## Boundary Value Problems

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# Solvability of the Dirichlet Problem for Elliptic Equations in Weighted Sobolev Spaces on Unbounded Domains

Boundary Value Problems20082008:901503

DOI: 10.1155/2008/901503

Accepted: 14 August 2008

Published: 20 August 2008

## Abstract

This paper is concerned with the study of the Dirichlet problem for a class of second-order linear elliptic equations in weighted Sobolev spaces on unbounded domains of , . We state a regularity result and we can deduce an existence and uniqueness theorem.

## 1. Introduction

Consider the Dirichlet problem
(1.1)
where is a sufficiently regular open subset of , , is the uniformly elliptic second-order linear differential operator defined by
(1.2)

with coefficients .

It is well known that if is bounded, the above problem has been largely studied by several authors under various hypotheses of discontinuity on the leading coefficients and considering the case . In particular, some -bounds for the solutions of the problem (1.1) and related existence and uniqueness theorems have been obtained. Among the other results on this subject, we quote here the classical result of [1], where the author assumed that the 's belong to . This result was later generalized in different ways, supposing that the derivatives of the leading coefficients belong to some wider spaces. More recently, a relevant contribution to the theory has been given in [25], where the coefficients are assumed to be in the class VMO and ; observe here that VMO contains the class .

If the set is unbounded, under assumptions similar to those required in [1], problem (1.1) has for instance been studied in [6] with , and in [7] with . Instead, in [8, 9], the leading coefficients satisfy restrictions similar to those in [2, 3].

In [10], we extended some results of [8, 9] to a weighted case. More precisely, we denoted by a weight function belonging to a suitable class and such that
(1.3)
Then we considered the problem
(1.4)

where and are some weighted Sobolev spaces and the weight functions are a suitable power of . We obtained that the operator has closed range and that for the problem (1.4) a uniqueness result holds.

In this paper, we study again the problem (1.4). We state a regularity result which allows us to obtain the solvability of the problem.

A similar weighted case was studied in [11] 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 . Let . Denote by the Lebesgue measure of , by the characteristic function of 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 . Finally, for any and , we put , and

Let be an open subset of . We introduce a class of weight functions defined on . Denote by the set of all measurable functions such that
(2.1)

where is independent of and .

We note that the class of all functions which are Lipschitz continuous in with Lipschitz coefficient is contained in (see [12]).

For , we put
(2.2)
It is known that
(2.3)

(see [12, 13]).

We assign an unbounded open subset of .

From now on, let be a function such that and
(2.4)
For example,
(2.5)
We put
(2.6)

and note that .

For any and , we set
(2.7)
and note that
(2.8)

where depend only on (see [12]).

If is a real function defined in , we denote by the zero extension of in .

We begin to prove the following.

Lemma 2.1.

If are two nonnegative functions in , respectively, then for any
(2.9)
and for the following also hold:
(2.10)

Proof.

The equality (2.9) follows by
(2.11)
Prove now the inequality (2.10). We observe that:
(2.12)
where we have put
(2.13)
On the other hand,
(2.14)
Therefore, from (2.12) and (2.14), we deduce that
(2.15)
By (2.15), it obviously follows
(2.16)

and (2.16) yields the inequality (2.10).

If , and , consider the space of distributions on such that for , equipped with the norm
(2.17)

Moreover, 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 [14].

From Lemma 2.1 we can deduce another lemma which we will need in the proof of our regularity result.

Lemma 2.2.

Let and . Then if and only if and the function belongs to . In addition, there exist such that
(2.18)
where and depend only on and . Moreover, if and , then the function belongs to and the following estimate holds:
(2.19)

with dependent only on and .

Proof.

The first part of the lemma follows from (2.9) for and , if one uses (2.1) and (2.8). The second part of the lemma follows in a similar way from the inequality (2.10), if one puts , , and .

## 3. An Embedding Lemma

We now recall the definitions of the function spaces in which the coefficients of the operator will be chosen. If has the property
(3.1)
where is a positive constant independent of and , it is possible to consider the space ( ) of functions such that
(3.2)
where
(3.3)
If , where
(3.4)
we will say that if for . A function
(3.5)
is called a modulus of continuity of in if
(3.6)
For and , we denote by the set of all functions in such that
(3.7)
endowed with the norm defined by (3.7). Then we define as the closure of in and 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:
(3.8)
where
(3.9)
Thus the modulus of continuity of is a function
(3.10)
such that
(3.11)

A more detailed account of properties of the above defined function spaces can be found in [6, 15, 16].

We consider the following condition:

(h0) has the cone property, are numbers such that
(3.12)

From [17, Theorem 3.1] we have the following.

Lemma 3.1.

If the assumption (h0) holds, then for any , it results that?? and

(3.13)

with dependent only on and

## 4. A Regularity Result

Assume that is an unbounded open subset of , with the uniform -regularity property, and let be the function defined by (2.6). Moreover, let and . Consider in the differential operator
(4.1)

with the following conditions on the coefficients:

(h1)
(4.2)

there exist functions , , and such that

(h2)
(4.3)
(h3)
(4.4)
where
(4.5)

Observe that under the assumptions (h1)–(h3), it follows that the operator is bounded from Lemma 3.1.

Theorem 4.1.

Suppose that the assumptions (h1), (h2), and (h3) hold, and let be a solution of the problem
(4.6)

where and . Then belongs to .

Proof.

By [8, Lemma 4.1], we have
(4.7)
We choose , with and a function such that
(4.8)

where depends only on .

We fix and put
(4.9)
Clearly, we have
(4.10)
Since , from [8, Theorem 3.1] it follows that
(4.11)
with depending on , , , , where depends on and is a function in such that
(4.12)
Since
(4.13)
from (4.11) and (4.13), we have
(4.14)

with depending on the same parameters of .

From Lemma 3.1 with , we have that
(4.15)

with dependent on and .

Using [18, Corollary 4.5], we can obtain the following interpolation estimates:
(4.16)

where depends on and the constants and depend on .

Thus by (4.14)–(4.16), with easy computations, we deduce the following bound:
(4.17)

where depends on , , , , .

By a well-known lemma of monotonicity of Miranda (see [19, Lemma 3.1]), it follows from (4.17) that
(4.18)
and then, using Young's inequality, we deduce from (4.18) that
(4.19)

with dependent on the same parameters of .

From (4.19) it follows that
(4.20)

where depends on the same parameters of .

By (4.20) and by Lemma 2.2, we have that
(4.21)

with dependent on the same parameters of and on .

Therefore, from (4.21), we have the result.

## 5. Existence and Uniqueness Results

In this section, we will prove our existence and uniqueness theorem. To this aim, we need two preliminary lemmas.

Observe that it is possible to find a function which is equivalent to and such that
(5.1)

where is independent of (see [12]).

Lemma 5.1.

The Dirichlet problem

(5.2)
where
(5.3)

is uniquely solvable. Moreover, if , then the solution belongs to for all in .

Proof.

Note that is a solution of problem (5.2) if and only if is a solution of the problem
(5.4)
But for any
(5.5)
then (5.4) is equivalent to the problem
(5.6)
where we have put
(5.7)

Using [7, Theorem 5.2], [6, Equation (1.6)], and (5.1), we obtain that (5.6) is uniquely solvable and then problem (5.2) is uniquely solvable too.

Moreover, if , then also . Therefore, using the theorem in [20], we have that the solution of (5.6) belongs to for all , and so the solution of (5.2) lies in for all .

Lemma 5.2.

The Dirichlet problem
(5.8)

is uniquely solvable, where is defined by (5.3).

Proof.

Let be a function in . Then, by Lemma 5.1, there exists a unique (for all ) such that .

Firstly, suppose that . It follows from Theorem 4.1 that belongs to . Moreover, by [10, Lemma 2.2], lies in .

Suppose now . Then (for all ) and then, using again Theorem 4.1, belongs to . Moreover, by [10, Lemma 2.2], lies in .

Therefore, in both cases, and it is a solution of the equation , so that . Since is dense in (see [14, Proposition 1.1]) and is a closed subspace of by [10, Theorem 4.1], we obtain that . The uniqueness of the solution follows from [10, Theorem 5.2].

Finally, adding the following assumption on the coefficients of :

(h4)
(5.9)

we are now in position to state the following uniqueness and existence result.

Theorem 5.3.

Suppose that conditions (h1)–(h4) hold. In addition, assume that a.e. in . Then the problem
(5.10)

is uniquely solvable.

Proof.

For each put
(5.11)
The function
(5.12)
is clearly continuous; moreover, it is easy to show that the coefficients of each operator satisfy the hypotheses of [10, Theorem 5.2] (see also [16, Lemma 3.2]), and hence . On the other hand, it follows from [10, Theorem 4.1] that is closed for any , so that [9, Lemma 4.1] can be used to obtain the existence of such that
(5.13)
By Lemma 5.2, the problem
(5.14)

is uniquely solvable.

Therefore, this latter result and estimate (5.13) allow to use the method of continuity along a parameter in order to prove that the problem
(5.15)

is likewise uniquely solvable. The proof is complete.

## Authors’ Affiliations

(1)
Dipartimento di Matematica e Informatica, Università di Salerno

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