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International Journal of Differential Equations
Volume 2011 (2011), Article ID 356356, 17 pages
http://dx.doi.org/10.1155/2011/356356
Research Article

-Solutions for Some Nonlinear Degenerate Elliptic Equations

Department of Mathematics, State University of Londrina, 86051-990 Londrina, PR, Brazil

Received 17 May 2011; Revised 6 October 2011; Accepted 6 October 2011

Academic Editor: Toka Diagana

Copyright © 2011 Albo Carlos Cavalheiro. 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 are interested in the existence of solutions for Dirichlet problem associated to the degenerate quasilinear elliptic equations in the setting of the weighted Sobolev spaces .

1. Introduction

In this paper we prove the existence of (weak) solutions in the weighted Sobolev spaces for the homogeneous Dirichlet problem: where is the partial differential operator: where is a bounded open set in (), and are two weight functions, and the functions , , and are Carathéodory functions.

By a weight, we will mean a locally integrable function on such that for a.e. . Every weight () gives rise to a measure on the measurable subsets on through integration. This measure will be denoted by . Thus, () for measurable sets .

In general, the Sobolev spaces without weights occur as spaces of solutions for elliptic and parabolic partial differential equations. For degenerate partial differential equations, that is, equations with various types of singularities in the coefficients, it is natural to look for solutions in weighted Sobolev spaces (see [14]).

A class of weights, which is particularly well understood, is the class of -weights (or Muckenhoupt class) that was introduced by Muckenhoupt (see [5]). These classes have found many useful applications in harmonic analysis (see [6, 7]). Another reason for studying -weights is the fact that powers of distance to submanifolds of often belong to (see [8]). There are, in fact, many interesting examples of weights (see [4] for -admissible weights).

Equations like (1.1) have been studied by many authors in the nondegenerate case (i.e., with ) (see, e.g., [9] and the references therein). The degenerate case with different conditions has been studied by many authors. In [2] Drábek et al. proved that under certain condition, the Dirichlet problem associated with the equation , has at least one solution , and in [1] the author proved the existence of solution when the nonlinear term is equal to zero.

Firstly, we prove an estimate for the bounded solutions of : we assume that , with (where as in Theorem 2.5), and we prove that any that solves satisfies , where depends only on the data, that is, , and . After that, we prove the existence of solution for problem if , with .

Note that, in the proof of our main result, many ideas have been adapted from [911].

The following theorem will be proved in Section 3.

Theorem 1.1. Let and be -weights, , with . Suppose the following. (H1) is measurable in for all : is continuous in for almost all . (H2), whenever , .(H3), with , where .(H4), where , and are positive functions, with and , and ().(H5) is measurable in for all is continuous in for almost all .(H6), where and are positive functions, with and .(H7), for all , where .(H8) is measurable in for all : is continuous in for almost all .(H9), where and are positive constants.(H10), with (where as in Theorem 2.5) and .
Let be a solution of problem . Then there exists a constant , which depends only on , and , such that .

The main result of this article is given in the next theorem, which is proved in Section 4.

Theorem 1.2. Assume that (H1)–(H9) hold true and suppose that
(H11), with ;(H12), for all .
Then there exists at least one solution of the problem .

Theorem 1.2 will be proved by approximating problem with the following problems: where , for . Note that and that .

2. Definitions and Basic Results

Let be a locally integrable nonnegative function in and assume that almost everywhere. We say that belongs to the Muckenhoupt class , , or that is an -weight, if there is a constant such that for all balls , where denotes the -dimensional Lebesgue measure in . If , then (see [4, 7, 12] or [13] for more information about -weights). The weight satisfies the doubling condition if , for all balls , where and denotes the ball with the same center as which is twice as large. If , then is doubling (see Corollary  15.7 in [4]).

As an example of -weight, the function , , is in if and only if (see Corollary  4.4, Chapter IX in [7]). If , then for some (see [6]).

Definition 2.1. Let be a weight, and let be open. For , we define as the set of measurable functions on such that

Remark 2.2. If , , then since is locally integrable, we have for every open set (see Remark  1.2.4 in [13]). It thus makes sense to talk about weak derivatives of functions in .

Definition 2.3. Let be open, , and let and be -weights, . We define the weighted Sobolev space as the set of functions with weak derivatives , for . The norm of in is given by The space is the closure of with respect to the norm . The dual space of is the space

Remark 2.4. (a) If , , then is dense in (see Corollary  2.16 in [13]).
(b) If , then

In this paper we use the following four results.

Theorem 2.5 (The Weighted Sobolev Inequality). Let be an open bounded set in and . There exist constants and positive such that for all and all satisfying , where .

Proof. See Theorem  1.3 in [3].

The following lemma is due to Stampacchia (see [14], Lemma  4.1).

Lemma 2.6. Let , , , and be real positive numbers, where .
Let be a decreasing function such that for all . Then , where .

Lemma 2.7. If , then , whenever is a ball in and is a measurable subset of .

Proof. See Theorem  15.5 Strong doubling of -weights in [4].

By Lemma 2.7, if , then .

Lemma 2.8. Let and be -weights, , , and a sequence , satisfies the following: (i) in and -a.e. in ;(ii) with .
Then in .

Proof. The proof of this lemma follows the line of Lemma  5 in [10].

Definition 2.9. We say that is a (weak) solution of problem if for all .

3. Proof of Theorem 1.1

Set and define for the functions and by

If is a solution of problem , define the set . We will use the test functions . Since and is a bounded set in , then and where is the characteristic function of the set (by we mean the function equal to for and to −1 for ).

Since , we have that .

Using the function in (2.8) we obtain We have the following estimates.(i) By (H3) we obtain(ii)By (H7) we obtain(iii) Using (H9) we obtain And we also have Hence in (3.3) we obtain

Since , we have for Hence in (3.8) we obtain Using (3.9) and if , we obtain Let us define the function by . We have that and We have the following.(a)For all , .(b)There exists, a constant () such that for all

This implies the following.(I1) a.e. on .(I2) If , then Combining (I1) and (I2) with (3.11) and (3.12) we obtain Define the function . Since and , we have that . Hence If , we have . Hence and we obtain

By Theorem 2.5, (3.15), and (3.18) we have Therefore, there exist positive constants and (depending only on , , , , , and ) such that Since , then and . For such that , using an interpolation inequality, Young’s inequality (with ), and Hölder’s inequality with exponents and we thus obtain Hence in (3.20) we obtain Now, we can choose in order to have and such that We obtain, from (3.22), that for every it results where . Hence for all we have Let us now take . Then we have Therefore for all we obtain (by (3.25) and (3.26)) that is, .

Let . Since , by Lemma 2.6 there exists a constant such that

Using Lemma 2.7 we have for all . Therefore any solution of problem satisfies the estimate .

4. Proof of Theorem 1.2

Step 1. Let us define for the approximation We have that , , and satisfies the conditions (H9) and (H12). We consider the approximate problem We say that is (weak) solution of problem if for all . We will prove that there exists at least one solution of the problem . For we define Then is a (weak) solution of problem if Let .(i)Using (H4) we obtain(ii) Using (H6) and , we obtain Hence, For each we have that is linear and continuous. Hence, there exists a linear and continuous operator such that . We set The operator is semimonotone; that is, by similar arguments as in the proof of Theorem  2 in [11] we have the following.(i) for all .(ii)For each , the operator is hemicontinuous and bounded from to and,for each the operator is hemicontinuous and bounded from to .(iii) If in and , then in as for all .(iv)If , in , and in , then as .(v)The operator is bounded.Hence the operator is pseudomonotone (see [15]).(vi)By (H3), (H7), and (H12) we have where . Since , we have that is, the operator is coercive. Then, by Theorem  27.B in [15], for each , the equation has a solution. Therefore, the problem , has a solution .

Step 2. We will show that and , where is independent of . If is a solution of problem , we define We have if . For , let us define the function . We have .
Now consider the function Since is a Lipschitz function and , then . Moreover, and We also have, for all measurable subset , By applying Vitali’s Convergence Theorem, with , we obtain Since , we obtain Hence in . Since is a solution of problem and , we have Using (H4), (H6), , (4.17), and (4.18), we obtain in (4.19) as Using in (4.2) (where ) we obtain Since we obtain the following.(i) By (H7) we have for all , and(ii) Using (H12) we have for all , and We have in . Using (H3), (i), and (ii) we obtain in (4.21) By Theorem  2.1.14 in [13] there is a positive constant such that Then we obtain Using (4.27) and Young’s inequality we obtain in (4.25) (for all ) where . We can choose so that , and there exists a constant such that Using Sobolev’s inequality (Theorem 2.5) and Hölder’s inequality with exponents and we obtain (since ) Let us now take and observe that . Then, from the previous inequality, it follows that Hence we obtain Since , by Lemma 2.6 there exists a constant such that for all , and using Lemma 2.7 we obtain . Therefore if is a solution of problem , we have and is independent of .

Step 3. Since and , then the sequence is relative compact in the strong topology of (by apply the analogous results of [10] and Lemma 2.8). Then, by extracting a subsequence which strongly converges in (i.e., there exists such that in ), we have for all Therefore is the solution of problem .

Example 4.1. Let , and consider the weights and ( and ), and the functions , , and defined by where . Let us consider the partial differential operator and , with . Therefore, by Theorem 1.2, the problem has a solution .

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