#### Abstract

We are concerned with determining values of , for which there exist nodal solutions of the boundary value problem , where is a sign-changing function, with . The proof of our main results is based upon global bifurcation techniques.

#### 1. Introduction

In [1], Ma and Thompson considered determining values of , for which there exist nodal solutions of the boundary value problem under the following assumptions:() with for ;() is continuous and does not vanish identically on any subinterval of ;()there exist , such that

Using the bifurcation theory of Rabinowitz [2, 3], they proved the following.

Theorem 1. *Let (), (), and () hold. Assume that, for some , either
**
Then (1) has two solutions and such that has exactly zeros in and is positive near 0 and has exactly zeros in and is negative near 0.*

The results of Theorem 1 have been extended to the case that the weight function changes its sign by Ma and Han [4]. Bifurcation methods have been applied to study the existence of nodal solutions of nonlinear two-point, multipoint, and periodic boundary value problems; see [5–9] and the references therein. The results they obtained extend some well-known theorems of the existence of positive solutions for the related problems [10].

However, no results on the existence of nodal solutions, even positive solutions, have been established for one-dimensional -Laplacian equation with sign-changing weight . It is the purpose of this paper to establish a similar result to Theorem 1 for one-dimensional -Laplacian equation with sign-changing weight. Problem with sign-changing weight arises from the selection-migration model in population genetics. In this model, changes sign corresponding to the fact that an allele holds an advantage over a rival allele at the same points and is at a disadvantage at others; the parameter corresponds to the reciprocal of diffusion; for details see [11].

If , Del Pino et al. [12] established the global bifurcation theory for one-dimensional -Laplacian eigenvalue problem. Peral [13] got the global bifurcation theory for -Laplacian eigenvalue problem on the unite ball. In [14], Del Pino and Manásevich obtained the global bifurcation from the principal eigenvalue for -Laplacian eigenvalue problem on the general domain. If and is singular at or , Lee and Sim [15] also established the bifurcation theory for one-dimensional -Laplacian eigenvalue problem. However, if changes sign, there are a few papers dealing with the -Laplacian eigenvalue problem via bifurcation techniques. In [16], Drábek and Huang established the global bifurcation from the principal eigenvalue for -Laplacian eigenvalue problem in .

The purpose of this paper is to study the bifurcation behavior of one-dimensional -Laplacian eigenvalue problem as follows: under the condition and() changes sign and ()there exists such that where with ;()there exists such that Moreover, based on our global bifurcation theorem, we will prove the existence of nodal solutions for the corresponding nonlinear problem with a parameter (see Theorem 11).

The main tool is the global bifurcation techniques in [17].

The rest of this paper is arranged as follows. In Section 2, we establish the global bifurcation theory for one-dimensional -Laplacian eigenvalue problem with sign-changing weight. In Section 3, we state and prove the main results of this paper.

#### 2. Some Preliminaries

Let be the Banach space with the norm

Let with its usual normal .

We start by considering the following auxiliary problem: for a given . By a solution of problem (9), we understand a function with absolutely continuous which satisfies (9). Problem (9) is equivalently written to where is a continuous function satisfying It is known that is continuous and maps equi-integrable sets of into relatively compacts of . One may refer to Lee and Sim [15] for details.

Since the bifurcation points of is related to the eigenvalues of the problem

We define the operator by Then is completely continuous and problem (13) is equivalent to

The following spectrum result plays a fundamental role in our study.

Lemma 2 (see [18, 19]). *Let () hold. Then*(i)*the set of all eigenvalues of the problem (13) is two infinite sequences of simple eigenvalues as follows:
*(ii)*for and , Ker is a space of with dimensional ;*(iii)*the eigenfunction corresponding to has exactly simple zeros in .*

*Remark 3. *Using the Gronwall inequality, we can easily show that all zeros of eigenfunction corresponding to eigenvalue are simple.

It is very known that is completely continuous in . Thus, the Leray-Schauder degree is well-defined for arbitrary -ball and and .

Lemma 4. *For , we have
*

*Proof. *We divide the proof into two cases.*Case 1*. . Since is compact and linear, by [20, Theorem 8.10] and Lemma 2 (ii) with ,
where is the sum of algebraic multiplicity of the eigenvalues of (13) satisfying .

If , then there are no such at all; then

If for some , then

This together with Lemma 2 (ii) implies the following:
*Case 2*. . In this case, we consider a new sign-changing eigenvalue problem as follows
where , . It is easy to check that
Thus, we may use the result obtained in Case 1 to deduce the desired result.

We first show that the principle eigenvalue function is continuous.

Proposition 5. *The eigenvalue function is continuous.*

*Proof. *We only show that is continuous since the case of is similar. In the following proof, we will shorten to . From the variational characterization of , it follows that

Let be a sequence in convergent to . We will show that

To do this, let . Then, from (24),

On applying the Dominated Convergence Theorem, we find that

Relation (27), the fact that is arbitrary and (24) yield

Thus, to prove (25), it suffices to show that
Let be a subsequence of such that .

Let us fix so that and, for each , is compactly embedded into . For , let us choose such that
For , there exists such that for any . Thus, for , (30) and Hölder's inequality imply that
This shows that is a bounded sequence in . Passing to a subsequence if necessary, we can assume that in and hence that in . Furthermore, and in for . It follows that
as . It is clear that

Thus,

Similarly, we can also obtain that
where and . Therefore,

We note that (30) and (31) imply that
for all . Thus, letting go to in (38) and using (37), we find that

On the other hand, since in , from (32) we obtain that

Now, letting and applying Fatou's Lemma, we find that
Hence, ; here denotes the radially symmetric subspace of . We claim that actually . Indeed, we know that for each . For , it is easy to see that
Then, letting , we obtain that
where . Since is arbitrary, from Proposition IX-18 of [21], we find that , as desired.

Finally, combining (39) and (41), we obtain that

This and the variational characterization of imply (29) and hence (25). This concludes the proof of the lemma.

Using Remark 3, Lemma 2, and Proposition 5, we will show that all eigenvalue functions , are continuous.

Lemma 6. *For fixed and , as a function of is continuous.*

*Proof. *Let be an eigenfunction corresponding to . By Lemma 2 and Remark 3, we know that has exactly simple zeros in ; that is, there exist such that . For convenience, we set , , and for . Let denote the first positive or negative eigenvalue of the restriction of problem (13) on for . Lemma 3 of [18] follows that for . Using a similar proof to Proposition 5, we can show that is continuous with respect to for . Therefore, is also continuous with respect to .

Lemma 7. *(i) Let be the sequence of positive eigenvalues of (13). Let be a constant with for all . Then, for arbitrary ,
**
where is the number of eigenvalues of problem (13) less than .**(ii) Let be the sequence of negative eigenvalues of (13). Consider , ; then
**
where is the number of eigenvalues of problem (25) larger than .*

*Proof. *We will only prove the case since the proof for the other cases is similar. We also only give the proof for the case . Proof for the case is similar. Assume that for some . Since the eigenvalues depend continuously on , there exists a continuous function and such that and . Define

It is easy to show that is a compact perturbation of the identity such that, for all , by definition of , , for all . Hence, the invariance of the degree under homotopology and the classical result for imply

For the existence of bifurcation branches for (12), we will make use of the following global bifurcation theorem results.

Lemma 8 (see [17]). *Let be a Banach space. Let be completely continuous such that for all . Suppose that there exist constants , , with , such that and are not bifurcation points for the equation
**
Furthermore, assume that
**
where is an isolating neighborhood of the trivial solution for both constants and . Let
**
and let be the component of containing . Then, either *(i)* is unbounded in or*(ii)*.*

Define the Nemytskii operators by

Then, it is clear that is continuous operator which sends bounded sets of into an equi-integrable sets of and problem (12) can be equivalently written as is completely continuous in and , for all .

Notice that (12) with has only the trivial solution. Applying this fact and Lemma 8 and the same method to prove [15, Theorem 2.1] with obvious changes, we may obtain the following.

Lemma 9. *Assume that (), (), and () hold. Then, for fixed and for fixed , each is a bifurcation point of (12) and the associated bifurcation branch satisfies the following; *(1)* is unbounded in ;*(2)*, where is the set of function which has exact simple zeros in , and is positive near .*

Finally, we give a key lemma that will be used in Section 3. Let

Lemma 10. *Let hold. Let be such that and
**Let be such that
**Let be a solution of the equation
**
Then, the number of zeros of goes to infinity as .*

*Proof. *After taking a subsequence if necessary, we may assume that
as . It is easy to check that the distance between any two consecutive zeros of any nontrivial solution of the equation
goes to zero as . Using this with [21, Lemma 2.5], it follows the desired results.

#### 3. Main Results and Its Proof

Let be the th positive or negative eigenvalue of (13). By applying Lemma 9, we will establish the main results as follows.

Theorem 11. *Let (), (), (), and () hold. Assume that, for some , either
**
or
**Then, (4) has two solutions and such that has exactly zeros in and is positive near 0 and has exactly zeros in and is negative near 0.*

*Proof. *We only prove the case of . The case of is similar. Consider the problem

Considering the results of Lemma 9, we have that, for each integer , , there exists a continuum of solutions of (62) joining to infinity in . Moreover, .

It is clear that any solution of (62) of the form yields a solution of (4). We will show that crosses the hyperplane in . To this end, it will be enough to show that joins to . Let satisfy
We note that for all since is the only solution of (62) for and .*Case 1*. . In this case, we only need to show that
We divide the proof into two steps. *Step 1*. We show that, if there exists a constant number such that
for large enough, then joins to .

In this case, it follows that

Let be such that
Then,
Let
Then, is nondecreasing and

We divide the equation
by and set . Since is bounded in , after taking a subsequence if necessary, we have for some and in with . Moreover, from (70) and the fact that is nondecreasing, we have
since

By the continuity and compactness of , it follows that
where , again choosing a subsequence and relabeling if necessary.

We claim that

Suppose on the contrary that . Since is a solution of (74) and all zeros of in are simple, it follows that for some and .

By the openness of , we have that there exists a neighborhood such that
which contradicts the facts that in and . Therefore, . Moreover, by Lemma 2, , so that
Therefore, joins to .*Step 2*. We show that there exists a constant such that for large enough.

On the contrary, we suppose that

Since , it follows that

Let
be the zeros of in . Then, after taking a subsequence if necessary,
Notice that Lemma 10 and the fact that has exactly simple zeros in yield
which implies that

However, this contradicts (): .*Case 2*. . In this case, we have that

Assume that is such that

If , then we are done!

If there exists , such that, for sufficiently large,

Applying the same method used in Step 1 of Case 1, after taking a subsequence and relabeling if necessary, it follows that

Thus, joins to .

#### Acknowledgments

This paper was supported by the NSFC (nos. 11061030, 11361047, and 11201378), SRFDP (no. 20126203110004), and Gansu Provincial National Science Foundation of China (no. 1208RJZA258).