research article optimal control strategies in an...
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Research ArticleOptimal Control Strategies in an Alcoholism Model
Xun-Yang Wang12 Hai-Feng Huo12 Qing-Kai Kong3 and Wei-Xuan Shi2
1 College of Electrical and Information Engineering Lanzhou University of Technology Lanzhou Gansu 730050 China2Department of Applied Mathematics Lanzhou University of Technology Lanzhou Gansu 730050 China3 School of Electrical and Information Engineering Anhui Industrial University Maanshan Anhui 243032 China
Correspondence should be addressed to Hai-Feng Huo huohf1970gmailcom
Received 10 December 2013 Accepted 9 January 2014 Published 2 March 2014
Academic Editor Weiming Wang
Copyright copy 2014 Xun-Yang Wang et alThis is an open access article distributed under theCreativeCommonsAttribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
This paper presents a deterministic SATQ-type mathematical model (including susceptible alcoholism treating and quittingcompartments) for the spread of alcoholismwith two control strategies to gain insights into this increasingly concerned about healthand social phenomenon Some properties of the solutions to themodel including positivity existence and stability are analyzedTheoptimal control strategies are derived by proposing an objective functional and using Pontryaginrsquos Maximum Principle Numericalsimulations are also conducted in the analytic results
1 Introduction
Alcoholism also known as alcohol dependence is a diseasethat includes the desire for alcohol and continuing to drink itdespite its negative effect on individualrsquos health relationshipsand social status [1] Similar to all other drug addictionsalcoholism can be regarded as a treatable disease The WorldHealth Organization estimates that about 140 million peoplethroughout the world suffer from alcohol dependence withrelated problems such as being sick losing a job among ahost of other things [2] Particularly young peoplersquos alco-holism problem is a major concern to public health USsurveys indicate that approximately 90 of college studentshave consumed alcohol at least once [3] and more than40 of college students have engaged in binge drinking [45] Unfortunately the biological mechanisms underpinningalcoholism are not known however risk factors includesocial environment stress mental health genetic sensitivityage ethnic group and sex [6 7] Long-term alcohol abusewill produce negative changes in the brain such as toleranceand physical dependenceThe subtle changesmake it difficultfor the alcoholics to stop drinking and result in alcoholwithdrawal symptoms upon discontinuation of alcohol con-sumption Alcohol damages almost all parts of the body andcontribute to a number of human diseases including butnot limited to liver cirrhosis pancreatitis heart disease and
sexual dysfunction and can eventually be deadly [8] Damageto the central and peripheral nervous systems can take placefrom sustained alcohol consumption [9ndash13]
Although alcoholism is becoming more and more dan-gerous and serious as well as a widespread social phe-nomenon only much less work has been done in the mathe-maticalmodelling of alcoholism as a growing health problemincluding a few studies which offered some mathematicalapproaches to understand the growing burden of alcoholism[10 14ndash19] In [10] a SIR-type model was proposed theauthors used standard contact rate between susceptiblesand alcoholism getting alcoholism reproductive numberand discussing the existence and stability of two equilibriaIn [14] a framework where drinking was modeled as asocially contagious process in low- and high-risk connectedenvironments was introduced they found that high levelsof social interaction between light and moderate drinkers inlow-risk environments can diminish the importance of thedistribution of relative drinking times on the prevalence ofheavy drinking In [15] neurophysiological examinations of100 long-term alcohol dependent patients who were havingneuropsychiatric treatment showed symptoms of polytopicdamage of the peripheral and central nervous system Theresults showed that for recognition of the damage an exten-sive diagnostic programmemust be used In [16] the authorsconsidered a kind of binge drinking model with two equal
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 954069 18 pageshttpdxdoiorg1011552014954069
2 Abstract and Applied Analysis
infectivity drunk states mathematical analyses establishedthat the global dynamics of themodel were determined by thebasic reproduction number In [17] the authors modified themodel from [16] that is they considered different infectivityof two drunk states and a SEIR-type model of alcoholismwas thus presented in which two alcohol related states wereinvolved namely no alcohol dependent consumers 119863(119905)and alcohol dependent consumers 119860(119905) In [18] the authorsformulated a deterministic model for evaluating the impactof heavy alcohol drinking on the reemerging gonorrheaepidemic and both analytical and numerical results wereprovided to ascertain whether heavy alcohol drinking hadan impact on the transmission dynamics of gonorrhea Theapproach of the literature [18] was very meaningful sinceit provided a new direction of thinking when the cross-infection between alcoholism and other pathological diseasesoccurs In recent monograph [19] the authors also proposeda SIR-type model to investigate alcohol abuse phenomenonand generated some useful insights for example the basicreproductive number was not always the key to controllingdrinking within the population For other papers that studythe model of giving up smoking or quitting drinking pleasesee [20 21] and references cited therein
As living standard and health awareness get improvedmore and more people who fall into binge drinking stateare actively seeking the quitting alcoholism measures andtreatment methods [1 11 22] In [22] treatment strategy wasintroduced into a simple SIR-type alcoholics quitting modelin which the authors used the bilinear incidence to depict theldquoinfectionrdquo between the occasional drinkers 119878 and problemdrinkers 119863 Motivated by some aforementioned documents[10 19 22] in this paper we will formulate a more reasonablealcoholics quitting model The fact that our model is morereasonable is embodied from the following three aspects
(1) Taking into account that alcoholism is a widespreadsocial phenomenon so the standard incidence issuperior to bilinear incidence when we portray therelationship between the alcoholism and the suscep-tibles during the course of infection While in [22]the authors adopted bilinear incidence we will adoptstandard incidence in this paper
(2) Since alcohol is harmful to health moreover aswe all know alcoholism is treatable if we can takeapproximate measure in time for example artificialisolation from alcoholisms medications persuasionand education programing on alcoholism So it isnecessary to take effective measures to avoid alcoholor to treat after alcoholism Documents [10 19] havenot considered these aspects
(3) Since there is effective prevention and treatment indescribing the phenomenon of alcoholism there aresome people who will never drink due to successfulprevention or some people who no longer drink aftersuccessful treatment Therefore when we formulatethe model in this paper itrsquos reasonable to introducea new compartment119876 the people in which will neverdrink for ever Obviously themodels of [10 19 22] arenot involving the quitting compartment 119876
Based on the above considerations we will premeditatetwo treating methods namely prevention of susceptiblesfrom alcoholism and treatment on alcoholism as controlvariables hence we will derive a SATQ-type model Wenote the fact that many authors are interested in solvingoptimal control problems such as cost minimization andoptimal control of various disease especially with biologicalbackground and various mathematical models [22ndash24] Inthis paper we will propose an objective functional whichconsiders not only alcohol quitting effects but also the costof controlling alcohol Then we consider a range of issuesrelated to the optimal control with themethod of PontryaginrsquosMaximum Principle including optimal control existenceuniqueness and characterization
The organization of this paper is as follows In thenext section the alcoholism model with prevention for thesusceptibles and treatment for alcoholism is formulated InSection 3 the basic reproduction number and the existence ofequilibria are investigatedThe stability of the disease free andendemic equilibria is proved in Section 4 Optimal controlstrategies by the classic method of PMP (Pontryaginrsquos Maxi-mum Principle) are discussed in Section 5 In Section 6 wegive some numerical simulations We give some discussionsand conclusions in the last section
2 The Model Formulation and SomeFundamental Properties
In this section we introduce a mathematical model withprevention and treatment for the alcoholism and then studysome important properties such as the boundness and posi-tivity of its solutions
21 Model Formulation and Parameter Explanation Thetotal population is partitioned into four compartments thesusceptible compartment 119878 which refers to the personswho never drink or drink moderately without affecting thephysical health the alcoholism compartment 119860 which refersto the persons who binge drink and affect the physical healthseriously the treatment compartment 119879 which refers to thepersons who have been receiving treatments by taking pills orothermedical interventions after alcoholism and the quittingcompartment 119876 which refers to the persons who recoverfrom alcoholism after treatment and stay off alcohol hereafterIn this paper we focus on a closed environment such as acommunity a university or a village So the total number ofpopulation to be considered is a constant we denote it as119873The population flow among those compartments is shown inthe following diagram (Figure 1)
The schematic diagram leads to the following system ofordinary differential equations
1198781015840= 120583119873 minus (1 minus 119906
1)120573119878119860
119873minus 120583119878
1198601015840= (1 minus 119906
1)120573119878119860
119873+ 120585119879 minus (119906
2+ 120583)119860
Abstract and Applied Analysis 3
A
S Q
T
120583N
120583S
120585T
120583T
120583A
120583Q
u2A
120575T
(1 minus u1)120573SA
N
Figure 1 Transfer diagram for the dynamics of alcoholism model
1198791015840= 1199062119860 minus (120583 + 120585 + 120575) 119879
1198761015840= 120575119879 minus 120583119876
(1)
Here 120583119873 is the birth number of the population 120583 is thenatural death rate of the population 119906
1is the fraction of
the susceptible individuals who successfully avoid to stayoff the alcoholism 119906
2is the fraction of the alcoholics who
take part in treatments here 0 le 119906119894le 1 119894 = 1 2 and
they will be considered as two control variables in Section 5120573 is the transmission coefficient of the ldquoinfectionrdquo for thesusceptible individuals from the alcoholic individuals 120585 is therate coefficient of the person who fail to be treated and returnto the alcoholism compartmentmostly due to their ownweakwill 120575 is the rate coefficient of the person who have receivedeffective treatment and recovered from alcoholism forever
22 Boundedness of Solutions to System and Positively Invari-ant Region It is important to show positivity and bounded-ness for the system (1) as they represent populations Firstlywe present the positivity of the solutions System (1) can beput into the matrix form
1198831015840= 119866 (119883) (2)
where119883 = (119878 119860 119879 119876)119879isin 1198774 and 119866(119883) is given by
119866 (119883) = (
1198661(119883)
1198662(119883)
1198663(119883)
1198664(119883)
)
=(
120583119873 minus (1 minus 1199061)120573119878119860
119873minus 120583119878
(1 minus 1199061)120573119878119860
119873+ 120585119879 minus (119906
2+ 120583)119860
1199062119860 minus (120583 + 120585 + 120575) 119879
120575119879 minus 120583119876
)
(3)
It is easy to check that
119866119894(119883)
1003816100381610038161003816119883119894(119905)=0119883119905isin119862+ge 0 119894 = 1 2 3 4 (4)
Due to Lemma 2 in [25] any solution of (1) is 119883(119905) isin 1198774+for
all 119905 ge 0We denote 119873(119905) = 119878(119905) + 119860(119905) + 119879(119905) + 119876(119905) summing
equations in (1) yields
119889119873 (119905)
119889119905= 0 (5)
so 119873(119905) = 119878(119905) + 119860(119905) + 119879(119905) + 119876(119905) = constant (denoted as119873) and the set
Ω = (119878 119860 119879 119876) isin 1198774
+ 119878 + 119860 + 119879 + 119876 le 119873 (6)
is a positively invariant region for (1) Therefore we willconsider the global stability of (1) on the setΩ
3 The Basic Reproduction Number andExistence of Alcoholism Equilibria
31 The Basic Reproduction Number 1198770 In epidemiology
the basic reproduction number (sometimes called basicreproductive rate or basic reproductive ratio) of an infectionis the number of infectious cases that one infectious casegenerates on average over the course of its infectious periodWhile in this context it means the number of persons thatan alcoholic will ldquoinfectrdquo during his ldquoinfectiousrdquo period in thepure susceptible environment so that the infected personswillenter the alcoholism compartment It is easy to see that themodel has an alcohol free equilibrium 119864
0= (1198780 0 0 0) =
(119873 0 0 0) In the following the basic reproduction numberof system (1) will be obtained by the next generation matrixmethod formulated in [26]
Let 119909 = (119860 119879 119876 119878)119879 then system (1) can be written as
119889119909
119889119905= F (119909) minusV (119909) (7)
where
F (119909) = (
(1 minus 1199061)120573119878119860
1198730
0
0
)
V (119909) = (
(1199062+ 120583)119860 minus 120585119879
(120583 + 120585 + 120575) 119879 minus 1199062119860
120583119876 minus 120575119879
(1 minus 1199061)120573119878119860
119873+ 120583119878 minus 120583119873
)
(8)
The Jacobian matrices of F(119909) and V(119909) at the alcohol freeequilibrium 119864
0are respectively
119863F (1198640) = (
1198653times3
0
0 0)
119863V (1198640) = (
1198813times3
0
(1 minus 1199061) 120573 0 0 120583
)
(9)
where
119865 = (
(1 minus 1199061) 120573 0 0
0 0 0
0 0 0
) 119881 = (
1199062+ 120583 minus120585 0
minus1199062
120583 + 120585 + 120575 0
0 minus120575 120583
)
(10)
The basic reproduction number denoted by 1198770 is thus given
by
1198770= 120588 (119865119881
minus1) =
120573 (1 minus 1199061) (120583 + 120585 + 120575)
1199062(120583 + 120575) + 120583 (120583 + 120585 + 120575)
(11)
4 Abstract and Applied Analysis
It is easy to see that both of the control parameters con-tributed to reducing the alcoholism From this point thecontrol measures are meaningful
32 Existence of Alcoholism Equilibrium The endemic equi-librium 119864
lowast(119878lowast 119860lowast 119879lowast 119876lowast) of system (1) is determined by
equations
120583119873 minus (1 minus 1199061)120573119878119860
119873minus 120583119878 = 0
(1 minus 1199061)120573119878119860
119873+ 120585119879 minus (119906
2+ 120583)119860 = 0
1199062119860 minus (120583 + 120585 + 120575) 119879 = 0
120575119879 minus 120583119876 = 0
(12)
The third equation in (12) leads to
119860 =120583 + 120585 + 120575
1199062
119879 (13)
From the last equation in (12) we have
119876 =120575119879
120583 (14)
From the first equation of (12) and together with (13) we canget
119878 =120583119873
120583 + (((1 minus 1199061) 120573) 119873)119860
=12058311987321199062
1205831198731199062+ (1 minus 119906
1) 120573119879 (120583 + 120585 + 120575)
(15)
Substituting (13)ndash(15) into the second equation of (12) gives
120583119873 minus120583211987321199062
1205831198731199062+ (1 minus 119906
1) 120573119879 (120583 + 120585 + 120575)
+ 120585119879 minus (120583 + 1199062)120583 + 120585 + 120575
1199062
119879 = 0
(16)
By simplifying (16) we can get
119879 [1199062120585 (1 minus 119906
1) 120573 (120583 + 120585 + 120575)
minus (1 minus 1199061) 120573(120583 + 120585 + 120575)
2
(120583 + 1199062)] 119879 + 120590 = 0
(17)
where120590 = 119906
2120583119873 (1 minus 119906
1) 120573 (120583 + 120585 + 120575)
+ (1199062)2
120583119873120585 minus 1205831198731199062(120583 + 119906
2) (120583 + 120585 + 120575)
(18)
Hence we get two explicit solutions to (17) one is 1198790= 0
which is corresponding to the alcohol free equilibria and theother is119879lowast= (120590) ((1 minus 119906
1) 120573(120583 + 120585 + 120575)
2
(120583 + 1199062)
minus1199062120585 (1 minus 119906
1) 120573 (120583 + 120585 + 120575) )
minus1
=120590
(1 minus 1199061) 120573 (120583 + 120585 + 120575) [120583120585 + (120575 + 120583) (120583 + 119906
2)]
(19)
which should be corresponding to the alcoholism equilibriaon condition that 119879lowast gt 0 otherwise the alcoholismequilibria are nonexistent It is enough to show the positivityof 120590 tomake sure the existence of alcoholism equilibria on thecondition 119877
0ge 1 By some simple calculations we simplify
the expression of 120590 to be
120590 = 1205831198731199062(120583 + 120585 + 120575) (1 minus 119906
1) 120573
minus [(120583 + 1199062) (120575 + 120583) + 120583120585]
(20)
Since 1198770gt 1 is equivalent to
120573 (1 minus 1199061) (120583 + 120585 + 120575) gt 119906
2(120583 + 120575) + 120583 (120583 + 120585 + 120575) (21)
the right side of this inequality is exactly equal to (120583+1199062)(120575+
120583) + 120583120585 Hence we have proved the existence of 119879lowast gt 0so are the alcoholism equilibria We summarize this result inTheorem 1
Theorem 1 For system (1) there is always an alcohol freeequilibrium 119864
0= (119873 0 0 0) When 119877
0gt 1 besides alcohol
free equilibrium 1198640 system (1) also has a unique alcoholism
equilibrium 119864lowast(119878lowast 119860lowast 119879lowast 119876lowast) where
119878lowast=
12058311987321199062
1205831198731199062+ (1 minus 119906
1) 120573119879lowast (120583 + 120585 + 120575)
119860lowast=120583 + 120585 + 120575
1199062
119879lowast
119876lowast=120575119879lowast
120583
119879lowast=
120590
(1 minus 1199061) 120573 (120583 + 120585 + 120575) [120583120585 + (120575 + 120583) (120583 + 119906
2)]
(22)
4 Stability Analysis of Equilibria
For the convenience of subsequent proof we denote a vector119883 = (119860 119879 119876 119878)
119879 and
119891 (119883) =(
(1 minus 1199061)120573119878119860
119873+ 120585119879 minus (119906
2+ 120583)119860
1199062119860 minus (120583 + 120585 + 120575) 119879
120575119879 minus 120583119876
120583119873 minus (1 minus 1199061)120573119878119860
119873minus 120583119878
) (23)
So the Jacobian matrix of 119891(119909) about vector 119883 is as thefollowing
119869 =
120597119891 (119883)
120597119883
=(
(1 minus 1199061) 120573119878
119873
minus (120583 + 1199062) 120585 0
(1 minus 1199061) 120573119860
119873
1199062 minus (120583 + 120585 + 120575) 0 0
0 120575 minus120583 0
minus
(1 minus 1199061) 120573119878
119873
0 0 minus120583 minus
(1 minus 1199061) 120573119860
119873
)
(24)
Theorem 2 For system (1) the alcohol free equilibrium 1198640is
locally asymptotically stable if 1198770lt 1
Abstract and Applied Analysis 5
Proof Since
119869 (1198640)
= (
(1 minus 1199061) 120573 minus (120583 + 119906
2) 120585 0 0
1199062
minus (120583 + 120585 + 120575) 0 0
0 120575 minus120583 0
minus (1 minus 1199061) 120573 0 0 minus120583
)
(25)
we can easily get that two of the eigenvalues are 1205821= 1205822=
minus120583 lt 0 while 1205823 1205824satisfy
1205822+ [2120583 + 120585 + 120575 + 119906
2minus (1 minus 119906
1) 120573] 120582
+ (120583 + 120585 + 120575) (120583 + 1199062minus (1 minus 119906
1) 120573) minus 119906
2120585 = 0
(26)
Thus
1205823+ 1205824= (1 minus 119906
1) 120573 minus (120583 + 120585 + 120575) minus (120583 + 119906
2) (27)
Since 1198770lt 1 is equivalent to
120573 (1 minus 1199061) (120583 + 120585 + 120575) lt 119906
2(120583 + 120575) + 120583 (120583 + 120585 + 120575)
lt (120583 + 1199062) (120583 + 120585 + 120575)
(28)
so
120573 (1 minus 1199061) lt 120583 + 119906
2 (29)
and then
1205823+ 1205824lt minus (120583 + 120585 + 120575) lt 0 (30)
while
12058231205824= (120583 + 120585 + 120575) (120583 + 119906
2minus (1 minus 119906
1) 120573) minus 119906
2120585 (31)
Similarly from 1198770lt 1 we can derive the inequality
minus120573 (1 minus 1199061) (120583 + 120585 + 120575) gt minus119906
2(120583 + 120575) minus 120583 (120583 + 120585 + 120575) (32)
so12058231205824gt (120583 + 119906
2) (120583 + 120585 + 120575)
minus 1199062120585 minus 1199062(120583 + 120575) minus 120583 (120583 + 120585 + 120575)
(33)
It reduces to
12058231205824gt 0 (34)
Hence Re 1205823lt 0 Re 120582
4lt 0 The proof is complete
Next we will turn to investigate the global stability of 1198640
Theorem 3 For system (1) the alcohol free equilibrium 1198640is
globally asymptotically stable if 1198770lt 1
Proof Consider the subsystem of (1) as follows
1198601015840= (1 minus 119906
1)120573119878119860
119873+ 120585119879 minus (119906
2+ 120583)119860
1198791015840= 1199062119860 minus (120583 + 120585 + 120575) 119879
1198761015840= 120575119879 minus 120583119876
(35)
Equation (35) can be rewritten as
(
) = (119865 minus 119881)(
119860
119879
119876
) minus (1 minus119878
119873)
times (
120573 (1 minus 1199061) 0 0
0 0 0
0 0 0
)(
119860
119879
119876
)
(36)
Since 119878 le 119873 and 0 le 1199061le 1 then for all 119905 gt 0 we can get
(
) le (119865 minus 119881)(
119860
119879
119876
) (37)
According to Lemma 1 in [26] all the eigenvalues of matrix119865 minus 119881 have negative real parts so the solutions of this sub-system are stable whenever 119877
0lt 1 So (119860(119905) 119879(119905) 119876(119905)) rarr
(0 0 0) as 119905 rarr infin By the comparison theorem [27] andbased on the fact that the total population is constant 119873 itfollows that (119860(119905) 119879(119905) 119876(119905)) rarr (0 0 0) and 119878(119905) rarr 119873
as 119905 rarr infin So the alcohol free equilibrium 1198640is globally
asymptotically stable the proof is complete
Theorem 4 For system (1) the alcoholism equilibrium119864lowast(119878lowast 119860lowast 119879lowast 119876lowast) is globally asymptotically stable if 119877
0gt 1
Proof Since the total population in model (1) is a constantnumber119873 in order to prove the global stability of system (1)it is sufficed to prove the corresponding stability of subsystem(38)
1198781015840= 120583119873 minus (1 minus 119906
1)120573119878119860
119873minus 120583119878
1198601015840= (1 minus 119906
1)120573119878119860
119873+ 120585119879 minus (119906
2+ 120583)119860
1198791015840= 1199062119860 minus (120583 + 120585 + 120575) 119879
(38)
We make normalization transform and still use the samesymbols 119878 119860 119879 to denote the variables then (38) can betransformed into
1198781015840= 120583 minus (1 minus 119906
1) 120573119878119860 minus 120583119878
1199041198601015840= (1 minus 119906
1) 120573119878119860 + 120585119879 minus (119906
2+ 120583)119860
1198791015840= 1199062119860 minus (120583 + 120585 + 120575) 119879
(39)
From (39) we can easily know that the equilibria (119878lowast 119860lowast 119879lowast)satisfy the following three equalities to be used later
120583 = (1 minus 1199061) 120573119878lowast119860lowastminus 120583119878lowast
(1 minus 1199061) 120573119878lowast119860lowast+ 120585119879lowast= (1199062+ 120583)119860
lowast
1199062119860lowast= (120583 + 120585 + 120575) 119879
lowast
(40)
6 Abstract and Applied Analysis
Let 119881 = 1199091(119878 minus 119878
lowastminus 119878lowast ln(119878119878lowast)) + 119909
2(119860 minus 119860
lowastminus
119860lowast ln(119860119860lowast)) + 119909
3(119879 minus 119879
lowastminus 119879lowast ln(119879119879lowast)) then
119881101584010038161003816100381610038161003816(39)
= 1199091[120583 minus (1 minus 119906
1) 120573119878119860 minus 120583119878
minus119878lowast
119878120583 + (1 minus 119906
1) 120573119878lowast119860 + 120583119878
lowast]
+ 1199092[ (1 minus 119906
1) 120573119878119860 + 120585119879 minus (119906
2+ 120583)119860
minus (1 minus 1199061) 120573119878119860
lowastminus119860lowast
119860120585119879 + (119906
2+ 120583)119860
lowast]
+ 1199093[1199062119860 minus (120583 + 120585 + 120575) 119879
minus119879lowast
1198791199062119860 + (120583 + 120585 + 120575) 119879
lowast]
= 1199091[ (1 minus 119906
1) 120573119878lowast119860lowast+ 120583119878lowastminus (1 minus 119906
1) 120573119878119860 minus 120583119878
minus119878lowast
119878((1 minus 119906
1) 120573119878lowast119860lowast+ 120583119878lowast)
+ (1 minus 1199061) 120573119878lowast119860 + 120583119878
lowast]
+ 1199092[ (1 minus 119906
1) 120573119878119860 + 120585119879 minus (119906
2+ 120583)119860
minus (1 minus 1199061) 120573119878119860
lowastminus119860lowast
119860120585119879 + (119906
2+ 120583)119860
lowast]
+ 1199093[1199062119860 minus (120583 + 120585 + 120575) 119879
minus119879lowast
1198791199062119860 + (120583 + 120585 + 120575) 119879
lowast]
= 1199091120583119878lowast(2 minus
119878
119878lowastminus119878lowast
119878)
+ [1199091(1 minus 119906
1) 120573119878lowast119860lowast+ 1199092(1199062+ 120583)119860
lowast
+ 1199093(120583 + 120585 + 120575) 119879
lowast]
minus [1199091
(119878lowast)2
(1 minus 1199061) 120573119860lowast
119878+ 1199092(1 minus 119906
1) 120573119878119860
lowast
+ 1199092
119860lowast120585119879
119860+ 1199093
119879lowast
1198791199062119860]
+ 119878119860 [minus1199091(1 minus 119906
1) 120573 + 119909
2(1 minus 119906
1) 120573]
+ 119860 [minus (1199062+ 120583) 119909
2+ 1199091(1 minus 119906
1) 120573119878lowast+ 11990621199093]
+ 119879 [1205851199092minus (120583 + 120585 + 120575) 119909
3]
(41)
To eliminate the cross-term 119878119860 and two single-variable terms119860 and 119879 we let
minus1199091(1 minus 119906
1) 120573 + 119909
2(1 minus 119906
1) 120573 = 0
minus (1199062+ 120583) 119909
2+ 1199091(1 minus 119906
1) 120573119878lowast+ 11990621199093= 0
1205851199092minus (120583 + 120585 + 120575) 119909
3= 0
(42)
By solving them we can get
1199091= 1 119909
2= 1
1199093=
120585
120583 + 120585 + 120575=1199062+ 120583 minus (1 minus 119906
1) 120573119878lowast
1199062
(43)
Next we let
1198811015840
1= 1199091(1 minus 119906
1) 120573119878lowast119860lowast+ 1199092(1199062+ 120583)119860
lowast
+ 1199093(120583 + 120585 + 120575) 119879
lowast
1198811015840
2= minus[119909
1
(119878lowast)2
(1 minus 1199061) 120573119860lowast
119878
+ 1199092(1 minus 119906
1) 120573119878119860
lowast+ 1199092
119860lowast120585119879
119860+ 1199093
119879lowast
1198791199062119860]
(44)
and then
1198811015840= 1199091120583119878lowast(2 minus
119878
119878lowastminus119878lowast
119878) + 119881
1015840
1+ 1198811015840
2 (45)
Due to
(1199062+ 120583)119860
lowast= (1 minus 119906
1) 120573119878lowast119860lowast+ 120585119879lowast
1199093(120583 + 120585 + 120575) 119879
lowast= 120585119879lowast1199092= 120585119879lowast
(46)
Abstract and Applied Analysis 7
so
1198811015840
1= 2 (1 minus 119906
1) 120573119878lowast119860lowast+ 2120585119879
lowast
1198811015840
2= minus[
(119878lowast)2
(1 minus 1199061) 120573119860lowast
119878
+ (1 minus 1199061) 120573119878119860
lowast+119860lowast120585119879
119860+ 1199093
119879lowast
1198791199062119860]
le minus2[(119878lowast)2
(1 minus 1199061) 120573119860lowast
119878sdot (1 minus 119906
1) 120573119878119860
lowast]
12
minus 2 [119860lowast120585119879
119860sdot 1199093
119879lowast
1198791199062119860]
= minus2 (1 minus 1199061) 120573119878lowast119860lowastminus 2(1199093119860lowast1205851199062119879lowast)12
= minus2 (1 minus 1199061) 120573119878lowast119860lowastminus 2[120585119879
lowast(11990931199062119860lowast)]12
= minus2 (1 minus 1199061) 120573119878lowast119860lowast
minus 2120585119879lowast[(1199062+ 120583)119860
lowastminus (1 minus 119906
1) 120573119878lowast119860lowast]12
= minus2 (1 minus 1199061) 120573119878lowast119860lowastminus 2(120585119879
lowast120585119879lowast)12
= minus2 (1 minus 1199061) 120573119878lowast119860lowastminus 2120585119879
lowast
(47)
Hence
1198811015840= 120583119878lowast(2 minus
119878
119878lowastminus119878lowast
119878) + 119881
1015840
1+ 1198811015840
2
le 120583119878lowast(2 minus
119878
119878lowastminus119878lowast
119878) + 2 (1 minus 119906
1) 120573119878lowast119860lowast
+ 2120585119879lowastminus 2 (1 minus 119906
1) 120573119878lowast119860lowastminus 2120585119879
lowast
= 120583119878lowast(2 minus
119878
119878lowastminus119878lowast
119878) le 0
(48)
1198811015840= 0 if and only if 119878 = 119878
lowast 119860 = 119860lowast 119879 = 119879
lowast Accordingto LaSallersquos invariance principle [28] we can derive theconclusion that the alcoholism equilibria 119864lowast(119878lowast 119860lowast 119879lowast 119876lowast)are globally asymptotically stable the proof is complete
5 Optimal Control Problem
51 The Existence of Optimal Control In order to investigatean effective campaign to control alcoholism in a communitywhich pursue the goals of the minimized alcoholisms andmore recovered individuals we will reconsider the system (1)and use two control variables to reduce the numbers of alco-holics The difference is that we will change the parameters1199061 1199062into control variable 119906
1(119905) 1199062(119905) Their aforementioned
definitions allow us to do so 1199061(119905) is used to limit the
proportion of the susceptible individual to contact withalcoholism usually by propaganda and education so that thesusceptible individual can stay off alcoholism consciously andbe free of ldquoinfectionrdquo we can understand the effect of 119906
1(119905)
is to prevent the the susceptible from contacting with thealcoholism The control variable 119906
2(119905) is used to control the
alcoholism to take appropriate treatment measures such astaking pills or seeking other medical help However just as acoin has two sides there will be a lot of costs generated duringthe control process So it is advisable to balance between thecosts and the alcohol effects In view of this our optimalcontrol problem tominimize the objective functional is givenby
119869 (1199061 1199062) = int
119905119891
0
[119860 (119905) +1198881
21199062
1(119905) +
1198882
21199062
2(119905)] 119889119905 (49)
which subjects to system
1198781015840= 120583119873 minus (1 minus 119906
1(119905))
120573119878119860
119873minus 120583119878
1198601015840= (1 minus 119906
1(119905))
120573119878119860
119873+ 120585119879 minus (119906
2(119905) + 120583)119860
1198791015840= 1199062(119905) 119860 minus (120583 + 120585 + 120575) 119879
1198761015840= 120575119879 minus 120583119876
(50)
with initial conditions
119878 (0) = 1198780 119860 (0) = 119860
0
119879 (0) = 1198790 119876 (0) = 119876
0
(51)
Here 119906119894(119905) isin 119880 ≜ (119906
1 1199062) | 119906119894(119905) is measurable and 0 le
119906119894(119905) le 1 for all 119905 isin [0 119905
119891] 119905119891is the end time to be
controlled 119880 is an admissible control set 119888119894 and 119894 = 1 2 are
weight factors (positive constants) that adjust the intensity oftwo different control measures
Next we will investigate the existence of the optimalcontrol of the above-mentioned problem
Theorem 5 There exists an optimal control pair 119906lowast
=
(119906lowast
1 119906lowast
2) isin 119880 such that
119869 (119906lowast
1 119906lowast
2) = min 119869 (119906
1 1199062) 119906
1(119905) 1199062(119905) isin 119880 (52)
subjects to the control system (1) with initial conditions (50)
Proof Toprove the existence of an optimal control accordingto the classic literature [29] we have to show the following
(1) The control and state variables are nonnegative values(2) The control set 119880 is convex and closed(3) The right side of the state system is bounded by linear
function in the state and control variables(4) The integrand of the objective functional is concave
on 119880(5) There exist constants 119889
1 1198892gt 0 and 120572 gt 1 such
that the integrand 119871(119905 1199061 1199062) ≜ 119860(119905) + (119888
12)1199062
1(119905) +
(11988822)1199062
2(119905) of the objective functional satisfies
119871 (119905 1199061 1199062) ge 1198891(100381610038161003816100381611990611003816100381610038161003816
2
+100381610038161003816100381611990621003816100381610038161003816
2
)1205722
minus 1198892
(53)
8 Abstract and Applied Analysis
statements (1) (2) and (3) are obvious satisfied we only needto test and verify the latter two ones Since the four statevariables have been all proved to be up bounded by 119873 wewill get the following equalities
1198781015840le 120583119873 119860
1015840le (1 minus 119906
1(119905)) 120573119878 + 120585119879
1198791015840le 1199062(119905) 119860 119876
1015840le 120575119879
(54)
so the fourth condition is set up As for the last condition
119871 (119905 1199061 1199062) ge 1198891(100381610038161003816100381611990611003816100381610038161003816
2
+100381610038161003816100381611990621003816100381610038161003816
2
)1205722
minus 1198892
(55)
is also true when we choose 1198891= min119888
12 11988822 and for all
1198892isin 119877+ 120572 = 2 The proof is complete
We next come to the core of this section
52 The Characterization of the Optimal Control With theexistence of the optimal control pairs established we nowpresent the optimality system and use a result from [30]we can easily know the existence of the solutions to theoptimality system (71) which will be gotten later Firstly wecome to discuss the theorem that relates to the character-ization of the optimal control The optimality system canbe used to compute candidates for optimal control pairsTo do this we begin by defining an augmented Hamilto-nian 119867 with penalty terms for the control constraints asfollows
119867 = 119860 (119905) +1198881
21199062
1(119905) +
1198882
21199062
2(119905)
+ 1205821[120583119873 minus (1 minus 119906
1(119905))
120573119878119860
119873minus 120583119878]
+ 1205822[(1 minus 119906
1(119905))
120573119878119860
119873+ 120585119879 minus (119906
2(119905) + 120583)119860]
+ 1205823[1199062(119905) 119860 minus (120583 + 120585 + 120575) 119879] + 120582
4(120575119879 minus 120583119876)
minus 119908111199061(119905) minus 119908
12(1 minus 119906
1(119905))
minus 119908211199062(119905) minus 119908
22(1 minus 119906
2(119905))
(56)
where 119908119894119895(119905) ge 0 are the penalty multipliers satisfying
11990811(119905) 1199061(119905) = 119908
12(119905) (1 minus 119906
1(119905))
= 0 at optimal control 119906lowast1
11990821(119905) 1199062(119905) = 119908
22(119905) (1 minus 119906
2(119905))
= 0 at optimal control 119906lowast2
(57)
Theorem6 Given optimal control pairs (119906lowast1 119906lowast
2) and solutions
119878(119905) 119860(119905) 119879(119905) 119876(119905) of the corresponding state system (50)there exist adjoint variables 120582
119894 119894 = 1 2 3 4 satisfying
1205821015840
1= 1205821(1 minus 119906
1(119905))
120573119860
119873+ 1205831205821minus 1205822(1 minus 119906
1(119905))
120573119860
119873
1205821015840
2= minus1 + 120582
1(1 minus 119906
1(119905))
120573119878
119873minus 1205822(1 minus 119906
1(119905))
120573119878
119873
+ 1205822(120583 + 119906
2(119905)) minus 120582
31199062(119905)
1205821015840
3= minus1205822120585 + 1205823(120583 + 120585 + 120575) minus 120582
4120575
1205821015840
4= 1205831205824
(58)
with the terminal conditions
120582119894(119905119891) = 0 119894 = 1 2 3 4 (59)
Furthermore (119906lowast1 119906lowast
2) are represented by
119906lowast
1= min(1max(0
120573119878119860 (1205822minus 1205821)
1198881119873
))
119906lowast
2= min(1max(0
119860 (1205822minus 1205823)
1198882
))
(60)
Proof According to Pontryagin Maximum Principle [29ndash31] we first differentiate the Hamiltonian operator 119867 withrespect to states Then the adjoint system can be written as
1205821015840
1= minus
120597119867
120597119878 120582
1015840
2= minus
120597119867
120597119860
1205821015840
3= minus
120597119867
120597119879 120582
1015840
4= minus
120597119867
120597119876
(61)
The terminal condition (56) of adjoint equations is given by120582119894(119905119891) = 0 119894 = 1 2 3 4
To obtain the necessary conditions of optimality (59) wealso differentiate theHamiltonian operator119867 with respect to119880 = (119906
1 1199062) and set them equal to zero then
120597119867
1205971199061
= 11988811199061(119905) + 120582
1
120573119878119860
119873minus 1205822
120573119878119860
119873minus 11990811+ 11990812= 0
120597119867
1205971199062
= 11988821199062(119905) minus 120582
2119860 + 120582
3119860 minus 119908
21+ 11990822= 0
(62)
By solving the optimal control we obtain
119906lowast
1=1
1198881
[(1205822minus 1205821)120573119878119860
119873+ 11990811minus 11990812] (63)
To determine an explicit expression for the optimalcontrol without119908
11and119908
12 a standard optimality technique
is utilized [29] We consider the following three cases
(i) On the set 119905 | 0 lt 119906lowast
1(119905) lt 1 we have 119908
11(119905) =
Abstract and Applied Analysis 9
11990812(119905) = 0 Hence the optimal control is
119906lowast
1=1
1198881
[(1205822minus 1205821)120573119878119860
119873] (64)
(ii) On the set 119905 | 119906lowast1(119905) = 1 we have 119908
11(119905) = 0 Hence
1 = 119906lowast
1(119905) =
1
1198881
[(1205822minus 1205821)120573119878119860
119873minus 11990812] (65)
This implies that
1
1198881
(1205822minus 1205821)120573119878119860
119873ge 1 since 119908
12(119905) ge 0 (66)
(iii) On the set 119905 | 119906lowast1(119905) = 0 we have 119908
12(119905) = 0 Hence
0 = 119906lowast
1(119905) =
1
1198881
[(1205822minus 1205821)120573119878119860
119873+ 11990811] (67)
This implies that
1
1198881
(1205822minus 1205821)120573119878119860
119873le 0 since 119908
11(119905) ge 0 (68)
Combining these results the optimal control 119906lowast1(119905) is
characterized as
119906lowast
1= min1max0 1
1198881
[(1205822minus 1205821)120573119878119860
119873] (69)
Using the similar arguments we can also obtain the otheroptimal control function
119906lowast
2= min1max0
(1205822minus 1205823) 119860
1198882
(70)
The proof is complete
We point out that the optimality system consistsof the state system (50) with the initial conditions119878(0) 119860(0) 119879(0) 119876(0) the adjoint (or costate) system(58) with the terminal conditions (59) and the optimalitycondition (60) Any optimal control pairs must satisfythis optimality system For the convenience of subsequent
numerical simulation in Section 6 we give the optimalitysystem as follows
1198781015840= 120583119873 minus (1 minusmin1max0 1
1198881
[(1205822minus 1205821)120573119878119860
119873] (119905))
times120573119878119860
119873minus 120583119878
1198601015840= (1minusmin1max0 1
1198881
[(1205822minus1205821)120573119878119860
119873] (119905))
120573119878119860
119873
+ 120585119879 minus (min1max0(1205822minus 1205823) 119860
1198882
(119905) + 120583)119860
1198791015840= min1max0
(1205822minus 1205823) 119860
1198882
(119905) 119860
minus (120583 + 120585 + 120575) 119879
1198761015840= 120575119879 minus 120583119876
1205821015840
1= 1205821(1 minusmin1max0 1
1198881
[(1205822minus 1205821)120573119878119860
119873] (119905))
times120573119860
119873+ 1205831205821
minus 1205822(1minusmin1max0 1
1198881
[(1205822minus1205821)120573119878119860
119873](119905))
times120573119860
119873
1205821015840
2= minus1
+ 1205821(1minusmin1max0 1
1198881
[(1205822minus1205821)120573119878119860
119873] (119905))
times120573119878
119873minus 1205823sdotmin1max0
(1205822minus 1205823) 119860
1198882
(119905)
minus 1205822(1 minusmin1max0 1
1198881
[(1205822minus 1205821)120573119878119860
119873])
times120573119878
119873+ 1205822(120583+min1max0
(1205822minus 1205823) 119860
1198882
(119905))
1205821015840
3= minus1205822120585 + 1205823(120583 + 120585 + 120575) minus 120582
4120575
1205821015840
4= 1205831205824
119878 (0) = 1198780 119860 (0) = 119860
0 119879 (0) = 119879
0
119876 (0) = 1198760 120582
119894(119905119891) = 0 119894 = 1 2 3 4
(71)
53 The Uniqueness of Optimal Control Due to the a prioriboundedness of the state adjoint functions and the resultingLipschitz structure of the ODEs we can obtain the unique-ness of the optimal control
10 Abstract and Applied Analysis
Lemma 7 (see [23]) The function 119906lowast(119904) = min (119887max (119904 119886))is Lipschitz continuous in 119904 where 119886 lt 119887 are some fixed positiveconstants
Theorem 8 For all 119905 isin [0 119905119891] the solution to the optimality
system (71) is unique
Proof Suppose (119878 119860 119879 119876 1205821 1205822 1205823 1205824) and
(119878 119860 119879 119876 1205821 1205822 1205823 1205824) are two different solutions of
our optimality system (71) Let
119878 = 119890120582119905119898 119860 = 119890
120582119905119899 119879 = 119890
120582119905119901
119876 = 119890120582119905119902 120582
1= 119890minus120582119905119903 120582
2= 119890minus120582119905119904
1205823= 119890minus120582119905119908 120582
4= 119890minus120582119905V
119878 = 119890120582119905119898 119860 = 119890
120582119905119899 119879 = 119890
120582119905119901
119876 = 119890120582119905119902 120582
1= 119890minus120582119905119903 120582
2= 119890minus120582119905119904
1205823= 119890minus120582119905119908 120582
4= 119890minus120582119905V
(72)
where 120582 gt 0 is to be chosenAccordingly we have
119906lowast
1(119905) = min1max0
120573119898119899 (119904 minus 119903) 119890120582119905
1198881119873
119906lowast
2(119905) = min1max0 (119904 minus 119908) 119899
1198882
119906lowast
1(119905) = min1max0
120573119898119899 (119904 minus 119903) 119890120582119905
1198881119873
119906lowast
2(119905) = min1max0 (119904 minus 119908) 119899
1198882
(73)
Now we substitute 119878 = 119890120582119905119898 119860 = 119890
120582119905119899 119879 = 119890
120582119905119901 119876 = 119890
120582119905119902
1205821= 119890minus120582119905119903 1205822= 119890minus120582119905119904 1205823= 119890minus120582119905119908 1205824= 119890minus120582119905V and 119878 =
119890120582119905119898 119860 = 119890
120582119905119899 119879 = 119890
120582119905119901 119876 = 119890
120582119905119902 1205821= 119890minus120582119905119903 1205822=
119890minus120582119905119904 1205823= 119890minus120582119905119908 1205824= 119890minus120582119905V into the first ODE of (71)
respectively then we can obtain
+ 120582119898 = 120583119873119890minus120582119905
minus (1 minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)
times120573119898119899119890
120582119905
119873minus 120583119898
(74)
for119898 and119898 respectively Similarly we can derive
119899 + 120582119899
= (1 minusmin1max1120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)120573119898119899119890
120582119905
119873
+ 120585119901 minus (min1max0 119899 (119904 minus 119908)1198882
+ 120583) 119899
(75)
for 119899 and 119899 respectively
+ 120582119901 = min1max0 119899 (119904 minus 119908)1198882
119899
minus (120583 + 120585 + 120575) 119901
(76)
for 119901 and 119901 respectively
119902 + 120582119902 = 120575119901 minus 120583119902 (77)
for 119902 and 119902 respectively
119903 minus 120582119903 = 119903(1 minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)120573119899119890120582119905
119873
+ 120583119903minus119904(1minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)
times120573119899119890120582119905
119873
(78)
for 119903 and 119903 respectively
119904 minus 120582119904 = minus119890120582119905
+ 119903(1 minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)120573119898119890120582119905
119873
minus 119904(1 minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)
times120573119898119890120582119905
119873+ 119904(min1max0 119899 (119904 minus 119908)
1198882
+ 120583)
minus 119908(min1max0 119899 (119904 minus 119908)1198882
)
(79)
for 119904 and 119904 respectively
minus 120582119908 = minus119904120585 + (120583 + 120585 + 120575)119908 minus 120575V (80)
for 119908 and 119908 respectively
V minus 120582V = 120583V (81)
for V and V respectively
Abstract and Applied Analysis 11
By Lemma 7 we can obtain
1003816100381610038161003816119906lowast
1(119905) minus 119906
lowast
1(119905)1003816100381610038161003816 le
120573119890120582119905
1198881119873|119898119899 (119904 minus 119903) minus 119898119899 (119904 minus 119903)|
1003816100381610038161003816119906lowast
2(119905) minus 119906
lowast
2(119905)1003816100381610038161003816 le
1
1198882
|119899 (119904 minus 119908) minus 119899 (119904 minus 119908)|
(82)
The equations for 119898 119899 119901 119902 119903 119904 119908 V and the equationsfor 119898 119899 119901 119902 119903 119904 119908 V are subtracted respectively then wemultiply each equation by appropriate difference of functionsand integrate from 0 to 119905
119891 Next we add all eight integral
equations and some inequality techniques to obtain unique-ness The following calculation is similar for the sake ofsimplicity we only take119898 and119898 for an example
minus + (120583 + 120582) (119898 minus 119898)
=120573119890120582119905
119873[minus(1 minusmin1max0
120573119898119899 (119904 minus 119903) 119890120582119905
1198881119873
)119898119899
+(1 minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)119898119899]
(83)
Multiplying both sides of (83) by (119898minus119898) and integratingfrom 0 to 119905
119891gives
1
2(119898 minus 119898)
2(119905119891) + (120583 + 120582)int
119905119891
0
(119898 minus 119898)2119889119905
= int
119905119891
0
(119898 minus 119898)120573119890120582119905
119873
times [minus (1 minus 119906lowast
1)119898119899 + (1 minus 119906
lowast
1)119898119899] 119889119905
= int
119905119891
0
(119898 minus 119898)120573119890120582119905
119873[(119906lowast
1minus 1) (119898119899 minus 119898119899 + 119898119899 minus 119898119899)
+ 119898119899 (119906lowast
1minus 119906lowast
1)] 119889119905
le120573119890120582119905119891
119873int
119905119891
0
(119898 minus 119898)
times [119906lowast
1minus 1
100381610038161003816100381610038161003816100381610038161003816
(119898 |119899 minus 119899| + |119898 minus 119898| 119899) + 119898119899120573119890120582119905
1198881119873
100381610038161003816100381610038161003816100381610038161003816
times 119898119899 (119904 minus 119903) minus 119898119899 (119904 minus 119903) ] 119889119905
le120573119890120582119905119891
119873int
119905119891
0
(119898 minus 119898)
times [1003816100381610038161003816119906lowast
1minus 1
1003816100381610038161003816 (119898 |119899 minus 119899| + |119898 minus 119898| 119899) + 119898119899120573119890120582119905
1198881119873
|119898119899 (119904 minus 119904) + (119898119899 minus 119898119899) 119904
minus (119898119899 (119903 minus 119903) + (119898119899 minus 119898119899) 119903)| ] 119889119905
le120573119890120582119905119891
119873int
119905119891
0
[1003816100381610038161003816119906lowast
1minus 1
1003816100381610038161003816 |119898| |119898 minus 119898| |119899 minus 119899|
+ |119898 minus 119898|2
|119899| + |119861| |119904| |119899| |119898 minus 119898|2
+ |119861| |119898119899| |119898 minus 119898| |119903 minus 119903|
+ |119861| |119898119899| |119898 minus 119898| |119904 minus 119904|
+ |119861| |119904| |119898| |119898 minus 119898| |119899 minus 119899|
+ |119861| |119903| |119898| |119898 minus 119898| |119899 minus 119899|
+ |119861| |119903| |119899| |119898 minus 119898|2] 119889119905
le120573119890120582119905119891
119873int
119905119891
0
[
1003816100381610038161003816119906lowastminus 1
1003816100381610038161003816
2|119898| ((119898 minus 119898)
2+ (119899 minus 119899)
2)
+ |119899| (119898 minus 119898)2
+|119861| |119898119899|
2((119898 minus 119898)
2+ (119904 minus 119904)
2)
+|119861| |119904| |119898|
2((119898 minus 119898)
2+ (119899 minus 119899)
2)
+ |119861| |119904| |119899| (119898 minus 119898)2
+|119861| |119898119899|
2((119898 minus 119898)
2+ (119903 minus 119903)
2)
+|119861| |119898| |119903|
2((119898 minus 119898)
2+ (119899 minus 119899)
2)
+ |119861| |119903| |119899| (119898 minus 119898)2]119889119905
(84)
In the above derivation we use many scaling techniquesfor inequality or absolute inequality Particularly what shouldbe noted is that to get the first inequality of above derivationwe use the estimation of |119906lowast
1(119905) minus 119906
lowast
1(119905)| which has been given
before besides for the sake of convenience we note 119861 =
119898119899(120573119890120582119905119891119873) Furthermore we notice that the coefficients of
all the eight terms in the last formula (119898 minus 119898)2+ (119899 minus 119899)
2(119898 minus119898)
2 (119898 minus119898)2+ (119904 minus 119904)
2 (119898 minus119898)2+ (119899 minus 119899)
2 (119898 minus119898)2
(119898minus119898)2+(119903minus119903)
2 (119898minus119898)2+(119899minus119899)2 (119898minus119898)2 namely (|119906lowast1minus
1|2)|119898| |119899| (|119861||119898119899|)2 (|119861||119904||119898|)2 |119861||119904|119899 (|119861||119898119899|)2(|119861||119898||119903|)2 |119861||119903||119899| are nonnegative and bounded So thereexists a positive constant 119888
2such that
1
2(119898 minus 119898)
2(119905119891) + (120583 + 120582)int
119905119891
0
(119898 minus 119898)2119889119905
12 Abstract and Applied Analysis
le 1198882
120573119890120582119905119891
119873int
119905119891
0
((119898 minus 119898)2+ (119899 minus 119899)
2
+ (119904 minus 119904)2+ (119903 minus 119903)
2) 119889119905
(85)
Combining eight of these inequalities gives
1
2(119898 minus 119898) (119905
119891) +
1
2(119899 minus 119899) (119905
119891) +
1
2(119901 minus 119901) (119905
119891)
+1
2(119902 minus 119902) (119905
119891) +
1
2(119903 minus 119903) (0) +
1
2(119904 minus 119904) (0)
+1
2(119908 minus 119908) (0) +
1
2(V minus V) (0) + (120583 + 120582)
times int
119905119891
0
(119898 minus 119898)2+ (119899 minus 119899)
2+ (119901 minus 119901)
2
+ (119902 minus 119902)2
+ (119904 minus 119904)2
+ (119903 minus 119903)2+ (119908 minus 119908)
2+ (V minus V)2) 119889119905
le 119861int
119905119891
0
(119898 minus 119898)2+ (119899 minus 119899)
2+ (119901 minus 119901)
2
+ (119902 minus 119902)2
+ (119903 minus 119903)2+ (119904 minus 119904)
2
+ (119908 minus 119908)2+ (V minus V)2 119889119905
(86)
Thus from the above inequality we can conclude that
(120583 + 120582 minus 119861)int
119905119891
0
(119898 minus 119898)2+ (119899 minus 119899)
2
+ (119901 minus 119901)2
+ (119902 minus 119902)2
+ (119903 minus 119903)2
+ (119904 minus 119904)2+(119908 minus 119908)
2+ (V minus V)2 119889119905 le 0
(87)
where 119861 depends on the coefficients and the bounds dependon119898 119899 119901 119902 119903 119904 119908 V If we choose 120582 such that 120583+120582 gt 119861 then119898 = 119898 119899 = 119899 119901 = 119901 119902 = 119902 119903 = 119903 119904 = 119904 119908 = 119908 and V = VHence the solution to the optimality system is unique Theproof is complete
6 Numerical Simulation
61 The Simulation of State System (1) without ControlParameters For the sake of simplicity but without loss ofgenerality we will perform the numerical simulation of statesystem (1) with parameters 119906
1= 0 119906
2= 0 Before
illustrating the analytic properties of the alcoholism model(1) we will target the populations in the environment ofa community or a university for example the school ofmaterial science and engineering in our university that isLan zhou University of Technology (LUT for short) owing tothe accurate and available information we can obtain Refer-ring to the information provided by the admissions officeof LUT this school will enroll almost 1200 undergraduatesand almost 300 various postgraduates at the beginning of
fall semester at the same time there will be almost 1500various students graduated and left this school so the scaleof students in school remained almost 6000 we can take thetotal population 119873 = 6000 In this simulation we will takeSeptember as the initial time and units in one week periodin one year According to the investigations of the studentunion implemented in September every year we can takeinitial values as 119878(0) = 4500 119860(0) = 1000 119879(0) = 300and 119876(0) = 200 It seems that the alcoholism is a little bitmore but it is rather natural because many freshmen feelconfused when they are faced with the new environment anda new lifestyle many of them have no better choice but gathertogether to drink in small groups to mediate the anxiety andget to know each other over time some of them developthe habit of drinking To a certain extent for example thefrequent drinking badly affects their study we can classifythem into the alcoholism compartment Other initial valuesseem more reasonable so we need no more explanation Aswe know alcoholism death is seldom happen within oneyear so we omit mortality from alcoholism then how tounderstand the recruitment rate as well as natural death rate120583 We can treat the freshmen admission as the recruitmentpopulation and graduation students as the natural ldquodeathrdquoparts So we can take 120583 = 15006000 = 025 which is exactlyconsistent with the value in [16] As for the infection rate 120573and recovery rate 120575 we will let them be variables since thedrinking behaviors are related to many factors such as theseason and the pressure
According to the data we get from the student unionwe choose 120585 = 04 To summarize we list the values ofthe parameters in Table 1 Using the values of parameters inTable 1 we can plot Figures 2 and 3 which are on condition1198770lt 1 and 119877
0ge 1 respectively From Figure 2 we easily
know when 1198770lt 1 holds the solution of system (1) tends to
the alcohol free equilibrium1198640and verifies the global stability
of 1198640 While seen from Figure 3 we also know that if 119877
0ge
1 holds the solution of system (1) tends to the alcoholismequilibrium 119864
lowast and verifies the global stability of 119864lowast
62 The Sensitivity of 1198770about Two Control Parameters
Although from the expression of the model reproductionnumber 119877
0 we can easily find out the fact that the two
control variables that is 1199061and 119906
2 attribute to reducing the
severity of alcoholism we will still depict the graph between1198770and the two control variables to see more intuitiveness see
Figure 4 It seems from the figure that 1198770is a monotonically
decreasing function about two control parameters so it isadvisable to take two approaches simultaneously to controlthe alcoholics
63 The Simulation of Optimality System (71) In this subsec-tion we will investigate numerically the optimal solution tooptimality system (71) by numerical method from [32] theoptimality system is solved with a fourth-order Runge-Kuttascheme Beginning with a guess for the control variables thestate system is solved forward in time and then those values ofstate are used to solve the adjoint equations backward in timeThe controls are updated at the end of each iteration using
Abstract and Applied Analysis 13
0 10 20 30 40 50
0
1000
2000
3000
4000
5000
6000
7000
Time
Popu
latio
n siz
e
S(t)
A(t)
T(t)
Q(t)
Figure 2 When 1198770lt 1 the alcohol free equilibrium 119864
0is globally
asymptotically stable (120573 = 02 120575 = 03 and 1198770= 071698)
S(t)
A(t)
T(t)
Q(t)
0 10 20 30 40
48
2520
303540
49 50
500
1000
2000
3000
4000
5000
6000
7000
Time
Popu
latio
n siz
e
Figure 3 When 1198770
ge 1 the alcoholism equilibrium 119864lowast
=
(4466 476 32 26) corresponding to the given parameters is globallyasymptotically stable (120573 = 03 120575 = 02 and 119877
0= 108511)
the values of optimal controls obtained lastly The iterationscontinue until convergence takes place
In the simulations we choose the available variable valuesas Table 1 shows besides 120573 = 03 120575 = 02 The initial valueof model (1) is assumed to be 119878(0) = 4500 119860(0) = 1000119879(0) = 300 and 119876(0) = 200 as before
The ideal weights in objective functional are very difficultto obtain in reality it needs much work on data mining andfittingHence the acquisition of appropriate practical weights
Table 1 The parameters description of model (1)
Parameter Description Values120583 Natural birth rate or death rate 025
120573Transmission coefficient betweenalcoholism and susceptibles Variables
119873 The total populations to be considered 6000
120575The rate of populations quitting fromalcoholism permanently after treatment Variables
120585The rate of populations failed intreatment and returned to be alcoholic 04
0
02
04
06
08
1
0
05
1
0 02 04 06 08 1
u1
u2
R0
Figure 4 The relationship between 1198770and two control variables 119906
1
and 1199062
is still a difficult problem and remains for further investiga-tionsThe cost associated with119860(119905) and 119906
1(119905)mainly includes
the cost of dangerous behaviours during the alcoholism timeand educating the public while the cost associated with1199062(119905)mainly comes from health professional and the medical
resource includingmedicines andnursing care In viewof thisand taking the expressions of 119906
1 1199062into account after many
numerical simulations we finally give weighting coefficientsas 1198881= 102 1198882= 104 It should be pointed out that the
weights here are of only theoretical interest to reveal thecontrol strategies proposed in this paper Another point tonote is that themaximumcontrol is very difficult to achieve inreality so we will omit the situation of the maximum controlduring the series of simulations
Next we will make some necessary instructions andexplanations to the above simulation graphs Figures 5 67 and 8 depict the number of four compartments underdifferent control levels when we choose the weight coeffi-cients in objective function to be 119888
1= 102 1198882= 104 From
the four simulation graphs we can observe the followingsimple facts in reducing the total number of alcoholismsand increasing the number of susceptibles the effectivenessof various control measures is as follows optimal control isevidently better than middle control and middle control isbetter than single control 119906
2 single control 119906
2is better than
single control 1199061 while single control 119906
1is much better than
no controlFigures 9 and 10 depict the optimal control law of 119906
1 1199062
respectively In the beginning of the simulation the control
14 Abstract and Applied Analysis
0 10 20 30 40 504500
5000
5500
6000
6500
7000
S(n
umbe
r of s
usce
ptib
les)
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 5Number of the susceptibles whenwe choose theweights inobjective function are 119888
1= 102 1198882= 104 and under different optimal
control strategies that is (1) without control 1199061= 1199062= 0 (2) single
control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0 119906
2= 05 (4)
middle control 1199061= 1199062= 05 (5) optimal control 119906
1= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
500
1000
1500
A(n
umbe
r of t
he al
coho
lism
)
t (unit week)
Figure 6 Number of the alcoholics when we choose the weights inobjective function are 119888
1= 102 1198882= 104 and under different optimal
control strategies that is (1) without control 1199061= 1199062= 0 (2) single
control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0 119906
2= 05 (4)
middle control 1199061= 1199062= 05 (5) optimal control 119906
1= 119906lowast
1 1199062= 119906lowast
2
0
50
100
150
200
250
300
350
400
T(n
umbe
r of t
reat
men
t)
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 7 Number of the persons in treatment when we choose theweights in objective function are 119888
1= 10
2 1198882= 10
4 and underdifferent optimal control strategies that is (1) without control 119906
1=
1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control 119906
1= 0
1199062= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal control
1199061= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
50
100
150
200
250
300
350
400
Q(n
umbe
r of q
uitti
ng)
Figure 8 Number of people in quitting compartment when wechoose the weights in objective function are 119888
1= 102 1198882= 104 and
under different optimal control strategies that is (1) without control1199061= 1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control
1199061= 0 119906
2= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal
control 1199061= 119906lowast
1 1199062= 119906lowast
2
Abstract and Applied Analysis 15
0 10 20 30 40 50t (unit week)
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
Optimal control of u1
Figure 9 Figures of the optimal control 1199061when we choose the
weight in objective function are 1198881= 102 1198882= 104
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
0 10 20 30 40 50t (unit week)
Optimal control of u2
Figure 10 Figures of the optimal control 1199062when we choose the
weight in objective function are 1198881= 102 1198882= 104
effort of 1199061should be decreased from 065 to 05 within the
first month and over the next week it should be increased tothe maximum control until 50 weeks then rapidly decreasedto 0 at the end of the simulation As for the control 119906
2 it
should start from around 05 due to the initial alcoholics thenincrease to 055 within one week since the rapid infection andnext decrease to almost 0 since the effectiveness of treatmentin three weeks but with the infection going on the controleffort of 119906
2should gradually increase to the maximum and
maintain this level until the tenth week for the purpose ofconsolidation therapy and preventing rebound hereafter itshould be gradually decreased to the level of almost 043 untilthe fifty weeks then quickly decreased to 0 in the end
0 10 20 30 40 504500
5000
5500
6000
6500
7000
S(n
umbe
r of s
usce
ptib
les)
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 11 Number of the susceptibles when we choose the weightsin objective function are 119888
1= 104 1198882= 102 and under different
optimal control strategies that is (1) without control 1199061= 1199062= 0
(2) single control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0
1199062= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal control
1199061= 119906lowast
1 1199062= 119906lowast
2
In order to investigate the influence of different weightcoefficients in the objective functional on the effect ofcontrolling at the same time for a better comparison we willchange the weight coefficients in objective function into 119888
2=
104 1198881= 102 and we will list the corresponding numerical
simulation results as Figures 11ndash16 showWhen we change the weight coefficients in objective
function into 1198881= 104 1198882= 102 we find that the results
of simulations derived from the graphs are very similar tothe ones before We speculate that the most likely reasons ofthis result are due to three respects one is that the weightcoefficients are not too sensitive in the numerical simulationand another possible reason is that both of the two controlsare important in some sense equivalently importantThe lastbut not themost unlikely reason is that we have not found themost appropriate weight coefficients in the simulation whichis very difficult to find as previously mentioned
7 Conclusions
In this paper we formulate an alcoholics quitting modeland firstly investigate the variation discipline of variouspopulations from the perspective of global stability thenwe propose an objective functional to examine two differentcontrolmeasures (ie prevention and treatment) on the effectof alcohol The basic reproduction number of the model wasderived and the global stability of the two equilibria is givenFrom the expression of the basic reproductionnumber119877
0and
16 Abstract and Applied Analysis
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
500
1000
1500
A(n
umbe
r of t
he al
coho
lism
)
t (unit week)
Figure 12 Number of the alcoholics when we choose the weights inobjective function are 119888
1= 104 1198882= 102 and under different optimal
control strategies that is (1) without control 1199061= 1199062= 0 (2) single
control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0 119906
2= 05 (4)
middle control 1199061= 1199062= 05 (5) optimal control 119906
1= 119906lowast
1 1199062= 119906lowast
2
0
50
100
150
200
250
300
350
400
T(n
umbe
r of t
reat
men
t)
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 13 Number of the persons in treatment when we choosethe weights in objective function are 119888
1= 104 1198882= 102 and under
different optimal control strategies that is (1) without control 1199061=
1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control 119906
1= 0
1199062= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal control
1199061= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
50
100
150
200
250
300
350
400
Q(n
umbe
r of q
uitti
ng)
Figure 14 Number of people in quitting compartment when wechoose the weights in objective function are 119888
1= 104 1198882= 102 and
under different optimal control strategies that is (1) without control1199061= 1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control
1199061= 0 119906
2= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal
control 1199061= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50t (unit week)
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
Optimal control of u1
Figure 15 Figures of the optimal control 1199061when we choose the
weight in objective function are 1198881= 104 1198882= 102
related numerical simulation we can easily see that the twocontrol strategies are effective in the alcoholics process
Using Pontryaginrsquos Maximum Principle we firstly deter-mine the necessary conditions for existence of optimalcontrol pairsThe uniqueness of the solution to the optimalitysystem (71) is derived by the classical method of contradic-tion Numerical simulations of the model suggest that thetwo different groups of weights in the objective function have
Abstract and Applied Analysis 17
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
0 10 20 30 40 50t (unit week)
Optimal control of u2
Figure 16 Figures of the optimal control 1199062when we choose the
weight in objective function are 1198881= 104 1198882= 102
much similar effects on the transmission of the alcoholismfrom this point the two control measures are almost equallyimportant in controlling the alcoholism although they willprobably have great influences on the cost of the objectivefunction From the simulation figures it seems that the effectof optimal control which is measured by the reduction inthe number of alcoholics and the increase in the number ofsusceptibles is much better than other control strategies asnoted earlier in the simulation section According to the real-time curve of two optimal controls we point out the specificimplementation methods of optimal control which can beachieved in practice
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
This work was partially supported by the NSF ofGansu Province of China (2013GS09485) the NNSF ofChina (10961018) the NSF of Gansu Province of China(1107RJZA088) the NSF for Distinguished Young Scholarsof Gansu Province of China (1111RJDA003) the Special Fundfor the Basic Requirements in the Research of University ofGansu Province of China and the Development Programfor Hong Liu Distinguished Young Scholars in LanzhouUniversity of Technology
References
[1] R Room T Babor and J Rehm ldquoAlcohol and public healthrdquoThe Lancet vol 365 no 9458 pp 519ndash530 2005
[2] J B Saunders O G Aasland A Amundsen and M GrantldquoAlcohol consumption and related problems among primary
health care patients WHO collaborative project on early detec-tion of personswith harmful alcohol consumption IrdquoAddictionvol 88 no 3 pp 349ndash362 1993
[3] L D Johnston P M OMalley and J G Bachman NationalSurvey Results on Drug Use From the Monitoring the FutureStudy 1975ndash1992 National Institute on Drug Abuse RockvilleMd USA 1993
[4] H Wechsler J E Lee M Kuo and H Lee ldquoCollege bingedrinking in the 1990s a continuing problem Results of theHarvard School of Public Health 1999 College Alcohol StudyrdquoJournal of American College Health vol 48 no 5 pp 199ndash2102000
[5] P M OrsquoMalley and L D Johnston ldquoEpidemiology of alcoholand other drug use among American college studentsrdquo Journalof Studies on Alcohol vol 63 no 14 pp 23ndash39 2002
[6] K Agarwal-Kozlowski and D P Agarwal ldquoGenetic predisposi-tion to alcoholismrdquo Therapeutische Umschau vol 57 no 4 pp179ndash184 2000
[7] C Y Chen C L Storr and J C Anthony ldquoEarly-onset drug useand risk for drug dependence problemsrdquo Addictive Behaviorsvol 34 no 3 pp 319ndash322 2009
[8] M Glavas and J Weinberg ldquoStress alcohol consumption andthe hypothalamic-pituitary adrenal axisrdquo in Nutrients Stressand Medical Disorders pp 165ndash183 Humana Press New YorkNY USA 2006
[9] L N Blum N H Nielsen J A Riggs and L B BresolinldquoAlcoholism and alcohol abuse among women report of theCouncil on Scientific Affairsrdquo Journal of Womenrsquos Health vol7 no 7 pp 861ndash871 1998
[10] B Benedict ldquoModeling alcoholism as a contagious disease howldquoinfectedrdquo drinking buddies spread problem drinkingrdquo SIAMNews vol 40 no 3 2007
[11] H Walter K Gutierrez K Ramskogler I Hertling A Dvorakand O M Lesch ldquoGender-specific differences in alcoholismImplications for treatmentrdquo Archives of Womenrsquos Mental Healthvol 6 no 4 pp 253ndash258 2003
[12] D Muller R D Koch H von Specht w Volker and E MMunch ldquoNeurophisiologic findings in chronic alcohol abuserdquoPsychiatrie Neurologie und Medizinische Psychologie vol 37no 3 pp 129ndash132 1985
[13] G Testino ldquoAlcoholic diseases in hepato-gastroenterology apoint of viewrdquo Hepato-Gastroenterology vol 55 no 82-83 pp371ndash377 2008
[14] A Mubayi P E Greenwood C Castillo-Chavez P J Grue-newald and D M Gorman ldquoThe impact of relative residencetimes on the distribution of heavy drinkers in highly distinctenvironmentsrdquo Socio-Economic Planning Sciences vol 44 no 1pp 45ndash56 2010
[15] D Muller R D Koch H von Specht W Volker and E MMunch ldquoNeurophysiologic findings in chronic alcohol abuserdquoPsychiatrie Neurologie Und Medizinische Psychologie vol 37no 3 pp 129ndash132 1985 (Leipz)
[16] G Mulone and B Straughan ldquoModeling binge drinkingrdquoInternational Journal of Biomathematics vol 5 no 1 Article ID1250005 14 pages 2012
[17] H F Huo and N N Song ldquoGlobal stability for a binge drinkingmodel with two stagesrdquo Discrete Dynamics in Nature andSociety vol 2012 Article ID 829386 15 pages 2012
[18] S S Mushayabasa and C P Bhunu ldquoModelling the effects ofheavy alcohol consumption on the transmission dynamics ofgonorrheardquo Nonlinear Dynamics vol 66 no 4 pp 695ndash7062011
18 Abstract and Applied Analysis
[19] F S Sancheznchez XWang C Castillo-Chvez DM Gormanand P J Gruenewald ldquoDrinking as an epidemic a simplemathematical model with recovery and relapserdquo in TherapistsGuide to Evidence-Based Relapse Prevention K Witkiewitz andG Marlatt Eds Academic Press New York NY USA 2007
[20] H-F Huo and C-C Zhu ldquoInfluence of relapse in a givingup smoking modelrdquo Abstract and Applied Analysis vol 2013Article ID 525461 12 pages 2013
[21] H F Huo and Q Wang ldquoThe effects of awareness programs by17 media on the drinking dynamicsrdquo Submitted
[22] S Lee E Jung and C Castillo-Chavez ldquoOptimal controlintervention strategies in low- and high-risk problem drinkingpopulationsrdquo Socio-Economic Planning Sciences vol 44 no 4pp 258ndash265 2010
[23] K R Fister S Lenhart and J S McNally ldquoOptimizingchemotherapy in an HIV modelrdquo Electronic Journal of Differ-ential Equations vol 1998 no 32 pp 1ndash12 1998
[24] S Lenhart and J T Workman Optimal Control Applied toBiological Models Chapman amp HallCRC Mathematical andComputational Biology Series Chapman amp HallCRC BocaRaton Fla USA 2007
[25] X Yang L Chen and J Chen ldquoPermanence and positiveperiodic solution for the single-species nonautonomous delaydiffusive modelsrdquo Computers amp Mathematics with ApplicationsAn International Journal vol 32 no 4 pp 109ndash116 1996
[26] P van denDriessche and JWatmough ldquoReproduction numbersand sub-threshold endemic equilibria for compartmental mod-els of disease transmissionrdquoMathematical Biosciences vol 180pp 29ndash48 2002
[27] V Lakshmikantham S Leela and A A Martynyuk StabilityAnalysis of Nonlinear Systems vol 125 Marcel Dekker NewYork NY USA 1989
[28] J P LaSalle The Stability of Dynamical Systems vol 25 ofRegional Conference Series in Applied Mathematics Society forIndustrial and Applied Mathematics 1976
[29] W H Fleming and R W Rishel Deterministic and StochasticOptimal Control vol 1 ofApplications of Mathematics SpringerBerlin Germany 1975
[30] D L Lukes Differential Equations Classical to ControlledMathematics in Science and Engineering Academic Press NewYork NY USA 1982
[31] L S Pontryagin V G Boltyanskii R V Gamkrelidze and EF Mishchenko The Mathematical Theory of Optimal ProcessesJohn Wiley amp Sons New Jersey NJ USA 1962
[32] W Hackbusch ldquoA numerical method for solving parabolicequations with opposite orientationsrdquo Computing vol 20 no3 pp 229ndash240 1978
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
2 Abstract and Applied Analysis
infectivity drunk states mathematical analyses establishedthat the global dynamics of themodel were determined by thebasic reproduction number In [17] the authors modified themodel from [16] that is they considered different infectivityof two drunk states and a SEIR-type model of alcoholismwas thus presented in which two alcohol related states wereinvolved namely no alcohol dependent consumers 119863(119905)and alcohol dependent consumers 119860(119905) In [18] the authorsformulated a deterministic model for evaluating the impactof heavy alcohol drinking on the reemerging gonorrheaepidemic and both analytical and numerical results wereprovided to ascertain whether heavy alcohol drinking hadan impact on the transmission dynamics of gonorrhea Theapproach of the literature [18] was very meaningful sinceit provided a new direction of thinking when the cross-infection between alcoholism and other pathological diseasesoccurs In recent monograph [19] the authors also proposeda SIR-type model to investigate alcohol abuse phenomenonand generated some useful insights for example the basicreproductive number was not always the key to controllingdrinking within the population For other papers that studythe model of giving up smoking or quitting drinking pleasesee [20 21] and references cited therein
As living standard and health awareness get improvedmore and more people who fall into binge drinking stateare actively seeking the quitting alcoholism measures andtreatment methods [1 11 22] In [22] treatment strategy wasintroduced into a simple SIR-type alcoholics quitting modelin which the authors used the bilinear incidence to depict theldquoinfectionrdquo between the occasional drinkers 119878 and problemdrinkers 119863 Motivated by some aforementioned documents[10 19 22] in this paper we will formulate a more reasonablealcoholics quitting model The fact that our model is morereasonable is embodied from the following three aspects
(1) Taking into account that alcoholism is a widespreadsocial phenomenon so the standard incidence issuperior to bilinear incidence when we portray therelationship between the alcoholism and the suscep-tibles during the course of infection While in [22]the authors adopted bilinear incidence we will adoptstandard incidence in this paper
(2) Since alcohol is harmful to health moreover aswe all know alcoholism is treatable if we can takeapproximate measure in time for example artificialisolation from alcoholisms medications persuasionand education programing on alcoholism So it isnecessary to take effective measures to avoid alcoholor to treat after alcoholism Documents [10 19] havenot considered these aspects
(3) Since there is effective prevention and treatment indescribing the phenomenon of alcoholism there aresome people who will never drink due to successfulprevention or some people who no longer drink aftersuccessful treatment Therefore when we formulatethe model in this paper itrsquos reasonable to introducea new compartment119876 the people in which will neverdrink for ever Obviously themodels of [10 19 22] arenot involving the quitting compartment 119876
Based on the above considerations we will premeditatetwo treating methods namely prevention of susceptiblesfrom alcoholism and treatment on alcoholism as controlvariables hence we will derive a SATQ-type model Wenote the fact that many authors are interested in solvingoptimal control problems such as cost minimization andoptimal control of various disease especially with biologicalbackground and various mathematical models [22ndash24] Inthis paper we will propose an objective functional whichconsiders not only alcohol quitting effects but also the costof controlling alcohol Then we consider a range of issuesrelated to the optimal control with themethod of PontryaginrsquosMaximum Principle including optimal control existenceuniqueness and characterization
The organization of this paper is as follows In thenext section the alcoholism model with prevention for thesusceptibles and treatment for alcoholism is formulated InSection 3 the basic reproduction number and the existence ofequilibria are investigatedThe stability of the disease free andendemic equilibria is proved in Section 4 Optimal controlstrategies by the classic method of PMP (Pontryaginrsquos Maxi-mum Principle) are discussed in Section 5 In Section 6 wegive some numerical simulations We give some discussionsand conclusions in the last section
2 The Model Formulation and SomeFundamental Properties
In this section we introduce a mathematical model withprevention and treatment for the alcoholism and then studysome important properties such as the boundness and posi-tivity of its solutions
21 Model Formulation and Parameter Explanation Thetotal population is partitioned into four compartments thesusceptible compartment 119878 which refers to the personswho never drink or drink moderately without affecting thephysical health the alcoholism compartment 119860 which refersto the persons who binge drink and affect the physical healthseriously the treatment compartment 119879 which refers to thepersons who have been receiving treatments by taking pills orothermedical interventions after alcoholism and the quittingcompartment 119876 which refers to the persons who recoverfrom alcoholism after treatment and stay off alcohol hereafterIn this paper we focus on a closed environment such as acommunity a university or a village So the total number ofpopulation to be considered is a constant we denote it as119873The population flow among those compartments is shown inthe following diagram (Figure 1)
The schematic diagram leads to the following system ofordinary differential equations
1198781015840= 120583119873 minus (1 minus 119906
1)120573119878119860
119873minus 120583119878
1198601015840= (1 minus 119906
1)120573119878119860
119873+ 120585119879 minus (119906
2+ 120583)119860
Abstract and Applied Analysis 3
A
S Q
T
120583N
120583S
120585T
120583T
120583A
120583Q
u2A
120575T
(1 minus u1)120573SA
N
Figure 1 Transfer diagram for the dynamics of alcoholism model
1198791015840= 1199062119860 minus (120583 + 120585 + 120575) 119879
1198761015840= 120575119879 minus 120583119876
(1)
Here 120583119873 is the birth number of the population 120583 is thenatural death rate of the population 119906
1is the fraction of
the susceptible individuals who successfully avoid to stayoff the alcoholism 119906
2is the fraction of the alcoholics who
take part in treatments here 0 le 119906119894le 1 119894 = 1 2 and
they will be considered as two control variables in Section 5120573 is the transmission coefficient of the ldquoinfectionrdquo for thesusceptible individuals from the alcoholic individuals 120585 is therate coefficient of the person who fail to be treated and returnto the alcoholism compartmentmostly due to their ownweakwill 120575 is the rate coefficient of the person who have receivedeffective treatment and recovered from alcoholism forever
22 Boundedness of Solutions to System and Positively Invari-ant Region It is important to show positivity and bounded-ness for the system (1) as they represent populations Firstlywe present the positivity of the solutions System (1) can beput into the matrix form
1198831015840= 119866 (119883) (2)
where119883 = (119878 119860 119879 119876)119879isin 1198774 and 119866(119883) is given by
119866 (119883) = (
1198661(119883)
1198662(119883)
1198663(119883)
1198664(119883)
)
=(
120583119873 minus (1 minus 1199061)120573119878119860
119873minus 120583119878
(1 minus 1199061)120573119878119860
119873+ 120585119879 minus (119906
2+ 120583)119860
1199062119860 minus (120583 + 120585 + 120575) 119879
120575119879 minus 120583119876
)
(3)
It is easy to check that
119866119894(119883)
1003816100381610038161003816119883119894(119905)=0119883119905isin119862+ge 0 119894 = 1 2 3 4 (4)
Due to Lemma 2 in [25] any solution of (1) is 119883(119905) isin 1198774+for
all 119905 ge 0We denote 119873(119905) = 119878(119905) + 119860(119905) + 119879(119905) + 119876(119905) summing
equations in (1) yields
119889119873 (119905)
119889119905= 0 (5)
so 119873(119905) = 119878(119905) + 119860(119905) + 119879(119905) + 119876(119905) = constant (denoted as119873) and the set
Ω = (119878 119860 119879 119876) isin 1198774
+ 119878 + 119860 + 119879 + 119876 le 119873 (6)
is a positively invariant region for (1) Therefore we willconsider the global stability of (1) on the setΩ
3 The Basic Reproduction Number andExistence of Alcoholism Equilibria
31 The Basic Reproduction Number 1198770 In epidemiology
the basic reproduction number (sometimes called basicreproductive rate or basic reproductive ratio) of an infectionis the number of infectious cases that one infectious casegenerates on average over the course of its infectious periodWhile in this context it means the number of persons thatan alcoholic will ldquoinfectrdquo during his ldquoinfectiousrdquo period in thepure susceptible environment so that the infected personswillenter the alcoholism compartment It is easy to see that themodel has an alcohol free equilibrium 119864
0= (1198780 0 0 0) =
(119873 0 0 0) In the following the basic reproduction numberof system (1) will be obtained by the next generation matrixmethod formulated in [26]
Let 119909 = (119860 119879 119876 119878)119879 then system (1) can be written as
119889119909
119889119905= F (119909) minusV (119909) (7)
where
F (119909) = (
(1 minus 1199061)120573119878119860
1198730
0
0
)
V (119909) = (
(1199062+ 120583)119860 minus 120585119879
(120583 + 120585 + 120575) 119879 minus 1199062119860
120583119876 minus 120575119879
(1 minus 1199061)120573119878119860
119873+ 120583119878 minus 120583119873
)
(8)
The Jacobian matrices of F(119909) and V(119909) at the alcohol freeequilibrium 119864
0are respectively
119863F (1198640) = (
1198653times3
0
0 0)
119863V (1198640) = (
1198813times3
0
(1 minus 1199061) 120573 0 0 120583
)
(9)
where
119865 = (
(1 minus 1199061) 120573 0 0
0 0 0
0 0 0
) 119881 = (
1199062+ 120583 minus120585 0
minus1199062
120583 + 120585 + 120575 0
0 minus120575 120583
)
(10)
The basic reproduction number denoted by 1198770 is thus given
by
1198770= 120588 (119865119881
minus1) =
120573 (1 minus 1199061) (120583 + 120585 + 120575)
1199062(120583 + 120575) + 120583 (120583 + 120585 + 120575)
(11)
4 Abstract and Applied Analysis
It is easy to see that both of the control parameters con-tributed to reducing the alcoholism From this point thecontrol measures are meaningful
32 Existence of Alcoholism Equilibrium The endemic equi-librium 119864
lowast(119878lowast 119860lowast 119879lowast 119876lowast) of system (1) is determined by
equations
120583119873 minus (1 minus 1199061)120573119878119860
119873minus 120583119878 = 0
(1 minus 1199061)120573119878119860
119873+ 120585119879 minus (119906
2+ 120583)119860 = 0
1199062119860 minus (120583 + 120585 + 120575) 119879 = 0
120575119879 minus 120583119876 = 0
(12)
The third equation in (12) leads to
119860 =120583 + 120585 + 120575
1199062
119879 (13)
From the last equation in (12) we have
119876 =120575119879
120583 (14)
From the first equation of (12) and together with (13) we canget
119878 =120583119873
120583 + (((1 minus 1199061) 120573) 119873)119860
=12058311987321199062
1205831198731199062+ (1 minus 119906
1) 120573119879 (120583 + 120585 + 120575)
(15)
Substituting (13)ndash(15) into the second equation of (12) gives
120583119873 minus120583211987321199062
1205831198731199062+ (1 minus 119906
1) 120573119879 (120583 + 120585 + 120575)
+ 120585119879 minus (120583 + 1199062)120583 + 120585 + 120575
1199062
119879 = 0
(16)
By simplifying (16) we can get
119879 [1199062120585 (1 minus 119906
1) 120573 (120583 + 120585 + 120575)
minus (1 minus 1199061) 120573(120583 + 120585 + 120575)
2
(120583 + 1199062)] 119879 + 120590 = 0
(17)
where120590 = 119906
2120583119873 (1 minus 119906
1) 120573 (120583 + 120585 + 120575)
+ (1199062)2
120583119873120585 minus 1205831198731199062(120583 + 119906
2) (120583 + 120585 + 120575)
(18)
Hence we get two explicit solutions to (17) one is 1198790= 0
which is corresponding to the alcohol free equilibria and theother is119879lowast= (120590) ((1 minus 119906
1) 120573(120583 + 120585 + 120575)
2
(120583 + 1199062)
minus1199062120585 (1 minus 119906
1) 120573 (120583 + 120585 + 120575) )
minus1
=120590
(1 minus 1199061) 120573 (120583 + 120585 + 120575) [120583120585 + (120575 + 120583) (120583 + 119906
2)]
(19)
which should be corresponding to the alcoholism equilibriaon condition that 119879lowast gt 0 otherwise the alcoholismequilibria are nonexistent It is enough to show the positivityof 120590 tomake sure the existence of alcoholism equilibria on thecondition 119877
0ge 1 By some simple calculations we simplify
the expression of 120590 to be
120590 = 1205831198731199062(120583 + 120585 + 120575) (1 minus 119906
1) 120573
minus [(120583 + 1199062) (120575 + 120583) + 120583120585]
(20)
Since 1198770gt 1 is equivalent to
120573 (1 minus 1199061) (120583 + 120585 + 120575) gt 119906
2(120583 + 120575) + 120583 (120583 + 120585 + 120575) (21)
the right side of this inequality is exactly equal to (120583+1199062)(120575+
120583) + 120583120585 Hence we have proved the existence of 119879lowast gt 0so are the alcoholism equilibria We summarize this result inTheorem 1
Theorem 1 For system (1) there is always an alcohol freeequilibrium 119864
0= (119873 0 0 0) When 119877
0gt 1 besides alcohol
free equilibrium 1198640 system (1) also has a unique alcoholism
equilibrium 119864lowast(119878lowast 119860lowast 119879lowast 119876lowast) where
119878lowast=
12058311987321199062
1205831198731199062+ (1 minus 119906
1) 120573119879lowast (120583 + 120585 + 120575)
119860lowast=120583 + 120585 + 120575
1199062
119879lowast
119876lowast=120575119879lowast
120583
119879lowast=
120590
(1 minus 1199061) 120573 (120583 + 120585 + 120575) [120583120585 + (120575 + 120583) (120583 + 119906
2)]
(22)
4 Stability Analysis of Equilibria
For the convenience of subsequent proof we denote a vector119883 = (119860 119879 119876 119878)
119879 and
119891 (119883) =(
(1 minus 1199061)120573119878119860
119873+ 120585119879 minus (119906
2+ 120583)119860
1199062119860 minus (120583 + 120585 + 120575) 119879
120575119879 minus 120583119876
120583119873 minus (1 minus 1199061)120573119878119860
119873minus 120583119878
) (23)
So the Jacobian matrix of 119891(119909) about vector 119883 is as thefollowing
119869 =
120597119891 (119883)
120597119883
=(
(1 minus 1199061) 120573119878
119873
minus (120583 + 1199062) 120585 0
(1 minus 1199061) 120573119860
119873
1199062 minus (120583 + 120585 + 120575) 0 0
0 120575 minus120583 0
minus
(1 minus 1199061) 120573119878
119873
0 0 minus120583 minus
(1 minus 1199061) 120573119860
119873
)
(24)
Theorem 2 For system (1) the alcohol free equilibrium 1198640is
locally asymptotically stable if 1198770lt 1
Abstract and Applied Analysis 5
Proof Since
119869 (1198640)
= (
(1 minus 1199061) 120573 minus (120583 + 119906
2) 120585 0 0
1199062
minus (120583 + 120585 + 120575) 0 0
0 120575 minus120583 0
minus (1 minus 1199061) 120573 0 0 minus120583
)
(25)
we can easily get that two of the eigenvalues are 1205821= 1205822=
minus120583 lt 0 while 1205823 1205824satisfy
1205822+ [2120583 + 120585 + 120575 + 119906
2minus (1 minus 119906
1) 120573] 120582
+ (120583 + 120585 + 120575) (120583 + 1199062minus (1 minus 119906
1) 120573) minus 119906
2120585 = 0
(26)
Thus
1205823+ 1205824= (1 minus 119906
1) 120573 minus (120583 + 120585 + 120575) minus (120583 + 119906
2) (27)
Since 1198770lt 1 is equivalent to
120573 (1 minus 1199061) (120583 + 120585 + 120575) lt 119906
2(120583 + 120575) + 120583 (120583 + 120585 + 120575)
lt (120583 + 1199062) (120583 + 120585 + 120575)
(28)
so
120573 (1 minus 1199061) lt 120583 + 119906
2 (29)
and then
1205823+ 1205824lt minus (120583 + 120585 + 120575) lt 0 (30)
while
12058231205824= (120583 + 120585 + 120575) (120583 + 119906
2minus (1 minus 119906
1) 120573) minus 119906
2120585 (31)
Similarly from 1198770lt 1 we can derive the inequality
minus120573 (1 minus 1199061) (120583 + 120585 + 120575) gt minus119906
2(120583 + 120575) minus 120583 (120583 + 120585 + 120575) (32)
so12058231205824gt (120583 + 119906
2) (120583 + 120585 + 120575)
minus 1199062120585 minus 1199062(120583 + 120575) minus 120583 (120583 + 120585 + 120575)
(33)
It reduces to
12058231205824gt 0 (34)
Hence Re 1205823lt 0 Re 120582
4lt 0 The proof is complete
Next we will turn to investigate the global stability of 1198640
Theorem 3 For system (1) the alcohol free equilibrium 1198640is
globally asymptotically stable if 1198770lt 1
Proof Consider the subsystem of (1) as follows
1198601015840= (1 minus 119906
1)120573119878119860
119873+ 120585119879 minus (119906
2+ 120583)119860
1198791015840= 1199062119860 minus (120583 + 120585 + 120575) 119879
1198761015840= 120575119879 minus 120583119876
(35)
Equation (35) can be rewritten as
(
) = (119865 minus 119881)(
119860
119879
119876
) minus (1 minus119878
119873)
times (
120573 (1 minus 1199061) 0 0
0 0 0
0 0 0
)(
119860
119879
119876
)
(36)
Since 119878 le 119873 and 0 le 1199061le 1 then for all 119905 gt 0 we can get
(
) le (119865 minus 119881)(
119860
119879
119876
) (37)
According to Lemma 1 in [26] all the eigenvalues of matrix119865 minus 119881 have negative real parts so the solutions of this sub-system are stable whenever 119877
0lt 1 So (119860(119905) 119879(119905) 119876(119905)) rarr
(0 0 0) as 119905 rarr infin By the comparison theorem [27] andbased on the fact that the total population is constant 119873 itfollows that (119860(119905) 119879(119905) 119876(119905)) rarr (0 0 0) and 119878(119905) rarr 119873
as 119905 rarr infin So the alcohol free equilibrium 1198640is globally
asymptotically stable the proof is complete
Theorem 4 For system (1) the alcoholism equilibrium119864lowast(119878lowast 119860lowast 119879lowast 119876lowast) is globally asymptotically stable if 119877
0gt 1
Proof Since the total population in model (1) is a constantnumber119873 in order to prove the global stability of system (1)it is sufficed to prove the corresponding stability of subsystem(38)
1198781015840= 120583119873 minus (1 minus 119906
1)120573119878119860
119873minus 120583119878
1198601015840= (1 minus 119906
1)120573119878119860
119873+ 120585119879 minus (119906
2+ 120583)119860
1198791015840= 1199062119860 minus (120583 + 120585 + 120575) 119879
(38)
We make normalization transform and still use the samesymbols 119878 119860 119879 to denote the variables then (38) can betransformed into
1198781015840= 120583 minus (1 minus 119906
1) 120573119878119860 minus 120583119878
1199041198601015840= (1 minus 119906
1) 120573119878119860 + 120585119879 minus (119906
2+ 120583)119860
1198791015840= 1199062119860 minus (120583 + 120585 + 120575) 119879
(39)
From (39) we can easily know that the equilibria (119878lowast 119860lowast 119879lowast)satisfy the following three equalities to be used later
120583 = (1 minus 1199061) 120573119878lowast119860lowastminus 120583119878lowast
(1 minus 1199061) 120573119878lowast119860lowast+ 120585119879lowast= (1199062+ 120583)119860
lowast
1199062119860lowast= (120583 + 120585 + 120575) 119879
lowast
(40)
6 Abstract and Applied Analysis
Let 119881 = 1199091(119878 minus 119878
lowastminus 119878lowast ln(119878119878lowast)) + 119909
2(119860 minus 119860
lowastminus
119860lowast ln(119860119860lowast)) + 119909
3(119879 minus 119879
lowastminus 119879lowast ln(119879119879lowast)) then
119881101584010038161003816100381610038161003816(39)
= 1199091[120583 minus (1 minus 119906
1) 120573119878119860 minus 120583119878
minus119878lowast
119878120583 + (1 minus 119906
1) 120573119878lowast119860 + 120583119878
lowast]
+ 1199092[ (1 minus 119906
1) 120573119878119860 + 120585119879 minus (119906
2+ 120583)119860
minus (1 minus 1199061) 120573119878119860
lowastminus119860lowast
119860120585119879 + (119906
2+ 120583)119860
lowast]
+ 1199093[1199062119860 minus (120583 + 120585 + 120575) 119879
minus119879lowast
1198791199062119860 + (120583 + 120585 + 120575) 119879
lowast]
= 1199091[ (1 minus 119906
1) 120573119878lowast119860lowast+ 120583119878lowastminus (1 minus 119906
1) 120573119878119860 minus 120583119878
minus119878lowast
119878((1 minus 119906
1) 120573119878lowast119860lowast+ 120583119878lowast)
+ (1 minus 1199061) 120573119878lowast119860 + 120583119878
lowast]
+ 1199092[ (1 minus 119906
1) 120573119878119860 + 120585119879 minus (119906
2+ 120583)119860
minus (1 minus 1199061) 120573119878119860
lowastminus119860lowast
119860120585119879 + (119906
2+ 120583)119860
lowast]
+ 1199093[1199062119860 minus (120583 + 120585 + 120575) 119879
minus119879lowast
1198791199062119860 + (120583 + 120585 + 120575) 119879
lowast]
= 1199091120583119878lowast(2 minus
119878
119878lowastminus119878lowast
119878)
+ [1199091(1 minus 119906
1) 120573119878lowast119860lowast+ 1199092(1199062+ 120583)119860
lowast
+ 1199093(120583 + 120585 + 120575) 119879
lowast]
minus [1199091
(119878lowast)2
(1 minus 1199061) 120573119860lowast
119878+ 1199092(1 minus 119906
1) 120573119878119860
lowast
+ 1199092
119860lowast120585119879
119860+ 1199093
119879lowast
1198791199062119860]
+ 119878119860 [minus1199091(1 minus 119906
1) 120573 + 119909
2(1 minus 119906
1) 120573]
+ 119860 [minus (1199062+ 120583) 119909
2+ 1199091(1 minus 119906
1) 120573119878lowast+ 11990621199093]
+ 119879 [1205851199092minus (120583 + 120585 + 120575) 119909
3]
(41)
To eliminate the cross-term 119878119860 and two single-variable terms119860 and 119879 we let
minus1199091(1 minus 119906
1) 120573 + 119909
2(1 minus 119906
1) 120573 = 0
minus (1199062+ 120583) 119909
2+ 1199091(1 minus 119906
1) 120573119878lowast+ 11990621199093= 0
1205851199092minus (120583 + 120585 + 120575) 119909
3= 0
(42)
By solving them we can get
1199091= 1 119909
2= 1
1199093=
120585
120583 + 120585 + 120575=1199062+ 120583 minus (1 minus 119906
1) 120573119878lowast
1199062
(43)
Next we let
1198811015840
1= 1199091(1 minus 119906
1) 120573119878lowast119860lowast+ 1199092(1199062+ 120583)119860
lowast
+ 1199093(120583 + 120585 + 120575) 119879
lowast
1198811015840
2= minus[119909
1
(119878lowast)2
(1 minus 1199061) 120573119860lowast
119878
+ 1199092(1 minus 119906
1) 120573119878119860
lowast+ 1199092
119860lowast120585119879
119860+ 1199093
119879lowast
1198791199062119860]
(44)
and then
1198811015840= 1199091120583119878lowast(2 minus
119878
119878lowastminus119878lowast
119878) + 119881
1015840
1+ 1198811015840
2 (45)
Due to
(1199062+ 120583)119860
lowast= (1 minus 119906
1) 120573119878lowast119860lowast+ 120585119879lowast
1199093(120583 + 120585 + 120575) 119879
lowast= 120585119879lowast1199092= 120585119879lowast
(46)
Abstract and Applied Analysis 7
so
1198811015840
1= 2 (1 minus 119906
1) 120573119878lowast119860lowast+ 2120585119879
lowast
1198811015840
2= minus[
(119878lowast)2
(1 minus 1199061) 120573119860lowast
119878
+ (1 minus 1199061) 120573119878119860
lowast+119860lowast120585119879
119860+ 1199093
119879lowast
1198791199062119860]
le minus2[(119878lowast)2
(1 minus 1199061) 120573119860lowast
119878sdot (1 minus 119906
1) 120573119878119860
lowast]
12
minus 2 [119860lowast120585119879
119860sdot 1199093
119879lowast
1198791199062119860]
= minus2 (1 minus 1199061) 120573119878lowast119860lowastminus 2(1199093119860lowast1205851199062119879lowast)12
= minus2 (1 minus 1199061) 120573119878lowast119860lowastminus 2[120585119879
lowast(11990931199062119860lowast)]12
= minus2 (1 minus 1199061) 120573119878lowast119860lowast
minus 2120585119879lowast[(1199062+ 120583)119860
lowastminus (1 minus 119906
1) 120573119878lowast119860lowast]12
= minus2 (1 minus 1199061) 120573119878lowast119860lowastminus 2(120585119879
lowast120585119879lowast)12
= minus2 (1 minus 1199061) 120573119878lowast119860lowastminus 2120585119879
lowast
(47)
Hence
1198811015840= 120583119878lowast(2 minus
119878
119878lowastminus119878lowast
119878) + 119881
1015840
1+ 1198811015840
2
le 120583119878lowast(2 minus
119878
119878lowastminus119878lowast
119878) + 2 (1 minus 119906
1) 120573119878lowast119860lowast
+ 2120585119879lowastminus 2 (1 minus 119906
1) 120573119878lowast119860lowastminus 2120585119879
lowast
= 120583119878lowast(2 minus
119878
119878lowastminus119878lowast
119878) le 0
(48)
1198811015840= 0 if and only if 119878 = 119878
lowast 119860 = 119860lowast 119879 = 119879
lowast Accordingto LaSallersquos invariance principle [28] we can derive theconclusion that the alcoholism equilibria 119864lowast(119878lowast 119860lowast 119879lowast 119876lowast)are globally asymptotically stable the proof is complete
5 Optimal Control Problem
51 The Existence of Optimal Control In order to investigatean effective campaign to control alcoholism in a communitywhich pursue the goals of the minimized alcoholisms andmore recovered individuals we will reconsider the system (1)and use two control variables to reduce the numbers of alco-holics The difference is that we will change the parameters1199061 1199062into control variable 119906
1(119905) 1199062(119905) Their aforementioned
definitions allow us to do so 1199061(119905) is used to limit the
proportion of the susceptible individual to contact withalcoholism usually by propaganda and education so that thesusceptible individual can stay off alcoholism consciously andbe free of ldquoinfectionrdquo we can understand the effect of 119906
1(119905)
is to prevent the the susceptible from contacting with thealcoholism The control variable 119906
2(119905) is used to control the
alcoholism to take appropriate treatment measures such astaking pills or seeking other medical help However just as acoin has two sides there will be a lot of costs generated duringthe control process So it is advisable to balance between thecosts and the alcohol effects In view of this our optimalcontrol problem tominimize the objective functional is givenby
119869 (1199061 1199062) = int
119905119891
0
[119860 (119905) +1198881
21199062
1(119905) +
1198882
21199062
2(119905)] 119889119905 (49)
which subjects to system
1198781015840= 120583119873 minus (1 minus 119906
1(119905))
120573119878119860
119873minus 120583119878
1198601015840= (1 minus 119906
1(119905))
120573119878119860
119873+ 120585119879 minus (119906
2(119905) + 120583)119860
1198791015840= 1199062(119905) 119860 minus (120583 + 120585 + 120575) 119879
1198761015840= 120575119879 minus 120583119876
(50)
with initial conditions
119878 (0) = 1198780 119860 (0) = 119860
0
119879 (0) = 1198790 119876 (0) = 119876
0
(51)
Here 119906119894(119905) isin 119880 ≜ (119906
1 1199062) | 119906119894(119905) is measurable and 0 le
119906119894(119905) le 1 for all 119905 isin [0 119905
119891] 119905119891is the end time to be
controlled 119880 is an admissible control set 119888119894 and 119894 = 1 2 are
weight factors (positive constants) that adjust the intensity oftwo different control measures
Next we will investigate the existence of the optimalcontrol of the above-mentioned problem
Theorem 5 There exists an optimal control pair 119906lowast
=
(119906lowast
1 119906lowast
2) isin 119880 such that
119869 (119906lowast
1 119906lowast
2) = min 119869 (119906
1 1199062) 119906
1(119905) 1199062(119905) isin 119880 (52)
subjects to the control system (1) with initial conditions (50)
Proof Toprove the existence of an optimal control accordingto the classic literature [29] we have to show the following
(1) The control and state variables are nonnegative values(2) The control set 119880 is convex and closed(3) The right side of the state system is bounded by linear
function in the state and control variables(4) The integrand of the objective functional is concave
on 119880(5) There exist constants 119889
1 1198892gt 0 and 120572 gt 1 such
that the integrand 119871(119905 1199061 1199062) ≜ 119860(119905) + (119888
12)1199062
1(119905) +
(11988822)1199062
2(119905) of the objective functional satisfies
119871 (119905 1199061 1199062) ge 1198891(100381610038161003816100381611990611003816100381610038161003816
2
+100381610038161003816100381611990621003816100381610038161003816
2
)1205722
minus 1198892
(53)
8 Abstract and Applied Analysis
statements (1) (2) and (3) are obvious satisfied we only needto test and verify the latter two ones Since the four statevariables have been all proved to be up bounded by 119873 wewill get the following equalities
1198781015840le 120583119873 119860
1015840le (1 minus 119906
1(119905)) 120573119878 + 120585119879
1198791015840le 1199062(119905) 119860 119876
1015840le 120575119879
(54)
so the fourth condition is set up As for the last condition
119871 (119905 1199061 1199062) ge 1198891(100381610038161003816100381611990611003816100381610038161003816
2
+100381610038161003816100381611990621003816100381610038161003816
2
)1205722
minus 1198892
(55)
is also true when we choose 1198891= min119888
12 11988822 and for all
1198892isin 119877+ 120572 = 2 The proof is complete
We next come to the core of this section
52 The Characterization of the Optimal Control With theexistence of the optimal control pairs established we nowpresent the optimality system and use a result from [30]we can easily know the existence of the solutions to theoptimality system (71) which will be gotten later Firstly wecome to discuss the theorem that relates to the character-ization of the optimal control The optimality system canbe used to compute candidates for optimal control pairsTo do this we begin by defining an augmented Hamilto-nian 119867 with penalty terms for the control constraints asfollows
119867 = 119860 (119905) +1198881
21199062
1(119905) +
1198882
21199062
2(119905)
+ 1205821[120583119873 minus (1 minus 119906
1(119905))
120573119878119860
119873minus 120583119878]
+ 1205822[(1 minus 119906
1(119905))
120573119878119860
119873+ 120585119879 minus (119906
2(119905) + 120583)119860]
+ 1205823[1199062(119905) 119860 minus (120583 + 120585 + 120575) 119879] + 120582
4(120575119879 minus 120583119876)
minus 119908111199061(119905) minus 119908
12(1 minus 119906
1(119905))
minus 119908211199062(119905) minus 119908
22(1 minus 119906
2(119905))
(56)
where 119908119894119895(119905) ge 0 are the penalty multipliers satisfying
11990811(119905) 1199061(119905) = 119908
12(119905) (1 minus 119906
1(119905))
= 0 at optimal control 119906lowast1
11990821(119905) 1199062(119905) = 119908
22(119905) (1 minus 119906
2(119905))
= 0 at optimal control 119906lowast2
(57)
Theorem6 Given optimal control pairs (119906lowast1 119906lowast
2) and solutions
119878(119905) 119860(119905) 119879(119905) 119876(119905) of the corresponding state system (50)there exist adjoint variables 120582
119894 119894 = 1 2 3 4 satisfying
1205821015840
1= 1205821(1 minus 119906
1(119905))
120573119860
119873+ 1205831205821minus 1205822(1 minus 119906
1(119905))
120573119860
119873
1205821015840
2= minus1 + 120582
1(1 minus 119906
1(119905))
120573119878
119873minus 1205822(1 minus 119906
1(119905))
120573119878
119873
+ 1205822(120583 + 119906
2(119905)) minus 120582
31199062(119905)
1205821015840
3= minus1205822120585 + 1205823(120583 + 120585 + 120575) minus 120582
4120575
1205821015840
4= 1205831205824
(58)
with the terminal conditions
120582119894(119905119891) = 0 119894 = 1 2 3 4 (59)
Furthermore (119906lowast1 119906lowast
2) are represented by
119906lowast
1= min(1max(0
120573119878119860 (1205822minus 1205821)
1198881119873
))
119906lowast
2= min(1max(0
119860 (1205822minus 1205823)
1198882
))
(60)
Proof According to Pontryagin Maximum Principle [29ndash31] we first differentiate the Hamiltonian operator 119867 withrespect to states Then the adjoint system can be written as
1205821015840
1= minus
120597119867
120597119878 120582
1015840
2= minus
120597119867
120597119860
1205821015840
3= minus
120597119867
120597119879 120582
1015840
4= minus
120597119867
120597119876
(61)
The terminal condition (56) of adjoint equations is given by120582119894(119905119891) = 0 119894 = 1 2 3 4
To obtain the necessary conditions of optimality (59) wealso differentiate theHamiltonian operator119867 with respect to119880 = (119906
1 1199062) and set them equal to zero then
120597119867
1205971199061
= 11988811199061(119905) + 120582
1
120573119878119860
119873minus 1205822
120573119878119860
119873minus 11990811+ 11990812= 0
120597119867
1205971199062
= 11988821199062(119905) minus 120582
2119860 + 120582
3119860 minus 119908
21+ 11990822= 0
(62)
By solving the optimal control we obtain
119906lowast
1=1
1198881
[(1205822minus 1205821)120573119878119860
119873+ 11990811minus 11990812] (63)
To determine an explicit expression for the optimalcontrol without119908
11and119908
12 a standard optimality technique
is utilized [29] We consider the following three cases
(i) On the set 119905 | 0 lt 119906lowast
1(119905) lt 1 we have 119908
11(119905) =
Abstract and Applied Analysis 9
11990812(119905) = 0 Hence the optimal control is
119906lowast
1=1
1198881
[(1205822minus 1205821)120573119878119860
119873] (64)
(ii) On the set 119905 | 119906lowast1(119905) = 1 we have 119908
11(119905) = 0 Hence
1 = 119906lowast
1(119905) =
1
1198881
[(1205822minus 1205821)120573119878119860
119873minus 11990812] (65)
This implies that
1
1198881
(1205822minus 1205821)120573119878119860
119873ge 1 since 119908
12(119905) ge 0 (66)
(iii) On the set 119905 | 119906lowast1(119905) = 0 we have 119908
12(119905) = 0 Hence
0 = 119906lowast
1(119905) =
1
1198881
[(1205822minus 1205821)120573119878119860
119873+ 11990811] (67)
This implies that
1
1198881
(1205822minus 1205821)120573119878119860
119873le 0 since 119908
11(119905) ge 0 (68)
Combining these results the optimal control 119906lowast1(119905) is
characterized as
119906lowast
1= min1max0 1
1198881
[(1205822minus 1205821)120573119878119860
119873] (69)
Using the similar arguments we can also obtain the otheroptimal control function
119906lowast
2= min1max0
(1205822minus 1205823) 119860
1198882
(70)
The proof is complete
We point out that the optimality system consistsof the state system (50) with the initial conditions119878(0) 119860(0) 119879(0) 119876(0) the adjoint (or costate) system(58) with the terminal conditions (59) and the optimalitycondition (60) Any optimal control pairs must satisfythis optimality system For the convenience of subsequent
numerical simulation in Section 6 we give the optimalitysystem as follows
1198781015840= 120583119873 minus (1 minusmin1max0 1
1198881
[(1205822minus 1205821)120573119878119860
119873] (119905))
times120573119878119860
119873minus 120583119878
1198601015840= (1minusmin1max0 1
1198881
[(1205822minus1205821)120573119878119860
119873] (119905))
120573119878119860
119873
+ 120585119879 minus (min1max0(1205822minus 1205823) 119860
1198882
(119905) + 120583)119860
1198791015840= min1max0
(1205822minus 1205823) 119860
1198882
(119905) 119860
minus (120583 + 120585 + 120575) 119879
1198761015840= 120575119879 minus 120583119876
1205821015840
1= 1205821(1 minusmin1max0 1
1198881
[(1205822minus 1205821)120573119878119860
119873] (119905))
times120573119860
119873+ 1205831205821
minus 1205822(1minusmin1max0 1
1198881
[(1205822minus1205821)120573119878119860
119873](119905))
times120573119860
119873
1205821015840
2= minus1
+ 1205821(1minusmin1max0 1
1198881
[(1205822minus1205821)120573119878119860
119873] (119905))
times120573119878
119873minus 1205823sdotmin1max0
(1205822minus 1205823) 119860
1198882
(119905)
minus 1205822(1 minusmin1max0 1
1198881
[(1205822minus 1205821)120573119878119860
119873])
times120573119878
119873+ 1205822(120583+min1max0
(1205822minus 1205823) 119860
1198882
(119905))
1205821015840
3= minus1205822120585 + 1205823(120583 + 120585 + 120575) minus 120582
4120575
1205821015840
4= 1205831205824
119878 (0) = 1198780 119860 (0) = 119860
0 119879 (0) = 119879
0
119876 (0) = 1198760 120582
119894(119905119891) = 0 119894 = 1 2 3 4
(71)
53 The Uniqueness of Optimal Control Due to the a prioriboundedness of the state adjoint functions and the resultingLipschitz structure of the ODEs we can obtain the unique-ness of the optimal control
10 Abstract and Applied Analysis
Lemma 7 (see [23]) The function 119906lowast(119904) = min (119887max (119904 119886))is Lipschitz continuous in 119904 where 119886 lt 119887 are some fixed positiveconstants
Theorem 8 For all 119905 isin [0 119905119891] the solution to the optimality
system (71) is unique
Proof Suppose (119878 119860 119879 119876 1205821 1205822 1205823 1205824) and
(119878 119860 119879 119876 1205821 1205822 1205823 1205824) are two different solutions of
our optimality system (71) Let
119878 = 119890120582119905119898 119860 = 119890
120582119905119899 119879 = 119890
120582119905119901
119876 = 119890120582119905119902 120582
1= 119890minus120582119905119903 120582
2= 119890minus120582119905119904
1205823= 119890minus120582119905119908 120582
4= 119890minus120582119905V
119878 = 119890120582119905119898 119860 = 119890
120582119905119899 119879 = 119890
120582119905119901
119876 = 119890120582119905119902 120582
1= 119890minus120582119905119903 120582
2= 119890minus120582119905119904
1205823= 119890minus120582119905119908 120582
4= 119890minus120582119905V
(72)
where 120582 gt 0 is to be chosenAccordingly we have
119906lowast
1(119905) = min1max0
120573119898119899 (119904 minus 119903) 119890120582119905
1198881119873
119906lowast
2(119905) = min1max0 (119904 minus 119908) 119899
1198882
119906lowast
1(119905) = min1max0
120573119898119899 (119904 minus 119903) 119890120582119905
1198881119873
119906lowast
2(119905) = min1max0 (119904 minus 119908) 119899
1198882
(73)
Now we substitute 119878 = 119890120582119905119898 119860 = 119890
120582119905119899 119879 = 119890
120582119905119901 119876 = 119890
120582119905119902
1205821= 119890minus120582119905119903 1205822= 119890minus120582119905119904 1205823= 119890minus120582119905119908 1205824= 119890minus120582119905V and 119878 =
119890120582119905119898 119860 = 119890
120582119905119899 119879 = 119890
120582119905119901 119876 = 119890
120582119905119902 1205821= 119890minus120582119905119903 1205822=
119890minus120582119905119904 1205823= 119890minus120582119905119908 1205824= 119890minus120582119905V into the first ODE of (71)
respectively then we can obtain
+ 120582119898 = 120583119873119890minus120582119905
minus (1 minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)
times120573119898119899119890
120582119905
119873minus 120583119898
(74)
for119898 and119898 respectively Similarly we can derive
119899 + 120582119899
= (1 minusmin1max1120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)120573119898119899119890
120582119905
119873
+ 120585119901 minus (min1max0 119899 (119904 minus 119908)1198882
+ 120583) 119899
(75)
for 119899 and 119899 respectively
+ 120582119901 = min1max0 119899 (119904 minus 119908)1198882
119899
minus (120583 + 120585 + 120575) 119901
(76)
for 119901 and 119901 respectively
119902 + 120582119902 = 120575119901 minus 120583119902 (77)
for 119902 and 119902 respectively
119903 minus 120582119903 = 119903(1 minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)120573119899119890120582119905
119873
+ 120583119903minus119904(1minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)
times120573119899119890120582119905
119873
(78)
for 119903 and 119903 respectively
119904 minus 120582119904 = minus119890120582119905
+ 119903(1 minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)120573119898119890120582119905
119873
minus 119904(1 minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)
times120573119898119890120582119905
119873+ 119904(min1max0 119899 (119904 minus 119908)
1198882
+ 120583)
minus 119908(min1max0 119899 (119904 minus 119908)1198882
)
(79)
for 119904 and 119904 respectively
minus 120582119908 = minus119904120585 + (120583 + 120585 + 120575)119908 minus 120575V (80)
for 119908 and 119908 respectively
V minus 120582V = 120583V (81)
for V and V respectively
Abstract and Applied Analysis 11
By Lemma 7 we can obtain
1003816100381610038161003816119906lowast
1(119905) minus 119906
lowast
1(119905)1003816100381610038161003816 le
120573119890120582119905
1198881119873|119898119899 (119904 minus 119903) minus 119898119899 (119904 minus 119903)|
1003816100381610038161003816119906lowast
2(119905) minus 119906
lowast
2(119905)1003816100381610038161003816 le
1
1198882
|119899 (119904 minus 119908) minus 119899 (119904 minus 119908)|
(82)
The equations for 119898 119899 119901 119902 119903 119904 119908 V and the equationsfor 119898 119899 119901 119902 119903 119904 119908 V are subtracted respectively then wemultiply each equation by appropriate difference of functionsand integrate from 0 to 119905
119891 Next we add all eight integral
equations and some inequality techniques to obtain unique-ness The following calculation is similar for the sake ofsimplicity we only take119898 and119898 for an example
minus + (120583 + 120582) (119898 minus 119898)
=120573119890120582119905
119873[minus(1 minusmin1max0
120573119898119899 (119904 minus 119903) 119890120582119905
1198881119873
)119898119899
+(1 minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)119898119899]
(83)
Multiplying both sides of (83) by (119898minus119898) and integratingfrom 0 to 119905
119891gives
1
2(119898 minus 119898)
2(119905119891) + (120583 + 120582)int
119905119891
0
(119898 minus 119898)2119889119905
= int
119905119891
0
(119898 minus 119898)120573119890120582119905
119873
times [minus (1 minus 119906lowast
1)119898119899 + (1 minus 119906
lowast
1)119898119899] 119889119905
= int
119905119891
0
(119898 minus 119898)120573119890120582119905
119873[(119906lowast
1minus 1) (119898119899 minus 119898119899 + 119898119899 minus 119898119899)
+ 119898119899 (119906lowast
1minus 119906lowast
1)] 119889119905
le120573119890120582119905119891
119873int
119905119891
0
(119898 minus 119898)
times [119906lowast
1minus 1
100381610038161003816100381610038161003816100381610038161003816
(119898 |119899 minus 119899| + |119898 minus 119898| 119899) + 119898119899120573119890120582119905
1198881119873
100381610038161003816100381610038161003816100381610038161003816
times 119898119899 (119904 minus 119903) minus 119898119899 (119904 minus 119903) ] 119889119905
le120573119890120582119905119891
119873int
119905119891
0
(119898 minus 119898)
times [1003816100381610038161003816119906lowast
1minus 1
1003816100381610038161003816 (119898 |119899 minus 119899| + |119898 minus 119898| 119899) + 119898119899120573119890120582119905
1198881119873
|119898119899 (119904 minus 119904) + (119898119899 minus 119898119899) 119904
minus (119898119899 (119903 minus 119903) + (119898119899 minus 119898119899) 119903)| ] 119889119905
le120573119890120582119905119891
119873int
119905119891
0
[1003816100381610038161003816119906lowast
1minus 1
1003816100381610038161003816 |119898| |119898 minus 119898| |119899 minus 119899|
+ |119898 minus 119898|2
|119899| + |119861| |119904| |119899| |119898 minus 119898|2
+ |119861| |119898119899| |119898 minus 119898| |119903 minus 119903|
+ |119861| |119898119899| |119898 minus 119898| |119904 minus 119904|
+ |119861| |119904| |119898| |119898 minus 119898| |119899 minus 119899|
+ |119861| |119903| |119898| |119898 minus 119898| |119899 minus 119899|
+ |119861| |119903| |119899| |119898 minus 119898|2] 119889119905
le120573119890120582119905119891
119873int
119905119891
0
[
1003816100381610038161003816119906lowastminus 1
1003816100381610038161003816
2|119898| ((119898 minus 119898)
2+ (119899 minus 119899)
2)
+ |119899| (119898 minus 119898)2
+|119861| |119898119899|
2((119898 minus 119898)
2+ (119904 minus 119904)
2)
+|119861| |119904| |119898|
2((119898 minus 119898)
2+ (119899 minus 119899)
2)
+ |119861| |119904| |119899| (119898 minus 119898)2
+|119861| |119898119899|
2((119898 minus 119898)
2+ (119903 minus 119903)
2)
+|119861| |119898| |119903|
2((119898 minus 119898)
2+ (119899 minus 119899)
2)
+ |119861| |119903| |119899| (119898 minus 119898)2]119889119905
(84)
In the above derivation we use many scaling techniquesfor inequality or absolute inequality Particularly what shouldbe noted is that to get the first inequality of above derivationwe use the estimation of |119906lowast
1(119905) minus 119906
lowast
1(119905)| which has been given
before besides for the sake of convenience we note 119861 =
119898119899(120573119890120582119905119891119873) Furthermore we notice that the coefficients of
all the eight terms in the last formula (119898 minus 119898)2+ (119899 minus 119899)
2(119898 minus119898)
2 (119898 minus119898)2+ (119904 minus 119904)
2 (119898 minus119898)2+ (119899 minus 119899)
2 (119898 minus119898)2
(119898minus119898)2+(119903minus119903)
2 (119898minus119898)2+(119899minus119899)2 (119898minus119898)2 namely (|119906lowast1minus
1|2)|119898| |119899| (|119861||119898119899|)2 (|119861||119904||119898|)2 |119861||119904|119899 (|119861||119898119899|)2(|119861||119898||119903|)2 |119861||119903||119899| are nonnegative and bounded So thereexists a positive constant 119888
2such that
1
2(119898 minus 119898)
2(119905119891) + (120583 + 120582)int
119905119891
0
(119898 minus 119898)2119889119905
12 Abstract and Applied Analysis
le 1198882
120573119890120582119905119891
119873int
119905119891
0
((119898 minus 119898)2+ (119899 minus 119899)
2
+ (119904 minus 119904)2+ (119903 minus 119903)
2) 119889119905
(85)
Combining eight of these inequalities gives
1
2(119898 minus 119898) (119905
119891) +
1
2(119899 minus 119899) (119905
119891) +
1
2(119901 minus 119901) (119905
119891)
+1
2(119902 minus 119902) (119905
119891) +
1
2(119903 minus 119903) (0) +
1
2(119904 minus 119904) (0)
+1
2(119908 minus 119908) (0) +
1
2(V minus V) (0) + (120583 + 120582)
times int
119905119891
0
(119898 minus 119898)2+ (119899 minus 119899)
2+ (119901 minus 119901)
2
+ (119902 minus 119902)2
+ (119904 minus 119904)2
+ (119903 minus 119903)2+ (119908 minus 119908)
2+ (V minus V)2) 119889119905
le 119861int
119905119891
0
(119898 minus 119898)2+ (119899 minus 119899)
2+ (119901 minus 119901)
2
+ (119902 minus 119902)2
+ (119903 minus 119903)2+ (119904 minus 119904)
2
+ (119908 minus 119908)2+ (V minus V)2 119889119905
(86)
Thus from the above inequality we can conclude that
(120583 + 120582 minus 119861)int
119905119891
0
(119898 minus 119898)2+ (119899 minus 119899)
2
+ (119901 minus 119901)2
+ (119902 minus 119902)2
+ (119903 minus 119903)2
+ (119904 minus 119904)2+(119908 minus 119908)
2+ (V minus V)2 119889119905 le 0
(87)
where 119861 depends on the coefficients and the bounds dependon119898 119899 119901 119902 119903 119904 119908 V If we choose 120582 such that 120583+120582 gt 119861 then119898 = 119898 119899 = 119899 119901 = 119901 119902 = 119902 119903 = 119903 119904 = 119904 119908 = 119908 and V = VHence the solution to the optimality system is unique Theproof is complete
6 Numerical Simulation
61 The Simulation of State System (1) without ControlParameters For the sake of simplicity but without loss ofgenerality we will perform the numerical simulation of statesystem (1) with parameters 119906
1= 0 119906
2= 0 Before
illustrating the analytic properties of the alcoholism model(1) we will target the populations in the environment ofa community or a university for example the school ofmaterial science and engineering in our university that isLan zhou University of Technology (LUT for short) owing tothe accurate and available information we can obtain Refer-ring to the information provided by the admissions officeof LUT this school will enroll almost 1200 undergraduatesand almost 300 various postgraduates at the beginning of
fall semester at the same time there will be almost 1500various students graduated and left this school so the scaleof students in school remained almost 6000 we can take thetotal population 119873 = 6000 In this simulation we will takeSeptember as the initial time and units in one week periodin one year According to the investigations of the studentunion implemented in September every year we can takeinitial values as 119878(0) = 4500 119860(0) = 1000 119879(0) = 300and 119876(0) = 200 It seems that the alcoholism is a little bitmore but it is rather natural because many freshmen feelconfused when they are faced with the new environment anda new lifestyle many of them have no better choice but gathertogether to drink in small groups to mediate the anxiety andget to know each other over time some of them developthe habit of drinking To a certain extent for example thefrequent drinking badly affects their study we can classifythem into the alcoholism compartment Other initial valuesseem more reasonable so we need no more explanation Aswe know alcoholism death is seldom happen within oneyear so we omit mortality from alcoholism then how tounderstand the recruitment rate as well as natural death rate120583 We can treat the freshmen admission as the recruitmentpopulation and graduation students as the natural ldquodeathrdquoparts So we can take 120583 = 15006000 = 025 which is exactlyconsistent with the value in [16] As for the infection rate 120573and recovery rate 120575 we will let them be variables since thedrinking behaviors are related to many factors such as theseason and the pressure
According to the data we get from the student unionwe choose 120585 = 04 To summarize we list the values ofthe parameters in Table 1 Using the values of parameters inTable 1 we can plot Figures 2 and 3 which are on condition1198770lt 1 and 119877
0ge 1 respectively From Figure 2 we easily
know when 1198770lt 1 holds the solution of system (1) tends to
the alcohol free equilibrium1198640and verifies the global stability
of 1198640 While seen from Figure 3 we also know that if 119877
0ge
1 holds the solution of system (1) tends to the alcoholismequilibrium 119864
lowast and verifies the global stability of 119864lowast
62 The Sensitivity of 1198770about Two Control Parameters
Although from the expression of the model reproductionnumber 119877
0 we can easily find out the fact that the two
control variables that is 1199061and 119906
2 attribute to reducing the
severity of alcoholism we will still depict the graph between1198770and the two control variables to see more intuitiveness see
Figure 4 It seems from the figure that 1198770is a monotonically
decreasing function about two control parameters so it isadvisable to take two approaches simultaneously to controlthe alcoholics
63 The Simulation of Optimality System (71) In this subsec-tion we will investigate numerically the optimal solution tooptimality system (71) by numerical method from [32] theoptimality system is solved with a fourth-order Runge-Kuttascheme Beginning with a guess for the control variables thestate system is solved forward in time and then those values ofstate are used to solve the adjoint equations backward in timeThe controls are updated at the end of each iteration using
Abstract and Applied Analysis 13
0 10 20 30 40 50
0
1000
2000
3000
4000
5000
6000
7000
Time
Popu
latio
n siz
e
S(t)
A(t)
T(t)
Q(t)
Figure 2 When 1198770lt 1 the alcohol free equilibrium 119864
0is globally
asymptotically stable (120573 = 02 120575 = 03 and 1198770= 071698)
S(t)
A(t)
T(t)
Q(t)
0 10 20 30 40
48
2520
303540
49 50
500
1000
2000
3000
4000
5000
6000
7000
Time
Popu
latio
n siz
e
Figure 3 When 1198770
ge 1 the alcoholism equilibrium 119864lowast
=
(4466 476 32 26) corresponding to the given parameters is globallyasymptotically stable (120573 = 03 120575 = 02 and 119877
0= 108511)
the values of optimal controls obtained lastly The iterationscontinue until convergence takes place
In the simulations we choose the available variable valuesas Table 1 shows besides 120573 = 03 120575 = 02 The initial valueof model (1) is assumed to be 119878(0) = 4500 119860(0) = 1000119879(0) = 300 and 119876(0) = 200 as before
The ideal weights in objective functional are very difficultto obtain in reality it needs much work on data mining andfittingHence the acquisition of appropriate practical weights
Table 1 The parameters description of model (1)
Parameter Description Values120583 Natural birth rate or death rate 025
120573Transmission coefficient betweenalcoholism and susceptibles Variables
119873 The total populations to be considered 6000
120575The rate of populations quitting fromalcoholism permanently after treatment Variables
120585The rate of populations failed intreatment and returned to be alcoholic 04
0
02
04
06
08
1
0
05
1
0 02 04 06 08 1
u1
u2
R0
Figure 4 The relationship between 1198770and two control variables 119906
1
and 1199062
is still a difficult problem and remains for further investiga-tionsThe cost associated with119860(119905) and 119906
1(119905)mainly includes
the cost of dangerous behaviours during the alcoholism timeand educating the public while the cost associated with1199062(119905)mainly comes from health professional and the medical
resource includingmedicines andnursing care In viewof thisand taking the expressions of 119906
1 1199062into account after many
numerical simulations we finally give weighting coefficientsas 1198881= 102 1198882= 104 It should be pointed out that the
weights here are of only theoretical interest to reveal thecontrol strategies proposed in this paper Another point tonote is that themaximumcontrol is very difficult to achieve inreality so we will omit the situation of the maximum controlduring the series of simulations
Next we will make some necessary instructions andexplanations to the above simulation graphs Figures 5 67 and 8 depict the number of four compartments underdifferent control levels when we choose the weight coeffi-cients in objective function to be 119888
1= 102 1198882= 104 From
the four simulation graphs we can observe the followingsimple facts in reducing the total number of alcoholismsand increasing the number of susceptibles the effectivenessof various control measures is as follows optimal control isevidently better than middle control and middle control isbetter than single control 119906
2 single control 119906
2is better than
single control 1199061 while single control 119906
1is much better than
no controlFigures 9 and 10 depict the optimal control law of 119906
1 1199062
respectively In the beginning of the simulation the control
14 Abstract and Applied Analysis
0 10 20 30 40 504500
5000
5500
6000
6500
7000
S(n
umbe
r of s
usce
ptib
les)
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 5Number of the susceptibles whenwe choose theweights inobjective function are 119888
1= 102 1198882= 104 and under different optimal
control strategies that is (1) without control 1199061= 1199062= 0 (2) single
control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0 119906
2= 05 (4)
middle control 1199061= 1199062= 05 (5) optimal control 119906
1= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
500
1000
1500
A(n
umbe
r of t
he al
coho
lism
)
t (unit week)
Figure 6 Number of the alcoholics when we choose the weights inobjective function are 119888
1= 102 1198882= 104 and under different optimal
control strategies that is (1) without control 1199061= 1199062= 0 (2) single
control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0 119906
2= 05 (4)
middle control 1199061= 1199062= 05 (5) optimal control 119906
1= 119906lowast
1 1199062= 119906lowast
2
0
50
100
150
200
250
300
350
400
T(n
umbe
r of t
reat
men
t)
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 7 Number of the persons in treatment when we choose theweights in objective function are 119888
1= 10
2 1198882= 10
4 and underdifferent optimal control strategies that is (1) without control 119906
1=
1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control 119906
1= 0
1199062= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal control
1199061= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
50
100
150
200
250
300
350
400
Q(n
umbe
r of q
uitti
ng)
Figure 8 Number of people in quitting compartment when wechoose the weights in objective function are 119888
1= 102 1198882= 104 and
under different optimal control strategies that is (1) without control1199061= 1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control
1199061= 0 119906
2= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal
control 1199061= 119906lowast
1 1199062= 119906lowast
2
Abstract and Applied Analysis 15
0 10 20 30 40 50t (unit week)
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
Optimal control of u1
Figure 9 Figures of the optimal control 1199061when we choose the
weight in objective function are 1198881= 102 1198882= 104
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
0 10 20 30 40 50t (unit week)
Optimal control of u2
Figure 10 Figures of the optimal control 1199062when we choose the
weight in objective function are 1198881= 102 1198882= 104
effort of 1199061should be decreased from 065 to 05 within the
first month and over the next week it should be increased tothe maximum control until 50 weeks then rapidly decreasedto 0 at the end of the simulation As for the control 119906
2 it
should start from around 05 due to the initial alcoholics thenincrease to 055 within one week since the rapid infection andnext decrease to almost 0 since the effectiveness of treatmentin three weeks but with the infection going on the controleffort of 119906
2should gradually increase to the maximum and
maintain this level until the tenth week for the purpose ofconsolidation therapy and preventing rebound hereafter itshould be gradually decreased to the level of almost 043 untilthe fifty weeks then quickly decreased to 0 in the end
0 10 20 30 40 504500
5000
5500
6000
6500
7000
S(n
umbe
r of s
usce
ptib
les)
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 11 Number of the susceptibles when we choose the weightsin objective function are 119888
1= 104 1198882= 102 and under different
optimal control strategies that is (1) without control 1199061= 1199062= 0
(2) single control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0
1199062= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal control
1199061= 119906lowast
1 1199062= 119906lowast
2
In order to investigate the influence of different weightcoefficients in the objective functional on the effect ofcontrolling at the same time for a better comparison we willchange the weight coefficients in objective function into 119888
2=
104 1198881= 102 and we will list the corresponding numerical
simulation results as Figures 11ndash16 showWhen we change the weight coefficients in objective
function into 1198881= 104 1198882= 102 we find that the results
of simulations derived from the graphs are very similar tothe ones before We speculate that the most likely reasons ofthis result are due to three respects one is that the weightcoefficients are not too sensitive in the numerical simulationand another possible reason is that both of the two controlsare important in some sense equivalently importantThe lastbut not themost unlikely reason is that we have not found themost appropriate weight coefficients in the simulation whichis very difficult to find as previously mentioned
7 Conclusions
In this paper we formulate an alcoholics quitting modeland firstly investigate the variation discipline of variouspopulations from the perspective of global stability thenwe propose an objective functional to examine two differentcontrolmeasures (ie prevention and treatment) on the effectof alcohol The basic reproduction number of the model wasderived and the global stability of the two equilibria is givenFrom the expression of the basic reproductionnumber119877
0and
16 Abstract and Applied Analysis
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
500
1000
1500
A(n
umbe
r of t
he al
coho
lism
)
t (unit week)
Figure 12 Number of the alcoholics when we choose the weights inobjective function are 119888
1= 104 1198882= 102 and under different optimal
control strategies that is (1) without control 1199061= 1199062= 0 (2) single
control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0 119906
2= 05 (4)
middle control 1199061= 1199062= 05 (5) optimal control 119906
1= 119906lowast
1 1199062= 119906lowast
2
0
50
100
150
200
250
300
350
400
T(n
umbe
r of t
reat
men
t)
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 13 Number of the persons in treatment when we choosethe weights in objective function are 119888
1= 104 1198882= 102 and under
different optimal control strategies that is (1) without control 1199061=
1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control 119906
1= 0
1199062= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal control
1199061= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
50
100
150
200
250
300
350
400
Q(n
umbe
r of q
uitti
ng)
Figure 14 Number of people in quitting compartment when wechoose the weights in objective function are 119888
1= 104 1198882= 102 and
under different optimal control strategies that is (1) without control1199061= 1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control
1199061= 0 119906
2= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal
control 1199061= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50t (unit week)
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
Optimal control of u1
Figure 15 Figures of the optimal control 1199061when we choose the
weight in objective function are 1198881= 104 1198882= 102
related numerical simulation we can easily see that the twocontrol strategies are effective in the alcoholics process
Using Pontryaginrsquos Maximum Principle we firstly deter-mine the necessary conditions for existence of optimalcontrol pairsThe uniqueness of the solution to the optimalitysystem (71) is derived by the classical method of contradic-tion Numerical simulations of the model suggest that thetwo different groups of weights in the objective function have
Abstract and Applied Analysis 17
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
0 10 20 30 40 50t (unit week)
Optimal control of u2
Figure 16 Figures of the optimal control 1199062when we choose the
weight in objective function are 1198881= 104 1198882= 102
much similar effects on the transmission of the alcoholismfrom this point the two control measures are almost equallyimportant in controlling the alcoholism although they willprobably have great influences on the cost of the objectivefunction From the simulation figures it seems that the effectof optimal control which is measured by the reduction inthe number of alcoholics and the increase in the number ofsusceptibles is much better than other control strategies asnoted earlier in the simulation section According to the real-time curve of two optimal controls we point out the specificimplementation methods of optimal control which can beachieved in practice
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
This work was partially supported by the NSF ofGansu Province of China (2013GS09485) the NNSF ofChina (10961018) the NSF of Gansu Province of China(1107RJZA088) the NSF for Distinguished Young Scholarsof Gansu Province of China (1111RJDA003) the Special Fundfor the Basic Requirements in the Research of University ofGansu Province of China and the Development Programfor Hong Liu Distinguished Young Scholars in LanzhouUniversity of Technology
References
[1] R Room T Babor and J Rehm ldquoAlcohol and public healthrdquoThe Lancet vol 365 no 9458 pp 519ndash530 2005
[2] J B Saunders O G Aasland A Amundsen and M GrantldquoAlcohol consumption and related problems among primary
health care patients WHO collaborative project on early detec-tion of personswith harmful alcohol consumption IrdquoAddictionvol 88 no 3 pp 349ndash362 1993
[3] L D Johnston P M OMalley and J G Bachman NationalSurvey Results on Drug Use From the Monitoring the FutureStudy 1975ndash1992 National Institute on Drug Abuse RockvilleMd USA 1993
[4] H Wechsler J E Lee M Kuo and H Lee ldquoCollege bingedrinking in the 1990s a continuing problem Results of theHarvard School of Public Health 1999 College Alcohol StudyrdquoJournal of American College Health vol 48 no 5 pp 199ndash2102000
[5] P M OrsquoMalley and L D Johnston ldquoEpidemiology of alcoholand other drug use among American college studentsrdquo Journalof Studies on Alcohol vol 63 no 14 pp 23ndash39 2002
[6] K Agarwal-Kozlowski and D P Agarwal ldquoGenetic predisposi-tion to alcoholismrdquo Therapeutische Umschau vol 57 no 4 pp179ndash184 2000
[7] C Y Chen C L Storr and J C Anthony ldquoEarly-onset drug useand risk for drug dependence problemsrdquo Addictive Behaviorsvol 34 no 3 pp 319ndash322 2009
[8] M Glavas and J Weinberg ldquoStress alcohol consumption andthe hypothalamic-pituitary adrenal axisrdquo in Nutrients Stressand Medical Disorders pp 165ndash183 Humana Press New YorkNY USA 2006
[9] L N Blum N H Nielsen J A Riggs and L B BresolinldquoAlcoholism and alcohol abuse among women report of theCouncil on Scientific Affairsrdquo Journal of Womenrsquos Health vol7 no 7 pp 861ndash871 1998
[10] B Benedict ldquoModeling alcoholism as a contagious disease howldquoinfectedrdquo drinking buddies spread problem drinkingrdquo SIAMNews vol 40 no 3 2007
[11] H Walter K Gutierrez K Ramskogler I Hertling A Dvorakand O M Lesch ldquoGender-specific differences in alcoholismImplications for treatmentrdquo Archives of Womenrsquos Mental Healthvol 6 no 4 pp 253ndash258 2003
[12] D Muller R D Koch H von Specht w Volker and E MMunch ldquoNeurophisiologic findings in chronic alcohol abuserdquoPsychiatrie Neurologie und Medizinische Psychologie vol 37no 3 pp 129ndash132 1985
[13] G Testino ldquoAlcoholic diseases in hepato-gastroenterology apoint of viewrdquo Hepato-Gastroenterology vol 55 no 82-83 pp371ndash377 2008
[14] A Mubayi P E Greenwood C Castillo-Chavez P J Grue-newald and D M Gorman ldquoThe impact of relative residencetimes on the distribution of heavy drinkers in highly distinctenvironmentsrdquo Socio-Economic Planning Sciences vol 44 no 1pp 45ndash56 2010
[15] D Muller R D Koch H von Specht W Volker and E MMunch ldquoNeurophysiologic findings in chronic alcohol abuserdquoPsychiatrie Neurologie Und Medizinische Psychologie vol 37no 3 pp 129ndash132 1985 (Leipz)
[16] G Mulone and B Straughan ldquoModeling binge drinkingrdquoInternational Journal of Biomathematics vol 5 no 1 Article ID1250005 14 pages 2012
[17] H F Huo and N N Song ldquoGlobal stability for a binge drinkingmodel with two stagesrdquo Discrete Dynamics in Nature andSociety vol 2012 Article ID 829386 15 pages 2012
[18] S S Mushayabasa and C P Bhunu ldquoModelling the effects ofheavy alcohol consumption on the transmission dynamics ofgonorrheardquo Nonlinear Dynamics vol 66 no 4 pp 695ndash7062011
18 Abstract and Applied Analysis
[19] F S Sancheznchez XWang C Castillo-Chvez DM Gormanand P J Gruenewald ldquoDrinking as an epidemic a simplemathematical model with recovery and relapserdquo in TherapistsGuide to Evidence-Based Relapse Prevention K Witkiewitz andG Marlatt Eds Academic Press New York NY USA 2007
[20] H-F Huo and C-C Zhu ldquoInfluence of relapse in a givingup smoking modelrdquo Abstract and Applied Analysis vol 2013Article ID 525461 12 pages 2013
[21] H F Huo and Q Wang ldquoThe effects of awareness programs by17 media on the drinking dynamicsrdquo Submitted
[22] S Lee E Jung and C Castillo-Chavez ldquoOptimal controlintervention strategies in low- and high-risk problem drinkingpopulationsrdquo Socio-Economic Planning Sciences vol 44 no 4pp 258ndash265 2010
[23] K R Fister S Lenhart and J S McNally ldquoOptimizingchemotherapy in an HIV modelrdquo Electronic Journal of Differ-ential Equations vol 1998 no 32 pp 1ndash12 1998
[24] S Lenhart and J T Workman Optimal Control Applied toBiological Models Chapman amp HallCRC Mathematical andComputational Biology Series Chapman amp HallCRC BocaRaton Fla USA 2007
[25] X Yang L Chen and J Chen ldquoPermanence and positiveperiodic solution for the single-species nonautonomous delaydiffusive modelsrdquo Computers amp Mathematics with ApplicationsAn International Journal vol 32 no 4 pp 109ndash116 1996
[26] P van denDriessche and JWatmough ldquoReproduction numbersand sub-threshold endemic equilibria for compartmental mod-els of disease transmissionrdquoMathematical Biosciences vol 180pp 29ndash48 2002
[27] V Lakshmikantham S Leela and A A Martynyuk StabilityAnalysis of Nonlinear Systems vol 125 Marcel Dekker NewYork NY USA 1989
[28] J P LaSalle The Stability of Dynamical Systems vol 25 ofRegional Conference Series in Applied Mathematics Society forIndustrial and Applied Mathematics 1976
[29] W H Fleming and R W Rishel Deterministic and StochasticOptimal Control vol 1 ofApplications of Mathematics SpringerBerlin Germany 1975
[30] D L Lukes Differential Equations Classical to ControlledMathematics in Science and Engineering Academic Press NewYork NY USA 1982
[31] L S Pontryagin V G Boltyanskii R V Gamkrelidze and EF Mishchenko The Mathematical Theory of Optimal ProcessesJohn Wiley amp Sons New Jersey NJ USA 1962
[32] W Hackbusch ldquoA numerical method for solving parabolicequations with opposite orientationsrdquo Computing vol 20 no3 pp 229ndash240 1978
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
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Discrete Dynamics in Nature and Society
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Decision SciencesAdvances in
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 3
A
S Q
T
120583N
120583S
120585T
120583T
120583A
120583Q
u2A
120575T
(1 minus u1)120573SA
N
Figure 1 Transfer diagram for the dynamics of alcoholism model
1198791015840= 1199062119860 minus (120583 + 120585 + 120575) 119879
1198761015840= 120575119879 minus 120583119876
(1)
Here 120583119873 is the birth number of the population 120583 is thenatural death rate of the population 119906
1is the fraction of
the susceptible individuals who successfully avoid to stayoff the alcoholism 119906
2is the fraction of the alcoholics who
take part in treatments here 0 le 119906119894le 1 119894 = 1 2 and
they will be considered as two control variables in Section 5120573 is the transmission coefficient of the ldquoinfectionrdquo for thesusceptible individuals from the alcoholic individuals 120585 is therate coefficient of the person who fail to be treated and returnto the alcoholism compartmentmostly due to their ownweakwill 120575 is the rate coefficient of the person who have receivedeffective treatment and recovered from alcoholism forever
22 Boundedness of Solutions to System and Positively Invari-ant Region It is important to show positivity and bounded-ness for the system (1) as they represent populations Firstlywe present the positivity of the solutions System (1) can beput into the matrix form
1198831015840= 119866 (119883) (2)
where119883 = (119878 119860 119879 119876)119879isin 1198774 and 119866(119883) is given by
119866 (119883) = (
1198661(119883)
1198662(119883)
1198663(119883)
1198664(119883)
)
=(
120583119873 minus (1 minus 1199061)120573119878119860
119873minus 120583119878
(1 minus 1199061)120573119878119860
119873+ 120585119879 minus (119906
2+ 120583)119860
1199062119860 minus (120583 + 120585 + 120575) 119879
120575119879 minus 120583119876
)
(3)
It is easy to check that
119866119894(119883)
1003816100381610038161003816119883119894(119905)=0119883119905isin119862+ge 0 119894 = 1 2 3 4 (4)
Due to Lemma 2 in [25] any solution of (1) is 119883(119905) isin 1198774+for
all 119905 ge 0We denote 119873(119905) = 119878(119905) + 119860(119905) + 119879(119905) + 119876(119905) summing
equations in (1) yields
119889119873 (119905)
119889119905= 0 (5)
so 119873(119905) = 119878(119905) + 119860(119905) + 119879(119905) + 119876(119905) = constant (denoted as119873) and the set
Ω = (119878 119860 119879 119876) isin 1198774
+ 119878 + 119860 + 119879 + 119876 le 119873 (6)
is a positively invariant region for (1) Therefore we willconsider the global stability of (1) on the setΩ
3 The Basic Reproduction Number andExistence of Alcoholism Equilibria
31 The Basic Reproduction Number 1198770 In epidemiology
the basic reproduction number (sometimes called basicreproductive rate or basic reproductive ratio) of an infectionis the number of infectious cases that one infectious casegenerates on average over the course of its infectious periodWhile in this context it means the number of persons thatan alcoholic will ldquoinfectrdquo during his ldquoinfectiousrdquo period in thepure susceptible environment so that the infected personswillenter the alcoholism compartment It is easy to see that themodel has an alcohol free equilibrium 119864
0= (1198780 0 0 0) =
(119873 0 0 0) In the following the basic reproduction numberof system (1) will be obtained by the next generation matrixmethod formulated in [26]
Let 119909 = (119860 119879 119876 119878)119879 then system (1) can be written as
119889119909
119889119905= F (119909) minusV (119909) (7)
where
F (119909) = (
(1 minus 1199061)120573119878119860
1198730
0
0
)
V (119909) = (
(1199062+ 120583)119860 minus 120585119879
(120583 + 120585 + 120575) 119879 minus 1199062119860
120583119876 minus 120575119879
(1 minus 1199061)120573119878119860
119873+ 120583119878 minus 120583119873
)
(8)
The Jacobian matrices of F(119909) and V(119909) at the alcohol freeequilibrium 119864
0are respectively
119863F (1198640) = (
1198653times3
0
0 0)
119863V (1198640) = (
1198813times3
0
(1 minus 1199061) 120573 0 0 120583
)
(9)
where
119865 = (
(1 minus 1199061) 120573 0 0
0 0 0
0 0 0
) 119881 = (
1199062+ 120583 minus120585 0
minus1199062
120583 + 120585 + 120575 0
0 minus120575 120583
)
(10)
The basic reproduction number denoted by 1198770 is thus given
by
1198770= 120588 (119865119881
minus1) =
120573 (1 minus 1199061) (120583 + 120585 + 120575)
1199062(120583 + 120575) + 120583 (120583 + 120585 + 120575)
(11)
4 Abstract and Applied Analysis
It is easy to see that both of the control parameters con-tributed to reducing the alcoholism From this point thecontrol measures are meaningful
32 Existence of Alcoholism Equilibrium The endemic equi-librium 119864
lowast(119878lowast 119860lowast 119879lowast 119876lowast) of system (1) is determined by
equations
120583119873 minus (1 minus 1199061)120573119878119860
119873minus 120583119878 = 0
(1 minus 1199061)120573119878119860
119873+ 120585119879 minus (119906
2+ 120583)119860 = 0
1199062119860 minus (120583 + 120585 + 120575) 119879 = 0
120575119879 minus 120583119876 = 0
(12)
The third equation in (12) leads to
119860 =120583 + 120585 + 120575
1199062
119879 (13)
From the last equation in (12) we have
119876 =120575119879
120583 (14)
From the first equation of (12) and together with (13) we canget
119878 =120583119873
120583 + (((1 minus 1199061) 120573) 119873)119860
=12058311987321199062
1205831198731199062+ (1 minus 119906
1) 120573119879 (120583 + 120585 + 120575)
(15)
Substituting (13)ndash(15) into the second equation of (12) gives
120583119873 minus120583211987321199062
1205831198731199062+ (1 minus 119906
1) 120573119879 (120583 + 120585 + 120575)
+ 120585119879 minus (120583 + 1199062)120583 + 120585 + 120575
1199062
119879 = 0
(16)
By simplifying (16) we can get
119879 [1199062120585 (1 minus 119906
1) 120573 (120583 + 120585 + 120575)
minus (1 minus 1199061) 120573(120583 + 120585 + 120575)
2
(120583 + 1199062)] 119879 + 120590 = 0
(17)
where120590 = 119906
2120583119873 (1 minus 119906
1) 120573 (120583 + 120585 + 120575)
+ (1199062)2
120583119873120585 minus 1205831198731199062(120583 + 119906
2) (120583 + 120585 + 120575)
(18)
Hence we get two explicit solutions to (17) one is 1198790= 0
which is corresponding to the alcohol free equilibria and theother is119879lowast= (120590) ((1 minus 119906
1) 120573(120583 + 120585 + 120575)
2
(120583 + 1199062)
minus1199062120585 (1 minus 119906
1) 120573 (120583 + 120585 + 120575) )
minus1
=120590
(1 minus 1199061) 120573 (120583 + 120585 + 120575) [120583120585 + (120575 + 120583) (120583 + 119906
2)]
(19)
which should be corresponding to the alcoholism equilibriaon condition that 119879lowast gt 0 otherwise the alcoholismequilibria are nonexistent It is enough to show the positivityof 120590 tomake sure the existence of alcoholism equilibria on thecondition 119877
0ge 1 By some simple calculations we simplify
the expression of 120590 to be
120590 = 1205831198731199062(120583 + 120585 + 120575) (1 minus 119906
1) 120573
minus [(120583 + 1199062) (120575 + 120583) + 120583120585]
(20)
Since 1198770gt 1 is equivalent to
120573 (1 minus 1199061) (120583 + 120585 + 120575) gt 119906
2(120583 + 120575) + 120583 (120583 + 120585 + 120575) (21)
the right side of this inequality is exactly equal to (120583+1199062)(120575+
120583) + 120583120585 Hence we have proved the existence of 119879lowast gt 0so are the alcoholism equilibria We summarize this result inTheorem 1
Theorem 1 For system (1) there is always an alcohol freeequilibrium 119864
0= (119873 0 0 0) When 119877
0gt 1 besides alcohol
free equilibrium 1198640 system (1) also has a unique alcoholism
equilibrium 119864lowast(119878lowast 119860lowast 119879lowast 119876lowast) where
119878lowast=
12058311987321199062
1205831198731199062+ (1 minus 119906
1) 120573119879lowast (120583 + 120585 + 120575)
119860lowast=120583 + 120585 + 120575
1199062
119879lowast
119876lowast=120575119879lowast
120583
119879lowast=
120590
(1 minus 1199061) 120573 (120583 + 120585 + 120575) [120583120585 + (120575 + 120583) (120583 + 119906
2)]
(22)
4 Stability Analysis of Equilibria
For the convenience of subsequent proof we denote a vector119883 = (119860 119879 119876 119878)
119879 and
119891 (119883) =(
(1 minus 1199061)120573119878119860
119873+ 120585119879 minus (119906
2+ 120583)119860
1199062119860 minus (120583 + 120585 + 120575) 119879
120575119879 minus 120583119876
120583119873 minus (1 minus 1199061)120573119878119860
119873minus 120583119878
) (23)
So the Jacobian matrix of 119891(119909) about vector 119883 is as thefollowing
119869 =
120597119891 (119883)
120597119883
=(
(1 minus 1199061) 120573119878
119873
minus (120583 + 1199062) 120585 0
(1 minus 1199061) 120573119860
119873
1199062 minus (120583 + 120585 + 120575) 0 0
0 120575 minus120583 0
minus
(1 minus 1199061) 120573119878
119873
0 0 minus120583 minus
(1 minus 1199061) 120573119860
119873
)
(24)
Theorem 2 For system (1) the alcohol free equilibrium 1198640is
locally asymptotically stable if 1198770lt 1
Abstract and Applied Analysis 5
Proof Since
119869 (1198640)
= (
(1 minus 1199061) 120573 minus (120583 + 119906
2) 120585 0 0
1199062
minus (120583 + 120585 + 120575) 0 0
0 120575 minus120583 0
minus (1 minus 1199061) 120573 0 0 minus120583
)
(25)
we can easily get that two of the eigenvalues are 1205821= 1205822=
minus120583 lt 0 while 1205823 1205824satisfy
1205822+ [2120583 + 120585 + 120575 + 119906
2minus (1 minus 119906
1) 120573] 120582
+ (120583 + 120585 + 120575) (120583 + 1199062minus (1 minus 119906
1) 120573) minus 119906
2120585 = 0
(26)
Thus
1205823+ 1205824= (1 minus 119906
1) 120573 minus (120583 + 120585 + 120575) minus (120583 + 119906
2) (27)
Since 1198770lt 1 is equivalent to
120573 (1 minus 1199061) (120583 + 120585 + 120575) lt 119906
2(120583 + 120575) + 120583 (120583 + 120585 + 120575)
lt (120583 + 1199062) (120583 + 120585 + 120575)
(28)
so
120573 (1 minus 1199061) lt 120583 + 119906
2 (29)
and then
1205823+ 1205824lt minus (120583 + 120585 + 120575) lt 0 (30)
while
12058231205824= (120583 + 120585 + 120575) (120583 + 119906
2minus (1 minus 119906
1) 120573) minus 119906
2120585 (31)
Similarly from 1198770lt 1 we can derive the inequality
minus120573 (1 minus 1199061) (120583 + 120585 + 120575) gt minus119906
2(120583 + 120575) minus 120583 (120583 + 120585 + 120575) (32)
so12058231205824gt (120583 + 119906
2) (120583 + 120585 + 120575)
minus 1199062120585 minus 1199062(120583 + 120575) minus 120583 (120583 + 120585 + 120575)
(33)
It reduces to
12058231205824gt 0 (34)
Hence Re 1205823lt 0 Re 120582
4lt 0 The proof is complete
Next we will turn to investigate the global stability of 1198640
Theorem 3 For system (1) the alcohol free equilibrium 1198640is
globally asymptotically stable if 1198770lt 1
Proof Consider the subsystem of (1) as follows
1198601015840= (1 minus 119906
1)120573119878119860
119873+ 120585119879 minus (119906
2+ 120583)119860
1198791015840= 1199062119860 minus (120583 + 120585 + 120575) 119879
1198761015840= 120575119879 minus 120583119876
(35)
Equation (35) can be rewritten as
(
) = (119865 minus 119881)(
119860
119879
119876
) minus (1 minus119878
119873)
times (
120573 (1 minus 1199061) 0 0
0 0 0
0 0 0
)(
119860
119879
119876
)
(36)
Since 119878 le 119873 and 0 le 1199061le 1 then for all 119905 gt 0 we can get
(
) le (119865 minus 119881)(
119860
119879
119876
) (37)
According to Lemma 1 in [26] all the eigenvalues of matrix119865 minus 119881 have negative real parts so the solutions of this sub-system are stable whenever 119877
0lt 1 So (119860(119905) 119879(119905) 119876(119905)) rarr
(0 0 0) as 119905 rarr infin By the comparison theorem [27] andbased on the fact that the total population is constant 119873 itfollows that (119860(119905) 119879(119905) 119876(119905)) rarr (0 0 0) and 119878(119905) rarr 119873
as 119905 rarr infin So the alcohol free equilibrium 1198640is globally
asymptotically stable the proof is complete
Theorem 4 For system (1) the alcoholism equilibrium119864lowast(119878lowast 119860lowast 119879lowast 119876lowast) is globally asymptotically stable if 119877
0gt 1
Proof Since the total population in model (1) is a constantnumber119873 in order to prove the global stability of system (1)it is sufficed to prove the corresponding stability of subsystem(38)
1198781015840= 120583119873 minus (1 minus 119906
1)120573119878119860
119873minus 120583119878
1198601015840= (1 minus 119906
1)120573119878119860
119873+ 120585119879 minus (119906
2+ 120583)119860
1198791015840= 1199062119860 minus (120583 + 120585 + 120575) 119879
(38)
We make normalization transform and still use the samesymbols 119878 119860 119879 to denote the variables then (38) can betransformed into
1198781015840= 120583 minus (1 minus 119906
1) 120573119878119860 minus 120583119878
1199041198601015840= (1 minus 119906
1) 120573119878119860 + 120585119879 minus (119906
2+ 120583)119860
1198791015840= 1199062119860 minus (120583 + 120585 + 120575) 119879
(39)
From (39) we can easily know that the equilibria (119878lowast 119860lowast 119879lowast)satisfy the following three equalities to be used later
120583 = (1 minus 1199061) 120573119878lowast119860lowastminus 120583119878lowast
(1 minus 1199061) 120573119878lowast119860lowast+ 120585119879lowast= (1199062+ 120583)119860
lowast
1199062119860lowast= (120583 + 120585 + 120575) 119879
lowast
(40)
6 Abstract and Applied Analysis
Let 119881 = 1199091(119878 minus 119878
lowastminus 119878lowast ln(119878119878lowast)) + 119909
2(119860 minus 119860
lowastminus
119860lowast ln(119860119860lowast)) + 119909
3(119879 minus 119879
lowastminus 119879lowast ln(119879119879lowast)) then
119881101584010038161003816100381610038161003816(39)
= 1199091[120583 minus (1 minus 119906
1) 120573119878119860 minus 120583119878
minus119878lowast
119878120583 + (1 minus 119906
1) 120573119878lowast119860 + 120583119878
lowast]
+ 1199092[ (1 minus 119906
1) 120573119878119860 + 120585119879 minus (119906
2+ 120583)119860
minus (1 minus 1199061) 120573119878119860
lowastminus119860lowast
119860120585119879 + (119906
2+ 120583)119860
lowast]
+ 1199093[1199062119860 minus (120583 + 120585 + 120575) 119879
minus119879lowast
1198791199062119860 + (120583 + 120585 + 120575) 119879
lowast]
= 1199091[ (1 minus 119906
1) 120573119878lowast119860lowast+ 120583119878lowastminus (1 minus 119906
1) 120573119878119860 minus 120583119878
minus119878lowast
119878((1 minus 119906
1) 120573119878lowast119860lowast+ 120583119878lowast)
+ (1 minus 1199061) 120573119878lowast119860 + 120583119878
lowast]
+ 1199092[ (1 minus 119906
1) 120573119878119860 + 120585119879 minus (119906
2+ 120583)119860
minus (1 minus 1199061) 120573119878119860
lowastminus119860lowast
119860120585119879 + (119906
2+ 120583)119860
lowast]
+ 1199093[1199062119860 minus (120583 + 120585 + 120575) 119879
minus119879lowast
1198791199062119860 + (120583 + 120585 + 120575) 119879
lowast]
= 1199091120583119878lowast(2 minus
119878
119878lowastminus119878lowast
119878)
+ [1199091(1 minus 119906
1) 120573119878lowast119860lowast+ 1199092(1199062+ 120583)119860
lowast
+ 1199093(120583 + 120585 + 120575) 119879
lowast]
minus [1199091
(119878lowast)2
(1 minus 1199061) 120573119860lowast
119878+ 1199092(1 minus 119906
1) 120573119878119860
lowast
+ 1199092
119860lowast120585119879
119860+ 1199093
119879lowast
1198791199062119860]
+ 119878119860 [minus1199091(1 minus 119906
1) 120573 + 119909
2(1 minus 119906
1) 120573]
+ 119860 [minus (1199062+ 120583) 119909
2+ 1199091(1 minus 119906
1) 120573119878lowast+ 11990621199093]
+ 119879 [1205851199092minus (120583 + 120585 + 120575) 119909
3]
(41)
To eliminate the cross-term 119878119860 and two single-variable terms119860 and 119879 we let
minus1199091(1 minus 119906
1) 120573 + 119909
2(1 minus 119906
1) 120573 = 0
minus (1199062+ 120583) 119909
2+ 1199091(1 minus 119906
1) 120573119878lowast+ 11990621199093= 0
1205851199092minus (120583 + 120585 + 120575) 119909
3= 0
(42)
By solving them we can get
1199091= 1 119909
2= 1
1199093=
120585
120583 + 120585 + 120575=1199062+ 120583 minus (1 minus 119906
1) 120573119878lowast
1199062
(43)
Next we let
1198811015840
1= 1199091(1 minus 119906
1) 120573119878lowast119860lowast+ 1199092(1199062+ 120583)119860
lowast
+ 1199093(120583 + 120585 + 120575) 119879
lowast
1198811015840
2= minus[119909
1
(119878lowast)2
(1 minus 1199061) 120573119860lowast
119878
+ 1199092(1 minus 119906
1) 120573119878119860
lowast+ 1199092
119860lowast120585119879
119860+ 1199093
119879lowast
1198791199062119860]
(44)
and then
1198811015840= 1199091120583119878lowast(2 minus
119878
119878lowastminus119878lowast
119878) + 119881
1015840
1+ 1198811015840
2 (45)
Due to
(1199062+ 120583)119860
lowast= (1 minus 119906
1) 120573119878lowast119860lowast+ 120585119879lowast
1199093(120583 + 120585 + 120575) 119879
lowast= 120585119879lowast1199092= 120585119879lowast
(46)
Abstract and Applied Analysis 7
so
1198811015840
1= 2 (1 minus 119906
1) 120573119878lowast119860lowast+ 2120585119879
lowast
1198811015840
2= minus[
(119878lowast)2
(1 minus 1199061) 120573119860lowast
119878
+ (1 minus 1199061) 120573119878119860
lowast+119860lowast120585119879
119860+ 1199093
119879lowast
1198791199062119860]
le minus2[(119878lowast)2
(1 minus 1199061) 120573119860lowast
119878sdot (1 minus 119906
1) 120573119878119860
lowast]
12
minus 2 [119860lowast120585119879
119860sdot 1199093
119879lowast
1198791199062119860]
= minus2 (1 minus 1199061) 120573119878lowast119860lowastminus 2(1199093119860lowast1205851199062119879lowast)12
= minus2 (1 minus 1199061) 120573119878lowast119860lowastminus 2[120585119879
lowast(11990931199062119860lowast)]12
= minus2 (1 minus 1199061) 120573119878lowast119860lowast
minus 2120585119879lowast[(1199062+ 120583)119860
lowastminus (1 minus 119906
1) 120573119878lowast119860lowast]12
= minus2 (1 minus 1199061) 120573119878lowast119860lowastminus 2(120585119879
lowast120585119879lowast)12
= minus2 (1 minus 1199061) 120573119878lowast119860lowastminus 2120585119879
lowast
(47)
Hence
1198811015840= 120583119878lowast(2 minus
119878
119878lowastminus119878lowast
119878) + 119881
1015840
1+ 1198811015840
2
le 120583119878lowast(2 minus
119878
119878lowastminus119878lowast
119878) + 2 (1 minus 119906
1) 120573119878lowast119860lowast
+ 2120585119879lowastminus 2 (1 minus 119906
1) 120573119878lowast119860lowastminus 2120585119879
lowast
= 120583119878lowast(2 minus
119878
119878lowastminus119878lowast
119878) le 0
(48)
1198811015840= 0 if and only if 119878 = 119878
lowast 119860 = 119860lowast 119879 = 119879
lowast Accordingto LaSallersquos invariance principle [28] we can derive theconclusion that the alcoholism equilibria 119864lowast(119878lowast 119860lowast 119879lowast 119876lowast)are globally asymptotically stable the proof is complete
5 Optimal Control Problem
51 The Existence of Optimal Control In order to investigatean effective campaign to control alcoholism in a communitywhich pursue the goals of the minimized alcoholisms andmore recovered individuals we will reconsider the system (1)and use two control variables to reduce the numbers of alco-holics The difference is that we will change the parameters1199061 1199062into control variable 119906
1(119905) 1199062(119905) Their aforementioned
definitions allow us to do so 1199061(119905) is used to limit the
proportion of the susceptible individual to contact withalcoholism usually by propaganda and education so that thesusceptible individual can stay off alcoholism consciously andbe free of ldquoinfectionrdquo we can understand the effect of 119906
1(119905)
is to prevent the the susceptible from contacting with thealcoholism The control variable 119906
2(119905) is used to control the
alcoholism to take appropriate treatment measures such astaking pills or seeking other medical help However just as acoin has two sides there will be a lot of costs generated duringthe control process So it is advisable to balance between thecosts and the alcohol effects In view of this our optimalcontrol problem tominimize the objective functional is givenby
119869 (1199061 1199062) = int
119905119891
0
[119860 (119905) +1198881
21199062
1(119905) +
1198882
21199062
2(119905)] 119889119905 (49)
which subjects to system
1198781015840= 120583119873 minus (1 minus 119906
1(119905))
120573119878119860
119873minus 120583119878
1198601015840= (1 minus 119906
1(119905))
120573119878119860
119873+ 120585119879 minus (119906
2(119905) + 120583)119860
1198791015840= 1199062(119905) 119860 minus (120583 + 120585 + 120575) 119879
1198761015840= 120575119879 minus 120583119876
(50)
with initial conditions
119878 (0) = 1198780 119860 (0) = 119860
0
119879 (0) = 1198790 119876 (0) = 119876
0
(51)
Here 119906119894(119905) isin 119880 ≜ (119906
1 1199062) | 119906119894(119905) is measurable and 0 le
119906119894(119905) le 1 for all 119905 isin [0 119905
119891] 119905119891is the end time to be
controlled 119880 is an admissible control set 119888119894 and 119894 = 1 2 are
weight factors (positive constants) that adjust the intensity oftwo different control measures
Next we will investigate the existence of the optimalcontrol of the above-mentioned problem
Theorem 5 There exists an optimal control pair 119906lowast
=
(119906lowast
1 119906lowast
2) isin 119880 such that
119869 (119906lowast
1 119906lowast
2) = min 119869 (119906
1 1199062) 119906
1(119905) 1199062(119905) isin 119880 (52)
subjects to the control system (1) with initial conditions (50)
Proof Toprove the existence of an optimal control accordingto the classic literature [29] we have to show the following
(1) The control and state variables are nonnegative values(2) The control set 119880 is convex and closed(3) The right side of the state system is bounded by linear
function in the state and control variables(4) The integrand of the objective functional is concave
on 119880(5) There exist constants 119889
1 1198892gt 0 and 120572 gt 1 such
that the integrand 119871(119905 1199061 1199062) ≜ 119860(119905) + (119888
12)1199062
1(119905) +
(11988822)1199062
2(119905) of the objective functional satisfies
119871 (119905 1199061 1199062) ge 1198891(100381610038161003816100381611990611003816100381610038161003816
2
+100381610038161003816100381611990621003816100381610038161003816
2
)1205722
minus 1198892
(53)
8 Abstract and Applied Analysis
statements (1) (2) and (3) are obvious satisfied we only needto test and verify the latter two ones Since the four statevariables have been all proved to be up bounded by 119873 wewill get the following equalities
1198781015840le 120583119873 119860
1015840le (1 minus 119906
1(119905)) 120573119878 + 120585119879
1198791015840le 1199062(119905) 119860 119876
1015840le 120575119879
(54)
so the fourth condition is set up As for the last condition
119871 (119905 1199061 1199062) ge 1198891(100381610038161003816100381611990611003816100381610038161003816
2
+100381610038161003816100381611990621003816100381610038161003816
2
)1205722
minus 1198892
(55)
is also true when we choose 1198891= min119888
12 11988822 and for all
1198892isin 119877+ 120572 = 2 The proof is complete
We next come to the core of this section
52 The Characterization of the Optimal Control With theexistence of the optimal control pairs established we nowpresent the optimality system and use a result from [30]we can easily know the existence of the solutions to theoptimality system (71) which will be gotten later Firstly wecome to discuss the theorem that relates to the character-ization of the optimal control The optimality system canbe used to compute candidates for optimal control pairsTo do this we begin by defining an augmented Hamilto-nian 119867 with penalty terms for the control constraints asfollows
119867 = 119860 (119905) +1198881
21199062
1(119905) +
1198882
21199062
2(119905)
+ 1205821[120583119873 minus (1 minus 119906
1(119905))
120573119878119860
119873minus 120583119878]
+ 1205822[(1 minus 119906
1(119905))
120573119878119860
119873+ 120585119879 minus (119906
2(119905) + 120583)119860]
+ 1205823[1199062(119905) 119860 minus (120583 + 120585 + 120575) 119879] + 120582
4(120575119879 minus 120583119876)
minus 119908111199061(119905) minus 119908
12(1 minus 119906
1(119905))
minus 119908211199062(119905) minus 119908
22(1 minus 119906
2(119905))
(56)
where 119908119894119895(119905) ge 0 are the penalty multipliers satisfying
11990811(119905) 1199061(119905) = 119908
12(119905) (1 minus 119906
1(119905))
= 0 at optimal control 119906lowast1
11990821(119905) 1199062(119905) = 119908
22(119905) (1 minus 119906
2(119905))
= 0 at optimal control 119906lowast2
(57)
Theorem6 Given optimal control pairs (119906lowast1 119906lowast
2) and solutions
119878(119905) 119860(119905) 119879(119905) 119876(119905) of the corresponding state system (50)there exist adjoint variables 120582
119894 119894 = 1 2 3 4 satisfying
1205821015840
1= 1205821(1 minus 119906
1(119905))
120573119860
119873+ 1205831205821minus 1205822(1 minus 119906
1(119905))
120573119860
119873
1205821015840
2= minus1 + 120582
1(1 minus 119906
1(119905))
120573119878
119873minus 1205822(1 minus 119906
1(119905))
120573119878
119873
+ 1205822(120583 + 119906
2(119905)) minus 120582
31199062(119905)
1205821015840
3= minus1205822120585 + 1205823(120583 + 120585 + 120575) minus 120582
4120575
1205821015840
4= 1205831205824
(58)
with the terminal conditions
120582119894(119905119891) = 0 119894 = 1 2 3 4 (59)
Furthermore (119906lowast1 119906lowast
2) are represented by
119906lowast
1= min(1max(0
120573119878119860 (1205822minus 1205821)
1198881119873
))
119906lowast
2= min(1max(0
119860 (1205822minus 1205823)
1198882
))
(60)
Proof According to Pontryagin Maximum Principle [29ndash31] we first differentiate the Hamiltonian operator 119867 withrespect to states Then the adjoint system can be written as
1205821015840
1= minus
120597119867
120597119878 120582
1015840
2= minus
120597119867
120597119860
1205821015840
3= minus
120597119867
120597119879 120582
1015840
4= minus
120597119867
120597119876
(61)
The terminal condition (56) of adjoint equations is given by120582119894(119905119891) = 0 119894 = 1 2 3 4
To obtain the necessary conditions of optimality (59) wealso differentiate theHamiltonian operator119867 with respect to119880 = (119906
1 1199062) and set them equal to zero then
120597119867
1205971199061
= 11988811199061(119905) + 120582
1
120573119878119860
119873minus 1205822
120573119878119860
119873minus 11990811+ 11990812= 0
120597119867
1205971199062
= 11988821199062(119905) minus 120582
2119860 + 120582
3119860 minus 119908
21+ 11990822= 0
(62)
By solving the optimal control we obtain
119906lowast
1=1
1198881
[(1205822minus 1205821)120573119878119860
119873+ 11990811minus 11990812] (63)
To determine an explicit expression for the optimalcontrol without119908
11and119908
12 a standard optimality technique
is utilized [29] We consider the following three cases
(i) On the set 119905 | 0 lt 119906lowast
1(119905) lt 1 we have 119908
11(119905) =
Abstract and Applied Analysis 9
11990812(119905) = 0 Hence the optimal control is
119906lowast
1=1
1198881
[(1205822minus 1205821)120573119878119860
119873] (64)
(ii) On the set 119905 | 119906lowast1(119905) = 1 we have 119908
11(119905) = 0 Hence
1 = 119906lowast
1(119905) =
1
1198881
[(1205822minus 1205821)120573119878119860
119873minus 11990812] (65)
This implies that
1
1198881
(1205822minus 1205821)120573119878119860
119873ge 1 since 119908
12(119905) ge 0 (66)
(iii) On the set 119905 | 119906lowast1(119905) = 0 we have 119908
12(119905) = 0 Hence
0 = 119906lowast
1(119905) =
1
1198881
[(1205822minus 1205821)120573119878119860
119873+ 11990811] (67)
This implies that
1
1198881
(1205822minus 1205821)120573119878119860
119873le 0 since 119908
11(119905) ge 0 (68)
Combining these results the optimal control 119906lowast1(119905) is
characterized as
119906lowast
1= min1max0 1
1198881
[(1205822minus 1205821)120573119878119860
119873] (69)
Using the similar arguments we can also obtain the otheroptimal control function
119906lowast
2= min1max0
(1205822minus 1205823) 119860
1198882
(70)
The proof is complete
We point out that the optimality system consistsof the state system (50) with the initial conditions119878(0) 119860(0) 119879(0) 119876(0) the adjoint (or costate) system(58) with the terminal conditions (59) and the optimalitycondition (60) Any optimal control pairs must satisfythis optimality system For the convenience of subsequent
numerical simulation in Section 6 we give the optimalitysystem as follows
1198781015840= 120583119873 minus (1 minusmin1max0 1
1198881
[(1205822minus 1205821)120573119878119860
119873] (119905))
times120573119878119860
119873minus 120583119878
1198601015840= (1minusmin1max0 1
1198881
[(1205822minus1205821)120573119878119860
119873] (119905))
120573119878119860
119873
+ 120585119879 minus (min1max0(1205822minus 1205823) 119860
1198882
(119905) + 120583)119860
1198791015840= min1max0
(1205822minus 1205823) 119860
1198882
(119905) 119860
minus (120583 + 120585 + 120575) 119879
1198761015840= 120575119879 minus 120583119876
1205821015840
1= 1205821(1 minusmin1max0 1
1198881
[(1205822minus 1205821)120573119878119860
119873] (119905))
times120573119860
119873+ 1205831205821
minus 1205822(1minusmin1max0 1
1198881
[(1205822minus1205821)120573119878119860
119873](119905))
times120573119860
119873
1205821015840
2= minus1
+ 1205821(1minusmin1max0 1
1198881
[(1205822minus1205821)120573119878119860
119873] (119905))
times120573119878
119873minus 1205823sdotmin1max0
(1205822minus 1205823) 119860
1198882
(119905)
minus 1205822(1 minusmin1max0 1
1198881
[(1205822minus 1205821)120573119878119860
119873])
times120573119878
119873+ 1205822(120583+min1max0
(1205822minus 1205823) 119860
1198882
(119905))
1205821015840
3= minus1205822120585 + 1205823(120583 + 120585 + 120575) minus 120582
4120575
1205821015840
4= 1205831205824
119878 (0) = 1198780 119860 (0) = 119860
0 119879 (0) = 119879
0
119876 (0) = 1198760 120582
119894(119905119891) = 0 119894 = 1 2 3 4
(71)
53 The Uniqueness of Optimal Control Due to the a prioriboundedness of the state adjoint functions and the resultingLipschitz structure of the ODEs we can obtain the unique-ness of the optimal control
10 Abstract and Applied Analysis
Lemma 7 (see [23]) The function 119906lowast(119904) = min (119887max (119904 119886))is Lipschitz continuous in 119904 where 119886 lt 119887 are some fixed positiveconstants
Theorem 8 For all 119905 isin [0 119905119891] the solution to the optimality
system (71) is unique
Proof Suppose (119878 119860 119879 119876 1205821 1205822 1205823 1205824) and
(119878 119860 119879 119876 1205821 1205822 1205823 1205824) are two different solutions of
our optimality system (71) Let
119878 = 119890120582119905119898 119860 = 119890
120582119905119899 119879 = 119890
120582119905119901
119876 = 119890120582119905119902 120582
1= 119890minus120582119905119903 120582
2= 119890minus120582119905119904
1205823= 119890minus120582119905119908 120582
4= 119890minus120582119905V
119878 = 119890120582119905119898 119860 = 119890
120582119905119899 119879 = 119890
120582119905119901
119876 = 119890120582119905119902 120582
1= 119890minus120582119905119903 120582
2= 119890minus120582119905119904
1205823= 119890minus120582119905119908 120582
4= 119890minus120582119905V
(72)
where 120582 gt 0 is to be chosenAccordingly we have
119906lowast
1(119905) = min1max0
120573119898119899 (119904 minus 119903) 119890120582119905
1198881119873
119906lowast
2(119905) = min1max0 (119904 minus 119908) 119899
1198882
119906lowast
1(119905) = min1max0
120573119898119899 (119904 minus 119903) 119890120582119905
1198881119873
119906lowast
2(119905) = min1max0 (119904 minus 119908) 119899
1198882
(73)
Now we substitute 119878 = 119890120582119905119898 119860 = 119890
120582119905119899 119879 = 119890
120582119905119901 119876 = 119890
120582119905119902
1205821= 119890minus120582119905119903 1205822= 119890minus120582119905119904 1205823= 119890minus120582119905119908 1205824= 119890minus120582119905V and 119878 =
119890120582119905119898 119860 = 119890
120582119905119899 119879 = 119890
120582119905119901 119876 = 119890
120582119905119902 1205821= 119890minus120582119905119903 1205822=
119890minus120582119905119904 1205823= 119890minus120582119905119908 1205824= 119890minus120582119905V into the first ODE of (71)
respectively then we can obtain
+ 120582119898 = 120583119873119890minus120582119905
minus (1 minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)
times120573119898119899119890
120582119905
119873minus 120583119898
(74)
for119898 and119898 respectively Similarly we can derive
119899 + 120582119899
= (1 minusmin1max1120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)120573119898119899119890
120582119905
119873
+ 120585119901 minus (min1max0 119899 (119904 minus 119908)1198882
+ 120583) 119899
(75)
for 119899 and 119899 respectively
+ 120582119901 = min1max0 119899 (119904 minus 119908)1198882
119899
minus (120583 + 120585 + 120575) 119901
(76)
for 119901 and 119901 respectively
119902 + 120582119902 = 120575119901 minus 120583119902 (77)
for 119902 and 119902 respectively
119903 minus 120582119903 = 119903(1 minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)120573119899119890120582119905
119873
+ 120583119903minus119904(1minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)
times120573119899119890120582119905
119873
(78)
for 119903 and 119903 respectively
119904 minus 120582119904 = minus119890120582119905
+ 119903(1 minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)120573119898119890120582119905
119873
minus 119904(1 minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)
times120573119898119890120582119905
119873+ 119904(min1max0 119899 (119904 minus 119908)
1198882
+ 120583)
minus 119908(min1max0 119899 (119904 minus 119908)1198882
)
(79)
for 119904 and 119904 respectively
minus 120582119908 = minus119904120585 + (120583 + 120585 + 120575)119908 minus 120575V (80)
for 119908 and 119908 respectively
V minus 120582V = 120583V (81)
for V and V respectively
Abstract and Applied Analysis 11
By Lemma 7 we can obtain
1003816100381610038161003816119906lowast
1(119905) minus 119906
lowast
1(119905)1003816100381610038161003816 le
120573119890120582119905
1198881119873|119898119899 (119904 minus 119903) minus 119898119899 (119904 minus 119903)|
1003816100381610038161003816119906lowast
2(119905) minus 119906
lowast
2(119905)1003816100381610038161003816 le
1
1198882
|119899 (119904 minus 119908) minus 119899 (119904 minus 119908)|
(82)
The equations for 119898 119899 119901 119902 119903 119904 119908 V and the equationsfor 119898 119899 119901 119902 119903 119904 119908 V are subtracted respectively then wemultiply each equation by appropriate difference of functionsand integrate from 0 to 119905
119891 Next we add all eight integral
equations and some inequality techniques to obtain unique-ness The following calculation is similar for the sake ofsimplicity we only take119898 and119898 for an example
minus + (120583 + 120582) (119898 minus 119898)
=120573119890120582119905
119873[minus(1 minusmin1max0
120573119898119899 (119904 minus 119903) 119890120582119905
1198881119873
)119898119899
+(1 minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)119898119899]
(83)
Multiplying both sides of (83) by (119898minus119898) and integratingfrom 0 to 119905
119891gives
1
2(119898 minus 119898)
2(119905119891) + (120583 + 120582)int
119905119891
0
(119898 minus 119898)2119889119905
= int
119905119891
0
(119898 minus 119898)120573119890120582119905
119873
times [minus (1 minus 119906lowast
1)119898119899 + (1 minus 119906
lowast
1)119898119899] 119889119905
= int
119905119891
0
(119898 minus 119898)120573119890120582119905
119873[(119906lowast
1minus 1) (119898119899 minus 119898119899 + 119898119899 minus 119898119899)
+ 119898119899 (119906lowast
1minus 119906lowast
1)] 119889119905
le120573119890120582119905119891
119873int
119905119891
0
(119898 minus 119898)
times [119906lowast
1minus 1
100381610038161003816100381610038161003816100381610038161003816
(119898 |119899 minus 119899| + |119898 minus 119898| 119899) + 119898119899120573119890120582119905
1198881119873
100381610038161003816100381610038161003816100381610038161003816
times 119898119899 (119904 minus 119903) minus 119898119899 (119904 minus 119903) ] 119889119905
le120573119890120582119905119891
119873int
119905119891
0
(119898 minus 119898)
times [1003816100381610038161003816119906lowast
1minus 1
1003816100381610038161003816 (119898 |119899 minus 119899| + |119898 minus 119898| 119899) + 119898119899120573119890120582119905
1198881119873
|119898119899 (119904 minus 119904) + (119898119899 minus 119898119899) 119904
minus (119898119899 (119903 minus 119903) + (119898119899 minus 119898119899) 119903)| ] 119889119905
le120573119890120582119905119891
119873int
119905119891
0
[1003816100381610038161003816119906lowast
1minus 1
1003816100381610038161003816 |119898| |119898 minus 119898| |119899 minus 119899|
+ |119898 minus 119898|2
|119899| + |119861| |119904| |119899| |119898 minus 119898|2
+ |119861| |119898119899| |119898 minus 119898| |119903 minus 119903|
+ |119861| |119898119899| |119898 minus 119898| |119904 minus 119904|
+ |119861| |119904| |119898| |119898 minus 119898| |119899 minus 119899|
+ |119861| |119903| |119898| |119898 minus 119898| |119899 minus 119899|
+ |119861| |119903| |119899| |119898 minus 119898|2] 119889119905
le120573119890120582119905119891
119873int
119905119891
0
[
1003816100381610038161003816119906lowastminus 1
1003816100381610038161003816
2|119898| ((119898 minus 119898)
2+ (119899 minus 119899)
2)
+ |119899| (119898 minus 119898)2
+|119861| |119898119899|
2((119898 minus 119898)
2+ (119904 minus 119904)
2)
+|119861| |119904| |119898|
2((119898 minus 119898)
2+ (119899 minus 119899)
2)
+ |119861| |119904| |119899| (119898 minus 119898)2
+|119861| |119898119899|
2((119898 minus 119898)
2+ (119903 minus 119903)
2)
+|119861| |119898| |119903|
2((119898 minus 119898)
2+ (119899 minus 119899)
2)
+ |119861| |119903| |119899| (119898 minus 119898)2]119889119905
(84)
In the above derivation we use many scaling techniquesfor inequality or absolute inequality Particularly what shouldbe noted is that to get the first inequality of above derivationwe use the estimation of |119906lowast
1(119905) minus 119906
lowast
1(119905)| which has been given
before besides for the sake of convenience we note 119861 =
119898119899(120573119890120582119905119891119873) Furthermore we notice that the coefficients of
all the eight terms in the last formula (119898 minus 119898)2+ (119899 minus 119899)
2(119898 minus119898)
2 (119898 minus119898)2+ (119904 minus 119904)
2 (119898 minus119898)2+ (119899 minus 119899)
2 (119898 minus119898)2
(119898minus119898)2+(119903minus119903)
2 (119898minus119898)2+(119899minus119899)2 (119898minus119898)2 namely (|119906lowast1minus
1|2)|119898| |119899| (|119861||119898119899|)2 (|119861||119904||119898|)2 |119861||119904|119899 (|119861||119898119899|)2(|119861||119898||119903|)2 |119861||119903||119899| are nonnegative and bounded So thereexists a positive constant 119888
2such that
1
2(119898 minus 119898)
2(119905119891) + (120583 + 120582)int
119905119891
0
(119898 minus 119898)2119889119905
12 Abstract and Applied Analysis
le 1198882
120573119890120582119905119891
119873int
119905119891
0
((119898 minus 119898)2+ (119899 minus 119899)
2
+ (119904 minus 119904)2+ (119903 minus 119903)
2) 119889119905
(85)
Combining eight of these inequalities gives
1
2(119898 minus 119898) (119905
119891) +
1
2(119899 minus 119899) (119905
119891) +
1
2(119901 minus 119901) (119905
119891)
+1
2(119902 minus 119902) (119905
119891) +
1
2(119903 minus 119903) (0) +
1
2(119904 minus 119904) (0)
+1
2(119908 minus 119908) (0) +
1
2(V minus V) (0) + (120583 + 120582)
times int
119905119891
0
(119898 minus 119898)2+ (119899 minus 119899)
2+ (119901 minus 119901)
2
+ (119902 minus 119902)2
+ (119904 minus 119904)2
+ (119903 minus 119903)2+ (119908 minus 119908)
2+ (V minus V)2) 119889119905
le 119861int
119905119891
0
(119898 minus 119898)2+ (119899 minus 119899)
2+ (119901 minus 119901)
2
+ (119902 minus 119902)2
+ (119903 minus 119903)2+ (119904 minus 119904)
2
+ (119908 minus 119908)2+ (V minus V)2 119889119905
(86)
Thus from the above inequality we can conclude that
(120583 + 120582 minus 119861)int
119905119891
0
(119898 minus 119898)2+ (119899 minus 119899)
2
+ (119901 minus 119901)2
+ (119902 minus 119902)2
+ (119903 minus 119903)2
+ (119904 minus 119904)2+(119908 minus 119908)
2+ (V minus V)2 119889119905 le 0
(87)
where 119861 depends on the coefficients and the bounds dependon119898 119899 119901 119902 119903 119904 119908 V If we choose 120582 such that 120583+120582 gt 119861 then119898 = 119898 119899 = 119899 119901 = 119901 119902 = 119902 119903 = 119903 119904 = 119904 119908 = 119908 and V = VHence the solution to the optimality system is unique Theproof is complete
6 Numerical Simulation
61 The Simulation of State System (1) without ControlParameters For the sake of simplicity but without loss ofgenerality we will perform the numerical simulation of statesystem (1) with parameters 119906
1= 0 119906
2= 0 Before
illustrating the analytic properties of the alcoholism model(1) we will target the populations in the environment ofa community or a university for example the school ofmaterial science and engineering in our university that isLan zhou University of Technology (LUT for short) owing tothe accurate and available information we can obtain Refer-ring to the information provided by the admissions officeof LUT this school will enroll almost 1200 undergraduatesand almost 300 various postgraduates at the beginning of
fall semester at the same time there will be almost 1500various students graduated and left this school so the scaleof students in school remained almost 6000 we can take thetotal population 119873 = 6000 In this simulation we will takeSeptember as the initial time and units in one week periodin one year According to the investigations of the studentunion implemented in September every year we can takeinitial values as 119878(0) = 4500 119860(0) = 1000 119879(0) = 300and 119876(0) = 200 It seems that the alcoholism is a little bitmore but it is rather natural because many freshmen feelconfused when they are faced with the new environment anda new lifestyle many of them have no better choice but gathertogether to drink in small groups to mediate the anxiety andget to know each other over time some of them developthe habit of drinking To a certain extent for example thefrequent drinking badly affects their study we can classifythem into the alcoholism compartment Other initial valuesseem more reasonable so we need no more explanation Aswe know alcoholism death is seldom happen within oneyear so we omit mortality from alcoholism then how tounderstand the recruitment rate as well as natural death rate120583 We can treat the freshmen admission as the recruitmentpopulation and graduation students as the natural ldquodeathrdquoparts So we can take 120583 = 15006000 = 025 which is exactlyconsistent with the value in [16] As for the infection rate 120573and recovery rate 120575 we will let them be variables since thedrinking behaviors are related to many factors such as theseason and the pressure
According to the data we get from the student unionwe choose 120585 = 04 To summarize we list the values ofthe parameters in Table 1 Using the values of parameters inTable 1 we can plot Figures 2 and 3 which are on condition1198770lt 1 and 119877
0ge 1 respectively From Figure 2 we easily
know when 1198770lt 1 holds the solution of system (1) tends to
the alcohol free equilibrium1198640and verifies the global stability
of 1198640 While seen from Figure 3 we also know that if 119877
0ge
1 holds the solution of system (1) tends to the alcoholismequilibrium 119864
lowast and verifies the global stability of 119864lowast
62 The Sensitivity of 1198770about Two Control Parameters
Although from the expression of the model reproductionnumber 119877
0 we can easily find out the fact that the two
control variables that is 1199061and 119906
2 attribute to reducing the
severity of alcoholism we will still depict the graph between1198770and the two control variables to see more intuitiveness see
Figure 4 It seems from the figure that 1198770is a monotonically
decreasing function about two control parameters so it isadvisable to take two approaches simultaneously to controlthe alcoholics
63 The Simulation of Optimality System (71) In this subsec-tion we will investigate numerically the optimal solution tooptimality system (71) by numerical method from [32] theoptimality system is solved with a fourth-order Runge-Kuttascheme Beginning with a guess for the control variables thestate system is solved forward in time and then those values ofstate are used to solve the adjoint equations backward in timeThe controls are updated at the end of each iteration using
Abstract and Applied Analysis 13
0 10 20 30 40 50
0
1000
2000
3000
4000
5000
6000
7000
Time
Popu
latio
n siz
e
S(t)
A(t)
T(t)
Q(t)
Figure 2 When 1198770lt 1 the alcohol free equilibrium 119864
0is globally
asymptotically stable (120573 = 02 120575 = 03 and 1198770= 071698)
S(t)
A(t)
T(t)
Q(t)
0 10 20 30 40
48
2520
303540
49 50
500
1000
2000
3000
4000
5000
6000
7000
Time
Popu
latio
n siz
e
Figure 3 When 1198770
ge 1 the alcoholism equilibrium 119864lowast
=
(4466 476 32 26) corresponding to the given parameters is globallyasymptotically stable (120573 = 03 120575 = 02 and 119877
0= 108511)
the values of optimal controls obtained lastly The iterationscontinue until convergence takes place
In the simulations we choose the available variable valuesas Table 1 shows besides 120573 = 03 120575 = 02 The initial valueof model (1) is assumed to be 119878(0) = 4500 119860(0) = 1000119879(0) = 300 and 119876(0) = 200 as before
The ideal weights in objective functional are very difficultto obtain in reality it needs much work on data mining andfittingHence the acquisition of appropriate practical weights
Table 1 The parameters description of model (1)
Parameter Description Values120583 Natural birth rate or death rate 025
120573Transmission coefficient betweenalcoholism and susceptibles Variables
119873 The total populations to be considered 6000
120575The rate of populations quitting fromalcoholism permanently after treatment Variables
120585The rate of populations failed intreatment and returned to be alcoholic 04
0
02
04
06
08
1
0
05
1
0 02 04 06 08 1
u1
u2
R0
Figure 4 The relationship between 1198770and two control variables 119906
1
and 1199062
is still a difficult problem and remains for further investiga-tionsThe cost associated with119860(119905) and 119906
1(119905)mainly includes
the cost of dangerous behaviours during the alcoholism timeand educating the public while the cost associated with1199062(119905)mainly comes from health professional and the medical
resource includingmedicines andnursing care In viewof thisand taking the expressions of 119906
1 1199062into account after many
numerical simulations we finally give weighting coefficientsas 1198881= 102 1198882= 104 It should be pointed out that the
weights here are of only theoretical interest to reveal thecontrol strategies proposed in this paper Another point tonote is that themaximumcontrol is very difficult to achieve inreality so we will omit the situation of the maximum controlduring the series of simulations
Next we will make some necessary instructions andexplanations to the above simulation graphs Figures 5 67 and 8 depict the number of four compartments underdifferent control levels when we choose the weight coeffi-cients in objective function to be 119888
1= 102 1198882= 104 From
the four simulation graphs we can observe the followingsimple facts in reducing the total number of alcoholismsand increasing the number of susceptibles the effectivenessof various control measures is as follows optimal control isevidently better than middle control and middle control isbetter than single control 119906
2 single control 119906
2is better than
single control 1199061 while single control 119906
1is much better than
no controlFigures 9 and 10 depict the optimal control law of 119906
1 1199062
respectively In the beginning of the simulation the control
14 Abstract and Applied Analysis
0 10 20 30 40 504500
5000
5500
6000
6500
7000
S(n
umbe
r of s
usce
ptib
les)
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 5Number of the susceptibles whenwe choose theweights inobjective function are 119888
1= 102 1198882= 104 and under different optimal
control strategies that is (1) without control 1199061= 1199062= 0 (2) single
control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0 119906
2= 05 (4)
middle control 1199061= 1199062= 05 (5) optimal control 119906
1= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
500
1000
1500
A(n
umbe
r of t
he al
coho
lism
)
t (unit week)
Figure 6 Number of the alcoholics when we choose the weights inobjective function are 119888
1= 102 1198882= 104 and under different optimal
control strategies that is (1) without control 1199061= 1199062= 0 (2) single
control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0 119906
2= 05 (4)
middle control 1199061= 1199062= 05 (5) optimal control 119906
1= 119906lowast
1 1199062= 119906lowast
2
0
50
100
150
200
250
300
350
400
T(n
umbe
r of t
reat
men
t)
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 7 Number of the persons in treatment when we choose theweights in objective function are 119888
1= 10
2 1198882= 10
4 and underdifferent optimal control strategies that is (1) without control 119906
1=
1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control 119906
1= 0
1199062= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal control
1199061= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
50
100
150
200
250
300
350
400
Q(n
umbe
r of q
uitti
ng)
Figure 8 Number of people in quitting compartment when wechoose the weights in objective function are 119888
1= 102 1198882= 104 and
under different optimal control strategies that is (1) without control1199061= 1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control
1199061= 0 119906
2= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal
control 1199061= 119906lowast
1 1199062= 119906lowast
2
Abstract and Applied Analysis 15
0 10 20 30 40 50t (unit week)
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
Optimal control of u1
Figure 9 Figures of the optimal control 1199061when we choose the
weight in objective function are 1198881= 102 1198882= 104
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
0 10 20 30 40 50t (unit week)
Optimal control of u2
Figure 10 Figures of the optimal control 1199062when we choose the
weight in objective function are 1198881= 102 1198882= 104
effort of 1199061should be decreased from 065 to 05 within the
first month and over the next week it should be increased tothe maximum control until 50 weeks then rapidly decreasedto 0 at the end of the simulation As for the control 119906
2 it
should start from around 05 due to the initial alcoholics thenincrease to 055 within one week since the rapid infection andnext decrease to almost 0 since the effectiveness of treatmentin three weeks but with the infection going on the controleffort of 119906
2should gradually increase to the maximum and
maintain this level until the tenth week for the purpose ofconsolidation therapy and preventing rebound hereafter itshould be gradually decreased to the level of almost 043 untilthe fifty weeks then quickly decreased to 0 in the end
0 10 20 30 40 504500
5000
5500
6000
6500
7000
S(n
umbe
r of s
usce
ptib
les)
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 11 Number of the susceptibles when we choose the weightsin objective function are 119888
1= 104 1198882= 102 and under different
optimal control strategies that is (1) without control 1199061= 1199062= 0
(2) single control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0
1199062= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal control
1199061= 119906lowast
1 1199062= 119906lowast
2
In order to investigate the influence of different weightcoefficients in the objective functional on the effect ofcontrolling at the same time for a better comparison we willchange the weight coefficients in objective function into 119888
2=
104 1198881= 102 and we will list the corresponding numerical
simulation results as Figures 11ndash16 showWhen we change the weight coefficients in objective
function into 1198881= 104 1198882= 102 we find that the results
of simulations derived from the graphs are very similar tothe ones before We speculate that the most likely reasons ofthis result are due to three respects one is that the weightcoefficients are not too sensitive in the numerical simulationand another possible reason is that both of the two controlsare important in some sense equivalently importantThe lastbut not themost unlikely reason is that we have not found themost appropriate weight coefficients in the simulation whichis very difficult to find as previously mentioned
7 Conclusions
In this paper we formulate an alcoholics quitting modeland firstly investigate the variation discipline of variouspopulations from the perspective of global stability thenwe propose an objective functional to examine two differentcontrolmeasures (ie prevention and treatment) on the effectof alcohol The basic reproduction number of the model wasderived and the global stability of the two equilibria is givenFrom the expression of the basic reproductionnumber119877
0and
16 Abstract and Applied Analysis
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
500
1000
1500
A(n
umbe
r of t
he al
coho
lism
)
t (unit week)
Figure 12 Number of the alcoholics when we choose the weights inobjective function are 119888
1= 104 1198882= 102 and under different optimal
control strategies that is (1) without control 1199061= 1199062= 0 (2) single
control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0 119906
2= 05 (4)
middle control 1199061= 1199062= 05 (5) optimal control 119906
1= 119906lowast
1 1199062= 119906lowast
2
0
50
100
150
200
250
300
350
400
T(n
umbe
r of t
reat
men
t)
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 13 Number of the persons in treatment when we choosethe weights in objective function are 119888
1= 104 1198882= 102 and under
different optimal control strategies that is (1) without control 1199061=
1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control 119906
1= 0
1199062= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal control
1199061= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
50
100
150
200
250
300
350
400
Q(n
umbe
r of q
uitti
ng)
Figure 14 Number of people in quitting compartment when wechoose the weights in objective function are 119888
1= 104 1198882= 102 and
under different optimal control strategies that is (1) without control1199061= 1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control
1199061= 0 119906
2= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal
control 1199061= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50t (unit week)
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
Optimal control of u1
Figure 15 Figures of the optimal control 1199061when we choose the
weight in objective function are 1198881= 104 1198882= 102
related numerical simulation we can easily see that the twocontrol strategies are effective in the alcoholics process
Using Pontryaginrsquos Maximum Principle we firstly deter-mine the necessary conditions for existence of optimalcontrol pairsThe uniqueness of the solution to the optimalitysystem (71) is derived by the classical method of contradic-tion Numerical simulations of the model suggest that thetwo different groups of weights in the objective function have
Abstract and Applied Analysis 17
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
0 10 20 30 40 50t (unit week)
Optimal control of u2
Figure 16 Figures of the optimal control 1199062when we choose the
weight in objective function are 1198881= 104 1198882= 102
much similar effects on the transmission of the alcoholismfrom this point the two control measures are almost equallyimportant in controlling the alcoholism although they willprobably have great influences on the cost of the objectivefunction From the simulation figures it seems that the effectof optimal control which is measured by the reduction inthe number of alcoholics and the increase in the number ofsusceptibles is much better than other control strategies asnoted earlier in the simulation section According to the real-time curve of two optimal controls we point out the specificimplementation methods of optimal control which can beachieved in practice
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
This work was partially supported by the NSF ofGansu Province of China (2013GS09485) the NNSF ofChina (10961018) the NSF of Gansu Province of China(1107RJZA088) the NSF for Distinguished Young Scholarsof Gansu Province of China (1111RJDA003) the Special Fundfor the Basic Requirements in the Research of University ofGansu Province of China and the Development Programfor Hong Liu Distinguished Young Scholars in LanzhouUniversity of Technology
References
[1] R Room T Babor and J Rehm ldquoAlcohol and public healthrdquoThe Lancet vol 365 no 9458 pp 519ndash530 2005
[2] J B Saunders O G Aasland A Amundsen and M GrantldquoAlcohol consumption and related problems among primary
health care patients WHO collaborative project on early detec-tion of personswith harmful alcohol consumption IrdquoAddictionvol 88 no 3 pp 349ndash362 1993
[3] L D Johnston P M OMalley and J G Bachman NationalSurvey Results on Drug Use From the Monitoring the FutureStudy 1975ndash1992 National Institute on Drug Abuse RockvilleMd USA 1993
[4] H Wechsler J E Lee M Kuo and H Lee ldquoCollege bingedrinking in the 1990s a continuing problem Results of theHarvard School of Public Health 1999 College Alcohol StudyrdquoJournal of American College Health vol 48 no 5 pp 199ndash2102000
[5] P M OrsquoMalley and L D Johnston ldquoEpidemiology of alcoholand other drug use among American college studentsrdquo Journalof Studies on Alcohol vol 63 no 14 pp 23ndash39 2002
[6] K Agarwal-Kozlowski and D P Agarwal ldquoGenetic predisposi-tion to alcoholismrdquo Therapeutische Umschau vol 57 no 4 pp179ndash184 2000
[7] C Y Chen C L Storr and J C Anthony ldquoEarly-onset drug useand risk for drug dependence problemsrdquo Addictive Behaviorsvol 34 no 3 pp 319ndash322 2009
[8] M Glavas and J Weinberg ldquoStress alcohol consumption andthe hypothalamic-pituitary adrenal axisrdquo in Nutrients Stressand Medical Disorders pp 165ndash183 Humana Press New YorkNY USA 2006
[9] L N Blum N H Nielsen J A Riggs and L B BresolinldquoAlcoholism and alcohol abuse among women report of theCouncil on Scientific Affairsrdquo Journal of Womenrsquos Health vol7 no 7 pp 861ndash871 1998
[10] B Benedict ldquoModeling alcoholism as a contagious disease howldquoinfectedrdquo drinking buddies spread problem drinkingrdquo SIAMNews vol 40 no 3 2007
[11] H Walter K Gutierrez K Ramskogler I Hertling A Dvorakand O M Lesch ldquoGender-specific differences in alcoholismImplications for treatmentrdquo Archives of Womenrsquos Mental Healthvol 6 no 4 pp 253ndash258 2003
[12] D Muller R D Koch H von Specht w Volker and E MMunch ldquoNeurophisiologic findings in chronic alcohol abuserdquoPsychiatrie Neurologie und Medizinische Psychologie vol 37no 3 pp 129ndash132 1985
[13] G Testino ldquoAlcoholic diseases in hepato-gastroenterology apoint of viewrdquo Hepato-Gastroenterology vol 55 no 82-83 pp371ndash377 2008
[14] A Mubayi P E Greenwood C Castillo-Chavez P J Grue-newald and D M Gorman ldquoThe impact of relative residencetimes on the distribution of heavy drinkers in highly distinctenvironmentsrdquo Socio-Economic Planning Sciences vol 44 no 1pp 45ndash56 2010
[15] D Muller R D Koch H von Specht W Volker and E MMunch ldquoNeurophysiologic findings in chronic alcohol abuserdquoPsychiatrie Neurologie Und Medizinische Psychologie vol 37no 3 pp 129ndash132 1985 (Leipz)
[16] G Mulone and B Straughan ldquoModeling binge drinkingrdquoInternational Journal of Biomathematics vol 5 no 1 Article ID1250005 14 pages 2012
[17] H F Huo and N N Song ldquoGlobal stability for a binge drinkingmodel with two stagesrdquo Discrete Dynamics in Nature andSociety vol 2012 Article ID 829386 15 pages 2012
[18] S S Mushayabasa and C P Bhunu ldquoModelling the effects ofheavy alcohol consumption on the transmission dynamics ofgonorrheardquo Nonlinear Dynamics vol 66 no 4 pp 695ndash7062011
18 Abstract and Applied Analysis
[19] F S Sancheznchez XWang C Castillo-Chvez DM Gormanand P J Gruenewald ldquoDrinking as an epidemic a simplemathematical model with recovery and relapserdquo in TherapistsGuide to Evidence-Based Relapse Prevention K Witkiewitz andG Marlatt Eds Academic Press New York NY USA 2007
[20] H-F Huo and C-C Zhu ldquoInfluence of relapse in a givingup smoking modelrdquo Abstract and Applied Analysis vol 2013Article ID 525461 12 pages 2013
[21] H F Huo and Q Wang ldquoThe effects of awareness programs by17 media on the drinking dynamicsrdquo Submitted
[22] S Lee E Jung and C Castillo-Chavez ldquoOptimal controlintervention strategies in low- and high-risk problem drinkingpopulationsrdquo Socio-Economic Planning Sciences vol 44 no 4pp 258ndash265 2010
[23] K R Fister S Lenhart and J S McNally ldquoOptimizingchemotherapy in an HIV modelrdquo Electronic Journal of Differ-ential Equations vol 1998 no 32 pp 1ndash12 1998
[24] S Lenhart and J T Workman Optimal Control Applied toBiological Models Chapman amp HallCRC Mathematical andComputational Biology Series Chapman amp HallCRC BocaRaton Fla USA 2007
[25] X Yang L Chen and J Chen ldquoPermanence and positiveperiodic solution for the single-species nonautonomous delaydiffusive modelsrdquo Computers amp Mathematics with ApplicationsAn International Journal vol 32 no 4 pp 109ndash116 1996
[26] P van denDriessche and JWatmough ldquoReproduction numbersand sub-threshold endemic equilibria for compartmental mod-els of disease transmissionrdquoMathematical Biosciences vol 180pp 29ndash48 2002
[27] V Lakshmikantham S Leela and A A Martynyuk StabilityAnalysis of Nonlinear Systems vol 125 Marcel Dekker NewYork NY USA 1989
[28] J P LaSalle The Stability of Dynamical Systems vol 25 ofRegional Conference Series in Applied Mathematics Society forIndustrial and Applied Mathematics 1976
[29] W H Fleming and R W Rishel Deterministic and StochasticOptimal Control vol 1 ofApplications of Mathematics SpringerBerlin Germany 1975
[30] D L Lukes Differential Equations Classical to ControlledMathematics in Science and Engineering Academic Press NewYork NY USA 1982
[31] L S Pontryagin V G Boltyanskii R V Gamkrelidze and EF Mishchenko The Mathematical Theory of Optimal ProcessesJohn Wiley amp Sons New Jersey NJ USA 1962
[32] W Hackbusch ldquoA numerical method for solving parabolicequations with opposite orientationsrdquo Computing vol 20 no3 pp 229ndash240 1978
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Differential EquationsInternational Journal of
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Applied MathematicsJournal of
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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
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Mathematical PhysicsAdvances in
Complex AnalysisJournal of
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OptimizationJournal of
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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
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Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
4 Abstract and Applied Analysis
It is easy to see that both of the control parameters con-tributed to reducing the alcoholism From this point thecontrol measures are meaningful
32 Existence of Alcoholism Equilibrium The endemic equi-librium 119864
lowast(119878lowast 119860lowast 119879lowast 119876lowast) of system (1) is determined by
equations
120583119873 minus (1 minus 1199061)120573119878119860
119873minus 120583119878 = 0
(1 minus 1199061)120573119878119860
119873+ 120585119879 minus (119906
2+ 120583)119860 = 0
1199062119860 minus (120583 + 120585 + 120575) 119879 = 0
120575119879 minus 120583119876 = 0
(12)
The third equation in (12) leads to
119860 =120583 + 120585 + 120575
1199062
119879 (13)
From the last equation in (12) we have
119876 =120575119879
120583 (14)
From the first equation of (12) and together with (13) we canget
119878 =120583119873
120583 + (((1 minus 1199061) 120573) 119873)119860
=12058311987321199062
1205831198731199062+ (1 minus 119906
1) 120573119879 (120583 + 120585 + 120575)
(15)
Substituting (13)ndash(15) into the second equation of (12) gives
120583119873 minus120583211987321199062
1205831198731199062+ (1 minus 119906
1) 120573119879 (120583 + 120585 + 120575)
+ 120585119879 minus (120583 + 1199062)120583 + 120585 + 120575
1199062
119879 = 0
(16)
By simplifying (16) we can get
119879 [1199062120585 (1 minus 119906
1) 120573 (120583 + 120585 + 120575)
minus (1 minus 1199061) 120573(120583 + 120585 + 120575)
2
(120583 + 1199062)] 119879 + 120590 = 0
(17)
where120590 = 119906
2120583119873 (1 minus 119906
1) 120573 (120583 + 120585 + 120575)
+ (1199062)2
120583119873120585 minus 1205831198731199062(120583 + 119906
2) (120583 + 120585 + 120575)
(18)
Hence we get two explicit solutions to (17) one is 1198790= 0
which is corresponding to the alcohol free equilibria and theother is119879lowast= (120590) ((1 minus 119906
1) 120573(120583 + 120585 + 120575)
2
(120583 + 1199062)
minus1199062120585 (1 minus 119906
1) 120573 (120583 + 120585 + 120575) )
minus1
=120590
(1 minus 1199061) 120573 (120583 + 120585 + 120575) [120583120585 + (120575 + 120583) (120583 + 119906
2)]
(19)
which should be corresponding to the alcoholism equilibriaon condition that 119879lowast gt 0 otherwise the alcoholismequilibria are nonexistent It is enough to show the positivityof 120590 tomake sure the existence of alcoholism equilibria on thecondition 119877
0ge 1 By some simple calculations we simplify
the expression of 120590 to be
120590 = 1205831198731199062(120583 + 120585 + 120575) (1 minus 119906
1) 120573
minus [(120583 + 1199062) (120575 + 120583) + 120583120585]
(20)
Since 1198770gt 1 is equivalent to
120573 (1 minus 1199061) (120583 + 120585 + 120575) gt 119906
2(120583 + 120575) + 120583 (120583 + 120585 + 120575) (21)
the right side of this inequality is exactly equal to (120583+1199062)(120575+
120583) + 120583120585 Hence we have proved the existence of 119879lowast gt 0so are the alcoholism equilibria We summarize this result inTheorem 1
Theorem 1 For system (1) there is always an alcohol freeequilibrium 119864
0= (119873 0 0 0) When 119877
0gt 1 besides alcohol
free equilibrium 1198640 system (1) also has a unique alcoholism
equilibrium 119864lowast(119878lowast 119860lowast 119879lowast 119876lowast) where
119878lowast=
12058311987321199062
1205831198731199062+ (1 minus 119906
1) 120573119879lowast (120583 + 120585 + 120575)
119860lowast=120583 + 120585 + 120575
1199062
119879lowast
119876lowast=120575119879lowast
120583
119879lowast=
120590
(1 minus 1199061) 120573 (120583 + 120585 + 120575) [120583120585 + (120575 + 120583) (120583 + 119906
2)]
(22)
4 Stability Analysis of Equilibria
For the convenience of subsequent proof we denote a vector119883 = (119860 119879 119876 119878)
119879 and
119891 (119883) =(
(1 minus 1199061)120573119878119860
119873+ 120585119879 minus (119906
2+ 120583)119860
1199062119860 minus (120583 + 120585 + 120575) 119879
120575119879 minus 120583119876
120583119873 minus (1 minus 1199061)120573119878119860
119873minus 120583119878
) (23)
So the Jacobian matrix of 119891(119909) about vector 119883 is as thefollowing
119869 =
120597119891 (119883)
120597119883
=(
(1 minus 1199061) 120573119878
119873
minus (120583 + 1199062) 120585 0
(1 minus 1199061) 120573119860
119873
1199062 minus (120583 + 120585 + 120575) 0 0
0 120575 minus120583 0
minus
(1 minus 1199061) 120573119878
119873
0 0 minus120583 minus
(1 minus 1199061) 120573119860
119873
)
(24)
Theorem 2 For system (1) the alcohol free equilibrium 1198640is
locally asymptotically stable if 1198770lt 1
Abstract and Applied Analysis 5
Proof Since
119869 (1198640)
= (
(1 minus 1199061) 120573 minus (120583 + 119906
2) 120585 0 0
1199062
minus (120583 + 120585 + 120575) 0 0
0 120575 minus120583 0
minus (1 minus 1199061) 120573 0 0 minus120583
)
(25)
we can easily get that two of the eigenvalues are 1205821= 1205822=
minus120583 lt 0 while 1205823 1205824satisfy
1205822+ [2120583 + 120585 + 120575 + 119906
2minus (1 minus 119906
1) 120573] 120582
+ (120583 + 120585 + 120575) (120583 + 1199062minus (1 minus 119906
1) 120573) minus 119906
2120585 = 0
(26)
Thus
1205823+ 1205824= (1 minus 119906
1) 120573 minus (120583 + 120585 + 120575) minus (120583 + 119906
2) (27)
Since 1198770lt 1 is equivalent to
120573 (1 minus 1199061) (120583 + 120585 + 120575) lt 119906
2(120583 + 120575) + 120583 (120583 + 120585 + 120575)
lt (120583 + 1199062) (120583 + 120585 + 120575)
(28)
so
120573 (1 minus 1199061) lt 120583 + 119906
2 (29)
and then
1205823+ 1205824lt minus (120583 + 120585 + 120575) lt 0 (30)
while
12058231205824= (120583 + 120585 + 120575) (120583 + 119906
2minus (1 minus 119906
1) 120573) minus 119906
2120585 (31)
Similarly from 1198770lt 1 we can derive the inequality
minus120573 (1 minus 1199061) (120583 + 120585 + 120575) gt minus119906
2(120583 + 120575) minus 120583 (120583 + 120585 + 120575) (32)
so12058231205824gt (120583 + 119906
2) (120583 + 120585 + 120575)
minus 1199062120585 minus 1199062(120583 + 120575) minus 120583 (120583 + 120585 + 120575)
(33)
It reduces to
12058231205824gt 0 (34)
Hence Re 1205823lt 0 Re 120582
4lt 0 The proof is complete
Next we will turn to investigate the global stability of 1198640
Theorem 3 For system (1) the alcohol free equilibrium 1198640is
globally asymptotically stable if 1198770lt 1
Proof Consider the subsystem of (1) as follows
1198601015840= (1 minus 119906
1)120573119878119860
119873+ 120585119879 minus (119906
2+ 120583)119860
1198791015840= 1199062119860 minus (120583 + 120585 + 120575) 119879
1198761015840= 120575119879 minus 120583119876
(35)
Equation (35) can be rewritten as
(
) = (119865 minus 119881)(
119860
119879
119876
) minus (1 minus119878
119873)
times (
120573 (1 minus 1199061) 0 0
0 0 0
0 0 0
)(
119860
119879
119876
)
(36)
Since 119878 le 119873 and 0 le 1199061le 1 then for all 119905 gt 0 we can get
(
) le (119865 minus 119881)(
119860
119879
119876
) (37)
According to Lemma 1 in [26] all the eigenvalues of matrix119865 minus 119881 have negative real parts so the solutions of this sub-system are stable whenever 119877
0lt 1 So (119860(119905) 119879(119905) 119876(119905)) rarr
(0 0 0) as 119905 rarr infin By the comparison theorem [27] andbased on the fact that the total population is constant 119873 itfollows that (119860(119905) 119879(119905) 119876(119905)) rarr (0 0 0) and 119878(119905) rarr 119873
as 119905 rarr infin So the alcohol free equilibrium 1198640is globally
asymptotically stable the proof is complete
Theorem 4 For system (1) the alcoholism equilibrium119864lowast(119878lowast 119860lowast 119879lowast 119876lowast) is globally asymptotically stable if 119877
0gt 1
Proof Since the total population in model (1) is a constantnumber119873 in order to prove the global stability of system (1)it is sufficed to prove the corresponding stability of subsystem(38)
1198781015840= 120583119873 minus (1 minus 119906
1)120573119878119860
119873minus 120583119878
1198601015840= (1 minus 119906
1)120573119878119860
119873+ 120585119879 minus (119906
2+ 120583)119860
1198791015840= 1199062119860 minus (120583 + 120585 + 120575) 119879
(38)
We make normalization transform and still use the samesymbols 119878 119860 119879 to denote the variables then (38) can betransformed into
1198781015840= 120583 minus (1 minus 119906
1) 120573119878119860 minus 120583119878
1199041198601015840= (1 minus 119906
1) 120573119878119860 + 120585119879 minus (119906
2+ 120583)119860
1198791015840= 1199062119860 minus (120583 + 120585 + 120575) 119879
(39)
From (39) we can easily know that the equilibria (119878lowast 119860lowast 119879lowast)satisfy the following three equalities to be used later
120583 = (1 minus 1199061) 120573119878lowast119860lowastminus 120583119878lowast
(1 minus 1199061) 120573119878lowast119860lowast+ 120585119879lowast= (1199062+ 120583)119860
lowast
1199062119860lowast= (120583 + 120585 + 120575) 119879
lowast
(40)
6 Abstract and Applied Analysis
Let 119881 = 1199091(119878 minus 119878
lowastminus 119878lowast ln(119878119878lowast)) + 119909
2(119860 minus 119860
lowastminus
119860lowast ln(119860119860lowast)) + 119909
3(119879 minus 119879
lowastminus 119879lowast ln(119879119879lowast)) then
119881101584010038161003816100381610038161003816(39)
= 1199091[120583 minus (1 minus 119906
1) 120573119878119860 minus 120583119878
minus119878lowast
119878120583 + (1 minus 119906
1) 120573119878lowast119860 + 120583119878
lowast]
+ 1199092[ (1 minus 119906
1) 120573119878119860 + 120585119879 minus (119906
2+ 120583)119860
minus (1 minus 1199061) 120573119878119860
lowastminus119860lowast
119860120585119879 + (119906
2+ 120583)119860
lowast]
+ 1199093[1199062119860 minus (120583 + 120585 + 120575) 119879
minus119879lowast
1198791199062119860 + (120583 + 120585 + 120575) 119879
lowast]
= 1199091[ (1 minus 119906
1) 120573119878lowast119860lowast+ 120583119878lowastminus (1 minus 119906
1) 120573119878119860 minus 120583119878
minus119878lowast
119878((1 minus 119906
1) 120573119878lowast119860lowast+ 120583119878lowast)
+ (1 minus 1199061) 120573119878lowast119860 + 120583119878
lowast]
+ 1199092[ (1 minus 119906
1) 120573119878119860 + 120585119879 minus (119906
2+ 120583)119860
minus (1 minus 1199061) 120573119878119860
lowastminus119860lowast
119860120585119879 + (119906
2+ 120583)119860
lowast]
+ 1199093[1199062119860 minus (120583 + 120585 + 120575) 119879
minus119879lowast
1198791199062119860 + (120583 + 120585 + 120575) 119879
lowast]
= 1199091120583119878lowast(2 minus
119878
119878lowastminus119878lowast
119878)
+ [1199091(1 minus 119906
1) 120573119878lowast119860lowast+ 1199092(1199062+ 120583)119860
lowast
+ 1199093(120583 + 120585 + 120575) 119879
lowast]
minus [1199091
(119878lowast)2
(1 minus 1199061) 120573119860lowast
119878+ 1199092(1 minus 119906
1) 120573119878119860
lowast
+ 1199092
119860lowast120585119879
119860+ 1199093
119879lowast
1198791199062119860]
+ 119878119860 [minus1199091(1 minus 119906
1) 120573 + 119909
2(1 minus 119906
1) 120573]
+ 119860 [minus (1199062+ 120583) 119909
2+ 1199091(1 minus 119906
1) 120573119878lowast+ 11990621199093]
+ 119879 [1205851199092minus (120583 + 120585 + 120575) 119909
3]
(41)
To eliminate the cross-term 119878119860 and two single-variable terms119860 and 119879 we let
minus1199091(1 minus 119906
1) 120573 + 119909
2(1 minus 119906
1) 120573 = 0
minus (1199062+ 120583) 119909
2+ 1199091(1 minus 119906
1) 120573119878lowast+ 11990621199093= 0
1205851199092minus (120583 + 120585 + 120575) 119909
3= 0
(42)
By solving them we can get
1199091= 1 119909
2= 1
1199093=
120585
120583 + 120585 + 120575=1199062+ 120583 minus (1 minus 119906
1) 120573119878lowast
1199062
(43)
Next we let
1198811015840
1= 1199091(1 minus 119906
1) 120573119878lowast119860lowast+ 1199092(1199062+ 120583)119860
lowast
+ 1199093(120583 + 120585 + 120575) 119879
lowast
1198811015840
2= minus[119909
1
(119878lowast)2
(1 minus 1199061) 120573119860lowast
119878
+ 1199092(1 minus 119906
1) 120573119878119860
lowast+ 1199092
119860lowast120585119879
119860+ 1199093
119879lowast
1198791199062119860]
(44)
and then
1198811015840= 1199091120583119878lowast(2 minus
119878
119878lowastminus119878lowast
119878) + 119881
1015840
1+ 1198811015840
2 (45)
Due to
(1199062+ 120583)119860
lowast= (1 minus 119906
1) 120573119878lowast119860lowast+ 120585119879lowast
1199093(120583 + 120585 + 120575) 119879
lowast= 120585119879lowast1199092= 120585119879lowast
(46)
Abstract and Applied Analysis 7
so
1198811015840
1= 2 (1 minus 119906
1) 120573119878lowast119860lowast+ 2120585119879
lowast
1198811015840
2= minus[
(119878lowast)2
(1 minus 1199061) 120573119860lowast
119878
+ (1 minus 1199061) 120573119878119860
lowast+119860lowast120585119879
119860+ 1199093
119879lowast
1198791199062119860]
le minus2[(119878lowast)2
(1 minus 1199061) 120573119860lowast
119878sdot (1 minus 119906
1) 120573119878119860
lowast]
12
minus 2 [119860lowast120585119879
119860sdot 1199093
119879lowast
1198791199062119860]
= minus2 (1 minus 1199061) 120573119878lowast119860lowastminus 2(1199093119860lowast1205851199062119879lowast)12
= minus2 (1 minus 1199061) 120573119878lowast119860lowastminus 2[120585119879
lowast(11990931199062119860lowast)]12
= minus2 (1 minus 1199061) 120573119878lowast119860lowast
minus 2120585119879lowast[(1199062+ 120583)119860
lowastminus (1 minus 119906
1) 120573119878lowast119860lowast]12
= minus2 (1 minus 1199061) 120573119878lowast119860lowastminus 2(120585119879
lowast120585119879lowast)12
= minus2 (1 minus 1199061) 120573119878lowast119860lowastminus 2120585119879
lowast
(47)
Hence
1198811015840= 120583119878lowast(2 minus
119878
119878lowastminus119878lowast
119878) + 119881
1015840
1+ 1198811015840
2
le 120583119878lowast(2 minus
119878
119878lowastminus119878lowast
119878) + 2 (1 minus 119906
1) 120573119878lowast119860lowast
+ 2120585119879lowastminus 2 (1 minus 119906
1) 120573119878lowast119860lowastminus 2120585119879
lowast
= 120583119878lowast(2 minus
119878
119878lowastminus119878lowast
119878) le 0
(48)
1198811015840= 0 if and only if 119878 = 119878
lowast 119860 = 119860lowast 119879 = 119879
lowast Accordingto LaSallersquos invariance principle [28] we can derive theconclusion that the alcoholism equilibria 119864lowast(119878lowast 119860lowast 119879lowast 119876lowast)are globally asymptotically stable the proof is complete
5 Optimal Control Problem
51 The Existence of Optimal Control In order to investigatean effective campaign to control alcoholism in a communitywhich pursue the goals of the minimized alcoholisms andmore recovered individuals we will reconsider the system (1)and use two control variables to reduce the numbers of alco-holics The difference is that we will change the parameters1199061 1199062into control variable 119906
1(119905) 1199062(119905) Their aforementioned
definitions allow us to do so 1199061(119905) is used to limit the
proportion of the susceptible individual to contact withalcoholism usually by propaganda and education so that thesusceptible individual can stay off alcoholism consciously andbe free of ldquoinfectionrdquo we can understand the effect of 119906
1(119905)
is to prevent the the susceptible from contacting with thealcoholism The control variable 119906
2(119905) is used to control the
alcoholism to take appropriate treatment measures such astaking pills or seeking other medical help However just as acoin has two sides there will be a lot of costs generated duringthe control process So it is advisable to balance between thecosts and the alcohol effects In view of this our optimalcontrol problem tominimize the objective functional is givenby
119869 (1199061 1199062) = int
119905119891
0
[119860 (119905) +1198881
21199062
1(119905) +
1198882
21199062
2(119905)] 119889119905 (49)
which subjects to system
1198781015840= 120583119873 minus (1 minus 119906
1(119905))
120573119878119860
119873minus 120583119878
1198601015840= (1 minus 119906
1(119905))
120573119878119860
119873+ 120585119879 minus (119906
2(119905) + 120583)119860
1198791015840= 1199062(119905) 119860 minus (120583 + 120585 + 120575) 119879
1198761015840= 120575119879 minus 120583119876
(50)
with initial conditions
119878 (0) = 1198780 119860 (0) = 119860
0
119879 (0) = 1198790 119876 (0) = 119876
0
(51)
Here 119906119894(119905) isin 119880 ≜ (119906
1 1199062) | 119906119894(119905) is measurable and 0 le
119906119894(119905) le 1 for all 119905 isin [0 119905
119891] 119905119891is the end time to be
controlled 119880 is an admissible control set 119888119894 and 119894 = 1 2 are
weight factors (positive constants) that adjust the intensity oftwo different control measures
Next we will investigate the existence of the optimalcontrol of the above-mentioned problem
Theorem 5 There exists an optimal control pair 119906lowast
=
(119906lowast
1 119906lowast
2) isin 119880 such that
119869 (119906lowast
1 119906lowast
2) = min 119869 (119906
1 1199062) 119906
1(119905) 1199062(119905) isin 119880 (52)
subjects to the control system (1) with initial conditions (50)
Proof Toprove the existence of an optimal control accordingto the classic literature [29] we have to show the following
(1) The control and state variables are nonnegative values(2) The control set 119880 is convex and closed(3) The right side of the state system is bounded by linear
function in the state and control variables(4) The integrand of the objective functional is concave
on 119880(5) There exist constants 119889
1 1198892gt 0 and 120572 gt 1 such
that the integrand 119871(119905 1199061 1199062) ≜ 119860(119905) + (119888
12)1199062
1(119905) +
(11988822)1199062
2(119905) of the objective functional satisfies
119871 (119905 1199061 1199062) ge 1198891(100381610038161003816100381611990611003816100381610038161003816
2
+100381610038161003816100381611990621003816100381610038161003816
2
)1205722
minus 1198892
(53)
8 Abstract and Applied Analysis
statements (1) (2) and (3) are obvious satisfied we only needto test and verify the latter two ones Since the four statevariables have been all proved to be up bounded by 119873 wewill get the following equalities
1198781015840le 120583119873 119860
1015840le (1 minus 119906
1(119905)) 120573119878 + 120585119879
1198791015840le 1199062(119905) 119860 119876
1015840le 120575119879
(54)
so the fourth condition is set up As for the last condition
119871 (119905 1199061 1199062) ge 1198891(100381610038161003816100381611990611003816100381610038161003816
2
+100381610038161003816100381611990621003816100381610038161003816
2
)1205722
minus 1198892
(55)
is also true when we choose 1198891= min119888
12 11988822 and for all
1198892isin 119877+ 120572 = 2 The proof is complete
We next come to the core of this section
52 The Characterization of the Optimal Control With theexistence of the optimal control pairs established we nowpresent the optimality system and use a result from [30]we can easily know the existence of the solutions to theoptimality system (71) which will be gotten later Firstly wecome to discuss the theorem that relates to the character-ization of the optimal control The optimality system canbe used to compute candidates for optimal control pairsTo do this we begin by defining an augmented Hamilto-nian 119867 with penalty terms for the control constraints asfollows
119867 = 119860 (119905) +1198881
21199062
1(119905) +
1198882
21199062
2(119905)
+ 1205821[120583119873 minus (1 minus 119906
1(119905))
120573119878119860
119873minus 120583119878]
+ 1205822[(1 minus 119906
1(119905))
120573119878119860
119873+ 120585119879 minus (119906
2(119905) + 120583)119860]
+ 1205823[1199062(119905) 119860 minus (120583 + 120585 + 120575) 119879] + 120582
4(120575119879 minus 120583119876)
minus 119908111199061(119905) minus 119908
12(1 minus 119906
1(119905))
minus 119908211199062(119905) minus 119908
22(1 minus 119906
2(119905))
(56)
where 119908119894119895(119905) ge 0 are the penalty multipliers satisfying
11990811(119905) 1199061(119905) = 119908
12(119905) (1 minus 119906
1(119905))
= 0 at optimal control 119906lowast1
11990821(119905) 1199062(119905) = 119908
22(119905) (1 minus 119906
2(119905))
= 0 at optimal control 119906lowast2
(57)
Theorem6 Given optimal control pairs (119906lowast1 119906lowast
2) and solutions
119878(119905) 119860(119905) 119879(119905) 119876(119905) of the corresponding state system (50)there exist adjoint variables 120582
119894 119894 = 1 2 3 4 satisfying
1205821015840
1= 1205821(1 minus 119906
1(119905))
120573119860
119873+ 1205831205821minus 1205822(1 minus 119906
1(119905))
120573119860
119873
1205821015840
2= minus1 + 120582
1(1 minus 119906
1(119905))
120573119878
119873minus 1205822(1 minus 119906
1(119905))
120573119878
119873
+ 1205822(120583 + 119906
2(119905)) minus 120582
31199062(119905)
1205821015840
3= minus1205822120585 + 1205823(120583 + 120585 + 120575) minus 120582
4120575
1205821015840
4= 1205831205824
(58)
with the terminal conditions
120582119894(119905119891) = 0 119894 = 1 2 3 4 (59)
Furthermore (119906lowast1 119906lowast
2) are represented by
119906lowast
1= min(1max(0
120573119878119860 (1205822minus 1205821)
1198881119873
))
119906lowast
2= min(1max(0
119860 (1205822minus 1205823)
1198882
))
(60)
Proof According to Pontryagin Maximum Principle [29ndash31] we first differentiate the Hamiltonian operator 119867 withrespect to states Then the adjoint system can be written as
1205821015840
1= minus
120597119867
120597119878 120582
1015840
2= minus
120597119867
120597119860
1205821015840
3= minus
120597119867
120597119879 120582
1015840
4= minus
120597119867
120597119876
(61)
The terminal condition (56) of adjoint equations is given by120582119894(119905119891) = 0 119894 = 1 2 3 4
To obtain the necessary conditions of optimality (59) wealso differentiate theHamiltonian operator119867 with respect to119880 = (119906
1 1199062) and set them equal to zero then
120597119867
1205971199061
= 11988811199061(119905) + 120582
1
120573119878119860
119873minus 1205822
120573119878119860
119873minus 11990811+ 11990812= 0
120597119867
1205971199062
= 11988821199062(119905) minus 120582
2119860 + 120582
3119860 minus 119908
21+ 11990822= 0
(62)
By solving the optimal control we obtain
119906lowast
1=1
1198881
[(1205822minus 1205821)120573119878119860
119873+ 11990811minus 11990812] (63)
To determine an explicit expression for the optimalcontrol without119908
11and119908
12 a standard optimality technique
is utilized [29] We consider the following three cases
(i) On the set 119905 | 0 lt 119906lowast
1(119905) lt 1 we have 119908
11(119905) =
Abstract and Applied Analysis 9
11990812(119905) = 0 Hence the optimal control is
119906lowast
1=1
1198881
[(1205822minus 1205821)120573119878119860
119873] (64)
(ii) On the set 119905 | 119906lowast1(119905) = 1 we have 119908
11(119905) = 0 Hence
1 = 119906lowast
1(119905) =
1
1198881
[(1205822minus 1205821)120573119878119860
119873minus 11990812] (65)
This implies that
1
1198881
(1205822minus 1205821)120573119878119860
119873ge 1 since 119908
12(119905) ge 0 (66)
(iii) On the set 119905 | 119906lowast1(119905) = 0 we have 119908
12(119905) = 0 Hence
0 = 119906lowast
1(119905) =
1
1198881
[(1205822minus 1205821)120573119878119860
119873+ 11990811] (67)
This implies that
1
1198881
(1205822minus 1205821)120573119878119860
119873le 0 since 119908
11(119905) ge 0 (68)
Combining these results the optimal control 119906lowast1(119905) is
characterized as
119906lowast
1= min1max0 1
1198881
[(1205822minus 1205821)120573119878119860
119873] (69)
Using the similar arguments we can also obtain the otheroptimal control function
119906lowast
2= min1max0
(1205822minus 1205823) 119860
1198882
(70)
The proof is complete
We point out that the optimality system consistsof the state system (50) with the initial conditions119878(0) 119860(0) 119879(0) 119876(0) the adjoint (or costate) system(58) with the terminal conditions (59) and the optimalitycondition (60) Any optimal control pairs must satisfythis optimality system For the convenience of subsequent
numerical simulation in Section 6 we give the optimalitysystem as follows
1198781015840= 120583119873 minus (1 minusmin1max0 1
1198881
[(1205822minus 1205821)120573119878119860
119873] (119905))
times120573119878119860
119873minus 120583119878
1198601015840= (1minusmin1max0 1
1198881
[(1205822minus1205821)120573119878119860
119873] (119905))
120573119878119860
119873
+ 120585119879 minus (min1max0(1205822minus 1205823) 119860
1198882
(119905) + 120583)119860
1198791015840= min1max0
(1205822minus 1205823) 119860
1198882
(119905) 119860
minus (120583 + 120585 + 120575) 119879
1198761015840= 120575119879 minus 120583119876
1205821015840
1= 1205821(1 minusmin1max0 1
1198881
[(1205822minus 1205821)120573119878119860
119873] (119905))
times120573119860
119873+ 1205831205821
minus 1205822(1minusmin1max0 1
1198881
[(1205822minus1205821)120573119878119860
119873](119905))
times120573119860
119873
1205821015840
2= minus1
+ 1205821(1minusmin1max0 1
1198881
[(1205822minus1205821)120573119878119860
119873] (119905))
times120573119878
119873minus 1205823sdotmin1max0
(1205822minus 1205823) 119860
1198882
(119905)
minus 1205822(1 minusmin1max0 1
1198881
[(1205822minus 1205821)120573119878119860
119873])
times120573119878
119873+ 1205822(120583+min1max0
(1205822minus 1205823) 119860
1198882
(119905))
1205821015840
3= minus1205822120585 + 1205823(120583 + 120585 + 120575) minus 120582
4120575
1205821015840
4= 1205831205824
119878 (0) = 1198780 119860 (0) = 119860
0 119879 (0) = 119879
0
119876 (0) = 1198760 120582
119894(119905119891) = 0 119894 = 1 2 3 4
(71)
53 The Uniqueness of Optimal Control Due to the a prioriboundedness of the state adjoint functions and the resultingLipschitz structure of the ODEs we can obtain the unique-ness of the optimal control
10 Abstract and Applied Analysis
Lemma 7 (see [23]) The function 119906lowast(119904) = min (119887max (119904 119886))is Lipschitz continuous in 119904 where 119886 lt 119887 are some fixed positiveconstants
Theorem 8 For all 119905 isin [0 119905119891] the solution to the optimality
system (71) is unique
Proof Suppose (119878 119860 119879 119876 1205821 1205822 1205823 1205824) and
(119878 119860 119879 119876 1205821 1205822 1205823 1205824) are two different solutions of
our optimality system (71) Let
119878 = 119890120582119905119898 119860 = 119890
120582119905119899 119879 = 119890
120582119905119901
119876 = 119890120582119905119902 120582
1= 119890minus120582119905119903 120582
2= 119890minus120582119905119904
1205823= 119890minus120582119905119908 120582
4= 119890minus120582119905V
119878 = 119890120582119905119898 119860 = 119890
120582119905119899 119879 = 119890
120582119905119901
119876 = 119890120582119905119902 120582
1= 119890minus120582119905119903 120582
2= 119890minus120582119905119904
1205823= 119890minus120582119905119908 120582
4= 119890minus120582119905V
(72)
where 120582 gt 0 is to be chosenAccordingly we have
119906lowast
1(119905) = min1max0
120573119898119899 (119904 minus 119903) 119890120582119905
1198881119873
119906lowast
2(119905) = min1max0 (119904 minus 119908) 119899
1198882
119906lowast
1(119905) = min1max0
120573119898119899 (119904 minus 119903) 119890120582119905
1198881119873
119906lowast
2(119905) = min1max0 (119904 minus 119908) 119899
1198882
(73)
Now we substitute 119878 = 119890120582119905119898 119860 = 119890
120582119905119899 119879 = 119890
120582119905119901 119876 = 119890
120582119905119902
1205821= 119890minus120582119905119903 1205822= 119890minus120582119905119904 1205823= 119890minus120582119905119908 1205824= 119890minus120582119905V and 119878 =
119890120582119905119898 119860 = 119890
120582119905119899 119879 = 119890
120582119905119901 119876 = 119890
120582119905119902 1205821= 119890minus120582119905119903 1205822=
119890minus120582119905119904 1205823= 119890minus120582119905119908 1205824= 119890minus120582119905V into the first ODE of (71)
respectively then we can obtain
+ 120582119898 = 120583119873119890minus120582119905
minus (1 minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)
times120573119898119899119890
120582119905
119873minus 120583119898
(74)
for119898 and119898 respectively Similarly we can derive
119899 + 120582119899
= (1 minusmin1max1120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)120573119898119899119890
120582119905
119873
+ 120585119901 minus (min1max0 119899 (119904 minus 119908)1198882
+ 120583) 119899
(75)
for 119899 and 119899 respectively
+ 120582119901 = min1max0 119899 (119904 minus 119908)1198882
119899
minus (120583 + 120585 + 120575) 119901
(76)
for 119901 and 119901 respectively
119902 + 120582119902 = 120575119901 minus 120583119902 (77)
for 119902 and 119902 respectively
119903 minus 120582119903 = 119903(1 minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)120573119899119890120582119905
119873
+ 120583119903minus119904(1minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)
times120573119899119890120582119905
119873
(78)
for 119903 and 119903 respectively
119904 minus 120582119904 = minus119890120582119905
+ 119903(1 minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)120573119898119890120582119905
119873
minus 119904(1 minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)
times120573119898119890120582119905
119873+ 119904(min1max0 119899 (119904 minus 119908)
1198882
+ 120583)
minus 119908(min1max0 119899 (119904 minus 119908)1198882
)
(79)
for 119904 and 119904 respectively
minus 120582119908 = minus119904120585 + (120583 + 120585 + 120575)119908 minus 120575V (80)
for 119908 and 119908 respectively
V minus 120582V = 120583V (81)
for V and V respectively
Abstract and Applied Analysis 11
By Lemma 7 we can obtain
1003816100381610038161003816119906lowast
1(119905) minus 119906
lowast
1(119905)1003816100381610038161003816 le
120573119890120582119905
1198881119873|119898119899 (119904 minus 119903) minus 119898119899 (119904 minus 119903)|
1003816100381610038161003816119906lowast
2(119905) minus 119906
lowast
2(119905)1003816100381610038161003816 le
1
1198882
|119899 (119904 minus 119908) minus 119899 (119904 minus 119908)|
(82)
The equations for 119898 119899 119901 119902 119903 119904 119908 V and the equationsfor 119898 119899 119901 119902 119903 119904 119908 V are subtracted respectively then wemultiply each equation by appropriate difference of functionsand integrate from 0 to 119905
119891 Next we add all eight integral
equations and some inequality techniques to obtain unique-ness The following calculation is similar for the sake ofsimplicity we only take119898 and119898 for an example
minus + (120583 + 120582) (119898 minus 119898)
=120573119890120582119905
119873[minus(1 minusmin1max0
120573119898119899 (119904 minus 119903) 119890120582119905
1198881119873
)119898119899
+(1 minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)119898119899]
(83)
Multiplying both sides of (83) by (119898minus119898) and integratingfrom 0 to 119905
119891gives
1
2(119898 minus 119898)
2(119905119891) + (120583 + 120582)int
119905119891
0
(119898 minus 119898)2119889119905
= int
119905119891
0
(119898 minus 119898)120573119890120582119905
119873
times [minus (1 minus 119906lowast
1)119898119899 + (1 minus 119906
lowast
1)119898119899] 119889119905
= int
119905119891
0
(119898 minus 119898)120573119890120582119905
119873[(119906lowast
1minus 1) (119898119899 minus 119898119899 + 119898119899 minus 119898119899)
+ 119898119899 (119906lowast
1minus 119906lowast
1)] 119889119905
le120573119890120582119905119891
119873int
119905119891
0
(119898 minus 119898)
times [119906lowast
1minus 1
100381610038161003816100381610038161003816100381610038161003816
(119898 |119899 minus 119899| + |119898 minus 119898| 119899) + 119898119899120573119890120582119905
1198881119873
100381610038161003816100381610038161003816100381610038161003816
times 119898119899 (119904 minus 119903) minus 119898119899 (119904 minus 119903) ] 119889119905
le120573119890120582119905119891
119873int
119905119891
0
(119898 minus 119898)
times [1003816100381610038161003816119906lowast
1minus 1
1003816100381610038161003816 (119898 |119899 minus 119899| + |119898 minus 119898| 119899) + 119898119899120573119890120582119905
1198881119873
|119898119899 (119904 minus 119904) + (119898119899 minus 119898119899) 119904
minus (119898119899 (119903 minus 119903) + (119898119899 minus 119898119899) 119903)| ] 119889119905
le120573119890120582119905119891
119873int
119905119891
0
[1003816100381610038161003816119906lowast
1minus 1
1003816100381610038161003816 |119898| |119898 minus 119898| |119899 minus 119899|
+ |119898 minus 119898|2
|119899| + |119861| |119904| |119899| |119898 minus 119898|2
+ |119861| |119898119899| |119898 minus 119898| |119903 minus 119903|
+ |119861| |119898119899| |119898 minus 119898| |119904 minus 119904|
+ |119861| |119904| |119898| |119898 minus 119898| |119899 minus 119899|
+ |119861| |119903| |119898| |119898 minus 119898| |119899 minus 119899|
+ |119861| |119903| |119899| |119898 minus 119898|2] 119889119905
le120573119890120582119905119891
119873int
119905119891
0
[
1003816100381610038161003816119906lowastminus 1
1003816100381610038161003816
2|119898| ((119898 minus 119898)
2+ (119899 minus 119899)
2)
+ |119899| (119898 minus 119898)2
+|119861| |119898119899|
2((119898 minus 119898)
2+ (119904 minus 119904)
2)
+|119861| |119904| |119898|
2((119898 minus 119898)
2+ (119899 minus 119899)
2)
+ |119861| |119904| |119899| (119898 minus 119898)2
+|119861| |119898119899|
2((119898 minus 119898)
2+ (119903 minus 119903)
2)
+|119861| |119898| |119903|
2((119898 minus 119898)
2+ (119899 minus 119899)
2)
+ |119861| |119903| |119899| (119898 minus 119898)2]119889119905
(84)
In the above derivation we use many scaling techniquesfor inequality or absolute inequality Particularly what shouldbe noted is that to get the first inequality of above derivationwe use the estimation of |119906lowast
1(119905) minus 119906
lowast
1(119905)| which has been given
before besides for the sake of convenience we note 119861 =
119898119899(120573119890120582119905119891119873) Furthermore we notice that the coefficients of
all the eight terms in the last formula (119898 minus 119898)2+ (119899 minus 119899)
2(119898 minus119898)
2 (119898 minus119898)2+ (119904 minus 119904)
2 (119898 minus119898)2+ (119899 minus 119899)
2 (119898 minus119898)2
(119898minus119898)2+(119903minus119903)
2 (119898minus119898)2+(119899minus119899)2 (119898minus119898)2 namely (|119906lowast1minus
1|2)|119898| |119899| (|119861||119898119899|)2 (|119861||119904||119898|)2 |119861||119904|119899 (|119861||119898119899|)2(|119861||119898||119903|)2 |119861||119903||119899| are nonnegative and bounded So thereexists a positive constant 119888
2such that
1
2(119898 minus 119898)
2(119905119891) + (120583 + 120582)int
119905119891
0
(119898 minus 119898)2119889119905
12 Abstract and Applied Analysis
le 1198882
120573119890120582119905119891
119873int
119905119891
0
((119898 minus 119898)2+ (119899 minus 119899)
2
+ (119904 minus 119904)2+ (119903 minus 119903)
2) 119889119905
(85)
Combining eight of these inequalities gives
1
2(119898 minus 119898) (119905
119891) +
1
2(119899 minus 119899) (119905
119891) +
1
2(119901 minus 119901) (119905
119891)
+1
2(119902 minus 119902) (119905
119891) +
1
2(119903 minus 119903) (0) +
1
2(119904 minus 119904) (0)
+1
2(119908 minus 119908) (0) +
1
2(V minus V) (0) + (120583 + 120582)
times int
119905119891
0
(119898 minus 119898)2+ (119899 minus 119899)
2+ (119901 minus 119901)
2
+ (119902 minus 119902)2
+ (119904 minus 119904)2
+ (119903 minus 119903)2+ (119908 minus 119908)
2+ (V minus V)2) 119889119905
le 119861int
119905119891
0
(119898 minus 119898)2+ (119899 minus 119899)
2+ (119901 minus 119901)
2
+ (119902 minus 119902)2
+ (119903 minus 119903)2+ (119904 minus 119904)
2
+ (119908 minus 119908)2+ (V minus V)2 119889119905
(86)
Thus from the above inequality we can conclude that
(120583 + 120582 minus 119861)int
119905119891
0
(119898 minus 119898)2+ (119899 minus 119899)
2
+ (119901 minus 119901)2
+ (119902 minus 119902)2
+ (119903 minus 119903)2
+ (119904 minus 119904)2+(119908 minus 119908)
2+ (V minus V)2 119889119905 le 0
(87)
where 119861 depends on the coefficients and the bounds dependon119898 119899 119901 119902 119903 119904 119908 V If we choose 120582 such that 120583+120582 gt 119861 then119898 = 119898 119899 = 119899 119901 = 119901 119902 = 119902 119903 = 119903 119904 = 119904 119908 = 119908 and V = VHence the solution to the optimality system is unique Theproof is complete
6 Numerical Simulation
61 The Simulation of State System (1) without ControlParameters For the sake of simplicity but without loss ofgenerality we will perform the numerical simulation of statesystem (1) with parameters 119906
1= 0 119906
2= 0 Before
illustrating the analytic properties of the alcoholism model(1) we will target the populations in the environment ofa community or a university for example the school ofmaterial science and engineering in our university that isLan zhou University of Technology (LUT for short) owing tothe accurate and available information we can obtain Refer-ring to the information provided by the admissions officeof LUT this school will enroll almost 1200 undergraduatesand almost 300 various postgraduates at the beginning of
fall semester at the same time there will be almost 1500various students graduated and left this school so the scaleof students in school remained almost 6000 we can take thetotal population 119873 = 6000 In this simulation we will takeSeptember as the initial time and units in one week periodin one year According to the investigations of the studentunion implemented in September every year we can takeinitial values as 119878(0) = 4500 119860(0) = 1000 119879(0) = 300and 119876(0) = 200 It seems that the alcoholism is a little bitmore but it is rather natural because many freshmen feelconfused when they are faced with the new environment anda new lifestyle many of them have no better choice but gathertogether to drink in small groups to mediate the anxiety andget to know each other over time some of them developthe habit of drinking To a certain extent for example thefrequent drinking badly affects their study we can classifythem into the alcoholism compartment Other initial valuesseem more reasonable so we need no more explanation Aswe know alcoholism death is seldom happen within oneyear so we omit mortality from alcoholism then how tounderstand the recruitment rate as well as natural death rate120583 We can treat the freshmen admission as the recruitmentpopulation and graduation students as the natural ldquodeathrdquoparts So we can take 120583 = 15006000 = 025 which is exactlyconsistent with the value in [16] As for the infection rate 120573and recovery rate 120575 we will let them be variables since thedrinking behaviors are related to many factors such as theseason and the pressure
According to the data we get from the student unionwe choose 120585 = 04 To summarize we list the values ofthe parameters in Table 1 Using the values of parameters inTable 1 we can plot Figures 2 and 3 which are on condition1198770lt 1 and 119877
0ge 1 respectively From Figure 2 we easily
know when 1198770lt 1 holds the solution of system (1) tends to
the alcohol free equilibrium1198640and verifies the global stability
of 1198640 While seen from Figure 3 we also know that if 119877
0ge
1 holds the solution of system (1) tends to the alcoholismequilibrium 119864
lowast and verifies the global stability of 119864lowast
62 The Sensitivity of 1198770about Two Control Parameters
Although from the expression of the model reproductionnumber 119877
0 we can easily find out the fact that the two
control variables that is 1199061and 119906
2 attribute to reducing the
severity of alcoholism we will still depict the graph between1198770and the two control variables to see more intuitiveness see
Figure 4 It seems from the figure that 1198770is a monotonically
decreasing function about two control parameters so it isadvisable to take two approaches simultaneously to controlthe alcoholics
63 The Simulation of Optimality System (71) In this subsec-tion we will investigate numerically the optimal solution tooptimality system (71) by numerical method from [32] theoptimality system is solved with a fourth-order Runge-Kuttascheme Beginning with a guess for the control variables thestate system is solved forward in time and then those values ofstate are used to solve the adjoint equations backward in timeThe controls are updated at the end of each iteration using
Abstract and Applied Analysis 13
0 10 20 30 40 50
0
1000
2000
3000
4000
5000
6000
7000
Time
Popu
latio
n siz
e
S(t)
A(t)
T(t)
Q(t)
Figure 2 When 1198770lt 1 the alcohol free equilibrium 119864
0is globally
asymptotically stable (120573 = 02 120575 = 03 and 1198770= 071698)
S(t)
A(t)
T(t)
Q(t)
0 10 20 30 40
48
2520
303540
49 50
500
1000
2000
3000
4000
5000
6000
7000
Time
Popu
latio
n siz
e
Figure 3 When 1198770
ge 1 the alcoholism equilibrium 119864lowast
=
(4466 476 32 26) corresponding to the given parameters is globallyasymptotically stable (120573 = 03 120575 = 02 and 119877
0= 108511)
the values of optimal controls obtained lastly The iterationscontinue until convergence takes place
In the simulations we choose the available variable valuesas Table 1 shows besides 120573 = 03 120575 = 02 The initial valueof model (1) is assumed to be 119878(0) = 4500 119860(0) = 1000119879(0) = 300 and 119876(0) = 200 as before
The ideal weights in objective functional are very difficultto obtain in reality it needs much work on data mining andfittingHence the acquisition of appropriate practical weights
Table 1 The parameters description of model (1)
Parameter Description Values120583 Natural birth rate or death rate 025
120573Transmission coefficient betweenalcoholism and susceptibles Variables
119873 The total populations to be considered 6000
120575The rate of populations quitting fromalcoholism permanently after treatment Variables
120585The rate of populations failed intreatment and returned to be alcoholic 04
0
02
04
06
08
1
0
05
1
0 02 04 06 08 1
u1
u2
R0
Figure 4 The relationship between 1198770and two control variables 119906
1
and 1199062
is still a difficult problem and remains for further investiga-tionsThe cost associated with119860(119905) and 119906
1(119905)mainly includes
the cost of dangerous behaviours during the alcoholism timeand educating the public while the cost associated with1199062(119905)mainly comes from health professional and the medical
resource includingmedicines andnursing care In viewof thisand taking the expressions of 119906
1 1199062into account after many
numerical simulations we finally give weighting coefficientsas 1198881= 102 1198882= 104 It should be pointed out that the
weights here are of only theoretical interest to reveal thecontrol strategies proposed in this paper Another point tonote is that themaximumcontrol is very difficult to achieve inreality so we will omit the situation of the maximum controlduring the series of simulations
Next we will make some necessary instructions andexplanations to the above simulation graphs Figures 5 67 and 8 depict the number of four compartments underdifferent control levels when we choose the weight coeffi-cients in objective function to be 119888
1= 102 1198882= 104 From
the four simulation graphs we can observe the followingsimple facts in reducing the total number of alcoholismsand increasing the number of susceptibles the effectivenessof various control measures is as follows optimal control isevidently better than middle control and middle control isbetter than single control 119906
2 single control 119906
2is better than
single control 1199061 while single control 119906
1is much better than
no controlFigures 9 and 10 depict the optimal control law of 119906
1 1199062
respectively In the beginning of the simulation the control
14 Abstract and Applied Analysis
0 10 20 30 40 504500
5000
5500
6000
6500
7000
S(n
umbe
r of s
usce
ptib
les)
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 5Number of the susceptibles whenwe choose theweights inobjective function are 119888
1= 102 1198882= 104 and under different optimal
control strategies that is (1) without control 1199061= 1199062= 0 (2) single
control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0 119906
2= 05 (4)
middle control 1199061= 1199062= 05 (5) optimal control 119906
1= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
500
1000
1500
A(n
umbe
r of t
he al
coho
lism
)
t (unit week)
Figure 6 Number of the alcoholics when we choose the weights inobjective function are 119888
1= 102 1198882= 104 and under different optimal
control strategies that is (1) without control 1199061= 1199062= 0 (2) single
control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0 119906
2= 05 (4)
middle control 1199061= 1199062= 05 (5) optimal control 119906
1= 119906lowast
1 1199062= 119906lowast
2
0
50
100
150
200
250
300
350
400
T(n
umbe
r of t
reat
men
t)
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 7 Number of the persons in treatment when we choose theweights in objective function are 119888
1= 10
2 1198882= 10
4 and underdifferent optimal control strategies that is (1) without control 119906
1=
1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control 119906
1= 0
1199062= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal control
1199061= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
50
100
150
200
250
300
350
400
Q(n
umbe
r of q
uitti
ng)
Figure 8 Number of people in quitting compartment when wechoose the weights in objective function are 119888
1= 102 1198882= 104 and
under different optimal control strategies that is (1) without control1199061= 1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control
1199061= 0 119906
2= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal
control 1199061= 119906lowast
1 1199062= 119906lowast
2
Abstract and Applied Analysis 15
0 10 20 30 40 50t (unit week)
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
Optimal control of u1
Figure 9 Figures of the optimal control 1199061when we choose the
weight in objective function are 1198881= 102 1198882= 104
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
0 10 20 30 40 50t (unit week)
Optimal control of u2
Figure 10 Figures of the optimal control 1199062when we choose the
weight in objective function are 1198881= 102 1198882= 104
effort of 1199061should be decreased from 065 to 05 within the
first month and over the next week it should be increased tothe maximum control until 50 weeks then rapidly decreasedto 0 at the end of the simulation As for the control 119906
2 it
should start from around 05 due to the initial alcoholics thenincrease to 055 within one week since the rapid infection andnext decrease to almost 0 since the effectiveness of treatmentin three weeks but with the infection going on the controleffort of 119906
2should gradually increase to the maximum and
maintain this level until the tenth week for the purpose ofconsolidation therapy and preventing rebound hereafter itshould be gradually decreased to the level of almost 043 untilthe fifty weeks then quickly decreased to 0 in the end
0 10 20 30 40 504500
5000
5500
6000
6500
7000
S(n
umbe
r of s
usce
ptib
les)
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 11 Number of the susceptibles when we choose the weightsin objective function are 119888
1= 104 1198882= 102 and under different
optimal control strategies that is (1) without control 1199061= 1199062= 0
(2) single control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0
1199062= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal control
1199061= 119906lowast
1 1199062= 119906lowast
2
In order to investigate the influence of different weightcoefficients in the objective functional on the effect ofcontrolling at the same time for a better comparison we willchange the weight coefficients in objective function into 119888
2=
104 1198881= 102 and we will list the corresponding numerical
simulation results as Figures 11ndash16 showWhen we change the weight coefficients in objective
function into 1198881= 104 1198882= 102 we find that the results
of simulations derived from the graphs are very similar tothe ones before We speculate that the most likely reasons ofthis result are due to three respects one is that the weightcoefficients are not too sensitive in the numerical simulationand another possible reason is that both of the two controlsare important in some sense equivalently importantThe lastbut not themost unlikely reason is that we have not found themost appropriate weight coefficients in the simulation whichis very difficult to find as previously mentioned
7 Conclusions
In this paper we formulate an alcoholics quitting modeland firstly investigate the variation discipline of variouspopulations from the perspective of global stability thenwe propose an objective functional to examine two differentcontrolmeasures (ie prevention and treatment) on the effectof alcohol The basic reproduction number of the model wasderived and the global stability of the two equilibria is givenFrom the expression of the basic reproductionnumber119877
0and
16 Abstract and Applied Analysis
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
500
1000
1500
A(n
umbe
r of t
he al
coho
lism
)
t (unit week)
Figure 12 Number of the alcoholics when we choose the weights inobjective function are 119888
1= 104 1198882= 102 and under different optimal
control strategies that is (1) without control 1199061= 1199062= 0 (2) single
control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0 119906
2= 05 (4)
middle control 1199061= 1199062= 05 (5) optimal control 119906
1= 119906lowast
1 1199062= 119906lowast
2
0
50
100
150
200
250
300
350
400
T(n
umbe
r of t
reat
men
t)
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 13 Number of the persons in treatment when we choosethe weights in objective function are 119888
1= 104 1198882= 102 and under
different optimal control strategies that is (1) without control 1199061=
1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control 119906
1= 0
1199062= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal control
1199061= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
50
100
150
200
250
300
350
400
Q(n
umbe
r of q
uitti
ng)
Figure 14 Number of people in quitting compartment when wechoose the weights in objective function are 119888
1= 104 1198882= 102 and
under different optimal control strategies that is (1) without control1199061= 1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control
1199061= 0 119906
2= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal
control 1199061= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50t (unit week)
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
Optimal control of u1
Figure 15 Figures of the optimal control 1199061when we choose the
weight in objective function are 1198881= 104 1198882= 102
related numerical simulation we can easily see that the twocontrol strategies are effective in the alcoholics process
Using Pontryaginrsquos Maximum Principle we firstly deter-mine the necessary conditions for existence of optimalcontrol pairsThe uniqueness of the solution to the optimalitysystem (71) is derived by the classical method of contradic-tion Numerical simulations of the model suggest that thetwo different groups of weights in the objective function have
Abstract and Applied Analysis 17
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
0 10 20 30 40 50t (unit week)
Optimal control of u2
Figure 16 Figures of the optimal control 1199062when we choose the
weight in objective function are 1198881= 104 1198882= 102
much similar effects on the transmission of the alcoholismfrom this point the two control measures are almost equallyimportant in controlling the alcoholism although they willprobably have great influences on the cost of the objectivefunction From the simulation figures it seems that the effectof optimal control which is measured by the reduction inthe number of alcoholics and the increase in the number ofsusceptibles is much better than other control strategies asnoted earlier in the simulation section According to the real-time curve of two optimal controls we point out the specificimplementation methods of optimal control which can beachieved in practice
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
This work was partially supported by the NSF ofGansu Province of China (2013GS09485) the NNSF ofChina (10961018) the NSF of Gansu Province of China(1107RJZA088) the NSF for Distinguished Young Scholarsof Gansu Province of China (1111RJDA003) the Special Fundfor the Basic Requirements in the Research of University ofGansu Province of China and the Development Programfor Hong Liu Distinguished Young Scholars in LanzhouUniversity of Technology
References
[1] R Room T Babor and J Rehm ldquoAlcohol and public healthrdquoThe Lancet vol 365 no 9458 pp 519ndash530 2005
[2] J B Saunders O G Aasland A Amundsen and M GrantldquoAlcohol consumption and related problems among primary
health care patients WHO collaborative project on early detec-tion of personswith harmful alcohol consumption IrdquoAddictionvol 88 no 3 pp 349ndash362 1993
[3] L D Johnston P M OMalley and J G Bachman NationalSurvey Results on Drug Use From the Monitoring the FutureStudy 1975ndash1992 National Institute on Drug Abuse RockvilleMd USA 1993
[4] H Wechsler J E Lee M Kuo and H Lee ldquoCollege bingedrinking in the 1990s a continuing problem Results of theHarvard School of Public Health 1999 College Alcohol StudyrdquoJournal of American College Health vol 48 no 5 pp 199ndash2102000
[5] P M OrsquoMalley and L D Johnston ldquoEpidemiology of alcoholand other drug use among American college studentsrdquo Journalof Studies on Alcohol vol 63 no 14 pp 23ndash39 2002
[6] K Agarwal-Kozlowski and D P Agarwal ldquoGenetic predisposi-tion to alcoholismrdquo Therapeutische Umschau vol 57 no 4 pp179ndash184 2000
[7] C Y Chen C L Storr and J C Anthony ldquoEarly-onset drug useand risk for drug dependence problemsrdquo Addictive Behaviorsvol 34 no 3 pp 319ndash322 2009
[8] M Glavas and J Weinberg ldquoStress alcohol consumption andthe hypothalamic-pituitary adrenal axisrdquo in Nutrients Stressand Medical Disorders pp 165ndash183 Humana Press New YorkNY USA 2006
[9] L N Blum N H Nielsen J A Riggs and L B BresolinldquoAlcoholism and alcohol abuse among women report of theCouncil on Scientific Affairsrdquo Journal of Womenrsquos Health vol7 no 7 pp 861ndash871 1998
[10] B Benedict ldquoModeling alcoholism as a contagious disease howldquoinfectedrdquo drinking buddies spread problem drinkingrdquo SIAMNews vol 40 no 3 2007
[11] H Walter K Gutierrez K Ramskogler I Hertling A Dvorakand O M Lesch ldquoGender-specific differences in alcoholismImplications for treatmentrdquo Archives of Womenrsquos Mental Healthvol 6 no 4 pp 253ndash258 2003
[12] D Muller R D Koch H von Specht w Volker and E MMunch ldquoNeurophisiologic findings in chronic alcohol abuserdquoPsychiatrie Neurologie und Medizinische Psychologie vol 37no 3 pp 129ndash132 1985
[13] G Testino ldquoAlcoholic diseases in hepato-gastroenterology apoint of viewrdquo Hepato-Gastroenterology vol 55 no 82-83 pp371ndash377 2008
[14] A Mubayi P E Greenwood C Castillo-Chavez P J Grue-newald and D M Gorman ldquoThe impact of relative residencetimes on the distribution of heavy drinkers in highly distinctenvironmentsrdquo Socio-Economic Planning Sciences vol 44 no 1pp 45ndash56 2010
[15] D Muller R D Koch H von Specht W Volker and E MMunch ldquoNeurophysiologic findings in chronic alcohol abuserdquoPsychiatrie Neurologie Und Medizinische Psychologie vol 37no 3 pp 129ndash132 1985 (Leipz)
[16] G Mulone and B Straughan ldquoModeling binge drinkingrdquoInternational Journal of Biomathematics vol 5 no 1 Article ID1250005 14 pages 2012
[17] H F Huo and N N Song ldquoGlobal stability for a binge drinkingmodel with two stagesrdquo Discrete Dynamics in Nature andSociety vol 2012 Article ID 829386 15 pages 2012
[18] S S Mushayabasa and C P Bhunu ldquoModelling the effects ofheavy alcohol consumption on the transmission dynamics ofgonorrheardquo Nonlinear Dynamics vol 66 no 4 pp 695ndash7062011
18 Abstract and Applied Analysis
[19] F S Sancheznchez XWang C Castillo-Chvez DM Gormanand P J Gruenewald ldquoDrinking as an epidemic a simplemathematical model with recovery and relapserdquo in TherapistsGuide to Evidence-Based Relapse Prevention K Witkiewitz andG Marlatt Eds Academic Press New York NY USA 2007
[20] H-F Huo and C-C Zhu ldquoInfluence of relapse in a givingup smoking modelrdquo Abstract and Applied Analysis vol 2013Article ID 525461 12 pages 2013
[21] H F Huo and Q Wang ldquoThe effects of awareness programs by17 media on the drinking dynamicsrdquo Submitted
[22] S Lee E Jung and C Castillo-Chavez ldquoOptimal controlintervention strategies in low- and high-risk problem drinkingpopulationsrdquo Socio-Economic Planning Sciences vol 44 no 4pp 258ndash265 2010
[23] K R Fister S Lenhart and J S McNally ldquoOptimizingchemotherapy in an HIV modelrdquo Electronic Journal of Differ-ential Equations vol 1998 no 32 pp 1ndash12 1998
[24] S Lenhart and J T Workman Optimal Control Applied toBiological Models Chapman amp HallCRC Mathematical andComputational Biology Series Chapman amp HallCRC BocaRaton Fla USA 2007
[25] X Yang L Chen and J Chen ldquoPermanence and positiveperiodic solution for the single-species nonautonomous delaydiffusive modelsrdquo Computers amp Mathematics with ApplicationsAn International Journal vol 32 no 4 pp 109ndash116 1996
[26] P van denDriessche and JWatmough ldquoReproduction numbersand sub-threshold endemic equilibria for compartmental mod-els of disease transmissionrdquoMathematical Biosciences vol 180pp 29ndash48 2002
[27] V Lakshmikantham S Leela and A A Martynyuk StabilityAnalysis of Nonlinear Systems vol 125 Marcel Dekker NewYork NY USA 1989
[28] J P LaSalle The Stability of Dynamical Systems vol 25 ofRegional Conference Series in Applied Mathematics Society forIndustrial and Applied Mathematics 1976
[29] W H Fleming and R W Rishel Deterministic and StochasticOptimal Control vol 1 ofApplications of Mathematics SpringerBerlin Germany 1975
[30] D L Lukes Differential Equations Classical to ControlledMathematics in Science and Engineering Academic Press NewYork NY USA 1982
[31] L S Pontryagin V G Boltyanskii R V Gamkrelidze and EF Mishchenko The Mathematical Theory of Optimal ProcessesJohn Wiley amp Sons New Jersey NJ USA 1962
[32] W Hackbusch ldquoA numerical method for solving parabolicequations with opposite orientationsrdquo Computing vol 20 no3 pp 229ndash240 1978
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 5
Proof Since
119869 (1198640)
= (
(1 minus 1199061) 120573 minus (120583 + 119906
2) 120585 0 0
1199062
minus (120583 + 120585 + 120575) 0 0
0 120575 minus120583 0
minus (1 minus 1199061) 120573 0 0 minus120583
)
(25)
we can easily get that two of the eigenvalues are 1205821= 1205822=
minus120583 lt 0 while 1205823 1205824satisfy
1205822+ [2120583 + 120585 + 120575 + 119906
2minus (1 minus 119906
1) 120573] 120582
+ (120583 + 120585 + 120575) (120583 + 1199062minus (1 minus 119906
1) 120573) minus 119906
2120585 = 0
(26)
Thus
1205823+ 1205824= (1 minus 119906
1) 120573 minus (120583 + 120585 + 120575) minus (120583 + 119906
2) (27)
Since 1198770lt 1 is equivalent to
120573 (1 minus 1199061) (120583 + 120585 + 120575) lt 119906
2(120583 + 120575) + 120583 (120583 + 120585 + 120575)
lt (120583 + 1199062) (120583 + 120585 + 120575)
(28)
so
120573 (1 minus 1199061) lt 120583 + 119906
2 (29)
and then
1205823+ 1205824lt minus (120583 + 120585 + 120575) lt 0 (30)
while
12058231205824= (120583 + 120585 + 120575) (120583 + 119906
2minus (1 minus 119906
1) 120573) minus 119906
2120585 (31)
Similarly from 1198770lt 1 we can derive the inequality
minus120573 (1 minus 1199061) (120583 + 120585 + 120575) gt minus119906
2(120583 + 120575) minus 120583 (120583 + 120585 + 120575) (32)
so12058231205824gt (120583 + 119906
2) (120583 + 120585 + 120575)
minus 1199062120585 minus 1199062(120583 + 120575) minus 120583 (120583 + 120585 + 120575)
(33)
It reduces to
12058231205824gt 0 (34)
Hence Re 1205823lt 0 Re 120582
4lt 0 The proof is complete
Next we will turn to investigate the global stability of 1198640
Theorem 3 For system (1) the alcohol free equilibrium 1198640is
globally asymptotically stable if 1198770lt 1
Proof Consider the subsystem of (1) as follows
1198601015840= (1 minus 119906
1)120573119878119860
119873+ 120585119879 minus (119906
2+ 120583)119860
1198791015840= 1199062119860 minus (120583 + 120585 + 120575) 119879
1198761015840= 120575119879 minus 120583119876
(35)
Equation (35) can be rewritten as
(
) = (119865 minus 119881)(
119860
119879
119876
) minus (1 minus119878
119873)
times (
120573 (1 minus 1199061) 0 0
0 0 0
0 0 0
)(
119860
119879
119876
)
(36)
Since 119878 le 119873 and 0 le 1199061le 1 then for all 119905 gt 0 we can get
(
) le (119865 minus 119881)(
119860
119879
119876
) (37)
According to Lemma 1 in [26] all the eigenvalues of matrix119865 minus 119881 have negative real parts so the solutions of this sub-system are stable whenever 119877
0lt 1 So (119860(119905) 119879(119905) 119876(119905)) rarr
(0 0 0) as 119905 rarr infin By the comparison theorem [27] andbased on the fact that the total population is constant 119873 itfollows that (119860(119905) 119879(119905) 119876(119905)) rarr (0 0 0) and 119878(119905) rarr 119873
as 119905 rarr infin So the alcohol free equilibrium 1198640is globally
asymptotically stable the proof is complete
Theorem 4 For system (1) the alcoholism equilibrium119864lowast(119878lowast 119860lowast 119879lowast 119876lowast) is globally asymptotically stable if 119877
0gt 1
Proof Since the total population in model (1) is a constantnumber119873 in order to prove the global stability of system (1)it is sufficed to prove the corresponding stability of subsystem(38)
1198781015840= 120583119873 minus (1 minus 119906
1)120573119878119860
119873minus 120583119878
1198601015840= (1 minus 119906
1)120573119878119860
119873+ 120585119879 minus (119906
2+ 120583)119860
1198791015840= 1199062119860 minus (120583 + 120585 + 120575) 119879
(38)
We make normalization transform and still use the samesymbols 119878 119860 119879 to denote the variables then (38) can betransformed into
1198781015840= 120583 minus (1 minus 119906
1) 120573119878119860 minus 120583119878
1199041198601015840= (1 minus 119906
1) 120573119878119860 + 120585119879 minus (119906
2+ 120583)119860
1198791015840= 1199062119860 minus (120583 + 120585 + 120575) 119879
(39)
From (39) we can easily know that the equilibria (119878lowast 119860lowast 119879lowast)satisfy the following three equalities to be used later
120583 = (1 minus 1199061) 120573119878lowast119860lowastminus 120583119878lowast
(1 minus 1199061) 120573119878lowast119860lowast+ 120585119879lowast= (1199062+ 120583)119860
lowast
1199062119860lowast= (120583 + 120585 + 120575) 119879
lowast
(40)
6 Abstract and Applied Analysis
Let 119881 = 1199091(119878 minus 119878
lowastminus 119878lowast ln(119878119878lowast)) + 119909
2(119860 minus 119860
lowastminus
119860lowast ln(119860119860lowast)) + 119909
3(119879 minus 119879
lowastminus 119879lowast ln(119879119879lowast)) then
119881101584010038161003816100381610038161003816(39)
= 1199091[120583 minus (1 minus 119906
1) 120573119878119860 minus 120583119878
minus119878lowast
119878120583 + (1 minus 119906
1) 120573119878lowast119860 + 120583119878
lowast]
+ 1199092[ (1 minus 119906
1) 120573119878119860 + 120585119879 minus (119906
2+ 120583)119860
minus (1 minus 1199061) 120573119878119860
lowastminus119860lowast
119860120585119879 + (119906
2+ 120583)119860
lowast]
+ 1199093[1199062119860 minus (120583 + 120585 + 120575) 119879
minus119879lowast
1198791199062119860 + (120583 + 120585 + 120575) 119879
lowast]
= 1199091[ (1 minus 119906
1) 120573119878lowast119860lowast+ 120583119878lowastminus (1 minus 119906
1) 120573119878119860 minus 120583119878
minus119878lowast
119878((1 minus 119906
1) 120573119878lowast119860lowast+ 120583119878lowast)
+ (1 minus 1199061) 120573119878lowast119860 + 120583119878
lowast]
+ 1199092[ (1 minus 119906
1) 120573119878119860 + 120585119879 minus (119906
2+ 120583)119860
minus (1 minus 1199061) 120573119878119860
lowastminus119860lowast
119860120585119879 + (119906
2+ 120583)119860
lowast]
+ 1199093[1199062119860 minus (120583 + 120585 + 120575) 119879
minus119879lowast
1198791199062119860 + (120583 + 120585 + 120575) 119879
lowast]
= 1199091120583119878lowast(2 minus
119878
119878lowastminus119878lowast
119878)
+ [1199091(1 minus 119906
1) 120573119878lowast119860lowast+ 1199092(1199062+ 120583)119860
lowast
+ 1199093(120583 + 120585 + 120575) 119879
lowast]
minus [1199091
(119878lowast)2
(1 minus 1199061) 120573119860lowast
119878+ 1199092(1 minus 119906
1) 120573119878119860
lowast
+ 1199092
119860lowast120585119879
119860+ 1199093
119879lowast
1198791199062119860]
+ 119878119860 [minus1199091(1 minus 119906
1) 120573 + 119909
2(1 minus 119906
1) 120573]
+ 119860 [minus (1199062+ 120583) 119909
2+ 1199091(1 minus 119906
1) 120573119878lowast+ 11990621199093]
+ 119879 [1205851199092minus (120583 + 120585 + 120575) 119909
3]
(41)
To eliminate the cross-term 119878119860 and two single-variable terms119860 and 119879 we let
minus1199091(1 minus 119906
1) 120573 + 119909
2(1 minus 119906
1) 120573 = 0
minus (1199062+ 120583) 119909
2+ 1199091(1 minus 119906
1) 120573119878lowast+ 11990621199093= 0
1205851199092minus (120583 + 120585 + 120575) 119909
3= 0
(42)
By solving them we can get
1199091= 1 119909
2= 1
1199093=
120585
120583 + 120585 + 120575=1199062+ 120583 minus (1 minus 119906
1) 120573119878lowast
1199062
(43)
Next we let
1198811015840
1= 1199091(1 minus 119906
1) 120573119878lowast119860lowast+ 1199092(1199062+ 120583)119860
lowast
+ 1199093(120583 + 120585 + 120575) 119879
lowast
1198811015840
2= minus[119909
1
(119878lowast)2
(1 minus 1199061) 120573119860lowast
119878
+ 1199092(1 minus 119906
1) 120573119878119860
lowast+ 1199092
119860lowast120585119879
119860+ 1199093
119879lowast
1198791199062119860]
(44)
and then
1198811015840= 1199091120583119878lowast(2 minus
119878
119878lowastminus119878lowast
119878) + 119881
1015840
1+ 1198811015840
2 (45)
Due to
(1199062+ 120583)119860
lowast= (1 minus 119906
1) 120573119878lowast119860lowast+ 120585119879lowast
1199093(120583 + 120585 + 120575) 119879
lowast= 120585119879lowast1199092= 120585119879lowast
(46)
Abstract and Applied Analysis 7
so
1198811015840
1= 2 (1 minus 119906
1) 120573119878lowast119860lowast+ 2120585119879
lowast
1198811015840
2= minus[
(119878lowast)2
(1 minus 1199061) 120573119860lowast
119878
+ (1 minus 1199061) 120573119878119860
lowast+119860lowast120585119879
119860+ 1199093
119879lowast
1198791199062119860]
le minus2[(119878lowast)2
(1 minus 1199061) 120573119860lowast
119878sdot (1 minus 119906
1) 120573119878119860
lowast]
12
minus 2 [119860lowast120585119879
119860sdot 1199093
119879lowast
1198791199062119860]
= minus2 (1 minus 1199061) 120573119878lowast119860lowastminus 2(1199093119860lowast1205851199062119879lowast)12
= minus2 (1 minus 1199061) 120573119878lowast119860lowastminus 2[120585119879
lowast(11990931199062119860lowast)]12
= minus2 (1 minus 1199061) 120573119878lowast119860lowast
minus 2120585119879lowast[(1199062+ 120583)119860
lowastminus (1 minus 119906
1) 120573119878lowast119860lowast]12
= minus2 (1 minus 1199061) 120573119878lowast119860lowastminus 2(120585119879
lowast120585119879lowast)12
= minus2 (1 minus 1199061) 120573119878lowast119860lowastminus 2120585119879
lowast
(47)
Hence
1198811015840= 120583119878lowast(2 minus
119878
119878lowastminus119878lowast
119878) + 119881
1015840
1+ 1198811015840
2
le 120583119878lowast(2 minus
119878
119878lowastminus119878lowast
119878) + 2 (1 minus 119906
1) 120573119878lowast119860lowast
+ 2120585119879lowastminus 2 (1 minus 119906
1) 120573119878lowast119860lowastminus 2120585119879
lowast
= 120583119878lowast(2 minus
119878
119878lowastminus119878lowast
119878) le 0
(48)
1198811015840= 0 if and only if 119878 = 119878
lowast 119860 = 119860lowast 119879 = 119879
lowast Accordingto LaSallersquos invariance principle [28] we can derive theconclusion that the alcoholism equilibria 119864lowast(119878lowast 119860lowast 119879lowast 119876lowast)are globally asymptotically stable the proof is complete
5 Optimal Control Problem
51 The Existence of Optimal Control In order to investigatean effective campaign to control alcoholism in a communitywhich pursue the goals of the minimized alcoholisms andmore recovered individuals we will reconsider the system (1)and use two control variables to reduce the numbers of alco-holics The difference is that we will change the parameters1199061 1199062into control variable 119906
1(119905) 1199062(119905) Their aforementioned
definitions allow us to do so 1199061(119905) is used to limit the
proportion of the susceptible individual to contact withalcoholism usually by propaganda and education so that thesusceptible individual can stay off alcoholism consciously andbe free of ldquoinfectionrdquo we can understand the effect of 119906
1(119905)
is to prevent the the susceptible from contacting with thealcoholism The control variable 119906
2(119905) is used to control the
alcoholism to take appropriate treatment measures such astaking pills or seeking other medical help However just as acoin has two sides there will be a lot of costs generated duringthe control process So it is advisable to balance between thecosts and the alcohol effects In view of this our optimalcontrol problem tominimize the objective functional is givenby
119869 (1199061 1199062) = int
119905119891
0
[119860 (119905) +1198881
21199062
1(119905) +
1198882
21199062
2(119905)] 119889119905 (49)
which subjects to system
1198781015840= 120583119873 minus (1 minus 119906
1(119905))
120573119878119860
119873minus 120583119878
1198601015840= (1 minus 119906
1(119905))
120573119878119860
119873+ 120585119879 minus (119906
2(119905) + 120583)119860
1198791015840= 1199062(119905) 119860 minus (120583 + 120585 + 120575) 119879
1198761015840= 120575119879 minus 120583119876
(50)
with initial conditions
119878 (0) = 1198780 119860 (0) = 119860
0
119879 (0) = 1198790 119876 (0) = 119876
0
(51)
Here 119906119894(119905) isin 119880 ≜ (119906
1 1199062) | 119906119894(119905) is measurable and 0 le
119906119894(119905) le 1 for all 119905 isin [0 119905
119891] 119905119891is the end time to be
controlled 119880 is an admissible control set 119888119894 and 119894 = 1 2 are
weight factors (positive constants) that adjust the intensity oftwo different control measures
Next we will investigate the existence of the optimalcontrol of the above-mentioned problem
Theorem 5 There exists an optimal control pair 119906lowast
=
(119906lowast
1 119906lowast
2) isin 119880 such that
119869 (119906lowast
1 119906lowast
2) = min 119869 (119906
1 1199062) 119906
1(119905) 1199062(119905) isin 119880 (52)
subjects to the control system (1) with initial conditions (50)
Proof Toprove the existence of an optimal control accordingto the classic literature [29] we have to show the following
(1) The control and state variables are nonnegative values(2) The control set 119880 is convex and closed(3) The right side of the state system is bounded by linear
function in the state and control variables(4) The integrand of the objective functional is concave
on 119880(5) There exist constants 119889
1 1198892gt 0 and 120572 gt 1 such
that the integrand 119871(119905 1199061 1199062) ≜ 119860(119905) + (119888
12)1199062
1(119905) +
(11988822)1199062
2(119905) of the objective functional satisfies
119871 (119905 1199061 1199062) ge 1198891(100381610038161003816100381611990611003816100381610038161003816
2
+100381610038161003816100381611990621003816100381610038161003816
2
)1205722
minus 1198892
(53)
8 Abstract and Applied Analysis
statements (1) (2) and (3) are obvious satisfied we only needto test and verify the latter two ones Since the four statevariables have been all proved to be up bounded by 119873 wewill get the following equalities
1198781015840le 120583119873 119860
1015840le (1 minus 119906
1(119905)) 120573119878 + 120585119879
1198791015840le 1199062(119905) 119860 119876
1015840le 120575119879
(54)
so the fourth condition is set up As for the last condition
119871 (119905 1199061 1199062) ge 1198891(100381610038161003816100381611990611003816100381610038161003816
2
+100381610038161003816100381611990621003816100381610038161003816
2
)1205722
minus 1198892
(55)
is also true when we choose 1198891= min119888
12 11988822 and for all
1198892isin 119877+ 120572 = 2 The proof is complete
We next come to the core of this section
52 The Characterization of the Optimal Control With theexistence of the optimal control pairs established we nowpresent the optimality system and use a result from [30]we can easily know the existence of the solutions to theoptimality system (71) which will be gotten later Firstly wecome to discuss the theorem that relates to the character-ization of the optimal control The optimality system canbe used to compute candidates for optimal control pairsTo do this we begin by defining an augmented Hamilto-nian 119867 with penalty terms for the control constraints asfollows
119867 = 119860 (119905) +1198881
21199062
1(119905) +
1198882
21199062
2(119905)
+ 1205821[120583119873 minus (1 minus 119906
1(119905))
120573119878119860
119873minus 120583119878]
+ 1205822[(1 minus 119906
1(119905))
120573119878119860
119873+ 120585119879 minus (119906
2(119905) + 120583)119860]
+ 1205823[1199062(119905) 119860 minus (120583 + 120585 + 120575) 119879] + 120582
4(120575119879 minus 120583119876)
minus 119908111199061(119905) minus 119908
12(1 minus 119906
1(119905))
minus 119908211199062(119905) minus 119908
22(1 minus 119906
2(119905))
(56)
where 119908119894119895(119905) ge 0 are the penalty multipliers satisfying
11990811(119905) 1199061(119905) = 119908
12(119905) (1 minus 119906
1(119905))
= 0 at optimal control 119906lowast1
11990821(119905) 1199062(119905) = 119908
22(119905) (1 minus 119906
2(119905))
= 0 at optimal control 119906lowast2
(57)
Theorem6 Given optimal control pairs (119906lowast1 119906lowast
2) and solutions
119878(119905) 119860(119905) 119879(119905) 119876(119905) of the corresponding state system (50)there exist adjoint variables 120582
119894 119894 = 1 2 3 4 satisfying
1205821015840
1= 1205821(1 minus 119906
1(119905))
120573119860
119873+ 1205831205821minus 1205822(1 minus 119906
1(119905))
120573119860
119873
1205821015840
2= minus1 + 120582
1(1 minus 119906
1(119905))
120573119878
119873minus 1205822(1 minus 119906
1(119905))
120573119878
119873
+ 1205822(120583 + 119906
2(119905)) minus 120582
31199062(119905)
1205821015840
3= minus1205822120585 + 1205823(120583 + 120585 + 120575) minus 120582
4120575
1205821015840
4= 1205831205824
(58)
with the terminal conditions
120582119894(119905119891) = 0 119894 = 1 2 3 4 (59)
Furthermore (119906lowast1 119906lowast
2) are represented by
119906lowast
1= min(1max(0
120573119878119860 (1205822minus 1205821)
1198881119873
))
119906lowast
2= min(1max(0
119860 (1205822minus 1205823)
1198882
))
(60)
Proof According to Pontryagin Maximum Principle [29ndash31] we first differentiate the Hamiltonian operator 119867 withrespect to states Then the adjoint system can be written as
1205821015840
1= minus
120597119867
120597119878 120582
1015840
2= minus
120597119867
120597119860
1205821015840
3= minus
120597119867
120597119879 120582
1015840
4= minus
120597119867
120597119876
(61)
The terminal condition (56) of adjoint equations is given by120582119894(119905119891) = 0 119894 = 1 2 3 4
To obtain the necessary conditions of optimality (59) wealso differentiate theHamiltonian operator119867 with respect to119880 = (119906
1 1199062) and set them equal to zero then
120597119867
1205971199061
= 11988811199061(119905) + 120582
1
120573119878119860
119873minus 1205822
120573119878119860
119873minus 11990811+ 11990812= 0
120597119867
1205971199062
= 11988821199062(119905) minus 120582
2119860 + 120582
3119860 minus 119908
21+ 11990822= 0
(62)
By solving the optimal control we obtain
119906lowast
1=1
1198881
[(1205822minus 1205821)120573119878119860
119873+ 11990811minus 11990812] (63)
To determine an explicit expression for the optimalcontrol without119908
11and119908
12 a standard optimality technique
is utilized [29] We consider the following three cases
(i) On the set 119905 | 0 lt 119906lowast
1(119905) lt 1 we have 119908
11(119905) =
Abstract and Applied Analysis 9
11990812(119905) = 0 Hence the optimal control is
119906lowast
1=1
1198881
[(1205822minus 1205821)120573119878119860
119873] (64)
(ii) On the set 119905 | 119906lowast1(119905) = 1 we have 119908
11(119905) = 0 Hence
1 = 119906lowast
1(119905) =
1
1198881
[(1205822minus 1205821)120573119878119860
119873minus 11990812] (65)
This implies that
1
1198881
(1205822minus 1205821)120573119878119860
119873ge 1 since 119908
12(119905) ge 0 (66)
(iii) On the set 119905 | 119906lowast1(119905) = 0 we have 119908
12(119905) = 0 Hence
0 = 119906lowast
1(119905) =
1
1198881
[(1205822minus 1205821)120573119878119860
119873+ 11990811] (67)
This implies that
1
1198881
(1205822minus 1205821)120573119878119860
119873le 0 since 119908
11(119905) ge 0 (68)
Combining these results the optimal control 119906lowast1(119905) is
characterized as
119906lowast
1= min1max0 1
1198881
[(1205822minus 1205821)120573119878119860
119873] (69)
Using the similar arguments we can also obtain the otheroptimal control function
119906lowast
2= min1max0
(1205822minus 1205823) 119860
1198882
(70)
The proof is complete
We point out that the optimality system consistsof the state system (50) with the initial conditions119878(0) 119860(0) 119879(0) 119876(0) the adjoint (or costate) system(58) with the terminal conditions (59) and the optimalitycondition (60) Any optimal control pairs must satisfythis optimality system For the convenience of subsequent
numerical simulation in Section 6 we give the optimalitysystem as follows
1198781015840= 120583119873 minus (1 minusmin1max0 1
1198881
[(1205822minus 1205821)120573119878119860
119873] (119905))
times120573119878119860
119873minus 120583119878
1198601015840= (1minusmin1max0 1
1198881
[(1205822minus1205821)120573119878119860
119873] (119905))
120573119878119860
119873
+ 120585119879 minus (min1max0(1205822minus 1205823) 119860
1198882
(119905) + 120583)119860
1198791015840= min1max0
(1205822minus 1205823) 119860
1198882
(119905) 119860
minus (120583 + 120585 + 120575) 119879
1198761015840= 120575119879 minus 120583119876
1205821015840
1= 1205821(1 minusmin1max0 1
1198881
[(1205822minus 1205821)120573119878119860
119873] (119905))
times120573119860
119873+ 1205831205821
minus 1205822(1minusmin1max0 1
1198881
[(1205822minus1205821)120573119878119860
119873](119905))
times120573119860
119873
1205821015840
2= minus1
+ 1205821(1minusmin1max0 1
1198881
[(1205822minus1205821)120573119878119860
119873] (119905))
times120573119878
119873minus 1205823sdotmin1max0
(1205822minus 1205823) 119860
1198882
(119905)
minus 1205822(1 minusmin1max0 1
1198881
[(1205822minus 1205821)120573119878119860
119873])
times120573119878
119873+ 1205822(120583+min1max0
(1205822minus 1205823) 119860
1198882
(119905))
1205821015840
3= minus1205822120585 + 1205823(120583 + 120585 + 120575) minus 120582
4120575
1205821015840
4= 1205831205824
119878 (0) = 1198780 119860 (0) = 119860
0 119879 (0) = 119879
0
119876 (0) = 1198760 120582
119894(119905119891) = 0 119894 = 1 2 3 4
(71)
53 The Uniqueness of Optimal Control Due to the a prioriboundedness of the state adjoint functions and the resultingLipschitz structure of the ODEs we can obtain the unique-ness of the optimal control
10 Abstract and Applied Analysis
Lemma 7 (see [23]) The function 119906lowast(119904) = min (119887max (119904 119886))is Lipschitz continuous in 119904 where 119886 lt 119887 are some fixed positiveconstants
Theorem 8 For all 119905 isin [0 119905119891] the solution to the optimality
system (71) is unique
Proof Suppose (119878 119860 119879 119876 1205821 1205822 1205823 1205824) and
(119878 119860 119879 119876 1205821 1205822 1205823 1205824) are two different solutions of
our optimality system (71) Let
119878 = 119890120582119905119898 119860 = 119890
120582119905119899 119879 = 119890
120582119905119901
119876 = 119890120582119905119902 120582
1= 119890minus120582119905119903 120582
2= 119890minus120582119905119904
1205823= 119890minus120582119905119908 120582
4= 119890minus120582119905V
119878 = 119890120582119905119898 119860 = 119890
120582119905119899 119879 = 119890
120582119905119901
119876 = 119890120582119905119902 120582
1= 119890minus120582119905119903 120582
2= 119890minus120582119905119904
1205823= 119890minus120582119905119908 120582
4= 119890minus120582119905V
(72)
where 120582 gt 0 is to be chosenAccordingly we have
119906lowast
1(119905) = min1max0
120573119898119899 (119904 minus 119903) 119890120582119905
1198881119873
119906lowast
2(119905) = min1max0 (119904 minus 119908) 119899
1198882
119906lowast
1(119905) = min1max0
120573119898119899 (119904 minus 119903) 119890120582119905
1198881119873
119906lowast
2(119905) = min1max0 (119904 minus 119908) 119899
1198882
(73)
Now we substitute 119878 = 119890120582119905119898 119860 = 119890
120582119905119899 119879 = 119890
120582119905119901 119876 = 119890
120582119905119902
1205821= 119890minus120582119905119903 1205822= 119890minus120582119905119904 1205823= 119890minus120582119905119908 1205824= 119890minus120582119905V and 119878 =
119890120582119905119898 119860 = 119890
120582119905119899 119879 = 119890
120582119905119901 119876 = 119890
120582119905119902 1205821= 119890minus120582119905119903 1205822=
119890minus120582119905119904 1205823= 119890minus120582119905119908 1205824= 119890minus120582119905V into the first ODE of (71)
respectively then we can obtain
+ 120582119898 = 120583119873119890minus120582119905
minus (1 minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)
times120573119898119899119890
120582119905
119873minus 120583119898
(74)
for119898 and119898 respectively Similarly we can derive
119899 + 120582119899
= (1 minusmin1max1120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)120573119898119899119890
120582119905
119873
+ 120585119901 minus (min1max0 119899 (119904 minus 119908)1198882
+ 120583) 119899
(75)
for 119899 and 119899 respectively
+ 120582119901 = min1max0 119899 (119904 minus 119908)1198882
119899
minus (120583 + 120585 + 120575) 119901
(76)
for 119901 and 119901 respectively
119902 + 120582119902 = 120575119901 minus 120583119902 (77)
for 119902 and 119902 respectively
119903 minus 120582119903 = 119903(1 minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)120573119899119890120582119905
119873
+ 120583119903minus119904(1minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)
times120573119899119890120582119905
119873
(78)
for 119903 and 119903 respectively
119904 minus 120582119904 = minus119890120582119905
+ 119903(1 minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)120573119898119890120582119905
119873
minus 119904(1 minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)
times120573119898119890120582119905
119873+ 119904(min1max0 119899 (119904 minus 119908)
1198882
+ 120583)
minus 119908(min1max0 119899 (119904 minus 119908)1198882
)
(79)
for 119904 and 119904 respectively
minus 120582119908 = minus119904120585 + (120583 + 120585 + 120575)119908 minus 120575V (80)
for 119908 and 119908 respectively
V minus 120582V = 120583V (81)
for V and V respectively
Abstract and Applied Analysis 11
By Lemma 7 we can obtain
1003816100381610038161003816119906lowast
1(119905) minus 119906
lowast
1(119905)1003816100381610038161003816 le
120573119890120582119905
1198881119873|119898119899 (119904 minus 119903) minus 119898119899 (119904 minus 119903)|
1003816100381610038161003816119906lowast
2(119905) minus 119906
lowast
2(119905)1003816100381610038161003816 le
1
1198882
|119899 (119904 minus 119908) minus 119899 (119904 minus 119908)|
(82)
The equations for 119898 119899 119901 119902 119903 119904 119908 V and the equationsfor 119898 119899 119901 119902 119903 119904 119908 V are subtracted respectively then wemultiply each equation by appropriate difference of functionsand integrate from 0 to 119905
119891 Next we add all eight integral
equations and some inequality techniques to obtain unique-ness The following calculation is similar for the sake ofsimplicity we only take119898 and119898 for an example
minus + (120583 + 120582) (119898 minus 119898)
=120573119890120582119905
119873[minus(1 minusmin1max0
120573119898119899 (119904 minus 119903) 119890120582119905
1198881119873
)119898119899
+(1 minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)119898119899]
(83)
Multiplying both sides of (83) by (119898minus119898) and integratingfrom 0 to 119905
119891gives
1
2(119898 minus 119898)
2(119905119891) + (120583 + 120582)int
119905119891
0
(119898 minus 119898)2119889119905
= int
119905119891
0
(119898 minus 119898)120573119890120582119905
119873
times [minus (1 minus 119906lowast
1)119898119899 + (1 minus 119906
lowast
1)119898119899] 119889119905
= int
119905119891
0
(119898 minus 119898)120573119890120582119905
119873[(119906lowast
1minus 1) (119898119899 minus 119898119899 + 119898119899 minus 119898119899)
+ 119898119899 (119906lowast
1minus 119906lowast
1)] 119889119905
le120573119890120582119905119891
119873int
119905119891
0
(119898 minus 119898)
times [119906lowast
1minus 1
100381610038161003816100381610038161003816100381610038161003816
(119898 |119899 minus 119899| + |119898 minus 119898| 119899) + 119898119899120573119890120582119905
1198881119873
100381610038161003816100381610038161003816100381610038161003816
times 119898119899 (119904 minus 119903) minus 119898119899 (119904 minus 119903) ] 119889119905
le120573119890120582119905119891
119873int
119905119891
0
(119898 minus 119898)
times [1003816100381610038161003816119906lowast
1minus 1
1003816100381610038161003816 (119898 |119899 minus 119899| + |119898 minus 119898| 119899) + 119898119899120573119890120582119905
1198881119873
|119898119899 (119904 minus 119904) + (119898119899 minus 119898119899) 119904
minus (119898119899 (119903 minus 119903) + (119898119899 minus 119898119899) 119903)| ] 119889119905
le120573119890120582119905119891
119873int
119905119891
0
[1003816100381610038161003816119906lowast
1minus 1
1003816100381610038161003816 |119898| |119898 minus 119898| |119899 minus 119899|
+ |119898 minus 119898|2
|119899| + |119861| |119904| |119899| |119898 minus 119898|2
+ |119861| |119898119899| |119898 minus 119898| |119903 minus 119903|
+ |119861| |119898119899| |119898 minus 119898| |119904 minus 119904|
+ |119861| |119904| |119898| |119898 minus 119898| |119899 minus 119899|
+ |119861| |119903| |119898| |119898 minus 119898| |119899 minus 119899|
+ |119861| |119903| |119899| |119898 minus 119898|2] 119889119905
le120573119890120582119905119891
119873int
119905119891
0
[
1003816100381610038161003816119906lowastminus 1
1003816100381610038161003816
2|119898| ((119898 minus 119898)
2+ (119899 minus 119899)
2)
+ |119899| (119898 minus 119898)2
+|119861| |119898119899|
2((119898 minus 119898)
2+ (119904 minus 119904)
2)
+|119861| |119904| |119898|
2((119898 minus 119898)
2+ (119899 minus 119899)
2)
+ |119861| |119904| |119899| (119898 minus 119898)2
+|119861| |119898119899|
2((119898 minus 119898)
2+ (119903 minus 119903)
2)
+|119861| |119898| |119903|
2((119898 minus 119898)
2+ (119899 minus 119899)
2)
+ |119861| |119903| |119899| (119898 minus 119898)2]119889119905
(84)
In the above derivation we use many scaling techniquesfor inequality or absolute inequality Particularly what shouldbe noted is that to get the first inequality of above derivationwe use the estimation of |119906lowast
1(119905) minus 119906
lowast
1(119905)| which has been given
before besides for the sake of convenience we note 119861 =
119898119899(120573119890120582119905119891119873) Furthermore we notice that the coefficients of
all the eight terms in the last formula (119898 minus 119898)2+ (119899 minus 119899)
2(119898 minus119898)
2 (119898 minus119898)2+ (119904 minus 119904)
2 (119898 minus119898)2+ (119899 minus 119899)
2 (119898 minus119898)2
(119898minus119898)2+(119903minus119903)
2 (119898minus119898)2+(119899minus119899)2 (119898minus119898)2 namely (|119906lowast1minus
1|2)|119898| |119899| (|119861||119898119899|)2 (|119861||119904||119898|)2 |119861||119904|119899 (|119861||119898119899|)2(|119861||119898||119903|)2 |119861||119903||119899| are nonnegative and bounded So thereexists a positive constant 119888
2such that
1
2(119898 minus 119898)
2(119905119891) + (120583 + 120582)int
119905119891
0
(119898 minus 119898)2119889119905
12 Abstract and Applied Analysis
le 1198882
120573119890120582119905119891
119873int
119905119891
0
((119898 minus 119898)2+ (119899 minus 119899)
2
+ (119904 minus 119904)2+ (119903 minus 119903)
2) 119889119905
(85)
Combining eight of these inequalities gives
1
2(119898 minus 119898) (119905
119891) +
1
2(119899 minus 119899) (119905
119891) +
1
2(119901 minus 119901) (119905
119891)
+1
2(119902 minus 119902) (119905
119891) +
1
2(119903 minus 119903) (0) +
1
2(119904 minus 119904) (0)
+1
2(119908 minus 119908) (0) +
1
2(V minus V) (0) + (120583 + 120582)
times int
119905119891
0
(119898 minus 119898)2+ (119899 minus 119899)
2+ (119901 minus 119901)
2
+ (119902 minus 119902)2
+ (119904 minus 119904)2
+ (119903 minus 119903)2+ (119908 minus 119908)
2+ (V minus V)2) 119889119905
le 119861int
119905119891
0
(119898 minus 119898)2+ (119899 minus 119899)
2+ (119901 minus 119901)
2
+ (119902 minus 119902)2
+ (119903 minus 119903)2+ (119904 minus 119904)
2
+ (119908 minus 119908)2+ (V minus V)2 119889119905
(86)
Thus from the above inequality we can conclude that
(120583 + 120582 minus 119861)int
119905119891
0
(119898 minus 119898)2+ (119899 minus 119899)
2
+ (119901 minus 119901)2
+ (119902 minus 119902)2
+ (119903 minus 119903)2
+ (119904 minus 119904)2+(119908 minus 119908)
2+ (V minus V)2 119889119905 le 0
(87)
where 119861 depends on the coefficients and the bounds dependon119898 119899 119901 119902 119903 119904 119908 V If we choose 120582 such that 120583+120582 gt 119861 then119898 = 119898 119899 = 119899 119901 = 119901 119902 = 119902 119903 = 119903 119904 = 119904 119908 = 119908 and V = VHence the solution to the optimality system is unique Theproof is complete
6 Numerical Simulation
61 The Simulation of State System (1) without ControlParameters For the sake of simplicity but without loss ofgenerality we will perform the numerical simulation of statesystem (1) with parameters 119906
1= 0 119906
2= 0 Before
illustrating the analytic properties of the alcoholism model(1) we will target the populations in the environment ofa community or a university for example the school ofmaterial science and engineering in our university that isLan zhou University of Technology (LUT for short) owing tothe accurate and available information we can obtain Refer-ring to the information provided by the admissions officeof LUT this school will enroll almost 1200 undergraduatesand almost 300 various postgraduates at the beginning of
fall semester at the same time there will be almost 1500various students graduated and left this school so the scaleof students in school remained almost 6000 we can take thetotal population 119873 = 6000 In this simulation we will takeSeptember as the initial time and units in one week periodin one year According to the investigations of the studentunion implemented in September every year we can takeinitial values as 119878(0) = 4500 119860(0) = 1000 119879(0) = 300and 119876(0) = 200 It seems that the alcoholism is a little bitmore but it is rather natural because many freshmen feelconfused when they are faced with the new environment anda new lifestyle many of them have no better choice but gathertogether to drink in small groups to mediate the anxiety andget to know each other over time some of them developthe habit of drinking To a certain extent for example thefrequent drinking badly affects their study we can classifythem into the alcoholism compartment Other initial valuesseem more reasonable so we need no more explanation Aswe know alcoholism death is seldom happen within oneyear so we omit mortality from alcoholism then how tounderstand the recruitment rate as well as natural death rate120583 We can treat the freshmen admission as the recruitmentpopulation and graduation students as the natural ldquodeathrdquoparts So we can take 120583 = 15006000 = 025 which is exactlyconsistent with the value in [16] As for the infection rate 120573and recovery rate 120575 we will let them be variables since thedrinking behaviors are related to many factors such as theseason and the pressure
According to the data we get from the student unionwe choose 120585 = 04 To summarize we list the values ofthe parameters in Table 1 Using the values of parameters inTable 1 we can plot Figures 2 and 3 which are on condition1198770lt 1 and 119877
0ge 1 respectively From Figure 2 we easily
know when 1198770lt 1 holds the solution of system (1) tends to
the alcohol free equilibrium1198640and verifies the global stability
of 1198640 While seen from Figure 3 we also know that if 119877
0ge
1 holds the solution of system (1) tends to the alcoholismequilibrium 119864
lowast and verifies the global stability of 119864lowast
62 The Sensitivity of 1198770about Two Control Parameters
Although from the expression of the model reproductionnumber 119877
0 we can easily find out the fact that the two
control variables that is 1199061and 119906
2 attribute to reducing the
severity of alcoholism we will still depict the graph between1198770and the two control variables to see more intuitiveness see
Figure 4 It seems from the figure that 1198770is a monotonically
decreasing function about two control parameters so it isadvisable to take two approaches simultaneously to controlthe alcoholics
63 The Simulation of Optimality System (71) In this subsec-tion we will investigate numerically the optimal solution tooptimality system (71) by numerical method from [32] theoptimality system is solved with a fourth-order Runge-Kuttascheme Beginning with a guess for the control variables thestate system is solved forward in time and then those values ofstate are used to solve the adjoint equations backward in timeThe controls are updated at the end of each iteration using
Abstract and Applied Analysis 13
0 10 20 30 40 50
0
1000
2000
3000
4000
5000
6000
7000
Time
Popu
latio
n siz
e
S(t)
A(t)
T(t)
Q(t)
Figure 2 When 1198770lt 1 the alcohol free equilibrium 119864
0is globally
asymptotically stable (120573 = 02 120575 = 03 and 1198770= 071698)
S(t)
A(t)
T(t)
Q(t)
0 10 20 30 40
48
2520
303540
49 50
500
1000
2000
3000
4000
5000
6000
7000
Time
Popu
latio
n siz
e
Figure 3 When 1198770
ge 1 the alcoholism equilibrium 119864lowast
=
(4466 476 32 26) corresponding to the given parameters is globallyasymptotically stable (120573 = 03 120575 = 02 and 119877
0= 108511)
the values of optimal controls obtained lastly The iterationscontinue until convergence takes place
In the simulations we choose the available variable valuesas Table 1 shows besides 120573 = 03 120575 = 02 The initial valueof model (1) is assumed to be 119878(0) = 4500 119860(0) = 1000119879(0) = 300 and 119876(0) = 200 as before
The ideal weights in objective functional are very difficultto obtain in reality it needs much work on data mining andfittingHence the acquisition of appropriate practical weights
Table 1 The parameters description of model (1)
Parameter Description Values120583 Natural birth rate or death rate 025
120573Transmission coefficient betweenalcoholism and susceptibles Variables
119873 The total populations to be considered 6000
120575The rate of populations quitting fromalcoholism permanently after treatment Variables
120585The rate of populations failed intreatment and returned to be alcoholic 04
0
02
04
06
08
1
0
05
1
0 02 04 06 08 1
u1
u2
R0
Figure 4 The relationship between 1198770and two control variables 119906
1
and 1199062
is still a difficult problem and remains for further investiga-tionsThe cost associated with119860(119905) and 119906
1(119905)mainly includes
the cost of dangerous behaviours during the alcoholism timeand educating the public while the cost associated with1199062(119905)mainly comes from health professional and the medical
resource includingmedicines andnursing care In viewof thisand taking the expressions of 119906
1 1199062into account after many
numerical simulations we finally give weighting coefficientsas 1198881= 102 1198882= 104 It should be pointed out that the
weights here are of only theoretical interest to reveal thecontrol strategies proposed in this paper Another point tonote is that themaximumcontrol is very difficult to achieve inreality so we will omit the situation of the maximum controlduring the series of simulations
Next we will make some necessary instructions andexplanations to the above simulation graphs Figures 5 67 and 8 depict the number of four compartments underdifferent control levels when we choose the weight coeffi-cients in objective function to be 119888
1= 102 1198882= 104 From
the four simulation graphs we can observe the followingsimple facts in reducing the total number of alcoholismsand increasing the number of susceptibles the effectivenessof various control measures is as follows optimal control isevidently better than middle control and middle control isbetter than single control 119906
2 single control 119906
2is better than
single control 1199061 while single control 119906
1is much better than
no controlFigures 9 and 10 depict the optimal control law of 119906
1 1199062
respectively In the beginning of the simulation the control
14 Abstract and Applied Analysis
0 10 20 30 40 504500
5000
5500
6000
6500
7000
S(n
umbe
r of s
usce
ptib
les)
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 5Number of the susceptibles whenwe choose theweights inobjective function are 119888
1= 102 1198882= 104 and under different optimal
control strategies that is (1) without control 1199061= 1199062= 0 (2) single
control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0 119906
2= 05 (4)
middle control 1199061= 1199062= 05 (5) optimal control 119906
1= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
500
1000
1500
A(n
umbe
r of t
he al
coho
lism
)
t (unit week)
Figure 6 Number of the alcoholics when we choose the weights inobjective function are 119888
1= 102 1198882= 104 and under different optimal
control strategies that is (1) without control 1199061= 1199062= 0 (2) single
control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0 119906
2= 05 (4)
middle control 1199061= 1199062= 05 (5) optimal control 119906
1= 119906lowast
1 1199062= 119906lowast
2
0
50
100
150
200
250
300
350
400
T(n
umbe
r of t
reat
men
t)
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 7 Number of the persons in treatment when we choose theweights in objective function are 119888
1= 10
2 1198882= 10
4 and underdifferent optimal control strategies that is (1) without control 119906
1=
1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control 119906
1= 0
1199062= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal control
1199061= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
50
100
150
200
250
300
350
400
Q(n
umbe
r of q
uitti
ng)
Figure 8 Number of people in quitting compartment when wechoose the weights in objective function are 119888
1= 102 1198882= 104 and
under different optimal control strategies that is (1) without control1199061= 1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control
1199061= 0 119906
2= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal
control 1199061= 119906lowast
1 1199062= 119906lowast
2
Abstract and Applied Analysis 15
0 10 20 30 40 50t (unit week)
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
Optimal control of u1
Figure 9 Figures of the optimal control 1199061when we choose the
weight in objective function are 1198881= 102 1198882= 104
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
0 10 20 30 40 50t (unit week)
Optimal control of u2
Figure 10 Figures of the optimal control 1199062when we choose the
weight in objective function are 1198881= 102 1198882= 104
effort of 1199061should be decreased from 065 to 05 within the
first month and over the next week it should be increased tothe maximum control until 50 weeks then rapidly decreasedto 0 at the end of the simulation As for the control 119906
2 it
should start from around 05 due to the initial alcoholics thenincrease to 055 within one week since the rapid infection andnext decrease to almost 0 since the effectiveness of treatmentin three weeks but with the infection going on the controleffort of 119906
2should gradually increase to the maximum and
maintain this level until the tenth week for the purpose ofconsolidation therapy and preventing rebound hereafter itshould be gradually decreased to the level of almost 043 untilthe fifty weeks then quickly decreased to 0 in the end
0 10 20 30 40 504500
5000
5500
6000
6500
7000
S(n
umbe
r of s
usce
ptib
les)
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 11 Number of the susceptibles when we choose the weightsin objective function are 119888
1= 104 1198882= 102 and under different
optimal control strategies that is (1) without control 1199061= 1199062= 0
(2) single control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0
1199062= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal control
1199061= 119906lowast
1 1199062= 119906lowast
2
In order to investigate the influence of different weightcoefficients in the objective functional on the effect ofcontrolling at the same time for a better comparison we willchange the weight coefficients in objective function into 119888
2=
104 1198881= 102 and we will list the corresponding numerical
simulation results as Figures 11ndash16 showWhen we change the weight coefficients in objective
function into 1198881= 104 1198882= 102 we find that the results
of simulations derived from the graphs are very similar tothe ones before We speculate that the most likely reasons ofthis result are due to three respects one is that the weightcoefficients are not too sensitive in the numerical simulationand another possible reason is that both of the two controlsare important in some sense equivalently importantThe lastbut not themost unlikely reason is that we have not found themost appropriate weight coefficients in the simulation whichis very difficult to find as previously mentioned
7 Conclusions
In this paper we formulate an alcoholics quitting modeland firstly investigate the variation discipline of variouspopulations from the perspective of global stability thenwe propose an objective functional to examine two differentcontrolmeasures (ie prevention and treatment) on the effectof alcohol The basic reproduction number of the model wasderived and the global stability of the two equilibria is givenFrom the expression of the basic reproductionnumber119877
0and
16 Abstract and Applied Analysis
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
500
1000
1500
A(n
umbe
r of t
he al
coho
lism
)
t (unit week)
Figure 12 Number of the alcoholics when we choose the weights inobjective function are 119888
1= 104 1198882= 102 and under different optimal
control strategies that is (1) without control 1199061= 1199062= 0 (2) single
control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0 119906
2= 05 (4)
middle control 1199061= 1199062= 05 (5) optimal control 119906
1= 119906lowast
1 1199062= 119906lowast
2
0
50
100
150
200
250
300
350
400
T(n
umbe
r of t
reat
men
t)
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 13 Number of the persons in treatment when we choosethe weights in objective function are 119888
1= 104 1198882= 102 and under
different optimal control strategies that is (1) without control 1199061=
1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control 119906
1= 0
1199062= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal control
1199061= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
50
100
150
200
250
300
350
400
Q(n
umbe
r of q
uitti
ng)
Figure 14 Number of people in quitting compartment when wechoose the weights in objective function are 119888
1= 104 1198882= 102 and
under different optimal control strategies that is (1) without control1199061= 1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control
1199061= 0 119906
2= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal
control 1199061= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50t (unit week)
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
Optimal control of u1
Figure 15 Figures of the optimal control 1199061when we choose the
weight in objective function are 1198881= 104 1198882= 102
related numerical simulation we can easily see that the twocontrol strategies are effective in the alcoholics process
Using Pontryaginrsquos Maximum Principle we firstly deter-mine the necessary conditions for existence of optimalcontrol pairsThe uniqueness of the solution to the optimalitysystem (71) is derived by the classical method of contradic-tion Numerical simulations of the model suggest that thetwo different groups of weights in the objective function have
Abstract and Applied Analysis 17
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
0 10 20 30 40 50t (unit week)
Optimal control of u2
Figure 16 Figures of the optimal control 1199062when we choose the
weight in objective function are 1198881= 104 1198882= 102
much similar effects on the transmission of the alcoholismfrom this point the two control measures are almost equallyimportant in controlling the alcoholism although they willprobably have great influences on the cost of the objectivefunction From the simulation figures it seems that the effectof optimal control which is measured by the reduction inthe number of alcoholics and the increase in the number ofsusceptibles is much better than other control strategies asnoted earlier in the simulation section According to the real-time curve of two optimal controls we point out the specificimplementation methods of optimal control which can beachieved in practice
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
This work was partially supported by the NSF ofGansu Province of China (2013GS09485) the NNSF ofChina (10961018) the NSF of Gansu Province of China(1107RJZA088) the NSF for Distinguished Young Scholarsof Gansu Province of China (1111RJDA003) the Special Fundfor the Basic Requirements in the Research of University ofGansu Province of China and the Development Programfor Hong Liu Distinguished Young Scholars in LanzhouUniversity of Technology
References
[1] R Room T Babor and J Rehm ldquoAlcohol and public healthrdquoThe Lancet vol 365 no 9458 pp 519ndash530 2005
[2] J B Saunders O G Aasland A Amundsen and M GrantldquoAlcohol consumption and related problems among primary
health care patients WHO collaborative project on early detec-tion of personswith harmful alcohol consumption IrdquoAddictionvol 88 no 3 pp 349ndash362 1993
[3] L D Johnston P M OMalley and J G Bachman NationalSurvey Results on Drug Use From the Monitoring the FutureStudy 1975ndash1992 National Institute on Drug Abuse RockvilleMd USA 1993
[4] H Wechsler J E Lee M Kuo and H Lee ldquoCollege bingedrinking in the 1990s a continuing problem Results of theHarvard School of Public Health 1999 College Alcohol StudyrdquoJournal of American College Health vol 48 no 5 pp 199ndash2102000
[5] P M OrsquoMalley and L D Johnston ldquoEpidemiology of alcoholand other drug use among American college studentsrdquo Journalof Studies on Alcohol vol 63 no 14 pp 23ndash39 2002
[6] K Agarwal-Kozlowski and D P Agarwal ldquoGenetic predisposi-tion to alcoholismrdquo Therapeutische Umschau vol 57 no 4 pp179ndash184 2000
[7] C Y Chen C L Storr and J C Anthony ldquoEarly-onset drug useand risk for drug dependence problemsrdquo Addictive Behaviorsvol 34 no 3 pp 319ndash322 2009
[8] M Glavas and J Weinberg ldquoStress alcohol consumption andthe hypothalamic-pituitary adrenal axisrdquo in Nutrients Stressand Medical Disorders pp 165ndash183 Humana Press New YorkNY USA 2006
[9] L N Blum N H Nielsen J A Riggs and L B BresolinldquoAlcoholism and alcohol abuse among women report of theCouncil on Scientific Affairsrdquo Journal of Womenrsquos Health vol7 no 7 pp 861ndash871 1998
[10] B Benedict ldquoModeling alcoholism as a contagious disease howldquoinfectedrdquo drinking buddies spread problem drinkingrdquo SIAMNews vol 40 no 3 2007
[11] H Walter K Gutierrez K Ramskogler I Hertling A Dvorakand O M Lesch ldquoGender-specific differences in alcoholismImplications for treatmentrdquo Archives of Womenrsquos Mental Healthvol 6 no 4 pp 253ndash258 2003
[12] D Muller R D Koch H von Specht w Volker and E MMunch ldquoNeurophisiologic findings in chronic alcohol abuserdquoPsychiatrie Neurologie und Medizinische Psychologie vol 37no 3 pp 129ndash132 1985
[13] G Testino ldquoAlcoholic diseases in hepato-gastroenterology apoint of viewrdquo Hepato-Gastroenterology vol 55 no 82-83 pp371ndash377 2008
[14] A Mubayi P E Greenwood C Castillo-Chavez P J Grue-newald and D M Gorman ldquoThe impact of relative residencetimes on the distribution of heavy drinkers in highly distinctenvironmentsrdquo Socio-Economic Planning Sciences vol 44 no 1pp 45ndash56 2010
[15] D Muller R D Koch H von Specht W Volker and E MMunch ldquoNeurophysiologic findings in chronic alcohol abuserdquoPsychiatrie Neurologie Und Medizinische Psychologie vol 37no 3 pp 129ndash132 1985 (Leipz)
[16] G Mulone and B Straughan ldquoModeling binge drinkingrdquoInternational Journal of Biomathematics vol 5 no 1 Article ID1250005 14 pages 2012
[17] H F Huo and N N Song ldquoGlobal stability for a binge drinkingmodel with two stagesrdquo Discrete Dynamics in Nature andSociety vol 2012 Article ID 829386 15 pages 2012
[18] S S Mushayabasa and C P Bhunu ldquoModelling the effects ofheavy alcohol consumption on the transmission dynamics ofgonorrheardquo Nonlinear Dynamics vol 66 no 4 pp 695ndash7062011
18 Abstract and Applied Analysis
[19] F S Sancheznchez XWang C Castillo-Chvez DM Gormanand P J Gruenewald ldquoDrinking as an epidemic a simplemathematical model with recovery and relapserdquo in TherapistsGuide to Evidence-Based Relapse Prevention K Witkiewitz andG Marlatt Eds Academic Press New York NY USA 2007
[20] H-F Huo and C-C Zhu ldquoInfluence of relapse in a givingup smoking modelrdquo Abstract and Applied Analysis vol 2013Article ID 525461 12 pages 2013
[21] H F Huo and Q Wang ldquoThe effects of awareness programs by17 media on the drinking dynamicsrdquo Submitted
[22] S Lee E Jung and C Castillo-Chavez ldquoOptimal controlintervention strategies in low- and high-risk problem drinkingpopulationsrdquo Socio-Economic Planning Sciences vol 44 no 4pp 258ndash265 2010
[23] K R Fister S Lenhart and J S McNally ldquoOptimizingchemotherapy in an HIV modelrdquo Electronic Journal of Differ-ential Equations vol 1998 no 32 pp 1ndash12 1998
[24] S Lenhart and J T Workman Optimal Control Applied toBiological Models Chapman amp HallCRC Mathematical andComputational Biology Series Chapman amp HallCRC BocaRaton Fla USA 2007
[25] X Yang L Chen and J Chen ldquoPermanence and positiveperiodic solution for the single-species nonautonomous delaydiffusive modelsrdquo Computers amp Mathematics with ApplicationsAn International Journal vol 32 no 4 pp 109ndash116 1996
[26] P van denDriessche and JWatmough ldquoReproduction numbersand sub-threshold endemic equilibria for compartmental mod-els of disease transmissionrdquoMathematical Biosciences vol 180pp 29ndash48 2002
[27] V Lakshmikantham S Leela and A A Martynyuk StabilityAnalysis of Nonlinear Systems vol 125 Marcel Dekker NewYork NY USA 1989
[28] J P LaSalle The Stability of Dynamical Systems vol 25 ofRegional Conference Series in Applied Mathematics Society forIndustrial and Applied Mathematics 1976
[29] W H Fleming and R W Rishel Deterministic and StochasticOptimal Control vol 1 ofApplications of Mathematics SpringerBerlin Germany 1975
[30] D L Lukes Differential Equations Classical to ControlledMathematics in Science and Engineering Academic Press NewYork NY USA 1982
[31] L S Pontryagin V G Boltyanskii R V Gamkrelidze and EF Mishchenko The Mathematical Theory of Optimal ProcessesJohn Wiley amp Sons New Jersey NJ USA 1962
[32] W Hackbusch ldquoA numerical method for solving parabolicequations with opposite orientationsrdquo Computing vol 20 no3 pp 229ndash240 1978
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International Journal of Mathematics and Mathematical Sciences
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Decision SciencesAdvances in
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
6 Abstract and Applied Analysis
Let 119881 = 1199091(119878 minus 119878
lowastminus 119878lowast ln(119878119878lowast)) + 119909
2(119860 minus 119860
lowastminus
119860lowast ln(119860119860lowast)) + 119909
3(119879 minus 119879
lowastminus 119879lowast ln(119879119879lowast)) then
119881101584010038161003816100381610038161003816(39)
= 1199091[120583 minus (1 minus 119906
1) 120573119878119860 minus 120583119878
minus119878lowast
119878120583 + (1 minus 119906
1) 120573119878lowast119860 + 120583119878
lowast]
+ 1199092[ (1 minus 119906
1) 120573119878119860 + 120585119879 minus (119906
2+ 120583)119860
minus (1 minus 1199061) 120573119878119860
lowastminus119860lowast
119860120585119879 + (119906
2+ 120583)119860
lowast]
+ 1199093[1199062119860 minus (120583 + 120585 + 120575) 119879
minus119879lowast
1198791199062119860 + (120583 + 120585 + 120575) 119879
lowast]
= 1199091[ (1 minus 119906
1) 120573119878lowast119860lowast+ 120583119878lowastminus (1 minus 119906
1) 120573119878119860 minus 120583119878
minus119878lowast
119878((1 minus 119906
1) 120573119878lowast119860lowast+ 120583119878lowast)
+ (1 minus 1199061) 120573119878lowast119860 + 120583119878
lowast]
+ 1199092[ (1 minus 119906
1) 120573119878119860 + 120585119879 minus (119906
2+ 120583)119860
minus (1 minus 1199061) 120573119878119860
lowastminus119860lowast
119860120585119879 + (119906
2+ 120583)119860
lowast]
+ 1199093[1199062119860 minus (120583 + 120585 + 120575) 119879
minus119879lowast
1198791199062119860 + (120583 + 120585 + 120575) 119879
lowast]
= 1199091120583119878lowast(2 minus
119878
119878lowastminus119878lowast
119878)
+ [1199091(1 minus 119906
1) 120573119878lowast119860lowast+ 1199092(1199062+ 120583)119860
lowast
+ 1199093(120583 + 120585 + 120575) 119879
lowast]
minus [1199091
(119878lowast)2
(1 minus 1199061) 120573119860lowast
119878+ 1199092(1 minus 119906
1) 120573119878119860
lowast
+ 1199092
119860lowast120585119879
119860+ 1199093
119879lowast
1198791199062119860]
+ 119878119860 [minus1199091(1 minus 119906
1) 120573 + 119909
2(1 minus 119906
1) 120573]
+ 119860 [minus (1199062+ 120583) 119909
2+ 1199091(1 minus 119906
1) 120573119878lowast+ 11990621199093]
+ 119879 [1205851199092minus (120583 + 120585 + 120575) 119909
3]
(41)
To eliminate the cross-term 119878119860 and two single-variable terms119860 and 119879 we let
minus1199091(1 minus 119906
1) 120573 + 119909
2(1 minus 119906
1) 120573 = 0
minus (1199062+ 120583) 119909
2+ 1199091(1 minus 119906
1) 120573119878lowast+ 11990621199093= 0
1205851199092minus (120583 + 120585 + 120575) 119909
3= 0
(42)
By solving them we can get
1199091= 1 119909
2= 1
1199093=
120585
120583 + 120585 + 120575=1199062+ 120583 minus (1 minus 119906
1) 120573119878lowast
1199062
(43)
Next we let
1198811015840
1= 1199091(1 minus 119906
1) 120573119878lowast119860lowast+ 1199092(1199062+ 120583)119860
lowast
+ 1199093(120583 + 120585 + 120575) 119879
lowast
1198811015840
2= minus[119909
1
(119878lowast)2
(1 minus 1199061) 120573119860lowast
119878
+ 1199092(1 minus 119906
1) 120573119878119860
lowast+ 1199092
119860lowast120585119879
119860+ 1199093
119879lowast
1198791199062119860]
(44)
and then
1198811015840= 1199091120583119878lowast(2 minus
119878
119878lowastminus119878lowast
119878) + 119881
1015840
1+ 1198811015840
2 (45)
Due to
(1199062+ 120583)119860
lowast= (1 minus 119906
1) 120573119878lowast119860lowast+ 120585119879lowast
1199093(120583 + 120585 + 120575) 119879
lowast= 120585119879lowast1199092= 120585119879lowast
(46)
Abstract and Applied Analysis 7
so
1198811015840
1= 2 (1 minus 119906
1) 120573119878lowast119860lowast+ 2120585119879
lowast
1198811015840
2= minus[
(119878lowast)2
(1 minus 1199061) 120573119860lowast
119878
+ (1 minus 1199061) 120573119878119860
lowast+119860lowast120585119879
119860+ 1199093
119879lowast
1198791199062119860]
le minus2[(119878lowast)2
(1 minus 1199061) 120573119860lowast
119878sdot (1 minus 119906
1) 120573119878119860
lowast]
12
minus 2 [119860lowast120585119879
119860sdot 1199093
119879lowast
1198791199062119860]
= minus2 (1 minus 1199061) 120573119878lowast119860lowastminus 2(1199093119860lowast1205851199062119879lowast)12
= minus2 (1 minus 1199061) 120573119878lowast119860lowastminus 2[120585119879
lowast(11990931199062119860lowast)]12
= minus2 (1 minus 1199061) 120573119878lowast119860lowast
minus 2120585119879lowast[(1199062+ 120583)119860
lowastminus (1 minus 119906
1) 120573119878lowast119860lowast]12
= minus2 (1 minus 1199061) 120573119878lowast119860lowastminus 2(120585119879
lowast120585119879lowast)12
= minus2 (1 minus 1199061) 120573119878lowast119860lowastminus 2120585119879
lowast
(47)
Hence
1198811015840= 120583119878lowast(2 minus
119878
119878lowastminus119878lowast
119878) + 119881
1015840
1+ 1198811015840
2
le 120583119878lowast(2 minus
119878
119878lowastminus119878lowast
119878) + 2 (1 minus 119906
1) 120573119878lowast119860lowast
+ 2120585119879lowastminus 2 (1 minus 119906
1) 120573119878lowast119860lowastminus 2120585119879
lowast
= 120583119878lowast(2 minus
119878
119878lowastminus119878lowast
119878) le 0
(48)
1198811015840= 0 if and only if 119878 = 119878
lowast 119860 = 119860lowast 119879 = 119879
lowast Accordingto LaSallersquos invariance principle [28] we can derive theconclusion that the alcoholism equilibria 119864lowast(119878lowast 119860lowast 119879lowast 119876lowast)are globally asymptotically stable the proof is complete
5 Optimal Control Problem
51 The Existence of Optimal Control In order to investigatean effective campaign to control alcoholism in a communitywhich pursue the goals of the minimized alcoholisms andmore recovered individuals we will reconsider the system (1)and use two control variables to reduce the numbers of alco-holics The difference is that we will change the parameters1199061 1199062into control variable 119906
1(119905) 1199062(119905) Their aforementioned
definitions allow us to do so 1199061(119905) is used to limit the
proportion of the susceptible individual to contact withalcoholism usually by propaganda and education so that thesusceptible individual can stay off alcoholism consciously andbe free of ldquoinfectionrdquo we can understand the effect of 119906
1(119905)
is to prevent the the susceptible from contacting with thealcoholism The control variable 119906
2(119905) is used to control the
alcoholism to take appropriate treatment measures such astaking pills or seeking other medical help However just as acoin has two sides there will be a lot of costs generated duringthe control process So it is advisable to balance between thecosts and the alcohol effects In view of this our optimalcontrol problem tominimize the objective functional is givenby
119869 (1199061 1199062) = int
119905119891
0
[119860 (119905) +1198881
21199062
1(119905) +
1198882
21199062
2(119905)] 119889119905 (49)
which subjects to system
1198781015840= 120583119873 minus (1 minus 119906
1(119905))
120573119878119860
119873minus 120583119878
1198601015840= (1 minus 119906
1(119905))
120573119878119860
119873+ 120585119879 minus (119906
2(119905) + 120583)119860
1198791015840= 1199062(119905) 119860 minus (120583 + 120585 + 120575) 119879
1198761015840= 120575119879 minus 120583119876
(50)
with initial conditions
119878 (0) = 1198780 119860 (0) = 119860
0
119879 (0) = 1198790 119876 (0) = 119876
0
(51)
Here 119906119894(119905) isin 119880 ≜ (119906
1 1199062) | 119906119894(119905) is measurable and 0 le
119906119894(119905) le 1 for all 119905 isin [0 119905
119891] 119905119891is the end time to be
controlled 119880 is an admissible control set 119888119894 and 119894 = 1 2 are
weight factors (positive constants) that adjust the intensity oftwo different control measures
Next we will investigate the existence of the optimalcontrol of the above-mentioned problem
Theorem 5 There exists an optimal control pair 119906lowast
=
(119906lowast
1 119906lowast
2) isin 119880 such that
119869 (119906lowast
1 119906lowast
2) = min 119869 (119906
1 1199062) 119906
1(119905) 1199062(119905) isin 119880 (52)
subjects to the control system (1) with initial conditions (50)
Proof Toprove the existence of an optimal control accordingto the classic literature [29] we have to show the following
(1) The control and state variables are nonnegative values(2) The control set 119880 is convex and closed(3) The right side of the state system is bounded by linear
function in the state and control variables(4) The integrand of the objective functional is concave
on 119880(5) There exist constants 119889
1 1198892gt 0 and 120572 gt 1 such
that the integrand 119871(119905 1199061 1199062) ≜ 119860(119905) + (119888
12)1199062
1(119905) +
(11988822)1199062
2(119905) of the objective functional satisfies
119871 (119905 1199061 1199062) ge 1198891(100381610038161003816100381611990611003816100381610038161003816
2
+100381610038161003816100381611990621003816100381610038161003816
2
)1205722
minus 1198892
(53)
8 Abstract and Applied Analysis
statements (1) (2) and (3) are obvious satisfied we only needto test and verify the latter two ones Since the four statevariables have been all proved to be up bounded by 119873 wewill get the following equalities
1198781015840le 120583119873 119860
1015840le (1 minus 119906
1(119905)) 120573119878 + 120585119879
1198791015840le 1199062(119905) 119860 119876
1015840le 120575119879
(54)
so the fourth condition is set up As for the last condition
119871 (119905 1199061 1199062) ge 1198891(100381610038161003816100381611990611003816100381610038161003816
2
+100381610038161003816100381611990621003816100381610038161003816
2
)1205722
minus 1198892
(55)
is also true when we choose 1198891= min119888
12 11988822 and for all
1198892isin 119877+ 120572 = 2 The proof is complete
We next come to the core of this section
52 The Characterization of the Optimal Control With theexistence of the optimal control pairs established we nowpresent the optimality system and use a result from [30]we can easily know the existence of the solutions to theoptimality system (71) which will be gotten later Firstly wecome to discuss the theorem that relates to the character-ization of the optimal control The optimality system canbe used to compute candidates for optimal control pairsTo do this we begin by defining an augmented Hamilto-nian 119867 with penalty terms for the control constraints asfollows
119867 = 119860 (119905) +1198881
21199062
1(119905) +
1198882
21199062
2(119905)
+ 1205821[120583119873 minus (1 minus 119906
1(119905))
120573119878119860
119873minus 120583119878]
+ 1205822[(1 minus 119906
1(119905))
120573119878119860
119873+ 120585119879 minus (119906
2(119905) + 120583)119860]
+ 1205823[1199062(119905) 119860 minus (120583 + 120585 + 120575) 119879] + 120582
4(120575119879 minus 120583119876)
minus 119908111199061(119905) minus 119908
12(1 minus 119906
1(119905))
minus 119908211199062(119905) minus 119908
22(1 minus 119906
2(119905))
(56)
where 119908119894119895(119905) ge 0 are the penalty multipliers satisfying
11990811(119905) 1199061(119905) = 119908
12(119905) (1 minus 119906
1(119905))
= 0 at optimal control 119906lowast1
11990821(119905) 1199062(119905) = 119908
22(119905) (1 minus 119906
2(119905))
= 0 at optimal control 119906lowast2
(57)
Theorem6 Given optimal control pairs (119906lowast1 119906lowast
2) and solutions
119878(119905) 119860(119905) 119879(119905) 119876(119905) of the corresponding state system (50)there exist adjoint variables 120582
119894 119894 = 1 2 3 4 satisfying
1205821015840
1= 1205821(1 minus 119906
1(119905))
120573119860
119873+ 1205831205821minus 1205822(1 minus 119906
1(119905))
120573119860
119873
1205821015840
2= minus1 + 120582
1(1 minus 119906
1(119905))
120573119878
119873minus 1205822(1 minus 119906
1(119905))
120573119878
119873
+ 1205822(120583 + 119906
2(119905)) minus 120582
31199062(119905)
1205821015840
3= minus1205822120585 + 1205823(120583 + 120585 + 120575) minus 120582
4120575
1205821015840
4= 1205831205824
(58)
with the terminal conditions
120582119894(119905119891) = 0 119894 = 1 2 3 4 (59)
Furthermore (119906lowast1 119906lowast
2) are represented by
119906lowast
1= min(1max(0
120573119878119860 (1205822minus 1205821)
1198881119873
))
119906lowast
2= min(1max(0
119860 (1205822minus 1205823)
1198882
))
(60)
Proof According to Pontryagin Maximum Principle [29ndash31] we first differentiate the Hamiltonian operator 119867 withrespect to states Then the adjoint system can be written as
1205821015840
1= minus
120597119867
120597119878 120582
1015840
2= minus
120597119867
120597119860
1205821015840
3= minus
120597119867
120597119879 120582
1015840
4= minus
120597119867
120597119876
(61)
The terminal condition (56) of adjoint equations is given by120582119894(119905119891) = 0 119894 = 1 2 3 4
To obtain the necessary conditions of optimality (59) wealso differentiate theHamiltonian operator119867 with respect to119880 = (119906
1 1199062) and set them equal to zero then
120597119867
1205971199061
= 11988811199061(119905) + 120582
1
120573119878119860
119873minus 1205822
120573119878119860
119873minus 11990811+ 11990812= 0
120597119867
1205971199062
= 11988821199062(119905) minus 120582
2119860 + 120582
3119860 minus 119908
21+ 11990822= 0
(62)
By solving the optimal control we obtain
119906lowast
1=1
1198881
[(1205822minus 1205821)120573119878119860
119873+ 11990811minus 11990812] (63)
To determine an explicit expression for the optimalcontrol without119908
11and119908
12 a standard optimality technique
is utilized [29] We consider the following three cases
(i) On the set 119905 | 0 lt 119906lowast
1(119905) lt 1 we have 119908
11(119905) =
Abstract and Applied Analysis 9
11990812(119905) = 0 Hence the optimal control is
119906lowast
1=1
1198881
[(1205822minus 1205821)120573119878119860
119873] (64)
(ii) On the set 119905 | 119906lowast1(119905) = 1 we have 119908
11(119905) = 0 Hence
1 = 119906lowast
1(119905) =
1
1198881
[(1205822minus 1205821)120573119878119860
119873minus 11990812] (65)
This implies that
1
1198881
(1205822minus 1205821)120573119878119860
119873ge 1 since 119908
12(119905) ge 0 (66)
(iii) On the set 119905 | 119906lowast1(119905) = 0 we have 119908
12(119905) = 0 Hence
0 = 119906lowast
1(119905) =
1
1198881
[(1205822minus 1205821)120573119878119860
119873+ 11990811] (67)
This implies that
1
1198881
(1205822minus 1205821)120573119878119860
119873le 0 since 119908
11(119905) ge 0 (68)
Combining these results the optimal control 119906lowast1(119905) is
characterized as
119906lowast
1= min1max0 1
1198881
[(1205822minus 1205821)120573119878119860
119873] (69)
Using the similar arguments we can also obtain the otheroptimal control function
119906lowast
2= min1max0
(1205822minus 1205823) 119860
1198882
(70)
The proof is complete
We point out that the optimality system consistsof the state system (50) with the initial conditions119878(0) 119860(0) 119879(0) 119876(0) the adjoint (or costate) system(58) with the terminal conditions (59) and the optimalitycondition (60) Any optimal control pairs must satisfythis optimality system For the convenience of subsequent
numerical simulation in Section 6 we give the optimalitysystem as follows
1198781015840= 120583119873 minus (1 minusmin1max0 1
1198881
[(1205822minus 1205821)120573119878119860
119873] (119905))
times120573119878119860
119873minus 120583119878
1198601015840= (1minusmin1max0 1
1198881
[(1205822minus1205821)120573119878119860
119873] (119905))
120573119878119860
119873
+ 120585119879 minus (min1max0(1205822minus 1205823) 119860
1198882
(119905) + 120583)119860
1198791015840= min1max0
(1205822minus 1205823) 119860
1198882
(119905) 119860
minus (120583 + 120585 + 120575) 119879
1198761015840= 120575119879 minus 120583119876
1205821015840
1= 1205821(1 minusmin1max0 1
1198881
[(1205822minus 1205821)120573119878119860
119873] (119905))
times120573119860
119873+ 1205831205821
minus 1205822(1minusmin1max0 1
1198881
[(1205822minus1205821)120573119878119860
119873](119905))
times120573119860
119873
1205821015840
2= minus1
+ 1205821(1minusmin1max0 1
1198881
[(1205822minus1205821)120573119878119860
119873] (119905))
times120573119878
119873minus 1205823sdotmin1max0
(1205822minus 1205823) 119860
1198882
(119905)
minus 1205822(1 minusmin1max0 1
1198881
[(1205822minus 1205821)120573119878119860
119873])
times120573119878
119873+ 1205822(120583+min1max0
(1205822minus 1205823) 119860
1198882
(119905))
1205821015840
3= minus1205822120585 + 1205823(120583 + 120585 + 120575) minus 120582
4120575
1205821015840
4= 1205831205824
119878 (0) = 1198780 119860 (0) = 119860
0 119879 (0) = 119879
0
119876 (0) = 1198760 120582
119894(119905119891) = 0 119894 = 1 2 3 4
(71)
53 The Uniqueness of Optimal Control Due to the a prioriboundedness of the state adjoint functions and the resultingLipschitz structure of the ODEs we can obtain the unique-ness of the optimal control
10 Abstract and Applied Analysis
Lemma 7 (see [23]) The function 119906lowast(119904) = min (119887max (119904 119886))is Lipschitz continuous in 119904 where 119886 lt 119887 are some fixed positiveconstants
Theorem 8 For all 119905 isin [0 119905119891] the solution to the optimality
system (71) is unique
Proof Suppose (119878 119860 119879 119876 1205821 1205822 1205823 1205824) and
(119878 119860 119879 119876 1205821 1205822 1205823 1205824) are two different solutions of
our optimality system (71) Let
119878 = 119890120582119905119898 119860 = 119890
120582119905119899 119879 = 119890
120582119905119901
119876 = 119890120582119905119902 120582
1= 119890minus120582119905119903 120582
2= 119890minus120582119905119904
1205823= 119890minus120582119905119908 120582
4= 119890minus120582119905V
119878 = 119890120582119905119898 119860 = 119890
120582119905119899 119879 = 119890
120582119905119901
119876 = 119890120582119905119902 120582
1= 119890minus120582119905119903 120582
2= 119890minus120582119905119904
1205823= 119890minus120582119905119908 120582
4= 119890minus120582119905V
(72)
where 120582 gt 0 is to be chosenAccordingly we have
119906lowast
1(119905) = min1max0
120573119898119899 (119904 minus 119903) 119890120582119905
1198881119873
119906lowast
2(119905) = min1max0 (119904 minus 119908) 119899
1198882
119906lowast
1(119905) = min1max0
120573119898119899 (119904 minus 119903) 119890120582119905
1198881119873
119906lowast
2(119905) = min1max0 (119904 minus 119908) 119899
1198882
(73)
Now we substitute 119878 = 119890120582119905119898 119860 = 119890
120582119905119899 119879 = 119890
120582119905119901 119876 = 119890
120582119905119902
1205821= 119890minus120582119905119903 1205822= 119890minus120582119905119904 1205823= 119890minus120582119905119908 1205824= 119890minus120582119905V and 119878 =
119890120582119905119898 119860 = 119890
120582119905119899 119879 = 119890
120582119905119901 119876 = 119890
120582119905119902 1205821= 119890minus120582119905119903 1205822=
119890minus120582119905119904 1205823= 119890minus120582119905119908 1205824= 119890minus120582119905V into the first ODE of (71)
respectively then we can obtain
+ 120582119898 = 120583119873119890minus120582119905
minus (1 minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)
times120573119898119899119890
120582119905
119873minus 120583119898
(74)
for119898 and119898 respectively Similarly we can derive
119899 + 120582119899
= (1 minusmin1max1120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)120573119898119899119890
120582119905
119873
+ 120585119901 minus (min1max0 119899 (119904 minus 119908)1198882
+ 120583) 119899
(75)
for 119899 and 119899 respectively
+ 120582119901 = min1max0 119899 (119904 minus 119908)1198882
119899
minus (120583 + 120585 + 120575) 119901
(76)
for 119901 and 119901 respectively
119902 + 120582119902 = 120575119901 minus 120583119902 (77)
for 119902 and 119902 respectively
119903 minus 120582119903 = 119903(1 minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)120573119899119890120582119905
119873
+ 120583119903minus119904(1minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)
times120573119899119890120582119905
119873
(78)
for 119903 and 119903 respectively
119904 minus 120582119904 = minus119890120582119905
+ 119903(1 minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)120573119898119890120582119905
119873
minus 119904(1 minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)
times120573119898119890120582119905
119873+ 119904(min1max0 119899 (119904 minus 119908)
1198882
+ 120583)
minus 119908(min1max0 119899 (119904 minus 119908)1198882
)
(79)
for 119904 and 119904 respectively
minus 120582119908 = minus119904120585 + (120583 + 120585 + 120575)119908 minus 120575V (80)
for 119908 and 119908 respectively
V minus 120582V = 120583V (81)
for V and V respectively
Abstract and Applied Analysis 11
By Lemma 7 we can obtain
1003816100381610038161003816119906lowast
1(119905) minus 119906
lowast
1(119905)1003816100381610038161003816 le
120573119890120582119905
1198881119873|119898119899 (119904 minus 119903) minus 119898119899 (119904 minus 119903)|
1003816100381610038161003816119906lowast
2(119905) minus 119906
lowast
2(119905)1003816100381610038161003816 le
1
1198882
|119899 (119904 minus 119908) minus 119899 (119904 minus 119908)|
(82)
The equations for 119898 119899 119901 119902 119903 119904 119908 V and the equationsfor 119898 119899 119901 119902 119903 119904 119908 V are subtracted respectively then wemultiply each equation by appropriate difference of functionsand integrate from 0 to 119905
119891 Next we add all eight integral
equations and some inequality techniques to obtain unique-ness The following calculation is similar for the sake ofsimplicity we only take119898 and119898 for an example
minus + (120583 + 120582) (119898 minus 119898)
=120573119890120582119905
119873[minus(1 minusmin1max0
120573119898119899 (119904 minus 119903) 119890120582119905
1198881119873
)119898119899
+(1 minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)119898119899]
(83)
Multiplying both sides of (83) by (119898minus119898) and integratingfrom 0 to 119905
119891gives
1
2(119898 minus 119898)
2(119905119891) + (120583 + 120582)int
119905119891
0
(119898 minus 119898)2119889119905
= int
119905119891
0
(119898 minus 119898)120573119890120582119905
119873
times [minus (1 minus 119906lowast
1)119898119899 + (1 minus 119906
lowast
1)119898119899] 119889119905
= int
119905119891
0
(119898 minus 119898)120573119890120582119905
119873[(119906lowast
1minus 1) (119898119899 minus 119898119899 + 119898119899 minus 119898119899)
+ 119898119899 (119906lowast
1minus 119906lowast
1)] 119889119905
le120573119890120582119905119891
119873int
119905119891
0
(119898 minus 119898)
times [119906lowast
1minus 1
100381610038161003816100381610038161003816100381610038161003816
(119898 |119899 minus 119899| + |119898 minus 119898| 119899) + 119898119899120573119890120582119905
1198881119873
100381610038161003816100381610038161003816100381610038161003816
times 119898119899 (119904 minus 119903) minus 119898119899 (119904 minus 119903) ] 119889119905
le120573119890120582119905119891
119873int
119905119891
0
(119898 minus 119898)
times [1003816100381610038161003816119906lowast
1minus 1
1003816100381610038161003816 (119898 |119899 minus 119899| + |119898 minus 119898| 119899) + 119898119899120573119890120582119905
1198881119873
|119898119899 (119904 minus 119904) + (119898119899 minus 119898119899) 119904
minus (119898119899 (119903 minus 119903) + (119898119899 minus 119898119899) 119903)| ] 119889119905
le120573119890120582119905119891
119873int
119905119891
0
[1003816100381610038161003816119906lowast
1minus 1
1003816100381610038161003816 |119898| |119898 minus 119898| |119899 minus 119899|
+ |119898 minus 119898|2
|119899| + |119861| |119904| |119899| |119898 minus 119898|2
+ |119861| |119898119899| |119898 minus 119898| |119903 minus 119903|
+ |119861| |119898119899| |119898 minus 119898| |119904 minus 119904|
+ |119861| |119904| |119898| |119898 minus 119898| |119899 minus 119899|
+ |119861| |119903| |119898| |119898 minus 119898| |119899 minus 119899|
+ |119861| |119903| |119899| |119898 minus 119898|2] 119889119905
le120573119890120582119905119891
119873int
119905119891
0
[
1003816100381610038161003816119906lowastminus 1
1003816100381610038161003816
2|119898| ((119898 minus 119898)
2+ (119899 minus 119899)
2)
+ |119899| (119898 minus 119898)2
+|119861| |119898119899|
2((119898 minus 119898)
2+ (119904 minus 119904)
2)
+|119861| |119904| |119898|
2((119898 minus 119898)
2+ (119899 minus 119899)
2)
+ |119861| |119904| |119899| (119898 minus 119898)2
+|119861| |119898119899|
2((119898 minus 119898)
2+ (119903 minus 119903)
2)
+|119861| |119898| |119903|
2((119898 minus 119898)
2+ (119899 minus 119899)
2)
+ |119861| |119903| |119899| (119898 minus 119898)2]119889119905
(84)
In the above derivation we use many scaling techniquesfor inequality or absolute inequality Particularly what shouldbe noted is that to get the first inequality of above derivationwe use the estimation of |119906lowast
1(119905) minus 119906
lowast
1(119905)| which has been given
before besides for the sake of convenience we note 119861 =
119898119899(120573119890120582119905119891119873) Furthermore we notice that the coefficients of
all the eight terms in the last formula (119898 minus 119898)2+ (119899 minus 119899)
2(119898 minus119898)
2 (119898 minus119898)2+ (119904 minus 119904)
2 (119898 minus119898)2+ (119899 minus 119899)
2 (119898 minus119898)2
(119898minus119898)2+(119903minus119903)
2 (119898minus119898)2+(119899minus119899)2 (119898minus119898)2 namely (|119906lowast1minus
1|2)|119898| |119899| (|119861||119898119899|)2 (|119861||119904||119898|)2 |119861||119904|119899 (|119861||119898119899|)2(|119861||119898||119903|)2 |119861||119903||119899| are nonnegative and bounded So thereexists a positive constant 119888
2such that
1
2(119898 minus 119898)
2(119905119891) + (120583 + 120582)int
119905119891
0
(119898 minus 119898)2119889119905
12 Abstract and Applied Analysis
le 1198882
120573119890120582119905119891
119873int
119905119891
0
((119898 minus 119898)2+ (119899 minus 119899)
2
+ (119904 minus 119904)2+ (119903 minus 119903)
2) 119889119905
(85)
Combining eight of these inequalities gives
1
2(119898 minus 119898) (119905
119891) +
1
2(119899 minus 119899) (119905
119891) +
1
2(119901 minus 119901) (119905
119891)
+1
2(119902 minus 119902) (119905
119891) +
1
2(119903 minus 119903) (0) +
1
2(119904 minus 119904) (0)
+1
2(119908 minus 119908) (0) +
1
2(V minus V) (0) + (120583 + 120582)
times int
119905119891
0
(119898 minus 119898)2+ (119899 minus 119899)
2+ (119901 minus 119901)
2
+ (119902 minus 119902)2
+ (119904 minus 119904)2
+ (119903 minus 119903)2+ (119908 minus 119908)
2+ (V minus V)2) 119889119905
le 119861int
119905119891
0
(119898 minus 119898)2+ (119899 minus 119899)
2+ (119901 minus 119901)
2
+ (119902 minus 119902)2
+ (119903 minus 119903)2+ (119904 minus 119904)
2
+ (119908 minus 119908)2+ (V minus V)2 119889119905
(86)
Thus from the above inequality we can conclude that
(120583 + 120582 minus 119861)int
119905119891
0
(119898 minus 119898)2+ (119899 minus 119899)
2
+ (119901 minus 119901)2
+ (119902 minus 119902)2
+ (119903 minus 119903)2
+ (119904 minus 119904)2+(119908 minus 119908)
2+ (V minus V)2 119889119905 le 0
(87)
where 119861 depends on the coefficients and the bounds dependon119898 119899 119901 119902 119903 119904 119908 V If we choose 120582 such that 120583+120582 gt 119861 then119898 = 119898 119899 = 119899 119901 = 119901 119902 = 119902 119903 = 119903 119904 = 119904 119908 = 119908 and V = VHence the solution to the optimality system is unique Theproof is complete
6 Numerical Simulation
61 The Simulation of State System (1) without ControlParameters For the sake of simplicity but without loss ofgenerality we will perform the numerical simulation of statesystem (1) with parameters 119906
1= 0 119906
2= 0 Before
illustrating the analytic properties of the alcoholism model(1) we will target the populations in the environment ofa community or a university for example the school ofmaterial science and engineering in our university that isLan zhou University of Technology (LUT for short) owing tothe accurate and available information we can obtain Refer-ring to the information provided by the admissions officeof LUT this school will enroll almost 1200 undergraduatesand almost 300 various postgraduates at the beginning of
fall semester at the same time there will be almost 1500various students graduated and left this school so the scaleof students in school remained almost 6000 we can take thetotal population 119873 = 6000 In this simulation we will takeSeptember as the initial time and units in one week periodin one year According to the investigations of the studentunion implemented in September every year we can takeinitial values as 119878(0) = 4500 119860(0) = 1000 119879(0) = 300and 119876(0) = 200 It seems that the alcoholism is a little bitmore but it is rather natural because many freshmen feelconfused when they are faced with the new environment anda new lifestyle many of them have no better choice but gathertogether to drink in small groups to mediate the anxiety andget to know each other over time some of them developthe habit of drinking To a certain extent for example thefrequent drinking badly affects their study we can classifythem into the alcoholism compartment Other initial valuesseem more reasonable so we need no more explanation Aswe know alcoholism death is seldom happen within oneyear so we omit mortality from alcoholism then how tounderstand the recruitment rate as well as natural death rate120583 We can treat the freshmen admission as the recruitmentpopulation and graduation students as the natural ldquodeathrdquoparts So we can take 120583 = 15006000 = 025 which is exactlyconsistent with the value in [16] As for the infection rate 120573and recovery rate 120575 we will let them be variables since thedrinking behaviors are related to many factors such as theseason and the pressure
According to the data we get from the student unionwe choose 120585 = 04 To summarize we list the values ofthe parameters in Table 1 Using the values of parameters inTable 1 we can plot Figures 2 and 3 which are on condition1198770lt 1 and 119877
0ge 1 respectively From Figure 2 we easily
know when 1198770lt 1 holds the solution of system (1) tends to
the alcohol free equilibrium1198640and verifies the global stability
of 1198640 While seen from Figure 3 we also know that if 119877
0ge
1 holds the solution of system (1) tends to the alcoholismequilibrium 119864
lowast and verifies the global stability of 119864lowast
62 The Sensitivity of 1198770about Two Control Parameters
Although from the expression of the model reproductionnumber 119877
0 we can easily find out the fact that the two
control variables that is 1199061and 119906
2 attribute to reducing the
severity of alcoholism we will still depict the graph between1198770and the two control variables to see more intuitiveness see
Figure 4 It seems from the figure that 1198770is a monotonically
decreasing function about two control parameters so it isadvisable to take two approaches simultaneously to controlthe alcoholics
63 The Simulation of Optimality System (71) In this subsec-tion we will investigate numerically the optimal solution tooptimality system (71) by numerical method from [32] theoptimality system is solved with a fourth-order Runge-Kuttascheme Beginning with a guess for the control variables thestate system is solved forward in time and then those values ofstate are used to solve the adjoint equations backward in timeThe controls are updated at the end of each iteration using
Abstract and Applied Analysis 13
0 10 20 30 40 50
0
1000
2000
3000
4000
5000
6000
7000
Time
Popu
latio
n siz
e
S(t)
A(t)
T(t)
Q(t)
Figure 2 When 1198770lt 1 the alcohol free equilibrium 119864
0is globally
asymptotically stable (120573 = 02 120575 = 03 and 1198770= 071698)
S(t)
A(t)
T(t)
Q(t)
0 10 20 30 40
48
2520
303540
49 50
500
1000
2000
3000
4000
5000
6000
7000
Time
Popu
latio
n siz
e
Figure 3 When 1198770
ge 1 the alcoholism equilibrium 119864lowast
=
(4466 476 32 26) corresponding to the given parameters is globallyasymptotically stable (120573 = 03 120575 = 02 and 119877
0= 108511)
the values of optimal controls obtained lastly The iterationscontinue until convergence takes place
In the simulations we choose the available variable valuesas Table 1 shows besides 120573 = 03 120575 = 02 The initial valueof model (1) is assumed to be 119878(0) = 4500 119860(0) = 1000119879(0) = 300 and 119876(0) = 200 as before
The ideal weights in objective functional are very difficultto obtain in reality it needs much work on data mining andfittingHence the acquisition of appropriate practical weights
Table 1 The parameters description of model (1)
Parameter Description Values120583 Natural birth rate or death rate 025
120573Transmission coefficient betweenalcoholism and susceptibles Variables
119873 The total populations to be considered 6000
120575The rate of populations quitting fromalcoholism permanently after treatment Variables
120585The rate of populations failed intreatment and returned to be alcoholic 04
0
02
04
06
08
1
0
05
1
0 02 04 06 08 1
u1
u2
R0
Figure 4 The relationship between 1198770and two control variables 119906
1
and 1199062
is still a difficult problem and remains for further investiga-tionsThe cost associated with119860(119905) and 119906
1(119905)mainly includes
the cost of dangerous behaviours during the alcoholism timeand educating the public while the cost associated with1199062(119905)mainly comes from health professional and the medical
resource includingmedicines andnursing care In viewof thisand taking the expressions of 119906
1 1199062into account after many
numerical simulations we finally give weighting coefficientsas 1198881= 102 1198882= 104 It should be pointed out that the
weights here are of only theoretical interest to reveal thecontrol strategies proposed in this paper Another point tonote is that themaximumcontrol is very difficult to achieve inreality so we will omit the situation of the maximum controlduring the series of simulations
Next we will make some necessary instructions andexplanations to the above simulation graphs Figures 5 67 and 8 depict the number of four compartments underdifferent control levels when we choose the weight coeffi-cients in objective function to be 119888
1= 102 1198882= 104 From
the four simulation graphs we can observe the followingsimple facts in reducing the total number of alcoholismsand increasing the number of susceptibles the effectivenessof various control measures is as follows optimal control isevidently better than middle control and middle control isbetter than single control 119906
2 single control 119906
2is better than
single control 1199061 while single control 119906
1is much better than
no controlFigures 9 and 10 depict the optimal control law of 119906
1 1199062
respectively In the beginning of the simulation the control
14 Abstract and Applied Analysis
0 10 20 30 40 504500
5000
5500
6000
6500
7000
S(n
umbe
r of s
usce
ptib
les)
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 5Number of the susceptibles whenwe choose theweights inobjective function are 119888
1= 102 1198882= 104 and under different optimal
control strategies that is (1) without control 1199061= 1199062= 0 (2) single
control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0 119906
2= 05 (4)
middle control 1199061= 1199062= 05 (5) optimal control 119906
1= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
500
1000
1500
A(n
umbe
r of t
he al
coho
lism
)
t (unit week)
Figure 6 Number of the alcoholics when we choose the weights inobjective function are 119888
1= 102 1198882= 104 and under different optimal
control strategies that is (1) without control 1199061= 1199062= 0 (2) single
control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0 119906
2= 05 (4)
middle control 1199061= 1199062= 05 (5) optimal control 119906
1= 119906lowast
1 1199062= 119906lowast
2
0
50
100
150
200
250
300
350
400
T(n
umbe
r of t
reat
men
t)
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 7 Number of the persons in treatment when we choose theweights in objective function are 119888
1= 10
2 1198882= 10
4 and underdifferent optimal control strategies that is (1) without control 119906
1=
1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control 119906
1= 0
1199062= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal control
1199061= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
50
100
150
200
250
300
350
400
Q(n
umbe
r of q
uitti
ng)
Figure 8 Number of people in quitting compartment when wechoose the weights in objective function are 119888
1= 102 1198882= 104 and
under different optimal control strategies that is (1) without control1199061= 1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control
1199061= 0 119906
2= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal
control 1199061= 119906lowast
1 1199062= 119906lowast
2
Abstract and Applied Analysis 15
0 10 20 30 40 50t (unit week)
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
Optimal control of u1
Figure 9 Figures of the optimal control 1199061when we choose the
weight in objective function are 1198881= 102 1198882= 104
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
0 10 20 30 40 50t (unit week)
Optimal control of u2
Figure 10 Figures of the optimal control 1199062when we choose the
weight in objective function are 1198881= 102 1198882= 104
effort of 1199061should be decreased from 065 to 05 within the
first month and over the next week it should be increased tothe maximum control until 50 weeks then rapidly decreasedto 0 at the end of the simulation As for the control 119906
2 it
should start from around 05 due to the initial alcoholics thenincrease to 055 within one week since the rapid infection andnext decrease to almost 0 since the effectiveness of treatmentin three weeks but with the infection going on the controleffort of 119906
2should gradually increase to the maximum and
maintain this level until the tenth week for the purpose ofconsolidation therapy and preventing rebound hereafter itshould be gradually decreased to the level of almost 043 untilthe fifty weeks then quickly decreased to 0 in the end
0 10 20 30 40 504500
5000
5500
6000
6500
7000
S(n
umbe
r of s
usce
ptib
les)
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 11 Number of the susceptibles when we choose the weightsin objective function are 119888
1= 104 1198882= 102 and under different
optimal control strategies that is (1) without control 1199061= 1199062= 0
(2) single control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0
1199062= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal control
1199061= 119906lowast
1 1199062= 119906lowast
2
In order to investigate the influence of different weightcoefficients in the objective functional on the effect ofcontrolling at the same time for a better comparison we willchange the weight coefficients in objective function into 119888
2=
104 1198881= 102 and we will list the corresponding numerical
simulation results as Figures 11ndash16 showWhen we change the weight coefficients in objective
function into 1198881= 104 1198882= 102 we find that the results
of simulations derived from the graphs are very similar tothe ones before We speculate that the most likely reasons ofthis result are due to three respects one is that the weightcoefficients are not too sensitive in the numerical simulationand another possible reason is that both of the two controlsare important in some sense equivalently importantThe lastbut not themost unlikely reason is that we have not found themost appropriate weight coefficients in the simulation whichis very difficult to find as previously mentioned
7 Conclusions
In this paper we formulate an alcoholics quitting modeland firstly investigate the variation discipline of variouspopulations from the perspective of global stability thenwe propose an objective functional to examine two differentcontrolmeasures (ie prevention and treatment) on the effectof alcohol The basic reproduction number of the model wasderived and the global stability of the two equilibria is givenFrom the expression of the basic reproductionnumber119877
0and
16 Abstract and Applied Analysis
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
500
1000
1500
A(n
umbe
r of t
he al
coho
lism
)
t (unit week)
Figure 12 Number of the alcoholics when we choose the weights inobjective function are 119888
1= 104 1198882= 102 and under different optimal
control strategies that is (1) without control 1199061= 1199062= 0 (2) single
control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0 119906
2= 05 (4)
middle control 1199061= 1199062= 05 (5) optimal control 119906
1= 119906lowast
1 1199062= 119906lowast
2
0
50
100
150
200
250
300
350
400
T(n
umbe
r of t
reat
men
t)
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 13 Number of the persons in treatment when we choosethe weights in objective function are 119888
1= 104 1198882= 102 and under
different optimal control strategies that is (1) without control 1199061=
1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control 119906
1= 0
1199062= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal control
1199061= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
50
100
150
200
250
300
350
400
Q(n
umbe
r of q
uitti
ng)
Figure 14 Number of people in quitting compartment when wechoose the weights in objective function are 119888
1= 104 1198882= 102 and
under different optimal control strategies that is (1) without control1199061= 1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control
1199061= 0 119906
2= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal
control 1199061= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50t (unit week)
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
Optimal control of u1
Figure 15 Figures of the optimal control 1199061when we choose the
weight in objective function are 1198881= 104 1198882= 102
related numerical simulation we can easily see that the twocontrol strategies are effective in the alcoholics process
Using Pontryaginrsquos Maximum Principle we firstly deter-mine the necessary conditions for existence of optimalcontrol pairsThe uniqueness of the solution to the optimalitysystem (71) is derived by the classical method of contradic-tion Numerical simulations of the model suggest that thetwo different groups of weights in the objective function have
Abstract and Applied Analysis 17
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
0 10 20 30 40 50t (unit week)
Optimal control of u2
Figure 16 Figures of the optimal control 1199062when we choose the
weight in objective function are 1198881= 104 1198882= 102
much similar effects on the transmission of the alcoholismfrom this point the two control measures are almost equallyimportant in controlling the alcoholism although they willprobably have great influences on the cost of the objectivefunction From the simulation figures it seems that the effectof optimal control which is measured by the reduction inthe number of alcoholics and the increase in the number ofsusceptibles is much better than other control strategies asnoted earlier in the simulation section According to the real-time curve of two optimal controls we point out the specificimplementation methods of optimal control which can beachieved in practice
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
This work was partially supported by the NSF ofGansu Province of China (2013GS09485) the NNSF ofChina (10961018) the NSF of Gansu Province of China(1107RJZA088) the NSF for Distinguished Young Scholarsof Gansu Province of China (1111RJDA003) the Special Fundfor the Basic Requirements in the Research of University ofGansu Province of China and the Development Programfor Hong Liu Distinguished Young Scholars in LanzhouUniversity of Technology
References
[1] R Room T Babor and J Rehm ldquoAlcohol and public healthrdquoThe Lancet vol 365 no 9458 pp 519ndash530 2005
[2] J B Saunders O G Aasland A Amundsen and M GrantldquoAlcohol consumption and related problems among primary
health care patients WHO collaborative project on early detec-tion of personswith harmful alcohol consumption IrdquoAddictionvol 88 no 3 pp 349ndash362 1993
[3] L D Johnston P M OMalley and J G Bachman NationalSurvey Results on Drug Use From the Monitoring the FutureStudy 1975ndash1992 National Institute on Drug Abuse RockvilleMd USA 1993
[4] H Wechsler J E Lee M Kuo and H Lee ldquoCollege bingedrinking in the 1990s a continuing problem Results of theHarvard School of Public Health 1999 College Alcohol StudyrdquoJournal of American College Health vol 48 no 5 pp 199ndash2102000
[5] P M OrsquoMalley and L D Johnston ldquoEpidemiology of alcoholand other drug use among American college studentsrdquo Journalof Studies on Alcohol vol 63 no 14 pp 23ndash39 2002
[6] K Agarwal-Kozlowski and D P Agarwal ldquoGenetic predisposi-tion to alcoholismrdquo Therapeutische Umschau vol 57 no 4 pp179ndash184 2000
[7] C Y Chen C L Storr and J C Anthony ldquoEarly-onset drug useand risk for drug dependence problemsrdquo Addictive Behaviorsvol 34 no 3 pp 319ndash322 2009
[8] M Glavas and J Weinberg ldquoStress alcohol consumption andthe hypothalamic-pituitary adrenal axisrdquo in Nutrients Stressand Medical Disorders pp 165ndash183 Humana Press New YorkNY USA 2006
[9] L N Blum N H Nielsen J A Riggs and L B BresolinldquoAlcoholism and alcohol abuse among women report of theCouncil on Scientific Affairsrdquo Journal of Womenrsquos Health vol7 no 7 pp 861ndash871 1998
[10] B Benedict ldquoModeling alcoholism as a contagious disease howldquoinfectedrdquo drinking buddies spread problem drinkingrdquo SIAMNews vol 40 no 3 2007
[11] H Walter K Gutierrez K Ramskogler I Hertling A Dvorakand O M Lesch ldquoGender-specific differences in alcoholismImplications for treatmentrdquo Archives of Womenrsquos Mental Healthvol 6 no 4 pp 253ndash258 2003
[12] D Muller R D Koch H von Specht w Volker and E MMunch ldquoNeurophisiologic findings in chronic alcohol abuserdquoPsychiatrie Neurologie und Medizinische Psychologie vol 37no 3 pp 129ndash132 1985
[13] G Testino ldquoAlcoholic diseases in hepato-gastroenterology apoint of viewrdquo Hepato-Gastroenterology vol 55 no 82-83 pp371ndash377 2008
[14] A Mubayi P E Greenwood C Castillo-Chavez P J Grue-newald and D M Gorman ldquoThe impact of relative residencetimes on the distribution of heavy drinkers in highly distinctenvironmentsrdquo Socio-Economic Planning Sciences vol 44 no 1pp 45ndash56 2010
[15] D Muller R D Koch H von Specht W Volker and E MMunch ldquoNeurophysiologic findings in chronic alcohol abuserdquoPsychiatrie Neurologie Und Medizinische Psychologie vol 37no 3 pp 129ndash132 1985 (Leipz)
[16] G Mulone and B Straughan ldquoModeling binge drinkingrdquoInternational Journal of Biomathematics vol 5 no 1 Article ID1250005 14 pages 2012
[17] H F Huo and N N Song ldquoGlobal stability for a binge drinkingmodel with two stagesrdquo Discrete Dynamics in Nature andSociety vol 2012 Article ID 829386 15 pages 2012
[18] S S Mushayabasa and C P Bhunu ldquoModelling the effects ofheavy alcohol consumption on the transmission dynamics ofgonorrheardquo Nonlinear Dynamics vol 66 no 4 pp 695ndash7062011
18 Abstract and Applied Analysis
[19] F S Sancheznchez XWang C Castillo-Chvez DM Gormanand P J Gruenewald ldquoDrinking as an epidemic a simplemathematical model with recovery and relapserdquo in TherapistsGuide to Evidence-Based Relapse Prevention K Witkiewitz andG Marlatt Eds Academic Press New York NY USA 2007
[20] H-F Huo and C-C Zhu ldquoInfluence of relapse in a givingup smoking modelrdquo Abstract and Applied Analysis vol 2013Article ID 525461 12 pages 2013
[21] H F Huo and Q Wang ldquoThe effects of awareness programs by17 media on the drinking dynamicsrdquo Submitted
[22] S Lee E Jung and C Castillo-Chavez ldquoOptimal controlintervention strategies in low- and high-risk problem drinkingpopulationsrdquo Socio-Economic Planning Sciences vol 44 no 4pp 258ndash265 2010
[23] K R Fister S Lenhart and J S McNally ldquoOptimizingchemotherapy in an HIV modelrdquo Electronic Journal of Differ-ential Equations vol 1998 no 32 pp 1ndash12 1998
[24] S Lenhart and J T Workman Optimal Control Applied toBiological Models Chapman amp HallCRC Mathematical andComputational Biology Series Chapman amp HallCRC BocaRaton Fla USA 2007
[25] X Yang L Chen and J Chen ldquoPermanence and positiveperiodic solution for the single-species nonautonomous delaydiffusive modelsrdquo Computers amp Mathematics with ApplicationsAn International Journal vol 32 no 4 pp 109ndash116 1996
[26] P van denDriessche and JWatmough ldquoReproduction numbersand sub-threshold endemic equilibria for compartmental mod-els of disease transmissionrdquoMathematical Biosciences vol 180pp 29ndash48 2002
[27] V Lakshmikantham S Leela and A A Martynyuk StabilityAnalysis of Nonlinear Systems vol 125 Marcel Dekker NewYork NY USA 1989
[28] J P LaSalle The Stability of Dynamical Systems vol 25 ofRegional Conference Series in Applied Mathematics Society forIndustrial and Applied Mathematics 1976
[29] W H Fleming and R W Rishel Deterministic and StochasticOptimal Control vol 1 ofApplications of Mathematics SpringerBerlin Germany 1975
[30] D L Lukes Differential Equations Classical to ControlledMathematics in Science and Engineering Academic Press NewYork NY USA 1982
[31] L S Pontryagin V G Boltyanskii R V Gamkrelidze and EF Mishchenko The Mathematical Theory of Optimal ProcessesJohn Wiley amp Sons New Jersey NJ USA 1962
[32] W Hackbusch ldquoA numerical method for solving parabolicequations with opposite orientationsrdquo Computing vol 20 no3 pp 229ndash240 1978
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 7
so
1198811015840
1= 2 (1 minus 119906
1) 120573119878lowast119860lowast+ 2120585119879
lowast
1198811015840
2= minus[
(119878lowast)2
(1 minus 1199061) 120573119860lowast
119878
+ (1 minus 1199061) 120573119878119860
lowast+119860lowast120585119879
119860+ 1199093
119879lowast
1198791199062119860]
le minus2[(119878lowast)2
(1 minus 1199061) 120573119860lowast
119878sdot (1 minus 119906
1) 120573119878119860
lowast]
12
minus 2 [119860lowast120585119879
119860sdot 1199093
119879lowast
1198791199062119860]
= minus2 (1 minus 1199061) 120573119878lowast119860lowastminus 2(1199093119860lowast1205851199062119879lowast)12
= minus2 (1 minus 1199061) 120573119878lowast119860lowastminus 2[120585119879
lowast(11990931199062119860lowast)]12
= minus2 (1 minus 1199061) 120573119878lowast119860lowast
minus 2120585119879lowast[(1199062+ 120583)119860
lowastminus (1 minus 119906
1) 120573119878lowast119860lowast]12
= minus2 (1 minus 1199061) 120573119878lowast119860lowastminus 2(120585119879
lowast120585119879lowast)12
= minus2 (1 minus 1199061) 120573119878lowast119860lowastminus 2120585119879
lowast
(47)
Hence
1198811015840= 120583119878lowast(2 minus
119878
119878lowastminus119878lowast
119878) + 119881
1015840
1+ 1198811015840
2
le 120583119878lowast(2 minus
119878
119878lowastminus119878lowast
119878) + 2 (1 minus 119906
1) 120573119878lowast119860lowast
+ 2120585119879lowastminus 2 (1 minus 119906
1) 120573119878lowast119860lowastminus 2120585119879
lowast
= 120583119878lowast(2 minus
119878
119878lowastminus119878lowast
119878) le 0
(48)
1198811015840= 0 if and only if 119878 = 119878
lowast 119860 = 119860lowast 119879 = 119879
lowast Accordingto LaSallersquos invariance principle [28] we can derive theconclusion that the alcoholism equilibria 119864lowast(119878lowast 119860lowast 119879lowast 119876lowast)are globally asymptotically stable the proof is complete
5 Optimal Control Problem
51 The Existence of Optimal Control In order to investigatean effective campaign to control alcoholism in a communitywhich pursue the goals of the minimized alcoholisms andmore recovered individuals we will reconsider the system (1)and use two control variables to reduce the numbers of alco-holics The difference is that we will change the parameters1199061 1199062into control variable 119906
1(119905) 1199062(119905) Their aforementioned
definitions allow us to do so 1199061(119905) is used to limit the
proportion of the susceptible individual to contact withalcoholism usually by propaganda and education so that thesusceptible individual can stay off alcoholism consciously andbe free of ldquoinfectionrdquo we can understand the effect of 119906
1(119905)
is to prevent the the susceptible from contacting with thealcoholism The control variable 119906
2(119905) is used to control the
alcoholism to take appropriate treatment measures such astaking pills or seeking other medical help However just as acoin has two sides there will be a lot of costs generated duringthe control process So it is advisable to balance between thecosts and the alcohol effects In view of this our optimalcontrol problem tominimize the objective functional is givenby
119869 (1199061 1199062) = int
119905119891
0
[119860 (119905) +1198881
21199062
1(119905) +
1198882
21199062
2(119905)] 119889119905 (49)
which subjects to system
1198781015840= 120583119873 minus (1 minus 119906
1(119905))
120573119878119860
119873minus 120583119878
1198601015840= (1 minus 119906
1(119905))
120573119878119860
119873+ 120585119879 minus (119906
2(119905) + 120583)119860
1198791015840= 1199062(119905) 119860 minus (120583 + 120585 + 120575) 119879
1198761015840= 120575119879 minus 120583119876
(50)
with initial conditions
119878 (0) = 1198780 119860 (0) = 119860
0
119879 (0) = 1198790 119876 (0) = 119876
0
(51)
Here 119906119894(119905) isin 119880 ≜ (119906
1 1199062) | 119906119894(119905) is measurable and 0 le
119906119894(119905) le 1 for all 119905 isin [0 119905
119891] 119905119891is the end time to be
controlled 119880 is an admissible control set 119888119894 and 119894 = 1 2 are
weight factors (positive constants) that adjust the intensity oftwo different control measures
Next we will investigate the existence of the optimalcontrol of the above-mentioned problem
Theorem 5 There exists an optimal control pair 119906lowast
=
(119906lowast
1 119906lowast
2) isin 119880 such that
119869 (119906lowast
1 119906lowast
2) = min 119869 (119906
1 1199062) 119906
1(119905) 1199062(119905) isin 119880 (52)
subjects to the control system (1) with initial conditions (50)
Proof Toprove the existence of an optimal control accordingto the classic literature [29] we have to show the following
(1) The control and state variables are nonnegative values(2) The control set 119880 is convex and closed(3) The right side of the state system is bounded by linear
function in the state and control variables(4) The integrand of the objective functional is concave
on 119880(5) There exist constants 119889
1 1198892gt 0 and 120572 gt 1 such
that the integrand 119871(119905 1199061 1199062) ≜ 119860(119905) + (119888
12)1199062
1(119905) +
(11988822)1199062
2(119905) of the objective functional satisfies
119871 (119905 1199061 1199062) ge 1198891(100381610038161003816100381611990611003816100381610038161003816
2
+100381610038161003816100381611990621003816100381610038161003816
2
)1205722
minus 1198892
(53)
8 Abstract and Applied Analysis
statements (1) (2) and (3) are obvious satisfied we only needto test and verify the latter two ones Since the four statevariables have been all proved to be up bounded by 119873 wewill get the following equalities
1198781015840le 120583119873 119860
1015840le (1 minus 119906
1(119905)) 120573119878 + 120585119879
1198791015840le 1199062(119905) 119860 119876
1015840le 120575119879
(54)
so the fourth condition is set up As for the last condition
119871 (119905 1199061 1199062) ge 1198891(100381610038161003816100381611990611003816100381610038161003816
2
+100381610038161003816100381611990621003816100381610038161003816
2
)1205722
minus 1198892
(55)
is also true when we choose 1198891= min119888
12 11988822 and for all
1198892isin 119877+ 120572 = 2 The proof is complete
We next come to the core of this section
52 The Characterization of the Optimal Control With theexistence of the optimal control pairs established we nowpresent the optimality system and use a result from [30]we can easily know the existence of the solutions to theoptimality system (71) which will be gotten later Firstly wecome to discuss the theorem that relates to the character-ization of the optimal control The optimality system canbe used to compute candidates for optimal control pairsTo do this we begin by defining an augmented Hamilto-nian 119867 with penalty terms for the control constraints asfollows
119867 = 119860 (119905) +1198881
21199062
1(119905) +
1198882
21199062
2(119905)
+ 1205821[120583119873 minus (1 minus 119906
1(119905))
120573119878119860
119873minus 120583119878]
+ 1205822[(1 minus 119906
1(119905))
120573119878119860
119873+ 120585119879 minus (119906
2(119905) + 120583)119860]
+ 1205823[1199062(119905) 119860 minus (120583 + 120585 + 120575) 119879] + 120582
4(120575119879 minus 120583119876)
minus 119908111199061(119905) minus 119908
12(1 minus 119906
1(119905))
minus 119908211199062(119905) minus 119908
22(1 minus 119906
2(119905))
(56)
where 119908119894119895(119905) ge 0 are the penalty multipliers satisfying
11990811(119905) 1199061(119905) = 119908
12(119905) (1 minus 119906
1(119905))
= 0 at optimal control 119906lowast1
11990821(119905) 1199062(119905) = 119908
22(119905) (1 minus 119906
2(119905))
= 0 at optimal control 119906lowast2
(57)
Theorem6 Given optimal control pairs (119906lowast1 119906lowast
2) and solutions
119878(119905) 119860(119905) 119879(119905) 119876(119905) of the corresponding state system (50)there exist adjoint variables 120582
119894 119894 = 1 2 3 4 satisfying
1205821015840
1= 1205821(1 minus 119906
1(119905))
120573119860
119873+ 1205831205821minus 1205822(1 minus 119906
1(119905))
120573119860
119873
1205821015840
2= minus1 + 120582
1(1 minus 119906
1(119905))
120573119878
119873minus 1205822(1 minus 119906
1(119905))
120573119878
119873
+ 1205822(120583 + 119906
2(119905)) minus 120582
31199062(119905)
1205821015840
3= minus1205822120585 + 1205823(120583 + 120585 + 120575) minus 120582
4120575
1205821015840
4= 1205831205824
(58)
with the terminal conditions
120582119894(119905119891) = 0 119894 = 1 2 3 4 (59)
Furthermore (119906lowast1 119906lowast
2) are represented by
119906lowast
1= min(1max(0
120573119878119860 (1205822minus 1205821)
1198881119873
))
119906lowast
2= min(1max(0
119860 (1205822minus 1205823)
1198882
))
(60)
Proof According to Pontryagin Maximum Principle [29ndash31] we first differentiate the Hamiltonian operator 119867 withrespect to states Then the adjoint system can be written as
1205821015840
1= minus
120597119867
120597119878 120582
1015840
2= minus
120597119867
120597119860
1205821015840
3= minus
120597119867
120597119879 120582
1015840
4= minus
120597119867
120597119876
(61)
The terminal condition (56) of adjoint equations is given by120582119894(119905119891) = 0 119894 = 1 2 3 4
To obtain the necessary conditions of optimality (59) wealso differentiate theHamiltonian operator119867 with respect to119880 = (119906
1 1199062) and set them equal to zero then
120597119867
1205971199061
= 11988811199061(119905) + 120582
1
120573119878119860
119873minus 1205822
120573119878119860
119873minus 11990811+ 11990812= 0
120597119867
1205971199062
= 11988821199062(119905) minus 120582
2119860 + 120582
3119860 minus 119908
21+ 11990822= 0
(62)
By solving the optimal control we obtain
119906lowast
1=1
1198881
[(1205822minus 1205821)120573119878119860
119873+ 11990811minus 11990812] (63)
To determine an explicit expression for the optimalcontrol without119908
11and119908
12 a standard optimality technique
is utilized [29] We consider the following three cases
(i) On the set 119905 | 0 lt 119906lowast
1(119905) lt 1 we have 119908
11(119905) =
Abstract and Applied Analysis 9
11990812(119905) = 0 Hence the optimal control is
119906lowast
1=1
1198881
[(1205822minus 1205821)120573119878119860
119873] (64)
(ii) On the set 119905 | 119906lowast1(119905) = 1 we have 119908
11(119905) = 0 Hence
1 = 119906lowast
1(119905) =
1
1198881
[(1205822minus 1205821)120573119878119860
119873minus 11990812] (65)
This implies that
1
1198881
(1205822minus 1205821)120573119878119860
119873ge 1 since 119908
12(119905) ge 0 (66)
(iii) On the set 119905 | 119906lowast1(119905) = 0 we have 119908
12(119905) = 0 Hence
0 = 119906lowast
1(119905) =
1
1198881
[(1205822minus 1205821)120573119878119860
119873+ 11990811] (67)
This implies that
1
1198881
(1205822minus 1205821)120573119878119860
119873le 0 since 119908
11(119905) ge 0 (68)
Combining these results the optimal control 119906lowast1(119905) is
characterized as
119906lowast
1= min1max0 1
1198881
[(1205822minus 1205821)120573119878119860
119873] (69)
Using the similar arguments we can also obtain the otheroptimal control function
119906lowast
2= min1max0
(1205822minus 1205823) 119860
1198882
(70)
The proof is complete
We point out that the optimality system consistsof the state system (50) with the initial conditions119878(0) 119860(0) 119879(0) 119876(0) the adjoint (or costate) system(58) with the terminal conditions (59) and the optimalitycondition (60) Any optimal control pairs must satisfythis optimality system For the convenience of subsequent
numerical simulation in Section 6 we give the optimalitysystem as follows
1198781015840= 120583119873 minus (1 minusmin1max0 1
1198881
[(1205822minus 1205821)120573119878119860
119873] (119905))
times120573119878119860
119873minus 120583119878
1198601015840= (1minusmin1max0 1
1198881
[(1205822minus1205821)120573119878119860
119873] (119905))
120573119878119860
119873
+ 120585119879 minus (min1max0(1205822minus 1205823) 119860
1198882
(119905) + 120583)119860
1198791015840= min1max0
(1205822minus 1205823) 119860
1198882
(119905) 119860
minus (120583 + 120585 + 120575) 119879
1198761015840= 120575119879 minus 120583119876
1205821015840
1= 1205821(1 minusmin1max0 1
1198881
[(1205822minus 1205821)120573119878119860
119873] (119905))
times120573119860
119873+ 1205831205821
minus 1205822(1minusmin1max0 1
1198881
[(1205822minus1205821)120573119878119860
119873](119905))
times120573119860
119873
1205821015840
2= minus1
+ 1205821(1minusmin1max0 1
1198881
[(1205822minus1205821)120573119878119860
119873] (119905))
times120573119878
119873minus 1205823sdotmin1max0
(1205822minus 1205823) 119860
1198882
(119905)
minus 1205822(1 minusmin1max0 1
1198881
[(1205822minus 1205821)120573119878119860
119873])
times120573119878
119873+ 1205822(120583+min1max0
(1205822minus 1205823) 119860
1198882
(119905))
1205821015840
3= minus1205822120585 + 1205823(120583 + 120585 + 120575) minus 120582
4120575
1205821015840
4= 1205831205824
119878 (0) = 1198780 119860 (0) = 119860
0 119879 (0) = 119879
0
119876 (0) = 1198760 120582
119894(119905119891) = 0 119894 = 1 2 3 4
(71)
53 The Uniqueness of Optimal Control Due to the a prioriboundedness of the state adjoint functions and the resultingLipschitz structure of the ODEs we can obtain the unique-ness of the optimal control
10 Abstract and Applied Analysis
Lemma 7 (see [23]) The function 119906lowast(119904) = min (119887max (119904 119886))is Lipschitz continuous in 119904 where 119886 lt 119887 are some fixed positiveconstants
Theorem 8 For all 119905 isin [0 119905119891] the solution to the optimality
system (71) is unique
Proof Suppose (119878 119860 119879 119876 1205821 1205822 1205823 1205824) and
(119878 119860 119879 119876 1205821 1205822 1205823 1205824) are two different solutions of
our optimality system (71) Let
119878 = 119890120582119905119898 119860 = 119890
120582119905119899 119879 = 119890
120582119905119901
119876 = 119890120582119905119902 120582
1= 119890minus120582119905119903 120582
2= 119890minus120582119905119904
1205823= 119890minus120582119905119908 120582
4= 119890minus120582119905V
119878 = 119890120582119905119898 119860 = 119890
120582119905119899 119879 = 119890
120582119905119901
119876 = 119890120582119905119902 120582
1= 119890minus120582119905119903 120582
2= 119890minus120582119905119904
1205823= 119890minus120582119905119908 120582
4= 119890minus120582119905V
(72)
where 120582 gt 0 is to be chosenAccordingly we have
119906lowast
1(119905) = min1max0
120573119898119899 (119904 minus 119903) 119890120582119905
1198881119873
119906lowast
2(119905) = min1max0 (119904 minus 119908) 119899
1198882
119906lowast
1(119905) = min1max0
120573119898119899 (119904 minus 119903) 119890120582119905
1198881119873
119906lowast
2(119905) = min1max0 (119904 minus 119908) 119899
1198882
(73)
Now we substitute 119878 = 119890120582119905119898 119860 = 119890
120582119905119899 119879 = 119890
120582119905119901 119876 = 119890
120582119905119902
1205821= 119890minus120582119905119903 1205822= 119890minus120582119905119904 1205823= 119890minus120582119905119908 1205824= 119890minus120582119905V and 119878 =
119890120582119905119898 119860 = 119890
120582119905119899 119879 = 119890
120582119905119901 119876 = 119890
120582119905119902 1205821= 119890minus120582119905119903 1205822=
119890minus120582119905119904 1205823= 119890minus120582119905119908 1205824= 119890minus120582119905V into the first ODE of (71)
respectively then we can obtain
+ 120582119898 = 120583119873119890minus120582119905
minus (1 minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)
times120573119898119899119890
120582119905
119873minus 120583119898
(74)
for119898 and119898 respectively Similarly we can derive
119899 + 120582119899
= (1 minusmin1max1120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)120573119898119899119890
120582119905
119873
+ 120585119901 minus (min1max0 119899 (119904 minus 119908)1198882
+ 120583) 119899
(75)
for 119899 and 119899 respectively
+ 120582119901 = min1max0 119899 (119904 minus 119908)1198882
119899
minus (120583 + 120585 + 120575) 119901
(76)
for 119901 and 119901 respectively
119902 + 120582119902 = 120575119901 minus 120583119902 (77)
for 119902 and 119902 respectively
119903 minus 120582119903 = 119903(1 minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)120573119899119890120582119905
119873
+ 120583119903minus119904(1minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)
times120573119899119890120582119905
119873
(78)
for 119903 and 119903 respectively
119904 minus 120582119904 = minus119890120582119905
+ 119903(1 minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)120573119898119890120582119905
119873
minus 119904(1 minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)
times120573119898119890120582119905
119873+ 119904(min1max0 119899 (119904 minus 119908)
1198882
+ 120583)
minus 119908(min1max0 119899 (119904 minus 119908)1198882
)
(79)
for 119904 and 119904 respectively
minus 120582119908 = minus119904120585 + (120583 + 120585 + 120575)119908 minus 120575V (80)
for 119908 and 119908 respectively
V minus 120582V = 120583V (81)
for V and V respectively
Abstract and Applied Analysis 11
By Lemma 7 we can obtain
1003816100381610038161003816119906lowast
1(119905) minus 119906
lowast
1(119905)1003816100381610038161003816 le
120573119890120582119905
1198881119873|119898119899 (119904 minus 119903) minus 119898119899 (119904 minus 119903)|
1003816100381610038161003816119906lowast
2(119905) minus 119906
lowast
2(119905)1003816100381610038161003816 le
1
1198882
|119899 (119904 minus 119908) minus 119899 (119904 minus 119908)|
(82)
The equations for 119898 119899 119901 119902 119903 119904 119908 V and the equationsfor 119898 119899 119901 119902 119903 119904 119908 V are subtracted respectively then wemultiply each equation by appropriate difference of functionsand integrate from 0 to 119905
119891 Next we add all eight integral
equations and some inequality techniques to obtain unique-ness The following calculation is similar for the sake ofsimplicity we only take119898 and119898 for an example
minus + (120583 + 120582) (119898 minus 119898)
=120573119890120582119905
119873[minus(1 minusmin1max0
120573119898119899 (119904 minus 119903) 119890120582119905
1198881119873
)119898119899
+(1 minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)119898119899]
(83)
Multiplying both sides of (83) by (119898minus119898) and integratingfrom 0 to 119905
119891gives
1
2(119898 minus 119898)
2(119905119891) + (120583 + 120582)int
119905119891
0
(119898 minus 119898)2119889119905
= int
119905119891
0
(119898 minus 119898)120573119890120582119905
119873
times [minus (1 minus 119906lowast
1)119898119899 + (1 minus 119906
lowast
1)119898119899] 119889119905
= int
119905119891
0
(119898 minus 119898)120573119890120582119905
119873[(119906lowast
1minus 1) (119898119899 minus 119898119899 + 119898119899 minus 119898119899)
+ 119898119899 (119906lowast
1minus 119906lowast
1)] 119889119905
le120573119890120582119905119891
119873int
119905119891
0
(119898 minus 119898)
times [119906lowast
1minus 1
100381610038161003816100381610038161003816100381610038161003816
(119898 |119899 minus 119899| + |119898 minus 119898| 119899) + 119898119899120573119890120582119905
1198881119873
100381610038161003816100381610038161003816100381610038161003816
times 119898119899 (119904 minus 119903) minus 119898119899 (119904 minus 119903) ] 119889119905
le120573119890120582119905119891
119873int
119905119891
0
(119898 minus 119898)
times [1003816100381610038161003816119906lowast
1minus 1
1003816100381610038161003816 (119898 |119899 minus 119899| + |119898 minus 119898| 119899) + 119898119899120573119890120582119905
1198881119873
|119898119899 (119904 minus 119904) + (119898119899 minus 119898119899) 119904
minus (119898119899 (119903 minus 119903) + (119898119899 minus 119898119899) 119903)| ] 119889119905
le120573119890120582119905119891
119873int
119905119891
0
[1003816100381610038161003816119906lowast
1minus 1
1003816100381610038161003816 |119898| |119898 minus 119898| |119899 minus 119899|
+ |119898 minus 119898|2
|119899| + |119861| |119904| |119899| |119898 minus 119898|2
+ |119861| |119898119899| |119898 minus 119898| |119903 minus 119903|
+ |119861| |119898119899| |119898 minus 119898| |119904 minus 119904|
+ |119861| |119904| |119898| |119898 minus 119898| |119899 minus 119899|
+ |119861| |119903| |119898| |119898 minus 119898| |119899 minus 119899|
+ |119861| |119903| |119899| |119898 minus 119898|2] 119889119905
le120573119890120582119905119891
119873int
119905119891
0
[
1003816100381610038161003816119906lowastminus 1
1003816100381610038161003816
2|119898| ((119898 minus 119898)
2+ (119899 minus 119899)
2)
+ |119899| (119898 minus 119898)2
+|119861| |119898119899|
2((119898 minus 119898)
2+ (119904 minus 119904)
2)
+|119861| |119904| |119898|
2((119898 minus 119898)
2+ (119899 minus 119899)
2)
+ |119861| |119904| |119899| (119898 minus 119898)2
+|119861| |119898119899|
2((119898 minus 119898)
2+ (119903 minus 119903)
2)
+|119861| |119898| |119903|
2((119898 minus 119898)
2+ (119899 minus 119899)
2)
+ |119861| |119903| |119899| (119898 minus 119898)2]119889119905
(84)
In the above derivation we use many scaling techniquesfor inequality or absolute inequality Particularly what shouldbe noted is that to get the first inequality of above derivationwe use the estimation of |119906lowast
1(119905) minus 119906
lowast
1(119905)| which has been given
before besides for the sake of convenience we note 119861 =
119898119899(120573119890120582119905119891119873) Furthermore we notice that the coefficients of
all the eight terms in the last formula (119898 minus 119898)2+ (119899 minus 119899)
2(119898 minus119898)
2 (119898 minus119898)2+ (119904 minus 119904)
2 (119898 minus119898)2+ (119899 minus 119899)
2 (119898 minus119898)2
(119898minus119898)2+(119903minus119903)
2 (119898minus119898)2+(119899minus119899)2 (119898minus119898)2 namely (|119906lowast1minus
1|2)|119898| |119899| (|119861||119898119899|)2 (|119861||119904||119898|)2 |119861||119904|119899 (|119861||119898119899|)2(|119861||119898||119903|)2 |119861||119903||119899| are nonnegative and bounded So thereexists a positive constant 119888
2such that
1
2(119898 minus 119898)
2(119905119891) + (120583 + 120582)int
119905119891
0
(119898 minus 119898)2119889119905
12 Abstract and Applied Analysis
le 1198882
120573119890120582119905119891
119873int
119905119891
0
((119898 minus 119898)2+ (119899 minus 119899)
2
+ (119904 minus 119904)2+ (119903 minus 119903)
2) 119889119905
(85)
Combining eight of these inequalities gives
1
2(119898 minus 119898) (119905
119891) +
1
2(119899 minus 119899) (119905
119891) +
1
2(119901 minus 119901) (119905
119891)
+1
2(119902 minus 119902) (119905
119891) +
1
2(119903 minus 119903) (0) +
1
2(119904 minus 119904) (0)
+1
2(119908 minus 119908) (0) +
1
2(V minus V) (0) + (120583 + 120582)
times int
119905119891
0
(119898 minus 119898)2+ (119899 minus 119899)
2+ (119901 minus 119901)
2
+ (119902 minus 119902)2
+ (119904 minus 119904)2
+ (119903 minus 119903)2+ (119908 minus 119908)
2+ (V minus V)2) 119889119905
le 119861int
119905119891
0
(119898 minus 119898)2+ (119899 minus 119899)
2+ (119901 minus 119901)
2
+ (119902 minus 119902)2
+ (119903 minus 119903)2+ (119904 minus 119904)
2
+ (119908 minus 119908)2+ (V minus V)2 119889119905
(86)
Thus from the above inequality we can conclude that
(120583 + 120582 minus 119861)int
119905119891
0
(119898 minus 119898)2+ (119899 minus 119899)
2
+ (119901 minus 119901)2
+ (119902 minus 119902)2
+ (119903 minus 119903)2
+ (119904 minus 119904)2+(119908 minus 119908)
2+ (V minus V)2 119889119905 le 0
(87)
where 119861 depends on the coefficients and the bounds dependon119898 119899 119901 119902 119903 119904 119908 V If we choose 120582 such that 120583+120582 gt 119861 then119898 = 119898 119899 = 119899 119901 = 119901 119902 = 119902 119903 = 119903 119904 = 119904 119908 = 119908 and V = VHence the solution to the optimality system is unique Theproof is complete
6 Numerical Simulation
61 The Simulation of State System (1) without ControlParameters For the sake of simplicity but without loss ofgenerality we will perform the numerical simulation of statesystem (1) with parameters 119906
1= 0 119906
2= 0 Before
illustrating the analytic properties of the alcoholism model(1) we will target the populations in the environment ofa community or a university for example the school ofmaterial science and engineering in our university that isLan zhou University of Technology (LUT for short) owing tothe accurate and available information we can obtain Refer-ring to the information provided by the admissions officeof LUT this school will enroll almost 1200 undergraduatesand almost 300 various postgraduates at the beginning of
fall semester at the same time there will be almost 1500various students graduated and left this school so the scaleof students in school remained almost 6000 we can take thetotal population 119873 = 6000 In this simulation we will takeSeptember as the initial time and units in one week periodin one year According to the investigations of the studentunion implemented in September every year we can takeinitial values as 119878(0) = 4500 119860(0) = 1000 119879(0) = 300and 119876(0) = 200 It seems that the alcoholism is a little bitmore but it is rather natural because many freshmen feelconfused when they are faced with the new environment anda new lifestyle many of them have no better choice but gathertogether to drink in small groups to mediate the anxiety andget to know each other over time some of them developthe habit of drinking To a certain extent for example thefrequent drinking badly affects their study we can classifythem into the alcoholism compartment Other initial valuesseem more reasonable so we need no more explanation Aswe know alcoholism death is seldom happen within oneyear so we omit mortality from alcoholism then how tounderstand the recruitment rate as well as natural death rate120583 We can treat the freshmen admission as the recruitmentpopulation and graduation students as the natural ldquodeathrdquoparts So we can take 120583 = 15006000 = 025 which is exactlyconsistent with the value in [16] As for the infection rate 120573and recovery rate 120575 we will let them be variables since thedrinking behaviors are related to many factors such as theseason and the pressure
According to the data we get from the student unionwe choose 120585 = 04 To summarize we list the values ofthe parameters in Table 1 Using the values of parameters inTable 1 we can plot Figures 2 and 3 which are on condition1198770lt 1 and 119877
0ge 1 respectively From Figure 2 we easily
know when 1198770lt 1 holds the solution of system (1) tends to
the alcohol free equilibrium1198640and verifies the global stability
of 1198640 While seen from Figure 3 we also know that if 119877
0ge
1 holds the solution of system (1) tends to the alcoholismequilibrium 119864
lowast and verifies the global stability of 119864lowast
62 The Sensitivity of 1198770about Two Control Parameters
Although from the expression of the model reproductionnumber 119877
0 we can easily find out the fact that the two
control variables that is 1199061and 119906
2 attribute to reducing the
severity of alcoholism we will still depict the graph between1198770and the two control variables to see more intuitiveness see
Figure 4 It seems from the figure that 1198770is a monotonically
decreasing function about two control parameters so it isadvisable to take two approaches simultaneously to controlthe alcoholics
63 The Simulation of Optimality System (71) In this subsec-tion we will investigate numerically the optimal solution tooptimality system (71) by numerical method from [32] theoptimality system is solved with a fourth-order Runge-Kuttascheme Beginning with a guess for the control variables thestate system is solved forward in time and then those values ofstate are used to solve the adjoint equations backward in timeThe controls are updated at the end of each iteration using
Abstract and Applied Analysis 13
0 10 20 30 40 50
0
1000
2000
3000
4000
5000
6000
7000
Time
Popu
latio
n siz
e
S(t)
A(t)
T(t)
Q(t)
Figure 2 When 1198770lt 1 the alcohol free equilibrium 119864
0is globally
asymptotically stable (120573 = 02 120575 = 03 and 1198770= 071698)
S(t)
A(t)
T(t)
Q(t)
0 10 20 30 40
48
2520
303540
49 50
500
1000
2000
3000
4000
5000
6000
7000
Time
Popu
latio
n siz
e
Figure 3 When 1198770
ge 1 the alcoholism equilibrium 119864lowast
=
(4466 476 32 26) corresponding to the given parameters is globallyasymptotically stable (120573 = 03 120575 = 02 and 119877
0= 108511)
the values of optimal controls obtained lastly The iterationscontinue until convergence takes place
In the simulations we choose the available variable valuesas Table 1 shows besides 120573 = 03 120575 = 02 The initial valueof model (1) is assumed to be 119878(0) = 4500 119860(0) = 1000119879(0) = 300 and 119876(0) = 200 as before
The ideal weights in objective functional are very difficultto obtain in reality it needs much work on data mining andfittingHence the acquisition of appropriate practical weights
Table 1 The parameters description of model (1)
Parameter Description Values120583 Natural birth rate or death rate 025
120573Transmission coefficient betweenalcoholism and susceptibles Variables
119873 The total populations to be considered 6000
120575The rate of populations quitting fromalcoholism permanently after treatment Variables
120585The rate of populations failed intreatment and returned to be alcoholic 04
0
02
04
06
08
1
0
05
1
0 02 04 06 08 1
u1
u2
R0
Figure 4 The relationship between 1198770and two control variables 119906
1
and 1199062
is still a difficult problem and remains for further investiga-tionsThe cost associated with119860(119905) and 119906
1(119905)mainly includes
the cost of dangerous behaviours during the alcoholism timeand educating the public while the cost associated with1199062(119905)mainly comes from health professional and the medical
resource includingmedicines andnursing care In viewof thisand taking the expressions of 119906
1 1199062into account after many
numerical simulations we finally give weighting coefficientsas 1198881= 102 1198882= 104 It should be pointed out that the
weights here are of only theoretical interest to reveal thecontrol strategies proposed in this paper Another point tonote is that themaximumcontrol is very difficult to achieve inreality so we will omit the situation of the maximum controlduring the series of simulations
Next we will make some necessary instructions andexplanations to the above simulation graphs Figures 5 67 and 8 depict the number of four compartments underdifferent control levels when we choose the weight coeffi-cients in objective function to be 119888
1= 102 1198882= 104 From
the four simulation graphs we can observe the followingsimple facts in reducing the total number of alcoholismsand increasing the number of susceptibles the effectivenessof various control measures is as follows optimal control isevidently better than middle control and middle control isbetter than single control 119906
2 single control 119906
2is better than
single control 1199061 while single control 119906
1is much better than
no controlFigures 9 and 10 depict the optimal control law of 119906
1 1199062
respectively In the beginning of the simulation the control
14 Abstract and Applied Analysis
0 10 20 30 40 504500
5000
5500
6000
6500
7000
S(n
umbe
r of s
usce
ptib
les)
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 5Number of the susceptibles whenwe choose theweights inobjective function are 119888
1= 102 1198882= 104 and under different optimal
control strategies that is (1) without control 1199061= 1199062= 0 (2) single
control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0 119906
2= 05 (4)
middle control 1199061= 1199062= 05 (5) optimal control 119906
1= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
500
1000
1500
A(n
umbe
r of t
he al
coho
lism
)
t (unit week)
Figure 6 Number of the alcoholics when we choose the weights inobjective function are 119888
1= 102 1198882= 104 and under different optimal
control strategies that is (1) without control 1199061= 1199062= 0 (2) single
control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0 119906
2= 05 (4)
middle control 1199061= 1199062= 05 (5) optimal control 119906
1= 119906lowast
1 1199062= 119906lowast
2
0
50
100
150
200
250
300
350
400
T(n
umbe
r of t
reat
men
t)
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 7 Number of the persons in treatment when we choose theweights in objective function are 119888
1= 10
2 1198882= 10
4 and underdifferent optimal control strategies that is (1) without control 119906
1=
1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control 119906
1= 0
1199062= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal control
1199061= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
50
100
150
200
250
300
350
400
Q(n
umbe
r of q
uitti
ng)
Figure 8 Number of people in quitting compartment when wechoose the weights in objective function are 119888
1= 102 1198882= 104 and
under different optimal control strategies that is (1) without control1199061= 1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control
1199061= 0 119906
2= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal
control 1199061= 119906lowast
1 1199062= 119906lowast
2
Abstract and Applied Analysis 15
0 10 20 30 40 50t (unit week)
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
Optimal control of u1
Figure 9 Figures of the optimal control 1199061when we choose the
weight in objective function are 1198881= 102 1198882= 104
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
0 10 20 30 40 50t (unit week)
Optimal control of u2
Figure 10 Figures of the optimal control 1199062when we choose the
weight in objective function are 1198881= 102 1198882= 104
effort of 1199061should be decreased from 065 to 05 within the
first month and over the next week it should be increased tothe maximum control until 50 weeks then rapidly decreasedto 0 at the end of the simulation As for the control 119906
2 it
should start from around 05 due to the initial alcoholics thenincrease to 055 within one week since the rapid infection andnext decrease to almost 0 since the effectiveness of treatmentin three weeks but with the infection going on the controleffort of 119906
2should gradually increase to the maximum and
maintain this level until the tenth week for the purpose ofconsolidation therapy and preventing rebound hereafter itshould be gradually decreased to the level of almost 043 untilthe fifty weeks then quickly decreased to 0 in the end
0 10 20 30 40 504500
5000
5500
6000
6500
7000
S(n
umbe
r of s
usce
ptib
les)
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 11 Number of the susceptibles when we choose the weightsin objective function are 119888
1= 104 1198882= 102 and under different
optimal control strategies that is (1) without control 1199061= 1199062= 0
(2) single control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0
1199062= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal control
1199061= 119906lowast
1 1199062= 119906lowast
2
In order to investigate the influence of different weightcoefficients in the objective functional on the effect ofcontrolling at the same time for a better comparison we willchange the weight coefficients in objective function into 119888
2=
104 1198881= 102 and we will list the corresponding numerical
simulation results as Figures 11ndash16 showWhen we change the weight coefficients in objective
function into 1198881= 104 1198882= 102 we find that the results
of simulations derived from the graphs are very similar tothe ones before We speculate that the most likely reasons ofthis result are due to three respects one is that the weightcoefficients are not too sensitive in the numerical simulationand another possible reason is that both of the two controlsare important in some sense equivalently importantThe lastbut not themost unlikely reason is that we have not found themost appropriate weight coefficients in the simulation whichis very difficult to find as previously mentioned
7 Conclusions
In this paper we formulate an alcoholics quitting modeland firstly investigate the variation discipline of variouspopulations from the perspective of global stability thenwe propose an objective functional to examine two differentcontrolmeasures (ie prevention and treatment) on the effectof alcohol The basic reproduction number of the model wasderived and the global stability of the two equilibria is givenFrom the expression of the basic reproductionnumber119877
0and
16 Abstract and Applied Analysis
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
500
1000
1500
A(n
umbe
r of t
he al
coho
lism
)
t (unit week)
Figure 12 Number of the alcoholics when we choose the weights inobjective function are 119888
1= 104 1198882= 102 and under different optimal
control strategies that is (1) without control 1199061= 1199062= 0 (2) single
control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0 119906
2= 05 (4)
middle control 1199061= 1199062= 05 (5) optimal control 119906
1= 119906lowast
1 1199062= 119906lowast
2
0
50
100
150
200
250
300
350
400
T(n
umbe
r of t
reat
men
t)
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 13 Number of the persons in treatment when we choosethe weights in objective function are 119888
1= 104 1198882= 102 and under
different optimal control strategies that is (1) without control 1199061=
1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control 119906
1= 0
1199062= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal control
1199061= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
50
100
150
200
250
300
350
400
Q(n
umbe
r of q
uitti
ng)
Figure 14 Number of people in quitting compartment when wechoose the weights in objective function are 119888
1= 104 1198882= 102 and
under different optimal control strategies that is (1) without control1199061= 1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control
1199061= 0 119906
2= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal
control 1199061= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50t (unit week)
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
Optimal control of u1
Figure 15 Figures of the optimal control 1199061when we choose the
weight in objective function are 1198881= 104 1198882= 102
related numerical simulation we can easily see that the twocontrol strategies are effective in the alcoholics process
Using Pontryaginrsquos Maximum Principle we firstly deter-mine the necessary conditions for existence of optimalcontrol pairsThe uniqueness of the solution to the optimalitysystem (71) is derived by the classical method of contradic-tion Numerical simulations of the model suggest that thetwo different groups of weights in the objective function have
Abstract and Applied Analysis 17
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
0 10 20 30 40 50t (unit week)
Optimal control of u2
Figure 16 Figures of the optimal control 1199062when we choose the
weight in objective function are 1198881= 104 1198882= 102
much similar effects on the transmission of the alcoholismfrom this point the two control measures are almost equallyimportant in controlling the alcoholism although they willprobably have great influences on the cost of the objectivefunction From the simulation figures it seems that the effectof optimal control which is measured by the reduction inthe number of alcoholics and the increase in the number ofsusceptibles is much better than other control strategies asnoted earlier in the simulation section According to the real-time curve of two optimal controls we point out the specificimplementation methods of optimal control which can beachieved in practice
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
This work was partially supported by the NSF ofGansu Province of China (2013GS09485) the NNSF ofChina (10961018) the NSF of Gansu Province of China(1107RJZA088) the NSF for Distinguished Young Scholarsof Gansu Province of China (1111RJDA003) the Special Fundfor the Basic Requirements in the Research of University ofGansu Province of China and the Development Programfor Hong Liu Distinguished Young Scholars in LanzhouUniversity of Technology
References
[1] R Room T Babor and J Rehm ldquoAlcohol and public healthrdquoThe Lancet vol 365 no 9458 pp 519ndash530 2005
[2] J B Saunders O G Aasland A Amundsen and M GrantldquoAlcohol consumption and related problems among primary
health care patients WHO collaborative project on early detec-tion of personswith harmful alcohol consumption IrdquoAddictionvol 88 no 3 pp 349ndash362 1993
[3] L D Johnston P M OMalley and J G Bachman NationalSurvey Results on Drug Use From the Monitoring the FutureStudy 1975ndash1992 National Institute on Drug Abuse RockvilleMd USA 1993
[4] H Wechsler J E Lee M Kuo and H Lee ldquoCollege bingedrinking in the 1990s a continuing problem Results of theHarvard School of Public Health 1999 College Alcohol StudyrdquoJournal of American College Health vol 48 no 5 pp 199ndash2102000
[5] P M OrsquoMalley and L D Johnston ldquoEpidemiology of alcoholand other drug use among American college studentsrdquo Journalof Studies on Alcohol vol 63 no 14 pp 23ndash39 2002
[6] K Agarwal-Kozlowski and D P Agarwal ldquoGenetic predisposi-tion to alcoholismrdquo Therapeutische Umschau vol 57 no 4 pp179ndash184 2000
[7] C Y Chen C L Storr and J C Anthony ldquoEarly-onset drug useand risk for drug dependence problemsrdquo Addictive Behaviorsvol 34 no 3 pp 319ndash322 2009
[8] M Glavas and J Weinberg ldquoStress alcohol consumption andthe hypothalamic-pituitary adrenal axisrdquo in Nutrients Stressand Medical Disorders pp 165ndash183 Humana Press New YorkNY USA 2006
[9] L N Blum N H Nielsen J A Riggs and L B BresolinldquoAlcoholism and alcohol abuse among women report of theCouncil on Scientific Affairsrdquo Journal of Womenrsquos Health vol7 no 7 pp 861ndash871 1998
[10] B Benedict ldquoModeling alcoholism as a contagious disease howldquoinfectedrdquo drinking buddies spread problem drinkingrdquo SIAMNews vol 40 no 3 2007
[11] H Walter K Gutierrez K Ramskogler I Hertling A Dvorakand O M Lesch ldquoGender-specific differences in alcoholismImplications for treatmentrdquo Archives of Womenrsquos Mental Healthvol 6 no 4 pp 253ndash258 2003
[12] D Muller R D Koch H von Specht w Volker and E MMunch ldquoNeurophisiologic findings in chronic alcohol abuserdquoPsychiatrie Neurologie und Medizinische Psychologie vol 37no 3 pp 129ndash132 1985
[13] G Testino ldquoAlcoholic diseases in hepato-gastroenterology apoint of viewrdquo Hepato-Gastroenterology vol 55 no 82-83 pp371ndash377 2008
[14] A Mubayi P E Greenwood C Castillo-Chavez P J Grue-newald and D M Gorman ldquoThe impact of relative residencetimes on the distribution of heavy drinkers in highly distinctenvironmentsrdquo Socio-Economic Planning Sciences vol 44 no 1pp 45ndash56 2010
[15] D Muller R D Koch H von Specht W Volker and E MMunch ldquoNeurophysiologic findings in chronic alcohol abuserdquoPsychiatrie Neurologie Und Medizinische Psychologie vol 37no 3 pp 129ndash132 1985 (Leipz)
[16] G Mulone and B Straughan ldquoModeling binge drinkingrdquoInternational Journal of Biomathematics vol 5 no 1 Article ID1250005 14 pages 2012
[17] H F Huo and N N Song ldquoGlobal stability for a binge drinkingmodel with two stagesrdquo Discrete Dynamics in Nature andSociety vol 2012 Article ID 829386 15 pages 2012
[18] S S Mushayabasa and C P Bhunu ldquoModelling the effects ofheavy alcohol consumption on the transmission dynamics ofgonorrheardquo Nonlinear Dynamics vol 66 no 4 pp 695ndash7062011
18 Abstract and Applied Analysis
[19] F S Sancheznchez XWang C Castillo-Chvez DM Gormanand P J Gruenewald ldquoDrinking as an epidemic a simplemathematical model with recovery and relapserdquo in TherapistsGuide to Evidence-Based Relapse Prevention K Witkiewitz andG Marlatt Eds Academic Press New York NY USA 2007
[20] H-F Huo and C-C Zhu ldquoInfluence of relapse in a givingup smoking modelrdquo Abstract and Applied Analysis vol 2013Article ID 525461 12 pages 2013
[21] H F Huo and Q Wang ldquoThe effects of awareness programs by17 media on the drinking dynamicsrdquo Submitted
[22] S Lee E Jung and C Castillo-Chavez ldquoOptimal controlintervention strategies in low- and high-risk problem drinkingpopulationsrdquo Socio-Economic Planning Sciences vol 44 no 4pp 258ndash265 2010
[23] K R Fister S Lenhart and J S McNally ldquoOptimizingchemotherapy in an HIV modelrdquo Electronic Journal of Differ-ential Equations vol 1998 no 32 pp 1ndash12 1998
[24] S Lenhart and J T Workman Optimal Control Applied toBiological Models Chapman amp HallCRC Mathematical andComputational Biology Series Chapman amp HallCRC BocaRaton Fla USA 2007
[25] X Yang L Chen and J Chen ldquoPermanence and positiveperiodic solution for the single-species nonautonomous delaydiffusive modelsrdquo Computers amp Mathematics with ApplicationsAn International Journal vol 32 no 4 pp 109ndash116 1996
[26] P van denDriessche and JWatmough ldquoReproduction numbersand sub-threshold endemic equilibria for compartmental mod-els of disease transmissionrdquoMathematical Biosciences vol 180pp 29ndash48 2002
[27] V Lakshmikantham S Leela and A A Martynyuk StabilityAnalysis of Nonlinear Systems vol 125 Marcel Dekker NewYork NY USA 1989
[28] J P LaSalle The Stability of Dynamical Systems vol 25 ofRegional Conference Series in Applied Mathematics Society forIndustrial and Applied Mathematics 1976
[29] W H Fleming and R W Rishel Deterministic and StochasticOptimal Control vol 1 ofApplications of Mathematics SpringerBerlin Germany 1975
[30] D L Lukes Differential Equations Classical to ControlledMathematics in Science and Engineering Academic Press NewYork NY USA 1982
[31] L S Pontryagin V G Boltyanskii R V Gamkrelidze and EF Mishchenko The Mathematical Theory of Optimal ProcessesJohn Wiley amp Sons New Jersey NJ USA 1962
[32] W Hackbusch ldquoA numerical method for solving parabolicequations with opposite orientationsrdquo Computing vol 20 no3 pp 229ndash240 1978
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
8 Abstract and Applied Analysis
statements (1) (2) and (3) are obvious satisfied we only needto test and verify the latter two ones Since the four statevariables have been all proved to be up bounded by 119873 wewill get the following equalities
1198781015840le 120583119873 119860
1015840le (1 minus 119906
1(119905)) 120573119878 + 120585119879
1198791015840le 1199062(119905) 119860 119876
1015840le 120575119879
(54)
so the fourth condition is set up As for the last condition
119871 (119905 1199061 1199062) ge 1198891(100381610038161003816100381611990611003816100381610038161003816
2
+100381610038161003816100381611990621003816100381610038161003816
2
)1205722
minus 1198892
(55)
is also true when we choose 1198891= min119888
12 11988822 and for all
1198892isin 119877+ 120572 = 2 The proof is complete
We next come to the core of this section
52 The Characterization of the Optimal Control With theexistence of the optimal control pairs established we nowpresent the optimality system and use a result from [30]we can easily know the existence of the solutions to theoptimality system (71) which will be gotten later Firstly wecome to discuss the theorem that relates to the character-ization of the optimal control The optimality system canbe used to compute candidates for optimal control pairsTo do this we begin by defining an augmented Hamilto-nian 119867 with penalty terms for the control constraints asfollows
119867 = 119860 (119905) +1198881
21199062
1(119905) +
1198882
21199062
2(119905)
+ 1205821[120583119873 minus (1 minus 119906
1(119905))
120573119878119860
119873minus 120583119878]
+ 1205822[(1 minus 119906
1(119905))
120573119878119860
119873+ 120585119879 minus (119906
2(119905) + 120583)119860]
+ 1205823[1199062(119905) 119860 minus (120583 + 120585 + 120575) 119879] + 120582
4(120575119879 minus 120583119876)
minus 119908111199061(119905) minus 119908
12(1 minus 119906
1(119905))
minus 119908211199062(119905) minus 119908
22(1 minus 119906
2(119905))
(56)
where 119908119894119895(119905) ge 0 are the penalty multipliers satisfying
11990811(119905) 1199061(119905) = 119908
12(119905) (1 minus 119906
1(119905))
= 0 at optimal control 119906lowast1
11990821(119905) 1199062(119905) = 119908
22(119905) (1 minus 119906
2(119905))
= 0 at optimal control 119906lowast2
(57)
Theorem6 Given optimal control pairs (119906lowast1 119906lowast
2) and solutions
119878(119905) 119860(119905) 119879(119905) 119876(119905) of the corresponding state system (50)there exist adjoint variables 120582
119894 119894 = 1 2 3 4 satisfying
1205821015840
1= 1205821(1 minus 119906
1(119905))
120573119860
119873+ 1205831205821minus 1205822(1 minus 119906
1(119905))
120573119860
119873
1205821015840
2= minus1 + 120582
1(1 minus 119906
1(119905))
120573119878
119873minus 1205822(1 minus 119906
1(119905))
120573119878
119873
+ 1205822(120583 + 119906
2(119905)) minus 120582
31199062(119905)
1205821015840
3= minus1205822120585 + 1205823(120583 + 120585 + 120575) minus 120582
4120575
1205821015840
4= 1205831205824
(58)
with the terminal conditions
120582119894(119905119891) = 0 119894 = 1 2 3 4 (59)
Furthermore (119906lowast1 119906lowast
2) are represented by
119906lowast
1= min(1max(0
120573119878119860 (1205822minus 1205821)
1198881119873
))
119906lowast
2= min(1max(0
119860 (1205822minus 1205823)
1198882
))
(60)
Proof According to Pontryagin Maximum Principle [29ndash31] we first differentiate the Hamiltonian operator 119867 withrespect to states Then the adjoint system can be written as
1205821015840
1= minus
120597119867
120597119878 120582
1015840
2= minus
120597119867
120597119860
1205821015840
3= minus
120597119867
120597119879 120582
1015840
4= minus
120597119867
120597119876
(61)
The terminal condition (56) of adjoint equations is given by120582119894(119905119891) = 0 119894 = 1 2 3 4
To obtain the necessary conditions of optimality (59) wealso differentiate theHamiltonian operator119867 with respect to119880 = (119906
1 1199062) and set them equal to zero then
120597119867
1205971199061
= 11988811199061(119905) + 120582
1
120573119878119860
119873minus 1205822
120573119878119860
119873minus 11990811+ 11990812= 0
120597119867
1205971199062
= 11988821199062(119905) minus 120582
2119860 + 120582
3119860 minus 119908
21+ 11990822= 0
(62)
By solving the optimal control we obtain
119906lowast
1=1
1198881
[(1205822minus 1205821)120573119878119860
119873+ 11990811minus 11990812] (63)
To determine an explicit expression for the optimalcontrol without119908
11and119908
12 a standard optimality technique
is utilized [29] We consider the following three cases
(i) On the set 119905 | 0 lt 119906lowast
1(119905) lt 1 we have 119908
11(119905) =
Abstract and Applied Analysis 9
11990812(119905) = 0 Hence the optimal control is
119906lowast
1=1
1198881
[(1205822minus 1205821)120573119878119860
119873] (64)
(ii) On the set 119905 | 119906lowast1(119905) = 1 we have 119908
11(119905) = 0 Hence
1 = 119906lowast
1(119905) =
1
1198881
[(1205822minus 1205821)120573119878119860
119873minus 11990812] (65)
This implies that
1
1198881
(1205822minus 1205821)120573119878119860
119873ge 1 since 119908
12(119905) ge 0 (66)
(iii) On the set 119905 | 119906lowast1(119905) = 0 we have 119908
12(119905) = 0 Hence
0 = 119906lowast
1(119905) =
1
1198881
[(1205822minus 1205821)120573119878119860
119873+ 11990811] (67)
This implies that
1
1198881
(1205822minus 1205821)120573119878119860
119873le 0 since 119908
11(119905) ge 0 (68)
Combining these results the optimal control 119906lowast1(119905) is
characterized as
119906lowast
1= min1max0 1
1198881
[(1205822minus 1205821)120573119878119860
119873] (69)
Using the similar arguments we can also obtain the otheroptimal control function
119906lowast
2= min1max0
(1205822minus 1205823) 119860
1198882
(70)
The proof is complete
We point out that the optimality system consistsof the state system (50) with the initial conditions119878(0) 119860(0) 119879(0) 119876(0) the adjoint (or costate) system(58) with the terminal conditions (59) and the optimalitycondition (60) Any optimal control pairs must satisfythis optimality system For the convenience of subsequent
numerical simulation in Section 6 we give the optimalitysystem as follows
1198781015840= 120583119873 minus (1 minusmin1max0 1
1198881
[(1205822minus 1205821)120573119878119860
119873] (119905))
times120573119878119860
119873minus 120583119878
1198601015840= (1minusmin1max0 1
1198881
[(1205822minus1205821)120573119878119860
119873] (119905))
120573119878119860
119873
+ 120585119879 minus (min1max0(1205822minus 1205823) 119860
1198882
(119905) + 120583)119860
1198791015840= min1max0
(1205822minus 1205823) 119860
1198882
(119905) 119860
minus (120583 + 120585 + 120575) 119879
1198761015840= 120575119879 minus 120583119876
1205821015840
1= 1205821(1 minusmin1max0 1
1198881
[(1205822minus 1205821)120573119878119860
119873] (119905))
times120573119860
119873+ 1205831205821
minus 1205822(1minusmin1max0 1
1198881
[(1205822minus1205821)120573119878119860
119873](119905))
times120573119860
119873
1205821015840
2= minus1
+ 1205821(1minusmin1max0 1
1198881
[(1205822minus1205821)120573119878119860
119873] (119905))
times120573119878
119873minus 1205823sdotmin1max0
(1205822minus 1205823) 119860
1198882
(119905)
minus 1205822(1 minusmin1max0 1
1198881
[(1205822minus 1205821)120573119878119860
119873])
times120573119878
119873+ 1205822(120583+min1max0
(1205822minus 1205823) 119860
1198882
(119905))
1205821015840
3= minus1205822120585 + 1205823(120583 + 120585 + 120575) minus 120582
4120575
1205821015840
4= 1205831205824
119878 (0) = 1198780 119860 (0) = 119860
0 119879 (0) = 119879
0
119876 (0) = 1198760 120582
119894(119905119891) = 0 119894 = 1 2 3 4
(71)
53 The Uniqueness of Optimal Control Due to the a prioriboundedness of the state adjoint functions and the resultingLipschitz structure of the ODEs we can obtain the unique-ness of the optimal control
10 Abstract and Applied Analysis
Lemma 7 (see [23]) The function 119906lowast(119904) = min (119887max (119904 119886))is Lipschitz continuous in 119904 where 119886 lt 119887 are some fixed positiveconstants
Theorem 8 For all 119905 isin [0 119905119891] the solution to the optimality
system (71) is unique
Proof Suppose (119878 119860 119879 119876 1205821 1205822 1205823 1205824) and
(119878 119860 119879 119876 1205821 1205822 1205823 1205824) are two different solutions of
our optimality system (71) Let
119878 = 119890120582119905119898 119860 = 119890
120582119905119899 119879 = 119890
120582119905119901
119876 = 119890120582119905119902 120582
1= 119890minus120582119905119903 120582
2= 119890minus120582119905119904
1205823= 119890minus120582119905119908 120582
4= 119890minus120582119905V
119878 = 119890120582119905119898 119860 = 119890
120582119905119899 119879 = 119890
120582119905119901
119876 = 119890120582119905119902 120582
1= 119890minus120582119905119903 120582
2= 119890minus120582119905119904
1205823= 119890minus120582119905119908 120582
4= 119890minus120582119905V
(72)
where 120582 gt 0 is to be chosenAccordingly we have
119906lowast
1(119905) = min1max0
120573119898119899 (119904 minus 119903) 119890120582119905
1198881119873
119906lowast
2(119905) = min1max0 (119904 minus 119908) 119899
1198882
119906lowast
1(119905) = min1max0
120573119898119899 (119904 minus 119903) 119890120582119905
1198881119873
119906lowast
2(119905) = min1max0 (119904 minus 119908) 119899
1198882
(73)
Now we substitute 119878 = 119890120582119905119898 119860 = 119890
120582119905119899 119879 = 119890
120582119905119901 119876 = 119890
120582119905119902
1205821= 119890minus120582119905119903 1205822= 119890minus120582119905119904 1205823= 119890minus120582119905119908 1205824= 119890minus120582119905V and 119878 =
119890120582119905119898 119860 = 119890
120582119905119899 119879 = 119890
120582119905119901 119876 = 119890
120582119905119902 1205821= 119890minus120582119905119903 1205822=
119890minus120582119905119904 1205823= 119890minus120582119905119908 1205824= 119890minus120582119905V into the first ODE of (71)
respectively then we can obtain
+ 120582119898 = 120583119873119890minus120582119905
minus (1 minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)
times120573119898119899119890
120582119905
119873minus 120583119898
(74)
for119898 and119898 respectively Similarly we can derive
119899 + 120582119899
= (1 minusmin1max1120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)120573119898119899119890
120582119905
119873
+ 120585119901 minus (min1max0 119899 (119904 minus 119908)1198882
+ 120583) 119899
(75)
for 119899 and 119899 respectively
+ 120582119901 = min1max0 119899 (119904 minus 119908)1198882
119899
minus (120583 + 120585 + 120575) 119901
(76)
for 119901 and 119901 respectively
119902 + 120582119902 = 120575119901 minus 120583119902 (77)
for 119902 and 119902 respectively
119903 minus 120582119903 = 119903(1 minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)120573119899119890120582119905
119873
+ 120583119903minus119904(1minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)
times120573119899119890120582119905
119873
(78)
for 119903 and 119903 respectively
119904 minus 120582119904 = minus119890120582119905
+ 119903(1 minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)120573119898119890120582119905
119873
minus 119904(1 minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)
times120573119898119890120582119905
119873+ 119904(min1max0 119899 (119904 minus 119908)
1198882
+ 120583)
minus 119908(min1max0 119899 (119904 minus 119908)1198882
)
(79)
for 119904 and 119904 respectively
minus 120582119908 = minus119904120585 + (120583 + 120585 + 120575)119908 minus 120575V (80)
for 119908 and 119908 respectively
V minus 120582V = 120583V (81)
for V and V respectively
Abstract and Applied Analysis 11
By Lemma 7 we can obtain
1003816100381610038161003816119906lowast
1(119905) minus 119906
lowast
1(119905)1003816100381610038161003816 le
120573119890120582119905
1198881119873|119898119899 (119904 minus 119903) minus 119898119899 (119904 minus 119903)|
1003816100381610038161003816119906lowast
2(119905) minus 119906
lowast
2(119905)1003816100381610038161003816 le
1
1198882
|119899 (119904 minus 119908) minus 119899 (119904 minus 119908)|
(82)
The equations for 119898 119899 119901 119902 119903 119904 119908 V and the equationsfor 119898 119899 119901 119902 119903 119904 119908 V are subtracted respectively then wemultiply each equation by appropriate difference of functionsand integrate from 0 to 119905
119891 Next we add all eight integral
equations and some inequality techniques to obtain unique-ness The following calculation is similar for the sake ofsimplicity we only take119898 and119898 for an example
minus + (120583 + 120582) (119898 minus 119898)
=120573119890120582119905
119873[minus(1 minusmin1max0
120573119898119899 (119904 minus 119903) 119890120582119905
1198881119873
)119898119899
+(1 minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)119898119899]
(83)
Multiplying both sides of (83) by (119898minus119898) and integratingfrom 0 to 119905
119891gives
1
2(119898 minus 119898)
2(119905119891) + (120583 + 120582)int
119905119891
0
(119898 minus 119898)2119889119905
= int
119905119891
0
(119898 minus 119898)120573119890120582119905
119873
times [minus (1 minus 119906lowast
1)119898119899 + (1 minus 119906
lowast
1)119898119899] 119889119905
= int
119905119891
0
(119898 minus 119898)120573119890120582119905
119873[(119906lowast
1minus 1) (119898119899 minus 119898119899 + 119898119899 minus 119898119899)
+ 119898119899 (119906lowast
1minus 119906lowast
1)] 119889119905
le120573119890120582119905119891
119873int
119905119891
0
(119898 minus 119898)
times [119906lowast
1minus 1
100381610038161003816100381610038161003816100381610038161003816
(119898 |119899 minus 119899| + |119898 minus 119898| 119899) + 119898119899120573119890120582119905
1198881119873
100381610038161003816100381610038161003816100381610038161003816
times 119898119899 (119904 minus 119903) minus 119898119899 (119904 minus 119903) ] 119889119905
le120573119890120582119905119891
119873int
119905119891
0
(119898 minus 119898)
times [1003816100381610038161003816119906lowast
1minus 1
1003816100381610038161003816 (119898 |119899 minus 119899| + |119898 minus 119898| 119899) + 119898119899120573119890120582119905
1198881119873
|119898119899 (119904 minus 119904) + (119898119899 minus 119898119899) 119904
minus (119898119899 (119903 minus 119903) + (119898119899 minus 119898119899) 119903)| ] 119889119905
le120573119890120582119905119891
119873int
119905119891
0
[1003816100381610038161003816119906lowast
1minus 1
1003816100381610038161003816 |119898| |119898 minus 119898| |119899 minus 119899|
+ |119898 minus 119898|2
|119899| + |119861| |119904| |119899| |119898 minus 119898|2
+ |119861| |119898119899| |119898 minus 119898| |119903 minus 119903|
+ |119861| |119898119899| |119898 minus 119898| |119904 minus 119904|
+ |119861| |119904| |119898| |119898 minus 119898| |119899 minus 119899|
+ |119861| |119903| |119898| |119898 minus 119898| |119899 minus 119899|
+ |119861| |119903| |119899| |119898 minus 119898|2] 119889119905
le120573119890120582119905119891
119873int
119905119891
0
[
1003816100381610038161003816119906lowastminus 1
1003816100381610038161003816
2|119898| ((119898 minus 119898)
2+ (119899 minus 119899)
2)
+ |119899| (119898 minus 119898)2
+|119861| |119898119899|
2((119898 minus 119898)
2+ (119904 minus 119904)
2)
+|119861| |119904| |119898|
2((119898 minus 119898)
2+ (119899 minus 119899)
2)
+ |119861| |119904| |119899| (119898 minus 119898)2
+|119861| |119898119899|
2((119898 minus 119898)
2+ (119903 minus 119903)
2)
+|119861| |119898| |119903|
2((119898 minus 119898)
2+ (119899 minus 119899)
2)
+ |119861| |119903| |119899| (119898 minus 119898)2]119889119905
(84)
In the above derivation we use many scaling techniquesfor inequality or absolute inequality Particularly what shouldbe noted is that to get the first inequality of above derivationwe use the estimation of |119906lowast
1(119905) minus 119906
lowast
1(119905)| which has been given
before besides for the sake of convenience we note 119861 =
119898119899(120573119890120582119905119891119873) Furthermore we notice that the coefficients of
all the eight terms in the last formula (119898 minus 119898)2+ (119899 minus 119899)
2(119898 minus119898)
2 (119898 minus119898)2+ (119904 minus 119904)
2 (119898 minus119898)2+ (119899 minus 119899)
2 (119898 minus119898)2
(119898minus119898)2+(119903minus119903)
2 (119898minus119898)2+(119899minus119899)2 (119898minus119898)2 namely (|119906lowast1minus
1|2)|119898| |119899| (|119861||119898119899|)2 (|119861||119904||119898|)2 |119861||119904|119899 (|119861||119898119899|)2(|119861||119898||119903|)2 |119861||119903||119899| are nonnegative and bounded So thereexists a positive constant 119888
2such that
1
2(119898 minus 119898)
2(119905119891) + (120583 + 120582)int
119905119891
0
(119898 minus 119898)2119889119905
12 Abstract and Applied Analysis
le 1198882
120573119890120582119905119891
119873int
119905119891
0
((119898 minus 119898)2+ (119899 minus 119899)
2
+ (119904 minus 119904)2+ (119903 minus 119903)
2) 119889119905
(85)
Combining eight of these inequalities gives
1
2(119898 minus 119898) (119905
119891) +
1
2(119899 minus 119899) (119905
119891) +
1
2(119901 minus 119901) (119905
119891)
+1
2(119902 minus 119902) (119905
119891) +
1
2(119903 minus 119903) (0) +
1
2(119904 minus 119904) (0)
+1
2(119908 minus 119908) (0) +
1
2(V minus V) (0) + (120583 + 120582)
times int
119905119891
0
(119898 minus 119898)2+ (119899 minus 119899)
2+ (119901 minus 119901)
2
+ (119902 minus 119902)2
+ (119904 minus 119904)2
+ (119903 minus 119903)2+ (119908 minus 119908)
2+ (V minus V)2) 119889119905
le 119861int
119905119891
0
(119898 minus 119898)2+ (119899 minus 119899)
2+ (119901 minus 119901)
2
+ (119902 minus 119902)2
+ (119903 minus 119903)2+ (119904 minus 119904)
2
+ (119908 minus 119908)2+ (V minus V)2 119889119905
(86)
Thus from the above inequality we can conclude that
(120583 + 120582 minus 119861)int
119905119891
0
(119898 minus 119898)2+ (119899 minus 119899)
2
+ (119901 minus 119901)2
+ (119902 minus 119902)2
+ (119903 minus 119903)2
+ (119904 minus 119904)2+(119908 minus 119908)
2+ (V minus V)2 119889119905 le 0
(87)
where 119861 depends on the coefficients and the bounds dependon119898 119899 119901 119902 119903 119904 119908 V If we choose 120582 such that 120583+120582 gt 119861 then119898 = 119898 119899 = 119899 119901 = 119901 119902 = 119902 119903 = 119903 119904 = 119904 119908 = 119908 and V = VHence the solution to the optimality system is unique Theproof is complete
6 Numerical Simulation
61 The Simulation of State System (1) without ControlParameters For the sake of simplicity but without loss ofgenerality we will perform the numerical simulation of statesystem (1) with parameters 119906
1= 0 119906
2= 0 Before
illustrating the analytic properties of the alcoholism model(1) we will target the populations in the environment ofa community or a university for example the school ofmaterial science and engineering in our university that isLan zhou University of Technology (LUT for short) owing tothe accurate and available information we can obtain Refer-ring to the information provided by the admissions officeof LUT this school will enroll almost 1200 undergraduatesand almost 300 various postgraduates at the beginning of
fall semester at the same time there will be almost 1500various students graduated and left this school so the scaleof students in school remained almost 6000 we can take thetotal population 119873 = 6000 In this simulation we will takeSeptember as the initial time and units in one week periodin one year According to the investigations of the studentunion implemented in September every year we can takeinitial values as 119878(0) = 4500 119860(0) = 1000 119879(0) = 300and 119876(0) = 200 It seems that the alcoholism is a little bitmore but it is rather natural because many freshmen feelconfused when they are faced with the new environment anda new lifestyle many of them have no better choice but gathertogether to drink in small groups to mediate the anxiety andget to know each other over time some of them developthe habit of drinking To a certain extent for example thefrequent drinking badly affects their study we can classifythem into the alcoholism compartment Other initial valuesseem more reasonable so we need no more explanation Aswe know alcoholism death is seldom happen within oneyear so we omit mortality from alcoholism then how tounderstand the recruitment rate as well as natural death rate120583 We can treat the freshmen admission as the recruitmentpopulation and graduation students as the natural ldquodeathrdquoparts So we can take 120583 = 15006000 = 025 which is exactlyconsistent with the value in [16] As for the infection rate 120573and recovery rate 120575 we will let them be variables since thedrinking behaviors are related to many factors such as theseason and the pressure
According to the data we get from the student unionwe choose 120585 = 04 To summarize we list the values ofthe parameters in Table 1 Using the values of parameters inTable 1 we can plot Figures 2 and 3 which are on condition1198770lt 1 and 119877
0ge 1 respectively From Figure 2 we easily
know when 1198770lt 1 holds the solution of system (1) tends to
the alcohol free equilibrium1198640and verifies the global stability
of 1198640 While seen from Figure 3 we also know that if 119877
0ge
1 holds the solution of system (1) tends to the alcoholismequilibrium 119864
lowast and verifies the global stability of 119864lowast
62 The Sensitivity of 1198770about Two Control Parameters
Although from the expression of the model reproductionnumber 119877
0 we can easily find out the fact that the two
control variables that is 1199061and 119906
2 attribute to reducing the
severity of alcoholism we will still depict the graph between1198770and the two control variables to see more intuitiveness see
Figure 4 It seems from the figure that 1198770is a monotonically
decreasing function about two control parameters so it isadvisable to take two approaches simultaneously to controlthe alcoholics
63 The Simulation of Optimality System (71) In this subsec-tion we will investigate numerically the optimal solution tooptimality system (71) by numerical method from [32] theoptimality system is solved with a fourth-order Runge-Kuttascheme Beginning with a guess for the control variables thestate system is solved forward in time and then those values ofstate are used to solve the adjoint equations backward in timeThe controls are updated at the end of each iteration using
Abstract and Applied Analysis 13
0 10 20 30 40 50
0
1000
2000
3000
4000
5000
6000
7000
Time
Popu
latio
n siz
e
S(t)
A(t)
T(t)
Q(t)
Figure 2 When 1198770lt 1 the alcohol free equilibrium 119864
0is globally
asymptotically stable (120573 = 02 120575 = 03 and 1198770= 071698)
S(t)
A(t)
T(t)
Q(t)
0 10 20 30 40
48
2520
303540
49 50
500
1000
2000
3000
4000
5000
6000
7000
Time
Popu
latio
n siz
e
Figure 3 When 1198770
ge 1 the alcoholism equilibrium 119864lowast
=
(4466 476 32 26) corresponding to the given parameters is globallyasymptotically stable (120573 = 03 120575 = 02 and 119877
0= 108511)
the values of optimal controls obtained lastly The iterationscontinue until convergence takes place
In the simulations we choose the available variable valuesas Table 1 shows besides 120573 = 03 120575 = 02 The initial valueof model (1) is assumed to be 119878(0) = 4500 119860(0) = 1000119879(0) = 300 and 119876(0) = 200 as before
The ideal weights in objective functional are very difficultto obtain in reality it needs much work on data mining andfittingHence the acquisition of appropriate practical weights
Table 1 The parameters description of model (1)
Parameter Description Values120583 Natural birth rate or death rate 025
120573Transmission coefficient betweenalcoholism and susceptibles Variables
119873 The total populations to be considered 6000
120575The rate of populations quitting fromalcoholism permanently after treatment Variables
120585The rate of populations failed intreatment and returned to be alcoholic 04
0
02
04
06
08
1
0
05
1
0 02 04 06 08 1
u1
u2
R0
Figure 4 The relationship between 1198770and two control variables 119906
1
and 1199062
is still a difficult problem and remains for further investiga-tionsThe cost associated with119860(119905) and 119906
1(119905)mainly includes
the cost of dangerous behaviours during the alcoholism timeand educating the public while the cost associated with1199062(119905)mainly comes from health professional and the medical
resource includingmedicines andnursing care In viewof thisand taking the expressions of 119906
1 1199062into account after many
numerical simulations we finally give weighting coefficientsas 1198881= 102 1198882= 104 It should be pointed out that the
weights here are of only theoretical interest to reveal thecontrol strategies proposed in this paper Another point tonote is that themaximumcontrol is very difficult to achieve inreality so we will omit the situation of the maximum controlduring the series of simulations
Next we will make some necessary instructions andexplanations to the above simulation graphs Figures 5 67 and 8 depict the number of four compartments underdifferent control levels when we choose the weight coeffi-cients in objective function to be 119888
1= 102 1198882= 104 From
the four simulation graphs we can observe the followingsimple facts in reducing the total number of alcoholismsand increasing the number of susceptibles the effectivenessof various control measures is as follows optimal control isevidently better than middle control and middle control isbetter than single control 119906
2 single control 119906
2is better than
single control 1199061 while single control 119906
1is much better than
no controlFigures 9 and 10 depict the optimal control law of 119906
1 1199062
respectively In the beginning of the simulation the control
14 Abstract and Applied Analysis
0 10 20 30 40 504500
5000
5500
6000
6500
7000
S(n
umbe
r of s
usce
ptib
les)
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 5Number of the susceptibles whenwe choose theweights inobjective function are 119888
1= 102 1198882= 104 and under different optimal
control strategies that is (1) without control 1199061= 1199062= 0 (2) single
control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0 119906
2= 05 (4)
middle control 1199061= 1199062= 05 (5) optimal control 119906
1= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
500
1000
1500
A(n
umbe
r of t
he al
coho
lism
)
t (unit week)
Figure 6 Number of the alcoholics when we choose the weights inobjective function are 119888
1= 102 1198882= 104 and under different optimal
control strategies that is (1) without control 1199061= 1199062= 0 (2) single
control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0 119906
2= 05 (4)
middle control 1199061= 1199062= 05 (5) optimal control 119906
1= 119906lowast
1 1199062= 119906lowast
2
0
50
100
150
200
250
300
350
400
T(n
umbe
r of t
reat
men
t)
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 7 Number of the persons in treatment when we choose theweights in objective function are 119888
1= 10
2 1198882= 10
4 and underdifferent optimal control strategies that is (1) without control 119906
1=
1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control 119906
1= 0
1199062= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal control
1199061= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
50
100
150
200
250
300
350
400
Q(n
umbe
r of q
uitti
ng)
Figure 8 Number of people in quitting compartment when wechoose the weights in objective function are 119888
1= 102 1198882= 104 and
under different optimal control strategies that is (1) without control1199061= 1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control
1199061= 0 119906
2= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal
control 1199061= 119906lowast
1 1199062= 119906lowast
2
Abstract and Applied Analysis 15
0 10 20 30 40 50t (unit week)
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
Optimal control of u1
Figure 9 Figures of the optimal control 1199061when we choose the
weight in objective function are 1198881= 102 1198882= 104
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
0 10 20 30 40 50t (unit week)
Optimal control of u2
Figure 10 Figures of the optimal control 1199062when we choose the
weight in objective function are 1198881= 102 1198882= 104
effort of 1199061should be decreased from 065 to 05 within the
first month and over the next week it should be increased tothe maximum control until 50 weeks then rapidly decreasedto 0 at the end of the simulation As for the control 119906
2 it
should start from around 05 due to the initial alcoholics thenincrease to 055 within one week since the rapid infection andnext decrease to almost 0 since the effectiveness of treatmentin three weeks but with the infection going on the controleffort of 119906
2should gradually increase to the maximum and
maintain this level until the tenth week for the purpose ofconsolidation therapy and preventing rebound hereafter itshould be gradually decreased to the level of almost 043 untilthe fifty weeks then quickly decreased to 0 in the end
0 10 20 30 40 504500
5000
5500
6000
6500
7000
S(n
umbe
r of s
usce
ptib
les)
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 11 Number of the susceptibles when we choose the weightsin objective function are 119888
1= 104 1198882= 102 and under different
optimal control strategies that is (1) without control 1199061= 1199062= 0
(2) single control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0
1199062= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal control
1199061= 119906lowast
1 1199062= 119906lowast
2
In order to investigate the influence of different weightcoefficients in the objective functional on the effect ofcontrolling at the same time for a better comparison we willchange the weight coefficients in objective function into 119888
2=
104 1198881= 102 and we will list the corresponding numerical
simulation results as Figures 11ndash16 showWhen we change the weight coefficients in objective
function into 1198881= 104 1198882= 102 we find that the results
of simulations derived from the graphs are very similar tothe ones before We speculate that the most likely reasons ofthis result are due to three respects one is that the weightcoefficients are not too sensitive in the numerical simulationand another possible reason is that both of the two controlsare important in some sense equivalently importantThe lastbut not themost unlikely reason is that we have not found themost appropriate weight coefficients in the simulation whichis very difficult to find as previously mentioned
7 Conclusions
In this paper we formulate an alcoholics quitting modeland firstly investigate the variation discipline of variouspopulations from the perspective of global stability thenwe propose an objective functional to examine two differentcontrolmeasures (ie prevention and treatment) on the effectof alcohol The basic reproduction number of the model wasderived and the global stability of the two equilibria is givenFrom the expression of the basic reproductionnumber119877
0and
16 Abstract and Applied Analysis
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
500
1000
1500
A(n
umbe
r of t
he al
coho
lism
)
t (unit week)
Figure 12 Number of the alcoholics when we choose the weights inobjective function are 119888
1= 104 1198882= 102 and under different optimal
control strategies that is (1) without control 1199061= 1199062= 0 (2) single
control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0 119906
2= 05 (4)
middle control 1199061= 1199062= 05 (5) optimal control 119906
1= 119906lowast
1 1199062= 119906lowast
2
0
50
100
150
200
250
300
350
400
T(n
umbe
r of t
reat
men
t)
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 13 Number of the persons in treatment when we choosethe weights in objective function are 119888
1= 104 1198882= 102 and under
different optimal control strategies that is (1) without control 1199061=
1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control 119906
1= 0
1199062= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal control
1199061= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
50
100
150
200
250
300
350
400
Q(n
umbe
r of q
uitti
ng)
Figure 14 Number of people in quitting compartment when wechoose the weights in objective function are 119888
1= 104 1198882= 102 and
under different optimal control strategies that is (1) without control1199061= 1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control
1199061= 0 119906
2= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal
control 1199061= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50t (unit week)
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
Optimal control of u1
Figure 15 Figures of the optimal control 1199061when we choose the
weight in objective function are 1198881= 104 1198882= 102
related numerical simulation we can easily see that the twocontrol strategies are effective in the alcoholics process
Using Pontryaginrsquos Maximum Principle we firstly deter-mine the necessary conditions for existence of optimalcontrol pairsThe uniqueness of the solution to the optimalitysystem (71) is derived by the classical method of contradic-tion Numerical simulations of the model suggest that thetwo different groups of weights in the objective function have
Abstract and Applied Analysis 17
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
0 10 20 30 40 50t (unit week)
Optimal control of u2
Figure 16 Figures of the optimal control 1199062when we choose the
weight in objective function are 1198881= 104 1198882= 102
much similar effects on the transmission of the alcoholismfrom this point the two control measures are almost equallyimportant in controlling the alcoholism although they willprobably have great influences on the cost of the objectivefunction From the simulation figures it seems that the effectof optimal control which is measured by the reduction inthe number of alcoholics and the increase in the number ofsusceptibles is much better than other control strategies asnoted earlier in the simulation section According to the real-time curve of two optimal controls we point out the specificimplementation methods of optimal control which can beachieved in practice
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
This work was partially supported by the NSF ofGansu Province of China (2013GS09485) the NNSF ofChina (10961018) the NSF of Gansu Province of China(1107RJZA088) the NSF for Distinguished Young Scholarsof Gansu Province of China (1111RJDA003) the Special Fundfor the Basic Requirements in the Research of University ofGansu Province of China and the Development Programfor Hong Liu Distinguished Young Scholars in LanzhouUniversity of Technology
References
[1] R Room T Babor and J Rehm ldquoAlcohol and public healthrdquoThe Lancet vol 365 no 9458 pp 519ndash530 2005
[2] J B Saunders O G Aasland A Amundsen and M GrantldquoAlcohol consumption and related problems among primary
health care patients WHO collaborative project on early detec-tion of personswith harmful alcohol consumption IrdquoAddictionvol 88 no 3 pp 349ndash362 1993
[3] L D Johnston P M OMalley and J G Bachman NationalSurvey Results on Drug Use From the Monitoring the FutureStudy 1975ndash1992 National Institute on Drug Abuse RockvilleMd USA 1993
[4] H Wechsler J E Lee M Kuo and H Lee ldquoCollege bingedrinking in the 1990s a continuing problem Results of theHarvard School of Public Health 1999 College Alcohol StudyrdquoJournal of American College Health vol 48 no 5 pp 199ndash2102000
[5] P M OrsquoMalley and L D Johnston ldquoEpidemiology of alcoholand other drug use among American college studentsrdquo Journalof Studies on Alcohol vol 63 no 14 pp 23ndash39 2002
[6] K Agarwal-Kozlowski and D P Agarwal ldquoGenetic predisposi-tion to alcoholismrdquo Therapeutische Umschau vol 57 no 4 pp179ndash184 2000
[7] C Y Chen C L Storr and J C Anthony ldquoEarly-onset drug useand risk for drug dependence problemsrdquo Addictive Behaviorsvol 34 no 3 pp 319ndash322 2009
[8] M Glavas and J Weinberg ldquoStress alcohol consumption andthe hypothalamic-pituitary adrenal axisrdquo in Nutrients Stressand Medical Disorders pp 165ndash183 Humana Press New YorkNY USA 2006
[9] L N Blum N H Nielsen J A Riggs and L B BresolinldquoAlcoholism and alcohol abuse among women report of theCouncil on Scientific Affairsrdquo Journal of Womenrsquos Health vol7 no 7 pp 861ndash871 1998
[10] B Benedict ldquoModeling alcoholism as a contagious disease howldquoinfectedrdquo drinking buddies spread problem drinkingrdquo SIAMNews vol 40 no 3 2007
[11] H Walter K Gutierrez K Ramskogler I Hertling A Dvorakand O M Lesch ldquoGender-specific differences in alcoholismImplications for treatmentrdquo Archives of Womenrsquos Mental Healthvol 6 no 4 pp 253ndash258 2003
[12] D Muller R D Koch H von Specht w Volker and E MMunch ldquoNeurophisiologic findings in chronic alcohol abuserdquoPsychiatrie Neurologie und Medizinische Psychologie vol 37no 3 pp 129ndash132 1985
[13] G Testino ldquoAlcoholic diseases in hepato-gastroenterology apoint of viewrdquo Hepato-Gastroenterology vol 55 no 82-83 pp371ndash377 2008
[14] A Mubayi P E Greenwood C Castillo-Chavez P J Grue-newald and D M Gorman ldquoThe impact of relative residencetimes on the distribution of heavy drinkers in highly distinctenvironmentsrdquo Socio-Economic Planning Sciences vol 44 no 1pp 45ndash56 2010
[15] D Muller R D Koch H von Specht W Volker and E MMunch ldquoNeurophysiologic findings in chronic alcohol abuserdquoPsychiatrie Neurologie Und Medizinische Psychologie vol 37no 3 pp 129ndash132 1985 (Leipz)
[16] G Mulone and B Straughan ldquoModeling binge drinkingrdquoInternational Journal of Biomathematics vol 5 no 1 Article ID1250005 14 pages 2012
[17] H F Huo and N N Song ldquoGlobal stability for a binge drinkingmodel with two stagesrdquo Discrete Dynamics in Nature andSociety vol 2012 Article ID 829386 15 pages 2012
[18] S S Mushayabasa and C P Bhunu ldquoModelling the effects ofheavy alcohol consumption on the transmission dynamics ofgonorrheardquo Nonlinear Dynamics vol 66 no 4 pp 695ndash7062011
18 Abstract and Applied Analysis
[19] F S Sancheznchez XWang C Castillo-Chvez DM Gormanand P J Gruenewald ldquoDrinking as an epidemic a simplemathematical model with recovery and relapserdquo in TherapistsGuide to Evidence-Based Relapse Prevention K Witkiewitz andG Marlatt Eds Academic Press New York NY USA 2007
[20] H-F Huo and C-C Zhu ldquoInfluence of relapse in a givingup smoking modelrdquo Abstract and Applied Analysis vol 2013Article ID 525461 12 pages 2013
[21] H F Huo and Q Wang ldquoThe effects of awareness programs by17 media on the drinking dynamicsrdquo Submitted
[22] S Lee E Jung and C Castillo-Chavez ldquoOptimal controlintervention strategies in low- and high-risk problem drinkingpopulationsrdquo Socio-Economic Planning Sciences vol 44 no 4pp 258ndash265 2010
[23] K R Fister S Lenhart and J S McNally ldquoOptimizingchemotherapy in an HIV modelrdquo Electronic Journal of Differ-ential Equations vol 1998 no 32 pp 1ndash12 1998
[24] S Lenhart and J T Workman Optimal Control Applied toBiological Models Chapman amp HallCRC Mathematical andComputational Biology Series Chapman amp HallCRC BocaRaton Fla USA 2007
[25] X Yang L Chen and J Chen ldquoPermanence and positiveperiodic solution for the single-species nonautonomous delaydiffusive modelsrdquo Computers amp Mathematics with ApplicationsAn International Journal vol 32 no 4 pp 109ndash116 1996
[26] P van denDriessche and JWatmough ldquoReproduction numbersand sub-threshold endemic equilibria for compartmental mod-els of disease transmissionrdquoMathematical Biosciences vol 180pp 29ndash48 2002
[27] V Lakshmikantham S Leela and A A Martynyuk StabilityAnalysis of Nonlinear Systems vol 125 Marcel Dekker NewYork NY USA 1989
[28] J P LaSalle The Stability of Dynamical Systems vol 25 ofRegional Conference Series in Applied Mathematics Society forIndustrial and Applied Mathematics 1976
[29] W H Fleming and R W Rishel Deterministic and StochasticOptimal Control vol 1 ofApplications of Mathematics SpringerBerlin Germany 1975
[30] D L Lukes Differential Equations Classical to ControlledMathematics in Science and Engineering Academic Press NewYork NY USA 1982
[31] L S Pontryagin V G Boltyanskii R V Gamkrelidze and EF Mishchenko The Mathematical Theory of Optimal ProcessesJohn Wiley amp Sons New Jersey NJ USA 1962
[32] W Hackbusch ldquoA numerical method for solving parabolicequations with opposite orientationsrdquo Computing vol 20 no3 pp 229ndash240 1978
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 9
11990812(119905) = 0 Hence the optimal control is
119906lowast
1=1
1198881
[(1205822minus 1205821)120573119878119860
119873] (64)
(ii) On the set 119905 | 119906lowast1(119905) = 1 we have 119908
11(119905) = 0 Hence
1 = 119906lowast
1(119905) =
1
1198881
[(1205822minus 1205821)120573119878119860
119873minus 11990812] (65)
This implies that
1
1198881
(1205822minus 1205821)120573119878119860
119873ge 1 since 119908
12(119905) ge 0 (66)
(iii) On the set 119905 | 119906lowast1(119905) = 0 we have 119908
12(119905) = 0 Hence
0 = 119906lowast
1(119905) =
1
1198881
[(1205822minus 1205821)120573119878119860
119873+ 11990811] (67)
This implies that
1
1198881
(1205822minus 1205821)120573119878119860
119873le 0 since 119908
11(119905) ge 0 (68)
Combining these results the optimal control 119906lowast1(119905) is
characterized as
119906lowast
1= min1max0 1
1198881
[(1205822minus 1205821)120573119878119860
119873] (69)
Using the similar arguments we can also obtain the otheroptimal control function
119906lowast
2= min1max0
(1205822minus 1205823) 119860
1198882
(70)
The proof is complete
We point out that the optimality system consistsof the state system (50) with the initial conditions119878(0) 119860(0) 119879(0) 119876(0) the adjoint (or costate) system(58) with the terminal conditions (59) and the optimalitycondition (60) Any optimal control pairs must satisfythis optimality system For the convenience of subsequent
numerical simulation in Section 6 we give the optimalitysystem as follows
1198781015840= 120583119873 minus (1 minusmin1max0 1
1198881
[(1205822minus 1205821)120573119878119860
119873] (119905))
times120573119878119860
119873minus 120583119878
1198601015840= (1minusmin1max0 1
1198881
[(1205822minus1205821)120573119878119860
119873] (119905))
120573119878119860
119873
+ 120585119879 minus (min1max0(1205822minus 1205823) 119860
1198882
(119905) + 120583)119860
1198791015840= min1max0
(1205822minus 1205823) 119860
1198882
(119905) 119860
minus (120583 + 120585 + 120575) 119879
1198761015840= 120575119879 minus 120583119876
1205821015840
1= 1205821(1 minusmin1max0 1
1198881
[(1205822minus 1205821)120573119878119860
119873] (119905))
times120573119860
119873+ 1205831205821
minus 1205822(1minusmin1max0 1
1198881
[(1205822minus1205821)120573119878119860
119873](119905))
times120573119860
119873
1205821015840
2= minus1
+ 1205821(1minusmin1max0 1
1198881
[(1205822minus1205821)120573119878119860
119873] (119905))
times120573119878
119873minus 1205823sdotmin1max0
(1205822minus 1205823) 119860
1198882
(119905)
minus 1205822(1 minusmin1max0 1
1198881
[(1205822minus 1205821)120573119878119860
119873])
times120573119878
119873+ 1205822(120583+min1max0
(1205822minus 1205823) 119860
1198882
(119905))
1205821015840
3= minus1205822120585 + 1205823(120583 + 120585 + 120575) minus 120582
4120575
1205821015840
4= 1205831205824
119878 (0) = 1198780 119860 (0) = 119860
0 119879 (0) = 119879
0
119876 (0) = 1198760 120582
119894(119905119891) = 0 119894 = 1 2 3 4
(71)
53 The Uniqueness of Optimal Control Due to the a prioriboundedness of the state adjoint functions and the resultingLipschitz structure of the ODEs we can obtain the unique-ness of the optimal control
10 Abstract and Applied Analysis
Lemma 7 (see [23]) The function 119906lowast(119904) = min (119887max (119904 119886))is Lipschitz continuous in 119904 where 119886 lt 119887 are some fixed positiveconstants
Theorem 8 For all 119905 isin [0 119905119891] the solution to the optimality
system (71) is unique
Proof Suppose (119878 119860 119879 119876 1205821 1205822 1205823 1205824) and
(119878 119860 119879 119876 1205821 1205822 1205823 1205824) are two different solutions of
our optimality system (71) Let
119878 = 119890120582119905119898 119860 = 119890
120582119905119899 119879 = 119890
120582119905119901
119876 = 119890120582119905119902 120582
1= 119890minus120582119905119903 120582
2= 119890minus120582119905119904
1205823= 119890minus120582119905119908 120582
4= 119890minus120582119905V
119878 = 119890120582119905119898 119860 = 119890
120582119905119899 119879 = 119890
120582119905119901
119876 = 119890120582119905119902 120582
1= 119890minus120582119905119903 120582
2= 119890minus120582119905119904
1205823= 119890minus120582119905119908 120582
4= 119890minus120582119905V
(72)
where 120582 gt 0 is to be chosenAccordingly we have
119906lowast
1(119905) = min1max0
120573119898119899 (119904 minus 119903) 119890120582119905
1198881119873
119906lowast
2(119905) = min1max0 (119904 minus 119908) 119899
1198882
119906lowast
1(119905) = min1max0
120573119898119899 (119904 minus 119903) 119890120582119905
1198881119873
119906lowast
2(119905) = min1max0 (119904 minus 119908) 119899
1198882
(73)
Now we substitute 119878 = 119890120582119905119898 119860 = 119890
120582119905119899 119879 = 119890
120582119905119901 119876 = 119890
120582119905119902
1205821= 119890minus120582119905119903 1205822= 119890minus120582119905119904 1205823= 119890minus120582119905119908 1205824= 119890minus120582119905V and 119878 =
119890120582119905119898 119860 = 119890
120582119905119899 119879 = 119890
120582119905119901 119876 = 119890
120582119905119902 1205821= 119890minus120582119905119903 1205822=
119890minus120582119905119904 1205823= 119890minus120582119905119908 1205824= 119890minus120582119905V into the first ODE of (71)
respectively then we can obtain
+ 120582119898 = 120583119873119890minus120582119905
minus (1 minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)
times120573119898119899119890
120582119905
119873minus 120583119898
(74)
for119898 and119898 respectively Similarly we can derive
119899 + 120582119899
= (1 minusmin1max1120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)120573119898119899119890
120582119905
119873
+ 120585119901 minus (min1max0 119899 (119904 minus 119908)1198882
+ 120583) 119899
(75)
for 119899 and 119899 respectively
+ 120582119901 = min1max0 119899 (119904 minus 119908)1198882
119899
minus (120583 + 120585 + 120575) 119901
(76)
for 119901 and 119901 respectively
119902 + 120582119902 = 120575119901 minus 120583119902 (77)
for 119902 and 119902 respectively
119903 minus 120582119903 = 119903(1 minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)120573119899119890120582119905
119873
+ 120583119903minus119904(1minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)
times120573119899119890120582119905
119873
(78)
for 119903 and 119903 respectively
119904 minus 120582119904 = minus119890120582119905
+ 119903(1 minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)120573119898119890120582119905
119873
minus 119904(1 minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)
times120573119898119890120582119905
119873+ 119904(min1max0 119899 (119904 minus 119908)
1198882
+ 120583)
minus 119908(min1max0 119899 (119904 minus 119908)1198882
)
(79)
for 119904 and 119904 respectively
minus 120582119908 = minus119904120585 + (120583 + 120585 + 120575)119908 minus 120575V (80)
for 119908 and 119908 respectively
V minus 120582V = 120583V (81)
for V and V respectively
Abstract and Applied Analysis 11
By Lemma 7 we can obtain
1003816100381610038161003816119906lowast
1(119905) minus 119906
lowast
1(119905)1003816100381610038161003816 le
120573119890120582119905
1198881119873|119898119899 (119904 minus 119903) minus 119898119899 (119904 minus 119903)|
1003816100381610038161003816119906lowast
2(119905) minus 119906
lowast
2(119905)1003816100381610038161003816 le
1
1198882
|119899 (119904 minus 119908) minus 119899 (119904 minus 119908)|
(82)
The equations for 119898 119899 119901 119902 119903 119904 119908 V and the equationsfor 119898 119899 119901 119902 119903 119904 119908 V are subtracted respectively then wemultiply each equation by appropriate difference of functionsand integrate from 0 to 119905
119891 Next we add all eight integral
equations and some inequality techniques to obtain unique-ness The following calculation is similar for the sake ofsimplicity we only take119898 and119898 for an example
minus + (120583 + 120582) (119898 minus 119898)
=120573119890120582119905
119873[minus(1 minusmin1max0
120573119898119899 (119904 minus 119903) 119890120582119905
1198881119873
)119898119899
+(1 minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)119898119899]
(83)
Multiplying both sides of (83) by (119898minus119898) and integratingfrom 0 to 119905
119891gives
1
2(119898 minus 119898)
2(119905119891) + (120583 + 120582)int
119905119891
0
(119898 minus 119898)2119889119905
= int
119905119891
0
(119898 minus 119898)120573119890120582119905
119873
times [minus (1 minus 119906lowast
1)119898119899 + (1 minus 119906
lowast
1)119898119899] 119889119905
= int
119905119891
0
(119898 minus 119898)120573119890120582119905
119873[(119906lowast
1minus 1) (119898119899 minus 119898119899 + 119898119899 minus 119898119899)
+ 119898119899 (119906lowast
1minus 119906lowast
1)] 119889119905
le120573119890120582119905119891
119873int
119905119891
0
(119898 minus 119898)
times [119906lowast
1minus 1
100381610038161003816100381610038161003816100381610038161003816
(119898 |119899 minus 119899| + |119898 minus 119898| 119899) + 119898119899120573119890120582119905
1198881119873
100381610038161003816100381610038161003816100381610038161003816
times 119898119899 (119904 minus 119903) minus 119898119899 (119904 minus 119903) ] 119889119905
le120573119890120582119905119891
119873int
119905119891
0
(119898 minus 119898)
times [1003816100381610038161003816119906lowast
1minus 1
1003816100381610038161003816 (119898 |119899 minus 119899| + |119898 minus 119898| 119899) + 119898119899120573119890120582119905
1198881119873
|119898119899 (119904 minus 119904) + (119898119899 minus 119898119899) 119904
minus (119898119899 (119903 minus 119903) + (119898119899 minus 119898119899) 119903)| ] 119889119905
le120573119890120582119905119891
119873int
119905119891
0
[1003816100381610038161003816119906lowast
1minus 1
1003816100381610038161003816 |119898| |119898 minus 119898| |119899 minus 119899|
+ |119898 minus 119898|2
|119899| + |119861| |119904| |119899| |119898 minus 119898|2
+ |119861| |119898119899| |119898 minus 119898| |119903 minus 119903|
+ |119861| |119898119899| |119898 minus 119898| |119904 minus 119904|
+ |119861| |119904| |119898| |119898 minus 119898| |119899 minus 119899|
+ |119861| |119903| |119898| |119898 minus 119898| |119899 minus 119899|
+ |119861| |119903| |119899| |119898 minus 119898|2] 119889119905
le120573119890120582119905119891
119873int
119905119891
0
[
1003816100381610038161003816119906lowastminus 1
1003816100381610038161003816
2|119898| ((119898 minus 119898)
2+ (119899 minus 119899)
2)
+ |119899| (119898 minus 119898)2
+|119861| |119898119899|
2((119898 minus 119898)
2+ (119904 minus 119904)
2)
+|119861| |119904| |119898|
2((119898 minus 119898)
2+ (119899 minus 119899)
2)
+ |119861| |119904| |119899| (119898 minus 119898)2
+|119861| |119898119899|
2((119898 minus 119898)
2+ (119903 minus 119903)
2)
+|119861| |119898| |119903|
2((119898 minus 119898)
2+ (119899 minus 119899)
2)
+ |119861| |119903| |119899| (119898 minus 119898)2]119889119905
(84)
In the above derivation we use many scaling techniquesfor inequality or absolute inequality Particularly what shouldbe noted is that to get the first inequality of above derivationwe use the estimation of |119906lowast
1(119905) minus 119906
lowast
1(119905)| which has been given
before besides for the sake of convenience we note 119861 =
119898119899(120573119890120582119905119891119873) Furthermore we notice that the coefficients of
all the eight terms in the last formula (119898 minus 119898)2+ (119899 minus 119899)
2(119898 minus119898)
2 (119898 minus119898)2+ (119904 minus 119904)
2 (119898 minus119898)2+ (119899 minus 119899)
2 (119898 minus119898)2
(119898minus119898)2+(119903minus119903)
2 (119898minus119898)2+(119899minus119899)2 (119898minus119898)2 namely (|119906lowast1minus
1|2)|119898| |119899| (|119861||119898119899|)2 (|119861||119904||119898|)2 |119861||119904|119899 (|119861||119898119899|)2(|119861||119898||119903|)2 |119861||119903||119899| are nonnegative and bounded So thereexists a positive constant 119888
2such that
1
2(119898 minus 119898)
2(119905119891) + (120583 + 120582)int
119905119891
0
(119898 minus 119898)2119889119905
12 Abstract and Applied Analysis
le 1198882
120573119890120582119905119891
119873int
119905119891
0
((119898 minus 119898)2+ (119899 minus 119899)
2
+ (119904 minus 119904)2+ (119903 minus 119903)
2) 119889119905
(85)
Combining eight of these inequalities gives
1
2(119898 minus 119898) (119905
119891) +
1
2(119899 minus 119899) (119905
119891) +
1
2(119901 minus 119901) (119905
119891)
+1
2(119902 minus 119902) (119905
119891) +
1
2(119903 minus 119903) (0) +
1
2(119904 minus 119904) (0)
+1
2(119908 minus 119908) (0) +
1
2(V minus V) (0) + (120583 + 120582)
times int
119905119891
0
(119898 minus 119898)2+ (119899 minus 119899)
2+ (119901 minus 119901)
2
+ (119902 minus 119902)2
+ (119904 minus 119904)2
+ (119903 minus 119903)2+ (119908 minus 119908)
2+ (V minus V)2) 119889119905
le 119861int
119905119891
0
(119898 minus 119898)2+ (119899 minus 119899)
2+ (119901 minus 119901)
2
+ (119902 minus 119902)2
+ (119903 minus 119903)2+ (119904 minus 119904)
2
+ (119908 minus 119908)2+ (V minus V)2 119889119905
(86)
Thus from the above inequality we can conclude that
(120583 + 120582 minus 119861)int
119905119891
0
(119898 minus 119898)2+ (119899 minus 119899)
2
+ (119901 minus 119901)2
+ (119902 minus 119902)2
+ (119903 minus 119903)2
+ (119904 minus 119904)2+(119908 minus 119908)
2+ (V minus V)2 119889119905 le 0
(87)
where 119861 depends on the coefficients and the bounds dependon119898 119899 119901 119902 119903 119904 119908 V If we choose 120582 such that 120583+120582 gt 119861 then119898 = 119898 119899 = 119899 119901 = 119901 119902 = 119902 119903 = 119903 119904 = 119904 119908 = 119908 and V = VHence the solution to the optimality system is unique Theproof is complete
6 Numerical Simulation
61 The Simulation of State System (1) without ControlParameters For the sake of simplicity but without loss ofgenerality we will perform the numerical simulation of statesystem (1) with parameters 119906
1= 0 119906
2= 0 Before
illustrating the analytic properties of the alcoholism model(1) we will target the populations in the environment ofa community or a university for example the school ofmaterial science and engineering in our university that isLan zhou University of Technology (LUT for short) owing tothe accurate and available information we can obtain Refer-ring to the information provided by the admissions officeof LUT this school will enroll almost 1200 undergraduatesand almost 300 various postgraduates at the beginning of
fall semester at the same time there will be almost 1500various students graduated and left this school so the scaleof students in school remained almost 6000 we can take thetotal population 119873 = 6000 In this simulation we will takeSeptember as the initial time and units in one week periodin one year According to the investigations of the studentunion implemented in September every year we can takeinitial values as 119878(0) = 4500 119860(0) = 1000 119879(0) = 300and 119876(0) = 200 It seems that the alcoholism is a little bitmore but it is rather natural because many freshmen feelconfused when they are faced with the new environment anda new lifestyle many of them have no better choice but gathertogether to drink in small groups to mediate the anxiety andget to know each other over time some of them developthe habit of drinking To a certain extent for example thefrequent drinking badly affects their study we can classifythem into the alcoholism compartment Other initial valuesseem more reasonable so we need no more explanation Aswe know alcoholism death is seldom happen within oneyear so we omit mortality from alcoholism then how tounderstand the recruitment rate as well as natural death rate120583 We can treat the freshmen admission as the recruitmentpopulation and graduation students as the natural ldquodeathrdquoparts So we can take 120583 = 15006000 = 025 which is exactlyconsistent with the value in [16] As for the infection rate 120573and recovery rate 120575 we will let them be variables since thedrinking behaviors are related to many factors such as theseason and the pressure
According to the data we get from the student unionwe choose 120585 = 04 To summarize we list the values ofthe parameters in Table 1 Using the values of parameters inTable 1 we can plot Figures 2 and 3 which are on condition1198770lt 1 and 119877
0ge 1 respectively From Figure 2 we easily
know when 1198770lt 1 holds the solution of system (1) tends to
the alcohol free equilibrium1198640and verifies the global stability
of 1198640 While seen from Figure 3 we also know that if 119877
0ge
1 holds the solution of system (1) tends to the alcoholismequilibrium 119864
lowast and verifies the global stability of 119864lowast
62 The Sensitivity of 1198770about Two Control Parameters
Although from the expression of the model reproductionnumber 119877
0 we can easily find out the fact that the two
control variables that is 1199061and 119906
2 attribute to reducing the
severity of alcoholism we will still depict the graph between1198770and the two control variables to see more intuitiveness see
Figure 4 It seems from the figure that 1198770is a monotonically
decreasing function about two control parameters so it isadvisable to take two approaches simultaneously to controlthe alcoholics
63 The Simulation of Optimality System (71) In this subsec-tion we will investigate numerically the optimal solution tooptimality system (71) by numerical method from [32] theoptimality system is solved with a fourth-order Runge-Kuttascheme Beginning with a guess for the control variables thestate system is solved forward in time and then those values ofstate are used to solve the adjoint equations backward in timeThe controls are updated at the end of each iteration using
Abstract and Applied Analysis 13
0 10 20 30 40 50
0
1000
2000
3000
4000
5000
6000
7000
Time
Popu
latio
n siz
e
S(t)
A(t)
T(t)
Q(t)
Figure 2 When 1198770lt 1 the alcohol free equilibrium 119864
0is globally
asymptotically stable (120573 = 02 120575 = 03 and 1198770= 071698)
S(t)
A(t)
T(t)
Q(t)
0 10 20 30 40
48
2520
303540
49 50
500
1000
2000
3000
4000
5000
6000
7000
Time
Popu
latio
n siz
e
Figure 3 When 1198770
ge 1 the alcoholism equilibrium 119864lowast
=
(4466 476 32 26) corresponding to the given parameters is globallyasymptotically stable (120573 = 03 120575 = 02 and 119877
0= 108511)
the values of optimal controls obtained lastly The iterationscontinue until convergence takes place
In the simulations we choose the available variable valuesas Table 1 shows besides 120573 = 03 120575 = 02 The initial valueof model (1) is assumed to be 119878(0) = 4500 119860(0) = 1000119879(0) = 300 and 119876(0) = 200 as before
The ideal weights in objective functional are very difficultto obtain in reality it needs much work on data mining andfittingHence the acquisition of appropriate practical weights
Table 1 The parameters description of model (1)
Parameter Description Values120583 Natural birth rate or death rate 025
120573Transmission coefficient betweenalcoholism and susceptibles Variables
119873 The total populations to be considered 6000
120575The rate of populations quitting fromalcoholism permanently after treatment Variables
120585The rate of populations failed intreatment and returned to be alcoholic 04
0
02
04
06
08
1
0
05
1
0 02 04 06 08 1
u1
u2
R0
Figure 4 The relationship between 1198770and two control variables 119906
1
and 1199062
is still a difficult problem and remains for further investiga-tionsThe cost associated with119860(119905) and 119906
1(119905)mainly includes
the cost of dangerous behaviours during the alcoholism timeand educating the public while the cost associated with1199062(119905)mainly comes from health professional and the medical
resource includingmedicines andnursing care In viewof thisand taking the expressions of 119906
1 1199062into account after many
numerical simulations we finally give weighting coefficientsas 1198881= 102 1198882= 104 It should be pointed out that the
weights here are of only theoretical interest to reveal thecontrol strategies proposed in this paper Another point tonote is that themaximumcontrol is very difficult to achieve inreality so we will omit the situation of the maximum controlduring the series of simulations
Next we will make some necessary instructions andexplanations to the above simulation graphs Figures 5 67 and 8 depict the number of four compartments underdifferent control levels when we choose the weight coeffi-cients in objective function to be 119888
1= 102 1198882= 104 From
the four simulation graphs we can observe the followingsimple facts in reducing the total number of alcoholismsand increasing the number of susceptibles the effectivenessof various control measures is as follows optimal control isevidently better than middle control and middle control isbetter than single control 119906
2 single control 119906
2is better than
single control 1199061 while single control 119906
1is much better than
no controlFigures 9 and 10 depict the optimal control law of 119906
1 1199062
respectively In the beginning of the simulation the control
14 Abstract and Applied Analysis
0 10 20 30 40 504500
5000
5500
6000
6500
7000
S(n
umbe
r of s
usce
ptib
les)
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 5Number of the susceptibles whenwe choose theweights inobjective function are 119888
1= 102 1198882= 104 and under different optimal
control strategies that is (1) without control 1199061= 1199062= 0 (2) single
control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0 119906
2= 05 (4)
middle control 1199061= 1199062= 05 (5) optimal control 119906
1= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
500
1000
1500
A(n
umbe
r of t
he al
coho
lism
)
t (unit week)
Figure 6 Number of the alcoholics when we choose the weights inobjective function are 119888
1= 102 1198882= 104 and under different optimal
control strategies that is (1) without control 1199061= 1199062= 0 (2) single
control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0 119906
2= 05 (4)
middle control 1199061= 1199062= 05 (5) optimal control 119906
1= 119906lowast
1 1199062= 119906lowast
2
0
50
100
150
200
250
300
350
400
T(n
umbe
r of t
reat
men
t)
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 7 Number of the persons in treatment when we choose theweights in objective function are 119888
1= 10
2 1198882= 10
4 and underdifferent optimal control strategies that is (1) without control 119906
1=
1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control 119906
1= 0
1199062= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal control
1199061= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
50
100
150
200
250
300
350
400
Q(n
umbe
r of q
uitti
ng)
Figure 8 Number of people in quitting compartment when wechoose the weights in objective function are 119888
1= 102 1198882= 104 and
under different optimal control strategies that is (1) without control1199061= 1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control
1199061= 0 119906
2= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal
control 1199061= 119906lowast
1 1199062= 119906lowast
2
Abstract and Applied Analysis 15
0 10 20 30 40 50t (unit week)
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
Optimal control of u1
Figure 9 Figures of the optimal control 1199061when we choose the
weight in objective function are 1198881= 102 1198882= 104
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
0 10 20 30 40 50t (unit week)
Optimal control of u2
Figure 10 Figures of the optimal control 1199062when we choose the
weight in objective function are 1198881= 102 1198882= 104
effort of 1199061should be decreased from 065 to 05 within the
first month and over the next week it should be increased tothe maximum control until 50 weeks then rapidly decreasedto 0 at the end of the simulation As for the control 119906
2 it
should start from around 05 due to the initial alcoholics thenincrease to 055 within one week since the rapid infection andnext decrease to almost 0 since the effectiveness of treatmentin three weeks but with the infection going on the controleffort of 119906
2should gradually increase to the maximum and
maintain this level until the tenth week for the purpose ofconsolidation therapy and preventing rebound hereafter itshould be gradually decreased to the level of almost 043 untilthe fifty weeks then quickly decreased to 0 in the end
0 10 20 30 40 504500
5000
5500
6000
6500
7000
S(n
umbe
r of s
usce
ptib
les)
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 11 Number of the susceptibles when we choose the weightsin objective function are 119888
1= 104 1198882= 102 and under different
optimal control strategies that is (1) without control 1199061= 1199062= 0
(2) single control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0
1199062= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal control
1199061= 119906lowast
1 1199062= 119906lowast
2
In order to investigate the influence of different weightcoefficients in the objective functional on the effect ofcontrolling at the same time for a better comparison we willchange the weight coefficients in objective function into 119888
2=
104 1198881= 102 and we will list the corresponding numerical
simulation results as Figures 11ndash16 showWhen we change the weight coefficients in objective
function into 1198881= 104 1198882= 102 we find that the results
of simulations derived from the graphs are very similar tothe ones before We speculate that the most likely reasons ofthis result are due to three respects one is that the weightcoefficients are not too sensitive in the numerical simulationand another possible reason is that both of the two controlsare important in some sense equivalently importantThe lastbut not themost unlikely reason is that we have not found themost appropriate weight coefficients in the simulation whichis very difficult to find as previously mentioned
7 Conclusions
In this paper we formulate an alcoholics quitting modeland firstly investigate the variation discipline of variouspopulations from the perspective of global stability thenwe propose an objective functional to examine two differentcontrolmeasures (ie prevention and treatment) on the effectof alcohol The basic reproduction number of the model wasderived and the global stability of the two equilibria is givenFrom the expression of the basic reproductionnumber119877
0and
16 Abstract and Applied Analysis
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
500
1000
1500
A(n
umbe
r of t
he al
coho
lism
)
t (unit week)
Figure 12 Number of the alcoholics when we choose the weights inobjective function are 119888
1= 104 1198882= 102 and under different optimal
control strategies that is (1) without control 1199061= 1199062= 0 (2) single
control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0 119906
2= 05 (4)
middle control 1199061= 1199062= 05 (5) optimal control 119906
1= 119906lowast
1 1199062= 119906lowast
2
0
50
100
150
200
250
300
350
400
T(n
umbe
r of t
reat
men
t)
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 13 Number of the persons in treatment when we choosethe weights in objective function are 119888
1= 104 1198882= 102 and under
different optimal control strategies that is (1) without control 1199061=
1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control 119906
1= 0
1199062= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal control
1199061= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
50
100
150
200
250
300
350
400
Q(n
umbe
r of q
uitti
ng)
Figure 14 Number of people in quitting compartment when wechoose the weights in objective function are 119888
1= 104 1198882= 102 and
under different optimal control strategies that is (1) without control1199061= 1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control
1199061= 0 119906
2= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal
control 1199061= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50t (unit week)
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
Optimal control of u1
Figure 15 Figures of the optimal control 1199061when we choose the
weight in objective function are 1198881= 104 1198882= 102
related numerical simulation we can easily see that the twocontrol strategies are effective in the alcoholics process
Using Pontryaginrsquos Maximum Principle we firstly deter-mine the necessary conditions for existence of optimalcontrol pairsThe uniqueness of the solution to the optimalitysystem (71) is derived by the classical method of contradic-tion Numerical simulations of the model suggest that thetwo different groups of weights in the objective function have
Abstract and Applied Analysis 17
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
0 10 20 30 40 50t (unit week)
Optimal control of u2
Figure 16 Figures of the optimal control 1199062when we choose the
weight in objective function are 1198881= 104 1198882= 102
much similar effects on the transmission of the alcoholismfrom this point the two control measures are almost equallyimportant in controlling the alcoholism although they willprobably have great influences on the cost of the objectivefunction From the simulation figures it seems that the effectof optimal control which is measured by the reduction inthe number of alcoholics and the increase in the number ofsusceptibles is much better than other control strategies asnoted earlier in the simulation section According to the real-time curve of two optimal controls we point out the specificimplementation methods of optimal control which can beachieved in practice
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
This work was partially supported by the NSF ofGansu Province of China (2013GS09485) the NNSF ofChina (10961018) the NSF of Gansu Province of China(1107RJZA088) the NSF for Distinguished Young Scholarsof Gansu Province of China (1111RJDA003) the Special Fundfor the Basic Requirements in the Research of University ofGansu Province of China and the Development Programfor Hong Liu Distinguished Young Scholars in LanzhouUniversity of Technology
References
[1] R Room T Babor and J Rehm ldquoAlcohol and public healthrdquoThe Lancet vol 365 no 9458 pp 519ndash530 2005
[2] J B Saunders O G Aasland A Amundsen and M GrantldquoAlcohol consumption and related problems among primary
health care patients WHO collaborative project on early detec-tion of personswith harmful alcohol consumption IrdquoAddictionvol 88 no 3 pp 349ndash362 1993
[3] L D Johnston P M OMalley and J G Bachman NationalSurvey Results on Drug Use From the Monitoring the FutureStudy 1975ndash1992 National Institute on Drug Abuse RockvilleMd USA 1993
[4] H Wechsler J E Lee M Kuo and H Lee ldquoCollege bingedrinking in the 1990s a continuing problem Results of theHarvard School of Public Health 1999 College Alcohol StudyrdquoJournal of American College Health vol 48 no 5 pp 199ndash2102000
[5] P M OrsquoMalley and L D Johnston ldquoEpidemiology of alcoholand other drug use among American college studentsrdquo Journalof Studies on Alcohol vol 63 no 14 pp 23ndash39 2002
[6] K Agarwal-Kozlowski and D P Agarwal ldquoGenetic predisposi-tion to alcoholismrdquo Therapeutische Umschau vol 57 no 4 pp179ndash184 2000
[7] C Y Chen C L Storr and J C Anthony ldquoEarly-onset drug useand risk for drug dependence problemsrdquo Addictive Behaviorsvol 34 no 3 pp 319ndash322 2009
[8] M Glavas and J Weinberg ldquoStress alcohol consumption andthe hypothalamic-pituitary adrenal axisrdquo in Nutrients Stressand Medical Disorders pp 165ndash183 Humana Press New YorkNY USA 2006
[9] L N Blum N H Nielsen J A Riggs and L B BresolinldquoAlcoholism and alcohol abuse among women report of theCouncil on Scientific Affairsrdquo Journal of Womenrsquos Health vol7 no 7 pp 861ndash871 1998
[10] B Benedict ldquoModeling alcoholism as a contagious disease howldquoinfectedrdquo drinking buddies spread problem drinkingrdquo SIAMNews vol 40 no 3 2007
[11] H Walter K Gutierrez K Ramskogler I Hertling A Dvorakand O M Lesch ldquoGender-specific differences in alcoholismImplications for treatmentrdquo Archives of Womenrsquos Mental Healthvol 6 no 4 pp 253ndash258 2003
[12] D Muller R D Koch H von Specht w Volker and E MMunch ldquoNeurophisiologic findings in chronic alcohol abuserdquoPsychiatrie Neurologie und Medizinische Psychologie vol 37no 3 pp 129ndash132 1985
[13] G Testino ldquoAlcoholic diseases in hepato-gastroenterology apoint of viewrdquo Hepato-Gastroenterology vol 55 no 82-83 pp371ndash377 2008
[14] A Mubayi P E Greenwood C Castillo-Chavez P J Grue-newald and D M Gorman ldquoThe impact of relative residencetimes on the distribution of heavy drinkers in highly distinctenvironmentsrdquo Socio-Economic Planning Sciences vol 44 no 1pp 45ndash56 2010
[15] D Muller R D Koch H von Specht W Volker and E MMunch ldquoNeurophysiologic findings in chronic alcohol abuserdquoPsychiatrie Neurologie Und Medizinische Psychologie vol 37no 3 pp 129ndash132 1985 (Leipz)
[16] G Mulone and B Straughan ldquoModeling binge drinkingrdquoInternational Journal of Biomathematics vol 5 no 1 Article ID1250005 14 pages 2012
[17] H F Huo and N N Song ldquoGlobal stability for a binge drinkingmodel with two stagesrdquo Discrete Dynamics in Nature andSociety vol 2012 Article ID 829386 15 pages 2012
[18] S S Mushayabasa and C P Bhunu ldquoModelling the effects ofheavy alcohol consumption on the transmission dynamics ofgonorrheardquo Nonlinear Dynamics vol 66 no 4 pp 695ndash7062011
18 Abstract and Applied Analysis
[19] F S Sancheznchez XWang C Castillo-Chvez DM Gormanand P J Gruenewald ldquoDrinking as an epidemic a simplemathematical model with recovery and relapserdquo in TherapistsGuide to Evidence-Based Relapse Prevention K Witkiewitz andG Marlatt Eds Academic Press New York NY USA 2007
[20] H-F Huo and C-C Zhu ldquoInfluence of relapse in a givingup smoking modelrdquo Abstract and Applied Analysis vol 2013Article ID 525461 12 pages 2013
[21] H F Huo and Q Wang ldquoThe effects of awareness programs by17 media on the drinking dynamicsrdquo Submitted
[22] S Lee E Jung and C Castillo-Chavez ldquoOptimal controlintervention strategies in low- and high-risk problem drinkingpopulationsrdquo Socio-Economic Planning Sciences vol 44 no 4pp 258ndash265 2010
[23] K R Fister S Lenhart and J S McNally ldquoOptimizingchemotherapy in an HIV modelrdquo Electronic Journal of Differ-ential Equations vol 1998 no 32 pp 1ndash12 1998
[24] S Lenhart and J T Workman Optimal Control Applied toBiological Models Chapman amp HallCRC Mathematical andComputational Biology Series Chapman amp HallCRC BocaRaton Fla USA 2007
[25] X Yang L Chen and J Chen ldquoPermanence and positiveperiodic solution for the single-species nonautonomous delaydiffusive modelsrdquo Computers amp Mathematics with ApplicationsAn International Journal vol 32 no 4 pp 109ndash116 1996
[26] P van denDriessche and JWatmough ldquoReproduction numbersand sub-threshold endemic equilibria for compartmental mod-els of disease transmissionrdquoMathematical Biosciences vol 180pp 29ndash48 2002
[27] V Lakshmikantham S Leela and A A Martynyuk StabilityAnalysis of Nonlinear Systems vol 125 Marcel Dekker NewYork NY USA 1989
[28] J P LaSalle The Stability of Dynamical Systems vol 25 ofRegional Conference Series in Applied Mathematics Society forIndustrial and Applied Mathematics 1976
[29] W H Fleming and R W Rishel Deterministic and StochasticOptimal Control vol 1 ofApplications of Mathematics SpringerBerlin Germany 1975
[30] D L Lukes Differential Equations Classical to ControlledMathematics in Science and Engineering Academic Press NewYork NY USA 1982
[31] L S Pontryagin V G Boltyanskii R V Gamkrelidze and EF Mishchenko The Mathematical Theory of Optimal ProcessesJohn Wiley amp Sons New Jersey NJ USA 1962
[32] W Hackbusch ldquoA numerical method for solving parabolicequations with opposite orientationsrdquo Computing vol 20 no3 pp 229ndash240 1978
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
10 Abstract and Applied Analysis
Lemma 7 (see [23]) The function 119906lowast(119904) = min (119887max (119904 119886))is Lipschitz continuous in 119904 where 119886 lt 119887 are some fixed positiveconstants
Theorem 8 For all 119905 isin [0 119905119891] the solution to the optimality
system (71) is unique
Proof Suppose (119878 119860 119879 119876 1205821 1205822 1205823 1205824) and
(119878 119860 119879 119876 1205821 1205822 1205823 1205824) are two different solutions of
our optimality system (71) Let
119878 = 119890120582119905119898 119860 = 119890
120582119905119899 119879 = 119890
120582119905119901
119876 = 119890120582119905119902 120582
1= 119890minus120582119905119903 120582
2= 119890minus120582119905119904
1205823= 119890minus120582119905119908 120582
4= 119890minus120582119905V
119878 = 119890120582119905119898 119860 = 119890
120582119905119899 119879 = 119890
120582119905119901
119876 = 119890120582119905119902 120582
1= 119890minus120582119905119903 120582
2= 119890minus120582119905119904
1205823= 119890minus120582119905119908 120582
4= 119890minus120582119905V
(72)
where 120582 gt 0 is to be chosenAccordingly we have
119906lowast
1(119905) = min1max0
120573119898119899 (119904 minus 119903) 119890120582119905
1198881119873
119906lowast
2(119905) = min1max0 (119904 minus 119908) 119899
1198882
119906lowast
1(119905) = min1max0
120573119898119899 (119904 minus 119903) 119890120582119905
1198881119873
119906lowast
2(119905) = min1max0 (119904 minus 119908) 119899
1198882
(73)
Now we substitute 119878 = 119890120582119905119898 119860 = 119890
120582119905119899 119879 = 119890
120582119905119901 119876 = 119890
120582119905119902
1205821= 119890minus120582119905119903 1205822= 119890minus120582119905119904 1205823= 119890minus120582119905119908 1205824= 119890minus120582119905V and 119878 =
119890120582119905119898 119860 = 119890
120582119905119899 119879 = 119890
120582119905119901 119876 = 119890
120582119905119902 1205821= 119890minus120582119905119903 1205822=
119890minus120582119905119904 1205823= 119890minus120582119905119908 1205824= 119890minus120582119905V into the first ODE of (71)
respectively then we can obtain
+ 120582119898 = 120583119873119890minus120582119905
minus (1 minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)
times120573119898119899119890
120582119905
119873minus 120583119898
(74)
for119898 and119898 respectively Similarly we can derive
119899 + 120582119899
= (1 minusmin1max1120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)120573119898119899119890
120582119905
119873
+ 120585119901 minus (min1max0 119899 (119904 minus 119908)1198882
+ 120583) 119899
(75)
for 119899 and 119899 respectively
+ 120582119901 = min1max0 119899 (119904 minus 119908)1198882
119899
minus (120583 + 120585 + 120575) 119901
(76)
for 119901 and 119901 respectively
119902 + 120582119902 = 120575119901 minus 120583119902 (77)
for 119902 and 119902 respectively
119903 minus 120582119903 = 119903(1 minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)120573119899119890120582119905
119873
+ 120583119903minus119904(1minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)
times120573119899119890120582119905
119873
(78)
for 119903 and 119903 respectively
119904 minus 120582119904 = minus119890120582119905
+ 119903(1 minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)120573119898119890120582119905
119873
minus 119904(1 minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)
times120573119898119890120582119905
119873+ 119904(min1max0 119899 (119904 minus 119908)
1198882
+ 120583)
minus 119908(min1max0 119899 (119904 minus 119908)1198882
)
(79)
for 119904 and 119904 respectively
minus 120582119908 = minus119904120585 + (120583 + 120585 + 120575)119908 minus 120575V (80)
for 119908 and 119908 respectively
V minus 120582V = 120583V (81)
for V and V respectively
Abstract and Applied Analysis 11
By Lemma 7 we can obtain
1003816100381610038161003816119906lowast
1(119905) minus 119906
lowast
1(119905)1003816100381610038161003816 le
120573119890120582119905
1198881119873|119898119899 (119904 minus 119903) minus 119898119899 (119904 minus 119903)|
1003816100381610038161003816119906lowast
2(119905) minus 119906
lowast
2(119905)1003816100381610038161003816 le
1
1198882
|119899 (119904 minus 119908) minus 119899 (119904 minus 119908)|
(82)
The equations for 119898 119899 119901 119902 119903 119904 119908 V and the equationsfor 119898 119899 119901 119902 119903 119904 119908 V are subtracted respectively then wemultiply each equation by appropriate difference of functionsand integrate from 0 to 119905
119891 Next we add all eight integral
equations and some inequality techniques to obtain unique-ness The following calculation is similar for the sake ofsimplicity we only take119898 and119898 for an example
minus + (120583 + 120582) (119898 minus 119898)
=120573119890120582119905
119873[minus(1 minusmin1max0
120573119898119899 (119904 minus 119903) 119890120582119905
1198881119873
)119898119899
+(1 minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)119898119899]
(83)
Multiplying both sides of (83) by (119898minus119898) and integratingfrom 0 to 119905
119891gives
1
2(119898 minus 119898)
2(119905119891) + (120583 + 120582)int
119905119891
0
(119898 minus 119898)2119889119905
= int
119905119891
0
(119898 minus 119898)120573119890120582119905
119873
times [minus (1 minus 119906lowast
1)119898119899 + (1 minus 119906
lowast
1)119898119899] 119889119905
= int
119905119891
0
(119898 minus 119898)120573119890120582119905
119873[(119906lowast
1minus 1) (119898119899 minus 119898119899 + 119898119899 minus 119898119899)
+ 119898119899 (119906lowast
1minus 119906lowast
1)] 119889119905
le120573119890120582119905119891
119873int
119905119891
0
(119898 minus 119898)
times [119906lowast
1minus 1
100381610038161003816100381610038161003816100381610038161003816
(119898 |119899 minus 119899| + |119898 minus 119898| 119899) + 119898119899120573119890120582119905
1198881119873
100381610038161003816100381610038161003816100381610038161003816
times 119898119899 (119904 minus 119903) minus 119898119899 (119904 minus 119903) ] 119889119905
le120573119890120582119905119891
119873int
119905119891
0
(119898 minus 119898)
times [1003816100381610038161003816119906lowast
1minus 1
1003816100381610038161003816 (119898 |119899 minus 119899| + |119898 minus 119898| 119899) + 119898119899120573119890120582119905
1198881119873
|119898119899 (119904 minus 119904) + (119898119899 minus 119898119899) 119904
minus (119898119899 (119903 minus 119903) + (119898119899 minus 119898119899) 119903)| ] 119889119905
le120573119890120582119905119891
119873int
119905119891
0
[1003816100381610038161003816119906lowast
1minus 1
1003816100381610038161003816 |119898| |119898 minus 119898| |119899 minus 119899|
+ |119898 minus 119898|2
|119899| + |119861| |119904| |119899| |119898 minus 119898|2
+ |119861| |119898119899| |119898 minus 119898| |119903 minus 119903|
+ |119861| |119898119899| |119898 minus 119898| |119904 minus 119904|
+ |119861| |119904| |119898| |119898 minus 119898| |119899 minus 119899|
+ |119861| |119903| |119898| |119898 minus 119898| |119899 minus 119899|
+ |119861| |119903| |119899| |119898 minus 119898|2] 119889119905
le120573119890120582119905119891
119873int
119905119891
0
[
1003816100381610038161003816119906lowastminus 1
1003816100381610038161003816
2|119898| ((119898 minus 119898)
2+ (119899 minus 119899)
2)
+ |119899| (119898 minus 119898)2
+|119861| |119898119899|
2((119898 minus 119898)
2+ (119904 minus 119904)
2)
+|119861| |119904| |119898|
2((119898 minus 119898)
2+ (119899 minus 119899)
2)
+ |119861| |119904| |119899| (119898 minus 119898)2
+|119861| |119898119899|
2((119898 minus 119898)
2+ (119903 minus 119903)
2)
+|119861| |119898| |119903|
2((119898 minus 119898)
2+ (119899 minus 119899)
2)
+ |119861| |119903| |119899| (119898 minus 119898)2]119889119905
(84)
In the above derivation we use many scaling techniquesfor inequality or absolute inequality Particularly what shouldbe noted is that to get the first inequality of above derivationwe use the estimation of |119906lowast
1(119905) minus 119906
lowast
1(119905)| which has been given
before besides for the sake of convenience we note 119861 =
119898119899(120573119890120582119905119891119873) Furthermore we notice that the coefficients of
all the eight terms in the last formula (119898 minus 119898)2+ (119899 minus 119899)
2(119898 minus119898)
2 (119898 minus119898)2+ (119904 minus 119904)
2 (119898 minus119898)2+ (119899 minus 119899)
2 (119898 minus119898)2
(119898minus119898)2+(119903minus119903)
2 (119898minus119898)2+(119899minus119899)2 (119898minus119898)2 namely (|119906lowast1minus
1|2)|119898| |119899| (|119861||119898119899|)2 (|119861||119904||119898|)2 |119861||119904|119899 (|119861||119898119899|)2(|119861||119898||119903|)2 |119861||119903||119899| are nonnegative and bounded So thereexists a positive constant 119888
2such that
1
2(119898 minus 119898)
2(119905119891) + (120583 + 120582)int
119905119891
0
(119898 minus 119898)2119889119905
12 Abstract and Applied Analysis
le 1198882
120573119890120582119905119891
119873int
119905119891
0
((119898 minus 119898)2+ (119899 minus 119899)
2
+ (119904 minus 119904)2+ (119903 minus 119903)
2) 119889119905
(85)
Combining eight of these inequalities gives
1
2(119898 minus 119898) (119905
119891) +
1
2(119899 minus 119899) (119905
119891) +
1
2(119901 minus 119901) (119905
119891)
+1
2(119902 minus 119902) (119905
119891) +
1
2(119903 minus 119903) (0) +
1
2(119904 minus 119904) (0)
+1
2(119908 minus 119908) (0) +
1
2(V minus V) (0) + (120583 + 120582)
times int
119905119891
0
(119898 minus 119898)2+ (119899 minus 119899)
2+ (119901 minus 119901)
2
+ (119902 minus 119902)2
+ (119904 minus 119904)2
+ (119903 minus 119903)2+ (119908 minus 119908)
2+ (V minus V)2) 119889119905
le 119861int
119905119891
0
(119898 minus 119898)2+ (119899 minus 119899)
2+ (119901 minus 119901)
2
+ (119902 minus 119902)2
+ (119903 minus 119903)2+ (119904 minus 119904)
2
+ (119908 minus 119908)2+ (V minus V)2 119889119905
(86)
Thus from the above inequality we can conclude that
(120583 + 120582 minus 119861)int
119905119891
0
(119898 minus 119898)2+ (119899 minus 119899)
2
+ (119901 minus 119901)2
+ (119902 minus 119902)2
+ (119903 minus 119903)2
+ (119904 minus 119904)2+(119908 minus 119908)
2+ (V minus V)2 119889119905 le 0
(87)
where 119861 depends on the coefficients and the bounds dependon119898 119899 119901 119902 119903 119904 119908 V If we choose 120582 such that 120583+120582 gt 119861 then119898 = 119898 119899 = 119899 119901 = 119901 119902 = 119902 119903 = 119903 119904 = 119904 119908 = 119908 and V = VHence the solution to the optimality system is unique Theproof is complete
6 Numerical Simulation
61 The Simulation of State System (1) without ControlParameters For the sake of simplicity but without loss ofgenerality we will perform the numerical simulation of statesystem (1) with parameters 119906
1= 0 119906
2= 0 Before
illustrating the analytic properties of the alcoholism model(1) we will target the populations in the environment ofa community or a university for example the school ofmaterial science and engineering in our university that isLan zhou University of Technology (LUT for short) owing tothe accurate and available information we can obtain Refer-ring to the information provided by the admissions officeof LUT this school will enroll almost 1200 undergraduatesand almost 300 various postgraduates at the beginning of
fall semester at the same time there will be almost 1500various students graduated and left this school so the scaleof students in school remained almost 6000 we can take thetotal population 119873 = 6000 In this simulation we will takeSeptember as the initial time and units in one week periodin one year According to the investigations of the studentunion implemented in September every year we can takeinitial values as 119878(0) = 4500 119860(0) = 1000 119879(0) = 300and 119876(0) = 200 It seems that the alcoholism is a little bitmore but it is rather natural because many freshmen feelconfused when they are faced with the new environment anda new lifestyle many of them have no better choice but gathertogether to drink in small groups to mediate the anxiety andget to know each other over time some of them developthe habit of drinking To a certain extent for example thefrequent drinking badly affects their study we can classifythem into the alcoholism compartment Other initial valuesseem more reasonable so we need no more explanation Aswe know alcoholism death is seldom happen within oneyear so we omit mortality from alcoholism then how tounderstand the recruitment rate as well as natural death rate120583 We can treat the freshmen admission as the recruitmentpopulation and graduation students as the natural ldquodeathrdquoparts So we can take 120583 = 15006000 = 025 which is exactlyconsistent with the value in [16] As for the infection rate 120573and recovery rate 120575 we will let them be variables since thedrinking behaviors are related to many factors such as theseason and the pressure
According to the data we get from the student unionwe choose 120585 = 04 To summarize we list the values ofthe parameters in Table 1 Using the values of parameters inTable 1 we can plot Figures 2 and 3 which are on condition1198770lt 1 and 119877
0ge 1 respectively From Figure 2 we easily
know when 1198770lt 1 holds the solution of system (1) tends to
the alcohol free equilibrium1198640and verifies the global stability
of 1198640 While seen from Figure 3 we also know that if 119877
0ge
1 holds the solution of system (1) tends to the alcoholismequilibrium 119864
lowast and verifies the global stability of 119864lowast
62 The Sensitivity of 1198770about Two Control Parameters
Although from the expression of the model reproductionnumber 119877
0 we can easily find out the fact that the two
control variables that is 1199061and 119906
2 attribute to reducing the
severity of alcoholism we will still depict the graph between1198770and the two control variables to see more intuitiveness see
Figure 4 It seems from the figure that 1198770is a monotonically
decreasing function about two control parameters so it isadvisable to take two approaches simultaneously to controlthe alcoholics
63 The Simulation of Optimality System (71) In this subsec-tion we will investigate numerically the optimal solution tooptimality system (71) by numerical method from [32] theoptimality system is solved with a fourth-order Runge-Kuttascheme Beginning with a guess for the control variables thestate system is solved forward in time and then those values ofstate are used to solve the adjoint equations backward in timeThe controls are updated at the end of each iteration using
Abstract and Applied Analysis 13
0 10 20 30 40 50
0
1000
2000
3000
4000
5000
6000
7000
Time
Popu
latio
n siz
e
S(t)
A(t)
T(t)
Q(t)
Figure 2 When 1198770lt 1 the alcohol free equilibrium 119864
0is globally
asymptotically stable (120573 = 02 120575 = 03 and 1198770= 071698)
S(t)
A(t)
T(t)
Q(t)
0 10 20 30 40
48
2520
303540
49 50
500
1000
2000
3000
4000
5000
6000
7000
Time
Popu
latio
n siz
e
Figure 3 When 1198770
ge 1 the alcoholism equilibrium 119864lowast
=
(4466 476 32 26) corresponding to the given parameters is globallyasymptotically stable (120573 = 03 120575 = 02 and 119877
0= 108511)
the values of optimal controls obtained lastly The iterationscontinue until convergence takes place
In the simulations we choose the available variable valuesas Table 1 shows besides 120573 = 03 120575 = 02 The initial valueof model (1) is assumed to be 119878(0) = 4500 119860(0) = 1000119879(0) = 300 and 119876(0) = 200 as before
The ideal weights in objective functional are very difficultto obtain in reality it needs much work on data mining andfittingHence the acquisition of appropriate practical weights
Table 1 The parameters description of model (1)
Parameter Description Values120583 Natural birth rate or death rate 025
120573Transmission coefficient betweenalcoholism and susceptibles Variables
119873 The total populations to be considered 6000
120575The rate of populations quitting fromalcoholism permanently after treatment Variables
120585The rate of populations failed intreatment and returned to be alcoholic 04
0
02
04
06
08
1
0
05
1
0 02 04 06 08 1
u1
u2
R0
Figure 4 The relationship between 1198770and two control variables 119906
1
and 1199062
is still a difficult problem and remains for further investiga-tionsThe cost associated with119860(119905) and 119906
1(119905)mainly includes
the cost of dangerous behaviours during the alcoholism timeand educating the public while the cost associated with1199062(119905)mainly comes from health professional and the medical
resource includingmedicines andnursing care In viewof thisand taking the expressions of 119906
1 1199062into account after many
numerical simulations we finally give weighting coefficientsas 1198881= 102 1198882= 104 It should be pointed out that the
weights here are of only theoretical interest to reveal thecontrol strategies proposed in this paper Another point tonote is that themaximumcontrol is very difficult to achieve inreality so we will omit the situation of the maximum controlduring the series of simulations
Next we will make some necessary instructions andexplanations to the above simulation graphs Figures 5 67 and 8 depict the number of four compartments underdifferent control levels when we choose the weight coeffi-cients in objective function to be 119888
1= 102 1198882= 104 From
the four simulation graphs we can observe the followingsimple facts in reducing the total number of alcoholismsand increasing the number of susceptibles the effectivenessof various control measures is as follows optimal control isevidently better than middle control and middle control isbetter than single control 119906
2 single control 119906
2is better than
single control 1199061 while single control 119906
1is much better than
no controlFigures 9 and 10 depict the optimal control law of 119906
1 1199062
respectively In the beginning of the simulation the control
14 Abstract and Applied Analysis
0 10 20 30 40 504500
5000
5500
6000
6500
7000
S(n
umbe
r of s
usce
ptib
les)
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 5Number of the susceptibles whenwe choose theweights inobjective function are 119888
1= 102 1198882= 104 and under different optimal
control strategies that is (1) without control 1199061= 1199062= 0 (2) single
control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0 119906
2= 05 (4)
middle control 1199061= 1199062= 05 (5) optimal control 119906
1= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
500
1000
1500
A(n
umbe
r of t
he al
coho
lism
)
t (unit week)
Figure 6 Number of the alcoholics when we choose the weights inobjective function are 119888
1= 102 1198882= 104 and under different optimal
control strategies that is (1) without control 1199061= 1199062= 0 (2) single
control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0 119906
2= 05 (4)
middle control 1199061= 1199062= 05 (5) optimal control 119906
1= 119906lowast
1 1199062= 119906lowast
2
0
50
100
150
200
250
300
350
400
T(n
umbe
r of t
reat
men
t)
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 7 Number of the persons in treatment when we choose theweights in objective function are 119888
1= 10
2 1198882= 10
4 and underdifferent optimal control strategies that is (1) without control 119906
1=
1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control 119906
1= 0
1199062= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal control
1199061= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
50
100
150
200
250
300
350
400
Q(n
umbe
r of q
uitti
ng)
Figure 8 Number of people in quitting compartment when wechoose the weights in objective function are 119888
1= 102 1198882= 104 and
under different optimal control strategies that is (1) without control1199061= 1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control
1199061= 0 119906
2= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal
control 1199061= 119906lowast
1 1199062= 119906lowast
2
Abstract and Applied Analysis 15
0 10 20 30 40 50t (unit week)
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
Optimal control of u1
Figure 9 Figures of the optimal control 1199061when we choose the
weight in objective function are 1198881= 102 1198882= 104
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
0 10 20 30 40 50t (unit week)
Optimal control of u2
Figure 10 Figures of the optimal control 1199062when we choose the
weight in objective function are 1198881= 102 1198882= 104
effort of 1199061should be decreased from 065 to 05 within the
first month and over the next week it should be increased tothe maximum control until 50 weeks then rapidly decreasedto 0 at the end of the simulation As for the control 119906
2 it
should start from around 05 due to the initial alcoholics thenincrease to 055 within one week since the rapid infection andnext decrease to almost 0 since the effectiveness of treatmentin three weeks but with the infection going on the controleffort of 119906
2should gradually increase to the maximum and
maintain this level until the tenth week for the purpose ofconsolidation therapy and preventing rebound hereafter itshould be gradually decreased to the level of almost 043 untilthe fifty weeks then quickly decreased to 0 in the end
0 10 20 30 40 504500
5000
5500
6000
6500
7000
S(n
umbe
r of s
usce
ptib
les)
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 11 Number of the susceptibles when we choose the weightsin objective function are 119888
1= 104 1198882= 102 and under different
optimal control strategies that is (1) without control 1199061= 1199062= 0
(2) single control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0
1199062= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal control
1199061= 119906lowast
1 1199062= 119906lowast
2
In order to investigate the influence of different weightcoefficients in the objective functional on the effect ofcontrolling at the same time for a better comparison we willchange the weight coefficients in objective function into 119888
2=
104 1198881= 102 and we will list the corresponding numerical
simulation results as Figures 11ndash16 showWhen we change the weight coefficients in objective
function into 1198881= 104 1198882= 102 we find that the results
of simulations derived from the graphs are very similar tothe ones before We speculate that the most likely reasons ofthis result are due to three respects one is that the weightcoefficients are not too sensitive in the numerical simulationand another possible reason is that both of the two controlsare important in some sense equivalently importantThe lastbut not themost unlikely reason is that we have not found themost appropriate weight coefficients in the simulation whichis very difficult to find as previously mentioned
7 Conclusions
In this paper we formulate an alcoholics quitting modeland firstly investigate the variation discipline of variouspopulations from the perspective of global stability thenwe propose an objective functional to examine two differentcontrolmeasures (ie prevention and treatment) on the effectof alcohol The basic reproduction number of the model wasderived and the global stability of the two equilibria is givenFrom the expression of the basic reproductionnumber119877
0and
16 Abstract and Applied Analysis
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
500
1000
1500
A(n
umbe
r of t
he al
coho
lism
)
t (unit week)
Figure 12 Number of the alcoholics when we choose the weights inobjective function are 119888
1= 104 1198882= 102 and under different optimal
control strategies that is (1) without control 1199061= 1199062= 0 (2) single
control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0 119906
2= 05 (4)
middle control 1199061= 1199062= 05 (5) optimal control 119906
1= 119906lowast
1 1199062= 119906lowast
2
0
50
100
150
200
250
300
350
400
T(n
umbe
r of t
reat
men
t)
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 13 Number of the persons in treatment when we choosethe weights in objective function are 119888
1= 104 1198882= 102 and under
different optimal control strategies that is (1) without control 1199061=
1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control 119906
1= 0
1199062= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal control
1199061= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
50
100
150
200
250
300
350
400
Q(n
umbe
r of q
uitti
ng)
Figure 14 Number of people in quitting compartment when wechoose the weights in objective function are 119888
1= 104 1198882= 102 and
under different optimal control strategies that is (1) without control1199061= 1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control
1199061= 0 119906
2= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal
control 1199061= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50t (unit week)
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
Optimal control of u1
Figure 15 Figures of the optimal control 1199061when we choose the
weight in objective function are 1198881= 104 1198882= 102
related numerical simulation we can easily see that the twocontrol strategies are effective in the alcoholics process
Using Pontryaginrsquos Maximum Principle we firstly deter-mine the necessary conditions for existence of optimalcontrol pairsThe uniqueness of the solution to the optimalitysystem (71) is derived by the classical method of contradic-tion Numerical simulations of the model suggest that thetwo different groups of weights in the objective function have
Abstract and Applied Analysis 17
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
0 10 20 30 40 50t (unit week)
Optimal control of u2
Figure 16 Figures of the optimal control 1199062when we choose the
weight in objective function are 1198881= 104 1198882= 102
much similar effects on the transmission of the alcoholismfrom this point the two control measures are almost equallyimportant in controlling the alcoholism although they willprobably have great influences on the cost of the objectivefunction From the simulation figures it seems that the effectof optimal control which is measured by the reduction inthe number of alcoholics and the increase in the number ofsusceptibles is much better than other control strategies asnoted earlier in the simulation section According to the real-time curve of two optimal controls we point out the specificimplementation methods of optimal control which can beachieved in practice
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
This work was partially supported by the NSF ofGansu Province of China (2013GS09485) the NNSF ofChina (10961018) the NSF of Gansu Province of China(1107RJZA088) the NSF for Distinguished Young Scholarsof Gansu Province of China (1111RJDA003) the Special Fundfor the Basic Requirements in the Research of University ofGansu Province of China and the Development Programfor Hong Liu Distinguished Young Scholars in LanzhouUniversity of Technology
References
[1] R Room T Babor and J Rehm ldquoAlcohol and public healthrdquoThe Lancet vol 365 no 9458 pp 519ndash530 2005
[2] J B Saunders O G Aasland A Amundsen and M GrantldquoAlcohol consumption and related problems among primary
health care patients WHO collaborative project on early detec-tion of personswith harmful alcohol consumption IrdquoAddictionvol 88 no 3 pp 349ndash362 1993
[3] L D Johnston P M OMalley and J G Bachman NationalSurvey Results on Drug Use From the Monitoring the FutureStudy 1975ndash1992 National Institute on Drug Abuse RockvilleMd USA 1993
[4] H Wechsler J E Lee M Kuo and H Lee ldquoCollege bingedrinking in the 1990s a continuing problem Results of theHarvard School of Public Health 1999 College Alcohol StudyrdquoJournal of American College Health vol 48 no 5 pp 199ndash2102000
[5] P M OrsquoMalley and L D Johnston ldquoEpidemiology of alcoholand other drug use among American college studentsrdquo Journalof Studies on Alcohol vol 63 no 14 pp 23ndash39 2002
[6] K Agarwal-Kozlowski and D P Agarwal ldquoGenetic predisposi-tion to alcoholismrdquo Therapeutische Umschau vol 57 no 4 pp179ndash184 2000
[7] C Y Chen C L Storr and J C Anthony ldquoEarly-onset drug useand risk for drug dependence problemsrdquo Addictive Behaviorsvol 34 no 3 pp 319ndash322 2009
[8] M Glavas and J Weinberg ldquoStress alcohol consumption andthe hypothalamic-pituitary adrenal axisrdquo in Nutrients Stressand Medical Disorders pp 165ndash183 Humana Press New YorkNY USA 2006
[9] L N Blum N H Nielsen J A Riggs and L B BresolinldquoAlcoholism and alcohol abuse among women report of theCouncil on Scientific Affairsrdquo Journal of Womenrsquos Health vol7 no 7 pp 861ndash871 1998
[10] B Benedict ldquoModeling alcoholism as a contagious disease howldquoinfectedrdquo drinking buddies spread problem drinkingrdquo SIAMNews vol 40 no 3 2007
[11] H Walter K Gutierrez K Ramskogler I Hertling A Dvorakand O M Lesch ldquoGender-specific differences in alcoholismImplications for treatmentrdquo Archives of Womenrsquos Mental Healthvol 6 no 4 pp 253ndash258 2003
[12] D Muller R D Koch H von Specht w Volker and E MMunch ldquoNeurophisiologic findings in chronic alcohol abuserdquoPsychiatrie Neurologie und Medizinische Psychologie vol 37no 3 pp 129ndash132 1985
[13] G Testino ldquoAlcoholic diseases in hepato-gastroenterology apoint of viewrdquo Hepato-Gastroenterology vol 55 no 82-83 pp371ndash377 2008
[14] A Mubayi P E Greenwood C Castillo-Chavez P J Grue-newald and D M Gorman ldquoThe impact of relative residencetimes on the distribution of heavy drinkers in highly distinctenvironmentsrdquo Socio-Economic Planning Sciences vol 44 no 1pp 45ndash56 2010
[15] D Muller R D Koch H von Specht W Volker and E MMunch ldquoNeurophysiologic findings in chronic alcohol abuserdquoPsychiatrie Neurologie Und Medizinische Psychologie vol 37no 3 pp 129ndash132 1985 (Leipz)
[16] G Mulone and B Straughan ldquoModeling binge drinkingrdquoInternational Journal of Biomathematics vol 5 no 1 Article ID1250005 14 pages 2012
[17] H F Huo and N N Song ldquoGlobal stability for a binge drinkingmodel with two stagesrdquo Discrete Dynamics in Nature andSociety vol 2012 Article ID 829386 15 pages 2012
[18] S S Mushayabasa and C P Bhunu ldquoModelling the effects ofheavy alcohol consumption on the transmission dynamics ofgonorrheardquo Nonlinear Dynamics vol 66 no 4 pp 695ndash7062011
18 Abstract and Applied Analysis
[19] F S Sancheznchez XWang C Castillo-Chvez DM Gormanand P J Gruenewald ldquoDrinking as an epidemic a simplemathematical model with recovery and relapserdquo in TherapistsGuide to Evidence-Based Relapse Prevention K Witkiewitz andG Marlatt Eds Academic Press New York NY USA 2007
[20] H-F Huo and C-C Zhu ldquoInfluence of relapse in a givingup smoking modelrdquo Abstract and Applied Analysis vol 2013Article ID 525461 12 pages 2013
[21] H F Huo and Q Wang ldquoThe effects of awareness programs by17 media on the drinking dynamicsrdquo Submitted
[22] S Lee E Jung and C Castillo-Chavez ldquoOptimal controlintervention strategies in low- and high-risk problem drinkingpopulationsrdquo Socio-Economic Planning Sciences vol 44 no 4pp 258ndash265 2010
[23] K R Fister S Lenhart and J S McNally ldquoOptimizingchemotherapy in an HIV modelrdquo Electronic Journal of Differ-ential Equations vol 1998 no 32 pp 1ndash12 1998
[24] S Lenhart and J T Workman Optimal Control Applied toBiological Models Chapman amp HallCRC Mathematical andComputational Biology Series Chapman amp HallCRC BocaRaton Fla USA 2007
[25] X Yang L Chen and J Chen ldquoPermanence and positiveperiodic solution for the single-species nonautonomous delaydiffusive modelsrdquo Computers amp Mathematics with ApplicationsAn International Journal vol 32 no 4 pp 109ndash116 1996
[26] P van denDriessche and JWatmough ldquoReproduction numbersand sub-threshold endemic equilibria for compartmental mod-els of disease transmissionrdquoMathematical Biosciences vol 180pp 29ndash48 2002
[27] V Lakshmikantham S Leela and A A Martynyuk StabilityAnalysis of Nonlinear Systems vol 125 Marcel Dekker NewYork NY USA 1989
[28] J P LaSalle The Stability of Dynamical Systems vol 25 ofRegional Conference Series in Applied Mathematics Society forIndustrial and Applied Mathematics 1976
[29] W H Fleming and R W Rishel Deterministic and StochasticOptimal Control vol 1 ofApplications of Mathematics SpringerBerlin Germany 1975
[30] D L Lukes Differential Equations Classical to ControlledMathematics in Science and Engineering Academic Press NewYork NY USA 1982
[31] L S Pontryagin V G Boltyanskii R V Gamkrelidze and EF Mishchenko The Mathematical Theory of Optimal ProcessesJohn Wiley amp Sons New Jersey NJ USA 1962
[32] W Hackbusch ldquoA numerical method for solving parabolicequations with opposite orientationsrdquo Computing vol 20 no3 pp 229ndash240 1978
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 11
By Lemma 7 we can obtain
1003816100381610038161003816119906lowast
1(119905) minus 119906
lowast
1(119905)1003816100381610038161003816 le
120573119890120582119905
1198881119873|119898119899 (119904 minus 119903) minus 119898119899 (119904 minus 119903)|
1003816100381610038161003816119906lowast
2(119905) minus 119906
lowast
2(119905)1003816100381610038161003816 le
1
1198882
|119899 (119904 minus 119908) minus 119899 (119904 minus 119908)|
(82)
The equations for 119898 119899 119901 119902 119903 119904 119908 V and the equationsfor 119898 119899 119901 119902 119903 119904 119908 V are subtracted respectively then wemultiply each equation by appropriate difference of functionsand integrate from 0 to 119905
119891 Next we add all eight integral
equations and some inequality techniques to obtain unique-ness The following calculation is similar for the sake ofsimplicity we only take119898 and119898 for an example
minus + (120583 + 120582) (119898 minus 119898)
=120573119890120582119905
119873[minus(1 minusmin1max0
120573119898119899 (119904 minus 119903) 119890120582119905
1198881119873
)119898119899
+(1 minusmin1max0120573119898119899 (119904 minus 119903) 119890
120582119905
1198881119873
)119898119899]
(83)
Multiplying both sides of (83) by (119898minus119898) and integratingfrom 0 to 119905
119891gives
1
2(119898 minus 119898)
2(119905119891) + (120583 + 120582)int
119905119891
0
(119898 minus 119898)2119889119905
= int
119905119891
0
(119898 minus 119898)120573119890120582119905
119873
times [minus (1 minus 119906lowast
1)119898119899 + (1 minus 119906
lowast
1)119898119899] 119889119905
= int
119905119891
0
(119898 minus 119898)120573119890120582119905
119873[(119906lowast
1minus 1) (119898119899 minus 119898119899 + 119898119899 minus 119898119899)
+ 119898119899 (119906lowast
1minus 119906lowast
1)] 119889119905
le120573119890120582119905119891
119873int
119905119891
0
(119898 minus 119898)
times [119906lowast
1minus 1
100381610038161003816100381610038161003816100381610038161003816
(119898 |119899 minus 119899| + |119898 minus 119898| 119899) + 119898119899120573119890120582119905
1198881119873
100381610038161003816100381610038161003816100381610038161003816
times 119898119899 (119904 minus 119903) minus 119898119899 (119904 minus 119903) ] 119889119905
le120573119890120582119905119891
119873int
119905119891
0
(119898 minus 119898)
times [1003816100381610038161003816119906lowast
1minus 1
1003816100381610038161003816 (119898 |119899 minus 119899| + |119898 minus 119898| 119899) + 119898119899120573119890120582119905
1198881119873
|119898119899 (119904 minus 119904) + (119898119899 minus 119898119899) 119904
minus (119898119899 (119903 minus 119903) + (119898119899 minus 119898119899) 119903)| ] 119889119905
le120573119890120582119905119891
119873int
119905119891
0
[1003816100381610038161003816119906lowast
1minus 1
1003816100381610038161003816 |119898| |119898 minus 119898| |119899 minus 119899|
+ |119898 minus 119898|2
|119899| + |119861| |119904| |119899| |119898 minus 119898|2
+ |119861| |119898119899| |119898 minus 119898| |119903 minus 119903|
+ |119861| |119898119899| |119898 minus 119898| |119904 minus 119904|
+ |119861| |119904| |119898| |119898 minus 119898| |119899 minus 119899|
+ |119861| |119903| |119898| |119898 minus 119898| |119899 minus 119899|
+ |119861| |119903| |119899| |119898 minus 119898|2] 119889119905
le120573119890120582119905119891
119873int
119905119891
0
[
1003816100381610038161003816119906lowastminus 1
1003816100381610038161003816
2|119898| ((119898 minus 119898)
2+ (119899 minus 119899)
2)
+ |119899| (119898 minus 119898)2
+|119861| |119898119899|
2((119898 minus 119898)
2+ (119904 minus 119904)
2)
+|119861| |119904| |119898|
2((119898 minus 119898)
2+ (119899 minus 119899)
2)
+ |119861| |119904| |119899| (119898 minus 119898)2
+|119861| |119898119899|
2((119898 minus 119898)
2+ (119903 minus 119903)
2)
+|119861| |119898| |119903|
2((119898 minus 119898)
2+ (119899 minus 119899)
2)
+ |119861| |119903| |119899| (119898 minus 119898)2]119889119905
(84)
In the above derivation we use many scaling techniquesfor inequality or absolute inequality Particularly what shouldbe noted is that to get the first inequality of above derivationwe use the estimation of |119906lowast
1(119905) minus 119906
lowast
1(119905)| which has been given
before besides for the sake of convenience we note 119861 =
119898119899(120573119890120582119905119891119873) Furthermore we notice that the coefficients of
all the eight terms in the last formula (119898 minus 119898)2+ (119899 minus 119899)
2(119898 minus119898)
2 (119898 minus119898)2+ (119904 minus 119904)
2 (119898 minus119898)2+ (119899 minus 119899)
2 (119898 minus119898)2
(119898minus119898)2+(119903minus119903)
2 (119898minus119898)2+(119899minus119899)2 (119898minus119898)2 namely (|119906lowast1minus
1|2)|119898| |119899| (|119861||119898119899|)2 (|119861||119904||119898|)2 |119861||119904|119899 (|119861||119898119899|)2(|119861||119898||119903|)2 |119861||119903||119899| are nonnegative and bounded So thereexists a positive constant 119888
2such that
1
2(119898 minus 119898)
2(119905119891) + (120583 + 120582)int
119905119891
0
(119898 minus 119898)2119889119905
12 Abstract and Applied Analysis
le 1198882
120573119890120582119905119891
119873int
119905119891
0
((119898 minus 119898)2+ (119899 minus 119899)
2
+ (119904 minus 119904)2+ (119903 minus 119903)
2) 119889119905
(85)
Combining eight of these inequalities gives
1
2(119898 minus 119898) (119905
119891) +
1
2(119899 minus 119899) (119905
119891) +
1
2(119901 minus 119901) (119905
119891)
+1
2(119902 minus 119902) (119905
119891) +
1
2(119903 minus 119903) (0) +
1
2(119904 minus 119904) (0)
+1
2(119908 minus 119908) (0) +
1
2(V minus V) (0) + (120583 + 120582)
times int
119905119891
0
(119898 minus 119898)2+ (119899 minus 119899)
2+ (119901 minus 119901)
2
+ (119902 minus 119902)2
+ (119904 minus 119904)2
+ (119903 minus 119903)2+ (119908 minus 119908)
2+ (V minus V)2) 119889119905
le 119861int
119905119891
0
(119898 minus 119898)2+ (119899 minus 119899)
2+ (119901 minus 119901)
2
+ (119902 minus 119902)2
+ (119903 minus 119903)2+ (119904 minus 119904)
2
+ (119908 minus 119908)2+ (V minus V)2 119889119905
(86)
Thus from the above inequality we can conclude that
(120583 + 120582 minus 119861)int
119905119891
0
(119898 minus 119898)2+ (119899 minus 119899)
2
+ (119901 minus 119901)2
+ (119902 minus 119902)2
+ (119903 minus 119903)2
+ (119904 minus 119904)2+(119908 minus 119908)
2+ (V minus V)2 119889119905 le 0
(87)
where 119861 depends on the coefficients and the bounds dependon119898 119899 119901 119902 119903 119904 119908 V If we choose 120582 such that 120583+120582 gt 119861 then119898 = 119898 119899 = 119899 119901 = 119901 119902 = 119902 119903 = 119903 119904 = 119904 119908 = 119908 and V = VHence the solution to the optimality system is unique Theproof is complete
6 Numerical Simulation
61 The Simulation of State System (1) without ControlParameters For the sake of simplicity but without loss ofgenerality we will perform the numerical simulation of statesystem (1) with parameters 119906
1= 0 119906
2= 0 Before
illustrating the analytic properties of the alcoholism model(1) we will target the populations in the environment ofa community or a university for example the school ofmaterial science and engineering in our university that isLan zhou University of Technology (LUT for short) owing tothe accurate and available information we can obtain Refer-ring to the information provided by the admissions officeof LUT this school will enroll almost 1200 undergraduatesand almost 300 various postgraduates at the beginning of
fall semester at the same time there will be almost 1500various students graduated and left this school so the scaleof students in school remained almost 6000 we can take thetotal population 119873 = 6000 In this simulation we will takeSeptember as the initial time and units in one week periodin one year According to the investigations of the studentunion implemented in September every year we can takeinitial values as 119878(0) = 4500 119860(0) = 1000 119879(0) = 300and 119876(0) = 200 It seems that the alcoholism is a little bitmore but it is rather natural because many freshmen feelconfused when they are faced with the new environment anda new lifestyle many of them have no better choice but gathertogether to drink in small groups to mediate the anxiety andget to know each other over time some of them developthe habit of drinking To a certain extent for example thefrequent drinking badly affects their study we can classifythem into the alcoholism compartment Other initial valuesseem more reasonable so we need no more explanation Aswe know alcoholism death is seldom happen within oneyear so we omit mortality from alcoholism then how tounderstand the recruitment rate as well as natural death rate120583 We can treat the freshmen admission as the recruitmentpopulation and graduation students as the natural ldquodeathrdquoparts So we can take 120583 = 15006000 = 025 which is exactlyconsistent with the value in [16] As for the infection rate 120573and recovery rate 120575 we will let them be variables since thedrinking behaviors are related to many factors such as theseason and the pressure
According to the data we get from the student unionwe choose 120585 = 04 To summarize we list the values ofthe parameters in Table 1 Using the values of parameters inTable 1 we can plot Figures 2 and 3 which are on condition1198770lt 1 and 119877
0ge 1 respectively From Figure 2 we easily
know when 1198770lt 1 holds the solution of system (1) tends to
the alcohol free equilibrium1198640and verifies the global stability
of 1198640 While seen from Figure 3 we also know that if 119877
0ge
1 holds the solution of system (1) tends to the alcoholismequilibrium 119864
lowast and verifies the global stability of 119864lowast
62 The Sensitivity of 1198770about Two Control Parameters
Although from the expression of the model reproductionnumber 119877
0 we can easily find out the fact that the two
control variables that is 1199061and 119906
2 attribute to reducing the
severity of alcoholism we will still depict the graph between1198770and the two control variables to see more intuitiveness see
Figure 4 It seems from the figure that 1198770is a monotonically
decreasing function about two control parameters so it isadvisable to take two approaches simultaneously to controlthe alcoholics
63 The Simulation of Optimality System (71) In this subsec-tion we will investigate numerically the optimal solution tooptimality system (71) by numerical method from [32] theoptimality system is solved with a fourth-order Runge-Kuttascheme Beginning with a guess for the control variables thestate system is solved forward in time and then those values ofstate are used to solve the adjoint equations backward in timeThe controls are updated at the end of each iteration using
Abstract and Applied Analysis 13
0 10 20 30 40 50
0
1000
2000
3000
4000
5000
6000
7000
Time
Popu
latio
n siz
e
S(t)
A(t)
T(t)
Q(t)
Figure 2 When 1198770lt 1 the alcohol free equilibrium 119864
0is globally
asymptotically stable (120573 = 02 120575 = 03 and 1198770= 071698)
S(t)
A(t)
T(t)
Q(t)
0 10 20 30 40
48
2520
303540
49 50
500
1000
2000
3000
4000
5000
6000
7000
Time
Popu
latio
n siz
e
Figure 3 When 1198770
ge 1 the alcoholism equilibrium 119864lowast
=
(4466 476 32 26) corresponding to the given parameters is globallyasymptotically stable (120573 = 03 120575 = 02 and 119877
0= 108511)
the values of optimal controls obtained lastly The iterationscontinue until convergence takes place
In the simulations we choose the available variable valuesas Table 1 shows besides 120573 = 03 120575 = 02 The initial valueof model (1) is assumed to be 119878(0) = 4500 119860(0) = 1000119879(0) = 300 and 119876(0) = 200 as before
The ideal weights in objective functional are very difficultto obtain in reality it needs much work on data mining andfittingHence the acquisition of appropriate practical weights
Table 1 The parameters description of model (1)
Parameter Description Values120583 Natural birth rate or death rate 025
120573Transmission coefficient betweenalcoholism and susceptibles Variables
119873 The total populations to be considered 6000
120575The rate of populations quitting fromalcoholism permanently after treatment Variables
120585The rate of populations failed intreatment and returned to be alcoholic 04
0
02
04
06
08
1
0
05
1
0 02 04 06 08 1
u1
u2
R0
Figure 4 The relationship between 1198770and two control variables 119906
1
and 1199062
is still a difficult problem and remains for further investiga-tionsThe cost associated with119860(119905) and 119906
1(119905)mainly includes
the cost of dangerous behaviours during the alcoholism timeand educating the public while the cost associated with1199062(119905)mainly comes from health professional and the medical
resource includingmedicines andnursing care In viewof thisand taking the expressions of 119906
1 1199062into account after many
numerical simulations we finally give weighting coefficientsas 1198881= 102 1198882= 104 It should be pointed out that the
weights here are of only theoretical interest to reveal thecontrol strategies proposed in this paper Another point tonote is that themaximumcontrol is very difficult to achieve inreality so we will omit the situation of the maximum controlduring the series of simulations
Next we will make some necessary instructions andexplanations to the above simulation graphs Figures 5 67 and 8 depict the number of four compartments underdifferent control levels when we choose the weight coeffi-cients in objective function to be 119888
1= 102 1198882= 104 From
the four simulation graphs we can observe the followingsimple facts in reducing the total number of alcoholismsand increasing the number of susceptibles the effectivenessof various control measures is as follows optimal control isevidently better than middle control and middle control isbetter than single control 119906
2 single control 119906
2is better than
single control 1199061 while single control 119906
1is much better than
no controlFigures 9 and 10 depict the optimal control law of 119906
1 1199062
respectively In the beginning of the simulation the control
14 Abstract and Applied Analysis
0 10 20 30 40 504500
5000
5500
6000
6500
7000
S(n
umbe
r of s
usce
ptib
les)
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 5Number of the susceptibles whenwe choose theweights inobjective function are 119888
1= 102 1198882= 104 and under different optimal
control strategies that is (1) without control 1199061= 1199062= 0 (2) single
control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0 119906
2= 05 (4)
middle control 1199061= 1199062= 05 (5) optimal control 119906
1= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
500
1000
1500
A(n
umbe
r of t
he al
coho
lism
)
t (unit week)
Figure 6 Number of the alcoholics when we choose the weights inobjective function are 119888
1= 102 1198882= 104 and under different optimal
control strategies that is (1) without control 1199061= 1199062= 0 (2) single
control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0 119906
2= 05 (4)
middle control 1199061= 1199062= 05 (5) optimal control 119906
1= 119906lowast
1 1199062= 119906lowast
2
0
50
100
150
200
250
300
350
400
T(n
umbe
r of t
reat
men
t)
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 7 Number of the persons in treatment when we choose theweights in objective function are 119888
1= 10
2 1198882= 10
4 and underdifferent optimal control strategies that is (1) without control 119906
1=
1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control 119906
1= 0
1199062= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal control
1199061= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
50
100
150
200
250
300
350
400
Q(n
umbe
r of q
uitti
ng)
Figure 8 Number of people in quitting compartment when wechoose the weights in objective function are 119888
1= 102 1198882= 104 and
under different optimal control strategies that is (1) without control1199061= 1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control
1199061= 0 119906
2= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal
control 1199061= 119906lowast
1 1199062= 119906lowast
2
Abstract and Applied Analysis 15
0 10 20 30 40 50t (unit week)
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
Optimal control of u1
Figure 9 Figures of the optimal control 1199061when we choose the
weight in objective function are 1198881= 102 1198882= 104
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
0 10 20 30 40 50t (unit week)
Optimal control of u2
Figure 10 Figures of the optimal control 1199062when we choose the
weight in objective function are 1198881= 102 1198882= 104
effort of 1199061should be decreased from 065 to 05 within the
first month and over the next week it should be increased tothe maximum control until 50 weeks then rapidly decreasedto 0 at the end of the simulation As for the control 119906
2 it
should start from around 05 due to the initial alcoholics thenincrease to 055 within one week since the rapid infection andnext decrease to almost 0 since the effectiveness of treatmentin three weeks but with the infection going on the controleffort of 119906
2should gradually increase to the maximum and
maintain this level until the tenth week for the purpose ofconsolidation therapy and preventing rebound hereafter itshould be gradually decreased to the level of almost 043 untilthe fifty weeks then quickly decreased to 0 in the end
0 10 20 30 40 504500
5000
5500
6000
6500
7000
S(n
umbe
r of s
usce
ptib
les)
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 11 Number of the susceptibles when we choose the weightsin objective function are 119888
1= 104 1198882= 102 and under different
optimal control strategies that is (1) without control 1199061= 1199062= 0
(2) single control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0
1199062= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal control
1199061= 119906lowast
1 1199062= 119906lowast
2
In order to investigate the influence of different weightcoefficients in the objective functional on the effect ofcontrolling at the same time for a better comparison we willchange the weight coefficients in objective function into 119888
2=
104 1198881= 102 and we will list the corresponding numerical
simulation results as Figures 11ndash16 showWhen we change the weight coefficients in objective
function into 1198881= 104 1198882= 102 we find that the results
of simulations derived from the graphs are very similar tothe ones before We speculate that the most likely reasons ofthis result are due to three respects one is that the weightcoefficients are not too sensitive in the numerical simulationand another possible reason is that both of the two controlsare important in some sense equivalently importantThe lastbut not themost unlikely reason is that we have not found themost appropriate weight coefficients in the simulation whichis very difficult to find as previously mentioned
7 Conclusions
In this paper we formulate an alcoholics quitting modeland firstly investigate the variation discipline of variouspopulations from the perspective of global stability thenwe propose an objective functional to examine two differentcontrolmeasures (ie prevention and treatment) on the effectof alcohol The basic reproduction number of the model wasderived and the global stability of the two equilibria is givenFrom the expression of the basic reproductionnumber119877
0and
16 Abstract and Applied Analysis
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
500
1000
1500
A(n
umbe
r of t
he al
coho
lism
)
t (unit week)
Figure 12 Number of the alcoholics when we choose the weights inobjective function are 119888
1= 104 1198882= 102 and under different optimal
control strategies that is (1) without control 1199061= 1199062= 0 (2) single
control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0 119906
2= 05 (4)
middle control 1199061= 1199062= 05 (5) optimal control 119906
1= 119906lowast
1 1199062= 119906lowast
2
0
50
100
150
200
250
300
350
400
T(n
umbe
r of t
reat
men
t)
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 13 Number of the persons in treatment when we choosethe weights in objective function are 119888
1= 104 1198882= 102 and under
different optimal control strategies that is (1) without control 1199061=
1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control 119906
1= 0
1199062= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal control
1199061= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
50
100
150
200
250
300
350
400
Q(n
umbe
r of q
uitti
ng)
Figure 14 Number of people in quitting compartment when wechoose the weights in objective function are 119888
1= 104 1198882= 102 and
under different optimal control strategies that is (1) without control1199061= 1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control
1199061= 0 119906
2= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal
control 1199061= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50t (unit week)
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
Optimal control of u1
Figure 15 Figures of the optimal control 1199061when we choose the
weight in objective function are 1198881= 104 1198882= 102
related numerical simulation we can easily see that the twocontrol strategies are effective in the alcoholics process
Using Pontryaginrsquos Maximum Principle we firstly deter-mine the necessary conditions for existence of optimalcontrol pairsThe uniqueness of the solution to the optimalitysystem (71) is derived by the classical method of contradic-tion Numerical simulations of the model suggest that thetwo different groups of weights in the objective function have
Abstract and Applied Analysis 17
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
0 10 20 30 40 50t (unit week)
Optimal control of u2
Figure 16 Figures of the optimal control 1199062when we choose the
weight in objective function are 1198881= 104 1198882= 102
much similar effects on the transmission of the alcoholismfrom this point the two control measures are almost equallyimportant in controlling the alcoholism although they willprobably have great influences on the cost of the objectivefunction From the simulation figures it seems that the effectof optimal control which is measured by the reduction inthe number of alcoholics and the increase in the number ofsusceptibles is much better than other control strategies asnoted earlier in the simulation section According to the real-time curve of two optimal controls we point out the specificimplementation methods of optimal control which can beachieved in practice
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
This work was partially supported by the NSF ofGansu Province of China (2013GS09485) the NNSF ofChina (10961018) the NSF of Gansu Province of China(1107RJZA088) the NSF for Distinguished Young Scholarsof Gansu Province of China (1111RJDA003) the Special Fundfor the Basic Requirements in the Research of University ofGansu Province of China and the Development Programfor Hong Liu Distinguished Young Scholars in LanzhouUniversity of Technology
References
[1] R Room T Babor and J Rehm ldquoAlcohol and public healthrdquoThe Lancet vol 365 no 9458 pp 519ndash530 2005
[2] J B Saunders O G Aasland A Amundsen and M GrantldquoAlcohol consumption and related problems among primary
health care patients WHO collaborative project on early detec-tion of personswith harmful alcohol consumption IrdquoAddictionvol 88 no 3 pp 349ndash362 1993
[3] L D Johnston P M OMalley and J G Bachman NationalSurvey Results on Drug Use From the Monitoring the FutureStudy 1975ndash1992 National Institute on Drug Abuse RockvilleMd USA 1993
[4] H Wechsler J E Lee M Kuo and H Lee ldquoCollege bingedrinking in the 1990s a continuing problem Results of theHarvard School of Public Health 1999 College Alcohol StudyrdquoJournal of American College Health vol 48 no 5 pp 199ndash2102000
[5] P M OrsquoMalley and L D Johnston ldquoEpidemiology of alcoholand other drug use among American college studentsrdquo Journalof Studies on Alcohol vol 63 no 14 pp 23ndash39 2002
[6] K Agarwal-Kozlowski and D P Agarwal ldquoGenetic predisposi-tion to alcoholismrdquo Therapeutische Umschau vol 57 no 4 pp179ndash184 2000
[7] C Y Chen C L Storr and J C Anthony ldquoEarly-onset drug useand risk for drug dependence problemsrdquo Addictive Behaviorsvol 34 no 3 pp 319ndash322 2009
[8] M Glavas and J Weinberg ldquoStress alcohol consumption andthe hypothalamic-pituitary adrenal axisrdquo in Nutrients Stressand Medical Disorders pp 165ndash183 Humana Press New YorkNY USA 2006
[9] L N Blum N H Nielsen J A Riggs and L B BresolinldquoAlcoholism and alcohol abuse among women report of theCouncil on Scientific Affairsrdquo Journal of Womenrsquos Health vol7 no 7 pp 861ndash871 1998
[10] B Benedict ldquoModeling alcoholism as a contagious disease howldquoinfectedrdquo drinking buddies spread problem drinkingrdquo SIAMNews vol 40 no 3 2007
[11] H Walter K Gutierrez K Ramskogler I Hertling A Dvorakand O M Lesch ldquoGender-specific differences in alcoholismImplications for treatmentrdquo Archives of Womenrsquos Mental Healthvol 6 no 4 pp 253ndash258 2003
[12] D Muller R D Koch H von Specht w Volker and E MMunch ldquoNeurophisiologic findings in chronic alcohol abuserdquoPsychiatrie Neurologie und Medizinische Psychologie vol 37no 3 pp 129ndash132 1985
[13] G Testino ldquoAlcoholic diseases in hepato-gastroenterology apoint of viewrdquo Hepato-Gastroenterology vol 55 no 82-83 pp371ndash377 2008
[14] A Mubayi P E Greenwood C Castillo-Chavez P J Grue-newald and D M Gorman ldquoThe impact of relative residencetimes on the distribution of heavy drinkers in highly distinctenvironmentsrdquo Socio-Economic Planning Sciences vol 44 no 1pp 45ndash56 2010
[15] D Muller R D Koch H von Specht W Volker and E MMunch ldquoNeurophysiologic findings in chronic alcohol abuserdquoPsychiatrie Neurologie Und Medizinische Psychologie vol 37no 3 pp 129ndash132 1985 (Leipz)
[16] G Mulone and B Straughan ldquoModeling binge drinkingrdquoInternational Journal of Biomathematics vol 5 no 1 Article ID1250005 14 pages 2012
[17] H F Huo and N N Song ldquoGlobal stability for a binge drinkingmodel with two stagesrdquo Discrete Dynamics in Nature andSociety vol 2012 Article ID 829386 15 pages 2012
[18] S S Mushayabasa and C P Bhunu ldquoModelling the effects ofheavy alcohol consumption on the transmission dynamics ofgonorrheardquo Nonlinear Dynamics vol 66 no 4 pp 695ndash7062011
18 Abstract and Applied Analysis
[19] F S Sancheznchez XWang C Castillo-Chvez DM Gormanand P J Gruenewald ldquoDrinking as an epidemic a simplemathematical model with recovery and relapserdquo in TherapistsGuide to Evidence-Based Relapse Prevention K Witkiewitz andG Marlatt Eds Academic Press New York NY USA 2007
[20] H-F Huo and C-C Zhu ldquoInfluence of relapse in a givingup smoking modelrdquo Abstract and Applied Analysis vol 2013Article ID 525461 12 pages 2013
[21] H F Huo and Q Wang ldquoThe effects of awareness programs by17 media on the drinking dynamicsrdquo Submitted
[22] S Lee E Jung and C Castillo-Chavez ldquoOptimal controlintervention strategies in low- and high-risk problem drinkingpopulationsrdquo Socio-Economic Planning Sciences vol 44 no 4pp 258ndash265 2010
[23] K R Fister S Lenhart and J S McNally ldquoOptimizingchemotherapy in an HIV modelrdquo Electronic Journal of Differ-ential Equations vol 1998 no 32 pp 1ndash12 1998
[24] S Lenhart and J T Workman Optimal Control Applied toBiological Models Chapman amp HallCRC Mathematical andComputational Biology Series Chapman amp HallCRC BocaRaton Fla USA 2007
[25] X Yang L Chen and J Chen ldquoPermanence and positiveperiodic solution for the single-species nonautonomous delaydiffusive modelsrdquo Computers amp Mathematics with ApplicationsAn International Journal vol 32 no 4 pp 109ndash116 1996
[26] P van denDriessche and JWatmough ldquoReproduction numbersand sub-threshold endemic equilibria for compartmental mod-els of disease transmissionrdquoMathematical Biosciences vol 180pp 29ndash48 2002
[27] V Lakshmikantham S Leela and A A Martynyuk StabilityAnalysis of Nonlinear Systems vol 125 Marcel Dekker NewYork NY USA 1989
[28] J P LaSalle The Stability of Dynamical Systems vol 25 ofRegional Conference Series in Applied Mathematics Society forIndustrial and Applied Mathematics 1976
[29] W H Fleming and R W Rishel Deterministic and StochasticOptimal Control vol 1 ofApplications of Mathematics SpringerBerlin Germany 1975
[30] D L Lukes Differential Equations Classical to ControlledMathematics in Science and Engineering Academic Press NewYork NY USA 1982
[31] L S Pontryagin V G Boltyanskii R V Gamkrelidze and EF Mishchenko The Mathematical Theory of Optimal ProcessesJohn Wiley amp Sons New Jersey NJ USA 1962
[32] W Hackbusch ldquoA numerical method for solving parabolicequations with opposite orientationsrdquo Computing vol 20 no3 pp 229ndash240 1978
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
12 Abstract and Applied Analysis
le 1198882
120573119890120582119905119891
119873int
119905119891
0
((119898 minus 119898)2+ (119899 minus 119899)
2
+ (119904 minus 119904)2+ (119903 minus 119903)
2) 119889119905
(85)
Combining eight of these inequalities gives
1
2(119898 minus 119898) (119905
119891) +
1
2(119899 minus 119899) (119905
119891) +
1
2(119901 minus 119901) (119905
119891)
+1
2(119902 minus 119902) (119905
119891) +
1
2(119903 minus 119903) (0) +
1
2(119904 minus 119904) (0)
+1
2(119908 minus 119908) (0) +
1
2(V minus V) (0) + (120583 + 120582)
times int
119905119891
0
(119898 minus 119898)2+ (119899 minus 119899)
2+ (119901 minus 119901)
2
+ (119902 minus 119902)2
+ (119904 minus 119904)2
+ (119903 minus 119903)2+ (119908 minus 119908)
2+ (V minus V)2) 119889119905
le 119861int
119905119891
0
(119898 minus 119898)2+ (119899 minus 119899)
2+ (119901 minus 119901)
2
+ (119902 minus 119902)2
+ (119903 minus 119903)2+ (119904 minus 119904)
2
+ (119908 minus 119908)2+ (V minus V)2 119889119905
(86)
Thus from the above inequality we can conclude that
(120583 + 120582 minus 119861)int
119905119891
0
(119898 minus 119898)2+ (119899 minus 119899)
2
+ (119901 minus 119901)2
+ (119902 minus 119902)2
+ (119903 minus 119903)2
+ (119904 minus 119904)2+(119908 minus 119908)
2+ (V minus V)2 119889119905 le 0
(87)
where 119861 depends on the coefficients and the bounds dependon119898 119899 119901 119902 119903 119904 119908 V If we choose 120582 such that 120583+120582 gt 119861 then119898 = 119898 119899 = 119899 119901 = 119901 119902 = 119902 119903 = 119903 119904 = 119904 119908 = 119908 and V = VHence the solution to the optimality system is unique Theproof is complete
6 Numerical Simulation
61 The Simulation of State System (1) without ControlParameters For the sake of simplicity but without loss ofgenerality we will perform the numerical simulation of statesystem (1) with parameters 119906
1= 0 119906
2= 0 Before
illustrating the analytic properties of the alcoholism model(1) we will target the populations in the environment ofa community or a university for example the school ofmaterial science and engineering in our university that isLan zhou University of Technology (LUT for short) owing tothe accurate and available information we can obtain Refer-ring to the information provided by the admissions officeof LUT this school will enroll almost 1200 undergraduatesand almost 300 various postgraduates at the beginning of
fall semester at the same time there will be almost 1500various students graduated and left this school so the scaleof students in school remained almost 6000 we can take thetotal population 119873 = 6000 In this simulation we will takeSeptember as the initial time and units in one week periodin one year According to the investigations of the studentunion implemented in September every year we can takeinitial values as 119878(0) = 4500 119860(0) = 1000 119879(0) = 300and 119876(0) = 200 It seems that the alcoholism is a little bitmore but it is rather natural because many freshmen feelconfused when they are faced with the new environment anda new lifestyle many of them have no better choice but gathertogether to drink in small groups to mediate the anxiety andget to know each other over time some of them developthe habit of drinking To a certain extent for example thefrequent drinking badly affects their study we can classifythem into the alcoholism compartment Other initial valuesseem more reasonable so we need no more explanation Aswe know alcoholism death is seldom happen within oneyear so we omit mortality from alcoholism then how tounderstand the recruitment rate as well as natural death rate120583 We can treat the freshmen admission as the recruitmentpopulation and graduation students as the natural ldquodeathrdquoparts So we can take 120583 = 15006000 = 025 which is exactlyconsistent with the value in [16] As for the infection rate 120573and recovery rate 120575 we will let them be variables since thedrinking behaviors are related to many factors such as theseason and the pressure
According to the data we get from the student unionwe choose 120585 = 04 To summarize we list the values ofthe parameters in Table 1 Using the values of parameters inTable 1 we can plot Figures 2 and 3 which are on condition1198770lt 1 and 119877
0ge 1 respectively From Figure 2 we easily
know when 1198770lt 1 holds the solution of system (1) tends to
the alcohol free equilibrium1198640and verifies the global stability
of 1198640 While seen from Figure 3 we also know that if 119877
0ge
1 holds the solution of system (1) tends to the alcoholismequilibrium 119864
lowast and verifies the global stability of 119864lowast
62 The Sensitivity of 1198770about Two Control Parameters
Although from the expression of the model reproductionnumber 119877
0 we can easily find out the fact that the two
control variables that is 1199061and 119906
2 attribute to reducing the
severity of alcoholism we will still depict the graph between1198770and the two control variables to see more intuitiveness see
Figure 4 It seems from the figure that 1198770is a monotonically
decreasing function about two control parameters so it isadvisable to take two approaches simultaneously to controlthe alcoholics
63 The Simulation of Optimality System (71) In this subsec-tion we will investigate numerically the optimal solution tooptimality system (71) by numerical method from [32] theoptimality system is solved with a fourth-order Runge-Kuttascheme Beginning with a guess for the control variables thestate system is solved forward in time and then those values ofstate are used to solve the adjoint equations backward in timeThe controls are updated at the end of each iteration using
Abstract and Applied Analysis 13
0 10 20 30 40 50
0
1000
2000
3000
4000
5000
6000
7000
Time
Popu
latio
n siz
e
S(t)
A(t)
T(t)
Q(t)
Figure 2 When 1198770lt 1 the alcohol free equilibrium 119864
0is globally
asymptotically stable (120573 = 02 120575 = 03 and 1198770= 071698)
S(t)
A(t)
T(t)
Q(t)
0 10 20 30 40
48
2520
303540
49 50
500
1000
2000
3000
4000
5000
6000
7000
Time
Popu
latio
n siz
e
Figure 3 When 1198770
ge 1 the alcoholism equilibrium 119864lowast
=
(4466 476 32 26) corresponding to the given parameters is globallyasymptotically stable (120573 = 03 120575 = 02 and 119877
0= 108511)
the values of optimal controls obtained lastly The iterationscontinue until convergence takes place
In the simulations we choose the available variable valuesas Table 1 shows besides 120573 = 03 120575 = 02 The initial valueof model (1) is assumed to be 119878(0) = 4500 119860(0) = 1000119879(0) = 300 and 119876(0) = 200 as before
The ideal weights in objective functional are very difficultto obtain in reality it needs much work on data mining andfittingHence the acquisition of appropriate practical weights
Table 1 The parameters description of model (1)
Parameter Description Values120583 Natural birth rate or death rate 025
120573Transmission coefficient betweenalcoholism and susceptibles Variables
119873 The total populations to be considered 6000
120575The rate of populations quitting fromalcoholism permanently after treatment Variables
120585The rate of populations failed intreatment and returned to be alcoholic 04
0
02
04
06
08
1
0
05
1
0 02 04 06 08 1
u1
u2
R0
Figure 4 The relationship between 1198770and two control variables 119906
1
and 1199062
is still a difficult problem and remains for further investiga-tionsThe cost associated with119860(119905) and 119906
1(119905)mainly includes
the cost of dangerous behaviours during the alcoholism timeand educating the public while the cost associated with1199062(119905)mainly comes from health professional and the medical
resource includingmedicines andnursing care In viewof thisand taking the expressions of 119906
1 1199062into account after many
numerical simulations we finally give weighting coefficientsas 1198881= 102 1198882= 104 It should be pointed out that the
weights here are of only theoretical interest to reveal thecontrol strategies proposed in this paper Another point tonote is that themaximumcontrol is very difficult to achieve inreality so we will omit the situation of the maximum controlduring the series of simulations
Next we will make some necessary instructions andexplanations to the above simulation graphs Figures 5 67 and 8 depict the number of four compartments underdifferent control levels when we choose the weight coeffi-cients in objective function to be 119888
1= 102 1198882= 104 From
the four simulation graphs we can observe the followingsimple facts in reducing the total number of alcoholismsand increasing the number of susceptibles the effectivenessof various control measures is as follows optimal control isevidently better than middle control and middle control isbetter than single control 119906
2 single control 119906
2is better than
single control 1199061 while single control 119906
1is much better than
no controlFigures 9 and 10 depict the optimal control law of 119906
1 1199062
respectively In the beginning of the simulation the control
14 Abstract and Applied Analysis
0 10 20 30 40 504500
5000
5500
6000
6500
7000
S(n
umbe
r of s
usce
ptib
les)
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 5Number of the susceptibles whenwe choose theweights inobjective function are 119888
1= 102 1198882= 104 and under different optimal
control strategies that is (1) without control 1199061= 1199062= 0 (2) single
control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0 119906
2= 05 (4)
middle control 1199061= 1199062= 05 (5) optimal control 119906
1= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
500
1000
1500
A(n
umbe
r of t
he al
coho
lism
)
t (unit week)
Figure 6 Number of the alcoholics when we choose the weights inobjective function are 119888
1= 102 1198882= 104 and under different optimal
control strategies that is (1) without control 1199061= 1199062= 0 (2) single
control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0 119906
2= 05 (4)
middle control 1199061= 1199062= 05 (5) optimal control 119906
1= 119906lowast
1 1199062= 119906lowast
2
0
50
100
150
200
250
300
350
400
T(n
umbe
r of t
reat
men
t)
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 7 Number of the persons in treatment when we choose theweights in objective function are 119888
1= 10
2 1198882= 10
4 and underdifferent optimal control strategies that is (1) without control 119906
1=
1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control 119906
1= 0
1199062= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal control
1199061= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
50
100
150
200
250
300
350
400
Q(n
umbe
r of q
uitti
ng)
Figure 8 Number of people in quitting compartment when wechoose the weights in objective function are 119888
1= 102 1198882= 104 and
under different optimal control strategies that is (1) without control1199061= 1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control
1199061= 0 119906
2= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal
control 1199061= 119906lowast
1 1199062= 119906lowast
2
Abstract and Applied Analysis 15
0 10 20 30 40 50t (unit week)
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
Optimal control of u1
Figure 9 Figures of the optimal control 1199061when we choose the
weight in objective function are 1198881= 102 1198882= 104
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
0 10 20 30 40 50t (unit week)
Optimal control of u2
Figure 10 Figures of the optimal control 1199062when we choose the
weight in objective function are 1198881= 102 1198882= 104
effort of 1199061should be decreased from 065 to 05 within the
first month and over the next week it should be increased tothe maximum control until 50 weeks then rapidly decreasedto 0 at the end of the simulation As for the control 119906
2 it
should start from around 05 due to the initial alcoholics thenincrease to 055 within one week since the rapid infection andnext decrease to almost 0 since the effectiveness of treatmentin three weeks but with the infection going on the controleffort of 119906
2should gradually increase to the maximum and
maintain this level until the tenth week for the purpose ofconsolidation therapy and preventing rebound hereafter itshould be gradually decreased to the level of almost 043 untilthe fifty weeks then quickly decreased to 0 in the end
0 10 20 30 40 504500
5000
5500
6000
6500
7000
S(n
umbe
r of s
usce
ptib
les)
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 11 Number of the susceptibles when we choose the weightsin objective function are 119888
1= 104 1198882= 102 and under different
optimal control strategies that is (1) without control 1199061= 1199062= 0
(2) single control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0
1199062= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal control
1199061= 119906lowast
1 1199062= 119906lowast
2
In order to investigate the influence of different weightcoefficients in the objective functional on the effect ofcontrolling at the same time for a better comparison we willchange the weight coefficients in objective function into 119888
2=
104 1198881= 102 and we will list the corresponding numerical
simulation results as Figures 11ndash16 showWhen we change the weight coefficients in objective
function into 1198881= 104 1198882= 102 we find that the results
of simulations derived from the graphs are very similar tothe ones before We speculate that the most likely reasons ofthis result are due to three respects one is that the weightcoefficients are not too sensitive in the numerical simulationand another possible reason is that both of the two controlsare important in some sense equivalently importantThe lastbut not themost unlikely reason is that we have not found themost appropriate weight coefficients in the simulation whichis very difficult to find as previously mentioned
7 Conclusions
In this paper we formulate an alcoholics quitting modeland firstly investigate the variation discipline of variouspopulations from the perspective of global stability thenwe propose an objective functional to examine two differentcontrolmeasures (ie prevention and treatment) on the effectof alcohol The basic reproduction number of the model wasderived and the global stability of the two equilibria is givenFrom the expression of the basic reproductionnumber119877
0and
16 Abstract and Applied Analysis
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
500
1000
1500
A(n
umbe
r of t
he al
coho
lism
)
t (unit week)
Figure 12 Number of the alcoholics when we choose the weights inobjective function are 119888
1= 104 1198882= 102 and under different optimal
control strategies that is (1) without control 1199061= 1199062= 0 (2) single
control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0 119906
2= 05 (4)
middle control 1199061= 1199062= 05 (5) optimal control 119906
1= 119906lowast
1 1199062= 119906lowast
2
0
50
100
150
200
250
300
350
400
T(n
umbe
r of t
reat
men
t)
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 13 Number of the persons in treatment when we choosethe weights in objective function are 119888
1= 104 1198882= 102 and under
different optimal control strategies that is (1) without control 1199061=
1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control 119906
1= 0
1199062= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal control
1199061= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
50
100
150
200
250
300
350
400
Q(n
umbe
r of q
uitti
ng)
Figure 14 Number of people in quitting compartment when wechoose the weights in objective function are 119888
1= 104 1198882= 102 and
under different optimal control strategies that is (1) without control1199061= 1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control
1199061= 0 119906
2= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal
control 1199061= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50t (unit week)
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
Optimal control of u1
Figure 15 Figures of the optimal control 1199061when we choose the
weight in objective function are 1198881= 104 1198882= 102
related numerical simulation we can easily see that the twocontrol strategies are effective in the alcoholics process
Using Pontryaginrsquos Maximum Principle we firstly deter-mine the necessary conditions for existence of optimalcontrol pairsThe uniqueness of the solution to the optimalitysystem (71) is derived by the classical method of contradic-tion Numerical simulations of the model suggest that thetwo different groups of weights in the objective function have
Abstract and Applied Analysis 17
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
0 10 20 30 40 50t (unit week)
Optimal control of u2
Figure 16 Figures of the optimal control 1199062when we choose the
weight in objective function are 1198881= 104 1198882= 102
much similar effects on the transmission of the alcoholismfrom this point the two control measures are almost equallyimportant in controlling the alcoholism although they willprobably have great influences on the cost of the objectivefunction From the simulation figures it seems that the effectof optimal control which is measured by the reduction inthe number of alcoholics and the increase in the number ofsusceptibles is much better than other control strategies asnoted earlier in the simulation section According to the real-time curve of two optimal controls we point out the specificimplementation methods of optimal control which can beachieved in practice
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
This work was partially supported by the NSF ofGansu Province of China (2013GS09485) the NNSF ofChina (10961018) the NSF of Gansu Province of China(1107RJZA088) the NSF for Distinguished Young Scholarsof Gansu Province of China (1111RJDA003) the Special Fundfor the Basic Requirements in the Research of University ofGansu Province of China and the Development Programfor Hong Liu Distinguished Young Scholars in LanzhouUniversity of Technology
References
[1] R Room T Babor and J Rehm ldquoAlcohol and public healthrdquoThe Lancet vol 365 no 9458 pp 519ndash530 2005
[2] J B Saunders O G Aasland A Amundsen and M GrantldquoAlcohol consumption and related problems among primary
health care patients WHO collaborative project on early detec-tion of personswith harmful alcohol consumption IrdquoAddictionvol 88 no 3 pp 349ndash362 1993
[3] L D Johnston P M OMalley and J G Bachman NationalSurvey Results on Drug Use From the Monitoring the FutureStudy 1975ndash1992 National Institute on Drug Abuse RockvilleMd USA 1993
[4] H Wechsler J E Lee M Kuo and H Lee ldquoCollege bingedrinking in the 1990s a continuing problem Results of theHarvard School of Public Health 1999 College Alcohol StudyrdquoJournal of American College Health vol 48 no 5 pp 199ndash2102000
[5] P M OrsquoMalley and L D Johnston ldquoEpidemiology of alcoholand other drug use among American college studentsrdquo Journalof Studies on Alcohol vol 63 no 14 pp 23ndash39 2002
[6] K Agarwal-Kozlowski and D P Agarwal ldquoGenetic predisposi-tion to alcoholismrdquo Therapeutische Umschau vol 57 no 4 pp179ndash184 2000
[7] C Y Chen C L Storr and J C Anthony ldquoEarly-onset drug useand risk for drug dependence problemsrdquo Addictive Behaviorsvol 34 no 3 pp 319ndash322 2009
[8] M Glavas and J Weinberg ldquoStress alcohol consumption andthe hypothalamic-pituitary adrenal axisrdquo in Nutrients Stressand Medical Disorders pp 165ndash183 Humana Press New YorkNY USA 2006
[9] L N Blum N H Nielsen J A Riggs and L B BresolinldquoAlcoholism and alcohol abuse among women report of theCouncil on Scientific Affairsrdquo Journal of Womenrsquos Health vol7 no 7 pp 861ndash871 1998
[10] B Benedict ldquoModeling alcoholism as a contagious disease howldquoinfectedrdquo drinking buddies spread problem drinkingrdquo SIAMNews vol 40 no 3 2007
[11] H Walter K Gutierrez K Ramskogler I Hertling A Dvorakand O M Lesch ldquoGender-specific differences in alcoholismImplications for treatmentrdquo Archives of Womenrsquos Mental Healthvol 6 no 4 pp 253ndash258 2003
[12] D Muller R D Koch H von Specht w Volker and E MMunch ldquoNeurophisiologic findings in chronic alcohol abuserdquoPsychiatrie Neurologie und Medizinische Psychologie vol 37no 3 pp 129ndash132 1985
[13] G Testino ldquoAlcoholic diseases in hepato-gastroenterology apoint of viewrdquo Hepato-Gastroenterology vol 55 no 82-83 pp371ndash377 2008
[14] A Mubayi P E Greenwood C Castillo-Chavez P J Grue-newald and D M Gorman ldquoThe impact of relative residencetimes on the distribution of heavy drinkers in highly distinctenvironmentsrdquo Socio-Economic Planning Sciences vol 44 no 1pp 45ndash56 2010
[15] D Muller R D Koch H von Specht W Volker and E MMunch ldquoNeurophysiologic findings in chronic alcohol abuserdquoPsychiatrie Neurologie Und Medizinische Psychologie vol 37no 3 pp 129ndash132 1985 (Leipz)
[16] G Mulone and B Straughan ldquoModeling binge drinkingrdquoInternational Journal of Biomathematics vol 5 no 1 Article ID1250005 14 pages 2012
[17] H F Huo and N N Song ldquoGlobal stability for a binge drinkingmodel with two stagesrdquo Discrete Dynamics in Nature andSociety vol 2012 Article ID 829386 15 pages 2012
[18] S S Mushayabasa and C P Bhunu ldquoModelling the effects ofheavy alcohol consumption on the transmission dynamics ofgonorrheardquo Nonlinear Dynamics vol 66 no 4 pp 695ndash7062011
18 Abstract and Applied Analysis
[19] F S Sancheznchez XWang C Castillo-Chvez DM Gormanand P J Gruenewald ldquoDrinking as an epidemic a simplemathematical model with recovery and relapserdquo in TherapistsGuide to Evidence-Based Relapse Prevention K Witkiewitz andG Marlatt Eds Academic Press New York NY USA 2007
[20] H-F Huo and C-C Zhu ldquoInfluence of relapse in a givingup smoking modelrdquo Abstract and Applied Analysis vol 2013Article ID 525461 12 pages 2013
[21] H F Huo and Q Wang ldquoThe effects of awareness programs by17 media on the drinking dynamicsrdquo Submitted
[22] S Lee E Jung and C Castillo-Chavez ldquoOptimal controlintervention strategies in low- and high-risk problem drinkingpopulationsrdquo Socio-Economic Planning Sciences vol 44 no 4pp 258ndash265 2010
[23] K R Fister S Lenhart and J S McNally ldquoOptimizingchemotherapy in an HIV modelrdquo Electronic Journal of Differ-ential Equations vol 1998 no 32 pp 1ndash12 1998
[24] S Lenhart and J T Workman Optimal Control Applied toBiological Models Chapman amp HallCRC Mathematical andComputational Biology Series Chapman amp HallCRC BocaRaton Fla USA 2007
[25] X Yang L Chen and J Chen ldquoPermanence and positiveperiodic solution for the single-species nonautonomous delaydiffusive modelsrdquo Computers amp Mathematics with ApplicationsAn International Journal vol 32 no 4 pp 109ndash116 1996
[26] P van denDriessche and JWatmough ldquoReproduction numbersand sub-threshold endemic equilibria for compartmental mod-els of disease transmissionrdquoMathematical Biosciences vol 180pp 29ndash48 2002
[27] V Lakshmikantham S Leela and A A Martynyuk StabilityAnalysis of Nonlinear Systems vol 125 Marcel Dekker NewYork NY USA 1989
[28] J P LaSalle The Stability of Dynamical Systems vol 25 ofRegional Conference Series in Applied Mathematics Society forIndustrial and Applied Mathematics 1976
[29] W H Fleming and R W Rishel Deterministic and StochasticOptimal Control vol 1 ofApplications of Mathematics SpringerBerlin Germany 1975
[30] D L Lukes Differential Equations Classical to ControlledMathematics in Science and Engineering Academic Press NewYork NY USA 1982
[31] L S Pontryagin V G Boltyanskii R V Gamkrelidze and EF Mishchenko The Mathematical Theory of Optimal ProcessesJohn Wiley amp Sons New Jersey NJ USA 1962
[32] W Hackbusch ldquoA numerical method for solving parabolicequations with opposite orientationsrdquo Computing vol 20 no3 pp 229ndash240 1978
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 13
0 10 20 30 40 50
0
1000
2000
3000
4000
5000
6000
7000
Time
Popu
latio
n siz
e
S(t)
A(t)
T(t)
Q(t)
Figure 2 When 1198770lt 1 the alcohol free equilibrium 119864
0is globally
asymptotically stable (120573 = 02 120575 = 03 and 1198770= 071698)
S(t)
A(t)
T(t)
Q(t)
0 10 20 30 40
48
2520
303540
49 50
500
1000
2000
3000
4000
5000
6000
7000
Time
Popu
latio
n siz
e
Figure 3 When 1198770
ge 1 the alcoholism equilibrium 119864lowast
=
(4466 476 32 26) corresponding to the given parameters is globallyasymptotically stable (120573 = 03 120575 = 02 and 119877
0= 108511)
the values of optimal controls obtained lastly The iterationscontinue until convergence takes place
In the simulations we choose the available variable valuesas Table 1 shows besides 120573 = 03 120575 = 02 The initial valueof model (1) is assumed to be 119878(0) = 4500 119860(0) = 1000119879(0) = 300 and 119876(0) = 200 as before
The ideal weights in objective functional are very difficultto obtain in reality it needs much work on data mining andfittingHence the acquisition of appropriate practical weights
Table 1 The parameters description of model (1)
Parameter Description Values120583 Natural birth rate or death rate 025
120573Transmission coefficient betweenalcoholism and susceptibles Variables
119873 The total populations to be considered 6000
120575The rate of populations quitting fromalcoholism permanently after treatment Variables
120585The rate of populations failed intreatment and returned to be alcoholic 04
0
02
04
06
08
1
0
05
1
0 02 04 06 08 1
u1
u2
R0
Figure 4 The relationship between 1198770and two control variables 119906
1
and 1199062
is still a difficult problem and remains for further investiga-tionsThe cost associated with119860(119905) and 119906
1(119905)mainly includes
the cost of dangerous behaviours during the alcoholism timeand educating the public while the cost associated with1199062(119905)mainly comes from health professional and the medical
resource includingmedicines andnursing care In viewof thisand taking the expressions of 119906
1 1199062into account after many
numerical simulations we finally give weighting coefficientsas 1198881= 102 1198882= 104 It should be pointed out that the
weights here are of only theoretical interest to reveal thecontrol strategies proposed in this paper Another point tonote is that themaximumcontrol is very difficult to achieve inreality so we will omit the situation of the maximum controlduring the series of simulations
Next we will make some necessary instructions andexplanations to the above simulation graphs Figures 5 67 and 8 depict the number of four compartments underdifferent control levels when we choose the weight coeffi-cients in objective function to be 119888
1= 102 1198882= 104 From
the four simulation graphs we can observe the followingsimple facts in reducing the total number of alcoholismsand increasing the number of susceptibles the effectivenessof various control measures is as follows optimal control isevidently better than middle control and middle control isbetter than single control 119906
2 single control 119906
2is better than
single control 1199061 while single control 119906
1is much better than
no controlFigures 9 and 10 depict the optimal control law of 119906
1 1199062
respectively In the beginning of the simulation the control
14 Abstract and Applied Analysis
0 10 20 30 40 504500
5000
5500
6000
6500
7000
S(n
umbe
r of s
usce
ptib
les)
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 5Number of the susceptibles whenwe choose theweights inobjective function are 119888
1= 102 1198882= 104 and under different optimal
control strategies that is (1) without control 1199061= 1199062= 0 (2) single
control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0 119906
2= 05 (4)
middle control 1199061= 1199062= 05 (5) optimal control 119906
1= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
500
1000
1500
A(n
umbe
r of t
he al
coho
lism
)
t (unit week)
Figure 6 Number of the alcoholics when we choose the weights inobjective function are 119888
1= 102 1198882= 104 and under different optimal
control strategies that is (1) without control 1199061= 1199062= 0 (2) single
control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0 119906
2= 05 (4)
middle control 1199061= 1199062= 05 (5) optimal control 119906
1= 119906lowast
1 1199062= 119906lowast
2
0
50
100
150
200
250
300
350
400
T(n
umbe
r of t
reat
men
t)
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 7 Number of the persons in treatment when we choose theweights in objective function are 119888
1= 10
2 1198882= 10
4 and underdifferent optimal control strategies that is (1) without control 119906
1=
1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control 119906
1= 0
1199062= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal control
1199061= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
50
100
150
200
250
300
350
400
Q(n
umbe
r of q
uitti
ng)
Figure 8 Number of people in quitting compartment when wechoose the weights in objective function are 119888
1= 102 1198882= 104 and
under different optimal control strategies that is (1) without control1199061= 1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control
1199061= 0 119906
2= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal
control 1199061= 119906lowast
1 1199062= 119906lowast
2
Abstract and Applied Analysis 15
0 10 20 30 40 50t (unit week)
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
Optimal control of u1
Figure 9 Figures of the optimal control 1199061when we choose the
weight in objective function are 1198881= 102 1198882= 104
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
0 10 20 30 40 50t (unit week)
Optimal control of u2
Figure 10 Figures of the optimal control 1199062when we choose the
weight in objective function are 1198881= 102 1198882= 104
effort of 1199061should be decreased from 065 to 05 within the
first month and over the next week it should be increased tothe maximum control until 50 weeks then rapidly decreasedto 0 at the end of the simulation As for the control 119906
2 it
should start from around 05 due to the initial alcoholics thenincrease to 055 within one week since the rapid infection andnext decrease to almost 0 since the effectiveness of treatmentin three weeks but with the infection going on the controleffort of 119906
2should gradually increase to the maximum and
maintain this level until the tenth week for the purpose ofconsolidation therapy and preventing rebound hereafter itshould be gradually decreased to the level of almost 043 untilthe fifty weeks then quickly decreased to 0 in the end
0 10 20 30 40 504500
5000
5500
6000
6500
7000
S(n
umbe
r of s
usce
ptib
les)
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 11 Number of the susceptibles when we choose the weightsin objective function are 119888
1= 104 1198882= 102 and under different
optimal control strategies that is (1) without control 1199061= 1199062= 0
(2) single control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0
1199062= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal control
1199061= 119906lowast
1 1199062= 119906lowast
2
In order to investigate the influence of different weightcoefficients in the objective functional on the effect ofcontrolling at the same time for a better comparison we willchange the weight coefficients in objective function into 119888
2=
104 1198881= 102 and we will list the corresponding numerical
simulation results as Figures 11ndash16 showWhen we change the weight coefficients in objective
function into 1198881= 104 1198882= 102 we find that the results
of simulations derived from the graphs are very similar tothe ones before We speculate that the most likely reasons ofthis result are due to three respects one is that the weightcoefficients are not too sensitive in the numerical simulationand another possible reason is that both of the two controlsare important in some sense equivalently importantThe lastbut not themost unlikely reason is that we have not found themost appropriate weight coefficients in the simulation whichis very difficult to find as previously mentioned
7 Conclusions
In this paper we formulate an alcoholics quitting modeland firstly investigate the variation discipline of variouspopulations from the perspective of global stability thenwe propose an objective functional to examine two differentcontrolmeasures (ie prevention and treatment) on the effectof alcohol The basic reproduction number of the model wasderived and the global stability of the two equilibria is givenFrom the expression of the basic reproductionnumber119877
0and
16 Abstract and Applied Analysis
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
500
1000
1500
A(n
umbe
r of t
he al
coho
lism
)
t (unit week)
Figure 12 Number of the alcoholics when we choose the weights inobjective function are 119888
1= 104 1198882= 102 and under different optimal
control strategies that is (1) without control 1199061= 1199062= 0 (2) single
control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0 119906
2= 05 (4)
middle control 1199061= 1199062= 05 (5) optimal control 119906
1= 119906lowast
1 1199062= 119906lowast
2
0
50
100
150
200
250
300
350
400
T(n
umbe
r of t
reat
men
t)
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 13 Number of the persons in treatment when we choosethe weights in objective function are 119888
1= 104 1198882= 102 and under
different optimal control strategies that is (1) without control 1199061=
1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control 119906
1= 0
1199062= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal control
1199061= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
50
100
150
200
250
300
350
400
Q(n
umbe
r of q
uitti
ng)
Figure 14 Number of people in quitting compartment when wechoose the weights in objective function are 119888
1= 104 1198882= 102 and
under different optimal control strategies that is (1) without control1199061= 1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control
1199061= 0 119906
2= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal
control 1199061= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50t (unit week)
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
Optimal control of u1
Figure 15 Figures of the optimal control 1199061when we choose the
weight in objective function are 1198881= 104 1198882= 102
related numerical simulation we can easily see that the twocontrol strategies are effective in the alcoholics process
Using Pontryaginrsquos Maximum Principle we firstly deter-mine the necessary conditions for existence of optimalcontrol pairsThe uniqueness of the solution to the optimalitysystem (71) is derived by the classical method of contradic-tion Numerical simulations of the model suggest that thetwo different groups of weights in the objective function have
Abstract and Applied Analysis 17
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
0 10 20 30 40 50t (unit week)
Optimal control of u2
Figure 16 Figures of the optimal control 1199062when we choose the
weight in objective function are 1198881= 104 1198882= 102
much similar effects on the transmission of the alcoholismfrom this point the two control measures are almost equallyimportant in controlling the alcoholism although they willprobably have great influences on the cost of the objectivefunction From the simulation figures it seems that the effectof optimal control which is measured by the reduction inthe number of alcoholics and the increase in the number ofsusceptibles is much better than other control strategies asnoted earlier in the simulation section According to the real-time curve of two optimal controls we point out the specificimplementation methods of optimal control which can beachieved in practice
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
This work was partially supported by the NSF ofGansu Province of China (2013GS09485) the NNSF ofChina (10961018) the NSF of Gansu Province of China(1107RJZA088) the NSF for Distinguished Young Scholarsof Gansu Province of China (1111RJDA003) the Special Fundfor the Basic Requirements in the Research of University ofGansu Province of China and the Development Programfor Hong Liu Distinguished Young Scholars in LanzhouUniversity of Technology
References
[1] R Room T Babor and J Rehm ldquoAlcohol and public healthrdquoThe Lancet vol 365 no 9458 pp 519ndash530 2005
[2] J B Saunders O G Aasland A Amundsen and M GrantldquoAlcohol consumption and related problems among primary
health care patients WHO collaborative project on early detec-tion of personswith harmful alcohol consumption IrdquoAddictionvol 88 no 3 pp 349ndash362 1993
[3] L D Johnston P M OMalley and J G Bachman NationalSurvey Results on Drug Use From the Monitoring the FutureStudy 1975ndash1992 National Institute on Drug Abuse RockvilleMd USA 1993
[4] H Wechsler J E Lee M Kuo and H Lee ldquoCollege bingedrinking in the 1990s a continuing problem Results of theHarvard School of Public Health 1999 College Alcohol StudyrdquoJournal of American College Health vol 48 no 5 pp 199ndash2102000
[5] P M OrsquoMalley and L D Johnston ldquoEpidemiology of alcoholand other drug use among American college studentsrdquo Journalof Studies on Alcohol vol 63 no 14 pp 23ndash39 2002
[6] K Agarwal-Kozlowski and D P Agarwal ldquoGenetic predisposi-tion to alcoholismrdquo Therapeutische Umschau vol 57 no 4 pp179ndash184 2000
[7] C Y Chen C L Storr and J C Anthony ldquoEarly-onset drug useand risk for drug dependence problemsrdquo Addictive Behaviorsvol 34 no 3 pp 319ndash322 2009
[8] M Glavas and J Weinberg ldquoStress alcohol consumption andthe hypothalamic-pituitary adrenal axisrdquo in Nutrients Stressand Medical Disorders pp 165ndash183 Humana Press New YorkNY USA 2006
[9] L N Blum N H Nielsen J A Riggs and L B BresolinldquoAlcoholism and alcohol abuse among women report of theCouncil on Scientific Affairsrdquo Journal of Womenrsquos Health vol7 no 7 pp 861ndash871 1998
[10] B Benedict ldquoModeling alcoholism as a contagious disease howldquoinfectedrdquo drinking buddies spread problem drinkingrdquo SIAMNews vol 40 no 3 2007
[11] H Walter K Gutierrez K Ramskogler I Hertling A Dvorakand O M Lesch ldquoGender-specific differences in alcoholismImplications for treatmentrdquo Archives of Womenrsquos Mental Healthvol 6 no 4 pp 253ndash258 2003
[12] D Muller R D Koch H von Specht w Volker and E MMunch ldquoNeurophisiologic findings in chronic alcohol abuserdquoPsychiatrie Neurologie und Medizinische Psychologie vol 37no 3 pp 129ndash132 1985
[13] G Testino ldquoAlcoholic diseases in hepato-gastroenterology apoint of viewrdquo Hepato-Gastroenterology vol 55 no 82-83 pp371ndash377 2008
[14] A Mubayi P E Greenwood C Castillo-Chavez P J Grue-newald and D M Gorman ldquoThe impact of relative residencetimes on the distribution of heavy drinkers in highly distinctenvironmentsrdquo Socio-Economic Planning Sciences vol 44 no 1pp 45ndash56 2010
[15] D Muller R D Koch H von Specht W Volker and E MMunch ldquoNeurophysiologic findings in chronic alcohol abuserdquoPsychiatrie Neurologie Und Medizinische Psychologie vol 37no 3 pp 129ndash132 1985 (Leipz)
[16] G Mulone and B Straughan ldquoModeling binge drinkingrdquoInternational Journal of Biomathematics vol 5 no 1 Article ID1250005 14 pages 2012
[17] H F Huo and N N Song ldquoGlobal stability for a binge drinkingmodel with two stagesrdquo Discrete Dynamics in Nature andSociety vol 2012 Article ID 829386 15 pages 2012
[18] S S Mushayabasa and C P Bhunu ldquoModelling the effects ofheavy alcohol consumption on the transmission dynamics ofgonorrheardquo Nonlinear Dynamics vol 66 no 4 pp 695ndash7062011
18 Abstract and Applied Analysis
[19] F S Sancheznchez XWang C Castillo-Chvez DM Gormanand P J Gruenewald ldquoDrinking as an epidemic a simplemathematical model with recovery and relapserdquo in TherapistsGuide to Evidence-Based Relapse Prevention K Witkiewitz andG Marlatt Eds Academic Press New York NY USA 2007
[20] H-F Huo and C-C Zhu ldquoInfluence of relapse in a givingup smoking modelrdquo Abstract and Applied Analysis vol 2013Article ID 525461 12 pages 2013
[21] H F Huo and Q Wang ldquoThe effects of awareness programs by17 media on the drinking dynamicsrdquo Submitted
[22] S Lee E Jung and C Castillo-Chavez ldquoOptimal controlintervention strategies in low- and high-risk problem drinkingpopulationsrdquo Socio-Economic Planning Sciences vol 44 no 4pp 258ndash265 2010
[23] K R Fister S Lenhart and J S McNally ldquoOptimizingchemotherapy in an HIV modelrdquo Electronic Journal of Differ-ential Equations vol 1998 no 32 pp 1ndash12 1998
[24] S Lenhart and J T Workman Optimal Control Applied toBiological Models Chapman amp HallCRC Mathematical andComputational Biology Series Chapman amp HallCRC BocaRaton Fla USA 2007
[25] X Yang L Chen and J Chen ldquoPermanence and positiveperiodic solution for the single-species nonautonomous delaydiffusive modelsrdquo Computers amp Mathematics with ApplicationsAn International Journal vol 32 no 4 pp 109ndash116 1996
[26] P van denDriessche and JWatmough ldquoReproduction numbersand sub-threshold endemic equilibria for compartmental mod-els of disease transmissionrdquoMathematical Biosciences vol 180pp 29ndash48 2002
[27] V Lakshmikantham S Leela and A A Martynyuk StabilityAnalysis of Nonlinear Systems vol 125 Marcel Dekker NewYork NY USA 1989
[28] J P LaSalle The Stability of Dynamical Systems vol 25 ofRegional Conference Series in Applied Mathematics Society forIndustrial and Applied Mathematics 1976
[29] W H Fleming and R W Rishel Deterministic and StochasticOptimal Control vol 1 ofApplications of Mathematics SpringerBerlin Germany 1975
[30] D L Lukes Differential Equations Classical to ControlledMathematics in Science and Engineering Academic Press NewYork NY USA 1982
[31] L S Pontryagin V G Boltyanskii R V Gamkrelidze and EF Mishchenko The Mathematical Theory of Optimal ProcessesJohn Wiley amp Sons New Jersey NJ USA 1962
[32] W Hackbusch ldquoA numerical method for solving parabolicequations with opposite orientationsrdquo Computing vol 20 no3 pp 229ndash240 1978
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
14 Abstract and Applied Analysis
0 10 20 30 40 504500
5000
5500
6000
6500
7000
S(n
umbe
r of s
usce
ptib
les)
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 5Number of the susceptibles whenwe choose theweights inobjective function are 119888
1= 102 1198882= 104 and under different optimal
control strategies that is (1) without control 1199061= 1199062= 0 (2) single
control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0 119906
2= 05 (4)
middle control 1199061= 1199062= 05 (5) optimal control 119906
1= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
500
1000
1500
A(n
umbe
r of t
he al
coho
lism
)
t (unit week)
Figure 6 Number of the alcoholics when we choose the weights inobjective function are 119888
1= 102 1198882= 104 and under different optimal
control strategies that is (1) without control 1199061= 1199062= 0 (2) single
control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0 119906
2= 05 (4)
middle control 1199061= 1199062= 05 (5) optimal control 119906
1= 119906lowast
1 1199062= 119906lowast
2
0
50
100
150
200
250
300
350
400
T(n
umbe
r of t
reat
men
t)
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 7 Number of the persons in treatment when we choose theweights in objective function are 119888
1= 10
2 1198882= 10
4 and underdifferent optimal control strategies that is (1) without control 119906
1=
1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control 119906
1= 0
1199062= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal control
1199061= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
50
100
150
200
250
300
350
400
Q(n
umbe
r of q
uitti
ng)
Figure 8 Number of people in quitting compartment when wechoose the weights in objective function are 119888
1= 102 1198882= 104 and
under different optimal control strategies that is (1) without control1199061= 1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control
1199061= 0 119906
2= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal
control 1199061= 119906lowast
1 1199062= 119906lowast
2
Abstract and Applied Analysis 15
0 10 20 30 40 50t (unit week)
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
Optimal control of u1
Figure 9 Figures of the optimal control 1199061when we choose the
weight in objective function are 1198881= 102 1198882= 104
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
0 10 20 30 40 50t (unit week)
Optimal control of u2
Figure 10 Figures of the optimal control 1199062when we choose the
weight in objective function are 1198881= 102 1198882= 104
effort of 1199061should be decreased from 065 to 05 within the
first month and over the next week it should be increased tothe maximum control until 50 weeks then rapidly decreasedto 0 at the end of the simulation As for the control 119906
2 it
should start from around 05 due to the initial alcoholics thenincrease to 055 within one week since the rapid infection andnext decrease to almost 0 since the effectiveness of treatmentin three weeks but with the infection going on the controleffort of 119906
2should gradually increase to the maximum and
maintain this level until the tenth week for the purpose ofconsolidation therapy and preventing rebound hereafter itshould be gradually decreased to the level of almost 043 untilthe fifty weeks then quickly decreased to 0 in the end
0 10 20 30 40 504500
5000
5500
6000
6500
7000
S(n
umbe
r of s
usce
ptib
les)
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 11 Number of the susceptibles when we choose the weightsin objective function are 119888
1= 104 1198882= 102 and under different
optimal control strategies that is (1) without control 1199061= 1199062= 0
(2) single control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0
1199062= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal control
1199061= 119906lowast
1 1199062= 119906lowast
2
In order to investigate the influence of different weightcoefficients in the objective functional on the effect ofcontrolling at the same time for a better comparison we willchange the weight coefficients in objective function into 119888
2=
104 1198881= 102 and we will list the corresponding numerical
simulation results as Figures 11ndash16 showWhen we change the weight coefficients in objective
function into 1198881= 104 1198882= 102 we find that the results
of simulations derived from the graphs are very similar tothe ones before We speculate that the most likely reasons ofthis result are due to three respects one is that the weightcoefficients are not too sensitive in the numerical simulationand another possible reason is that both of the two controlsare important in some sense equivalently importantThe lastbut not themost unlikely reason is that we have not found themost appropriate weight coefficients in the simulation whichis very difficult to find as previously mentioned
7 Conclusions
In this paper we formulate an alcoholics quitting modeland firstly investigate the variation discipline of variouspopulations from the perspective of global stability thenwe propose an objective functional to examine two differentcontrolmeasures (ie prevention and treatment) on the effectof alcohol The basic reproduction number of the model wasderived and the global stability of the two equilibria is givenFrom the expression of the basic reproductionnumber119877
0and
16 Abstract and Applied Analysis
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
500
1000
1500
A(n
umbe
r of t
he al
coho
lism
)
t (unit week)
Figure 12 Number of the alcoholics when we choose the weights inobjective function are 119888
1= 104 1198882= 102 and under different optimal
control strategies that is (1) without control 1199061= 1199062= 0 (2) single
control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0 119906
2= 05 (4)
middle control 1199061= 1199062= 05 (5) optimal control 119906
1= 119906lowast
1 1199062= 119906lowast
2
0
50
100
150
200
250
300
350
400
T(n
umbe
r of t
reat
men
t)
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 13 Number of the persons in treatment when we choosethe weights in objective function are 119888
1= 104 1198882= 102 and under
different optimal control strategies that is (1) without control 1199061=
1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control 119906
1= 0
1199062= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal control
1199061= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
50
100
150
200
250
300
350
400
Q(n
umbe
r of q
uitti
ng)
Figure 14 Number of people in quitting compartment when wechoose the weights in objective function are 119888
1= 104 1198882= 102 and
under different optimal control strategies that is (1) without control1199061= 1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control
1199061= 0 119906
2= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal
control 1199061= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50t (unit week)
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
Optimal control of u1
Figure 15 Figures of the optimal control 1199061when we choose the
weight in objective function are 1198881= 104 1198882= 102
related numerical simulation we can easily see that the twocontrol strategies are effective in the alcoholics process
Using Pontryaginrsquos Maximum Principle we firstly deter-mine the necessary conditions for existence of optimalcontrol pairsThe uniqueness of the solution to the optimalitysystem (71) is derived by the classical method of contradic-tion Numerical simulations of the model suggest that thetwo different groups of weights in the objective function have
Abstract and Applied Analysis 17
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
0 10 20 30 40 50t (unit week)
Optimal control of u2
Figure 16 Figures of the optimal control 1199062when we choose the
weight in objective function are 1198881= 104 1198882= 102
much similar effects on the transmission of the alcoholismfrom this point the two control measures are almost equallyimportant in controlling the alcoholism although they willprobably have great influences on the cost of the objectivefunction From the simulation figures it seems that the effectof optimal control which is measured by the reduction inthe number of alcoholics and the increase in the number ofsusceptibles is much better than other control strategies asnoted earlier in the simulation section According to the real-time curve of two optimal controls we point out the specificimplementation methods of optimal control which can beachieved in practice
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
This work was partially supported by the NSF ofGansu Province of China (2013GS09485) the NNSF ofChina (10961018) the NSF of Gansu Province of China(1107RJZA088) the NSF for Distinguished Young Scholarsof Gansu Province of China (1111RJDA003) the Special Fundfor the Basic Requirements in the Research of University ofGansu Province of China and the Development Programfor Hong Liu Distinguished Young Scholars in LanzhouUniversity of Technology
References
[1] R Room T Babor and J Rehm ldquoAlcohol and public healthrdquoThe Lancet vol 365 no 9458 pp 519ndash530 2005
[2] J B Saunders O G Aasland A Amundsen and M GrantldquoAlcohol consumption and related problems among primary
health care patients WHO collaborative project on early detec-tion of personswith harmful alcohol consumption IrdquoAddictionvol 88 no 3 pp 349ndash362 1993
[3] L D Johnston P M OMalley and J G Bachman NationalSurvey Results on Drug Use From the Monitoring the FutureStudy 1975ndash1992 National Institute on Drug Abuse RockvilleMd USA 1993
[4] H Wechsler J E Lee M Kuo and H Lee ldquoCollege bingedrinking in the 1990s a continuing problem Results of theHarvard School of Public Health 1999 College Alcohol StudyrdquoJournal of American College Health vol 48 no 5 pp 199ndash2102000
[5] P M OrsquoMalley and L D Johnston ldquoEpidemiology of alcoholand other drug use among American college studentsrdquo Journalof Studies on Alcohol vol 63 no 14 pp 23ndash39 2002
[6] K Agarwal-Kozlowski and D P Agarwal ldquoGenetic predisposi-tion to alcoholismrdquo Therapeutische Umschau vol 57 no 4 pp179ndash184 2000
[7] C Y Chen C L Storr and J C Anthony ldquoEarly-onset drug useand risk for drug dependence problemsrdquo Addictive Behaviorsvol 34 no 3 pp 319ndash322 2009
[8] M Glavas and J Weinberg ldquoStress alcohol consumption andthe hypothalamic-pituitary adrenal axisrdquo in Nutrients Stressand Medical Disorders pp 165ndash183 Humana Press New YorkNY USA 2006
[9] L N Blum N H Nielsen J A Riggs and L B BresolinldquoAlcoholism and alcohol abuse among women report of theCouncil on Scientific Affairsrdquo Journal of Womenrsquos Health vol7 no 7 pp 861ndash871 1998
[10] B Benedict ldquoModeling alcoholism as a contagious disease howldquoinfectedrdquo drinking buddies spread problem drinkingrdquo SIAMNews vol 40 no 3 2007
[11] H Walter K Gutierrez K Ramskogler I Hertling A Dvorakand O M Lesch ldquoGender-specific differences in alcoholismImplications for treatmentrdquo Archives of Womenrsquos Mental Healthvol 6 no 4 pp 253ndash258 2003
[12] D Muller R D Koch H von Specht w Volker and E MMunch ldquoNeurophisiologic findings in chronic alcohol abuserdquoPsychiatrie Neurologie und Medizinische Psychologie vol 37no 3 pp 129ndash132 1985
[13] G Testino ldquoAlcoholic diseases in hepato-gastroenterology apoint of viewrdquo Hepato-Gastroenterology vol 55 no 82-83 pp371ndash377 2008
[14] A Mubayi P E Greenwood C Castillo-Chavez P J Grue-newald and D M Gorman ldquoThe impact of relative residencetimes on the distribution of heavy drinkers in highly distinctenvironmentsrdquo Socio-Economic Planning Sciences vol 44 no 1pp 45ndash56 2010
[15] D Muller R D Koch H von Specht W Volker and E MMunch ldquoNeurophysiologic findings in chronic alcohol abuserdquoPsychiatrie Neurologie Und Medizinische Psychologie vol 37no 3 pp 129ndash132 1985 (Leipz)
[16] G Mulone and B Straughan ldquoModeling binge drinkingrdquoInternational Journal of Biomathematics vol 5 no 1 Article ID1250005 14 pages 2012
[17] H F Huo and N N Song ldquoGlobal stability for a binge drinkingmodel with two stagesrdquo Discrete Dynamics in Nature andSociety vol 2012 Article ID 829386 15 pages 2012
[18] S S Mushayabasa and C P Bhunu ldquoModelling the effects ofheavy alcohol consumption on the transmission dynamics ofgonorrheardquo Nonlinear Dynamics vol 66 no 4 pp 695ndash7062011
18 Abstract and Applied Analysis
[19] F S Sancheznchez XWang C Castillo-Chvez DM Gormanand P J Gruenewald ldquoDrinking as an epidemic a simplemathematical model with recovery and relapserdquo in TherapistsGuide to Evidence-Based Relapse Prevention K Witkiewitz andG Marlatt Eds Academic Press New York NY USA 2007
[20] H-F Huo and C-C Zhu ldquoInfluence of relapse in a givingup smoking modelrdquo Abstract and Applied Analysis vol 2013Article ID 525461 12 pages 2013
[21] H F Huo and Q Wang ldquoThe effects of awareness programs by17 media on the drinking dynamicsrdquo Submitted
[22] S Lee E Jung and C Castillo-Chavez ldquoOptimal controlintervention strategies in low- and high-risk problem drinkingpopulationsrdquo Socio-Economic Planning Sciences vol 44 no 4pp 258ndash265 2010
[23] K R Fister S Lenhart and J S McNally ldquoOptimizingchemotherapy in an HIV modelrdquo Electronic Journal of Differ-ential Equations vol 1998 no 32 pp 1ndash12 1998
[24] S Lenhart and J T Workman Optimal Control Applied toBiological Models Chapman amp HallCRC Mathematical andComputational Biology Series Chapman amp HallCRC BocaRaton Fla USA 2007
[25] X Yang L Chen and J Chen ldquoPermanence and positiveperiodic solution for the single-species nonautonomous delaydiffusive modelsrdquo Computers amp Mathematics with ApplicationsAn International Journal vol 32 no 4 pp 109ndash116 1996
[26] P van denDriessche and JWatmough ldquoReproduction numbersand sub-threshold endemic equilibria for compartmental mod-els of disease transmissionrdquoMathematical Biosciences vol 180pp 29ndash48 2002
[27] V Lakshmikantham S Leela and A A Martynyuk StabilityAnalysis of Nonlinear Systems vol 125 Marcel Dekker NewYork NY USA 1989
[28] J P LaSalle The Stability of Dynamical Systems vol 25 ofRegional Conference Series in Applied Mathematics Society forIndustrial and Applied Mathematics 1976
[29] W H Fleming and R W Rishel Deterministic and StochasticOptimal Control vol 1 ofApplications of Mathematics SpringerBerlin Germany 1975
[30] D L Lukes Differential Equations Classical to ControlledMathematics in Science and Engineering Academic Press NewYork NY USA 1982
[31] L S Pontryagin V G Boltyanskii R V Gamkrelidze and EF Mishchenko The Mathematical Theory of Optimal ProcessesJohn Wiley amp Sons New Jersey NJ USA 1962
[32] W Hackbusch ldquoA numerical method for solving parabolicequations with opposite orientationsrdquo Computing vol 20 no3 pp 229ndash240 1978
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 15
0 10 20 30 40 50t (unit week)
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
Optimal control of u1
Figure 9 Figures of the optimal control 1199061when we choose the
weight in objective function are 1198881= 102 1198882= 104
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
0 10 20 30 40 50t (unit week)
Optimal control of u2
Figure 10 Figures of the optimal control 1199062when we choose the
weight in objective function are 1198881= 102 1198882= 104
effort of 1199061should be decreased from 065 to 05 within the
first month and over the next week it should be increased tothe maximum control until 50 weeks then rapidly decreasedto 0 at the end of the simulation As for the control 119906
2 it
should start from around 05 due to the initial alcoholics thenincrease to 055 within one week since the rapid infection andnext decrease to almost 0 since the effectiveness of treatmentin three weeks but with the infection going on the controleffort of 119906
2should gradually increase to the maximum and
maintain this level until the tenth week for the purpose ofconsolidation therapy and preventing rebound hereafter itshould be gradually decreased to the level of almost 043 untilthe fifty weeks then quickly decreased to 0 in the end
0 10 20 30 40 504500
5000
5500
6000
6500
7000
S(n
umbe
r of s
usce
ptib
les)
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 11 Number of the susceptibles when we choose the weightsin objective function are 119888
1= 104 1198882= 102 and under different
optimal control strategies that is (1) without control 1199061= 1199062= 0
(2) single control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0
1199062= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal control
1199061= 119906lowast
1 1199062= 119906lowast
2
In order to investigate the influence of different weightcoefficients in the objective functional on the effect ofcontrolling at the same time for a better comparison we willchange the weight coefficients in objective function into 119888
2=
104 1198881= 102 and we will list the corresponding numerical
simulation results as Figures 11ndash16 showWhen we change the weight coefficients in objective
function into 1198881= 104 1198882= 102 we find that the results
of simulations derived from the graphs are very similar tothe ones before We speculate that the most likely reasons ofthis result are due to three respects one is that the weightcoefficients are not too sensitive in the numerical simulationand another possible reason is that both of the two controlsare important in some sense equivalently importantThe lastbut not themost unlikely reason is that we have not found themost appropriate weight coefficients in the simulation whichis very difficult to find as previously mentioned
7 Conclusions
In this paper we formulate an alcoholics quitting modeland firstly investigate the variation discipline of variouspopulations from the perspective of global stability thenwe propose an objective functional to examine two differentcontrolmeasures (ie prevention and treatment) on the effectof alcohol The basic reproduction number of the model wasderived and the global stability of the two equilibria is givenFrom the expression of the basic reproductionnumber119877
0and
16 Abstract and Applied Analysis
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
500
1000
1500
A(n
umbe
r of t
he al
coho
lism
)
t (unit week)
Figure 12 Number of the alcoholics when we choose the weights inobjective function are 119888
1= 104 1198882= 102 and under different optimal
control strategies that is (1) without control 1199061= 1199062= 0 (2) single
control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0 119906
2= 05 (4)
middle control 1199061= 1199062= 05 (5) optimal control 119906
1= 119906lowast
1 1199062= 119906lowast
2
0
50
100
150
200
250
300
350
400
T(n
umbe
r of t
reat
men
t)
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 13 Number of the persons in treatment when we choosethe weights in objective function are 119888
1= 104 1198882= 102 and under
different optimal control strategies that is (1) without control 1199061=
1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control 119906
1= 0
1199062= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal control
1199061= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
50
100
150
200
250
300
350
400
Q(n
umbe
r of q
uitti
ng)
Figure 14 Number of people in quitting compartment when wechoose the weights in objective function are 119888
1= 104 1198882= 102 and
under different optimal control strategies that is (1) without control1199061= 1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control
1199061= 0 119906
2= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal
control 1199061= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50t (unit week)
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
Optimal control of u1
Figure 15 Figures of the optimal control 1199061when we choose the
weight in objective function are 1198881= 104 1198882= 102
related numerical simulation we can easily see that the twocontrol strategies are effective in the alcoholics process
Using Pontryaginrsquos Maximum Principle we firstly deter-mine the necessary conditions for existence of optimalcontrol pairsThe uniqueness of the solution to the optimalitysystem (71) is derived by the classical method of contradic-tion Numerical simulations of the model suggest that thetwo different groups of weights in the objective function have
Abstract and Applied Analysis 17
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
0 10 20 30 40 50t (unit week)
Optimal control of u2
Figure 16 Figures of the optimal control 1199062when we choose the
weight in objective function are 1198881= 104 1198882= 102
much similar effects on the transmission of the alcoholismfrom this point the two control measures are almost equallyimportant in controlling the alcoholism although they willprobably have great influences on the cost of the objectivefunction From the simulation figures it seems that the effectof optimal control which is measured by the reduction inthe number of alcoholics and the increase in the number ofsusceptibles is much better than other control strategies asnoted earlier in the simulation section According to the real-time curve of two optimal controls we point out the specificimplementation methods of optimal control which can beachieved in practice
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
This work was partially supported by the NSF ofGansu Province of China (2013GS09485) the NNSF ofChina (10961018) the NSF of Gansu Province of China(1107RJZA088) the NSF for Distinguished Young Scholarsof Gansu Province of China (1111RJDA003) the Special Fundfor the Basic Requirements in the Research of University ofGansu Province of China and the Development Programfor Hong Liu Distinguished Young Scholars in LanzhouUniversity of Technology
References
[1] R Room T Babor and J Rehm ldquoAlcohol and public healthrdquoThe Lancet vol 365 no 9458 pp 519ndash530 2005
[2] J B Saunders O G Aasland A Amundsen and M GrantldquoAlcohol consumption and related problems among primary
health care patients WHO collaborative project on early detec-tion of personswith harmful alcohol consumption IrdquoAddictionvol 88 no 3 pp 349ndash362 1993
[3] L D Johnston P M OMalley and J G Bachman NationalSurvey Results on Drug Use From the Monitoring the FutureStudy 1975ndash1992 National Institute on Drug Abuse RockvilleMd USA 1993
[4] H Wechsler J E Lee M Kuo and H Lee ldquoCollege bingedrinking in the 1990s a continuing problem Results of theHarvard School of Public Health 1999 College Alcohol StudyrdquoJournal of American College Health vol 48 no 5 pp 199ndash2102000
[5] P M OrsquoMalley and L D Johnston ldquoEpidemiology of alcoholand other drug use among American college studentsrdquo Journalof Studies on Alcohol vol 63 no 14 pp 23ndash39 2002
[6] K Agarwal-Kozlowski and D P Agarwal ldquoGenetic predisposi-tion to alcoholismrdquo Therapeutische Umschau vol 57 no 4 pp179ndash184 2000
[7] C Y Chen C L Storr and J C Anthony ldquoEarly-onset drug useand risk for drug dependence problemsrdquo Addictive Behaviorsvol 34 no 3 pp 319ndash322 2009
[8] M Glavas and J Weinberg ldquoStress alcohol consumption andthe hypothalamic-pituitary adrenal axisrdquo in Nutrients Stressand Medical Disorders pp 165ndash183 Humana Press New YorkNY USA 2006
[9] L N Blum N H Nielsen J A Riggs and L B BresolinldquoAlcoholism and alcohol abuse among women report of theCouncil on Scientific Affairsrdquo Journal of Womenrsquos Health vol7 no 7 pp 861ndash871 1998
[10] B Benedict ldquoModeling alcoholism as a contagious disease howldquoinfectedrdquo drinking buddies spread problem drinkingrdquo SIAMNews vol 40 no 3 2007
[11] H Walter K Gutierrez K Ramskogler I Hertling A Dvorakand O M Lesch ldquoGender-specific differences in alcoholismImplications for treatmentrdquo Archives of Womenrsquos Mental Healthvol 6 no 4 pp 253ndash258 2003
[12] D Muller R D Koch H von Specht w Volker and E MMunch ldquoNeurophisiologic findings in chronic alcohol abuserdquoPsychiatrie Neurologie und Medizinische Psychologie vol 37no 3 pp 129ndash132 1985
[13] G Testino ldquoAlcoholic diseases in hepato-gastroenterology apoint of viewrdquo Hepato-Gastroenterology vol 55 no 82-83 pp371ndash377 2008
[14] A Mubayi P E Greenwood C Castillo-Chavez P J Grue-newald and D M Gorman ldquoThe impact of relative residencetimes on the distribution of heavy drinkers in highly distinctenvironmentsrdquo Socio-Economic Planning Sciences vol 44 no 1pp 45ndash56 2010
[15] D Muller R D Koch H von Specht W Volker and E MMunch ldquoNeurophysiologic findings in chronic alcohol abuserdquoPsychiatrie Neurologie Und Medizinische Psychologie vol 37no 3 pp 129ndash132 1985 (Leipz)
[16] G Mulone and B Straughan ldquoModeling binge drinkingrdquoInternational Journal of Biomathematics vol 5 no 1 Article ID1250005 14 pages 2012
[17] H F Huo and N N Song ldquoGlobal stability for a binge drinkingmodel with two stagesrdquo Discrete Dynamics in Nature andSociety vol 2012 Article ID 829386 15 pages 2012
[18] S S Mushayabasa and C P Bhunu ldquoModelling the effects ofheavy alcohol consumption on the transmission dynamics ofgonorrheardquo Nonlinear Dynamics vol 66 no 4 pp 695ndash7062011
18 Abstract and Applied Analysis
[19] F S Sancheznchez XWang C Castillo-Chvez DM Gormanand P J Gruenewald ldquoDrinking as an epidemic a simplemathematical model with recovery and relapserdquo in TherapistsGuide to Evidence-Based Relapse Prevention K Witkiewitz andG Marlatt Eds Academic Press New York NY USA 2007
[20] H-F Huo and C-C Zhu ldquoInfluence of relapse in a givingup smoking modelrdquo Abstract and Applied Analysis vol 2013Article ID 525461 12 pages 2013
[21] H F Huo and Q Wang ldquoThe effects of awareness programs by17 media on the drinking dynamicsrdquo Submitted
[22] S Lee E Jung and C Castillo-Chavez ldquoOptimal controlintervention strategies in low- and high-risk problem drinkingpopulationsrdquo Socio-Economic Planning Sciences vol 44 no 4pp 258ndash265 2010
[23] K R Fister S Lenhart and J S McNally ldquoOptimizingchemotherapy in an HIV modelrdquo Electronic Journal of Differ-ential Equations vol 1998 no 32 pp 1ndash12 1998
[24] S Lenhart and J T Workman Optimal Control Applied toBiological Models Chapman amp HallCRC Mathematical andComputational Biology Series Chapman amp HallCRC BocaRaton Fla USA 2007
[25] X Yang L Chen and J Chen ldquoPermanence and positiveperiodic solution for the single-species nonautonomous delaydiffusive modelsrdquo Computers amp Mathematics with ApplicationsAn International Journal vol 32 no 4 pp 109ndash116 1996
[26] P van denDriessche and JWatmough ldquoReproduction numbersand sub-threshold endemic equilibria for compartmental mod-els of disease transmissionrdquoMathematical Biosciences vol 180pp 29ndash48 2002
[27] V Lakshmikantham S Leela and A A Martynyuk StabilityAnalysis of Nonlinear Systems vol 125 Marcel Dekker NewYork NY USA 1989
[28] J P LaSalle The Stability of Dynamical Systems vol 25 ofRegional Conference Series in Applied Mathematics Society forIndustrial and Applied Mathematics 1976
[29] W H Fleming and R W Rishel Deterministic and StochasticOptimal Control vol 1 ofApplications of Mathematics SpringerBerlin Germany 1975
[30] D L Lukes Differential Equations Classical to ControlledMathematics in Science and Engineering Academic Press NewYork NY USA 1982
[31] L S Pontryagin V G Boltyanskii R V Gamkrelidze and EF Mishchenko The Mathematical Theory of Optimal ProcessesJohn Wiley amp Sons New Jersey NJ USA 1962
[32] W Hackbusch ldquoA numerical method for solving parabolicequations with opposite orientationsrdquo Computing vol 20 no3 pp 229ndash240 1978
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
16 Abstract and Applied Analysis
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
500
1000
1500
A(n
umbe
r of t
he al
coho
lism
)
t (unit week)
Figure 12 Number of the alcoholics when we choose the weights inobjective function are 119888
1= 104 1198882= 102 and under different optimal
control strategies that is (1) without control 1199061= 1199062= 0 (2) single
control 1199061= 05 119906
2= 0 (3) single control 119906
1= 0 119906
2= 05 (4)
middle control 1199061= 1199062= 05 (5) optimal control 119906
1= 119906lowast
1 1199062= 119906lowast
2
0
50
100
150
200
250
300
350
400
T(n
umbe
r of t
reat
men
t)
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
Figure 13 Number of the persons in treatment when we choosethe weights in objective function are 119888
1= 104 1198882= 102 and under
different optimal control strategies that is (1) without control 1199061=
1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control 119906
1= 0
1199062= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal control
1199061= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50
Constant control measure u1 = 0 u2 = 0
Single control measure u1 = 05 u2 = 0
Single control measure u1 = 0 u2 = 05
Middle control measure u1 = u2 = 05
Optimal control
0
50
100
150
200
250
300
350
400
Q(n
umbe
r of q
uitti
ng)
Figure 14 Number of people in quitting compartment when wechoose the weights in objective function are 119888
1= 104 1198882= 102 and
under different optimal control strategies that is (1) without control1199061= 1199062= 0 (2) single control 119906
1= 05 119906
2= 0 (3) single control
1199061= 0 119906
2= 05 (4) middle control 119906
1= 1199062= 05 (5) optimal
control 1199061= 119906lowast
1 1199062= 119906lowast
2
0 10 20 30 40 50t (unit week)
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
Optimal control of u1
Figure 15 Figures of the optimal control 1199061when we choose the
weight in objective function are 1198881= 104 1198882= 102
related numerical simulation we can easily see that the twocontrol strategies are effective in the alcoholics process
Using Pontryaginrsquos Maximum Principle we firstly deter-mine the necessary conditions for existence of optimalcontrol pairsThe uniqueness of the solution to the optimalitysystem (71) is derived by the classical method of contradic-tion Numerical simulations of the model suggest that thetwo different groups of weights in the objective function have
Abstract and Applied Analysis 17
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
0 10 20 30 40 50t (unit week)
Optimal control of u2
Figure 16 Figures of the optimal control 1199062when we choose the
weight in objective function are 1198881= 104 1198882= 102
much similar effects on the transmission of the alcoholismfrom this point the two control measures are almost equallyimportant in controlling the alcoholism although they willprobably have great influences on the cost of the objectivefunction From the simulation figures it seems that the effectof optimal control which is measured by the reduction inthe number of alcoholics and the increase in the number ofsusceptibles is much better than other control strategies asnoted earlier in the simulation section According to the real-time curve of two optimal controls we point out the specificimplementation methods of optimal control which can beachieved in practice
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
This work was partially supported by the NSF ofGansu Province of China (2013GS09485) the NNSF ofChina (10961018) the NSF of Gansu Province of China(1107RJZA088) the NSF for Distinguished Young Scholarsof Gansu Province of China (1111RJDA003) the Special Fundfor the Basic Requirements in the Research of University ofGansu Province of China and the Development Programfor Hong Liu Distinguished Young Scholars in LanzhouUniversity of Technology
References
[1] R Room T Babor and J Rehm ldquoAlcohol and public healthrdquoThe Lancet vol 365 no 9458 pp 519ndash530 2005
[2] J B Saunders O G Aasland A Amundsen and M GrantldquoAlcohol consumption and related problems among primary
health care patients WHO collaborative project on early detec-tion of personswith harmful alcohol consumption IrdquoAddictionvol 88 no 3 pp 349ndash362 1993
[3] L D Johnston P M OMalley and J G Bachman NationalSurvey Results on Drug Use From the Monitoring the FutureStudy 1975ndash1992 National Institute on Drug Abuse RockvilleMd USA 1993
[4] H Wechsler J E Lee M Kuo and H Lee ldquoCollege bingedrinking in the 1990s a continuing problem Results of theHarvard School of Public Health 1999 College Alcohol StudyrdquoJournal of American College Health vol 48 no 5 pp 199ndash2102000
[5] P M OrsquoMalley and L D Johnston ldquoEpidemiology of alcoholand other drug use among American college studentsrdquo Journalof Studies on Alcohol vol 63 no 14 pp 23ndash39 2002
[6] K Agarwal-Kozlowski and D P Agarwal ldquoGenetic predisposi-tion to alcoholismrdquo Therapeutische Umschau vol 57 no 4 pp179ndash184 2000
[7] C Y Chen C L Storr and J C Anthony ldquoEarly-onset drug useand risk for drug dependence problemsrdquo Addictive Behaviorsvol 34 no 3 pp 319ndash322 2009
[8] M Glavas and J Weinberg ldquoStress alcohol consumption andthe hypothalamic-pituitary adrenal axisrdquo in Nutrients Stressand Medical Disorders pp 165ndash183 Humana Press New YorkNY USA 2006
[9] L N Blum N H Nielsen J A Riggs and L B BresolinldquoAlcoholism and alcohol abuse among women report of theCouncil on Scientific Affairsrdquo Journal of Womenrsquos Health vol7 no 7 pp 861ndash871 1998
[10] B Benedict ldquoModeling alcoholism as a contagious disease howldquoinfectedrdquo drinking buddies spread problem drinkingrdquo SIAMNews vol 40 no 3 2007
[11] H Walter K Gutierrez K Ramskogler I Hertling A Dvorakand O M Lesch ldquoGender-specific differences in alcoholismImplications for treatmentrdquo Archives of Womenrsquos Mental Healthvol 6 no 4 pp 253ndash258 2003
[12] D Muller R D Koch H von Specht w Volker and E MMunch ldquoNeurophisiologic findings in chronic alcohol abuserdquoPsychiatrie Neurologie und Medizinische Psychologie vol 37no 3 pp 129ndash132 1985
[13] G Testino ldquoAlcoholic diseases in hepato-gastroenterology apoint of viewrdquo Hepato-Gastroenterology vol 55 no 82-83 pp371ndash377 2008
[14] A Mubayi P E Greenwood C Castillo-Chavez P J Grue-newald and D M Gorman ldquoThe impact of relative residencetimes on the distribution of heavy drinkers in highly distinctenvironmentsrdquo Socio-Economic Planning Sciences vol 44 no 1pp 45ndash56 2010
[15] D Muller R D Koch H von Specht W Volker and E MMunch ldquoNeurophysiologic findings in chronic alcohol abuserdquoPsychiatrie Neurologie Und Medizinische Psychologie vol 37no 3 pp 129ndash132 1985 (Leipz)
[16] G Mulone and B Straughan ldquoModeling binge drinkingrdquoInternational Journal of Biomathematics vol 5 no 1 Article ID1250005 14 pages 2012
[17] H F Huo and N N Song ldquoGlobal stability for a binge drinkingmodel with two stagesrdquo Discrete Dynamics in Nature andSociety vol 2012 Article ID 829386 15 pages 2012
[18] S S Mushayabasa and C P Bhunu ldquoModelling the effects ofheavy alcohol consumption on the transmission dynamics ofgonorrheardquo Nonlinear Dynamics vol 66 no 4 pp 695ndash7062011
18 Abstract and Applied Analysis
[19] F S Sancheznchez XWang C Castillo-Chvez DM Gormanand P J Gruenewald ldquoDrinking as an epidemic a simplemathematical model with recovery and relapserdquo in TherapistsGuide to Evidence-Based Relapse Prevention K Witkiewitz andG Marlatt Eds Academic Press New York NY USA 2007
[20] H-F Huo and C-C Zhu ldquoInfluence of relapse in a givingup smoking modelrdquo Abstract and Applied Analysis vol 2013Article ID 525461 12 pages 2013
[21] H F Huo and Q Wang ldquoThe effects of awareness programs by17 media on the drinking dynamicsrdquo Submitted
[22] S Lee E Jung and C Castillo-Chavez ldquoOptimal controlintervention strategies in low- and high-risk problem drinkingpopulationsrdquo Socio-Economic Planning Sciences vol 44 no 4pp 258ndash265 2010
[23] K R Fister S Lenhart and J S McNally ldquoOptimizingchemotherapy in an HIV modelrdquo Electronic Journal of Differ-ential Equations vol 1998 no 32 pp 1ndash12 1998
[24] S Lenhart and J T Workman Optimal Control Applied toBiological Models Chapman amp HallCRC Mathematical andComputational Biology Series Chapman amp HallCRC BocaRaton Fla USA 2007
[25] X Yang L Chen and J Chen ldquoPermanence and positiveperiodic solution for the single-species nonautonomous delaydiffusive modelsrdquo Computers amp Mathematics with ApplicationsAn International Journal vol 32 no 4 pp 109ndash116 1996
[26] P van denDriessche and JWatmough ldquoReproduction numbersand sub-threshold endemic equilibria for compartmental mod-els of disease transmissionrdquoMathematical Biosciences vol 180pp 29ndash48 2002
[27] V Lakshmikantham S Leela and A A Martynyuk StabilityAnalysis of Nonlinear Systems vol 125 Marcel Dekker NewYork NY USA 1989
[28] J P LaSalle The Stability of Dynamical Systems vol 25 ofRegional Conference Series in Applied Mathematics Society forIndustrial and Applied Mathematics 1976
[29] W H Fleming and R W Rishel Deterministic and StochasticOptimal Control vol 1 ofApplications of Mathematics SpringerBerlin Germany 1975
[30] D L Lukes Differential Equations Classical to ControlledMathematics in Science and Engineering Academic Press NewYork NY USA 1982
[31] L S Pontryagin V G Boltyanskii R V Gamkrelidze and EF Mishchenko The Mathematical Theory of Optimal ProcessesJohn Wiley amp Sons New Jersey NJ USA 1962
[32] W Hackbusch ldquoA numerical method for solving parabolicequations with opposite orientationsrdquo Computing vol 20 no3 pp 229ndash240 1978
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 17
0
02
04
06
08
1
12
Deg
ree o
f the
cont
rol m
easu
re
0 10 20 30 40 50t (unit week)
Optimal control of u2
Figure 16 Figures of the optimal control 1199062when we choose the
weight in objective function are 1198881= 104 1198882= 102
much similar effects on the transmission of the alcoholismfrom this point the two control measures are almost equallyimportant in controlling the alcoholism although they willprobably have great influences on the cost of the objectivefunction From the simulation figures it seems that the effectof optimal control which is measured by the reduction inthe number of alcoholics and the increase in the number ofsusceptibles is much better than other control strategies asnoted earlier in the simulation section According to the real-time curve of two optimal controls we point out the specificimplementation methods of optimal control which can beachieved in practice
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
This work was partially supported by the NSF ofGansu Province of China (2013GS09485) the NNSF ofChina (10961018) the NSF of Gansu Province of China(1107RJZA088) the NSF for Distinguished Young Scholarsof Gansu Province of China (1111RJDA003) the Special Fundfor the Basic Requirements in the Research of University ofGansu Province of China and the Development Programfor Hong Liu Distinguished Young Scholars in LanzhouUniversity of Technology
References
[1] R Room T Babor and J Rehm ldquoAlcohol and public healthrdquoThe Lancet vol 365 no 9458 pp 519ndash530 2005
[2] J B Saunders O G Aasland A Amundsen and M GrantldquoAlcohol consumption and related problems among primary
health care patients WHO collaborative project on early detec-tion of personswith harmful alcohol consumption IrdquoAddictionvol 88 no 3 pp 349ndash362 1993
[3] L D Johnston P M OMalley and J G Bachman NationalSurvey Results on Drug Use From the Monitoring the FutureStudy 1975ndash1992 National Institute on Drug Abuse RockvilleMd USA 1993
[4] H Wechsler J E Lee M Kuo and H Lee ldquoCollege bingedrinking in the 1990s a continuing problem Results of theHarvard School of Public Health 1999 College Alcohol StudyrdquoJournal of American College Health vol 48 no 5 pp 199ndash2102000
[5] P M OrsquoMalley and L D Johnston ldquoEpidemiology of alcoholand other drug use among American college studentsrdquo Journalof Studies on Alcohol vol 63 no 14 pp 23ndash39 2002
[6] K Agarwal-Kozlowski and D P Agarwal ldquoGenetic predisposi-tion to alcoholismrdquo Therapeutische Umschau vol 57 no 4 pp179ndash184 2000
[7] C Y Chen C L Storr and J C Anthony ldquoEarly-onset drug useand risk for drug dependence problemsrdquo Addictive Behaviorsvol 34 no 3 pp 319ndash322 2009
[8] M Glavas and J Weinberg ldquoStress alcohol consumption andthe hypothalamic-pituitary adrenal axisrdquo in Nutrients Stressand Medical Disorders pp 165ndash183 Humana Press New YorkNY USA 2006
[9] L N Blum N H Nielsen J A Riggs and L B BresolinldquoAlcoholism and alcohol abuse among women report of theCouncil on Scientific Affairsrdquo Journal of Womenrsquos Health vol7 no 7 pp 861ndash871 1998
[10] B Benedict ldquoModeling alcoholism as a contagious disease howldquoinfectedrdquo drinking buddies spread problem drinkingrdquo SIAMNews vol 40 no 3 2007
[11] H Walter K Gutierrez K Ramskogler I Hertling A Dvorakand O M Lesch ldquoGender-specific differences in alcoholismImplications for treatmentrdquo Archives of Womenrsquos Mental Healthvol 6 no 4 pp 253ndash258 2003
[12] D Muller R D Koch H von Specht w Volker and E MMunch ldquoNeurophisiologic findings in chronic alcohol abuserdquoPsychiatrie Neurologie und Medizinische Psychologie vol 37no 3 pp 129ndash132 1985
[13] G Testino ldquoAlcoholic diseases in hepato-gastroenterology apoint of viewrdquo Hepato-Gastroenterology vol 55 no 82-83 pp371ndash377 2008
[14] A Mubayi P E Greenwood C Castillo-Chavez P J Grue-newald and D M Gorman ldquoThe impact of relative residencetimes on the distribution of heavy drinkers in highly distinctenvironmentsrdquo Socio-Economic Planning Sciences vol 44 no 1pp 45ndash56 2010
[15] D Muller R D Koch H von Specht W Volker and E MMunch ldquoNeurophysiologic findings in chronic alcohol abuserdquoPsychiatrie Neurologie Und Medizinische Psychologie vol 37no 3 pp 129ndash132 1985 (Leipz)
[16] G Mulone and B Straughan ldquoModeling binge drinkingrdquoInternational Journal of Biomathematics vol 5 no 1 Article ID1250005 14 pages 2012
[17] H F Huo and N N Song ldquoGlobal stability for a binge drinkingmodel with two stagesrdquo Discrete Dynamics in Nature andSociety vol 2012 Article ID 829386 15 pages 2012
[18] S S Mushayabasa and C P Bhunu ldquoModelling the effects ofheavy alcohol consumption on the transmission dynamics ofgonorrheardquo Nonlinear Dynamics vol 66 no 4 pp 695ndash7062011
18 Abstract and Applied Analysis
[19] F S Sancheznchez XWang C Castillo-Chvez DM Gormanand P J Gruenewald ldquoDrinking as an epidemic a simplemathematical model with recovery and relapserdquo in TherapistsGuide to Evidence-Based Relapse Prevention K Witkiewitz andG Marlatt Eds Academic Press New York NY USA 2007
[20] H-F Huo and C-C Zhu ldquoInfluence of relapse in a givingup smoking modelrdquo Abstract and Applied Analysis vol 2013Article ID 525461 12 pages 2013
[21] H F Huo and Q Wang ldquoThe effects of awareness programs by17 media on the drinking dynamicsrdquo Submitted
[22] S Lee E Jung and C Castillo-Chavez ldquoOptimal controlintervention strategies in low- and high-risk problem drinkingpopulationsrdquo Socio-Economic Planning Sciences vol 44 no 4pp 258ndash265 2010
[23] K R Fister S Lenhart and J S McNally ldquoOptimizingchemotherapy in an HIV modelrdquo Electronic Journal of Differ-ential Equations vol 1998 no 32 pp 1ndash12 1998
[24] S Lenhart and J T Workman Optimal Control Applied toBiological Models Chapman amp HallCRC Mathematical andComputational Biology Series Chapman amp HallCRC BocaRaton Fla USA 2007
[25] X Yang L Chen and J Chen ldquoPermanence and positiveperiodic solution for the single-species nonautonomous delaydiffusive modelsrdquo Computers amp Mathematics with ApplicationsAn International Journal vol 32 no 4 pp 109ndash116 1996
[26] P van denDriessche and JWatmough ldquoReproduction numbersand sub-threshold endemic equilibria for compartmental mod-els of disease transmissionrdquoMathematical Biosciences vol 180pp 29ndash48 2002
[27] V Lakshmikantham S Leela and A A Martynyuk StabilityAnalysis of Nonlinear Systems vol 125 Marcel Dekker NewYork NY USA 1989
[28] J P LaSalle The Stability of Dynamical Systems vol 25 ofRegional Conference Series in Applied Mathematics Society forIndustrial and Applied Mathematics 1976
[29] W H Fleming and R W Rishel Deterministic and StochasticOptimal Control vol 1 ofApplications of Mathematics SpringerBerlin Germany 1975
[30] D L Lukes Differential Equations Classical to ControlledMathematics in Science and Engineering Academic Press NewYork NY USA 1982
[31] L S Pontryagin V G Boltyanskii R V Gamkrelidze and EF Mishchenko The Mathematical Theory of Optimal ProcessesJohn Wiley amp Sons New Jersey NJ USA 1962
[32] W Hackbusch ldquoA numerical method for solving parabolicequations with opposite orientationsrdquo Computing vol 20 no3 pp 229ndash240 1978
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
18 Abstract and Applied Analysis
[19] F S Sancheznchez XWang C Castillo-Chvez DM Gormanand P J Gruenewald ldquoDrinking as an epidemic a simplemathematical model with recovery and relapserdquo in TherapistsGuide to Evidence-Based Relapse Prevention K Witkiewitz andG Marlatt Eds Academic Press New York NY USA 2007
[20] H-F Huo and C-C Zhu ldquoInfluence of relapse in a givingup smoking modelrdquo Abstract and Applied Analysis vol 2013Article ID 525461 12 pages 2013
[21] H F Huo and Q Wang ldquoThe effects of awareness programs by17 media on the drinking dynamicsrdquo Submitted
[22] S Lee E Jung and C Castillo-Chavez ldquoOptimal controlintervention strategies in low- and high-risk problem drinkingpopulationsrdquo Socio-Economic Planning Sciences vol 44 no 4pp 258ndash265 2010
[23] K R Fister S Lenhart and J S McNally ldquoOptimizingchemotherapy in an HIV modelrdquo Electronic Journal of Differ-ential Equations vol 1998 no 32 pp 1ndash12 1998
[24] S Lenhart and J T Workman Optimal Control Applied toBiological Models Chapman amp HallCRC Mathematical andComputational Biology Series Chapman amp HallCRC BocaRaton Fla USA 2007
[25] X Yang L Chen and J Chen ldquoPermanence and positiveperiodic solution for the single-species nonautonomous delaydiffusive modelsrdquo Computers amp Mathematics with ApplicationsAn International Journal vol 32 no 4 pp 109ndash116 1996
[26] P van denDriessche and JWatmough ldquoReproduction numbersand sub-threshold endemic equilibria for compartmental mod-els of disease transmissionrdquoMathematical Biosciences vol 180pp 29ndash48 2002
[27] V Lakshmikantham S Leela and A A Martynyuk StabilityAnalysis of Nonlinear Systems vol 125 Marcel Dekker NewYork NY USA 1989
[28] J P LaSalle The Stability of Dynamical Systems vol 25 ofRegional Conference Series in Applied Mathematics Society forIndustrial and Applied Mathematics 1976
[29] W H Fleming and R W Rishel Deterministic and StochasticOptimal Control vol 1 ofApplications of Mathematics SpringerBerlin Germany 1975
[30] D L Lukes Differential Equations Classical to ControlledMathematics in Science and Engineering Academic Press NewYork NY USA 1982
[31] L S Pontryagin V G Boltyanskii R V Gamkrelidze and EF Mishchenko The Mathematical Theory of Optimal ProcessesJohn Wiley amp Sons New Jersey NJ USA 1962
[32] W Hackbusch ldquoA numerical method for solving parabolicequations with opposite orientationsrdquo Computing vol 20 no3 pp 229ndash240 1978
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of