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# Nontrivial Solutions of the Asymmetric Beam System with Jumping Nonlinear Terms

*Boundary Value Problems*
**volume 2010**, Article number: 728101 (2010)

## Abstract

We investigate the existence of multiple nontrivial solutions for perturbations and of the beam system with Dirichlet boundary condition in , in , where , and are nonzero constants. Here is the beam operator in , and the nonlinearity crosses the eigenvalues of the beam operator.

## 1. Introduction

Let be the beam operator in , In this paper, we investigate the existence of multiple nontrivial solutions for perturbations of the beam system with Dirichlet boundary condition

where and the nonlinearity crosses the eigenvalues of the beam operator. This system represents a bending beam supported by cables in the two directions.

In [1, 2], the authors investigated the multiplicity of solutions of a nonlinear suspension bridge equation in an interval

where the nonlinearity crosses an eigenvalue. This equation represents a bending beam supported by cables under a load The constant represents the restoring force if the cables stretch. The nonlinearity models the fact that cables resist expansion but do not resist compression.

In [2] Lazer and McKenna point out that the kind of nonlinearity

can furnish a model to study travelling waves in suspension bridges. This is a one-dimensional beam equation that represents only the up and down travelling waves of the beam. But the beam has also the right and left travelling waves. Hence we can consider two-dimensional beam equation (1.1).

The nonlinear equation with jumping nonlinearity has been extensively studied by many authors. For the fourth order elliptic equation, Taratello [3] and Micheletti and Pistoia [4, 5] proved the existence of nontrivial solutions, by using degree theory and critical point theory, separately. For one-dimensional case, Lazer and McKenna [6] proved the existence of nontrivial solution by the global bifurcation method. For this jumping nonlinearity, we are interested in the multiple nontrivial solutions of the equation. Here we used variational reduction method to find the nontrivial solutions of problem (1.1).

In Section 2, we investigate some properties of the Hilbert space spanned by eigenfunctions of the beam operator. We show that only the trivial solution exists for problem (1.4) when , and . In Section 3, we state the Mountain Pass Theorem. In Section 4, we investigate the existence of nontrivial solutions for a perturbation of the asymmetric beam equation

where , and are constants. This equation satisfies Dirichlet boundary condition on the interval and periodic condition on the variable . We use the variational reduction method to apply mountain pass theorem in order to get the main result that for (1.2) has at least three periodic solutions, two of which are nontrivial. In Section 5, we investigate the existence of multiple nontrivial solutions for perturbations of beam system (1.1). We also prove that for (1.1) has only the trivial solution.

## 2. Preliminaries

Let be the differential operator and Then the eigenvalue problem

has infinitely many eigenvalues and corresponding normalized eigenfunctions given by

We note that all eigenvalues in the interval are given by

Let be the square and the Hilbert space defined by

Then the set of functions is an orthonormal basis in . Let us denote an element in as

and we define a subspace of as

Then this is a complete normed space with a norm

Since for all , we have that

(), where denotes the norm of ;

() if and only if .

Define . Then we have the following lemma (cf. [7]).

Lemma 2.1.

Let , . Then we have that

Theorem 2.2.

Let , and . Then the equation, with Dirichlet boundary condition,

has only the trivial solution in .

Proof.

Since and , let . The equation is equivalent to

By Lemma 2.1, is a compact linear map from into . Therefore, it is norm . We note that

So the right-hand side of (2.10) defines a Lipschitz mapping of into with Lipschitz constant . Therefore, by the contraction mapping principle, there exists a unique solution . Since is a solution of (2.10), is the unique solution.

## 3. Mountain Pass Theorem

The mountain pass theorem concerns itself with proving the existence of critical points of functional which satisfy the Palais-Smale (PS) condition, which occurs repeatedly in critical point theory.

Definition 3.1.

We say that satisfies the Palais-Smale condition if any sequence for which is bounded and as possesses a convergent sequence.

The following deformation theorem is stated in [8].

Theorem 3.2.

Let be a real Banach space and . Suppose satisfies Palais-Smale condition. Let be a given neighborhood of the set of the critical points of at a given level . Then there exists , as small as we want, and a deformation such that we denote by the set :

(i),

(ii),

(iii).

We state the Mountain Pass Theorem.

Theorem 3.3.

Let be a real Banach space and satisfy condition. Suppose that

() there are constants such that , and

() there is an such that .

Then possesses a critical value . Moreover, can be characterized as

where

## 4. Critical Point Theory and Multiple Nontrivial Solutions

We investigate the existence of multiple solutions of (1.1) when . We define a functional on by

Then the functional is well defined in and the solutions of (1.4) coincide with the critical points of . Now we investigate the property of functional .

Lemma 4.1 (cf. [7]).

is continuous and Frechet differentiable at each with

We will use a variational reduction method to apply the mountain pass theorem.

Let be the two-dimensional subspace of . Both of them have the same eigenvalue . Then for . Let be the orthogonal complement of in . Let denote onto and denote onto . Then every element is expressed by

where , .

Lemma 4.2.

Let , and . Let be given. Then we have that there exists a unique solution of equation

Let . Then satisfies a uniform Lipschitz continuous on with respect to the norm (also the norm ).

Proof.

Choose and let Then (4.4) can be written as

Since is a self-adjoint, compact, linear map from into itself, the eigenvalues of in are , where or . Therefore, is . Since

the right-hand side of (4.5) defines a Lipschitz mapping because for fixed maps into itself. By the contraction mapping principle, there exists a unique (also ) for fixed . Since is bounded from to there exists a unique solution of (4.4) for given .

Let

Then . If and for any , , then

Hence

Since

Therefore, is continuous on with respect to norm (also, to ).

Lemma 4.3.

If is defined by , then is a continuous Frechet derivative with respect to and

If is a critical point of , then is a solution of (1.4) and conversely every solution of (1.4) is of this form.

Proof.

Let and set . If , then from (4.4)

Since

Let be the two subspaces of defined as follows:

Given and considering the function : defined by

the function has continuous partial Fréchet derivatives and with respect to its first and second variables given by

Therefore, let with and . Then by Lemma 4.2

If and , then

Since for any and , it is easy to know that

And

It follows that

Therefore, is strictly convex with respect to the second variable.

Similarly, using the fact that for any , if and are in and , then

where . Therefore, is strictly concave with respect to the first variable. From (4.17), it follows that

with equality if and only if .

Since is strictly concave (convex) with respect to its first (second) variable, [9, Theorem ] implies that is with respect to and

Suppose that there exists such that . From (4.24), it follows that for all . Then by Lemma 4.2, it follows that for any . Therefore, is a solution of (1.4).

Conversely, if is a solution of (1.4) and , then for any .

Lemma 4.4.

Let , and . Then there exists a small open neighborhood of 0 in such that is a strict local minimum of .

Proof.

For , and , problem (1.4) has a trivial solution . Thus we have . Since the subspace is orthogonal complement of subspace , we get and . Furthermore, is the unique solution of (4.4) in for . The trivial solution is of the form and , where is an identity map on , is continuous, it follows that there exists a small open neighborhood of 0 in such that if then , . By Lemma 4.2, is the solution of (4.5) for any . Therefore, if , then for we have . Thus

If , then . Therefore, in ,

where . It follows that is a strict local point of minimum of .

Proposition 4.5.

If , then the equation admits only the trivial solution in .

Proof.

is invariant under and under the map . So the spectrum of restricted to contains in . The spectrum of restricted to contains in . From the symmetry theorem in [10], any solution of this equation satisfies . This nontrivial periodic solution is periodic with periodic . This shows that there is no nontrivial solution of

Lemma 4.6.

Let and . Then the functional , defined on , satisfies the Palais-Smale condition.

Proof.

Let be a Palais-Smale sequence that is is bounded and in . Since is two-dimensional, it is enough to prove that is bounded in .

Let be the solution of (1.4) with where . So

By contradiction, we suppose that , also . Dividing by and taking , we get

Since , we get weakly in . Since is a compact operator, passing to a subsequence, we get strongly in . Taking the limit of both sides of (4.28), it follows that

with . This contradicts to the fact that for the following equation

has only the trivial solution by Proposition 4.5. Hence is bounded in .

We now define the functional on , for ,

The critical points of coincide with solutions of the equation

The above equation () has only the trivial solution and hence has only one critical point .

Given , let be the unique solution of the equation

where . Let us define the reduced functional on by . We note that we can obtain the same results as Lemmas 4.1 and 4.2 when we replace and by and . We also note that, for has only the critical point .

Lemma 4.7.

Let , , and . Then we have for all with .

The proof of the lemma can be found in [1].

Lemma 4.8.

Let , , and . Then we have

for all (certainly for also the norm ).

Proof.

Suppose that it is not true that

Then there exists a sequence in and a constant such that

Given , let be the unique solution of the equation

Let . Then . By dividing , we have

By Lemma 4.2, is Lipschitz continuous on . So the sequence is bounded in . Since (), it follows that and are bounded in . Since is a compact operator, there is a subsequence of converging to some in , denoted by itself. Since is a two-dimensional space, assume that sequence converges to with . Therefore, we can get that the sequence converges to an element in .

On the other hand, since , dividing this inequality by , we get

By Lemma 4.2, it follows that for any

If we set in (4.40) and divide by , then we obtain

Let be arbitrary. Dividing (4.40) by and letting , we obtain

where Then (4.42) can be written in the form for all . Put . Letting in (4.41), we obtain

where we have used (4.42). Hence

Letting in (4.39), we obtain

Since , this contradicts to the fact that for all . This proves that .

Now we state the main result in this paper.

Theorem 4.9.

Let , , and . Then there exist at least three solutions of the equation

two of which are nontrivial solutions.

Proof.

We remark that is the trivial solution of problem (1.4). Then is a critical point of functional . Next we want to find others critical points of which are corresponding to the solutions of problem (1.4).

By Lemma 4.4, there exists a small open neighborhood of 0 in such that is a strict local point of minimum of . Since from Lemma 4.8 and is a two-dimensional space, there exists a critical point of such that

Let be an open neighborhood of in such that . Since , we can choose such that . Since satisfies the Palais-Smale condition, by the Mountain Pass Theorem (Theorem 3.3), there is a critical value

where

If , then there exists a critical point of at level such that , 0 ( since and ). Therefore, in case , the functional has also at least 3 critical points .

If , then define

where . Hence,

That is . By contradiction, assume . Use the functional for the deformation theorem (Theorem 4.9) and taking . We choose such that . From the deformation theorem (Theorem 3.2), and

which is a contradiction. Therefore, there exists a critical point of at level such that , 0, which means that (1.4) has at least three critical points. Since , these two critical points coincide with two nontrivial period solutions of problem (1.4).

## 5. Nontrivial Solutions for the Beam System

In this section, we investigate the existence of multiple nontrivial solutions for perturbations of the beam system with Dirichlet boundary condition

where and the nonlinearity crosses the eigenvalues of the beam operator.

Theorem 5.1.

Let , , and . Then beam system (5.1) has at least three solutions , two of which are nontrivial solutions.

Proof.

From problem (5.1), we get the equation

where the nonlinearity

Let . Then the above equation is equivalent to

Since , , and , the above equation has at least three solutions, two of which are nontrivial solutions, say . Hence we get the solutions of problem (5.1) from the following systems:

where and . When , from the above equation, we get the trivial solution . When , from the above equation, we get the nontrivial solutions , .Therefore, system(5.1) has at least three solutions , two of which are nontrivial solutions.

Theorem 5.2.

Let , and . Then system (5.1) has only the trivial solution .

Proof.

From problem (5.1), we get the equation

where the nonlinearity

Let . Then the above equation is equivalent to

Since , and , by Theorem 2.2, the above equation has the trivial solution. Hence we have the trivial solution of problem (5.1) from the following system:

From (5.9), we get the trivial solution .

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## Acknowledgments

This work (Choi) was supported by Inha University Research Grant. The authors appreciate very much the referee's corrections and revisions.

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### Keywords

- Dirichlet Boundary Condition
- Nontrivial Solution
- Trivial Solution
- Suspension Bridge
- Critical Point Theory