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Stability of weak solutions to obstacle problem in Clifford analysis

Advances in Difference Equations20132013:250

https://doi.org/10.1186/1687-1847-2013-250

Received: 24 May 2013

Accepted: 6 August 2013

Published: 20 August 2013

Abstract

This paper is concerned with the regularity properties of weak solutions to the obstacle problem for Clifford-valued functions. Our main results are a global reverse Hölder inequality and stability of the weak solutions to the obstacle problem.

Keywords

  • Clifford-valued function
  • obstacle problem
  • higher integrability
  • stability

1 Introduction

Let e 1 , e 2 , , e n be the standard basis of R n with the relation e i e j + e j e i = 2 δ i j . For k = 0 , 1 , , n , we denote by C n k = C n k ( R n ) the linear space of all k-multivectors, spanned by the reduced products e I = e i 1 e i 2 e i k , corresponding to all ordered k-tuples I = ( i 1 , i 2 , , i k ) , 1 i 1 < i 2 < < i k n . Thus, Clifford algebra C n = k C n k is a graded algebra, especially C n 0 = R and C n 1 = R n . R C H C n 3 is an increasing chain, where is the Hamilton’s algebra of quaternions. For u C n , u can be written as
u = I u I e I = 1 i 1 < < i k n u i 1 , , i k e i 1 e i k ,

where 1 k n .

The norm of u C n is given by | u | = ( I u I 2 ) 1 / 2 . Clifford conjugation e α 1 e α k ¯ = ( 1 ) k e α k e α 1 . For u = I u I e I C n , v = J v J e J C n ,
u , v = I u I e I , J v J e J = I u I v I

defines the corresponding inner product on C n . Denote by Sc ( u ) the scalar part of u, the coefficient of the element e 0 , and we also have u , v = Sc ( u ¯ v ) .

The Dirac operator used in this paper is given by
D = j = i n e j x j .
(1)
Throughout this paper, we write D ( Ω , C n ) for the space of Clifford-valued functions whose coefficients are Schwartz distributions on Ω. For p > 0 , we write L p ( Ω , C n k ) for the space of Clifford-valued functions u, whose coefficients belong to the usual L p ( Ω ) space. It is a reflexive Banach space endowed with the norm
u p = ( Ω | u | p d x ) 1 p .
Also, W 1 , p ( Ω , C n k ) is the Sobolev space of Clifford-valued functions u, whose coefficients belong to W 1 , p ( Ω ) . It is a reflexive Banach space endowed with the norm
u 1 , p = u p + u p ,

where u = ( u x 1 , , u x n ) , u x i = I u I x i e I .

Let Ω be a bounded domain. This paper is concerned with the A-Dirac equation
D A p ( x , D u ) = 0 ,
(2)
where 1 < p < , A p : Ω × C n C n preserves the grading of the Clifford algebra and satisfies the following conditions for some constants 0 < a b < :
  1. (i)

    The mapping x A p ( x , ξ ) is measurable for all ξ C n k .

     
  2. (ii)

    The mapping ξ A p ( x , ξ ) is continuous for a.e. x Ω .

     
  3. (iii)

    | A p ( x , ξ ) | b | ξ | p 1 .

     
  4. (iv)

    A p ( x , ξ ) A p ( x , η ) , ξ η a | ξ η | p for all x Ω with ξ η .

     

Definition 1.1 [1]

A Clifford-valued function u W 1 , p ( Ω , C n k ) for k = 0 , 1 , , n , is a weak solution to equation (2) if for all φ W 1 , p ( Ω , C n k ) with compact support, we have
Ω A p ( x , D u ) ¯ D φ = 0 .
(3)
In [1, 2], Nolder explained how quasi-linear elliptic equations
div A p ( x , u ) = 0
(4)

arise as components of Dirac systems (2) and discussed some properties of the weak solutions to the scalar parts of equations, such as the Caccioppoli estimate and the removability theorem. In [3], Heinonen et al. studied the quasi-linear elliptic equations (4) by means of potential theory systematically. Many other mathematicians also work on properties of solutions to equations (4), such as the regularity, stability, convergence and so on, see [49].

This paper is organized as follows. In Section 2, some preliminary results about Clifford-valued functions are presented. In Section 3, higher integrability of weak solutions to obstacle problem for Clifford-valued functions are obtained. Section 4 is concerned with the stability of the weak solutions to obstacle problem. For other works about Clifford analysis and Dirac equations, see [1, 2, 1012].

2 Preliminary results

Lemma 2.1 C 0 ( Ω , C n k ) is dense in W 0 1 , p ( Ω , C n k ) for 1 p < .

Proof Let u = I u I ( x ) e I W 0 1 , p ( Ω , C n k ) , then u I W 0 1 , p ( Ω ) . Since C 0 ( Ω ) is dense in W 0 1 , p ( Ω ) , we can find a sequence { u I k } C 0 ( Ω ) converges to u I in W 0 1 , p ( Ω ) . Let u k = I u I k e I , then { u k } C 0 ( Ω , C n k ) .
Ω | u k u | p d x + Ω | u k u | p d x Ω ( I | u I k u I | 2 ) p 2 d x + Ω ( i = 1 n | u k x i u x i | 2 ) p 2 d x c ( n , p ) I Ω | u I k u I | p d x + Ω ( i = 1 n I ( u I k x i u I x i ) 2 ) p 2 d x c 1 ( n , p ) I Ω | u I k u I | p d x + c 2 ( n , p ) I Ω | u I k u I | p d x .
Since { u I k } converges to u I in W 0 1 , p ( Ω ) . Then Ω | u I k u I | p d x 0 , Ω | u I k u I | d x 0 . Thus,
Ω | u k u | p d x + Ω | u k u | p d x 0 ( k ) .

This means u k converges to u in W 0 1 , p ( Ω , C n k ) . □

Here, we denote the Grassmann algebra = k k , d is the exterior derivative operator, and d is the formal adjoint operator. For more details about differential form, see [1]. From [13], we know that the linear mapping
λ : e i 1 e i k e i 1 e i k ( k 0 )

to all of ( R n ) defines a linear isomorphism form ( R n ) onto C n ( R n ) independently of the choice of orthonormal basis { e j } for R n .

For a Clifford-valued function u, we write u for λ 1 u . Via λ, the operator d + d is mapped to D, it means
( D u ) = d u + d u .
(5)
Then, if u D ( Ω , C n k ) , there exists a constant C, such that
max { | d u | , | d u | } C | D u | .
(6)

In [4], Iwaniec gave the following Poincaré inequality for differential forms.

Theorem 2.2 For each u C 0 ( Ω , k ) , there exits a constant C ( n , p ) , 1 < p < such that
u p C ( n , p ) ( d u p + d u p ) .
(7)

Combining (6), (7) and Lemma 2.1, we have the following Poincaré-Sobolev inequality.

Proposition 2.3 For each u W 0 1 , p ( Ω , C n k ) , 1 < p < , there exists a constant C, such that
Ω | u | p c ( n , p ) Ω | D u | p d x .
(8)

Using the Poincaré inequality of the real function and | | u | | c ( n ) | u | , we have

Corollary 2.4 For each u W 0 1 , p ( Ω , C n k ) , 1 < p < n , there exists a constant C such that
( Ω | u | n p n p d x ) n p n p C ( Ω | D u | p d x ) 1 p .
(9)

Similar to the process of the Ponicaré inequality for differential forms in [14], the following inequalities for Clifford-valued function can be obtained.

Corollary 2.5 Let u W 0 1 , s ( Ω , C n k ) , 1 < s < , then, there exists a constant C such that
(10)
Proof If 1 < s < n n 1 , then n s n 1 { s , n n 1 } . From the Hölder inequality and Corollary 2.4, we have
If s > n n 1 , then
( Ω | u | s d x ) 1 s C ( Ω | D u | n s n + s d x ) n + s n s C | Ω | n + s n s ( Ω | D u | n s n + s 1 d x ) n + s 1 n s .
Thus, we have

 □

3 Higher integrability

In this section, we will prove the higher integrability of the weak solutions to the obstacle problem K ψ f , p ( Ω , C n k ) . Here, we assume that the complement of Ω satisfies the measure density condition, it means that there exists a positive constant C > 0 such that
| Ω c B | C | B | ,
(11)

where B is a ball in R n . In order to prove the higher integrability, we need the following Gehring’s lemma, which appeared in [5].

Lemma 3.1 Let B 0 be a ball in R n , and let g , h L p ( B 0 ) be a nonnegative function satisfying
(12)
for all balls B 2 B B 0 , 1 < p < . Then, for each 0 < σ < 1 , p < s < p + p 1 10 n p 4 n k p , we have
(13)

Definition 3.2 For Clifford-valued functions a = I a I ( x ) e I and b = I b I ( x ) e I , we say that a b in Ω if a I ( x ) b I ( x ) for a.e. x Ω and all ordered tuple I.

Let ψ = I ψ I e I be a Clifford-valued function in W 1 , p ( Ω , C n k ) , where ψ I : Ω [ , + ) . f W 1 , p ( Ω , C n k ) is a function, which gives the boundary. We consider the obstacle problem
K ψ f , p ( Ω , C n k ) = { v W 1 , p ( Ω , C n k ) : v f W 0 1 , p ( Ω , C n k ) , v ψ  a.e.  Ω }
(14)

for Clifford-valued functions. To avoid trivialities, we always assume that the set K ψ f , p ( Ω , C n k ) is not empty.

Definition 3.3 We say that a function u W loc 1 , p ( Ω , C n k ) is a weak solution to the obstacle problem K ψ f , p ( Ω , C n k ) , if
Ω A p ( x , D u ) , D v D u d x 0 ,
(15)

whenever v K ψ f , p ( Ω , C n k ) .

Remark 3.4 If u is a weak solution to obstacle problem (15), then u holds for the scalar part of (3), i.e.,
Ω Sc ( A ( x , D u ) ¯ D φ ) d x = 0 .
(16)
Remark 3.5 We assume f ψ in Ω. Indeed, denote f = I f I e I , ψ = I ψ I e I . Then, since u f W 0 1 , p ( Ω , C n k ) ,
0 ( ψ I f I ) + ( u f I ) + .

Thus, we have ( ψ I f I ) + W 0 1 , p ( Ω ) and f I 1 = ( ψ I f I ) + + f I ψ I in Ω. Let f = I f I 1 e I , then f ψ in Ω.

Theorem 3.6 Suppose that the complement of Ω satisfies the measure density condition (11), and let u W loc 1 , p ( Ω , C n k ) be a weak solution to the obstacle problem (14), where f W 1 , p + δ ( Ω , C n k ) for some δ > 0 . Then, there exists a positive number ε 0 > 0 and a constant C, independent of u, such that | D u | L p + ε ( Ω ) , whenever 0 < ε < ε 0 , and
Ω | D u | p + ε d x C [ ( Ω | D u | p d x ) 1 + ε p + Ω | D f | p + ε d x ] .
(17)
Proof First, let B 0 be a ball with Ω 1 2 B 0 , 2 B B 0 be a ball such that Ω B and η C 0 ( 2 B ) be the cut-off function such that η | B ¯ = 1 , 0 η 1 , | η | c ( n ) | B | 1 n . Denote by D = 2 B Ω , now we test (15) with v = u η p ( u f ) . Since u f W 0 1 , p ( Ω , C n k ) , then v f W 0 1 , p ( Ω , C n k ) . For any I, we get
v I ψ I = ( 1 η p ) ( u I ψ I ) + η p ( f I ψ I ) 0
a.e. Ω. Thus, v is an admissible test function, and we have
0 Ω A p ( x , D u ) , D v D u d x D A p ( x , D u ) , η p D u + η p D f p η p 1 D η ( u f ) d x .
(18)
Since A satisfies (iii), (iv), combining the Hölder inequality and the Young inequality, it follows from (18) that
a D η p | D u | p d x b D η p | D u | p 1 | D f | d x + p D | D u | p 1 η p 1 | D η | | u f | d x b ( D η p | D u | p d x ) p 1 p ( D η p | D f | p d x ) 1 p + b p ( D η p | D u | p d x ) p 1 p ( D η p | u f | p d x ) 1 p b ϵ 1 ( p 1 ) p D η p | D u | p d x + b c ( p , ϵ 1 ) D η p | D f | p d x + b ϵ 2 ( p 1 ) p D η p | D u | p d x + b c ( p , ϵ 2 ) D | D η | p | u f | p d x .
Choose ϵ 1 , ϵ 2 such that
b ϵ 1 ( p 1 ) p + b ϵ 2 ( p 1 ) p a 2 .
Since u f W 0 1 , p ( Ω , C n k ) , using Lemma 2.5, we have
B Ω | D u | p d x c 1 2 B Ω | D f | p d x + c 2 | B | 1 n 2 B Ω | u f | p d x c 1 2 B Ω | D f | p d x + c 2 | B | 1 n | 2 B Ω | p n ( 2 B Ω | D u D f | p θ d x ) θ ,
(19)
where θ = n + p 1 n > 1 . By means of condition (11), we have | 2 B Ω | ( 1 C ) | 2 B | , then (19) becomes
1 | B | B Ω | D u | p d x c 3 1 | 2 B | 2 B Ω | D f | p d x + c 4 1 | 2 B | θ ( 2 B Ω | D u | p θ + | D f | p θ d x ) θ .
(20)
Let
g ( x ) = { | D u | p θ , if  x Ω , 0 , otherwise ,
and
h ( x ) = { | D f | p θ , if  x Ω , 0 , otherwise .
Then, the following reverse Hölder estimate
(21)
holds when 2 B B 0 . According to Lemma 3.1, (21) becomes
That is,
B | D u | p + ε d x C [ ( Ω | D u | p d x ) 1 + ε p + Ω | D f | p + ε d x ] .

Since Ω is a bounded domain, Ω ¯ can be covered by a finite number of balls such that the previous inequality holds, then the estimates (17) follows immediately. □

4 Stability

In this section, we will show that the weak solutions to obstacle problem are stable under some suitable assumptions.

Suppose that { p i } is a sequence such that p i p and for each i = 0 , 1 ,  , there is an operator A p i satisfying (i)-(iv) for p = p i and for a.e. x Ω ,
A p i ( x , ξ ) A p ( x , ξ )
(22)
uniformly on compact subset of R n . We consider the obstacle problem
K ψ f , p i ( Ω , C n k ) = { v W 1 , p i ( Ω , C n k ) : v f W 0 1 , p i ( Ω , C n k ) , v ψ  a.e.  Ω } .
(23)
Assume that A p i preserves the grading of the Clifford algebra and converges to A p ( x , ξ ) uniformly on compact subset of R n . Suppose that u i W 1 , p i ( Ω , C n k ) is the weak solution to the following K ψ f , p i ( Ω , C n k ) -obstacle problem
Ω A p i ( x , D u i ) , D v D u i d x 0
(24)
for each v K ψ f , p i ( Ω , C n k ) and u 0 W 1 , p ( Ω , C n k ) is the weak solution to
Ω A p ( x , D u 0 ) , D v D u 0 d x 0
(25)

for all v K ψ f , p ( Ω , C n k ) .

Definition 4.1 We call a sequence u j L p ( Ω , C n ) converges weakly in L p ( Ω , C n ) to u if
Ω v u j d x Ω v u d x ,
(26)

whenever v L p p 1 ( Ω , C n ) . Denote it by u j u .

Remark 4.2 Suppose that u i weakly converges to u in L p ( Ω , C n k ) , then
Ω v , u i d x Ω v , u d x ,
(27)

whenever v W 1 , p ( Ω , C n k ) .

Now, we start with the main result of this section.

Theorem 4.3 Suppose that the complement of Ω satisfies the measure density condition (11). Let p i , A p i ( x , ξ ) , A p ( x , ξ ) be defined as described above. Let the boundary value function f be in W 1 , p + τ ( Ω , C n k ) for some τ > 0 . Then, there is a small number δ 0 such that the sequence { u i } of weak solutions to the obstacle problem (24) has a subsequence, which converges to the weak solution u 0 of (25) in W 1 , p + δ ( Ω , C n k ) for any δ < δ 0 .

Before proving Theorem 4.3, we need the following lemmas.

Lemma 4.4 [6]

Suppose that a bounded open set Ω satisfies (11). Let p i > 1 and u i W 1 , p i ( Ω ) , i = 0 , 1 ,  , and suppose that p i p and u i u 0 a.e in Ω. If θ W 1 , p + ϵ ( Ω ) , ϵ > 0 and if u i θ W 0 1 , p i ( Ω ) with
Ω | ( u i θ ) | p i d x M
(28)

i = 1 , 2 ,  , where M < is independent of i, then u 0 W 0 1 , p ( Ω ) .

Lemma 4.5 Let u i be described in Theorem  4.3. There exists a constant γ 0 > 0 such that for each γ [ 0 , γ 0 ] ,
Ω | D u i | p + γ 2 d x C ,
(29)

where i is sufficiently large and C is independent of i.

Proof By virtue of Remark 3.5, we can use f as a test function in (24), so that
Ω A p i ( x , D u i ) , D f D u i d x 0 .
From the structure conditions on A i and the Hölder inequality, it follows
a Ω | D u i | p i d x b ( Ω | D u i | p i d x ) p i 1 p i ( Ω | D f | p i d x ) 1 p i .
Then we have
Ω | D u i | p i d x C Ω | D f | p i d x .
(30)
By the Hölder inequality and Young’s inequality, (30) becomes
Ω | D u i | p i d x C ( | Ω | + Ω | D f | p + σ d x ) ,
(31)
when i is sufficiently large and σ > 0 . According to Theorem 3.6, there exists a constant γ 0 > 0 such that for each γ [ 0 , γ 0 ] ,
( Ω | D u i | p i + γ d x ) p i p i + γ C [ Ω | D u i | p i d x + ( Ω | D f | p i + γ d x ) p i p i + γ ] .
(32)
Since p i p , choose i sufficiently large such that p γ 2 p i p + γ 2 . Then p + γ 2 p + γ . So, there exists a constant C, such that
Ω | D u i | p + γ 2 d x C .

 □

Proof of Theorem 4.3 Let κ = n n 1 . Since u i f W 0 1 , p ( Ω , C n k ) , we choose i so large that p + γ 2 κ p i and p i p + γ 2 , using Lemma 2.4, we get
u i f p + γ 2 C u i f κ p i C u i f n p i n p i C D u i D f p i C ( D u i p + γ 2 + D f p + γ 2 ) .
Implied by the Minkowski inequality that
u i p + γ 2 u i f p + γ 2 + f p + γ 2 C ( D u i p + γ 2 + D f p + γ 2 + f p + γ 2 ) .
Then it follows from Lemma 4.5 that
u i W 1 , p + γ 2 ( Ω , C n k ) C .
(33)
Write δ = γ 2 , u i = I u I i e I , according to (33), we have u i W 1 , p + δ ( Ω , C n k ) . So, we can extract a subsequence, still denote by u I i , such that
{ u I i u I , in  W 1 , p + δ ( Ω ) , u I i u I , in  L p + δ ( Ω ) , u I i u I , pointwise a.e. in  Ω .
(34)
Then u I i u I . Let u = I u I e I , then u i u in L p + δ ( Ω ) ; u i u pointwise a.e. in Ω. Since u I i u I , for each j = 1 , 2 , , n , we have
Ω y α u I i x j d x Ω y α u I x j d x ,

whenever y α W 1 , p ( Ω ) . Then, Ω y D u i d x Ω y D u d x for each y W 1 , p ( Ω , C n ) . This yields D u i D u in L p ( Ω , C n ) .

The next stage is to extract a further subsequence, so that D u i D u pointwise almost everywhere in Ω.

Because of u i f W 0 1 , p i ( Ω , C n k ) and u i f u f , we get u f W 0 1 , p i ( Ω , C n k ) a.e. in Ω, that is to say, u K ψ f , p i ( Ω , C n k ) . So,
Ω A p i ( x , D u i ) , D u D u i d x 0 .
(35)
Moreover, by the convergence assumption, we obtain
A p i ( x , D u ) A p ( x , D u )

almost everywhere in Ω.

From this, if follows that u i u in W 1 , p + δ ( Ω , C n k ) for all δ < δ .

At last, we will show that u = u 0 . Using Lemma (4.4), we get u f W 0 1 , p ( Ω , C n k ) , then u K ψ f , p ( Ω , C n k ) . This yields
Ω A p ( x , D u 0 ) , D u D u 0 d x 0 .
(36)
On the other hand,
Ω A p i ( x , D u i ) , D u 0 D u i d x 0 .
Let i , since u i u in W 1 , p ( Ω , C n k ) , we get
Ω A p ( x , D u ) , D u 0 D u d x 0 .
(37)
Combining (36) and (37), we have
0 Ω A p ( x , D u ) A p ( x , D u 0 ) , D u D u 0 d x 0 .

It follows that u = u 0 . This proof is completed. □

Declarations

Acknowledgements

The authors would like to thank the anonymous referees for their time and thoughtful suggestions. The research is supported by the National Science Foundation of China (#11071048).

Authors’ Affiliations

(1)
Department of Mathematics, Harbin Institute of Technology, Harbin, China

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© Lu and Bao; licensee Springer. 2013

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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