Abstract
We investigate the spectrum structure of the eigenvalue problem . As for the application of the spectrum structure, we show the existence of solutions of the fourth-order boundary value problem at resonance , which models a statically elastic beam with both end-points being cantilevered or fixed, where is the first eigenvalue of the corresponding eigenvalue problem and nonlinearity may be unbounded.
1. Introduction
Starting from the seminal paper of Landesman and Lazer [1], the existence and multiplicity of solutions of nonlinear second-order boundary value problem at resonance, and its general case have been extensively studied; see Gupta [2, 3], Iannacci and Nkashama [4, 5], Costa and Goncalves [6], Ambrosetti and Mancini [7], Fonda and Habets [8], Các [9], and Ahmad [10] and the references therein. Because of the linear operator , is not reversible; this kind of problems as (1) is of problems at resonance.
In the past twenty years, the existence and multiplicity of solutions (or positive solutions) of nonlinear fourth-order boundary value problems at nonresonance case have been investigated by many authors. Especially, many works address the nonlinear fourth-order differential equation of the following form: with one of the following sets of boundary conditions:(i)Both end-points simply supported conditions: (ii)Both end-points cantilevered or fixed conditions: (iii)One end simply supported and the other end sliding clamped conditions: See Rynne [11], Korman [12], Ma et al. [13–15], Cabada et al. [16, 17], Vrabel [18], Schröder [19], Drábek and Holubová [20], Webb et al. [21], Chu and O’Regan [22], and Cid et al. [23] for references along this line.
However, relatively little is known about the fourth-order problem at resonance; see Gupta et al. [24, 25], Jurkiewicz [26], and Iannacci and Nkashama [27]. The likely reason is that the spectrum theory of fourth-order operators is not available.
The purpose of this paper is to show the existence of solutions of the fourth-order boundary value problem at resonance: which models a statically elastic beam with both end-points being cantilevered or fixed, where is a -Carathéodory function, , and is the first eigenvalue of the corresponding linear eigenvalue problem. More precisely, we provide a sufficient condition for the solvability of problem (7) in which nonlinearity is not necessarily needed to satisfy the Landesman-Lazer type condition or the monotonicity assumption.
To do this, we investigate the spectrum structure of the linear eigenvalue problem: We will show that the eigenvalues of (8) form a sequence: Moreover, for each , (, is the simple root of the equation ) is simple, and the corresponding eigenfunction is , which forms (with a suitable normalization) an orthogonal system of . Sincehas nontrivial solution , problem (7) is a problem in resonance.
We refer, for motivations and results, to the classical papers of [4] for the second-order boundary value problems at resonance and [24] for the fourth-order boundary value problems which are simply supported at both ends and are at resonance. In this paper, we will use the classical spaces , , and ; we shall make use, in what follows, of the Sobolev spaces and ; we refer the reader to see [28] for their definitions and properties.
The rest of the paper is arranged as follows. In Section 2, we investigate the spectrum structure of eigenvalue problem (8). In Section 3, we give some preliminary results that are needed to apply Leray-Schauder continuation method to obtain the existence of solutions for problem (7). Finally, Section 4 is devoted to stating and proving our main result.
2. The Eigenvalue Problem
In this section, we consider the linear eigenvalue problem:
Lemma 1. The equationhas infinitely many simple roots Moreover, for .
Proof. Let It is easy to check that, for ,We claim that has exactly one root ; moreover, for any , is simple. Assume that the claim is not true. Then, the following two cases must occur.
Case 1. There are three zeros in for some . In this case, we may find such that However, this contradicts the fact that Case 2. There is a double zero for some . In this case, we only deal with case being odd. Case is even and can be treated similarly.
Since we may assume that there exists such that Combining this with the fact , it concludes that in some left neighborhood of . However, this is a contradiction.
Lemma 2. The linear eigenvalue problem (11) has infinitely many eigenvalues: and the eigenfunction corresponding to is given by Moreover, , where denote the set of such that(i) has only simple zeros in and has exactly such zeros;(ii) and .
Proof. By [11, P. 308], we know problem (11) has a sequence of eigenvalues with . For any given , each eigenvalue is simple and has a corresponding eigenfunction satisfying (i) and (ii). By a direct calculation, we have (21) and (22).
3. Preliminaries
Throughout the paper, we assume that(H0) such that, for a.e. , ; moreover on a subset of of positive measure.
Define a linear operator bywhere . Then is the linear self-adjoint operator, and thus admits the orthogonal direct sum decomposition , where is the one-dimensional null space of and is the range space of , namely, Therefore each has a unique decomposition: where , , so that, with obvious notations, .
Since , it follows that has the Fourier series expansion:By (25), we observe that
Lemma 3. Assume that satisfies (H0). Let and satisfy Then there exists a constant such that, for all , we have
Proof. We will divide the proof into three steps.
Step 1. It follows from Lemma 2 that, for all , Multiplying both sides of the equation in (30) by and integrating from to , we get that This together with the orthogonality of and in implies thatSubsequently, by (28), we haveStep 2. We show that if , then there exists a constant satisfyingFirstly, by (27), we observe thatBy Lemma 2, is a solution of (11). So that, substituting into (11) and multiplying both sides of the equation by and integrating from to , we get that, for ,This fact together with (35) and using Parseval identity yields thatTherefore, by (H0), we find that with equality if and only if for all and . Therefore, for , one has , and, by using the series expansion, reduces to . But then, we haveIt follows from (H0) that , and hence .
Next we will prove that (34) is true. Suppose, on the contrary, that there exists a sequence and such thatIt follows from the compact embedding of into that Now (41) implies thatAt the same time by (40) and (41), we obtain This together with (42) implies thatthat is, . By the fact that with equality if and only if , we know ; this contradicts the fact that .
Step 3. By a direct observation of (33) and (34), we obtain the desired results.
Lemma 4. Let be fixed constant. Define a linear operator by Then is completely continuous.
Proof. By the theory of linear fourth-order differential equations, the operator defined by is one-to-one and continuous obviously. It follows that is completely continuous.
4. The Main Result and the Proof
The main result of the paper addresses the existence of solutions of fourth-order problem (7), when the nonlinearity is unbounded. For the sake of simplicity, we assume the following:(H1) is a -Carathéodory function; namely, is measurable on for every , is continuous on for a.e. , for any constant , and there exists a function such that for a.e. and all with .(H2) for a.e. and all .(H3)For all constant , there exist a constant and a function such that for a.e. and all with , where has given by (H0).
Theorem 5. Assume that (H0)–(H3) hold. Then problem (7) has at least one solution for any provided:
Proof. Let be associated with function and be a fixed constant with . To study problem (7) using Leray-Schauder continuation method, we prove that each of the possible solutions of the homotopy has a priori bound. Therefore, we claim that if is a solution of (50), then there exists a constant independently of such that If we assume on the contrary that there exists a sequence and a sequence with for all such that Let . ThenObviously, by Lemma 4, (53) is equivalent to (47) together with (48) yields that there exists a function depending only on such that Subsequently, the right-hand member of (54) is bounded in independently of . By Lemma 4, there exists such that in . Moreover, .
On the other hand, (H3) yields that there exist and such that for a.e. and all with , where is chosen such that . Let us define a function by Then, this together with (H2) and (56) yields thatMoreover, is a -Carathéodory function. Define by By (H1), it yields that, for a.e. and all , there exists , such that Observe that depend only on and .
Thus, problem (50) is equivalent to The fact with together with (58) yields that Therefore, by Lemma 3, (60), and the compact embedding of into , we have for some constant .
By (63), we deduce immediately that in . Therefore we can write . Since , we shall suppose thatNow, using Lemma 2, we can get that there exists such that, for , on . So that, for ,Multiplying both sides of the equation in (53) by and integrating from to , by (31), (49), and the fact , we haveSo that . By (H2) and (65), we conclude a contradiction.
Conflicts of Interest
All of the authors of this article declare that there are no conflicts of interest regarding the publication of this paper.
Authors’ Contributions
Man Xu and Ruyun Ma completed the main study together and Man Xu wrote the manuscript; Ruyun Ma checked the proofs process and verified the calculation. Moreover, all the authors read and approved the last version of the manuscript.
Acknowledgments
This work was supported by NSFC (no. 11671322) and NSFC (no. 11361054).