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Uniqueness results for the Dirichlet problem for higher order elliptic equations in polyhedral angles

Boundary Value Problems20142014:232

https://doi.org/10.1186/s13661-014-0232-1

Received: 23 June 2014

Accepted: 8 October 2014

Published: 4 November 2014

Abstract

We consider the Dirichlet boundary value problem for higher order elliptic equations in divergence form with discontinuous coefficients in polyhedral angles. Some uniqueness results are proved.

MSC: 35J30, 35J40.

Keywords

higher order elliptic equationsDirichlet problemdiscontinuous coefficients

1 Introduction

The Dirichlet problem for the polyharmonic equation in a bounded domain of R n has been studied by Sobolev in [1]. Later on, different problems (Dirichlet, Neumann and Riquier problems) for harmonic, biharmonic and meta-harmonic functions have been considered by Vekua in [2] and [3], both in the cases of bounded and unbounded domains of R n . Successively, many authors studied analogous problems in more general cases and with different methods (see, for instance, [4]–[10], the general survey on this subject [11] and the references quoted therein). In particular, in [10], the author obtains the uniqueness of the solution of the Dirichlet problem,
{ Δ m u ( x ) = 0 in  R l n , j u ( x ) ν j | R l n = 0 , j = 0 , , m 1 ,
(1.1)

where Δ m denotes the polyharmonic operator of order m, Δ is the Laplace operator and R l n is a polyhedral angle of R n , defined in Section 2. We explicitly observe that for l = 0 the above mentioned definition gives the half-space R + n . We note that, due to the tools used in the proof, some restrictions on the dimension n of the space are required.

Our aim, in this paper, is to generalize the uniqueness result of [10]. More precisely, we are concerned with the following Dirichlet problem for a homogeneous equation in divergence form of order 2m:
{ | α | = | β | = m D α ( a α β ( x ) D β u ( x ) ) = 0 in  R l n , j u ( x ) ν j | R l n = 0 , j = 0 , , m 1 ,
(1.2)

where the discontinuous coefficients a α β are bounded and measurable functions satisfying the uniform ellipticity condition.

Let us remark that if we take α = β and if the coefficients of the equation are constants a α β ( x ) = m ! α ! , then the left-hand side of the equation in (1.2) is exactly the polyharmonic operator Δ m in (1.1).

Our main results consist in two uniqueness theorems obtained for some particular cases of problem (1.2). More precisely, in Section 4 we consider problem (1.2) in the case m = 1 and in Section 5 we assume that α = β . The main tool in our analysis is a generalization of the Hardy inequality proved by Kondrat’ev and Oleinik in [5] (see Section 3).

2 Notation

Throughout this work we use the following notation:

n N is the dimension of the considered space;

Greek letters denote n-dimensional multi-indices, for instance α = ( α 1 , α 2 , , α n ) , where α i N { 0 } , i = 1 , , n ;

| α | = α 1 + + α n is the module of the multi-index α;

α ! = α 1 ! α n ! is the factorial of the multi-index α;

φ , i ( x ) = φ ( x ) x i , i = 1 , , n ;

D i α i = α i ( x i ) α i , i = 1 , , n ;

D α = D 1 α 1 D n α n ;

for ξ = ( ξ 1 , , ξ n ) R n we set ξ α = ξ 1 α 1 ξ n α n ;

for every l { 0 , , n 1 } ,
R l n = { x = ( x 1 , x 2 , , x n ) R n : x i > 0 , i = n l , , n }

is the ‘polyhedral angle’ with vertex in the origin;

for l = 0 the above definition gives the half-space R + n ;

for ρ > 0 we denote by Q ρ = { x R l n : | x | < ρ } .

3 Setting of the problem

We want to consider the following differential equation in divergence form of order 2m, m N , in certain unbounded domains of R n , n > 2 :
| α | = | β | = m D β ( a α β ( x ) D α u ( x ) ) = f ( x ) ,
(3.1)
where f ( x ) is a given datum and the coefficients a α β ( x ) are bounded measurable functions satisfying the uniform ellipticity condition, i.e. there exist two positive constants λ 1 and λ 2 such that for each nonzero vector ξ R n one has
λ 1 | ξ | 2 m | α | = | β | = m a α β ( x ) ξ α ξ β λ 2 | ξ | 2 m a.e.
(3.2)

Let us mention that if we take α = β in (3.1) and if the coefficients of the equation are constants a α ( x ) = m ! α ! , then left-hand side of this equation is the polyharmonic operator Δ m , where Δ denotes, as usual, the Laplace operator.

For every sufficiently differentiable functions u and v let us set
E m a ( u , v ) = | α | = | β | = m a α β ( x ) D α u D β v , E m ( u , v ) = | α | = m m ! α ! D α u D α v , E m a ( u ) = E m a ( u , u ) , E m ( u ) = E m ( u , u ) .

Definition 3.1

We say that the function u is a generalized solution of (3.1) in R l n with homogeneous Dirichlet boundary conditions if u W m , 2 ( R l n ) and it satisfies the integral identity
( 1 ) m Q ρ E m a ( u , v ) d x = Q ρ f ( x ) v ( x ) d x ,
(3.3)

for any ρ > 0 and any function v W 0 m , 2 ( Q ρ ) , where f L 2 ( R l n ) .

To prove our main results, consisting in two uniqueness theorems, we will essentially use the following generalized Hardy inequality.

Lemma 3.2

(Generalized Hardy inequality)

Let p > 1 , j, and n be such that j + n p 0 . Assume that for a sufficiently smooth function g the following condition is fulfilled in a cone V R n with vertex in the origin of coordinates:
V | x | j | g ( x ) | p d x < ,
(3.4)
where g = ( g x 1 , , g x n ) is the gradient of the function g. Then there exist two constants M , K > 0 such that
V | x | j p | g ( x ) M | p d x < K V | x | j | g ( x ) | p d x ,
(3.5)

where the constant K does not depend on the function g. If, in addition, g ( 0 ) = 0 then M = 0 .

Remark 3.3

The previous lemma, which was proved by Kondrat’ev and Oleinik in [5], holds also if we replace (3.5) with the following inequality:
V R 2 V R 1 | x | j p | g ( x ) M | p d x < K V R 2 V R 1 | x | j | g ( x ) | p d x ,
(3.6)

with 0 < R 1 < R 2 , where V R , R > 0 , denotes the intersection between the cone V and the open ball of center in the origin and radius R.

This result can be deduced by the proof of Lemma 3.2, with slight modifications. We point out that in this proof it is also well rendered that the constant K does not depend on R 1 and R 2 .

Remark 3.4

As evidenced in many works about different variants of Hardy or Caffarelli-Kohn-Nirenberg type inequalities (see for instance [5], [12]–[16]), there are always very important restrictions on the dimension of the space n, the order of ‘singularity’ j and the order of the integral norm p.

4 Dirichlet problem for second order elliptic equations

In this section we consider, for m = 1 , the homogeneous equation (3.1) in the polyhedral angle R l n , 0 l n 1 , with the homogeneous Dirichlet boundary condition, namely
{ i , j = 1 n x j ( a i j ( x ) u x i ) = 0 in  R l n , u | R l n = 0 .
(4.1)
Let us observe that by Definition 3.1 it follows that every generalized solution is such that
R l n E 1 ( u ) d x < .
(4.2)

Now we prove our first uniqueness result.

Theorem 4.1

Let n > 2 . Assume that (3.2) is satisfied, with m = 1 . If u is a generalized solution of problem (4.1), then u 0 in R l n .

Proof

Let Θ ( s ) be an auxiliary function in C 0 ( [ 0 , [ ) defined by
Θ ( s ) { 1 0 s 1 , θ ( s ) 1 s 2 , 0 s 2 ,
(4.3)
with 0 θ ( s ) 1 , and such that there exists a positive constant K 0 such that
| Θ ( s ) | 2 K 0 Θ ( s ) .
(4.4)

We note that in order to obtain a cut-off function Θ ( s ) of the above mentioned type one can consider a classical mollifier and modify it suitably near to s = 1 and s = 2 .

Set, for any R > 0 ,
Θ R ( x ) = Θ ( | x | R ) .
(4.5)
Note that the function Θ R is such that, for any j = 1 , , n , one has
Θ R , j ( x ) = Θ ( | x | R ) x j R | x | .
(4.6)
Let us now consider the function
v ( x ) = u ( x ) Θ R ( x ) .
Clearly, by definition of Θ R and as a consequence of our boundary condition, one has v ( x ) W 0 1 , 2 ( Q 2 R ) . Thus, substituting this function in the integral identity (3.3), we get
Q 2 R i , j = 1 n a i j ( x ) u , i u , j Θ R ( x ) d x + Q 2 R Q R i , j = 1 n a i j ( x ) u , i u Θ R , j ( x ) d x = 0 .
Hence, by (3.2) we have
Q 2 R E 1 a ( u ) Θ R ( x ) d x = | Q 2 R Q R i , j = 1 n a i j ( x ) u , i u Θ ( | x | R ) x j R | x | d x | .
Recalling that for any ε > 0 one has
a b ε a 2 2 + b 2 2 ε , a , b 0 ,
(4.7)
in view of the boundedness of the coefficients and of (4.4), one gets
Q 2 R E 1 a ( u ) Θ R ( x ) d x ε 2 K 1 K 0 Q 2 R Q R E 1 ( u ) Θ R d x + K 1 2 ε Q 2 R Q R u 2 | x | 2 d x ,
where K 0 is the constant in (4.4) and K 1 = K 1 ( n , λ 2 ) . Thus, taking into account the ellipticity of the coefficients, we obtain
Q 2 R E 1 ( u ) Θ R ( x ) d x ε 2 K 2 K 0 Q 2 R Q R E 1 ( u ) Θ R d x + K 2 2 ε Q 2 R Q R u 2 | x | 2 d x ,
(4.8)

with K 2 = K 2 ( n , λ 1 , λ 2 ) .

Therefore, if we choose ε = 1 K 2 K 0 and we apply the generalized Hardy inequality (3.6) (with p = 2 and j = 0 ) to the second term in the right-hand side of (4.8), we deduce that
Q 2 R E 1 ( u ) Θ R ( x ) d x K 3 Q 2 R Q R E 1 ( u ) d x ,

where the constant K 3 does not depend on the radius R and on the function u (see Remark 3.3).

Now, observe that clearly for any P > 0 there exists R > P such that Q P Q R and therefore, in view of the definition of Θ R , by the former inequality we obtain
Q P E 1 ( u ) d x K 3 Q 2 R Q R E 1 ( u ) d x .
(4.9)
Condition (4.2) being satisfied, the right-hand side of (4.9) tends to zero when R . Now, since the left-hand side of (4.9) is independent of the radius R, we have, for any P > 0 ,
Q P E 1 ( u ) d x = Q P i = 1 n u , i u , i d x = 0 .

This means that the function u ( x ) is a constant and, according to the boundary condition in (4.1), this constant is zero. This concludes our proof. □

Remark 4.2

Note that our proof do not provide any uniqueness result for n = 2 , since in this case the generalized Hardy inequality in Lemma 3.2 does not apply, as a consequence of our choice of p and j.

5 Dirichlet problem for higher order elliptic equations

Here, we consider the following homogeneous equation of order 2m with homogeneous Dirichlet boundary conditions in the polyhedral angle R l n , l { 0 , , n 1 } :
{ | α | = m D α ( a α ( x ) D α u ( x ) ) = 0 in  R l n , j u ( x ) ν j | R l n = 0 , j = 0 , , m 1 .
(5.1)
Note that, again, in view of Definition 3.1 one finds that every generalized solution of problem (5.1) is such that
R l n E m ( u ) d x < .
(5.2)

Theorem 5.1

Let n > 2 m or n = 2 k + 1 , with k N . Assume that (3.2) is satisfied. If u is a generalized solution of problem (5.1), then u ( x ) 0 in R l n .

Proof

Let us use again the function Θ R introduced in the proof of Theorem 4.1.

It is easy to check that
D α Θ R ( x ) = i = 1 | α | Θ ( i ) ( | x | R ) P | α | ( x ) R i | x | ( 2 | α | i ) ,

where Θ ( i ) denotes the derivative of order i of the function Θ and P | α | ( x ) is a polynomial of order | α | .

Moreover, if we assume that there exist some positive constants K i , such that
| Θ ( i ) ( s ) | 2 K i Θ ( s ) , i = 1 , , | α | ,
then, for R < | x | < 2 R , one has
| D α Θ R ( x ) | 2 K α Θ R ( x ) | x | 2 | α | ,
(5.3)

where the constant K α depends only on α.

Note that function v ( x ) = u ( x ) Θ R ( x ) W 0 m , 2 ( Q 2 R ) , thus, substituting it in the integral identity (3.3), we deduce
Q 2 R E m a ( u , u Θ R ) d x = Q 2 R E m a ( u ) Θ R d x + Q 2 R Q R | α | = m a α ( x ) D α u [ γ + ι = α , ι 0 ( | γ | + | ι | ) ! γ ! ι ! D γ u D ι Θ R ] d x = 0 .
Therefore
Q 2 R E m a ( u ) Θ R d x = | Q 2 R Q R | α | = m a α ( x ) D α u [ γ + ι = α , ι 0 ( | γ | + | ι | ) ! γ ! ι ! D γ u D ι Θ R ] d x | .
From (4.7) and (5.3), arguing as in the proof of Theorem 4.1, we deduce that
Q 2 R E m ( u ) Θ R ( x ) d x | α | = m K α ε α Q 2 R Q R E m ( u ) Θ R ( x ) d x + | α | = m K α ε α Q 2 R Q R [ | γ | = 0 m 1 E | γ | ( u ) 1 | x | 2 ( m | γ | ) ] d x ,
(5.4)

with K α = K α ( n , α , λ 1 , λ 2 , K α ) and K α = K α ( n , α , λ 1 , λ 2 ) .

Now, we apply repeatedly the Hardy inequality (3.6) to the single summands of the second term in the right-hand side of (5.4) until the order of the partial derivatives achieves m. Thus, after an appropriate selection of ε α , we get
Q 2 R E m ( u ) Θ R ( x ) d x K ˜ Q 2 R Q R E m ( u ) d x ,

where the constant K ˜ is independent of the radius R and of the function u.

Finally, following the same argument used in Theorem 4.1, for any P > 0 we find R > P such that Q P Q R and therefore, taking into account the definition of Θ R , we obtain
Q P E m ( u ) d x K ˜ Q 2 R Q R E m ( u ) d x .
(5.5)
In view of condition (5.2), the right-hand side of (5.5) goes to zero when R , and, since the left-hand side of (5.5) is independent of R, we have, for any P > 0 ,
Q P E m ( u ) d x = Q P [ | α | = m m ! α ! D α u D α u ] d x = 0 .

Therefore, the partial derivatives of any order of the solution are equal to zero, thus, as a consequence of the boundary conditions in (5.1), we deduce that u ( x ) 0 in R l n . □

Remark 5.2

Clearly also in this case the repeated application of the Hardy inequality yields the restrictions n > 2 m or n = 2 k + 1 , k N , on the space dimension.

Authors’ information

Sara Monsurrò and Maria Transirico are members of the Gruppo Nazionale per l’Analisi Matematica, la Probabilità e le loro Applicazioni (GNAMPA) of the Istituto Nazionale di Alta Matematica (INdAM).

Declarations

Authors’ Affiliations

(1)
Dipartimento di Matematica, Università di Salerno, Fisciano, Italy
(2)
Iv. Javakhishvili Tbilisi State University, Tbilisi, Georgia

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Copyright

© Monsurrò et al.; licensee Springer. 2014

This article is published under license to BioMed Central Ltd.Open Access This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly credited.

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