#### Abstract

We study the existence of nodal solutions for the following problem: , , , where is a parameter, with on any subinterval of , , and ; , for , and , where and . We use bifurcation techniques to prove our main results.

#### 1. Introduction

By applying the bifurcation techniques of Rabinowitz [1, 2], Im et al. [3] studied the existence of positive solutions of the problems and some authors [4–7] studied the existence of nodal solutions of the problems. Recently, Dai and Ma [8] established the unilateral global bifurcation theory for the problems by applying the bifurcation techniques of Dancer [9]. Later, Dai [10] also considered the existence of nodal solutions for the problems with nonasymptotic nonlinearity at or by applying the unilateral global bifurcation theory of [8]. For the abstract unilateral global bifurcation theory, we refer the reader to [1, 2, 8, 9, 11–13] and the references therein. On the other hand, problems involving nondifferentiable nonlinearity have also been investigated by applying bifurcation techniques; see [13–17]. In particular, Berestycki [14] established the Rabinowitz-type global interval bifurcation result and Ma and Dai [13] established the unilateral global interval bifurcation theorem. Meanwhile, half-linear or half-quasilinear boundary value problems have attracted the attention of many specialists in different equations because of their interesting applications; see [13, 14, 16–18]. Among them, Berestycki [14] (or see [13]) established the spectrum for the following half-linear eigenvalue problem:where , , , and satisfies the following condition:) with on any subinterval of .

Let with the norm . Let with the norm .

We define the linear operator with . Then is a closed operator and is completely continuous.

Let under the product topology. Let denote the set of function in which have exactly interior nodal (i.e., nondegenerate) zeros in and are positive near , set , and . They are disjoint and open in . Finally, let and .

By Theorem 1 of [14], Berestycki [14] obtained the following result.

Lemma 1 (see [14, Theorem 2]). *Let hold. There exist two sequences of simple half eigenvalues for (1)The corresponding half-lines solutions are in and . Further, aside from these solutions and the trivial ones, there are no other solutions of (1).*

Furthermore, following Lemma 1, Ma and Dai [13] (or see [14]) considered the existence of nodal solutions for the following half-linear eigenvalue problem:where is a parameter and satisfies the following assumptions:() for .()There exist such that

Firstly, we show that the existence and uniqueness theorem is valid for (4).

Lemma 2 (see [13, Lemma 2.2]). *If is a nontrivial solution of (4) under assumptions and has a double zero, then .*

Under assumptions ()–(), they obtained the following result.

Theorem 3 (see [13, Theorem 4.1]). *Let , , and hold. For , either or . Then problem (4) has a solution such that has exactly zeros in and is positive near .*

Of course, the natural question is that of what would happen if or . Obviously, the previous results cannot deal with this case. The purpose of this work is to establish several results similar to those of [13]. The main methods used in this work are global bifurcation techniques and the approximation of connected components. Moreover, we consider the cases of , while the authors of [13, 16, 17] only studied the cases of .

In this paper, we will investigate the existence of nodal solutions for problem (4), where satisfies condition . Throughout this paper, we assume that satisfies and the following assumptions:() and .() and .() and .() and .() and .() and .() and .() and .

The rest of this paper is arranged as follows. In Section 2, we gave some preliminaries. In Section 3, we give the interval for the parameter which ensure the existence of single or multiple nodal solutions for half-linear problem (4) under assumptions and ()–() for the nonlinearity .

#### 2. Preliminaries

Let denote the closure of the set of nontrivial solutions of (4) in , with denoting the subset of with and .

In order to prove our main results, we need the following Sturm type comparison result.

Lemma 4. *Let for and , . Also let , be solutions of the following differential equations: respectively. If , and , in , then either there exists such that or or or and for some constant .*

*Proof. *We discuss four cases.*Case 1 (**, ** in **).* By the Picone identity [19] (or (0.4) of [20]), we havewhere , . It follows thatThe left-hand side of (8) equalsWe prove that . If , then . If , noting the conclusion of Lemma 2, then . By L’Hospital rule, we have thatSimilarly, we can obtain that . Therefore, the left-hand side of (8) equals zero. Hence, the right-hand side of (8) also equals zero. It follows that there exists a constant such that and .*Case 2 (**, ** in **).* Similar to (8), we can getThe above argument is still valid for this case.*Case 3 (**, ** in **).* Similar to Case 1, we can get the result.*Case 4 (**, ** in **)*. Similar to (8), we can getSimilar to the proof of Case 1, we can obtain the result.

By Lemma 4, we obtain the following result that will be used later.

Lemma 5. *Let be such that and **Let be such that**Let be a solution of the equationThen must change sign on as .*

*Proof. *Set and . By simple computation, we can show that After taking a subsequence if necessary, we may assume that as , where is th eigenvalue of the following problem:Set , . By some simple computations, we can showLet be the corresponding eigenfunction of . Since as , Lemma 2.1 of [4] implies that must change sign on . Note that the conclusion of Lemma 4 also is valid if . Using these facts and Lemma 4, we can obtain the desired result.

*Definition 6 (see [21]). *Let be a Banach space and be a family of subsets of . Then the superior limit of is defined by

Lemma 7 (see [22]). *Let be a Banach space and let be a family of closed connected subsets of . Assume that*(i)*there exist , , and , such that ;*(ii)*;*(iii)*for all , is a relative compact set of , where**Then there exists an unbounded component in and .*

#### 3. Nodal Solutions for Half-Linear Eigenvalue Problems

In this section, we will study the existence of nodal solutions for problem (4) with or .

We start this section by studying the following eigenvalue problem:where is a parameter. Firstly, under the conditions ()–(), let be such thatwith . Let us consideras a bifurcation problem from the trivial solution . By using Theorem 3 of [13] or Ma and Dai [13, Lemma 4.1] obtained the following Lemma.

Lemma 8 (see [13, Lemma 4.1]). *For , is a bifurcation point for problem (24). Moreover, there exists an unbounded continuum of solutions of problem (24), such that .*

*Remark 9. *Any solution of (22) of the form yields a solution of (4). In order to prove our main results, one will only show that crosses the hyperplane in .

Clearly, implies . Hence, is always the solution of (4). Applying Lemma 8 (or Lemma 4.1 of [13]), we will establish the existence of nodal solutions of (4) as follows.

Theorem 10. *Let , , and hold. For , assume that one of the following conditions holds:*(i)* for .*(ii)

*(iii)*

*for*.*(iv)*

*for*.

*for*.*Then problem (4) possesses 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 (i) since the proofs of the cases for (ii), (iii), and (iv) can be given similarly.

In view of the proof to prove [13, Theorem 4.1], we only need to show that joins to . To do this, it is enough to prove that .

Assume on the contrary that ; then there exists a sequence such thatfor some positive constant not depending on .

By , let , then is nondecreasing andWe consider the equationLet , and should be the solutions of problemSince is bounded in , choosing a subsequence and relabelling if necessary, we have that for some and .

Furthermore, from (26) and the fact that is nondecreasing, we have thatsinceBy (29) and the compactness of , we obtain that By , there exists such that . Without loss of generality, it follows that .

By the Gronwall-Bellman inequality [3, Lemma 2.1], we get , .

This contradicts .

Theorem 11. *Let , , and hold. For , assume that one of the following conditions holds:*(i)* for .*(ii)

*(iii)*

*for*.*(iv)*

*for*.

*for*.*Then problem (4) possesses two solutions and such that has exactly zeros in and is positive near and has exactly zeros in and is negative near 0.*

*Proof. *We will only prove the case of (i) since the proofs of the cases for (ii), (iii), and (iv) are completely analogous.

Inspired by the idea of [23] or see [10], we define the cut-off function of as follows:We consider the following problem:Clearly, we can see that , , and .

Similar to the proof of [13, Theorem 4.1], by Lemma 8 and Remark 9, there exists an unbounded continuum of solutions of problem (34) emanating from , such that , and joins to .

Taking and , we have that .

So condition (i) in Lemma 7 is satisfied with .

Obviouslyand, accordingly (ii) in Lemma 7 holds. (iii) in Lemma 7 can be deduced directly from the Arzela-Ascoli Theorem and the definition of .

Therefore, by Lemma 7, contains an unbounded connected component with .

From , (34) can be converted to equivalent equation (22). Since , we conclude . Moreover, by (4).

In the following, we show that .

Let satisfy .

Let be such thatWith , let , then is nondecreasing andWe divide the equationLet , and should be the solutions of problemSince is bounded in , choosing a subsequence and relabelling if necessary, we have that for some .

Furthermore, from (37) and the fact that is nondecreasing, we have thatsinceBy (40) and the compactness of , we obtain that It is clear that and since is closed in . Moreover, by Lemma 1, , so thatThus joins to .

Theorem 12. *Let , , and hold. For , assume that one of the following conditions holds:*(i)* for .*(ii)

*(iii)*

*for*.*(iv)*

*for*.

*for*.*Then problem (4) possesses 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. *DefineWe consider the following problem:Clearly, we can see that , , and .

Similar to the proof of [13, Theorem 4.1], by Lemma 8 and Remark 9, there exists an unbounded continuum of solutions of problem (45) emanating from , such that , and joins to .

Taking and , we have that .

So condition (i) in Lemma 7 is satisfied with .

Obviouslyand accordingly (ii) in Lemma 7 holds. (iii) in Lemma 7 can be deduced directly from the Arzela-Ascoli Theorem and the definition of .

Therefore, by Lemma 7, contains an unbounded connected component with .

Similar to Theorem 11, we obtain .

Similar to the method of the proof of Case 2 of [13, Theorem 4.1], we can obtain that .

Theorem 13. *Let , , and hold. For , assume that one of the following conditions holds:*(i)* for .*(ii)

*(iii)*

*for**or*.

*for*.*Then problem (4) possesses 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. *DefineWe consider the following problem:Clearly, we can see that , , and .

Applying a similar method used in the proof of Theorem 12, we obtain an unbounded connected component with .

Similar to the method of the proof of Theorem 10, we can obtain that . It follows that the result is obtained.

Theorem 14. *Let , , and hold. For , assume that one of the following conditions holds:*(i)*There exists for , such that .*(ii)

*There exists*(iii)

*for*, such that .*There exists*(iv)

*for*, such that .*There exists*

*for*, such that .*Proof. *DefineWe consider the following problem:Clearly, we can see that , , and .

Applying a similar method used in the proof of Theorem 12, we obtain an unbounded connected component with .

In the following, we can show that .

Assume that there exists a sequence such thatWe divide the proof into two steps.*Step 1.* We show that .

If is bounded, then there exists a constant not depending on such that . It follows that Letdenote the simple zeros of in . Then, after choosing a subsequence and relabelling if necessary, We claim that there exists such that Otherwise, we haveThis is a contradiction.

Let , with . For all sufficiently large, we have . So does not change its sign in . Without loss of generality, we assume that , . Moreover, we have . Combining this with the fact and with the relationFurthermore, we obtain that for any . Let be the corresponding eigenfunction of of (1). If for any , applying Lemma 4 to and on , we can get that must change its sign in for large enough. While, this is impossible. So we have that for any . Applying Lemma 5 to on , we can get that must change its sign in for large enough. This is a contradiction.*Step 2.* We show that .

Assume on the contrary that . We consider the problem