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Abstract and Applied Analysis
Volume 2013 (2013), Article ID 908062, 10 pages
Existence of Solution for Impulsive Differential Equations with Nonlinear Derivative Dependence via Variational Methods
1Department of Mathematics, Hunan Normal University, Changsha, Hunan 410081, China
2Hunan Normal University Press, Changsha, Hunan 410081, China
3School of Economics and Management, Changsha University of Science and Technology, Changsha, Hunan 410004, China
Received 30 May 2013; Revised 1 August 2013; Accepted 22 August 2013
Academic Editor: M. Victoria Otero-Espinar
Copyright © 2013 Lizhao Yan et al. 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.
We use variational methods and iterative methods to investigate the solutions of impulsive differential equations with nonlinear derivative dependence. The conditions for the existence of solutions are established. The main results are also demonstrated with examples.
Many dynamical systems have an impulsive dynamical behavior due to abrupt changes at certain instants during the evolution process. The mathematical description of these phenomena leads to impulsive differential equations. Recent development in this field has been motivated by many applied problems, such as control theory, population dynamics, and medicine [1–9].
We consider the following nonlinear Dirichlet boundary value problems for impulsive differential equations: where , , , is continuous, and , , are continuous.
The characteristic of (1) is the presence of the first order derivative in the nonlinearity term. Most of the results concerning the existence of solutions of these equations are obtained using upper and lower solutions methods, coincidence degree theory, and fixed point theorems [10–14]. However, to the best of our knowledge, there are few papers concerned with the existence of solutions for impulsive boundary value problems like problem (1) by using variational methods. Motivated by [15, 16], in this paper we will fill the gap in this area.
When there is no derivative in the nonlinearity term, problem (1) has been extensively studied by [17–23], using variational methods. We know, contrary to these equations, (1) is not variational and the well-developed critical point theory is of no avail for, at least, a direct attack to problem (1). The technique used in this paper consists of, associating with problem (1), a family of the following Dirichlet boundary value problems with no dependence on the derivative of the solution. Namely, for each , we consider the problem Now problem (2) is variational and we can treat it by variational methods.
In this paper, we need the following conditions.) is measurable in for every and continuous in for a.e. .() as uniformly for and and for and .() There exist constants and such that () There exist constants and such that where .() There exist constants such that () The function satisfies the following conditions: where is a constant.() are odd and nondecreasing, and there exist constants , , and such that () There exists such that (), for all , .
Firstly, we recall some facts which will be used in the proof of our main result. It has been shown, for instance, in  that the set of all eigenvalues of the following problem is given by the sequence of positive numbers where Each eigenvalue is simple with the associated eigenfunction We recall the well-known characterization of as the best constant in the Poincaré inequality
Let be the Sobolev space endowed with the norm Throughout the paper, it will be assumed that . We also consider the norm
We need the following Lemmas.
Lemma 1 (see [24, Lemma 2.4]). If , then the norms and are equivalent.
Lemma 2 (see [24, Lemma 2.5]). There exists such that if , then
Lemma 3 (see [24, Lemma 2.6]). Let ; then there exists such that
For , we have that and are both absolutely continuous. Hence for any . If , then is absolutely continuous. In this case, may not hold for some . It leads to the impulsive effects. As a consequence, we need to introduce a different concept of solution.
Definition 4. A function is said to be a classical solution of problem (1) if satisfies the equation a.e. on and the limits , and , exist and satisfy the impulsive condition and the boundary condition .
We have the following fact. Take and multiply the equation in problem (1) by and integrate from 0 to : The first term is now Hence
Definition 5. We say that a function is a weak solution of problem (1) if the identity holds for any .
Proposition 6. Under the hypotheses and , the functional defined by is continuous and differentiable and for any . Moreover, the critical point of is a classical solutions of problem (2).
Proof. Using the assumptions , we can obtain the continuity and differentiability of and that is defined by
for any . It follows that the critical point of is the weak solution of (2). Moreover, it is a classical solution of problem (2).
Evidently, since . By the definition of weak solution, we have Choose with for every ; then This implies that Hence, for every . The impulsive condition in (2) is satisfied. This completes the proof.
We will obtain the critical points of by using the Mountain Pass Theorem. Therefore, we state this theorem precisely.
Lemma 7 (see ). Let be a real Banach space and satisfy (PS)-condition. Suppose that satisfies the following conditions:
there exists constants such that ;
there exists such that .
Then possesses a critical value given by where is an open ball in of radius centered at 0 and
3. The Solvability of (2)
Theorem 8. Suppose that and hold; then there exist positive constants and such that, for each , problem (2) has one solution such that .
Proof. (I) We show that satisfies the (PS)-condition.
Assume that is a sequence such that is bounded and as . We will prove that the sequence is bounded. Obviously, there exists a constant such that We set From (17), (25), (26), and (33), , and , we have By , we obtain that is bounded in .
Since is a reflexive Banach space, passing to a subsequence if necessary, we can assume that Hence Notice that From [26, Lemma 4.2], we see that there exists , for any : Combining this inequality with (39), we have It follows from (37)–(39) that in . Hence, satisfies (PS)-condition.
(II) We verify assumption (ii) of Lemma 7.
By (), there exists such that Since , it follows that Using , we have Hence, from (18), (25), (35), , and , for , we have Set , . Equation (45) shows that implies that ; that is, satisfies assumption (ii) of Lemma 7.
(III) We verify assumption (iii) of Lemma 7.
By and , we know that for Take such that . Since , , and , (46) implies that there exists such that , and if we set . By Lemma 7, possesses a critical value given by where . Obviously, , so according to Lemma 7, there exists and , such that
(IV) We prove that .
Since is the solution of problem (2), we have So It follows from and that, given , there exists a positive constant , independent of , such that Hence, using , we have By Sobolev embedding theorem, we obtain Set and ; then which implies From the inf max characterization of in (III), we obtain with chosen in (III). We estimate using and : whose maximum is achieved at some and the value can be taken as . Clearly it is independent of . Which implies This completes the proof.
Lemma 9. Suppose that hold and , , and hold only for positive ; then there exist positive constants , and such that, for each , problem (2) has a positive solution such that .
Proof. Set Consider the function Obviously, satisfies , and satisfies , so using Theorem 8, we obtained a solution of (60). Multiplying the equation by and integrating by parks, we conclude that . So is positive.
Theorem 10. Suppose that and hold; then there exist positive constants and such that, for each , problem (2) has one solution such that .
Proof. (I) We show that satisfies the (PS)-condition.
Assume that is a sequence such that is bounded and as . We will prove that the sequence is bounded.
It follows from , , and that Hence, is bounded in . The following proof of (PS)-condition is similar to that in (I) of Theorem 8.
(II) We verify assumption (ii) of Lemma 7.
Using , we have The proof is similar to that in (II) of Theorem 8.
(III) We verify assumption (iii) of Lemma 7.
Take such that . By and , using Sobolev embedding theorem, we have Since , (63) implies that there exists such that and if we set . By Lemma 7, possess a critical value , given by where . Obviously, , so according to Lemma 7, there exists and , such that
(IV) We prove that .
Like (IV) of Theorem 8, we can obtain that there exist such that It follows from satisfying (65) that Let Since , then can achieve its maximum at some , which implies there exist such that This completes the proof.
Lemma 11. Suppose that hold and , , and hold only for positive , then there exist positive constants , and such that, for each , problem (2) has a positive solution such that .
Proof. The proof is similar to that of Lemma 9, so we omit it.
Remark 12. By the same method, we can discuss the negative solution of (2).
4. The Solvability of (1)
Theorem 13. Assume that and and hold; then problem (1) has one nontrivial solution provided satisfying .
Proof. We construct a sequence as solutions of the following problem: obtained in Theorem 8, starting with an arbitrary . It follows from the Sobolev embedding theorem that . Using and , we obtain Hence, Using (17) and (18) and by Hölder inequality, we have that is, Since is less than 1, then it follows that strongly converges in , as it easily follows proving that is a Cauchy sequence in . By Theorem 8, we know that . In this way we obtain a nontrivial solution of (1).
Example 14. Let and ; consider the following nonlinear Dirichlet impulsive problem:
Compared with (1), and . We can take and . Then by simple computation, it is easy to verify that all conditions of Theorem 13 are satisfied. Hence, by Theorem 13, problem (75) has one nontrivial solution.
Theorem 15. Assume that and and hold; then problem (1) has one nontrivial solution provided satisfying .
Example 16. Let and ; consider the following nonlinear Dirichlet impulsive problem:
Compared with (1), and . We can take and . Then by simple computation, it is easy to verify that all conditions of Theorem 15 are satisfied. Hence, by Theorem 15, problem (77) has one nontrivial solution.
The authors are grateful to the referees for their useful suggestions. This work is partially supported by the National Natural Science Foundation of China (no. 71201013), the Humanities and Social Sciences Project of the Ministry of Education of China (no. 12YJC630118), the Innovation Platform Open Funds for Universities in Hunan Province (no. 13K059), and the Provincial Natural Science Foundation of Hunan (no. 11JJ3012).
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