Research ArticleReversed S-Shaped Bifurcation Curve for a Neumann Problem

Hui Xing 1 Hongbin Chen2 and Ruofei Yao2

1Department of Mathematics Xirsquoan Polytechnic University Xirsquoan 710048 China2School of Mathematics and Statistics Xirsquoan Jiaotong University Xirsquoan 710049 China

Correspondence should be addressed to Hui Xing xinghui210163com

Received 24 April 2018 Revised 21 June 2018 Accepted 4 July 2018 Published 1 August 2018

Academic Editor Douglas R Anderson

Copyright copy 2018 Hui Xing et alThis is an open access article distributed under the Creative Commons Attribution License whichpermits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

We study the bifurcation and the exact multiplicity of solutions for a class of Neumann boundary value problem with indefiniteweight We prove that all the solutions obtained form a smooth reversed S-shaped curve by topological degree theory Crandall-Rabinowitz bifurcation theorem and the uniform antimaximum principle in terms of eigenvalues Moreover we obtain that theequation has exactly either one two or three solutions depending on the real parameter The stability is obtained by the eigenvaluecomparison principle

1 Introduction

The existence and multiplicity of solutions of Neumannproblems have been investigated by many authors see forexample [1ndash5] It is well known that determining the exactnumber of solutions of semilinear equations is usually a verydifficult and challenging taskThe results on exactmultiplicityof solutions for Neumann problems are very few in theprevious literature

In this paper we study the bifurcation and the exactmultiplicity of solutions for the Neumann problem

11990610158401015840 + 119886 (119909) 119906 minus 119887 (119909) 1199063 = 120582119891 (119909) 119909 isin (0 1) 1199061015840 (0) = 1199061015840 (1) = 0

(1)

depending on a real parameter 120582 where 119886(119909) 119887(119909) and 119891(119909)are given continuous functions and 119887(119909) gt 0 and 119886(119909) maychange sign

Llibre and Roberto [6] studied the existence and thestability of periodic solutions of Duffing equation

11990910158401015840 + 1198881199091015840 + 119886 (119905) 119909 + 119887 (119905) 1199093 = 120582ℎ (119905) (2)

Chen and Li [7] also studied the Duffing equation (2) in avery particular case ie 119886(119905) equiv 119886 gt 0 119887(119905) = 1 and 119886 119888constantsThey obtained that (2) has exactly three T-periodicsolutions Lomtatidze et al [8] also studied the existence ofperiodic solutions of Duffing type equations

Tzeng et al [9] also studied the global bifurcation andexact multiplicity of positive solutions of Dirichlet problemwith cubic nonlinearity

11990610158401015840 + 120582 (minus1205761199063 + 1205901199062 minus 120581119906 + 120588) = 0 minus 1 lt 119909 lt 1119906 (minus1) = 119906 (1) = 0

(3)

using the time-map method where 120576 120590 120581 120588 are constantsEquation (3) is an autonomous system and the time-mapmethod has been successfully employed to solve the problem(3) but it is not applicable to study the nonautonomoussystem (1) There are many results on exact multiplicity ofsolutions for the Dirichlet problems see [9ndash12]

In [13] under Neumann boundary value conditions thebifurcation of solutions to a logistic equation with harvest-ing has been investigated using the uniform antimaximumprinciple and Crandall-Rabinowitz bifurcation theoremTheuniform antimaximum principle plays an important role inproving the main results More theories and applications ofantimaximum principle can be seen for example [14 15]As continuation of [13] in this paper the bifurcation ofsolutions for a Neumann problem with cubic nonlinearityis investigated using the uniform antimaximum principleCrandall-Rabinowitz bifurcation theorem the topologicaldegree theory and the continuation method More detailedresults on bifurcation theory can be seen in [16 17] Thetopological degree theory and the uniform antimaximum

HindawiDiscrete Dynamics in Nature and SocietyVolume 2018 Article ID 5376075 8 pageshttpsdoiorg10115520185376075

2 Discrete Dynamics in Nature and Society

principle play an important role in proving the main resultsin this paper

2 Preliminaries

Definition 1 (see [18]) We call a solution 119906 of the equation

11990610158401015840 + 119892 (119909 119906) = 119891 (119909) 119909 isin (0 1) 1199061015840 (0) = 1199061015840 (1) = 0

(4)

a stable solution if the principal eigenvalue 1205831(119892119906(119909 119906)) of theequation

11990810158401015840 + 119892119906 (119909 119906) 119908 = minus120583119908 119909 isin (0 1) 1199081015840 (0) = 1199081015840 (1) = 0

(5)

is strictly positive The solution 119906 is unstable if the principaleigenvalue 1205831(119892119906(119909 119906)) is negativeDefinition 2 (see [18]) We call a solution 119906 of the equation

11990610158401015840 + 119892 (119909 119906) = 119891 (119909) 119909 isin (0 1) 1199061015840 (0) = 1199061015840 (1) = 0

(6)

a nondegenerate solution if the linearized equation

11990810158401015840 + 119892119906 (119909 119906) 119908 = 0 119909 isin (0 1) 1199081015840 (0) = 1199081015840 (1) = 0

(7)

does not admit any nontrivial solutions The solution 119906is degenerate (singular) if the linearized equation (7) hasnontrivial solutions

Definition 3 (see [19]) A mapping 119866 119883 997888rarr 119884 (119883 and 119884 aretopological spaces) is said to be proper if for every compactset 119870 sub 119884 the set 119866minus1(119870) is compact in119883

Lemma 4 (see [19]) Let 119883 be a real Banach space and 119871 alinear compact map in 119883 Suppose that 120582 = 0 and 120582minus1 is notan eigenvalue of 119871 LetΩ sub 119883 be an open bounded and 0 isin ΩThen

deg (119868 minus 120582119871Ω 0) = (minus1)119898(120582) (8)

where 119898(120582) is the sum of the algebraic multiplicities of all theeigenvalues 120583 satisfying 120583120582 gt 1 and 119898(120582) = 0 if 119871 has noeigenvalues 120583 of this kind

Lemma 5 Consider the eigenvalue problem

minus11990610158401015840 (119909) + 119906 (119909) = 120583119906 0 lt 119909 lt 11199061015840 (0) = 1199061015840 (1) = 0

(9)

Then (9) has the Green function

119866 (119909 119904) =

cosh (1 minus 119909) cosh 119904sinh 1 0 le 119904 le 119909 le 1

cosh (1 minus 119904) cosh119909sinh 1 0 le 119909 le 119904 le 1

(10)

and

deg (119868 minus 120583119871 119861119877 0)

= 1 for minus infin lt 120583 lt 1(minus1)119899+1 for 11989921205872 + 1 lt 120583 lt (119899 + 1)2 1205872 + 1

(11)

where for any 119877 gt 0 119861119877 = 119906 isin 119862[0 1] 119906 le 119877 denotes aball of radius 119877 and (119871119906)(119909) = int1

0119866(119909 119904)119906(119904)119889119904

Proof It is easy to obtain (10) and the proof is omitted Theeigenvalue problem (9) is equivalent to the equation119906minus120583119871119906 =0 The equation 119906 minus 120583119871119906 = 0 has nontrivial solutions if andonly if 120583minus1 is the eigenvalue of the operator 119871 For 120583 gt 1 thegeneral solution of equation (9) is

119906 (119909) = 119888 cos (radic120583 minus 1119909) + 119889 sin (radic120583 minus 1119909) (12)

By Neumann boundary value condition we have

minus119888radic120583 minus 1 sin (radic120583 minus 1) = 0 (13)

When 119888 = 0 we have 120583 = 11989921205872 + 1 where 119899 ge 0 119899 isin NTherefore 120582119899 = (11989921205872 + 1)minus1 is the eigenvalue of 119871 where119899 ge 0119873(119871 minus 120582119899) is the one-dimensional subspace of Banachspace119883 spanned by 119906119899 = cos(119899120587119909)

For 119899 ge 0 Lemma 4 implies that

deg (119868 minus 120583119871 119861119877 0)

= 1 for minus infin lt 120583 lt 1(minus1)119899+1 for 11989921205872 + 1 lt 120583 lt (119899 + 1)2 1205872 + 1

(14)

This completes the proof

Consider the eigenvalue problem

11990610158401015840 + 119886 (119909) 119906 + 120583119906 = 0 119909 isin (0 1) 1199061015840 (0) = 1199061015840 (1) = 0

(15)

We recall some propositions listed as follows

Proposition 6 It is well known that the problem (15) has theeigenvalues 1205831(119886(119909)) lt 1205832(119886(119909)) le 1205833(119886(119909)) le sdot sdot sdot The firsteigenvalue 1205831(119886(119909)) is real and simple and the correspondingeigenfunction does not change sign When 119886(119909) equiv 0 we denote120583119894(0) by 120583119894 It is obvious that 1205831 = 0 is the first eigenvalue and1205832 = 1205872 is the second eigenvalue of the eigenvalue equation

11990610158401015840 + 120583119906 = 0 119909 isin (0 1) 1199061015840 (0) = 1199061015840 (1) = 0

(16)

Proposition 7 The comparison theorem of eigenvalues canbe stated as follows 120583119894(119886(119909)) is strictly decreasing in the sensethat 1198861(119909) ≪ 1198862(119909) implies that 120583119894(1198861(119909)) gt 120583119894(1198862(119909))where 1198861(119909) ≪ 1198862(119909) namely 1198861(119909) le 1198862(119909) with the strictinequality on a set of positive measure

Discrete Dynamics in Nature and Society 3

Proposition 8 Suppose that 120583119896 ≪ 119886(119909) ≪ 120583119896+1 119896 = 1 2 Then the equation

11990610158401015840 + 119886 (119909) 119906 = 0 119909 isin (0 1) 1199061015840 (0) = 1199061015840 (1) = 0

(17)

does not admit any nontrivial solutions

Proof Argue by contradiction (17) admits nontrivial solu-tions Let 119906(119909) be a nontrivial solution of the equation (17)It follows from 120583119896 ≪ 119886(119909) ≪ 120583119896+1 that 120583119896(119886(119909)) lt 120583119896(120583119896) = 0and 120583119896+1(119886(119909)) gt 120583119896+1(120583119896+1) = 0 for 119896 = 1 2 3 sdot sdot sdot It followsfrom (17) that 0 is the eigenvalue contradiction Therefore(17) admits only trivial solution 0 Equation (17) does notadmit any nontrivial solutionsThis completes the proof

In order to prove our main theorem using the topologicaldegree theory the following lemma is essential In thefollowing lemma the weight function 119886(119909)may change sign

Lemma 9 Suppose that 119886(119909) ≪ 1205872 satisfies the equation119871119886119906 = 11990610158401015840 + 119886 (119909) 119906 = 0 119909 isin (0 1)

1199061015840 (0) = 1199061015840 (1) = 0(18)

Then

deg (119868 minus 119860 119861119877 0) = 1 for 1205831 (119886 (119909)) gt 0minus1 for 1205831 (119886 (119909)) lt 0 (19)

where (119860119906)(119909) = int10119866(119909 119904)(119886(119904) + 1)119906(119904)119889119904 119866(119909 119904) is defined

in (10) for any 119877 gt 0 119861119877 = 119906 isin 119862[0 1] 119906 le 119877 denotes aball of radius 119877Proof Equation (18) is equivalent to the equation

119906 (119909) = int1

0119866 (119909 119904) (119886 (119904) + 1) 119906 (119904) 119889119904 (20)

In (11) let 119899 = 0 and 120583 minus 1 = 120573 we havedeg (119868 minus (120573 + 1) 119871 119861119877 0)

= 1 for minus infin lt 120573 lt 0minus1 for 0 lt 120573 lt 1205872

(21)

Next we calculate deg(119868minus119860 119861119877 0) Let119867(119905 119906) = 119906minus(1minus119905)(120573+1)119871119906minus 119905119860119906 Now we prove that119867(119905 119906) = 0 for forall119905 isin [0 1] and119909 isin 120597119861119877 Argue by contradiction and assume that119867(119905 119906) = 0that is for forall119905 isin [0 1] the equation

11990610158401015840 + (1 minus 119905) 120573119906 + 119905119886 (119909) 119906 = 0 119909 isin (0 1) 1199061015840 (0) = 1199061015840 (1) = 0

(22)

has solution 119906 satisfying 119906 = 119877 on 120597119861119877

Case 1 When 119886(119909) does not change sign(i) For 119886(119909) ≪ 0 choose 120573 lt 0 that is for forall119905 isin [0 1]

((1 minus 119905)120573 + 119905119886(119909)) lt 0 by Proposition 7 we have 1205831((1 minus119905)120573 + 119905119886(119909)) gt 1205831(0) = 0 It follows that (22) has only trivialsolution 119906 equiv 0 Contradiction therefore we have119867(119905 119906) = 0for 119905 isin [0 1] 119909 isin 120597119861119877 Using the homotopy invariance we get

deg (119868 minus 119860 119861119877 0) = deg (119867 (1 sdot) 119861119877 0)= deg (119867 (0 sdot) 119861119877 0)= deg (119868 minus (120573 + 1) 119871 119861119877 0) = 1

(23)

Therefore for 119886(119909) ≪ 0 that is 1205831(119886(119909)) gt 0deg (119868 minus 119860 119861119877 0) = 1 (24)

(ii) For 0 ≪ 119886(119909) ≪ 1205872 choose 120573 isin (0 1205872) that is forforall119905 isin [0 1] 0 ≪ (1 minus 119905)120573 + 119905119886(119909) ≪ 1205872 we have 1205831((1 minus 119905)120573 +119905119886(119909)) lt 1205831(0) = 0 and 1205832((1 minus 119905)120573 + 119905119886(119909)) gt 1205832(1205872) = 0 ByProposition 8 it follows that (22) has only trivial solution 119906 equiv0 contradictionTherefore we have119867(119905 119906) = 0 for 119905 isin [0 1]119909 isin 120597119861119877 Using the homotopy invariance we get

deg (119868 minus 119860 119861119877 0) = deg (119867 (1 sdot) 119861119877 0)= deg (119867 (0 sdot) 119861119877 0)= deg (119868 minus (120573 + 1) 119871 119861119877 0) = minus1

(25)

Therefore for 1205831(119886(119909)) lt 0 119886(119909) ≪ 1205872 we havedeg (119868 minus 119860 119861119877 0) = minus1 (26)

Case 2 When 119886(119909) changes sign let 119886(119909) = 119886+(119909) minus 119886minus(119909)where 119886+(119909) = max119886(119909) 0 119886minus(119909) = maxminus119886(119909) 0

(i) For 1205831(119886(119909)) gt 0 consider the following equation11990610158401015840 + (119905119886+ (119909) minus 119886minus (119909)) 119906 = 0 119909 isin (0 1)

1199061015840 (0) = 1199061015840 (1) = 0(27)

where 119905 isin [0 1] Obviously119905119886+ (119909) minus 119886minus (119909) le 119886+ (119909) minus 119886minus (119909) = 119886 (119909)

forall119905 isin [0 1] (28)

By Proposition 7 and (28) we have

1205831 (119905119886+ (119909) minus 119886minus (119909)) ge 1205831 (119886 (119909)) gt 0 forall119905 isin [0 1] (29)

When 119905 = 0 the solution of (27) is

119906 (119909) = int1

0119866 (119909 119904) (minus119886minus (119904) + 1) 119906 (119904) 119889119904 (30)

Let (1198711119906)(119909) = int10119866(119909 119904)(minus119886minus(119904) + 1)119906(119904)119889119904 since minus119886minus(119904) ≪

0 we havedeg (119868 minus 1198711 119861119877 0) = 1 (31)

4 Discrete Dynamics in Nature and Society

The solution of (27) is

119906 (119909) = int1

0119866 (119909 119904) (119905119886+ (119904) minus 119886minus (119904) + 1) 119906 (119904) 119889119904 (32)

Let ℎ1(119905 119906) = 119906 minus 119905119860119906 minus (1 minus 119905)1198711119906 It is easy to prove thatℎ1(119905 119906) = 0 for forall119905 isin [0 1] 119906 isin 120597119861119877 Therefore using thehomotopy invariance we get

deg (119868 minus 119860 119861119877 0) = deg (ℎ1 (1 sdot) 119861119877 0)= deg (ℎ1 (0 sdot) 119861119877 0)= deg (119868 minus 1198711 119861119877 0) = 1

(33)

(ii) For 1205831(119886(119909)) lt 0 119886(119909) ≪ 1205872 by Proposition 71205832(119886(119909)) gt 0 similarly let

(1198712119906) (119909) = int1

0119866 (119909 119904) (119886+ (119904) + 1) 119906 (119904) 119889119904 (34)

Since 119886+(119904) ≫ 0 we havedeg (119868 minus 1198712 119861119877 0) = minus1 (35)

Let ℎ2(119905 119906) = 119906 minus 119905119860119906 minus (1 minus 119905)1198712119906 It is easy to provethat ℎ2(119905 119906) = 0 for forall119905 isin [0 1] 119906 isin 120597119861119877 Therefore using thehomotopy invariance we get

deg (119868 minus 119860 119861119877 0) = deg (ℎ2 (1 sdot) 119861119877 0)= deg (ℎ2 (0 sdot) 119861119877 0)= deg (119868 minus 1198712 119861119877 0) = minus1

(36)

This completes the proof

Definition 10 (see [20]) Let Ω be open set in Banach space119864 Suppose 119860 Ω 997888rarr 119864 is completely continuous operator119891 = 119868minus119860 Let 1199090 isin Ω be an isolated fixed point of119891 inΩ thatis there exists 119903 gt 0 such that 119861119903 = 119909 119909 minus 1199090 lt 119903 sub Ωand 1199090 is only fixed point of 119891 in 119861119903 Define

ind (119868 minus 119860 1199090) = deg (119868 minus 119860 119861119903 0) (37)

Lemma 11 (see [20]) Let Ω be open set in Banach space 119864Suppose 119860 Ω 997888rarr 119864 is completely continuous operator 119860 hasno fixed point on 120597Ω Suppose that there are finite isolated fixedpoints 1199091 1199092 119909119898 Then

deg (119868 minus 119860Ω 0) =119898

sum119894=1

ind (119868 minus 119860 119909119894) (38)

Lemma 12 (see [20]) Let Ω be open set in Banach space 119864Suppose 119860 Ω 997888rarr 119864 is completely continuous operator 1199090 isinΩ1198601199090 = 1199090 Suppose that119860 is Frechet differentiable at 1199090 and1 is not the eigenvalue of derived operator 1198601015840(1199090) Then 1199090 isan isolated fixed point of 119860 and

ind (119868 minus 119860 1199090) = ind (119868 minus 1198601015840 (1199090) 0) = (minus1)120573 (39)

where 120573 is the sum of the algebraic multiplicities of all theeigenvalues 1198601015840(1199090) in (0 1)

Now we state three known results which plays a key rolein proving the main results in this paper We do not providetheir proof which can be seen in [21]

Lemma 13 Suppose that 119886(119909) 1198861(119909) and 1198862(119909) isin 119862[0 1] suchthat

119886 (119909) 1198861 (119909) 1198862 (119909) ≪ 1205872 (40)

Then

(1) the possible solution 119906 of (18) is either 119906(119909) equiv 0 or119906(119909) = 0 for each 119909 isin [0 1]

(2) 119871119886119894119906 = 0 (119894 = 1 2) cannot both admit nontrivialsolutions if 1198861(119909) ≪ 1198862(119909)

Lemma 14 Let 119891(119909) gt 0 on [0 1] and 119886(119909) satisfy

119886 (119909) ≪ 12058724 (41)

If 119906(119909) is a solution of the nonhomogeneous differentialequation

11990610158401015840 + 119886 (119909) 119906 = 119891 (119909) 119909 isin (0 1) 1199061015840 (0) = 1199061015840 (1) = 0

(42)

then the following statements hold

(1) either 119906(119909) gt 0 or 119906(119909) lt 0 for all 119909 isin [0 1](2) maximum principle 119906(119909) lt 0 if 1205831(119886(119909)) gt 0(3) uniform antimaximum principle 119906(119909) gt 0 if

1205831(119886(119909)) lt 0For the readerrsquos convenience we denote the continuous

function 119892(119909 119906) = 119886(119909)119906 minus 119887(119909)1199063 in (1) The assumptions onfunction 119892(119909 119906) are listed as follows

(f1) 119892(119909 119906) is locally differentiable with respect to thesecond variable with 119892119906(119909 119906) ≪ 1205872

(f2) lim inf119906997888rarrplusmninfin119906(119892(119909 119906) minus 120582119891(119909)) lt 0 uniformly in 119909 isin[0 1]

Lemma 15 Let119883 be an order Banach space and [119908 V] = 119906 isin119883 | 119908 le 119906 le V an order set Assume that the given continuousfunction 119892(119909 119906) satisfies (f1) Then

(1) the solutions of (1) are totally ordered(2) (1) cannot admit three distinct solutions in [119908 V] if

119892119906(119909 119906) is strictly increasing or strictly decreasing in[119908 V]

For the convenience we recast (1) in the operator form

119865 (120582 119906) fl 11990610158401015840 + 119892 (119909 119906) minus 120582119891 (119909) R times 1198622 [0 1] 997888rarr 119862 [0 1]

(43)

Discrete Dynamics in Nature and Society 5

Lemma 16 (see [16]) Let119883 and 119885 be Banach spaces Assumethat 119865 Rtimes119880 997888rarr 119885 is continuously differentiable on119881times119880 subR times 119883 satisfying the following three conditions

(1) 119865(120582lowast 119906lowast) = 0 for some (120582lowast 119906lowast) isin 119881 times119880 dim119873(119863119906119865(120582lowast 119906lowast)) = 1 and the null space119873(119863119906119865(120582lowast 119906lowast)) = span119908

(2) the Fredholm index of 119863119906119865(120582lowast 119906lowast) is zerocodim119877(119863119906119865(120582lowast 119906lowast)) = 1

(3) 119863120582119865(120582lowast 119906lowast)isin119877(119863119906119865(120582lowast 119906lowast))Then there is a continuously differentiable curve through(120582lowast 119906lowast) that is there exists (120582(119904) 119906(119904)) = (120582lowast + 120591(119904) 119906lowast +119904119908 + 119911(119904)) | 119904 isin (minus120575 120575) (120582(0) 119906(0)) = (120582lowast 119906lowast) such that

119865 (120582 (119904) 119906 (119904)) = 0 for 119904 isin (minus120575 120575) (44)

and all solutions of 119865(120582 119906) = 0 in a neighborhood of (120582lowast 119906lowast)belong to the curve 120591(0) = 1205911015840(0) = 119911(0) = 1199111015840(0) = 0

3 Main Results

Lemma 17 Assume (f2) holds Then for every fixed 120582 (1) hasat least a solution and

deg (119868 minus 119860 119861119877 0) = 1 (45)

for 119877 large enough where (119860119906)(119909) = int10119866(119909 119904)(119892(119904 119906(119904)) +

119906(119904) minus 120582119891(119904))119889119904 and 119866 is defined in (10)

Proof It follows from (f2) that there exists119877 gt 0 large enoughsuch that 119892(119909 119877) minus 120582119891(119909) lt 0 and 119892(119909 minus119877) minus 120582119891(119909) gt 0Therefore minus119877 and 119877 are subsolution and supersolution of (1)respectively Therefore there exists at least one solution of (1)between minus119877 and 119877 Next we show that all the solutions of(1) are between minus119877 and 119877 for 119877 gt 0 large enough Argueby contradiction suppose that there exists 1199090 such that 119906(119909)attains its maximum value and 119906(1199090) gt 119877 We have

11990610158401015840 (1199090) + 119886 (1199090) 119906 (1199090) minus 119887 (1199090) 1199063 (1199090) minus 120582119891 (1199090)= 0 (46)

It follows from 1199061015840(1199090) = 0 and 11990610158401015840(1199090) lt 0 that119886 (1199090) 119906 (1199090) minus 119887 (1199090) 1199063 (1199090) minus 120582119891 (1199090) ge 0 (47)

From the above there exists 119877 gt 0 large enough such that119892(119909 119877) minus 120582119891(119909) lt 0 contradiction Similar suppose thatthere exists 1199091 such that 119906(119909) attains its minimum value and119906(1199091) lt minus119877 we will obtain contradiction Therefore thereexists 119877 gt 0 large enough such that 119906 lt 119877 for all thesolutions 119906(119909) (1) is equivalent to the following equation

119906 (119909) = (119860119906) (119909) (48)

Next we calculate deg(119868minus119860 119861119877 0) Let ℎ(119905 119906) = 119906minus119905119860119906 forall119905 isin[0 1] The equation ℎ(119905 119906) = 0 is equivalent to the followingequation

11990610158401015840 + 119905119892 (119909 119906) minus (1 minus 119905) 119906 minus 119905120582119891 (119909) = 0 (49)

It is evident that minus119877 and 119877 are subsolution and supersolutionof (49) respectively Therefore all the solutions of (49) mustbe between minus119877 and 119877 that is the equation ℎ(119905 119906) = 0 has nosolution on 120597119861119877 Therefore ℎ(119905 119906) = 0 for 119905 isin [0 1] 119906 isin 120597119861119877By the homotopy invariance properties of the topologicaldegree we have

deg (119868 minus 119860 119861119877 0) = deg (ℎ (sdot 1) 119861119877 0)= deg (ℎ (sdot 0) 119861119877 0)= deg (119868 119861119877 0) = 1

(50)

This completes the proof

Lemma 18 Suppose that 1205831(119886(119909)) lt 0 119886(119909) ≪ 1205872 and119891(119909) gt 0 for all 119909 isin [0 1] Then (1) has exactly three solutions1199060(119909) 0 V0(119909) for 120582 = 0 where 1199060(119909) is the unique positivestable solution and V0(119909) is the unique negative stable solutionProof For the mapping 119865 defined in (43) let 120582 = 0 and wehave

119865 (119906) = 11990610158401015840 + 119886 (119909) 119906 minus 119887 (119909) 1199063 (51)

Next we prove that any solution 119906(119909) of 119865(119906) = 0 isnondegenerate

First we have the Frechet derivative of (51)

119865119906 (119906) [119908] = 11990810158401015840 + (119886 (119909) minus 3119887 (119909) 1199062)119908 (52)

It is obvious that 119906 = 0 is the solution of 119865(119906) = 0 andtherefore when 119906 = 0 119886(119909)minus3119887(119909)1199062 = 119886(119909) in (52) It followsfrom 1205831(119886(119909)) lt 0 that 0 is nondegenerate

Next we prove that any nontrivial solution 119906(119909) of119865(119906) =0 is nondegenerate 119865(119906) = 0 is equivalent to the followingequation

119906 (119909) = (119860119906) (119909) (53)

where (119860119906)(119909) = int10119866(119909 119904)(119892(119904 119906(119904)) + 119906(119904))119889119904 and 119866 is

defined in (10) The equation

119908 minus 1198601015840 (119906) 119908 = 0 (54)

is equivalent to 119865119906(119906)[119908] = 0 where 1198601015840(119906) denotes thederivative operator of 119860 When 1205831(119886(119909)) lt 0 119886(119909) ≪ 1205872by Lemma 9 Definition 10 and Lemma 12 there exists 119877 gt 0and we have

ind (119868 minus 119860 0) = ind (119868 minus 1198601015840 (0) 0)= deg (119868 minus 119860 119861119877 0) = minus1

(55)

If 119906(119909) is a nontrivial solution of 119865(119906) = 0 it follows fromLemma 13 that 119906(119909) = 0 for all 119909 isin [0 1] Therefore

119892119906 (119909 119906) = 119886 (119909) minus 3119887 (119909) 1199062 (119909) lt 119892 (119909 119906)119906

= 119886 (119909) minus 119887 (119909) 1199062 (119909) ≪ 1205872(56)

6 Discrete Dynamics in Nature and Society

Let 1199021(119909) = 119892(119909 119906)119906 and 1199022(119909) = 119892119906(119909 119906) It follows fromthe above hypothesis that 119906(119909) is a nontrivial solution of1198711199021119906 = 0 By the second conclusion of Lemma 13 we havethat1198711199022119906 = 0does not have nontrivial solutionwhich impliesthat 119906(119909) is a nondegenerate solution

Finally we prove the positive solution is unique andstable We denote by 1199060 the nontrivial solution of (51) It isobvious that minus1199060(119909) is also the nontrivial solution of (51)It follows from (56) and the comparison of eigenvalues that1205831(1199022(119909)) gt 1205831(1199021(119909)) = 0 By Lemma 9 Definition 10 andLemma 12 we have

ind (119868 minus 119860 119906) = ind (119868 minus 1198601015840 (119906) 0)= deg (119868 minus 119860 119861119877 0) = 1

(57)

Let 119898 be the number of nontrivial solutions of 119865(119906) = 0Hence 119865 is proper Since 0 is a regular value of 119865119898 must befinite According to Lemma 17 we have deg(119868 minus 119860 119861119877 0) = 1for sufficiently large 119877 By Lemmas 9 and 11 and the indexformula we have that

1 = deg (119868 minus 119860 119861119877 0)= ind (119868 minus 119860 0) + sum

119906 =0

ind (119868 minus 119860 119906119894) = minus1 + 119898 (58)

It is obvious that 119898 = 2 Let V0(119909) equiv minus1199060(119909) for 119909 isin [0 1]Therefore 119865(119906) = 0 has exactly three solutions 1199060 0 and V0Since 1205831(1199022) gt 1205831(1199021) = 0 we have that the positive solution1199060(119909) is unique and stable The negative solution V0(119909) is alsostable This completes the proof

Lemma 19 Suppose that 119886(119909) ≪ 12058724 For some 120582 lt 0 (1)has a unique positive solution 1199061(119909)with 1199061(119909) gt 1199060(119909) where1199060(119909) is the unique positive solution of (1) for 120582 = 0Proof By Lemma 18 for 120582 = 0 (1) has the unique positivesolution 1199060 which provided a subsolution of (1) for 120582 lt 0There exists 119877 gt 0 large enough such that 119892(119909 119877) minus 120582119891(119909) lt0 Thus 119877 is a supersolution of (1) Therefore we prove thatthere exists a positive solution 1199061(119909) of (1) for 120582 lt 0 such that1199060(119909) lt 1199061(119909) lt 119877

Next we will prove that the positive solution 1199061 is uniqueAssume by contradiction that (1) has another positive solu-tion 119906(119909) for 120582 lt 0 Let V(119909) = 119906(119909) minus 1199061(119909) = 0 then 119908satisfies the following equation

V10158401015840 + 119886 (119909) V minus 119887 (119909) [1199062 (119909) + 119906 (119909) 1199061 (119909) + 11990621 (119909)] V= 0

(59)

Clearly 119886(119909) minus 119887(119909)[1199062(119909) + 119906(119909)1199061(119909) + 11990621(119909)] lt 119886(119909) ≪12058724 By Lemma 13 V(119909) equiv 0 is the solution of (59) which isa contradiction Therefore the solution 1199061 obtained above isonly positive solution For 120582 lt 0 1199061 satisfies (1) and

119906101584010158401 + 119902 (119909) 1199061 = 120582119891 (119909) (60)

where 119902(119909) = 119886(119909) minus 119887(119909)11990621(119909) Again since 119891(119909) gt 0 and120582 lt 0 it follows from Lemma 14 that 1205831(119902(119909)) gt 0 Considerthe linearization associated with (1)

11990810158401015840 + (119886 (119909) minus 3119887 (119909) 1199062 (119909))119908 = 0 (61)

where 119892119906(119909 119906) = 119886(119909) minus 3119887(119909)1199062(119909) lt 119902(119909) Thus1205831(119892119906(119909 119906)) gt 1205831(119902(119909)) gt 0Therefore by Lemma 13 we have1199061(119909) is nondegenerate This completes the proof

Lemma20 For 120582 gt 0 (1) has a unique negative solution V1(119909)with V1(119909) lt V0(119909) where V0(119909) is the unique negative solutionof (1) for 120582 = 0Proof The proof is similar to the proof of Lemma 19

Theorem 21 Assume that the first eigenvalue 1205831(119886(119909)) lt 0and 119891(119909) gt 0 for all 119909 isin [0 1] Suppose that 119886(119909) ≪12058724 Then all the solutions of (1) are of one sign and lie ona unique reversed S-shaped solution curve which is symmetricwith respect to the origin More precisely there exists 120582lowast gt 0such that

(i) For 120582 gt 120582lowast (1) has no positive solution and has aunique negative solution which is stable

(ii) For 120582 = plusmn120582lowast (1) has exactly two solutions Moreoverwhen 120582 = 120582lowast the negative solution is stable and thepositive solution is degenerate When 120582 = minus120582lowast thepositive solution is stable and the negative solution isdegenerate

(iii) For minus120582lowast lt 120582 lt 120582lowast (1) has exactly three orderedsolutions at the same 120582 and the middle solution isunstable and the remaining two are stable Moreoverwhen 120582 lt 0 the maximal solution is positive and theother two are negative When 120582 gt 0 the minimalsolution is negative and the other two are positive

(iv) For 120582 lt minus120582lowast (1) has no negative solution and has aunique positive solution which is stable

Proof It follows from Lemma 19 that (1) has a uniquenondegenerate positive solution 1199061(119909) for some 120582 lt 0 Thesolution curve can be continued a little bit such that 1199061(119909)remains positive for increasing 120582 when 120582 lt 0 By Lemma 18(1) has a unique nondegenerate positive solution 1199060(119909) for120582 = 0The positive solution curve can pass through 1199061(119909) and1199060(119909) and can be continued further for increasing 120582 until thelinearized equation (61) admits the nontrivial solutions Weclaim that the curve of positive solutions cannot be continuedfor 120582 gt 120582lowast Next we will prove the existence of 120582lowast Since1205831(119886(119909)) lt 0 let V(119909) gt 0 be the first eigenfunction of theproblem

V10158401015840 + 119886 (119909) V + 120583V = 0 119909 isin (0 1) V1015840 (0) = V1015840 (1) = 0

(62)

Discrete Dynamics in Nature and Society 7

Multiplying (1) by V(119909) and subtracting from (62) multipliedby 119906 after that integrating over [0 1] we obtain

int1

0[119887 (119909) 1199063 (119909) + 120583119906 (119909)] V (119909) 119889119909

= minus120582int1

0119891 (119909) V (119909) 119889119909

(63)

Applying the mean-value theorem for (63) there exists 120585 isin[0 1] such that

ℎ (119906 (120585) 119887 (120585)) = 119887 (120585) 1199063 (120585) + 1205831119906 (120585)

= minus120582int10119891 (119909) V (119909) 119889119909int10V (119909) 119889119909

(64)

For all 119887(120585) le 1198870 = max119909isin[01]119887(119909) the only root ofℎ(119906(120585) 119887(120585)) = 119898 is negative for 119898 lt ℎ(radicminus120583131198870 1198870) lt 0Let119872 = ℎ(radicminus120583131198870 1198870) and

120582lowast = minus 119872int10V (119909) 119889119909

int10119891 (119909) V (119909) 119889119909 (65)

and it follows from (64) that 119906(120585) lt 0 for 120582 gt 120582lowast Thereforethere exists a 120582lowast gt 0 such that (1) has no positive solution for120582 gt 120582lowast

By applying Lemma 14 we obtain that 119906lowast(119909) is still apositive solution of (1) for 120582 = 120582lowast We denote the degeneratesolution (120582lowast 119906lowast) At (120582lowast 119906lowast) we verify that Lemma 16 canbe applied here It follows from (43) that 119863119906119865(120582lowast 119906lowast)[V] =V10158401015840 + [119886(119909) minus 3119887(119909)1199062lowast]V = 0 In fact 0 is simple and theprincipal eigenvalue of (5) and the first eigenfunction V(119909) gt0 therefore dim119873(119863119906119865(120582lowast 119906lowast)) = codim119877(119863119906119865(120582lowast 119906lowast)) =1 and the null space 119873(119863119906119865(120582lowast 119906lowast)) = spanV Thereforethe Fredholm index of 119863119906119865(120582lowast 119906lowast) is zero Conditions (1)and (2) of Lemma 16 are satisfied Next we verify Condition(3) of Lemma 16 Suppose on the contrary that119863120582119865(120582lowast 119906lowast) =minus119891(119909) isin 119877(119863119906119865(120582lowast 119906lowast)) namely there is a continuousfunction 119908 satisfying

11990810158401015840 + [119886 (119909) minus 3119887 (119909) 1199062lowast]119908 = minus119891 (119909) (66)

Consider the linearized equation of (1) at (120582lowast 119906lowast)V10158401015840 + [119886 (119909) minus 3119887 (119909) 1199062lowast] V = 0 (67)

Since 119865 is singular at (120582lowast 119906lowast) that is (67) has a nontrivialsolution V(119909) such that V(119909) gt 0 Multiplying (66) by Vsubtracting from (67) multiplied by 119908 and integrating byparts on [0 1] we have

0 = int1

0V10158401015840119908 minus 11990810158401015840V 119889119909 = int

1

0V (119909) 119891 (119909) 119889119909 (68)

a contradiction since both V(119909) and 119891(119909) are positiveTherefore Condition (3) of Lemma 16 is satisfied Near(120582lowast 119906lowast) the solutions of

119865 (120582 (119904) 119906 (119904)) = 0 for all 119904 isin [minus120575 120575] (69)

form a curve

(120582 (119904) 119906 (119904)) = (120582lowast + 120591 (119904) 119906lowast + 119904V + 119911 (119904)) | 119904isin [minus120575 120575] (120582 (0) 119906 (0)) = (120582lowast 119906lowast)

(70)

Differentiating (69) twice in 119904 setting 119904 = 0 and 119906119904|119904=0 = V(119909)and evaluating at (120582lowast 119906lowast) we have

11990610158401015840119904119904 + [119886 (119909) minus 3119887 (119909) 1199062lowast] 119906119904119904= 6119887 (119909) 119906lowastV2 + 120591119904119904 (0) 119891 (119909)

(71)

Multiplying the linearized equation (67) by 119906119904119904 subtractingfrom (71) multiplied by V and integrating by parts over [0 1]we obtain

120591119904119904 (0) = minus6 int10119887 (119909) 119906lowastV3119889119909

int10119891 (119909) V (119909) 119889119909 lt 0 (72)

Therefore (120582lowast 119906lowast) is a fold point of 119865(120582 119906) to the left Itfollows from the above that the curve of positive solutionscannot be continued to the right indefinitely for all 120582 gt 0Hence the positive solution curve will make a left turn at(120582lowast 119906lowast) Near the critical point (120582lowast 119906lowast) by the Crandall-Rabinowitz bifurcation theorem there are two branches ofpositive solutions denoted by the upper branch 119906+(119909 120582)and the lower branch 119906minus(119909 120582) with 119906minus(119909 120582) lt 119906+(119909 120582)It follows from Lemmas 14 15 and 18 that the fold point(120582lowast 119906lowast) is unique and the upper branch119906+(119909 120582) ismonotonedecreasing for all 120582 lt 120582lowast and lower branch 119906minus(119909 120582) ismonotone increasing for all 0 lt 120582 lt 120582lowast Therefore the lowerbranch curve ismonotone increasing and continues to the leftwithout any turnings Rewriting (1) in the following form

11990610158401015840 + (119886 (119909) minus 119887 (119909) 1199062) 119906 = 120582119891 (119909) (73)

since 119886(119909) minus 119887(119909)1199062 lt 119886(119909) ≪ 12058724 it follows fromLemma 14 that 119906(119909) gt 0 or 119906(119909) lt 0 This shows that thesolution changes its sign only at (120582 119906) = (0 0) Since thenonlinearity 119892(119909 119906) = 119886(119909)119906 minus 119887(119909)1199063 is an odd function in119906 it follows from the symmetry that if (120582 119906(119909)) is a solutionof (1) so is (minus120582 minus119906(119909)) Thus the component of solutions weconstructed above forms a smooth reversed S-shaped curvewith exactly two turning points (120582lowast 119906lowast) and (minus120582lowast minus119906lowast)

It follows from Lemma 18 that the positive solution 1199060is stable For 120582 lt 120582lowast the upper branch 119906+(119909 120582) remainsstable until it reaches the degenerate solution (120582lowast 119906lowast) Nextwe prove the lower branch 119906minus(119909 120582) is unstable Let 119906119898 beany one solution of lower branch 119906minus(119909 120582) such that 0 lt119906119898 lt 119906lowast Since 119886(119909) minus 3119887(119909)1199062119898 gt 119886(119909) minus 3119887(119909)1199062lowast we have1205831(119886(119909) minus 3119887(119909)1199062119898) lt 1205831(119886(119909) minus 3119887(119909)1199062lowast) = 0 Thereforethe lower branch 119906minus(119909 120582) is unstable Since the solution set issymmetric with respect to the origin the stability of negativesolutions 119906(119909) is obtained by using the property of symmetryThe upper branch of the negative solutions is unstable andthe lower branch of negative solutions is stable Therefore allsolutions of (1) lie on a unique reversed S-shaped solutioncurve This completes the proof

8 Discrete Dynamics in Nature and Society

Data Availability

No data were used to support this study

Conflicts of Interest

The authors declare that there are no conflicts of interestregarding the publication of this paper

Acknowledgments

This work was completed with the support of Tian YuanSpecial Funds of the National Science Foundation of China(no 11626182)

References

[1] J Chu X Lin D Jiang D OrsquoRegan and R P Agarwal ldquoPositivesolutions for second-order superlinear repulsive singular Neu-mann boundary value problemsrdquo Positivity An InternationalMathematics Journal Devoted to Theory and Applications ofPositivity vol 12 no 3 pp 555ndash569 2008

[2] Z Li ldquoExistence of positive solutions of superlinear second-order Neumann boundary value problemrdquo Nonlinear AnalysisTheory Methods amp Applications An International Multidisci-plinary Journal vol 72 no 6 pp 3216ndash3221 2010

[3] G Bonanno and A Sciammetta ldquoExistence and multiplicityresults to Neumann problems for elliptic equations involvingthe p-Laplacianrdquo Journal of Mathematical Analysis and Appli-cations vol 390 no 1 pp 59ndash67 2012

[4] E Sovrano and F Zanolin ldquoIndefinite weight nonlinear prob-lems with Neumann boundary conditionsrdquo Journal of Mathe-matical Analysis and Applications vol 452 no 1 pp 126ndash1472017

[5] G Feltrin and E Sovrano ldquoThree positive solutions to anindefinite Neumann problem a shooting methodrdquo NonlinearAnalysis Theory Methods amp Applications An InternationalMultidisciplinary Journal vol 166 pp 87ndash101 2018

[6] J Llibre and L A Roberto ldquoOn the periodic solutions of aclass of Duffing differential equationsrdquoDiscrete and ContinuousDynamical Systems - Series A vol 33 no 1 pp 277ndash282 2013

[7] H Chen and Y Li ldquoStability and exact multiplicity of periodicsolutions of Duffing equations with cubic nonlinearitiesrdquo Pro-ceedings of the American Mathematical Society vol 135 no 12pp 3925ndash3932 2007

[8] A Lomtatidze and J Sremr ldquoOn periodic solutions to second-order Duffing type equationsrdquo Nonlinear Analysis Real WorldApplications vol 40 pp 215ndash242 2018

[9] C-C Tzeng K-C Hung and S-H Wang ldquoGlobal bifurcationand exact multiplicity of positive solutions for a positone prob-lem with cubic nonlinearityrdquo Journal of Differential Equationsvol 252 no 12 pp 6250ndash6274 2012

[10] S-H Wang and T-S Yeh ldquoA complete classification of bifurca-tion diagrams of a Dirichlet problemwith concave-convex non-linearitiesrdquo Journal of Mathematical Analysis and Applicationsvol 291 no 1 pp 128ndash153 2004

[11] K-C Hung and S-H Wang ldquoA theorem on S-shaped bifur-cation curve for a positone problem with convex-concavenonlinearity and its applications to the perturbed Gelfandproblemrdquo Journal of Differential Equations vol 251 no 2 pp223ndash237 2011

[12] H Pan and R Xing ldquoTime maps and exact multiplicity resultsfor one-dimensional prescribed mean curvature equationsrdquoNonlinear Analysis Theory Methods amp Applications An Inter-national Multidisciplinary Journal vol 74 no 4 pp 1234ndash12602011

[13] H Chen H Xing and X He ldquoBifurcation and stability ofsolutions to a logistic equation with harvestingrdquo MathematicalMethods in the Applied Sciences vol 38 no 8 pp 1623ndash16302015

[14] P Clrsquoement andG Sweers ldquoUniform anti-maximumprinciplesrdquoJournal of Differential Equations vol 164 no 1 pp 118ndash1542000

[15] W Reichel ldquoSharp parameter ranges in the uniform anti-maximum principle for second-order ordinary diffrential oper-atorsrdquo Zeitschrift fur Angewandte Mathematik und Physik vol54 no 5 pp 822ndash838 2003

[16] MG Crandall and PH Rabinowitz ldquoBifurcation perturbationof simple eigenvalues and linearized stabilityrdquo Archive forRational Mechanics and Analysis vol 52 pp 161ndash180 1973

[17] H Kielhofer Bifurcation Theory An Introduction with Applica-tions to PDEs Springer New York NY USA 2003

[18] S Oruganti J Shi and R Shivaji ldquoDiffusive logistic equationwith constant yield harvesting I Steady statesrdquo Transactions ofthe American Mathematical Society vol 354 no 9 pp 3601ndash3619 2002

[19] K Deimling Nonlinear Functional Analysis Springer Berlin Germany 1985

[20] J X Sun Nonlinear Functional Analysis and Its ApplicationsScience Press Beijing China 2007

[21] H Xing H Chen and X He ldquoExact multiplicity and stability ofsolutions of second-order Neumann boundary value problemrdquoApplied Mathematics and Computation vol 232 pp 1104ndash11112014

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Discrete Dynamics in Nature and Society 3

Proposition 8 Suppose that 120583119896 ≪ 119886(119909) ≪ 120583119896+1 119896 = 1 2 Then the equation

11990610158401015840 + 119886 (119909) 119906 = 0 119909 isin (0 1) 1199061015840 (0) = 1199061015840 (1) = 0

(17)

does not admit any nontrivial solutions

Proof Argue by contradiction (17) admits nontrivial solu-tions Let 119906(119909) be a nontrivial solution of the equation (17)It follows from 120583119896 ≪ 119886(119909) ≪ 120583119896+1 that 120583119896(119886(119909)) lt 120583119896(120583119896) = 0and 120583119896+1(119886(119909)) gt 120583119896+1(120583119896+1) = 0 for 119896 = 1 2 3 sdot sdot sdot It followsfrom (17) that 0 is the eigenvalue contradiction Therefore(17) admits only trivial solution 0 Equation (17) does notadmit any nontrivial solutionsThis completes the proof

In order to prove our main theorem using the topologicaldegree theory the following lemma is essential In thefollowing lemma the weight function 119886(119909)may change sign

Lemma 9 Suppose that 119886(119909) ≪ 1205872 satisfies the equation119871119886119906 = 11990610158401015840 + 119886 (119909) 119906 = 0 119909 isin (0 1)

1199061015840 (0) = 1199061015840 (1) = 0(18)

Then

deg (119868 minus 119860 119861119877 0) = 1 for 1205831 (119886 (119909)) gt 0minus1 for 1205831 (119886 (119909)) lt 0 (19)

where (119860119906)(119909) = int10119866(119909 119904)(119886(119904) + 1)119906(119904)119889119904 119866(119909 119904) is defined

in (10) for any 119877 gt 0 119861119877 = 119906 isin 119862[0 1] 119906 le 119877 denotes aball of radius 119877Proof Equation (18) is equivalent to the equation

119906 (119909) = int1

0119866 (119909 119904) (119886 (119904) + 1) 119906 (119904) 119889119904 (20)

In (11) let 119899 = 0 and 120583 minus 1 = 120573 we havedeg (119868 minus (120573 + 1) 119871 119861119877 0)

= 1 for minus infin lt 120573 lt 0minus1 for 0 lt 120573 lt 1205872

(21)

Next we calculate deg(119868minus119860 119861119877 0) Let119867(119905 119906) = 119906minus(1minus119905)(120573+1)119871119906minus 119905119860119906 Now we prove that119867(119905 119906) = 0 for forall119905 isin [0 1] and119909 isin 120597119861119877 Argue by contradiction and assume that119867(119905 119906) = 0that is for forall119905 isin [0 1] the equation

11990610158401015840 + (1 minus 119905) 120573119906 + 119905119886 (119909) 119906 = 0 119909 isin (0 1) 1199061015840 (0) = 1199061015840 (1) = 0

(22)

has solution 119906 satisfying 119906 = 119877 on 120597119861119877

Case 1 When 119886(119909) does not change sign(i) For 119886(119909) ≪ 0 choose 120573 lt 0 that is for forall119905 isin [0 1]

((1 minus 119905)120573 + 119905119886(119909)) lt 0 by Proposition 7 we have 1205831((1 minus119905)120573 + 119905119886(119909)) gt 1205831(0) = 0 It follows that (22) has only trivialsolution 119906 equiv 0 Contradiction therefore we have119867(119905 119906) = 0for 119905 isin [0 1] 119909 isin 120597119861119877 Using the homotopy invariance we get

deg (119868 minus 119860 119861119877 0) = deg (119867 (1 sdot) 119861119877 0)= deg (119867 (0 sdot) 119861119877 0)= deg (119868 minus (120573 + 1) 119871 119861119877 0) = 1

(23)

Therefore for 119886(119909) ≪ 0 that is 1205831(119886(119909)) gt 0deg (119868 minus 119860 119861119877 0) = 1 (24)

(ii) For 0 ≪ 119886(119909) ≪ 1205872 choose 120573 isin (0 1205872) that is forforall119905 isin [0 1] 0 ≪ (1 minus 119905)120573 + 119905119886(119909) ≪ 1205872 we have 1205831((1 minus 119905)120573 +119905119886(119909)) lt 1205831(0) = 0 and 1205832((1 minus 119905)120573 + 119905119886(119909)) gt 1205832(1205872) = 0 ByProposition 8 it follows that (22) has only trivial solution 119906 equiv0 contradictionTherefore we have119867(119905 119906) = 0 for 119905 isin [0 1]119909 isin 120597119861119877 Using the homotopy invariance we get

deg (119868 minus 119860 119861119877 0) = deg (119867 (1 sdot) 119861119877 0)= deg (119867 (0 sdot) 119861119877 0)= deg (119868 minus (120573 + 1) 119871 119861119877 0) = minus1

(25)

Therefore for 1205831(119886(119909)) lt 0 119886(119909) ≪ 1205872 we havedeg (119868 minus 119860 119861119877 0) = minus1 (26)

Case 2 When 119886(119909) changes sign let 119886(119909) = 119886+(119909) minus 119886minus(119909)where 119886+(119909) = max119886(119909) 0 119886minus(119909) = maxminus119886(119909) 0

(i) For 1205831(119886(119909)) gt 0 consider the following equation11990610158401015840 + (119905119886+ (119909) minus 119886minus (119909)) 119906 = 0 119909 isin (0 1)

1199061015840 (0) = 1199061015840 (1) = 0(27)

where 119905 isin [0 1] Obviously119905119886+ (119909) minus 119886minus (119909) le 119886+ (119909) minus 119886minus (119909) = 119886 (119909)

forall119905 isin [0 1] (28)

By Proposition 7 and (28) we have

1205831 (119905119886+ (119909) minus 119886minus (119909)) ge 1205831 (119886 (119909)) gt 0 forall119905 isin [0 1] (29)

When 119905 = 0 the solution of (27) is

119906 (119909) = int1

0119866 (119909 119904) (minus119886minus (119904) + 1) 119906 (119904) 119889119904 (30)

Let (1198711119906)(119909) = int10119866(119909 119904)(minus119886minus(119904) + 1)119906(119904)119889119904 since minus119886minus(119904) ≪

0 we havedeg (119868 minus 1198711 119861119877 0) = 1 (31)

4 Discrete Dynamics in Nature and Society

The solution of (27) is

119906 (119909) = int1

0119866 (119909 119904) (119905119886+ (119904) minus 119886minus (119904) + 1) 119906 (119904) 119889119904 (32)

Let ℎ1(119905 119906) = 119906 minus 119905119860119906 minus (1 minus 119905)1198711119906 It is easy to prove thatℎ1(119905 119906) = 0 for forall119905 isin [0 1] 119906 isin 120597119861119877 Therefore using thehomotopy invariance we get

deg (119868 minus 119860 119861119877 0) = deg (ℎ1 (1 sdot) 119861119877 0)= deg (ℎ1 (0 sdot) 119861119877 0)= deg (119868 minus 1198711 119861119877 0) = 1

(33)

(ii) For 1205831(119886(119909)) lt 0 119886(119909) ≪ 1205872 by Proposition 71205832(119886(119909)) gt 0 similarly let

(1198712119906) (119909) = int1

0119866 (119909 119904) (119886+ (119904) + 1) 119906 (119904) 119889119904 (34)

Since 119886+(119904) ≫ 0 we havedeg (119868 minus 1198712 119861119877 0) = minus1 (35)

Let ℎ2(119905 119906) = 119906 minus 119905119860119906 minus (1 minus 119905)1198712119906 It is easy to provethat ℎ2(119905 119906) = 0 for forall119905 isin [0 1] 119906 isin 120597119861119877 Therefore using thehomotopy invariance we get

deg (119868 minus 119860 119861119877 0) = deg (ℎ2 (1 sdot) 119861119877 0)= deg (ℎ2 (0 sdot) 119861119877 0)= deg (119868 minus 1198712 119861119877 0) = minus1

(36)

This completes the proof

Definition 10 (see [20]) Let Ω be open set in Banach space119864 Suppose 119860 Ω 997888rarr 119864 is completely continuous operator119891 = 119868minus119860 Let 1199090 isin Ω be an isolated fixed point of119891 inΩ thatis there exists 119903 gt 0 such that 119861119903 = 119909 119909 minus 1199090 lt 119903 sub Ωand 1199090 is only fixed point of 119891 in 119861119903 Define

ind (119868 minus 119860 1199090) = deg (119868 minus 119860 119861119903 0) (37)

Lemma 11 (see [20]) Let Ω be open set in Banach space 119864Suppose 119860 Ω 997888rarr 119864 is completely continuous operator 119860 hasno fixed point on 120597Ω Suppose that there are finite isolated fixedpoints 1199091 1199092 119909119898 Then

deg (119868 minus 119860Ω 0) =119898

sum119894=1

ind (119868 minus 119860 119909119894) (38)

Lemma 12 (see [20]) Let Ω be open set in Banach space 119864Suppose 119860 Ω 997888rarr 119864 is completely continuous operator 1199090 isinΩ1198601199090 = 1199090 Suppose that119860 is Frechet differentiable at 1199090 and1 is not the eigenvalue of derived operator 1198601015840(1199090) Then 1199090 isan isolated fixed point of 119860 and

ind (119868 minus 119860 1199090) = ind (119868 minus 1198601015840 (1199090) 0) = (minus1)120573 (39)

where 120573 is the sum of the algebraic multiplicities of all theeigenvalues 1198601015840(1199090) in (0 1)

Now we state three known results which plays a key rolein proving the main results in this paper We do not providetheir proof which can be seen in [21]

Lemma 13 Suppose that 119886(119909) 1198861(119909) and 1198862(119909) isin 119862[0 1] suchthat

119886 (119909) 1198861 (119909) 1198862 (119909) ≪ 1205872 (40)

Then

(1) the possible solution 119906 of (18) is either 119906(119909) equiv 0 or119906(119909) = 0 for each 119909 isin [0 1]

(2) 119871119886119894119906 = 0 (119894 = 1 2) cannot both admit nontrivialsolutions if 1198861(119909) ≪ 1198862(119909)

Lemma 14 Let 119891(119909) gt 0 on [0 1] and 119886(119909) satisfy

119886 (119909) ≪ 12058724 (41)

If 119906(119909) is a solution of the nonhomogeneous differentialequation

11990610158401015840 + 119886 (119909) 119906 = 119891 (119909) 119909 isin (0 1) 1199061015840 (0) = 1199061015840 (1) = 0

(42)

then the following statements hold

(1) either 119906(119909) gt 0 or 119906(119909) lt 0 for all 119909 isin [0 1](2) maximum principle 119906(119909) lt 0 if 1205831(119886(119909)) gt 0(3) uniform antimaximum principle 119906(119909) gt 0 if

1205831(119886(119909)) lt 0For the readerrsquos convenience we denote the continuous

function 119892(119909 119906) = 119886(119909)119906 minus 119887(119909)1199063 in (1) The assumptions onfunction 119892(119909 119906) are listed as follows

(f1) 119892(119909 119906) is locally differentiable with respect to thesecond variable with 119892119906(119909 119906) ≪ 1205872

(f2) lim inf119906997888rarrplusmninfin119906(119892(119909 119906) minus 120582119891(119909)) lt 0 uniformly in 119909 isin[0 1]

Lemma 15 Let119883 be an order Banach space and [119908 V] = 119906 isin119883 | 119908 le 119906 le V an order set Assume that the given continuousfunction 119892(119909 119906) satisfies (f1) Then

(1) the solutions of (1) are totally ordered(2) (1) cannot admit three distinct solutions in [119908 V] if

119892119906(119909 119906) is strictly increasing or strictly decreasing in[119908 V]

For the convenience we recast (1) in the operator form

119865 (120582 119906) fl 11990610158401015840 + 119892 (119909 119906) minus 120582119891 (119909) R times 1198622 [0 1] 997888rarr 119862 [0 1]

(43)

Discrete Dynamics in Nature and Society 5

Lemma 16 (see [16]) Let119883 and 119885 be Banach spaces Assumethat 119865 Rtimes119880 997888rarr 119885 is continuously differentiable on119881times119880 subR times 119883 satisfying the following three conditions

(1) 119865(120582lowast 119906lowast) = 0 for some (120582lowast 119906lowast) isin 119881 times119880 dim119873(119863119906119865(120582lowast 119906lowast)) = 1 and the null space119873(119863119906119865(120582lowast 119906lowast)) = span119908

(2) the Fredholm index of 119863119906119865(120582lowast 119906lowast) is zerocodim119877(119863119906119865(120582lowast 119906lowast)) = 1

(3) 119863120582119865(120582lowast 119906lowast)isin119877(119863119906119865(120582lowast 119906lowast))Then there is a continuously differentiable curve through(120582lowast 119906lowast) that is there exists (120582(119904) 119906(119904)) = (120582lowast + 120591(119904) 119906lowast +119904119908 + 119911(119904)) | 119904 isin (minus120575 120575) (120582(0) 119906(0)) = (120582lowast 119906lowast) such that

119865 (120582 (119904) 119906 (119904)) = 0 for 119904 isin (minus120575 120575) (44)

and all solutions of 119865(120582 119906) = 0 in a neighborhood of (120582lowast 119906lowast)belong to the curve 120591(0) = 1205911015840(0) = 119911(0) = 1199111015840(0) = 0

3 Main Results

Lemma 17 Assume (f2) holds Then for every fixed 120582 (1) hasat least a solution and

deg (119868 minus 119860 119861119877 0) = 1 (45)

for 119877 large enough where (119860119906)(119909) = int10119866(119909 119904)(119892(119904 119906(119904)) +

119906(119904) minus 120582119891(119904))119889119904 and 119866 is defined in (10)

Proof It follows from (f2) that there exists119877 gt 0 large enoughsuch that 119892(119909 119877) minus 120582119891(119909) lt 0 and 119892(119909 minus119877) minus 120582119891(119909) gt 0Therefore minus119877 and 119877 are subsolution and supersolution of (1)respectively Therefore there exists at least one solution of (1)between minus119877 and 119877 Next we show that all the solutions of(1) are between minus119877 and 119877 for 119877 gt 0 large enough Argueby contradiction suppose that there exists 1199090 such that 119906(119909)attains its maximum value and 119906(1199090) gt 119877 We have

11990610158401015840 (1199090) + 119886 (1199090) 119906 (1199090) minus 119887 (1199090) 1199063 (1199090) minus 120582119891 (1199090)= 0 (46)

It follows from 1199061015840(1199090) = 0 and 11990610158401015840(1199090) lt 0 that119886 (1199090) 119906 (1199090) minus 119887 (1199090) 1199063 (1199090) minus 120582119891 (1199090) ge 0 (47)

From the above there exists 119877 gt 0 large enough such that119892(119909 119877) minus 120582119891(119909) lt 0 contradiction Similar suppose thatthere exists 1199091 such that 119906(119909) attains its minimum value and119906(1199091) lt minus119877 we will obtain contradiction Therefore thereexists 119877 gt 0 large enough such that 119906 lt 119877 for all thesolutions 119906(119909) (1) is equivalent to the following equation

119906 (119909) = (119860119906) (119909) (48)

Next we calculate deg(119868minus119860 119861119877 0) Let ℎ(119905 119906) = 119906minus119905119860119906 forall119905 isin[0 1] The equation ℎ(119905 119906) = 0 is equivalent to the followingequation

11990610158401015840 + 119905119892 (119909 119906) minus (1 minus 119905) 119906 minus 119905120582119891 (119909) = 0 (49)

It is evident that minus119877 and 119877 are subsolution and supersolutionof (49) respectively Therefore all the solutions of (49) mustbe between minus119877 and 119877 that is the equation ℎ(119905 119906) = 0 has nosolution on 120597119861119877 Therefore ℎ(119905 119906) = 0 for 119905 isin [0 1] 119906 isin 120597119861119877By the homotopy invariance properties of the topologicaldegree we have

deg (119868 minus 119860 119861119877 0) = deg (ℎ (sdot 1) 119861119877 0)= deg (ℎ (sdot 0) 119861119877 0)= deg (119868 119861119877 0) = 1

(50)

This completes the proof

Lemma 18 Suppose that 1205831(119886(119909)) lt 0 119886(119909) ≪ 1205872 and119891(119909) gt 0 for all 119909 isin [0 1] Then (1) has exactly three solutions1199060(119909) 0 V0(119909) for 120582 = 0 where 1199060(119909) is the unique positivestable solution and V0(119909) is the unique negative stable solutionProof For the mapping 119865 defined in (43) let 120582 = 0 and wehave

119865 (119906) = 11990610158401015840 + 119886 (119909) 119906 minus 119887 (119909) 1199063 (51)

Next we prove that any solution 119906(119909) of 119865(119906) = 0 isnondegenerate

First we have the Frechet derivative of (51)

119865119906 (119906) [119908] = 11990810158401015840 + (119886 (119909) minus 3119887 (119909) 1199062)119908 (52)

It is obvious that 119906 = 0 is the solution of 119865(119906) = 0 andtherefore when 119906 = 0 119886(119909)minus3119887(119909)1199062 = 119886(119909) in (52) It followsfrom 1205831(119886(119909)) lt 0 that 0 is nondegenerate

Next we prove that any nontrivial solution 119906(119909) of119865(119906) =0 is nondegenerate 119865(119906) = 0 is equivalent to the followingequation

119906 (119909) = (119860119906) (119909) (53)

where (119860119906)(119909) = int10119866(119909 119904)(119892(119904 119906(119904)) + 119906(119904))119889119904 and 119866 is

defined in (10) The equation

119908 minus 1198601015840 (119906) 119908 = 0 (54)

is equivalent to 119865119906(119906)[119908] = 0 where 1198601015840(119906) denotes thederivative operator of 119860 When 1205831(119886(119909)) lt 0 119886(119909) ≪ 1205872by Lemma 9 Definition 10 and Lemma 12 there exists 119877 gt 0and we have

ind (119868 minus 119860 0) = ind (119868 minus 1198601015840 (0) 0)= deg (119868 minus 119860 119861119877 0) = minus1

(55)

If 119906(119909) is a nontrivial solution of 119865(119906) = 0 it follows fromLemma 13 that 119906(119909) = 0 for all 119909 isin [0 1] Therefore

119892119906 (119909 119906) = 119886 (119909) minus 3119887 (119909) 1199062 (119909) lt 119892 (119909 119906)119906

= 119886 (119909) minus 119887 (119909) 1199062 (119909) ≪ 1205872(56)

6 Discrete Dynamics in Nature and Society

Let 1199021(119909) = 119892(119909 119906)119906 and 1199022(119909) = 119892119906(119909 119906) It follows fromthe above hypothesis that 119906(119909) is a nontrivial solution of1198711199021119906 = 0 By the second conclusion of Lemma 13 we havethat1198711199022119906 = 0does not have nontrivial solutionwhich impliesthat 119906(119909) is a nondegenerate solution

Finally we prove the positive solution is unique andstable We denote by 1199060 the nontrivial solution of (51) It isobvious that minus1199060(119909) is also the nontrivial solution of (51)It follows from (56) and the comparison of eigenvalues that1205831(1199022(119909)) gt 1205831(1199021(119909)) = 0 By Lemma 9 Definition 10 andLemma 12 we have

ind (119868 minus 119860 119906) = ind (119868 minus 1198601015840 (119906) 0)= deg (119868 minus 119860 119861119877 0) = 1

(57)

Let 119898 be the number of nontrivial solutions of 119865(119906) = 0Hence 119865 is proper Since 0 is a regular value of 119865119898 must befinite According to Lemma 17 we have deg(119868 minus 119860 119861119877 0) = 1for sufficiently large 119877 By Lemmas 9 and 11 and the indexformula we have that

1 = deg (119868 minus 119860 119861119877 0)= ind (119868 minus 119860 0) + sum

119906 =0

ind (119868 minus 119860 119906119894) = minus1 + 119898 (58)

It is obvious that 119898 = 2 Let V0(119909) equiv minus1199060(119909) for 119909 isin [0 1]Therefore 119865(119906) = 0 has exactly three solutions 1199060 0 and V0Since 1205831(1199022) gt 1205831(1199021) = 0 we have that the positive solution1199060(119909) is unique and stable The negative solution V0(119909) is alsostable This completes the proof

Lemma 19 Suppose that 119886(119909) ≪ 12058724 For some 120582 lt 0 (1)has a unique positive solution 1199061(119909)with 1199061(119909) gt 1199060(119909) where1199060(119909) is the unique positive solution of (1) for 120582 = 0Proof By Lemma 18 for 120582 = 0 (1) has the unique positivesolution 1199060 which provided a subsolution of (1) for 120582 lt 0There exists 119877 gt 0 large enough such that 119892(119909 119877) minus 120582119891(119909) lt0 Thus 119877 is a supersolution of (1) Therefore we prove thatthere exists a positive solution 1199061(119909) of (1) for 120582 lt 0 such that1199060(119909) lt 1199061(119909) lt 119877

Next we will prove that the positive solution 1199061 is uniqueAssume by contradiction that (1) has another positive solu-tion 119906(119909) for 120582 lt 0 Let V(119909) = 119906(119909) minus 1199061(119909) = 0 then 119908satisfies the following equation

V10158401015840 + 119886 (119909) V minus 119887 (119909) [1199062 (119909) + 119906 (119909) 1199061 (119909) + 11990621 (119909)] V= 0

(59)

Clearly 119886(119909) minus 119887(119909)[1199062(119909) + 119906(119909)1199061(119909) + 11990621(119909)] lt 119886(119909) ≪12058724 By Lemma 13 V(119909) equiv 0 is the solution of (59) which isa contradiction Therefore the solution 1199061 obtained above isonly positive solution For 120582 lt 0 1199061 satisfies (1) and

119906101584010158401 + 119902 (119909) 1199061 = 120582119891 (119909) (60)

where 119902(119909) = 119886(119909) minus 119887(119909)11990621(119909) Again since 119891(119909) gt 0 and120582 lt 0 it follows from Lemma 14 that 1205831(119902(119909)) gt 0 Considerthe linearization associated with (1)

11990810158401015840 + (119886 (119909) minus 3119887 (119909) 1199062 (119909))119908 = 0 (61)

where 119892119906(119909 119906) = 119886(119909) minus 3119887(119909)1199062(119909) lt 119902(119909) Thus1205831(119892119906(119909 119906)) gt 1205831(119902(119909)) gt 0Therefore by Lemma 13 we have1199061(119909) is nondegenerate This completes the proof

Lemma20 For 120582 gt 0 (1) has a unique negative solution V1(119909)with V1(119909) lt V0(119909) where V0(119909) is the unique negative solutionof (1) for 120582 = 0Proof The proof is similar to the proof of Lemma 19

Theorem 21 Assume that the first eigenvalue 1205831(119886(119909)) lt 0and 119891(119909) gt 0 for all 119909 isin [0 1] Suppose that 119886(119909) ≪12058724 Then all the solutions of (1) are of one sign and lie ona unique reversed S-shaped solution curve which is symmetricwith respect to the origin More precisely there exists 120582lowast gt 0such that

(i) For 120582 gt 120582lowast (1) has no positive solution and has aunique negative solution which is stable

(ii) For 120582 = plusmn120582lowast (1) has exactly two solutions Moreoverwhen 120582 = 120582lowast the negative solution is stable and thepositive solution is degenerate When 120582 = minus120582lowast thepositive solution is stable and the negative solution isdegenerate

(iii) For minus120582lowast lt 120582 lt 120582lowast (1) has exactly three orderedsolutions at the same 120582 and the middle solution isunstable and the remaining two are stable Moreoverwhen 120582 lt 0 the maximal solution is positive and theother two are negative When 120582 gt 0 the minimalsolution is negative and the other two are positive

(iv) For 120582 lt minus120582lowast (1) has no negative solution and has aunique positive solution which is stable

Proof It follows from Lemma 19 that (1) has a uniquenondegenerate positive solution 1199061(119909) for some 120582 lt 0 Thesolution curve can be continued a little bit such that 1199061(119909)remains positive for increasing 120582 when 120582 lt 0 By Lemma 18(1) has a unique nondegenerate positive solution 1199060(119909) for120582 = 0The positive solution curve can pass through 1199061(119909) and1199060(119909) and can be continued further for increasing 120582 until thelinearized equation (61) admits the nontrivial solutions Weclaim that the curve of positive solutions cannot be continuedfor 120582 gt 120582lowast Next we will prove the existence of 120582lowast Since1205831(119886(119909)) lt 0 let V(119909) gt 0 be the first eigenfunction of theproblem

V10158401015840 + 119886 (119909) V + 120583V = 0 119909 isin (0 1) V1015840 (0) = V1015840 (1) = 0

(62)

Discrete Dynamics in Nature and Society 7

Multiplying (1) by V(119909) and subtracting from (62) multipliedby 119906 after that integrating over [0 1] we obtain

int1

0[119887 (119909) 1199063 (119909) + 120583119906 (119909)] V (119909) 119889119909

= minus120582int1

0119891 (119909) V (119909) 119889119909

(63)

Applying the mean-value theorem for (63) there exists 120585 isin[0 1] such that

ℎ (119906 (120585) 119887 (120585)) = 119887 (120585) 1199063 (120585) + 1205831119906 (120585)

= minus120582int10119891 (119909) V (119909) 119889119909int10V (119909) 119889119909

(64)

For all 119887(120585) le 1198870 = max119909isin[01]119887(119909) the only root ofℎ(119906(120585) 119887(120585)) = 119898 is negative for 119898 lt ℎ(radicminus120583131198870 1198870) lt 0Let119872 = ℎ(radicminus120583131198870 1198870) and

120582lowast = minus 119872int10V (119909) 119889119909

int10119891 (119909) V (119909) 119889119909 (65)

and it follows from (64) that 119906(120585) lt 0 for 120582 gt 120582lowast Thereforethere exists a 120582lowast gt 0 such that (1) has no positive solution for120582 gt 120582lowast

By applying Lemma 14 we obtain that 119906lowast(119909) is still apositive solution of (1) for 120582 = 120582lowast We denote the degeneratesolution (120582lowast 119906lowast) At (120582lowast 119906lowast) we verify that Lemma 16 canbe applied here It follows from (43) that 119863119906119865(120582lowast 119906lowast)[V] =V10158401015840 + [119886(119909) minus 3119887(119909)1199062lowast]V = 0 In fact 0 is simple and theprincipal eigenvalue of (5) and the first eigenfunction V(119909) gt0 therefore dim119873(119863119906119865(120582lowast 119906lowast)) = codim119877(119863119906119865(120582lowast 119906lowast)) =1 and the null space 119873(119863119906119865(120582lowast 119906lowast)) = spanV Thereforethe Fredholm index of 119863119906119865(120582lowast 119906lowast) is zero Conditions (1)and (2) of Lemma 16 are satisfied Next we verify Condition(3) of Lemma 16 Suppose on the contrary that119863120582119865(120582lowast 119906lowast) =minus119891(119909) isin 119877(119863119906119865(120582lowast 119906lowast)) namely there is a continuousfunction 119908 satisfying

11990810158401015840 + [119886 (119909) minus 3119887 (119909) 1199062lowast]119908 = minus119891 (119909) (66)

Consider the linearized equation of (1) at (120582lowast 119906lowast)V10158401015840 + [119886 (119909) minus 3119887 (119909) 1199062lowast] V = 0 (67)

Since 119865 is singular at (120582lowast 119906lowast) that is (67) has a nontrivialsolution V(119909) such that V(119909) gt 0 Multiplying (66) by Vsubtracting from (67) multiplied by 119908 and integrating byparts on [0 1] we have

0 = int1

0V10158401015840119908 minus 11990810158401015840V 119889119909 = int

1

0V (119909) 119891 (119909) 119889119909 (68)

a contradiction since both V(119909) and 119891(119909) are positiveTherefore Condition (3) of Lemma 16 is satisfied Near(120582lowast 119906lowast) the solutions of

119865 (120582 (119904) 119906 (119904)) = 0 for all 119904 isin [minus120575 120575] (69)

form a curve

(120582 (119904) 119906 (119904)) = (120582lowast + 120591 (119904) 119906lowast + 119904V + 119911 (119904)) | 119904isin [minus120575 120575] (120582 (0) 119906 (0)) = (120582lowast 119906lowast)

(70)

Differentiating (69) twice in 119904 setting 119904 = 0 and 119906119904|119904=0 = V(119909)and evaluating at (120582lowast 119906lowast) we have

11990610158401015840119904119904 + [119886 (119909) minus 3119887 (119909) 1199062lowast] 119906119904119904= 6119887 (119909) 119906lowastV2 + 120591119904119904 (0) 119891 (119909)

(71)

Multiplying the linearized equation (67) by 119906119904119904 subtractingfrom (71) multiplied by V and integrating by parts over [0 1]we obtain

120591119904119904 (0) = minus6 int10119887 (119909) 119906lowastV3119889119909

int10119891 (119909) V (119909) 119889119909 lt 0 (72)

Therefore (120582lowast 119906lowast) is a fold point of 119865(120582 119906) to the left Itfollows from the above that the curve of positive solutionscannot be continued to the right indefinitely for all 120582 gt 0Hence the positive solution curve will make a left turn at(120582lowast 119906lowast) Near the critical point (120582lowast 119906lowast) by the Crandall-Rabinowitz bifurcation theorem there are two branches ofpositive solutions denoted by the upper branch 119906+(119909 120582)and the lower branch 119906minus(119909 120582) with 119906minus(119909 120582) lt 119906+(119909 120582)It follows from Lemmas 14 15 and 18 that the fold point(120582lowast 119906lowast) is unique and the upper branch119906+(119909 120582) ismonotonedecreasing for all 120582 lt 120582lowast and lower branch 119906minus(119909 120582) ismonotone increasing for all 0 lt 120582 lt 120582lowast Therefore the lowerbranch curve ismonotone increasing and continues to the leftwithout any turnings Rewriting (1) in the following form

11990610158401015840 + (119886 (119909) minus 119887 (119909) 1199062) 119906 = 120582119891 (119909) (73)

since 119886(119909) minus 119887(119909)1199062 lt 119886(119909) ≪ 12058724 it follows fromLemma 14 that 119906(119909) gt 0 or 119906(119909) lt 0 This shows that thesolution changes its sign only at (120582 119906) = (0 0) Since thenonlinearity 119892(119909 119906) = 119886(119909)119906 minus 119887(119909)1199063 is an odd function in119906 it follows from the symmetry that if (120582 119906(119909)) is a solutionof (1) so is (minus120582 minus119906(119909)) Thus the component of solutions weconstructed above forms a smooth reversed S-shaped curvewith exactly two turning points (120582lowast 119906lowast) and (minus120582lowast minus119906lowast)

It follows from Lemma 18 that the positive solution 1199060is stable For 120582 lt 120582lowast the upper branch 119906+(119909 120582) remainsstable until it reaches the degenerate solution (120582lowast 119906lowast) Nextwe prove the lower branch 119906minus(119909 120582) is unstable Let 119906119898 beany one solution of lower branch 119906minus(119909 120582) such that 0 lt119906119898 lt 119906lowast Since 119886(119909) minus 3119887(119909)1199062119898 gt 119886(119909) minus 3119887(119909)1199062lowast we have1205831(119886(119909) minus 3119887(119909)1199062119898) lt 1205831(119886(119909) minus 3119887(119909)1199062lowast) = 0 Thereforethe lower branch 119906minus(119909 120582) is unstable Since the solution set issymmetric with respect to the origin the stability of negativesolutions 119906(119909) is obtained by using the property of symmetryThe upper branch of the negative solutions is unstable andthe lower branch of negative solutions is stable Therefore allsolutions of (1) lie on a unique reversed S-shaped solutioncurve This completes the proof

8 Discrete Dynamics in Nature and Society

Data Availability

No data were used to support this study

Conflicts of Interest

The authors declare that there are no conflicts of interestregarding the publication of this paper

Acknowledgments

This work was completed with the support of Tian YuanSpecial Funds of the National Science Foundation of China(no 11626182)

References

[1] J Chu X Lin D Jiang D OrsquoRegan and R P Agarwal ldquoPositivesolutions for second-order superlinear repulsive singular Neu-mann boundary value problemsrdquo Positivity An InternationalMathematics Journal Devoted to Theory and Applications ofPositivity vol 12 no 3 pp 555ndash569 2008

[2] Z Li ldquoExistence of positive solutions of superlinear second-order Neumann boundary value problemrdquo Nonlinear AnalysisTheory Methods amp Applications An International Multidisci-plinary Journal vol 72 no 6 pp 3216ndash3221 2010

[3] G Bonanno and A Sciammetta ldquoExistence and multiplicityresults to Neumann problems for elliptic equations involvingthe p-Laplacianrdquo Journal of Mathematical Analysis and Appli-cations vol 390 no 1 pp 59ndash67 2012

[4] E Sovrano and F Zanolin ldquoIndefinite weight nonlinear prob-lems with Neumann boundary conditionsrdquo Journal of Mathe-matical Analysis and Applications vol 452 no 1 pp 126ndash1472017

[5] G Feltrin and E Sovrano ldquoThree positive solutions to anindefinite Neumann problem a shooting methodrdquo NonlinearAnalysis Theory Methods amp Applications An InternationalMultidisciplinary Journal vol 166 pp 87ndash101 2018

[6] J Llibre and L A Roberto ldquoOn the periodic solutions of aclass of Duffing differential equationsrdquoDiscrete and ContinuousDynamical Systems - Series A vol 33 no 1 pp 277ndash282 2013

[7] H Chen and Y Li ldquoStability and exact multiplicity of periodicsolutions of Duffing equations with cubic nonlinearitiesrdquo Pro-ceedings of the American Mathematical Society vol 135 no 12pp 3925ndash3932 2007

[8] A Lomtatidze and J Sremr ldquoOn periodic solutions to second-order Duffing type equationsrdquo Nonlinear Analysis Real WorldApplications vol 40 pp 215ndash242 2018

[9] C-C Tzeng K-C Hung and S-H Wang ldquoGlobal bifurcationand exact multiplicity of positive solutions for a positone prob-lem with cubic nonlinearityrdquo Journal of Differential Equationsvol 252 no 12 pp 6250ndash6274 2012

[10] S-H Wang and T-S Yeh ldquoA complete classification of bifurca-tion diagrams of a Dirichlet problemwith concave-convex non-linearitiesrdquo Journal of Mathematical Analysis and Applicationsvol 291 no 1 pp 128ndash153 2004

[11] K-C Hung and S-H Wang ldquoA theorem on S-shaped bifur-cation curve for a positone problem with convex-concavenonlinearity and its applications to the perturbed Gelfandproblemrdquo Journal of Differential Equations vol 251 no 2 pp223ndash237 2011

[12] H Pan and R Xing ldquoTime maps and exact multiplicity resultsfor one-dimensional prescribed mean curvature equationsrdquoNonlinear Analysis Theory Methods amp Applications An Inter-national Multidisciplinary Journal vol 74 no 4 pp 1234ndash12602011

[13] H Chen H Xing and X He ldquoBifurcation and stability ofsolutions to a logistic equation with harvestingrdquo MathematicalMethods in the Applied Sciences vol 38 no 8 pp 1623ndash16302015

[14] P Clrsquoement andG Sweers ldquoUniform anti-maximumprinciplesrdquoJournal of Differential Equations vol 164 no 1 pp 118ndash1542000

[15] W Reichel ldquoSharp parameter ranges in the uniform anti-maximum principle for second-order ordinary diffrential oper-atorsrdquo Zeitschrift fur Angewandte Mathematik und Physik vol54 no 5 pp 822ndash838 2003

[16] MG Crandall and PH Rabinowitz ldquoBifurcation perturbationof simple eigenvalues and linearized stabilityrdquo Archive forRational Mechanics and Analysis vol 52 pp 161ndash180 1973

[17] H Kielhofer Bifurcation Theory An Introduction with Applica-tions to PDEs Springer New York NY USA 2003

[18] S Oruganti J Shi and R Shivaji ldquoDiffusive logistic equationwith constant yield harvesting I Steady statesrdquo Transactions ofthe American Mathematical Society vol 354 no 9 pp 3601ndash3619 2002

[19] K Deimling Nonlinear Functional Analysis Springer Berlin Germany 1985

[20] J X Sun Nonlinear Functional Analysis and Its ApplicationsScience Press Beijing China 2007

[21] H Xing H Chen and X He ldquoExact multiplicity and stability ofsolutions of second-order Neumann boundary value problemrdquoApplied Mathematics and Computation vol 232 pp 1104ndash11112014

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4 Discrete Dynamics in Nature and Society

The solution of (27) is

119906 (119909) = int1

0119866 (119909 119904) (119905119886+ (119904) minus 119886minus (119904) + 1) 119906 (119904) 119889119904 (32)

Let ℎ1(119905 119906) = 119906 minus 119905119860119906 minus (1 minus 119905)1198711119906 It is easy to prove thatℎ1(119905 119906) = 0 for forall119905 isin [0 1] 119906 isin 120597119861119877 Therefore using thehomotopy invariance we get

deg (119868 minus 119860 119861119877 0) = deg (ℎ1 (1 sdot) 119861119877 0)= deg (ℎ1 (0 sdot) 119861119877 0)= deg (119868 minus 1198711 119861119877 0) = 1

(33)

(ii) For 1205831(119886(119909)) lt 0 119886(119909) ≪ 1205872 by Proposition 71205832(119886(119909)) gt 0 similarly let

(1198712119906) (119909) = int1

0119866 (119909 119904) (119886+ (119904) + 1) 119906 (119904) 119889119904 (34)

Since 119886+(119904) ≫ 0 we havedeg (119868 minus 1198712 119861119877 0) = minus1 (35)

Let ℎ2(119905 119906) = 119906 minus 119905119860119906 minus (1 minus 119905)1198712119906 It is easy to provethat ℎ2(119905 119906) = 0 for forall119905 isin [0 1] 119906 isin 120597119861119877 Therefore using thehomotopy invariance we get

deg (119868 minus 119860 119861119877 0) = deg (ℎ2 (1 sdot) 119861119877 0)= deg (ℎ2 (0 sdot) 119861119877 0)= deg (119868 minus 1198712 119861119877 0) = minus1

(36)

This completes the proof

Definition 10 (see [20]) Let Ω be open set in Banach space119864 Suppose 119860 Ω 997888rarr 119864 is completely continuous operator119891 = 119868minus119860 Let 1199090 isin Ω be an isolated fixed point of119891 inΩ thatis there exists 119903 gt 0 such that 119861119903 = 119909 119909 minus 1199090 lt 119903 sub Ωand 1199090 is only fixed point of 119891 in 119861119903 Define

ind (119868 minus 119860 1199090) = deg (119868 minus 119860 119861119903 0) (37)

Lemma 11 (see [20]) Let Ω be open set in Banach space 119864Suppose 119860 Ω 997888rarr 119864 is completely continuous operator 119860 hasno fixed point on 120597Ω Suppose that there are finite isolated fixedpoints 1199091 1199092 119909119898 Then

deg (119868 minus 119860Ω 0) =119898

sum119894=1

ind (119868 minus 119860 119909119894) (38)

Lemma 12 (see [20]) Let Ω be open set in Banach space 119864Suppose 119860 Ω 997888rarr 119864 is completely continuous operator 1199090 isinΩ1198601199090 = 1199090 Suppose that119860 is Frechet differentiable at 1199090 and1 is not the eigenvalue of derived operator 1198601015840(1199090) Then 1199090 isan isolated fixed point of 119860 and

ind (119868 minus 119860 1199090) = ind (119868 minus 1198601015840 (1199090) 0) = (minus1)120573 (39)

where 120573 is the sum of the algebraic multiplicities of all theeigenvalues 1198601015840(1199090) in (0 1)

Now we state three known results which plays a key rolein proving the main results in this paper We do not providetheir proof which can be seen in [21]

Lemma 13 Suppose that 119886(119909) 1198861(119909) and 1198862(119909) isin 119862[0 1] suchthat

119886 (119909) 1198861 (119909) 1198862 (119909) ≪ 1205872 (40)

Then

(1) the possible solution 119906 of (18) is either 119906(119909) equiv 0 or119906(119909) = 0 for each 119909 isin [0 1]

(2) 119871119886119894119906 = 0 (119894 = 1 2) cannot both admit nontrivialsolutions if 1198861(119909) ≪ 1198862(119909)

Lemma 14 Let 119891(119909) gt 0 on [0 1] and 119886(119909) satisfy

119886 (119909) ≪ 12058724 (41)

If 119906(119909) is a solution of the nonhomogeneous differentialequation

11990610158401015840 + 119886 (119909) 119906 = 119891 (119909) 119909 isin (0 1) 1199061015840 (0) = 1199061015840 (1) = 0

(42)

then the following statements hold

(1) either 119906(119909) gt 0 or 119906(119909) lt 0 for all 119909 isin [0 1](2) maximum principle 119906(119909) lt 0 if 1205831(119886(119909)) gt 0(3) uniform antimaximum principle 119906(119909) gt 0 if

1205831(119886(119909)) lt 0For the readerrsquos convenience we denote the continuous

function 119892(119909 119906) = 119886(119909)119906 minus 119887(119909)1199063 in (1) The assumptions onfunction 119892(119909 119906) are listed as follows

(f1) 119892(119909 119906) is locally differentiable with respect to thesecond variable with 119892119906(119909 119906) ≪ 1205872

(f2) lim inf119906997888rarrplusmninfin119906(119892(119909 119906) minus 120582119891(119909)) lt 0 uniformly in 119909 isin[0 1]

Lemma 15 Let119883 be an order Banach space and [119908 V] = 119906 isin119883 | 119908 le 119906 le V an order set Assume that the given continuousfunction 119892(119909 119906) satisfies (f1) Then

(1) the solutions of (1) are totally ordered(2) (1) cannot admit three distinct solutions in [119908 V] if

119892119906(119909 119906) is strictly increasing or strictly decreasing in[119908 V]

For the convenience we recast (1) in the operator form

119865 (120582 119906) fl 11990610158401015840 + 119892 (119909 119906) minus 120582119891 (119909) R times 1198622 [0 1] 997888rarr 119862 [0 1]

(43)

Discrete Dynamics in Nature and Society 5

Lemma 16 (see [16]) Let119883 and 119885 be Banach spaces Assumethat 119865 Rtimes119880 997888rarr 119885 is continuously differentiable on119881times119880 subR times 119883 satisfying the following three conditions

(1) 119865(120582lowast 119906lowast) = 0 for some (120582lowast 119906lowast) isin 119881 times119880 dim119873(119863119906119865(120582lowast 119906lowast)) = 1 and the null space119873(119863119906119865(120582lowast 119906lowast)) = span119908

(2) the Fredholm index of 119863119906119865(120582lowast 119906lowast) is zerocodim119877(119863119906119865(120582lowast 119906lowast)) = 1

(3) 119863120582119865(120582lowast 119906lowast)isin119877(119863119906119865(120582lowast 119906lowast))Then there is a continuously differentiable curve through(120582lowast 119906lowast) that is there exists (120582(119904) 119906(119904)) = (120582lowast + 120591(119904) 119906lowast +119904119908 + 119911(119904)) | 119904 isin (minus120575 120575) (120582(0) 119906(0)) = (120582lowast 119906lowast) such that

119865 (120582 (119904) 119906 (119904)) = 0 for 119904 isin (minus120575 120575) (44)

and all solutions of 119865(120582 119906) = 0 in a neighborhood of (120582lowast 119906lowast)belong to the curve 120591(0) = 1205911015840(0) = 119911(0) = 1199111015840(0) = 0

3 Main Results

Lemma 17 Assume (f2) holds Then for every fixed 120582 (1) hasat least a solution and

deg (119868 minus 119860 119861119877 0) = 1 (45)

for 119877 large enough where (119860119906)(119909) = int10119866(119909 119904)(119892(119904 119906(119904)) +

119906(119904) minus 120582119891(119904))119889119904 and 119866 is defined in (10)

Proof It follows from (f2) that there exists119877 gt 0 large enoughsuch that 119892(119909 119877) minus 120582119891(119909) lt 0 and 119892(119909 minus119877) minus 120582119891(119909) gt 0Therefore minus119877 and 119877 are subsolution and supersolution of (1)respectively Therefore there exists at least one solution of (1)between minus119877 and 119877 Next we show that all the solutions of(1) are between minus119877 and 119877 for 119877 gt 0 large enough Argueby contradiction suppose that there exists 1199090 such that 119906(119909)attains its maximum value and 119906(1199090) gt 119877 We have

11990610158401015840 (1199090) + 119886 (1199090) 119906 (1199090) minus 119887 (1199090) 1199063 (1199090) minus 120582119891 (1199090)= 0 (46)

It follows from 1199061015840(1199090) = 0 and 11990610158401015840(1199090) lt 0 that119886 (1199090) 119906 (1199090) minus 119887 (1199090) 1199063 (1199090) minus 120582119891 (1199090) ge 0 (47)

From the above there exists 119877 gt 0 large enough such that119892(119909 119877) minus 120582119891(119909) lt 0 contradiction Similar suppose thatthere exists 1199091 such that 119906(119909) attains its minimum value and119906(1199091) lt minus119877 we will obtain contradiction Therefore thereexists 119877 gt 0 large enough such that 119906 lt 119877 for all thesolutions 119906(119909) (1) is equivalent to the following equation

119906 (119909) = (119860119906) (119909) (48)

Next we calculate deg(119868minus119860 119861119877 0) Let ℎ(119905 119906) = 119906minus119905119860119906 forall119905 isin[0 1] The equation ℎ(119905 119906) = 0 is equivalent to the followingequation

11990610158401015840 + 119905119892 (119909 119906) minus (1 minus 119905) 119906 minus 119905120582119891 (119909) = 0 (49)

It is evident that minus119877 and 119877 are subsolution and supersolutionof (49) respectively Therefore all the solutions of (49) mustbe between minus119877 and 119877 that is the equation ℎ(119905 119906) = 0 has nosolution on 120597119861119877 Therefore ℎ(119905 119906) = 0 for 119905 isin [0 1] 119906 isin 120597119861119877By the homotopy invariance properties of the topologicaldegree we have

deg (119868 minus 119860 119861119877 0) = deg (ℎ (sdot 1) 119861119877 0)= deg (ℎ (sdot 0) 119861119877 0)= deg (119868 119861119877 0) = 1

(50)

This completes the proof

Lemma 18 Suppose that 1205831(119886(119909)) lt 0 119886(119909) ≪ 1205872 and119891(119909) gt 0 for all 119909 isin [0 1] Then (1) has exactly three solutions1199060(119909) 0 V0(119909) for 120582 = 0 where 1199060(119909) is the unique positivestable solution and V0(119909) is the unique negative stable solutionProof For the mapping 119865 defined in (43) let 120582 = 0 and wehave

119865 (119906) = 11990610158401015840 + 119886 (119909) 119906 minus 119887 (119909) 1199063 (51)

Next we prove that any solution 119906(119909) of 119865(119906) = 0 isnondegenerate

First we have the Frechet derivative of (51)

119865119906 (119906) [119908] = 11990810158401015840 + (119886 (119909) minus 3119887 (119909) 1199062)119908 (52)

It is obvious that 119906 = 0 is the solution of 119865(119906) = 0 andtherefore when 119906 = 0 119886(119909)minus3119887(119909)1199062 = 119886(119909) in (52) It followsfrom 1205831(119886(119909)) lt 0 that 0 is nondegenerate

Next we prove that any nontrivial solution 119906(119909) of119865(119906) =0 is nondegenerate 119865(119906) = 0 is equivalent to the followingequation

119906 (119909) = (119860119906) (119909) (53)

where (119860119906)(119909) = int10119866(119909 119904)(119892(119904 119906(119904)) + 119906(119904))119889119904 and 119866 is

defined in (10) The equation

119908 minus 1198601015840 (119906) 119908 = 0 (54)

is equivalent to 119865119906(119906)[119908] = 0 where 1198601015840(119906) denotes thederivative operator of 119860 When 1205831(119886(119909)) lt 0 119886(119909) ≪ 1205872by Lemma 9 Definition 10 and Lemma 12 there exists 119877 gt 0and we have

ind (119868 minus 119860 0) = ind (119868 minus 1198601015840 (0) 0)= deg (119868 minus 119860 119861119877 0) = minus1

(55)

If 119906(119909) is a nontrivial solution of 119865(119906) = 0 it follows fromLemma 13 that 119906(119909) = 0 for all 119909 isin [0 1] Therefore

119892119906 (119909 119906) = 119886 (119909) minus 3119887 (119909) 1199062 (119909) lt 119892 (119909 119906)119906

= 119886 (119909) minus 119887 (119909) 1199062 (119909) ≪ 1205872(56)

6 Discrete Dynamics in Nature and Society

Let 1199021(119909) = 119892(119909 119906)119906 and 1199022(119909) = 119892119906(119909 119906) It follows fromthe above hypothesis that 119906(119909) is a nontrivial solution of1198711199021119906 = 0 By the second conclusion of Lemma 13 we havethat1198711199022119906 = 0does not have nontrivial solutionwhich impliesthat 119906(119909) is a nondegenerate solution

Finally we prove the positive solution is unique andstable We denote by 1199060 the nontrivial solution of (51) It isobvious that minus1199060(119909) is also the nontrivial solution of (51)It follows from (56) and the comparison of eigenvalues that1205831(1199022(119909)) gt 1205831(1199021(119909)) = 0 By Lemma 9 Definition 10 andLemma 12 we have

ind (119868 minus 119860 119906) = ind (119868 minus 1198601015840 (119906) 0)= deg (119868 minus 119860 119861119877 0) = 1

(57)

Let 119898 be the number of nontrivial solutions of 119865(119906) = 0Hence 119865 is proper Since 0 is a regular value of 119865119898 must befinite According to Lemma 17 we have deg(119868 minus 119860 119861119877 0) = 1for sufficiently large 119877 By Lemmas 9 and 11 and the indexformula we have that

1 = deg (119868 minus 119860 119861119877 0)= ind (119868 minus 119860 0) + sum

119906 =0

ind (119868 minus 119860 119906119894) = minus1 + 119898 (58)

It is obvious that 119898 = 2 Let V0(119909) equiv minus1199060(119909) for 119909 isin [0 1]Therefore 119865(119906) = 0 has exactly three solutions 1199060 0 and V0Since 1205831(1199022) gt 1205831(1199021) = 0 we have that the positive solution1199060(119909) is unique and stable The negative solution V0(119909) is alsostable This completes the proof

Lemma 19 Suppose that 119886(119909) ≪ 12058724 For some 120582 lt 0 (1)has a unique positive solution 1199061(119909)with 1199061(119909) gt 1199060(119909) where1199060(119909) is the unique positive solution of (1) for 120582 = 0Proof By Lemma 18 for 120582 = 0 (1) has the unique positivesolution 1199060 which provided a subsolution of (1) for 120582 lt 0There exists 119877 gt 0 large enough such that 119892(119909 119877) minus 120582119891(119909) lt0 Thus 119877 is a supersolution of (1) Therefore we prove thatthere exists a positive solution 1199061(119909) of (1) for 120582 lt 0 such that1199060(119909) lt 1199061(119909) lt 119877

Next we will prove that the positive solution 1199061 is uniqueAssume by contradiction that (1) has another positive solu-tion 119906(119909) for 120582 lt 0 Let V(119909) = 119906(119909) minus 1199061(119909) = 0 then 119908satisfies the following equation

V10158401015840 + 119886 (119909) V minus 119887 (119909) [1199062 (119909) + 119906 (119909) 1199061 (119909) + 11990621 (119909)] V= 0

(59)

Clearly 119886(119909) minus 119887(119909)[1199062(119909) + 119906(119909)1199061(119909) + 11990621(119909)] lt 119886(119909) ≪12058724 By Lemma 13 V(119909) equiv 0 is the solution of (59) which isa contradiction Therefore the solution 1199061 obtained above isonly positive solution For 120582 lt 0 1199061 satisfies (1) and

119906101584010158401 + 119902 (119909) 1199061 = 120582119891 (119909) (60)

where 119902(119909) = 119886(119909) minus 119887(119909)11990621(119909) Again since 119891(119909) gt 0 and120582 lt 0 it follows from Lemma 14 that 1205831(119902(119909)) gt 0 Considerthe linearization associated with (1)

11990810158401015840 + (119886 (119909) minus 3119887 (119909) 1199062 (119909))119908 = 0 (61)

where 119892119906(119909 119906) = 119886(119909) minus 3119887(119909)1199062(119909) lt 119902(119909) Thus1205831(119892119906(119909 119906)) gt 1205831(119902(119909)) gt 0Therefore by Lemma 13 we have1199061(119909) is nondegenerate This completes the proof

Lemma20 For 120582 gt 0 (1) has a unique negative solution V1(119909)with V1(119909) lt V0(119909) where V0(119909) is the unique negative solutionof (1) for 120582 = 0Proof The proof is similar to the proof of Lemma 19

Theorem 21 Assume that the first eigenvalue 1205831(119886(119909)) lt 0and 119891(119909) gt 0 for all 119909 isin [0 1] Suppose that 119886(119909) ≪12058724 Then all the solutions of (1) are of one sign and lie ona unique reversed S-shaped solution curve which is symmetricwith respect to the origin More precisely there exists 120582lowast gt 0such that

(i) For 120582 gt 120582lowast (1) has no positive solution and has aunique negative solution which is stable

(ii) For 120582 = plusmn120582lowast (1) has exactly two solutions Moreoverwhen 120582 = 120582lowast the negative solution is stable and thepositive solution is degenerate When 120582 = minus120582lowast thepositive solution is stable and the negative solution isdegenerate

(iii) For minus120582lowast lt 120582 lt 120582lowast (1) has exactly three orderedsolutions at the same 120582 and the middle solution isunstable and the remaining two are stable Moreoverwhen 120582 lt 0 the maximal solution is positive and theother two are negative When 120582 gt 0 the minimalsolution is negative and the other two are positive

(iv) For 120582 lt minus120582lowast (1) has no negative solution and has aunique positive solution which is stable

Proof It follows from Lemma 19 that (1) has a uniquenondegenerate positive solution 1199061(119909) for some 120582 lt 0 Thesolution curve can be continued a little bit such that 1199061(119909)remains positive for increasing 120582 when 120582 lt 0 By Lemma 18(1) has a unique nondegenerate positive solution 1199060(119909) for120582 = 0The positive solution curve can pass through 1199061(119909) and1199060(119909) and can be continued further for increasing 120582 until thelinearized equation (61) admits the nontrivial solutions Weclaim that the curve of positive solutions cannot be continuedfor 120582 gt 120582lowast Next we will prove the existence of 120582lowast Since1205831(119886(119909)) lt 0 let V(119909) gt 0 be the first eigenfunction of theproblem

V10158401015840 + 119886 (119909) V + 120583V = 0 119909 isin (0 1) V1015840 (0) = V1015840 (1) = 0

(62)

Discrete Dynamics in Nature and Society 7

Multiplying (1) by V(119909) and subtracting from (62) multipliedby 119906 after that integrating over [0 1] we obtain

int1

0[119887 (119909) 1199063 (119909) + 120583119906 (119909)] V (119909) 119889119909

= minus120582int1

0119891 (119909) V (119909) 119889119909

(63)

Applying the mean-value theorem for (63) there exists 120585 isin[0 1] such that

ℎ (119906 (120585) 119887 (120585)) = 119887 (120585) 1199063 (120585) + 1205831119906 (120585)

= minus120582int10119891 (119909) V (119909) 119889119909int10V (119909) 119889119909

(64)

For all 119887(120585) le 1198870 = max119909isin[01]119887(119909) the only root ofℎ(119906(120585) 119887(120585)) = 119898 is negative for 119898 lt ℎ(radicminus120583131198870 1198870) lt 0Let119872 = ℎ(radicminus120583131198870 1198870) and

120582lowast = minus 119872int10V (119909) 119889119909

int10119891 (119909) V (119909) 119889119909 (65)

and it follows from (64) that 119906(120585) lt 0 for 120582 gt 120582lowast Thereforethere exists a 120582lowast gt 0 such that (1) has no positive solution for120582 gt 120582lowast

By applying Lemma 14 we obtain that 119906lowast(119909) is still apositive solution of (1) for 120582 = 120582lowast We denote the degeneratesolution (120582lowast 119906lowast) At (120582lowast 119906lowast) we verify that Lemma 16 canbe applied here It follows from (43) that 119863119906119865(120582lowast 119906lowast)[V] =V10158401015840 + [119886(119909) minus 3119887(119909)1199062lowast]V = 0 In fact 0 is simple and theprincipal eigenvalue of (5) and the first eigenfunction V(119909) gt0 therefore dim119873(119863119906119865(120582lowast 119906lowast)) = codim119877(119863119906119865(120582lowast 119906lowast)) =1 and the null space 119873(119863119906119865(120582lowast 119906lowast)) = spanV Thereforethe Fredholm index of 119863119906119865(120582lowast 119906lowast) is zero Conditions (1)and (2) of Lemma 16 are satisfied Next we verify Condition(3) of Lemma 16 Suppose on the contrary that119863120582119865(120582lowast 119906lowast) =minus119891(119909) isin 119877(119863119906119865(120582lowast 119906lowast)) namely there is a continuousfunction 119908 satisfying

11990810158401015840 + [119886 (119909) minus 3119887 (119909) 1199062lowast]119908 = minus119891 (119909) (66)

Consider the linearized equation of (1) at (120582lowast 119906lowast)V10158401015840 + [119886 (119909) minus 3119887 (119909) 1199062lowast] V = 0 (67)

Since 119865 is singular at (120582lowast 119906lowast) that is (67) has a nontrivialsolution V(119909) such that V(119909) gt 0 Multiplying (66) by Vsubtracting from (67) multiplied by 119908 and integrating byparts on [0 1] we have

0 = int1

0V10158401015840119908 minus 11990810158401015840V 119889119909 = int

1

0V (119909) 119891 (119909) 119889119909 (68)

a contradiction since both V(119909) and 119891(119909) are positiveTherefore Condition (3) of Lemma 16 is satisfied Near(120582lowast 119906lowast) the solutions of

119865 (120582 (119904) 119906 (119904)) = 0 for all 119904 isin [minus120575 120575] (69)

form a curve

(120582 (119904) 119906 (119904)) = (120582lowast + 120591 (119904) 119906lowast + 119904V + 119911 (119904)) | 119904isin [minus120575 120575] (120582 (0) 119906 (0)) = (120582lowast 119906lowast)

(70)

Differentiating (69) twice in 119904 setting 119904 = 0 and 119906119904|119904=0 = V(119909)and evaluating at (120582lowast 119906lowast) we have

11990610158401015840119904119904 + [119886 (119909) minus 3119887 (119909) 1199062lowast] 119906119904119904= 6119887 (119909) 119906lowastV2 + 120591119904119904 (0) 119891 (119909)

(71)

Multiplying the linearized equation (67) by 119906119904119904 subtractingfrom (71) multiplied by V and integrating by parts over [0 1]we obtain

120591119904119904 (0) = minus6 int10119887 (119909) 119906lowastV3119889119909

int10119891 (119909) V (119909) 119889119909 lt 0 (72)

Therefore (120582lowast 119906lowast) is a fold point of 119865(120582 119906) to the left Itfollows from the above that the curve of positive solutionscannot be continued to the right indefinitely for all 120582 gt 0Hence the positive solution curve will make a left turn at(120582lowast 119906lowast) Near the critical point (120582lowast 119906lowast) by the Crandall-Rabinowitz bifurcation theorem there are two branches ofpositive solutions denoted by the upper branch 119906+(119909 120582)and the lower branch 119906minus(119909 120582) with 119906minus(119909 120582) lt 119906+(119909 120582)It follows from Lemmas 14 15 and 18 that the fold point(120582lowast 119906lowast) is unique and the upper branch119906+(119909 120582) ismonotonedecreasing for all 120582 lt 120582lowast and lower branch 119906minus(119909 120582) ismonotone increasing for all 0 lt 120582 lt 120582lowast Therefore the lowerbranch curve ismonotone increasing and continues to the leftwithout any turnings Rewriting (1) in the following form

11990610158401015840 + (119886 (119909) minus 119887 (119909) 1199062) 119906 = 120582119891 (119909) (73)

since 119886(119909) minus 119887(119909)1199062 lt 119886(119909) ≪ 12058724 it follows fromLemma 14 that 119906(119909) gt 0 or 119906(119909) lt 0 This shows that thesolution changes its sign only at (120582 119906) = (0 0) Since thenonlinearity 119892(119909 119906) = 119886(119909)119906 minus 119887(119909)1199063 is an odd function in119906 it follows from the symmetry that if (120582 119906(119909)) is a solutionof (1) so is (minus120582 minus119906(119909)) Thus the component of solutions weconstructed above forms a smooth reversed S-shaped curvewith exactly two turning points (120582lowast 119906lowast) and (minus120582lowast minus119906lowast)

It follows from Lemma 18 that the positive solution 1199060is stable For 120582 lt 120582lowast the upper branch 119906+(119909 120582) remainsstable until it reaches the degenerate solution (120582lowast 119906lowast) Nextwe prove the lower branch 119906minus(119909 120582) is unstable Let 119906119898 beany one solution of lower branch 119906minus(119909 120582) such that 0 lt119906119898 lt 119906lowast Since 119886(119909) minus 3119887(119909)1199062119898 gt 119886(119909) minus 3119887(119909)1199062lowast we have1205831(119886(119909) minus 3119887(119909)1199062119898) lt 1205831(119886(119909) minus 3119887(119909)1199062lowast) = 0 Thereforethe lower branch 119906minus(119909 120582) is unstable Since the solution set issymmetric with respect to the origin the stability of negativesolutions 119906(119909) is obtained by using the property of symmetryThe upper branch of the negative solutions is unstable andthe lower branch of negative solutions is stable Therefore allsolutions of (1) lie on a unique reversed S-shaped solutioncurve This completes the proof

8 Discrete Dynamics in Nature and Society

Data Availability

No data were used to support this study

Conflicts of Interest

The authors declare that there are no conflicts of interestregarding the publication of this paper

Acknowledgments

This work was completed with the support of Tian YuanSpecial Funds of the National Science Foundation of China(no 11626182)

References

[1] J Chu X Lin D Jiang D OrsquoRegan and R P Agarwal ldquoPositivesolutions for second-order superlinear repulsive singular Neu-mann boundary value problemsrdquo Positivity An InternationalMathematics Journal Devoted to Theory and Applications ofPositivity vol 12 no 3 pp 555ndash569 2008

[2] Z Li ldquoExistence of positive solutions of superlinear second-order Neumann boundary value problemrdquo Nonlinear AnalysisTheory Methods amp Applications An International Multidisci-plinary Journal vol 72 no 6 pp 3216ndash3221 2010

[3] G Bonanno and A Sciammetta ldquoExistence and multiplicityresults to Neumann problems for elliptic equations involvingthe p-Laplacianrdquo Journal of Mathematical Analysis and Appli-cations vol 390 no 1 pp 59ndash67 2012

[4] E Sovrano and F Zanolin ldquoIndefinite weight nonlinear prob-lems with Neumann boundary conditionsrdquo Journal of Mathe-matical Analysis and Applications vol 452 no 1 pp 126ndash1472017

[5] G Feltrin and E Sovrano ldquoThree positive solutions to anindefinite Neumann problem a shooting methodrdquo NonlinearAnalysis Theory Methods amp Applications An InternationalMultidisciplinary Journal vol 166 pp 87ndash101 2018

[6] J Llibre and L A Roberto ldquoOn the periodic solutions of aclass of Duffing differential equationsrdquoDiscrete and ContinuousDynamical Systems - Series A vol 33 no 1 pp 277ndash282 2013

[7] H Chen and Y Li ldquoStability and exact multiplicity of periodicsolutions of Duffing equations with cubic nonlinearitiesrdquo Pro-ceedings of the American Mathematical Society vol 135 no 12pp 3925ndash3932 2007

[8] A Lomtatidze and J Sremr ldquoOn periodic solutions to second-order Duffing type equationsrdquo Nonlinear Analysis Real WorldApplications vol 40 pp 215ndash242 2018

[9] C-C Tzeng K-C Hung and S-H Wang ldquoGlobal bifurcationand exact multiplicity of positive solutions for a positone prob-lem with cubic nonlinearityrdquo Journal of Differential Equationsvol 252 no 12 pp 6250ndash6274 2012

[10] S-H Wang and T-S Yeh ldquoA complete classification of bifurca-tion diagrams of a Dirichlet problemwith concave-convex non-linearitiesrdquo Journal of Mathematical Analysis and Applicationsvol 291 no 1 pp 128ndash153 2004

[11] K-C Hung and S-H Wang ldquoA theorem on S-shaped bifur-cation curve for a positone problem with convex-concavenonlinearity and its applications to the perturbed Gelfandproblemrdquo Journal of Differential Equations vol 251 no 2 pp223ndash237 2011

[12] H Pan and R Xing ldquoTime maps and exact multiplicity resultsfor one-dimensional prescribed mean curvature equationsrdquoNonlinear Analysis Theory Methods amp Applications An Inter-national Multidisciplinary Journal vol 74 no 4 pp 1234ndash12602011

[13] H Chen H Xing and X He ldquoBifurcation and stability ofsolutions to a logistic equation with harvestingrdquo MathematicalMethods in the Applied Sciences vol 38 no 8 pp 1623ndash16302015

[14] P Clrsquoement andG Sweers ldquoUniform anti-maximumprinciplesrdquoJournal of Differential Equations vol 164 no 1 pp 118ndash1542000

[15] W Reichel ldquoSharp parameter ranges in the uniform anti-maximum principle for second-order ordinary diffrential oper-atorsrdquo Zeitschrift fur Angewandte Mathematik und Physik vol54 no 5 pp 822ndash838 2003

[16] MG Crandall and PH Rabinowitz ldquoBifurcation perturbationof simple eigenvalues and linearized stabilityrdquo Archive forRational Mechanics and Analysis vol 52 pp 161ndash180 1973

[17] H Kielhofer Bifurcation Theory An Introduction with Applica-tions to PDEs Springer New York NY USA 2003

[18] S Oruganti J Shi and R Shivaji ldquoDiffusive logistic equationwith constant yield harvesting I Steady statesrdquo Transactions ofthe American Mathematical Society vol 354 no 9 pp 3601ndash3619 2002

[19] K Deimling Nonlinear Functional Analysis Springer Berlin Germany 1985

[20] J X Sun Nonlinear Functional Analysis and Its ApplicationsScience Press Beijing China 2007

[21] H Xing H Chen and X He ldquoExact multiplicity and stability ofsolutions of second-order Neumann boundary value problemrdquoApplied Mathematics and Computation vol 232 pp 1104ndash11112014

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Discrete Dynamics in Nature and Society 5

Lemma 16 (see [16]) Let119883 and 119885 be Banach spaces Assumethat 119865 Rtimes119880 997888rarr 119885 is continuously differentiable on119881times119880 subR times 119883 satisfying the following three conditions

(1) 119865(120582lowast 119906lowast) = 0 for some (120582lowast 119906lowast) isin 119881 times119880 dim119873(119863119906119865(120582lowast 119906lowast)) = 1 and the null space119873(119863119906119865(120582lowast 119906lowast)) = span119908

(2) the Fredholm index of 119863119906119865(120582lowast 119906lowast) is zerocodim119877(119863119906119865(120582lowast 119906lowast)) = 1

(3) 119863120582119865(120582lowast 119906lowast)isin119877(119863119906119865(120582lowast 119906lowast))Then there is a continuously differentiable curve through(120582lowast 119906lowast) that is there exists (120582(119904) 119906(119904)) = (120582lowast + 120591(119904) 119906lowast +119904119908 + 119911(119904)) | 119904 isin (minus120575 120575) (120582(0) 119906(0)) = (120582lowast 119906lowast) such that

119865 (120582 (119904) 119906 (119904)) = 0 for 119904 isin (minus120575 120575) (44)

and all solutions of 119865(120582 119906) = 0 in a neighborhood of (120582lowast 119906lowast)belong to the curve 120591(0) = 1205911015840(0) = 119911(0) = 1199111015840(0) = 0

3 Main Results

Lemma 17 Assume (f2) holds Then for every fixed 120582 (1) hasat least a solution and

deg (119868 minus 119860 119861119877 0) = 1 (45)

for 119877 large enough where (119860119906)(119909) = int10119866(119909 119904)(119892(119904 119906(119904)) +

119906(119904) minus 120582119891(119904))119889119904 and 119866 is defined in (10)

Proof It follows from (f2) that there exists119877 gt 0 large enoughsuch that 119892(119909 119877) minus 120582119891(119909) lt 0 and 119892(119909 minus119877) minus 120582119891(119909) gt 0Therefore minus119877 and 119877 are subsolution and supersolution of (1)respectively Therefore there exists at least one solution of (1)between minus119877 and 119877 Next we show that all the solutions of(1) are between minus119877 and 119877 for 119877 gt 0 large enough Argueby contradiction suppose that there exists 1199090 such that 119906(119909)attains its maximum value and 119906(1199090) gt 119877 We have

11990610158401015840 (1199090) + 119886 (1199090) 119906 (1199090) minus 119887 (1199090) 1199063 (1199090) minus 120582119891 (1199090)= 0 (46)

It follows from 1199061015840(1199090) = 0 and 11990610158401015840(1199090) lt 0 that119886 (1199090) 119906 (1199090) minus 119887 (1199090) 1199063 (1199090) minus 120582119891 (1199090) ge 0 (47)

From the above there exists 119877 gt 0 large enough such that119892(119909 119877) minus 120582119891(119909) lt 0 contradiction Similar suppose thatthere exists 1199091 such that 119906(119909) attains its minimum value and119906(1199091) lt minus119877 we will obtain contradiction Therefore thereexists 119877 gt 0 large enough such that 119906 lt 119877 for all thesolutions 119906(119909) (1) is equivalent to the following equation

119906 (119909) = (119860119906) (119909) (48)

Next we calculate deg(119868minus119860 119861119877 0) Let ℎ(119905 119906) = 119906minus119905119860119906 forall119905 isin[0 1] The equation ℎ(119905 119906) = 0 is equivalent to the followingequation

11990610158401015840 + 119905119892 (119909 119906) minus (1 minus 119905) 119906 minus 119905120582119891 (119909) = 0 (49)

It is evident that minus119877 and 119877 are subsolution and supersolutionof (49) respectively Therefore all the solutions of (49) mustbe between minus119877 and 119877 that is the equation ℎ(119905 119906) = 0 has nosolution on 120597119861119877 Therefore ℎ(119905 119906) = 0 for 119905 isin [0 1] 119906 isin 120597119861119877By the homotopy invariance properties of the topologicaldegree we have

deg (119868 minus 119860 119861119877 0) = deg (ℎ (sdot 1) 119861119877 0)= deg (ℎ (sdot 0) 119861119877 0)= deg (119868 119861119877 0) = 1

(50)

This completes the proof

Lemma 18 Suppose that 1205831(119886(119909)) lt 0 119886(119909) ≪ 1205872 and119891(119909) gt 0 for all 119909 isin [0 1] Then (1) has exactly three solutions1199060(119909) 0 V0(119909) for 120582 = 0 where 1199060(119909) is the unique positivestable solution and V0(119909) is the unique negative stable solutionProof For the mapping 119865 defined in (43) let 120582 = 0 and wehave

119865 (119906) = 11990610158401015840 + 119886 (119909) 119906 minus 119887 (119909) 1199063 (51)

Next we prove that any solution 119906(119909) of 119865(119906) = 0 isnondegenerate

First we have the Frechet derivative of (51)

119865119906 (119906) [119908] = 11990810158401015840 + (119886 (119909) minus 3119887 (119909) 1199062)119908 (52)

It is obvious that 119906 = 0 is the solution of 119865(119906) = 0 andtherefore when 119906 = 0 119886(119909)minus3119887(119909)1199062 = 119886(119909) in (52) It followsfrom 1205831(119886(119909)) lt 0 that 0 is nondegenerate

Next we prove that any nontrivial solution 119906(119909) of119865(119906) =0 is nondegenerate 119865(119906) = 0 is equivalent to the followingequation

119906 (119909) = (119860119906) (119909) (53)

where (119860119906)(119909) = int10119866(119909 119904)(119892(119904 119906(119904)) + 119906(119904))119889119904 and 119866 is

defined in (10) The equation

119908 minus 1198601015840 (119906) 119908 = 0 (54)

is equivalent to 119865119906(119906)[119908] = 0 where 1198601015840(119906) denotes thederivative operator of 119860 When 1205831(119886(119909)) lt 0 119886(119909) ≪ 1205872by Lemma 9 Definition 10 and Lemma 12 there exists 119877 gt 0and we have

ind (119868 minus 119860 0) = ind (119868 minus 1198601015840 (0) 0)= deg (119868 minus 119860 119861119877 0) = minus1

(55)

If 119906(119909) is a nontrivial solution of 119865(119906) = 0 it follows fromLemma 13 that 119906(119909) = 0 for all 119909 isin [0 1] Therefore

119892119906 (119909 119906) = 119886 (119909) minus 3119887 (119909) 1199062 (119909) lt 119892 (119909 119906)119906

= 119886 (119909) minus 119887 (119909) 1199062 (119909) ≪ 1205872(56)

6 Discrete Dynamics in Nature and Society

Let 1199021(119909) = 119892(119909 119906)119906 and 1199022(119909) = 119892119906(119909 119906) It follows fromthe above hypothesis that 119906(119909) is a nontrivial solution of1198711199021119906 = 0 By the second conclusion of Lemma 13 we havethat1198711199022119906 = 0does not have nontrivial solutionwhich impliesthat 119906(119909) is a nondegenerate solution

Finally we prove the positive solution is unique andstable We denote by 1199060 the nontrivial solution of (51) It isobvious that minus1199060(119909) is also the nontrivial solution of (51)It follows from (56) and the comparison of eigenvalues that1205831(1199022(119909)) gt 1205831(1199021(119909)) = 0 By Lemma 9 Definition 10 andLemma 12 we have

ind (119868 minus 119860 119906) = ind (119868 minus 1198601015840 (119906) 0)= deg (119868 minus 119860 119861119877 0) = 1

(57)

Let 119898 be the number of nontrivial solutions of 119865(119906) = 0Hence 119865 is proper Since 0 is a regular value of 119865119898 must befinite According to Lemma 17 we have deg(119868 minus 119860 119861119877 0) = 1for sufficiently large 119877 By Lemmas 9 and 11 and the indexformula we have that

1 = deg (119868 minus 119860 119861119877 0)= ind (119868 minus 119860 0) + sum

119906 =0

ind (119868 minus 119860 119906119894) = minus1 + 119898 (58)

It is obvious that 119898 = 2 Let V0(119909) equiv minus1199060(119909) for 119909 isin [0 1]Therefore 119865(119906) = 0 has exactly three solutions 1199060 0 and V0Since 1205831(1199022) gt 1205831(1199021) = 0 we have that the positive solution1199060(119909) is unique and stable The negative solution V0(119909) is alsostable This completes the proof

Lemma 19 Suppose that 119886(119909) ≪ 12058724 For some 120582 lt 0 (1)has a unique positive solution 1199061(119909)with 1199061(119909) gt 1199060(119909) where1199060(119909) is the unique positive solution of (1) for 120582 = 0Proof By Lemma 18 for 120582 = 0 (1) has the unique positivesolution 1199060 which provided a subsolution of (1) for 120582 lt 0There exists 119877 gt 0 large enough such that 119892(119909 119877) minus 120582119891(119909) lt0 Thus 119877 is a supersolution of (1) Therefore we prove thatthere exists a positive solution 1199061(119909) of (1) for 120582 lt 0 such that1199060(119909) lt 1199061(119909) lt 119877

Next we will prove that the positive solution 1199061 is uniqueAssume by contradiction that (1) has another positive solu-tion 119906(119909) for 120582 lt 0 Let V(119909) = 119906(119909) minus 1199061(119909) = 0 then 119908satisfies the following equation

V10158401015840 + 119886 (119909) V minus 119887 (119909) [1199062 (119909) + 119906 (119909) 1199061 (119909) + 11990621 (119909)] V= 0

(59)

Clearly 119886(119909) minus 119887(119909)[1199062(119909) + 119906(119909)1199061(119909) + 11990621(119909)] lt 119886(119909) ≪12058724 By Lemma 13 V(119909) equiv 0 is the solution of (59) which isa contradiction Therefore the solution 1199061 obtained above isonly positive solution For 120582 lt 0 1199061 satisfies (1) and

119906101584010158401 + 119902 (119909) 1199061 = 120582119891 (119909) (60)

where 119902(119909) = 119886(119909) minus 119887(119909)11990621(119909) Again since 119891(119909) gt 0 and120582 lt 0 it follows from Lemma 14 that 1205831(119902(119909)) gt 0 Considerthe linearization associated with (1)

11990810158401015840 + (119886 (119909) minus 3119887 (119909) 1199062 (119909))119908 = 0 (61)

where 119892119906(119909 119906) = 119886(119909) minus 3119887(119909)1199062(119909) lt 119902(119909) Thus1205831(119892119906(119909 119906)) gt 1205831(119902(119909)) gt 0Therefore by Lemma 13 we have1199061(119909) is nondegenerate This completes the proof

Lemma20 For 120582 gt 0 (1) has a unique negative solution V1(119909)with V1(119909) lt V0(119909) where V0(119909) is the unique negative solutionof (1) for 120582 = 0Proof The proof is similar to the proof of Lemma 19

Theorem 21 Assume that the first eigenvalue 1205831(119886(119909)) lt 0and 119891(119909) gt 0 for all 119909 isin [0 1] Suppose that 119886(119909) ≪12058724 Then all the solutions of (1) are of one sign and lie ona unique reversed S-shaped solution curve which is symmetricwith respect to the origin More precisely there exists 120582lowast gt 0such that

(i) For 120582 gt 120582lowast (1) has no positive solution and has aunique negative solution which is stable

(ii) For 120582 = plusmn120582lowast (1) has exactly two solutions Moreoverwhen 120582 = 120582lowast the negative solution is stable and thepositive solution is degenerate When 120582 = minus120582lowast thepositive solution is stable and the negative solution isdegenerate

(iii) For minus120582lowast lt 120582 lt 120582lowast (1) has exactly three orderedsolutions at the same 120582 and the middle solution isunstable and the remaining two are stable Moreoverwhen 120582 lt 0 the maximal solution is positive and theother two are negative When 120582 gt 0 the minimalsolution is negative and the other two are positive

(iv) For 120582 lt minus120582lowast (1) has no negative solution and has aunique positive solution which is stable

Proof It follows from Lemma 19 that (1) has a uniquenondegenerate positive solution 1199061(119909) for some 120582 lt 0 Thesolution curve can be continued a little bit such that 1199061(119909)remains positive for increasing 120582 when 120582 lt 0 By Lemma 18(1) has a unique nondegenerate positive solution 1199060(119909) for120582 = 0The positive solution curve can pass through 1199061(119909) and1199060(119909) and can be continued further for increasing 120582 until thelinearized equation (61) admits the nontrivial solutions Weclaim that the curve of positive solutions cannot be continuedfor 120582 gt 120582lowast Next we will prove the existence of 120582lowast Since1205831(119886(119909)) lt 0 let V(119909) gt 0 be the first eigenfunction of theproblem

V10158401015840 + 119886 (119909) V + 120583V = 0 119909 isin (0 1) V1015840 (0) = V1015840 (1) = 0

(62)

Discrete Dynamics in Nature and Society 7

Multiplying (1) by V(119909) and subtracting from (62) multipliedby 119906 after that integrating over [0 1] we obtain

int1

0[119887 (119909) 1199063 (119909) + 120583119906 (119909)] V (119909) 119889119909

= minus120582int1

0119891 (119909) V (119909) 119889119909

(63)

Applying the mean-value theorem for (63) there exists 120585 isin[0 1] such that

ℎ (119906 (120585) 119887 (120585)) = 119887 (120585) 1199063 (120585) + 1205831119906 (120585)

= minus120582int10119891 (119909) V (119909) 119889119909int10V (119909) 119889119909

(64)

For all 119887(120585) le 1198870 = max119909isin[01]119887(119909) the only root ofℎ(119906(120585) 119887(120585)) = 119898 is negative for 119898 lt ℎ(radicminus120583131198870 1198870) lt 0Let119872 = ℎ(radicminus120583131198870 1198870) and

120582lowast = minus 119872int10V (119909) 119889119909

int10119891 (119909) V (119909) 119889119909 (65)

and it follows from (64) that 119906(120585) lt 0 for 120582 gt 120582lowast Thereforethere exists a 120582lowast gt 0 such that (1) has no positive solution for120582 gt 120582lowast

By applying Lemma 14 we obtain that 119906lowast(119909) is still apositive solution of (1) for 120582 = 120582lowast We denote the degeneratesolution (120582lowast 119906lowast) At (120582lowast 119906lowast) we verify that Lemma 16 canbe applied here It follows from (43) that 119863119906119865(120582lowast 119906lowast)[V] =V10158401015840 + [119886(119909) minus 3119887(119909)1199062lowast]V = 0 In fact 0 is simple and theprincipal eigenvalue of (5) and the first eigenfunction V(119909) gt0 therefore dim119873(119863119906119865(120582lowast 119906lowast)) = codim119877(119863119906119865(120582lowast 119906lowast)) =1 and the null space 119873(119863119906119865(120582lowast 119906lowast)) = spanV Thereforethe Fredholm index of 119863119906119865(120582lowast 119906lowast) is zero Conditions (1)and (2) of Lemma 16 are satisfied Next we verify Condition(3) of Lemma 16 Suppose on the contrary that119863120582119865(120582lowast 119906lowast) =minus119891(119909) isin 119877(119863119906119865(120582lowast 119906lowast)) namely there is a continuousfunction 119908 satisfying

11990810158401015840 + [119886 (119909) minus 3119887 (119909) 1199062lowast]119908 = minus119891 (119909) (66)

Consider the linearized equation of (1) at (120582lowast 119906lowast)V10158401015840 + [119886 (119909) minus 3119887 (119909) 1199062lowast] V = 0 (67)

Since 119865 is singular at (120582lowast 119906lowast) that is (67) has a nontrivialsolution V(119909) such that V(119909) gt 0 Multiplying (66) by Vsubtracting from (67) multiplied by 119908 and integrating byparts on [0 1] we have

0 = int1

0V10158401015840119908 minus 11990810158401015840V 119889119909 = int

1

0V (119909) 119891 (119909) 119889119909 (68)

a contradiction since both V(119909) and 119891(119909) are positiveTherefore Condition (3) of Lemma 16 is satisfied Near(120582lowast 119906lowast) the solutions of

119865 (120582 (119904) 119906 (119904)) = 0 for all 119904 isin [minus120575 120575] (69)

form a curve

(120582 (119904) 119906 (119904)) = (120582lowast + 120591 (119904) 119906lowast + 119904V + 119911 (119904)) | 119904isin [minus120575 120575] (120582 (0) 119906 (0)) = (120582lowast 119906lowast)

(70)

Differentiating (69) twice in 119904 setting 119904 = 0 and 119906119904|119904=0 = V(119909)and evaluating at (120582lowast 119906lowast) we have

11990610158401015840119904119904 + [119886 (119909) minus 3119887 (119909) 1199062lowast] 119906119904119904= 6119887 (119909) 119906lowastV2 + 120591119904119904 (0) 119891 (119909)

(71)

Multiplying the linearized equation (67) by 119906119904119904 subtractingfrom (71) multiplied by V and integrating by parts over [0 1]we obtain

120591119904119904 (0) = minus6 int10119887 (119909) 119906lowastV3119889119909

int10119891 (119909) V (119909) 119889119909 lt 0 (72)

Therefore (120582lowast 119906lowast) is a fold point of 119865(120582 119906) to the left Itfollows from the above that the curve of positive solutionscannot be continued to the right indefinitely for all 120582 gt 0Hence the positive solution curve will make a left turn at(120582lowast 119906lowast) Near the critical point (120582lowast 119906lowast) by the Crandall-Rabinowitz bifurcation theorem there are two branches ofpositive solutions denoted by the upper branch 119906+(119909 120582)and the lower branch 119906minus(119909 120582) with 119906minus(119909 120582) lt 119906+(119909 120582)It follows from Lemmas 14 15 and 18 that the fold point(120582lowast 119906lowast) is unique and the upper branch119906+(119909 120582) ismonotonedecreasing for all 120582 lt 120582lowast and lower branch 119906minus(119909 120582) ismonotone increasing for all 0 lt 120582 lt 120582lowast Therefore the lowerbranch curve ismonotone increasing and continues to the leftwithout any turnings Rewriting (1) in the following form

11990610158401015840 + (119886 (119909) minus 119887 (119909) 1199062) 119906 = 120582119891 (119909) (73)

since 119886(119909) minus 119887(119909)1199062 lt 119886(119909) ≪ 12058724 it follows fromLemma 14 that 119906(119909) gt 0 or 119906(119909) lt 0 This shows that thesolution changes its sign only at (120582 119906) = (0 0) Since thenonlinearity 119892(119909 119906) = 119886(119909)119906 minus 119887(119909)1199063 is an odd function in119906 it follows from the symmetry that if (120582 119906(119909)) is a solutionof (1) so is (minus120582 minus119906(119909)) Thus the component of solutions weconstructed above forms a smooth reversed S-shaped curvewith exactly two turning points (120582lowast 119906lowast) and (minus120582lowast minus119906lowast)

It follows from Lemma 18 that the positive solution 1199060is stable For 120582 lt 120582lowast the upper branch 119906+(119909 120582) remainsstable until it reaches the degenerate solution (120582lowast 119906lowast) Nextwe prove the lower branch 119906minus(119909 120582) is unstable Let 119906119898 beany one solution of lower branch 119906minus(119909 120582) such that 0 lt119906119898 lt 119906lowast Since 119886(119909) minus 3119887(119909)1199062119898 gt 119886(119909) minus 3119887(119909)1199062lowast we have1205831(119886(119909) minus 3119887(119909)1199062119898) lt 1205831(119886(119909) minus 3119887(119909)1199062lowast) = 0 Thereforethe lower branch 119906minus(119909 120582) is unstable Since the solution set issymmetric with respect to the origin the stability of negativesolutions 119906(119909) is obtained by using the property of symmetryThe upper branch of the negative solutions is unstable andthe lower branch of negative solutions is stable Therefore allsolutions of (1) lie on a unique reversed S-shaped solutioncurve This completes the proof

8 Discrete Dynamics in Nature and Society

Data Availability

No data were used to support this study

Conflicts of Interest

The authors declare that there are no conflicts of interestregarding the publication of this paper

Acknowledgments

This work was completed with the support of Tian YuanSpecial Funds of the National Science Foundation of China(no 11626182)

References

[1] J Chu X Lin D Jiang D OrsquoRegan and R P Agarwal ldquoPositivesolutions for second-order superlinear repulsive singular Neu-mann boundary value problemsrdquo Positivity An InternationalMathematics Journal Devoted to Theory and Applications ofPositivity vol 12 no 3 pp 555ndash569 2008

[2] Z Li ldquoExistence of positive solutions of superlinear second-order Neumann boundary value problemrdquo Nonlinear AnalysisTheory Methods amp Applications An International Multidisci-plinary Journal vol 72 no 6 pp 3216ndash3221 2010

[3] G Bonanno and A Sciammetta ldquoExistence and multiplicityresults to Neumann problems for elliptic equations involvingthe p-Laplacianrdquo Journal of Mathematical Analysis and Appli-cations vol 390 no 1 pp 59ndash67 2012

[4] E Sovrano and F Zanolin ldquoIndefinite weight nonlinear prob-lems with Neumann boundary conditionsrdquo Journal of Mathe-matical Analysis and Applications vol 452 no 1 pp 126ndash1472017

[5] G Feltrin and E Sovrano ldquoThree positive solutions to anindefinite Neumann problem a shooting methodrdquo NonlinearAnalysis Theory Methods amp Applications An InternationalMultidisciplinary Journal vol 166 pp 87ndash101 2018

[6] J Llibre and L A Roberto ldquoOn the periodic solutions of aclass of Duffing differential equationsrdquoDiscrete and ContinuousDynamical Systems - Series A vol 33 no 1 pp 277ndash282 2013

[7] H Chen and Y Li ldquoStability and exact multiplicity of periodicsolutions of Duffing equations with cubic nonlinearitiesrdquo Pro-ceedings of the American Mathematical Society vol 135 no 12pp 3925ndash3932 2007

[8] A Lomtatidze and J Sremr ldquoOn periodic solutions to second-order Duffing type equationsrdquo Nonlinear Analysis Real WorldApplications vol 40 pp 215ndash242 2018

[9] C-C Tzeng K-C Hung and S-H Wang ldquoGlobal bifurcationand exact multiplicity of positive solutions for a positone prob-lem with cubic nonlinearityrdquo Journal of Differential Equationsvol 252 no 12 pp 6250ndash6274 2012

[10] S-H Wang and T-S Yeh ldquoA complete classification of bifurca-tion diagrams of a Dirichlet problemwith concave-convex non-linearitiesrdquo Journal of Mathematical Analysis and Applicationsvol 291 no 1 pp 128ndash153 2004

[11] K-C Hung and S-H Wang ldquoA theorem on S-shaped bifur-cation curve for a positone problem with convex-concavenonlinearity and its applications to the perturbed Gelfandproblemrdquo Journal of Differential Equations vol 251 no 2 pp223ndash237 2011

[12] H Pan and R Xing ldquoTime maps and exact multiplicity resultsfor one-dimensional prescribed mean curvature equationsrdquoNonlinear Analysis Theory Methods amp Applications An Inter-national Multidisciplinary Journal vol 74 no 4 pp 1234ndash12602011

[13] H Chen H Xing and X He ldquoBifurcation and stability ofsolutions to a logistic equation with harvestingrdquo MathematicalMethods in the Applied Sciences vol 38 no 8 pp 1623ndash16302015

[14] P Clrsquoement andG Sweers ldquoUniform anti-maximumprinciplesrdquoJournal of Differential Equations vol 164 no 1 pp 118ndash1542000

[15] W Reichel ldquoSharp parameter ranges in the uniform anti-maximum principle for second-order ordinary diffrential oper-atorsrdquo Zeitschrift fur Angewandte Mathematik und Physik vol54 no 5 pp 822ndash838 2003

[16] MG Crandall and PH Rabinowitz ldquoBifurcation perturbationof simple eigenvalues and linearized stabilityrdquo Archive forRational Mechanics and Analysis vol 52 pp 161ndash180 1973

[17] H Kielhofer Bifurcation Theory An Introduction with Applica-tions to PDEs Springer New York NY USA 2003

[18] S Oruganti J Shi and R Shivaji ldquoDiffusive logistic equationwith constant yield harvesting I Steady statesrdquo Transactions ofthe American Mathematical Society vol 354 no 9 pp 3601ndash3619 2002

[19] K Deimling Nonlinear Functional Analysis Springer Berlin Germany 1985

[20] J X Sun Nonlinear Functional Analysis and Its ApplicationsScience Press Beijing China 2007

[21] H Xing H Chen and X He ldquoExact multiplicity and stability ofsolutions of second-order Neumann boundary value problemrdquoApplied Mathematics and Computation vol 232 pp 1104ndash11112014

Hindawiwwwhindawicom Volume 2018

MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Mathematical Problems in Engineering

Applied MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Probability and StatisticsHindawiwwwhindawicom Volume 2018

Journal of

Hindawiwwwhindawicom Volume 2018

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawiwwwhindawicom Volume 2018

OptimizationJournal of

Hindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom Volume 2018

Engineering Mathematics

International Journal of

Hindawiwwwhindawicom Volume 2018

Operations ResearchAdvances in

Journal of

Hindawiwwwhindawicom Volume 2018

Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018

International Journal of Mathematics and Mathematical Sciences

Hindawiwwwhindawicom Volume 2018

Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom

The Scientific World Journal

Volume 2018

Hindawiwwwhindawicom Volume 2018Volume 2018

Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in

Nature and SocietyHindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom

Dierential EquationsInternational Journal of

Volume 2018

Hindawiwwwhindawicom Volume 2018

Decision SciencesAdvances in

Hindawiwwwhindawicom Volume 2018

AnalysisInternational Journal of

Hindawiwwwhindawicom Volume 2018

Stochastic AnalysisInternational Journal of

Submit your manuscripts atwwwhindawicom

6 Discrete Dynamics in Nature and Society

Let 1199021(119909) = 119892(119909 119906)119906 and 1199022(119909) = 119892119906(119909 119906) It follows fromthe above hypothesis that 119906(119909) is a nontrivial solution of1198711199021119906 = 0 By the second conclusion of Lemma 13 we havethat1198711199022119906 = 0does not have nontrivial solutionwhich impliesthat 119906(119909) is a nondegenerate solution

Finally we prove the positive solution is unique andstable We denote by 1199060 the nontrivial solution of (51) It isobvious that minus1199060(119909) is also the nontrivial solution of (51)It follows from (56) and the comparison of eigenvalues that1205831(1199022(119909)) gt 1205831(1199021(119909)) = 0 By Lemma 9 Definition 10 andLemma 12 we have

ind (119868 minus 119860 119906) = ind (119868 minus 1198601015840 (119906) 0)= deg (119868 minus 119860 119861119877 0) = 1

(57)

Let 119898 be the number of nontrivial solutions of 119865(119906) = 0Hence 119865 is proper Since 0 is a regular value of 119865119898 must befinite According to Lemma 17 we have deg(119868 minus 119860 119861119877 0) = 1for sufficiently large 119877 By Lemmas 9 and 11 and the indexformula we have that

1 = deg (119868 minus 119860 119861119877 0)= ind (119868 minus 119860 0) + sum

119906 =0

ind (119868 minus 119860 119906119894) = minus1 + 119898 (58)

It is obvious that 119898 = 2 Let V0(119909) equiv minus1199060(119909) for 119909 isin [0 1]Therefore 119865(119906) = 0 has exactly three solutions 1199060 0 and V0Since 1205831(1199022) gt 1205831(1199021) = 0 we have that the positive solution1199060(119909) is unique and stable The negative solution V0(119909) is alsostable This completes the proof

Lemma 19 Suppose that 119886(119909) ≪ 12058724 For some 120582 lt 0 (1)has a unique positive solution 1199061(119909)with 1199061(119909) gt 1199060(119909) where1199060(119909) is the unique positive solution of (1) for 120582 = 0Proof By Lemma 18 for 120582 = 0 (1) has the unique positivesolution 1199060 which provided a subsolution of (1) for 120582 lt 0There exists 119877 gt 0 large enough such that 119892(119909 119877) minus 120582119891(119909) lt0 Thus 119877 is a supersolution of (1) Therefore we prove thatthere exists a positive solution 1199061(119909) of (1) for 120582 lt 0 such that1199060(119909) lt 1199061(119909) lt 119877

Next we will prove that the positive solution 1199061 is uniqueAssume by contradiction that (1) has another positive solu-tion 119906(119909) for 120582 lt 0 Let V(119909) = 119906(119909) minus 1199061(119909) = 0 then 119908satisfies the following equation

V10158401015840 + 119886 (119909) V minus 119887 (119909) [1199062 (119909) + 119906 (119909) 1199061 (119909) + 11990621 (119909)] V= 0

(59)

Clearly 119886(119909) minus 119887(119909)[1199062(119909) + 119906(119909)1199061(119909) + 11990621(119909)] lt 119886(119909) ≪12058724 By Lemma 13 V(119909) equiv 0 is the solution of (59) which isa contradiction Therefore the solution 1199061 obtained above isonly positive solution For 120582 lt 0 1199061 satisfies (1) and

119906101584010158401 + 119902 (119909) 1199061 = 120582119891 (119909) (60)

where 119902(119909) = 119886(119909) minus 119887(119909)11990621(119909) Again since 119891(119909) gt 0 and120582 lt 0 it follows from Lemma 14 that 1205831(119902(119909)) gt 0 Considerthe linearization associated with (1)

11990810158401015840 + (119886 (119909) minus 3119887 (119909) 1199062 (119909))119908 = 0 (61)

where 119892119906(119909 119906) = 119886(119909) minus 3119887(119909)1199062(119909) lt 119902(119909) Thus1205831(119892119906(119909 119906)) gt 1205831(119902(119909)) gt 0Therefore by Lemma 13 we have1199061(119909) is nondegenerate This completes the proof

Lemma20 For 120582 gt 0 (1) has a unique negative solution V1(119909)with V1(119909) lt V0(119909) where V0(119909) is the unique negative solutionof (1) for 120582 = 0Proof The proof is similar to the proof of Lemma 19

Theorem 21 Assume that the first eigenvalue 1205831(119886(119909)) lt 0and 119891(119909) gt 0 for all 119909 isin [0 1] Suppose that 119886(119909) ≪12058724 Then all the solutions of (1) are of one sign and lie ona unique reversed S-shaped solution curve which is symmetricwith respect to the origin More precisely there exists 120582lowast gt 0such that

(i) For 120582 gt 120582lowast (1) has no positive solution and has aunique negative solution which is stable

(ii) For 120582 = plusmn120582lowast (1) has exactly two solutions Moreoverwhen 120582 = 120582lowast the negative solution is stable and thepositive solution is degenerate When 120582 = minus120582lowast thepositive solution is stable and the negative solution isdegenerate

(iii) For minus120582lowast lt 120582 lt 120582lowast (1) has exactly three orderedsolutions at the same 120582 and the middle solution isunstable and the remaining two are stable Moreoverwhen 120582 lt 0 the maximal solution is positive and theother two are negative When 120582 gt 0 the minimalsolution is negative and the other two are positive

(iv) For 120582 lt minus120582lowast (1) has no negative solution and has aunique positive solution which is stable

Proof It follows from Lemma 19 that (1) has a uniquenondegenerate positive solution 1199061(119909) for some 120582 lt 0 Thesolution curve can be continued a little bit such that 1199061(119909)remains positive for increasing 120582 when 120582 lt 0 By Lemma 18(1) has a unique nondegenerate positive solution 1199060(119909) for120582 = 0The positive solution curve can pass through 1199061(119909) and1199060(119909) and can be continued further for increasing 120582 until thelinearized equation (61) admits the nontrivial solutions Weclaim that the curve of positive solutions cannot be continuedfor 120582 gt 120582lowast Next we will prove the existence of 120582lowast Since1205831(119886(119909)) lt 0 let V(119909) gt 0 be the first eigenfunction of theproblem

V10158401015840 + 119886 (119909) V + 120583V = 0 119909 isin (0 1) V1015840 (0) = V1015840 (1) = 0

(62)

Discrete Dynamics in Nature and Society 7

Multiplying (1) by V(119909) and subtracting from (62) multipliedby 119906 after that integrating over [0 1] we obtain

int1

0[119887 (119909) 1199063 (119909) + 120583119906 (119909)] V (119909) 119889119909

= minus120582int1

0119891 (119909) V (119909) 119889119909

(63)

Applying the mean-value theorem for (63) there exists 120585 isin[0 1] such that

ℎ (119906 (120585) 119887 (120585)) = 119887 (120585) 1199063 (120585) + 1205831119906 (120585)

= minus120582int10119891 (119909) V (119909) 119889119909int10V (119909) 119889119909

(64)

For all 119887(120585) le 1198870 = max119909isin[01]119887(119909) the only root ofℎ(119906(120585) 119887(120585)) = 119898 is negative for 119898 lt ℎ(radicminus120583131198870 1198870) lt 0Let119872 = ℎ(radicminus120583131198870 1198870) and

120582lowast = minus 119872int10V (119909) 119889119909

int10119891 (119909) V (119909) 119889119909 (65)

and it follows from (64) that 119906(120585) lt 0 for 120582 gt 120582lowast Thereforethere exists a 120582lowast gt 0 such that (1) has no positive solution for120582 gt 120582lowast

By applying Lemma 14 we obtain that 119906lowast(119909) is still apositive solution of (1) for 120582 = 120582lowast We denote the degeneratesolution (120582lowast 119906lowast) At (120582lowast 119906lowast) we verify that Lemma 16 canbe applied here It follows from (43) that 119863119906119865(120582lowast 119906lowast)[V] =V10158401015840 + [119886(119909) minus 3119887(119909)1199062lowast]V = 0 In fact 0 is simple and theprincipal eigenvalue of (5) and the first eigenfunction V(119909) gt0 therefore dim119873(119863119906119865(120582lowast 119906lowast)) = codim119877(119863119906119865(120582lowast 119906lowast)) =1 and the null space 119873(119863119906119865(120582lowast 119906lowast)) = spanV Thereforethe Fredholm index of 119863119906119865(120582lowast 119906lowast) is zero Conditions (1)and (2) of Lemma 16 are satisfied Next we verify Condition(3) of Lemma 16 Suppose on the contrary that119863120582119865(120582lowast 119906lowast) =minus119891(119909) isin 119877(119863119906119865(120582lowast 119906lowast)) namely there is a continuousfunction 119908 satisfying

11990810158401015840 + [119886 (119909) minus 3119887 (119909) 1199062lowast]119908 = minus119891 (119909) (66)

Consider the linearized equation of (1) at (120582lowast 119906lowast)V10158401015840 + [119886 (119909) minus 3119887 (119909) 1199062lowast] V = 0 (67)

Since 119865 is singular at (120582lowast 119906lowast) that is (67) has a nontrivialsolution V(119909) such that V(119909) gt 0 Multiplying (66) by Vsubtracting from (67) multiplied by 119908 and integrating byparts on [0 1] we have

0 = int1

0V10158401015840119908 minus 11990810158401015840V 119889119909 = int

1

0V (119909) 119891 (119909) 119889119909 (68)

a contradiction since both V(119909) and 119891(119909) are positiveTherefore Condition (3) of Lemma 16 is satisfied Near(120582lowast 119906lowast) the solutions of

119865 (120582 (119904) 119906 (119904)) = 0 for all 119904 isin [minus120575 120575] (69)

form a curve

(120582 (119904) 119906 (119904)) = (120582lowast + 120591 (119904) 119906lowast + 119904V + 119911 (119904)) | 119904isin [minus120575 120575] (120582 (0) 119906 (0)) = (120582lowast 119906lowast)

(70)

Differentiating (69) twice in 119904 setting 119904 = 0 and 119906119904|119904=0 = V(119909)and evaluating at (120582lowast 119906lowast) we have

11990610158401015840119904119904 + [119886 (119909) minus 3119887 (119909) 1199062lowast] 119906119904119904= 6119887 (119909) 119906lowastV2 + 120591119904119904 (0) 119891 (119909)

(71)

Multiplying the linearized equation (67) by 119906119904119904 subtractingfrom (71) multiplied by V and integrating by parts over [0 1]we obtain

120591119904119904 (0) = minus6 int10119887 (119909) 119906lowastV3119889119909

int10119891 (119909) V (119909) 119889119909 lt 0 (72)

Therefore (120582lowast 119906lowast) is a fold point of 119865(120582 119906) to the left Itfollows from the above that the curve of positive solutionscannot be continued to the right indefinitely for all 120582 gt 0Hence the positive solution curve will make a left turn at(120582lowast 119906lowast) Near the critical point (120582lowast 119906lowast) by the Crandall-Rabinowitz bifurcation theorem there are two branches ofpositive solutions denoted by the upper branch 119906+(119909 120582)and the lower branch 119906minus(119909 120582) with 119906minus(119909 120582) lt 119906+(119909 120582)It follows from Lemmas 14 15 and 18 that the fold point(120582lowast 119906lowast) is unique and the upper branch119906+(119909 120582) ismonotonedecreasing for all 120582 lt 120582lowast and lower branch 119906minus(119909 120582) ismonotone increasing for all 0 lt 120582 lt 120582lowast Therefore the lowerbranch curve ismonotone increasing and continues to the leftwithout any turnings Rewriting (1) in the following form

11990610158401015840 + (119886 (119909) minus 119887 (119909) 1199062) 119906 = 120582119891 (119909) (73)

since 119886(119909) minus 119887(119909)1199062 lt 119886(119909) ≪ 12058724 it follows fromLemma 14 that 119906(119909) gt 0 or 119906(119909) lt 0 This shows that thesolution changes its sign only at (120582 119906) = (0 0) Since thenonlinearity 119892(119909 119906) = 119886(119909)119906 minus 119887(119909)1199063 is an odd function in119906 it follows from the symmetry that if (120582 119906(119909)) is a solutionof (1) so is (minus120582 minus119906(119909)) Thus the component of solutions weconstructed above forms a smooth reversed S-shaped curvewith exactly two turning points (120582lowast 119906lowast) and (minus120582lowast minus119906lowast)

It follows from Lemma 18 that the positive solution 1199060is stable For 120582 lt 120582lowast the upper branch 119906+(119909 120582) remainsstable until it reaches the degenerate solution (120582lowast 119906lowast) Nextwe prove the lower branch 119906minus(119909 120582) is unstable Let 119906119898 beany one solution of lower branch 119906minus(119909 120582) such that 0 lt119906119898 lt 119906lowast Since 119886(119909) minus 3119887(119909)1199062119898 gt 119886(119909) minus 3119887(119909)1199062lowast we have1205831(119886(119909) minus 3119887(119909)1199062119898) lt 1205831(119886(119909) minus 3119887(119909)1199062lowast) = 0 Thereforethe lower branch 119906minus(119909 120582) is unstable Since the solution set issymmetric with respect to the origin the stability of negativesolutions 119906(119909) is obtained by using the property of symmetryThe upper branch of the negative solutions is unstable andthe lower branch of negative solutions is stable Therefore allsolutions of (1) lie on a unique reversed S-shaped solutioncurve This completes the proof

8 Discrete Dynamics in Nature and Society

Data Availability

No data were used to support this study

Conflicts of Interest

The authors declare that there are no conflicts of interestregarding the publication of this paper

Acknowledgments

This work was completed with the support of Tian YuanSpecial Funds of the National Science Foundation of China(no 11626182)

References

[1] J Chu X Lin D Jiang D OrsquoRegan and R P Agarwal ldquoPositivesolutions for second-order superlinear repulsive singular Neu-mann boundary value problemsrdquo Positivity An InternationalMathematics Journal Devoted to Theory and Applications ofPositivity vol 12 no 3 pp 555ndash569 2008

[2] Z Li ldquoExistence of positive solutions of superlinear second-order Neumann boundary value problemrdquo Nonlinear AnalysisTheory Methods amp Applications An International Multidisci-plinary Journal vol 72 no 6 pp 3216ndash3221 2010

[3] G Bonanno and A Sciammetta ldquoExistence and multiplicityresults to Neumann problems for elliptic equations involvingthe p-Laplacianrdquo Journal of Mathematical Analysis and Appli-cations vol 390 no 1 pp 59ndash67 2012

[4] E Sovrano and F Zanolin ldquoIndefinite weight nonlinear prob-lems with Neumann boundary conditionsrdquo Journal of Mathe-matical Analysis and Applications vol 452 no 1 pp 126ndash1472017

[5] G Feltrin and E Sovrano ldquoThree positive solutions to anindefinite Neumann problem a shooting methodrdquo NonlinearAnalysis Theory Methods amp Applications An InternationalMultidisciplinary Journal vol 166 pp 87ndash101 2018

[6] J Llibre and L A Roberto ldquoOn the periodic solutions of aclass of Duffing differential equationsrdquoDiscrete and ContinuousDynamical Systems - Series A vol 33 no 1 pp 277ndash282 2013

[7] H Chen and Y Li ldquoStability and exact multiplicity of periodicsolutions of Duffing equations with cubic nonlinearitiesrdquo Pro-ceedings of the American Mathematical Society vol 135 no 12pp 3925ndash3932 2007

[8] A Lomtatidze and J Sremr ldquoOn periodic solutions to second-order Duffing type equationsrdquo Nonlinear Analysis Real WorldApplications vol 40 pp 215ndash242 2018

[9] C-C Tzeng K-C Hung and S-H Wang ldquoGlobal bifurcationand exact multiplicity of positive solutions for a positone prob-lem with cubic nonlinearityrdquo Journal of Differential Equationsvol 252 no 12 pp 6250ndash6274 2012

[10] S-H Wang and T-S Yeh ldquoA complete classification of bifurca-tion diagrams of a Dirichlet problemwith concave-convex non-linearitiesrdquo Journal of Mathematical Analysis and Applicationsvol 291 no 1 pp 128ndash153 2004

[11] K-C Hung and S-H Wang ldquoA theorem on S-shaped bifur-cation curve for a positone problem with convex-concavenonlinearity and its applications to the perturbed Gelfandproblemrdquo Journal of Differential Equations vol 251 no 2 pp223ndash237 2011

[12] H Pan and R Xing ldquoTime maps and exact multiplicity resultsfor one-dimensional prescribed mean curvature equationsrdquoNonlinear Analysis Theory Methods amp Applications An Inter-national Multidisciplinary Journal vol 74 no 4 pp 1234ndash12602011

[13] H Chen H Xing and X He ldquoBifurcation and stability ofsolutions to a logistic equation with harvestingrdquo MathematicalMethods in the Applied Sciences vol 38 no 8 pp 1623ndash16302015

[14] P Clrsquoement andG Sweers ldquoUniform anti-maximumprinciplesrdquoJournal of Differential Equations vol 164 no 1 pp 118ndash1542000

[15] W Reichel ldquoSharp parameter ranges in the uniform anti-maximum principle for second-order ordinary diffrential oper-atorsrdquo Zeitschrift fur Angewandte Mathematik und Physik vol54 no 5 pp 822ndash838 2003

[16] MG Crandall and PH Rabinowitz ldquoBifurcation perturbationof simple eigenvalues and linearized stabilityrdquo Archive forRational Mechanics and Analysis vol 52 pp 161ndash180 1973

[17] H Kielhofer Bifurcation Theory An Introduction with Applica-tions to PDEs Springer New York NY USA 2003

[18] S Oruganti J Shi and R Shivaji ldquoDiffusive logistic equationwith constant yield harvesting I Steady statesrdquo Transactions ofthe American Mathematical Society vol 354 no 9 pp 3601ndash3619 2002

[19] K Deimling Nonlinear Functional Analysis Springer Berlin Germany 1985

[20] J X Sun Nonlinear Functional Analysis and Its ApplicationsScience Press Beijing China 2007

[21] H Xing H Chen and X He ldquoExact multiplicity and stability ofsolutions of second-order Neumann boundary value problemrdquoApplied Mathematics and Computation vol 232 pp 1104ndash11112014

Hindawiwwwhindawicom Volume 2018

MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Mathematical Problems in Engineering

Applied MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Probability and StatisticsHindawiwwwhindawicom Volume 2018

Journal of

Hindawiwwwhindawicom Volume 2018

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawiwwwhindawicom Volume 2018

OptimizationJournal of

Hindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom Volume 2018

Engineering Mathematics

International Journal of

Hindawiwwwhindawicom Volume 2018

Operations ResearchAdvances in

Journal of

Hindawiwwwhindawicom Volume 2018

Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018

International Journal of Mathematics and Mathematical Sciences

Hindawiwwwhindawicom Volume 2018

Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom

The Scientific World Journal

Volume 2018

Hindawiwwwhindawicom Volume 2018Volume 2018

Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in

Nature and SocietyHindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom

Dierential EquationsInternational Journal of

Volume 2018

Hindawiwwwhindawicom Volume 2018

Decision SciencesAdvances in

Hindawiwwwhindawicom Volume 2018

AnalysisInternational Journal of

Hindawiwwwhindawicom Volume 2018

Stochastic AnalysisInternational Journal of

Submit your manuscripts atwwwhindawicom

Discrete Dynamics in Nature and Society 7

Multiplying (1) by V(119909) and subtracting from (62) multipliedby 119906 after that integrating over [0 1] we obtain

int1

0[119887 (119909) 1199063 (119909) + 120583119906 (119909)] V (119909) 119889119909

= minus120582int1

0119891 (119909) V (119909) 119889119909

(63)

Applying the mean-value theorem for (63) there exists 120585 isin[0 1] such that

ℎ (119906 (120585) 119887 (120585)) = 119887 (120585) 1199063 (120585) + 1205831119906 (120585)

= minus120582int10119891 (119909) V (119909) 119889119909int10V (119909) 119889119909

(64)

For all 119887(120585) le 1198870 = max119909isin[01]119887(119909) the only root ofℎ(119906(120585) 119887(120585)) = 119898 is negative for 119898 lt ℎ(radicminus120583131198870 1198870) lt 0Let119872 = ℎ(radicminus120583131198870 1198870) and

120582lowast = minus 119872int10V (119909) 119889119909

int10119891 (119909) V (119909) 119889119909 (65)

and it follows from (64) that 119906(120585) lt 0 for 120582 gt 120582lowast Thereforethere exists a 120582lowast gt 0 such that (1) has no positive solution for120582 gt 120582lowast

By applying Lemma 14 we obtain that 119906lowast(119909) is still apositive solution of (1) for 120582 = 120582lowast We denote the degeneratesolution (120582lowast 119906lowast) At (120582lowast 119906lowast) we verify that Lemma 16 canbe applied here It follows from (43) that 119863119906119865(120582lowast 119906lowast)[V] =V10158401015840 + [119886(119909) minus 3119887(119909)1199062lowast]V = 0 In fact 0 is simple and theprincipal eigenvalue of (5) and the first eigenfunction V(119909) gt0 therefore dim119873(119863119906119865(120582lowast 119906lowast)) = codim119877(119863119906119865(120582lowast 119906lowast)) =1 and the null space 119873(119863119906119865(120582lowast 119906lowast)) = spanV Thereforethe Fredholm index of 119863119906119865(120582lowast 119906lowast) is zero Conditions (1)and (2) of Lemma 16 are satisfied Next we verify Condition(3) of Lemma 16 Suppose on the contrary that119863120582119865(120582lowast 119906lowast) =minus119891(119909) isin 119877(119863119906119865(120582lowast 119906lowast)) namely there is a continuousfunction 119908 satisfying

11990810158401015840 + [119886 (119909) minus 3119887 (119909) 1199062lowast]119908 = minus119891 (119909) (66)

Consider the linearized equation of (1) at (120582lowast 119906lowast)V10158401015840 + [119886 (119909) minus 3119887 (119909) 1199062lowast] V = 0 (67)

Since 119865 is singular at (120582lowast 119906lowast) that is (67) has a nontrivialsolution V(119909) such that V(119909) gt 0 Multiplying (66) by Vsubtracting from (67) multiplied by 119908 and integrating byparts on [0 1] we have

0 = int1

0V10158401015840119908 minus 11990810158401015840V 119889119909 = int

1

0V (119909) 119891 (119909) 119889119909 (68)

a contradiction since both V(119909) and 119891(119909) are positiveTherefore Condition (3) of Lemma 16 is satisfied Near(120582lowast 119906lowast) the solutions of

119865 (120582 (119904) 119906 (119904)) = 0 for all 119904 isin [minus120575 120575] (69)

form a curve

(120582 (119904) 119906 (119904)) = (120582lowast + 120591 (119904) 119906lowast + 119904V + 119911 (119904)) | 119904isin [minus120575 120575] (120582 (0) 119906 (0)) = (120582lowast 119906lowast)

(70)

Differentiating (69) twice in 119904 setting 119904 = 0 and 119906119904|119904=0 = V(119909)and evaluating at (120582lowast 119906lowast) we have

11990610158401015840119904119904 + [119886 (119909) minus 3119887 (119909) 1199062lowast] 119906119904119904= 6119887 (119909) 119906lowastV2 + 120591119904119904 (0) 119891 (119909)

(71)

Multiplying the linearized equation (67) by 119906119904119904 subtractingfrom (71) multiplied by V and integrating by parts over [0 1]we obtain

120591119904119904 (0) = minus6 int10119887 (119909) 119906lowastV3119889119909

int10119891 (119909) V (119909) 119889119909 lt 0 (72)

Therefore (120582lowast 119906lowast) is a fold point of 119865(120582 119906) to the left Itfollows from the above that the curve of positive solutionscannot be continued to the right indefinitely for all 120582 gt 0Hence the positive solution curve will make a left turn at(120582lowast 119906lowast) Near the critical point (120582lowast 119906lowast) by the Crandall-Rabinowitz bifurcation theorem there are two branches ofpositive solutions denoted by the upper branch 119906+(119909 120582)and the lower branch 119906minus(119909 120582) with 119906minus(119909 120582) lt 119906+(119909 120582)It follows from Lemmas 14 15 and 18 that the fold point(120582lowast 119906lowast) is unique and the upper branch119906+(119909 120582) ismonotonedecreasing for all 120582 lt 120582lowast and lower branch 119906minus(119909 120582) ismonotone increasing for all 0 lt 120582 lt 120582lowast Therefore the lowerbranch curve ismonotone increasing and continues to the leftwithout any turnings Rewriting (1) in the following form

11990610158401015840 + (119886 (119909) minus 119887 (119909) 1199062) 119906 = 120582119891 (119909) (73)

since 119886(119909) minus 119887(119909)1199062 lt 119886(119909) ≪ 12058724 it follows fromLemma 14 that 119906(119909) gt 0 or 119906(119909) lt 0 This shows that thesolution changes its sign only at (120582 119906) = (0 0) Since thenonlinearity 119892(119909 119906) = 119886(119909)119906 minus 119887(119909)1199063 is an odd function in119906 it follows from the symmetry that if (120582 119906(119909)) is a solutionof (1) so is (minus120582 minus119906(119909)) Thus the component of solutions weconstructed above forms a smooth reversed S-shaped curvewith exactly two turning points (120582lowast 119906lowast) and (minus120582lowast minus119906lowast)

It follows from Lemma 18 that the positive solution 1199060is stable For 120582 lt 120582lowast the upper branch 119906+(119909 120582) remainsstable until it reaches the degenerate solution (120582lowast 119906lowast) Nextwe prove the lower branch 119906minus(119909 120582) is unstable Let 119906119898 beany one solution of lower branch 119906minus(119909 120582) such that 0 lt119906119898 lt 119906lowast Since 119886(119909) minus 3119887(119909)1199062119898 gt 119886(119909) minus 3119887(119909)1199062lowast we have1205831(119886(119909) minus 3119887(119909)1199062119898) lt 1205831(119886(119909) minus 3119887(119909)1199062lowast) = 0 Thereforethe lower branch 119906minus(119909 120582) is unstable Since the solution set issymmetric with respect to the origin the stability of negativesolutions 119906(119909) is obtained by using the property of symmetryThe upper branch of the negative solutions is unstable andthe lower branch of negative solutions is stable Therefore allsolutions of (1) lie on a unique reversed S-shaped solutioncurve This completes the proof

8 Discrete Dynamics in Nature and Society

Data Availability

No data were used to support this study

Conflicts of Interest

The authors declare that there are no conflicts of interestregarding the publication of this paper

Acknowledgments

This work was completed with the support of Tian YuanSpecial Funds of the National Science Foundation of China(no 11626182)

References

[1] J Chu X Lin D Jiang D OrsquoRegan and R P Agarwal ldquoPositivesolutions for second-order superlinear repulsive singular Neu-mann boundary value problemsrdquo Positivity An InternationalMathematics Journal Devoted to Theory and Applications ofPositivity vol 12 no 3 pp 555ndash569 2008

[2] Z Li ldquoExistence of positive solutions of superlinear second-order Neumann boundary value problemrdquo Nonlinear AnalysisTheory Methods amp Applications An International Multidisci-plinary Journal vol 72 no 6 pp 3216ndash3221 2010

[3] G Bonanno and A Sciammetta ldquoExistence and multiplicityresults to Neumann problems for elliptic equations involvingthe p-Laplacianrdquo Journal of Mathematical Analysis and Appli-cations vol 390 no 1 pp 59ndash67 2012

[4] E Sovrano and F Zanolin ldquoIndefinite weight nonlinear prob-lems with Neumann boundary conditionsrdquo Journal of Mathe-matical Analysis and Applications vol 452 no 1 pp 126ndash1472017

[5] G Feltrin and E Sovrano ldquoThree positive solutions to anindefinite Neumann problem a shooting methodrdquo NonlinearAnalysis Theory Methods amp Applications An InternationalMultidisciplinary Journal vol 166 pp 87ndash101 2018

[6] J Llibre and L A Roberto ldquoOn the periodic solutions of aclass of Duffing differential equationsrdquoDiscrete and ContinuousDynamical Systems - Series A vol 33 no 1 pp 277ndash282 2013

[7] H Chen and Y Li ldquoStability and exact multiplicity of periodicsolutions of Duffing equations with cubic nonlinearitiesrdquo Pro-ceedings of the American Mathematical Society vol 135 no 12pp 3925ndash3932 2007

[8] A Lomtatidze and J Sremr ldquoOn periodic solutions to second-order Duffing type equationsrdquo Nonlinear Analysis Real WorldApplications vol 40 pp 215ndash242 2018

[9] C-C Tzeng K-C Hung and S-H Wang ldquoGlobal bifurcationand exact multiplicity of positive solutions for a positone prob-lem with cubic nonlinearityrdquo Journal of Differential Equationsvol 252 no 12 pp 6250ndash6274 2012

[10] S-H Wang and T-S Yeh ldquoA complete classification of bifurca-tion diagrams of a Dirichlet problemwith concave-convex non-linearitiesrdquo Journal of Mathematical Analysis and Applicationsvol 291 no 1 pp 128ndash153 2004

[11] K-C Hung and S-H Wang ldquoA theorem on S-shaped bifur-cation curve for a positone problem with convex-concavenonlinearity and its applications to the perturbed Gelfandproblemrdquo Journal of Differential Equations vol 251 no 2 pp223ndash237 2011

[12] H Pan and R Xing ldquoTime maps and exact multiplicity resultsfor one-dimensional prescribed mean curvature equationsrdquoNonlinear Analysis Theory Methods amp Applications An Inter-national Multidisciplinary Journal vol 74 no 4 pp 1234ndash12602011

[13] H Chen H Xing and X He ldquoBifurcation and stability ofsolutions to a logistic equation with harvestingrdquo MathematicalMethods in the Applied Sciences vol 38 no 8 pp 1623ndash16302015

[14] P Clrsquoement andG Sweers ldquoUniform anti-maximumprinciplesrdquoJournal of Differential Equations vol 164 no 1 pp 118ndash1542000

[15] W Reichel ldquoSharp parameter ranges in the uniform anti-maximum principle for second-order ordinary diffrential oper-atorsrdquo Zeitschrift fur Angewandte Mathematik und Physik vol54 no 5 pp 822ndash838 2003

[16] MG Crandall and PH Rabinowitz ldquoBifurcation perturbationof simple eigenvalues and linearized stabilityrdquo Archive forRational Mechanics and Analysis vol 52 pp 161ndash180 1973

[17] H Kielhofer Bifurcation Theory An Introduction with Applica-tions to PDEs Springer New York NY USA 2003

[18] S Oruganti J Shi and R Shivaji ldquoDiffusive logistic equationwith constant yield harvesting I Steady statesrdquo Transactions ofthe American Mathematical Society vol 354 no 9 pp 3601ndash3619 2002

[19] K Deimling Nonlinear Functional Analysis Springer Berlin Germany 1985

[20] J X Sun Nonlinear Functional Analysis and Its ApplicationsScience Press Beijing China 2007

[21] H Xing H Chen and X He ldquoExact multiplicity and stability ofsolutions of second-order Neumann boundary value problemrdquoApplied Mathematics and Computation vol 232 pp 1104ndash11112014

Hindawiwwwhindawicom Volume 2018

MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Mathematical Problems in Engineering

Applied MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Probability and StatisticsHindawiwwwhindawicom Volume 2018

Journal of

Hindawiwwwhindawicom Volume 2018

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawiwwwhindawicom Volume 2018

OptimizationJournal of

Hindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom Volume 2018

Engineering Mathematics

International Journal of

Hindawiwwwhindawicom Volume 2018

Operations ResearchAdvances in

Journal of

Hindawiwwwhindawicom Volume 2018

Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018

International Journal of Mathematics and Mathematical Sciences

Hindawiwwwhindawicom Volume 2018

Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom

The Scientific World Journal

Volume 2018

Hindawiwwwhindawicom Volume 2018Volume 2018

Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in

Nature and SocietyHindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom

Dierential EquationsInternational Journal of

Volume 2018

Hindawiwwwhindawicom Volume 2018

Decision SciencesAdvances in

Hindawiwwwhindawicom Volume 2018

AnalysisInternational Journal of

Hindawiwwwhindawicom Volume 2018

Stochastic AnalysisInternational Journal of

Submit your manuscripts atwwwhindawicom

8 Discrete Dynamics in Nature and Society

Data Availability

No data were used to support this study

Conflicts of Interest

The authors declare that there are no conflicts of interestregarding the publication of this paper

Acknowledgments

This work was completed with the support of Tian YuanSpecial Funds of the National Science Foundation of China(no 11626182)

References

[1] J Chu X Lin D Jiang D OrsquoRegan and R P Agarwal ldquoPositivesolutions for second-order superlinear repulsive singular Neu-mann boundary value problemsrdquo Positivity An InternationalMathematics Journal Devoted to Theory and Applications ofPositivity vol 12 no 3 pp 555ndash569 2008

[2] Z Li ldquoExistence of positive solutions of superlinear second-order Neumann boundary value problemrdquo Nonlinear AnalysisTheory Methods amp Applications An International Multidisci-plinary Journal vol 72 no 6 pp 3216ndash3221 2010

[3] G Bonanno and A Sciammetta ldquoExistence and multiplicityresults to Neumann problems for elliptic equations involvingthe p-Laplacianrdquo Journal of Mathematical Analysis and Appli-cations vol 390 no 1 pp 59ndash67 2012

[4] E Sovrano and F Zanolin ldquoIndefinite weight nonlinear prob-lems with Neumann boundary conditionsrdquo Journal of Mathe-matical Analysis and Applications vol 452 no 1 pp 126ndash1472017

[5] G Feltrin and E Sovrano ldquoThree positive solutions to anindefinite Neumann problem a shooting methodrdquo NonlinearAnalysis Theory Methods amp Applications An InternationalMultidisciplinary Journal vol 166 pp 87ndash101 2018

[6] J Llibre and L A Roberto ldquoOn the periodic solutions of aclass of Duffing differential equationsrdquoDiscrete and ContinuousDynamical Systems - Series A vol 33 no 1 pp 277ndash282 2013

[7] H Chen and Y Li ldquoStability and exact multiplicity of periodicsolutions of Duffing equations with cubic nonlinearitiesrdquo Pro-ceedings of the American Mathematical Society vol 135 no 12pp 3925ndash3932 2007

[8] A Lomtatidze and J Sremr ldquoOn periodic solutions to second-order Duffing type equationsrdquo Nonlinear Analysis Real WorldApplications vol 40 pp 215ndash242 2018

[9] C-C Tzeng K-C Hung and S-H Wang ldquoGlobal bifurcationand exact multiplicity of positive solutions for a positone prob-lem with cubic nonlinearityrdquo Journal of Differential Equationsvol 252 no 12 pp 6250ndash6274 2012

[10] S-H Wang and T-S Yeh ldquoA complete classification of bifurca-tion diagrams of a Dirichlet problemwith concave-convex non-linearitiesrdquo Journal of Mathematical Analysis and Applicationsvol 291 no 1 pp 128ndash153 2004

[11] K-C Hung and S-H Wang ldquoA theorem on S-shaped bifur-cation curve for a positone problem with convex-concavenonlinearity and its applications to the perturbed Gelfandproblemrdquo Journal of Differential Equations vol 251 no 2 pp223ndash237 2011

[12] H Pan and R Xing ldquoTime maps and exact multiplicity resultsfor one-dimensional prescribed mean curvature equationsrdquoNonlinear Analysis Theory Methods amp Applications An Inter-national Multidisciplinary Journal vol 74 no 4 pp 1234ndash12602011

[13] H Chen H Xing and X He ldquoBifurcation and stability ofsolutions to a logistic equation with harvestingrdquo MathematicalMethods in the Applied Sciences vol 38 no 8 pp 1623ndash16302015

[14] P Clrsquoement andG Sweers ldquoUniform anti-maximumprinciplesrdquoJournal of Differential Equations vol 164 no 1 pp 118ndash1542000

[15] W Reichel ldquoSharp parameter ranges in the uniform anti-maximum principle for second-order ordinary diffrential oper-atorsrdquo Zeitschrift fur Angewandte Mathematik und Physik vol54 no 5 pp 822ndash838 2003

[16] MG Crandall and PH Rabinowitz ldquoBifurcation perturbationof simple eigenvalues and linearized stabilityrdquo Archive forRational Mechanics and Analysis vol 52 pp 161ndash180 1973

[17] H Kielhofer Bifurcation Theory An Introduction with Applica-tions to PDEs Springer New York NY USA 2003

[18] S Oruganti J Shi and R Shivaji ldquoDiffusive logistic equationwith constant yield harvesting I Steady statesrdquo Transactions ofthe American Mathematical Society vol 354 no 9 pp 3601ndash3619 2002

[19] K Deimling Nonlinear Functional Analysis Springer Berlin Germany 1985

[20] J X Sun Nonlinear Functional Analysis and Its ApplicationsScience Press Beijing China 2007

[21] H Xing H Chen and X He ldquoExact multiplicity and stability ofsolutions of second-order Neumann boundary value problemrdquoApplied Mathematics and Computation vol 232 pp 1104ndash11112014

Hindawiwwwhindawicom Volume 2018

MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Mathematical Problems in Engineering

Applied MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Probability and StatisticsHindawiwwwhindawicom Volume 2018

Journal of

Hindawiwwwhindawicom Volume 2018

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawiwwwhindawicom Volume 2018

OptimizationJournal of

Hindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom Volume 2018

Engineering Mathematics

International Journal of

Hindawiwwwhindawicom Volume 2018

Operations ResearchAdvances in

Journal of

Hindawiwwwhindawicom Volume 2018

Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018

International Journal of Mathematics and Mathematical Sciences

Hindawiwwwhindawicom Volume 2018

Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom

The Scientific World Journal

Volume 2018

Hindawiwwwhindawicom Volume 2018Volume 2018

Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in

Nature and SocietyHindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom

Dierential EquationsInternational Journal of

Volume 2018

Hindawiwwwhindawicom Volume 2018

Decision SciencesAdvances in

Hindawiwwwhindawicom Volume 2018

AnalysisInternational Journal of

Hindawiwwwhindawicom Volume 2018

Stochastic AnalysisInternational Journal of

Submit your manuscripts atwwwhindawicom

Hindawiwwwhindawicom Volume 2018

MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Mathematical Problems in Engineering

Applied MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Probability and StatisticsHindawiwwwhindawicom Volume 2018

Journal of

Hindawiwwwhindawicom Volume 2018

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawiwwwhindawicom Volume 2018

OptimizationJournal of

Hindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom Volume 2018

Engineering Mathematics

International Journal of

Hindawiwwwhindawicom Volume 2018

Operations ResearchAdvances in

Journal of

Hindawiwwwhindawicom Volume 2018

Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018

International Journal of Mathematics and Mathematical Sciences

Hindawiwwwhindawicom Volume 2018

Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom

The Scientific World Journal

Volume 2018

Hindawiwwwhindawicom Volume 2018Volume 2018

Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in

Nature and SocietyHindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom

Dierential EquationsInternational Journal of

Volume 2018

Hindawiwwwhindawicom Volume 2018

Decision SciencesAdvances in

Hindawiwwwhindawicom Volume 2018

AnalysisInternational Journal of

Hindawiwwwhindawicom Volume 2018

Stochastic AnalysisInternational Journal of

Submit your manuscripts atwwwhindawicom