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Abstract and Applied Analysis
Volume 2014 (2014), Article ID 697974, 10 pages
http://dx.doi.org/10.1155/2014/697974
Research Article

Variational Integrals of a Class of Nonhomogeneous -Harmonic Equations

Department of Mathematics, Harbin Institute of Technology, Harbin 150001, China

Received 20 September 2013; Accepted 10 January 2014; Published 27 February 2014

Academic Editor: Shusen Ding

Copyright © 2014 Guanfeng Li 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.

Abstract

We introduce a class of variational integrals whose Euler equations are nonhomogeneous -harmonic equations. We investigate the relationship between the minimization problem and the Euler equation and give a simple proof of the existence of some nonhomogeneous -harmonic equations by applying direct methods of the calculus of variations. Besides, we establish some interesting results on variational integrals.

1. Introduction

In this paper, we study the variational integral of the form whose Euler equations are nonhomogeneous -harmonic equations where , , , and are operators satisfying some assumptions. There are many literatures on (2) and a large of useful results have been established; see [13] and their references. We investigate the relationship between the minimization of and solutions of the Euler equation. Based on that, we give a simple proof of the existence of some nonhomogeneous -harmonic equations by applying direct methods of the calculus of variations. Besides, we establish some interesting results on variational integrals. The results of this paper make the theory on (2) easier to comprehend.

We recall the weighted Sobolev spaces which are adopted in [4].

Let be the real Euclidean space with the dimension , . Throughout this paper, will denote an open subset of and . Let be a locally integrable, nonnegative function in . A Radon measure is canonically associated with the weight , Thus , where is the -dimensional Lebesgue measure. In this paper, unless otherwise stated, we always assume that is a -admissible measure and ; see [4].

Let and . Denote the norm of and by , where (or ).

For , let where is the gradient of . The Sobolev space is defined to be the completion of the set with respect to the norm . In other words, if and only if and there is a function and a sequence , such that We call the gradient of in and write .

The space is the closure of in . Obviously, and are Banach space with respect to the norm . Moreover, is uniformly convex and the Sobolev space and are reflexive; see [5] for details.

The corresponding local Sobolev space is defined in the obvious manner: a function is in if and only if is in each open set .

2. Variational Integrals

Suppose that is a measurable set and that for an open neighborhood of . Then, we have the following variational integral: where is a variational kernel satisfying the following assumptions for some constants : for a.e. ; and is also a variational kernel satisfying the following assumptions for some constants : for a.e. ;

Remark 1. Note that a convex function is differential if and only if it is continuously differentiable; see [6]. Thus, by assumptions (10) and (14), mappings and are continuously differentiable for a.e. . Denote by the usual gradient of with respect to the second variable and by the usual derivative of with respect to the second variable. Obviously, and exist for a.e. .
The value lies in the interval and by assumptions (9) and (13), if and only if and ; that is, .
The convexity assumptions (10) and (14) can imply the following useful inequalities.

Lemma 2. For a.e. , whenever , , and .

Proof. The proof is based on assumptions (10) and (14) and the definition of directional derivative. Here, we only show the proof of the first inequality (15) and the other is similar.
Fix such that the mapping is strictly convex and differentiable. Then, for , Setting , we can get Dividing by and subtracting from both sides, we obtain that By the definition of directional derivative, we have that Then, we can get that .
Suppose that there exist , , such that , let , and then we can obtain that On the other hand, since , we have that Then, (22) contradicts (21) and the lemma follows.

3. Nonhomogeneous -Harmonic Equations and the Obstacles Problem

The following nonlinear elliptic equation: is called the nonhomogeneous -harmonic equation, where is an operator satisfying the following assumptions for some constants : for all and almost all , whenever , , and whenever , , and is also an operator satisfying the following similar assumptions for some constants : for all and almost all , whenever .

Definition 3. A function is a (weak) solution of (2) in if for all . A function is a supersolution of (2) in if whenever is nonnegative. A function is a subsolution of (2) in if whenever is nonnegative.

Next is the obstacles problem associated with nonhomogeneous -harmonic equations (2).

Suppose that is bounded in is a function, and . Let If , write .

The problem is to find a function in such that whenever . We call the function an obstacle.

Definition 4. If a function satisfies (37) for all , we say that is a solution to the obstacle problem with obstacle and boundary values or a solution to the obstacle problem in .
If is a solution to the obstacle problem in , we say that is a solution to the obstacle problem with obstacle .

Proposition 5. (1) A solution to the obstacle problem is always a supersolution to (2) in .
(2) If is a supersolution to (2) in , is a solution to the obstacle problem in for each open sets . Moreover, if is bounded and , is a solution to the obstacle problem in .
(3) A solution to the obstacle problem in is a solution to (2) in .
(4) If is a solution to (2) in , is a solution to the obstacle problem in for each open set . Moreover, if is bounded , is a solution to the obstacle problem in .

Proof. By the definition of supersolution and solution to (2) and the definition of solution to the obstacle problem, it is easy to prove that Proposition 5 is true. Here, we only give a proof of (1).
Suppose is a solution to the obstacle problem in , and is obviously in . For all nonnegative , , a.e. in , and Then, . By (37), we can get Therefore, is a supersolution to (2) in .

Theorem 6 (see [3]). Suppose is a solution to the obstacle problem in . If is a supersolution to (2) in , such that , then a.e. in .

4. Relationship between the Minimization Problem and the Euler Equation

In this section, we establish that the variational integral (7) gives rise to an equation of the type (2) as its Euler equation, where the mappings and satisfy the structural assumptions (24)–(32).

Theorem 7. Suppose that and are two variational kernels satisfying (8)–(14) and let and . Then, and satisfy assumptions (24)–(32) with , , , and .

Proof. For points for which (9), (10), (11), (13), and (14) do not hold, we are free to define and arbitrarily. Fix such that satisfies (9)–(11) and satisfies (13) and (14).
(i) By the definition of partial derivative, the th coordinate of equals Then, the mapping is measurable. Moreover, by (10), is continuously differentiable. Then, is continuous and satisfies (24).
(ii) If , then . By (15), we have that Since and , we can obtain that . Then, If , . Therefore, satisfies (25).
(iii) If , then by (25). Write and . Then, and . Applying (15) with and , we can obtain that Since , we have that .
If , we just need to verify that . If not, for each , write and . By (15), . Therefore, Thus, for . The right hand side goes to zero as and . Then, satisfies (26).
(iv) For , , by (15), we have that Combining (47) with (48), we obtain that Then, and satisfies (27).
(v) For , , . Taking partial derivative from both sides with respect to yields Then, .
By the similar argument, we can obtain that satisfies assumptions (29)–(32).

The next theorem shows that minimizers of the variational integral are solutions to the corresponding Euler equation and vice versa.

Theorem 8. Suppose that is a convex set and . Then, if and only if for all .

Proof. (i) By Lemma 2, we have that for all . Then, That is, . Therefore, .
(ii) Fix and let . Then, since is convex and by (51), we have that for all and . Therefore, By assumptions (10) and (14) and the definition of directional derivative, for a.e. . By the mean value theorem, there exists a real number , such that Then, we have By Theorem 7, the following inequalities hold: Write and by , we have that that is, . Then, we can get the following conditions: as . By the Lebesgue’s dominated convergence theorem, we obtain that The theorem is proved.

5. -Extremals and -Superextremals with Obstacles

Definition 9. A function is called an -extremal in with boundary values if and whenever . A function is called a (free) -extremal in if is an -extremal with boundary values in each open set .

It is immediate that an -extremal with boundary values is a free -extremal.

Theorem 10. Suppose that is a free -extremal in . Then, whenever .

Proof. For , has compact support. Choose an open set such that . Then, and is an -extremal with boundary values in . Since , we have that Since vanishes outside , we can obtain that whenever .
Fix with and let be a sequence with converging to in . By Lemma 2, we can get that By the inequality (67), we obtain that Since converges to in , letting converge to in inequality (69), it follows that The theorem follows.

Theorem 11. A function is an (free) -extremal in if and only if in , that is, for all .

Proof. Write for each open set .
(i) Fix and let be an open set such that . Then, and . Thus, , , , and    is a convex set. Since is an (free) -extremal in and , we have that By Theorem 8, it follows that Then, Since and vanish outside , it follows that
(ii) Fix the open set and . Then, and . Choose a sequence with converging to in . By Theorem 7, we obtain that as . Therefore, it follows that By Theorem 8, we have that . Then, is a free -extremal in .

Based on the proof of Theorem 11, we easily infer the following corollary.

Corollary 12. Suppose that a sequence converges to in , and then

Next, we formulate the obstacle problem in terms of variational integrals. This makes the essence of the problem clearer.

Definition 13. Suppose that is bounded. Let be an arbitrary function and call it an obstacle. For , write A function is called an -superextremal with obstacle and boundary values if for all .
A function is called a (free) -superextremal in if is an -superextremal with obstacle and boundary values in each open set .

Remark 14. (1) The -superextremal with obstacle and boundary values minimizes the variational integral among all functions which, roughly speaking, coincide with on the boundary and lie above the obstacle . Naturally, this problem makes sense only if is nonempty. Moreover, we always assume that the notation that includes the assumptions is bounded in this paper.
(2) -extremal can be interpreted as -superextremal with identically .

Theorem 15. Suppose that and . Then, a function is an -superextremal with obstacle and boundary values if and only if is a solution to the obstacle problem in with and .

Proof. It is easy to see that is a convex subset of and . Then, is a -superextremal with obstacle and boundary values if and only if for all .
By Theorem 8, we can get that for all if and only if for all .
Thus, the theorem follows by the definition of the solution to the obstacle problem.

6. Existence of -Superextremals

In this section, we establish the existence of -superextremals by the direct methods of the calculus of variations.

First, we show a lemma, which is a direct corollary of Mazur lemma.

Lemma 16. If is a normed space with the norm and converges weakly in to , then there exists a sequence of convex combinations of , such that converges to in the norm topology of .

Proof. Fix and it is easily to see that the subsequence , , of converges weakly in to . By the Mazur lemma, we can get that converges to in the norm topology of , where Then, there exists a number , such that for all . Let and the lemma follows.

Theorem 17. Suppose that is a nonempty convex closed set. Then there is such that

Proof. Let be a minimizing sequence, that is, as . Since , , and we can assume that for all . By assumptions (9) and (13), we have that Therefore, is a bounded sequence in . Thus a subsequence which we still denote by converges weakly in to a function . By Lemma 16, there exists a sequence of convex combinations of , such that converges to in . Since is closed and convex, . By Corollary 12, we have that For each , since as , there exists a number such that for all . By assumptions (10) and (14), we obtain that whenever . By (93) and (95), it follows that Then, and is the desired minimizer.

Theorem 18. Suppose that is bounded, that , and that . If there exists a unique -superextremal with obstacle and boundary values .

Proof. Since the set is nonempty convex sub