RESEARCH Open Access
The existence of solutions to the
nonhomogeneous
A
-harmonic equation
Guanfeng Li, Yong Wang
*
and Gejun Bao
* Correspondence: mathwy@hit.
edu.cn
Department of Mathematics,
Harbin Institute of Technology,
Harbin 150001, People’s Republic
of China
Abstract
In this paper, we introduce the obstacl e 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 problem, the nonhomogeneous A-harmonic equation, exis-
tence and uniqueness of solutions
1 Introduction
In this paper, we study the nonhomogeneous
A
-harmonic equation
−divA
(
x, ∇u
(

¯
E
is a compact
subset of F. C(Ω) is the set of all continuous functions u : Ω ® ℝ. sptu is the s mal-
lest closed set such that u vanishes outside sptu.
C
k
()={ϕ :  → R : the kth - derivative of ϕ is continuous},
C
k
0
()={ϕ ∈ C
k
():sptϕ  },
C
∞
()=
∞

k=1
C
k
()
and
C
∞
0
()={ϕ ∈ C
∞
():sptϕ  }

provided the original work is properly cited .
|
|φ||
p
=
⎛
⎝


|φ|
p
dx
⎞
⎠
1/p
,
where j Î L
p
(Ω)(or L
p
(Ω; ℝ
n
)).
For  Î C
∞
(Ω), let
|
|ϕ||
1,p
=(

(Ω)ifandonlyifu Î L
p
(Ω)andthereisafunctionv Î
L
p
(Ω; ℝ
n
) and a sequence 
i
Î C
∞
(Ω), such that


|ϕ
i
− u|
p
dx → 0 and


|∇ϕ
i
− v|
p
dx → 0, i →∞
.
We call v the gradient of u in H
1,p
(Ω) and write v = ▽u.

( Ω)and
H
1
,p
0
(
)
are reflexive; see [1] for
details.
u
∈ H
1,p
loc
() if and only if u ∈ H
1,p
(

) for each open set 

 
.
The Dirichlet space L
1,p
( Ω)and
L
1,p
0
(
)
are defined as follows: u Î L

L
1
,p
0
(
)
is the set of all functions u
Î L
1,p
(Ω), for which there is a sequence
ϕ
j
∈ C
∞
0
(
)
such that ▽
j
® ▽u in L
p
(Ω; ℝ
n
).
Lemma 1.1 [2]Let 1 <p<∞ and f
i
be a bounded sequence in L
p
(Ω), i.e. f
i

(Ω), then min{u, v} and max{u, v} are in H
1,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
1,p
0
(
)
,thenmin{u, v} and max{u, v} are in
H
1,p
0
(
)
. Moreover, if
u
∈ H
1

-harmonic equation, where
A
: R
n
× R
n
→ R
n
is an
operator satisfying the following assumptions for some constants 0 < a ≤ b <∞:
Li et al. Journal of Inequalities and Applications 2011, 2011:80
/>Page 2 of 13
(I)
the mapping x → A(x, ξ) is measurable for all ξ ∈ R
n
an
d
the mapping ξ → A
(
x, ξ
)
is continuous for a.e. x ∈ R
n
;
for all ξ Î ℝ
n
and almost all x Î ℝ
n
,
(II)

1
− ξ
2
)
> 0
,
whenever ξ
1
, ξ
2
Î ℝ
n
, ξ
1
≠ ξ
2
; and
(V)
A(
x, λξ
)
= λ|λ|
p−2
A
(
x, ξ
)
whenever l Î ℝ, l ≠ 0, and f is a function satisfying f Î L
p/(p-1)
(Ω).

x, ∇u
)
= f
weakly in Ω, i.e.


(A(x, ∇u) ·∇ϕ − f ϕ)dx =
0
for all
ϕ ∈ C
∞
0
(
)
.
Afunction
u ∈ H
1
,p
loc
(
)
is a supersolution to (2.1) in Ω, if
−divA
(
x, ∇u
)
≥
f
in

(A(x, ∇u) ·∇ϕ − f ϕ)dx ≤
0
whenever
ϕ ∈ C
∞
0
(
)
is nonnegative.
Remark: If u is a solution (a supersolution or a subsolution), then u+τ is also a solu-
tion (a supersolution or a subsolution), but lu + τ, l, τ Î ℝ 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.Wejustgivetheproofinthecasethatu is a solution and the others are
similar.
(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.
(ii) Let

=

λ
∈
I

λ
and u be the solution to (2.1) in Ω
l
for each l Î I,whereI is an index

m
}, subordinate to the cover-
ing Ω
i
,suchthat
g
i
∈ C
∞
0
(
i
)
,0≤ g
i
≤ 1and
m

i
=1
g
i
≡
1
in D; see Lemma 2.3.1 in [11]. Thus,


(A(x, ∇u) ·∇ϕ − f ϕ)dx =

D

g
i
∈ C
∞
0
(
i
)
and
ϕ ∈ C
∞
0
(
)
, it is easy to see that
g
i
ϕ ∈ C
∞
0
(
i
)
. Since u is
solution in Ω
i
, we have

D
(A(x, ∇u) ·∇(g

(A(x, ∇u) ·∇ϕ − f ϕ)dx =0(respectively, ≥ 0 or ≤ 0
)
for all
ϕ ∈ H
1,p
0
(
)
(respectively, for all nonnegative
ϕ ∈ H
1,p
0
(
)
or for all nonnegative
ϕ ∈ H
1
,p
0
(
)
).
Proof.Forall
ϕ ∈ H
1
,p
0
(
)
, there is a sequence


(A(x, ∇u) ·∇ϕ
i
− fϕ
i
)dx






=








(A(x, ∇u) · (∇ϕ −∇ϕ
i
) − f(ϕ − ϕ
i
))dx





p
dx)
1−
1
p
(


|∇ϕ −∇ϕ
i
|
p
dx)
1
p
+(


|f |
p/(p−1)
dx)
1−
1
p
(


|ϕ − ϕ
i
|

→ 0,
Li et al. Journal of Inequalities and Applications 2011, 2011:80
/>Page 4 of 13
where
M =max{β(


|∇u|
p
dx)
1−
1
p
,(


|f |
p/(p−1)
dx)
1−
1
p
} <
∞
.
Since u is a solution,


(A(x, ∇u) ·∇ϕ − f ϕ)dx = lim
i→∞

1
,p
0
(
)
with compact support.
Prop osition 2.2 A function u is a solut ion 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 conver se, for each
ϕ ∈ C
∞
0
(
)
,let
+
be the positive part and 
-
be
the negative part of . Then, both 
+
and 
-
are in
H
1
,p
0
(

and


(A(x, ∇u) ·∇(−ϕ
−
) − f(−ϕ
−
))dx ≤ 0
.
By the above inequalities,


(A(x, ∇u) ·∇ϕ
+
− f ϕ
+
)dx =0and


(A(x, ∇u) ·∇ϕ
−
− f ϕ
−
)dx =0
.
Then,


(A(x, ∇u) ·∇ϕ − fϕ)dx =


∇
η =

∇u −∇v, u <
v
0, u ≥ v
.
Since u Î H
1,p
(Ω)isasupersolutionandv Î H
1,p
(Ω) is a subsolution, the following
inequalities hold,
−


(A(x, ∇u) ·∇η − f η)dx =


(A(x, ∇u) ·∇(−η) − f (−η))dx ≥ 0
,
and


(A(x, ∇v) ·∇η − f η)dx ≥ 0
.
Then, by the assumption (IV),
0 ≤



)
and Lemma 1.2, h = 0 a.e. in Ω. Thus, u ≥ v a.e. in Ω.
3 The obstacle problem
Suppose that Ω is bounded in ℝ
n
, ψ : Ω ® [-∞,∞] is a function and ϑ Î H
1,p
(Ω)). Let
K
ψ,ϑ
= K
ψ,ϑ
()={v ∈ H
1,p
():v ≥ ψ a.e. in  and v − ϑ ∈ H
1,p
0
()}
.
If ψ = ϑ, write
K
ψ
,
ψ
()=K
ψ
(
)
.
The problem is to find a function u in K

solution to the obstacle problem in
K
ψ
,ϑ
(
)
.
If u is a solution to the obstacle problem in
K
ψ
,u
(
)
, 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.
Li et al. Journal of Inequalities and Applications 2011, 2011:80
/>Page 6 of 13
(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
(

. If v Î H
1,p
(Ω) is a supersolution to (2.1) in Ω, such that
min{u, v}∈
K
ψ
,ϑ
(
)
, then v ≥ ua.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 theoreti cal basis for our work,
which is a general result in the theory of monotone operators; see [12]. Let X be a
refl exive 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
weakly in X’, i.e.
L u
j
, v→L u, v for each v ∈ X
,
(4:3)
whenever u
j
Î K converges to u Î KinX.
Proposition 4.1 Let K be a nonempty closed convex subset of X and let
L
: K ® X
’
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 Î K.
Lemma 4.1 Let x
i
be a sequence of X. For any subsequence
x
i
j
of x
i

0
weakly in X. Then, there exist ε
0
>0,
y
0
Î X’ and a subsequence
x
i
j
of x
i
, such that
Li et al. Journal of Inequalities and Applications 2011, 2011:80
/>Page 7 of 13

y
0
, x
i
j
− x
0

≥ ε
0
for each j Î N.
Obviou sly, for any su bsequence
x
i

(Ω; ℝ
n
). The norm of X is
|
|g|| = ||g
1
||
p
+ ||g
2
||
p
for all g =(g
1
, g
2
) Î X. 〈·, ·〉 is the usual pairing between X’ and X,
h, g =


(h
1
g
1
+ h
2
· g
2
)dx
,

,
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)Supposethat(v, ▽v ) Î K. Because
v ∈ K
ψ
,
ϑ
, v is in H
1.p
(Ω). Then, v Î L
p
(Ω) and ▽v Î L
p
(Ω). That means (v, ▽v) Î X. Therefore, K ⊂ X.
(ii) If (v
i
, ▽v
i
) Î K is a sequence which converges to (v, )inX, where  =(
1
, , 
n


|∇v
i
− ϕ|
p
dx → 0
.
Since
v
i
− ϑ ∈ H
1,p
0
()
,
v − ϑ ∈ H
1,p
0
(
)
and ▽ν = .
Since v
i
® v in L
p
(Ω), there exists a subsequence
v
i
j
, such that

)
+
(
1 − λ
)(
v, ∇v
)
=
(
λu +
(
1 − λ
)
v, ∇
(
λu +
(
1 − λ
)
v
))
∈ K
.
Li et al. Journal of Inequalities and Applications 2011, 2011:80
/>Page 8 of 13
Therefore, K is convex in X.
Define a mapping
L :
K → X


2
)dx =


(A(x, ∇v) · h
2
− fh
1
)dx
.
Since f Î L
p/(p-1)
(Ω), by the assumption (III) and the Hölder inequality, we have
|


(A(x, ∇v) · h
2
− fh
1
)dx|
≤


|A(x, ∇v)||h
2
|dx +


|f ||h

1
p
dx)
1
p
+(


|f |
p
p − 1
dx)
1−
1
p
(


|h
1
|
1
p
dx)
1
p
≤ M[(


|h

p
dx)
1−
1
p
,(


|f |
p
p − 1
dx)
1−
1
p
} <
∞
.
Bytheinequality(4.5),
L v
∈
X

for each (v, ▽ v) Î K. The mapping
L
is well
defined.
The following three lemmas show that
L
is monotone, coercive and weakly continu-

x, ∇u
))
−
(
−f , A
(
x, ∇v
))
=
(
0, A
(
x, ∇u
)
− A
(
x, ∇v
))
.
Since (u - v, ▽u - ▽v) Î X, by the assumption (IV), we have
L u − L v, u − v =


(A(x, ∇u) − A(x, ∇v)) · (∇u −∇v)dx ≥ 0
.
This proves the lemma.
Lemma 4.4
L
is coercive on K, i.e. there exists  Î K such that


p
p
) − β(||∇u||
p−1
p
||∇ϕ||
p
+ ||∇u||
p
||∇ϕ||
p−1
p
)
.
(4:6)
Using the inequ ality (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−1
p
)
and
|
|∇u||
p
≤ ||∇u||
p
+ ||∇ϕ −∇u||
p
.
Putting the above inequalities into (4.6), we get
L u − L ϕ, u − ϕn ≥ α2
−p
||∇u −∇ϕ||
p
p
− β2
p−1
||∇ϕ||
p
(||∇ϕ||
p−1
p
+ ||∇u −∇ϕ||
p−1
p
)
− β||∇ϕ||

Then, we have
L u − L ϕ, u − ϕ
||∇u −∇ϕ||
p
≥α2
−p
||∇u −∇ϕ||
p−1
p
− β2
p−1
||∇ϕ||
p
||∇u −∇ϕ||
p−
2
p
− β||∇ϕ||
p−1
p
− β(2
p−1
+1)||∇ϕ||
p
p
1
||∇u −∇ϕ||
p
.
(4:7)

+ ||∇u −∇ϕ||
p
= ||u − ϕ|| ≤ (C diam +1))||∇u −∇ϕ||
p
.
(4:9)
Combining the inequality ||u
j
- || ≥ ||u
j
||-|||| and (4.9), we have
(C diam +1)||∇u
j
−∇ϕ||
p
≥||u
j
− ϕ|| ≥ ||u
j
|| − ||ϕ||
.
For each sequence (u
j
, ▽u
j
) Î K with ||u
j
|| ® ∞,||▽u
j
- ▽||

p−1
||∇ϕ||
p
1
||∇u
j
−∇ϕ||
p
) →∞
,
β(2
p−1
+1)||∇ϕ||
p
p
1
||∇u −∇ϕ||
p
→ 0.
(4:10)
Li et al. Journal of Inequalities and Applications 2011, 2011:80
/>Page 10 of 13
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

j
, ▽u
j
) Î K converges to (u, ▽u) Î KinX.
Proof. Let (u
j
, ▽u
j
) Î K be any sequence that converges to an element (u, ▽u) Î 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→0forallv =(v
1
, v
2
) ∈ X
.
By the definition of
L
,
L u
j
− L u, v =


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


|A(x, ∇u
j
)|
p/(p−1)
dx ≤


(β|∇u
j
|
p−1
)
p/(p−1)
dx = β
p/(p−1)


|∇u

x, ∇u
)
weakly in L
p/(p-1)
(Ω; ℝ
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
,suchthat
A(x, ∇u

p/(p-1)
(Ω; ℝ
n
).
Consequently, for all v =(v
1
, v
2
) Î X,
L u
j
− L u, v =


(A(x, ∇u
j
) − A(x, ∇u)) · v
2
dx → 0
.
Then,
L u
j
, v→L u, v

for allv Î X. Hence,
L
is weakly continuous on K.
Based on the above lemmas, we can prove our main results.
Theorem 4.1 Let Ω ⊂ ℝ

,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 Î K.
This means that there is a function u in
K
ψ
,
ϑ
, such that


(A(x, ∇u) · (∇v −∇u) − f (v − u))dx ≥
0
Whenever

≥ u
2
a.e. in Ω and u
2
≥ u
1
a.e. in Ω. Thus, u
1
= u
2
a.e. in Ω and the uniqueness
is proved.
Theorem 4.2 Let Ω ⊂ ℝ
n
be a bounded open set and ϑ Î H
1,p
(Ω). There is a unique
function u Î H
1·p
(Ω) with
u
− ϑ ∈ H
1
,p
0
(
)
such that



ψ
,
ϑ
such that


(A(x, ∇u) · (∇v −∇u) − f (v − u))dx ≥
0
whenever
v ∈ K
ψ
,
ϑ
.
For each
ϕ ∈ H
1
,p
0
(
)
,
u
+ ϕ ≥−∞= ψ,
u
− ϕ ≥−∞= ψ,
u
+ ϕ − ϑ =(u − ϑ)+ϕ ∈ H
1,p
0

Thus,


(A(x, ∇u) ·∇ϕ − f ϕ)dx =0
.
(ii) Let u
1
and u
2
are two solutions to (2.1) with
u
i
− ϑ ∈ H
1,p
0
(
)
, i = 1, 2. Since ϑ Î
H
1,p
(Ω), u
1
, u
2
Î H
1,p
(Ω)and
u
1
− u

is a supersolution and u
2
is a subsolution. By Lemma 2.2, u
1
≥ u
2
a.e in Ω.
Similarly, u
2
≥ u
1
a.e in Ω. Thus, u
1
= u
2
a.e. in Ω and the uniqueness is proved.
Acknowledgements
This work was supported by the National Natural Science Foundation of China (Grant No. 11071048).
Authors’ contributions
All authors contributed equally in this paper. They read and approved the final manuscript.
Competing interests
The authors declare that they have no competing interests.
Received: 13 April 2011 Accepted: 7 October 2011 Published: 7 October 2011
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