#### Abstract

We will establish unilateral global bifurcation result for a class of fourth-order problems. Under some natural hypotheses on perturbation function, we show that is a bifurcation point of the above problems and there are two distinct unbounded continua, and , consisting of the bifurcation branch from , where is the th eigenvalue of the linear problem corresponding to the above problems. As the applications of the above result, we study the existence of nodal solutions for the following problems: , , where is a parameter and are given constants; with on any subinterval of ; and is continuous with for We give the intervals for the parameter which ensure the existence of nodal solutions for the above fourth-order Dirichlet problems if or where and We use unilateral global bifurcation techniques and the approximation of connected components to prove our main results.

#### 1. Introduction

The deformations of an elastic beam in equilibrium state with fixed both endpoints can be described by the fourth-order boundary value problem where is a parameter, , are given constants, and is continuous. When , since problem (1) cannot transform into a system of second-order equation, the treatment method of second-order system does not apply to problem (1). Thus, there exists some difficulty studying problem (1) even in the case of .

In recent years, there has been considerable interest in the above BVP (1) mainly because of their interesting applications. For example, Agarwal and Chow [1] () first investigated the existence of the solutions of problem (1) by contraction mapping and iterative methods. Subsequently, when , by fixed point theory on cones, Ma et al. [2, 3], Yao [4, 5], Zhai et al. [6], and Webb et al. [7] studied the existence of positive solutions of problem (1).

On the other hand, by applying the bifurcation techniques of Rabinowitz [8, 9], Gupta and Mawhin [10], Lazer and McKenna [11], Liu and O’Regan [12], and Ma et al. [13–15] studied the existence of nodal solutions for the fourth-order BVP where both ends were simply supported, and Rynne [16] investigated the nodal properties of the solutions for a general th-order problem.

Meanwhile, it is well known that the spectrum structure of the linear eigenvalue problems according to (1) plays a key role to study problem (1) by the bifurcation techniques. Kratochvíl and Nečas [17] first studied the spectrum of the -biharmonic operator together with . Subsequently, Benedikt [18–21] also studied the spectral properties of the corresponding eigenvalue problem of the same problems as [17], and Benedikt [22] studied existence and global bifurcation of solutions for the above problems. When , by applying the bifurcation techniques, Korman [23] investigated the uniqueness of positive solutions and Rynne [24] studied nodal properties of the solutions for problem (1), respectively. By Elias’s theory [25, 26], Xu and Han [27] (), Ma et al. [28] (), and Ma and Gao [29] () established the spectrum structure of the linear eigenvalue problems according to (1) and studied the existence of nodal solutions of problem (1) using bifurcation theory [8]. In 2012, Shen [30, 31] established the following spectrum structure by applying disconjugate operator theory [25, 26].

Lemma 1 (see [30, 31]). *Let and hold. The linear eigenvalue problem has a unique infinite number of positive eigenvalues Moreover, each eigenvalue is simple. The eigenfunction corresponding to has exactly simple zeros in . For each , the algebraic multiplicity of is , where () one of following conditions holds:**(i) , satisfying , are given constants with **(ii) satisfying , are given constants with ** with on any subinterval of .*

On the basis of Lemma 1, Shen [30, 31] studied the existence of nodal solutions of problem (1) by applying Rabinowitz’s global bifurcation theorem [8].

In 2013, when , satisfy and , Shen and He [32] also studied bifurcation from interval and the existence of positive solutions for problem (1) by applying Rabinowitz’s global bifurcation theorem [9].

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 eigenvalue of has multiplicity , Dancer [33] has shown that there are two distinct unbounded continua and , consisting of the bifurcation branch of emanating from , which satisfy either that and are both unbounded or . This result has been extended to one-dimensional -Laplacian problem by Dai and Ma [34]. The above results [34] have been improved partially by Dai [35] with nonasymptotic nonlinearity at or . Later, Dancer’s result [33] has been also extended to the periodic -Laplacian problems by Dai et al. [36]. In 2013, Dai and Han [37] established Dancer-type unilateral global bifurcation results for fourth-order problems of the deformations of an elastic beam in equilibrium state where both ends are simply supported by Dancer [33].

In this paper, based the spectral theory of [30, 31], we will establish Dancer-type unilateral global bifurcation results about the continuum of solutions for the following fourth-order eigenvalue problem:where satisfies , and the perturbation function is continuous with and satisfies the following hypotheses uniformly for and on bounded sets.

Let with the norm and with the norm . Let denote the set of functions in which have exactly interior nodal (i.e., nondegenerate) zeros in and are positive near , set , and They are disjoint and open in . Let , , and under the product topology. Let denote the closure in of the set of nontrivial solutions of (1) and let denote the subset of with and .

Under condition (9), we will show that is a bifurcation point of (8) and there are two distinct unbounded continua, and , consisting of the bifurcation branch from , where is the th eigenvalue of problem (2). Based on the above result, we investigate the existence of nodal solutions for problem (1).

*Remark 2. *By applying disconjugate operator theory [25, 26], the authors [13, 14, 16] also established the spectrum structure of the corresponding linear eigenvalue problems. On the basis of the above spectrum structure, the authors [13, 14, 16] studied the existence of nodal solutions of the above problem by applying Rabinowitz’s global bifurcation theorem [8].

The rest of this paper is arranged as follows. In Section 2, we will establish unilateral global bifurcation results. In Section 3, we will investigate the existence of nodal solutions for problem (1) under the linear growth condition on .

#### 2. Unilateral Global Bifurcation Results

We define the linear operator with

From [31, p. 93], we consider the following auxiliary problem: for a given . We can get that problem (11) can be equivalently written as where was given in of [31, p. 93].

Then is a closed operator and is completely continuous.

Define the operator by

Furthermore, it is clear that problem (8) can be equivalently written as Clearly, is completely continuous from and

Let and then is nondecreasing and uniformly for and on bounded sets. Further it follows from (16) thatuniformly for and on bounded sets.

By (17), we have that as uniformly for and on bounded sets. Furthermore, Applying Theorem of [33], we may obtain the following result.

Theorem 3. *Assume that , , and (9) hold. Then is a bifurcation point of problem (8) and there exist two distinct unbounded continua and of problem (8) emanating from such that either they are both unbounded or .*

Next, we prove that the first choice of the alternative of Theorem 3 is the only possibility. To do it, we give the following lemma.

Lemma 4. *Let . If , then cannot contain a pair and .*

*Proof. *Suppose on the contrary that there exists when with and Let ; then should be a solution of problemBy (17), (18), and the compactness of we obtain that for some convenient subsequence as . Now verifies the equationand Hence which is an open set in , and as a consequence for some large enough, , and this is a contradiction.

Lemma 5. *If is a solution of (9) and , then .*

*Proof. *By the proof of Theorem in [16, p. 467] (see also Corollary and the proof of Theorem , together with the remark following that proof, in [16]), we easily obtain the result.

Connecting Theorem 3 with Lemma 4, we can easily deduce the following Dancer-type unilateral global bifurcation result.

Theorem 6. *Assume that , , and (9) hold; then and are unbounded continua. Moreover, we have*

*Proof. *By Theorem 3 with Lemma 4, we only prove for . In the following, we only prove the case of since the proof of is similar.

We claim that there exists a neighborhood of such that . Suppose on the contrary that there exists when with and . Let ; then should be a solution of problemBy (17), (21), and the compactness of , we obtain that for some convenient subsequence as . Now verifies the equationand Hence which is an open set in , and as a consequence for some large enough, , and this is a contradiction.

Suppose that . Then there exists such that and with . Since , by Lemma 8, . Let ; then should be a solution of problem By (17), (23), and the compactness of , we obtain that for some convenient subsequence as . Now verifies the equationand . Hence , for some . Therefore, with . This contradicts Lemma 4.

In order to treat the case or , we will need the following results.

*Definition 7 (see [38]). *Let be a Banach space and let be a family of subsets of . Then the superior limit of is defined by

Lemma 8 (see [38]). *Each connected subset of metric space is contained in a component, and each connected component of is closed.*

Lemma 9 (see [39]). *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 *

Lemma 10. *Assume and . Let . Assume that is a subset of with , and let uniformly on . Let be a solution of the equation and then must change sign on as is large enough.*

*Proof. *After taking a subsequence if necessary, we may assume thatfor large enough. By [32, Lemma ], is disconjugate on , which is a key condition in Elias [25]. Obviously, have the property . (For the definition of property , see [25, p. 36].) Now, from the proof of [25, Lemma ] (see also the remarks in the final paragraph in [25, p. 43]; or see the proof of [16, Lemma ]), it follows that, for all sufficiently large, must change sign on .

#### 3. Main Results

In this section, we first study the following eigenvalue problem: where is a parameter.

In the section, satisfy the following conditions: for and and and and and and and and ,where

Let be such that with Let us consideras a bifurcation problem from the trivial solution and as a bifurcation problem from infinity.

We add the points to space . By [40], we note that problem (34) and problem (35) are the same, and each of them is equivalent to problem (30). By Theorems 3 and 6 and the results of Rabinowitz [41], we have the following Lemma.

Lemma 11. *Let , , , and hold. and are bifurcation points for problem (30). Moreover, there are two distinct unbounded subcontinua of solutions to problem (30), and , consisting of the bifurcation branch emanating from or . For , joins to , such that and .*

*Remark 12. *Any solution of the problem (30) of the form yields a solution of the problem (1). In order to prove our main results, one will only show that crosses the hyperplane in .

Theorem 13. *Let , , , and hold, and either or Then problem (1) has two solutions and , has exactly simple zeros in and is positive near , and has exactly simple zeros in and is negative near .*

*Proof of Theorem 13. *By Lemma 11 and Remark 12, we only prove and crosses the hyperplane in . We only prove the case of since the case of is similar.*Case 1*. (i) Consider .

In this case, we only need to show thatWe divide the proof into two steps.

Let satisfy We note that for all , since is the only solution of the problem (30) for and .*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 By (32) and (33), let and then is nondecreasing and We divide the equation by and set . Since is bounded in , after taking a subsequence if necessary, we have that for some with . Moreover, from (41) and the fact that is nondecreasing, we have that since By the continuity and compactness of , it follows thatwhere , again choosing a subsequence and relabeling if necessary.

We claim that

It is clear that since is closed in . Thus, , so that Therefore, joins to .*Step 2*. We show that there exists a constant such that for all . On the contrary, choosing a subsequence and relabeling if necessary, it follows that Since , it follows thatLet denote the simple zeros of in . Let and Then, after taking a subsequence if necessary, We claim that there exists such that Otherwise, we have This is a contradiction. Let with . For all sufficiently large, we have . So does not change its sign in .

On the other hand, letwhere Conditions and imply that there exists a positive constant such that for any and all . By Lemma 10, we get that must change its sign in for large enough, which is the contradicts. Therefore, for some constant number and sufficiently large.*Case 2*. (ii) Consider

The proof is similar to that for Case 1, so we omit it.

Theorem 14. *Let , , , and hold. Assume condition holds for some . Then problem (1) has two solutions and , has exactly simple zeros in and is positive near , and has exactly simple zeros in and is negative near .*

*Proof. *Inspired by the idea of [42], we define the cut-off function of as the following:We consider the following problem: Clearly, we can see that , , and .

Similar to the proof of Theorem 13, by Lemma 11 and Remark 12, there are two distinct unbounded subcontinua of solutions to problem (57), and emanating from , and joins to .

Taking and , we have that

So condition (i) in Lemma 9 is satisfied with

Obviously and accordingly, (ii) in Lemma 9 holds. (iii) in Lemma 9 can be deduced directly from the Arzela-Ascoli Theorem and the definition of .

Therefore, by Lemma 9, contains an unbounded component emanating from , and joins to .

From , (57) can be converted to the equivalent equation (30). Thus, is an unbounded component of solutions of problem (30) emanating from , and joins to . We can prove the result.

Theorem 15. *Let , , , and hold. Assume that condition holds for some . Then problem (1) has two solutions and , has exactly simple zeros in and is positive near , and has exactly simple zeros in and is negative near .*

*Proof. *If is any nontrivial solution of problem (30), dividing problem (30) by and setting yieldDefine Evidently, problem (59) is equivalent toIt is obvious that is always the solution of problem (59). By simple computation, we can show that and . Now, applying Theorem 13, there are two distinct unbounded subcontinua of solutions to problem (61), and emanating from , and joins to .

Under the inversion , we obtain being an unbounded component of solutions of problem (30) emanating from , and joins to .

Moreover, by Remark 12 and the problem (1), we can obtain that .

Thus, is an unbounded component of solutions of problem (1) such that joins to .

Theorem 16. *Let , , , and hold. Assume that condition holds. Then problem (1) has two solutions and , has exactly simple zeros in and is positive near , and has exactly simple zeros in and is negative near .*

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

Assume on the contrary that , and then there exists a sequence such that for some positive constant depending not on .

By , let , and then is nondecreasing andWe consider the equation Let , and should be the solutions of problem Since is bounded in , choosing a subsequence and relabeling if necessary, we have that for some and

Furthermore, from (63) and the fact that is nondecreasing, we have that since By (12), (65), (66), and the compactness of , we obtain that .

This contradicts

Theorem 17. *Let , , , and hold. Assume that condition holds. Then problem (1) has two solutions and , has exactly simple zeros in and is positive near , and has exactly simple zeros in and is negative near .*

*Proof. *Similar to the method of the proof of Theorem 15 and the conclusions of Theorem 16, we can prove the conclusion.

Theorem 18. *Let , , , and hold. Assume that condition holds. Then problem (1) has two solutions and , has exactly simple zeros in and is positive near , and has exactly simple zeros in and is negative near .*

*Proof. *Define