BOUNDARY REGULARITY OF WEAK SOLUTIONS TO
NONLINEAR ELLIPTIC OBSTACLE PROBLEMS
MENG JUNXIA AND CHU YUMING
Received 25 April 2005; Revised 10 Septe mber 2005; Accepted 14 September 2005
We study the boundary regularity of weak solutions to nonlinear obstacle problem with
C
1,β
-obstacle function, and obtain the C
loc
1,α
boundary regularity.
Copyright © 2006 M. Junxia and C. Yuming. This is an open access article distributed
under the Creative Commons Attribution License, which permits unrestricted use, dis-
tribution, and reproduction in any medium, provided the original work is properly cited.
1. Introduction
We consider the following variational inequality:
u
∈:

Ω
A(x, ∇u) · (∇v −∇u)dx
≥

Ω
H(x,u,∇u)(v − u)dx +

Ω
F(x, u) · (∇v −∇u)dx
(1.1)
for all v
∈={v ∈W

i
(x, h)/∂x
j
|≤τ
1
|h|
p−1
;
(v)

N
i, j
=1
a
ij
ξ
i
ξ
j
≥ τ
2
|h|
p−2
|ξ|
2
;
(vi)
|A(x, ξ) − A(y,ξ)|≤b
1
(1 + |ξ|

Volume 2006, Article ID 72012, Pages 1–15
DOI 10.1155/BVP/2006/72012
2 Boundary regularity
We assume that H(x,u, λ), F(x,u)
={F
i
(x, u)}
1≤i≤N
in (1.1) are of the form:


H(x,u,∇u)


≤
c

|∇
u|
p/r

+ |u|
r−1
+ g(x)

, (1.2)


F(x, u)


K(ψ)
=

v ∈ W
1,p
(Ω):v ≥ ψ a.e.

, (1.5)
has been studied by various authors. In the case when ψ is assumed to have only minimal
regularity properties, it was shown by [8, 11] that the solution of (1.1)iscontinuous.In
particular, if ψ
∈ C
0,α
(Ω), then the s olution u is also an element of C
0,α

(Ω). In the case
when ψ
∈ C
2
(Ω), papers [4, 6, 10, 12] employed different techniques to prove interior
C
1,α
(Ω) regularity for the solution u to (1.4). Reference [1] gave an interesting result: the
condition for
 to be nonempty is just that ψ should have finite capacity. This implies,
among other things, that ψ
+
= max(ψ,0) must vanish on ∂Ω, C—almost everywhere.
This condition is important for the existence of weak solutions to obstacle problem.

B
R
(z) =

x ∈ R
N
: |x − z| <R

, Γ
R
(z) =

x ∈ B
R
(z):x
n
= 0

,
B
+
R
(z) =

x ∈ B
R
(z):x
n
> 0


E its closure, and by
|E| its Lebesgue measure. ( f )
R
= (1/|B
R
|)

B
R
f (x)dx.The
M. Junxia and C. Yuming 3
letter c is used throughout to denote a positive constant, not necessarily the same at each
occurrence.
Since
Ω is compact, ∂Ω can be covered by a finite number of neighbourhoods V of its
points. It is enough to prove the better regularity of u holds true in V
∩ Ω.Since∂Ω is a
Lipschitz boundary, one can find T which is an invertible Lipschitz mapping such that
T(V)
= B, T(V ∩ Ω) = B
+
, T(V\Ω) = B
−
, T(V ∩ ∂Ω) = Γ. (2.2)
Under the mapping T the variational inequality in Ω is transfor med to a variational
inequality of the same form in B
+
,for
¯
u = u ◦ T

¯
F(x,
¯
u)
· (∇v −∇
¯
u)dx,
∀v ∈
¯
,
(2.3)
where
¯
={v ∈ W
1,p
0
(B
+
), v ≥ ψ,a.e.inB
+
},
¯
A,
¯
H,
¯
F satisfy assumptions of type (i)–
(vii), (1.2), (1.3)withdifferent constants.
In order to simplify the notations, we still denote
¯

, ,x
n−1
,−x
n

+4u

x
1
, ,x
n−1
,−
x
n
2

,ifx ∈ B
−
.
(2.4)
In light of Extension theorem [5, page 254], we only need to prove a better regularity of
u in B
+
.
Definit ion 2.1. The function u
∈that satisfies (2.3)forallv ∈is called a weak solu-
tion to the obstacle problem with obstacle ψ.
Definit ion 2.2. Call f
∈ C
0,α

∈ Γ to be understood.
In the following, we will use some lemmas which we state below.
Lemma 2.3. Let w
∈ W
1,p
(B
+
R
) beasolutionoftheDirichletproblem

B
+
R
A(x, ∇w)∇φdx= 0 ∀φ ∈ W
1,p
0

B
+
R

,
w
− u ∈ W
1,p
0

B
+
R


N

B
+
R
|∇w|
p
dx, (2.7)

B
+
ρ


∇
w − (∇w)
ρ


p
dx ≤ c

ρ
R

N+σ

B
+

A(x, ∇ψ)∇φdx,
w
− v ∈ W
1,p
0

B
+
R

, ∀φ ∈ W
1,p
0

B
+
R

,
(2.9)
then

B
+
R
|∇w|
p
dx ≤ c

B

|∇v −∇w|
p
dx ≤ c

B
+
R

A(x, ∇v) − A(x,∇w)

·
(∇v −∇w)dx
= c

B
+
R
A(x, ∇ψ) · (∇v −∇w)dx
≤ c

B
+
R
|∇ψ|
p−1
|∇v −∇w|dx
≤ c


B

= u on ∂B
+
R
, u ∈,thatv ≥ ψ on ∂B
+
R
.Letξ = min(v,ψ), ξ = ψ
on ∂B
+
R
, ξ − ψ ∈ W
1,p
0
(B
+
R
). As test functions in (2.9)wetakeφ = ξ − ψ,from(2.9)and
M. Junxia and C. Yuming 5
monotony inequality (iii), we have
0
=

B
+
R
A(x, ∇v) − A(x,∇ψ) ·∇(ξ − ψ)dx
=

B
+

+
R
. 
This lemma is a useful comparison principle, it can be used to obtain the existence or
regularity of solutions to elliptic equation or variational inequalit y.
We ex tend v to B
+
by setting v = u on B
+
\B
+
R
, and hence v ∈. We have t he following
corollary.
Corollary 2.6. Suppose u is a weak solution to the obstacle problem (2.3), v
∈ W
1,p
(B
+
R
)
is a solution of the D irichlet problem (2.9), then v
∈satisfies the variational inequality

B
+
A(x, ∇u) · (∇v −∇u)dx ≥

B
+

+
R
|∇u|
p
dx +

B
+
R
|∇ψ|
p
dx

, (2.15)

B
+
R
|∇u −∇v|
p
dx ≤ c


B
+
R

|∇
u|
p

+ R
N(1− p

/s)
+ R
N(1− p/m)

,
(2.16)
where δ
= (r − p)/r(p − 1) > 0.
Proof. By inserting φ
= v − u in (2.9), an application of H
¨
older’s inequality and Young’s
inequality y ields

B
+
R
|∇v|
p
dx ≤ c

B
+
R
A(x, ∇v) ·∇vdx
= c


c



B
+
R
|∇v|
p
dx

(p−1)/p


B
+
R
|∇u|
p
dx

1/p
+


B
+
R
|∇ψ|
p

B
+
R
|∇u|
p
dx + c
2

B
+
R
|∇u −∇v|
p
dx
+ c


2
, p


B
+
R
|∇ψ|
p
dx
≤

c


2
, p


B
+
R
|∇ψ|
p
dx
(2.17)
for (c

1
+ c
2
)sufficiently small (c
1
+ c
2
< 1), we can get (2.15). 
By ψ ∈ W
1,m
(Ω), m>N,wehave

B
R
|∇ψ|
p

+
R

H(x,u,∇u)(u− v)+

F(x, u) − A(x, ∇ψ)

·
(∇u −∇v)

dx
≤ c

B
+
R

|∇
u|
p(1−1/r)
+ |u|
r−1
+ |g|

|u − v|dx
+ c

B
+
R


dx

1−1/r


B
+
R
|u − v|
r
dx

1/r
+


B
+
R

|
u|
q
+ |h|
p


dx


p
+ |u|
r

dx

1−1/r
+ g
t
R
N(1−1/r−1/t)

×
R
1−N(1/p−1/r)


B
+
R
|∇u −∇v|
p
dx

1/p
+ c


B
+


B
+
R
|∇u −∇v|
p
dx

1/p
(2.19)
since 0 <R
≤ 1, by (2.18), H
¨
older inequality, Young inequality, we have

B
+
R
|∇u −∇v|
p
dx
≤ c


B
+
R

|∇
u|

p/(p−1)
t
R
Np(1−1/r−1/t)/(p−1)
+ h
p/(p−1)
s
R
N(1− p

/s)
+

B
+
R
|∇ψ|
p
dx

≤
c


B
+
R

|∇
u|

Np(1−1/r−1/t)/(p−1)
+ R
N(1− p

/s)
+ R
N(1− p/m)

≤
c


B
+
R

|∇
u|
p
+ |u|
r

dx

1+δ
+


B
+

+ R
N(1− p/m)

≤
c


B
+
R

|∇
u|
p
+ |u|
r

dx

1+δ
+


B
+
R
|∇u|
p
dx



|∇
u|
p
+ |u|
r

dx

1+δ
+


B
+
R
|∇u|
p
dx

q/p
+


B
+
R
|u|
r
dx

(B
+
) with m>N.Ifu ∈
makes (2.3)hold,thenu ∈ C
0,λ
(Γ) with λ = min{1 − N(1/t +1/r − 1/p)/(p − 1),1 − N/
s(p
− 1),1 − N/m}.
Before proceeding with the formal proof, we make an impor tant observation. It is a
well-known result.
Proposition 3.2. If f
∈ W
1,p
(Ω), then for all constants k ∈ R
N
,

B
ρ


∇
f − (∇ f )
ρ,x
0


p
dx ≤ C(p)


dx +

B
ρ


k − (∇ f )
ρ,x
0


p
dx

. (3.2)
Moreover

B
ρ


k − (∇ f )
ρ,x
0


p
dx =





B
ρ
∇ fdx




p
=


B
ρ






1


B
ρ



B


B
ρ


1−p

B
ρ
|∇ f − k|
p
dx


B
ρ


p(1−1/p)
=

B
ρ
|∇ f − k|
p
dx.
(3.3)
Therefore (3.1)holdsforanyk
∈ R
N

+
, we c an deduce the same result for u in B
−
, so we need not care about
this situation.
M. Junxia and C. Yuming 9
(3) If B
R
(y
0
) ∩ B
+
= Ø, we also g ive three different situations as follows:
(a) y
0
∈ Γ,
(b) y
0
∈ B
−
,
(c) y
0
∈ B
+
.
We only prove the situation (a), since the others can be transformed into the situation
(a) or the interior regularity situation by applying the finitely covered theorem, see [13].
Assume h
∈ L

ψ|
p−2
∇ψ −

|∇ψ|
p−2
∇ψ

R



p/(p−1)
dx ≤ c

B
+
R
|∇ψ|
p
dx ≤ c∇ψ
p
m
R
N(1− p/m)
.
(3.5)
Combining (2.7), (2.10), (2.11), (2.16), and (3.5), we have

B

|∇u|
p
dx + c

B
+
R
|∇u −∇v|
p
dx + c

B
+
R
|∇v −∇w|
p
dx
≤ c

ρ
R

N

B
+
R
|∇u|
p
dx

B
+
R
|u|
r
dx

q/p
+ R
Np(1−1/r−1/t)/(p−1)
+ R
N[1− p/s(p−1)]
+ R
N(1− p/m)

≤
c

ρ
R

N

B
+
R
|∇u|
p
dx + c


R
|u|
r
dx

q/p
+ cR
N− p+pλ
,
(3.6)
where λ
= min{1 − N(1/t +1/r − 1/p)/(p − 1),1 − N/s(p − 1),1 − N/m}.
By t>N/p, we have the following.
(i) If 2
≤ p<N,then1/t< p/N,1/t+1/r−1/p<p/N+(N − p)/N p−1/p= (p − 1)/N,
N(1/t +1/r
− 1/p)/(p − 1) < 1.
(ii) If p
= N,byt>1, we can assume that r is a positive number sufficiently large,
such that: 1/t+1/r<1, so N(1/t+1/r
−1/N)/(N −1)<(N/(N −1))(1−1/N)=1.
Hence, if 2
≤ p ≤ N, we always have N(1/t +1/r − 1/p)/(p − 1) < 1.
Using s>N/(p
− 1), m>N, by the definition of λ,weseethat:0<λ<1.
10 Boundary regularity
In the meantime, by Poincare’s inequality and H
¨
older’s inequality, we also have


≤ c

ρ
R

N

B
+
R
|u|
r
dx + cR
r[1−N(1/p−1/r)]


B
+
R
|∇u|
p
dx

r/p
,
(3.7)
where
r

1 − N

=
0, if 2 ≤ p<N;
r

1 − N

1
N
−
1
r

=
N,ifp = N.
(3.8)
Adding (3.7)to(3.6) and setting
φ(R)
=

B
+
R

|∇
u|
p
+ |u|
r

dx, (3.9)


δ
+


B
+
R
|∇u|
p
dx

(q−p)/p
+


B
+
R
|u|
r
dx

(q−p)/p
+
⎧
⎪
⎪
⎪
⎪

(3.11)
We can always get χ(R)
→ 0asR → 0
+
.Applying[7, page 86, Lemma 2.1], we deduce
that for ρ sufficiently small,

B
+
ρ
|∇u|
p
dx ≤ φ(ρ) ≤ cρ
N− p+pλ
. (3.12)

By Dirichlet growth theorem (see [7, page 64, Theorem 1.1]), u ∈ C
0,λ
loc
(Γ).
M. Junxia and C. Yuming 11
4. C
1,α
1
regularity
Theorem 4.1. Assume that H(x,u,
∇u) satisfies (1.2), g ∈ L
t
(B
+

dx ≤ cρ
N+pα
1
. (4.1)
It is easy to see that
|∇ψ|
p−2
∇ψ ∈ C
0,γ
(B
+
)ifψ ∈ C
1,γ
(B
+
)and2≤ p ≤ N.
Utilizing the conditions of Theorem 4.1,weseethat:

B
+
R


F − F
R


p/(p−1)
dx ≤F
p/(p−1)


|∇
ψ|
p−2
∇ψ


C
0,γ
(B
+
)
|x − y|
γ
, ∀x, y ∈ B
+
.
(4.3)
By ψ
∈ C
1,γ
(B
+
), we can get |∇ψ|
p−1
≤ c, so combining condition (vi) we have


A(x, ∇ψ) − A(y,∇ψ)



·∇
φdx
=

B
+
R

B
R
−

A(x, ∇ψ(x)

−
A

y,∇ψ(y)

·∇
φ(x)dydx
=

B
+
R

B
R

−

|x − y|
α
0
+





∇
ψ(x)


p−2
∇ψ(x)−


∇ψ(y)


p−2
∇ψ(y)







A(x, ∇ψ) ·∇φdx≤ c

B
+
R

R
α
0
+ R
γ



∇
φ(x)


dx. (4.6)
In the following, we give two lemmas which will b e used in the proof of Theorem 4.1.
12 Boundary regularity
Lemma 4.2. Assume that ψ
∈ C
1,γ
(B
+
), w,v ∈ W
1,p
(B
+

dx ≤ c

B
+
R

A(x, ∇v) − A(x,∇w)

·
(∇v −∇w)dx
= c

B
+
R
A(x, ∇ψ) · (∇v −∇w)dx
≤ c

B
+
R

R
α
0
+ R
γ

|∇
v −∇w|dx

0,β
(B
+
), β>0; ψ ∈ C
1,γ
(B
+
),
γ>0. v
∈ W
1,p
(B
+
R
) solves the Dirichlet problem (2.9), then there holds

B
+
R
|∇u −∇v|
p
dx ≤ c


B
+
R

|∇
u|

|∇u −∇v|
p
dx ≤ c

B
+
R

A(x, ∇u) − A(x, ∇v)

·
(∇u −∇v)dx
≤ c

B
+
R

H(x,u,∇u)(u− v)+

F(x, u) − A(x, ∇ψ)

·
(∇u −∇v)

dx
≤ c

B
+

0
+ R
γ

|∇
u −∇v|dx
M. Junxia and C. Yuming 13
≤ c


B
+
R

|∇
u|
p
+ |u|
r
+ |g|
r/(r−1)

dx

1−1/r


B
+
R

dx

1/p
+

B
+
R

R
α
0
+ R
γ

|∇
u −∇v|dx

≤
c


B
+
R

|∇
u|
p
+ |u|


F − F
R


p

dx

1/p

+ R
γ+N(p−1)/p
+ R
α
0
+N(p−1)/p

×


B
+
R
|∇u −∇v|
p
dx

1/p
.

N+γp/(p−1)
+ R
N+α
0
.p/(p−1)

.
(4.11)
Hence (4.9)holds.

Similar to the proof of (3.12), we get for any μ ∈ (0,1)

B
+
ρ
|∇u|
p
dx,

B
+
ρ
|∇v|
p
dx,

B
+
ρ
|u|

p
dx
≤ c

B
+
ρ


∇
u − (∇w)
ρ


p
dx
≤ c

B
+
ρ


∇
w − (∇w)
ρ


p
dx + c


B
+
R

|∇
u|
p
+ |u|
r

dx

1+δ
+ R
Np(1−1/r−1/t)/(p−1)
+ R
N+βp/(p−1)
+ R
N+γp/(p−1)
+ R
N+α
0
p/(p−1)

+ c

R
N+γp/(p−1)
+ R


|∇
u|
p
+ |u|
r

dx

1+δ
+ R
Np(1−1/r−1/t)/(p−1)
+ R
N+βp/(p−1)
+ R
N+γp/(P−1)
+ R
N+α
0
p/(p−1)

≤
c

ρ
R

N+σ

B


B
+
ρ


∇
u − (∇u)
ρ


p
dx ≤ c

ρ
R

N+σ

B
R


∇
u − (∇u)
R


p
dx + cR

Acknowledgments
We would like to thank professor Fang AiNong for many useful comments on the sub-
ject, the project supported by the Natural Science Foundation of Hebei University, the
National Natural Science Foundation of China under Grant 10471039, and the Zhejiang
Province Natural Science Foundation of China under Grant M103087.
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