- Research Article
- Open Access

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

- Serena Boccia
^{1}, - Sara Monsurrò
^{1}and - Maria Transirico
^{1}Email author

**2008**:901503

https://doi.org/10.1155/2008/901503

© Serena Boccia et al. 2008

**Received:**2 March 2008**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.

## Keywords

- Weight Function
- Differential Operator
- Open Subset
- Dirichlet Problem
- Detailed Account

## 1. Introduction

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 [2–5], 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].

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

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]).

We assign an unbounded open subset of .

and note that .

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.

Proof.

and (2.16) yields the inequality (2.10).

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.

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

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

We consider the following condition:

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

Lemma 3.1.

If the assumption (h_{0}) holds, then for any
, it results that??
and

with dependent only on and

## 4. A Regularity Result

with the following conditions on the coefficients:

there exist functions , , and such that

Observe that under the assumptions (h_{1})–(h_{3}), it follows that the operator
is bounded from Lemma 3.1.

Theorem 4.1.

_{1}), (h

_{2}), and (h

_{3}) hold, and let be a solution of the problem

where and . Then belongs to .

Proof.

where depends only on .

with depending on the same parameters of .

with dependent on and .

where depends on and the constants and depend on .

where depends on , , , , .

with dependent on the same parameters of .

where depends on the same parameters of .

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.

where is independent of (see [12]).

Lemma 5.1.

The Dirichlet problem

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

Proof.

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.

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 :

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

Theorem 5.3.

_{1})–(h

_{4}) hold. In addition, assume that a.e. in . Then the problem

is uniquely solvable.

Proof.

is uniquely solvable.

is likewise uniquely solvable. The proof is complete.

## Authors’ Affiliations

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