#### Abstract

Timoshenko beam equations with external damping and internal damping terms and forcing terms are investigated, and boundary conditions (end conditions) to be considered are hinged ends (pinned ends), hinged-sliding ends, and sliding ends. Unboundedness of solutions of boundary value problems for Timoshenko beam equations is studied, and it is shown that the magnitude of the displacement of the beam grows up to ∞ as under some assumptions on the forcing term. Our approach is to reduce the multidimensional problems to one-dimensional problems for fourth-order ordinary differential inequalities.

#### 1. Introduction

The most fundamental beam equations are of the following form: with the length , the mass density , the cross-sectional area , the modulus of elasticity , and the flexural rigidity (see [1, page 416]). Taking account of the rotary inertia and the deflection due to shear, we obtain the following fourth-order beam equation for the transverse vibrations of prismatic beams on elastic foundations: (see [1, page 433] and Wang and Stephens [2, page 150]).

Dividing the above equation by , letting , and taking into account the nonlinear term , the external damping term and the internal damping terms we obtain the Timoshenko beam equation where , and are positive constants.

Let and we assume throughout this paper that(H1) is a real-valued continuous function in ;(H2) and for ;(H3) is a nondecreasing function in ;(H4) is a real-valued continuous function on .

Definition 1. By a solution of (5), one means a function such that the partial derivatives exist and are continuous on .

Oscillations of beam equations have been treated by numerous authors; see, for example, Feireisl and Herrmann [3], Herrmann [4], Kopáčková [5], Kusano and Yoshida [6], Yoshida [710], and the references therein. In particular, we mention the paper [4] by Herrmann which deals with the Euler-Bernoulli beam equations that is similar to (5). We note that the oscillation of (5) was studied by Yoshida [10]. We refer to Ball [11], Fitzgibbon [12], and Narazaki [13] for stability and existence results for beam equations.

However, there appears to be no known unboundedness results for beam equations. The objective of this paper is to provide unboundedness results for (5) by reducing the multi-dimensional problems to one-dimensional problems for ordinary differential inequalities of fourth-order.

In Section 2 we treat the hinged ends and reduce unboundedness problem for (5) to that for ordinary differential inequalities. Sections 3 and 4 are devoted to the hinged-sliding ends and sliding ends, respectively. In Section 5, we study fourth-order differential inequalities, and we derive unboundedness results for (5) in Section 6.

#### 2. Hinged Ends

In this section, we treat the case where the ends of the beam are hinged, so that solutions are required to satisfy the boundary condition

Theorem 2. Every solution of (5) satisfying is unbounded on if for any constant , all solutions of the fourth-order differential inequalities are not bounded from below, where

Proof. Suppose to the contrary that there exists a solution of the boundary value problem (5), which is bounded on . Then, there exists a constant such that that is, First we consider the case where on . It follows from the hypotheses (H2) and (H3) that and therefore we see from (5) that on . Multiplying (13) by and then integrating over , we derive Integrating by parts and using , we obtain Combining (14) and (15) yields where . It is easy to check that that is, is bounded from below. Hence, we conclude that is a solution of (7) which is bounded from below. This contradicts the hypothesis. In the case where on , satisfies in view of the hypothesis (H2). It is readily verified that and . By the same arguments as in the case where , we conclude that is a solution of (8) which is bounded from below in light of . This contradicts the hypothesis and the proof is complete.

#### 3. Hinged-Sliding Ends

In this section, we deal with the case of hinged-sliding ends, for which the boundary condition takes the form

Theorem 3. Every solution of (5) satisfying is unbounded on if for any constant , all solutions of the fourth-order differential inequalities are not bounded from below, where

Proof. Suppose that there is a solution of the boundary value problem (5), which is bounded on . Let on for some . First, consider the case where on . Proceeding as in the proof of Theorem 2, we find that the inequality (13) holds. Multiplying (13) by and then integrating over , we obtain the inequality (14) with replaced by . Integration by parts yields As in the proof of Theorem 2, we observe that is a solution of (19) which is bounded from below. This contradicts the hypothesis. The case where on can be treated similarly, and we find that is a solution of (20) which is bounded from below. The contradiction establishes the theorem.

#### 4. Sliding Ends

We study the case of sliding ends for which the boundary condition takes the form

Theorem 4. Every solution of (5) satisfying is unbounded on if for any constant , all solutions of the fourth-order differential inequalities are not bounded from below, where

Proof. Suppose that the boundary value problem (5), has a solution which is bounded on . There exists a constant such that on . First, consider the case where on . Arguing as in the proof of Theorem 2, we see that the inequality (13) holds. Integrating (13) over , we obtain the inequality (14) with replaced by and . It is easy to see that Hence, we have the inequality where . Therefore, we conclude that is a solution of (23) which is bounded from below. This contradicts the hypothesis. The case where on can be treated analogously, and we observe that is a solution of (24) which is bounded from below. This is a contradiction and the proof is complete.

#### 5. Fourth-Order Ordinary Differential Inequalities

We deal with the ordinary differential inequality of the fourth order and derive sufficient condition for every solution of (28) to be unbounded from below. It is assumed that , and are nonnegative constants, and is a continuous function on .

Theorem 5. Every solution of (28) is not bounded from below if

Proof. Assume, on the contrary, that there exists a solution of (28) which is bounded from below. Let on for some constant . Then we obtain from (28) the inequality Integrating (30) over , we get and hence where . Integration of (32) over yields that is, where . Integrating (34) over twice, we obtain and therefore in view of the identity where , and . Dividing (36) by yields The left hand side of (38) is bounded from below, whereas the right hand side of (38) is not bounded from below from the condition (29). This is a contradiction and the proof is complete.

#### 6. Unboundedness Results for Timoshenko Beam Equations

Combining Theorems 24 with Theorem 5, we present unboundedness results for the three types of boundary value problems for (5) under consideration.

Theorem 6. Every solution of (5) satisfying is unbounded on if

Proof. The hypothesis (39) implies that every solution of (7) is not bounded from below via Theorem 5. Since the hypothesis (40) implies that we observe that every solution of (8) is not bounded from below. The conclusion follows from Theorem 2. The proof is complete.

We combine Theorems 3 and 4 with Theorem 5 to establish the following two theorems.

Theorem 7. Every solution of (5) satisfying is unbounded on if

Theorem 8. Every solution of (5) satisfying is unbounded on if

Remark 9. If a solution of (5) is unbounded on , then there exists a sequence such that

Remark 10. In the case where , and being constants, we see that satisfies the hypotheses (H1)–(H3).

Example 11. We consider the Timoshenko beam equation with the hinged ends , where Here , and . An easy computation shows that Since we observe that where are constants, and are bounded functions on . Hence, the conditions (39) and (40) are satisfied. It follows from Theorem 6 that every solution of the problem (46), is unbounded on . For example, is such a solution.