Research Article | Open Access

Wandong Lou, "Weak Solutions for a Second Order Dirichlet Boundary Value Problem on Time Scale", *Journal of Applied Mathematics*, vol. 2013, Article ID 684038, 7 pages, 2013. https://doi.org/10.1155/2013/684038

# Weak Solutions for a Second Order Dirichlet Boundary Value Problem on Time Scale

**Academic Editor:**Jin Liang

#### Abstract

We adopt the Leray-Schauder degree theory and critical point theory to consider a second order Dirichlet boundary value problem on time scales and obtain some existence theorems of weak solutions for the previous problem.

#### 1. Introduction

In recent years, differential equations on time scales have been studied extensively in the literature. There have been many approaches to study the existence and multiplicity of solutions for differential equations on time scales. The variational method is, to the best of our knowledge, novel, and it may open a new approach to deal with nonlinear problems on time scales. For more details about recent development in the direction, we refer the reader to [1–12] and the references therein. The two books [13, 14] by Bohner and Peterson summarize some excellent results on time scales.

In [1, 2], the authors utilized variational techniques and critical point theory to derive some sufficient conditions for the existence of positive solutions for the following second order dynamic equation with Dirichlet boundary conditions:

In [8], Zhou and Li studied Sobolev's spaces on time scales, and given their properties, as applications, they presented variational methods and critical point theory to obtain the existence of solutions for the second order Hamiltonian systems on time scales as follows: where is -measurable in for every and continuously differentiable in for -a.e. .

Motivated by the works cited previously, in this paper, we study the second order Dirichlet boundary value problem on time scales. Consider where we say that a property holds for -almost every or -almost everywhere on , -a.e., whenever there exists a set with null Lebesgue -measure such that this property holds for every , and is an arbitrary bounded time scale such that and .

We assume that is a parameter, , , where means that fulfils the Carathéodory conditions (see [15, Definition 3.2.22] and [2, (i) and (ii) in ()]). By virtue of the Leray-Schauder degree theory, a result involving the existence of weak solutions is established. Next, we utilize two critical point theorems to investigate that problem (3) has at least one weak solution and infinitely many weak solutions with the parameter , respectively. The results obtained here improve some existing results in the literature.

#### 2. Preliminaries and an Existence Theorem via Leray-Schauder Degree Theory

We first offer some related preliminaries concerning the basic definitions and results on time scales. For convenience, for , we denote . Let Sobolev's space be defined by (see [1–3]) where (see [3, Definition 2.9]) denotes the set of absolutely continuous functions on . Then is a Hilbert space with the inner product and let be the norm induced by the inner product (see [1, page 1265] and [2, page 371]).

Lemma 1 (see [3, Proposition 3.7]). *The immersion is compact.**We firstly need to consider the following eigenvalue problem:
**
For each , multiply by on both sides of the previous equation in (6) and integrate over to obtain
**By (7) of [8], we obtain
*

Lemma 2 (see [9, Lemma 3.4] and [13, Theorem 4.95]). *The eigenvalues of (6) may be arranged as , and can be expressed by
**
By Lemma 2, if there exists and , then we have
**
Choosing , we can easily obtain
**
Moreover, we also have
*

Lemma 3 (see [16, Corollary 3.3]). *Let . Then the Wirtinger-type inequality is
**
where is defined by Lemma 2.**Denote an operator as follows
**
By Hölders inequality and Lemma 3, we have
**
Then is a bounded linear operator. By Lemma 1, is compact. Clearly, is also symmetric, for . Consequently, the supremum is achieved by Theorem 6.3.12 in [15]. Hence, (10)–(12) are true.*

In what follows, we offer two lemmas involving Leray-Schauder degree.

Lemma 4 (see [15, Proposition 5.2.22]). *Let be an open bounded set in a Banach space and . Let be a unique solution in of the equation . Assume that the Fréchet derivative exists and is continuously invertible. Then
**
and is the multiplicity of the eigenvalue of the operator .*

Lemma 5 (see [15, Theorem 5.2.13]). *Let be an open bounded set in a Banach space . There exists a mapping defined for all and such that , . This mapping has the homotopy invariance property: if and , , and are such that , for every and , then .*

Now, we list our hypotheses for (3).(H1) is a parameter and , where are determined by Lemma 2.(H2); there exist , and such that (H3) is -Lipschitz continuous with respect to the second variable; that is, there exists a constant , such that

*Remark 6. *We can take , where are as in (H2). Clearly, it also satisfies (H3). In fact,

It is clear that (3) is equivalent to the following integral equation:
We define an operator as follows:

Lemma 7. * is compact on .*

*Proof. *We first prove that there exists a ball () such that maps into itself if is large enough. Indeed, (H2) leads to
Let , and . For any ,
Therefore, the above claim is true. Meanwhile, we also arrive at immediately is uniformly bounded on . On the other hand, we shall prove that is equicontinuous on . By (H3) and Hölders inequality, we have
We have from Arzelà-Ascoli theorem is compact on .

Theorem 8. *Assume that (H1)–(H3) hold; problem (3) has at least a weak solution.*

*Proof. *We will utilize Leray-Schauder degree to prove the result. One is invited to verify that the existence of a solution of (3) is equivalent to the existence of a solution of the operator equation
where and are defined by (14) and (21), respectively. For the reason that is bounded, linear, and compact on , we easily see that Fréchet derivative exists and is continuously invertible by (H1). Consequently, Lemma 4 implies
So, to complete the proof, we have to find an admissible homotopy connecting and . Define
We shall prove that there exists such that for all , , and , we obtain
If the claim is false, we can find sequences and such that and
Set and divide (29) by to get
This is equivalent to
for each . Now, passing to suitable subsequences, without loss of generality, we may assume that and in . Note that, similar with (22), we find
On the other hand, by the compactness of , we have (see [15, Proposition 2.2.4(iii)]). In (30), let ; we have , and satisfies . However, this contradicts our assumption , . This implies that (28) holds. By Lemma 5 and (26), we have
Therefore, (3) has at least a weak solution.

#### 3. Two Existence Theorems via Critical Point Theory

We still use the Sobolev’s space defined by (4), which is equipped with the inner product and the corresponding norm Now, we will establish the corresponding variational formulations for problem (3) as follows: where .

We now list our hypotheses for (3).(H4) is -measurable in for every and continuously differentiable in for , and there exist and for all and -a.e. .(H5) There exist and such that (H6) There exist such that (H7) uniformly for .(H8) , , .

*Remark 9. *Let
where is a fixed positive integer. Then
Direct computation shows
Clearly, (H4)–(H8) hold.

By (H4), we find (see [11, Theorem 2.27]), and for ,
Clearly, the existence of weak solutions for (3) is equivalent to the existence of critical points for . In what follows, we take , .

Lemma 10 (see [17, Theorem 1.2]). *Suppose is a reflexive Banach space with norm , and is coercive and weak (sequentially) lower semicontinuous; that is, the following conditions are fulfilled as follows:*(1)* as , .*(2)* For any , any sequence such that , there holds
* *Then is bounded from below and attains its infimum on .*

Theorem 11. *If (H4) and (H5) with hold, (3) has at least a weak solution.*

*Proof. *Our working space is a Hilbert space, so it is reflexive. By Lemma 2.1 and Theorem 3.3 in [8], we see is weakly lower semicontinuous on . On the other hand, by (H5), we have
and thus as . Lemma 10 implies can attain its infimum in ; that is, (3) has at least a weak solution. This completes the proof.

If is a Hilbert space, there exist (see [18]) and such that , and . For , denote , , and . Clearly, with for all . Since is a Hilbert space, we can choose an orthonormal basis such that

*Definition 12 (see [19, Definition 1.1]). *Assume that is a Banach space with norm ; we say that satisfies Cerami condition (C) if for all ,(i)any bounded sequence satisfying , possesses a convergent subsequence;(ii)there exist such that for any with , . Denote . We will introduce the following Fountain theorem under condition (C).

Lemma 13 (see [19, Proposition 1.2]). *Assume that satisfies condition (C), and . For each , there exists such that*(i)*, ;*(ii)*.**Then has a sequence of critical points , such that as .*

Lemma 14. *Suppose that (H4), (H5), and (H6) hold; then satisfies the condition (C).*

*Proof. *For all , we assume that is bounded and
Going, if necessary, to a subsequence, we can assume that in ; then
Lemma 1 leads to
It follows that in and , we see

Clearly, it is equivalent to , and then we have
Hence, condition (i) of Definition 12 holds. Next, we prove condition (ii) of Definition 12, suppose the contrary, there exists a sequence such that
By (52), there exists a constant such that
On the other hand, (H6) implies
This, together with (54), leads to there is a constant such that . By (H5), similar with (45), we have
and thus if (53) holds, which contradicts in (52). This proves that satisfies condition (C).

Theorem 15. *Under assumptions (H4)–(H8) with in (H5), problem (3) has infinitely many solutions.*

*Proof. *(H8) and Lemma 14 enable us to obtain that and satisfies the condition (C). For any , let
and it is easy to verify that defined by (57) is a norm of . Since all the norms of a finite dimensional normed space are equivalent, so there exists positive constant such that . In view of (H7), there exist and such that
This implies that
Since ; then there exists positive constant such that
Let , . Then by Lemma 3.8 of [20] and Lemma 1 we obtain , as . For any , note that , in view of (H5), we find that
Choosing , then as , then we have;
Hence, as . Combining this and (60), we can take , and thus . Up until now, we have proved that the functional satisfies all the conditions of Lemma 13; then has infinitely many solutions.

#### Acknowledgment

This research is supported by the Project of Shandong Province Higher Educational Science and Technology Program (Grant no. J13LI51).

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#### Copyright

Copyright © 2013 Wandong Lou. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.