Journal of Function Spaces

Volume 2017 (2017), Article ID 1069491, 8 pages

https://doi.org/10.1155/2017/1069491

*S*-Shaped Connected Component for Nonlinear Fourth-Order Problem of Elastic Beam Equation

^{1}Department of Mathematics, Northwest Normal University, Lanzhou 730070, China^{2}Department of Applied Mathematics, Lanzhou University of Technology, Lanzhou, China

Correspondence should be addressed to Ruyun Ma

Received 1 December 2016; Accepted 20 June 2017; Published 24 July 2017

Academic Editor: Gennaro Infante

Copyright © 2017 Jinxiang Wang et al. 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 investigate the existence of -shaped connected component in the set of positive solutions of the fourth-order boundary value problem: , , where is a parameter, , and with . We develop a bifurcation approach to deal with this extreme situation by constructing a sequence of functions satisfying and By studying the auxiliary problems, we get a sequence of unbounded connected components , and, then, we find an unbounded connected component in the set of positive solutions of the fourth-order boundary value problem which satisfies and is -shaped.

#### 1. Introduction

The fourth-order boundary value problem describes the deformations of an elastic beam with simple support at the end; see Gupta [1]. This kind of problems has been extensively studied by using topological degree theory, fixed point theorems, lower and upper solutions method, and critical point theory; see [2–17] and the references therein.

Let In this case, , the global structure of solution set of (1) has been studied extensively by several authors via the well-known Rabinowitz global bifurcation theorem since studying an unbounded connected component bifurcating from the trivial solution at ; see Rynne [18], Ma et al. [19], and Dai and Han [20]. Very recently, Wang and Ma [21] considered the more general problem under the following assumption.

(A) There exist constants with and , such that Since , the Rabinowitz global bifurcation theorem can be used to guarantee an unbounded connected component bifurcating from the trivial solution at . However, in the extreme situation , the Rabinowitz global bifurcation theorem cannot be directly used to get a connected component bifurcating from the trivial solution anymore.

To overcome this difficulty, we have to consider a special problem where . We will develop a bifurcation approach to deal with this extreme situation by constructing a sequence of functions satisfying By studying the auxiliary problems we get a sequence of unbounded connected components of the set of positive solutions of , and, then, we find an unbounded component in the set of positive solutions of (5) which satisfies .

More precisely, we will prove the following.

Theorem 1. *Assume that*(H1) *, in , and in ;*(H2) *, for all , , and ;*(H3) *;*(H4) *there exists such that implies that *(H5) *there exists such that **Then there exist and such that*(i)*(5) has at least one positive solution if ;*(ii)*(5) has at least two positive solutions if ;*(iii)*(5) has at least three positive solutions if ;*(iv)*(5) has at least two positive solutions if ;*(v)*(5) has at least one positive solution if .*

*For other results on the shape of the connected component of solution set, see Wang and Ma [21] and Sim and Tanaka [22].*

*The rest of this paper is arranged as follows. In Section 2, we give some preliminaries. In Section 3, we consider the corresponding auxiliary problems and obtain a sequence of unbounded components with rightward direction near the initial point. Section 4 is devoted to showing the direction turns of the component, and we complete the proof of Theorem 1.*

*2. Preliminaries*

*In this section, we state some preliminary results.*

*Definition 2 (see [23]). *Let be a Banach space and be a family of subsets of . Then the superior limit of is defined by

*Definition 3 (see [23]). *A component of a set means a maximal connected subset of .

*Lemma 4 (see [24]). Let be a Banach space, and let be a family of connected subsets of . Assume that(a)there exist , , and , such that ;(b), where ;(c)for every is a relatively compact set of , where .Then there exists an unbounded component in and .*

*Lemma 5 (see [19, Theorem 2.1]). Assume that (H1) holds. Then the linear problem has a positive simple principal eigenvalue Moreover, the corresponding eigenfunction is positive in .Let ; it is well known that the fourth-order linear problem has a unique solution where Moreover, if and , then that is, is concave.Since Green’s function has the properties (i), ,(ii), , ,then, for any , we have *

*3. Auxiliary Problems and Rightward Bifurcation*

*3. Auxiliary Problems and Rightward Bifurcation*

*For each , we define by Then with for all and By (H2), it follows that .*

*We extend to an odd function by Similarly we may extend to an odd function for each .*

*Now let us consider the auxiliary family of the equations We rewrite by Since (18) implies , then, using Rabinowitz’s global bifurcation theorem and following the similar arguments in the proof of Theorem1.1 in [19] or Theorem2.2 in [20], we have the following.*

*Lemma 6. Assume that (H1) and (H2) hold; then, for each fixed , from there emanates an unbounded subcontinuum of positive solutions of in the set , where with the norm .*

*Lemma 7. Assume that (H1) and (H2) hold. For each fixed , let be a sequence of positive solutions to which satisfies and as . Let be the first eigenfunction of (10) which satisfies Then there exists a subsequence of , again denoted by , such that converges uniformly to on .*

*Proof. *Set Then . For every , we have From the boundary condition , there exists such that . Integrating (21) on , we obtain Dividing both sides of (22) by , we get Since implies , then, by (18), there exists a constant such that From , it follows that there exists a constant such that Then, for , (23) implies that that is, Since is bounded, by the Ascoli-Arzela theorem, a subsequence of uniformly converges to a limit with , and we again denote by the subsequence.

For every , we have Dividing both sides of (28) by , we get Since , we conclude that for each fixed . Then Lebesgue’s dominated convergence theorem shows that which means that is a nontrivial solution of (10) with , and hence .

*Lemma 8. Assume that (H1) and (H2) hold. For each fixed , let be as in Lemma 6. Then there exists such that and imply .*

*Proof. *Assume to the contrary that there exists a sequence such that , , and . By Lemma 7, there exists a subsequence of , again denoted by , such that converges uniformly to on . Multiplying the equation of with by and integrating it over , we have By simple computation, one has that Combining (31) with (32), we obtain that is Since implies , then, from the definition of and (17), we have Then Lebesgue’s dominated convergence theorem, Lemma 7, and (35) imply that Similarly, by Lebesgue’s dominated convergence theorem and Lemma 7 again, we conclude that this contradicts (34).

*4. Direction Turns of Component and Proof of Theorem 1*

*4. Direction Turns of Component and Proof of Theorem 1*

*Lemma 9. Assume that (H1) and (H2) hold; then there exists an unbounded connected component with in the solutions set of (5). Moreover, the component grows to the right near .*

*Proof. *Let us verify that satisfy all of the conditions of Lemma 4.

Since Condition (a) in Lemma 4 is satisfied with . Obviously and, accordingly, (b) holds. (c) can be deduced directly from the Ascoli-Arzela theorem and the definition of . Therefore, the superior limit of , that is, , contains an unbounded connected component with . From Lemma 8, the component grows to right near .

*Lemma 10. Assume that (H1), (H2), and (H3) hold. Let be a sequence of positive solutions to (5); then implies .*

*Proof. *Assume on the contrary that is bounded; we divide the proof into two cases.*Case 1 *. By recalling (22) and (21), we have that and are bounded.

From the boundary condition , there exists such that . Integrating the equation of (5) on , we obtain then is bounded too. Finally, we conclude that is bounded; this deduces a contradiction.*Case 2 *. Since is bounded, then, by (H2) and (H3), there exists constant such that Since , combining (42) with (16) we have which yields that is bounded; this deduces a contradiction.

*Lemma 11. Assume that (H1), (H2), and (H3) hold. Then, joins to in .*

*Proof. *We divide the proof into two steps.*Step 1*. We show that

Assume on the contrary that Let be such that ; then , and from Lemma 10,

Since , we have that Set , and then , and From (H3), we have that is bounded uniformly; then is bounded. By Ascoli-Arzela theorem, choosing a subsequence and relabelling it if necessary, it follows that there exists with such that Letthen is nondecreasing and (H3) implies that Since this together with (48) and implies thatNotice that (45) is equivalent to Combining this with (50) and using (46) and Lebesgue dominated convergence theorem, it follows that This contradicts with . Therefore, *Step 2*. We show that

Assume on the contrary that Let be such that Since implies , then, following the same arguments in the proof of Case2 in Lemma 10, we can get a contradiction.

*Lemma 12. Assume that (H1), (H2), (H3), and (H4) hold. Let be a solution of (5) with ; then .*

*Proof. *Let be a solution of (5) with ; then, by Condition (H4) and the property of , we have then .

*Lemma 13. Assume that (H1), (H2), (H3), (H4), and (H5) hold. Let be a solution of (5) with ; then .*

*Proof. *Let be a solution of (5) with ; then (16) implies that Suppose on the contrary that ; then, by (H5) we have Multiplying inequality (56) by and integrating it over , we have On the other hand, by simple computation, one has that since is concave, then ; this deduces a contradiction.

*Proof of Theorem 1. *From Lemma 9, there exists an unbounded connected component in the positive solutions set of (5); moreover, with and it grows to the right near . From Lemma 11, there exists a sequence such that and . Lemma 10 implies that ; then there exist and such that and ; Lemmas 12 and 13 imply that and , respectively.

By Lemmas 11, 12, and 13, there exist and which satisfy and , such that the component turns to the left at and to the right at ; that is, is an -shaped component; this together with Lemma 9 completes the proof of Theorem 1.

*Remark 14. *Let us take then Condition (A) in [21] (see (4)) implies which means is in linear growth at zero. Equation (60) guarantees that Rabinowitz global bifurcation theorem can be directly used to bifurcate a connected component from . However, in this paper, we deal with (5) under the superlinear growth condition at zero; that is, (H2) In the situation, , the Rabinowitz global bifurcation theorem cannot be directly used to get a connected component joining with infinity anymore. To overcome this difficulty, we have to construct a sequence of functions which is in linear growth at zero and satisfies By studying the corresponding auxiliary problems , we obtain a sequence of unbounded connected components via Rabinowitz global bifurcation theorem. Now, by using the fact that the superior limit of certain infinity collection of connected components contains an unbounded connected component (see Ma and An in [24]), we get a connected component : which joins with infinity.

*Therefore, the key conditions, the conclusion, and the proofs of the main results in this paper and in [21] are very different.*

*Conflicts of Interest*

*Conflicts of Interest*

*The authors declare that there are no conflicts of interest regarding the publication of this paper.*

*Authors’ Contributions*

*Authors’ Contributions*

*The authors contributed equally to this paper. All authors read and approved the final manuscript.*

*Acknowledgments*

*Acknowledgments*

*This work was supported by the NSFC (no. 11361054, no. 11671322).*

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