International Journal of Differential Equations

Volume 2018 (2018), Article ID 6259307, 6 pages

https://doi.org/10.1155/2018/6259307

## Uniqueness Results for Higher Order Elliptic Equations in Weighted Sobolev Spaces

University of Salerno, Via Giovanni Paolo II, No. 132, 84084 Fisciano, Italy

Correspondence should be addressed to Sara Monsurrò; ti.asinu@orrusnoms

Received 28 November 2017; Accepted 11 January 2018; Published 1 March 2018

Academic Editor: P. A. Krutitskii

Copyright © 2018 Loredana Caso 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 prove some uniqueness results for the solution of two kinds of Dirichlet boundary value problems for second- and fourth-order linear elliptic differential equations with discontinuous coefficients in polyhedral angles, in weighted Sobolev spaces.

#### 1. Introduction

The Dirichlet problem for polyharmonic equations in bounded domains of has been studied, among the first, by Sobolev in [1].

The problem was developed in various directions. For instance, Vekua in [2, 3] considers different boundary value problems in not necessarily bounded domains for harmonic, biharmonic, and metaharmonic functions. Successively, analogous problems in more general cases, for what concerns domains and operators, have been studied with different methods by many authors (see, e.g., [4–7]).

In particular, in [7], the author obtains a uniqueness result for the Dirichlet problem for polyharmonic operators of order in polyhedral angles of . This result has been later on generalized, in [5], to the case of operators in divergence form of order with discontinuous bounded measurable elliptic coefficients.

In [6] the authors study a boundary value problem for biharmonic functions in presence of nonregular points on the boundary of the domain. It is well known that in the neighborhood of these singular points (corners or edges) the solution of the problem presents a singularity that can be characterized by the presence of a suitable weight.

Uniqueness results for different Dirichlet problems in weighted Sobolev spaces for different classes of weights can be found in [8–12]. Studies of Dirichlet problems in the framework of weighted Sobolev spaces and in the case of unbounded domains can be found in [13–22].

In this paper, we extend the results of [5, 7] to the case of weighted Sobolev spaces. More precisely, we prove some uniqueness results for the solution of two kinds of Dirichlet boundary value problems for second- and fourth-order linear elliptic differential equations with discontinuous coefficients in the polyhedral angle , , , in weighted Sobolev spaces.

The first problem we consider is the following:where, for and , denotes a weighted Sobolev space where the weight is a power of the distance from the origin, is the closure of in , and ; see Section 2 for details.

The second problem we study isIn both cases the coefficients belong to some weighted Sobolev spaces.

The main tool in our analysis is a generalization of the Hardy’s inequality proved by Kondrat’ev and Olènik in [23].

#### 2. Preliminary Results

Let be an open subset of with , whose boundary contains . For and , denotes the space of all functions such that for , normed byFrom [24] and Propositions 6.3 and 6.5, we get the following.

Proposition 1. *If is a bounded open subset in with , then Furthermore, for each there exists such that *

In the present paper we use the following notation:(i) is a cone with vertex in the origin of coordinates;(ii), , is the open ball of center in the origin and radius ;(iii);(iv)for every , is the “polyhedral angle” with vertex in the origin;(v) is the half-space;(vi).

To prove our main results, consisting in two uniqueness theorems, we will use the following inequality. We observe that this is a slightly modified version of a generalized Hardy’s inequality that was proved by Kondrat’ev and Olènik in [23], adapted to our needs (see also [5]).

Lemma 2 (generalized Hardy’s inequality). *Let and be such that . Assume that for a sufficiently smooth function the following condition is fulfilled: where is the gradient of the function and . Then, there exist two constants such thatwhere does not depend on the function , , and . If, in addition, then .*

*Remark 3. *We remark that there are always important restrictions on the dimension of the space, the order of “singularity” , and the summability exponent (see, e.g., [23, 25–29], where different variants of Hardy or Caffarelli-Kohn-Nirenberg type inequalities are proved).

#### 3. Dirichlet Problem for Second-Order Elliptic Equations

We consider the following differential operator in divergence form in the polyhedral angle , :where the coefficients are measurable functions such that there exist two positive constants and such that

We study the Dirichlet problemwhere .

*Definition 4. *We say that a function is a generalized solution of problem (11) if it satisfies the integral identityfor any and any function .

Now we prove our first uniqueness result.

Theorem 5. *Let be a generalized solution of problem (11), with . Then there exists such that if and one has in .*

*Proof. *Let be an auxiliary function in defined bywhere is such that . Let us also assume that there exists a positive constant such thatSet, for any ,Note that the function is such that, for any , one hasLet be a generalized solution of problem (11), with . We put Clearly, by definition of and as a consequence of our boundary condition, one has that .

Thus, using as test function in (12), we getFrom (10), (16), and (18) we deduce that there exists a positive constant such that where denotes the modulus of the gradient of .

By applying Young’s inequality one gets that for any Thus, taking into account (14) and applying the generalized Hardy’s inequality (8) (with and ) to the second term in the right-hand side of (20), we deduce that if , From the ellipticity condition in (10) and for , we have where the constant .

Thus for any and for any we obtainSince is a generalized solution of problem (11), with , and the constant does not depend on the radius and on the solution , the right-hand side of (23) tends to zero when and then This implies that thereforeBy Proposition 1 we deduce that if the solution with , then , for any . On the other hand, if for any there exists such that if , then for any . Thus, by (26) the function is a constant in , and since one concludes that in .

#### 4. Dirichlet Problem for 4th-Order Elliptic Equations

Let us now consider the following differential operator of 4th order in the polyhedral angle , , where are measurable symmetric coefficients and there exist two positive constants and such that We want to prove a uniqueness result for the solution of the Dirichlet problemwhere .

*Definition 6. *We say that a function is a generalized solution of problem (29) if it satisfies the integral identityfor any and any function .

The result is the following.

Theorem 7. *Let be a generalized solution of problem (29), with . Then there exists such that if and one has in .*

*Proof. *We shall rely on the methods developed in [5, 7]. We consider the function defined in (13) and satisfying (14). Furthermore, we assume that there exists a positive constant such thatNote that the function is such that, for any , one has (16) andwhere denotes the Kronecker delta.

Again we put where is a generalized solution of problem (29), with .

Observe that the definition of together with the boundary condition satisfied by gives that . Hence, by the symmetry of , if we take as test function in (30) we getFrom (28) and (34) we deduce that By applying (16), (32), and Young’s inequality one gets that there exist two positive constants and such that for any Thus, applying repeatedly the generalized Hardy’s inequality (8) (with and to the third integral on the right-hand side and with and to the last integral on the right-hand side and then again with and ), we deduce that if , where the constant .

Thus for any and for any we obtainNow, arguing as in the proof of Theorem 5, since is a generalized solution of problem (29), with , the right-hand side of (38) tends to zero when and then This implies that thereforeIn view of Proposition 1 we obtain that if the solution with , then , for any , while if for any there exists such that if , then for any . Thus, by (41) the function is constant a.e. in , and since one concludes that a.e. in , for any . The thesis follows then as the one of Theorem 5.

#### Conflicts of Interest

The authors declare that they have no conflicts of interest.

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