Hindawi Publishing CorporationInternational Journal of Differential EquationsVolume 2013 Article ID 297085 11 pageshttpdxdoiorg1011552013297085

Research ArticleOscillations of a Class of Forced Second-Order DifferentialEquations with Possible Discontinuous Coefficients

Siniša MiliIiT Mervan PašiT and Darko CubriniT

Department of Mathematics Faculty of Electrical Engeneering and Computing University of Zagreb 10000 Zagreb Croatia

Correspondence should be addressed to Mervan Pasic mervanpasicgmailcom

Received 29 January 2013 Accepted 16 April 2013

Academic Editor Ondrej Dosly

Copyright copy 2013 Sinisa Milicic et al This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

We study the oscillation of all solutions of a general class of forced second-order differential equations where their second derivativeis not necessarily a continuous function and the coefficients of the main equation may be discontinuous Our main results are notincluded in the previously published known oscillation criteria of interval type Many examples and consequences are presentedillustrating the main results

1 Introduction

Let 1199050gt 0 and let 119860119862loc([1199050infin)R) denote the set of all real

functions absolutely continuous on every bounded interval[119886 119887] sub [119905

0infin) We study the oscillatory behaviour of all

solutions 119909 = 119909(119905) of the following class of forced second-order differential equations

(119903 (119905) Φ (119909 (119905) 1199091015840(119905)))1015840

+ 119902 (119905) 119891 (119909 (119905)) = 119890 (119905)

ae in [1199050infin)

119909 119903Φ (119909 1199091015840) isin 119860119862loc ([1199050infin) R)

(1)

where the functions Φ R2 rarr R Φ = Φ(119906 V) 119891

R rarr R and 119891 = 119891(119906) satisfy some general conditionsgiven in Section 2 A continuous function 119909 = 119909(119905) is saidto be oscillatory if there is a sequence 119905

119899isin [1199050infin) such that

119909(119905119899) = 0 for all 119899 isin N and 119905

119899rarr infin as 119899 rarr infin A differential

equation is oscillatory if all its solutions are oscillatoryThe forcing term 119890(119905) is a sign-changing function (possi-

bly discontinuous) This can be formulated by the followinghypothesis for every 119879 ge 119905

0there exist two intervals (119886

1 1198871)

and (1198862 1198872) 119879 le 119886

1lt 1198871le 1198862lt 1198872 such that

119890 (119905) ge 0 119905 isin (1198861 1198871)

119890 (119905) le 0 119905 isin (1198862 1198872)

(2)

The coefficient 119902(119905) may be a discontinuous function on[1199050infin) and the case 119909 notin 119862

2((1199050infin)R) occurs in our

main results and examples too Two important classes offunctions Φ(119906 V) are included in the differential operator(119903(119905)Φ(119909 119909

1015840))1015840 as

Φ (119906 V) = 120601 (119906) V Φ (119906 V) =120601 (119906) V

radic1 + V2 (119906 V) isin R

2 (3)

The first one is the classic second-order differential operatorwhich is linear in 119909

1015840 and the second one is the so-calledone-dimensional mean curvature differential operator seeExamples 1 and 2

Depending on 119902(119905) we propose the following four simplemodels for (1)

(i) 119902(119905) is strictly positive and continuous on [1199050infin) as

11990910158401015840+ 41198982119891 (119909) = ℎ (sin (119898119905)) ae in [119905

0infin)

119909 1199091015840isin 119860119862loc ([1199050infin) R) 119909 notin 119862

2((1199050infin) R)

(4)

(ii) 119902(119905) is nonnegative and continuous on [1199050infin) as

11990910158401015840+ 11989821205872[cos (119898119905)]+119891 (119909) = ℎ (sin (119898119905))

ae in [1199050infin)

119909 1199091015840isin 119860119862loc ([1199050infin) R)

(5)

2 International Journal of Differential Equations

5 10 15

051

minus1

minus05

Figure 1 Function 119909(119905) = |sin(119905)| sin(119905) is a solution of (4)

5 10 15

051

minus1

minus05

Figure 21199091015840(119905) = 2|sin(119905)| cos(119905) and hence11990910158401015840(119905) is not a continuousfunction

(iii) 119902(119905) is nonnegative and discontinuous on [1199050infin) as

11990910158401015840+ 41198982[sign (cos (119898119905))]+119891 (119909) = ℎ (sin (119898119905))

ae in [1199050infin)

119909 1199091015840isin 119860119862loc ([1199050infin) R)

(6)

(iv) 119902(119905) is sign changing and discontinuous on [1199050infin) as

11990910158401015840+ 41198982 sign (cos (119898119905)) 119891 (119909) = ℎ (sin (119898119905))

ae in [1199050infin)

119909 1199091015840isin 119860119862loc ([1199050infin) R)

(7)

where 119898 isin N and ℎ(119904) is an arbitrary function such thatℎ(119904) 119904 gt 0 for all 119904 = 0 for instance ℎ(119904) = 119904 or ℎ(119904) = sign(119904)According to Corollaries 7 and 10 we will show that (4)ndash(7) are oscillatory provided the function 119891 = 119891(119906) satisfies119891(119906)119906 ge 119870 ge 1 for all 119906 = 0 see Examples 8ndash13 It isinteresting that in particular for 119891(119906) = 119906 and ℎ(119904) =

21198982 sign(119904) (4) allows an explicit oscillatory solution 119909(119905) =

|sin(119898119905)| sin(119898119905) as shown in Figures 1 and 2Moreover as a consequence of Corollary 7 one can

show that all solutions of (4) are oscillatory for details seeExample 8 The main goal of this paper is to give somesufficient conditions on functions Φ(119906 V) 119891(119906) and thecoefficients 119903(119905) 119902(119905) and 119890(119905) such that (1) is oscillatory seeTheorems 3 and 4 It will also cover the model equations (4)ndash(7) as well as some other examples presented in Section 2

To the best of our knowledge it seems that there are onlyfew papers which study the oscillation of the second-orderdifferential equations with nonsmooth (local integrable)coefficients see [1ndash3]More precisely in [1] the author studiedthe interval oscillation criteria for the following second-orderhalf-linear differential equation

(119903 (119905)100381610038161003816100381610038161199091015840(119905)10038161003816100381610038161003816

120590minus1

1199091015840(119905))

1015840

+ 119902 (119905) |119909 (119905)|120590minus1

119909 (119905) = 0

ae in (0infin)

119909 11990310038161003816100381610038161003816119909101584010038161003816100381610038161003816

120590minus1

1199091015840isin 119860119862loc ((0infin) R)

(8)

where 120590 gt 1 and 1119903 119902 isin 119871 loc((0infin)R) such thatintinfin

0119903minus1120590

(119905)119889119903 = infin See also [2] but with the solutionspace1198621((0infin)R) instead of119860119862loc((0infin)R) that is 119909 and119903|1199091015840|120590minus1

1199091015840isin 1198621((0infin)R)

In [3] the authors consider the following second-orderdifferential equation

(119903 (119905) 1199091015840(119905))1015840

+ 119876 (119905 119909 (119905) 1199091015840(119905)) = 0 119905 ge 119905

0 (9)

where 119903(119905) gt 0 ae in [1199050infin) 1119903 isin 119871 loc([1199050infin)R) and

119876(119905 119910 119911) is locally integrable function in 119905 and continuous in(119910 119911) Equation (9) allows the forcing term 119890(119905) in the nextsense as follows

119910119876 (119905 119910 119911) ge 119902 (119905) 119910119891 (119910) minus 119890 (119905) 119910

forall (119905 119910 119911) isin [1199050infin) timesR

2

(10)

where 119890 = 119890(119905) satisfies (2) but the functions 119902 = 119902(119905) and119891 = 119891(119910) are smooth enough in their variables that is 119902 isin

119862([1199050infin)R) and 119891 isin 119862

1(RR)

On certain oscillation criteria for various classes offorced second-order differential equations with continuouscoefficients we refer the reader to [4ndash13] Our methodmodifies a recently used one in [14 15] and it containsthe classic Riccati transformation of the main equation ablow-up argument and pointwise comparison principle Thecomparison principle applies to all sub- and supersolutions ofa class of the generalized Riccati differential equations withnonlinear terms that are supposed to be locally integrablein the first variable and locally Lipschitz continuous in thesecond variable

2 Hypotheses Results and Consequences

First of all the functionΦ(119906 V)which appears in the second-order differential operator of (1) satisfies

|119906|120574minus2

VΦ (119906 V) ge 119892 (|Φ (119906 V)|) forall119906 V isin R (11)

where 120574 ge 2 and 119892 R+rarr R+is a locally Lipschitz function

1198920 R+rarr R+satisfying

119892 (119888119904) ge 1198881205741198920 (119904) forall119888 gt 0 119904 gt 0

1198920(119904) + 119872

0ge 1199042 for some 119872

0ge 0 and all 119904 isin R

+

(12)

In most cases 119892(119904) = 1198920(119904)1198921(119904) 119904 gt 0 where 119892

0(119904) = 119904

120574120574 ge 2 and 119892

1(119904) is an arbitrary function satisfying 119892

1(119904) ge 1

Thus for such 119892(119904) with 1198921(119904) equiv 1 condition (11) became

|119906|120574minus2

VΦ (119906 V) ge |Φ (119906 V)|120574

forall119906 V isin R (13)

It is not difficult to check that if 1198920isin 1198621(R+) or 119892

0(119904) is a

convex function then it is locally Lipschitz onR+too see for

instance [16 Theorem 133]Two essential classes of the second-order differential

operators (119903(119905)Φ(119909 1199091015840))1015840 satisfy condition (13) as is shown inthe next examples

International Journal of Differential Equations 3

Example 1 We consider the second-order differential opera-tor which is linear in 1199091015840 as follows

(119903 (119905) Φ (119909 1199091015840))1015840

= 120572(120601 (119909) 1199091015840)1015840

(14)

where 120572 gt 0 and 0 le 120601(119906) le 1 for all 119906 isin R Obviously thefunctionΦ(119906 V) = 120601(119906)V satisfies condition (13) in particularfor 120574 = 2 Two usual choices for 120601(119906) are 120601(119906) = |sin 119906| and120601(119906) = |119906|(1 + |119906|)

Example 2 We consider a quasilinear differential operator(the so-called one-dimensional prescribed mean curvatureoperator) as follows

(119903 (119905) Φ (119909 1199091015840))1015840

= 120572(120601 (119909)1199091015840

radic1 + 11990910158402)

1015840

(15)

where 120572 gt 0 and 0 le 120601120574minus1(119906) le |119906|

120574minus2 for all 119906 isin R It is notdifficult to check that condition (13) is satisfied in particularfor Φ(119906 V) = 120601(119906)V(1 + V2)12 and for any 120574 gt 1 For 120601(119906)we can take the same choice as in the previous example

Next we suppose the existence of a constant119870 such that

119891 (119906)

119906ge 119870 gt 0 forall119906 = 0 (16)

In order to simplify our consideration here inmany exampleswe often use 119891(119906) = 119870119906

Condition (2) means that there exists a sequence of pairsof intervals 119869

1119895= [1198861119895 1198871119895] and 119869

2119895= [1198862119895 1198872119895] 119895 isin N

contained in (1199050infin) such that the sequences (119886

1119895)119895ge1

(1198871119895)119895ge1

(1198862119895)119895ge1

and (1198872119895)119895ge1

are increasing 1198861119895lt 1198871119895le 1198862119895lt 1198872119895for

each 119895 and

119890 (119905) ge 0 on 1198691119895 119890 (119905) le 0 on 119869

2119895

for each 119895 isin N

lim119895rarrinfin

1198861119895= infin

(17)

On the intervals 1198691119895and 1198692119895 the coefficient 119903(119905) satisfies

119903 (119905) gt 0 on 119869119894119895 1199031minus120574

isin 1198711(119869119894119895)

forall119894 isin 1 2 119895 isin N(18)

Let there be a real function 119862 = 119862(119905) 119862 isin

1198711

loc((1199050infin)R) and let there exist a sequence of positive realnumbers (120582

119895)119895isinN such that

119862 (119905) ge 0 on 119869119894119895 119888119894119895= int

119869119894119895

119862 (120591) 119889120591 gt 0

forall119894 isin 1 2 119895 isin N

1

119888119894119895

119862 (119905) le1

120587min(120582

119895119903 (119905))1minus120574

119870

1198720+ 1

120582119895119902 (119905)

forall119905 isin 119869119894119895 119894 isin 1 2 119895 isin N

(19)

where 120574 1198720 and 119870 are constants defined in (11) (12) and

(16) respectivelyThe proof of the following main result will be presented

in Section 4

Theorem 3 Let the functions Φ(119906 V) 119891(119906) 119890(119905) and 119903(119905)

satisfy (11) (12) (16) (17) and (18) respectively Let 119902(119905) ge 0

and 119902(119905) equiv 0 on each interval 119869119894119895 119894 isin 1 2 119895 isin N If (19) is

fulfilled then (1) is oscillatory

Condition (19) can be replaced by an equivalent onewhich has a more practical value and takes a simpler formsince we do not need a sequence of auxiliary parameters(120582119895)119895isinN let there be a real function 119862 = 119862(119905) 119862 isin

1198711

loc((1199050infin)R) such that

119862 (119905) ge 0 on 119869119894119895 119888119894119895= int

119868119894119895

119862 (120591) 119889120591 gt 0 forall119894 isin 1 2

119895 isin N

sup119905isin119869119894119895

[119862 (119905)

119902 (119905)] sup119905isin119869119894119895

[119903 (119905) 119862(119905)1(120574minus1)

] le119870

1198720+ 1

(

119888119894119895

120587)

120574(120574minus1)

forall119894 isin 1 2 119895 isin N

(20)

where 120574 1198720 and 119870 are constants defined in (11) (12) and

(16) respectively Since we will show that (19) and (20)are equivalent see page 8 the next oscillation criterionimmediately follows fromTheorem 3

Theorem 4 Let the functions Φ(119906 V) 119891(119906) 119890(119905) and 119903(119905)

satisfy (11) (12) (16) (17) and (18) respectively Let 119902(119905) ge 0

and 119902(119905) equiv 0 on each interval 119869119894119895 119894 isin 1 2 119895 isin N If (20) is

fulfilled then (1) is oscillatory

Remark 5 Assuming that 119871 = lim119895rarrinfin

1198861119895

lt infin wecan ensure the oscillation in the point 119871 Note that 119871 =

lim119895rarrinfin

1198862119895

since 1198861119895

lt 1198862119895

lt 1198861119895+1

Thus we can generatea one-sided (right) limit

Now we consider some consequences of Theorem 4which depend on the qualitative properties of the coefficient119902(119905)

Substituting 119862(119905) equiv 1 in (20) Theorem 4 implies thefollowing result involving lower bounds on the lengths ofintervals |119869

119894119895| = 119887119894119895minus 119886119894119895

Corollary 6 (119902(119905) is positive) Let the functionsΦ(119906 V) 119891(119906)119890(119905) and 119903(119905) satisfy (11) (12) (16) (17) and (18) respectivelyLet inf

119905isin119869119894119895119902(119905) gt 0 for each 119894 isin 1 2 119895 isin N If

10038161003816100381610038161003816119869119894119895

10038161003816100381610038161003816ge 120587(

(1198720+ 1) sup

119905isin119869119894119895119903 (119905)

119870inf119905isin119869119894119895

119902 (119905))

(120574minus1)120574

forall119894 isin 1 2 119895 isin N

(21)

then (1) is oscillatory

4 International Journal of Differential Equations

Corollary 7 (119902(119905) is bounded from below by a positiveconstant) Let the functionsΦ(119906 V)119891(119906) 119890(119905) and 119903(119905) satisfy(11) (12) (16) (17) and (18) respectively Let there be twoconstants 119903

0 1199020satisfying

0 lt 119903 (119905) le 1199030 119902 (119905) ge 119902

0gt 0 forall119894 isin 1 2 119895 isin N (22)

If

10038161003816100381610038161003816119869119894119895

10038161003816100381610038161003816ge 120587(

(1198720+ 1) 1199030

1198701199020

)

(120574minus1)120574

forall119894 isin 1 2 119895 isin N (23)

then (1) is oscillatory where 1205741198720 and119870 are constants defined

in (11) (12) and (16) respectively

Example 8 (oscillation of (4)) We know that 119909(119905) =

|sin(119898119905)| sin(119898119905) is an oscillatory solution of (4) Howeveraccording to Corollary 7 we can show that all solutions of (4)are oscillatory Indeed since Φ(119906 V) equiv V the conditions (11)and (12) are satisfied especially for 120574 = 2 119892(119904) = 119892

0(119904) = 119904

2

and 1198720= 0 Next 119891(119906) equiv 119906 implies that condition (16) is

satisfied especially for 119870 = 1 Since 119903(119905) equiv 1 and 119902(119905) equiv 41198982

it is clear that conditions (18) and (22) are also satisfied inparticular for 119903

0= 1 and 119902

0= 4119898

2 Moreover since 119890(119905) =ℎ(sin(119898119905)) and ℎ(119904)119904 gt 0 119904 = 0 we have that (17) is fulfilled for1198861119895= 2119895120587119898 119887

1119895= (2119895 + 1)120587119898 = 119886

2119895and 1198872119895= (2119895 + 2)120587119898

Moreover10038161003816100381610038161003816119869119894119895

10038161003816100381610038161003816= 119887119894minus 119886119894=120587

119898gt 0 119894 isin 1 2 (24)

Hence we conclude that the required condition (23) isfulfilled that is10038161003816100381610038161003816119869119894119895

10038161003816100381610038161003816=120587

119898ge

120587

radic41198982

= 120587((1198720+ 1) 1199030

1199020

)

(120574minus1)120574

ge 120587((1198720+ 1) 1199030

1198701199020

)

(120574minus1)120574

(25)

Thus all conditions of Corollary 7 are satisfied and hence (4)is oscillatory

Example 9 We consider the following class of equations

11990910158401015840minus 2

11987810158401015840(119905)

119878 (119905)119909 = 2 sign (119878 (119905)) 1198781015840

2

(119905) ae in [1199050infin)

119909 1199091015840isin 119860119862loc ([1199050infin) R) 119909 notin 119862

2((1199050infin) R)

(26)

where 119878 = 119878(119905) 119878 isin 1198622(R) is an oscillatory function such thatthe zeros 119905

119899of the function sign(119878(119905))11987810158402(119905) satisfy 119905

119899rarr infin

there is a 1205910isin R such that 119905

119899+1minus 119905119899ge 1205910gt 0 for all 119899 isin N and

119878(119905) = 0 on (119905119899 119905119899+1) This equation allows an explicitly given

oscillatory solution 119909(119905) = |119878(119905)|119878(119905) Moreover if there is aconstant 119904

0gt 0 such that

minus211987810158401015840(119905)

119878 (119905)ge 1199040 119905 isin (119905

119899 119905119899+1) 120591

0ge

120587

radic1199040

(27)

then by Corollary 7 we conclude that (26) is oscillatoryIndeed conditions (11) (12) and (16) are satisfied by the samereasons as in Example 8 Condition (18) is satisfied becauseof 119903(119905) equiv 1 Also from (27) it follows that (22) and (23) arefulfilled in particular for 120574 = 2119872

0= 0 119903

0= 1 119902

0= 1199040 and

119870 ge 1 that is10038161003816100381610038161003816119869119894119895

10038161003816100381610038161003816=10038161003816100381610038161003816119905119895+1

minus 119905119895

10038161003816100381610038161003816ge 1205910ge

120587

radic1199040

= 120587((1198720+ 1) 1199030

1199020

)

(120574minus1)120574

ge 120587((1198720+ 1) 1199030

1198701199020

)

(120574minus1)120574

(28)

Hence Corollary 7 proves this result

As the second consequence ofTheorem3 is unlike the firstone we consider the case when the coefficient 119902(119905) is not astrictly positive function Here by 119902 = 0 we denote the setof all 119905 isin R such that 119902(119905) = 0

Corollary 10 (119902(119905) is nonnegative but not equiv 0) Let thefunctions Φ(119906 V) 119891(119906) 119890(119905) and 119903(119905) satisfy (11) (12) (16)(17) and (18) respectively Let 119902(119905) ge 0 on each interval 119869

119894119895

119894 isin 1 2 119895 isin N such that

119902119894119895= int

119869119894119895

119902 (120591) 119889120591 gt 0 119894 isin 1 2 119895 isin N (29)

If

119902120574

119894119895

119902 (119905)ge 120587120574((1198720+ 1) 119903 (119905)

119870)

120574minus1

119905 isin 119869119894119895 119902 = 0 119894 isin 1 2 119895 isin N

(30)

then (1) is oscillatory

Proof It suffices to show that (30) is equivalent to theexistence of a real number 120582

119895such that

120582119895ge120587 (1198720+ 1)

119870

1

119902119894119895

1

119902119894119895

119902 (119905) le1

120587(120582119895119903 (119905))1minus120574

119905 isin 119869119894119895 119894 isin 1 2 119895 isin N

(31)

The claim will then follow fromTheorem 3 Inequality (31) isfor any 119905 isin 119869

119894119895 119902 = 0 equivalent to

120587 (1198720+ 1)

119870

1

119902119894119895

le 120582119895le (

119902119894119895

120587119902 (119905))

1(120574minus1)1

119903 (119905) (32)

that is to

120587 (1198720+ 1)

119870

1

119902119894119895

le (

119902119894119895

120587119902 (119905))

1(120574minus1)1

119903 (119905) (33)

This inequality is easily seen to be equivalent to (30) for any119905 isin 119869119894119895 119902 = 0 Note that if 119905 isin 119902 = 0 then the second

inequality in (31) is trivially satisfied

International Journal of Differential Equations 5

The following three examples are simple consequences ofCorollary 10

Example 11 (oscillation of (5)) Equation (5) as well asequation

11990910158401015840+ 11989821205872[sin (119898119905)]+119891 (119909) = ℎ (cos (119898119905))

ae in [1199050infin)

119909 1199091015840isin 119860119862loc ([1199050infin) R)

(34)

are oscillatory where 119891(119906) satisfies (16) with 119870 ge 1 andℎ(119904)119904 gt 0 119904 = 0 Indeed in (5) we have 119903(119905) equiv 1 119902(119905) =

11989821205872[cos(119898119905)] and 119890(119905) = ℎ(sin(119898119905)) and so (17) and (18) are

fulfilled and 119902(119905) ge 0 119902(119905) equiv 0 on each interval 119869119894119895 119894 isin 1 2

119895 isin N where 1198691119895= [1198861119895 1198871119895] 1198692119895= [1198862119895 1198872119895] and

1198861119895=2119895120587

119898 119887

1119895=(2119895 + 12) 120587

119898

1198862119895=(2119895 + 32) 120587

119898 119887

2119895=(2119895 + 2) 120587

119898

(35)

Moreover

119902119894119895= int

119869119894119895

119902 (120591) 119889120591

= 11989821205872int

119887119894119895

119886119894119895

[cos (119898120591)]+119889120591 = 1198981205872 119894 isin 1 2 119895 isin N

(36)

and since 120574 = 2 119903(119905) equiv 1119870 ge 1 and1198720= 0 for 120582

119895= 1(119898120587)

we have

120582119895=

1

119898120587ge1

119870

1

119898120587=120587 (1198720+ 1)

119870

1

1198981205872

=120587 (1198720+ 1)

119870

1

119902119894119895

119894 isin 1 2 119895 isin N

1

119902119894119895

119902 (119905) =1

119898120587211989821205872[cos (119898119905)]+ le 119898

=1

120587

1

(1119898120587)=1

120587

1

(120582119895119903 (119905))120574minus1

119905 isin 119869119894119895 119894 isin 1 2 119895 isin N

(37)

Thus the required condition (30) is fulfilled in this case andhence we may apply Corollary 10 to (5) to verify that thisequation is oscillatory Analogously we can show that (34)is oscillatory too

In the next example the coefficient 119902(119905) is a discontinuousfunction on [119905

0infin)

Example 12 (oscillation of (6)) If 119891(119906) satisfies (16) with119870 ge

1 then (6) is oscillatory since the required condition (30) issatisfied especially for 119902(119905) = 4119898

2 sign(cos(119898119905)) on 119869119894119895 120582119895=

1(2119898) and 119869119894119895as in (35) 120574 = 2 119903(119905) equiv 1119870 ge 1 and119872

0= 0

Indeed

119902119894119895= int

119869119894119895

119902 (120591) 119889120591 = 41198982int

119869119894119895

[sign (cos (119898120591))]+119889120591

= 41198982 120587

2119898= 2119898120587 119894 isin 1 2 119895 isin N

120582119895=

1

2119898=

120587

2119898120587= 120587 (119872

0+ 1)

1

119902119894119895

ge120587 (1198720+ 1)

119870

1

119902119894119895

119894 isin 1 2 119895 isin N

1

119902119894

119902 (119905) =1

211989812058741198982[sign (cos (119898119905))]+

le2119898

120587=1

120587

1

radic1 (41198982)

=1

120587(120582119895119903 (119905))120574minus1

119905 isin 119869119894119895 119894 isin 1 2 119895 isin N

(38)

which shows that (30) is satisfied

At the end of this section we consider the case when 119902(119905)changes sign on [119905

0infin) but with the help of Corollary 10 since

119902(119905) is strictly positive on all intervals 119869119894119895

Example 13 (oscillation of (7)) If 119891(119906) satisfies (16) with119870 ge

1 then model equation (7) is oscillatory In fact let 119869119894119895be as

in (35) If ℎ(119904)119904 gt 0 119904 = 0 it is clear that 119890(119905) = ℎ(sin(119898119905))satisfies (2) that is 119890(119905) ge 0 on 119869

1119895and 119890(119905) le 0 on 119869

2119895 On

the other hand we have 119902(119905) = 41198982 sign(cos(119898119905)) on 119869

119894119895and

so 119902(119905) ge 0 119902(119905) equiv 0 on 119869119894119895for all 119894 isin 1 2 119895 isin N Hence

the proof of the fact that all assumptions of Corollary 7 arefulfilled is the same as in the preceding example

3 Proofs of Theorems 3 and 4

Let 119860119862loc([119886 119887)R) denote the set of all real functions whichare absolutely continuous on every interval [119886 119887 minus 120576] where120576 isin (0 119887 minus 119886) For arbitrary numbers 119879

0lt 119879lowast and functions

119866 = 119866(119905 119906) and 119887 = 119887(119905) we consider the ordinary differentialequation

1199081015840= 119866 (119905 119908) + 119887 (119905) ae in [119879

0 119879lowast) (39)

which generalizes the classic Riccati equation 1199081015840 = 119886(119905)1199082+

119887(119905) where 119886(119905) is an arbitrary function We associate to(39) the corresponding sub- and supersolutions 119908119908 isin

119860119862loc([1198790 119879lowast)R) which satisfy respectively

1199081015840le 119866 (119905 119908) + 119887 (119905) 119908

1015840ge 119866 (119905 119908) + 119887 (119905)

ae in [1198790 119879lowast)

(40)

We are interested in studying the following property

119908 (1198790) le 119908 (119879

0) implies 119908 (119905) le 119908 (119905)

forall119905 isin [T0 119879lowast)

(41)

In this way we introduce the following definition

6 International Journal of Differential Equations

Definition 14 We say that comparison principle holds for (39)on an interval [119879

0 119879lowast) 119879 le 119879

0lt 119879lowast if property (41)

is fulfilled for all sub- and supersolutions 119908119908 isin

119860119862loc([1198790 119879lowast)R) of (39) on [119879

0 119879lowast)

Now we are able to state the next result

Lemma 15 Let 119879 le 1198790lt 119879lowast Let 119866 = 119866(119905 119906) and 119887 = 119887(119905) be

two arbitrary function For any119872 gt 0 let there be a function119871 = 119871(119905) gt 0 depending on 119879

0 119879lowast119872 such that

119871 (119905) gt 0 on [1198790 119879lowast) 119871 isin 119871

1(1198790 119879lowast)

1003816100381610038161003816119866 (119905 119906

1) minus 119866 (119905 119906

2)1003816100381610038161003816le 119871 (119905)

10038161003816100381610038161199061minus 1199062

1003816100381610038161003816

forall119905 isin [1198790 119879lowast) 1199061 1199062isin [minus119872119872]

(42)

Then comparison principle (41) holds for (39) on an interval[1198790 119879lowast) 119879 le 119879

0lt 119879lowast

Proof It is enough to use the idea of the proof of [14 Lemma41] Hence this proof is left to the reader

Corollary 16 Let 120582 gt 0 and let 1198920 R+rarr R+be a locally

Lipschitz function on R+ Let 119870 be an arbitrary real number

and 119902(119905) an arbitrary function Let 119879 le 1198790lt 119879lowast If 119903(119905) gt 0

on [1198790 119879lowast) and 119903minus1+120574 isin 1198711(119879

0 119879lowast) then comparison principle

(41) holds for the Riccati differential equation

1199081015840=

1

(120582119903 (119905))120574minus1

1198920 (|119908|) + 119870120582119902 (119905) ae in [119879

0 119879lowast)

(43)

Proof It is clear that (43) is a particular case of (39) inparticular for

119866 (119905 119906) =1

(120582119903 (119905))120574minus1

1198920 (|119906|) 119887 (119905) = 119870120582119902 (119905) (44)

Since1198920is a locally Lipschitz function onR

+ for every119872 gt 0

there is an 1198710gt 0 depending on119872 such that

10038161003816100381610038161198920(1199041) minus 1198920(1199042)1003816100381610038161003816le 1198710

10038161003816100381610038161199041minus 1199042

1003816100381610038161003816 forall119904

1 1199041isin [minus119872119872]

(45)

Hence for any119872 gt 0 and all 1199061 1199062isin [minus119872119872] we obtain

1003816100381610038161003816119866 (119905 119906

1) minus 119866 (119905 119906

2)1003816100381610038161003816=

1

(120582119903 (119905))120574minus1

10038161003816100381610038161198920(10038161003816100381610038161199061

1003816100381610038161003816) minus 1198920(10038161003816100381610038161199062

1003816100381610038161003816)1003816100381610038161003816

le1198710

(120582119903 (119905))120574minus1

1003816100381610038161003816

10038161003816100381610038161199061

1003816100381610038161003816minus10038161003816100381610038161199062

1003816100381610038161003816

1003816100381610038161003816

le1198710

(120582119903 (119905))120574minus1

10038161003816100381610038161199061minus 1199062

1003816100381610038161003816

(46)

Thus according to assumption 119903minus1+120574

isin 1198711(1198790 119879lowast) the

required condition (42) is fulfilled in particular for 119871(119905) =

1198710(120582119903(119905))

minus120574+1 and so Lemma 15 proves this corollary

Before we give the proof ofTheorem 3 we state and provethe next two propositions In the first one by a nonoscillatorysolution 119909(119905) of the main equation (1) we get the existenceof a supersolution 119908(119905) of the Riccati differential equation(43) on the interval (119886

1 1198871) or (119886

2 1198872) In the second one we

construct two subsolutions 1199081(119905) and 119908

2(119905) of (43) which

blow up on intervals (1198861 119879lowast

1) sube (119886

1 1198871) and (119886

2 119879lowast

2) sube (119886

2 1198872)

respectively

Proposition 17 Let Φ(119906 V) and 119891(119906) satisfy (11) (12) and(16) respectively and let 119890(119905) satisfy (2) Let 119903(119905) gt 0 119902(119905) ge0 and 119902(119905) equiv 0 on [119886

1 1198871] cup [119886

2 1198872] Let 119909 = 119909(119905) be a

nonoscillatory solution of (1) and let for some 119879 ge 1199050and

120582 gt 0 the function 119908 = 119908(119905) be defined by

119908 (119905) = minus

120582119903 (119905)Φ (119909 (119905) 1199091015840(119905))

119909 (119905) 119905 ge 119879 (47)

Then 119908 isin 119860119862loc([119879infin)R) and 119908(119905) satisfies the inequality

1199081015840ge (120582119903 (119905))

1minus1205741198920(|119908|) + 119870120582119902 (119905) ae in 119869 (48)

where 119869 = (1198861 1198871) if 119909(119905) lt 0 and 119869 = (119886

2 1198872) if 119909(119905) gt 0

Proof Since 119909 = 119909(119905) is a nonoscillatory solution of (1) thereis a 119879 ge 119905

0such that 119909(119905) = 0 on [119879infin) Hence 119908(119905) is well

defined by (47)Because of (2) we have 119890(119905)119909(119905) le 0 for all 119905 isin 119869 where

119869 = (1198861 1198871) if 119909(119905) lt 0 and 119869 = (119886

2 1198872) if 119909(119905) gt 0 and since

120582 gt 0 we have

minus120582119890 (119905)

119909 (119905)ge 0 in 119869 (49)

Next since 119909 and 119903(119905)Φ(119909(119905) 1199091015840(119905)) are from

119860119862loc([1199050infin)R) we can take the first derivative of 119908(119905)for almost everywhere in 119869 which together with 119903(119905) gt 0119902(119905) ge 0 and 119902(119905) equiv 0 on 119869 gives

1199081015840(119905) = minus

120582(119903 (119905) Φ (119909 (119905) 1199091015840(119905)))1015840

119909 (119905)

+

120582119903 (119905)Φ (119909 (119905) 1199091015840(119905))

1199092(119905)

1199091015840(119905)

=120582119903 (119905)

|119909 (119905)|120574Φ(119909 (119905) 119909

1015840(119905)) 1199091015840(119905) |119909 (119905)|

120574minus2

+ 120582119902 (119905)119891 (119909 (119905))

119909 (119905)minus 120582

119890 (119905)

119909 (119905)

International Journal of Differential Equations 7

ge120582119903 (119905)

|119909 (119905)|120574119892 (10038161003816100381610038161003816Φ (119909 (119905) 119909

1015840(119905))

10038161003816100381610038161003816) + 119870120582119902 (119905) minus 120582

119890 (119905)

119909 (119905)

=120582119903 (119905)

|119909 (119905)|120574119892(

|119909 (119905)|

120582119903 (119905)|119908 (119905)|) + 119870120582119902 (119905) minus 120582

119890 (119905)

119909 (119905)

ge120582119903 (119905)

|119909 (119905)|120574

|119909 (119905)|120574

(120582119903 (119905))1205741198920 (|119908 (119905)|)

+ 119870120582119902 (119905) minus 120582119890 (119905)

119909 (119905)ae in 119869

(50)

that is

1199081015840ge (120582119903 (119905))

1minus1205741198920(|119908|) + 119870120582119902 (119905) minus 120582

119890 (119905)

119909 (119905)ae in 119869

(51)

Hence (49) and (51) prove the desired inequality (48)

Proposition 18 Let (19) be satisfied Let 1198771and 119877

2be two

arbitrary real numbers Then there are two points 119879lowast1isin (1198861 1198871)

and 119879lowast

2isin (1198862 1198872) and two continuous functions 119908

1(119905) and

1199082(119905) such that

119908119894(119886119894) le 119877119894 lim119905rarr119879

lowast

119894

119908119894(119905) = infin 119894 isin 1 2

1199081015840

119894le (120582119903 (119905))

1minus1205741198920(1003816100381610038161003816119908119894

1003816100381610038161003816) + 119870120582119902 (119905)

for ae 119905 isin (119886119894 119879lowast

119894) 119894 isin 1 2

(52)

Proof Since the function 119910(119904) = tan(119904) is a bijection fromthe interval (minus1205872 1205872) intoR we observe that there are two1199041 1199042isin (minus1205872 1205872) such that

tan (119904119894) le 119877119894 119894 isin 1 2 (53)

Next we define two functions 1198811(119905) and 119881

2(119905) by

119881119894 (119905) = 119904119894 +

120587

119888119894

int

119905

119886119894

119862 (120591) 119889120591 119905 isin [119886119894 119887119894] 119894 isin 1 2 (54)

where the real numbers 1198881 1198882and the function 119862(119905) are

defined in (19) From (19) and (54) one can immediatelyconclude that

119881119894(119886119894) = 119904119894lt120587

2 119881

119894(119887119894) = 119904119894+ 120587 gt

120587

2 119894 isin 1 2 (55)

Since 119862 isin 1198711

loc((1199050infin)R) it follows that 119862 isin 1198711((1199050infin)R)

which implies that 119881119894isin 119860119862([119886

119894 119887119894]R) 119894 isin 1 2 Therefore

exploiting the fact that in (55) we have 119881119894isin 119862([119886

119894 119887119894]R) we

obtain the existence of two points 119879lowast1isin (1198861 1198871) and 119879

lowast

2isin

(1198862 1198872) such that

119881119894(119879lowast

119894)=

120587

2 minus

120587

2lt119881119894(119905)lt

120587

2

forall119905 isin [119886119894 119879lowast

119894) 119894 isin 1 2

(56)

Now we are able to define the following two functions 1205961(119905)

and 1205962(119905) by

120596119894(119905) = tan (119881i (119905)) 119905 isin [119886

119894 119879lowast

119894) 119894 isin 1 2 (57)

That are well defined because of (56) Moreover from (53)(54) and (56) we obtain

120596119894(119886119894) = tan (119904

119894) le 119877119894

lim119905rarr119879

lowast

119898

120596119894(119905) = tan (119881

119894(119879lowast

119894)) = tan(120587

2) = infin

(58)

Also since119881119894isin 119860119862([119886

119894 119887119894]R) we can take the first derivative

of 120596119894(119905) and hence from (19) and (57) we obtain

1199081015840

119894(119905) =

120587

119888119894

119862 (119905)1

cos2 (119881119894 (119905))

=120587

119888119894

119862 (119905) (1003816100381610038161003816120596119894(119905)1003816100381610038161003816

2+ 1)

le120587

119888119894

119862 (119905) [1198920(1003816100381610038161003816120596119894(119905)1003816100381610038161003816) + 119872

0+ 1]

le120587

119888119894

119862 (119905) 1198920(1003816100381610038161003816120596119894(119905)1003816100381610038161003816) +

120587

119888119894

119862 (119905) (1198720+ 1)

le (120582119903 (119905))1minus1205741198920(1003816100381610038161003816120596119894(119905)1003816100381610038161003816) + 119870120582119902 (119905) ae in (119886

119894 119879lowast

119894)

(59)

which together with (58) proves this proposition

Now we are able to present the proof of Theorem 3 basedon Lemma 15 and Propositions 17 and 18

Proof of Theorem 3 Assuming the contrary then there is anonoscillatory solution 119909(119905) and 119879 ge 119905

0such that 119909(119905) = 0 for

all 119905 ge 119879 119909(119905) and 119903(119905)Φ(119909 1199091015840) are from 119860119862loc([1199050infin)R)In order to simplify notation let every 119895 isin N be fixed andomitted in the notation For instance instead of (119886

1119895 1198871119895) and

(1198862119895 1198872119895) we write (119886

1 1198871) and (119886

2 1198872) respectively and so on

On one hand we observe by Proposition 17 that thefunction 119908(119905) defined by (47) is a supersolution of thefollowing Riccati differential equation

1199081015840= (120582119903 (119905))

1minus1205741198920(|119908|) + 119870120582119902 (119905) ae in 119869 (60)

where 119908 isin 119860119862loc(119869R) and 119869 = (1198861 1198871) if 119909(119905) lt 0 and 119869 =

(1198862 1198872) if 119909(119905) gt 0 see the proof of Proposition 17 Let for

instance 119909(119905) lt 0 and thus 119869 = (1198861 1198871)

On the other hand by Proposition 18 there are a number119879lowast

1isin (1198861 1198871) and subsolutions 119908

1(119905) 1199081isin 119860119862([119886

1 119879lowast

1)R) of

the Riccati differential equation (60) such that

1199081(1198861) le 119908 (119886

1) lim

119905rarr119879lowast

1

1199081(119905) = infin (61)

(in the case when 119869 = (1198862 1198872) then we work with119879lowast

2isin (1198862 1198872)

and the other subsolution 1199082(119905) 119908

2isin 119860119862([119886

2 119879lowast

2)R))

Hence by Corollary 16 and (61) we conclude that1199081(119905) le 119908(119905)

for all 119905 isin [1198861 119879lowast

1) and therefore

infin = lim119905rarr119879

lowast

1

1199081(119905) le lim119905rarr119879

lowast

1

119908 (119905) (62)

which contradicts 119908 isin 119860119862loc([119879infin)R) Thus the assump-tion that 119909(119905) is nonoscillatory is not possible and hence everysolution of (1) is oscillatory

8 International Journal of Differential Equations

Proof of Theorem 4 There is only one difference betweenTheorems 3 and 4 and it is the difference between conditions(19) and (20) Hence Theorem 4 immediately follows fromTheorem 3 provided that we prove the equivalence between(19) and (20)

First of all we need the following two lemmas

Lemma 19 Let 119886 119887 and 119888 be positive real numbers and 120574 gt 1Then the existence of a positive real number 120582 such that

119888 le min 119886

120582120574minus1

119887120582 (63)

is equivalent to

119888 le 119886112057411988711205741015840

(64)

Here 1205741015840 = 120574(120574 minus 1) is the conjugate exponent of 120574

Proof Let us define 119892(120582) = min119886120582120574minus1 119887120582 Since 120582 997891rarr

119886120582120574minus1 is decreasing and 120582 997891rarr 119887120582 is increasing there exists

a unique point of maximum of 119892 It is achieved at 120582 = 1205820

which is a solution of 119886120582120574minus10

= 1198871205820 hence 120582

0= (119886119887)

1120574If there is a positive real number 120582 such that 119888 le 119892(120582)

then 119888 le 119892(1205820) = 119887120582

0= 119887(119886119887)

1120574= 119886112057411988711205741015840

Converselyif 119888 le 119886

112057411988711205741015840

then 119888 le 119892(1205820) and the equivalence in the

lemma is proved

The preceding lemma is a special case of the followingmore general statement

Lemma 20 Assume that 119886(119905) 119887(119905) and 119888(119905) are positive realfunctions defined on a subset 119869 sube R Then the existence of apositive real number 120582 such that for all 119905 isin 119869

119888 (119905) le min 119886 (119905)120582120574minus1

119887 (119905) 120582 (65)

is equivalent to

sup119905isin119869

119888 (119905)

119887 (119905)le (inf119905isin119869

119886 (119905)

119888 (119905))

1(120574minus1)

(66)

Proof The condition that there exists 120582 gt 0 such that (65)is true for all 119905 isin 119869 is equivalent with the following twoinequalities which have to hold simultaneously

119888 (119905) le119886 (119905)

120582120574minus1

119888 (119905) le 119887 (119905) 120582 (67)

that is with

119888 (119905)

119887 (119905)le 120582 le (

119886 (119905)

119888 (119905))

1(120574minus1)

forall119905 isin 119869 (68)

Taking the supremum of the left-hand side and the infimumof the right-hand side we obtain (66) Conversely if (66)is satisfied then since the left-hand side in (66) cannotbe equal to zero while the right-hand side is less than infinthen inequality (66) defines a nonempty closed interval in R

(possibly reducing to just one point) in which we can chooseany positive 120582 Hence (67) is satisfied for all 119905 isin 119869 andtherefore (65) as well

Remark 21 If 119886 = 119886(119905) 119887 = 119887(119905) and 119888 = 119888(119905) appearingin Lemma 20 are constant functions then condition (66) isequivalent to 119888119887 = (119886119888)1(120574minus1) that is to (64)

Proof of Equivalence of (19) and (20) Wewill use Lemma 20Let us fix 119869

119894119895 where 119894 isin 1 2 and 119895 isin N We see that that the

inequality appearing in the second line of condition (19) is ofthe form (65) where

119888 (119905) =1

119888119894119895

119862 (119905) 119886 (119905) =1

120587119903(119905)1minus120574

119887 (119905) =119870119902 (119905)

120587 (1198720+ 1)

(69)

Condition (66) is equivalent to

sup119905isin119869119894119895

119862 (119905)

119902 (119905)le

1198701198881205741015840

119894119895

1205871205741015840

(1198720+ 1)

inf119905isin119869119894119895

1

119903 (119905) 119862(119905)120574minus1

=

1198701198881205741015840

119894119895

(1198720+ 1) sup

119905isin119869119894119895119903 (119905) 119862(119905)

120574minus1(

119888119894119895

120587)

1205741015840

(70)

This inequality is equivalent to the corresponding one in (19)and the claim follows from Lemma 19

Proof of Corollary 6 Substituting inf119905isin119869119894119895

119902(119905) instead of 119902(119905)and sup

119905isin119869119894119895119903(119905) instead of 119903(119905) in the inequality appearing in

the second line of (20) after a short computation we obtain astronger inequality than (20) as

1

119888119894119895

sup119905isin119869119894119895

119862 (119905)

le1

120587(

119870inf119905isin119869119894119895

119902 (119905)

(1198720+ 1) sup

119905isin119869119894119895119903 (119905)

)

(120574minus1)120574

forall119894 isin 1 2 119895 isin N

(71)

Substituting119862(119905) equiv 1we obtain that the left-hand side of (71)is equal to |119869

119894119895|minus1 and the resulting inequality is equivalent to

(21) Since (21) implies (20) the claim is proved

Remark 22 Here we show that the choice of 119862(119905) equiv 1 inthe proof of Corollary 6 is the best possible To prove thisnote that on the left-hand side of (71) we have the expressiondepending on an auxiliary function 119862(119905) and this functionis absent on the right-hand side Therefore it has sense totry to find a function 119862(119905) 119905 isin 119869

119894119895 such that the value of

119876(119862) = (1119888119894119895)sup119905isin119869119894119895

119862(119905) is minimal (note that 119888119894119895depends

on the function 119862 = 119862(119905) as well) It is easy to see that theminimum is achieved for 119862(119905) equiv 1 (or any positive constant)Indeed since 119876(119862) = 119876(120583119888) for any 120583 gt 0 by taking 120583 =

(sup119905isin119869119894119895

119862(119905))minus1 it suffices to assume that sup

119905isin119869119894119895119862(119905) = 1

The value of 119876(119862) is minimal if 119888119894119895= int119869119894119895

119862(120591) 119889120591 is maximalpossible and since 0 le 119862(119905) le 1 it is clear that the maximumof 119888119894119895= 119888119894119895(119862) is achieved when 119862(119905) equiv 1

International Journal of Differential Equations 9

Proof of Corollary 7 It is clear that the condition (23) implies(21) Furthermore since the lengths |119869

119894119895| are uniformly

bounded from below by a positive constant see (23) thenobviously 119886

1119895rarr infin as 119895 rarr infin Thus the claim follows

immediately from Corollary 6

Proof of Corollary 10 Let 119862(119905) = 119902(119905) Then the requiredcondition (19) is clearly satisfied because of assumption(30) Hence this corollary immediately follows from Theo-rem 3

4 An Extension of Condition (12) tothe Case of 120574 gt 1

In this section we consider the oscillation of (1) in the casewhen 120574 isin (1 2) is allowed in the assumption (12) In this waythe assumption (11) is slightly modified by a real number 119901 gt1 such that

|119906|(119901minus1)120574minus119901

VΦ (119906 V) ge 119892 (|Φ (119906 V)|) (72)

where (12) is supposed that is 119892(119888119904) ge 1198881205741198920(119904) for every 119888 119904 gt

0 and for a locally Lipschitz function 1198920 [0infin) rarr [0infin)

for which there exists a constant1198720such that 119892

0(119904)+119872

0ge 119904119901

for all 119904 isin [0infin)For the function 119891 we assume that there exists a constant

119870 gt 0 such that

119891 (119906) sign (119906) ge 119870|119906|119901minus1 for every 119906 isin R (73)

We will need the following two lemmas

Lemma 23 Let 119886 lt 119887 and let 119891 [119886 119887] rarr [0infin) be a meas-urable function Then

int

119887

119886

119891 (119905) 119889119905 ge 1 (74)

if and only if there exists a function 119892 isin 1198711([119886 119887] [0infin))

1198921= 1 such that

119891 (119905) ge 119892 (119905) forall119905 isin [119886 119887] (75)

Proof We assume (74) If int119887119886119891(119905)119889119905 is finite we may choose

119892(119909) = 119891(119909)1198911

For int119887119886119891(119905)119889119905 = infin we may define for every natural

number 119899 a function

119891119899(119905) =

119891 (119905) 119891 (119905) le 119899

119899 119891 (119905) gt 119899(76)

Since 119891119899(119905) le 119899 we have 119891

119899isin 1198711([119886 119887]R) Obviously

lim119899int119887

119886119891119899(119905)119889119905 is also infiniteThis implies that there exists an

1198990such that int119887

1198861198911198990(119905)119889119905 ge 1 We define 119892(119905) = 119891

1198990(119905)119891

11989901

Now we assume (75) Integrating over the whole intervalwe get

int

119887

119886

119891 (119905) 119889119905 ge int

119887

119886

119892 (119905) 119889119905 = 1 (77)

Lemma 24 Let119901 gt 1There exists a unique function 119910 = 119910(119904)that satisfies the following Cauchy problem

1199101015840(119904) = 1 +

1003816100381610038161003816119910 (119904)

1003816100381610038161003816

119901 119904 isin R (78)

119910 (0) = 0 (79)

Furthermore there exists a finite real number 119878lowast such that

lim119904rarrplusmn119878

lowast119910 (119904) = plusmninfin (80)

and 119910 isin 1198621((minus119878lowast 119878lowast)R) is injective and monotonous We

have 119878lowast = (120587119901)(1(sin(120587119901))) and 119910 is odd

We will denote by tan119901the solution of (78)-(79)

Proof Obviously every solution to (78) is injective andmonotonous on a connected set since 1199101015840 = 1 + |119910|119901 ge 1 gt 0

Let 119891(119909) = 1(1 + |119909|119901) Obviously 1 ge 119891(119909) gt 0 for all

119909 isin R and 119862119901gt |1198911015840(119909)| gt 0 for some constants 119862

119901isin R

We may assume 119862119901gt 1 Hence 1199111015840 = 119891(119905) 119911(0) = 0 has (by

[17Theorem 31]) a locally unique solution on (minus1119862119901 1119862119901)

By the same argument 119911 can be extended to the whole of Rinterval by interval of type ((119899 minus 1)119862

119901 (119899 + 1)119862

119901) for all

119899 isin ZWe know from [18] that the for the integral of the left-

hand side we have

int

infin

0

119891 (119909) 119889119909 = int

infin

0

119889119909

1 + 119909119901=120587

119901

1

sin (120587119901)= 119878lowast (81)

so lim119905rarrplusmninfin

119911(119905) = plusmn119878lowast

Since 1199111015840(119905) = 0 119911 is injective monotonous and of class1198621(R (minus119878lowast 119878lowast)) and so we define 119910(119904) = 119911

minus1(119904) It imme-

diately follows that 119910 isin 1198621((minus119878lowast 119878lowast)R) and that 119910 is

injective and monotonous Also by continuity of 119911 we havelim119904rarrplusmn119878

lowast119910(119904) = lim119908rarrplusmninfin

119910(119911(119908)) = lim119908rarrplusmninfin

119908 = plusmninfinDifferentiating it also follows that 1199101015840 = 1 + |119910|

119901 and bydefinition of 119910 119910(0) = 119911minus1(0) = 0 So 119910 is a solution of (78)-(79) Since any other such solution must have 119911 as its inverse119910 is unique on its domain

Proposition 25 Let 119886 lt 119887 and 120574 gt 119901 gt 1 Assume (72) and(73) and let 119902(119905) ge 0 and 119890(119905) le 0 (or 119890(119905) ge 0 ) for all 119905 isin [119886 119887]If there exists a constant 120582 gt 0 such that

119901

2120587sin(120587

119901)int

119887

119886

min119901 minus 1

119903120574minus1

(119905) 120582120574minus1

119870120582119902 (119905)

1 +1198720

119889119905 ge 1 (82)

then for any solution 119909(119905) of (1) with 119909(119886) ge 0 (or 119909(119886) le 0)there exists a number 119905lowast isin [119886 119887] such that 119909(119905lowast) = 0

Proof We assume that 119909(119905) = 0 for all 119905 isin [119886 119887] Then we maydefine the function

120596 (119905) = minus

120582119903 (119905) Φ (119909 (119905) 1199091015840(119905))

sign (119909 (119905)) |119909 (119905)|119901minus1 (83)

10 International Journal of Differential Equations

This function is well defined on [119886 119887] and its derivative is

1199081015840(119905) = minus

120582(119903 (119905) Φ (119909 (119905) 1199091015840(119905)))1015840

sign (119909 (119905)) |119909 (119905)|119901minus1

+

(119901 minus 1) 120582119903 (119905) Φ (119909 (119905) 1199091015840(119905))

|119909 (119905)|119901

1199091015840(119905)

(84)

Using the fact that 119909(119905) is a solution of (1) and omitting 119905 from119909 and 1199091015840 for clarity we get

1199081015840(119905) =

120582 (119902 (119905) 119891 (119909) minus 119890 (119905))

sign (119909) |119909|119901minus1+

(119901 minus 1) 120582119903 (119905) Φ (119909 1199091015840)

119909119901

1199091015840

=

(119901 minus 1) 120582119903 (119905) Φ (119909 1199091015840)

|119909|119901

1199091015840+ 120582119902 (119905)

119891 (119909)

sign (119909) |119909|119901minus1

minus 120582119890 (119905)

sign (119909) |119909|119901minus1

=(119901 minus 1) 120582119903 (119905)

|119909|120574(119901minus1)

Φ(119909 1199091015840) 1199091015840|119909|(119901minus1)120574minus119901

+ 120582119902 (119905)119891 (119909)

sign (119909) |119909|119901minus1minus 120582

119890 (119905)

sign (119909) |119909|119901minus1

(72)

ge(119901 minus 1) 120582119903 (119905)

|119909|120574(119901minus1)

119892 (10038161003816100381610038161003816Φ (119909 119909

1015840)10038161003816100381610038161003816)

+ 120582119902 (119905)119891 (119909)

sign (119909) |119909|119901minus1

minus 120582119890 (119905)

sign (119909) |119909|119901minus1

=(119901 minus 1) 120582119903 (119905)

|119909|120574(119901minus1)

119892(|119909|119901minus1

120582119903 (119905)|120596 (119905)|)

+ 120582119902 (119905)119891 (119909)

sign (119909) |119909|119901minus1minus 120582

119890 (119905)

sign (119909) |119909|119901minus1

(12)

ge(119901 minus 1) 120582119903 (119905)

|119909|120574(119901minus1)

(|119909|119901minus1

120582119903 (119905))

120574

1198920 (|120596 (119905)|)

+ 120582119902 (119905)119891 (119909)

sign (119909) |119909|119901minus1minus 120582

119890 (119905)

sign (119909) |119909|119901minus1

=119901 minus 1

(120582119903 (119905))120574minus1

1198920 (|120596 (119905)|)+120582119902 (119905)

119891 (119909)

sign (119909) |119909|119901minus1

minus 120582119890 (119905)

sign (119909) |119909|119901minus1

(73)

ge119901 minus 1

(120582119903 (119905))120574minus1

1198920 (|120596 (119905)|) + 120582119902 (119905) 119870

minus 120582119890 (119905)

sign (119909) |119909|119901minus1

(85)

Since 119890(119905)(sign(119909)|119909|119901minus1) le 0 we have

1199081015840(119905) ge

119901 minus 1

(120582119903 (119905))120574minus1

1198920(|120596 (119905)|) + 120582119870119902 (119905) (86)

We apply Lemma 23 to (82) to obtain a function119862(119905) suchthat

int

119887

119886

119862 (119905) 119889119905 = 1 (87)

and by (75)

0 le 119862 (119905) le119901

2120587sin(120587

119901)min

119901 minus 1

119903120574minus1

(119905) 120582120574minus1

119870120582119902 (119905)

1 +1198720

(88)

Let 1199040isin (minus(120587119901)(1 sin(120587119901)) (120587119901)(1 sin(120587119901))) be such

that tan119901(1199040) le 120596(119886) We define the function

119881 (119905) = 1199040+2120587

119901

1

sin (120587119901)int

119905

119886

119862 (120591) 119889120591 (89)

Note that (minus(120587119901)(1 sin(120587119901))) lt 1199040

= 119881(119886) lt

(120587119901)(1 sin(120587119901)) and 119881(119887) = 1199040+ (2120587119901)(1 sin(120587119901)) gt

(120587119901)(1 sin(120587119901)) Since119881 is continuous there exists a 119879lowast isin(119886 119887) such that 119881(119905) lt (120587119901)(1 sin(120587119901)) for 119905 isin [119886 119879lowast) and119881(119879lowast) = (120587119901)(1 sin(120587119901))

We define

120596 (119905) = tan119901(119881 (119905)) 119905 isin [119886 119879

lowast) (90)

and note that by Lemma 24 lim119905rarr119879

lowast120596(119905) = +infin wheretan119901(119904) = 119910(119904) and 119910(119904) is determined by Lemma 24The derivative of 120596 then satisfies

1205961015840(119905) = tan1015840

119901(119881 (119905)) 119881

1015840(119905)

= (1 + tan119901119901(119881 (119905))) 119881

1015840(119905)

=2120587

119901

1

sin (120587119901)(1 + 120596

119901(119905)) 119862 (119905)

le (1 +1198720+ 1198920(1003816100381610038161003816120596 (119905)

1003816100381610038161003816))min

119901 minus 1

119903120574minus1

(119905) 120582120574minus1

119870120582119902 (119905)

1 +1198720

le119901 minus 1

119903120574minus1

(119905) 120582120574minus1

1198920(1003816100381610038161003816120596 (119905)

1003816100381610038161003816) + 119870120582119902 (119905)

(91)

Thus we may apply the comparison principle ofLemma 15 to 120596(119886) le 120596(119886) and their respective differentialinequalities to conclude that 120596(119905) le 120596(119905) for all 119905 isin [119886 119887]But since 120596 rarr +infin as 119905 rarr 119879

lowastlt 119887 and 120596(119905) is well defined

on the whole segment [119886 119887] we arrive at a contradictionHence 119909(119905) gt 0 cannot hold for all 119905 isin [119886 119887] and since 119909(119905) iscontinuous there exists a 119905lowast isin [119886 119887] such that 119909(119905lowast) = 0

Theorem 26 Let 120574 gt 119901 gt 1 Assume (72) and (73) and let forany 119879 gt 119905

0there exist 119879 le 119886

1lt 1198871le 1198862lt 1198872such that 119892(119905) ge 0

on [1198861 1198871] cup [119886

2 1198872] and 119890(119905) le 0 on [119886

1 1198871] and 119890(119905) ge 0 on

International Journal of Differential Equations 11

[1198862 1198872] If for both 119894 isin 1 2 there exist constants 120582

119894gt 0 such

that

119901

2120587sin (120587119901)int

119887119894

119886119894

min119901 minus 1

119903120574minus1

(119905) 120582120574minus1

119894

119870120582119894119902 (119905)

1 +1198720

119889119905 ge 1

(92)

then (1) is oscillatory

Proof For (1) to be oscillatory it is sufficient that for every119899 isin N there exists a 119905

119899gt 119899 such that 119909(119905

119899) = 0

Let 119899 isin N and let 119899 lt 1198861lt 1198871le 1198862lt 1198872as assumed in the

theorem It is enough to show that every solution 119909(119905) of (1)has a zero on [119886

1 1198872] If 119909(119886

1) ge 0 we apply Proposition 25 to

the segment [1198861 1198871] to obtain the sought zero If 119909(119886

2) le 0 we

again apply Proposition 25 to the segment [1198862 1198872] Suppose

119909(1198861) lt 0 and 119909(119886

2) gt 0 Then since 119909 is continuous there

exists a number 1199051isin [1198861 1198862] such that 119909(119905

1) = 0

Acknowledgment

This work is supported by the Ministry of Science of theRepublic of Croatia under Grant no 036-0361621-1291

References

[1] Q Kong ldquoNonoscillation and oscillation of second order half-linear differential equationsrdquo Journal of Mathematical Analysisand Applications vol 332 no 1 pp 512ndash522 2007

[2] Q-R Wang and Q-G Yang ldquoInterval criteria for oscillationof second-order half-linear differential equationsrdquo Journal ofMathematical Analysis and Applications vol 291 no 1 pp 224ndash236 2004

[3] Q Long and Q-R Wang ldquoNew oscillation criteria of second-order nonlinear differential equationsrdquo Applied Mathematicsand Computation vol 212 no 2 pp 357ndash365 2009

[4] Z Zheng and F Meng ldquoOscillation criteria for forced second-order quasi-linear differential equationsrdquo Mathematical andComputer Modelling vol 45 no 1-2 pp 215ndash220 2007

[5] D Cakmak and A Tiryaki ldquoOscillation criteria for certainforced second-order nonlinear differential equationsrdquo AppliedMathematics Letters vol 17 no 3 pp 275ndash279 2004

[6] F Jiang and F Meng ldquoNew oscillation criteria for a class ofsecond-order nonlinear forced differential equationsrdquo Journalof Mathematical Analysis and Applications vol 336 no 2 pp1476ndash1485 2007

[7] Q Yang ldquoInterval oscillation criteria for a forced secondorder nonlinear ordinary differential equations with oscillatorypotentialrdquo Applied Mathematics and Computation vol 135 no1 pp 49ndash64 2003

[8] A Tiryaki and B Ayanlar ldquoOscillation theorems for certainnonlinear differential equations of second orderrdquo Computers ampMathematics with Applications vol 47 no 1 pp 149ndash159 2004

[9] J Jaros T Kusano and N Yoshida ldquoGeneralized Piconersquosformula and forced oscillations in quasilinear differential equa-tions of the second orderrdquoArchivumMathematicum vol 38 no1 pp 53ndash59 2002

[10] R P Agarwal S R Grace and D OrsquoRegan Oscillation Theoryfor Second Order Linear Half-Linear Superlinear and SublinearDynamic Equations Kluwer Academic Publishers DordrechtThe Netherlands 2002

[11] S Jing ldquoA new oscillation criterion for forced second-orderquasilinear differential equationsrdquoDiscrete Dynamics in Natureand Society vol 2011 Article ID 428976 8 pages 2011

[12] W-T Li and S S Cheng ldquoAn oscillation criterion for nonhomo-geneous half-linear differential equationsrdquoAppliedMathematicsLetters vol 15 no 3 pp 259ndash263 2002

[13] J S W Wong ldquoOscillation criteria for a forced second-orderlinear differential equationrdquo Journal of Mathematical Analysisand Applications vol 231 no 1 pp 235ndash240 1999

[14] M Pasic ldquoFite-Wintner-Leighton type oscillation criteria forsecond-order differential equations with nonlinear dampingrdquoAbstract and Applied Analysis vol 2013 Article ID 852180 10pages 2013

[15] M Pasic ldquoNew interval oscillation criteria for forced second-order differential equations with nonlinear dampingrdquo Interna-tional Journal of Mathematical Analysis vol 7 no 25 pp 1239ndash1255 2013

[16] L Gasinski and N S Papageorgiou Nonsmooth Critical PointTheory and Nonlinear Boundary Value Problems vol 8 of Seriesin Mathematical Analysis and Applications Edited by R PAgarwal and D OrsquoRegan Chapman amp HallCRC Boca RatonFla USA 2005

[17] S Lang Real and Functional Analysis vol 142 of GraduateTexts in Mathematics Springer-Verlag New York NY USA 3rdedition 1993

[18] I N Bronshtein K A Semendyayev G Musiol and HMuehligHandbook of Mathematics Springer Berlin Germany5th edition 2007

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

2 International Journal of Differential Equations

5 10 15

051

minus1

minus05

Figure 1 Function 119909(119905) = |sin(119905)| sin(119905) is a solution of (4)

5 10 15

051

minus1

minus05

Figure 21199091015840(119905) = 2|sin(119905)| cos(119905) and hence11990910158401015840(119905) is not a continuousfunction

(iii) 119902(119905) is nonnegative and discontinuous on [1199050infin) as

11990910158401015840+ 41198982[sign (cos (119898119905))]+119891 (119909) = ℎ (sin (119898119905))

ae in [1199050infin)

119909 1199091015840isin 119860119862loc ([1199050infin) R)

(6)

(iv) 119902(119905) is sign changing and discontinuous on [1199050infin) as

11990910158401015840+ 41198982 sign (cos (119898119905)) 119891 (119909) = ℎ (sin (119898119905))

ae in [1199050infin)

119909 1199091015840isin 119860119862loc ([1199050infin) R)

(7)

where 119898 isin N and ℎ(119904) is an arbitrary function such thatℎ(119904) 119904 gt 0 for all 119904 = 0 for instance ℎ(119904) = 119904 or ℎ(119904) = sign(119904)According to Corollaries 7 and 10 we will show that (4)ndash(7) are oscillatory provided the function 119891 = 119891(119906) satisfies119891(119906)119906 ge 119870 ge 1 for all 119906 = 0 see Examples 8ndash13 It isinteresting that in particular for 119891(119906) = 119906 and ℎ(119904) =

21198982 sign(119904) (4) allows an explicit oscillatory solution 119909(119905) =

|sin(119898119905)| sin(119898119905) as shown in Figures 1 and 2Moreover as a consequence of Corollary 7 one can

show that all solutions of (4) are oscillatory for details seeExample 8 The main goal of this paper is to give somesufficient conditions on functions Φ(119906 V) 119891(119906) and thecoefficients 119903(119905) 119902(119905) and 119890(119905) such that (1) is oscillatory seeTheorems 3 and 4 It will also cover the model equations (4)ndash(7) as well as some other examples presented in Section 2

To the best of our knowledge it seems that there are onlyfew papers which study the oscillation of the second-orderdifferential equations with nonsmooth (local integrable)coefficients see [1ndash3]More precisely in [1] the author studiedthe interval oscillation criteria for the following second-orderhalf-linear differential equation

(119903 (119905)100381610038161003816100381610038161199091015840(119905)10038161003816100381610038161003816

120590minus1

1199091015840(119905))

1015840

+ 119902 (119905) |119909 (119905)|120590minus1

119909 (119905) = 0

ae in (0infin)

119909 11990310038161003816100381610038161003816119909101584010038161003816100381610038161003816

120590minus1

1199091015840isin 119860119862loc ((0infin) R)

(8)

where 120590 gt 1 and 1119903 119902 isin 119871 loc((0infin)R) such thatintinfin

0119903minus1120590

(119905)119889119903 = infin See also [2] but with the solutionspace1198621((0infin)R) instead of119860119862loc((0infin)R) that is 119909 and119903|1199091015840|120590minus1

1199091015840isin 1198621((0infin)R)

In [3] the authors consider the following second-orderdifferential equation

(119903 (119905) 1199091015840(119905))1015840

+ 119876 (119905 119909 (119905) 1199091015840(119905)) = 0 119905 ge 119905

0 (9)

where 119903(119905) gt 0 ae in [1199050infin) 1119903 isin 119871 loc([1199050infin)R) and

119876(119905 119910 119911) is locally integrable function in 119905 and continuous in(119910 119911) Equation (9) allows the forcing term 119890(119905) in the nextsense as follows

119910119876 (119905 119910 119911) ge 119902 (119905) 119910119891 (119910) minus 119890 (119905) 119910

forall (119905 119910 119911) isin [1199050infin) timesR

2

(10)

where 119890 = 119890(119905) satisfies (2) but the functions 119902 = 119902(119905) and119891 = 119891(119910) are smooth enough in their variables that is 119902 isin

119862([1199050infin)R) and 119891 isin 119862

1(RR)

On certain oscillation criteria for various classes offorced second-order differential equations with continuouscoefficients we refer the reader to [4ndash13] Our methodmodifies a recently used one in [14 15] and it containsthe classic Riccati transformation of the main equation ablow-up argument and pointwise comparison principle Thecomparison principle applies to all sub- and supersolutions ofa class of the generalized Riccati differential equations withnonlinear terms that are supposed to be locally integrablein the first variable and locally Lipschitz continuous in thesecond variable

2 Hypotheses Results and Consequences

First of all the functionΦ(119906 V)which appears in the second-order differential operator of (1) satisfies

|119906|120574minus2

VΦ (119906 V) ge 119892 (|Φ (119906 V)|) forall119906 V isin R (11)

where 120574 ge 2 and 119892 R+rarr R+is a locally Lipschitz function

1198920 R+rarr R+satisfying

119892 (119888119904) ge 1198881205741198920 (119904) forall119888 gt 0 119904 gt 0

1198920(119904) + 119872

0ge 1199042 for some 119872

0ge 0 and all 119904 isin R

+

(12)

In most cases 119892(119904) = 1198920(119904)1198921(119904) 119904 gt 0 where 119892

0(119904) = 119904

120574120574 ge 2 and 119892

1(119904) is an arbitrary function satisfying 119892

1(119904) ge 1

Thus for such 119892(119904) with 1198921(119904) equiv 1 condition (11) became

|119906|120574minus2

VΦ (119906 V) ge |Φ (119906 V)|120574

forall119906 V isin R (13)

It is not difficult to check that if 1198920isin 1198621(R+) or 119892

0(119904) is a

convex function then it is locally Lipschitz onR+too see for

instance [16 Theorem 133]Two essential classes of the second-order differential

operators (119903(119905)Φ(119909 1199091015840))1015840 satisfy condition (13) as is shown inthe next examples

International Journal of Differential Equations 3

Example 1 We consider the second-order differential opera-tor which is linear in 1199091015840 as follows

(119903 (119905) Φ (119909 1199091015840))1015840

= 120572(120601 (119909) 1199091015840)1015840

(14)

where 120572 gt 0 and 0 le 120601(119906) le 1 for all 119906 isin R Obviously thefunctionΦ(119906 V) = 120601(119906)V satisfies condition (13) in particularfor 120574 = 2 Two usual choices for 120601(119906) are 120601(119906) = |sin 119906| and120601(119906) = |119906|(1 + |119906|)

Example 2 We consider a quasilinear differential operator(the so-called one-dimensional prescribed mean curvatureoperator) as follows

(119903 (119905) Φ (119909 1199091015840))1015840

= 120572(120601 (119909)1199091015840

radic1 + 11990910158402)

1015840

(15)

where 120572 gt 0 and 0 le 120601120574minus1(119906) le |119906|

120574minus2 for all 119906 isin R It is notdifficult to check that condition (13) is satisfied in particularfor Φ(119906 V) = 120601(119906)V(1 + V2)12 and for any 120574 gt 1 For 120601(119906)we can take the same choice as in the previous example

Next we suppose the existence of a constant119870 such that

119891 (119906)

119906ge 119870 gt 0 forall119906 = 0 (16)

In order to simplify our consideration here inmany exampleswe often use 119891(119906) = 119870119906

Condition (2) means that there exists a sequence of pairsof intervals 119869

1119895= [1198861119895 1198871119895] and 119869

2119895= [1198862119895 1198872119895] 119895 isin N

contained in (1199050infin) such that the sequences (119886

1119895)119895ge1

(1198871119895)119895ge1

(1198862119895)119895ge1

and (1198872119895)119895ge1

are increasing 1198861119895lt 1198871119895le 1198862119895lt 1198872119895for

each 119895 and

119890 (119905) ge 0 on 1198691119895 119890 (119905) le 0 on 119869

2119895

for each 119895 isin N

lim119895rarrinfin

1198861119895= infin

(17)

On the intervals 1198691119895and 1198692119895 the coefficient 119903(119905) satisfies

119903 (119905) gt 0 on 119869119894119895 1199031minus120574

isin 1198711(119869119894119895)

forall119894 isin 1 2 119895 isin N(18)

Let there be a real function 119862 = 119862(119905) 119862 isin

1198711

loc((1199050infin)R) and let there exist a sequence of positive realnumbers (120582

119895)119895isinN such that

119862 (119905) ge 0 on 119869119894119895 119888119894119895= int

119869119894119895

119862 (120591) 119889120591 gt 0

forall119894 isin 1 2 119895 isin N

1

119888119894119895

119862 (119905) le1

120587min(120582

119895119903 (119905))1minus120574

119870

1198720+ 1

120582119895119902 (119905)

forall119905 isin 119869119894119895 119894 isin 1 2 119895 isin N

(19)

where 120574 1198720 and 119870 are constants defined in (11) (12) and

(16) respectivelyThe proof of the following main result will be presented

in Section 4

Theorem 3 Let the functions Φ(119906 V) 119891(119906) 119890(119905) and 119903(119905)

satisfy (11) (12) (16) (17) and (18) respectively Let 119902(119905) ge 0

and 119902(119905) equiv 0 on each interval 119869119894119895 119894 isin 1 2 119895 isin N If (19) is

fulfilled then (1) is oscillatory

Condition (19) can be replaced by an equivalent onewhich has a more practical value and takes a simpler formsince we do not need a sequence of auxiliary parameters(120582119895)119895isinN let there be a real function 119862 = 119862(119905) 119862 isin

1198711

loc((1199050infin)R) such that

119862 (119905) ge 0 on 119869119894119895 119888119894119895= int

119868119894119895

119862 (120591) 119889120591 gt 0 forall119894 isin 1 2

119895 isin N

sup119905isin119869119894119895

[119862 (119905)

119902 (119905)] sup119905isin119869119894119895

[119903 (119905) 119862(119905)1(120574minus1)

] le119870

1198720+ 1

(

119888119894119895

120587)

120574(120574minus1)

forall119894 isin 1 2 119895 isin N

(20)

where 120574 1198720 and 119870 are constants defined in (11) (12) and

(16) respectively Since we will show that (19) and (20)are equivalent see page 8 the next oscillation criterionimmediately follows fromTheorem 3

Theorem 4 Let the functions Φ(119906 V) 119891(119906) 119890(119905) and 119903(119905)

satisfy (11) (12) (16) (17) and (18) respectively Let 119902(119905) ge 0

and 119902(119905) equiv 0 on each interval 119869119894119895 119894 isin 1 2 119895 isin N If (20) is

fulfilled then (1) is oscillatory

Remark 5 Assuming that 119871 = lim119895rarrinfin

1198861119895

lt infin wecan ensure the oscillation in the point 119871 Note that 119871 =

lim119895rarrinfin

1198862119895

since 1198861119895

lt 1198862119895

lt 1198861119895+1

Thus we can generatea one-sided (right) limit

Now we consider some consequences of Theorem 4which depend on the qualitative properties of the coefficient119902(119905)

Substituting 119862(119905) equiv 1 in (20) Theorem 4 implies thefollowing result involving lower bounds on the lengths ofintervals |119869

119894119895| = 119887119894119895minus 119886119894119895

Corollary 6 (119902(119905) is positive) Let the functionsΦ(119906 V) 119891(119906)119890(119905) and 119903(119905) satisfy (11) (12) (16) (17) and (18) respectivelyLet inf

119905isin119869119894119895119902(119905) gt 0 for each 119894 isin 1 2 119895 isin N If

10038161003816100381610038161003816119869119894119895

10038161003816100381610038161003816ge 120587(

(1198720+ 1) sup

119905isin119869119894119895119903 (119905)

119870inf119905isin119869119894119895

119902 (119905))

(120574minus1)120574

forall119894 isin 1 2 119895 isin N

(21)

then (1) is oscillatory

4 International Journal of Differential Equations

Corollary 7 (119902(119905) is bounded from below by a positiveconstant) Let the functionsΦ(119906 V)119891(119906) 119890(119905) and 119903(119905) satisfy(11) (12) (16) (17) and (18) respectively Let there be twoconstants 119903

0 1199020satisfying

0 lt 119903 (119905) le 1199030 119902 (119905) ge 119902

0gt 0 forall119894 isin 1 2 119895 isin N (22)

If

10038161003816100381610038161003816119869119894119895

10038161003816100381610038161003816ge 120587(

(1198720+ 1) 1199030

1198701199020

)

(120574minus1)120574

forall119894 isin 1 2 119895 isin N (23)

then (1) is oscillatory where 1205741198720 and119870 are constants defined

in (11) (12) and (16) respectively

Example 8 (oscillation of (4)) We know that 119909(119905) =

|sin(119898119905)| sin(119898119905) is an oscillatory solution of (4) Howeveraccording to Corollary 7 we can show that all solutions of (4)are oscillatory Indeed since Φ(119906 V) equiv V the conditions (11)and (12) are satisfied especially for 120574 = 2 119892(119904) = 119892

0(119904) = 119904

2

and 1198720= 0 Next 119891(119906) equiv 119906 implies that condition (16) is

satisfied especially for 119870 = 1 Since 119903(119905) equiv 1 and 119902(119905) equiv 41198982

it is clear that conditions (18) and (22) are also satisfied inparticular for 119903

0= 1 and 119902

0= 4119898

2 Moreover since 119890(119905) =ℎ(sin(119898119905)) and ℎ(119904)119904 gt 0 119904 = 0 we have that (17) is fulfilled for1198861119895= 2119895120587119898 119887

1119895= (2119895 + 1)120587119898 = 119886

2119895and 1198872119895= (2119895 + 2)120587119898

Moreover10038161003816100381610038161003816119869119894119895

10038161003816100381610038161003816= 119887119894minus 119886119894=120587

119898gt 0 119894 isin 1 2 (24)

Hence we conclude that the required condition (23) isfulfilled that is10038161003816100381610038161003816119869119894119895

10038161003816100381610038161003816=120587

119898ge

120587

radic41198982

= 120587((1198720+ 1) 1199030

1199020

)

(120574minus1)120574

ge 120587((1198720+ 1) 1199030

1198701199020

)

(120574minus1)120574

(25)

Thus all conditions of Corollary 7 are satisfied and hence (4)is oscillatory

Example 9 We consider the following class of equations

11990910158401015840minus 2

11987810158401015840(119905)

119878 (119905)119909 = 2 sign (119878 (119905)) 1198781015840

2

(119905) ae in [1199050infin)

119909 1199091015840isin 119860119862loc ([1199050infin) R) 119909 notin 119862

2((1199050infin) R)

(26)

where 119878 = 119878(119905) 119878 isin 1198622(R) is an oscillatory function such thatthe zeros 119905

119899of the function sign(119878(119905))11987810158402(119905) satisfy 119905

119899rarr infin

there is a 1205910isin R such that 119905

119899+1minus 119905119899ge 1205910gt 0 for all 119899 isin N and

119878(119905) = 0 on (119905119899 119905119899+1) This equation allows an explicitly given

oscillatory solution 119909(119905) = |119878(119905)|119878(119905) Moreover if there is aconstant 119904

0gt 0 such that

minus211987810158401015840(119905)

119878 (119905)ge 1199040 119905 isin (119905

119899 119905119899+1) 120591

0ge

120587

radic1199040

(27)

then by Corollary 7 we conclude that (26) is oscillatoryIndeed conditions (11) (12) and (16) are satisfied by the samereasons as in Example 8 Condition (18) is satisfied becauseof 119903(119905) equiv 1 Also from (27) it follows that (22) and (23) arefulfilled in particular for 120574 = 2119872

0= 0 119903

0= 1 119902

0= 1199040 and

119870 ge 1 that is10038161003816100381610038161003816119869119894119895

10038161003816100381610038161003816=10038161003816100381610038161003816119905119895+1

minus 119905119895

10038161003816100381610038161003816ge 1205910ge

120587

radic1199040

= 120587((1198720+ 1) 1199030

1199020

)

(120574minus1)120574

ge 120587((1198720+ 1) 1199030

1198701199020

)

(120574minus1)120574

(28)

Hence Corollary 7 proves this result

As the second consequence ofTheorem3 is unlike the firstone we consider the case when the coefficient 119902(119905) is not astrictly positive function Here by 119902 = 0 we denote the setof all 119905 isin R such that 119902(119905) = 0

Corollary 10 (119902(119905) is nonnegative but not equiv 0) Let thefunctions Φ(119906 V) 119891(119906) 119890(119905) and 119903(119905) satisfy (11) (12) (16)(17) and (18) respectively Let 119902(119905) ge 0 on each interval 119869

119894119895

119894 isin 1 2 119895 isin N such that

119902119894119895= int

119869119894119895

119902 (120591) 119889120591 gt 0 119894 isin 1 2 119895 isin N (29)

If

119902120574

119894119895

119902 (119905)ge 120587120574((1198720+ 1) 119903 (119905)

119870)

120574minus1

119905 isin 119869119894119895 119902 = 0 119894 isin 1 2 119895 isin N

(30)

then (1) is oscillatory

Proof It suffices to show that (30) is equivalent to theexistence of a real number 120582

119895such that

120582119895ge120587 (1198720+ 1)

119870

1

119902119894119895

1

119902119894119895

119902 (119905) le1

120587(120582119895119903 (119905))1minus120574

119905 isin 119869119894119895 119894 isin 1 2 119895 isin N

(31)

The claim will then follow fromTheorem 3 Inequality (31) isfor any 119905 isin 119869

119894119895 119902 = 0 equivalent to

120587 (1198720+ 1)

119870

1

119902119894119895

le 120582119895le (

119902119894119895

120587119902 (119905))

1(120574minus1)1

119903 (119905) (32)

that is to

120587 (1198720+ 1)

119870

1

119902119894119895

le (

119902119894119895

120587119902 (119905))

1(120574minus1)1

119903 (119905) (33)

This inequality is easily seen to be equivalent to (30) for any119905 isin 119869119894119895 119902 = 0 Note that if 119905 isin 119902 = 0 then the second

inequality in (31) is trivially satisfied

International Journal of Differential Equations 5

The following three examples are simple consequences ofCorollary 10

Example 11 (oscillation of (5)) Equation (5) as well asequation

11990910158401015840+ 11989821205872[sin (119898119905)]+119891 (119909) = ℎ (cos (119898119905))

ae in [1199050infin)

119909 1199091015840isin 119860119862loc ([1199050infin) R)

(34)

are oscillatory where 119891(119906) satisfies (16) with 119870 ge 1 andℎ(119904)119904 gt 0 119904 = 0 Indeed in (5) we have 119903(119905) equiv 1 119902(119905) =

11989821205872[cos(119898119905)] and 119890(119905) = ℎ(sin(119898119905)) and so (17) and (18) are

fulfilled and 119902(119905) ge 0 119902(119905) equiv 0 on each interval 119869119894119895 119894 isin 1 2

119895 isin N where 1198691119895= [1198861119895 1198871119895] 1198692119895= [1198862119895 1198872119895] and

1198861119895=2119895120587

119898 119887

1119895=(2119895 + 12) 120587

119898

1198862119895=(2119895 + 32) 120587

119898 119887

2119895=(2119895 + 2) 120587

119898

(35)

Moreover

119902119894119895= int

119869119894119895

119902 (120591) 119889120591

= 11989821205872int

119887119894119895

119886119894119895

[cos (119898120591)]+119889120591 = 1198981205872 119894 isin 1 2 119895 isin N

(36)

and since 120574 = 2 119903(119905) equiv 1119870 ge 1 and1198720= 0 for 120582

119895= 1(119898120587)

we have

120582119895=

1

119898120587ge1

119870

1

119898120587=120587 (1198720+ 1)

119870

1

1198981205872

=120587 (1198720+ 1)

119870

1

119902119894119895

119894 isin 1 2 119895 isin N

1

119902119894119895

119902 (119905) =1

119898120587211989821205872[cos (119898119905)]+ le 119898

=1

120587

1

(1119898120587)=1

120587

1

(120582119895119903 (119905))120574minus1

119905 isin 119869119894119895 119894 isin 1 2 119895 isin N

(37)

Thus the required condition (30) is fulfilled in this case andhence we may apply Corollary 10 to (5) to verify that thisequation is oscillatory Analogously we can show that (34)is oscillatory too

In the next example the coefficient 119902(119905) is a discontinuousfunction on [119905

0infin)

Example 12 (oscillation of (6)) If 119891(119906) satisfies (16) with119870 ge

1 then (6) is oscillatory since the required condition (30) issatisfied especially for 119902(119905) = 4119898

2 sign(cos(119898119905)) on 119869119894119895 120582119895=

1(2119898) and 119869119894119895as in (35) 120574 = 2 119903(119905) equiv 1119870 ge 1 and119872

0= 0

Indeed

119902119894119895= int

119869119894119895

119902 (120591) 119889120591 = 41198982int

119869119894119895

[sign (cos (119898120591))]+119889120591

= 41198982 120587

2119898= 2119898120587 119894 isin 1 2 119895 isin N

120582119895=

1

2119898=

120587

2119898120587= 120587 (119872

0+ 1)

1

119902119894119895

ge120587 (1198720+ 1)

119870

1

119902119894119895

119894 isin 1 2 119895 isin N

1

119902119894

119902 (119905) =1

211989812058741198982[sign (cos (119898119905))]+

le2119898

120587=1

120587

1

radic1 (41198982)

=1

120587(120582119895119903 (119905))120574minus1

119905 isin 119869119894119895 119894 isin 1 2 119895 isin N

(38)

which shows that (30) is satisfied

At the end of this section we consider the case when 119902(119905)changes sign on [119905

0infin) but with the help of Corollary 10 since

119902(119905) is strictly positive on all intervals 119869119894119895

Example 13 (oscillation of (7)) If 119891(119906) satisfies (16) with119870 ge

1 then model equation (7) is oscillatory In fact let 119869119894119895be as

in (35) If ℎ(119904)119904 gt 0 119904 = 0 it is clear that 119890(119905) = ℎ(sin(119898119905))satisfies (2) that is 119890(119905) ge 0 on 119869

1119895and 119890(119905) le 0 on 119869

2119895 On

the other hand we have 119902(119905) = 41198982 sign(cos(119898119905)) on 119869

119894119895and

so 119902(119905) ge 0 119902(119905) equiv 0 on 119869119894119895for all 119894 isin 1 2 119895 isin N Hence

the proof of the fact that all assumptions of Corollary 7 arefulfilled is the same as in the preceding example

3 Proofs of Theorems 3 and 4

Let 119860119862loc([119886 119887)R) denote the set of all real functions whichare absolutely continuous on every interval [119886 119887 minus 120576] where120576 isin (0 119887 minus 119886) For arbitrary numbers 119879

0lt 119879lowast and functions

119866 = 119866(119905 119906) and 119887 = 119887(119905) we consider the ordinary differentialequation

1199081015840= 119866 (119905 119908) + 119887 (119905) ae in [119879

0 119879lowast) (39)

which generalizes the classic Riccati equation 1199081015840 = 119886(119905)1199082+

119887(119905) where 119886(119905) is an arbitrary function We associate to(39) the corresponding sub- and supersolutions 119908119908 isin

119860119862loc([1198790 119879lowast)R) which satisfy respectively

1199081015840le 119866 (119905 119908) + 119887 (119905) 119908

1015840ge 119866 (119905 119908) + 119887 (119905)

ae in [1198790 119879lowast)

(40)

We are interested in studying the following property

119908 (1198790) le 119908 (119879

0) implies 119908 (119905) le 119908 (119905)

forall119905 isin [T0 119879lowast)

(41)

In this way we introduce the following definition

6 International Journal of Differential Equations

Definition 14 We say that comparison principle holds for (39)on an interval [119879

0 119879lowast) 119879 le 119879

0lt 119879lowast if property (41)

is fulfilled for all sub- and supersolutions 119908119908 isin

119860119862loc([1198790 119879lowast)R) of (39) on [119879

0 119879lowast)

Now we are able to state the next result

Lemma 15 Let 119879 le 1198790lt 119879lowast Let 119866 = 119866(119905 119906) and 119887 = 119887(119905) be

two arbitrary function For any119872 gt 0 let there be a function119871 = 119871(119905) gt 0 depending on 119879

0 119879lowast119872 such that

119871 (119905) gt 0 on [1198790 119879lowast) 119871 isin 119871

1(1198790 119879lowast)

1003816100381610038161003816119866 (119905 119906

1) minus 119866 (119905 119906

2)1003816100381610038161003816le 119871 (119905)

10038161003816100381610038161199061minus 1199062

1003816100381610038161003816

forall119905 isin [1198790 119879lowast) 1199061 1199062isin [minus119872119872]

(42)

Then comparison principle (41) holds for (39) on an interval[1198790 119879lowast) 119879 le 119879

0lt 119879lowast

Proof It is enough to use the idea of the proof of [14 Lemma41] Hence this proof is left to the reader

Corollary 16 Let 120582 gt 0 and let 1198920 R+rarr R+be a locally

Lipschitz function on R+ Let 119870 be an arbitrary real number

and 119902(119905) an arbitrary function Let 119879 le 1198790lt 119879lowast If 119903(119905) gt 0

on [1198790 119879lowast) and 119903minus1+120574 isin 1198711(119879

0 119879lowast) then comparison principle

(41) holds for the Riccati differential equation

1199081015840=

1

(120582119903 (119905))120574minus1

1198920 (|119908|) + 119870120582119902 (119905) ae in [119879

0 119879lowast)

(43)

Proof It is clear that (43) is a particular case of (39) inparticular for

119866 (119905 119906) =1

(120582119903 (119905))120574minus1

1198920 (|119906|) 119887 (119905) = 119870120582119902 (119905) (44)

Since1198920is a locally Lipschitz function onR

+ for every119872 gt 0

there is an 1198710gt 0 depending on119872 such that

10038161003816100381610038161198920(1199041) minus 1198920(1199042)1003816100381610038161003816le 1198710

10038161003816100381610038161199041minus 1199042

1003816100381610038161003816 forall119904

1 1199041isin [minus119872119872]

(45)

Hence for any119872 gt 0 and all 1199061 1199062isin [minus119872119872] we obtain

1003816100381610038161003816119866 (119905 119906

1) minus 119866 (119905 119906

2)1003816100381610038161003816=

1

(120582119903 (119905))120574minus1

10038161003816100381610038161198920(10038161003816100381610038161199061

1003816100381610038161003816) minus 1198920(10038161003816100381610038161199062

1003816100381610038161003816)1003816100381610038161003816

le1198710

(120582119903 (119905))120574minus1

1003816100381610038161003816

10038161003816100381610038161199061

1003816100381610038161003816minus10038161003816100381610038161199062

1003816100381610038161003816

1003816100381610038161003816

le1198710

(120582119903 (119905))120574minus1

10038161003816100381610038161199061minus 1199062

1003816100381610038161003816

(46)

Thus according to assumption 119903minus1+120574

isin 1198711(1198790 119879lowast) the

required condition (42) is fulfilled in particular for 119871(119905) =

1198710(120582119903(119905))

minus120574+1 and so Lemma 15 proves this corollary

Before we give the proof ofTheorem 3 we state and provethe next two propositions In the first one by a nonoscillatorysolution 119909(119905) of the main equation (1) we get the existenceof a supersolution 119908(119905) of the Riccati differential equation(43) on the interval (119886

1 1198871) or (119886

2 1198872) In the second one we

construct two subsolutions 1199081(119905) and 119908

2(119905) of (43) which

blow up on intervals (1198861 119879lowast

1) sube (119886

1 1198871) and (119886

2 119879lowast

2) sube (119886

2 1198872)

respectively

Proposition 17 Let Φ(119906 V) and 119891(119906) satisfy (11) (12) and(16) respectively and let 119890(119905) satisfy (2) Let 119903(119905) gt 0 119902(119905) ge0 and 119902(119905) equiv 0 on [119886

1 1198871] cup [119886

2 1198872] Let 119909 = 119909(119905) be a

nonoscillatory solution of (1) and let for some 119879 ge 1199050and

120582 gt 0 the function 119908 = 119908(119905) be defined by

119908 (119905) = minus

120582119903 (119905)Φ (119909 (119905) 1199091015840(119905))

119909 (119905) 119905 ge 119879 (47)

Then 119908 isin 119860119862loc([119879infin)R) and 119908(119905) satisfies the inequality

1199081015840ge (120582119903 (119905))

1minus1205741198920(|119908|) + 119870120582119902 (119905) ae in 119869 (48)

where 119869 = (1198861 1198871) if 119909(119905) lt 0 and 119869 = (119886

2 1198872) if 119909(119905) gt 0

Proof Since 119909 = 119909(119905) is a nonoscillatory solution of (1) thereis a 119879 ge 119905

0such that 119909(119905) = 0 on [119879infin) Hence 119908(119905) is well

defined by (47)Because of (2) we have 119890(119905)119909(119905) le 0 for all 119905 isin 119869 where

119869 = (1198861 1198871) if 119909(119905) lt 0 and 119869 = (119886

2 1198872) if 119909(119905) gt 0 and since

120582 gt 0 we have

minus120582119890 (119905)

119909 (119905)ge 0 in 119869 (49)

Next since 119909 and 119903(119905)Φ(119909(119905) 1199091015840(119905)) are from

119860119862loc([1199050infin)R) we can take the first derivative of 119908(119905)for almost everywhere in 119869 which together with 119903(119905) gt 0119902(119905) ge 0 and 119902(119905) equiv 0 on 119869 gives

1199081015840(119905) = minus

120582(119903 (119905) Φ (119909 (119905) 1199091015840(119905)))1015840

119909 (119905)

+

120582119903 (119905)Φ (119909 (119905) 1199091015840(119905))

1199092(119905)

1199091015840(119905)

=120582119903 (119905)

|119909 (119905)|120574Φ(119909 (119905) 119909

1015840(119905)) 1199091015840(119905) |119909 (119905)|

120574minus2

+ 120582119902 (119905)119891 (119909 (119905))

119909 (119905)minus 120582

119890 (119905)

119909 (119905)

International Journal of Differential Equations 7

ge120582119903 (119905)

|119909 (119905)|120574119892 (10038161003816100381610038161003816Φ (119909 (119905) 119909

1015840(119905))

10038161003816100381610038161003816) + 119870120582119902 (119905) minus 120582

119890 (119905)

119909 (119905)

=120582119903 (119905)

|119909 (119905)|120574119892(

|119909 (119905)|

120582119903 (119905)|119908 (119905)|) + 119870120582119902 (119905) minus 120582

119890 (119905)

119909 (119905)

ge120582119903 (119905)

|119909 (119905)|120574

|119909 (119905)|120574

(120582119903 (119905))1205741198920 (|119908 (119905)|)

+ 119870120582119902 (119905) minus 120582119890 (119905)

119909 (119905)ae in 119869

(50)

that is

1199081015840ge (120582119903 (119905))

1minus1205741198920(|119908|) + 119870120582119902 (119905) minus 120582

119890 (119905)

119909 (119905)ae in 119869

(51)

Hence (49) and (51) prove the desired inequality (48)

Proposition 18 Let (19) be satisfied Let 1198771and 119877

2be two

arbitrary real numbers Then there are two points 119879lowast1isin (1198861 1198871)

and 119879lowast

2isin (1198862 1198872) and two continuous functions 119908

1(119905) and

1199082(119905) such that

119908119894(119886119894) le 119877119894 lim119905rarr119879

lowast

119894

119908119894(119905) = infin 119894 isin 1 2

1199081015840

119894le (120582119903 (119905))

1minus1205741198920(1003816100381610038161003816119908119894

1003816100381610038161003816) + 119870120582119902 (119905)

for ae 119905 isin (119886119894 119879lowast

119894) 119894 isin 1 2

(52)

Proof Since the function 119910(119904) = tan(119904) is a bijection fromthe interval (minus1205872 1205872) intoR we observe that there are two1199041 1199042isin (minus1205872 1205872) such that

tan (119904119894) le 119877119894 119894 isin 1 2 (53)

Next we define two functions 1198811(119905) and 119881

2(119905) by

119881119894 (119905) = 119904119894 +

120587

119888119894

int

119905

119886119894

119862 (120591) 119889120591 119905 isin [119886119894 119887119894] 119894 isin 1 2 (54)

where the real numbers 1198881 1198882and the function 119862(119905) are

defined in (19) From (19) and (54) one can immediatelyconclude that

119881119894(119886119894) = 119904119894lt120587

2 119881

119894(119887119894) = 119904119894+ 120587 gt

120587

2 119894 isin 1 2 (55)

Since 119862 isin 1198711

loc((1199050infin)R) it follows that 119862 isin 1198711((1199050infin)R)

which implies that 119881119894isin 119860119862([119886

119894 119887119894]R) 119894 isin 1 2 Therefore

exploiting the fact that in (55) we have 119881119894isin 119862([119886

119894 119887119894]R) we

obtain the existence of two points 119879lowast1isin (1198861 1198871) and 119879

lowast

2isin

(1198862 1198872) such that

119881119894(119879lowast

119894)=

120587

2 minus

120587

2lt119881119894(119905)lt

120587

2

forall119905 isin [119886119894 119879lowast

119894) 119894 isin 1 2

(56)

Now we are able to define the following two functions 1205961(119905)

and 1205962(119905) by

120596119894(119905) = tan (119881i (119905)) 119905 isin [119886

119894 119879lowast

119894) 119894 isin 1 2 (57)

That are well defined because of (56) Moreover from (53)(54) and (56) we obtain

120596119894(119886119894) = tan (119904

119894) le 119877119894

lim119905rarr119879

lowast

119898

120596119894(119905) = tan (119881

119894(119879lowast

119894)) = tan(120587

2) = infin

(58)

Also since119881119894isin 119860119862([119886

119894 119887119894]R) we can take the first derivative

of 120596119894(119905) and hence from (19) and (57) we obtain

1199081015840

119894(119905) =

120587

119888119894

119862 (119905)1

cos2 (119881119894 (119905))

=120587

119888119894

119862 (119905) (1003816100381610038161003816120596119894(119905)1003816100381610038161003816

2+ 1)

le120587

119888119894

119862 (119905) [1198920(1003816100381610038161003816120596119894(119905)1003816100381610038161003816) + 119872

0+ 1]

le120587

119888119894

119862 (119905) 1198920(1003816100381610038161003816120596119894(119905)1003816100381610038161003816) +

120587

119888119894

119862 (119905) (1198720+ 1)

le (120582119903 (119905))1minus1205741198920(1003816100381610038161003816120596119894(119905)1003816100381610038161003816) + 119870120582119902 (119905) ae in (119886

119894 119879lowast

119894)

(59)

which together with (58) proves this proposition

Now we are able to present the proof of Theorem 3 basedon Lemma 15 and Propositions 17 and 18

Proof of Theorem 3 Assuming the contrary then there is anonoscillatory solution 119909(119905) and 119879 ge 119905

0such that 119909(119905) = 0 for

all 119905 ge 119879 119909(119905) and 119903(119905)Φ(119909 1199091015840) are from 119860119862loc([1199050infin)R)In order to simplify notation let every 119895 isin N be fixed andomitted in the notation For instance instead of (119886

1119895 1198871119895) and

(1198862119895 1198872119895) we write (119886

1 1198871) and (119886

2 1198872) respectively and so on

On one hand we observe by Proposition 17 that thefunction 119908(119905) defined by (47) is a supersolution of thefollowing Riccati differential equation

1199081015840= (120582119903 (119905))

1minus1205741198920(|119908|) + 119870120582119902 (119905) ae in 119869 (60)

where 119908 isin 119860119862loc(119869R) and 119869 = (1198861 1198871) if 119909(119905) lt 0 and 119869 =

(1198862 1198872) if 119909(119905) gt 0 see the proof of Proposition 17 Let for

instance 119909(119905) lt 0 and thus 119869 = (1198861 1198871)

On the other hand by Proposition 18 there are a number119879lowast

1isin (1198861 1198871) and subsolutions 119908

1(119905) 1199081isin 119860119862([119886

1 119879lowast

1)R) of

the Riccati differential equation (60) such that

1199081(1198861) le 119908 (119886

1) lim

119905rarr119879lowast

1

1199081(119905) = infin (61)

(in the case when 119869 = (1198862 1198872) then we work with119879lowast

2isin (1198862 1198872)

and the other subsolution 1199082(119905) 119908

2isin 119860119862([119886

2 119879lowast

2)R))

Hence by Corollary 16 and (61) we conclude that1199081(119905) le 119908(119905)

for all 119905 isin [1198861 119879lowast

1) and therefore

infin = lim119905rarr119879

lowast

1

1199081(119905) le lim119905rarr119879

lowast

1

119908 (119905) (62)

which contradicts 119908 isin 119860119862loc([119879infin)R) Thus the assump-tion that 119909(119905) is nonoscillatory is not possible and hence everysolution of (1) is oscillatory

8 International Journal of Differential Equations

Proof of Theorem 4 There is only one difference betweenTheorems 3 and 4 and it is the difference between conditions(19) and (20) Hence Theorem 4 immediately follows fromTheorem 3 provided that we prove the equivalence between(19) and (20)

First of all we need the following two lemmas

Lemma 19 Let 119886 119887 and 119888 be positive real numbers and 120574 gt 1Then the existence of a positive real number 120582 such that

119888 le min 119886

120582120574minus1

119887120582 (63)

is equivalent to

119888 le 119886112057411988711205741015840

(64)

Here 1205741015840 = 120574(120574 minus 1) is the conjugate exponent of 120574

Proof Let us define 119892(120582) = min119886120582120574minus1 119887120582 Since 120582 997891rarr

119886120582120574minus1 is decreasing and 120582 997891rarr 119887120582 is increasing there exists

a unique point of maximum of 119892 It is achieved at 120582 = 1205820

which is a solution of 119886120582120574minus10

= 1198871205820 hence 120582

0= (119886119887)

1120574If there is a positive real number 120582 such that 119888 le 119892(120582)

then 119888 le 119892(1205820) = 119887120582

0= 119887(119886119887)

1120574= 119886112057411988711205741015840

Converselyif 119888 le 119886

112057411988711205741015840

then 119888 le 119892(1205820) and the equivalence in the

lemma is proved

The preceding lemma is a special case of the followingmore general statement

Lemma 20 Assume that 119886(119905) 119887(119905) and 119888(119905) are positive realfunctions defined on a subset 119869 sube R Then the existence of apositive real number 120582 such that for all 119905 isin 119869

119888 (119905) le min 119886 (119905)120582120574minus1

119887 (119905) 120582 (65)

is equivalent to

sup119905isin119869

119888 (119905)

119887 (119905)le (inf119905isin119869

119886 (119905)

119888 (119905))

1(120574minus1)

(66)

Proof The condition that there exists 120582 gt 0 such that (65)is true for all 119905 isin 119869 is equivalent with the following twoinequalities which have to hold simultaneously

119888 (119905) le119886 (119905)

120582120574minus1

119888 (119905) le 119887 (119905) 120582 (67)

that is with

119888 (119905)

119887 (119905)le 120582 le (

119886 (119905)

119888 (119905))

1(120574minus1)

forall119905 isin 119869 (68)

Taking the supremum of the left-hand side and the infimumof the right-hand side we obtain (66) Conversely if (66)is satisfied then since the left-hand side in (66) cannotbe equal to zero while the right-hand side is less than infinthen inequality (66) defines a nonempty closed interval in R

(possibly reducing to just one point) in which we can chooseany positive 120582 Hence (67) is satisfied for all 119905 isin 119869 andtherefore (65) as well

Remark 21 If 119886 = 119886(119905) 119887 = 119887(119905) and 119888 = 119888(119905) appearingin Lemma 20 are constant functions then condition (66) isequivalent to 119888119887 = (119886119888)1(120574minus1) that is to (64)

Proof of Equivalence of (19) and (20) Wewill use Lemma 20Let us fix 119869

119894119895 where 119894 isin 1 2 and 119895 isin N We see that that the

inequality appearing in the second line of condition (19) is ofthe form (65) where

119888 (119905) =1

119888119894119895

119862 (119905) 119886 (119905) =1

120587119903(119905)1minus120574

119887 (119905) =119870119902 (119905)

120587 (1198720+ 1)

(69)

Condition (66) is equivalent to

sup119905isin119869119894119895

119862 (119905)

119902 (119905)le

1198701198881205741015840

119894119895

1205871205741015840

(1198720+ 1)

inf119905isin119869119894119895

1

119903 (119905) 119862(119905)120574minus1

=

1198701198881205741015840

119894119895

(1198720+ 1) sup

119905isin119869119894119895119903 (119905) 119862(119905)

120574minus1(

119888119894119895

120587)

1205741015840

(70)

This inequality is equivalent to the corresponding one in (19)and the claim follows from Lemma 19

Proof of Corollary 6 Substituting inf119905isin119869119894119895

119902(119905) instead of 119902(119905)and sup

119905isin119869119894119895119903(119905) instead of 119903(119905) in the inequality appearing in

the second line of (20) after a short computation we obtain astronger inequality than (20) as

1

119888119894119895

sup119905isin119869119894119895

119862 (119905)

le1

120587(

119870inf119905isin119869119894119895

119902 (119905)

(1198720+ 1) sup

119905isin119869119894119895119903 (119905)

)

(120574minus1)120574

forall119894 isin 1 2 119895 isin N

(71)

Substituting119862(119905) equiv 1we obtain that the left-hand side of (71)is equal to |119869

119894119895|minus1 and the resulting inequality is equivalent to

(21) Since (21) implies (20) the claim is proved

Remark 22 Here we show that the choice of 119862(119905) equiv 1 inthe proof of Corollary 6 is the best possible To prove thisnote that on the left-hand side of (71) we have the expressiondepending on an auxiliary function 119862(119905) and this functionis absent on the right-hand side Therefore it has sense totry to find a function 119862(119905) 119905 isin 119869

119894119895 such that the value of

119876(119862) = (1119888119894119895)sup119905isin119869119894119895

119862(119905) is minimal (note that 119888119894119895depends

on the function 119862 = 119862(119905) as well) It is easy to see that theminimum is achieved for 119862(119905) equiv 1 (or any positive constant)Indeed since 119876(119862) = 119876(120583119888) for any 120583 gt 0 by taking 120583 =

(sup119905isin119869119894119895

119862(119905))minus1 it suffices to assume that sup

119905isin119869119894119895119862(119905) = 1

The value of 119876(119862) is minimal if 119888119894119895= int119869119894119895

119862(120591) 119889120591 is maximalpossible and since 0 le 119862(119905) le 1 it is clear that the maximumof 119888119894119895= 119888119894119895(119862) is achieved when 119862(119905) equiv 1

International Journal of Differential Equations 9

Proof of Corollary 7 It is clear that the condition (23) implies(21) Furthermore since the lengths |119869

119894119895| are uniformly

bounded from below by a positive constant see (23) thenobviously 119886

1119895rarr infin as 119895 rarr infin Thus the claim follows

immediately from Corollary 6

Proof of Corollary 10 Let 119862(119905) = 119902(119905) Then the requiredcondition (19) is clearly satisfied because of assumption(30) Hence this corollary immediately follows from Theo-rem 3

4 An Extension of Condition (12) tothe Case of 120574 gt 1

In this section we consider the oscillation of (1) in the casewhen 120574 isin (1 2) is allowed in the assumption (12) In this waythe assumption (11) is slightly modified by a real number 119901 gt1 such that

|119906|(119901minus1)120574minus119901

VΦ (119906 V) ge 119892 (|Φ (119906 V)|) (72)

where (12) is supposed that is 119892(119888119904) ge 1198881205741198920(119904) for every 119888 119904 gt

0 and for a locally Lipschitz function 1198920 [0infin) rarr [0infin)

for which there exists a constant1198720such that 119892

0(119904)+119872

0ge 119904119901

for all 119904 isin [0infin)For the function 119891 we assume that there exists a constant

119870 gt 0 such that

119891 (119906) sign (119906) ge 119870|119906|119901minus1 for every 119906 isin R (73)

We will need the following two lemmas

Lemma 23 Let 119886 lt 119887 and let 119891 [119886 119887] rarr [0infin) be a meas-urable function Then

int

119887

119886

119891 (119905) 119889119905 ge 1 (74)

if and only if there exists a function 119892 isin 1198711([119886 119887] [0infin))

1198921= 1 such that

119891 (119905) ge 119892 (119905) forall119905 isin [119886 119887] (75)

Proof We assume (74) If int119887119886119891(119905)119889119905 is finite we may choose

119892(119909) = 119891(119909)1198911

For int119887119886119891(119905)119889119905 = infin we may define for every natural

number 119899 a function

119891119899(119905) =

119891 (119905) 119891 (119905) le 119899

119899 119891 (119905) gt 119899(76)

Since 119891119899(119905) le 119899 we have 119891

119899isin 1198711([119886 119887]R) Obviously

lim119899int119887

119886119891119899(119905)119889119905 is also infiniteThis implies that there exists an

1198990such that int119887

1198861198911198990(119905)119889119905 ge 1 We define 119892(119905) = 119891

1198990(119905)119891

11989901

Now we assume (75) Integrating over the whole intervalwe get

int

119887

119886

119891 (119905) 119889119905 ge int

119887

119886

119892 (119905) 119889119905 = 1 (77)

Lemma 24 Let119901 gt 1There exists a unique function 119910 = 119910(119904)that satisfies the following Cauchy problem

1199101015840(119904) = 1 +

1003816100381610038161003816119910 (119904)

1003816100381610038161003816

119901 119904 isin R (78)

119910 (0) = 0 (79)

Furthermore there exists a finite real number 119878lowast such that

lim119904rarrplusmn119878

lowast119910 (119904) = plusmninfin (80)

and 119910 isin 1198621((minus119878lowast 119878lowast)R) is injective and monotonous We

have 119878lowast = (120587119901)(1(sin(120587119901))) and 119910 is odd

We will denote by tan119901the solution of (78)-(79)

Proof Obviously every solution to (78) is injective andmonotonous on a connected set since 1199101015840 = 1 + |119910|119901 ge 1 gt 0

Let 119891(119909) = 1(1 + |119909|119901) Obviously 1 ge 119891(119909) gt 0 for all

119909 isin R and 119862119901gt |1198911015840(119909)| gt 0 for some constants 119862

119901isin R

We may assume 119862119901gt 1 Hence 1199111015840 = 119891(119905) 119911(0) = 0 has (by

[17Theorem 31]) a locally unique solution on (minus1119862119901 1119862119901)

By the same argument 119911 can be extended to the whole of Rinterval by interval of type ((119899 minus 1)119862

119901 (119899 + 1)119862

119901) for all

119899 isin ZWe know from [18] that the for the integral of the left-

hand side we have

int

infin

0

119891 (119909) 119889119909 = int

infin

0

119889119909

1 + 119909119901=120587

119901

1

sin (120587119901)= 119878lowast (81)

so lim119905rarrplusmninfin

119911(119905) = plusmn119878lowast

Since 1199111015840(119905) = 0 119911 is injective monotonous and of class1198621(R (minus119878lowast 119878lowast)) and so we define 119910(119904) = 119911

minus1(119904) It imme-

diately follows that 119910 isin 1198621((minus119878lowast 119878lowast)R) and that 119910 is

injective and monotonous Also by continuity of 119911 we havelim119904rarrplusmn119878

lowast119910(119904) = lim119908rarrplusmninfin

119910(119911(119908)) = lim119908rarrplusmninfin

119908 = plusmninfinDifferentiating it also follows that 1199101015840 = 1 + |119910|

119901 and bydefinition of 119910 119910(0) = 119911minus1(0) = 0 So 119910 is a solution of (78)-(79) Since any other such solution must have 119911 as its inverse119910 is unique on its domain

Proposition 25 Let 119886 lt 119887 and 120574 gt 119901 gt 1 Assume (72) and(73) and let 119902(119905) ge 0 and 119890(119905) le 0 (or 119890(119905) ge 0 ) for all 119905 isin [119886 119887]If there exists a constant 120582 gt 0 such that

119901

2120587sin(120587

119901)int

119887

119886

min119901 minus 1

119903120574minus1

(119905) 120582120574minus1

119870120582119902 (119905)

1 +1198720

119889119905 ge 1 (82)

then for any solution 119909(119905) of (1) with 119909(119886) ge 0 (or 119909(119886) le 0)there exists a number 119905lowast isin [119886 119887] such that 119909(119905lowast) = 0

Proof We assume that 119909(119905) = 0 for all 119905 isin [119886 119887] Then we maydefine the function

120596 (119905) = minus

120582119903 (119905) Φ (119909 (119905) 1199091015840(119905))

sign (119909 (119905)) |119909 (119905)|119901minus1 (83)

10 International Journal of Differential Equations

This function is well defined on [119886 119887] and its derivative is

1199081015840(119905) = minus

120582(119903 (119905) Φ (119909 (119905) 1199091015840(119905)))1015840

sign (119909 (119905)) |119909 (119905)|119901minus1

+

(119901 minus 1) 120582119903 (119905) Φ (119909 (119905) 1199091015840(119905))

|119909 (119905)|119901

1199091015840(119905)

(84)

Using the fact that 119909(119905) is a solution of (1) and omitting 119905 from119909 and 1199091015840 for clarity we get

1199081015840(119905) =

120582 (119902 (119905) 119891 (119909) minus 119890 (119905))

sign (119909) |119909|119901minus1+

(119901 minus 1) 120582119903 (119905) Φ (119909 1199091015840)

119909119901

1199091015840

=

(119901 minus 1) 120582119903 (119905) Φ (119909 1199091015840)

|119909|119901

1199091015840+ 120582119902 (119905)

119891 (119909)

sign (119909) |119909|119901minus1

minus 120582119890 (119905)

sign (119909) |119909|119901minus1

=(119901 minus 1) 120582119903 (119905)

|119909|120574(119901minus1)

Φ(119909 1199091015840) 1199091015840|119909|(119901minus1)120574minus119901

+ 120582119902 (119905)119891 (119909)

sign (119909) |119909|119901minus1minus 120582

119890 (119905)

sign (119909) |119909|119901minus1

(72)

ge(119901 minus 1) 120582119903 (119905)

|119909|120574(119901minus1)

119892 (10038161003816100381610038161003816Φ (119909 119909

1015840)10038161003816100381610038161003816)

+ 120582119902 (119905)119891 (119909)

sign (119909) |119909|119901minus1

minus 120582119890 (119905)

sign (119909) |119909|119901minus1

=(119901 minus 1) 120582119903 (119905)

|119909|120574(119901minus1)

119892(|119909|119901minus1

120582119903 (119905)|120596 (119905)|)

+ 120582119902 (119905)119891 (119909)

sign (119909) |119909|119901minus1minus 120582

119890 (119905)

sign (119909) |119909|119901minus1

(12)

ge(119901 minus 1) 120582119903 (119905)

|119909|120574(119901minus1)

(|119909|119901minus1

120582119903 (119905))

120574

1198920 (|120596 (119905)|)

+ 120582119902 (119905)119891 (119909)

sign (119909) |119909|119901minus1minus 120582

119890 (119905)

sign (119909) |119909|119901minus1

=119901 minus 1

(120582119903 (119905))120574minus1

1198920 (|120596 (119905)|)+120582119902 (119905)

119891 (119909)

sign (119909) |119909|119901minus1

minus 120582119890 (119905)

sign (119909) |119909|119901minus1

(73)

ge119901 minus 1

(120582119903 (119905))120574minus1

1198920 (|120596 (119905)|) + 120582119902 (119905) 119870

minus 120582119890 (119905)

sign (119909) |119909|119901minus1

(85)

Since 119890(119905)(sign(119909)|119909|119901minus1) le 0 we have

1199081015840(119905) ge

119901 minus 1

(120582119903 (119905))120574minus1

1198920(|120596 (119905)|) + 120582119870119902 (119905) (86)

We apply Lemma 23 to (82) to obtain a function119862(119905) suchthat

int

119887

119886

119862 (119905) 119889119905 = 1 (87)

and by (75)

0 le 119862 (119905) le119901

2120587sin(120587

119901)min

119901 minus 1

119903120574minus1

(119905) 120582120574minus1

119870120582119902 (119905)

1 +1198720

(88)

Let 1199040isin (minus(120587119901)(1 sin(120587119901)) (120587119901)(1 sin(120587119901))) be such

that tan119901(1199040) le 120596(119886) We define the function

119881 (119905) = 1199040+2120587

119901

1

sin (120587119901)int

119905

119886

119862 (120591) 119889120591 (89)

Note that (minus(120587119901)(1 sin(120587119901))) lt 1199040

= 119881(119886) lt

(120587119901)(1 sin(120587119901)) and 119881(119887) = 1199040+ (2120587119901)(1 sin(120587119901)) gt

(120587119901)(1 sin(120587119901)) Since119881 is continuous there exists a 119879lowast isin(119886 119887) such that 119881(119905) lt (120587119901)(1 sin(120587119901)) for 119905 isin [119886 119879lowast) and119881(119879lowast) = (120587119901)(1 sin(120587119901))

We define

120596 (119905) = tan119901(119881 (119905)) 119905 isin [119886 119879

lowast) (90)

and note that by Lemma 24 lim119905rarr119879

lowast120596(119905) = +infin wheretan119901(119904) = 119910(119904) and 119910(119904) is determined by Lemma 24The derivative of 120596 then satisfies

1205961015840(119905) = tan1015840

119901(119881 (119905)) 119881

1015840(119905)

= (1 + tan119901119901(119881 (119905))) 119881

1015840(119905)

=2120587

119901

1

sin (120587119901)(1 + 120596

119901(119905)) 119862 (119905)

le (1 +1198720+ 1198920(1003816100381610038161003816120596 (119905)

1003816100381610038161003816))min

119901 minus 1

119903120574minus1

(119905) 120582120574minus1

119870120582119902 (119905)

1 +1198720

le119901 minus 1

119903120574minus1

(119905) 120582120574minus1

1198920(1003816100381610038161003816120596 (119905)

1003816100381610038161003816) + 119870120582119902 (119905)

(91)

Thus we may apply the comparison principle ofLemma 15 to 120596(119886) le 120596(119886) and their respective differentialinequalities to conclude that 120596(119905) le 120596(119905) for all 119905 isin [119886 119887]But since 120596 rarr +infin as 119905 rarr 119879

lowastlt 119887 and 120596(119905) is well defined

on the whole segment [119886 119887] we arrive at a contradictionHence 119909(119905) gt 0 cannot hold for all 119905 isin [119886 119887] and since 119909(119905) iscontinuous there exists a 119905lowast isin [119886 119887] such that 119909(119905lowast) = 0

Theorem 26 Let 120574 gt 119901 gt 1 Assume (72) and (73) and let forany 119879 gt 119905

0there exist 119879 le 119886

1lt 1198871le 1198862lt 1198872such that 119892(119905) ge 0

on [1198861 1198871] cup [119886

2 1198872] and 119890(119905) le 0 on [119886

1 1198871] and 119890(119905) ge 0 on

International Journal of Differential Equations 11

[1198862 1198872] If for both 119894 isin 1 2 there exist constants 120582

119894gt 0 such

that

119901

2120587sin (120587119901)int

119887119894

119886119894

min119901 minus 1

119903120574minus1

(119905) 120582120574minus1

119894

119870120582119894119902 (119905)

1 +1198720

119889119905 ge 1

(92)

then (1) is oscillatory

Proof For (1) to be oscillatory it is sufficient that for every119899 isin N there exists a 119905

119899gt 119899 such that 119909(119905

119899) = 0

Let 119899 isin N and let 119899 lt 1198861lt 1198871le 1198862lt 1198872as assumed in the

theorem It is enough to show that every solution 119909(119905) of (1)has a zero on [119886

1 1198872] If 119909(119886

1) ge 0 we apply Proposition 25 to

the segment [1198861 1198871] to obtain the sought zero If 119909(119886

2) le 0 we

again apply Proposition 25 to the segment [1198862 1198872] Suppose

119909(1198861) lt 0 and 119909(119886

2) gt 0 Then since 119909 is continuous there

exists a number 1199051isin [1198861 1198862] such that 119909(119905

1) = 0

Acknowledgment

This work is supported by the Ministry of Science of theRepublic of Croatia under Grant no 036-0361621-1291

References

[1] Q Kong ldquoNonoscillation and oscillation of second order half-linear differential equationsrdquo Journal of Mathematical Analysisand Applications vol 332 no 1 pp 512ndash522 2007

[2] Q-R Wang and Q-G Yang ldquoInterval criteria for oscillationof second-order half-linear differential equationsrdquo Journal ofMathematical Analysis and Applications vol 291 no 1 pp 224ndash236 2004

[3] Q Long and Q-R Wang ldquoNew oscillation criteria of second-order nonlinear differential equationsrdquo Applied Mathematicsand Computation vol 212 no 2 pp 357ndash365 2009

[4] Z Zheng and F Meng ldquoOscillation criteria for forced second-order quasi-linear differential equationsrdquo Mathematical andComputer Modelling vol 45 no 1-2 pp 215ndash220 2007

[5] D Cakmak and A Tiryaki ldquoOscillation criteria for certainforced second-order nonlinear differential equationsrdquo AppliedMathematics Letters vol 17 no 3 pp 275ndash279 2004

[6] F Jiang and F Meng ldquoNew oscillation criteria for a class ofsecond-order nonlinear forced differential equationsrdquo Journalof Mathematical Analysis and Applications vol 336 no 2 pp1476ndash1485 2007

[7] Q Yang ldquoInterval oscillation criteria for a forced secondorder nonlinear ordinary differential equations with oscillatorypotentialrdquo Applied Mathematics and Computation vol 135 no1 pp 49ndash64 2003

[8] A Tiryaki and B Ayanlar ldquoOscillation theorems for certainnonlinear differential equations of second orderrdquo Computers ampMathematics with Applications vol 47 no 1 pp 149ndash159 2004

[9] J Jaros T Kusano and N Yoshida ldquoGeneralized Piconersquosformula and forced oscillations in quasilinear differential equa-tions of the second orderrdquoArchivumMathematicum vol 38 no1 pp 53ndash59 2002

[10] R P Agarwal S R Grace and D OrsquoRegan Oscillation Theoryfor Second Order Linear Half-Linear Superlinear and SublinearDynamic Equations Kluwer Academic Publishers DordrechtThe Netherlands 2002

[11] S Jing ldquoA new oscillation criterion for forced second-orderquasilinear differential equationsrdquoDiscrete Dynamics in Natureand Society vol 2011 Article ID 428976 8 pages 2011

[12] W-T Li and S S Cheng ldquoAn oscillation criterion for nonhomo-geneous half-linear differential equationsrdquoAppliedMathematicsLetters vol 15 no 3 pp 259ndash263 2002

[13] J S W Wong ldquoOscillation criteria for a forced second-orderlinear differential equationrdquo Journal of Mathematical Analysisand Applications vol 231 no 1 pp 235ndash240 1999

[14] M Pasic ldquoFite-Wintner-Leighton type oscillation criteria forsecond-order differential equations with nonlinear dampingrdquoAbstract and Applied Analysis vol 2013 Article ID 852180 10pages 2013

[15] M Pasic ldquoNew interval oscillation criteria for forced second-order differential equations with nonlinear dampingrdquo Interna-tional Journal of Mathematical Analysis vol 7 no 25 pp 1239ndash1255 2013

[16] L Gasinski and N S Papageorgiou Nonsmooth Critical PointTheory and Nonlinear Boundary Value Problems vol 8 of Seriesin Mathematical Analysis and Applications Edited by R PAgarwal and D OrsquoRegan Chapman amp HallCRC Boca RatonFla USA 2005

[17] S Lang Real and Functional Analysis vol 142 of GraduateTexts in Mathematics Springer-Verlag New York NY USA 3rdedition 1993

[18] I N Bronshtein K A Semendyayev G Musiol and HMuehligHandbook of Mathematics Springer Berlin Germany5th edition 2007

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

International Journal of Differential Equations 3

Example 1 We consider the second-order differential opera-tor which is linear in 1199091015840 as follows

(119903 (119905) Φ (119909 1199091015840))1015840

= 120572(120601 (119909) 1199091015840)1015840

(14)

where 120572 gt 0 and 0 le 120601(119906) le 1 for all 119906 isin R Obviously thefunctionΦ(119906 V) = 120601(119906)V satisfies condition (13) in particularfor 120574 = 2 Two usual choices for 120601(119906) are 120601(119906) = |sin 119906| and120601(119906) = |119906|(1 + |119906|)

Example 2 We consider a quasilinear differential operator(the so-called one-dimensional prescribed mean curvatureoperator) as follows

(119903 (119905) Φ (119909 1199091015840))1015840

= 120572(120601 (119909)1199091015840

radic1 + 11990910158402)

1015840

(15)

where 120572 gt 0 and 0 le 120601120574minus1(119906) le |119906|

120574minus2 for all 119906 isin R It is notdifficult to check that condition (13) is satisfied in particularfor Φ(119906 V) = 120601(119906)V(1 + V2)12 and for any 120574 gt 1 For 120601(119906)we can take the same choice as in the previous example

Next we suppose the existence of a constant119870 such that

119891 (119906)

119906ge 119870 gt 0 forall119906 = 0 (16)

In order to simplify our consideration here inmany exampleswe often use 119891(119906) = 119870119906

Condition (2) means that there exists a sequence of pairsof intervals 119869

1119895= [1198861119895 1198871119895] and 119869

2119895= [1198862119895 1198872119895] 119895 isin N

contained in (1199050infin) such that the sequences (119886

1119895)119895ge1

(1198871119895)119895ge1

(1198862119895)119895ge1

and (1198872119895)119895ge1

are increasing 1198861119895lt 1198871119895le 1198862119895lt 1198872119895for

each 119895 and

119890 (119905) ge 0 on 1198691119895 119890 (119905) le 0 on 119869

2119895

for each 119895 isin N

lim119895rarrinfin

1198861119895= infin

(17)

On the intervals 1198691119895and 1198692119895 the coefficient 119903(119905) satisfies

119903 (119905) gt 0 on 119869119894119895 1199031minus120574

isin 1198711(119869119894119895)

forall119894 isin 1 2 119895 isin N(18)

Let there be a real function 119862 = 119862(119905) 119862 isin

1198711

loc((1199050infin)R) and let there exist a sequence of positive realnumbers (120582

119895)119895isinN such that

119862 (119905) ge 0 on 119869119894119895 119888119894119895= int

119869119894119895

119862 (120591) 119889120591 gt 0

forall119894 isin 1 2 119895 isin N

1

119888119894119895

119862 (119905) le1

120587min(120582

119895119903 (119905))1minus120574

119870

1198720+ 1

120582119895119902 (119905)

forall119905 isin 119869119894119895 119894 isin 1 2 119895 isin N

(19)

where 120574 1198720 and 119870 are constants defined in (11) (12) and

(16) respectivelyThe proof of the following main result will be presented

in Section 4

Theorem 3 Let the functions Φ(119906 V) 119891(119906) 119890(119905) and 119903(119905)

satisfy (11) (12) (16) (17) and (18) respectively Let 119902(119905) ge 0

and 119902(119905) equiv 0 on each interval 119869119894119895 119894 isin 1 2 119895 isin N If (19) is

fulfilled then (1) is oscillatory

Condition (19) can be replaced by an equivalent onewhich has a more practical value and takes a simpler formsince we do not need a sequence of auxiliary parameters(120582119895)119895isinN let there be a real function 119862 = 119862(119905) 119862 isin

1198711

loc((1199050infin)R) such that

119862 (119905) ge 0 on 119869119894119895 119888119894119895= int

119868119894119895

119862 (120591) 119889120591 gt 0 forall119894 isin 1 2

119895 isin N

sup119905isin119869119894119895

[119862 (119905)

119902 (119905)] sup119905isin119869119894119895

[119903 (119905) 119862(119905)1(120574minus1)

] le119870

1198720+ 1

(

119888119894119895

120587)

120574(120574minus1)

forall119894 isin 1 2 119895 isin N

(20)

where 120574 1198720 and 119870 are constants defined in (11) (12) and

(16) respectively Since we will show that (19) and (20)are equivalent see page 8 the next oscillation criterionimmediately follows fromTheorem 3

Theorem 4 Let the functions Φ(119906 V) 119891(119906) 119890(119905) and 119903(119905)

satisfy (11) (12) (16) (17) and (18) respectively Let 119902(119905) ge 0

and 119902(119905) equiv 0 on each interval 119869119894119895 119894 isin 1 2 119895 isin N If (20) is

fulfilled then (1) is oscillatory

Remark 5 Assuming that 119871 = lim119895rarrinfin

1198861119895

lt infin wecan ensure the oscillation in the point 119871 Note that 119871 =

lim119895rarrinfin

1198862119895

since 1198861119895

lt 1198862119895

lt 1198861119895+1

Thus we can generatea one-sided (right) limit

Now we consider some consequences of Theorem 4which depend on the qualitative properties of the coefficient119902(119905)

Substituting 119862(119905) equiv 1 in (20) Theorem 4 implies thefollowing result involving lower bounds on the lengths ofintervals |119869

119894119895| = 119887119894119895minus 119886119894119895

Corollary 6 (119902(119905) is positive) Let the functionsΦ(119906 V) 119891(119906)119890(119905) and 119903(119905) satisfy (11) (12) (16) (17) and (18) respectivelyLet inf

119905isin119869119894119895119902(119905) gt 0 for each 119894 isin 1 2 119895 isin N If

10038161003816100381610038161003816119869119894119895

10038161003816100381610038161003816ge 120587(

(1198720+ 1) sup

119905isin119869119894119895119903 (119905)

119870inf119905isin119869119894119895

119902 (119905))

(120574minus1)120574

forall119894 isin 1 2 119895 isin N

(21)

then (1) is oscillatory

4 International Journal of Differential Equations

Corollary 7 (119902(119905) is bounded from below by a positiveconstant) Let the functionsΦ(119906 V)119891(119906) 119890(119905) and 119903(119905) satisfy(11) (12) (16) (17) and (18) respectively Let there be twoconstants 119903

0 1199020satisfying

0 lt 119903 (119905) le 1199030 119902 (119905) ge 119902

0gt 0 forall119894 isin 1 2 119895 isin N (22)

If

10038161003816100381610038161003816119869119894119895

10038161003816100381610038161003816ge 120587(

(1198720+ 1) 1199030

1198701199020

)

(120574minus1)120574

forall119894 isin 1 2 119895 isin N (23)

then (1) is oscillatory where 1205741198720 and119870 are constants defined

in (11) (12) and (16) respectively

Example 8 (oscillation of (4)) We know that 119909(119905) =

|sin(119898119905)| sin(119898119905) is an oscillatory solution of (4) Howeveraccording to Corollary 7 we can show that all solutions of (4)are oscillatory Indeed since Φ(119906 V) equiv V the conditions (11)and (12) are satisfied especially for 120574 = 2 119892(119904) = 119892

0(119904) = 119904

2

and 1198720= 0 Next 119891(119906) equiv 119906 implies that condition (16) is

satisfied especially for 119870 = 1 Since 119903(119905) equiv 1 and 119902(119905) equiv 41198982

it is clear that conditions (18) and (22) are also satisfied inparticular for 119903

0= 1 and 119902

0= 4119898

2 Moreover since 119890(119905) =ℎ(sin(119898119905)) and ℎ(119904)119904 gt 0 119904 = 0 we have that (17) is fulfilled for1198861119895= 2119895120587119898 119887

1119895= (2119895 + 1)120587119898 = 119886

2119895and 1198872119895= (2119895 + 2)120587119898

Moreover10038161003816100381610038161003816119869119894119895

10038161003816100381610038161003816= 119887119894minus 119886119894=120587

119898gt 0 119894 isin 1 2 (24)

Hence we conclude that the required condition (23) isfulfilled that is10038161003816100381610038161003816119869119894119895

10038161003816100381610038161003816=120587

119898ge

120587

radic41198982

= 120587((1198720+ 1) 1199030

1199020

)

(120574minus1)120574

ge 120587((1198720+ 1) 1199030

1198701199020

)

(120574minus1)120574

(25)

Thus all conditions of Corollary 7 are satisfied and hence (4)is oscillatory

Example 9 We consider the following class of equations

11990910158401015840minus 2

11987810158401015840(119905)

119878 (119905)119909 = 2 sign (119878 (119905)) 1198781015840

2

(119905) ae in [1199050infin)

119909 1199091015840isin 119860119862loc ([1199050infin) R) 119909 notin 119862

2((1199050infin) R)

(26)

where 119878 = 119878(119905) 119878 isin 1198622(R) is an oscillatory function such thatthe zeros 119905

119899of the function sign(119878(119905))11987810158402(119905) satisfy 119905

119899rarr infin

there is a 1205910isin R such that 119905

119899+1minus 119905119899ge 1205910gt 0 for all 119899 isin N and

119878(119905) = 0 on (119905119899 119905119899+1) This equation allows an explicitly given

oscillatory solution 119909(119905) = |119878(119905)|119878(119905) Moreover if there is aconstant 119904

0gt 0 such that

minus211987810158401015840(119905)

119878 (119905)ge 1199040 119905 isin (119905

119899 119905119899+1) 120591

0ge

120587

radic1199040

(27)

then by Corollary 7 we conclude that (26) is oscillatoryIndeed conditions (11) (12) and (16) are satisfied by the samereasons as in Example 8 Condition (18) is satisfied becauseof 119903(119905) equiv 1 Also from (27) it follows that (22) and (23) arefulfilled in particular for 120574 = 2119872

0= 0 119903

0= 1 119902

0= 1199040 and

119870 ge 1 that is10038161003816100381610038161003816119869119894119895

10038161003816100381610038161003816=10038161003816100381610038161003816119905119895+1

minus 119905119895

10038161003816100381610038161003816ge 1205910ge

120587

radic1199040

= 120587((1198720+ 1) 1199030

1199020

)

(120574minus1)120574

ge 120587((1198720+ 1) 1199030

1198701199020

)

(120574minus1)120574

(28)

Hence Corollary 7 proves this result

As the second consequence ofTheorem3 is unlike the firstone we consider the case when the coefficient 119902(119905) is not astrictly positive function Here by 119902 = 0 we denote the setof all 119905 isin R such that 119902(119905) = 0

Corollary 10 (119902(119905) is nonnegative but not equiv 0) Let thefunctions Φ(119906 V) 119891(119906) 119890(119905) and 119903(119905) satisfy (11) (12) (16)(17) and (18) respectively Let 119902(119905) ge 0 on each interval 119869

119894119895

119894 isin 1 2 119895 isin N such that

119902119894119895= int

119869119894119895

119902 (120591) 119889120591 gt 0 119894 isin 1 2 119895 isin N (29)

If

119902120574

119894119895

119902 (119905)ge 120587120574((1198720+ 1) 119903 (119905)

119870)

120574minus1

119905 isin 119869119894119895 119902 = 0 119894 isin 1 2 119895 isin N

(30)

then (1) is oscillatory

Proof It suffices to show that (30) is equivalent to theexistence of a real number 120582

119895such that

120582119895ge120587 (1198720+ 1)

119870

1

119902119894119895

1

119902119894119895

119902 (119905) le1

120587(120582119895119903 (119905))1minus120574

119905 isin 119869119894119895 119894 isin 1 2 119895 isin N

(31)

The claim will then follow fromTheorem 3 Inequality (31) isfor any 119905 isin 119869

119894119895 119902 = 0 equivalent to

120587 (1198720+ 1)

119870

1

119902119894119895

le 120582119895le (

119902119894119895

120587119902 (119905))

1(120574minus1)1

119903 (119905) (32)

that is to

120587 (1198720+ 1)

119870

1

119902119894119895

le (

119902119894119895

120587119902 (119905))

1(120574minus1)1

119903 (119905) (33)

This inequality is easily seen to be equivalent to (30) for any119905 isin 119869119894119895 119902 = 0 Note that if 119905 isin 119902 = 0 then the second

inequality in (31) is trivially satisfied

International Journal of Differential Equations 5

The following three examples are simple consequences ofCorollary 10

Example 11 (oscillation of (5)) Equation (5) as well asequation

11990910158401015840+ 11989821205872[sin (119898119905)]+119891 (119909) = ℎ (cos (119898119905))

ae in [1199050infin)

119909 1199091015840isin 119860119862loc ([1199050infin) R)

(34)

are oscillatory where 119891(119906) satisfies (16) with 119870 ge 1 andℎ(119904)119904 gt 0 119904 = 0 Indeed in (5) we have 119903(119905) equiv 1 119902(119905) =

11989821205872[cos(119898119905)] and 119890(119905) = ℎ(sin(119898119905)) and so (17) and (18) are

fulfilled and 119902(119905) ge 0 119902(119905) equiv 0 on each interval 119869119894119895 119894 isin 1 2

119895 isin N where 1198691119895= [1198861119895 1198871119895] 1198692119895= [1198862119895 1198872119895] and

1198861119895=2119895120587

119898 119887

1119895=(2119895 + 12) 120587

119898

1198862119895=(2119895 + 32) 120587

119898 119887

2119895=(2119895 + 2) 120587

119898

(35)

Moreover

119902119894119895= int

119869119894119895

119902 (120591) 119889120591

= 11989821205872int

119887119894119895

119886119894119895

[cos (119898120591)]+119889120591 = 1198981205872 119894 isin 1 2 119895 isin N

(36)

and since 120574 = 2 119903(119905) equiv 1119870 ge 1 and1198720= 0 for 120582

119895= 1(119898120587)

we have

120582119895=

1

119898120587ge1

119870

1

119898120587=120587 (1198720+ 1)

119870

1

1198981205872

=120587 (1198720+ 1)

119870

1

119902119894119895

119894 isin 1 2 119895 isin N

1

119902119894119895

119902 (119905) =1

119898120587211989821205872[cos (119898119905)]+ le 119898

=1

120587

1

(1119898120587)=1

120587

1

(120582119895119903 (119905))120574minus1

119905 isin 119869119894119895 119894 isin 1 2 119895 isin N

(37)

Thus the required condition (30) is fulfilled in this case andhence we may apply Corollary 10 to (5) to verify that thisequation is oscillatory Analogously we can show that (34)is oscillatory too

In the next example the coefficient 119902(119905) is a discontinuousfunction on [119905

0infin)

Example 12 (oscillation of (6)) If 119891(119906) satisfies (16) with119870 ge

1 then (6) is oscillatory since the required condition (30) issatisfied especially for 119902(119905) = 4119898

2 sign(cos(119898119905)) on 119869119894119895 120582119895=

1(2119898) and 119869119894119895as in (35) 120574 = 2 119903(119905) equiv 1119870 ge 1 and119872

0= 0

Indeed

119902119894119895= int

119869119894119895

119902 (120591) 119889120591 = 41198982int

119869119894119895

[sign (cos (119898120591))]+119889120591

= 41198982 120587

2119898= 2119898120587 119894 isin 1 2 119895 isin N

120582119895=

1

2119898=

120587

2119898120587= 120587 (119872

0+ 1)

1

119902119894119895

ge120587 (1198720+ 1)

119870

1

119902119894119895

119894 isin 1 2 119895 isin N

1

119902119894

119902 (119905) =1

211989812058741198982[sign (cos (119898119905))]+

le2119898

120587=1

120587

1

radic1 (41198982)

=1

120587(120582119895119903 (119905))120574minus1

119905 isin 119869119894119895 119894 isin 1 2 119895 isin N

(38)

which shows that (30) is satisfied

At the end of this section we consider the case when 119902(119905)changes sign on [119905

0infin) but with the help of Corollary 10 since

119902(119905) is strictly positive on all intervals 119869119894119895

Example 13 (oscillation of (7)) If 119891(119906) satisfies (16) with119870 ge

1 then model equation (7) is oscillatory In fact let 119869119894119895be as

in (35) If ℎ(119904)119904 gt 0 119904 = 0 it is clear that 119890(119905) = ℎ(sin(119898119905))satisfies (2) that is 119890(119905) ge 0 on 119869

1119895and 119890(119905) le 0 on 119869

2119895 On

the other hand we have 119902(119905) = 41198982 sign(cos(119898119905)) on 119869

119894119895and

so 119902(119905) ge 0 119902(119905) equiv 0 on 119869119894119895for all 119894 isin 1 2 119895 isin N Hence

the proof of the fact that all assumptions of Corollary 7 arefulfilled is the same as in the preceding example

3 Proofs of Theorems 3 and 4

Let 119860119862loc([119886 119887)R) denote the set of all real functions whichare absolutely continuous on every interval [119886 119887 minus 120576] where120576 isin (0 119887 minus 119886) For arbitrary numbers 119879

0lt 119879lowast and functions

119866 = 119866(119905 119906) and 119887 = 119887(119905) we consider the ordinary differentialequation

1199081015840= 119866 (119905 119908) + 119887 (119905) ae in [119879

0 119879lowast) (39)

which generalizes the classic Riccati equation 1199081015840 = 119886(119905)1199082+

119887(119905) where 119886(119905) is an arbitrary function We associate to(39) the corresponding sub- and supersolutions 119908119908 isin

119860119862loc([1198790 119879lowast)R) which satisfy respectively

1199081015840le 119866 (119905 119908) + 119887 (119905) 119908

1015840ge 119866 (119905 119908) + 119887 (119905)

ae in [1198790 119879lowast)

(40)

We are interested in studying the following property

119908 (1198790) le 119908 (119879

0) implies 119908 (119905) le 119908 (119905)

forall119905 isin [T0 119879lowast)

(41)

In this way we introduce the following definition

6 International Journal of Differential Equations

Definition 14 We say that comparison principle holds for (39)on an interval [119879

0 119879lowast) 119879 le 119879

0lt 119879lowast if property (41)

is fulfilled for all sub- and supersolutions 119908119908 isin

119860119862loc([1198790 119879lowast)R) of (39) on [119879

0 119879lowast)

Now we are able to state the next result

Lemma 15 Let 119879 le 1198790lt 119879lowast Let 119866 = 119866(119905 119906) and 119887 = 119887(119905) be

two arbitrary function For any119872 gt 0 let there be a function119871 = 119871(119905) gt 0 depending on 119879

0 119879lowast119872 such that

119871 (119905) gt 0 on [1198790 119879lowast) 119871 isin 119871

1(1198790 119879lowast)

1003816100381610038161003816119866 (119905 119906

1) minus 119866 (119905 119906

2)1003816100381610038161003816le 119871 (119905)

10038161003816100381610038161199061minus 1199062

1003816100381610038161003816

forall119905 isin [1198790 119879lowast) 1199061 1199062isin [minus119872119872]

(42)

Then comparison principle (41) holds for (39) on an interval[1198790 119879lowast) 119879 le 119879

0lt 119879lowast

Proof It is enough to use the idea of the proof of [14 Lemma41] Hence this proof is left to the reader

Corollary 16 Let 120582 gt 0 and let 1198920 R+rarr R+be a locally

Lipschitz function on R+ Let 119870 be an arbitrary real number

and 119902(119905) an arbitrary function Let 119879 le 1198790lt 119879lowast If 119903(119905) gt 0

on [1198790 119879lowast) and 119903minus1+120574 isin 1198711(119879

0 119879lowast) then comparison principle

(41) holds for the Riccati differential equation

1199081015840=

1

(120582119903 (119905))120574minus1

1198920 (|119908|) + 119870120582119902 (119905) ae in [119879

0 119879lowast)

(43)

Proof It is clear that (43) is a particular case of (39) inparticular for

119866 (119905 119906) =1

(120582119903 (119905))120574minus1

1198920 (|119906|) 119887 (119905) = 119870120582119902 (119905) (44)

Since1198920is a locally Lipschitz function onR

+ for every119872 gt 0

there is an 1198710gt 0 depending on119872 such that

10038161003816100381610038161198920(1199041) minus 1198920(1199042)1003816100381610038161003816le 1198710

10038161003816100381610038161199041minus 1199042

1003816100381610038161003816 forall119904

1 1199041isin [minus119872119872]

(45)

Hence for any119872 gt 0 and all 1199061 1199062isin [minus119872119872] we obtain

1003816100381610038161003816119866 (119905 119906

1) minus 119866 (119905 119906

2)1003816100381610038161003816=

1

(120582119903 (119905))120574minus1

10038161003816100381610038161198920(10038161003816100381610038161199061

1003816100381610038161003816) minus 1198920(10038161003816100381610038161199062

1003816100381610038161003816)1003816100381610038161003816

le1198710

(120582119903 (119905))120574minus1

1003816100381610038161003816

10038161003816100381610038161199061

1003816100381610038161003816minus10038161003816100381610038161199062

1003816100381610038161003816

1003816100381610038161003816

le1198710

(120582119903 (119905))120574minus1

10038161003816100381610038161199061minus 1199062

1003816100381610038161003816

(46)

Thus according to assumption 119903minus1+120574

isin 1198711(1198790 119879lowast) the

required condition (42) is fulfilled in particular for 119871(119905) =

1198710(120582119903(119905))

minus120574+1 and so Lemma 15 proves this corollary

Before we give the proof ofTheorem 3 we state and provethe next two propositions In the first one by a nonoscillatorysolution 119909(119905) of the main equation (1) we get the existenceof a supersolution 119908(119905) of the Riccati differential equation(43) on the interval (119886

1 1198871) or (119886

2 1198872) In the second one we

construct two subsolutions 1199081(119905) and 119908

2(119905) of (43) which

blow up on intervals (1198861 119879lowast

1) sube (119886

1 1198871) and (119886

2 119879lowast

2) sube (119886

2 1198872)

respectively

Proposition 17 Let Φ(119906 V) and 119891(119906) satisfy (11) (12) and(16) respectively and let 119890(119905) satisfy (2) Let 119903(119905) gt 0 119902(119905) ge0 and 119902(119905) equiv 0 on [119886

1 1198871] cup [119886

2 1198872] Let 119909 = 119909(119905) be a

nonoscillatory solution of (1) and let for some 119879 ge 1199050and

120582 gt 0 the function 119908 = 119908(119905) be defined by

119908 (119905) = minus

120582119903 (119905)Φ (119909 (119905) 1199091015840(119905))

119909 (119905) 119905 ge 119879 (47)

Then 119908 isin 119860119862loc([119879infin)R) and 119908(119905) satisfies the inequality

1199081015840ge (120582119903 (119905))

1minus1205741198920(|119908|) + 119870120582119902 (119905) ae in 119869 (48)

where 119869 = (1198861 1198871) if 119909(119905) lt 0 and 119869 = (119886

2 1198872) if 119909(119905) gt 0

Proof Since 119909 = 119909(119905) is a nonoscillatory solution of (1) thereis a 119879 ge 119905

0such that 119909(119905) = 0 on [119879infin) Hence 119908(119905) is well

defined by (47)Because of (2) we have 119890(119905)119909(119905) le 0 for all 119905 isin 119869 where

119869 = (1198861 1198871) if 119909(119905) lt 0 and 119869 = (119886

2 1198872) if 119909(119905) gt 0 and since

120582 gt 0 we have

minus120582119890 (119905)

119909 (119905)ge 0 in 119869 (49)

Next since 119909 and 119903(119905)Φ(119909(119905) 1199091015840(119905)) are from

119860119862loc([1199050infin)R) we can take the first derivative of 119908(119905)for almost everywhere in 119869 which together with 119903(119905) gt 0119902(119905) ge 0 and 119902(119905) equiv 0 on 119869 gives

1199081015840(119905) = minus

120582(119903 (119905) Φ (119909 (119905) 1199091015840(119905)))1015840

119909 (119905)

+

120582119903 (119905)Φ (119909 (119905) 1199091015840(119905))

1199092(119905)

1199091015840(119905)

=120582119903 (119905)

|119909 (119905)|120574Φ(119909 (119905) 119909

1015840(119905)) 1199091015840(119905) |119909 (119905)|

120574minus2

+ 120582119902 (119905)119891 (119909 (119905))

119909 (119905)minus 120582

119890 (119905)

119909 (119905)

International Journal of Differential Equations 7

ge120582119903 (119905)

|119909 (119905)|120574119892 (10038161003816100381610038161003816Φ (119909 (119905) 119909

1015840(119905))

10038161003816100381610038161003816) + 119870120582119902 (119905) minus 120582

119890 (119905)

119909 (119905)

=120582119903 (119905)

|119909 (119905)|120574119892(

|119909 (119905)|

120582119903 (119905)|119908 (119905)|) + 119870120582119902 (119905) minus 120582

119890 (119905)

119909 (119905)

ge120582119903 (119905)

|119909 (119905)|120574

|119909 (119905)|120574

(120582119903 (119905))1205741198920 (|119908 (119905)|)

+ 119870120582119902 (119905) minus 120582119890 (119905)

119909 (119905)ae in 119869

(50)

that is

1199081015840ge (120582119903 (119905))

1minus1205741198920(|119908|) + 119870120582119902 (119905) minus 120582

119890 (119905)

119909 (119905)ae in 119869

(51)

Hence (49) and (51) prove the desired inequality (48)

Proposition 18 Let (19) be satisfied Let 1198771and 119877

2be two

arbitrary real numbers Then there are two points 119879lowast1isin (1198861 1198871)

and 119879lowast

2isin (1198862 1198872) and two continuous functions 119908

1(119905) and

1199082(119905) such that

119908119894(119886119894) le 119877119894 lim119905rarr119879

lowast

119894

119908119894(119905) = infin 119894 isin 1 2

1199081015840

119894le (120582119903 (119905))

1minus1205741198920(1003816100381610038161003816119908119894

1003816100381610038161003816) + 119870120582119902 (119905)

for ae 119905 isin (119886119894 119879lowast

119894) 119894 isin 1 2

(52)

Proof Since the function 119910(119904) = tan(119904) is a bijection fromthe interval (minus1205872 1205872) intoR we observe that there are two1199041 1199042isin (minus1205872 1205872) such that

tan (119904119894) le 119877119894 119894 isin 1 2 (53)

Next we define two functions 1198811(119905) and 119881

2(119905) by

119881119894 (119905) = 119904119894 +

120587

119888119894

int

119905

119886119894

119862 (120591) 119889120591 119905 isin [119886119894 119887119894] 119894 isin 1 2 (54)

where the real numbers 1198881 1198882and the function 119862(119905) are

defined in (19) From (19) and (54) one can immediatelyconclude that

119881119894(119886119894) = 119904119894lt120587

2 119881

119894(119887119894) = 119904119894+ 120587 gt

120587

2 119894 isin 1 2 (55)

Since 119862 isin 1198711

loc((1199050infin)R) it follows that 119862 isin 1198711((1199050infin)R)

which implies that 119881119894isin 119860119862([119886

119894 119887119894]R) 119894 isin 1 2 Therefore

exploiting the fact that in (55) we have 119881119894isin 119862([119886

119894 119887119894]R) we

obtain the existence of two points 119879lowast1isin (1198861 1198871) and 119879

lowast

2isin

(1198862 1198872) such that

119881119894(119879lowast

119894)=

120587

2 minus

120587

2lt119881119894(119905)lt

120587

2

forall119905 isin [119886119894 119879lowast

119894) 119894 isin 1 2

(56)

Now we are able to define the following two functions 1205961(119905)

and 1205962(119905) by

120596119894(119905) = tan (119881i (119905)) 119905 isin [119886

119894 119879lowast

119894) 119894 isin 1 2 (57)

That are well defined because of (56) Moreover from (53)(54) and (56) we obtain

120596119894(119886119894) = tan (119904

119894) le 119877119894

lim119905rarr119879

lowast

119898

120596119894(119905) = tan (119881

119894(119879lowast

119894)) = tan(120587

2) = infin

(58)

Also since119881119894isin 119860119862([119886

119894 119887119894]R) we can take the first derivative

of 120596119894(119905) and hence from (19) and (57) we obtain

1199081015840

119894(119905) =

120587

119888119894

119862 (119905)1

cos2 (119881119894 (119905))

=120587

119888119894

119862 (119905) (1003816100381610038161003816120596119894(119905)1003816100381610038161003816

2+ 1)

le120587

119888119894

119862 (119905) [1198920(1003816100381610038161003816120596119894(119905)1003816100381610038161003816) + 119872

0+ 1]

le120587

119888119894

119862 (119905) 1198920(1003816100381610038161003816120596119894(119905)1003816100381610038161003816) +

120587

119888119894

119862 (119905) (1198720+ 1)

le (120582119903 (119905))1minus1205741198920(1003816100381610038161003816120596119894(119905)1003816100381610038161003816) + 119870120582119902 (119905) ae in (119886

119894 119879lowast

119894)

(59)

which together with (58) proves this proposition

Now we are able to present the proof of Theorem 3 basedon Lemma 15 and Propositions 17 and 18

Proof of Theorem 3 Assuming the contrary then there is anonoscillatory solution 119909(119905) and 119879 ge 119905

0such that 119909(119905) = 0 for

all 119905 ge 119879 119909(119905) and 119903(119905)Φ(119909 1199091015840) are from 119860119862loc([1199050infin)R)In order to simplify notation let every 119895 isin N be fixed andomitted in the notation For instance instead of (119886

1119895 1198871119895) and

(1198862119895 1198872119895) we write (119886

1 1198871) and (119886

2 1198872) respectively and so on

On one hand we observe by Proposition 17 that thefunction 119908(119905) defined by (47) is a supersolution of thefollowing Riccati differential equation

1199081015840= (120582119903 (119905))

1minus1205741198920(|119908|) + 119870120582119902 (119905) ae in 119869 (60)

where 119908 isin 119860119862loc(119869R) and 119869 = (1198861 1198871) if 119909(119905) lt 0 and 119869 =

(1198862 1198872) if 119909(119905) gt 0 see the proof of Proposition 17 Let for

instance 119909(119905) lt 0 and thus 119869 = (1198861 1198871)

On the other hand by Proposition 18 there are a number119879lowast

1isin (1198861 1198871) and subsolutions 119908

1(119905) 1199081isin 119860119862([119886

1 119879lowast

1)R) of

the Riccati differential equation (60) such that

1199081(1198861) le 119908 (119886

1) lim

119905rarr119879lowast

1

1199081(119905) = infin (61)

(in the case when 119869 = (1198862 1198872) then we work with119879lowast

2isin (1198862 1198872)

and the other subsolution 1199082(119905) 119908

2isin 119860119862([119886

2 119879lowast

2)R))

Hence by Corollary 16 and (61) we conclude that1199081(119905) le 119908(119905)

for all 119905 isin [1198861 119879lowast

1) and therefore

infin = lim119905rarr119879

lowast

1

1199081(119905) le lim119905rarr119879

lowast

1

119908 (119905) (62)

which contradicts 119908 isin 119860119862loc([119879infin)R) Thus the assump-tion that 119909(119905) is nonoscillatory is not possible and hence everysolution of (1) is oscillatory

8 International Journal of Differential Equations

Proof of Theorem 4 There is only one difference betweenTheorems 3 and 4 and it is the difference between conditions(19) and (20) Hence Theorem 4 immediately follows fromTheorem 3 provided that we prove the equivalence between(19) and (20)

First of all we need the following two lemmas

Lemma 19 Let 119886 119887 and 119888 be positive real numbers and 120574 gt 1Then the existence of a positive real number 120582 such that

119888 le min 119886

120582120574minus1

119887120582 (63)

is equivalent to

119888 le 119886112057411988711205741015840

(64)

Here 1205741015840 = 120574(120574 minus 1) is the conjugate exponent of 120574

Proof Let us define 119892(120582) = min119886120582120574minus1 119887120582 Since 120582 997891rarr

119886120582120574minus1 is decreasing and 120582 997891rarr 119887120582 is increasing there exists

a unique point of maximum of 119892 It is achieved at 120582 = 1205820

which is a solution of 119886120582120574minus10

= 1198871205820 hence 120582

0= (119886119887)

1120574If there is a positive real number 120582 such that 119888 le 119892(120582)

then 119888 le 119892(1205820) = 119887120582

0= 119887(119886119887)

1120574= 119886112057411988711205741015840

Converselyif 119888 le 119886

112057411988711205741015840

then 119888 le 119892(1205820) and the equivalence in the

lemma is proved

The preceding lemma is a special case of the followingmore general statement

Lemma 20 Assume that 119886(119905) 119887(119905) and 119888(119905) are positive realfunctions defined on a subset 119869 sube R Then the existence of apositive real number 120582 such that for all 119905 isin 119869

119888 (119905) le min 119886 (119905)120582120574minus1

119887 (119905) 120582 (65)

is equivalent to

sup119905isin119869

119888 (119905)

119887 (119905)le (inf119905isin119869

119886 (119905)

119888 (119905))

1(120574minus1)

(66)

Proof The condition that there exists 120582 gt 0 such that (65)is true for all 119905 isin 119869 is equivalent with the following twoinequalities which have to hold simultaneously

119888 (119905) le119886 (119905)

120582120574minus1

119888 (119905) le 119887 (119905) 120582 (67)

that is with

119888 (119905)

119887 (119905)le 120582 le (

119886 (119905)

119888 (119905))

1(120574minus1)

forall119905 isin 119869 (68)

Taking the supremum of the left-hand side and the infimumof the right-hand side we obtain (66) Conversely if (66)is satisfied then since the left-hand side in (66) cannotbe equal to zero while the right-hand side is less than infinthen inequality (66) defines a nonempty closed interval in R

(possibly reducing to just one point) in which we can chooseany positive 120582 Hence (67) is satisfied for all 119905 isin 119869 andtherefore (65) as well

Remark 21 If 119886 = 119886(119905) 119887 = 119887(119905) and 119888 = 119888(119905) appearingin Lemma 20 are constant functions then condition (66) isequivalent to 119888119887 = (119886119888)1(120574minus1) that is to (64)

Proof of Equivalence of (19) and (20) Wewill use Lemma 20Let us fix 119869

119894119895 where 119894 isin 1 2 and 119895 isin N We see that that the

inequality appearing in the second line of condition (19) is ofthe form (65) where

119888 (119905) =1

119888119894119895

119862 (119905) 119886 (119905) =1

120587119903(119905)1minus120574

119887 (119905) =119870119902 (119905)

120587 (1198720+ 1)

(69)

Condition (66) is equivalent to

sup119905isin119869119894119895

119862 (119905)

119902 (119905)le

1198701198881205741015840

119894119895

1205871205741015840

(1198720+ 1)

inf119905isin119869119894119895

1

119903 (119905) 119862(119905)120574minus1

=

1198701198881205741015840

119894119895

(1198720+ 1) sup

119905isin119869119894119895119903 (119905) 119862(119905)

120574minus1(

119888119894119895

120587)

1205741015840

(70)

This inequality is equivalent to the corresponding one in (19)and the claim follows from Lemma 19

Proof of Corollary 6 Substituting inf119905isin119869119894119895

119902(119905) instead of 119902(119905)and sup

119905isin119869119894119895119903(119905) instead of 119903(119905) in the inequality appearing in

the second line of (20) after a short computation we obtain astronger inequality than (20) as

1

119888119894119895

sup119905isin119869119894119895

119862 (119905)

le1

120587(

119870inf119905isin119869119894119895

119902 (119905)

(1198720+ 1) sup

119905isin119869119894119895119903 (119905)

)

(120574minus1)120574

forall119894 isin 1 2 119895 isin N

(71)

Substituting119862(119905) equiv 1we obtain that the left-hand side of (71)is equal to |119869

119894119895|minus1 and the resulting inequality is equivalent to

(21) Since (21) implies (20) the claim is proved

Remark 22 Here we show that the choice of 119862(119905) equiv 1 inthe proof of Corollary 6 is the best possible To prove thisnote that on the left-hand side of (71) we have the expressiondepending on an auxiliary function 119862(119905) and this functionis absent on the right-hand side Therefore it has sense totry to find a function 119862(119905) 119905 isin 119869

119894119895 such that the value of

119876(119862) = (1119888119894119895)sup119905isin119869119894119895

119862(119905) is minimal (note that 119888119894119895depends

on the function 119862 = 119862(119905) as well) It is easy to see that theminimum is achieved for 119862(119905) equiv 1 (or any positive constant)Indeed since 119876(119862) = 119876(120583119888) for any 120583 gt 0 by taking 120583 =

(sup119905isin119869119894119895

119862(119905))minus1 it suffices to assume that sup

119905isin119869119894119895119862(119905) = 1

The value of 119876(119862) is minimal if 119888119894119895= int119869119894119895

119862(120591) 119889120591 is maximalpossible and since 0 le 119862(119905) le 1 it is clear that the maximumof 119888119894119895= 119888119894119895(119862) is achieved when 119862(119905) equiv 1

International Journal of Differential Equations 9

Proof of Corollary 7 It is clear that the condition (23) implies(21) Furthermore since the lengths |119869

119894119895| are uniformly

bounded from below by a positive constant see (23) thenobviously 119886

1119895rarr infin as 119895 rarr infin Thus the claim follows

immediately from Corollary 6

Proof of Corollary 10 Let 119862(119905) = 119902(119905) Then the requiredcondition (19) is clearly satisfied because of assumption(30) Hence this corollary immediately follows from Theo-rem 3

4 An Extension of Condition (12) tothe Case of 120574 gt 1

In this section we consider the oscillation of (1) in the casewhen 120574 isin (1 2) is allowed in the assumption (12) In this waythe assumption (11) is slightly modified by a real number 119901 gt1 such that

|119906|(119901minus1)120574minus119901

VΦ (119906 V) ge 119892 (|Φ (119906 V)|) (72)

where (12) is supposed that is 119892(119888119904) ge 1198881205741198920(119904) for every 119888 119904 gt

0 and for a locally Lipschitz function 1198920 [0infin) rarr [0infin)

for which there exists a constant1198720such that 119892

0(119904)+119872

0ge 119904119901

for all 119904 isin [0infin)For the function 119891 we assume that there exists a constant

119870 gt 0 such that

119891 (119906) sign (119906) ge 119870|119906|119901minus1 for every 119906 isin R (73)

We will need the following two lemmas

Lemma 23 Let 119886 lt 119887 and let 119891 [119886 119887] rarr [0infin) be a meas-urable function Then

int

119887

119886

119891 (119905) 119889119905 ge 1 (74)

if and only if there exists a function 119892 isin 1198711([119886 119887] [0infin))

1198921= 1 such that

119891 (119905) ge 119892 (119905) forall119905 isin [119886 119887] (75)

Proof We assume (74) If int119887119886119891(119905)119889119905 is finite we may choose

119892(119909) = 119891(119909)1198911

For int119887119886119891(119905)119889119905 = infin we may define for every natural

number 119899 a function

119891119899(119905) =

119891 (119905) 119891 (119905) le 119899

119899 119891 (119905) gt 119899(76)

Since 119891119899(119905) le 119899 we have 119891

119899isin 1198711([119886 119887]R) Obviously

lim119899int119887

119886119891119899(119905)119889119905 is also infiniteThis implies that there exists an

1198990such that int119887

1198861198911198990(119905)119889119905 ge 1 We define 119892(119905) = 119891

1198990(119905)119891

11989901

Now we assume (75) Integrating over the whole intervalwe get

int

119887

119886

119891 (119905) 119889119905 ge int

119887

119886

119892 (119905) 119889119905 = 1 (77)

Lemma 24 Let119901 gt 1There exists a unique function 119910 = 119910(119904)that satisfies the following Cauchy problem

1199101015840(119904) = 1 +

1003816100381610038161003816119910 (119904)

1003816100381610038161003816

119901 119904 isin R (78)

119910 (0) = 0 (79)

Furthermore there exists a finite real number 119878lowast such that

lim119904rarrplusmn119878

lowast119910 (119904) = plusmninfin (80)

and 119910 isin 1198621((minus119878lowast 119878lowast)R) is injective and monotonous We

have 119878lowast = (120587119901)(1(sin(120587119901))) and 119910 is odd

We will denote by tan119901the solution of (78)-(79)

Proof Obviously every solution to (78) is injective andmonotonous on a connected set since 1199101015840 = 1 + |119910|119901 ge 1 gt 0

Let 119891(119909) = 1(1 + |119909|119901) Obviously 1 ge 119891(119909) gt 0 for all

119909 isin R and 119862119901gt |1198911015840(119909)| gt 0 for some constants 119862

119901isin R

We may assume 119862119901gt 1 Hence 1199111015840 = 119891(119905) 119911(0) = 0 has (by

[17Theorem 31]) a locally unique solution on (minus1119862119901 1119862119901)

By the same argument 119911 can be extended to the whole of Rinterval by interval of type ((119899 minus 1)119862

119901 (119899 + 1)119862

119901) for all

119899 isin ZWe know from [18] that the for the integral of the left-

hand side we have

int

infin

0

119891 (119909) 119889119909 = int

infin

0

119889119909

1 + 119909119901=120587

119901

1

sin (120587119901)= 119878lowast (81)

so lim119905rarrplusmninfin

119911(119905) = plusmn119878lowast

Since 1199111015840(119905) = 0 119911 is injective monotonous and of class1198621(R (minus119878lowast 119878lowast)) and so we define 119910(119904) = 119911

minus1(119904) It imme-

diately follows that 119910 isin 1198621((minus119878lowast 119878lowast)R) and that 119910 is

injective and monotonous Also by continuity of 119911 we havelim119904rarrplusmn119878

lowast119910(119904) = lim119908rarrplusmninfin

119910(119911(119908)) = lim119908rarrplusmninfin

119908 = plusmninfinDifferentiating it also follows that 1199101015840 = 1 + |119910|

119901 and bydefinition of 119910 119910(0) = 119911minus1(0) = 0 So 119910 is a solution of (78)-(79) Since any other such solution must have 119911 as its inverse119910 is unique on its domain

Proposition 25 Let 119886 lt 119887 and 120574 gt 119901 gt 1 Assume (72) and(73) and let 119902(119905) ge 0 and 119890(119905) le 0 (or 119890(119905) ge 0 ) for all 119905 isin [119886 119887]If there exists a constant 120582 gt 0 such that

119901

2120587sin(120587

119901)int

119887

119886

min119901 minus 1

119903120574minus1

(119905) 120582120574minus1

119870120582119902 (119905)

1 +1198720

119889119905 ge 1 (82)

then for any solution 119909(119905) of (1) with 119909(119886) ge 0 (or 119909(119886) le 0)there exists a number 119905lowast isin [119886 119887] such that 119909(119905lowast) = 0

Proof We assume that 119909(119905) = 0 for all 119905 isin [119886 119887] Then we maydefine the function

120596 (119905) = minus

120582119903 (119905) Φ (119909 (119905) 1199091015840(119905))

sign (119909 (119905)) |119909 (119905)|119901minus1 (83)

10 International Journal of Differential Equations

This function is well defined on [119886 119887] and its derivative is

1199081015840(119905) = minus

120582(119903 (119905) Φ (119909 (119905) 1199091015840(119905)))1015840

sign (119909 (119905)) |119909 (119905)|119901minus1

+

(119901 minus 1) 120582119903 (119905) Φ (119909 (119905) 1199091015840(119905))

|119909 (119905)|119901

1199091015840(119905)

(84)

Using the fact that 119909(119905) is a solution of (1) and omitting 119905 from119909 and 1199091015840 for clarity we get

1199081015840(119905) =

120582 (119902 (119905) 119891 (119909) minus 119890 (119905))

sign (119909) |119909|119901minus1+

(119901 minus 1) 120582119903 (119905) Φ (119909 1199091015840)

119909119901

1199091015840

=

(119901 minus 1) 120582119903 (119905) Φ (119909 1199091015840)

|119909|119901

1199091015840+ 120582119902 (119905)

119891 (119909)

sign (119909) |119909|119901minus1

minus 120582119890 (119905)

sign (119909) |119909|119901minus1

=(119901 minus 1) 120582119903 (119905)

|119909|120574(119901minus1)

Φ(119909 1199091015840) 1199091015840|119909|(119901minus1)120574minus119901

+ 120582119902 (119905)119891 (119909)

sign (119909) |119909|119901minus1minus 120582

119890 (119905)

sign (119909) |119909|119901minus1

(72)

ge(119901 minus 1) 120582119903 (119905)

|119909|120574(119901minus1)

119892 (10038161003816100381610038161003816Φ (119909 119909

1015840)10038161003816100381610038161003816)

+ 120582119902 (119905)119891 (119909)

sign (119909) |119909|119901minus1

minus 120582119890 (119905)

sign (119909) |119909|119901minus1

=(119901 minus 1) 120582119903 (119905)

|119909|120574(119901minus1)

119892(|119909|119901minus1

120582119903 (119905)|120596 (119905)|)

+ 120582119902 (119905)119891 (119909)

sign (119909) |119909|119901minus1minus 120582

119890 (119905)

sign (119909) |119909|119901minus1

(12)

ge(119901 minus 1) 120582119903 (119905)

|119909|120574(119901minus1)

(|119909|119901minus1

120582119903 (119905))

120574

1198920 (|120596 (119905)|)

+ 120582119902 (119905)119891 (119909)

sign (119909) |119909|119901minus1minus 120582

119890 (119905)

sign (119909) |119909|119901minus1

=119901 minus 1

(120582119903 (119905))120574minus1

1198920 (|120596 (119905)|)+120582119902 (119905)

119891 (119909)

sign (119909) |119909|119901minus1

minus 120582119890 (119905)

sign (119909) |119909|119901minus1

(73)

ge119901 minus 1

(120582119903 (119905))120574minus1

1198920 (|120596 (119905)|) + 120582119902 (119905) 119870

minus 120582119890 (119905)

sign (119909) |119909|119901minus1

(85)

Since 119890(119905)(sign(119909)|119909|119901minus1) le 0 we have

1199081015840(119905) ge

119901 minus 1

(120582119903 (119905))120574minus1

1198920(|120596 (119905)|) + 120582119870119902 (119905) (86)

We apply Lemma 23 to (82) to obtain a function119862(119905) suchthat

int

119887

119886

119862 (119905) 119889119905 = 1 (87)

and by (75)

0 le 119862 (119905) le119901

2120587sin(120587

119901)min

119901 minus 1

119903120574minus1

(119905) 120582120574minus1

119870120582119902 (119905)

1 +1198720

(88)

Let 1199040isin (minus(120587119901)(1 sin(120587119901)) (120587119901)(1 sin(120587119901))) be such

that tan119901(1199040) le 120596(119886) We define the function

119881 (119905) = 1199040+2120587

119901

1

sin (120587119901)int

119905

119886

119862 (120591) 119889120591 (89)

Note that (minus(120587119901)(1 sin(120587119901))) lt 1199040

= 119881(119886) lt

(120587119901)(1 sin(120587119901)) and 119881(119887) = 1199040+ (2120587119901)(1 sin(120587119901)) gt

(120587119901)(1 sin(120587119901)) Since119881 is continuous there exists a 119879lowast isin(119886 119887) such that 119881(119905) lt (120587119901)(1 sin(120587119901)) for 119905 isin [119886 119879lowast) and119881(119879lowast) = (120587119901)(1 sin(120587119901))

We define

120596 (119905) = tan119901(119881 (119905)) 119905 isin [119886 119879

lowast) (90)

and note that by Lemma 24 lim119905rarr119879

lowast120596(119905) = +infin wheretan119901(119904) = 119910(119904) and 119910(119904) is determined by Lemma 24The derivative of 120596 then satisfies

1205961015840(119905) = tan1015840

119901(119881 (119905)) 119881

1015840(119905)

= (1 + tan119901119901(119881 (119905))) 119881

1015840(119905)

=2120587

119901

1

sin (120587119901)(1 + 120596

119901(119905)) 119862 (119905)

le (1 +1198720+ 1198920(1003816100381610038161003816120596 (119905)

1003816100381610038161003816))min

119901 minus 1

119903120574minus1

(119905) 120582120574minus1

119870120582119902 (119905)

1 +1198720

le119901 minus 1

119903120574minus1

(119905) 120582120574minus1

1198920(1003816100381610038161003816120596 (119905)

1003816100381610038161003816) + 119870120582119902 (119905)

(91)

Thus we may apply the comparison principle ofLemma 15 to 120596(119886) le 120596(119886) and their respective differentialinequalities to conclude that 120596(119905) le 120596(119905) for all 119905 isin [119886 119887]But since 120596 rarr +infin as 119905 rarr 119879

lowastlt 119887 and 120596(119905) is well defined

on the whole segment [119886 119887] we arrive at a contradictionHence 119909(119905) gt 0 cannot hold for all 119905 isin [119886 119887] and since 119909(119905) iscontinuous there exists a 119905lowast isin [119886 119887] such that 119909(119905lowast) = 0

Theorem 26 Let 120574 gt 119901 gt 1 Assume (72) and (73) and let forany 119879 gt 119905

0there exist 119879 le 119886

1lt 1198871le 1198862lt 1198872such that 119892(119905) ge 0

on [1198861 1198871] cup [119886

2 1198872] and 119890(119905) le 0 on [119886

1 1198871] and 119890(119905) ge 0 on

International Journal of Differential Equations 11

[1198862 1198872] If for both 119894 isin 1 2 there exist constants 120582

119894gt 0 such

that

119901

2120587sin (120587119901)int

119887119894

119886119894

min119901 minus 1

119903120574minus1

(119905) 120582120574minus1

119894

119870120582119894119902 (119905)

1 +1198720

119889119905 ge 1

(92)

then (1) is oscillatory

Proof For (1) to be oscillatory it is sufficient that for every119899 isin N there exists a 119905

119899gt 119899 such that 119909(119905

119899) = 0

Let 119899 isin N and let 119899 lt 1198861lt 1198871le 1198862lt 1198872as assumed in the

theorem It is enough to show that every solution 119909(119905) of (1)has a zero on [119886

1 1198872] If 119909(119886

1) ge 0 we apply Proposition 25 to

the segment [1198861 1198871] to obtain the sought zero If 119909(119886

2) le 0 we

again apply Proposition 25 to the segment [1198862 1198872] Suppose

119909(1198861) lt 0 and 119909(119886

2) gt 0 Then since 119909 is continuous there

exists a number 1199051isin [1198861 1198862] such that 119909(119905

1) = 0

Acknowledgment

This work is supported by the Ministry of Science of theRepublic of Croatia under Grant no 036-0361621-1291

References

[1] Q Kong ldquoNonoscillation and oscillation of second order half-linear differential equationsrdquo Journal of Mathematical Analysisand Applications vol 332 no 1 pp 512ndash522 2007

[2] Q-R Wang and Q-G Yang ldquoInterval criteria for oscillationof second-order half-linear differential equationsrdquo Journal ofMathematical Analysis and Applications vol 291 no 1 pp 224ndash236 2004

[3] Q Long and Q-R Wang ldquoNew oscillation criteria of second-order nonlinear differential equationsrdquo Applied Mathematicsand Computation vol 212 no 2 pp 357ndash365 2009

[4] Z Zheng and F Meng ldquoOscillation criteria for forced second-order quasi-linear differential equationsrdquo Mathematical andComputer Modelling vol 45 no 1-2 pp 215ndash220 2007

[5] D Cakmak and A Tiryaki ldquoOscillation criteria for certainforced second-order nonlinear differential equationsrdquo AppliedMathematics Letters vol 17 no 3 pp 275ndash279 2004

[6] F Jiang and F Meng ldquoNew oscillation criteria for a class ofsecond-order nonlinear forced differential equationsrdquo Journalof Mathematical Analysis and Applications vol 336 no 2 pp1476ndash1485 2007

[7] Q Yang ldquoInterval oscillation criteria for a forced secondorder nonlinear ordinary differential equations with oscillatorypotentialrdquo Applied Mathematics and Computation vol 135 no1 pp 49ndash64 2003

[8] A Tiryaki and B Ayanlar ldquoOscillation theorems for certainnonlinear differential equations of second orderrdquo Computers ampMathematics with Applications vol 47 no 1 pp 149ndash159 2004

[9] J Jaros T Kusano and N Yoshida ldquoGeneralized Piconersquosformula and forced oscillations in quasilinear differential equa-tions of the second orderrdquoArchivumMathematicum vol 38 no1 pp 53ndash59 2002

[10] R P Agarwal S R Grace and D OrsquoRegan Oscillation Theoryfor Second Order Linear Half-Linear Superlinear and SublinearDynamic Equations Kluwer Academic Publishers DordrechtThe Netherlands 2002

[11] S Jing ldquoA new oscillation criterion for forced second-orderquasilinear differential equationsrdquoDiscrete Dynamics in Natureand Society vol 2011 Article ID 428976 8 pages 2011

[12] W-T Li and S S Cheng ldquoAn oscillation criterion for nonhomo-geneous half-linear differential equationsrdquoAppliedMathematicsLetters vol 15 no 3 pp 259ndash263 2002

[13] J S W Wong ldquoOscillation criteria for a forced second-orderlinear differential equationrdquo Journal of Mathematical Analysisand Applications vol 231 no 1 pp 235ndash240 1999

[14] M Pasic ldquoFite-Wintner-Leighton type oscillation criteria forsecond-order differential equations with nonlinear dampingrdquoAbstract and Applied Analysis vol 2013 Article ID 852180 10pages 2013

[15] M Pasic ldquoNew interval oscillation criteria for forced second-order differential equations with nonlinear dampingrdquo Interna-tional Journal of Mathematical Analysis vol 7 no 25 pp 1239ndash1255 2013

[16] L Gasinski and N S Papageorgiou Nonsmooth Critical PointTheory and Nonlinear Boundary Value Problems vol 8 of Seriesin Mathematical Analysis and Applications Edited by R PAgarwal and D OrsquoRegan Chapman amp HallCRC Boca RatonFla USA 2005

[17] S Lang Real and Functional Analysis vol 142 of GraduateTexts in Mathematics Springer-Verlag New York NY USA 3rdedition 1993

[18] I N Bronshtein K A Semendyayev G Musiol and HMuehligHandbook of Mathematics Springer Berlin Germany5th edition 2007

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

4 International Journal of Differential Equations

Corollary 7 (119902(119905) is bounded from below by a positiveconstant) Let the functionsΦ(119906 V)119891(119906) 119890(119905) and 119903(119905) satisfy(11) (12) (16) (17) and (18) respectively Let there be twoconstants 119903

0 1199020satisfying

0 lt 119903 (119905) le 1199030 119902 (119905) ge 119902

0gt 0 forall119894 isin 1 2 119895 isin N (22)

If

10038161003816100381610038161003816119869119894119895

10038161003816100381610038161003816ge 120587(

(1198720+ 1) 1199030

1198701199020

)

(120574minus1)120574

forall119894 isin 1 2 119895 isin N (23)

then (1) is oscillatory where 1205741198720 and119870 are constants defined

in (11) (12) and (16) respectively

Example 8 (oscillation of (4)) We know that 119909(119905) =

|sin(119898119905)| sin(119898119905) is an oscillatory solution of (4) Howeveraccording to Corollary 7 we can show that all solutions of (4)are oscillatory Indeed since Φ(119906 V) equiv V the conditions (11)and (12) are satisfied especially for 120574 = 2 119892(119904) = 119892

0(119904) = 119904

2

and 1198720= 0 Next 119891(119906) equiv 119906 implies that condition (16) is

satisfied especially for 119870 = 1 Since 119903(119905) equiv 1 and 119902(119905) equiv 41198982

it is clear that conditions (18) and (22) are also satisfied inparticular for 119903

0= 1 and 119902

0= 4119898

2 Moreover since 119890(119905) =ℎ(sin(119898119905)) and ℎ(119904)119904 gt 0 119904 = 0 we have that (17) is fulfilled for1198861119895= 2119895120587119898 119887

1119895= (2119895 + 1)120587119898 = 119886

2119895and 1198872119895= (2119895 + 2)120587119898

Moreover10038161003816100381610038161003816119869119894119895

10038161003816100381610038161003816= 119887119894minus 119886119894=120587

119898gt 0 119894 isin 1 2 (24)

Hence we conclude that the required condition (23) isfulfilled that is10038161003816100381610038161003816119869119894119895

10038161003816100381610038161003816=120587

119898ge

120587

radic41198982

= 120587((1198720+ 1) 1199030

1199020

)

(120574minus1)120574

ge 120587((1198720+ 1) 1199030

1198701199020

)

(120574minus1)120574

(25)

Thus all conditions of Corollary 7 are satisfied and hence (4)is oscillatory

Example 9 We consider the following class of equations

11990910158401015840minus 2

11987810158401015840(119905)

119878 (119905)119909 = 2 sign (119878 (119905)) 1198781015840

2

(119905) ae in [1199050infin)

119909 1199091015840isin 119860119862loc ([1199050infin) R) 119909 notin 119862

2((1199050infin) R)

(26)

where 119878 = 119878(119905) 119878 isin 1198622(R) is an oscillatory function such thatthe zeros 119905

119899of the function sign(119878(119905))11987810158402(119905) satisfy 119905

119899rarr infin

there is a 1205910isin R such that 119905

119899+1minus 119905119899ge 1205910gt 0 for all 119899 isin N and

119878(119905) = 0 on (119905119899 119905119899+1) This equation allows an explicitly given

oscillatory solution 119909(119905) = |119878(119905)|119878(119905) Moreover if there is aconstant 119904

0gt 0 such that

minus211987810158401015840(119905)

119878 (119905)ge 1199040 119905 isin (119905

119899 119905119899+1) 120591

0ge

120587

radic1199040

(27)

then by Corollary 7 we conclude that (26) is oscillatoryIndeed conditions (11) (12) and (16) are satisfied by the samereasons as in Example 8 Condition (18) is satisfied becauseof 119903(119905) equiv 1 Also from (27) it follows that (22) and (23) arefulfilled in particular for 120574 = 2119872

0= 0 119903

0= 1 119902

0= 1199040 and

119870 ge 1 that is10038161003816100381610038161003816119869119894119895

10038161003816100381610038161003816=10038161003816100381610038161003816119905119895+1

minus 119905119895

10038161003816100381610038161003816ge 1205910ge

120587

radic1199040

= 120587((1198720+ 1) 1199030

1199020

)

(120574minus1)120574

ge 120587((1198720+ 1) 1199030

1198701199020

)

(120574minus1)120574

(28)

Hence Corollary 7 proves this result

As the second consequence ofTheorem3 is unlike the firstone we consider the case when the coefficient 119902(119905) is not astrictly positive function Here by 119902 = 0 we denote the setof all 119905 isin R such that 119902(119905) = 0

Corollary 10 (119902(119905) is nonnegative but not equiv 0) Let thefunctions Φ(119906 V) 119891(119906) 119890(119905) and 119903(119905) satisfy (11) (12) (16)(17) and (18) respectively Let 119902(119905) ge 0 on each interval 119869

119894119895

119894 isin 1 2 119895 isin N such that

119902119894119895= int

119869119894119895

119902 (120591) 119889120591 gt 0 119894 isin 1 2 119895 isin N (29)

If

119902120574

119894119895

119902 (119905)ge 120587120574((1198720+ 1) 119903 (119905)

119870)

120574minus1

119905 isin 119869119894119895 119902 = 0 119894 isin 1 2 119895 isin N

(30)

then (1) is oscillatory

Proof It suffices to show that (30) is equivalent to theexistence of a real number 120582

119895such that

120582119895ge120587 (1198720+ 1)

119870

1

119902119894119895

1

119902119894119895

119902 (119905) le1

120587(120582119895119903 (119905))1minus120574

119905 isin 119869119894119895 119894 isin 1 2 119895 isin N

(31)

The claim will then follow fromTheorem 3 Inequality (31) isfor any 119905 isin 119869

119894119895 119902 = 0 equivalent to

120587 (1198720+ 1)

119870

1

119902119894119895

le 120582119895le (

119902119894119895

120587119902 (119905))

1(120574minus1)1

119903 (119905) (32)

that is to

120587 (1198720+ 1)

119870

1

119902119894119895

le (

119902119894119895

120587119902 (119905))

1(120574minus1)1

119903 (119905) (33)

This inequality is easily seen to be equivalent to (30) for any119905 isin 119869119894119895 119902 = 0 Note that if 119905 isin 119902 = 0 then the second

inequality in (31) is trivially satisfied

International Journal of Differential Equations 5

The following three examples are simple consequences ofCorollary 10

Example 11 (oscillation of (5)) Equation (5) as well asequation

11990910158401015840+ 11989821205872[sin (119898119905)]+119891 (119909) = ℎ (cos (119898119905))

ae in [1199050infin)

119909 1199091015840isin 119860119862loc ([1199050infin) R)

(34)

are oscillatory where 119891(119906) satisfies (16) with 119870 ge 1 andℎ(119904)119904 gt 0 119904 = 0 Indeed in (5) we have 119903(119905) equiv 1 119902(119905) =

11989821205872[cos(119898119905)] and 119890(119905) = ℎ(sin(119898119905)) and so (17) and (18) are

fulfilled and 119902(119905) ge 0 119902(119905) equiv 0 on each interval 119869119894119895 119894 isin 1 2

119895 isin N where 1198691119895= [1198861119895 1198871119895] 1198692119895= [1198862119895 1198872119895] and

1198861119895=2119895120587

119898 119887

1119895=(2119895 + 12) 120587

119898

1198862119895=(2119895 + 32) 120587

119898 119887

2119895=(2119895 + 2) 120587

119898

(35)

Moreover

119902119894119895= int

119869119894119895

119902 (120591) 119889120591

= 11989821205872int

119887119894119895

119886119894119895

[cos (119898120591)]+119889120591 = 1198981205872 119894 isin 1 2 119895 isin N

(36)

and since 120574 = 2 119903(119905) equiv 1119870 ge 1 and1198720= 0 for 120582

119895= 1(119898120587)

we have

120582119895=

1

119898120587ge1

119870

1

119898120587=120587 (1198720+ 1)

119870

1

1198981205872

=120587 (1198720+ 1)

119870

1

119902119894119895

119894 isin 1 2 119895 isin N

1

119902119894119895

119902 (119905) =1

119898120587211989821205872[cos (119898119905)]+ le 119898

=1

120587

1

(1119898120587)=1

120587

1

(120582119895119903 (119905))120574minus1

119905 isin 119869119894119895 119894 isin 1 2 119895 isin N

(37)

Thus the required condition (30) is fulfilled in this case andhence we may apply Corollary 10 to (5) to verify that thisequation is oscillatory Analogously we can show that (34)is oscillatory too

In the next example the coefficient 119902(119905) is a discontinuousfunction on [119905

0infin)

Example 12 (oscillation of (6)) If 119891(119906) satisfies (16) with119870 ge

1 then (6) is oscillatory since the required condition (30) issatisfied especially for 119902(119905) = 4119898

2 sign(cos(119898119905)) on 119869119894119895 120582119895=

1(2119898) and 119869119894119895as in (35) 120574 = 2 119903(119905) equiv 1119870 ge 1 and119872

0= 0

Indeed

119902119894119895= int

119869119894119895

119902 (120591) 119889120591 = 41198982int

119869119894119895

[sign (cos (119898120591))]+119889120591

= 41198982 120587

2119898= 2119898120587 119894 isin 1 2 119895 isin N

120582119895=

1

2119898=

120587

2119898120587= 120587 (119872

0+ 1)

1

119902119894119895

ge120587 (1198720+ 1)

119870

1

119902119894119895

119894 isin 1 2 119895 isin N

1

119902119894

119902 (119905) =1

211989812058741198982[sign (cos (119898119905))]+

le2119898

120587=1

120587

1

radic1 (41198982)

=1

120587(120582119895119903 (119905))120574minus1

119905 isin 119869119894119895 119894 isin 1 2 119895 isin N

(38)

which shows that (30) is satisfied

At the end of this section we consider the case when 119902(119905)changes sign on [119905

0infin) but with the help of Corollary 10 since

119902(119905) is strictly positive on all intervals 119869119894119895

Example 13 (oscillation of (7)) If 119891(119906) satisfies (16) with119870 ge

1 then model equation (7) is oscillatory In fact let 119869119894119895be as

in (35) If ℎ(119904)119904 gt 0 119904 = 0 it is clear that 119890(119905) = ℎ(sin(119898119905))satisfies (2) that is 119890(119905) ge 0 on 119869

1119895and 119890(119905) le 0 on 119869

2119895 On

the other hand we have 119902(119905) = 41198982 sign(cos(119898119905)) on 119869

119894119895and

so 119902(119905) ge 0 119902(119905) equiv 0 on 119869119894119895for all 119894 isin 1 2 119895 isin N Hence

the proof of the fact that all assumptions of Corollary 7 arefulfilled is the same as in the preceding example

3 Proofs of Theorems 3 and 4

Let 119860119862loc([119886 119887)R) denote the set of all real functions whichare absolutely continuous on every interval [119886 119887 minus 120576] where120576 isin (0 119887 minus 119886) For arbitrary numbers 119879

0lt 119879lowast and functions

119866 = 119866(119905 119906) and 119887 = 119887(119905) we consider the ordinary differentialequation

1199081015840= 119866 (119905 119908) + 119887 (119905) ae in [119879

0 119879lowast) (39)

which generalizes the classic Riccati equation 1199081015840 = 119886(119905)1199082+

119887(119905) where 119886(119905) is an arbitrary function We associate to(39) the corresponding sub- and supersolutions 119908119908 isin

119860119862loc([1198790 119879lowast)R) which satisfy respectively

1199081015840le 119866 (119905 119908) + 119887 (119905) 119908

1015840ge 119866 (119905 119908) + 119887 (119905)

ae in [1198790 119879lowast)

(40)

We are interested in studying the following property

119908 (1198790) le 119908 (119879

0) implies 119908 (119905) le 119908 (119905)

forall119905 isin [T0 119879lowast)

(41)

In this way we introduce the following definition

6 International Journal of Differential Equations

Definition 14 We say that comparison principle holds for (39)on an interval [119879

0 119879lowast) 119879 le 119879

0lt 119879lowast if property (41)

is fulfilled for all sub- and supersolutions 119908119908 isin

119860119862loc([1198790 119879lowast)R) of (39) on [119879

0 119879lowast)

Now we are able to state the next result

Lemma 15 Let 119879 le 1198790lt 119879lowast Let 119866 = 119866(119905 119906) and 119887 = 119887(119905) be

two arbitrary function For any119872 gt 0 let there be a function119871 = 119871(119905) gt 0 depending on 119879

0 119879lowast119872 such that

119871 (119905) gt 0 on [1198790 119879lowast) 119871 isin 119871

1(1198790 119879lowast)

1003816100381610038161003816119866 (119905 119906

1) minus 119866 (119905 119906

2)1003816100381610038161003816le 119871 (119905)

10038161003816100381610038161199061minus 1199062

1003816100381610038161003816

forall119905 isin [1198790 119879lowast) 1199061 1199062isin [minus119872119872]

(42)

Then comparison principle (41) holds for (39) on an interval[1198790 119879lowast) 119879 le 119879

0lt 119879lowast

Proof It is enough to use the idea of the proof of [14 Lemma41] Hence this proof is left to the reader

Corollary 16 Let 120582 gt 0 and let 1198920 R+rarr R+be a locally

Lipschitz function on R+ Let 119870 be an arbitrary real number

and 119902(119905) an arbitrary function Let 119879 le 1198790lt 119879lowast If 119903(119905) gt 0

on [1198790 119879lowast) and 119903minus1+120574 isin 1198711(119879

0 119879lowast) then comparison principle

(41) holds for the Riccati differential equation

1199081015840=

1

(120582119903 (119905))120574minus1

1198920 (|119908|) + 119870120582119902 (119905) ae in [119879

0 119879lowast)

(43)

Proof It is clear that (43) is a particular case of (39) inparticular for

119866 (119905 119906) =1

(120582119903 (119905))120574minus1

1198920 (|119906|) 119887 (119905) = 119870120582119902 (119905) (44)

Since1198920is a locally Lipschitz function onR

+ for every119872 gt 0

there is an 1198710gt 0 depending on119872 such that

10038161003816100381610038161198920(1199041) minus 1198920(1199042)1003816100381610038161003816le 1198710

10038161003816100381610038161199041minus 1199042

1003816100381610038161003816 forall119904

1 1199041isin [minus119872119872]

(45)

Hence for any119872 gt 0 and all 1199061 1199062isin [minus119872119872] we obtain

1003816100381610038161003816119866 (119905 119906

1) minus 119866 (119905 119906

2)1003816100381610038161003816=

1

(120582119903 (119905))120574minus1

10038161003816100381610038161198920(10038161003816100381610038161199061

1003816100381610038161003816) minus 1198920(10038161003816100381610038161199062

1003816100381610038161003816)1003816100381610038161003816

le1198710

(120582119903 (119905))120574minus1

1003816100381610038161003816

10038161003816100381610038161199061

1003816100381610038161003816minus10038161003816100381610038161199062

1003816100381610038161003816

1003816100381610038161003816

le1198710

(120582119903 (119905))120574minus1

10038161003816100381610038161199061minus 1199062

1003816100381610038161003816

(46)

Thus according to assumption 119903minus1+120574

isin 1198711(1198790 119879lowast) the

required condition (42) is fulfilled in particular for 119871(119905) =

1198710(120582119903(119905))

minus120574+1 and so Lemma 15 proves this corollary

Before we give the proof ofTheorem 3 we state and provethe next two propositions In the first one by a nonoscillatorysolution 119909(119905) of the main equation (1) we get the existenceof a supersolution 119908(119905) of the Riccati differential equation(43) on the interval (119886

1 1198871) or (119886

2 1198872) In the second one we

construct two subsolutions 1199081(119905) and 119908

2(119905) of (43) which

blow up on intervals (1198861 119879lowast

1) sube (119886

1 1198871) and (119886

2 119879lowast

2) sube (119886

2 1198872)

respectively

Proposition 17 Let Φ(119906 V) and 119891(119906) satisfy (11) (12) and(16) respectively and let 119890(119905) satisfy (2) Let 119903(119905) gt 0 119902(119905) ge0 and 119902(119905) equiv 0 on [119886

1 1198871] cup [119886

2 1198872] Let 119909 = 119909(119905) be a

nonoscillatory solution of (1) and let for some 119879 ge 1199050and

120582 gt 0 the function 119908 = 119908(119905) be defined by

119908 (119905) = minus

120582119903 (119905)Φ (119909 (119905) 1199091015840(119905))

119909 (119905) 119905 ge 119879 (47)

Then 119908 isin 119860119862loc([119879infin)R) and 119908(119905) satisfies the inequality

1199081015840ge (120582119903 (119905))

1minus1205741198920(|119908|) + 119870120582119902 (119905) ae in 119869 (48)

where 119869 = (1198861 1198871) if 119909(119905) lt 0 and 119869 = (119886

2 1198872) if 119909(119905) gt 0

Proof Since 119909 = 119909(119905) is a nonoscillatory solution of (1) thereis a 119879 ge 119905

0such that 119909(119905) = 0 on [119879infin) Hence 119908(119905) is well

defined by (47)Because of (2) we have 119890(119905)119909(119905) le 0 for all 119905 isin 119869 where

119869 = (1198861 1198871) if 119909(119905) lt 0 and 119869 = (119886

2 1198872) if 119909(119905) gt 0 and since

120582 gt 0 we have

minus120582119890 (119905)

119909 (119905)ge 0 in 119869 (49)

Next since 119909 and 119903(119905)Φ(119909(119905) 1199091015840(119905)) are from

119860119862loc([1199050infin)R) we can take the first derivative of 119908(119905)for almost everywhere in 119869 which together with 119903(119905) gt 0119902(119905) ge 0 and 119902(119905) equiv 0 on 119869 gives

1199081015840(119905) = minus

120582(119903 (119905) Φ (119909 (119905) 1199091015840(119905)))1015840

119909 (119905)

+

120582119903 (119905)Φ (119909 (119905) 1199091015840(119905))

1199092(119905)

1199091015840(119905)

=120582119903 (119905)

|119909 (119905)|120574Φ(119909 (119905) 119909

1015840(119905)) 1199091015840(119905) |119909 (119905)|

120574minus2

+ 120582119902 (119905)119891 (119909 (119905))

119909 (119905)minus 120582

119890 (119905)

119909 (119905)

International Journal of Differential Equations 7

ge120582119903 (119905)

|119909 (119905)|120574119892 (10038161003816100381610038161003816Φ (119909 (119905) 119909

1015840(119905))

10038161003816100381610038161003816) + 119870120582119902 (119905) minus 120582

119890 (119905)

119909 (119905)

=120582119903 (119905)

|119909 (119905)|120574119892(

|119909 (119905)|

120582119903 (119905)|119908 (119905)|) + 119870120582119902 (119905) minus 120582

119890 (119905)

119909 (119905)

ge120582119903 (119905)

|119909 (119905)|120574

|119909 (119905)|120574

(120582119903 (119905))1205741198920 (|119908 (119905)|)

+ 119870120582119902 (119905) minus 120582119890 (119905)

119909 (119905)ae in 119869

(50)

that is

1199081015840ge (120582119903 (119905))

1minus1205741198920(|119908|) + 119870120582119902 (119905) minus 120582

119890 (119905)

119909 (119905)ae in 119869

(51)

Hence (49) and (51) prove the desired inequality (48)

Proposition 18 Let (19) be satisfied Let 1198771and 119877

2be two

arbitrary real numbers Then there are two points 119879lowast1isin (1198861 1198871)

and 119879lowast

2isin (1198862 1198872) and two continuous functions 119908

1(119905) and

1199082(119905) such that

119908119894(119886119894) le 119877119894 lim119905rarr119879

lowast

119894

119908119894(119905) = infin 119894 isin 1 2

1199081015840

119894le (120582119903 (119905))

1minus1205741198920(1003816100381610038161003816119908119894

1003816100381610038161003816) + 119870120582119902 (119905)

for ae 119905 isin (119886119894 119879lowast

119894) 119894 isin 1 2

(52)

Proof Since the function 119910(119904) = tan(119904) is a bijection fromthe interval (minus1205872 1205872) intoR we observe that there are two1199041 1199042isin (minus1205872 1205872) such that

tan (119904119894) le 119877119894 119894 isin 1 2 (53)

Next we define two functions 1198811(119905) and 119881

2(119905) by

119881119894 (119905) = 119904119894 +

120587

119888119894

int

119905

119886119894

119862 (120591) 119889120591 119905 isin [119886119894 119887119894] 119894 isin 1 2 (54)

where the real numbers 1198881 1198882and the function 119862(119905) are

defined in (19) From (19) and (54) one can immediatelyconclude that

119881119894(119886119894) = 119904119894lt120587

2 119881

119894(119887119894) = 119904119894+ 120587 gt

120587

2 119894 isin 1 2 (55)

Since 119862 isin 1198711

loc((1199050infin)R) it follows that 119862 isin 1198711((1199050infin)R)

which implies that 119881119894isin 119860119862([119886

119894 119887119894]R) 119894 isin 1 2 Therefore

exploiting the fact that in (55) we have 119881119894isin 119862([119886

119894 119887119894]R) we

obtain the existence of two points 119879lowast1isin (1198861 1198871) and 119879

lowast

2isin

(1198862 1198872) such that

119881119894(119879lowast

119894)=

120587

2 minus

120587

2lt119881119894(119905)lt

120587

2

forall119905 isin [119886119894 119879lowast

119894) 119894 isin 1 2

(56)

Now we are able to define the following two functions 1205961(119905)

and 1205962(119905) by

120596119894(119905) = tan (119881i (119905)) 119905 isin [119886

119894 119879lowast

119894) 119894 isin 1 2 (57)

That are well defined because of (56) Moreover from (53)(54) and (56) we obtain

120596119894(119886119894) = tan (119904

119894) le 119877119894

lim119905rarr119879

lowast

119898

120596119894(119905) = tan (119881

119894(119879lowast

119894)) = tan(120587

2) = infin

(58)

Also since119881119894isin 119860119862([119886

119894 119887119894]R) we can take the first derivative

of 120596119894(119905) and hence from (19) and (57) we obtain

1199081015840

119894(119905) =

120587

119888119894

119862 (119905)1

cos2 (119881119894 (119905))

=120587

119888119894

119862 (119905) (1003816100381610038161003816120596119894(119905)1003816100381610038161003816

2+ 1)

le120587

119888119894

119862 (119905) [1198920(1003816100381610038161003816120596119894(119905)1003816100381610038161003816) + 119872

0+ 1]

le120587

119888119894

119862 (119905) 1198920(1003816100381610038161003816120596119894(119905)1003816100381610038161003816) +

120587

119888119894

119862 (119905) (1198720+ 1)

le (120582119903 (119905))1minus1205741198920(1003816100381610038161003816120596119894(119905)1003816100381610038161003816) + 119870120582119902 (119905) ae in (119886

119894 119879lowast

119894)

(59)

which together with (58) proves this proposition

Now we are able to present the proof of Theorem 3 basedon Lemma 15 and Propositions 17 and 18

Proof of Theorem 3 Assuming the contrary then there is anonoscillatory solution 119909(119905) and 119879 ge 119905

0such that 119909(119905) = 0 for

all 119905 ge 119879 119909(119905) and 119903(119905)Φ(119909 1199091015840) are from 119860119862loc([1199050infin)R)In order to simplify notation let every 119895 isin N be fixed andomitted in the notation For instance instead of (119886

1119895 1198871119895) and

(1198862119895 1198872119895) we write (119886

1 1198871) and (119886

2 1198872) respectively and so on

On one hand we observe by Proposition 17 that thefunction 119908(119905) defined by (47) is a supersolution of thefollowing Riccati differential equation

1199081015840= (120582119903 (119905))

1minus1205741198920(|119908|) + 119870120582119902 (119905) ae in 119869 (60)

where 119908 isin 119860119862loc(119869R) and 119869 = (1198861 1198871) if 119909(119905) lt 0 and 119869 =

(1198862 1198872) if 119909(119905) gt 0 see the proof of Proposition 17 Let for

instance 119909(119905) lt 0 and thus 119869 = (1198861 1198871)

On the other hand by Proposition 18 there are a number119879lowast

1isin (1198861 1198871) and subsolutions 119908

1(119905) 1199081isin 119860119862([119886

1 119879lowast

1)R) of

the Riccati differential equation (60) such that

1199081(1198861) le 119908 (119886

1) lim

119905rarr119879lowast

1

1199081(119905) = infin (61)

(in the case when 119869 = (1198862 1198872) then we work with119879lowast

2isin (1198862 1198872)

and the other subsolution 1199082(119905) 119908

2isin 119860119862([119886

2 119879lowast

2)R))

Hence by Corollary 16 and (61) we conclude that1199081(119905) le 119908(119905)

for all 119905 isin [1198861 119879lowast

1) and therefore

infin = lim119905rarr119879

lowast

1

1199081(119905) le lim119905rarr119879

lowast

1

119908 (119905) (62)

which contradicts 119908 isin 119860119862loc([119879infin)R) Thus the assump-tion that 119909(119905) is nonoscillatory is not possible and hence everysolution of (1) is oscillatory

8 International Journal of Differential Equations

Proof of Theorem 4 There is only one difference betweenTheorems 3 and 4 and it is the difference between conditions(19) and (20) Hence Theorem 4 immediately follows fromTheorem 3 provided that we prove the equivalence between(19) and (20)

First of all we need the following two lemmas

Lemma 19 Let 119886 119887 and 119888 be positive real numbers and 120574 gt 1Then the existence of a positive real number 120582 such that

119888 le min 119886

120582120574minus1

119887120582 (63)

is equivalent to

119888 le 119886112057411988711205741015840

(64)

Here 1205741015840 = 120574(120574 minus 1) is the conjugate exponent of 120574

Proof Let us define 119892(120582) = min119886120582120574minus1 119887120582 Since 120582 997891rarr

119886120582120574minus1 is decreasing and 120582 997891rarr 119887120582 is increasing there exists

a unique point of maximum of 119892 It is achieved at 120582 = 1205820

which is a solution of 119886120582120574minus10

= 1198871205820 hence 120582

0= (119886119887)

1120574If there is a positive real number 120582 such that 119888 le 119892(120582)

then 119888 le 119892(1205820) = 119887120582

0= 119887(119886119887)

1120574= 119886112057411988711205741015840

Converselyif 119888 le 119886

112057411988711205741015840

then 119888 le 119892(1205820) and the equivalence in the

lemma is proved

The preceding lemma is a special case of the followingmore general statement

Lemma 20 Assume that 119886(119905) 119887(119905) and 119888(119905) are positive realfunctions defined on a subset 119869 sube R Then the existence of apositive real number 120582 such that for all 119905 isin 119869

119888 (119905) le min 119886 (119905)120582120574minus1

119887 (119905) 120582 (65)

is equivalent to

sup119905isin119869

119888 (119905)

119887 (119905)le (inf119905isin119869

119886 (119905)

119888 (119905))

1(120574minus1)

(66)

Proof The condition that there exists 120582 gt 0 such that (65)is true for all 119905 isin 119869 is equivalent with the following twoinequalities which have to hold simultaneously

119888 (119905) le119886 (119905)

120582120574minus1

119888 (119905) le 119887 (119905) 120582 (67)

that is with

119888 (119905)

119887 (119905)le 120582 le (

119886 (119905)

119888 (119905))

1(120574minus1)

forall119905 isin 119869 (68)

Taking the supremum of the left-hand side and the infimumof the right-hand side we obtain (66) Conversely if (66)is satisfied then since the left-hand side in (66) cannotbe equal to zero while the right-hand side is less than infinthen inequality (66) defines a nonempty closed interval in R

(possibly reducing to just one point) in which we can chooseany positive 120582 Hence (67) is satisfied for all 119905 isin 119869 andtherefore (65) as well

Remark 21 If 119886 = 119886(119905) 119887 = 119887(119905) and 119888 = 119888(119905) appearingin Lemma 20 are constant functions then condition (66) isequivalent to 119888119887 = (119886119888)1(120574minus1) that is to (64)

Proof of Equivalence of (19) and (20) Wewill use Lemma 20Let us fix 119869

119894119895 where 119894 isin 1 2 and 119895 isin N We see that that the

inequality appearing in the second line of condition (19) is ofthe form (65) where

119888 (119905) =1

119888119894119895

119862 (119905) 119886 (119905) =1

120587119903(119905)1minus120574

119887 (119905) =119870119902 (119905)

120587 (1198720+ 1)

(69)

Condition (66) is equivalent to

sup119905isin119869119894119895

119862 (119905)

119902 (119905)le

1198701198881205741015840

119894119895

1205871205741015840

(1198720+ 1)

inf119905isin119869119894119895

1

119903 (119905) 119862(119905)120574minus1

=

1198701198881205741015840

119894119895

(1198720+ 1) sup

119905isin119869119894119895119903 (119905) 119862(119905)

120574minus1(

119888119894119895

120587)

1205741015840

(70)

This inequality is equivalent to the corresponding one in (19)and the claim follows from Lemma 19

Proof of Corollary 6 Substituting inf119905isin119869119894119895

119902(119905) instead of 119902(119905)and sup

119905isin119869119894119895119903(119905) instead of 119903(119905) in the inequality appearing in

the second line of (20) after a short computation we obtain astronger inequality than (20) as

1

119888119894119895

sup119905isin119869119894119895

119862 (119905)

le1

120587(

119870inf119905isin119869119894119895

119902 (119905)

(1198720+ 1) sup

119905isin119869119894119895119903 (119905)

)

(120574minus1)120574

forall119894 isin 1 2 119895 isin N

(71)

Substituting119862(119905) equiv 1we obtain that the left-hand side of (71)is equal to |119869

119894119895|minus1 and the resulting inequality is equivalent to

(21) Since (21) implies (20) the claim is proved

Remark 22 Here we show that the choice of 119862(119905) equiv 1 inthe proof of Corollary 6 is the best possible To prove thisnote that on the left-hand side of (71) we have the expressiondepending on an auxiliary function 119862(119905) and this functionis absent on the right-hand side Therefore it has sense totry to find a function 119862(119905) 119905 isin 119869

119894119895 such that the value of

119876(119862) = (1119888119894119895)sup119905isin119869119894119895

119862(119905) is minimal (note that 119888119894119895depends

on the function 119862 = 119862(119905) as well) It is easy to see that theminimum is achieved for 119862(119905) equiv 1 (or any positive constant)Indeed since 119876(119862) = 119876(120583119888) for any 120583 gt 0 by taking 120583 =

(sup119905isin119869119894119895

119862(119905))minus1 it suffices to assume that sup

119905isin119869119894119895119862(119905) = 1

The value of 119876(119862) is minimal if 119888119894119895= int119869119894119895

119862(120591) 119889120591 is maximalpossible and since 0 le 119862(119905) le 1 it is clear that the maximumof 119888119894119895= 119888119894119895(119862) is achieved when 119862(119905) equiv 1

International Journal of Differential Equations 9

Proof of Corollary 7 It is clear that the condition (23) implies(21) Furthermore since the lengths |119869

119894119895| are uniformly

bounded from below by a positive constant see (23) thenobviously 119886

1119895rarr infin as 119895 rarr infin Thus the claim follows

immediately from Corollary 6

Proof of Corollary 10 Let 119862(119905) = 119902(119905) Then the requiredcondition (19) is clearly satisfied because of assumption(30) Hence this corollary immediately follows from Theo-rem 3

4 An Extension of Condition (12) tothe Case of 120574 gt 1

In this section we consider the oscillation of (1) in the casewhen 120574 isin (1 2) is allowed in the assumption (12) In this waythe assumption (11) is slightly modified by a real number 119901 gt1 such that

|119906|(119901minus1)120574minus119901

VΦ (119906 V) ge 119892 (|Φ (119906 V)|) (72)

where (12) is supposed that is 119892(119888119904) ge 1198881205741198920(119904) for every 119888 119904 gt

0 and for a locally Lipschitz function 1198920 [0infin) rarr [0infin)

for which there exists a constant1198720such that 119892

0(119904)+119872

0ge 119904119901

for all 119904 isin [0infin)For the function 119891 we assume that there exists a constant

119870 gt 0 such that

119891 (119906) sign (119906) ge 119870|119906|119901minus1 for every 119906 isin R (73)

We will need the following two lemmas

Lemma 23 Let 119886 lt 119887 and let 119891 [119886 119887] rarr [0infin) be a meas-urable function Then

int

119887

119886

119891 (119905) 119889119905 ge 1 (74)

if and only if there exists a function 119892 isin 1198711([119886 119887] [0infin))

1198921= 1 such that

119891 (119905) ge 119892 (119905) forall119905 isin [119886 119887] (75)

Proof We assume (74) If int119887119886119891(119905)119889119905 is finite we may choose

119892(119909) = 119891(119909)1198911

For int119887119886119891(119905)119889119905 = infin we may define for every natural

number 119899 a function

119891119899(119905) =

119891 (119905) 119891 (119905) le 119899

119899 119891 (119905) gt 119899(76)

Since 119891119899(119905) le 119899 we have 119891

119899isin 1198711([119886 119887]R) Obviously

lim119899int119887

119886119891119899(119905)119889119905 is also infiniteThis implies that there exists an

1198990such that int119887

1198861198911198990(119905)119889119905 ge 1 We define 119892(119905) = 119891

1198990(119905)119891

11989901

Now we assume (75) Integrating over the whole intervalwe get

int

119887

119886

119891 (119905) 119889119905 ge int

119887

119886

119892 (119905) 119889119905 = 1 (77)

Lemma 24 Let119901 gt 1There exists a unique function 119910 = 119910(119904)that satisfies the following Cauchy problem

1199101015840(119904) = 1 +

1003816100381610038161003816119910 (119904)

1003816100381610038161003816

119901 119904 isin R (78)

119910 (0) = 0 (79)

Furthermore there exists a finite real number 119878lowast such that

lim119904rarrplusmn119878

lowast119910 (119904) = plusmninfin (80)

and 119910 isin 1198621((minus119878lowast 119878lowast)R) is injective and monotonous We

have 119878lowast = (120587119901)(1(sin(120587119901))) and 119910 is odd

We will denote by tan119901the solution of (78)-(79)

Proof Obviously every solution to (78) is injective andmonotonous on a connected set since 1199101015840 = 1 + |119910|119901 ge 1 gt 0

Let 119891(119909) = 1(1 + |119909|119901) Obviously 1 ge 119891(119909) gt 0 for all

119909 isin R and 119862119901gt |1198911015840(119909)| gt 0 for some constants 119862

119901isin R

We may assume 119862119901gt 1 Hence 1199111015840 = 119891(119905) 119911(0) = 0 has (by

[17Theorem 31]) a locally unique solution on (minus1119862119901 1119862119901)

By the same argument 119911 can be extended to the whole of Rinterval by interval of type ((119899 minus 1)119862

119901 (119899 + 1)119862

119901) for all

119899 isin ZWe know from [18] that the for the integral of the left-

hand side we have

int

infin

0

119891 (119909) 119889119909 = int

infin

0

119889119909

1 + 119909119901=120587

119901

1

sin (120587119901)= 119878lowast (81)

so lim119905rarrplusmninfin

119911(119905) = plusmn119878lowast

Since 1199111015840(119905) = 0 119911 is injective monotonous and of class1198621(R (minus119878lowast 119878lowast)) and so we define 119910(119904) = 119911

minus1(119904) It imme-

diately follows that 119910 isin 1198621((minus119878lowast 119878lowast)R) and that 119910 is

injective and monotonous Also by continuity of 119911 we havelim119904rarrplusmn119878

lowast119910(119904) = lim119908rarrplusmninfin

119910(119911(119908)) = lim119908rarrplusmninfin

119908 = plusmninfinDifferentiating it also follows that 1199101015840 = 1 + |119910|

119901 and bydefinition of 119910 119910(0) = 119911minus1(0) = 0 So 119910 is a solution of (78)-(79) Since any other such solution must have 119911 as its inverse119910 is unique on its domain

Proposition 25 Let 119886 lt 119887 and 120574 gt 119901 gt 1 Assume (72) and(73) and let 119902(119905) ge 0 and 119890(119905) le 0 (or 119890(119905) ge 0 ) for all 119905 isin [119886 119887]If there exists a constant 120582 gt 0 such that

119901

2120587sin(120587

119901)int

119887

119886

min119901 minus 1

119903120574minus1

(119905) 120582120574minus1

119870120582119902 (119905)

1 +1198720

119889119905 ge 1 (82)

then for any solution 119909(119905) of (1) with 119909(119886) ge 0 (or 119909(119886) le 0)there exists a number 119905lowast isin [119886 119887] such that 119909(119905lowast) = 0

Proof We assume that 119909(119905) = 0 for all 119905 isin [119886 119887] Then we maydefine the function

120596 (119905) = minus

120582119903 (119905) Φ (119909 (119905) 1199091015840(119905))

sign (119909 (119905)) |119909 (119905)|119901minus1 (83)

10 International Journal of Differential Equations

This function is well defined on [119886 119887] and its derivative is

1199081015840(119905) = minus

120582(119903 (119905) Φ (119909 (119905) 1199091015840(119905)))1015840

sign (119909 (119905)) |119909 (119905)|119901minus1

+

(119901 minus 1) 120582119903 (119905) Φ (119909 (119905) 1199091015840(119905))

|119909 (119905)|119901

1199091015840(119905)

(84)

Using the fact that 119909(119905) is a solution of (1) and omitting 119905 from119909 and 1199091015840 for clarity we get

1199081015840(119905) =

120582 (119902 (119905) 119891 (119909) minus 119890 (119905))

sign (119909) |119909|119901minus1+

(119901 minus 1) 120582119903 (119905) Φ (119909 1199091015840)

119909119901

1199091015840

=

(119901 minus 1) 120582119903 (119905) Φ (119909 1199091015840)

|119909|119901

1199091015840+ 120582119902 (119905)

119891 (119909)

sign (119909) |119909|119901minus1

minus 120582119890 (119905)

sign (119909) |119909|119901minus1

=(119901 minus 1) 120582119903 (119905)

|119909|120574(119901minus1)

Φ(119909 1199091015840) 1199091015840|119909|(119901minus1)120574minus119901

+ 120582119902 (119905)119891 (119909)

sign (119909) |119909|119901minus1minus 120582

119890 (119905)

sign (119909) |119909|119901minus1

(72)

ge(119901 minus 1) 120582119903 (119905)

|119909|120574(119901minus1)

119892 (10038161003816100381610038161003816Φ (119909 119909

1015840)10038161003816100381610038161003816)

+ 120582119902 (119905)119891 (119909)

sign (119909) |119909|119901minus1

minus 120582119890 (119905)

sign (119909) |119909|119901minus1

=(119901 minus 1) 120582119903 (119905)

|119909|120574(119901minus1)

119892(|119909|119901minus1

120582119903 (119905)|120596 (119905)|)

+ 120582119902 (119905)119891 (119909)

sign (119909) |119909|119901minus1minus 120582

119890 (119905)

sign (119909) |119909|119901minus1

(12)

ge(119901 minus 1) 120582119903 (119905)

|119909|120574(119901minus1)

(|119909|119901minus1

120582119903 (119905))

120574

1198920 (|120596 (119905)|)

+ 120582119902 (119905)119891 (119909)

sign (119909) |119909|119901minus1minus 120582

119890 (119905)

sign (119909) |119909|119901minus1

=119901 minus 1

(120582119903 (119905))120574minus1

1198920 (|120596 (119905)|)+120582119902 (119905)

119891 (119909)

sign (119909) |119909|119901minus1

minus 120582119890 (119905)

sign (119909) |119909|119901minus1

(73)

ge119901 minus 1

(120582119903 (119905))120574minus1

1198920 (|120596 (119905)|) + 120582119902 (119905) 119870

minus 120582119890 (119905)

sign (119909) |119909|119901minus1

(85)

Since 119890(119905)(sign(119909)|119909|119901minus1) le 0 we have

1199081015840(119905) ge

119901 minus 1

(120582119903 (119905))120574minus1

1198920(|120596 (119905)|) + 120582119870119902 (119905) (86)

We apply Lemma 23 to (82) to obtain a function119862(119905) suchthat

int

119887

119886

119862 (119905) 119889119905 = 1 (87)

and by (75)

0 le 119862 (119905) le119901

2120587sin(120587

119901)min

119901 minus 1

119903120574minus1

(119905) 120582120574minus1

119870120582119902 (119905)

1 +1198720

(88)

Let 1199040isin (minus(120587119901)(1 sin(120587119901)) (120587119901)(1 sin(120587119901))) be such

that tan119901(1199040) le 120596(119886) We define the function

119881 (119905) = 1199040+2120587

119901

1

sin (120587119901)int

119905

119886

119862 (120591) 119889120591 (89)

Note that (minus(120587119901)(1 sin(120587119901))) lt 1199040

= 119881(119886) lt

(120587119901)(1 sin(120587119901)) and 119881(119887) = 1199040+ (2120587119901)(1 sin(120587119901)) gt

(120587119901)(1 sin(120587119901)) Since119881 is continuous there exists a 119879lowast isin(119886 119887) such that 119881(119905) lt (120587119901)(1 sin(120587119901)) for 119905 isin [119886 119879lowast) and119881(119879lowast) = (120587119901)(1 sin(120587119901))

We define

120596 (119905) = tan119901(119881 (119905)) 119905 isin [119886 119879

lowast) (90)

and note that by Lemma 24 lim119905rarr119879

lowast120596(119905) = +infin wheretan119901(119904) = 119910(119904) and 119910(119904) is determined by Lemma 24The derivative of 120596 then satisfies

1205961015840(119905) = tan1015840

119901(119881 (119905)) 119881

1015840(119905)

= (1 + tan119901119901(119881 (119905))) 119881

1015840(119905)

=2120587

119901

1

sin (120587119901)(1 + 120596

119901(119905)) 119862 (119905)

le (1 +1198720+ 1198920(1003816100381610038161003816120596 (119905)

1003816100381610038161003816))min

119901 minus 1

119903120574minus1

(119905) 120582120574minus1

119870120582119902 (119905)

1 +1198720

le119901 minus 1

119903120574minus1

(119905) 120582120574minus1

1198920(1003816100381610038161003816120596 (119905)

1003816100381610038161003816) + 119870120582119902 (119905)

(91)

Thus we may apply the comparison principle ofLemma 15 to 120596(119886) le 120596(119886) and their respective differentialinequalities to conclude that 120596(119905) le 120596(119905) for all 119905 isin [119886 119887]But since 120596 rarr +infin as 119905 rarr 119879

lowastlt 119887 and 120596(119905) is well defined

on the whole segment [119886 119887] we arrive at a contradictionHence 119909(119905) gt 0 cannot hold for all 119905 isin [119886 119887] and since 119909(119905) iscontinuous there exists a 119905lowast isin [119886 119887] such that 119909(119905lowast) = 0

Theorem 26 Let 120574 gt 119901 gt 1 Assume (72) and (73) and let forany 119879 gt 119905

0there exist 119879 le 119886

1lt 1198871le 1198862lt 1198872such that 119892(119905) ge 0

on [1198861 1198871] cup [119886

2 1198872] and 119890(119905) le 0 on [119886

1 1198871] and 119890(119905) ge 0 on

International Journal of Differential Equations 11

[1198862 1198872] If for both 119894 isin 1 2 there exist constants 120582

119894gt 0 such

that

119901

2120587sin (120587119901)int

119887119894

119886119894

min119901 minus 1

119903120574minus1

(119905) 120582120574minus1

119894

119870120582119894119902 (119905)

1 +1198720

119889119905 ge 1

(92)

then (1) is oscillatory

Proof For (1) to be oscillatory it is sufficient that for every119899 isin N there exists a 119905

119899gt 119899 such that 119909(119905

119899) = 0

Let 119899 isin N and let 119899 lt 1198861lt 1198871le 1198862lt 1198872as assumed in the

theorem It is enough to show that every solution 119909(119905) of (1)has a zero on [119886

1 1198872] If 119909(119886

1) ge 0 we apply Proposition 25 to

the segment [1198861 1198871] to obtain the sought zero If 119909(119886

2) le 0 we

again apply Proposition 25 to the segment [1198862 1198872] Suppose

119909(1198861) lt 0 and 119909(119886

2) gt 0 Then since 119909 is continuous there

exists a number 1199051isin [1198861 1198862] such that 119909(119905

1) = 0

Acknowledgment

This work is supported by the Ministry of Science of theRepublic of Croatia under Grant no 036-0361621-1291

References

[1] Q Kong ldquoNonoscillation and oscillation of second order half-linear differential equationsrdquo Journal of Mathematical Analysisand Applications vol 332 no 1 pp 512ndash522 2007

[2] Q-R Wang and Q-G Yang ldquoInterval criteria for oscillationof second-order half-linear differential equationsrdquo Journal ofMathematical Analysis and Applications vol 291 no 1 pp 224ndash236 2004

[3] Q Long and Q-R Wang ldquoNew oscillation criteria of second-order nonlinear differential equationsrdquo Applied Mathematicsand Computation vol 212 no 2 pp 357ndash365 2009

[4] Z Zheng and F Meng ldquoOscillation criteria for forced second-order quasi-linear differential equationsrdquo Mathematical andComputer Modelling vol 45 no 1-2 pp 215ndash220 2007

[5] D Cakmak and A Tiryaki ldquoOscillation criteria for certainforced second-order nonlinear differential equationsrdquo AppliedMathematics Letters vol 17 no 3 pp 275ndash279 2004

[6] F Jiang and F Meng ldquoNew oscillation criteria for a class ofsecond-order nonlinear forced differential equationsrdquo Journalof Mathematical Analysis and Applications vol 336 no 2 pp1476ndash1485 2007

[7] Q Yang ldquoInterval oscillation criteria for a forced secondorder nonlinear ordinary differential equations with oscillatorypotentialrdquo Applied Mathematics and Computation vol 135 no1 pp 49ndash64 2003

[8] A Tiryaki and B Ayanlar ldquoOscillation theorems for certainnonlinear differential equations of second orderrdquo Computers ampMathematics with Applications vol 47 no 1 pp 149ndash159 2004

[9] J Jaros T Kusano and N Yoshida ldquoGeneralized Piconersquosformula and forced oscillations in quasilinear differential equa-tions of the second orderrdquoArchivumMathematicum vol 38 no1 pp 53ndash59 2002

[10] R P Agarwal S R Grace and D OrsquoRegan Oscillation Theoryfor Second Order Linear Half-Linear Superlinear and SublinearDynamic Equations Kluwer Academic Publishers DordrechtThe Netherlands 2002

[11] S Jing ldquoA new oscillation criterion for forced second-orderquasilinear differential equationsrdquoDiscrete Dynamics in Natureand Society vol 2011 Article ID 428976 8 pages 2011

[12] W-T Li and S S Cheng ldquoAn oscillation criterion for nonhomo-geneous half-linear differential equationsrdquoAppliedMathematicsLetters vol 15 no 3 pp 259ndash263 2002

[13] J S W Wong ldquoOscillation criteria for a forced second-orderlinear differential equationrdquo Journal of Mathematical Analysisand Applications vol 231 no 1 pp 235ndash240 1999

[14] M Pasic ldquoFite-Wintner-Leighton type oscillation criteria forsecond-order differential equations with nonlinear dampingrdquoAbstract and Applied Analysis vol 2013 Article ID 852180 10pages 2013

[15] M Pasic ldquoNew interval oscillation criteria for forced second-order differential equations with nonlinear dampingrdquo Interna-tional Journal of Mathematical Analysis vol 7 no 25 pp 1239ndash1255 2013

[16] L Gasinski and N S Papageorgiou Nonsmooth Critical PointTheory and Nonlinear Boundary Value Problems vol 8 of Seriesin Mathematical Analysis and Applications Edited by R PAgarwal and D OrsquoRegan Chapman amp HallCRC Boca RatonFla USA 2005

[17] S Lang Real and Functional Analysis vol 142 of GraduateTexts in Mathematics Springer-Verlag New York NY USA 3rdedition 1993

[18] I N Bronshtein K A Semendyayev G Musiol and HMuehligHandbook of Mathematics Springer Berlin Germany5th edition 2007

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

International Journal of Differential Equations 5

The following three examples are simple consequences ofCorollary 10

Example 11 (oscillation of (5)) Equation (5) as well asequation

11990910158401015840+ 11989821205872[sin (119898119905)]+119891 (119909) = ℎ (cos (119898119905))

ae in [1199050infin)

119909 1199091015840isin 119860119862loc ([1199050infin) R)

(34)

are oscillatory where 119891(119906) satisfies (16) with 119870 ge 1 andℎ(119904)119904 gt 0 119904 = 0 Indeed in (5) we have 119903(119905) equiv 1 119902(119905) =

11989821205872[cos(119898119905)] and 119890(119905) = ℎ(sin(119898119905)) and so (17) and (18) are

fulfilled and 119902(119905) ge 0 119902(119905) equiv 0 on each interval 119869119894119895 119894 isin 1 2

119895 isin N where 1198691119895= [1198861119895 1198871119895] 1198692119895= [1198862119895 1198872119895] and

1198861119895=2119895120587

119898 119887

1119895=(2119895 + 12) 120587

119898

1198862119895=(2119895 + 32) 120587

119898 119887

2119895=(2119895 + 2) 120587

119898

(35)

Moreover

119902119894119895= int

119869119894119895

119902 (120591) 119889120591

= 11989821205872int

119887119894119895

119886119894119895

[cos (119898120591)]+119889120591 = 1198981205872 119894 isin 1 2 119895 isin N

(36)

and since 120574 = 2 119903(119905) equiv 1119870 ge 1 and1198720= 0 for 120582

119895= 1(119898120587)

we have

120582119895=

1

119898120587ge1

119870

1

119898120587=120587 (1198720+ 1)

119870

1

1198981205872

=120587 (1198720+ 1)

119870

1

119902119894119895

119894 isin 1 2 119895 isin N

1

119902119894119895

119902 (119905) =1

119898120587211989821205872[cos (119898119905)]+ le 119898

=1

120587

1

(1119898120587)=1

120587

1

(120582119895119903 (119905))120574minus1

119905 isin 119869119894119895 119894 isin 1 2 119895 isin N

(37)

Thus the required condition (30) is fulfilled in this case andhence we may apply Corollary 10 to (5) to verify that thisequation is oscillatory Analogously we can show that (34)is oscillatory too

In the next example the coefficient 119902(119905) is a discontinuousfunction on [119905

0infin)

Example 12 (oscillation of (6)) If 119891(119906) satisfies (16) with119870 ge

1 then (6) is oscillatory since the required condition (30) issatisfied especially for 119902(119905) = 4119898

2 sign(cos(119898119905)) on 119869119894119895 120582119895=

1(2119898) and 119869119894119895as in (35) 120574 = 2 119903(119905) equiv 1119870 ge 1 and119872

0= 0

Indeed

119902119894119895= int

119869119894119895

119902 (120591) 119889120591 = 41198982int

119869119894119895

[sign (cos (119898120591))]+119889120591

= 41198982 120587

2119898= 2119898120587 119894 isin 1 2 119895 isin N

120582119895=

1

2119898=

120587

2119898120587= 120587 (119872

0+ 1)

1

119902119894119895

ge120587 (1198720+ 1)

119870

1

119902119894119895

119894 isin 1 2 119895 isin N

1

119902119894

119902 (119905) =1

211989812058741198982[sign (cos (119898119905))]+

le2119898

120587=1

120587

1

radic1 (41198982)

=1

120587(120582119895119903 (119905))120574minus1

119905 isin 119869119894119895 119894 isin 1 2 119895 isin N

(38)

which shows that (30) is satisfied

At the end of this section we consider the case when 119902(119905)changes sign on [119905

0infin) but with the help of Corollary 10 since

119902(119905) is strictly positive on all intervals 119869119894119895

Example 13 (oscillation of (7)) If 119891(119906) satisfies (16) with119870 ge

1 then model equation (7) is oscillatory In fact let 119869119894119895be as

in (35) If ℎ(119904)119904 gt 0 119904 = 0 it is clear that 119890(119905) = ℎ(sin(119898119905))satisfies (2) that is 119890(119905) ge 0 on 119869

1119895and 119890(119905) le 0 on 119869

2119895 On

the other hand we have 119902(119905) = 41198982 sign(cos(119898119905)) on 119869

119894119895and

so 119902(119905) ge 0 119902(119905) equiv 0 on 119869119894119895for all 119894 isin 1 2 119895 isin N Hence

the proof of the fact that all assumptions of Corollary 7 arefulfilled is the same as in the preceding example

3 Proofs of Theorems 3 and 4

Let 119860119862loc([119886 119887)R) denote the set of all real functions whichare absolutely continuous on every interval [119886 119887 minus 120576] where120576 isin (0 119887 minus 119886) For arbitrary numbers 119879

0lt 119879lowast and functions

119866 = 119866(119905 119906) and 119887 = 119887(119905) we consider the ordinary differentialequation

1199081015840= 119866 (119905 119908) + 119887 (119905) ae in [119879

0 119879lowast) (39)

which generalizes the classic Riccati equation 1199081015840 = 119886(119905)1199082+

119887(119905) where 119886(119905) is an arbitrary function We associate to(39) the corresponding sub- and supersolutions 119908119908 isin

119860119862loc([1198790 119879lowast)R) which satisfy respectively

1199081015840le 119866 (119905 119908) + 119887 (119905) 119908

1015840ge 119866 (119905 119908) + 119887 (119905)

ae in [1198790 119879lowast)

(40)

We are interested in studying the following property

119908 (1198790) le 119908 (119879

0) implies 119908 (119905) le 119908 (119905)

forall119905 isin [T0 119879lowast)

(41)

In this way we introduce the following definition

6 International Journal of Differential Equations

Definition 14 We say that comparison principle holds for (39)on an interval [119879

0 119879lowast) 119879 le 119879

0lt 119879lowast if property (41)

is fulfilled for all sub- and supersolutions 119908119908 isin

119860119862loc([1198790 119879lowast)R) of (39) on [119879

0 119879lowast)

Now we are able to state the next result

Lemma 15 Let 119879 le 1198790lt 119879lowast Let 119866 = 119866(119905 119906) and 119887 = 119887(119905) be

two arbitrary function For any119872 gt 0 let there be a function119871 = 119871(119905) gt 0 depending on 119879

0 119879lowast119872 such that

119871 (119905) gt 0 on [1198790 119879lowast) 119871 isin 119871

1(1198790 119879lowast)

1003816100381610038161003816119866 (119905 119906

1) minus 119866 (119905 119906

2)1003816100381610038161003816le 119871 (119905)

10038161003816100381610038161199061minus 1199062

1003816100381610038161003816

forall119905 isin [1198790 119879lowast) 1199061 1199062isin [minus119872119872]

(42)

Then comparison principle (41) holds for (39) on an interval[1198790 119879lowast) 119879 le 119879

0lt 119879lowast

Proof It is enough to use the idea of the proof of [14 Lemma41] Hence this proof is left to the reader

Corollary 16 Let 120582 gt 0 and let 1198920 R+rarr R+be a locally

Lipschitz function on R+ Let 119870 be an arbitrary real number

and 119902(119905) an arbitrary function Let 119879 le 1198790lt 119879lowast If 119903(119905) gt 0

on [1198790 119879lowast) and 119903minus1+120574 isin 1198711(119879

0 119879lowast) then comparison principle

(41) holds for the Riccati differential equation

1199081015840=

1

(120582119903 (119905))120574minus1

1198920 (|119908|) + 119870120582119902 (119905) ae in [119879

0 119879lowast)

(43)

Proof It is clear that (43) is a particular case of (39) inparticular for

119866 (119905 119906) =1

(120582119903 (119905))120574minus1

1198920 (|119906|) 119887 (119905) = 119870120582119902 (119905) (44)

Since1198920is a locally Lipschitz function onR

+ for every119872 gt 0

there is an 1198710gt 0 depending on119872 such that

10038161003816100381610038161198920(1199041) minus 1198920(1199042)1003816100381610038161003816le 1198710

10038161003816100381610038161199041minus 1199042

1003816100381610038161003816 forall119904

1 1199041isin [minus119872119872]

(45)

Hence for any119872 gt 0 and all 1199061 1199062isin [minus119872119872] we obtain

1003816100381610038161003816119866 (119905 119906

1) minus 119866 (119905 119906

2)1003816100381610038161003816=

1

(120582119903 (119905))120574minus1

10038161003816100381610038161198920(10038161003816100381610038161199061

1003816100381610038161003816) minus 1198920(10038161003816100381610038161199062

1003816100381610038161003816)1003816100381610038161003816

le1198710

(120582119903 (119905))120574minus1

1003816100381610038161003816

10038161003816100381610038161199061

1003816100381610038161003816minus10038161003816100381610038161199062

1003816100381610038161003816

1003816100381610038161003816

le1198710

(120582119903 (119905))120574minus1

10038161003816100381610038161199061minus 1199062

1003816100381610038161003816

(46)

Thus according to assumption 119903minus1+120574

isin 1198711(1198790 119879lowast) the

required condition (42) is fulfilled in particular for 119871(119905) =

1198710(120582119903(119905))

minus120574+1 and so Lemma 15 proves this corollary

Before we give the proof ofTheorem 3 we state and provethe next two propositions In the first one by a nonoscillatorysolution 119909(119905) of the main equation (1) we get the existenceof a supersolution 119908(119905) of the Riccati differential equation(43) on the interval (119886

1 1198871) or (119886

2 1198872) In the second one we

construct two subsolutions 1199081(119905) and 119908

2(119905) of (43) which

blow up on intervals (1198861 119879lowast

1) sube (119886

1 1198871) and (119886

2 119879lowast

2) sube (119886

2 1198872)

respectively

Proposition 17 Let Φ(119906 V) and 119891(119906) satisfy (11) (12) and(16) respectively and let 119890(119905) satisfy (2) Let 119903(119905) gt 0 119902(119905) ge0 and 119902(119905) equiv 0 on [119886

1 1198871] cup [119886

2 1198872] Let 119909 = 119909(119905) be a

nonoscillatory solution of (1) and let for some 119879 ge 1199050and

120582 gt 0 the function 119908 = 119908(119905) be defined by

119908 (119905) = minus

120582119903 (119905)Φ (119909 (119905) 1199091015840(119905))

119909 (119905) 119905 ge 119879 (47)

Then 119908 isin 119860119862loc([119879infin)R) and 119908(119905) satisfies the inequality

1199081015840ge (120582119903 (119905))

1minus1205741198920(|119908|) + 119870120582119902 (119905) ae in 119869 (48)

where 119869 = (1198861 1198871) if 119909(119905) lt 0 and 119869 = (119886

2 1198872) if 119909(119905) gt 0

Proof Since 119909 = 119909(119905) is a nonoscillatory solution of (1) thereis a 119879 ge 119905

0such that 119909(119905) = 0 on [119879infin) Hence 119908(119905) is well

defined by (47)Because of (2) we have 119890(119905)119909(119905) le 0 for all 119905 isin 119869 where

119869 = (1198861 1198871) if 119909(119905) lt 0 and 119869 = (119886

2 1198872) if 119909(119905) gt 0 and since

120582 gt 0 we have

minus120582119890 (119905)

119909 (119905)ge 0 in 119869 (49)

Next since 119909 and 119903(119905)Φ(119909(119905) 1199091015840(119905)) are from

119860119862loc([1199050infin)R) we can take the first derivative of 119908(119905)for almost everywhere in 119869 which together with 119903(119905) gt 0119902(119905) ge 0 and 119902(119905) equiv 0 on 119869 gives

1199081015840(119905) = minus

120582(119903 (119905) Φ (119909 (119905) 1199091015840(119905)))1015840

119909 (119905)

+

120582119903 (119905)Φ (119909 (119905) 1199091015840(119905))

1199092(119905)

1199091015840(119905)

=120582119903 (119905)

|119909 (119905)|120574Φ(119909 (119905) 119909

1015840(119905)) 1199091015840(119905) |119909 (119905)|

120574minus2

+ 120582119902 (119905)119891 (119909 (119905))

119909 (119905)minus 120582

119890 (119905)

119909 (119905)

International Journal of Differential Equations 7

ge120582119903 (119905)

|119909 (119905)|120574119892 (10038161003816100381610038161003816Φ (119909 (119905) 119909

1015840(119905))

10038161003816100381610038161003816) + 119870120582119902 (119905) minus 120582

119890 (119905)

119909 (119905)

=120582119903 (119905)

|119909 (119905)|120574119892(

|119909 (119905)|

120582119903 (119905)|119908 (119905)|) + 119870120582119902 (119905) minus 120582

119890 (119905)

119909 (119905)

ge120582119903 (119905)

|119909 (119905)|120574

|119909 (119905)|120574

(120582119903 (119905))1205741198920 (|119908 (119905)|)

+ 119870120582119902 (119905) minus 120582119890 (119905)

119909 (119905)ae in 119869

(50)

that is

1199081015840ge (120582119903 (119905))

1minus1205741198920(|119908|) + 119870120582119902 (119905) minus 120582

119890 (119905)

119909 (119905)ae in 119869

(51)

Hence (49) and (51) prove the desired inequality (48)

Proposition 18 Let (19) be satisfied Let 1198771and 119877

2be two

arbitrary real numbers Then there are two points 119879lowast1isin (1198861 1198871)

and 119879lowast

2isin (1198862 1198872) and two continuous functions 119908

1(119905) and

1199082(119905) such that

119908119894(119886119894) le 119877119894 lim119905rarr119879

lowast

119894

119908119894(119905) = infin 119894 isin 1 2

1199081015840

119894le (120582119903 (119905))

1minus1205741198920(1003816100381610038161003816119908119894

1003816100381610038161003816) + 119870120582119902 (119905)

for ae 119905 isin (119886119894 119879lowast

119894) 119894 isin 1 2

(52)

Proof Since the function 119910(119904) = tan(119904) is a bijection fromthe interval (minus1205872 1205872) intoR we observe that there are two1199041 1199042isin (minus1205872 1205872) such that

tan (119904119894) le 119877119894 119894 isin 1 2 (53)

Next we define two functions 1198811(119905) and 119881

2(119905) by

119881119894 (119905) = 119904119894 +

120587

119888119894

int

119905

119886119894

119862 (120591) 119889120591 119905 isin [119886119894 119887119894] 119894 isin 1 2 (54)

where the real numbers 1198881 1198882and the function 119862(119905) are

defined in (19) From (19) and (54) one can immediatelyconclude that

119881119894(119886119894) = 119904119894lt120587

2 119881

119894(119887119894) = 119904119894+ 120587 gt

120587

2 119894 isin 1 2 (55)

Since 119862 isin 1198711

loc((1199050infin)R) it follows that 119862 isin 1198711((1199050infin)R)

which implies that 119881119894isin 119860119862([119886

119894 119887119894]R) 119894 isin 1 2 Therefore

exploiting the fact that in (55) we have 119881119894isin 119862([119886

119894 119887119894]R) we

obtain the existence of two points 119879lowast1isin (1198861 1198871) and 119879

lowast

2isin

(1198862 1198872) such that

119881119894(119879lowast

119894)=

120587

2 minus

120587

2lt119881119894(119905)lt

120587

2

forall119905 isin [119886119894 119879lowast

119894) 119894 isin 1 2

(56)

Now we are able to define the following two functions 1205961(119905)

and 1205962(119905) by

120596119894(119905) = tan (119881i (119905)) 119905 isin [119886

119894 119879lowast

119894) 119894 isin 1 2 (57)

That are well defined because of (56) Moreover from (53)(54) and (56) we obtain

120596119894(119886119894) = tan (119904

119894) le 119877119894

lim119905rarr119879

lowast

119898

120596119894(119905) = tan (119881

119894(119879lowast

119894)) = tan(120587

2) = infin

(58)

Also since119881119894isin 119860119862([119886

119894 119887119894]R) we can take the first derivative

of 120596119894(119905) and hence from (19) and (57) we obtain

1199081015840

119894(119905) =

120587

119888119894

119862 (119905)1

cos2 (119881119894 (119905))

=120587

119888119894

119862 (119905) (1003816100381610038161003816120596119894(119905)1003816100381610038161003816

2+ 1)

le120587

119888119894

119862 (119905) [1198920(1003816100381610038161003816120596119894(119905)1003816100381610038161003816) + 119872

0+ 1]

le120587

119888119894

119862 (119905) 1198920(1003816100381610038161003816120596119894(119905)1003816100381610038161003816) +

120587

119888119894

119862 (119905) (1198720+ 1)

le (120582119903 (119905))1minus1205741198920(1003816100381610038161003816120596119894(119905)1003816100381610038161003816) + 119870120582119902 (119905) ae in (119886

119894 119879lowast

119894)

(59)

which together with (58) proves this proposition

Now we are able to present the proof of Theorem 3 basedon Lemma 15 and Propositions 17 and 18

Proof of Theorem 3 Assuming the contrary then there is anonoscillatory solution 119909(119905) and 119879 ge 119905

0such that 119909(119905) = 0 for

all 119905 ge 119879 119909(119905) and 119903(119905)Φ(119909 1199091015840) are from 119860119862loc([1199050infin)R)In order to simplify notation let every 119895 isin N be fixed andomitted in the notation For instance instead of (119886

1119895 1198871119895) and

(1198862119895 1198872119895) we write (119886

1 1198871) and (119886

2 1198872) respectively and so on

On one hand we observe by Proposition 17 that thefunction 119908(119905) defined by (47) is a supersolution of thefollowing Riccati differential equation

1199081015840= (120582119903 (119905))

1minus1205741198920(|119908|) + 119870120582119902 (119905) ae in 119869 (60)

where 119908 isin 119860119862loc(119869R) and 119869 = (1198861 1198871) if 119909(119905) lt 0 and 119869 =

(1198862 1198872) if 119909(119905) gt 0 see the proof of Proposition 17 Let for

instance 119909(119905) lt 0 and thus 119869 = (1198861 1198871)

On the other hand by Proposition 18 there are a number119879lowast

1isin (1198861 1198871) and subsolutions 119908

1(119905) 1199081isin 119860119862([119886

1 119879lowast

1)R) of

the Riccati differential equation (60) such that

1199081(1198861) le 119908 (119886

1) lim

119905rarr119879lowast

1

1199081(119905) = infin (61)

(in the case when 119869 = (1198862 1198872) then we work with119879lowast

2isin (1198862 1198872)

and the other subsolution 1199082(119905) 119908

2isin 119860119862([119886

2 119879lowast

2)R))

Hence by Corollary 16 and (61) we conclude that1199081(119905) le 119908(119905)

for all 119905 isin [1198861 119879lowast

1) and therefore

infin = lim119905rarr119879

lowast

1

1199081(119905) le lim119905rarr119879

lowast

1

119908 (119905) (62)

which contradicts 119908 isin 119860119862loc([119879infin)R) Thus the assump-tion that 119909(119905) is nonoscillatory is not possible and hence everysolution of (1) is oscillatory

8 International Journal of Differential Equations

Proof of Theorem 4 There is only one difference betweenTheorems 3 and 4 and it is the difference between conditions(19) and (20) Hence Theorem 4 immediately follows fromTheorem 3 provided that we prove the equivalence between(19) and (20)

First of all we need the following two lemmas

Lemma 19 Let 119886 119887 and 119888 be positive real numbers and 120574 gt 1Then the existence of a positive real number 120582 such that

119888 le min 119886

120582120574minus1

119887120582 (63)

is equivalent to

119888 le 119886112057411988711205741015840

(64)

Here 1205741015840 = 120574(120574 minus 1) is the conjugate exponent of 120574

Proof Let us define 119892(120582) = min119886120582120574minus1 119887120582 Since 120582 997891rarr

119886120582120574minus1 is decreasing and 120582 997891rarr 119887120582 is increasing there exists

a unique point of maximum of 119892 It is achieved at 120582 = 1205820

which is a solution of 119886120582120574minus10

= 1198871205820 hence 120582

0= (119886119887)

1120574If there is a positive real number 120582 such that 119888 le 119892(120582)

then 119888 le 119892(1205820) = 119887120582

0= 119887(119886119887)

1120574= 119886112057411988711205741015840

Converselyif 119888 le 119886

112057411988711205741015840

then 119888 le 119892(1205820) and the equivalence in the

lemma is proved

The preceding lemma is a special case of the followingmore general statement

Lemma 20 Assume that 119886(119905) 119887(119905) and 119888(119905) are positive realfunctions defined on a subset 119869 sube R Then the existence of apositive real number 120582 such that for all 119905 isin 119869

119888 (119905) le min 119886 (119905)120582120574minus1

119887 (119905) 120582 (65)

is equivalent to

sup119905isin119869

119888 (119905)

119887 (119905)le (inf119905isin119869

119886 (119905)

119888 (119905))

1(120574minus1)

(66)

Proof The condition that there exists 120582 gt 0 such that (65)is true for all 119905 isin 119869 is equivalent with the following twoinequalities which have to hold simultaneously

119888 (119905) le119886 (119905)

120582120574minus1

119888 (119905) le 119887 (119905) 120582 (67)

that is with

119888 (119905)

119887 (119905)le 120582 le (

119886 (119905)

119888 (119905))

1(120574minus1)

forall119905 isin 119869 (68)

Taking the supremum of the left-hand side and the infimumof the right-hand side we obtain (66) Conversely if (66)is satisfied then since the left-hand side in (66) cannotbe equal to zero while the right-hand side is less than infinthen inequality (66) defines a nonempty closed interval in R

(possibly reducing to just one point) in which we can chooseany positive 120582 Hence (67) is satisfied for all 119905 isin 119869 andtherefore (65) as well

Remark 21 If 119886 = 119886(119905) 119887 = 119887(119905) and 119888 = 119888(119905) appearingin Lemma 20 are constant functions then condition (66) isequivalent to 119888119887 = (119886119888)1(120574minus1) that is to (64)

Proof of Equivalence of (19) and (20) Wewill use Lemma 20Let us fix 119869

119894119895 where 119894 isin 1 2 and 119895 isin N We see that that the

inequality appearing in the second line of condition (19) is ofthe form (65) where

119888 (119905) =1

119888119894119895

119862 (119905) 119886 (119905) =1

120587119903(119905)1minus120574

119887 (119905) =119870119902 (119905)

120587 (1198720+ 1)

(69)

Condition (66) is equivalent to

sup119905isin119869119894119895

119862 (119905)

119902 (119905)le

1198701198881205741015840

119894119895

1205871205741015840

(1198720+ 1)

inf119905isin119869119894119895

1

119903 (119905) 119862(119905)120574minus1

=

1198701198881205741015840

119894119895

(1198720+ 1) sup

119905isin119869119894119895119903 (119905) 119862(119905)

120574minus1(

119888119894119895

120587)

1205741015840

(70)

This inequality is equivalent to the corresponding one in (19)and the claim follows from Lemma 19

Proof of Corollary 6 Substituting inf119905isin119869119894119895

119902(119905) instead of 119902(119905)and sup

119905isin119869119894119895119903(119905) instead of 119903(119905) in the inequality appearing in

the second line of (20) after a short computation we obtain astronger inequality than (20) as

1

119888119894119895

sup119905isin119869119894119895

119862 (119905)

le1

120587(

119870inf119905isin119869119894119895

119902 (119905)

(1198720+ 1) sup

119905isin119869119894119895119903 (119905)

)

(120574minus1)120574

forall119894 isin 1 2 119895 isin N

(71)

Substituting119862(119905) equiv 1we obtain that the left-hand side of (71)is equal to |119869

119894119895|minus1 and the resulting inequality is equivalent to

(21) Since (21) implies (20) the claim is proved

Remark 22 Here we show that the choice of 119862(119905) equiv 1 inthe proof of Corollary 6 is the best possible To prove thisnote that on the left-hand side of (71) we have the expressiondepending on an auxiliary function 119862(119905) and this functionis absent on the right-hand side Therefore it has sense totry to find a function 119862(119905) 119905 isin 119869

119894119895 such that the value of

119876(119862) = (1119888119894119895)sup119905isin119869119894119895

119862(119905) is minimal (note that 119888119894119895depends

on the function 119862 = 119862(119905) as well) It is easy to see that theminimum is achieved for 119862(119905) equiv 1 (or any positive constant)Indeed since 119876(119862) = 119876(120583119888) for any 120583 gt 0 by taking 120583 =

(sup119905isin119869119894119895

119862(119905))minus1 it suffices to assume that sup

119905isin119869119894119895119862(119905) = 1

The value of 119876(119862) is minimal if 119888119894119895= int119869119894119895

119862(120591) 119889120591 is maximalpossible and since 0 le 119862(119905) le 1 it is clear that the maximumof 119888119894119895= 119888119894119895(119862) is achieved when 119862(119905) equiv 1

International Journal of Differential Equations 9

Proof of Corollary 7 It is clear that the condition (23) implies(21) Furthermore since the lengths |119869

119894119895| are uniformly

bounded from below by a positive constant see (23) thenobviously 119886

1119895rarr infin as 119895 rarr infin Thus the claim follows

immediately from Corollary 6

Proof of Corollary 10 Let 119862(119905) = 119902(119905) Then the requiredcondition (19) is clearly satisfied because of assumption(30) Hence this corollary immediately follows from Theo-rem 3

4 An Extension of Condition (12) tothe Case of 120574 gt 1

In this section we consider the oscillation of (1) in the casewhen 120574 isin (1 2) is allowed in the assumption (12) In this waythe assumption (11) is slightly modified by a real number 119901 gt1 such that

|119906|(119901minus1)120574minus119901

VΦ (119906 V) ge 119892 (|Φ (119906 V)|) (72)

where (12) is supposed that is 119892(119888119904) ge 1198881205741198920(119904) for every 119888 119904 gt

0 and for a locally Lipschitz function 1198920 [0infin) rarr [0infin)

for which there exists a constant1198720such that 119892

0(119904)+119872

0ge 119904119901

for all 119904 isin [0infin)For the function 119891 we assume that there exists a constant

119870 gt 0 such that

119891 (119906) sign (119906) ge 119870|119906|119901minus1 for every 119906 isin R (73)

We will need the following two lemmas

Lemma 23 Let 119886 lt 119887 and let 119891 [119886 119887] rarr [0infin) be a meas-urable function Then

int

119887

119886

119891 (119905) 119889119905 ge 1 (74)

if and only if there exists a function 119892 isin 1198711([119886 119887] [0infin))

1198921= 1 such that

119891 (119905) ge 119892 (119905) forall119905 isin [119886 119887] (75)

Proof We assume (74) If int119887119886119891(119905)119889119905 is finite we may choose

119892(119909) = 119891(119909)1198911

For int119887119886119891(119905)119889119905 = infin we may define for every natural

number 119899 a function

119891119899(119905) =

119891 (119905) 119891 (119905) le 119899

119899 119891 (119905) gt 119899(76)

Since 119891119899(119905) le 119899 we have 119891

119899isin 1198711([119886 119887]R) Obviously

lim119899int119887

119886119891119899(119905)119889119905 is also infiniteThis implies that there exists an

1198990such that int119887

1198861198911198990(119905)119889119905 ge 1 We define 119892(119905) = 119891

1198990(119905)119891

11989901

Now we assume (75) Integrating over the whole intervalwe get

int

119887

119886

119891 (119905) 119889119905 ge int

119887

119886

119892 (119905) 119889119905 = 1 (77)

Lemma 24 Let119901 gt 1There exists a unique function 119910 = 119910(119904)that satisfies the following Cauchy problem

1199101015840(119904) = 1 +

1003816100381610038161003816119910 (119904)

1003816100381610038161003816

119901 119904 isin R (78)

119910 (0) = 0 (79)

Furthermore there exists a finite real number 119878lowast such that

lim119904rarrplusmn119878

lowast119910 (119904) = plusmninfin (80)

and 119910 isin 1198621((minus119878lowast 119878lowast)R) is injective and monotonous We

have 119878lowast = (120587119901)(1(sin(120587119901))) and 119910 is odd

We will denote by tan119901the solution of (78)-(79)

Proof Obviously every solution to (78) is injective andmonotonous on a connected set since 1199101015840 = 1 + |119910|119901 ge 1 gt 0

Let 119891(119909) = 1(1 + |119909|119901) Obviously 1 ge 119891(119909) gt 0 for all

119909 isin R and 119862119901gt |1198911015840(119909)| gt 0 for some constants 119862

119901isin R

We may assume 119862119901gt 1 Hence 1199111015840 = 119891(119905) 119911(0) = 0 has (by

[17Theorem 31]) a locally unique solution on (minus1119862119901 1119862119901)

By the same argument 119911 can be extended to the whole of Rinterval by interval of type ((119899 minus 1)119862

119901 (119899 + 1)119862

119901) for all

119899 isin ZWe know from [18] that the for the integral of the left-

hand side we have

int

infin

0

119891 (119909) 119889119909 = int

infin

0

119889119909

1 + 119909119901=120587

119901

1

sin (120587119901)= 119878lowast (81)

so lim119905rarrplusmninfin

119911(119905) = plusmn119878lowast

Since 1199111015840(119905) = 0 119911 is injective monotonous and of class1198621(R (minus119878lowast 119878lowast)) and so we define 119910(119904) = 119911

minus1(119904) It imme-

diately follows that 119910 isin 1198621((minus119878lowast 119878lowast)R) and that 119910 is

injective and monotonous Also by continuity of 119911 we havelim119904rarrplusmn119878

lowast119910(119904) = lim119908rarrplusmninfin

119910(119911(119908)) = lim119908rarrplusmninfin

119908 = plusmninfinDifferentiating it also follows that 1199101015840 = 1 + |119910|

119901 and bydefinition of 119910 119910(0) = 119911minus1(0) = 0 So 119910 is a solution of (78)-(79) Since any other such solution must have 119911 as its inverse119910 is unique on its domain

Proposition 25 Let 119886 lt 119887 and 120574 gt 119901 gt 1 Assume (72) and(73) and let 119902(119905) ge 0 and 119890(119905) le 0 (or 119890(119905) ge 0 ) for all 119905 isin [119886 119887]If there exists a constant 120582 gt 0 such that

119901

2120587sin(120587

119901)int

119887

119886

min119901 minus 1

119903120574minus1

(119905) 120582120574minus1

119870120582119902 (119905)

1 +1198720

119889119905 ge 1 (82)

then for any solution 119909(119905) of (1) with 119909(119886) ge 0 (or 119909(119886) le 0)there exists a number 119905lowast isin [119886 119887] such that 119909(119905lowast) = 0

Proof We assume that 119909(119905) = 0 for all 119905 isin [119886 119887] Then we maydefine the function

120596 (119905) = minus

120582119903 (119905) Φ (119909 (119905) 1199091015840(119905))

sign (119909 (119905)) |119909 (119905)|119901minus1 (83)

10 International Journal of Differential Equations

This function is well defined on [119886 119887] and its derivative is

1199081015840(119905) = minus

120582(119903 (119905) Φ (119909 (119905) 1199091015840(119905)))1015840

sign (119909 (119905)) |119909 (119905)|119901minus1

+

(119901 minus 1) 120582119903 (119905) Φ (119909 (119905) 1199091015840(119905))

|119909 (119905)|119901

1199091015840(119905)

(84)

Using the fact that 119909(119905) is a solution of (1) and omitting 119905 from119909 and 1199091015840 for clarity we get

1199081015840(119905) =

120582 (119902 (119905) 119891 (119909) minus 119890 (119905))

sign (119909) |119909|119901minus1+

(119901 minus 1) 120582119903 (119905) Φ (119909 1199091015840)

119909119901

1199091015840

=

(119901 minus 1) 120582119903 (119905) Φ (119909 1199091015840)

|119909|119901

1199091015840+ 120582119902 (119905)

119891 (119909)

sign (119909) |119909|119901minus1

minus 120582119890 (119905)

sign (119909) |119909|119901minus1

=(119901 minus 1) 120582119903 (119905)

|119909|120574(119901minus1)

Φ(119909 1199091015840) 1199091015840|119909|(119901minus1)120574minus119901

+ 120582119902 (119905)119891 (119909)

sign (119909) |119909|119901minus1minus 120582

119890 (119905)

sign (119909) |119909|119901minus1

(72)

ge(119901 minus 1) 120582119903 (119905)

|119909|120574(119901minus1)

119892 (10038161003816100381610038161003816Φ (119909 119909

1015840)10038161003816100381610038161003816)

+ 120582119902 (119905)119891 (119909)

sign (119909) |119909|119901minus1

minus 120582119890 (119905)

sign (119909) |119909|119901minus1

=(119901 minus 1) 120582119903 (119905)

|119909|120574(119901minus1)

119892(|119909|119901minus1

120582119903 (119905)|120596 (119905)|)

+ 120582119902 (119905)119891 (119909)

sign (119909) |119909|119901minus1minus 120582

119890 (119905)

sign (119909) |119909|119901minus1

(12)

ge(119901 minus 1) 120582119903 (119905)

|119909|120574(119901minus1)

(|119909|119901minus1

120582119903 (119905))

120574

1198920 (|120596 (119905)|)

+ 120582119902 (119905)119891 (119909)

sign (119909) |119909|119901minus1minus 120582

119890 (119905)

sign (119909) |119909|119901minus1

=119901 minus 1

(120582119903 (119905))120574minus1

1198920 (|120596 (119905)|)+120582119902 (119905)

119891 (119909)

sign (119909) |119909|119901minus1

minus 120582119890 (119905)

sign (119909) |119909|119901minus1

(73)

ge119901 minus 1

(120582119903 (119905))120574minus1

1198920 (|120596 (119905)|) + 120582119902 (119905) 119870

minus 120582119890 (119905)

sign (119909) |119909|119901minus1

(85)

Since 119890(119905)(sign(119909)|119909|119901minus1) le 0 we have

1199081015840(119905) ge

119901 minus 1

(120582119903 (119905))120574minus1

1198920(|120596 (119905)|) + 120582119870119902 (119905) (86)

We apply Lemma 23 to (82) to obtain a function119862(119905) suchthat

int

119887

119886

119862 (119905) 119889119905 = 1 (87)

and by (75)

0 le 119862 (119905) le119901

2120587sin(120587

119901)min

119901 minus 1

119903120574minus1

(119905) 120582120574minus1

119870120582119902 (119905)

1 +1198720

(88)

Let 1199040isin (minus(120587119901)(1 sin(120587119901)) (120587119901)(1 sin(120587119901))) be such

that tan119901(1199040) le 120596(119886) We define the function

119881 (119905) = 1199040+2120587

119901

1

sin (120587119901)int

119905

119886

119862 (120591) 119889120591 (89)

Note that (minus(120587119901)(1 sin(120587119901))) lt 1199040

= 119881(119886) lt

(120587119901)(1 sin(120587119901)) and 119881(119887) = 1199040+ (2120587119901)(1 sin(120587119901)) gt

(120587119901)(1 sin(120587119901)) Since119881 is continuous there exists a 119879lowast isin(119886 119887) such that 119881(119905) lt (120587119901)(1 sin(120587119901)) for 119905 isin [119886 119879lowast) and119881(119879lowast) = (120587119901)(1 sin(120587119901))

We define

120596 (119905) = tan119901(119881 (119905)) 119905 isin [119886 119879

lowast) (90)

and note that by Lemma 24 lim119905rarr119879

lowast120596(119905) = +infin wheretan119901(119904) = 119910(119904) and 119910(119904) is determined by Lemma 24The derivative of 120596 then satisfies

1205961015840(119905) = tan1015840

119901(119881 (119905)) 119881

1015840(119905)

= (1 + tan119901119901(119881 (119905))) 119881

1015840(119905)

=2120587

119901

1

sin (120587119901)(1 + 120596

119901(119905)) 119862 (119905)

le (1 +1198720+ 1198920(1003816100381610038161003816120596 (119905)

1003816100381610038161003816))min

119901 minus 1

119903120574minus1

(119905) 120582120574minus1

119870120582119902 (119905)

1 +1198720

le119901 minus 1

119903120574minus1

(119905) 120582120574minus1

1198920(1003816100381610038161003816120596 (119905)

1003816100381610038161003816) + 119870120582119902 (119905)

(91)

Thus we may apply the comparison principle ofLemma 15 to 120596(119886) le 120596(119886) and their respective differentialinequalities to conclude that 120596(119905) le 120596(119905) for all 119905 isin [119886 119887]But since 120596 rarr +infin as 119905 rarr 119879

lowastlt 119887 and 120596(119905) is well defined

on the whole segment [119886 119887] we arrive at a contradictionHence 119909(119905) gt 0 cannot hold for all 119905 isin [119886 119887] and since 119909(119905) iscontinuous there exists a 119905lowast isin [119886 119887] such that 119909(119905lowast) = 0

Theorem 26 Let 120574 gt 119901 gt 1 Assume (72) and (73) and let forany 119879 gt 119905

0there exist 119879 le 119886

1lt 1198871le 1198862lt 1198872such that 119892(119905) ge 0

on [1198861 1198871] cup [119886

2 1198872] and 119890(119905) le 0 on [119886

1 1198871] and 119890(119905) ge 0 on

International Journal of Differential Equations 11

[1198862 1198872] If for both 119894 isin 1 2 there exist constants 120582

119894gt 0 such

that

119901

2120587sin (120587119901)int

119887119894

119886119894

min119901 minus 1

119903120574minus1

(119905) 120582120574minus1

119894

119870120582119894119902 (119905)

1 +1198720

119889119905 ge 1

(92)

then (1) is oscillatory

Proof For (1) to be oscillatory it is sufficient that for every119899 isin N there exists a 119905

119899gt 119899 such that 119909(119905

119899) = 0

Let 119899 isin N and let 119899 lt 1198861lt 1198871le 1198862lt 1198872as assumed in the

theorem It is enough to show that every solution 119909(119905) of (1)has a zero on [119886

1 1198872] If 119909(119886

1) ge 0 we apply Proposition 25 to

the segment [1198861 1198871] to obtain the sought zero If 119909(119886

2) le 0 we

again apply Proposition 25 to the segment [1198862 1198872] Suppose

119909(1198861) lt 0 and 119909(119886

2) gt 0 Then since 119909 is continuous there

exists a number 1199051isin [1198861 1198862] such that 119909(119905

1) = 0

Acknowledgment

This work is supported by the Ministry of Science of theRepublic of Croatia under Grant no 036-0361621-1291

References

[1] Q Kong ldquoNonoscillation and oscillation of second order half-linear differential equationsrdquo Journal of Mathematical Analysisand Applications vol 332 no 1 pp 512ndash522 2007

[2] Q-R Wang and Q-G Yang ldquoInterval criteria for oscillationof second-order half-linear differential equationsrdquo Journal ofMathematical Analysis and Applications vol 291 no 1 pp 224ndash236 2004

[3] Q Long and Q-R Wang ldquoNew oscillation criteria of second-order nonlinear differential equationsrdquo Applied Mathematicsand Computation vol 212 no 2 pp 357ndash365 2009

[4] Z Zheng and F Meng ldquoOscillation criteria for forced second-order quasi-linear differential equationsrdquo Mathematical andComputer Modelling vol 45 no 1-2 pp 215ndash220 2007

[5] D Cakmak and A Tiryaki ldquoOscillation criteria for certainforced second-order nonlinear differential equationsrdquo AppliedMathematics Letters vol 17 no 3 pp 275ndash279 2004

[6] F Jiang and F Meng ldquoNew oscillation criteria for a class ofsecond-order nonlinear forced differential equationsrdquo Journalof Mathematical Analysis and Applications vol 336 no 2 pp1476ndash1485 2007

[7] Q Yang ldquoInterval oscillation criteria for a forced secondorder nonlinear ordinary differential equations with oscillatorypotentialrdquo Applied Mathematics and Computation vol 135 no1 pp 49ndash64 2003

[8] A Tiryaki and B Ayanlar ldquoOscillation theorems for certainnonlinear differential equations of second orderrdquo Computers ampMathematics with Applications vol 47 no 1 pp 149ndash159 2004

[9] J Jaros T Kusano and N Yoshida ldquoGeneralized Piconersquosformula and forced oscillations in quasilinear differential equa-tions of the second orderrdquoArchivumMathematicum vol 38 no1 pp 53ndash59 2002

[10] R P Agarwal S R Grace and D OrsquoRegan Oscillation Theoryfor Second Order Linear Half-Linear Superlinear and SublinearDynamic Equations Kluwer Academic Publishers DordrechtThe Netherlands 2002

[11] S Jing ldquoA new oscillation criterion for forced second-orderquasilinear differential equationsrdquoDiscrete Dynamics in Natureand Society vol 2011 Article ID 428976 8 pages 2011

[12] W-T Li and S S Cheng ldquoAn oscillation criterion for nonhomo-geneous half-linear differential equationsrdquoAppliedMathematicsLetters vol 15 no 3 pp 259ndash263 2002

[13] J S W Wong ldquoOscillation criteria for a forced second-orderlinear differential equationrdquo Journal of Mathematical Analysisand Applications vol 231 no 1 pp 235ndash240 1999

[14] M Pasic ldquoFite-Wintner-Leighton type oscillation criteria forsecond-order differential equations with nonlinear dampingrdquoAbstract and Applied Analysis vol 2013 Article ID 852180 10pages 2013

[15] M Pasic ldquoNew interval oscillation criteria for forced second-order differential equations with nonlinear dampingrdquo Interna-tional Journal of Mathematical Analysis vol 7 no 25 pp 1239ndash1255 2013

[16] L Gasinski and N S Papageorgiou Nonsmooth Critical PointTheory and Nonlinear Boundary Value Problems vol 8 of Seriesin Mathematical Analysis and Applications Edited by R PAgarwal and D OrsquoRegan Chapman amp HallCRC Boca RatonFla USA 2005

[17] S Lang Real and Functional Analysis vol 142 of GraduateTexts in Mathematics Springer-Verlag New York NY USA 3rdedition 1993

[18] I N Bronshtein K A Semendyayev G Musiol and HMuehligHandbook of Mathematics Springer Berlin Germany5th edition 2007

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

6 International Journal of Differential Equations

Definition 14 We say that comparison principle holds for (39)on an interval [119879

0 119879lowast) 119879 le 119879

0lt 119879lowast if property (41)

is fulfilled for all sub- and supersolutions 119908119908 isin

119860119862loc([1198790 119879lowast)R) of (39) on [119879

0 119879lowast)

Now we are able to state the next result

Lemma 15 Let 119879 le 1198790lt 119879lowast Let 119866 = 119866(119905 119906) and 119887 = 119887(119905) be

two arbitrary function For any119872 gt 0 let there be a function119871 = 119871(119905) gt 0 depending on 119879

0 119879lowast119872 such that

119871 (119905) gt 0 on [1198790 119879lowast) 119871 isin 119871

1(1198790 119879lowast)

1003816100381610038161003816119866 (119905 119906

1) minus 119866 (119905 119906

2)1003816100381610038161003816le 119871 (119905)

10038161003816100381610038161199061minus 1199062

1003816100381610038161003816

forall119905 isin [1198790 119879lowast) 1199061 1199062isin [minus119872119872]

(42)

Then comparison principle (41) holds for (39) on an interval[1198790 119879lowast) 119879 le 119879

0lt 119879lowast

Proof It is enough to use the idea of the proof of [14 Lemma41] Hence this proof is left to the reader

Corollary 16 Let 120582 gt 0 and let 1198920 R+rarr R+be a locally

Lipschitz function on R+ Let 119870 be an arbitrary real number

and 119902(119905) an arbitrary function Let 119879 le 1198790lt 119879lowast If 119903(119905) gt 0

on [1198790 119879lowast) and 119903minus1+120574 isin 1198711(119879

0 119879lowast) then comparison principle

(41) holds for the Riccati differential equation

1199081015840=

1

(120582119903 (119905))120574minus1

1198920 (|119908|) + 119870120582119902 (119905) ae in [119879

0 119879lowast)

(43)

Proof It is clear that (43) is a particular case of (39) inparticular for

119866 (119905 119906) =1

(120582119903 (119905))120574minus1

1198920 (|119906|) 119887 (119905) = 119870120582119902 (119905) (44)

Since1198920is a locally Lipschitz function onR

+ for every119872 gt 0

there is an 1198710gt 0 depending on119872 such that

10038161003816100381610038161198920(1199041) minus 1198920(1199042)1003816100381610038161003816le 1198710

10038161003816100381610038161199041minus 1199042

1003816100381610038161003816 forall119904

1 1199041isin [minus119872119872]

(45)

Hence for any119872 gt 0 and all 1199061 1199062isin [minus119872119872] we obtain

1003816100381610038161003816119866 (119905 119906

1) minus 119866 (119905 119906

2)1003816100381610038161003816=

1

(120582119903 (119905))120574minus1

10038161003816100381610038161198920(10038161003816100381610038161199061

1003816100381610038161003816) minus 1198920(10038161003816100381610038161199062

1003816100381610038161003816)1003816100381610038161003816

le1198710

(120582119903 (119905))120574minus1

1003816100381610038161003816

10038161003816100381610038161199061

1003816100381610038161003816minus10038161003816100381610038161199062

1003816100381610038161003816

1003816100381610038161003816

le1198710

(120582119903 (119905))120574minus1

10038161003816100381610038161199061minus 1199062

1003816100381610038161003816

(46)

Thus according to assumption 119903minus1+120574

isin 1198711(1198790 119879lowast) the

required condition (42) is fulfilled in particular for 119871(119905) =

1198710(120582119903(119905))

minus120574+1 and so Lemma 15 proves this corollary

Before we give the proof ofTheorem 3 we state and provethe next two propositions In the first one by a nonoscillatorysolution 119909(119905) of the main equation (1) we get the existenceof a supersolution 119908(119905) of the Riccati differential equation(43) on the interval (119886

1 1198871) or (119886

2 1198872) In the second one we

construct two subsolutions 1199081(119905) and 119908

2(119905) of (43) which

blow up on intervals (1198861 119879lowast

1) sube (119886

1 1198871) and (119886

2 119879lowast

2) sube (119886

2 1198872)

respectively

Proposition 17 Let Φ(119906 V) and 119891(119906) satisfy (11) (12) and(16) respectively and let 119890(119905) satisfy (2) Let 119903(119905) gt 0 119902(119905) ge0 and 119902(119905) equiv 0 on [119886

1 1198871] cup [119886

2 1198872] Let 119909 = 119909(119905) be a

nonoscillatory solution of (1) and let for some 119879 ge 1199050and

120582 gt 0 the function 119908 = 119908(119905) be defined by

119908 (119905) = minus

120582119903 (119905)Φ (119909 (119905) 1199091015840(119905))

119909 (119905) 119905 ge 119879 (47)

Then 119908 isin 119860119862loc([119879infin)R) and 119908(119905) satisfies the inequality

1199081015840ge (120582119903 (119905))

1minus1205741198920(|119908|) + 119870120582119902 (119905) ae in 119869 (48)

where 119869 = (1198861 1198871) if 119909(119905) lt 0 and 119869 = (119886

2 1198872) if 119909(119905) gt 0

Proof Since 119909 = 119909(119905) is a nonoscillatory solution of (1) thereis a 119879 ge 119905

0such that 119909(119905) = 0 on [119879infin) Hence 119908(119905) is well

defined by (47)Because of (2) we have 119890(119905)119909(119905) le 0 for all 119905 isin 119869 where

119869 = (1198861 1198871) if 119909(119905) lt 0 and 119869 = (119886

2 1198872) if 119909(119905) gt 0 and since

120582 gt 0 we have

minus120582119890 (119905)

119909 (119905)ge 0 in 119869 (49)

Next since 119909 and 119903(119905)Φ(119909(119905) 1199091015840(119905)) are from

119860119862loc([1199050infin)R) we can take the first derivative of 119908(119905)for almost everywhere in 119869 which together with 119903(119905) gt 0119902(119905) ge 0 and 119902(119905) equiv 0 on 119869 gives

1199081015840(119905) = minus

120582(119903 (119905) Φ (119909 (119905) 1199091015840(119905)))1015840

119909 (119905)

+

120582119903 (119905)Φ (119909 (119905) 1199091015840(119905))

1199092(119905)

1199091015840(119905)

=120582119903 (119905)

|119909 (119905)|120574Φ(119909 (119905) 119909

1015840(119905)) 1199091015840(119905) |119909 (119905)|

120574minus2

+ 120582119902 (119905)119891 (119909 (119905))

119909 (119905)minus 120582

119890 (119905)

119909 (119905)

International Journal of Differential Equations 7

ge120582119903 (119905)

|119909 (119905)|120574119892 (10038161003816100381610038161003816Φ (119909 (119905) 119909

1015840(119905))

10038161003816100381610038161003816) + 119870120582119902 (119905) minus 120582

119890 (119905)

119909 (119905)

=120582119903 (119905)

|119909 (119905)|120574119892(

|119909 (119905)|

120582119903 (119905)|119908 (119905)|) + 119870120582119902 (119905) minus 120582

119890 (119905)

119909 (119905)

ge120582119903 (119905)

|119909 (119905)|120574

|119909 (119905)|120574

(120582119903 (119905))1205741198920 (|119908 (119905)|)

+ 119870120582119902 (119905) minus 120582119890 (119905)

119909 (119905)ae in 119869

(50)

that is

1199081015840ge (120582119903 (119905))

1minus1205741198920(|119908|) + 119870120582119902 (119905) minus 120582

119890 (119905)

119909 (119905)ae in 119869

(51)

Hence (49) and (51) prove the desired inequality (48)

Proposition 18 Let (19) be satisfied Let 1198771and 119877

2be two

arbitrary real numbers Then there are two points 119879lowast1isin (1198861 1198871)

and 119879lowast

2isin (1198862 1198872) and two continuous functions 119908

1(119905) and

1199082(119905) such that

119908119894(119886119894) le 119877119894 lim119905rarr119879

lowast

119894

119908119894(119905) = infin 119894 isin 1 2

1199081015840

119894le (120582119903 (119905))

1minus1205741198920(1003816100381610038161003816119908119894

1003816100381610038161003816) + 119870120582119902 (119905)

for ae 119905 isin (119886119894 119879lowast

119894) 119894 isin 1 2

(52)

Proof Since the function 119910(119904) = tan(119904) is a bijection fromthe interval (minus1205872 1205872) intoR we observe that there are two1199041 1199042isin (minus1205872 1205872) such that

tan (119904119894) le 119877119894 119894 isin 1 2 (53)

Next we define two functions 1198811(119905) and 119881

2(119905) by

119881119894 (119905) = 119904119894 +

120587

119888119894

int

119905

119886119894

119862 (120591) 119889120591 119905 isin [119886119894 119887119894] 119894 isin 1 2 (54)

where the real numbers 1198881 1198882and the function 119862(119905) are

defined in (19) From (19) and (54) one can immediatelyconclude that

119881119894(119886119894) = 119904119894lt120587

2 119881

119894(119887119894) = 119904119894+ 120587 gt

120587

2 119894 isin 1 2 (55)

Since 119862 isin 1198711

loc((1199050infin)R) it follows that 119862 isin 1198711((1199050infin)R)

which implies that 119881119894isin 119860119862([119886

119894 119887119894]R) 119894 isin 1 2 Therefore

exploiting the fact that in (55) we have 119881119894isin 119862([119886

119894 119887119894]R) we

obtain the existence of two points 119879lowast1isin (1198861 1198871) and 119879

lowast

2isin

(1198862 1198872) such that

119881119894(119879lowast

119894)=

120587

2 minus

120587

2lt119881119894(119905)lt

120587

2

forall119905 isin [119886119894 119879lowast

119894) 119894 isin 1 2

(56)

Now we are able to define the following two functions 1205961(119905)

and 1205962(119905) by

120596119894(119905) = tan (119881i (119905)) 119905 isin [119886

119894 119879lowast

119894) 119894 isin 1 2 (57)

That are well defined because of (56) Moreover from (53)(54) and (56) we obtain

120596119894(119886119894) = tan (119904

119894) le 119877119894

lim119905rarr119879

lowast

119898

120596119894(119905) = tan (119881

119894(119879lowast

119894)) = tan(120587

2) = infin

(58)

Also since119881119894isin 119860119862([119886

119894 119887119894]R) we can take the first derivative

of 120596119894(119905) and hence from (19) and (57) we obtain

1199081015840

119894(119905) =

120587

119888119894

119862 (119905)1

cos2 (119881119894 (119905))

=120587

119888119894

119862 (119905) (1003816100381610038161003816120596119894(119905)1003816100381610038161003816

2+ 1)

le120587

119888119894

119862 (119905) [1198920(1003816100381610038161003816120596119894(119905)1003816100381610038161003816) + 119872

0+ 1]

le120587

119888119894

119862 (119905) 1198920(1003816100381610038161003816120596119894(119905)1003816100381610038161003816) +

120587

119888119894

119862 (119905) (1198720+ 1)

le (120582119903 (119905))1minus1205741198920(1003816100381610038161003816120596119894(119905)1003816100381610038161003816) + 119870120582119902 (119905) ae in (119886

119894 119879lowast

119894)

(59)

which together with (58) proves this proposition

Now we are able to present the proof of Theorem 3 basedon Lemma 15 and Propositions 17 and 18

Proof of Theorem 3 Assuming the contrary then there is anonoscillatory solution 119909(119905) and 119879 ge 119905

0such that 119909(119905) = 0 for

all 119905 ge 119879 119909(119905) and 119903(119905)Φ(119909 1199091015840) are from 119860119862loc([1199050infin)R)In order to simplify notation let every 119895 isin N be fixed andomitted in the notation For instance instead of (119886

1119895 1198871119895) and

(1198862119895 1198872119895) we write (119886

1 1198871) and (119886

2 1198872) respectively and so on

On one hand we observe by Proposition 17 that thefunction 119908(119905) defined by (47) is a supersolution of thefollowing Riccati differential equation

1199081015840= (120582119903 (119905))

1minus1205741198920(|119908|) + 119870120582119902 (119905) ae in 119869 (60)

where 119908 isin 119860119862loc(119869R) and 119869 = (1198861 1198871) if 119909(119905) lt 0 and 119869 =

(1198862 1198872) if 119909(119905) gt 0 see the proof of Proposition 17 Let for

instance 119909(119905) lt 0 and thus 119869 = (1198861 1198871)

On the other hand by Proposition 18 there are a number119879lowast

1isin (1198861 1198871) and subsolutions 119908

1(119905) 1199081isin 119860119862([119886

1 119879lowast

1)R) of

the Riccati differential equation (60) such that

1199081(1198861) le 119908 (119886

1) lim

119905rarr119879lowast

1

1199081(119905) = infin (61)

(in the case when 119869 = (1198862 1198872) then we work with119879lowast

2isin (1198862 1198872)

and the other subsolution 1199082(119905) 119908

2isin 119860119862([119886

2 119879lowast

2)R))

Hence by Corollary 16 and (61) we conclude that1199081(119905) le 119908(119905)

for all 119905 isin [1198861 119879lowast

1) and therefore

infin = lim119905rarr119879

lowast

1

1199081(119905) le lim119905rarr119879

lowast

1

119908 (119905) (62)

which contradicts 119908 isin 119860119862loc([119879infin)R) Thus the assump-tion that 119909(119905) is nonoscillatory is not possible and hence everysolution of (1) is oscillatory

8 International Journal of Differential Equations

Proof of Theorem 4 There is only one difference betweenTheorems 3 and 4 and it is the difference between conditions(19) and (20) Hence Theorem 4 immediately follows fromTheorem 3 provided that we prove the equivalence between(19) and (20)

First of all we need the following two lemmas

Lemma 19 Let 119886 119887 and 119888 be positive real numbers and 120574 gt 1Then the existence of a positive real number 120582 such that

119888 le min 119886

120582120574minus1

119887120582 (63)

is equivalent to

119888 le 119886112057411988711205741015840

(64)

Here 1205741015840 = 120574(120574 minus 1) is the conjugate exponent of 120574

Proof Let us define 119892(120582) = min119886120582120574minus1 119887120582 Since 120582 997891rarr

119886120582120574minus1 is decreasing and 120582 997891rarr 119887120582 is increasing there exists

a unique point of maximum of 119892 It is achieved at 120582 = 1205820

which is a solution of 119886120582120574minus10

= 1198871205820 hence 120582

0= (119886119887)

1120574If there is a positive real number 120582 such that 119888 le 119892(120582)

then 119888 le 119892(1205820) = 119887120582

0= 119887(119886119887)

1120574= 119886112057411988711205741015840

Converselyif 119888 le 119886

112057411988711205741015840

then 119888 le 119892(1205820) and the equivalence in the

lemma is proved

The preceding lemma is a special case of the followingmore general statement

Lemma 20 Assume that 119886(119905) 119887(119905) and 119888(119905) are positive realfunctions defined on a subset 119869 sube R Then the existence of apositive real number 120582 such that for all 119905 isin 119869

119888 (119905) le min 119886 (119905)120582120574minus1

119887 (119905) 120582 (65)

is equivalent to

sup119905isin119869

119888 (119905)

119887 (119905)le (inf119905isin119869

119886 (119905)

119888 (119905))

1(120574minus1)

(66)

Proof The condition that there exists 120582 gt 0 such that (65)is true for all 119905 isin 119869 is equivalent with the following twoinequalities which have to hold simultaneously

119888 (119905) le119886 (119905)

120582120574minus1

119888 (119905) le 119887 (119905) 120582 (67)

that is with

119888 (119905)

119887 (119905)le 120582 le (

119886 (119905)

119888 (119905))

1(120574minus1)

forall119905 isin 119869 (68)

Taking the supremum of the left-hand side and the infimumof the right-hand side we obtain (66) Conversely if (66)is satisfied then since the left-hand side in (66) cannotbe equal to zero while the right-hand side is less than infinthen inequality (66) defines a nonempty closed interval in R

(possibly reducing to just one point) in which we can chooseany positive 120582 Hence (67) is satisfied for all 119905 isin 119869 andtherefore (65) as well

Remark 21 If 119886 = 119886(119905) 119887 = 119887(119905) and 119888 = 119888(119905) appearingin Lemma 20 are constant functions then condition (66) isequivalent to 119888119887 = (119886119888)1(120574minus1) that is to (64)

Proof of Equivalence of (19) and (20) Wewill use Lemma 20Let us fix 119869

119894119895 where 119894 isin 1 2 and 119895 isin N We see that that the

inequality appearing in the second line of condition (19) is ofthe form (65) where

119888 (119905) =1

119888119894119895

119862 (119905) 119886 (119905) =1

120587119903(119905)1minus120574

119887 (119905) =119870119902 (119905)

120587 (1198720+ 1)

(69)

Condition (66) is equivalent to

sup119905isin119869119894119895

119862 (119905)

119902 (119905)le

1198701198881205741015840

119894119895

1205871205741015840

(1198720+ 1)

inf119905isin119869119894119895

1

119903 (119905) 119862(119905)120574minus1

=

1198701198881205741015840

119894119895

(1198720+ 1) sup

119905isin119869119894119895119903 (119905) 119862(119905)

120574minus1(

119888119894119895

120587)

1205741015840

(70)

This inequality is equivalent to the corresponding one in (19)and the claim follows from Lemma 19

Proof of Corollary 6 Substituting inf119905isin119869119894119895

119902(119905) instead of 119902(119905)and sup

119905isin119869119894119895119903(119905) instead of 119903(119905) in the inequality appearing in

the second line of (20) after a short computation we obtain astronger inequality than (20) as

1

119888119894119895

sup119905isin119869119894119895

119862 (119905)

le1

120587(

119870inf119905isin119869119894119895

119902 (119905)

(1198720+ 1) sup

119905isin119869119894119895119903 (119905)

)

(120574minus1)120574

forall119894 isin 1 2 119895 isin N

(71)

Substituting119862(119905) equiv 1we obtain that the left-hand side of (71)is equal to |119869

119894119895|minus1 and the resulting inequality is equivalent to

(21) Since (21) implies (20) the claim is proved

Remark 22 Here we show that the choice of 119862(119905) equiv 1 inthe proof of Corollary 6 is the best possible To prove thisnote that on the left-hand side of (71) we have the expressiondepending on an auxiliary function 119862(119905) and this functionis absent on the right-hand side Therefore it has sense totry to find a function 119862(119905) 119905 isin 119869

119894119895 such that the value of

119876(119862) = (1119888119894119895)sup119905isin119869119894119895

119862(119905) is minimal (note that 119888119894119895depends

on the function 119862 = 119862(119905) as well) It is easy to see that theminimum is achieved for 119862(119905) equiv 1 (or any positive constant)Indeed since 119876(119862) = 119876(120583119888) for any 120583 gt 0 by taking 120583 =

(sup119905isin119869119894119895

119862(119905))minus1 it suffices to assume that sup

119905isin119869119894119895119862(119905) = 1

The value of 119876(119862) is minimal if 119888119894119895= int119869119894119895

119862(120591) 119889120591 is maximalpossible and since 0 le 119862(119905) le 1 it is clear that the maximumof 119888119894119895= 119888119894119895(119862) is achieved when 119862(119905) equiv 1

International Journal of Differential Equations 9

Proof of Corollary 7 It is clear that the condition (23) implies(21) Furthermore since the lengths |119869

119894119895| are uniformly

bounded from below by a positive constant see (23) thenobviously 119886

1119895rarr infin as 119895 rarr infin Thus the claim follows

immediately from Corollary 6

Proof of Corollary 10 Let 119862(119905) = 119902(119905) Then the requiredcondition (19) is clearly satisfied because of assumption(30) Hence this corollary immediately follows from Theo-rem 3

4 An Extension of Condition (12) tothe Case of 120574 gt 1

In this section we consider the oscillation of (1) in the casewhen 120574 isin (1 2) is allowed in the assumption (12) In this waythe assumption (11) is slightly modified by a real number 119901 gt1 such that

|119906|(119901minus1)120574minus119901

VΦ (119906 V) ge 119892 (|Φ (119906 V)|) (72)

where (12) is supposed that is 119892(119888119904) ge 1198881205741198920(119904) for every 119888 119904 gt

0 and for a locally Lipschitz function 1198920 [0infin) rarr [0infin)

for which there exists a constant1198720such that 119892

0(119904)+119872

0ge 119904119901

for all 119904 isin [0infin)For the function 119891 we assume that there exists a constant

119870 gt 0 such that

119891 (119906) sign (119906) ge 119870|119906|119901minus1 for every 119906 isin R (73)

We will need the following two lemmas

Lemma 23 Let 119886 lt 119887 and let 119891 [119886 119887] rarr [0infin) be a meas-urable function Then

int

119887

119886

119891 (119905) 119889119905 ge 1 (74)

if and only if there exists a function 119892 isin 1198711([119886 119887] [0infin))

1198921= 1 such that

119891 (119905) ge 119892 (119905) forall119905 isin [119886 119887] (75)

Proof We assume (74) If int119887119886119891(119905)119889119905 is finite we may choose

119892(119909) = 119891(119909)1198911

For int119887119886119891(119905)119889119905 = infin we may define for every natural

number 119899 a function

119891119899(119905) =

119891 (119905) 119891 (119905) le 119899

119899 119891 (119905) gt 119899(76)

Since 119891119899(119905) le 119899 we have 119891

119899isin 1198711([119886 119887]R) Obviously

lim119899int119887

119886119891119899(119905)119889119905 is also infiniteThis implies that there exists an

1198990such that int119887

1198861198911198990(119905)119889119905 ge 1 We define 119892(119905) = 119891

1198990(119905)119891

11989901

Now we assume (75) Integrating over the whole intervalwe get

int

119887

119886

119891 (119905) 119889119905 ge int

119887

119886

119892 (119905) 119889119905 = 1 (77)

Lemma 24 Let119901 gt 1There exists a unique function 119910 = 119910(119904)that satisfies the following Cauchy problem

1199101015840(119904) = 1 +

1003816100381610038161003816119910 (119904)

1003816100381610038161003816

119901 119904 isin R (78)

119910 (0) = 0 (79)

Furthermore there exists a finite real number 119878lowast such that

lim119904rarrplusmn119878

lowast119910 (119904) = plusmninfin (80)

and 119910 isin 1198621((minus119878lowast 119878lowast)R) is injective and monotonous We

have 119878lowast = (120587119901)(1(sin(120587119901))) and 119910 is odd

We will denote by tan119901the solution of (78)-(79)

Proof Obviously every solution to (78) is injective andmonotonous on a connected set since 1199101015840 = 1 + |119910|119901 ge 1 gt 0

Let 119891(119909) = 1(1 + |119909|119901) Obviously 1 ge 119891(119909) gt 0 for all

119909 isin R and 119862119901gt |1198911015840(119909)| gt 0 for some constants 119862

119901isin R

We may assume 119862119901gt 1 Hence 1199111015840 = 119891(119905) 119911(0) = 0 has (by

[17Theorem 31]) a locally unique solution on (minus1119862119901 1119862119901)

By the same argument 119911 can be extended to the whole of Rinterval by interval of type ((119899 minus 1)119862

119901 (119899 + 1)119862

119901) for all

119899 isin ZWe know from [18] that the for the integral of the left-

hand side we have

int

infin

0

119891 (119909) 119889119909 = int

infin

0

119889119909

1 + 119909119901=120587

119901

1

sin (120587119901)= 119878lowast (81)

so lim119905rarrplusmninfin

119911(119905) = plusmn119878lowast

Since 1199111015840(119905) = 0 119911 is injective monotonous and of class1198621(R (minus119878lowast 119878lowast)) and so we define 119910(119904) = 119911

minus1(119904) It imme-

diately follows that 119910 isin 1198621((minus119878lowast 119878lowast)R) and that 119910 is

injective and monotonous Also by continuity of 119911 we havelim119904rarrplusmn119878

lowast119910(119904) = lim119908rarrplusmninfin

119910(119911(119908)) = lim119908rarrplusmninfin

119908 = plusmninfinDifferentiating it also follows that 1199101015840 = 1 + |119910|

119901 and bydefinition of 119910 119910(0) = 119911minus1(0) = 0 So 119910 is a solution of (78)-(79) Since any other such solution must have 119911 as its inverse119910 is unique on its domain

Proposition 25 Let 119886 lt 119887 and 120574 gt 119901 gt 1 Assume (72) and(73) and let 119902(119905) ge 0 and 119890(119905) le 0 (or 119890(119905) ge 0 ) for all 119905 isin [119886 119887]If there exists a constant 120582 gt 0 such that

119901

2120587sin(120587

119901)int

119887

119886

min119901 minus 1

119903120574minus1

(119905) 120582120574minus1

119870120582119902 (119905)

1 +1198720

119889119905 ge 1 (82)

then for any solution 119909(119905) of (1) with 119909(119886) ge 0 (or 119909(119886) le 0)there exists a number 119905lowast isin [119886 119887] such that 119909(119905lowast) = 0

Proof We assume that 119909(119905) = 0 for all 119905 isin [119886 119887] Then we maydefine the function

120596 (119905) = minus

120582119903 (119905) Φ (119909 (119905) 1199091015840(119905))

sign (119909 (119905)) |119909 (119905)|119901minus1 (83)

10 International Journal of Differential Equations

This function is well defined on [119886 119887] and its derivative is

1199081015840(119905) = minus

120582(119903 (119905) Φ (119909 (119905) 1199091015840(119905)))1015840

sign (119909 (119905)) |119909 (119905)|119901minus1

+

(119901 minus 1) 120582119903 (119905) Φ (119909 (119905) 1199091015840(119905))

|119909 (119905)|119901

1199091015840(119905)

(84)

Using the fact that 119909(119905) is a solution of (1) and omitting 119905 from119909 and 1199091015840 for clarity we get

1199081015840(119905) =

120582 (119902 (119905) 119891 (119909) minus 119890 (119905))

sign (119909) |119909|119901minus1+

(119901 minus 1) 120582119903 (119905) Φ (119909 1199091015840)

119909119901

1199091015840

=

(119901 minus 1) 120582119903 (119905) Φ (119909 1199091015840)

|119909|119901

1199091015840+ 120582119902 (119905)

119891 (119909)

sign (119909) |119909|119901minus1

minus 120582119890 (119905)

sign (119909) |119909|119901minus1

=(119901 minus 1) 120582119903 (119905)

|119909|120574(119901minus1)

Φ(119909 1199091015840) 1199091015840|119909|(119901minus1)120574minus119901

+ 120582119902 (119905)119891 (119909)

sign (119909) |119909|119901minus1minus 120582

119890 (119905)

sign (119909) |119909|119901minus1

(72)

ge(119901 minus 1) 120582119903 (119905)

|119909|120574(119901minus1)

119892 (10038161003816100381610038161003816Φ (119909 119909

1015840)10038161003816100381610038161003816)

+ 120582119902 (119905)119891 (119909)

sign (119909) |119909|119901minus1

minus 120582119890 (119905)

sign (119909) |119909|119901minus1

=(119901 minus 1) 120582119903 (119905)

|119909|120574(119901minus1)

119892(|119909|119901minus1

120582119903 (119905)|120596 (119905)|)

+ 120582119902 (119905)119891 (119909)

sign (119909) |119909|119901minus1minus 120582

119890 (119905)

sign (119909) |119909|119901minus1

(12)

ge(119901 minus 1) 120582119903 (119905)

|119909|120574(119901minus1)

(|119909|119901minus1

120582119903 (119905))

120574

1198920 (|120596 (119905)|)

+ 120582119902 (119905)119891 (119909)

sign (119909) |119909|119901minus1minus 120582

119890 (119905)

sign (119909) |119909|119901minus1

=119901 minus 1

(120582119903 (119905))120574minus1

1198920 (|120596 (119905)|)+120582119902 (119905)

119891 (119909)

sign (119909) |119909|119901minus1

minus 120582119890 (119905)

sign (119909) |119909|119901minus1

(73)

ge119901 minus 1

(120582119903 (119905))120574minus1

1198920 (|120596 (119905)|) + 120582119902 (119905) 119870

minus 120582119890 (119905)

sign (119909) |119909|119901minus1

(85)

Since 119890(119905)(sign(119909)|119909|119901minus1) le 0 we have

1199081015840(119905) ge

119901 minus 1

(120582119903 (119905))120574minus1

1198920(|120596 (119905)|) + 120582119870119902 (119905) (86)

We apply Lemma 23 to (82) to obtain a function119862(119905) suchthat

int

119887

119886

119862 (119905) 119889119905 = 1 (87)

and by (75)

0 le 119862 (119905) le119901

2120587sin(120587

119901)min

119901 minus 1

119903120574minus1

(119905) 120582120574minus1

119870120582119902 (119905)

1 +1198720

(88)

Let 1199040isin (minus(120587119901)(1 sin(120587119901)) (120587119901)(1 sin(120587119901))) be such

that tan119901(1199040) le 120596(119886) We define the function

119881 (119905) = 1199040+2120587

119901

1

sin (120587119901)int

119905

119886

119862 (120591) 119889120591 (89)

Note that (minus(120587119901)(1 sin(120587119901))) lt 1199040

= 119881(119886) lt

(120587119901)(1 sin(120587119901)) and 119881(119887) = 1199040+ (2120587119901)(1 sin(120587119901)) gt

(120587119901)(1 sin(120587119901)) Since119881 is continuous there exists a 119879lowast isin(119886 119887) such that 119881(119905) lt (120587119901)(1 sin(120587119901)) for 119905 isin [119886 119879lowast) and119881(119879lowast) = (120587119901)(1 sin(120587119901))

We define

120596 (119905) = tan119901(119881 (119905)) 119905 isin [119886 119879

lowast) (90)

and note that by Lemma 24 lim119905rarr119879

lowast120596(119905) = +infin wheretan119901(119904) = 119910(119904) and 119910(119904) is determined by Lemma 24The derivative of 120596 then satisfies

1205961015840(119905) = tan1015840

119901(119881 (119905)) 119881

1015840(119905)

= (1 + tan119901119901(119881 (119905))) 119881

1015840(119905)

=2120587

119901

1

sin (120587119901)(1 + 120596

119901(119905)) 119862 (119905)

le (1 +1198720+ 1198920(1003816100381610038161003816120596 (119905)

1003816100381610038161003816))min

119901 minus 1

119903120574minus1

(119905) 120582120574minus1

119870120582119902 (119905)

1 +1198720

le119901 minus 1

119903120574minus1

(119905) 120582120574minus1

1198920(1003816100381610038161003816120596 (119905)

1003816100381610038161003816) + 119870120582119902 (119905)

(91)

Thus we may apply the comparison principle ofLemma 15 to 120596(119886) le 120596(119886) and their respective differentialinequalities to conclude that 120596(119905) le 120596(119905) for all 119905 isin [119886 119887]But since 120596 rarr +infin as 119905 rarr 119879

lowastlt 119887 and 120596(119905) is well defined

on the whole segment [119886 119887] we arrive at a contradictionHence 119909(119905) gt 0 cannot hold for all 119905 isin [119886 119887] and since 119909(119905) iscontinuous there exists a 119905lowast isin [119886 119887] such that 119909(119905lowast) = 0

Theorem 26 Let 120574 gt 119901 gt 1 Assume (72) and (73) and let forany 119879 gt 119905

0there exist 119879 le 119886

1lt 1198871le 1198862lt 1198872such that 119892(119905) ge 0

on [1198861 1198871] cup [119886

2 1198872] and 119890(119905) le 0 on [119886

1 1198871] and 119890(119905) ge 0 on

International Journal of Differential Equations 11

[1198862 1198872] If for both 119894 isin 1 2 there exist constants 120582

119894gt 0 such

that

119901

2120587sin (120587119901)int

119887119894

119886119894

min119901 minus 1

119903120574minus1

(119905) 120582120574minus1

119894

119870120582119894119902 (119905)

1 +1198720

119889119905 ge 1

(92)

then (1) is oscillatory

Proof For (1) to be oscillatory it is sufficient that for every119899 isin N there exists a 119905

119899gt 119899 such that 119909(119905

119899) = 0

Let 119899 isin N and let 119899 lt 1198861lt 1198871le 1198862lt 1198872as assumed in the

theorem It is enough to show that every solution 119909(119905) of (1)has a zero on [119886

1 1198872] If 119909(119886

1) ge 0 we apply Proposition 25 to

the segment [1198861 1198871] to obtain the sought zero If 119909(119886

2) le 0 we

again apply Proposition 25 to the segment [1198862 1198872] Suppose

119909(1198861) lt 0 and 119909(119886

2) gt 0 Then since 119909 is continuous there

exists a number 1199051isin [1198861 1198862] such that 119909(119905

1) = 0

Acknowledgment

This work is supported by the Ministry of Science of theRepublic of Croatia under Grant no 036-0361621-1291

References

[1] Q Kong ldquoNonoscillation and oscillation of second order half-linear differential equationsrdquo Journal of Mathematical Analysisand Applications vol 332 no 1 pp 512ndash522 2007

[2] Q-R Wang and Q-G Yang ldquoInterval criteria for oscillationof second-order half-linear differential equationsrdquo Journal ofMathematical Analysis and Applications vol 291 no 1 pp 224ndash236 2004

[3] Q Long and Q-R Wang ldquoNew oscillation criteria of second-order nonlinear differential equationsrdquo Applied Mathematicsand Computation vol 212 no 2 pp 357ndash365 2009

[4] Z Zheng and F Meng ldquoOscillation criteria for forced second-order quasi-linear differential equationsrdquo Mathematical andComputer Modelling vol 45 no 1-2 pp 215ndash220 2007

[5] D Cakmak and A Tiryaki ldquoOscillation criteria for certainforced second-order nonlinear differential equationsrdquo AppliedMathematics Letters vol 17 no 3 pp 275ndash279 2004

[6] F Jiang and F Meng ldquoNew oscillation criteria for a class ofsecond-order nonlinear forced differential equationsrdquo Journalof Mathematical Analysis and Applications vol 336 no 2 pp1476ndash1485 2007

[7] Q Yang ldquoInterval oscillation criteria for a forced secondorder nonlinear ordinary differential equations with oscillatorypotentialrdquo Applied Mathematics and Computation vol 135 no1 pp 49ndash64 2003

[8] A Tiryaki and B Ayanlar ldquoOscillation theorems for certainnonlinear differential equations of second orderrdquo Computers ampMathematics with Applications vol 47 no 1 pp 149ndash159 2004

[9] J Jaros T Kusano and N Yoshida ldquoGeneralized Piconersquosformula and forced oscillations in quasilinear differential equa-tions of the second orderrdquoArchivumMathematicum vol 38 no1 pp 53ndash59 2002

[10] R P Agarwal S R Grace and D OrsquoRegan Oscillation Theoryfor Second Order Linear Half-Linear Superlinear and SublinearDynamic Equations Kluwer Academic Publishers DordrechtThe Netherlands 2002

[11] S Jing ldquoA new oscillation criterion for forced second-orderquasilinear differential equationsrdquoDiscrete Dynamics in Natureand Society vol 2011 Article ID 428976 8 pages 2011

[12] W-T Li and S S Cheng ldquoAn oscillation criterion for nonhomo-geneous half-linear differential equationsrdquoAppliedMathematicsLetters vol 15 no 3 pp 259ndash263 2002

[13] J S W Wong ldquoOscillation criteria for a forced second-orderlinear differential equationrdquo Journal of Mathematical Analysisand Applications vol 231 no 1 pp 235ndash240 1999

[14] M Pasic ldquoFite-Wintner-Leighton type oscillation criteria forsecond-order differential equations with nonlinear dampingrdquoAbstract and Applied Analysis vol 2013 Article ID 852180 10pages 2013

[15] M Pasic ldquoNew interval oscillation criteria for forced second-order differential equations with nonlinear dampingrdquo Interna-tional Journal of Mathematical Analysis vol 7 no 25 pp 1239ndash1255 2013

[16] L Gasinski and N S Papageorgiou Nonsmooth Critical PointTheory and Nonlinear Boundary Value Problems vol 8 of Seriesin Mathematical Analysis and Applications Edited by R PAgarwal and D OrsquoRegan Chapman amp HallCRC Boca RatonFla USA 2005

[17] S Lang Real and Functional Analysis vol 142 of GraduateTexts in Mathematics Springer-Verlag New York NY USA 3rdedition 1993

[18] I N Bronshtein K A Semendyayev G Musiol and HMuehligHandbook of Mathematics Springer Berlin Germany5th edition 2007

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

International Journal of Differential Equations 7

ge120582119903 (119905)

|119909 (119905)|120574119892 (10038161003816100381610038161003816Φ (119909 (119905) 119909

1015840(119905))

10038161003816100381610038161003816) + 119870120582119902 (119905) minus 120582

119890 (119905)

119909 (119905)

=120582119903 (119905)

|119909 (119905)|120574119892(

|119909 (119905)|

120582119903 (119905)|119908 (119905)|) + 119870120582119902 (119905) minus 120582

119890 (119905)

119909 (119905)

ge120582119903 (119905)

|119909 (119905)|120574

|119909 (119905)|120574

(120582119903 (119905))1205741198920 (|119908 (119905)|)

+ 119870120582119902 (119905) minus 120582119890 (119905)

119909 (119905)ae in 119869

(50)

that is

1199081015840ge (120582119903 (119905))

1minus1205741198920(|119908|) + 119870120582119902 (119905) minus 120582

119890 (119905)

119909 (119905)ae in 119869

(51)

Hence (49) and (51) prove the desired inequality (48)

Proposition 18 Let (19) be satisfied Let 1198771and 119877

2be two

arbitrary real numbers Then there are two points 119879lowast1isin (1198861 1198871)

and 119879lowast

2isin (1198862 1198872) and two continuous functions 119908

1(119905) and

1199082(119905) such that

119908119894(119886119894) le 119877119894 lim119905rarr119879

lowast

119894

119908119894(119905) = infin 119894 isin 1 2

1199081015840

119894le (120582119903 (119905))

1minus1205741198920(1003816100381610038161003816119908119894

1003816100381610038161003816) + 119870120582119902 (119905)

for ae 119905 isin (119886119894 119879lowast

119894) 119894 isin 1 2

(52)

Proof Since the function 119910(119904) = tan(119904) is a bijection fromthe interval (minus1205872 1205872) intoR we observe that there are two1199041 1199042isin (minus1205872 1205872) such that

tan (119904119894) le 119877119894 119894 isin 1 2 (53)

Next we define two functions 1198811(119905) and 119881

2(119905) by

119881119894 (119905) = 119904119894 +

120587

119888119894

int

119905

119886119894

119862 (120591) 119889120591 119905 isin [119886119894 119887119894] 119894 isin 1 2 (54)

where the real numbers 1198881 1198882and the function 119862(119905) are

defined in (19) From (19) and (54) one can immediatelyconclude that

119881119894(119886119894) = 119904119894lt120587

2 119881

119894(119887119894) = 119904119894+ 120587 gt

120587

2 119894 isin 1 2 (55)

Since 119862 isin 1198711

loc((1199050infin)R) it follows that 119862 isin 1198711((1199050infin)R)

which implies that 119881119894isin 119860119862([119886

119894 119887119894]R) 119894 isin 1 2 Therefore

exploiting the fact that in (55) we have 119881119894isin 119862([119886

119894 119887119894]R) we

obtain the existence of two points 119879lowast1isin (1198861 1198871) and 119879

lowast

2isin

(1198862 1198872) such that

119881119894(119879lowast

119894)=

120587

2 minus

120587

2lt119881119894(119905)lt

120587

2

forall119905 isin [119886119894 119879lowast

119894) 119894 isin 1 2

(56)

Now we are able to define the following two functions 1205961(119905)

and 1205962(119905) by

120596119894(119905) = tan (119881i (119905)) 119905 isin [119886

119894 119879lowast

119894) 119894 isin 1 2 (57)

That are well defined because of (56) Moreover from (53)(54) and (56) we obtain

120596119894(119886119894) = tan (119904

119894) le 119877119894

lim119905rarr119879

lowast

119898

120596119894(119905) = tan (119881

119894(119879lowast

119894)) = tan(120587

2) = infin

(58)

Also since119881119894isin 119860119862([119886

119894 119887119894]R) we can take the first derivative

of 120596119894(119905) and hence from (19) and (57) we obtain

1199081015840

119894(119905) =

120587

119888119894

119862 (119905)1

cos2 (119881119894 (119905))

=120587

119888119894

119862 (119905) (1003816100381610038161003816120596119894(119905)1003816100381610038161003816

2+ 1)

le120587

119888119894

119862 (119905) [1198920(1003816100381610038161003816120596119894(119905)1003816100381610038161003816) + 119872

0+ 1]

le120587

119888119894

119862 (119905) 1198920(1003816100381610038161003816120596119894(119905)1003816100381610038161003816) +

120587

119888119894

119862 (119905) (1198720+ 1)

le (120582119903 (119905))1minus1205741198920(1003816100381610038161003816120596119894(119905)1003816100381610038161003816) + 119870120582119902 (119905) ae in (119886

119894 119879lowast

119894)

(59)

which together with (58) proves this proposition

Now we are able to present the proof of Theorem 3 basedon Lemma 15 and Propositions 17 and 18

Proof of Theorem 3 Assuming the contrary then there is anonoscillatory solution 119909(119905) and 119879 ge 119905

0such that 119909(119905) = 0 for

all 119905 ge 119879 119909(119905) and 119903(119905)Φ(119909 1199091015840) are from 119860119862loc([1199050infin)R)In order to simplify notation let every 119895 isin N be fixed andomitted in the notation For instance instead of (119886

1119895 1198871119895) and

(1198862119895 1198872119895) we write (119886

1 1198871) and (119886

2 1198872) respectively and so on

On one hand we observe by Proposition 17 that thefunction 119908(119905) defined by (47) is a supersolution of thefollowing Riccati differential equation

1199081015840= (120582119903 (119905))

1minus1205741198920(|119908|) + 119870120582119902 (119905) ae in 119869 (60)

where 119908 isin 119860119862loc(119869R) and 119869 = (1198861 1198871) if 119909(119905) lt 0 and 119869 =

(1198862 1198872) if 119909(119905) gt 0 see the proof of Proposition 17 Let for

instance 119909(119905) lt 0 and thus 119869 = (1198861 1198871)

On the other hand by Proposition 18 there are a number119879lowast

1isin (1198861 1198871) and subsolutions 119908

1(119905) 1199081isin 119860119862([119886

1 119879lowast

1)R) of

the Riccati differential equation (60) such that

1199081(1198861) le 119908 (119886

1) lim

119905rarr119879lowast

1

1199081(119905) = infin (61)

(in the case when 119869 = (1198862 1198872) then we work with119879lowast

2isin (1198862 1198872)

and the other subsolution 1199082(119905) 119908

2isin 119860119862([119886

2 119879lowast

2)R))

Hence by Corollary 16 and (61) we conclude that1199081(119905) le 119908(119905)

for all 119905 isin [1198861 119879lowast

1) and therefore

infin = lim119905rarr119879

lowast

1

1199081(119905) le lim119905rarr119879

lowast

1

119908 (119905) (62)

which contradicts 119908 isin 119860119862loc([119879infin)R) Thus the assump-tion that 119909(119905) is nonoscillatory is not possible and hence everysolution of (1) is oscillatory

8 International Journal of Differential Equations

Proof of Theorem 4 There is only one difference betweenTheorems 3 and 4 and it is the difference between conditions(19) and (20) Hence Theorem 4 immediately follows fromTheorem 3 provided that we prove the equivalence between(19) and (20)

First of all we need the following two lemmas

Lemma 19 Let 119886 119887 and 119888 be positive real numbers and 120574 gt 1Then the existence of a positive real number 120582 such that

119888 le min 119886

120582120574minus1

119887120582 (63)

is equivalent to

119888 le 119886112057411988711205741015840

(64)

Here 1205741015840 = 120574(120574 minus 1) is the conjugate exponent of 120574

Proof Let us define 119892(120582) = min119886120582120574minus1 119887120582 Since 120582 997891rarr

119886120582120574minus1 is decreasing and 120582 997891rarr 119887120582 is increasing there exists

a unique point of maximum of 119892 It is achieved at 120582 = 1205820

which is a solution of 119886120582120574minus10

= 1198871205820 hence 120582

0= (119886119887)

1120574If there is a positive real number 120582 such that 119888 le 119892(120582)

then 119888 le 119892(1205820) = 119887120582

0= 119887(119886119887)

1120574= 119886112057411988711205741015840

Converselyif 119888 le 119886

112057411988711205741015840

then 119888 le 119892(1205820) and the equivalence in the

lemma is proved

The preceding lemma is a special case of the followingmore general statement

Lemma 20 Assume that 119886(119905) 119887(119905) and 119888(119905) are positive realfunctions defined on a subset 119869 sube R Then the existence of apositive real number 120582 such that for all 119905 isin 119869

119888 (119905) le min 119886 (119905)120582120574minus1

119887 (119905) 120582 (65)

is equivalent to

sup119905isin119869

119888 (119905)

119887 (119905)le (inf119905isin119869

119886 (119905)

119888 (119905))

1(120574minus1)

(66)

Proof The condition that there exists 120582 gt 0 such that (65)is true for all 119905 isin 119869 is equivalent with the following twoinequalities which have to hold simultaneously

119888 (119905) le119886 (119905)

120582120574minus1

119888 (119905) le 119887 (119905) 120582 (67)

that is with

119888 (119905)

119887 (119905)le 120582 le (

119886 (119905)

119888 (119905))

1(120574minus1)

forall119905 isin 119869 (68)

Taking the supremum of the left-hand side and the infimumof the right-hand side we obtain (66) Conversely if (66)is satisfied then since the left-hand side in (66) cannotbe equal to zero while the right-hand side is less than infinthen inequality (66) defines a nonempty closed interval in R

(possibly reducing to just one point) in which we can chooseany positive 120582 Hence (67) is satisfied for all 119905 isin 119869 andtherefore (65) as well

Remark 21 If 119886 = 119886(119905) 119887 = 119887(119905) and 119888 = 119888(119905) appearingin Lemma 20 are constant functions then condition (66) isequivalent to 119888119887 = (119886119888)1(120574minus1) that is to (64)

Proof of Equivalence of (19) and (20) Wewill use Lemma 20Let us fix 119869

119894119895 where 119894 isin 1 2 and 119895 isin N We see that that the

inequality appearing in the second line of condition (19) is ofthe form (65) where

119888 (119905) =1

119888119894119895

119862 (119905) 119886 (119905) =1

120587119903(119905)1minus120574

119887 (119905) =119870119902 (119905)

120587 (1198720+ 1)

(69)

Condition (66) is equivalent to

sup119905isin119869119894119895

119862 (119905)

119902 (119905)le

1198701198881205741015840

119894119895

1205871205741015840

(1198720+ 1)

inf119905isin119869119894119895

1

119903 (119905) 119862(119905)120574minus1

=

1198701198881205741015840

119894119895

(1198720+ 1) sup

119905isin119869119894119895119903 (119905) 119862(119905)

120574minus1(

119888119894119895

120587)

1205741015840

(70)

This inequality is equivalent to the corresponding one in (19)and the claim follows from Lemma 19

Proof of Corollary 6 Substituting inf119905isin119869119894119895

119902(119905) instead of 119902(119905)and sup

119905isin119869119894119895119903(119905) instead of 119903(119905) in the inequality appearing in

the second line of (20) after a short computation we obtain astronger inequality than (20) as

1

119888119894119895

sup119905isin119869119894119895

119862 (119905)

le1

120587(

119870inf119905isin119869119894119895

119902 (119905)

(1198720+ 1) sup

119905isin119869119894119895119903 (119905)

)

(120574minus1)120574

forall119894 isin 1 2 119895 isin N

(71)

Substituting119862(119905) equiv 1we obtain that the left-hand side of (71)is equal to |119869

119894119895|minus1 and the resulting inequality is equivalent to

(21) Since (21) implies (20) the claim is proved

Remark 22 Here we show that the choice of 119862(119905) equiv 1 inthe proof of Corollary 6 is the best possible To prove thisnote that on the left-hand side of (71) we have the expressiondepending on an auxiliary function 119862(119905) and this functionis absent on the right-hand side Therefore it has sense totry to find a function 119862(119905) 119905 isin 119869

119894119895 such that the value of

119876(119862) = (1119888119894119895)sup119905isin119869119894119895

119862(119905) is minimal (note that 119888119894119895depends

on the function 119862 = 119862(119905) as well) It is easy to see that theminimum is achieved for 119862(119905) equiv 1 (or any positive constant)Indeed since 119876(119862) = 119876(120583119888) for any 120583 gt 0 by taking 120583 =

(sup119905isin119869119894119895

119862(119905))minus1 it suffices to assume that sup

119905isin119869119894119895119862(119905) = 1

The value of 119876(119862) is minimal if 119888119894119895= int119869119894119895

119862(120591) 119889120591 is maximalpossible and since 0 le 119862(119905) le 1 it is clear that the maximumof 119888119894119895= 119888119894119895(119862) is achieved when 119862(119905) equiv 1

International Journal of Differential Equations 9

Proof of Corollary 7 It is clear that the condition (23) implies(21) Furthermore since the lengths |119869

119894119895| are uniformly

bounded from below by a positive constant see (23) thenobviously 119886

1119895rarr infin as 119895 rarr infin Thus the claim follows

immediately from Corollary 6

Proof of Corollary 10 Let 119862(119905) = 119902(119905) Then the requiredcondition (19) is clearly satisfied because of assumption(30) Hence this corollary immediately follows from Theo-rem 3

4 An Extension of Condition (12) tothe Case of 120574 gt 1

In this section we consider the oscillation of (1) in the casewhen 120574 isin (1 2) is allowed in the assumption (12) In this waythe assumption (11) is slightly modified by a real number 119901 gt1 such that

|119906|(119901minus1)120574minus119901

VΦ (119906 V) ge 119892 (|Φ (119906 V)|) (72)

where (12) is supposed that is 119892(119888119904) ge 1198881205741198920(119904) for every 119888 119904 gt

0 and for a locally Lipschitz function 1198920 [0infin) rarr [0infin)

for which there exists a constant1198720such that 119892

0(119904)+119872

0ge 119904119901

for all 119904 isin [0infin)For the function 119891 we assume that there exists a constant

119870 gt 0 such that

119891 (119906) sign (119906) ge 119870|119906|119901minus1 for every 119906 isin R (73)

We will need the following two lemmas

Lemma 23 Let 119886 lt 119887 and let 119891 [119886 119887] rarr [0infin) be a meas-urable function Then

int

119887

119886

119891 (119905) 119889119905 ge 1 (74)

if and only if there exists a function 119892 isin 1198711([119886 119887] [0infin))

1198921= 1 such that

119891 (119905) ge 119892 (119905) forall119905 isin [119886 119887] (75)

Proof We assume (74) If int119887119886119891(119905)119889119905 is finite we may choose

119892(119909) = 119891(119909)1198911

For int119887119886119891(119905)119889119905 = infin we may define for every natural

number 119899 a function

119891119899(119905) =

119891 (119905) 119891 (119905) le 119899

119899 119891 (119905) gt 119899(76)

Since 119891119899(119905) le 119899 we have 119891

119899isin 1198711([119886 119887]R) Obviously

lim119899int119887

119886119891119899(119905)119889119905 is also infiniteThis implies that there exists an

1198990such that int119887

1198861198911198990(119905)119889119905 ge 1 We define 119892(119905) = 119891

1198990(119905)119891

11989901

Now we assume (75) Integrating over the whole intervalwe get

int

119887

119886

119891 (119905) 119889119905 ge int

119887

119886

119892 (119905) 119889119905 = 1 (77)

Lemma 24 Let119901 gt 1There exists a unique function 119910 = 119910(119904)that satisfies the following Cauchy problem

1199101015840(119904) = 1 +

1003816100381610038161003816119910 (119904)

1003816100381610038161003816

119901 119904 isin R (78)

119910 (0) = 0 (79)

Furthermore there exists a finite real number 119878lowast such that

lim119904rarrplusmn119878

lowast119910 (119904) = plusmninfin (80)

and 119910 isin 1198621((minus119878lowast 119878lowast)R) is injective and monotonous We

have 119878lowast = (120587119901)(1(sin(120587119901))) and 119910 is odd

We will denote by tan119901the solution of (78)-(79)

Proof Obviously every solution to (78) is injective andmonotonous on a connected set since 1199101015840 = 1 + |119910|119901 ge 1 gt 0

Let 119891(119909) = 1(1 + |119909|119901) Obviously 1 ge 119891(119909) gt 0 for all

119909 isin R and 119862119901gt |1198911015840(119909)| gt 0 for some constants 119862

119901isin R

We may assume 119862119901gt 1 Hence 1199111015840 = 119891(119905) 119911(0) = 0 has (by

[17Theorem 31]) a locally unique solution on (minus1119862119901 1119862119901)

By the same argument 119911 can be extended to the whole of Rinterval by interval of type ((119899 minus 1)119862

119901 (119899 + 1)119862

119901) for all

119899 isin ZWe know from [18] that the for the integral of the left-

hand side we have

int

infin

0

119891 (119909) 119889119909 = int

infin

0

119889119909

1 + 119909119901=120587

119901

1

sin (120587119901)= 119878lowast (81)

so lim119905rarrplusmninfin

119911(119905) = plusmn119878lowast

Since 1199111015840(119905) = 0 119911 is injective monotonous and of class1198621(R (minus119878lowast 119878lowast)) and so we define 119910(119904) = 119911

minus1(119904) It imme-

diately follows that 119910 isin 1198621((minus119878lowast 119878lowast)R) and that 119910 is

injective and monotonous Also by continuity of 119911 we havelim119904rarrplusmn119878

lowast119910(119904) = lim119908rarrplusmninfin

119910(119911(119908)) = lim119908rarrplusmninfin

119908 = plusmninfinDifferentiating it also follows that 1199101015840 = 1 + |119910|

119901 and bydefinition of 119910 119910(0) = 119911minus1(0) = 0 So 119910 is a solution of (78)-(79) Since any other such solution must have 119911 as its inverse119910 is unique on its domain

Proposition 25 Let 119886 lt 119887 and 120574 gt 119901 gt 1 Assume (72) and(73) and let 119902(119905) ge 0 and 119890(119905) le 0 (or 119890(119905) ge 0 ) for all 119905 isin [119886 119887]If there exists a constant 120582 gt 0 such that

119901

2120587sin(120587

119901)int

119887

119886

min119901 minus 1

119903120574minus1

(119905) 120582120574minus1

119870120582119902 (119905)

1 +1198720

119889119905 ge 1 (82)

then for any solution 119909(119905) of (1) with 119909(119886) ge 0 (or 119909(119886) le 0)there exists a number 119905lowast isin [119886 119887] such that 119909(119905lowast) = 0

Proof We assume that 119909(119905) = 0 for all 119905 isin [119886 119887] Then we maydefine the function

120596 (119905) = minus

120582119903 (119905) Φ (119909 (119905) 1199091015840(119905))

sign (119909 (119905)) |119909 (119905)|119901minus1 (83)

10 International Journal of Differential Equations

This function is well defined on [119886 119887] and its derivative is

1199081015840(119905) = minus

120582(119903 (119905) Φ (119909 (119905) 1199091015840(119905)))1015840

sign (119909 (119905)) |119909 (119905)|119901minus1

+

(119901 minus 1) 120582119903 (119905) Φ (119909 (119905) 1199091015840(119905))

|119909 (119905)|119901

1199091015840(119905)

(84)

Using the fact that 119909(119905) is a solution of (1) and omitting 119905 from119909 and 1199091015840 for clarity we get

1199081015840(119905) =

120582 (119902 (119905) 119891 (119909) minus 119890 (119905))

sign (119909) |119909|119901minus1+

(119901 minus 1) 120582119903 (119905) Φ (119909 1199091015840)

119909119901

1199091015840

=

(119901 minus 1) 120582119903 (119905) Φ (119909 1199091015840)

|119909|119901

1199091015840+ 120582119902 (119905)

119891 (119909)

sign (119909) |119909|119901minus1

minus 120582119890 (119905)

sign (119909) |119909|119901minus1

=(119901 minus 1) 120582119903 (119905)

|119909|120574(119901minus1)

Φ(119909 1199091015840) 1199091015840|119909|(119901minus1)120574minus119901

+ 120582119902 (119905)119891 (119909)

sign (119909) |119909|119901minus1minus 120582

119890 (119905)

sign (119909) |119909|119901minus1

(72)

ge(119901 minus 1) 120582119903 (119905)

|119909|120574(119901minus1)

119892 (10038161003816100381610038161003816Φ (119909 119909

1015840)10038161003816100381610038161003816)

+ 120582119902 (119905)119891 (119909)

sign (119909) |119909|119901minus1

minus 120582119890 (119905)

sign (119909) |119909|119901minus1

=(119901 minus 1) 120582119903 (119905)

|119909|120574(119901minus1)

119892(|119909|119901minus1

120582119903 (119905)|120596 (119905)|)

+ 120582119902 (119905)119891 (119909)

sign (119909) |119909|119901minus1minus 120582

119890 (119905)

sign (119909) |119909|119901minus1

(12)

ge(119901 minus 1) 120582119903 (119905)

|119909|120574(119901minus1)

(|119909|119901minus1

120582119903 (119905))

120574

1198920 (|120596 (119905)|)

+ 120582119902 (119905)119891 (119909)

sign (119909) |119909|119901minus1minus 120582

119890 (119905)

sign (119909) |119909|119901minus1

=119901 minus 1

(120582119903 (119905))120574minus1

1198920 (|120596 (119905)|)+120582119902 (119905)

119891 (119909)

sign (119909) |119909|119901minus1

minus 120582119890 (119905)

sign (119909) |119909|119901minus1

(73)

ge119901 minus 1

(120582119903 (119905))120574minus1

1198920 (|120596 (119905)|) + 120582119902 (119905) 119870

minus 120582119890 (119905)

sign (119909) |119909|119901minus1

(85)

Since 119890(119905)(sign(119909)|119909|119901minus1) le 0 we have

1199081015840(119905) ge

119901 minus 1

(120582119903 (119905))120574minus1

1198920(|120596 (119905)|) + 120582119870119902 (119905) (86)

We apply Lemma 23 to (82) to obtain a function119862(119905) suchthat

int

119887

119886

119862 (119905) 119889119905 = 1 (87)

and by (75)

0 le 119862 (119905) le119901

2120587sin(120587

119901)min

119901 minus 1

119903120574minus1

(119905) 120582120574minus1

119870120582119902 (119905)

1 +1198720

(88)

Let 1199040isin (minus(120587119901)(1 sin(120587119901)) (120587119901)(1 sin(120587119901))) be such

that tan119901(1199040) le 120596(119886) We define the function

119881 (119905) = 1199040+2120587

119901

1

sin (120587119901)int

119905

119886

119862 (120591) 119889120591 (89)

Note that (minus(120587119901)(1 sin(120587119901))) lt 1199040

= 119881(119886) lt

(120587119901)(1 sin(120587119901)) and 119881(119887) = 1199040+ (2120587119901)(1 sin(120587119901)) gt

(120587119901)(1 sin(120587119901)) Since119881 is continuous there exists a 119879lowast isin(119886 119887) such that 119881(119905) lt (120587119901)(1 sin(120587119901)) for 119905 isin [119886 119879lowast) and119881(119879lowast) = (120587119901)(1 sin(120587119901))

We define

120596 (119905) = tan119901(119881 (119905)) 119905 isin [119886 119879

lowast) (90)

and note that by Lemma 24 lim119905rarr119879

lowast120596(119905) = +infin wheretan119901(119904) = 119910(119904) and 119910(119904) is determined by Lemma 24The derivative of 120596 then satisfies

1205961015840(119905) = tan1015840

119901(119881 (119905)) 119881

1015840(119905)

= (1 + tan119901119901(119881 (119905))) 119881

1015840(119905)

=2120587

119901

1

sin (120587119901)(1 + 120596

119901(119905)) 119862 (119905)

le (1 +1198720+ 1198920(1003816100381610038161003816120596 (119905)

1003816100381610038161003816))min

119901 minus 1

119903120574minus1

(119905) 120582120574minus1

119870120582119902 (119905)

1 +1198720

le119901 minus 1

119903120574minus1

(119905) 120582120574minus1

1198920(1003816100381610038161003816120596 (119905)

1003816100381610038161003816) + 119870120582119902 (119905)

(91)

Thus we may apply the comparison principle ofLemma 15 to 120596(119886) le 120596(119886) and their respective differentialinequalities to conclude that 120596(119905) le 120596(119905) for all 119905 isin [119886 119887]But since 120596 rarr +infin as 119905 rarr 119879

lowastlt 119887 and 120596(119905) is well defined

on the whole segment [119886 119887] we arrive at a contradictionHence 119909(119905) gt 0 cannot hold for all 119905 isin [119886 119887] and since 119909(119905) iscontinuous there exists a 119905lowast isin [119886 119887] such that 119909(119905lowast) = 0

Theorem 26 Let 120574 gt 119901 gt 1 Assume (72) and (73) and let forany 119879 gt 119905

0there exist 119879 le 119886

1lt 1198871le 1198862lt 1198872such that 119892(119905) ge 0

on [1198861 1198871] cup [119886

2 1198872] and 119890(119905) le 0 on [119886

1 1198871] and 119890(119905) ge 0 on

International Journal of Differential Equations 11

[1198862 1198872] If for both 119894 isin 1 2 there exist constants 120582

119894gt 0 such

that

119901

2120587sin (120587119901)int

119887119894

119886119894

min119901 minus 1

119903120574minus1

(119905) 120582120574minus1

119894

119870120582119894119902 (119905)

1 +1198720

119889119905 ge 1

(92)

then (1) is oscillatory

Proof For (1) to be oscillatory it is sufficient that for every119899 isin N there exists a 119905

119899gt 119899 such that 119909(119905

119899) = 0

Let 119899 isin N and let 119899 lt 1198861lt 1198871le 1198862lt 1198872as assumed in the

theorem It is enough to show that every solution 119909(119905) of (1)has a zero on [119886

1 1198872] If 119909(119886

1) ge 0 we apply Proposition 25 to

the segment [1198861 1198871] to obtain the sought zero If 119909(119886

2) le 0 we

again apply Proposition 25 to the segment [1198862 1198872] Suppose

119909(1198861) lt 0 and 119909(119886

2) gt 0 Then since 119909 is continuous there

exists a number 1199051isin [1198861 1198862] such that 119909(119905

1) = 0

Acknowledgment

This work is supported by the Ministry of Science of theRepublic of Croatia under Grant no 036-0361621-1291

References

[1] Q Kong ldquoNonoscillation and oscillation of second order half-linear differential equationsrdquo Journal of Mathematical Analysisand Applications vol 332 no 1 pp 512ndash522 2007

[2] Q-R Wang and Q-G Yang ldquoInterval criteria for oscillationof second-order half-linear differential equationsrdquo Journal ofMathematical Analysis and Applications vol 291 no 1 pp 224ndash236 2004

[3] Q Long and Q-R Wang ldquoNew oscillation criteria of second-order nonlinear differential equationsrdquo Applied Mathematicsand Computation vol 212 no 2 pp 357ndash365 2009

[4] Z Zheng and F Meng ldquoOscillation criteria for forced second-order quasi-linear differential equationsrdquo Mathematical andComputer Modelling vol 45 no 1-2 pp 215ndash220 2007

[5] D Cakmak and A Tiryaki ldquoOscillation criteria for certainforced second-order nonlinear differential equationsrdquo AppliedMathematics Letters vol 17 no 3 pp 275ndash279 2004

[6] F Jiang and F Meng ldquoNew oscillation criteria for a class ofsecond-order nonlinear forced differential equationsrdquo Journalof Mathematical Analysis and Applications vol 336 no 2 pp1476ndash1485 2007

[7] Q Yang ldquoInterval oscillation criteria for a forced secondorder nonlinear ordinary differential equations with oscillatorypotentialrdquo Applied Mathematics and Computation vol 135 no1 pp 49ndash64 2003

[8] A Tiryaki and B Ayanlar ldquoOscillation theorems for certainnonlinear differential equations of second orderrdquo Computers ampMathematics with Applications vol 47 no 1 pp 149ndash159 2004

[9] J Jaros T Kusano and N Yoshida ldquoGeneralized Piconersquosformula and forced oscillations in quasilinear differential equa-tions of the second orderrdquoArchivumMathematicum vol 38 no1 pp 53ndash59 2002

[10] R P Agarwal S R Grace and D OrsquoRegan Oscillation Theoryfor Second Order Linear Half-Linear Superlinear and SublinearDynamic Equations Kluwer Academic Publishers DordrechtThe Netherlands 2002

[11] S Jing ldquoA new oscillation criterion for forced second-orderquasilinear differential equationsrdquoDiscrete Dynamics in Natureand Society vol 2011 Article ID 428976 8 pages 2011

[12] W-T Li and S S Cheng ldquoAn oscillation criterion for nonhomo-geneous half-linear differential equationsrdquoAppliedMathematicsLetters vol 15 no 3 pp 259ndash263 2002

[13] J S W Wong ldquoOscillation criteria for a forced second-orderlinear differential equationrdquo Journal of Mathematical Analysisand Applications vol 231 no 1 pp 235ndash240 1999

[14] M Pasic ldquoFite-Wintner-Leighton type oscillation criteria forsecond-order differential equations with nonlinear dampingrdquoAbstract and Applied Analysis vol 2013 Article ID 852180 10pages 2013

[15] M Pasic ldquoNew interval oscillation criteria for forced second-order differential equations with nonlinear dampingrdquo Interna-tional Journal of Mathematical Analysis vol 7 no 25 pp 1239ndash1255 2013

[16] L Gasinski and N S Papageorgiou Nonsmooth Critical PointTheory and Nonlinear Boundary Value Problems vol 8 of Seriesin Mathematical Analysis and Applications Edited by R PAgarwal and D OrsquoRegan Chapman amp HallCRC Boca RatonFla USA 2005

[17] S Lang Real and Functional Analysis vol 142 of GraduateTexts in Mathematics Springer-Verlag New York NY USA 3rdedition 1993

[18] I N Bronshtein K A Semendyayev G Musiol and HMuehligHandbook of Mathematics Springer Berlin Germany5th edition 2007

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 International Journal of Differential Equations

Proof of Theorem 4 There is only one difference betweenTheorems 3 and 4 and it is the difference between conditions(19) and (20) Hence Theorem 4 immediately follows fromTheorem 3 provided that we prove the equivalence between(19) and (20)

First of all we need the following two lemmas

Lemma 19 Let 119886 119887 and 119888 be positive real numbers and 120574 gt 1Then the existence of a positive real number 120582 such that

119888 le min 119886

120582120574minus1

119887120582 (63)

is equivalent to

119888 le 119886112057411988711205741015840

(64)

Here 1205741015840 = 120574(120574 minus 1) is the conjugate exponent of 120574

Proof Let us define 119892(120582) = min119886120582120574minus1 119887120582 Since 120582 997891rarr

119886120582120574minus1 is decreasing and 120582 997891rarr 119887120582 is increasing there exists

a unique point of maximum of 119892 It is achieved at 120582 = 1205820

which is a solution of 119886120582120574minus10

= 1198871205820 hence 120582

0= (119886119887)

1120574If there is a positive real number 120582 such that 119888 le 119892(120582)

then 119888 le 119892(1205820) = 119887120582

0= 119887(119886119887)

1120574= 119886112057411988711205741015840

Converselyif 119888 le 119886

112057411988711205741015840

then 119888 le 119892(1205820) and the equivalence in the

lemma is proved

The preceding lemma is a special case of the followingmore general statement

Lemma 20 Assume that 119886(119905) 119887(119905) and 119888(119905) are positive realfunctions defined on a subset 119869 sube R Then the existence of apositive real number 120582 such that for all 119905 isin 119869

119888 (119905) le min 119886 (119905)120582120574minus1

119887 (119905) 120582 (65)

is equivalent to

sup119905isin119869

119888 (119905)

119887 (119905)le (inf119905isin119869

119886 (119905)

119888 (119905))

1(120574minus1)

(66)

Proof The condition that there exists 120582 gt 0 such that (65)is true for all 119905 isin 119869 is equivalent with the following twoinequalities which have to hold simultaneously

119888 (119905) le119886 (119905)

120582120574minus1

119888 (119905) le 119887 (119905) 120582 (67)

that is with

119888 (119905)

119887 (119905)le 120582 le (

119886 (119905)

119888 (119905))

1(120574minus1)

forall119905 isin 119869 (68)

Taking the supremum of the left-hand side and the infimumof the right-hand side we obtain (66) Conversely if (66)is satisfied then since the left-hand side in (66) cannotbe equal to zero while the right-hand side is less than infinthen inequality (66) defines a nonempty closed interval in R

(possibly reducing to just one point) in which we can chooseany positive 120582 Hence (67) is satisfied for all 119905 isin 119869 andtherefore (65) as well

Remark 21 If 119886 = 119886(119905) 119887 = 119887(119905) and 119888 = 119888(119905) appearingin Lemma 20 are constant functions then condition (66) isequivalent to 119888119887 = (119886119888)1(120574minus1) that is to (64)

Proof of Equivalence of (19) and (20) Wewill use Lemma 20Let us fix 119869

119894119895 where 119894 isin 1 2 and 119895 isin N We see that that the

inequality appearing in the second line of condition (19) is ofthe form (65) where

119888 (119905) =1

119888119894119895

119862 (119905) 119886 (119905) =1

120587119903(119905)1minus120574

119887 (119905) =119870119902 (119905)

120587 (1198720+ 1)

(69)

Condition (66) is equivalent to

sup119905isin119869119894119895

119862 (119905)

119902 (119905)le

1198701198881205741015840

119894119895

1205871205741015840

(1198720+ 1)

inf119905isin119869119894119895

1

119903 (119905) 119862(119905)120574minus1

=

1198701198881205741015840

119894119895

(1198720+ 1) sup

119905isin119869119894119895119903 (119905) 119862(119905)

120574minus1(

119888119894119895

120587)

1205741015840

(70)

This inequality is equivalent to the corresponding one in (19)and the claim follows from Lemma 19

Proof of Corollary 6 Substituting inf119905isin119869119894119895

119902(119905) instead of 119902(119905)and sup

119905isin119869119894119895119903(119905) instead of 119903(119905) in the inequality appearing in

the second line of (20) after a short computation we obtain astronger inequality than (20) as

1

119888119894119895

sup119905isin119869119894119895

119862 (119905)

le1

120587(

119870inf119905isin119869119894119895

119902 (119905)

(1198720+ 1) sup

119905isin119869119894119895119903 (119905)

)

(120574minus1)120574

forall119894 isin 1 2 119895 isin N

(71)

Substituting119862(119905) equiv 1we obtain that the left-hand side of (71)is equal to |119869

119894119895|minus1 and the resulting inequality is equivalent to

(21) Since (21) implies (20) the claim is proved

Remark 22 Here we show that the choice of 119862(119905) equiv 1 inthe proof of Corollary 6 is the best possible To prove thisnote that on the left-hand side of (71) we have the expressiondepending on an auxiliary function 119862(119905) and this functionis absent on the right-hand side Therefore it has sense totry to find a function 119862(119905) 119905 isin 119869

119894119895 such that the value of

119876(119862) = (1119888119894119895)sup119905isin119869119894119895

119862(119905) is minimal (note that 119888119894119895depends

on the function 119862 = 119862(119905) as well) It is easy to see that theminimum is achieved for 119862(119905) equiv 1 (or any positive constant)Indeed since 119876(119862) = 119876(120583119888) for any 120583 gt 0 by taking 120583 =

(sup119905isin119869119894119895

119862(119905))minus1 it suffices to assume that sup

119905isin119869119894119895119862(119905) = 1

The value of 119876(119862) is minimal if 119888119894119895= int119869119894119895

119862(120591) 119889120591 is maximalpossible and since 0 le 119862(119905) le 1 it is clear that the maximumof 119888119894119895= 119888119894119895(119862) is achieved when 119862(119905) equiv 1

International Journal of Differential Equations 9

Proof of Corollary 7 It is clear that the condition (23) implies(21) Furthermore since the lengths |119869

119894119895| are uniformly

bounded from below by a positive constant see (23) thenobviously 119886

1119895rarr infin as 119895 rarr infin Thus the claim follows

immediately from Corollary 6

Proof of Corollary 10 Let 119862(119905) = 119902(119905) Then the requiredcondition (19) is clearly satisfied because of assumption(30) Hence this corollary immediately follows from Theo-rem 3

4 An Extension of Condition (12) tothe Case of 120574 gt 1

In this section we consider the oscillation of (1) in the casewhen 120574 isin (1 2) is allowed in the assumption (12) In this waythe assumption (11) is slightly modified by a real number 119901 gt1 such that

|119906|(119901minus1)120574minus119901

VΦ (119906 V) ge 119892 (|Φ (119906 V)|) (72)

where (12) is supposed that is 119892(119888119904) ge 1198881205741198920(119904) for every 119888 119904 gt

0 and for a locally Lipschitz function 1198920 [0infin) rarr [0infin)

for which there exists a constant1198720such that 119892

0(119904)+119872

0ge 119904119901

for all 119904 isin [0infin)For the function 119891 we assume that there exists a constant

119870 gt 0 such that

119891 (119906) sign (119906) ge 119870|119906|119901minus1 for every 119906 isin R (73)

We will need the following two lemmas

Lemma 23 Let 119886 lt 119887 and let 119891 [119886 119887] rarr [0infin) be a meas-urable function Then

int

119887

119886

119891 (119905) 119889119905 ge 1 (74)

if and only if there exists a function 119892 isin 1198711([119886 119887] [0infin))

1198921= 1 such that

119891 (119905) ge 119892 (119905) forall119905 isin [119886 119887] (75)

Proof We assume (74) If int119887119886119891(119905)119889119905 is finite we may choose

119892(119909) = 119891(119909)1198911

For int119887119886119891(119905)119889119905 = infin we may define for every natural

number 119899 a function

119891119899(119905) =

119891 (119905) 119891 (119905) le 119899

119899 119891 (119905) gt 119899(76)

Since 119891119899(119905) le 119899 we have 119891

119899isin 1198711([119886 119887]R) Obviously

lim119899int119887

119886119891119899(119905)119889119905 is also infiniteThis implies that there exists an

1198990such that int119887

1198861198911198990(119905)119889119905 ge 1 We define 119892(119905) = 119891

1198990(119905)119891

11989901

Now we assume (75) Integrating over the whole intervalwe get

int

119887

119886

119891 (119905) 119889119905 ge int

119887

119886

119892 (119905) 119889119905 = 1 (77)

Lemma 24 Let119901 gt 1There exists a unique function 119910 = 119910(119904)that satisfies the following Cauchy problem

1199101015840(119904) = 1 +

1003816100381610038161003816119910 (119904)

1003816100381610038161003816

119901 119904 isin R (78)

119910 (0) = 0 (79)

Furthermore there exists a finite real number 119878lowast such that

lim119904rarrplusmn119878

lowast119910 (119904) = plusmninfin (80)

and 119910 isin 1198621((minus119878lowast 119878lowast)R) is injective and monotonous We

have 119878lowast = (120587119901)(1(sin(120587119901))) and 119910 is odd

We will denote by tan119901the solution of (78)-(79)

Proof Obviously every solution to (78) is injective andmonotonous on a connected set since 1199101015840 = 1 + |119910|119901 ge 1 gt 0

Let 119891(119909) = 1(1 + |119909|119901) Obviously 1 ge 119891(119909) gt 0 for all

119909 isin R and 119862119901gt |1198911015840(119909)| gt 0 for some constants 119862

119901isin R

We may assume 119862119901gt 1 Hence 1199111015840 = 119891(119905) 119911(0) = 0 has (by

[17Theorem 31]) a locally unique solution on (minus1119862119901 1119862119901)

By the same argument 119911 can be extended to the whole of Rinterval by interval of type ((119899 minus 1)119862

119901 (119899 + 1)119862

119901) for all

119899 isin ZWe know from [18] that the for the integral of the left-

hand side we have

int

infin

0

119891 (119909) 119889119909 = int

infin

0

119889119909

1 + 119909119901=120587

119901

1

sin (120587119901)= 119878lowast (81)

so lim119905rarrplusmninfin

119911(119905) = plusmn119878lowast

Since 1199111015840(119905) = 0 119911 is injective monotonous and of class1198621(R (minus119878lowast 119878lowast)) and so we define 119910(119904) = 119911

minus1(119904) It imme-

diately follows that 119910 isin 1198621((minus119878lowast 119878lowast)R) and that 119910 is

injective and monotonous Also by continuity of 119911 we havelim119904rarrplusmn119878

lowast119910(119904) = lim119908rarrplusmninfin

119910(119911(119908)) = lim119908rarrplusmninfin

119908 = plusmninfinDifferentiating it also follows that 1199101015840 = 1 + |119910|

119901 and bydefinition of 119910 119910(0) = 119911minus1(0) = 0 So 119910 is a solution of (78)-(79) Since any other such solution must have 119911 as its inverse119910 is unique on its domain

Proposition 25 Let 119886 lt 119887 and 120574 gt 119901 gt 1 Assume (72) and(73) and let 119902(119905) ge 0 and 119890(119905) le 0 (or 119890(119905) ge 0 ) for all 119905 isin [119886 119887]If there exists a constant 120582 gt 0 such that

119901

2120587sin(120587

119901)int

119887

119886

min119901 minus 1

119903120574minus1

(119905) 120582120574minus1

119870120582119902 (119905)

1 +1198720

119889119905 ge 1 (82)

then for any solution 119909(119905) of (1) with 119909(119886) ge 0 (or 119909(119886) le 0)there exists a number 119905lowast isin [119886 119887] such that 119909(119905lowast) = 0

Proof We assume that 119909(119905) = 0 for all 119905 isin [119886 119887] Then we maydefine the function

120596 (119905) = minus

120582119903 (119905) Φ (119909 (119905) 1199091015840(119905))

sign (119909 (119905)) |119909 (119905)|119901minus1 (83)

10 International Journal of Differential Equations

This function is well defined on [119886 119887] and its derivative is

1199081015840(119905) = minus

120582(119903 (119905) Φ (119909 (119905) 1199091015840(119905)))1015840

sign (119909 (119905)) |119909 (119905)|119901minus1

+

(119901 minus 1) 120582119903 (119905) Φ (119909 (119905) 1199091015840(119905))

|119909 (119905)|119901

1199091015840(119905)

(84)

Using the fact that 119909(119905) is a solution of (1) and omitting 119905 from119909 and 1199091015840 for clarity we get

1199081015840(119905) =

120582 (119902 (119905) 119891 (119909) minus 119890 (119905))

sign (119909) |119909|119901minus1+

(119901 minus 1) 120582119903 (119905) Φ (119909 1199091015840)

119909119901

1199091015840

=

(119901 minus 1) 120582119903 (119905) Φ (119909 1199091015840)

|119909|119901

1199091015840+ 120582119902 (119905)

119891 (119909)

sign (119909) |119909|119901minus1

minus 120582119890 (119905)

sign (119909) |119909|119901minus1

=(119901 minus 1) 120582119903 (119905)

|119909|120574(119901minus1)

Φ(119909 1199091015840) 1199091015840|119909|(119901minus1)120574minus119901

+ 120582119902 (119905)119891 (119909)

sign (119909) |119909|119901minus1minus 120582

119890 (119905)

sign (119909) |119909|119901minus1

(72)

ge(119901 minus 1) 120582119903 (119905)

|119909|120574(119901minus1)

119892 (10038161003816100381610038161003816Φ (119909 119909

1015840)10038161003816100381610038161003816)

+ 120582119902 (119905)119891 (119909)

sign (119909) |119909|119901minus1

minus 120582119890 (119905)

sign (119909) |119909|119901minus1

=(119901 minus 1) 120582119903 (119905)

|119909|120574(119901minus1)

119892(|119909|119901minus1

120582119903 (119905)|120596 (119905)|)

+ 120582119902 (119905)119891 (119909)

sign (119909) |119909|119901minus1minus 120582

119890 (119905)

sign (119909) |119909|119901minus1

(12)

ge(119901 minus 1) 120582119903 (119905)

|119909|120574(119901minus1)

(|119909|119901minus1

120582119903 (119905))

120574

1198920 (|120596 (119905)|)

+ 120582119902 (119905)119891 (119909)

sign (119909) |119909|119901minus1minus 120582

119890 (119905)

sign (119909) |119909|119901minus1

=119901 minus 1

(120582119903 (119905))120574minus1

1198920 (|120596 (119905)|)+120582119902 (119905)

119891 (119909)

sign (119909) |119909|119901minus1

minus 120582119890 (119905)

sign (119909) |119909|119901minus1

(73)

ge119901 minus 1

(120582119903 (119905))120574minus1

1198920 (|120596 (119905)|) + 120582119902 (119905) 119870

minus 120582119890 (119905)

sign (119909) |119909|119901minus1

(85)

Since 119890(119905)(sign(119909)|119909|119901minus1) le 0 we have

1199081015840(119905) ge

119901 minus 1

(120582119903 (119905))120574minus1

1198920(|120596 (119905)|) + 120582119870119902 (119905) (86)

We apply Lemma 23 to (82) to obtain a function119862(119905) suchthat

int

119887

119886

119862 (119905) 119889119905 = 1 (87)

and by (75)

0 le 119862 (119905) le119901

2120587sin(120587

119901)min

119901 minus 1

119903120574minus1

(119905) 120582120574minus1

119870120582119902 (119905)

1 +1198720

(88)

Let 1199040isin (minus(120587119901)(1 sin(120587119901)) (120587119901)(1 sin(120587119901))) be such

that tan119901(1199040) le 120596(119886) We define the function

119881 (119905) = 1199040+2120587

119901

1

sin (120587119901)int

119905

119886

119862 (120591) 119889120591 (89)

Note that (minus(120587119901)(1 sin(120587119901))) lt 1199040

= 119881(119886) lt

(120587119901)(1 sin(120587119901)) and 119881(119887) = 1199040+ (2120587119901)(1 sin(120587119901)) gt

(120587119901)(1 sin(120587119901)) Since119881 is continuous there exists a 119879lowast isin(119886 119887) such that 119881(119905) lt (120587119901)(1 sin(120587119901)) for 119905 isin [119886 119879lowast) and119881(119879lowast) = (120587119901)(1 sin(120587119901))

We define

120596 (119905) = tan119901(119881 (119905)) 119905 isin [119886 119879

lowast) (90)

and note that by Lemma 24 lim119905rarr119879

lowast120596(119905) = +infin wheretan119901(119904) = 119910(119904) and 119910(119904) is determined by Lemma 24The derivative of 120596 then satisfies

1205961015840(119905) = tan1015840

119901(119881 (119905)) 119881

1015840(119905)

= (1 + tan119901119901(119881 (119905))) 119881

1015840(119905)

=2120587

119901

1

sin (120587119901)(1 + 120596

119901(119905)) 119862 (119905)

le (1 +1198720+ 1198920(1003816100381610038161003816120596 (119905)

1003816100381610038161003816))min

119901 minus 1

119903120574minus1

(119905) 120582120574minus1

119870120582119902 (119905)

1 +1198720

le119901 minus 1

119903120574minus1

(119905) 120582120574minus1

1198920(1003816100381610038161003816120596 (119905)

1003816100381610038161003816) + 119870120582119902 (119905)

(91)

Thus we may apply the comparison principle ofLemma 15 to 120596(119886) le 120596(119886) and their respective differentialinequalities to conclude that 120596(119905) le 120596(119905) for all 119905 isin [119886 119887]But since 120596 rarr +infin as 119905 rarr 119879

lowastlt 119887 and 120596(119905) is well defined

on the whole segment [119886 119887] we arrive at a contradictionHence 119909(119905) gt 0 cannot hold for all 119905 isin [119886 119887] and since 119909(119905) iscontinuous there exists a 119905lowast isin [119886 119887] such that 119909(119905lowast) = 0

Theorem 26 Let 120574 gt 119901 gt 1 Assume (72) and (73) and let forany 119879 gt 119905

0there exist 119879 le 119886

1lt 1198871le 1198862lt 1198872such that 119892(119905) ge 0

on [1198861 1198871] cup [119886

2 1198872] and 119890(119905) le 0 on [119886

1 1198871] and 119890(119905) ge 0 on

International Journal of Differential Equations 11

[1198862 1198872] If for both 119894 isin 1 2 there exist constants 120582

119894gt 0 such

that

119901

2120587sin (120587119901)int

119887119894

119886119894

min119901 minus 1

119903120574minus1

(119905) 120582120574minus1

119894

119870120582119894119902 (119905)

1 +1198720

119889119905 ge 1

(92)

then (1) is oscillatory

Proof For (1) to be oscillatory it is sufficient that for every119899 isin N there exists a 119905

119899gt 119899 such that 119909(119905

119899) = 0

Let 119899 isin N and let 119899 lt 1198861lt 1198871le 1198862lt 1198872as assumed in the

theorem It is enough to show that every solution 119909(119905) of (1)has a zero on [119886

1 1198872] If 119909(119886

1) ge 0 we apply Proposition 25 to

the segment [1198861 1198871] to obtain the sought zero If 119909(119886

2) le 0 we

again apply Proposition 25 to the segment [1198862 1198872] Suppose

119909(1198861) lt 0 and 119909(119886

2) gt 0 Then since 119909 is continuous there

exists a number 1199051isin [1198861 1198862] such that 119909(119905

1) = 0

Acknowledgment

This work is supported by the Ministry of Science of theRepublic of Croatia under Grant no 036-0361621-1291

References

[1] Q Kong ldquoNonoscillation and oscillation of second order half-linear differential equationsrdquo Journal of Mathematical Analysisand Applications vol 332 no 1 pp 512ndash522 2007

[2] Q-R Wang and Q-G Yang ldquoInterval criteria for oscillationof second-order half-linear differential equationsrdquo Journal ofMathematical Analysis and Applications vol 291 no 1 pp 224ndash236 2004

[3] Q Long and Q-R Wang ldquoNew oscillation criteria of second-order nonlinear differential equationsrdquo Applied Mathematicsand Computation vol 212 no 2 pp 357ndash365 2009

[4] Z Zheng and F Meng ldquoOscillation criteria for forced second-order quasi-linear differential equationsrdquo Mathematical andComputer Modelling vol 45 no 1-2 pp 215ndash220 2007

[5] D Cakmak and A Tiryaki ldquoOscillation criteria for certainforced second-order nonlinear differential equationsrdquo AppliedMathematics Letters vol 17 no 3 pp 275ndash279 2004

[6] F Jiang and F Meng ldquoNew oscillation criteria for a class ofsecond-order nonlinear forced differential equationsrdquo Journalof Mathematical Analysis and Applications vol 336 no 2 pp1476ndash1485 2007

[7] Q Yang ldquoInterval oscillation criteria for a forced secondorder nonlinear ordinary differential equations with oscillatorypotentialrdquo Applied Mathematics and Computation vol 135 no1 pp 49ndash64 2003

[8] A Tiryaki and B Ayanlar ldquoOscillation theorems for certainnonlinear differential equations of second orderrdquo Computers ampMathematics with Applications vol 47 no 1 pp 149ndash159 2004

[9] J Jaros T Kusano and N Yoshida ldquoGeneralized Piconersquosformula and forced oscillations in quasilinear differential equa-tions of the second orderrdquoArchivumMathematicum vol 38 no1 pp 53ndash59 2002

[10] R P Agarwal S R Grace and D OrsquoRegan Oscillation Theoryfor Second Order Linear Half-Linear Superlinear and SublinearDynamic Equations Kluwer Academic Publishers DordrechtThe Netherlands 2002

[11] S Jing ldquoA new oscillation criterion for forced second-orderquasilinear differential equationsrdquoDiscrete Dynamics in Natureand Society vol 2011 Article ID 428976 8 pages 2011

[12] W-T Li and S S Cheng ldquoAn oscillation criterion for nonhomo-geneous half-linear differential equationsrdquoAppliedMathematicsLetters vol 15 no 3 pp 259ndash263 2002

[13] J S W Wong ldquoOscillation criteria for a forced second-orderlinear differential equationrdquo Journal of Mathematical Analysisand Applications vol 231 no 1 pp 235ndash240 1999

[14] M Pasic ldquoFite-Wintner-Leighton type oscillation criteria forsecond-order differential equations with nonlinear dampingrdquoAbstract and Applied Analysis vol 2013 Article ID 852180 10pages 2013

[15] M Pasic ldquoNew interval oscillation criteria for forced second-order differential equations with nonlinear dampingrdquo Interna-tional Journal of Mathematical Analysis vol 7 no 25 pp 1239ndash1255 2013

[16] L Gasinski and N S Papageorgiou Nonsmooth Critical PointTheory and Nonlinear Boundary Value Problems vol 8 of Seriesin Mathematical Analysis and Applications Edited by R PAgarwal and D OrsquoRegan Chapman amp HallCRC Boca RatonFla USA 2005

[17] S Lang Real and Functional Analysis vol 142 of GraduateTexts in Mathematics Springer-Verlag New York NY USA 3rdedition 1993

[18] I N Bronshtein K A Semendyayev G Musiol and HMuehligHandbook of Mathematics Springer Berlin Germany5th edition 2007

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

International Journal of Differential Equations 9

Proof of Corollary 7 It is clear that the condition (23) implies(21) Furthermore since the lengths |119869

119894119895| are uniformly

bounded from below by a positive constant see (23) thenobviously 119886

1119895rarr infin as 119895 rarr infin Thus the claim follows

immediately from Corollary 6

Proof of Corollary 10 Let 119862(119905) = 119902(119905) Then the requiredcondition (19) is clearly satisfied because of assumption(30) Hence this corollary immediately follows from Theo-rem 3

4 An Extension of Condition (12) tothe Case of 120574 gt 1

In this section we consider the oscillation of (1) in the casewhen 120574 isin (1 2) is allowed in the assumption (12) In this waythe assumption (11) is slightly modified by a real number 119901 gt1 such that

|119906|(119901minus1)120574minus119901

VΦ (119906 V) ge 119892 (|Φ (119906 V)|) (72)

where (12) is supposed that is 119892(119888119904) ge 1198881205741198920(119904) for every 119888 119904 gt

0 and for a locally Lipschitz function 1198920 [0infin) rarr [0infin)

for which there exists a constant1198720such that 119892

0(119904)+119872

0ge 119904119901

for all 119904 isin [0infin)For the function 119891 we assume that there exists a constant

119870 gt 0 such that

119891 (119906) sign (119906) ge 119870|119906|119901minus1 for every 119906 isin R (73)

We will need the following two lemmas

Lemma 23 Let 119886 lt 119887 and let 119891 [119886 119887] rarr [0infin) be a meas-urable function Then

int

119887

119886

119891 (119905) 119889119905 ge 1 (74)

if and only if there exists a function 119892 isin 1198711([119886 119887] [0infin))

1198921= 1 such that

119891 (119905) ge 119892 (119905) forall119905 isin [119886 119887] (75)

Proof We assume (74) If int119887119886119891(119905)119889119905 is finite we may choose

119892(119909) = 119891(119909)1198911

For int119887119886119891(119905)119889119905 = infin we may define for every natural

number 119899 a function

119891119899(119905) =

119891 (119905) 119891 (119905) le 119899

119899 119891 (119905) gt 119899(76)

Since 119891119899(119905) le 119899 we have 119891

119899isin 1198711([119886 119887]R) Obviously

lim119899int119887

119886119891119899(119905)119889119905 is also infiniteThis implies that there exists an

1198990such that int119887

1198861198911198990(119905)119889119905 ge 1 We define 119892(119905) = 119891

1198990(119905)119891

11989901

Now we assume (75) Integrating over the whole intervalwe get

int

119887

119886

119891 (119905) 119889119905 ge int

119887

119886

119892 (119905) 119889119905 = 1 (77)

Lemma 24 Let119901 gt 1There exists a unique function 119910 = 119910(119904)that satisfies the following Cauchy problem

1199101015840(119904) = 1 +

1003816100381610038161003816119910 (119904)

1003816100381610038161003816

119901 119904 isin R (78)

119910 (0) = 0 (79)

Furthermore there exists a finite real number 119878lowast such that

lim119904rarrplusmn119878

lowast119910 (119904) = plusmninfin (80)

and 119910 isin 1198621((minus119878lowast 119878lowast)R) is injective and monotonous We

have 119878lowast = (120587119901)(1(sin(120587119901))) and 119910 is odd

We will denote by tan119901the solution of (78)-(79)

Proof Obviously every solution to (78) is injective andmonotonous on a connected set since 1199101015840 = 1 + |119910|119901 ge 1 gt 0

Let 119891(119909) = 1(1 + |119909|119901) Obviously 1 ge 119891(119909) gt 0 for all

119909 isin R and 119862119901gt |1198911015840(119909)| gt 0 for some constants 119862

119901isin R

We may assume 119862119901gt 1 Hence 1199111015840 = 119891(119905) 119911(0) = 0 has (by

[17Theorem 31]) a locally unique solution on (minus1119862119901 1119862119901)

By the same argument 119911 can be extended to the whole of Rinterval by interval of type ((119899 minus 1)119862

119901 (119899 + 1)119862

119901) for all

119899 isin ZWe know from [18] that the for the integral of the left-

hand side we have

int

infin

0

119891 (119909) 119889119909 = int

infin

0

119889119909

1 + 119909119901=120587

119901

1

sin (120587119901)= 119878lowast (81)

so lim119905rarrplusmninfin

119911(119905) = plusmn119878lowast

Since 1199111015840(119905) = 0 119911 is injective monotonous and of class1198621(R (minus119878lowast 119878lowast)) and so we define 119910(119904) = 119911

minus1(119904) It imme-

diately follows that 119910 isin 1198621((minus119878lowast 119878lowast)R) and that 119910 is

injective and monotonous Also by continuity of 119911 we havelim119904rarrplusmn119878

lowast119910(119904) = lim119908rarrplusmninfin

119910(119911(119908)) = lim119908rarrplusmninfin

119908 = plusmninfinDifferentiating it also follows that 1199101015840 = 1 + |119910|

119901 and bydefinition of 119910 119910(0) = 119911minus1(0) = 0 So 119910 is a solution of (78)-(79) Since any other such solution must have 119911 as its inverse119910 is unique on its domain

Proposition 25 Let 119886 lt 119887 and 120574 gt 119901 gt 1 Assume (72) and(73) and let 119902(119905) ge 0 and 119890(119905) le 0 (or 119890(119905) ge 0 ) for all 119905 isin [119886 119887]If there exists a constant 120582 gt 0 such that

119901

2120587sin(120587

119901)int

119887

119886

min119901 minus 1

119903120574minus1

(119905) 120582120574minus1

119870120582119902 (119905)

1 +1198720

119889119905 ge 1 (82)

then for any solution 119909(119905) of (1) with 119909(119886) ge 0 (or 119909(119886) le 0)there exists a number 119905lowast isin [119886 119887] such that 119909(119905lowast) = 0

Proof We assume that 119909(119905) = 0 for all 119905 isin [119886 119887] Then we maydefine the function

120596 (119905) = minus

120582119903 (119905) Φ (119909 (119905) 1199091015840(119905))

sign (119909 (119905)) |119909 (119905)|119901minus1 (83)

10 International Journal of Differential Equations

This function is well defined on [119886 119887] and its derivative is

1199081015840(119905) = minus

120582(119903 (119905) Φ (119909 (119905) 1199091015840(119905)))1015840

sign (119909 (119905)) |119909 (119905)|119901minus1

+

(119901 minus 1) 120582119903 (119905) Φ (119909 (119905) 1199091015840(119905))

|119909 (119905)|119901

1199091015840(119905)

(84)

Using the fact that 119909(119905) is a solution of (1) and omitting 119905 from119909 and 1199091015840 for clarity we get

1199081015840(119905) =

120582 (119902 (119905) 119891 (119909) minus 119890 (119905))

sign (119909) |119909|119901minus1+

(119901 minus 1) 120582119903 (119905) Φ (119909 1199091015840)

119909119901

1199091015840

=

(119901 minus 1) 120582119903 (119905) Φ (119909 1199091015840)

|119909|119901

1199091015840+ 120582119902 (119905)

119891 (119909)

sign (119909) |119909|119901minus1

minus 120582119890 (119905)

sign (119909) |119909|119901minus1

=(119901 minus 1) 120582119903 (119905)

|119909|120574(119901minus1)

Φ(119909 1199091015840) 1199091015840|119909|(119901minus1)120574minus119901

+ 120582119902 (119905)119891 (119909)

sign (119909) |119909|119901minus1minus 120582

119890 (119905)

sign (119909) |119909|119901minus1

(72)

ge(119901 minus 1) 120582119903 (119905)

|119909|120574(119901minus1)

119892 (10038161003816100381610038161003816Φ (119909 119909

1015840)10038161003816100381610038161003816)

+ 120582119902 (119905)119891 (119909)

sign (119909) |119909|119901minus1

minus 120582119890 (119905)

sign (119909) |119909|119901minus1

=(119901 minus 1) 120582119903 (119905)

|119909|120574(119901minus1)

119892(|119909|119901minus1

120582119903 (119905)|120596 (119905)|)

+ 120582119902 (119905)119891 (119909)

sign (119909) |119909|119901minus1minus 120582

119890 (119905)

sign (119909) |119909|119901minus1

(12)

ge(119901 minus 1) 120582119903 (119905)

|119909|120574(119901minus1)

(|119909|119901minus1

120582119903 (119905))

120574

1198920 (|120596 (119905)|)

+ 120582119902 (119905)119891 (119909)

sign (119909) |119909|119901minus1minus 120582

119890 (119905)

sign (119909) |119909|119901minus1

=119901 minus 1

(120582119903 (119905))120574minus1

1198920 (|120596 (119905)|)+120582119902 (119905)

119891 (119909)

sign (119909) |119909|119901minus1

minus 120582119890 (119905)

sign (119909) |119909|119901minus1

(73)

ge119901 minus 1

(120582119903 (119905))120574minus1

1198920 (|120596 (119905)|) + 120582119902 (119905) 119870

minus 120582119890 (119905)

sign (119909) |119909|119901minus1

(85)

Since 119890(119905)(sign(119909)|119909|119901minus1) le 0 we have

1199081015840(119905) ge

119901 minus 1

(120582119903 (119905))120574minus1

1198920(|120596 (119905)|) + 120582119870119902 (119905) (86)

We apply Lemma 23 to (82) to obtain a function119862(119905) suchthat

int

119887

119886

119862 (119905) 119889119905 = 1 (87)

and by (75)

0 le 119862 (119905) le119901

2120587sin(120587

119901)min

119901 minus 1

119903120574minus1

(119905) 120582120574minus1

119870120582119902 (119905)

1 +1198720

(88)

Let 1199040isin (minus(120587119901)(1 sin(120587119901)) (120587119901)(1 sin(120587119901))) be such

that tan119901(1199040) le 120596(119886) We define the function

119881 (119905) = 1199040+2120587

119901

1

sin (120587119901)int

119905

119886

119862 (120591) 119889120591 (89)

Note that (minus(120587119901)(1 sin(120587119901))) lt 1199040

= 119881(119886) lt

(120587119901)(1 sin(120587119901)) and 119881(119887) = 1199040+ (2120587119901)(1 sin(120587119901)) gt

(120587119901)(1 sin(120587119901)) Since119881 is continuous there exists a 119879lowast isin(119886 119887) such that 119881(119905) lt (120587119901)(1 sin(120587119901)) for 119905 isin [119886 119879lowast) and119881(119879lowast) = (120587119901)(1 sin(120587119901))

We define

120596 (119905) = tan119901(119881 (119905)) 119905 isin [119886 119879

lowast) (90)

and note that by Lemma 24 lim119905rarr119879

lowast120596(119905) = +infin wheretan119901(119904) = 119910(119904) and 119910(119904) is determined by Lemma 24The derivative of 120596 then satisfies

1205961015840(119905) = tan1015840

119901(119881 (119905)) 119881

1015840(119905)

= (1 + tan119901119901(119881 (119905))) 119881

1015840(119905)

=2120587

119901

1

sin (120587119901)(1 + 120596

119901(119905)) 119862 (119905)

le (1 +1198720+ 1198920(1003816100381610038161003816120596 (119905)

1003816100381610038161003816))min

119901 minus 1

119903120574minus1

(119905) 120582120574minus1

119870120582119902 (119905)

1 +1198720

le119901 minus 1

119903120574minus1

(119905) 120582120574minus1

1198920(1003816100381610038161003816120596 (119905)

1003816100381610038161003816) + 119870120582119902 (119905)

(91)

Thus we may apply the comparison principle ofLemma 15 to 120596(119886) le 120596(119886) and their respective differentialinequalities to conclude that 120596(119905) le 120596(119905) for all 119905 isin [119886 119887]But since 120596 rarr +infin as 119905 rarr 119879

lowastlt 119887 and 120596(119905) is well defined

on the whole segment [119886 119887] we arrive at a contradictionHence 119909(119905) gt 0 cannot hold for all 119905 isin [119886 119887] and since 119909(119905) iscontinuous there exists a 119905lowast isin [119886 119887] such that 119909(119905lowast) = 0

Theorem 26 Let 120574 gt 119901 gt 1 Assume (72) and (73) and let forany 119879 gt 119905

0there exist 119879 le 119886

1lt 1198871le 1198862lt 1198872such that 119892(119905) ge 0

on [1198861 1198871] cup [119886

2 1198872] and 119890(119905) le 0 on [119886

1 1198871] and 119890(119905) ge 0 on

International Journal of Differential Equations 11

[1198862 1198872] If for both 119894 isin 1 2 there exist constants 120582

119894gt 0 such

that

119901

2120587sin (120587119901)int

119887119894

119886119894

min119901 minus 1

119903120574minus1

(119905) 120582120574minus1

119894

119870120582119894119902 (119905)

1 +1198720

119889119905 ge 1

(92)

then (1) is oscillatory

Proof For (1) to be oscillatory it is sufficient that for every119899 isin N there exists a 119905

119899gt 119899 such that 119909(119905

119899) = 0

Let 119899 isin N and let 119899 lt 1198861lt 1198871le 1198862lt 1198872as assumed in the

theorem It is enough to show that every solution 119909(119905) of (1)has a zero on [119886

1 1198872] If 119909(119886

1) ge 0 we apply Proposition 25 to

the segment [1198861 1198871] to obtain the sought zero If 119909(119886

2) le 0 we

again apply Proposition 25 to the segment [1198862 1198872] Suppose

119909(1198861) lt 0 and 119909(119886

2) gt 0 Then since 119909 is continuous there

exists a number 1199051isin [1198861 1198862] such that 119909(119905

1) = 0

Acknowledgment

This work is supported by the Ministry of Science of theRepublic of Croatia under Grant no 036-0361621-1291

References

[1] Q Kong ldquoNonoscillation and oscillation of second order half-linear differential equationsrdquo Journal of Mathematical Analysisand Applications vol 332 no 1 pp 512ndash522 2007

[2] Q-R Wang and Q-G Yang ldquoInterval criteria for oscillationof second-order half-linear differential equationsrdquo Journal ofMathematical Analysis and Applications vol 291 no 1 pp 224ndash236 2004

[3] Q Long and Q-R Wang ldquoNew oscillation criteria of second-order nonlinear differential equationsrdquo Applied Mathematicsand Computation vol 212 no 2 pp 357ndash365 2009

[4] Z Zheng and F Meng ldquoOscillation criteria for forced second-order quasi-linear differential equationsrdquo Mathematical andComputer Modelling vol 45 no 1-2 pp 215ndash220 2007

[5] D Cakmak and A Tiryaki ldquoOscillation criteria for certainforced second-order nonlinear differential equationsrdquo AppliedMathematics Letters vol 17 no 3 pp 275ndash279 2004

[6] F Jiang and F Meng ldquoNew oscillation criteria for a class ofsecond-order nonlinear forced differential equationsrdquo Journalof Mathematical Analysis and Applications vol 336 no 2 pp1476ndash1485 2007

[7] Q Yang ldquoInterval oscillation criteria for a forced secondorder nonlinear ordinary differential equations with oscillatorypotentialrdquo Applied Mathematics and Computation vol 135 no1 pp 49ndash64 2003

[8] A Tiryaki and B Ayanlar ldquoOscillation theorems for certainnonlinear differential equations of second orderrdquo Computers ampMathematics with Applications vol 47 no 1 pp 149ndash159 2004

[9] J Jaros T Kusano and N Yoshida ldquoGeneralized Piconersquosformula and forced oscillations in quasilinear differential equa-tions of the second orderrdquoArchivumMathematicum vol 38 no1 pp 53ndash59 2002

[10] R P Agarwal S R Grace and D OrsquoRegan Oscillation Theoryfor Second Order Linear Half-Linear Superlinear and SublinearDynamic Equations Kluwer Academic Publishers DordrechtThe Netherlands 2002

[11] S Jing ldquoA new oscillation criterion for forced second-orderquasilinear differential equationsrdquoDiscrete Dynamics in Natureand Society vol 2011 Article ID 428976 8 pages 2011

[12] W-T Li and S S Cheng ldquoAn oscillation criterion for nonhomo-geneous half-linear differential equationsrdquoAppliedMathematicsLetters vol 15 no 3 pp 259ndash263 2002

[13] J S W Wong ldquoOscillation criteria for a forced second-orderlinear differential equationrdquo Journal of Mathematical Analysisand Applications vol 231 no 1 pp 235ndash240 1999

[14] M Pasic ldquoFite-Wintner-Leighton type oscillation criteria forsecond-order differential equations with nonlinear dampingrdquoAbstract and Applied Analysis vol 2013 Article ID 852180 10pages 2013

[15] M Pasic ldquoNew interval oscillation criteria for forced second-order differential equations with nonlinear dampingrdquo Interna-tional Journal of Mathematical Analysis vol 7 no 25 pp 1239ndash1255 2013

[16] L Gasinski and N S Papageorgiou Nonsmooth Critical PointTheory and Nonlinear Boundary Value Problems vol 8 of Seriesin Mathematical Analysis and Applications Edited by R PAgarwal and D OrsquoRegan Chapman amp HallCRC Boca RatonFla USA 2005

[17] S Lang Real and Functional Analysis vol 142 of GraduateTexts in Mathematics Springer-Verlag New York NY USA 3rdedition 1993

[18] I N Bronshtein K A Semendyayev G Musiol and HMuehligHandbook of Mathematics Springer Berlin Germany5th edition 2007

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

10 International Journal of Differential Equations

This function is well defined on [119886 119887] and its derivative is

1199081015840(119905) = minus

120582(119903 (119905) Φ (119909 (119905) 1199091015840(119905)))1015840

sign (119909 (119905)) |119909 (119905)|119901minus1

+

(119901 minus 1) 120582119903 (119905) Φ (119909 (119905) 1199091015840(119905))

|119909 (119905)|119901

1199091015840(119905)

(84)

Using the fact that 119909(119905) is a solution of (1) and omitting 119905 from119909 and 1199091015840 for clarity we get

1199081015840(119905) =

120582 (119902 (119905) 119891 (119909) minus 119890 (119905))

sign (119909) |119909|119901minus1+

(119901 minus 1) 120582119903 (119905) Φ (119909 1199091015840)

119909119901

1199091015840

=

(119901 minus 1) 120582119903 (119905) Φ (119909 1199091015840)

|119909|119901

1199091015840+ 120582119902 (119905)

119891 (119909)

sign (119909) |119909|119901minus1

minus 120582119890 (119905)

sign (119909) |119909|119901minus1

=(119901 minus 1) 120582119903 (119905)

|119909|120574(119901minus1)

Φ(119909 1199091015840) 1199091015840|119909|(119901minus1)120574minus119901

+ 120582119902 (119905)119891 (119909)

sign (119909) |119909|119901minus1minus 120582

119890 (119905)

sign (119909) |119909|119901minus1

(72)

ge(119901 minus 1) 120582119903 (119905)

|119909|120574(119901minus1)

119892 (10038161003816100381610038161003816Φ (119909 119909

1015840)10038161003816100381610038161003816)

+ 120582119902 (119905)119891 (119909)

sign (119909) |119909|119901minus1

minus 120582119890 (119905)

sign (119909) |119909|119901minus1

=(119901 minus 1) 120582119903 (119905)

|119909|120574(119901minus1)

119892(|119909|119901minus1

120582119903 (119905)|120596 (119905)|)

+ 120582119902 (119905)119891 (119909)

sign (119909) |119909|119901minus1minus 120582

119890 (119905)

sign (119909) |119909|119901minus1

(12)

ge(119901 minus 1) 120582119903 (119905)

|119909|120574(119901minus1)

(|119909|119901minus1

120582119903 (119905))

120574

1198920 (|120596 (119905)|)

+ 120582119902 (119905)119891 (119909)

sign (119909) |119909|119901minus1minus 120582

119890 (119905)

sign (119909) |119909|119901minus1

=119901 minus 1

(120582119903 (119905))120574minus1

1198920 (|120596 (119905)|)+120582119902 (119905)

119891 (119909)

sign (119909) |119909|119901minus1

minus 120582119890 (119905)

sign (119909) |119909|119901minus1

(73)

ge119901 minus 1

(120582119903 (119905))120574minus1

1198920 (|120596 (119905)|) + 120582119902 (119905) 119870

minus 120582119890 (119905)

sign (119909) |119909|119901minus1

(85)

Since 119890(119905)(sign(119909)|119909|119901minus1) le 0 we have

1199081015840(119905) ge

119901 minus 1

(120582119903 (119905))120574minus1

1198920(|120596 (119905)|) + 120582119870119902 (119905) (86)

We apply Lemma 23 to (82) to obtain a function119862(119905) suchthat

int

119887

119886

119862 (119905) 119889119905 = 1 (87)

and by (75)

0 le 119862 (119905) le119901

2120587sin(120587

119901)min

119901 minus 1

119903120574minus1

(119905) 120582120574minus1

119870120582119902 (119905)

1 +1198720

(88)

Let 1199040isin (minus(120587119901)(1 sin(120587119901)) (120587119901)(1 sin(120587119901))) be such

that tan119901(1199040) le 120596(119886) We define the function

119881 (119905) = 1199040+2120587

119901

1

sin (120587119901)int

119905

119886

119862 (120591) 119889120591 (89)

Note that (minus(120587119901)(1 sin(120587119901))) lt 1199040

= 119881(119886) lt

(120587119901)(1 sin(120587119901)) and 119881(119887) = 1199040+ (2120587119901)(1 sin(120587119901)) gt

(120587119901)(1 sin(120587119901)) Since119881 is continuous there exists a 119879lowast isin(119886 119887) such that 119881(119905) lt (120587119901)(1 sin(120587119901)) for 119905 isin [119886 119879lowast) and119881(119879lowast) = (120587119901)(1 sin(120587119901))

We define

120596 (119905) = tan119901(119881 (119905)) 119905 isin [119886 119879

lowast) (90)

and note that by Lemma 24 lim119905rarr119879

lowast120596(119905) = +infin wheretan119901(119904) = 119910(119904) and 119910(119904) is determined by Lemma 24The derivative of 120596 then satisfies

1205961015840(119905) = tan1015840

119901(119881 (119905)) 119881

1015840(119905)

= (1 + tan119901119901(119881 (119905))) 119881

1015840(119905)

=2120587

119901

1

sin (120587119901)(1 + 120596

119901(119905)) 119862 (119905)

le (1 +1198720+ 1198920(1003816100381610038161003816120596 (119905)

1003816100381610038161003816))min

119901 minus 1

119903120574minus1

(119905) 120582120574minus1

119870120582119902 (119905)

1 +1198720

le119901 minus 1

119903120574minus1

(119905) 120582120574minus1

1198920(1003816100381610038161003816120596 (119905)

1003816100381610038161003816) + 119870120582119902 (119905)

(91)

Thus we may apply the comparison principle ofLemma 15 to 120596(119886) le 120596(119886) and their respective differentialinequalities to conclude that 120596(119905) le 120596(119905) for all 119905 isin [119886 119887]But since 120596 rarr +infin as 119905 rarr 119879

lowastlt 119887 and 120596(119905) is well defined

on the whole segment [119886 119887] we arrive at a contradictionHence 119909(119905) gt 0 cannot hold for all 119905 isin [119886 119887] and since 119909(119905) iscontinuous there exists a 119905lowast isin [119886 119887] such that 119909(119905lowast) = 0

Theorem 26 Let 120574 gt 119901 gt 1 Assume (72) and (73) and let forany 119879 gt 119905

0there exist 119879 le 119886

1lt 1198871le 1198862lt 1198872such that 119892(119905) ge 0

on [1198861 1198871] cup [119886

2 1198872] and 119890(119905) le 0 on [119886

1 1198871] and 119890(119905) ge 0 on

International Journal of Differential Equations 11

[1198862 1198872] If for both 119894 isin 1 2 there exist constants 120582

119894gt 0 such

that

119901

2120587sin (120587119901)int

119887119894

119886119894

min119901 minus 1

119903120574minus1

(119905) 120582120574minus1

119894

119870120582119894119902 (119905)

1 +1198720

119889119905 ge 1

(92)

then (1) is oscillatory

Proof For (1) to be oscillatory it is sufficient that for every119899 isin N there exists a 119905

119899gt 119899 such that 119909(119905

119899) = 0

Let 119899 isin N and let 119899 lt 1198861lt 1198871le 1198862lt 1198872as assumed in the

theorem It is enough to show that every solution 119909(119905) of (1)has a zero on [119886

1 1198872] If 119909(119886

1) ge 0 we apply Proposition 25 to

the segment [1198861 1198871] to obtain the sought zero If 119909(119886

2) le 0 we

again apply Proposition 25 to the segment [1198862 1198872] Suppose

119909(1198861) lt 0 and 119909(119886

2) gt 0 Then since 119909 is continuous there

exists a number 1199051isin [1198861 1198862] such that 119909(119905

1) = 0

Acknowledgment

This work is supported by the Ministry of Science of theRepublic of Croatia under Grant no 036-0361621-1291

References

[1] Q Kong ldquoNonoscillation and oscillation of second order half-linear differential equationsrdquo Journal of Mathematical Analysisand Applications vol 332 no 1 pp 512ndash522 2007

[2] Q-R Wang and Q-G Yang ldquoInterval criteria for oscillationof second-order half-linear differential equationsrdquo Journal ofMathematical Analysis and Applications vol 291 no 1 pp 224ndash236 2004

[3] Q Long and Q-R Wang ldquoNew oscillation criteria of second-order nonlinear differential equationsrdquo Applied Mathematicsand Computation vol 212 no 2 pp 357ndash365 2009

[4] Z Zheng and F Meng ldquoOscillation criteria for forced second-order quasi-linear differential equationsrdquo Mathematical andComputer Modelling vol 45 no 1-2 pp 215ndash220 2007

[5] D Cakmak and A Tiryaki ldquoOscillation criteria for certainforced second-order nonlinear differential equationsrdquo AppliedMathematics Letters vol 17 no 3 pp 275ndash279 2004

[6] F Jiang and F Meng ldquoNew oscillation criteria for a class ofsecond-order nonlinear forced differential equationsrdquo Journalof Mathematical Analysis and Applications vol 336 no 2 pp1476ndash1485 2007

[7] Q Yang ldquoInterval oscillation criteria for a forced secondorder nonlinear ordinary differential equations with oscillatorypotentialrdquo Applied Mathematics and Computation vol 135 no1 pp 49ndash64 2003

[8] A Tiryaki and B Ayanlar ldquoOscillation theorems for certainnonlinear differential equations of second orderrdquo Computers ampMathematics with Applications vol 47 no 1 pp 149ndash159 2004

[9] J Jaros T Kusano and N Yoshida ldquoGeneralized Piconersquosformula and forced oscillations in quasilinear differential equa-tions of the second orderrdquoArchivumMathematicum vol 38 no1 pp 53ndash59 2002

[10] R P Agarwal S R Grace and D OrsquoRegan Oscillation Theoryfor Second Order Linear Half-Linear Superlinear and SublinearDynamic Equations Kluwer Academic Publishers DordrechtThe Netherlands 2002

[11] S Jing ldquoA new oscillation criterion for forced second-orderquasilinear differential equationsrdquoDiscrete Dynamics in Natureand Society vol 2011 Article ID 428976 8 pages 2011

[12] W-T Li and S S Cheng ldquoAn oscillation criterion for nonhomo-geneous half-linear differential equationsrdquoAppliedMathematicsLetters vol 15 no 3 pp 259ndash263 2002

[13] J S W Wong ldquoOscillation criteria for a forced second-orderlinear differential equationrdquo Journal of Mathematical Analysisand Applications vol 231 no 1 pp 235ndash240 1999

[14] M Pasic ldquoFite-Wintner-Leighton type oscillation criteria forsecond-order differential equations with nonlinear dampingrdquoAbstract and Applied Analysis vol 2013 Article ID 852180 10pages 2013

[15] M Pasic ldquoNew interval oscillation criteria for forced second-order differential equations with nonlinear dampingrdquo Interna-tional Journal of Mathematical Analysis vol 7 no 25 pp 1239ndash1255 2013

[16] L Gasinski and N S Papageorgiou Nonsmooth Critical PointTheory and Nonlinear Boundary Value Problems vol 8 of Seriesin Mathematical Analysis and Applications Edited by R PAgarwal and D OrsquoRegan Chapman amp HallCRC Boca RatonFla USA 2005

[17] S Lang Real and Functional Analysis vol 142 of GraduateTexts in Mathematics Springer-Verlag New York NY USA 3rdedition 1993

[18] I N Bronshtein K A Semendyayev G Musiol and HMuehligHandbook of Mathematics Springer Berlin Germany5th edition 2007

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

International Journal of Differential Equations 11

[1198862 1198872] If for both 119894 isin 1 2 there exist constants 120582

119894gt 0 such

that

119901

2120587sin (120587119901)int

119887119894

119886119894

min119901 minus 1

119903120574minus1

(119905) 120582120574minus1

119894

119870120582119894119902 (119905)

1 +1198720

119889119905 ge 1

(92)

then (1) is oscillatory

Proof For (1) to be oscillatory it is sufficient that for every119899 isin N there exists a 119905

119899gt 119899 such that 119909(119905

119899) = 0

Let 119899 isin N and let 119899 lt 1198861lt 1198871le 1198862lt 1198872as assumed in the

theorem It is enough to show that every solution 119909(119905) of (1)has a zero on [119886

1 1198872] If 119909(119886

1) ge 0 we apply Proposition 25 to

the segment [1198861 1198871] to obtain the sought zero If 119909(119886

2) le 0 we

again apply Proposition 25 to the segment [1198862 1198872] Suppose

119909(1198861) lt 0 and 119909(119886

2) gt 0 Then since 119909 is continuous there

exists a number 1199051isin [1198861 1198862] such that 119909(119905

1) = 0

Acknowledgment

This work is supported by the Ministry of Science of theRepublic of Croatia under Grant no 036-0361621-1291

References

[1] Q Kong ldquoNonoscillation and oscillation of second order half-linear differential equationsrdquo Journal of Mathematical Analysisand Applications vol 332 no 1 pp 512ndash522 2007

[2] Q-R Wang and Q-G Yang ldquoInterval criteria for oscillationof second-order half-linear differential equationsrdquo Journal ofMathematical Analysis and Applications vol 291 no 1 pp 224ndash236 2004

[3] Q Long and Q-R Wang ldquoNew oscillation criteria of second-order nonlinear differential equationsrdquo Applied Mathematicsand Computation vol 212 no 2 pp 357ndash365 2009

[4] Z Zheng and F Meng ldquoOscillation criteria for forced second-order quasi-linear differential equationsrdquo Mathematical andComputer Modelling vol 45 no 1-2 pp 215ndash220 2007

[5] D Cakmak and A Tiryaki ldquoOscillation criteria for certainforced second-order nonlinear differential equationsrdquo AppliedMathematics Letters vol 17 no 3 pp 275ndash279 2004

[6] F Jiang and F Meng ldquoNew oscillation criteria for a class ofsecond-order nonlinear forced differential equationsrdquo Journalof Mathematical Analysis and Applications vol 336 no 2 pp1476ndash1485 2007

[7] Q Yang ldquoInterval oscillation criteria for a forced secondorder nonlinear ordinary differential equations with oscillatorypotentialrdquo Applied Mathematics and Computation vol 135 no1 pp 49ndash64 2003

[8] A Tiryaki and B Ayanlar ldquoOscillation theorems for certainnonlinear differential equations of second orderrdquo Computers ampMathematics with Applications vol 47 no 1 pp 149ndash159 2004

[9] J Jaros T Kusano and N Yoshida ldquoGeneralized Piconersquosformula and forced oscillations in quasilinear differential equa-tions of the second orderrdquoArchivumMathematicum vol 38 no1 pp 53ndash59 2002

[10] R P Agarwal S R Grace and D OrsquoRegan Oscillation Theoryfor Second Order Linear Half-Linear Superlinear and SublinearDynamic Equations Kluwer Academic Publishers DordrechtThe Netherlands 2002

[11] S Jing ldquoA new oscillation criterion for forced second-orderquasilinear differential equationsrdquoDiscrete Dynamics in Natureand Society vol 2011 Article ID 428976 8 pages 2011

[12] W-T Li and S S Cheng ldquoAn oscillation criterion for nonhomo-geneous half-linear differential equationsrdquoAppliedMathematicsLetters vol 15 no 3 pp 259ndash263 2002

[13] J S W Wong ldquoOscillation criteria for a forced second-orderlinear differential equationrdquo Journal of Mathematical Analysisand Applications vol 231 no 1 pp 235ndash240 1999

[14] M Pasic ldquoFite-Wintner-Leighton type oscillation criteria forsecond-order differential equations with nonlinear dampingrdquoAbstract and Applied Analysis vol 2013 Article ID 852180 10pages 2013

[15] M Pasic ldquoNew interval oscillation criteria for forced second-order differential equations with nonlinear dampingrdquo Interna-tional Journal of Mathematical Analysis vol 7 no 25 pp 1239ndash1255 2013

[16] L Gasinski and N S Papageorgiou Nonsmooth Critical PointTheory and Nonlinear Boundary Value Problems vol 8 of Seriesin Mathematical Analysis and Applications Edited by R PAgarwal and D OrsquoRegan Chapman amp HallCRC Boca RatonFla USA 2005

[17] S Lang Real and Functional Analysis vol 142 of GraduateTexts in Mathematics Springer-Verlag New York NY USA 3rdedition 1993

[18] I N Bronshtein K A Semendyayev G Musiol and HMuehligHandbook of Mathematics Springer Berlin Germany5th edition 2007

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