Research Article | Open Access

# Unilateral Global Bifurcation from Intervals for Fourth-Order Problems and Its Applications

**Academic Editor:**Kousuke Kuto

#### Abstract

We establish a unilateral global bifurcation result from interval for a class of fourth-order problems with nondifferentiable nonlinearity. By applying the above result, we firstly establish the spectrum for a class of half-linear fourth-order eigenvalue problems. Moreover, we also investigate the existence of nodal solutions for the following half-linear fourth-order problems: , , , where is a parameter, , , , , and , for . We give the intervals for the parameter which ensure the existence of nodal solutions for the above fourth-order half-linear problems if or where and . We use the unilateral global bifurcation techniques and the approximation of connected components to prove our main results.

#### 1. Introduction

In the past twenty years, fourth-order BVP have attracted the attention of many specialists in differential equations because of their interesting applications. For example, Bai and Wang [1], Ma and Wang [2], and Chu and O’Regan [3] have investigated the fourth-order BVP by the fixed point theory in cones. Meanwhile, by applying the bifurcation techniques of Rabinowitz [4, 5], Gupta and Mawhin [6], Lazer and McKenna [7], Rynne [8], Liu and O’Regan [9], Ma et. al [10, 11], Shen [12, 13], and Ma [14] studied the existence of nodal solutions for the fourth-order BVP.

Now, consider the following operator equation:where is a compact linear operator and is compact with at uniformly on bounded intervals, where is a real Banach space with the norm . If the characteristic value of has multiplicity and Dancer [15] has shown that there are two distinct unbounded continua and , consisting of the bifurcation branch of emanating from , where either and are both unbounded or . This result has been extended to the fourth-order problems by Dai and Han [16]. More specifically, Dai and Han [16] considered the following fourth-order problem:where is continuous satisfying , anduniformly on and on bounded sets.

Let with the norm Let with the norm . Let under the product topology. Let denote the set of functions in which have exactly generalized simple zeros in and are positive near , set , and They are disjoint and open in . Finally, let and . Let denote the closure of the set of nontrivial solutions of (3) in , and denote the subset of with for and

From Dancer [15], Dai and Han [16] obtained that problem (3) has two distinct unbounded subcontinua and , consisting of the bifurcation branch emanating from , which satisfy the following.

Lemma 1. *Either and are both unbounded or , and where and were defined as in [10] or [16].*

For the abstract unilateral global bifurcation theory, we refer the reader to [4, 5, 15, 17–19] and the references therein.

However, among the above papers, the nonlinearities are linear in the zeros and infinity. The problems involving nondifferentiable nonlinearities have also been investigated by applying bifurcation techniques; see Berestycki [20], Schmitt and Smith [19], Rynne [21], Ma and Dai [22], and Dai et al. [23–25] and references therein. Among them, in 1977, Berestycki [20] studied the differential equations involving nondifferentiable nonlinearity. The above Berestycki’s ([20]) result has been improved partially by Schmitt and Smith [19] by applying a set-valued version of Rabinowitz global bifurcation theorem. In 1998, Rynne [21] established the interval bifurcation from and and obtained sets of positive or negative solutions with the approximation technique from Berestycki [20]. Recently, Ma and Dai [22] established the global interval bifurcation for a Sturm-Liouville problem with a nonsmooth nonlinearity by [15]. Later, Dai et al. [23–25] studied the bifurcation from intervals for Sturm-Liouville problems and its applications and established the unilateral global interval bifurcation for -Laplacian with non--linearization nonlinearity, respectively.

On the other hand, half-linear or half-quasilinear problems have attracted the attention of some specialists; see [20, 22, 25]. Among them, Berestycki [20] studied the bifurcation structure for the half-linear equations. Recently, Ma and Dai [22] and Dai and Ma [25] studied the existence of nodal solutions for a class of half-linear or half-quasilinear eigenvalue problems and improved Berestycki’s result, respectively.

Motivated by the above papers, in this paper, we will firstly establish some Dancer-type unilateral global bifurcation results about the continuum of solutions for the fourth-order problems:where is a parameter, the nonlinear term has the form , where and are continuous functions on , and satisfy the following conditions:).(), for all , , , where is a positive constant.() near , uniformly in and on bounded sets.

Let denote the closure of the set of nontrivial solutions of (6) in , and denote the subset of with for and .

Under assumptions , we will show that is a bifurcation interval of problem (6) and there are two distinct unbounded subcontinua, and , consisting of the bifurcation branch from , where is given in Lemma 1.

On the basis of the above unilateral global interval bifurcation result, we study the following half-linear eigenvalue problem:where , and and satisfy the following:().

We will show that there exist two sequences of simple half-eigenvalues for problem (7): and The corresponding half-linear solutions are in and Furthermore, aside from these solutions and the trivial ones, there is no other solutions of problem (7).

Following the above eigenvalue theory (see Theorem 19), we will investigate the existence of nodal solutions for the following fourth-order problem:Let denote the closure of the set of nontrivial solutions of (8) in , with denoting the subset of with for and For , Dai and Han [16] have established the existence of nodal solutions for problem (8) with crossing nonlinearity. In this paper, we assume that satisfies the following assumptions:() for .() and .() and .() and .() and .() and .() and ,where

The rest of this paper is arranged as follows. In Section 2, we have given some preliminaries. In Section 3, we establish the unilateral global bifurcation result from the interval for problem (6). In Section 4, on the basis of the unilateral global interval bifurcation result, we will establish the spectrum for a class of the half-linear fourth-order eigenvalue problem (7) (see Theorem 19). In Section 5, following the above eigenvalue theory (see Theorem 19), we will investigate the existence of nodal solutions for a class of the half-linear fourth-order problem (8).

#### 2. Hypotheses and Lemmas

We define the linear operator with

From [26, p. 439-440], we consider the following auxiliary problem:for a given . We can get that problem (11) can be equivalently written as whereThen is a closed operator and is completely continuous.

Define the Nemytskii operator by

Then it is clear that is continuous operator and problem (3) can be equivalently written as Clearly, is completely continuous and , .

Letand then is nondecreasing with respect to , anduniformly for and on bounded sets. Further it follows from (17) thatuniformly for and on bounded sets.

In the following, we summarize some preliminary results from [10, 16].

*Definition 2 (see [16]). *Let and such that . We call that a generalized simple zero if or . Otherwise, we call that a generalized double zero. If there is no generalized double zero of , we call that a nodal solution.

Lemma 3 (see [10] or [16]). *Let hold. The linear eigenvalue problemhas an infinite sequence of positive eigenvalues: Moreover, each eigenvalue is simple. To each eigenvalue there corresponds an essential unique eigenfunction which has exactly generalized simple zeros in and is positive near .*

Lemma 4 (see [16]). *If is a nontrivial solution of (6) under assumptions and and has a generalized double zero, then .*

*Remark 5. *By Lemma 4, we can see that if is a nontrivial solution of (6) under assumptions and , then .

Lemma 6 (see [20, Lemma ]). *Let and be two integers such that . Suppose that there exist two families of real numbers:Then, if , there exist two integers and having the same parity: and , such that *

*Definition 7 (see [27]). *Let be a Banach space and let be a family of subsets of . Then the superior limit of is defined by

Lemma 8 (see [27]). *Each connected subset of metric space is contained in a component, and each connected component of is closed.*

Lemma 9 (see [28]). *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. Unilateral Global Bifurcation

The main result for problem (6) is the following theorem.

Theorem 10. *Let , , and hold. Let , where , and let for every . The component of , containing , is unbounded and lies in and the component of , containing , is unbounded and lies in .*

In order to prove Theorem 10, we need the following results.

Lemma 11. *If is bounded, we can find a neighborhood of such that , where () is given by Theorem 10.*

*Proof. *We only prove the case of since the case of is similar.

Let be a uniform neighborhood of in .

We discuss two cases.*Case 1 (if **)*. Since the solutions of problem (6) are bounded in , then is compact in . It follows that is compact metric space. Obviously, and are two disjoint closed subsets of . Because of the maximal connectedness of , there does not exist a component of such that , . By [27] or [4, Lemma ], there exist two disjoint compact subsets of , such that , , . Evidently, Let , and let be the -neighborhood of Set and then*Case 2 (if **)*. In this case, we take . It is obvious that the result holds.

Consider the following auxiliary approximate problem:For , it is easy to show that nonlinear term is differ-entiable at the origin. Let

By Lemma 1, there are two unbounded continua of and of bifurcating from , consisting of the bifurcation branch , which satisfy the following result.

Lemma 12. * and are both unbounded and *

*Proof. *By Lemma of [4], there exists a bounded open neighborhood of such that It follows that By Lemma 4, we show that cannot leave outside of a neighborhood of . Thus, we haveNext, we will prove that both and are unbounded.

Without loss of generality, we may suppose that is bounded. Therefore, in view of (32) and Lemma 1, there exists such that and . This contradicts the definitions of and .

To prove Theorem 10, the next lemma will play a key role.

Lemma 13. *Let , , be a sequence converging to . If there exists a sequence such that is a nontrivial solution of problem (27) corresponding to , and converges to in , then .*

*Proof. *Without loss of generality, we may assume that . Let ; then satisfies the problemwhereBy (18), it follows thatuniformly for and on bounded sets. Furthermore, implies thatfor all .

Note that implies . Using this fact with (35) and (36), we have that is bounded in for large enough. The compactness of implies that is convergence in . Without loss of generality, we may assume that in with . Clearly, we have .

We claim that . On the contrary, supposing that , by Lemma 4, then , which is a contradiction with .

Now, we deduce the boundedness of . Let be an eigenfunction of problem (19) corresponding to .

Let be, respectively, the sequences of generalized simple zeros of and .

Suppose ; then we deduce from Lemma 6 existence of integers and having the same parity such that Therefore, without loss of generality, we choose , , , and and since and have the same parity and do not vanish and have the same sign in both the intervals and . It follows that , and .

We can assume without loss of generality that and in . By the Picone identity in [29, Theorem ], we have thatThe left-hand side of (39) equalsWe prove that . If , then . If , noting the conclusion of Lemma 3, then . By L’Hospital rule, we have thatIn the following, we will prove thatLet Then and in For some small enough, let and be such that , , and . Then we haveSet and Let and be such that , in and and Set ; then should be a solution of the problemThe Strong Maximum Principle implies that in . This follows that in It follows that

Next, we will prove that . If , then . If , noting the conclusion of Lemma 3, then . By L’Hospital rule, we have thatBy (42), we can show that . Therefore, the left-hand side of (39) .

By (42), we have .

It follows thatSimilarly, we can also show thatBy (35), (46), and (47), taking the limit as , we can obtain thatBy (36), and have the same sign in ; one has So, by (48), we have thatFurthermore, it follows that if , if ,Therefore, we have that .

Lemma 14. *For , the component of satisfies , where is given by Theorem 10.*

*Proof. *We only prove the case of since the case of is similar. For any