Open Access

The existence of solutions to the nonhomogeneous A -harmonic equation

Journal of Inequalities and Applications20112011:80

https://doi.org/10.1186/1029-242X-2011-80

Received: 13 April 2011

Accepted: 7 October 2011

Published: 7 October 2011

Abstract

In this paper, we introduce the obstacle problem about the nonhomogeneous A -harmonic equation. Then, we prove the existence and uniqueness of solutions to the nonhomogeneous A -harmonic equation and the obstacle problem.

Keywords

the obstacle problemthe nonhomogeneous A -harmonic equationexistence and uniqueness of solutions

1 Introduction

In this paper, we study the nonhomogeneous A -harmonic equation
- div A ( x , u ( x ) ) = f ( x ) ,

where A : n × n n is an operator and f is a function satisfying some assumptions given in the next section. We give the definition of solutions to the nonhomogeneous A -harmonic equation and the obstacle problem. In the mean time, we show some properties of their solutions. Then, we prove the existence and uniqueness of solutions to the Dirichlet problem for the nonhomogeneous A -harmonic equation with Sobolev boundary values.

Let n be the real Euclidean space with the dimension n. Throughout this paper, all the topological notions are taken with respect to n . E F means that Ē is a compact subset of F. C(Ω) is the set of all continuous functions u : Ω → . spt u is the smallest closed set such that u vanishes outside spt u.
C k ( Ω ) = { φ : Ω : the k th - derivative of φ is continuous } , C 0 k ( Ω ) = { φ C k ( Ω ) : spt φ Ω } , C ( Ω ) = k = 1 C k ( Ω )
and
C 0 ( Ω ) = { φ C ( Ω ) : spt φ Ω } .
Let L p (Ω) = {φ: Ω → : ∫Ω |φ| p dx < ∞} and L p (Ω; n ) = {φ: Ω → n : ∫Ω |φ| p dx < ∞}, 1 < p < ∞. Denote the norm of L p (Ω) and L p (Ω; n ) by || · || p ,
| | ϕ | | p = Ω | ϕ | p d x 1 p ,

where ϕ L p (Ω)(or L p (Ω; n )).

For φ C(Ω), let
| | φ | | 1 , p = ( Ω | φ | p d x ) 1 p + ( Ω | φ | p d x ) 1 p ,
where φ = (∂1φ, ..., ∂ n φ) is the gradient of φ. The Sobolev space H1,p(Ω) is defined to be the completion of the set {φ C(Ω): ||φ||1,p< ∞} with respect to the norm || · ||1,p. In other words, u H1,p(Ω) if and only if u L p (Ω) and there is a function v L p (Ω; n ) and a sequence φ i C(Ω), such that
Ω | φ i - u | p d x 0 and Ω | φ i - v | p d x 0 , i .

We call v the gradient of u in H1,p(Ω) and write v = u.

The space H 0 1 , p ( Ω ) is the closure of C 0 ( Ω ) in H1,p(Ω). Obviously, H1,p(Ω) and H 0 1 , p ( Ω ) are Banach space with respect to the norm ||·||1,p. Moreover, ||·||1,pis uniformly convex and the Sobolev space H1,p(Ω) and H 0 1 , p ( Ω ) are reflexive; see [1] for details.
u H l o c 1 , p ( Ω ) if and only if  u H 1 , p ( Ω ) for each open set  Ω Ω .

The Dirichlet space L1,p(Ω) and L 0 1 , p ( Ω ) are defined as follows: u L1,p(Ω) if and only if u H l o c 1 , p ( Ω ) and u L p (Ω); L 0 1 , p ( Ω ) is the closure of C 0 ( Ω ) with respect to the semi-norm p(u) = (∫Ω|u| p )1/p. In other words, L 0 1 , p ( Ω ) is the set of all functions u L1,p(Ω), for which there is a sequence φ j C 0 ( Ω ) such that φ j u in L p (Ω; n ).

Lemma 1.1 [2] Let 1 < p <and f i be a bounded sequence in L p (Ω), i.e. f i L p (Ω) and sup i | | f i | | p < . If f i f a.e. in Ω, then f i converges to f weakly in L p (Ω).

Lemma 1.2 [3] (1) If u H 0 1 , p ( Ω ) withu = 0, then u = 0.

(2) If u, v H1,p(Ω), then min{u, v} and max{u, v} are in H1,p(Ω) with
max { u , v } = u , u v v , u v and min { u , v } = v , u v u , u v .

(3) If u , v H 0 1 , p ( Ω ) , then min{u, v} and max{u, v} are in H 0 1 , p ( Ω ) . Moreover, if u H 0 1 , p ( Ω ) is nonnegative, then there is a sequence of nonnegative functions φ i C 0 ( Ω ) converging to u in H1,p(Ω).

2 The nonhomogeneous A -harmonic equation

The following nonlinear elliptic equation
- div A ( x , u ) = f ( x )
(2.1)

is called the nonhomogeneous A -harmonic equation, where A : n × n n is an operator satisfying the following assumptions for some constants 0 < αβ < ∞:

(I)    the mapping  x A ( x , ξ ) is measurable for all  ξ n and the mapping  ξ A ( x , ξ ) is continuous for a .e x n ;

for all ξ n and almost all x n ,

(II)    A ( x , ξ ) ξ α | ξ | p ,

(III)    | A ( x , ξ ) | β | ξ | p - 1 ,

(IV)    ( A ( x , ξ 1 ) - A ( x , ξ 2 ) ) ( ξ 1 - ξ 2 ) > 0 ,

whenever ξ1, ξ2 n , ξ1ξ2; and

(V)    A ( x , λ ξ ) = λ | λ | p - 2 A ( x , ξ )

whenever λ , λ ≠ 0, and f is a function satisfying f Lp/(p-1)(Ω).

If f = 0, the equation (2.1) degenerates into the homogeneous A -harmonic equation
- div A ( x , u ( x ) ) = 0 .
(2.2)

A continuous solution to (2.2) in Ω is called A -harmonic function. Many well-known results have been developed about (2.2), especially as (2.2) is the corresponding A -harmonic equation of differential forms; see [410].

Definition 2.1 A function u H l o c 1 , p ( Ω ) is a (weak) solution to the equation (2.1) in Ω, if - div A ( x , u ) = f weakly in Ω, i.e.
Ω ( A ( x , u ) φ - f φ ) d x = 0

for all φ C 0 ( Ω ) .

A function u H l o c 1 , p ( Ω ) is a supersolution to (2.1) in Ω, if - div A ( x , u ) f in weakly Ω, i.e.
Ω ( A ( x , u ) φ - f φ ) d x 0

whenever φ C 0 ( Ω ) is nonnegative.

A function u H l o c 1 , p ( Ω ) is a subsolution to (2.1) in Ω, if - div A ( x , u ) f weakly in Ω, i.e.
Ω ( A ( x , u ) φ - f φ ) d x 0

whenever φ C 0 ( Ω ) is nonnegative.

Remark: If u is a solution (a supersolution or a subsolution), then u+τ is also a solution (a supersolution or a subsolution), but λu + τ, λ, τ may not.

Proposition 2.1 A function u is a solution (a supersolution or a subsolution) to (2.1) in Ω if and only if Ω can be covered by open sets where u is a solution (a supersolution or a subsolution).

Proof. We just give the proof in the case that u is a solution and the others are similar.
  1. (i)

    Since Ω is covered by itself, it is easy to know that Ω can be covered by open sets where u is a solution.

     
  2. (ii)
    Let Ω = λ I Ω λ and u be the solution to (2.1) in Ω λ for each λ I, where I is an index set. For each φ C 0 ( Ω ) , there is a subset {Ω1, ..., Ω m } of {Ω λ }λIsuch that spt φ i = 1 m Ω i = D . Choose a partition of unity of D, {g1, ..., g m }, subordinate to the covering Ω i , such that g i C 0 ( Ω i ) , 0 ≤ g i ≤ 1 and i = 1 m g i 1 in D; see Lemma 2.3.1 in [11]. Thus,
    Ω ( A ( x , u ) φ - f φ ) d x = D ( A ( x , u ) φ - f φ ) d x = D ( A ( x , u ) ( i = 1 m g i φ ) - f ( i = 1 m g i φ ) ) d x = i = 1 m D ( A ( x , u ) ( g i φ ) - g i φ f ) d x .
     
Note that g i C 0 ( Ω i ) and φ C 0 ( Ω ) , it is easy to see that g i φ C 0 ( Ω i ) . Since u is solution in Ω i , we have
D ( A ( x , u ) ( g i φ ) - g i φ f ) d x = Ω i ( A ( x , u ) ( g i φ ) - g i φ f ) d x = 0 .
Therefore,
Ω ( A ( x , u ) φ - f φ ) d x = 0 .

It means that u is a solution in Ω.

Lemma 2.1 If u L 1 , p ( Ω ) is a solution (respectively, a supersolution or a subsolution) to (2.1), then
Ω ( A ( x , u ) φ - f φ ) d x = 0 ( respectively , 0 or 0 )

for all φ H 0 1 , p ( Ω ) (respectively, for all nonnegative φ H 0 1 , p ( Ω ) or for all nonnegative φ H 0 1 , p ( Ω ) ).

Proof. For all φ H 0 1 , p ( Ω ) , there is a sequence φ i C 0 ( Ω ) , such that φ i φ in H1,p(Ω).

Since A satisfies the assumption (III), f Lp/(p-1)(Ω) and u Ω L1,p(Ω), it follows that
Ω ( A ( x , u ) φ - f φ ) d x - Ω ( A ( x , u ) φ i - f φ i ) d x = Ω ( A ( x , u ) ( φ - φ i ) - f ( φ - φ i ) ) d x Ω | A ( x , u ) | | φ - φ i | d x + Ω | f | | φ - φ i | d x β Ω | u | p - 1 | φ - φ i | d x + Ω | f | | φ - φ i | d x β ( Ω | u | p d x ) 1 - 1 p ( Ω | φ - φ i | p d x ) 1 p + ( Ω | f | p ( p - 1 ) d x ) 1 - 1 p ( Ω | φ - φ i | p d x ) 1 p M { ( Ω | φ - φ i | p d x ) 1 p + ( Ω | φ - φ i | p d x ) 1 p } = M | | φ - φ i | | 1 , p 0 ,

where M = max { β ( Ω | u | p d x ) 1 - 1 p , ( Ω | f | p ( p - 1 ) d x ) 1 - 1 p } < .

Since u is a solution,
Ω ( A ( x , u ) φ - f φ ) d x = lim i Ω ( A ( x , u ) φ i - f φ i ) d x = 0 .

If u L1,p(Ω) is a supersolution or a subsolution, by Lemma 1.2, there is a sequence of nonnegative functions φ i C 0 ( Ω ) converging to the nonnegative function φ in H1,p(Ω). By the same discussion, the lemma follows.

Remark: Using the similar method as above, it is easy to prove that, if u is a solution (a supersolution or a subsolution),
Ω ( A ( x , u ) φ - f φ ) d x = 0 ( 0 or 0 )

for all (nonegative) φ H 0 1 , p ( Ω ) with compact support.

Proposition 2.2 A function u is a solution to (2.1) if and only if u is a supersolution and a subsolution.

Proof. Obviously, u is both a supersolution and a subsolution if u is a solution.

To establish the converse, for each φ C 0 ( Ω ) , let φ+ be the positive part and φ- be the negative part of φ. Then, both φ+ and φ- are in H 0 1 , p ( Ω ) and have compact support. Since u is both a supersolution and a subsolution and φ+ ≥ 0, -φ- ≥ 0, the following inequalities hold,
Ω ( A ( x , u ) φ + - f φ + ) d x 0 , Ω ( A ( x , u ) ( - φ - ) - f ( - φ - ) ) d x 0 , Ω ( A ( x , u ) φ + - f φ + ) d x 0
and
Ω ( A ( x , u ) ( - φ - ) - f ( - φ - ) ) d x 0 .
By the above inequalities,
Ω ( A ( x , u ) φ + - f φ + ) d x = 0 and Ω ( A ( x , u ) φ - - f φ - ) d x = 0 .
Then,
Ω ( A ( x , u ) φ - f φ ) d x = Ω ( A ( x , u ) φ + - f φ + ) d x + Ω ( A ( x , u ) φ - - f φ - ) d x = 0

This proves that u is a solution to (2.1).

Lemma 2.2 (Comparison Lemma) Let u H1,p(Ω) be a supersolution and v H1,p(Ω) be a subsolution to (2.1). If η = min { u - v , 0 } H 0 1 , p ( Ω ) , then uv a.e. in Ω.

Proof. By η = min {u - v, 0} and Lemma 1.2, η ≤ 0 and η = u - v , u < v 0 , u v .

Since u H1,p(Ω) is a supersolution and v H1,p(Ω) is a subsolution, the following inequalities hold,
- Ω ( A ( x , u ) η - f η ) d x = Ω ( A ( x , u ) ( - η ) - f ( - η ) ) d x 0 ,
and
Ω ( A ( x , v ) η - f η ) d x 0 .
Then, by the assumption (IV),
0 Ω ( A ( x , v ) · η f η ) d x Ω ( A ( x , u ) η f η ) d x = Ω ( A ( x , v ) ( A ( x , u ) ) η d x = { u < v } ( A ( x , v ) ( A ( x , u ) ) ( u v ) d x = { u < v } ( A ( x , v ) ( A ( x , u ) ) ( v u ) d x = { u < v } { u v } ( A ( x , v ) ( A ( x , u ) ) ( v u ) d x 0

Therefore, the Lebesgue measure of the set {u < v} ∩ {uv} is zero. That is η = 0 a.e. in Ω By η H 0 1 , p ( Ω ) and Lemma 1.2, η = 0 a.e. in Ω. Thus, u ≥ v a.e. in Ω.

3 The obstacle problem

Suppose that Ω is bounded in n , ψ : Ω → [-∞,∞] is a function and ϑ H1,p(Ω)). Let
K ψ , ϑ = K ψ , ϑ ( Ω ) = { v H 1 , p ( Ω ) : v ψ a.e.  in Ω and v - ϑ H 0 1 , p ( Ω ) } .

If ψ = ϑ, write K ψ , ψ ( Ω ) = K ψ ( Ω ) .

The problem is to find a function u in K ψ , ϑ such that
Ω ( A ( x , u ) ( v - u ) - f ( v - u ) ) d x 0
(3.1)

whenever v K ψ , ϑ . We call the function ψ an obstacle.

Definition 3.1 If a function u K ψ , ϑ ( Ω ) satisfies (3.1) for all v K ψ , ϑ ( Ω ) , we say that u is a solution to the obstacle problem with obstacle ψ and boundary vales ϑ or a solution to the obstacle problem in K ψ , ϑ ( Ω ) .

If u is a solution to the obstacle problem in K ψ , u ( Ω ) , we say that u is a solution to the obstacle problem with obstacle ψ.

Proposition 3.1 (1) A solution u to the obstacle problem is always a supersolution to (2.1) in Ω.

(2) If u is a supersolution to (2.1) in Ω, u is a solution to the obstacle problem in K u , u ( D ) for each open sets D Ω. Moreover, if Ω is bounded, u is a solution to the obstacle problem in K u , u ( Ω ) .

(3) A solution u to the obstacle problem in K - , u ( Ω ) is a solution to (2.1) in Ω.

(4) If u is a solution to (2.1) in Ω, u is a solution to the obstacle problem in K - , u ( D ) for each open set D Ω. Moreover, if Ω is bounded, u is a solution to the obstacle problem in K - , u ( Ω ) .

Theorem 3.1 Suppose u is a solution to the obstacle problem in K ψ , ϑ ( Ω ) . If v H1,p(Ω) is a supersolution to (2.1) in Ω, such that min { u , v } K ψ , ϑ ( Ω ) , then v ≥ u a.e. in Ω.

The proof is similar to Lemma 2.2.

4 The existence of solutions

In this section, we introduce the main work of this paper, to prove the existence and the uniqueness of solutions to the nonhomogeneous A - harmonic equation. We can see this work for the A -harmonic equation (2.2) in [3, Chapter 3 and Appendix I] for details. We use the similar method to prove our results.

First, we introduce the following proposition as the theoretical basis for our work, which is a general result in the theory of monotone operators; see [12]. Let X be a reflexive Banach space and denote its dual by X'. Let || · || be the norm of X and 〈·, ·〉 be a pairing between X' and X. K is a closed convex subset of X.

Definition 4.1 A mapping L : K X is called monotone, if
L u - L v , u - v 0
(4.1)

for all u, v in K.

L
is called coercive on K, if there exists φ K such that
L u j - L φ , u j - φ | | u j - φ | |
(4.2)

for each sequence u j in K with ||u j || → ∞.

L
is called weakly continuous on K, if L u j converges to L u weakly in X', i.e.
L u j , v L u , v for each v X ,
(4.3)

whenever u j K converges to u K in X.

Proposition 4.1 Let K be a nonempty closed convex subset of X and let L : KX ' be monotone, coercive and weakly continuous on K. Then there exists an element u in K such that
L u , v - u 0
(4.4)

whenever v K.

Lemma 4.1 Let x i be a sequence of X. For any subsequence x i j of x i , there is a subsequence x i j k of, x i j such that x i j k converges to x0 weakly in X and the weak limit x0 is independent of the choice of the subsequence of x i . Then x i converges to x0 weakly in X.

Proof. Suppose that x i does not converge to x0 weakly in X. Then, there exist ε0 > 0, y0 X' and a subsequence x i j of x i , such that
y 0 , x i j - x 0 ε 0

for each j .

Obviously, for any subsequence x i j k of x i j , x i j k cannot converge to x0 weakly in X. This contradicts the condition of the lemma.

Therefore, x i converges to x0 weakly in X.

Now let X = L p (Ω) × L p (Ω; n ). Then, X is a reflexive Banach space and its dual X' = Lp/(p- 1)(Ω) × Lp/(p- 1)(Ω; n ). The norm of X is
| | g | | = | | g 1 | | p + | | g 2 | | p
for all g = (g1, g2) X. 〈·, ·〉 is the usual pairing between X' and X,
h , g = Ω ( h 1 g 1 + h 2 g 2 ) d x ,

where g = (g1, g2) is in X and h = (h1, h2) in X'.

Let Ω be a bounded open set in n , ϑ H1,p(Ω) and ψ: Ω [-∞,∞] be any function. Construct the obstacle set
K ψ , ϑ = K ψ , ϑ ( Ω ) = { v H 1 , p ( Ω ) : v ψ a.e.  in Ω and v - ϑ H 0 1 , p ( Ω ) }

and suppose that K ψ , ϑ is not empty.

Let K = { ( v , v ) : v K ψ , ϑ } . Then, K is also not empty.

Lemma 4.2 K is a nonempty closed convex subset of X.

Proof. (i) Suppose that (v, v) K. Because v K ψ , ϑ , v is in H1.p(Ω). Then, v L p (Ω) and v L p (Ω). That means (v, v) X. Therefore, K X.
  1. (ii)
    If (v i , v i ) K is a sequence which converges to (v, φ) in X, where φ = (φ1, ..., φ n ) L p (Ω; n ), it follows that
    Ω | ( v i - ϑ ) - ( v - ϑ ) | p d x = Ω | v i - v | p d x 0 ,
     
and
Ω | ( v i - ϑ ) - ( φ - ϑ ) | p d x = Ω | v i - φ | p d x 0 .

Since v i - ϑ H 0 1 , p ( Ω ) , v - ϑ H 0 1 , p ( Ω ) and ν = φ.

Since v i v in L p (Ω), there exists a subsequence v i j , such that v i j v a.e. in Ω. By v i ≥ ψ a.e. in Ω, v ≥ ψ a.e. in Ω.

By the argumentation above, we have v K ψ , ϑ and v = φ. So (v, φ) = (v, v) K. This means K is closed in X.
  1. (iii)
    Let (u, u) K, (v, v) K and λ [0, 1].
    λ u + ( 1 - λ ) v λ ψ + ( 1 - λ ) ψ = ψ a.e. in Ω , λ u + ( 1 - λ ) v - ϑ = λ ( u - ϑ ) + ( 1 - λ ) ( v - ϑ ) H 0 1 , p ( Ω ) .
     
It means that λu + (1 - λ)v K ψ , ϑ . Then,
λ ( u , u ) + ( 1 - λ ) ( v , v ) = ( λ u + ( 1 - λ ) v , ( λ u + ( 1 - λ ) v ) ) K .

Therefore, K is convex in X.

Define a mapping L : K X by L ( v , v ) = ( - f , A ( x , v ) ) for each (v, v) K. For convenience, we denote L ( v , v ) simply by L v . For any element h = (h1, h2) X,
L v , h = Ω ( ( - f ) h 1 + A ( x , v ) h 2 ) d x = Ω ( A ( x , v ) h 2 - f h 1 ) d x .
Since f Lp/(p- 1)(Ω), by the assumption (III) and the Hölder inequality, we have
| Ω ( A ( x , v ) h 2 - f h 1 ) d x | Ω | A ( x , v ) | | h 2 | d x + Ω | f | | h 1 | d x β Ω | v | p - 1 | h 2 | d x + Ω | f | | h 1 | d x β ( Ω | v | p d x ) 1 - 1 p ( Ω | h 2 | 1 p d x ) 1 p + ( Ω | f | p p - 1 d x ) 1 - 1 p ( Ω | h 1 | 1 p d x ) 1 p M [ ( Ω | h 2 | 1 p d x ) 1 p + ( Ω | h 1 | 1 p d x ) 1 p ] = M | | h | | ,
(4.5)

where M = max { β ( Ω | v | p d x ) 1 - 1 p , ( Ω | f | p p - 1 d x ) 1 - 1 p } < .

By the inequality (4.5), L v X for each (v, v) K. The mapping L is well defined.

The following three lemmas show that L is monotone, coercive and weakly continuous on K.

Lemma 4.3 L is monotone on K, i.e. L u - L v , u - v 0 for all(u, u), (v, v) in K.

Proof. For all (u, u), (v, v) in K, L u = ( - f , A ( x , u ) ) and L v = ( - f , A ( x , v ) ) .

Then, L u - L v = ( - f , A ( x , u ) ) - ( - f , A ( x , v ) ) = ( 0 , A ( x , u ) - A ( x , v ) ) .

Since (u - v, u - v) X, by the assumption (IV), we have
L u - L v , u - v = Ω ( A ( x , u ) - A ( x , v ) ) ( u - v ) d x 0 .

This proves the lemma.

Lemma 4.4 L is coercive on K, i.e. there exists φ K such that
L u j - L φ , u j - φ | | u j - φ | |

for each sequence u j in K with ||u j || → ∞.

Proof. Fix (φ, φ) K. For each (u, u) K, by assumptions (II), (III) and the Hölder inequality,
L u - L φ , u - φ = Ω ( A ( x , u ) - A ( x , φ ) ) ( u - φ ) d x α ( | | u | | p p + | | φ | | p p ) - β ( | | u | | p p - 1 | | φ | | p + | | u | | p | | φ | | p p - 1 ) .
(4.6)
Using the inequality (a + b) r ≤ 2 r (a r + b r ) for all a ≥ 0, b ≥ 0 and r > 0, the following inequalities hold.
| | u + φ | | p p ( | | u | | p + | | φ | | p ) p 2 p ( | | u | | p p + | | φ | | p p ) , | | u | | p p - 1 ( | | u | | p + | | φ - u | | p ) p - 1 2 p - 1 ( | | u | | p p - 1 + | | u - φ | | p p - 1 )
and
| | u | | p | | u | | p + | | φ - u | | p .
Putting the above inequalities into (4.6), we get
u φ , u φ n α 2 p | | u φ | | p p β 2 p 1 | | φ | | p ( | | φ | | p p 1 + | | u φ | | p p 1 ) β | | φ | | p p 1 ( | | φ | | p + | | u φ | | p ) = α 2 p | | u φ | | p p β 2 p 1 | | φ | | p | | u φ | | p p 1 β | | φ | | p p 1 | | u φ | | p β ( 2 p 1 + 1 ) | | φ | | p p .
Then, we have
L u - L φ , u - φ | | u - φ | | p α 2 - p | | u - φ | | p p - 1 - β 2 p - 1 | | φ | | p | | u - φ | | p p - 2 - β | | φ | | p p - 1 - β ( 2 p - 1 + 1 ) | | φ | | p p 1 | | u - φ | | p .
(4.7)

Since (u,u), (φ, φ) K, both u and φ are in K ψ , ϑ . Thus, u - φ = u - ϑ - ( φ - ϑ ) H 0 1 , p ( Ω ) .

By the Poincaré inequality,
| | u - φ | | p C diam Ω | | u - φ | | p ,
(4.8)

where C is a constant independent of u and φ.

By the definition of the norm of X and the inequality (4.8), we obtain
| | u - φ | | p | | u - φ | | p + | | u - φ | | p = | | u - φ | | ( C diam Ω + 1 ) ) | | u - φ | | p .
(4.9)
Combining the inequality ||u j - φ|| ≥ ||u j || - ||φ|| and (4.9), we have
( C diam Ω + 1 ) | | u j - φ | | p | | u j - φ | | | | u j | | - | | φ | | .

For each sequence (u j , u j ) K with ||u j || → ∞, ||u j - φ|| p → ∞.

Thus,
α 2 - p | | u j - φ | | p p - 1 - β 2 p - 1 | | φ | | p | | u j - φ | | p p - 2 = | | u j - φ | | p p - 1 ( α 2 - p - β 2 p - 1 | | φ | | p 1 | | u j - φ | | p ) , β ( 2 p - 1 + 1 ) | | φ | | p p 1 | | u - φ | | p 0 .
(4.10)
Combining (4.10) with (4.7), we obtain
L u - L φ , u - φ | | u - φ | | p .
Using (4.9), we conclude
L u - L φ , u - φ | | u - φ | | L u - L φ , u - φ ( C diam Ω + 1 ) | | u - φ | | p .

It follows that L is coercive on K.

Lemma 4.5 L is weakly continuous on K, that means L u j converges to L u weakly in X', i.e.
L u j , v L u , v for all v = ( v 1 , v 2 ) X ,
(4.11)

whenever (u j , u j ) K converges to (u, u) K in X.

Proof. Let (u j , u j ) K be any sequence that converges to an element (u, u) K in X. It suffices to prove that L u j converges to L u weakly in X', i.e.
L u j - L u , v 0 for all v = ( v 1 , v 2 ) X .
By the definition of L ,
L u j - L u , v = Ω ( A ( x , u j ) - A ( x , u ) ) v 2 d x .

By the definition of X and (u j , u j ) (u, u) in X, u j u in L p (Ω; n ). There exists a subsequence u j k such that u j k u a.e. in Ω.

Since A satisfies the assumption (I), A ( x , u j k ( x ) ) A ( x , u ( x ) ) a.e. in Ω.

By the assumption (III), we obtain
Ω | A ( x , u j ) | p / ( p 1 ) d x Ω ( β | u j | p 1 ) p / ( p 1 ) d x = β p / ( p 1 ) Ω | u j | p d x .
(4.12)

Since u j u in L p (Ω; n ), (4.12) shows Lp/(p-1)(Ω; n ) - norms of A ( x , u j ) are uniformly bounded. By Lemma 1.1, A ( x , u j k ) converges to A ( x , u ) weakly in Lp/(p-1)(Ω; n).

By the same discussion, we know that, for any subsequence u j k of u j , there exists a subsequence u j k l of u j k , such that A ( x , u j k l ) converges to A ( x , u ) weakly in Lp/(p-1)(Ω; n ).

Since the weak limit A ( x , u ) is independent of the choice of the subsequence and by Lemma 4.1, it follows that A ( x , u j ) converges to A ( x , u ) weakly in Lp/(p-1)(Ω; n).

Consequently, for all v = (v1, v2) X,
L u j - L u , v = Ω ( A ( x , u j ) - A ( x , u ) ) v 2 d x 0 .

Then, L u j , v L u , v for all v X. Hence, L is weakly continuous on K.

Based on the above lemmas, we can prove our main results.

Theorem 4.1 Let Ω n is a bounded open set, ϑ H1,p(Ω) and ψ: Ω [-∞, ∞] be any function. If K ψ , ϑ , then there is a unique function u in K ψ , ϑ , such that
Ω ( A ( x , u ) ( v - u ) - f ( v - u ) ) d x 0

whenever v K ψ , ϑ . That is, if K ψ , ϑ , there is a unique solution u to the obstacle problem in K ψ , ϑ .

Proof. (i) Construct X, K and L as Lemmas 4.2, 4.3, 4.4 and 4.5. By the proposition (4.1) and Lemmas 4.2, 4.3, 4.4 and 4.5, there exists an element u in K such that
L u , v - u 0

whenever ν K.

This means that there is a function u in K ψ , ϑ , such that
Ω ( A ( x , u ) ( v - u ) - f ( v - u ) ) d x 0
Whenever v K ψ , ϑ .
  1. (ii)

    Suppose that u1 and u2 are two solutions to the obstacle problem in K ψ , ϑ . Then min { u 1 , u 2 } K ψ , ϑ and both u1 and u2 are supersolutions to (2.1) in Ω. By Lemma 3.1, u1u2 a.e. in Ω and u2u1 a.e. in Ω. Thus, u1 = u2 a.e. in Ω and the uniqueness is proved.

     
Theorem 4.2 Let Ω n be a bounded open set and ϑ H1,p(Ω). There is a unique function u Hp(Ω) with u - ϑ H 0 1 , p ( Ω ) such that
Ω ( A ( x , u ) ( φ ) - f φ ) d x = 0

whenever φ H 0 1 , p ( Ω ) . That is u is the unique solution to (2.1) with u - ϑ H 0 1 , p ( Ω ) .

Proof. (i) Choose ψ ≡ -∞. Since ϑ ≥ -∞ = ψ and ϑ - ϑ = 0 H 0 1 , p ( Ω ) , ϑ K ψ , ϑ .

By Theorem 4.1, there is a function u in K ψ , ϑ such that
Ω ( A ( x , u ) ( v - u ) - f ( v - u ) ) d x 0

whenever v K ψ , ϑ .

For each φ H 0 1 , p ( Ω ) ,
u + φ - = ψ , u - φ - = ψ , u + φ - ϑ = ( u - ϑ ) + φ H 0 1 , p ( Ω ) , u - φ - ϑ = ( u - ϑ ) - φ H 0 1 , p ( Ω ) .
Therefore, both u + φ and u - ϑ are in K ψ , ϑ for all φ H 0 1 , p ( Ω ) Then,
Ω ( A ( x , u ) φ - f φ ) d x 0
and
Ω ( A ( x , u ) ( - φ ) - f ( - φ ) ) d x 0 .
Thus,
Ω ( A ( x , u ) φ - f φ ) d x = 0 .
  1. (ii)

    Let u1 and u2 are two solutions to (2.1) with u i - ϑ H 0 1 , p ( Ω ) , i = 1, 2. Since ϑ H1,p(Ω), u1, u2 H1,p(Ω) and u 1 - u 2 = ( u 1 - ϑ ) - ( u 2 - ϑ ) H 0 1 , p ( Ω ) . Then, η 1 = min { u 1 - u 2 , 0 } H 0 1 , p ( Ω ) . Since u1 and u2 are two solutions and by Proposition 2.2, u1 is a supersolution and u2 is a subsolution. By Lemma 2.2, u1u2 a.e in Ω. Similarly, u2u1 a.e in Ω. Thus, u1 = u2 a.e. in Ω and the uniqueness is proved.

     

Declarations

Acknowledgements

This work was supported by the National Natural Science Foundation of China (Grant No. 11071048).

Authors’ Affiliations

(1)
Department of Mathematics, Harbin Institute of Technology

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© Li et al; licensee Springer. 2011

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