Abstract

We study in this paper a class of second-order linear elliptic equations in weighted Sobolev spaces on unbounded domains of , . We obtain an a priori bound, and a regularity result from which we deduce a uniqueness theorem.

1. Introduction

Let be an open subset of , . Assign in the uniformly elliptic second-order linear differential operatorwith coefficients , , and consider the associate Dirichlet problem:where .

It is well known that if is a bounded and sufficiently regular set, the above problem has been widely investigated by several authors under various hypotheses of discontinuity on the leading coefficients, in the case or sufficiently close to 2. In particular, some -bounds for the solutions of the problem (1.2) and related existence and uniqueness results have been obtained. Among the other results on this subject, we quote here those proved in [1], where the author assumed that 's belong to (and considered the case ) and in [24] (where the coefficients belong to some classes wider than ). More recently, a relevant contribution has been given in [58], where the coefficients are assumed to be in the class and ; observe here that contains the space .

If the set is unbounded and regular enough, under assumptions similar to those required in [1], problem (1.2) has for instance been studied in [911] with , and in [12] with . Instead, in [13, 14] the leading coefficients satisfy restrictions similar to those in [5, 6].

In this paper, we extend some results of [13, 14] to a weighted case. More precisely, we denote by a weight function belonging to a suitable class such thatand consider the Dirichlet problem:where , , , and are some weighted Sobolev spaces and the weight functions are a suitable power of . We obtain an a priori bound for the solutions of (1.4). Moreover, we state a regularity result that allows us to deduce a uniqueness theorem for the problem (1.4). A similar weighted case was studied in [15] with the leading coefficients satisfying hypotheses of Miranda's type and when .

2. Weight Functions and Weighted Spaces

Let be any Lebesgue measurable subset of and let be the collection of all Lebesgue measurable subsets of . If , denote by the Lebesgue measure of , by the characteristic function of , by the intersection (, )—where is the open ball of radius centered at —and by  the class of restrictions to of functions with . Moreover, if is a space of functions defined on , we denote by the class of all functions , such that for any .

We introduce a class of weight functions defined on an open subset of . Denote by the set of all measurable functions , such thatwhere is independent of and . Examples of functions in are the functionand, if and is a nonempty subset of , the functionFor , we putIt is known that(see [16, 17]).

We assign an unbounded open subset of .

Let be a function, such that andWe putFor any and , we set

If , , , and , consider the space of distributions on , such that for , equipped with the normMoreover, denote by the closure of in and put . A more detailed account of properties of the above defined spaces can be found, for instance, in [18].

From [15, Lemmas 1.1 and 2.1], we deduce the following two lemmas, respectively.

Lemma 2.1. For any , , and , if and only if and the function belongs to . Moreover, there exist , such that where and depend on , , , , and .

Lemma 2.2. If has the segment property, then for any , , and one has

3. Some Embedding Lemmas

We now recall the definitions of the function spaces in which the coefficients of the operator will be chosen. If has the propertywhere is a positive constant independent of and , it is possible to consider the space () of functions such thatwhereIf , wherewe will say that if for . A function is called a modulus of continuity of in ifFor and , we denote by the set of all functions in such thatendowed with the norm defined by (3.6). Then, we define as the closure of in . In particular, we put , and In order to define the modulus of continuity of a function in , recall first that for a function the following characterization holds:whereand , , is a function in such thatwith the position . Thus, the modulus of continuity of is a functionsuch thatA more detailed account of properties of the above defined function spaces can be found in [9, 19, 20].

We consider the following condition: has the cone property, , , , , are numbers such that

From [21, Theorem 3.1] we have the following.

Lemma 3.1. If the assumption holds, then for any one has and with dependent only on , , , , , and .

From [21, Theorem 3.2] it follows Lemma 3.2.

Lemma 3.2. If the assumption is satisfied and in addition , then for any there exist a constant and a bounded open set , with the cone property, such that where , depend on , , , , , , , , , and .

4. An a Priori Bound

Assume that is an unbounded open subset of , , with the uniform -regularity property, and let be the function defined by (2.7). Moreover, let and . Consider in the differential operator:with the following conditions on the coefficients: there exist functions , , and such thatwhere

Observe that under the assumptions , it follows that the operator is bounded from Lemma 3.1.

Theorem 4.1. If the hypotheses , , and are verified, then there exist a constant and a bounded open subset , with the cone property, such that with and depending on , , , , , , , , , , , , , , , , , , , , where depends on , , , , , , , , , ,  

Proof. We consider a function , such that where depends only on , fix and putClearly we have Now, we putand fix Since , from [14, Theorem 3.3], it follows that there exist and a bounded open subset , with the cone property, such thatwith and depending on , , , , , , , , , , , , , , where depends on , , , , , , , , , , Sincefrom (4.11) and (4.12), we have with dependent on the same parameters of .
On the other hand, since , we have thatwhere depends only on .
Therefore, by (4.13) and (4.14), we deduce the bound: where depends on the same parameters of and on .
From (4.15) it follows where depends on the same parameters of .
Sincefrom (4.16) and from Lemma 2.1 we have thatwith dependent on the same parameters of and also on .
Moreover, from Lemma 3.2 it follows that for any , there exist , , and two bounded open sets , , both with the cone property, such that where , depend on , , , , , , and , depend on , , , , , , , , , and .
From (4.18) and (4.19) it follows (4.6) and then we have the result.

5. A Uniqueness Result

In this section, we will prove our uniqueness theorem. We begin to prove a regularity result.

Lemma 5.1. Suppose that the assumptions , , and (with and ) hold, and let be a solution of the problem where and . Then, belongs to .

Proof. By [13, Lemma 4.1] we haveWe choose , with and a function , such that where depends only on .
We fix and putClearly we have
Since , from [14, Theorem 3.3] it follows that there exist and a bounded open subset , with the cone property, such thatwith and depending on , , , , , , , , , , , , , , , , , , where depends on , , , , , , , , , ,
Since from (5.6) and (5.7), we have with dependent on the same parameters of .
From Lemma 3.1 with , we have thatwith dependent on , , , and .
Using [22, Corollary 4.5], we can obtain the following interpolation estimate: where the constant depends on , , .
Thus, by (5.8)–(5.10), with easy computations, we deduce the bound: where depends on , , , , , , , , , , , , , , , , , , , .
By a well-known lemma of monotonicity of Miranda (see [23, Lemma 3.1]), it follows from (5.11) that and then, using Young's inequality, we deduce from (5.12) thatwith dependent on the same parameters of .
From (5.13) it follows where depends on the same parameters of .
If , sincefrom (5.14) and from Lemma 2.1 we have thatwith dependent on the same parameters of and on . Therefore, belongs to .
If , we denote by the positive integer, such thatThen, for , we have thatTherefore, using (5.14) and (5.16) with , , instead of , we deduce that On the other hand, we have thatand then, since , (5.14) holds. Thus, satisfies (5.16) and then .

Theorem 5.2. If conditions , , and (with and ) hold, and a.e. in , then the problemadmits only the zero solution.

Proof. Fix , such that . From Lemma 5.1 it follows that . On the other hand, since , from Lemma 2.2 we have that . Thus, from [13, Theorem 5.2] we deduce that .