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Abstract and Applied Analysis

Volume 2013 (2013), Article ID 897307, 9 pages

http://dx.doi.org/10.1155/2013/897307

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

Department of Mathematics, Shanghai Institute of Technology, Shanghai 201418, China

Received 29 March 2013; Accepted 13 May 2013

Academic Editor: Maoan Han

Copyright © 2013 Yulian An. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### 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 , , , and , where satisfies and the limits , . Weight function satisfies on .

#### 1. Introduction

In this paper, we consider the existence and exact multiplicity of sign-changing solutions for the boundary value problem where satisfies and weight function satisfies on . 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 , 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 under the conditions(C1) satisfies , for ;(C2), if ;(C3) the limit .

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 . The set consists of some curves which bifurcate from the trivial solution line 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 . Bari and Rynne [15] considered the th order boundary value problem where is a positive parameter, and the function satisfies , , for , and . 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 . 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 , . On the other hand, we discuss the case that instead of . The main difficulty is to show any nontrivial solution of (1) to be nondegenerate. We will introduce an auxiliary function 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 and . In Section 4, we consider the exact multiplicity of constant sign solutions of (1) under the conditions and .

#### 2. Some Notations

For applying bifurcation theorem, we consider the auxiliary problem where is a parameter.

Clearly, any solution of (4) of the form corresponds to a solution of (1). From , we know is a solution of (4) for any , such solutions will be called trivial solutions.

We introduce some notations to describe the properties of solutions of (4). For a function , if , then is a simple zero of if . For any and any , we define sets consisting of the functions with (i), ; (ii) has only simple zeros in and has exactly zeros in .

Obviously, the sets are open in and disjoint.

Suppose that is a solution of (4). Then the corresponding linearized problem of (4) is
We call the solution 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

*Remark 1. *Note that and on . It is well known that the eigenvalues of (6) are given by
For each , is algebraic simple and the corresponding normalized eigenfunction can be chosen .

In this paper, we work in the following spaces:
where is the normal supnorm. Obviously, and are Banach spaces.

#### 3. The Main Results under ,

In the section, we assume(H1) satisfies , if ;(H2), if ;(H3) the limits and ;(H4) satisfies and , if .

Note that (H1) ensures that the solution of the initial value problem is not only existent but also unique on the whole interval for any and . 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 , where with , , for all, is a large enough constant. It is easy to check that satisfies (H4).

Now, we give some important lemmas.

Lemma 3. *Suppose is a nontrivial solution of (4). Then,* (i)* for some and ;* (ii)*the zeros of and the zeros of are separated as .*

*Proof. *(i) Since is nontrivial, “IVPU” implies that all the zeros of are simple. Thus, (i) is true.(ii) When , we have
This implies that the zeros of and the zeros of are separated. Since for some and , we get has zero on and has exact zero on .

Lemma 4. *Suppose and is a nontrivial solution of (4). Then is nondegenerate.*

*Proof. *we need to show that (5) has only trivial solution. Consider the initial problem
Clearly, (11) has a unique solution . For every solution of (5), there exists a unique constant such that . We claim that
If (12) holds, then we immediately have if and only if . That is, to say . Then we will finish the proof. Now, we prove (12) is true.

Firstly, we show that there is at least one zero of between consecutive zeros of .

Suppose are consecutive zeros of , that is, . Without loss of generality, suppose , . If has no zero on , we assume , . (When , , the proof is similar to the case .)

Note that and satisfy the following equations, respectively:
Multiply (14) by and subtract from it (13) multiplied by , with to be specified. Then integrate over
We denote the left side of (15) by and a constant by . Integrating by parts,
Let
on . By the above supposition, we have
However, the right side of (15) is zero. A contradiction. Hence, there is at least one zero of between consecutive zeros of .

Note that the functions , satisfy the following equations:
respectively. Since and , by the Sturm comparison lemma, there exists at least one zero of between any two consecutive zeros of . This implies that has at least zeros on .

Secondly, we show (12). On the contrary, assuming , then has at least zeros on since . We conclude that has at least zeros on . This contradicts the fact that has exact zeros on .

Finally, we give a proper function satisfying (17). Integrating the differential equation in (17), we can choose
In view of (H4), we conclude . So, the auxiliary function exists. This completes the proof.

Our main results are the following.

Theorem 5. *Let (H1)–(H4) hold. Then, for fixed and , all solutions of (4) belonging to (resp. ) lie on a unique continuous curve on which there is not any turning point. This curve starts from , tends to left, and passes through the hyperplane . Precisely, (4) has exactly one solution belonging to (resp. ) for and has no solution belonging to (resp. ) for , where is the th eigenvalue of (6). See Figure 1(a).*

From Theorem 5, we immediately obtain the following theorem.

Theorem 6. *Let (H1)–(H4) hold. Then (1) has a unique solution in for every and .*

We give some lemmas for proving Theorem 5.

Lemma 7. *Let be a closed and bounded interval. Suppose for some and is a sequence of solutions of (4). Then
*

*Proof. *Define by setting
Then is completely continuous. Noting that , consider
as a bifurcation problem from . 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 for some and is a solution of (4). Then .*

*Proof. *Note that and satisfy equations
respectively. On the contrary, supposing , then since if and . Then by the Sturm comparison lemma, between any two consecutive zeros of , there exists at least one zero of . This implies that has at least zeros on . It is impossible.

Lemma 9. *Let be a closed subinterval of . Then there is such that for every solution of (4).*

*Proof. *Note that when , we have
Suppose on the contrary that there exists a sequence of solutions of (4) with , and as . Let
denote the zeros of in . Then we can choose at least one subinterval which is of length at least , for some . Without loss of generality, we suppose on . Moreover, we claim that is an unbounded sequence.

Assume that is uniformly bounded. Since is concave on , has only one zero in . Integrating the equation in (4), for any ,
It implies that is uniformly bounded. Then we obtain are uniformly bounded. Consider consecutive intervals and . By convexity of on and and the uniform boundedness of , is uniformly bounded on and . So,
are uniformly bounded. Hence, is uniformly bounded on and . By some finite steps, this procedure shows that is uniformly bounded on . This is a contradiction.

Taking subsequences if necessary, put
Noting that , then . We may assume . The case or can be considered similarly. In the rest of the proof, denotes . Since is concave on , for any small enough, there exists a constant such that
By the condition in (H3), there exists a real number such that
where can be given by
Since , for sufficiently large . Thus
Moreover,
Hence,
It follows that
This contradicts (32). This ends the proof.

Suppose there exits a nontrivial solution of (4) with . It follows from Lemma 3 that for some and . Meanwhile, from the implicit function theorem and Lemma 4, we have that all solutions of (4) in near lie on a curve passing through and parameterized by . We denote the local curve by . Then, we have the following lemma.

Lemma 10. *.*

*Proof. *This can be obtained by Lemma 3 and the fact that is open.

Lemma 11. * can be continued on the interval . Moreover, there is a constant such that
*

*Proof. *Suppose that there exist a sequence and a small constant such that , , and for large enough . Then, after choosing a subsequence if necessary, there exists such that in . It then follows from the equation in (4) that and is a nontrivial solution of (4). Hence, by Lemma 4 and the implicit function theorem, the curve can still be continued. Meanwhile, we know from Lemma 9. Combining this with Lemma 8, we have can be continued on the interval . Moreover, by Lemma 7, there is a constant such that

Assume is the intersection of and hyperplane . From Lemma 4, is also a nondegenerate solution of (4) and . Hence, can still be continued to the direction of .

*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 exactly two local curves of nontrivial solutions bifurcate from the point in , one of which lies in and the other in . By the above discussion, each of these local curves can be continued at least on the interval . We will denote these particular curves by . Then by Lemma 10. From Lemmas 4 and 11, can pass through the hyperplane and go to the direction of .

Finally, we exclude the possibility that there exists another solution curve of (4) belonging to the set . Suppose that there exists a solution but . Similarly, by continuation, we can obtain another solution curve such that for . Since is the unique solution curves near in by the standard Crandall and Rabinowitz theorem, must coincide with . This completes the proof.

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

#### 4. The Main Results under ,

In this section, we study the exact multiplicity of constant sign, that is, positive (resp. negative) solution of (1) under the conditions(H1′), . When , exists and satisfies . is nondecreasing on and nonincreasing on . For any fixed , and .(H2′), if ;(H3′), ;(H4′) satisfies , .

*Remark 12. *There are many functions satisfying conditions (H1′)–(H3′). For example,
As before, we discuss the structure of the set of solutions of the auxiliary problem (4).

Lemma 13. *If is a solution of (4), then the zero of is simple and , . Moreover, for some and .*

*Proof. *From proportion 2.1 in [20], we can obtain, for every and , the initial value problem
has a unique solution. Particularly, it is true for the case . Thus, for some and and .

Lemma 14. *For , if is a solution of (4), then , where is the first eigenvalue of (6).*

*Proof. *Note that and satisfy, respectively, the equations
On the contrary, assume . Noting that for we have and , then . By the Sturm comparison lemma, there exists at least one zero of between any two consecutive zeros of . This implies that has at least one zeros on . It is impossible.

Lemma 15. *Let be a bounded and closed interval and . For , if are solutions of (4) and , then
*

*Proof. *Note that 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 be a closed interval. Then there exists such that
**
for every solution of (4).*

Lemma 17. *Let be a closed interval. Then there exists such that
**
for each solution of (4).*

*Proof. *Without loss of generality, we select to be discussed. When , since is a positive solution of (4), we have
On the contrary, suppose there exists a sequence of solutions of (4) satisfying

Noting that is concave on the interval , then has only one zero in , denoted by . Integrating
we get
Taking subsequences if necessary, let
Without loss of generality, suppose . For the case or , the proof is similar. Since is concave on the interval , for small enough, there exists a constant such that for large
Since , there exists constant such that
where satisfies
Since , we have . Thus, for large enough , . And for every , we have
Moreover,
Therefore,
From this, we have
It contradicts (52). This ends the proof.

Lemma 18. *Let and be a solution of (4). Then is nondegenerate. *

*Proof. *It is sufficient to prove that (5) has only trivial solution. Consider the initial value problem
where denotes the derivative of to . 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
It is sufficient to prove
Noting that for every , if , then ; if , then . Suppose (when , the proof is similar). Then there must exist such that , for all . By condition (H1′), we have
It follows that
From Lemma 3.2 in [3], (57) has a unique solution. When , we denote the solution of (57) by . Then, for every solution of (57), there exists a constant such that .

We claim that

If (62) holds, then we immediately have if and only if . That is to say, . Then we will finish the proof.

Note that satisfy the following equations: respectively. Since , , , by the Sturm comparison lemma, there exists at least one zero of between any two consecutive zeros of . If (62) does not hold, then Complying this with , we obtain there exists at least one zero of in the interval . This contradicts that is a positive (resp. negative) solution of (4). The proof is ended.

Suppose and is a positive (resp. negative) solution of (4). For , we have . By Lemma 18 and the implicit function theorem, all solutions of (4) near lie on a unique curve which passes through and is parameterized by . Denote the curve by . Then we have the following lemma.

Lemma 19. *.*

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

Lemma 20. * can be continued on the interval and there exists such that
*

*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 is the intersection point of and hyperplane . From Lemma 18, is also a nondegenarate solution of (4) and . Hence, can be continued to the direction of .

Theorem 21. *Let (H1′), (H2′), (H3′), and (H4′) hold. Then for all solutions of (4) belonging to (resp. ) lie on a unique continuous curve on which there is not any turning point. This curve bifurcates from , tends to left, and passes through the hyperplane . Precisely, (4) has exactly one positive (resp. negative) solution if and has no positive (resp. negative) solution for , where 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).

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