Discrete Dynamics in Nature and Society
Volume 2016 (2016), Article ID 5956713, 15 pages
http://dx.doi.org/10.1155/2016/5956713

Research Article

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

Wenguo Shen and Tao He

Department of Basic Courses, Lanzhou Institute of Technology, Lanzhou 730050, China

Received 9 November 2015; Accepted 11 January 2016

Academic Editor: Kousuke Kuto

Copyright © 2016 Wenguo Shen and Tao He. 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.

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 , and

uniformly 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

where

Then 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 , .

Let

and then is nondecreasing with respect to , and

uniformly for and on bounded sets. Further it follows from (17) that

uniformly 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 problem

has 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 have

Next, 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 problem

where

By (18), it follows that

uniformly for and on bounded sets. Furthermore, implies that

for 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 that

The left-hand side of (39) equals

We prove that . If , then . If , noting the conclusion of Lemma 3, then . By L’Hospital rule, we have that

In the following, we will prove that

Let Then and in For some small enough, let and be such that , , and . Then we have

Set and Let and be such that , in and and Set ; then should be a solution of the problem

The 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 that

By (42), we can show that . Therefore, the left-hand side of (39) .
By (42), we have .
It follows that

Similarly, we can also show that

By (35), (46), and (47), taking the limit as , we can obtain that

By (36), and have the same sign in ; one has

So, by (48), we have that

Furthermore, 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 , there are two possibilities: (i) or (ii) . It is obvious that in the case of (i). While case (ii) implies that has at least one double zero in , Lemma 4 follows that . Hence, there exists a sequence such that is a solution of problem (27) corresponding to , and converges to in . By Lemma 13, we have : that is, in the case of (ii). Hence, .

Proof of Theorem 10. We only prove the case of since the case of is similar. Let be the component of , containing . By Lemma 14, we can show that .
Suppose on the contrary that is bounded. By Lemma 11, we can find a neighborhood of such that .
In order to complete the proof of this theorem, we consider problem (27). By Lemma 1, there are two continua and , consisting of the bifurcation branch . By Lemma 12, we have that and are both unbounded and

So there exists for all . Since is bounded in , (27) shows that is bounded in independently of . By the compactness of , one can find a sequence such that converges to a solution of (6). So . If , then from Lemma 4 or Remark 5 it follows that . Note that since is closed subset of . By Lemma 13, , which contradicts the definition of . On the other hand, if , then which contradicts

From Theorem 10 and its proof, we can easily get the following two corollaries.

Corollary 15. There exist two unbounded subcontinua and of solutions of (6) in , bifurcating from , and for and .
We relax the assumption of as the following:

() is a sign-changing weight.

Corollary 16. Let hold and . Then there exist two unbounded subcontinua and of solutions of (6) in , bifurcating from , and for and .

4. Spectrum of Half-Quasi-Linear Problems

In this section, we consider the half-linear problem (7). Problem (7) is called half-linear because it is positive homogeneous in the cones and . Similar to that of [20], we say that is a half-eigenvalue of problem (7) if there exists a nontrivial solution . is said to be simple if , for all solutions of problem (7).

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

Lemma 17. Let for and , . Also let be solutions of the following differential equations:

respectively. If , , and or and , and or , and in , then either there exists such that or or or and for some constant .

Proof. We discuss four cases.
Case  1 ( , in ). Then an easy calculation shows that

Similar to the step of the proof of Lemma of [16], we can obtain the left-hand side of (55) . This is a contradiction.
Case  2 ( , in ). Similar to (55), we can get

Similar to the step of the proof of Lemma of [16], we can obtain that the left-hand side of (56) . This is a contradiction.
Case  3 ( , in ). Similar to Case , we can get the result.
Case  4 ( , in ). We can get

Similar to the proof of Lemma of [16], we can obtain the left-hand side of (57) . This is a contradiction.

By Lemma 17, we obtain the following result that will be used later.

Lemma 18. Let be such that and

Let be such that

Let be a solution of the equation

Then the number of zeros of in goes to infinity as .

Proof. Set and . By simple computation, we can show that

After taking a subsequence if necessary, we may assume that

as , where is the th eigenvalue of the following problem:

Let be the corresponding eigenfunction of . It is easy to check that the distance between any two consecutive zeros of is (also see [30]).
Hence, the number of zeros of goes to infinity as . Note that the conclusion of Lemma 17 also is valid if . Using these facts and Lemma 17, we can obtain the desired results.

On the basis of the unilateral global interval bifurcation result, we establish the spectrum of the half-linear problem (8). More precisely, we will use Theorem 10 to prove the following result.

Theorem 19. 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).

Proof of Theorem 19. By Theorem 10, we know that there exists at least one solution of problem (7), , for every , and . The positive of problem (7) implies that are half-linear solutions in . Lemma 4 implies that any nontrivial solution of problem (7) lies in some . We claim that, for any solution of problem (7) with , we have that and    for some positive constant .
Next, we only prove the case of since the case of is similar.
We divide the proof into three steps.
Step  1 (we show that ). We may assume without loss of generality that the first generalized simple zero of to occur in (0, 1) is a generalized simple zero of . That is, there exists such that and or , and and do not vanish and have the same sign in . It follows that

Similar to the proof of Lemma of [16], we can obtain the left-hand side of (64) . So, one has that .
On the other hand, similar to the proof of Lemma 13, by Lemma 6, there must exist an interval such that and do not vanish and have the same sign in , and . We have

Similar to the proof of Lemma of [16], we can obtain the left-hand side of (65) . So, one has that . Hence .
Step  2 (we will prove that for some positive constant ). Without loss of generality, we may assume that and are positive in . By the Picone identity in [29, Theorem ], noting , we have that

where

Using similar methods of the proof of Lemma 13, we can show that the left-hand side of (66) . Hence, . It follows that

Thus, . Furthermore, we obtain that on for some positive constant . We may assume without loss of generality that the first zero of to occur in is a generalized simple zero of . That is, there exists such that and or , and and do not vanish and have the same sign in . Using method similar to the above, we can show that on for some positive constant . Clearly, and Lemma 3 imply . Repeating the above process times, we can show that for some positive constant .
Step  3 (we prove that ( or ) are increasing). In fact, if and are the solutions of problem (7) with , , and , the first generalized simple zero of to occur in (0, 1) is a generalized simple zero of . Indeed, if this were not, by Lemma 6, using method similar to the proof in Step , we could obtain , which is impossible, since the half-eigenvalues were shown to be simple. Therefore, by Lemma 17, we can get .
Naturally, we can consider the bifurcation structure of the perturbation of problem (7) of the form

where satisfies (4).

Theorem 20. For , is a bifurcation point for problem (69). Moreover, there exists an unbounded continuum of solutions of problem (69) emanating from , such that .

Proof. Let and :

Corollary 15 shows that there exist two unbounded subcontinua and of solutions of (69) in , bifurcating from , and for and . Let us show that ; that is, is a bifurcation point for problem (69). Indeed, if there exists being a sequence of solutions of problem (69) converging to , let , and then should be a solution of problem

By (18) and the compactness of , we obtain that for some convenient subsequence as . Now verifies the equation

and This implies that for some and .

5. Nodal Solutions for Half-Linear Eigenvalue Problems

We start this section by studying the following eigenvalue problem:

where is a parameter.

Let be such that

with

Let us consider

as a bifurcation problem from the trivial solution , and

as a bifurcation problem from infinity.

Applying Theorem 20 to problem (76), we have the following result.

Lemma 21. Let , , , and hold. For , is a bifurcation point for problem (76). Moreover, there exists an unbounded continuum of solutions of problem (76) that joins to infinity, such that .
We add the points to space ; by the results of Rabinowitz [31], we have the following lemma.

Lemma 22. Let , , , and hold. For , is a bifurcation point for problem (77). Moreover, there exists an unbounded continuum of solutions of problem (77) meeting , such that .
We note that problem (76) and problem (77) are the same, and each of them is equivalent to problem (73). By Lemmas 21 and 22, we obtain the following lemma.

Lemma 23. Let , , , and hold. There exists an unbounded continuum of solutions of problem (73) emanating from or , such that and , and joins to .

Remark 24. Any solution of (73) of the form yields a solution of (8). In order to prove our main results, one will only show that crosses the hyperplane in .

Remark 25. From and , we can see that there exists a positive constant such that for all .

Theorem 26. Let , , , and hold. For , either or . Then problem (8) possesses a solution such that has exactly zeros in and is positive near .

Proof. By Remark 24, in order to prove Theorem 26, we only show that joins to . Let satisfy

We note that for all since is the only solution of (73) for and .
Case  1 ( ). In this case, we 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

Let be such that

with

Let , then is nondecreasing, and

Moreover, from (84) and the fact that is nondecreasing, we have that

since

We divide the equation

by and set . Since is bounded in , choosing a subsequence and relabeling if necessary, we have that for some with .
By the compactness of , we obtain that

where , again choosing a subsequence and relabeling if necessary.
It is clear that and since is closed in . Moreover, by Theorem 19, , so that

Thus, joins to .
Step  2. We show that there exists a constant number such that , for large enough.
On the contrary, we suppose that . Since , it follows from the compactness of that

where by

By Remark 25, we have

Let be an eigenfunction corresponding to But if , applying Lemma 17 to and , we have that must change sign for large enough, which is impossible. So . By Lemma 18, we get that must change sign for large enough, and this contradicts the fact that .
Case  2 ( ). In this case, if is such that

then

and, moreover,
Assume that there exists such that, for all ,

Applying a similar argument to that used in Step  1 of Case  1, after taking a subsequence and relabeling if necessary, it follows that

Again joins to and the result follows.

Theorem 27. Let , , , and hold. For , assume that one of the following conditions holds.

(i) for (ii) for (iii) for (iv) for

Then problem (8) 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 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 [32], 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 26, by Lemma 23 and Remark 24, there exists an unbounded continuum of solutions of problem (98) emanating from , such that , 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 and (iii) in Lemma 9 can be deduced directly from the Arezela-Ascoli Theorem and the definition of .
Therefore, by Lemma 9, contains an unbounded connected component with .
From , (98) can be converted to the equivalent equation (73). Since , we conclude that . Moreover, by (8).
In the following, we show that .
Let satisfy .
Let be such that

with , let , then is nondecreasing, and

We divide the equation

Let should be the solutions of problem

Since is bounded in , choosing a subsequence and relabeling if necessary, we have that for some .
Furthermore, from (101) and the fact that is nondecreasing, we have that

since

By (104) and the compactness of , we obtain that

It is clear that and since is closed in . Moreover, by Theorem 19, , so that

Thus, joins to .

Theorem 28. Let , , , and hold. For , assume that one of the following conditions holds.

(i) for (ii) for (iii) for (iv) for

Then problem (8) 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. Define

We consider the following problem:

Clearly, we can see that , , and .
Similar to the proof of Theorem 26, by Lemma 23 and Remark 24, there exists an unbounded continuum of solutions of problem (109) emanating from , such that , 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 and (iii) in Lemma 9 can be deduced directly from the Arezela-Ascoli Theorem and the definition of .
Therefore, by Lemma 9, contains an unbounded connected component with .
From , (109) can be converted to the equivalent equation (73). Since , we conclude . Moreover, by (8).
Similar to the proof of Theorem 27, we can obtain that .

Theorem 29. Let , , , and hold. For , assume that one of the following conditions holds.

(i) for (ii) for (iii) for (iv) for

Then problem (8) 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 will only prove the case of (i) since the proofs of the cases for (ii), (iii), and (iv) are completely analogous.
If is any nontrivial solution of problem (73), dividing problem (73) by and setting yields

Define

Evidently, problem (111) is equivalent to

It is obvious that is always the solution of problem (113). By simple computation, we can show that and . Similar to the proof of Theorem 28, there exists an unbounded continuum of solutions of problem (113) emanating from , such that , and joins to .
Under the inversion , we obtain being an unbounded component of solutions of problem (73) emanating from , such that , and joins to .
Moreover, by (8), we can obtain that .
Thus, is an unbounded component of solutions of problem (8) such that joins to .

Theorem 30. Let , , , and hold. For , assume that one of the following conditions holds.

(i) for (ii) for , or (iii) for

Then problem (8) 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. Define

We consider the following problem:

Clearly, we can see that , , and .
Applying the similar method used in the proof of Theorem 26, by Lemma 23 and Remark 24, there exists an unbounded continuum of solutions of problem (115) emanating from or , such that and , and joins to .
Taking and or and , we have that By Lemma 9, we obtain an unbounded component with .
From , (115) can be converted to the equivalent equation (73). Thus, is an unbounded component of solutions of problem (73) emanating from or , such that and and joins to .
Since , we conclude . Moreover, by Remark 24 and (8), we can obtain that .
Thus, is an unbounded component of solutions of problem (8) emanating from or , such that and , and joins to .
Applying the similar method used in the proof of Theorem 27, we obtain an unbounded connected component with .

Theorem 31. Let , , , and hold. For , assume that one of the following conditions holds.

(i)There exists a for , such that .(ii)There exists a for , such that . .(iii)There exists a for , such that .(iv)There exists a for , such that .

Then problem (8) 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. Define

We consider the following problem:

Clearly, we can see that , , and .
Applying the similar method used in the proof of Theorem 30, there exists an unbounded continuum of solutions of problem (117) emanating from or , such that and , and joins to .
Moreover, we can obtain the desired results.

Remark 32. If and , Ma et al. [10, 11] studied problem (8) under the conditions and ; in the situation, the results of Theorem 26 improve on Theorem of [10] and Theorem of [11], respectively.

Remark 33. When and , Dai and Han [16] investigated the existence of nodal solutions for problem (8). Thus, our results partially extend and improve the corresponding Theorem of [16].

Remark 34. When and , Lazer and McKenna [7] investigated the existence of nodal solutions for problem (8). Thus, our results partially extend and improve the corresponding Theorem of [7].

Remark 35. The nonlinear term of (8) is not necessarily homogeneous linearizable at the origin and infinity because of the influence of the term . Clearly, so the bifurcation results of [7, 8, 10–13, 16] cannot be applied directly to obtain our results.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

The authors were supported by the NSFC (no. 11561038) and Gansu Provincial National Science Foundation of China (no. 145RJZA087).

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