Abstract

We discuss the solvability of the fourth-order boundary value problem , , , which models a statically bending elastic beam whose two ends are simply supported, where is continuous. Under a condition allowing that is superlinear in and , we obtain an existence and uniqueness result. Our discussion is based on the Leray-Schauder fixed point theorem.

1. Introduction and Main Results

In this paper we deal with the existence of a solution of the fourth-order ordinary differential equation boundary value problem (BVP) where is continuous. This problem models deformations of an elastic beam in the equilibrium state, whose ends are simply supported. Owing to its importance in physics, the solvability of this problem has been studied by many authors; see [1–14].

In [1], Aftabizadeh showed the existence of a solution to BVP(1.1) under the restriction that is a bounded function. In [2, Theorem ], Yang extended Aftabizadeh's result and showed the existence for BVP(1.1) under the growth condition of the form where , , and are positive constants such that .

In [6], del Pino and Manásevich further extended the result of Yang and obtained the following existence theorem.

Theorem A. Assume that there is a pair satisfying and that there are positive constants , , and such that and satisfies the growth condition Then the BVP(1.1) possesses at least one solution.

Obviously, the result of Yang follows from Theorem A by just setting . Conditions (1.3)–(1.5) concern a nonresonance condition involving the two-parameter linear eigenvalue problem (LEVP) In [6] it was shown that is an eigenvalue pair of LEVP(1.6) if and only if for some . Hence, for the straight line is called an eigenline of LEVP(1.6). Conditions (1.3)-(1.4) trivially imply that It is easy to prove that condition (1.8) is equivalent to the fact that the rectangle does not intersect any of the eigenline of LEVP(1.6). In [6], del Pino and Manásevich conjecture that Theorem A is also valid if (1.8) is replaced by (1.4). Particularly, in the case that the partial derivatives and exist, the conjecture means that if for large the pair lies in a certain rectangle which does not intersect any of the eigenline of LEVP(1.6); then BVP(1.1) is solvable. But they could not prove the conjecture.

Recently, the present author [11] has partly answered this conjecture and shows that if the rectangle is replaced by the circle the conjecture is correct. In other words, the following result is obtained.

Theorem B. Assume that has partial derivatives and in . If there exists a circle , which does not intersect any of the eigenline of LEVP(1.6), such that for large , then the BVP(1.1) has at least one solution.

See [11, Theorem and Corollary ]. Condition (1.12) means that is linear growth on and . If is not linear growth on or , Theorem B is invalid.

In this paper, we will extend Theorem B to the case that the circle is replaced by an unbounded domain. Let be a positive constant; then we will use the parabolic sector to substitute the the circle in Theorem B. Noting that is contained in the parabolic sector and only contacts the first eigenline at , we see that does not intersect any of the eigenline . Our new result is as follows.

Theorem 1.1. Assume that has partial derivatives and in . If there is a positive constant such that then the BVP(1.1) has a unique solution.

In Theorem 1.1, Condition (1.15) allows to be superlinear in and , and an example will be showed at the end of the paper. The proof of Theorem 1.1 is based on Leray-Schauder fixed point theorem and a differential inequality, which will be given in the next section.

2. Proof of the Main Results

Let and be the usual Hilbert space with the interior product and the norm . For , let be the usual Sobolev space with the norm . which means that , is absolutely continuous on and .

Given , we consider the linear fourth-order boundary value problem (LBVP) Let be Green's function to the second-order linear boundary value problem which is explicitly expressed by For every given , it is easy to verify that the LBVP(2.1) has a unique solution in Carathéodory sense, which is given by If , the solution is in , and it is a classical solution. Moreover, the solution operator of LBVP(2.1), is a linearly bounded operator. By the compactness of the Sobolev embedding , we see that is a completely continuous operator. Hence the restriction is completely continuous.

Lemma 2.1. For every , the unique solution of LBVP(2.1) satisfies the inequalities

Proof. Since sine system is a complete orthogonal system of , every can be expressed by the Fourier series expansion where , , and the Parseval equality holds. Let , then is the unique solution of LBVP(2.1), and , ,  and can be expressed by the Fourier series expansion of the sine system. Since , by the integral formula of Fourier coefficient, we obtain that
On the other hand, since cosine system is another complete orthogonal system of , every can be expressed by the cosine series expansion where , . For the above , by the integral formula of the coefficient of cosine series, we obtain the cosine series expansions of and :
By (2.8)–(2.10) and Parseval equality, we have that This implies that (2.5) holds.

Proof of Theorem 1.1. We define a mapping by By the continuity of , is continuous and it maps every bounded set of into a bounded set of . Hence, the composite mapping is completely continuous. By the definition of the solution operator of LBVP(2.1), the solution of BVP(1.1) is equivalent to the fixed point of . We first use the Leray-Schauder fixed point theorem [15] to show that has a fixed point. For this, we consider the homotopic family of the operator equations We need to prove that the set of the solutions of (2.13) is bounded in .
Let be a solution of (2.13) for . Set . Since , by the definition of , is the unique solution of LBVP(2.1). Hence satisfies the differential equation Set . Multiplying the first formula of (2.14) by and by the theorem of differential mean value, we have where , for some . In the last step of this estimation we use the inequality which is derived from the inequality by choosing Since , it follows that Hence, we obtain that in which we use the inequality for by choosing and . Integrating inequality (2.19) on using integration by parts and Lemma 2.1, we have from which it follows that From this and Lemma 2.1, we obtain that Hence, by the continuity of the Sobolev embedding , we have where is the Sobolev embedding constant. This means that the set of the solutions of (2.13) is bounded in . By the Leray-Schauder fixed point theorem [15], has a fixed point in which is a solution of BVP(1.1).
Now, let be two solutions of BVP(1.1). Set and . Then is the solution of LBVP(2.1), and it satisfies the equation Multiplying this equality by and by the theorem of differential mean value and Condition (1.15), we have that where , for some . Integrating this inequality on and using Lemma 2.1, we obtain that This implies that , and hence we have . Thus BVP(1.1) has only one solution.
The proof of Theorem 1.1 is completed.

Example 2.2. Consider the fourth-order boundary value problem where . Noting that is upperlinear growth on , one can check that all the known results of [1–14] are not applicable to this equation. But, if , then for small enough . Hence, Condition (1.15) holds, and by Theorem 1.1, the boundary value problem (2.27) has a unique solution.

Acknowledgments

The research is supported by NNSF of China (10871160), the NSF of Gansu Province (0710RJZA103), and Project of NWNU-KJCXGC-3-47.