On a Fourth-Order Boundary Value Problem at Resonance

We investigate the spectrum structure of the eigenvalue problem {u(4)(x) = λu(x), x ∈ (0, 1); u(0) = u(1) = u󸀠(0) = u󸀠(1) = 0}. As for the application of the spectrum structure, we show the existence of solutions of the fourth-order boundary value problem at resonance {−u(4)(x) + λ1u(x) + g(x, u(x)) = h(x), x ∈ (0, 1); u(0) = u(1) = u󸀠(0) = u󸀠(1) = 0}, which models a statically elastic beam with both end-points being cantilevered or fixed, where λ1 is the first eigenvalue of the corresponding eigenvalue problem and nonlinearity gmay be unbounded.

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  ∈ N,   (  =  4  ,   is the simple root of the equation cos  cosh  − 1 = 0) is simple, and the corresponding eigenfunction is   , which forms (with a suitable normalization) an orthogonal system of  2 (0, 1).Since  (4) () =  4  1  () ,  ∈ (0, 1) , 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   [0, 1],   (0, 1), and  ∞ (0, 1); we shall make use, in what follows, of the Sobolev spaces   (0, 1) and   0 (0, 1); 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.

The Eigenvalue Problem
In this section, we consider the linear eigenvalue problem: has infinitely many simple roots Moreover, for  ∈ N.
It is easy to check that, for  ∈ N, We claim that () has exactly one root   ∈ [, ( + 1)]; moreover, for any  ∈ N,   is simple.Assume that the claim is not true.Then, the following two cases must occur.
In this case, we may find  ∈ ( 0 , ( 0 + 1)) such that However, this contradicts the fact that Case 2. There is a double zero t ∈ ( 0 , ( 0 + 1)) for some  0 ∈ N. In this case, we only deal with case  0 being odd.Case  0 is even and can be treated similarly.Since we may assume that there exists t ∈ ( 0 , ( 0 + 1)) such that  () < 0,  ∈ (, t) , Combining this with the fact   ( t) > 0, it concludes that () > 0 in some left neighborhood of t.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  ∈
Step 3. By a direct observation of (33) and (34), we obtain the desired results.
Proof.By the theory of linear fourth-order differential equations, the operator  :  →  2 (0, 1) defined by is one-to-one and continuous obviously.It follows that  −1 :  2 (0, 1) →  is completely continuous.