# Reversed S-Shaped Bifurcation Curve for a Neumann Problem

Authors: Hui Xing; Hongbin Chen; Ruofei Yao
Journal: Discrete Dynamics in Nature and Society
 (2018)

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## Abstract

We study the bifurcation and the exact multiplicity of solutions for a class of Neumann boundary value problem with indefinite weight. 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 the equation has exactly either one, two, or three solutions depending on the real parameter. The stability is obtained by the eigenvalue comparison principle.


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## Body

## 1. Introduction

The existence and multiplicity of solutions of Neumann problems have been investigated by many authors; see, for example, [1–5]. It is well known that determining the exact number of solutions of semilinear equations is usually a very difficult and challenging task. The results on exact multiplicity of solutions for Neumann problems are very few in the previous literature.In this paper, we study the bifurcation and the exact multiplicity of solutions for the Neumann problem(1)u′′+axu-bxu3=λfx,x∈0,1,u′0=u′1=0,depending on a real parameter λ, where a(x), b(x), and f(x) are given continuous functions and b(x)>0 and a(x) may change sign.Llibre and Roberto [6] studied the existence and the stability of periodic solutions of Duffing equation(2)x′′+cx′+atx+btx3=λht.Chen and Li [7] also studied the Duffing equation (2) in a very particular case, i.e., a(t)≡a>0,b(t)=1, and a,c constants. They obtained that (2) has exactly three T-periodic solutions. Lomtatidze et al. [8] also studied the existence of periodic solutions of Duffing type equations.Tzeng et al. [9] also studied the global bifurcation and exact multiplicity of positive solutions of Dirichlet problem with cubic nonlinearity(3)u′′+λ-εu3+σu2-κu+ρ=0,-1<x<1,u-1=u1=0,using the time-map method, where ε,σ,κ,ρ are constants. Equation (3) is an autonomous system, and the time-map method has been successfully employed to solve the problem (3), but it is not applicable to study the nonautonomous system (1). There are many results on exact multiplicity of solutions for the Dirichlet problems; see [9–12].In [13], under Neumann boundary value conditions, the bifurcation of solutions to a logistic equation with harvesting has been investigated using the uniform antimaximum principle and Crandall-Rabinowitz bifurcation theorem. The uniform antimaximum principle plays an important role in proving the main results. More theories and applications of antimaximum principle can be seen, for example, [14, 15]. As continuation of [13], in this paper, the bifurcation of solutions for a Neumann problem with cubic nonlinearity is investigated using the uniform antimaximum principle, Crandall-Rabinowitz bifurcation theorem, the topological degree theory, and the continuation method. More detailed results on bifurcation theory can be seen in [16, 17]. The topological degree theory and the uniform antimaximum principle play an important role in proving the main results in this paper.


## 2. Preliminaries

Definition 1 (see [18]).
We call a solutionu of the equation(4)u′′+gx,u=fx,x∈0,1,u′0=u′1=0a stable solution if the principal eigenvalue μ1(gu(x,u)) of the equation(5)w′′+gux,uw=-μw,x∈0,1,w′0=w′1=0is strictly positive. The solution u is unstable if the principal eigenvalue μ1(gu(x,u)) is negative.Definition 2 (see [18]).
We call a solutionu of the equation(6)u′′+gx,u=fx,x∈0,1,u′0=u′1=0a nondegenerate solution if the linearized equation(7)w′′+gux,uw=0,x∈0,1,w′0=w′1=0does not admit any nontrivial solutions. The solution u is degenerate (singular) if the linearized equation (7) has nontrivial solutions.Definition 3 (see [19]).
A mappingG:X→Y (X and Y are topological spaces) is said to be proper if for every compact set K⊂Y, the set G-1(K) is compact in X.Lemma 4 (see [19]).
LetX be a real Banach space and L a linear, compact map in X. Suppose that λ≠0 and λ-1 is not an eigenvalue of L. Let Ω⊂X be an open bounded and 0∈Ω. Then(8)degI-λL,Ω,0=-1mλ,where m(λ) is the sum of the algebraic multiplicities of all the eigenvalues μ satisfying μλ>1, and m(λ)=0 if L has no eigenvalues μ of this kind.Lemma 5.
Consider the eigenvalue problem(9)-u′′x+ux=μu,0<x<1,u′0=u′1=0.Then (9) has the Green function(10)Gx,s=cosh⁡1-xcosh⁡ssinh⁡1,0≤s≤x≤1,cosh⁡1-scosh⁡xsinh⁡1,0≤x≤s≤1,and(11)degI-μL,BR,0=1,for-∞<μ<1,-1n+1,forn2π2+1<μ<n+12π2+1,where for any R>0, BR={u∈C[0,1]:u≤R} denotes a ball of radius R, and (Lu)(x)=∫01G(x,s)u(s)ds.Proof.
It is easy to obtain (10) and the proof is omitted. The eigenvalue problem (9) is equivalent to the equation u-μLu=0. The equation u-μLu=0 has nontrivial solutions if and only if μ-1 is the eigenvalue of the operator L. For μ>1, the general solution of equation (9) is(12)ux=ccos⁡μ-1x+dsin⁡μ-1x.By Neumann boundary value condition, we have(13)-cμ-1sin⁡μ-1=0.When c≠0, we have μ=n2π2+1, where n≥0, n∈N. Therefore, λn=(n2π2+1)-1 is the eigenvalue of L, where n≥0, N(L-λn) is the one-dimensional subspace of Banach space X spanned by un=cos⁡(nπx).
Forn≥0, Lemma 4 implies that(14)degI-μL,BR,0=1,for-∞<μ<1,-1n+1,forn2π2+1<μ<n+12π2+1.This completes the proof.Consider the eigenvalue problem(15)u′′+axu+μu=0,x∈0,1,u′0=u′1=0.We recall some propositions listed as follows.Proposition 6.
It is well known that the problem (15) has the eigenvalues μ1(a(x))<μ2(a(x))≤μ3(a(x))≤⋯. The first eigenvalue μ1(a(x)) is real and simple, and the corresponding eigenfunction does not change sign. When a(x)≡0, we denote μi(0) by μi. It is obvious that μ1=0 is the first eigenvalue and μ2=π2 is the second eigenvalue of the eigenvalue equation(16)u′′+μu=0,x∈0,1,u′0=u′1=0.Proposition 7.
The comparison theorem of eigenvalues can be stated as follows:μi(a(x)) is strictly decreasing in the sense that a1(x)≪a2(x) implies that μi(a1(x))>μi(a2(x)), where a1(x)≪a2(x), namely, a1(x)≤a2(x), with the strict inequality on a set of positive measure.Proposition 8.
Suppose thatμk≪a(x)≪μk+1,k=1,2,…. Then the equation(17)u′′+axu=0,x∈0,1,u′0=u′1=0does not admit any nontrivial solutions.Proof.
Argue by contradiction, (17) admits nontrivial solutions. Let u(x) be a nontrivial solution of the equation (17). It follows from μk≪a(x)≪μk+1 that μk(a(x))<μk(μk)=0 and μk+1(a(x))>μk+1(μk+1)=0, for k=1,2,3⋯. It follows from (17) that 0 is the eigenvalue, contradiction. Therefore, (17) admits only trivial solution 0. Equation (17) does not admit any nontrivial solutions. This completes the proof.In order to prove our main theorem using the topological degree theory, the following lemma is essential. In the following lemma, the weight functiona(x) may change sign.Lemma 9.
Suppose thata(x)≪π2 satisfies the equation(18)Lau=u′′+axu=0,x∈0,1,u′0=u′1=0.Then(19)degI-A,BR,0=1,forμ1ax>0,-1,forμ1ax<0,where (Au)(x)=∫01‍G(x,s)(a(s)+1)u(s)ds, G(x,s) is defined in (10), for any R>0, BR={u∈C[0,1]:u≤R} denotes a ball of radius R.Proof.
Equation (18) is equivalent to the equation(20)ux=∫01Gx,sas+1usds.In (11), let n=0 and μ-1=β; we have(21)degI-β+1L,BR,0=1,for-∞<β<0,-1,for0<β<π2.Next we calculate deg(I-A,BR,0). Let H(t,u)=u-(1-t)(β+1)Lu-tAu. Now we prove that H(t,u)≠0 for ∀t∈[0,1] and x∈∂BR. Argue by contradiction and assume that H(t,u)=0; that is, for ∀t∈[0,1], the equation(22)u′′+1-tβu+taxu=0,x∈0,1,u′0=u′1=0has solution u satisfying u=R on ∂BR.
Case  1. When a(x) does not change sign:
(i) Fora(x)≪0, choose β<0; that is, for ∀t∈[0,1], ((1-t)β+ta(x))<0, by Proposition 7, we have μ1((1-t)β+ta(x))>μ1(0)=0. It follows that (22) has only trivial solution u≡0. Contradiction, therefore, we have H(t,u)≠0 for t∈[0,1], x∈∂BR. Using the homotopy invariance we get(23)deg⁡I-A,BR,0=deg⁡H1,·,BR,0=deg⁡H0,·,BR,0=deg⁡I-β+1L,BR,0=1.Therefore, for a(x)≪0, that is, μ1(a(x))>0,(24)degI-A,BR,0=1.
(ii) For0≪a(x)≪π2, choose β∈(0,π2), that is, for ∀t∈[0,1], 0≪(1-t)β+ta(x)≪π2, we have μ1((1-t)β+ta(x))<μ1(0)=0 and μ2((1-t)β+ta(x))>μ2(π2)=0. By Proposition 8, it follows that (22) has only trivial solution u≡0, contradiction. Therefore, we have H(t,u)≠0 for t∈[0,1], x∈∂BR. Using the homotopy invariance, we get(25)degI-A,BR,0=degH1,·,BR,0=degH0,·,BR,0=degI-β+1L,BR,0=-1.Therefore, for μ1(a(x))<0, a(x)≪π2, we have(26)degI-A,BR,0=-1.
Case  2. When a(x) changes sign, let a(x)=a+(x)-a-(x), where a+(x)=max{a(x),0}, a-(x)=max{-a(x),0}.
(i) Forμ1(a(x))>0, consider the following equation:(27)u′′+ta+x-a-xu=0,x∈0,1,u′0=u′1=0.where t∈[0,1]. Obviously,(28)ta+x-a-x≤a+x-a-x=ax,∀t∈0,1.By Proposition 7 and (28), we have(29)μ1ta+x-a-x≥μ1ax>0,∀t∈0,1.When t=0, the solution of (27) is(30)ux=∫01Gx,s-a-s+1usds.Let (L1u)(x)=∫01‍G(x,s)(-a-(s)+1)u(s)ds; since -a-(s)≪0, we have(31)degI-L1,BR,0=1.The solution of (27) is(32)ux=∫01Gx,sta+s-a-s+1usds.Let h1(t,u)=u-tAu-(1-t)L1u. It is easy to prove that h1(t,u)≠0 for ∀t∈[0,1], u∈∂BR. Therefore, using the homotopy invariance, we get(33)degI-A,BR,0=degh11,·,BR,0=degh10,·,BR,0=degI-L1,BR,0=1.
(ii) Forμ1(a(x))<0, a(x)≪π2, by Proposition 7, μ2(a(x))>0; similarly, let(34)L2ux=∫01Gx,sa+s+1usds.Since a+(s)≫0, we have(35)degI-L2,BR,0=-1.
Leth2(t,u)=u-tAu-(1-t)L2u. It is easy to prove that h2(t,u)≠0 for ∀t∈[0,1], u∈∂BR. Therefore, using the homotopy invariance we get(36)deg⁡I-A,BR,0=deg⁡h21,·,BR,0=deg⁡h20,·,BR,0=deg⁡I-L2,BR,0=-1.This completes the proof.Definition 10 (see [20]).
LetΩ be open set in Banach space E. Suppose A:Ω¯→E is completely continuous operator, f=I-A. Let x0∈Ω be an isolated fixed point of f in Ω; that is, there exists r>0, such that Br={x:x-x0<r}⊂Ω, and x0 is only fixed point of f in B¯r. Define(37)indI-A,x0=degI-A,Br,0.Lemma 11 (see [20]).
LetΩ be open set in Banach space E. Suppose A:Ω¯→E is completely continuous operator; A has no fixed point on ∂Ω. Suppose that there are finite isolated fixed points x1,x2,…,xm. Then(38)degI-A,Ω,0=∑i=1mindI-A,xi.Lemma 12 (see [20]).
LetΩ be open set in Banach space E. Suppose A:Ω¯→E is completely continuous operator, x0∈Ω, Ax0=x0. Suppose that A is Fréchet differentiable at x0 and 1 is not the eigenvalue of derived operator A′(x0). Then x0 is an isolated fixed point of A and(39)indI-A,x0=indI-A′x0,0=-1β,where β is the sum of the algebraic multiplicities of all the eigenvalues A′(x0) in (0,1).Now we state three known results, which plays a key role in proving the main results in this paper. We do not provide their proof which can be seen in [21].Lemma 13.
Suppose thata(x),a1(x), and a2(x)∈C[0,1] such that(40)ax,a1x,a2x≪π2.Then(1)
the possible solutionu of (18) is either u(x)≡0 or u(x)≠0 for each x∈[0,1];(2)
Laiu=0(i=1,2) cannot both admit nontrivial solutions if a1(x)≪a2(x).Lemma 14.
Letf(x)>0 on [0,1] and a(x) satisfy(41)ax≪π24.If u(x) is a solution of the nonhomogeneous differential equation(42)u′′+axu=fx,x∈0,1,u′0=u′1=0,then the following statements hold:(1)
eitheru(x)>0 or u(x)<0for allx∈[0,1];(2)
maximum principle:u(x)<0,ifμ1(a(x))>0;(3)
uniform antimaximum principle:u(x)>0,ifμ1(a(x))<0.For the reader’s convenience, we denote the continuous functiong(x,u)=a(x)u-b(x)u3 in (1). The assumptions on function g(x,u) are listed as follows:(f1)
g(x,u) is locally differentiable with respect to the second variable with gu(x,u)≪π2;(f2)
liminfu→±∞⁡u(g(x,u)-λf(x))<0,uniformly inx∈[0,1].Lemma 15.
LetX be an order Banach space and [w,v]={u∈X∣w≤u≤v} an order set. Assume that the given continuous function g(x,u) satisfies (f1). Then(1)
the solutions of (1) are totally ordered;(2)
(1) cannot admit three distinct solutions in [w,v] if gu(x,u) is strictly increasing or strictly decreasing in [w,v].For the convenience, we recast (1) in the operator form(43)Fλ,u≔u′′+gx,u-λfx,R×C20,1→C0,1.Lemma 16 (see [16]).
LetX and Z be Banach spaces. Assume that F:R×U→Z is continuously differentiable on V×U⊂R×X satisfying the following three conditions:(1)
F(λ∗,u∗)=0 for some (λ∗,u∗)∈V×U, dim⁡N(DuF(λ∗,u∗))=1 and the null space N(DuF(λ∗,u∗))=span{w};(2)
the Fredholm index ofDuF(λ∗,u∗) is zero, codimR(DuF(λ∗,u∗))=1;(3)
DλF(λ∗,u∗)∈R(DuF(λ∗,u∗)).
Then there is a continuously differentiable curve through(λ∗,u∗); that is, there exists {(λ(s),u(s))=(λ∗+τ(s),u∗+sw+z(s))∣s∈(-δ,δ),(λ(0),u(0))=(λ∗,u∗)} such that(44)Fλs,us=0,fors∈-δ,δ,and all solutions of F(λ,u)=0 in a neighborhood of (λ∗,u∗) belong to the curve, τ(0)=τ′(0)=z(0)=z′(0)=0.


## 3. Main Results

Lemma 17.
Assume (f2) holds. Then for every fixedλ, (1) has at least a solution and(45)degI-A,BR,0=1,for R large enough, where (Au)(x)=∫01‍G(x,s)(g(s,u(s))+u(s)-λf(s))ds and G is defined in (10).Proof.
It follows from (f2) that there existsR>0 large enough such that g(x,R)-λf(x)<0 and g(x,-R)-λf(x)>0. Therefore, -R and R are subsolution and supersolution of (1), respectively. Therefore, there exists at least one solution of (1) between -R and R. Next, we show that all the solutions of (1) are between -R and R for R>0 large enough. Argue by contradiction, suppose that there exists x0 such that u(x) attains its maximum value and u(x0)>R. We have(46)u′′x0+ax0ux0-bx0u3x0-λfx0=0.It follows from u′(x0)=0 and u′′(x0)<0 that(47)ax0ux0-bx0u3x0-λfx0≥0.From the above there exists R>0 large enough such that g(x,R)-λf(x)<0, contradiction. Similar, suppose that there exists x1 such that u(x) attains its minimum value and u(x1)<-R; we will obtain contradiction. Therefore, there exists R>0 large enough such that u<R for all the solutions u(x). (1) is equivalent to the following equation:(48)ux=Aux.Next we calculate deg(I-A,BR,0). Let h(t,u)=u-tAu,∀t∈[0,1]. The equation h(t,u)=0 is equivalent to the following equation:(49)u′′+tgx,u-1-tu-tλfx=0.It is evident that -R and R are subsolution and supersolution of (49), respectively. Therefore, all the solutions of (49) must be between -R and R; that is, the equation h(t,u)=0 has no solution on ∂BR. Therefore, h(t,u)≠0 for t∈[0,1], u∈∂BR. By the homotopy invariance properties of the topological degree we have(50)degI-A,BR,0=degh·,1,BR,0=degh·,0,BR,0=degI,BR,0=1.This completes the proof.Lemma 18.
Suppose thatμ1(a(x))<0, a(x)≪π2, and f(x)>0 for all x∈[0,1]. Then (1) has exactly three solutions u0(x), 0,v0(x) for λ=0, where u0(x) is the unique positive stable solution and v0(x) is the unique negative stable solution.Proof.
For the mappingF defined in (43), let λ=0, and we have(51)Fu=u′′+axu-bxu3.Next we prove that any solution u(x) of F(u)=0 is nondegenerate.
First, we have the Fréchet derivative of (51)(52)Fuuw=w′′+ax-3bxu2w.It is obvious that u=0 is the solution of F(u)=0, and therefore, when u=0, a(x)-3b(x)u2=a(x) in (52). It follows from μ1(a(x))<0 that 0 is nondegenerate.
Next, we prove that any nontrivial solutionu(x) of F(u)=0 is nondegenerate. F(u)=0 is equivalent to the following equation:(53)ux=Aux,where (Au)(x)=∫01‍G(x,s)(g(s,u(s))+u(s))ds and G is defined in (10). The equation(54)w-A′uw=0is equivalent to Fu(u)[w]=0, where A′(u) denotes the derivative operator of A. When μ1(a(x))<0, a(x)≪π2, by Lemma 9, Definition 10, and Lemma 12, there exists R>0, and we have(55)indI-A,0=indI-A′0,0=deg⁡I-A,BR,0=-1.If u(x) is a nontrivial solution of F(u)=0, it follows from Lemma 13 that u(x)≠0 for all x∈[0,1]. Therefore,(56)gux,u=ax-3bxu2x<gx,uu=ax-bxu2x≪π2.Let q1(x)=g(x,u)/u and q2(x)=gu(x,u). It follows from the above hypothesis that u(x) is a nontrivial solution of Lq1u=0. By the second conclusion of Lemma 13, we have that Lq2u=0 does not have nontrivial solution, which implies that u(x) is a nondegenerate solution.
Finally, we prove the positive solution is unique and stable. We denote byu0 the nontrivial solution of (51). It is obvious that -u0(x) is also the nontrivial solution of (51). It follows from (56) and the comparison of eigenvalues that μ1q2x>μ1q1x=0. By Lemma 9, Definition 10, and Lemma 12, we have(57)indI-A,u=indI-A′u,0=degI-A,BR,0=1.Let m be the number of nontrivial solutions of F(u)=0. Hence, F is proper. Since 0 is a regular value of F, m must be finite. According to Lemma 17, we have deg(I-A,BR,0)=1 for sufficiently large R. By Lemmas 9 and 11 and the index formula, we have that(58)1=degI-A,BR,0=indI-A,0+∑u≠0indI-A,ui=-1+m.It is obvious that m=2. Let v0(x)≡-u0(x) for x∈[0,1]. Therefore, F(u)=0 has exactly three solutions u0,0, and v0. Since μ1(q2)>μ1(q1)=0, we have that the positive solution u0(x) is unique and stable. The negative solution v0(x) is also stable. This completes the proof.Lemma 19.
Suppose thata(x)≪π2/4. For some λ<0, (1) has a unique positive solution u1(x) with u1(x)>u0(x), where u0(x) is the unique positive solution of (1) for λ=0.Proof.
By Lemma18, for λ=0 (1) has the unique positive solution u0, which provided a subsolution of (1) for λ<0. There exists R>0 large enough such that g(x,R)-λf(x)<0. Thus, R is a supersolution of (1). Therefore, we prove that there exists a positive solution u1(x) of (1) for λ<0 such that u0(x)<u1(x)<R.
Next we will prove that the positive solutionu1 is unique. Assume, by contradiction that (1) has another positive solution u(x) for λ<0. Let v(x)=u(x)-u1(x)≠0; then w satisfies the following equation:(59)v′′+axv-bxu2x+uxu1x+u12xv=0.Clearly, a(x)-b(x)[u2(x)+u(x)u1(x)+u12(x)]<a(x)≪π2/4. By Lemma 13, v(x)≡0 is the solution of (59), which is a contradiction. Therefore, the solution u1 obtained above is only positive solution. For λ<0, u1 satisfies (1) and(60)u1′′+qxu1=λfx,where q(x)=a(x)-b(x)u12(x). Again since f(x)>0 and λ<0, it follows from Lemma 14 that μ1(q(x))>0. Consider the linearization associated with (1)(61)w′′+ax-3bxu2xw=0,where gu(x,u)=a(x)-3b(x)u2(x)<q(x). Thus, μ1(gu(x,u))>μ1(q(x))>0. Therefore, by Lemma 13 we have u1(x) is nondegenerate. This completes the proof.Lemma 20.
Forλ>0, (1) has a unique negative solution v1(x) with v1(x)<v0(x), where v0(x) is the unique negative solution of (1) for λ=0.Proof.
The proof is similar to the proof of Lemma19.Theorem 21.
Assume that the first eigenvalueμ1(a(x))<0 and f(x)>0 for all x∈[0,1]. Suppose that a(x)≪π2/4. Then all the solutions of (1) are of one sign and lie on a unique reversed S-shaped solution curve, which is symmetric with respect to the origin. More precisely, there exists λ∗>0, such that(i)
Forλ>λ∗, (1) has no positive solution and has a unique negative solution which is stable.(ii)
Forλ=±λ∗, (1) has exactly two solutions. Moreover, when λ=λ∗, the negative solution is stable and the positive solution is degenerate. When λ=-λ∗, the positive solution is stable and the negative solution is degenerate.(iii)
For-λ∗<λ<λ∗, (1) has exactly three ordered solutions at the same λ and the middle solution is unstable and the remaining two are stable. Moreover, when λ<0, the maximal solution is positive and the other two are negative. When λ>0, the minimal solution is negative and the other two are positive.(iv)
Forλ<-λ∗, (1) has no negative solution and has a unique positive solution which is stable.Proof.
It follows from Lemma19 that (1) has a unique nondegenerate positive solution u1(x) for some λ<0. The solution curve can be continued a little bit such that u1(x) remains positive for increasing λ when λ<0. By Lemma 18, (1) has a unique nondegenerate positive solution u0(x) for λ=0. The positive solution curve can pass through u1(x) and u0(x) and can be continued further for increasing λ until the linearized equation (61) admits the nontrivial solutions. We claim that the curve of positive solutions cannot be continued for λ>λ∗. Next we will prove the existence of λ∗. Since μ1(a(x))<0, let v(x)>0 be the first eigenfunction of the problem(62)v′′+axv+μv=0,x∈0,1,v′0=v′1=0.Multiplying (1) by v(x) and subtracting from (62) multiplied by u, after that integrating over [0,1], we obtain(63)∫01bxu3x+μuxvxdx=-λ∫01fxvxdx.Applying the mean-value theorem for (63), there exists ξ∈[0,1] such that(64)huξ,bξ=bξu3ξ+μ1uξ=-λ∫01fxvxdx∫01vxdx.For all b(ξ)≤b0=maxx∈[0,1]⁡b(x), the only root of h(u(ξ),b(ξ))=m is negative for m<h(-μ1/3b0,b0)<0. Let M=h(-μ1/3b0,b0) and(65)λ∗=-M∫01vxdx∫01fxvxdx,and it follows from (64) that u(ξ)<0 for λ>λ∗. Therefore, there exists a λ∗>0 such that (1) has no positive solution for λ>λ∗.
By applying Lemma14, we obtain that u∗(x) is still a positive solution of (1) for λ=λ∗. We denote the degenerate solution (λ∗,u∗). At (λ∗,u∗), we verify that Lemma 16 can be applied here. It follows from (43) that DuF(λ∗,u∗)[v]=v′′+[a(x)-3b(x)u∗2]v=0. In fact, 0 is simple and the principal eigenvalue of (5), and the first eigenfunction v(x)>0; therefore, dimN(DuF(λ∗,u∗))=codimR(DuF(λ∗,u∗))=1 and the null space N(DuF(λ∗,u∗))=span{v}. Therefore, the Fredholm index of DuF(λ∗,u∗) is zero. Conditions (1) and (2) of Lemma 16 are satisfied. Next we verify Condition (3) of Lemma 16. Suppose on the contrary that DλF(λ∗,u∗)=-f(x)∈R(DuF(λ∗,u∗)); namely, there is a continuous function w satisfying(66)w′′+ax-3bxu∗2w=-fx.Consider the linearized equation of (1) at (λ∗,u∗)(67)v′′+ax-3bxu∗2v=0.Since F is singular at (λ∗,u∗), that is, (67) has a nontrivial solution v(x) such that v(x)>0. Multiplying (66) by v, subtracting from (67) multiplied by w and integrating by parts on [0,1], we have(68)0=∫01v′′w-w′′vdx=∫01vxfxdx,a contradiction, since both v(x) and f(x) are positive. Therefore, Condition (3) of Lemma 16 is satisfied. Near (λ∗,u∗), the solutions of(69)Fλs,us=0,for  alls∈-δ,δform a curve(70)λs,us=λ∗+τs,u∗+sv+zs∣s∈-δ,δ,λ0,u0=λ∗,u∗.Differentiating (69) twice in s, setting s=0 and uss=0=v(x), and evaluating at (λ∗,u∗), we have(71)uss′′+ax-3bxu∗2uss=6bxu∗v2+τss0fx.Multiplying the linearized equation (67) by uss, subtracting from (71) multiplied by v, and integrating by parts over [0,1], we obtain(72)τss0=-6∫01bxu∗v3dx∫01fxvxdx<0.Therefore, (λ∗,u∗) is a fold point of F(λ,u) to the left. It follows from the above that the curve of positive solutions cannot be continued to the right indefinitely for all λ>0. Hence, the positive solution curve will make a left turn at (λ∗,u∗). Near the critical point (λ∗,u∗), by the Crandall-Rabinowitz bifurcation theorem, there are two branches of positive solutions denoted by the upper branch u+(x,λ) and the lower branch u-(x,λ) with u-(x,λ)<u+(x,λ). It follows from Lemmas 14, 15, and 18 that the fold point (λ∗,u∗) is unique, and the upper branch u+(x,λ) is monotone decreasing for all λ<λ∗ and lower branch u-(x,λ) is monotone increasing for all 0<λ<λ∗. Therefore, the lower branch curve is monotone increasing and continues to the left without any turnings. Rewriting (1) in the following form:(73)u′′+ax-bxu2u=λfx,since a(x)-b(x)u2<a(x)≪π2/4, it follows from Lemma 14 that u(x)>0 or u(x)<0. This shows that the solution changes its sign only at (λ,u)=(0,0). Since the nonlinearity g(x,u)=a(x)u-b(x)u3 is an odd function in u, it follows from the symmetry that if (λ,u(x)) is a solution of (1), so is (-λ,-u(x)). Thus, the component of solutions we constructed above forms a smooth reversed S-shaped curve with exactly two turning points (λ∗,u∗) and (-λ∗,-u∗).
It follows from Lemma18 that the positive solution u0 is stable. For λ<λ∗, the upper branch u+(x,λ) remains stable until it reaches the degenerate solution (λ∗,u∗). Next we prove the lower branch u-(x,λ) is unstable. Let um be any one solution of lower branch u-(x,λ) such that 0<um<u∗. Since a(x)-3b(x)um2>a(x)-3b(x)u∗2, we have μ1(a(x)-3b(x)um2)<μ1(a(x)-3b(x)u∗2)=0. Therefore, the lower branch u-(x,λ) is unstable. Since the solution set is symmetric with respect to the origin, the stability of negative solutions u(x) is obtained by using the property of symmetry. The upper branch of the negative solutions is unstable and the lower branch of negative solutions is stable. Therefore, all solutions of (1) lie on a unique reversed S-shaped solution curve. This completes the proof.



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*Source: 5376075-2018-08-01.xml*