# Exact Multiplicity of Sign-Changing Solutions for a Class of Second-Order Dirichlet Boundary Value Problem with Weight Function

Authors: Yulian An
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

Using bifurcation techniques and Sturm comparison theorem, we establish exact multiplicity results of sign-changing or constant sign solutions for the boundary value problemsu″+a(t)f(u)=0, t∈(0,1), u(0)=0, and u(1)=0, where f∈C(ℝ,ℝ) satisfies f(0)=0 and the limits f∞=lim|s|→∞(f(s)/s), f0=lim|s|→0(f(s)/s)∈{0,∞}. Weight function a(t)∈C1[0,1] satisfies a(t)>0 on [0,1].


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## Body

## 1. Introduction

In this paper, we consider the existence and exact multiplicity of sign-changing solutions for the boundary value problem(1)u′′+a(t)f(u)=0,t∈(0,1),u(0)=0,u(1)=0,
where f∈C(ℝ,ℝ) satisfies f(0)=0 and weight function a(t)∈C1[0,1] satisfies a(t)>0 on [0,1]. The existence and multiplicity of positive or sign-changing solutions of boundary value problems have been extensively studied in the literature, see [1–7] and references therein.However, for most nonlinearitiesf, a full description for the positive or sign-changing solution sets of many boundary value problems remains open. For some different boundary value problems, [8, 9] studied the exact multiplicity of positive solutions by bifurcation techniques, and [10–12] discussed the bifurcation diagrams of positive solutions by analyzing corresponding time maps. Recently, multiplicity of positive solutions to boundary blow-up elliptic problems with sign-changing weights was considered by [13].As for exact multiplicity of sign-changing solutions, only few papers considered this problem. In [14], Shi studied the problem
(2)u′′+λf(u)=0,t∈(0,1),u(0)=0,u(1)=0,
under the conditions(C1)
f∈C1(ℝ,ℝ) satisfies f(0)=0, f′(u)>0 for u∈ℝ;(C2)
f′(u)>f(u)/u, if u≠0;(C3)
the limitf∞=lim|s|→∞(f(s)/s)∈(0,∞).By using the implicit function theorem and local bifurcation theorems, the author obtained a full description of the set of sign-changing solutions of (2) for all values of λ. The set consists of some curves which bifurcate from the trivial solution line u=0 and tend to infinity. Particularly, there is no any turning points on these curves. Thus, they obtained the exact number of sign-changing solutions of the problem (2) for every given λ∈ℝ. Bari and Rynne [15] considered the 2mth (m≥2) order boundary value problem
(3)(-1)mu(2m)(t)=λg(u(t))u(t),t∈(0,1),u(i)(-1)=u(i)(1)=0,i=0,…,m-1,
where λ is a positive parameter, and the function g∈C1(ℝ,ℝ) satisfies g(0)>0, ±g′(ξ)>0, for ±ξ>0, and lim|ξ|→∞g(ξ)=∞. They got results similar to those in [14].The basic steps developed in [14, 15] to prove exact multiplicity of sign-changing solutions involve: showing any nontrivial solution of (2) (or (3)) to be nondegenerate and proving uniqueness of solution curve on which any nontrivial solution of (2) (or (3)) has certain zero point number.For other works on the exact multiplicity of sign-changing, see [16–19, 23]. In [16, 18, 23], the main tools are also bifurcation techniques. Time maps and exact multiplicity results of sign-changing solutions for one-dimensional prescribed mean curvature equations were considered by [17, 19]. However, all equations that had been studied in these works do not contain weight function a(t). Reference [20] discussed the existence and multiplicity of sign-changing solutions of some boundary value problems with weight functions. This work was extended to more general cases by [21] by shooting method and [22] by bifurcation method.In this paper, we consider the exact multiplicity of sign-changing solutions of (1). Compared with [14–19, 23], this paper considers the case that the nonlinearity contains a weight function a(t)≠1, t∈[0,1]. On the other hand, we discuss the case that f0,f∞∈{0,∞} instead of f0,f∞∈(0,∞). The main difficulty is to show any nontrivial solution of (1) to be nondegenerate. We will introduce an auxiliary function g(·) to deal with it. The method is motivated by the proof of Lemma  2.6 in [24], where the authors study global positive curves for a class of two-point boundary value problems. Compared with [24], we discuss not only the exact multiplicity of positive solutions but also of sign-changing solutions of (1).The organization of this paper is as follows. In Section2, we introduce some notations needed in later sections. We prove our main results in Sections 3 and 4. In Section 3, we study the exact multiplicity of sign-changing solutions of (1) under the hypotheses f0=0 and f∞=∞. In Section 4, we consider the exact multiplicity of constant sign solutions of (1) under the conditions f0=∞ and f∞=0.


## 2. Some Notations

For applying bifurcation theorem, we consider the auxiliary problem(4)u′′+μa(t)u+a(t)f(u)=0,t∈(0,1),u(0)=0,u(1)=0,
where μ∈ℝ is a parameter.Clearly, any solution of (4) of the form (0,u) corresponds to a solution u of (1). From f(0)=0, we know u≡0 is a solution of (4) for any μ∈ℝ, such solutions will be called trivial solutions.We introduce some notations to describe the properties of solutions of (4). For a function u∈C1[0,1], if u(x0)=0, then x0 is a simple zero of u if u′(x0)≠0. For any k∈ℕ:={1,2,…} and any ν∈{+,-}, we define sets Skν⊂C2[0,1] consisting of the functions u∈C2[0,1] with(i)
u(0)=0, νu′(0)>0;(ii)
u has only simple zeros in [0,1] and has exactly k-1 zeros in (0,1).Obviously, the setsSkν are open in {u∈C1[0,1]|u(0)=u(1)=0} and disjoint.Suppose that(μ,u) is a solution of (4). Then the corresponding linearized problem of (4) is
(5)w′′+μa(t)w+a(t)f′(u)w=0,t∈(0,1),w(0)=0,w(1)=0.
We call the solution (μ,u) is nondegenerate if (5) has no nontrivial solution; otherwise it is degenerate. Sometimes, we call a degenerate solution a turning point.Consider the linear problem(6)φ′′+μa(t)φ=0,t∈(0,1),φ(0)=0,φ(1)=0.Remark 1.
Note thata(t)∈C1[0,1] and a(t)>0 on [0,1]. It is well known that the eigenvalues of (6) are given by
(7)0<λ1<λ2<⋯<λk<λk+1<⋯,limk→∞λk=∞.
For each k∈ℕ, λk is algebraic simple and the corresponding normalized eigenfunction can be chosen φk∈Sk+.
In this paper, we work in the following spaces:(8)X={u∈C2[0,1]:u(0)=u(1)=0},∥u∥X=max{∥u∥∞,∥u′∥∞,∥u′′∥∞},E={u∈C1[0,1]:u(0)=u(1)=0},∥u∥E=max{∥u∥∞,∥u′∥∞},Y=C[0,1],∥u∥Y=∥u∥∞,
where ∥·∥∞ is the normal supnorm. Obviously, (X,∥·∥X) and (E,∥·∥E) are Banach spaces.


## 3. The Main Results underf0=0, f∞=∞

In the section, we assume(H1)
f∈C1(ℝ,ℝ) satisfies uf(u)>0, if u≠0;(H2)
f′(u)>f(u)/u, if u≠0;(H3)
the limitsf0:=lim|s|→0(f(s)/s)=0 and f∞:=lim|s|→∞(f(s)/s)=∞;(H4)
a(t)∈C2[0,1] satisfies a(t)>0 and 3(a′(t))2<2a(t)a′′(t), if t∈[0,1].Note that (H1) ensures that the solution of the initial value problem(9)u′′+μa(t)u+a(t)f(u)=0,t∈(0,1),u(t0)=α,u′(t0)=β
is not only existent but also unique on the whole interval [0,1] for any t0∈[0,1] and α,β∈ℝ. This fact will be used repeatedly in the following proof so, for brevity, it will be abbreviated to “IVPU”.Remark 2.
The condition (H4) appeared firstly in [24]. There are many functions satisfying (H4). Let a(t)=b(t)+c, where b(·)∈C2[0,1] with b′′(t)>0, b′(t)<0, for allt∈[0,1], c is a large enough constant. It is easy to check that a(·) satisfies (H4).Now, we give some important lemmas.Lemma 3.
Suppose(μ,u) is a nontrivial solution of (4). Then,(i)
u∈Skν  for some k∈ℕ and ν∈{+,-};(ii)
the zeros ofu and the zeros of u′ are separated as μ≥0.Proof.
(i) Sinceu is nontrivial, “IVPU” implies that all the zeros of u are simple. Thus, (i) is true.(ii)
Whenμ≥0, we have
(10)u′′=-μa(t)u-a(t)f(u)≤0,ifu≥0,u′′=-μa(t)u-a(t)f(u)≥0,ifu≤0.This implies that the zeros of u and the zeros of u′ are separated. Since u∈Skν for some k∈ℕ and ν∈{+,-}, we get u has k-1 zero on (0,1) and u′ has exact k zero on (0,1).Lemma 4.
Supposeμ≥0 and (μ,u) is a nontrivial solution of (4). Then (μ,u) is nondegenerate.Proof.
we need to show that (5) has only trivial solution. Consider the initial problem
(11)w′′+μa(t)w+a(t)f′(u)w=0,t∈(0,1),w(0)=0,w′(0)=1.
Clearly, (11) has a unique solution w0. For every solution w of (5), there exists a unique constant c∈ℝ such that w=cw0. We claim that
(12)w0(1)≠0.
If (12) holds, then we immediately have w(1)=0 if and only if c=0. That is, to say w≡0. Then we will finish the proof. Now, we prove (12) is true.
Firstly, we show that there is at least one zero ofu′ between consecutive zeros of w0.
Supposeα,β are consecutive zeros of w0, that is, w0(α)=w0(β)=0. Without loss of generality, suppose w0(t)>0, t∈(α,β). If u′ has no zero on (α,β), we assume u′(t)>0, t∈(α,β). (When u′(t)<0, t∈(α,β), the proof is similar to the case u′(t)>0.)
Note thatw0(t) and u′(t) satisfy the following equations, respectively:
(13)w0′′+μa(t)w0+a(t)f′(u)w0=0,(14)(u′)′′+μa(t)u′+a(t)f′(u)u′=-μa′(t)u-a′(t)f(u).
Multiply (14) by g(t)w0 and subtract from it (13) multiplied by g(t)u′, with g(t)>0 to be specified. Then integrate over (α,β)(15)∫αβ‍(u′)′′gw0dt-∫αβ‍w0′′gu′dt+∫αβ‍(μu+f(u))a′gw0dt=0.
We denote the left side of (15) by I and a constant -w0′(β)g(β)u′(β)+w0′(α)g(α)u′(α) by A. Integrating by parts,
(16)I=u′′gw0|αβ-∫αβ‍u′′[g′w0+gw0′]dt-w0′gu′|αβ+∫αβ‍w0′[g′u′+gu′′]dt+∫αβ‍(μu+f(u))a′gw0dt=-w0′(β)g(β)u′(β)+w0′(α)g(α)u′(α)+∫αβ‍u′[g′′w0+g′w0′]dt+∫αβ‍w0′g′u′dt+∫αβ‍(μu+f(u))a′gw0dt=A+∫αβ‍u′g′′w0dt+2∫αβ‍u′g′w0′dt+∫αβ‍(μu+f(u))a′gw0dt=A+∫αβ‍u′g′′w0dt+2[u′g′w0|αβ-∫αβ‍w0[u′′g′+u′g′′]dt]+∫αβ‍(μu+f(u))a′gw0dt=A-∫αβ‍u′g′′w0dt-2∫αβ‍w0g′u′′dt+∫αβ‍(μu+f(u))a′gw0dt=A-∫αβ‍u′g′′w0dt-2∫αβ‍w0g′(-μau-af(u))dt+∫αβ‍(μu+f(u))a′gw0dt=A-∫αβ‍u′g′′w0dt+2∫αβ‍(μu+f(u))a′gw0dt+∫αβ‍(μu+f(u))a′gw0dt=A-∫αβ‍u′g′′w0dt+∫αβ‍(μu+f(u))(2g′a+a′g)w0dt.
Let
(17)2g′a+a′g=0,g′′<0
on (0,1). By the above supposition, we have
(18)I=-w0′(β)g(β)u′(β)+w0′(α)g(α)u′(α)-∫αβ‍u′g′′w0dt>0.
However, the right side of (15) is zero. A contradiction. Hence, there is at least one zero of u′ between consecutive zeros of w0.
Note that the functionsu(t), w0(t) satisfy the following equations:
(19)u′′+(μ+f(u)u)a(t)u=0,w0′′+(μ+f′(u))a(t)w0=0,
respectively. Since a(t)>0 and f′(u)>f(u)/u, by the Sturm comparison lemma, there exists at least one zero of w0 between any two consecutive zeros of u. This implies that w0 has at least k zeros on (0,1).
Secondly, we show (12). On the contrary, assuming w0(1)=0, then w0 has at least k+2 zeros on [0,1] since w0(0)=0. We conclude that u′ has at least k+1 zeros on (0,1). This contradicts the fact that u′ has exact k zeros on (0,1).
Finally, we give a proper functiong satisfying (17). Integrating the differential equation in (17), we can choose
(20)g(t)=a-1/2(t).
In view of (H4), we conclude g′′<0. So, the auxiliary function g exists. This completes the proof.Our main results are the following.Theorem 5.
Let (H1)–(H4) hold. Then, for fixedk∈ℕ and μ≥0, all solutions of (4) belonging to Sk+ (resp. Sk-) lie on a unique continuous curve on which there is not any turning point. This curve starts from (λk,0), tends to left, and passes through the hyperplane {0}×E. Precisely, (4) has exactly one solution belonging to Sk+ (resp. Sk-) for μ∈[0,λk) and has no solution belonging to Sk+ (resp. Sk-) for μ∈[λk,∞), where λk is the kth eigenvalue of (6). See Figure 1(a).The bifurcation diagrams of solutions of (4).
(a)
(b)From Theorem5, we immediately obtain the following theorem.Theorem 6.
Let (H1)–(H4) hold. Then (1) has a unique solution in Skν  for every k∈ℕ and ν∈{+,-}.We give some lemmas for proving Theorem5.Lemma 7.
LetI⊂ℝ be a closed and bounded interval. Suppose {(μn,un)}⊂I×Skν for some k≥1 and ν∈{+,-} is a sequence of solutions of (4). Then
(21)μn⟶λk,if∥un∥E⟶0.Proof.
DefineL:X→Y by setting
(22)Lu:=-u′′,u∈X.
Then L-1:Y→E is completely continuous. Noting that f0=0, consider
(23)Lu-μa(t)u=a(t)f(u)
as a bifurcation problem from u≡0. Remark 1, Crandall and Rabinowitz theorem on bifurcation from simple eigenvalues (see [25]) and the method in [5] ensure that the result is correct.Lemma 8.
Suppose(μ,u)∈ℝ×Skν for some k≥1 and ν∈{+,-} is a solution of (4). Then μ<λk.Proof.
Note thatu and φk satisfy equations
(24)u′′+(μ+f(u)u)a(t)u=0,φk′′+λka(t)φk=0,
respectively. On the contrary, supposing μ≥λk, then μ+f(u)/u≥λk since f(u)/u>0 if u≠0 and f0=0. Then by the Sturm comparison lemma, between any two consecutive zeros of φk, there exists at least one zero of u. This implies that u has at least k zeros on (0,1). It is impossible.Lemma 9.
LetJ be a closed subinterval of [0,λk). Then there is bJ>0 such that ∥u∥E≤bJ for every solution (μ,u)∈J×Skν of (4).Proof.
Note that whenμ∈[0,λk), we have
(25)u′′=-μa(t)u-a(t)f(u)≤0,ifu≥0,u′′=-μa(t)u-a(t)f(u)≥0,ifu≤0.
Suppose on the contrary that there exists a sequence {(μn,un)} of solutions of (4) with μn∈J, un∈Skν and ∥un∥E→∞ as n→∞. Let
(26)0=τn0<τn1<⋯<τnk=1
denote the zeros of un in [0,1]. Then we can choose at least one subinterval (τnj,τnj+1)≜Inj which is of length at least 1/k, for some j∈{0,1,2,…,k-1}. Without loss of generality, we suppose u′′≤0 on Inj. Moreover, we claim that {maxInj|un|}n=1∞ is an unbounded sequence.
Assume that{maxInj|un|}n=1∞ is uniformly bounded. Since un is concave on Inj, un′ has only one zero ξnj in Inj. Integrating the equation in (4), for any t∈Inj,
(27)un′(t)=-∫ξnjt‍(μnun+f(un))a(s)ds.
It implies that {maxInj|un′|}n=1∞ is uniformly bounded. Then we obtain un′(τnj),un′(τnj+1) are uniformly bounded. Consider consecutive intervals Inj-1 and Inj+1. By convexity of un on Inj-1 and Inj+1 and the uniform boundedness of un′(τnj),un′(τnj+1), un′ is uniformly bounded on [ξnj-1,τnj] and [τnj+1,ξnj+1]. So,
(28)un(ξnj-1)=∫τnjξnj-1‍un′(s)ds,un(ξnj+1)=∫τnj+1ξnj+1‍un′(s)ds
are uniformly bounded. Hence, {max|un|}n=1∞ is uniformly bounded on Inj-1 and Inj+1. By some finite steps, this procedure shows that {max|un|}n=1∞ is uniformly bounded on [0,1]. This is a contradiction.
Taking subsequences if necessary, put(29)limn→∞τnj=τ∞j,limn→∞τnj+1=τ∞j+1,limn→∞ξnj=ξ∞j.
Noting that τnj+1-τnj≥1/k, then τ∞j+1-τ∞j≥1/k. We may assume τ∞j<ξ∞j<τ∞j+1. The case ξ∞j=τ∞j or ξ∞j=τ∞j+1 can be considered similarly. In the rest of the proof, ∥un∥∞ denotes maxInj|un|. Since un is concave on Inj, for any 0<ϵ<(ξ∞j-τ∞j)/4 small enough, there exists a constant σ>0 such that
(30)un(t)≥σ∥un∥∞,∀t∈[τnj+ϵ,τnj+1-ϵ].
By the condition f∞=∞ in (H3), there exists a real number R1>0 such that
(31)f(u)≥ηu,foru≥R1,
where η>0 can be given by
(32)(μn+η)σ∫τ∞j+ϵτ∞j+2ϵ‍ds∫sτ∞j+2ϵ‍a(τ)dτ>1.
Since ∥un∥∞→∞, ∥un∥∞>R1/σ for sufficiently large n. Thus
(33)un(t)≥σ∥un∥∞>R1,∀t∈[τnj+ϵ,τnj+1-ϵ].
Moreover,
(34)un(ξnj)=∫τnjξnj‍(∫sξnj‍[μna(τ)un(τ)+a(τ)f(un(τ))]dτ)ds≥(μn+η)∫τ∞j+ϵτ∞j+2ϵ‍(∫sτ∞j+2ϵ‍[a(τ)un(τ)]dτ)ds≥∥un∥∞(μn+η)σ∫τ∞j+ϵτ∞j+2ϵ‍ds∫sτ∞j+2ϵ‍a(τ)dτ.
Hence,
(35)∥un∥∞≥∥un∥∞(μn+η)σ∫τ∞j+ϵτ∞j+2ϵ‍ds∫sτ∞j+2ϵ‍a(τ)dτ.
It follows that
(36)(μn+η)σ∫τ∞j+ϵτ∞j+2ϵ‍ds∫sτ∞j+2ϵ‍a(τ)dτ≤1.
This contradicts (32). This ends the proof.Suppose there exits a nontrivial solution(μ0,u0) of (4) with μ0≥0. It follows from Lemma 3 that u0∈Skν for some k≥1 and ν∈{+,-}. Meanwhile, from the implicit function theorem and Lemma 4, we have that all solutions of (4) in ℝ×E near (μ0,u0) lie on a C1 curve passing through (μ0,u0) and parameterized by μ. We denote the local curve by Γ0. Then, we have the following lemma.Lemma 10.
Γ
0
⊂
ℝ
×
S
k
ν.Proof.
This can be obtained by Lemma3 and the fact that Skν is open.Lemma 11.
Γ
0 can be continued on the interval [0,λk). Moreover, there is a constant c>0 such that
(37)limμ→0∥u∥E=c>0,limμ→λk∥u∥E=0.Proof.
Suppose that there exist a sequence{(μn,un)}⊂Γ0 and a small constant 0<δ<1/δ such that μn→e∈(0,λk), n→∞, and δ≤∥un∥X≤1/δ for large enough n. Then, after choosing a subsequence if necessary, there exists u∞∈E such that un→u∞ in E. It then follows from the equation in (4) that u∞∈X and (e,u∞) is a nontrivial solution of (4). Hence, by Lemma 4 and the implicit function theorem, the curve Γ0 can still be continued. Meanwhile, we know limμn→e∥un∥E≠∞ from Lemma 9. Combining this with Lemma 8, we have Γ0 can be continued on the interval [0,λk). Moreover, by Lemma 7, there is a constant c>0 such that
(38)limμ→0∥u∥E=c>0,limμ→λk∥u∥E=0.
Assume(0,u0) is the intersection of Γ0 and hyperplane {0}×E. From Lemma 4, (0,u0) is also a nondegenerate solution of (4) and ∥u0∥E>0. Hence, Γ0 can still be continued to the direction of μ<0.Proof of Theorem5.
Consider bifurcation problems (23), by the standard Crandall and Rabinowitz theorem on local bifurcation from simple eigenvalues (see [25]), for each k≥1 exactly two local curves of nontrivial solutions bifurcate from the point (λk,0) in ℝ×E, one of which lies in ℝ×Sk+ and the other in ℝ×Sk-. By the above discussion, each of these local curves can be continued at least on the interval [0,λk). We will denote these particular curves by 𝒞kν. Then 𝒞kν⊂ℝ×Skν by Lemma 10. From Lemmas 4 and 11, 𝒞kν can pass through the hyperplane {0}×E and go to the direction of μ<0.
Finally, we exclude the possibility that there exists another solution curve of (4) belonging to the set [0,λk)×Skν. Suppose that there exists a solution (μ,u)∈[0,λk)×Skν but (μ,u)∉𝒞kν. Similarly, by continuation, we can obtain another solution curve Γ⊂[0,λk)×Skν such that limμ→λk∥u∥E=0 for (μ,u)∈Γ. Since 𝒞kν is the unique solution curves near (λk,0) in ℝ×E by the standard Crandall and Rabinowitz theorem, Γ must coincide with 𝒞kν. This completes the proof.Proof of Theorem6.
From Theorem5 and Lemma 11, we obtain the result.


## 4. The Main Results underf0=∞, f∞=0

In this section, we study the exact multiplicity of constant sign, that is, positive (resp. negative) solution of (1) under the conditions(H1′)
f∈C(ℝ,ℝ), f(0)=0. When u≠0, f′(u) exists and satisfies f′(u)>0. f′(u) is nondecreasing on (-∞,0) and nonincreasing on (0,∞). For any fixed α>0, ∫01‍tf′(±αt(1-t))dt<∞ and ∫01‍(1-t)f′(±αt(1-t))dt<∞.(H2′)
f′(u)<f(u)/u, if u≠0;(H3′)
f0=lim|s|→0(f(s)/s)=∞, f∞=lim|s|→∞(f(s)/s)=0;(H4′)
a(t)∈C1[0,1] satisfies a(t)>0, t∈[0,1].Remark 12.
There are many functions satisfying conditions (H1′)–(H3′). For example,(39)f(u)={u,u≥0,--u,u<0.
As before, we discuss the structure of the set of solutions of the auxiliary problem (4).Lemma 13.
If(μ,u) is a solution of (4), then the zero of u is simple and u′(0)≠0, u′(1)≠0. Moreover, u∈Skν for some k≥1 and ν∈{+,-}.Proof.
From proportion  2.1 in [20], we can obtain, for every t0∈[0,1] and β∈ℝ, the initial value problem
(40)u′′+μa(t)u+a(t)f(u)=0,t∈(0,1),u(t0)=0,u′(t0)=β
has a unique solution. Particularly, it is true for the case f0=∞. Thus, u∈Skν for some k≥1 and ν∈{+,-} and u′(0)≠0,u′(1)≠0.Lemma 14.
Forν∈{+,-}, if (μ,u)∈S1ν is a solution of (4), then μ<λ1, where λ1 is the first eigenvalue of (6).Proof.
Note thatu and φ1 satisfy, respectively, the equations
(41)u′′+(μ+f(u)u)a(t)u=0,φ1′′+λ1a(t)φ1=0.
On the contrary, assume μ≥λ1. Noting that for u≠0 we have f(u)/u>0 and f0=∞, then μ+f(u)/u≥λ1. By the Sturm comparison lemma, there exists at least one zero of u between any two consecutive zeros of φ1. This implies that u has at least one zeros on (0,1). It is impossible.Lemma 15.
LetI⊂ℝ be a bounded and closed interval and λ1∈I. For ν∈{+,-}, if (μn,un)∈S1ν(n=1,2,…) are solutions of (4) and μn∈I, then
(42)μn⟶λ1as∥un∥E⟶∞.Proof.
Note thatf∞=0 consider the bifurcation problems (23) from infinity. By using Remark 1 and standard Rabinowitz bifurcation theorem from simple eigenvalue in [26] and the proof method in [5], we can obtain the result.From Lemma15, we have the following lemma.Lemma 16.
LetJ-⊂[0,λ1) be a closed interval. Then there exists bJ->0 such that
(43)∥u∥E≤bJ-,
for every solution (μ,u)∈J-×S1ν of (4).Lemma 17.
LetJ~⊂[0,λ1) be a closed interval. Then there exists cJ~>0 such that
(44)∥u∥E≥cJ~,
for each solution (μ,u)∈J~×S1ν of (4).Proof.
Without loss of generality, we selectS1+ to be discussed. When μ∈[0,λ1), since (μ,u) is a positive solution of (4), we have
(45)u′′=-μa(t)u-a(t)f(u)≤0.
On the contrary, suppose there exists a sequence of solutions of (4) {(μn,un)}⊂J~×S1+ satisfying
(46)μn∈J,∥un∥E⟶0,asn⟶∞.
Noting thatun is concave on the interval (0,1), then un′ has only one zero in (0,1), denoted by ξn. Integrating
(47)un′′+μna(t)un+a(t)f(un)=0,t∈(0,1),
we get
(48)un′(t)=-∫ξnt‍(μnun+f(un))a(s)ds.
Taking subsequences if necessary, let
(49)limn→∞ξn=ξ∞.
Without loss of generality, suppose 0<ξ∞<1. For the case ξ∞=0 or ξ∞=1, the proof is similar. Since un is concave on the interval (0,1), for 0<ϵ<ξ∞/4 small enough, there exists a constant σ->0 such that for large n(50)un(t)≥σ-∥un∥∞,∀t∈[ϵ,1-ϵ].
Since f0=∞, there exists constant R1>0 such that
(51)f(u)≥η-u,as|u|≤R1,
where η->0 satisfies
(52)(μn+η-)σ-∫ϵ2ϵ‍ds∫s2ϵ‍a(τ)dτ>1.
Since ∥un∥E→0, we have ∥un∥∞→0. Thus, for large enough n, ∥un∥∞<R1. And for every t∈[ϵ,1-ϵ], we have
(53)σ-∥un∥∞≤un(t)≤∥un∥∞.
Moreover,
(54)un(ξn)=∫0ξn‍(∫sξn‍[μna(τ)un(τ)+a(τ)f(un(τ))]dτ)ds≥(μn+η-)∫ϵ2ϵ‍(∫s2ϵ‍[a(τ)un(τ)]dτ)ds≥∥un∥∞(μn+η-)σ-∫ϵ2ϵ‍ds∫s2ϵ‍a(τ)dτ.
Therefore,
(55)∥un∥∞≥∥un∥∞(μn+η-)σ-∫ϵ2ϵ‍ds∫s2ϵ‍a(τ)dτ.
From this, we have
(56)(μn+η)σ∫ϵ2ϵ‍ds∫s2ϵ‍a(τ)dτ≤1.
It contradicts (52). This ends the proof.Lemma 18.
Letμ∈[0,λ1) and (μ,u)∈ℝ×S1ν be a solution of (4). Then (μ,u) is nondegenerate.Proof.
It is sufficient to prove that (5) has only trivial solution. Consider the initial value problem
(57)w′′+μa(t)w+a(t)fu(u(t))w=0,t∈(0,1),w(0)=0,w′(0)=β≠0,
where fu denotes the derivative of f to u. β∈ℝ is a constant. First, we prove (57) has a unique solution. From Lemma 3.1 and its proof, Lemma  3.2 in [3], we need to show
(58)∫01‍t|μ+fu(u(t))|a(t)dt<∞,∫01‍(1-t)|μ+fu(u(t))|a(t)dt<∞.
It is sufficient to prove
(59)∫01‍tfu(u(t))dt<∞,∫01‍(1-t)fu(u(t))dt<∞.
Noting that for every t∈(0,1), if u(t)>0, then u′′(t)<0; if u(t)<0, then u′′(t)>0. Suppose (μ,u)∈ℝ×S1+  (when (μ,u)∈ℝ×S1-, the proof is similar). Then there must exist α>0 such that u(t)>αt(1-t)>0,  for all  t∈(0,1). By condition (H1′), we have
(60)fu(u(t))<fu(αt(1-t)),∀t∈(0,1).
It follows that
(61)∫01‍tfu(u(t))dt<∫01‍tfu(αt(1-t))dt<∞,∫01‍(1-t)fu(u(t))dt<∫01‍(1-t)fu(αt(1-t))dt<∞.
From Lemma  3.2 in [3], (57) has a unique solution. When β=1, we denote the solution of (57) by w0. Then, for every solution w of (57), there exists a constant d∈ℝ such that w=dw0.
We claim that(62)w0(1)≠0.
If (62) holds, then we immediately have w(1)=0 if and only if d=0. That is to say, w≡0. Then we will finish the proof.Note thatu(t),w0(t) satisfy the following equations:
(63)u′′+(μ+f(u)u)a(t)u=0,w0′′+(μ+f′(u))a(t)w0=0,
respectively. Since a(t)>0, f′(u)<f(u)/u, u≠0, by the Sturm comparison lemma, there exists at least one zero of u between any two consecutive zeros of w0. If (62) does not hold, then
(64)w0(1)=0.
Complying this with w0(0)=0, we obtain there exists at least one zero of u in the interval (0,1). This contradicts that u is a positive (resp. negative) solution of (4). The proof is ended.Supposeμ0≥0 and (μ0,u0) is a positive (resp. negative) solution of (4). For ν∈{+,-}, we have u0∈S1ν. By Lemma 18 and the implicit function theorem, all solutions of (4) near (μ0,u0) lie on a unique curve which passes through (μ0,u0) and is parameterized by μ. Denote the curve by Γ~0. Then we have the following lemma.Lemma 19.
Γ
~
0
⊂
ℝ
×
S
1
ν.Proof.
Note thatS1ν is an open set. This together with Lemma 13 implies the conclusion.Lemma 20.
Γ
~
0 can be continued on the interval [0,λ1) and there exists h>0 such that
(65)limμ→0∥u∥E=h>0,limμ→λ1∥u∥E=∞.Proof.
From Lemmas14–18 and the implicit function theorem, we can prove the result by using similar method of proving Lemma 11. We omit it.
Suppose(0,u0) is the intersection point of Γ~0 and hyperplane {0}×E. From Lemma 18, (0,u0) is also a nondegenarate solution of (4) and ∥u∥E=h>0. Hence, Γ~0 can be continued to the direction of μ<0.Theorem 21.
Let (H1′), (H2′), (H3′), and (H4′) hold. Then forμ≥0 all solutions of (4) belonging to S1+ (resp. S1-) lie on a unique continuous curve on which there is not any turning point. This curve bifurcates from (λ1,∞), tends to left, and passes through the hyperplane {0}×E. Precisely, (4) has exactly one positive (resp. negative) solution if μ∈[0,λ1) and has no positive (resp. negative) solution for μ∈[λ1,∞), where λ1 is the first eigenvalue of the linear problem (6). See Figure 1(b).Proof.
With the uniqueness of local curve of solutions bifurcating from infinity [26], the proof is similar to that of Theorem 5. For simplicity, we omit it.From Lemma20 and Theorem 21, we immediately obtain the following.Theorem 22.
Let (H1′), (H2′), (H3′), and (H4′) hold. Then (1) has a unique positive (resp. negative) solution.



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*Source: 897307-2013-06-03.xml*