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

Abstract

Using bifurcation techniques and Sturm comparison theorem, we establish exact multiplicity results of sign-changing or constant sign solutions for the boundary value problems 𝑢^″+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].

1. Introduction

In this paper, we consider the existence and exact multiplicity of sign-changing solutions for the boundary value problem 𝑢^′′(𝑡)𝑓(𝑢)(),𝑢(0)(1)+𝑎=0,𝑡∈0,1=0,𝑢=0, (1) 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 nonlinearities 𝑓, 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 𝑢^′′(𝑢)(),𝑢(0)(1)+𝜆𝑓=0,𝑡∈0,1=0,𝑢=0, (2) 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 limit 𝑓_∞=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 (−1)^𝑚𝑢^(2𝑚)(𝑡)(𝑢(𝑡𝑢(𝑡)(),𝑢=𝜆𝑔)),𝑡∈0,1_(𝑖)(−1)=𝑢^(𝑖)(1)=0,𝑖=0,…,𝑚−1, (3) 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 Section 2, 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 𝑢^′′(𝑡)(𝑡)𝑓(𝑢)(),𝑢(0)(1)+𝜇𝑎𝑢+𝑎=0,𝑡∈0,1=0,𝑢=0, (4) 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 𝑆^𝜈_𝑘⊂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 sets 𝑆^𝜈_𝑘 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 𝑤^′′+𝜇𝑎(𝑡)𝑤+𝑎(𝑡)𝑓^′(𝑢)(),𝑤(0)(1)𝑤=0,𝑡∈0,1=0,𝑤=0. (5) 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 𝜑^′′(𝑡)(),𝜑(0)(1)+𝜇𝑎𝜑=0,𝑡∈0,1=0,𝜑=0. (6)

Remark 1. Note that 𝑎(𝑡)∈𝐶¹[0,1] and 𝑎(𝑡)>0 on [0,1]. It is well known that the eigenvalues of (6) are given by 0<𝜆₁<𝜆₂<⋯<𝜆_𝑘<𝜆_𝑘+1<⋯,lim_𝑘→∞𝜆_𝑘=∞. (7) For each 𝑘∈ℕ, 𝜆_𝑘 is algebraic simple and the corresponding normalized eigenfunction can be chosen 𝜑_𝑘∈𝑆⁺_𝑘.
In this paper, we work in the following spaces: 𝑋=𝑢∈𝐶²[](0)(1),‖𝑢‖0,1:𝑢=𝑢=0_𝑋=max‖𝑢‖_∞,𝑢^′_∞,𝑢^′′_∞,𝐸=𝑢∈𝐶¹[](0)(1),‖𝑢‖0,1∶𝑢=𝑢=0_𝐸=max‖𝑢‖_∞,𝑢^′_∞,𝑌=𝐶[0,1],‖𝑢‖_𝑌=‖𝑢‖_∞, (8) where ‖⋅‖_∞ is the normal supnorm. Obviously, (𝑋,‖⋅‖_𝑋) and (𝐸,‖⋅‖_𝐸) are Banach spaces.

3. The Main Results under 𝑓₀=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 limits 𝑓₀∶=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 𝑢^′′(𝑡)(𝑡)𝑓(𝑢)(),𝑢𝑡+𝜇𝑎𝑢+𝑎=0,𝑡∈0,1₀=𝛼,𝑢^′𝑡₀=𝛽 (9) 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∈𝑆^𝜈_𝑘  for some 𝑘∈ℕ and 𝜈∈{+,−}; (ii)the zeros of 𝑢 and the zeros of 𝑢^′ are separated as 𝜇≥0.

Proof. (i) Since 𝑢 is nontrivial, “IVPU” implies that all the zeros of 𝑢 are simple. Thus, (i) is true.(ii) When 𝜇≥0, we have 𝑢^′′(𝑡)(𝑡)𝑓(𝑢)𝑢=−𝜇𝑎𝑢−𝑎≤0,if𝑢≥0,_′′=−𝜇𝑎(𝑡)𝑢−𝑎(𝑡)𝑓(𝑢)≥0,if𝑢≤0. (10)This implies that the zeros of 𝑢 and the zeros of 𝑢^′ are separated. Since 𝑢∈𝑆^𝜈_𝑘 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. Suppose 𝜇≥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 𝑤^′′+𝜇𝑎(𝑡)𝑤+𝑎(𝑡)𝑓^′(𝑢)(),𝑤(0)𝑤=0,𝑡∈0,1=0,𝑤^′(0)=1. (11) 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 𝑤₀(1)≠0. (12) 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 of 𝑢^′ between consecutive zeros of 𝑤₀.
Suppose 𝛼,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 that 𝑤₀(𝑡) and 𝑢^′(𝑡) satisfy the following equations, respectively: 𝑤^′′₀+𝜇𝑎(𝑡)𝑤₀+𝑎(𝑡)𝑓^′(𝑢)𝑤₀𝑢=0,_′′′+𝜇𝑎(𝑡)𝑢^′+𝑎(𝑡)𝑓^′(𝑢)𝑢^′=−𝜇𝑎^′(𝑡)𝑢−𝑎^′(𝑡)𝑓(𝑢). (13)(14) 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)^𝛽_𝛼𝑢^′′′𝑔𝑤₀𝑑𝑡−^𝛽_𝛼𝑤^′′₀𝑔𝑢^′+𝑑𝑡_𝛽𝛼𝜇𝑢+𝑓(𝑢)𝑎^′𝑔𝑤₀𝑑𝑡=0. (15) We denote the left side of (15) by 𝐼 and a constant −𝑤^′₀(𝛽)𝑔(𝛽)𝑢^′(𝛽)+𝑤^′₀(𝛼)𝑔(𝛼)𝑢^′(𝛼) by 𝐴. Integrating by parts, 𝐼=𝑢^′′𝑔𝑤₀^𝛽_𝛼−^𝛽_𝛼𝑢^′′𝑔^′𝑤₀+𝑔𝑤^′₀𝑑𝑡−𝑤^′₀𝑔𝑢^′𝛽_𝛼+^𝛽_𝛼𝑤^′₀𝑔^′𝑢^′+𝑔𝑢^′′𝑑𝑡+^𝛽_𝛼𝜇𝑢+𝑓(𝑢)𝑎^′𝑔𝑤₀𝑑𝑡=−𝑤^′₀𝛽𝑔𝛽𝑢^′𝛽+𝑤^′₀(𝛼)𝑔(𝛼)𝑢^′(𝛼)+^𝛽_𝛼𝑢^′𝑔^′′𝑤₀+𝑔^′𝑤^′₀𝑑𝑡+^𝛽_𝛼𝑤^′₀𝑔^′𝑢^′+𝑑𝑡_𝛽𝛼𝜇𝑢+𝑓(𝑢)𝑎^′𝑔𝑤₀𝑑𝑡=𝐴+^𝛽_𝛼𝑢^′𝑔^′′𝑤₀𝑑𝑡+2^𝛽_𝛼𝑢^′𝑔^′𝑤^′₀+𝑑𝑡_𝛽𝛼𝜇𝑢+𝑓(𝑢)𝑎^′𝑔𝑤₀𝑑𝑡=𝐴+^𝛽_𝛼𝑢^′𝑔^′′𝑤₀𝑢𝑑𝑡+2^′𝑔^′𝑤₀^𝛽_𝛼−^𝛽_𝛼𝑤₀𝑢^′′𝑔^′+𝑢^′𝑔^′′𝑑𝑡+^𝛽_𝛼𝜇𝑢+𝑓(𝑢)𝑎^′𝑔𝑤₀𝑑𝑡=𝐴−^𝛽_𝛼𝑢^′𝑔^′′𝑤₀𝑑𝑡−2^𝛽_𝛼𝑤₀𝑔^′𝑢^′′+𝑑𝑡_𝛽𝛼𝜇𝑢+𝑓(𝑢)𝑎^′𝑔𝑤₀𝑑𝑡=𝐴−^𝛽_𝛼𝑢^′𝑔^′′𝑤₀𝑑𝑡−2^𝛽_𝛼𝑤₀𝑔^′(𝑢)+−𝜇𝑎𝑢−𝑎𝑓𝑑𝑡_𝛽𝛼𝜇𝑢+𝑓(𝑢)𝑎^′𝑔𝑤₀𝑑𝑡=𝐴−^𝛽_𝛼𝑢^′𝑔^′′𝑤₀𝑑𝑡+2^𝛽_𝛼𝜇𝑢+𝑓(𝑢)𝑎^′𝑔𝑤₀𝑑𝑡+^𝛽_𝛼𝜇𝑢+𝑓(𝑢)𝑎^′𝑔𝑤₀𝑑𝑡=𝐴−^𝛽_𝛼𝑢^′𝑔^′′𝑤₀+𝑑𝑡_𝛽𝛼𝜇𝑢+𝑓(𝑢)2𝑔^′𝑎+𝑎^′𝑔𝑤₀𝑑𝑡. (16) Let 2𝑔^′𝑎+𝑎^′𝑔=0,𝑔^′′<0 (17) on (0,1). By the above supposition, we have 𝐼=−𝑤^′₀𝛽𝑔𝛽𝑢^′𝛽+𝑤^′₀(𝛼)𝑔(𝛼)𝑢^′(𝛼)−^𝛽_𝛼𝑢^′𝑔^′′𝑤₀𝑑𝑡>0. (18) 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 functions 𝑢(𝑡), 𝑤₀(𝑡) satisfy the following equations: 𝑢^′′+𝑓(𝑢)𝜇+𝑢𝑎(𝑡)𝑤𝑢=0,_′′0+𝜇+𝑓^′(𝑢)𝑎(𝑡)𝑤₀=0, (19) 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 function 𝑔 satisfying (17). Integrating the differential equation in (17), we can choose 𝑔(𝑡)=𝑎^−1/2(𝑡). (20) 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 fixed 𝑘∈ℕ and 𝜇≥0, all solutions of (4) belonging to 𝑆⁺_𝑘 (resp. 𝑆⁻_𝑘) lie on a unique continuous curve on which there is not any turning point. This curve starts from (𝜆_𝑘,0), tends to left, and passes through the hyperplane {0}×𝐸. Precisely, (4) has exactly one solution belonging to 𝑆⁺_𝑘 (resp. 𝑆⁻_𝑘) for 𝜇∈[0,𝜆_𝑘) and has no solution belonging to 𝑆⁺_𝑘 (resp. 𝑆⁻_𝑘) for 𝜇∈[𝜆_𝑘,∞), where 𝜆_𝑘 is the 𝑘th eigenvalue of (6). See Figure 1(a).

(a)

(b)

Figure 1 

The bifurcation diagrams of solutions of (4).

From Theorem 5, we immediately obtain the following theorem.

Theorem 6. Let (H1)–(H4) hold. Then (1) has a unique solution in 𝑆^𝜈_𝑘  for every 𝑘∈ℕ and 𝜈∈{+,−}.

We give some lemmas for proving Theorem 5.

Lemma 7. Let 𝐼⊂ℝ be a closed and bounded interval. Suppose {(𝜇_𝑛,𝑢_𝑛)}⊂𝐼×𝑆^𝜈_𝑘 for some 𝑘≥1 and 𝜈∈{+,−} is a sequence of solutions of (4). Then 𝜇_𝑛⟶𝜆_𝑘,𝑖𝑓𝑢_𝑛𝐸⟶0. (21)

Proof. Define 𝐿∶𝑋→𝑌 by setting 𝐿𝑢∶=−𝑢^′′,𝑢∈𝑋. (22) Then 𝐿⁻¹∶𝑌→𝐸 is completely continuous. Noting that 𝑓₀=0, consider 𝐿𝑢−𝜇𝑎(𝑡)𝑢=𝑎(𝑡)𝑓(𝑢) (23) 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×𝑆^𝜈_𝑘 for some 𝑘≥1 and 𝜈∈{+,−} is a solution of (4). Then 𝜇<𝜆_𝑘.

Proof. Note that 𝑢 and 𝜑_𝑘 satisfy equations 𝑢^′′+𝑓(𝑢)𝜇+𝑢𝑎(𝑡)𝜑𝑢=0,_′′𝑘+𝜆_𝑘𝑎(𝑡)𝜑_𝑘=0, (24) respectively. On the contrary, supposing 𝜇≥𝜆_𝑘, then 𝜇+𝑓(𝑢)/𝑢≥𝜆_𝑘 since 𝑓(𝑢)/𝑢>0 if 𝑢≠0 and 𝑓₀=0. Then by the Sturm comparison lemma, between any two consecutive zeros of 𝜑_𝑘, 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. Let 𝐽 be a closed subinterval of [0,𝜆_𝑘). Then there is 𝑏_𝐽>0 such that ‖𝑢‖_𝐸≤𝑏_𝐽 for every solution (𝜇,𝑢)∈𝐽×𝑆^𝜈_𝑘 of (4).

Proof. Note that when 𝜇∈[0,𝜆_𝑘), we have 𝑢^′′(𝑡)(𝑡)𝑓(𝑢)𝑢=−𝜇𝑎𝑢−𝑎≤0,if𝑢≥0,_′′=−𝜇𝑎(𝑡)𝑢−𝑎(𝑡)𝑓(𝑢)≥0,if𝑢≤0. (25) Suppose on the contrary that there exists a sequence {(𝜇_𝑛,𝑢_𝑛)} of solutions of (4) with 𝜇_𝑛∈𝐽, 𝑢_𝑛∈𝑆^𝜈_𝑘 and ‖𝑢_𝑛‖_𝐸→∞ as 𝑛→∞. Let 0=𝜏⁰_𝑛<𝜏¹_𝑛<⋯<𝜏^𝑘_𝑛=1 (26) denote the zeros of 𝑢_𝑛 in [0,1]. Then we can choose at least one subinterval (𝜏^𝑗_𝑛,𝜏_𝑛^𝑗+1)≜𝐼^𝑗_𝑛 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 𝐼^𝑗_𝑛. Moreover, we claim that {max_𝐼^𝑗_𝑛|𝑢_𝑛|}^∞_𝑛=1 is an unbounded sequence.
Assume that {max_𝐼^𝑗_𝑛|𝑢_𝑛|}^∞_𝑛=1 is uniformly bounded. Since 𝑢_𝑛 is concave on 𝐼^𝑗_𝑛, 𝑢^′_𝑛 has only one zero 𝜉^𝑗_𝑛 in 𝐼^𝑗_𝑛. Integrating the equation in (4), for any 𝑡∈𝐼^𝑗_𝑛, 𝑢^′_𝑛(𝑡)=−^𝑡_𝜉𝑗𝑛𝜇_𝑛𝑢_𝑛+𝑓𝑢_𝑛𝑎(𝑠)𝑑𝑠. (27) It implies that {max_𝐼^𝑗_𝑛|𝑢^′_𝑛|}^∞_𝑛=1 is uniformly bounded. Then we obtain 𝑢^′_𝑛(𝜏^𝑗_𝑛),𝑢^′_𝑛(𝜏_𝑛^𝑗+1) are uniformly bounded. Consider consecutive intervals 𝐼_𝑛^𝑗−1 and 𝐼_𝑛^𝑗+1. By convexity of 𝑢_𝑛 on 𝐼_𝑛^𝑗−1 and 𝐼_𝑛^𝑗+1 and the uniform boundedness of 𝑢^′_𝑛(𝜏^𝑗_𝑛),𝑢^′_𝑛(𝜏_𝑛^𝑗+1), 𝑢^′_𝑛 is uniformly bounded on [𝜉_𝑛^𝑗−1,𝜏^𝑗_𝑛] and [𝜏_𝑛^𝑗+1,𝜉_𝑛^𝑗+1]. So, 𝑢_𝑛𝜉_𝑛^𝑗−1=^{𝜉𝑛𝑗−1}_𝜏𝑗𝑛𝑢^′_𝑛(𝑠)𝑢𝑑𝑠,_𝑛𝜉_𝑛^𝑗+1=^{𝜉𝑛𝑗+1}_{𝜏𝑛𝑗+1}𝑢^′_𝑛(𝑠)𝑑𝑠 (28) are uniformly bounded. Hence, {max|𝑢_𝑛|}^∞_𝑛=1 is uniformly bounded on 𝐼_𝑛^𝑗−1 and 𝐼_𝑛^𝑗+1. By some finite steps, this procedure shows that {max|𝑢_𝑛|}^∞_𝑛=1 is uniformly bounded on [0,1]. This is a contradiction.
Taking subsequences if necessary, put lim_𝑛→∞𝜏^𝑗_𝑛=𝜏^𝑗_∞,lim_𝑛→∞𝜏_𝑛^𝑗+1=𝜏_∞^𝑗+1,lim_𝑛→∞𝜉^𝑗_𝑛=𝜉^𝑗_∞. (29) Noting that 𝜏_𝑛^𝑗+1−𝜏^𝑗_𝑛≥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, ‖𝑢_𝑛‖_∞ denotes max_𝐼^𝑗_𝑛|𝑢_𝑛|. Since 𝑢_𝑛 is concave on 𝐼^𝑗_𝑛, for any 0<𝜖<(𝜉^𝑗_∞−𝜏^𝑗_∞)/4 small enough, there exists a constant 𝜎>0 such that 𝑢_𝑛(𝑡)≥𝜎𝑢_𝑛∞,∀𝑡∈𝜏^𝑗_𝑛+𝜖,𝜏_𝑛^𝑗+1−𝜖. (30) By the condition 𝑓_∞=∞ in (H3), there exists a real number 𝑅₁>0 such that 𝑓(𝑢)≥𝜂𝑢,for𝑢≥𝑅₁, (31) where 𝜂>0 can be given by 𝜇_𝑛+𝜂𝜎^𝜏𝑗∞_𝜏^+2𝜖_𝑗∞+𝜖𝑑𝑠^𝜏𝑗∞_𝑠^+2𝜖𝑎(𝜏)𝑑𝜏>1. (32) Since ‖𝑢_𝑛‖_∞→∞, ‖𝑢_𝑛‖_∞>𝑅₁/𝜎 for sufficiently large 𝑛. Thus 𝑢_𝑛(𝑡)≥𝜎𝑢_𝑛∞>𝑅₁,∀𝑡∈𝜏^𝑗_𝑛+𝜖,𝜏_𝑛^𝑗+1−𝜖. (33) Moreover, 𝑢_𝑛𝜉^𝑗_𝑛=^𝜉𝑗𝑛_𝜏𝑗𝑛^𝜉𝑗𝑛_𝑠𝜇_𝑛𝑎(𝜏)𝑢_𝑛(𝜏)+𝑎(𝜏)𝑓𝑢_𝑛(𝜏≥𝜇)𝑑𝜏𝑑𝑠_𝑛+𝜂^𝜏𝑗∞_𝜏^+2𝜖_𝑗∞+𝜖^𝜏𝑗∞_𝑠^+2𝜖𝑎(𝜏)𝑢_𝑛(𝜏≥𝑢)𝑑𝜏𝑑𝑠_𝑛∞𝜇_𝑛+𝜂𝜎^𝜏𝑗∞_𝜏^+2𝜖_𝑗∞+𝜖𝑑𝑠^𝜏𝑗∞_𝑠^+2𝜖𝑎(𝜏)𝑑𝜏. (34) Hence, 𝑢_𝑛∞≥𝑢_𝑛∞𝜇_𝑛+𝜂𝜎^𝜏𝑗∞_𝜏^+2𝜖_𝑗∞+𝜖𝑑𝑠^𝜏𝑗∞_𝑠^+2𝜖𝑎(𝜏)𝑑𝜏. (35) It follows that 𝜇_𝑛+𝜂𝜎^𝜏𝑗∞_𝜏^+2𝜖_𝑗∞+𝜖𝑑𝑠^𝜏𝑗∞_𝑠^+2𝜖𝑎(𝜏)𝑑𝜏≤1. (36) 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∈𝑆^𝜈_𝑘 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. Γ₀⊂ℝ×𝑆^𝜈_𝑘.

Proof. This can be obtained by Lemma 3 and the fact that 𝑆^𝜈_𝑘 is open.

Lemma 11. Γ₀ can be continued on the interval [0,𝜆_𝑘). Moreover, there is a constant 𝑐>0 such that lim_𝜇→0‖𝑢‖_𝐸=𝑐>0,lim_{𝜇→𝜆𝑘}‖𝑢‖_𝐸=0. (37)

Proof. Suppose that there exist a sequence {(𝜇_𝑛,𝑢_𝑛)}⊂Γ₀ and a small constant 0<𝛿<1/𝛿 such that 𝜇_𝑛→𝑒∈(0,𝜆_𝑘), 𝑛→∞, and 𝛿≤‖𝑢_𝑛‖_𝑋≤1/𝛿 for large enough 𝑛. Then, after choosing a subsequence if necessary, there exists 𝑢_∞∈𝐸 such that 𝑢_𝑛→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 lim_{𝜇𝑛→𝑒}‖𝑢_𝑛‖_𝐸≠∞ from Lemma 9. Combining this with Lemma 8, we have Γ₀ can be continued on the interval [0,𝜆_𝑘). Moreover, by Lemma 7, there is a constant 𝑐>0 such that lim_𝜇→0‖𝑢‖_𝐸=𝑐>0,lim_{𝜇→𝜆𝑘}‖𝑢‖_𝐸=0. (38)
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 Theorem 5. 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 (𝜆_𝑘,0) in ℝ×𝐸, one of which lies in ℝ×𝑆⁺_𝑘 and the other in ℝ×𝑆⁻_𝑘. By the above discussion, each of these local curves can be continued at least on the interval [0,𝜆_𝑘). We will denote these particular curves by 𝒞^𝜈_𝑘. Then 𝒞^𝜈_𝑘⊂ℝ×𝑆^𝜈_𝑘 by Lemma 10. From Lemmas 4 and 11, 𝒞^𝜈_𝑘 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,𝜆_𝑘)×𝑆^𝜈_𝑘. Suppose that there exists a solution (𝜇,𝑢)∈[0,𝜆_𝑘)×𝑆^𝜈_𝑘 but (𝜇,𝑢)∉𝒞^𝜈_𝑘. Similarly, by continuation, we can obtain another solution curve Γ⊂[0,𝜆_𝑘)×𝑆^𝜈_𝑘 such that lim_{𝜇→𝜆𝑘}‖𝑢‖_𝐸=0 for (𝜇,𝑢)∈Γ. Since 𝒞^𝜈_𝑘 is the unique solution curves near (𝜆_𝑘,0) in ℝ×𝐸 by the standard Crandall and Rabinowitz theorem, Γ must coincide with 𝒞^𝜈_𝑘. This completes the proof.

Proof of Theorem 6. From Theorem 5 and Lemma 11, we obtain the result.

4. The Main Results under 𝑓₀=∞, 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, ∫¹₀𝑡𝑓^′(±𝛼𝑡(1−𝑡))𝑑𝑡<∞ and ∫¹₀(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, 𝑓(𝑢)=√−√𝑢,𝑢≥0,−𝑢,𝑢<0. (39) 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∈𝑆^𝜈_𝑘 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 𝑢^′′(𝑡)(𝑡)𝑓(𝑢)(),𝑢𝑡+𝜇𝑎𝑢+𝑎=0,𝑡∈0,1₀=0,𝑢^′𝑡₀=𝛽 (40) has a unique solution. Particularly, it is true for the case 𝑓₀=∞. Thus, 𝑢∈𝑆^𝜈_𝑘 for some 𝑘≥1 and 𝜈∈{+,−} and 𝑢^′(0)≠0,𝑢^′(1)≠0.

Lemma 14. For 𝜈∈{+,−}, if (𝜇,𝑢)∈𝑆^𝜈₁ is a solution of (4), then 𝜇<𝜆₁, where 𝜆₁ is the first eigenvalue of (6).

Proof. Note that 𝑢 and 𝜑₁ satisfy, respectively, the equations 𝑢^′′+𝑓(𝑢)𝜇+𝑢𝑎(𝑡)𝜑𝑢=0,_′′1+𝜆₁𝑎(𝑡)𝜑₁=0. (41) 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. Let 𝐼⊂ℝ be a bounded and closed interval and 𝜆₁∈𝐼. For 𝜈∈{+,−}, if (𝜇_𝑛,𝑢_𝑛)∈𝑆^𝜈₁(𝑛=1,2,…) are solutions of (4) and 𝜇_𝑛∈𝐼, then 𝜇_𝑛⟶𝜆₁𝑎𝑠𝑢_𝑛𝐸⟶∞. (42)

Proof. Note that 𝑓_∞=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 Lemma 15, we have the following lemma.

Lemma 16. Let 𝐽⊂[0,𝜆₁) be a closed interval. Then there exists 𝑏_𝐽>0 such that ‖𝑢‖_𝐸≤𝑏_𝐽, (43) for every solution (𝜇,𝑢)∈𝐽×𝑆^𝜈₁ of (4).

Lemma 17. Let 𝐽̃⊂[0,𝜆₁) be a closed interval. Then there exists 𝑐𝐽̃>0 such that ‖𝑢‖_𝐸≥𝑐𝐽,̃ (44) for each solution (𝜇,𝑢)∈𝐽̃×𝑆^𝜈₁ of (4).

Proof. Without loss of generality, we select 𝑆⁺₁ to be discussed. When 𝜇∈[0,𝜆₁), since (𝜇,𝑢) is a positive solution of (4), we have 𝑢^′′=−𝜇𝑎(𝑡)𝑢−𝑎(𝑡)𝑓(𝑢)≤0. (45) On the contrary, suppose there exists a sequence of solutions of (4) {(𝜇_𝑛,𝑢_𝑛)}⊂𝐽̃×𝑆⁺₁ satisfying 𝜇_𝑛∈𝐽,𝑢_𝑛𝐸⟶0,as𝑛⟶∞. (46)
Noting that 𝑢_𝑛 is concave on the interval (0,1), then 𝑢^′_𝑛 has only one zero in (0,1), denoted by 𝜉_𝑛. Integrating 𝑢^′′_𝑛+𝜇_𝑛𝑎(𝑡)𝑢_𝑛+𝑎(𝑡)𝑓𝑢_𝑛=0,𝑡∈(0,1), (47) we get 𝑢^′_𝑛(𝑡)=−^𝑡_𝜉𝑛𝜇_𝑛𝑢_𝑛+𝑓𝑢_𝑛𝑎(𝑠)𝑑𝑠. (48) Taking subsequences if necessary, let lim_𝑛→∞𝜉_𝑛=𝜉_∞. (49) Without loss of generality, suppose 0<𝜉_∞<1. For the case 𝜉_∞=0 or 𝜉_∞=1, the proof is similar. Since 𝑢_𝑛 is concave on the interval (0,1), for 0<𝜖<𝜉_∞/4 small enough, there exists a constant 𝜎>0 such that for large 𝑛𝑢_𝑛(𝑡)≥𝜎𝑢_𝑛∞,∀𝑡∈[𝜖,1−𝜖]. (50) Since 𝑓₀=∞, there exists constant 𝑅₁>0 such that 𝑓(𝑢)≥𝜂𝑢,as|𝑢|≤𝑅₁, (51) where 𝜂>0 satisfies 𝜇_𝑛+𝜂𝜎_𝜖^2𝜖𝑑𝑠_𝑠^2𝜖𝑎(𝜏)𝑑𝜏>1. (52) Since ‖𝑢_𝑛‖_𝐸→0, we have ‖𝑢_𝑛‖_∞→0. Thus, for large enough 𝑛, ‖𝑢_𝑛‖_∞<𝑅₁. And for every 𝑡∈[𝜖,1−𝜖], we have 𝜎𝑢_𝑛∞≤𝑢_𝑛(𝑡)≤𝑢_𝑛∞. (53) Moreover, 𝑢_𝑛𝜉_𝑛=^𝜉𝑛₀^𝜉𝑛_𝑠𝜇_𝑛𝑎(𝜏)𝑢_𝑛(𝜏)+𝑎(𝜏)𝑓𝑢_𝑛(𝜏)≥𝜇𝑑𝜏𝑑𝑠_𝑛+𝜂_𝜖^2𝜖_𝑠^2𝜖𝑎(𝜏)𝑢_𝑛(𝜏≥𝑢)𝑑𝜏𝑑𝑠_𝑛∞𝜇_𝑛+𝜂𝜎_𝜖^2𝜖𝑑𝑠_𝑠^2𝜖𝑎(𝜏)𝑑𝜏. (54) Therefore, 𝑢_𝑛∞≥𝑢_𝑛∞𝜇_𝑛+𝜂𝜎_𝜖^2𝜖𝑑𝑠_𝑠^2𝜖𝑎(𝜏)𝑑𝜏. (55) From this, we have 𝜇_𝑛+𝜂𝜎_𝜖^2𝜖𝑑𝑠_𝑠^2𝜖𝑎(𝜏)𝑑𝜏≤1. (56) It contradicts (52). This ends the proof.

Lemma 18. Let 𝜇∈[0,𝜆₁) and (𝜇,𝑢)∈ℝ×𝑆^𝜈₁ 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 𝑤^′′+𝜇𝑎(𝑡)𝑤+𝑎(𝑡)𝑓_𝑢(𝑢(𝑡(),𝑤(0)))𝑤=0,𝑡∈0,1=0,𝑤^′(0)=𝛽≠0, (57) where 𝑓_𝑢 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 ¹₀𝑡𝜇+𝑓_𝑢(𝑢(𝑡))𝑎(𝑡)𝑑𝑡<∞,₁₀(1−𝑡)𝜇+𝑓_𝑢(𝑢(𝑡))𝑎(𝑡)𝑑𝑡<∞. (58) It is sufficient to prove ¹₀𝑡𝑓_𝑢(𝑢(𝑡))𝑑𝑡<∞,₁₀(1−𝑡)𝑓_𝑢(𝑢(𝑡))𝑑𝑡<∞. (59) 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×𝑆⁺₁  (when (𝜇,𝑢)∈ℝ×𝑆⁻₁, 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 𝑓_𝑢(𝑢(𝑡))<𝑓_𝑢(𝛼𝑡(1−𝑡)),∀𝑡∈(0,1). (60) It follows that ¹₀𝑡𝑓_𝑢(𝑢(𝑡))𝑑𝑡<¹₀𝑡𝑓_𝑢(𝛼𝑡(1−𝑡))𝑑𝑡<∞,₁₀(1−𝑡)𝑓_𝑢(𝑢(𝑡))𝑑𝑡<¹₀(1−𝑡)𝑓_𝑢(𝛼𝑡(1−𝑡))𝑑𝑡<∞. (61) 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 𝑤₀(1)≠0. (62)
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 that 𝑢(𝑡),𝑤₀(𝑡) satisfy the following equations: 𝑢^′′+𝑓(𝑢)𝜇+𝑢𝑎(𝑡)𝑤𝑢=0,_′′0+𝜇+𝑓^′(𝑢)𝑎(𝑡)𝑤₀=0, (63) 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 𝑤₀(1)=0. (64) 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.

Suppose 𝜇₀≥0 and (𝜇₀,𝑢₀) is a positive (resp. negative) solution of (4). For 𝜈∈{+,−}, we have 𝑢₀∈𝑆^𝜈₁. 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 Γ₀. Then we have the following lemma.

Lemma 19. Γ₀⊂ℝ×𝑆^𝜈₁.

Proof. Note that 𝑆^𝜈₁ is an open set. This together with Lemma 13 implies the conclusion.

Lemma 20. Γ₀ can be continued on the interval [0,𝜆₁) and there exists ℎ>0 such that lim_𝜇→0‖𝑢‖_𝐸=ℎ>0,lim_𝜇→𝜆1‖𝑢‖_𝐸=∞. (65)

Proof. From Lemmas 14–18 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 Γ₀ and hyperplane {0}×𝐸. From Lemma 18, (0,𝑢₀) is also a nondegenarate solution of (4) and ‖𝑢‖_𝐸=ℎ>0. Hence, Γ₀ can be continued to the direction of 𝜇<0.

Theorem 21. Let (H1′), (H2′), (H3′), and (H4′) hold. Then for 𝜇≥0 all solutions of (4) belonging to 𝑆⁺₁ (resp. 𝑆⁻₁) 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 Lemma 20 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.

Acknowledgments

This paper is supported by the NFSC (10971139), China Postdoctoral Fund (no. 2011M500615), Innovation Program of Shanghai Municipal Education Commission (no. 11YZ225).