Boundary regularity of weak solutions to nonlinear elliptic obstacle problems

NONLINEAR ELLIPTIC OBSTACLE PROBLEMS

Received 25 April 2005; Revised 10 September 2005; Accepted 14 September 2005

We study the boundary regularity of weak solutions to nonlinear obstacle problem with C1,β_{-obstacle function, and obtain theC}

loc1,αboundary 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∈ :

ΩA(x,∇u)·(∇v− ∇u)dx

≥

ΩH(x,u,∇u)(v−u)dx+

ΩF(x,u)·(∇v− ∇u)dx

(1.1)

for allv∈={v∈W01,p(Ω),v≥ψa.e. inΩ}. HereΩis a bounded domain inRN(N≥2)

with Lipschitz boundary, 2≤p≤N.

A(x,ξ) :Ω×RN_→RN_{satisfies the following conditions:}

(i)Ais a vector valued function, the mappingx→A(x,ξ) is measurable for allξ∈

RN_{,ξ→A(x,ξ) is continuous for a.e.x∈Ω;}

(ii) the homogeneity condition:A(x,tξ)=t|t|p−2_{A(x,ξ),t∈R,t =0;}

(iii) the monotone inequality: (A(x,ξ)−A(x,ζ))(ξ−ζ)≥a|ξ−ζ|p_; (iv)|h||ai j_|+|∂Ai_{(x,h)/∂xj| ≤τ1|h|}p−1_;

(v)Ni,j=1ai jξiξj≥τ2|h|p−2|ξ|2;

(vi)|A(x,ξ)−A(y,ξ)| ≤b1(1 +|ξ|p−1)|x−y|α0;

(vii)|A(x,ξ)−A(x,η)| ≤b2||ξ|p−2_{ξ− |η|}p−2_η|;

whereai j_=∂Ai_/∂hj,a,b

1,b2,τ1,τ2are positive constants.

Hindawi Publishing Corporation Boundary Value Problems

Volume 2006, Article ID 72012, Pages1–15

We assume thatH(x,u,λ),F(x,u)= {Fi(x,u)}1≤i≤Nin (1.1) are of the form: _{H(x,u,∇u)≤c}

|∇u|p/r_+|u|r−1_{+g(x), (1.2)}

_F(x,u)≤c

|u|q/ p_{+h(x), (1.3)}

wherep < q < r,r=r/(r−1),p=p/(p−1), and if 2≤p < N,r=N p/(N−p), while if p=N, thenrcan be some sufficiently large positive number.

Higher regularity of the weak solution to thep-Laplacian obstacle problem

I(u)=inf

Ω|∇u|

p_dx:u∈K(ψ)

, (1.4)

where

K(ψ)= v∈W1,p_{(Ω) :v≥ψa.e., (1.5)}

has been studied by various authors. In the case whenψis assumed to have only minimal regularity properties, it was shown by [8,11] that the solution of (1.1) is continuous. In particular, ifψ∈C0,α_{(Ω), then the solutionuis also an element ofC}0,α

(Ω). In the case whenψ∈C2_{(Ω), papers [4,6,10,12] employed different techniques to prove interior}

C1,α_{(Ω) regularity for the solutionuto (1.4). Reference [1] gave an interesting result: the} condition forto be nonempty is just thatψshould have finite capacity. This implies, among other things, thatψ+_{=max(ψ, 0) must vanish on ∂Ω,C—almost everywhere.}

This condition is important for the existence of weak solutions to obstacle problem. Whenψis smooth (sayC1,α_{(Ω)), the interior regularity of weak solutions to problem} (1.1) has been studied extensively by many authors ([3,13,14]).

In view of De Giorgi class, paper [2] obtainedC0,α_{interior regularity for solutions of} nonlinear elliptic obstacle problem with natural growth in the gradient by taking appro-priate test function.

The main concern of these papers is the question of the regularity of the solutionu in terms of the given regularity properties of the obstacleψ and relevant data. This is especially interesting in view of the fact that there is a limit to the amount of regularity thatucan inherit fromψ: it is possible forψto be real analytic, butuwill be at bestC1,1_,

that is, have bounded second derivatives.

This paper obtainsCloc1,αboundary regularity of weak solutions to the obstacle problem

withC1,β_{-obstacle function under controllable growth condition (1.2). We present a new} proof to a useful comparison principle.

2. Notations and preliminaries

Ωis an open bounded subset ofRN_{,N≥2;∂Ωis the boundary ofΩ. Ifz∈R}N_{, we put}

BR(z)= x∈RN_{:|x−z|< R, Γ}

R(z)= x∈BR(z) :xn=0, B+

R(z)= x∈BR(z) :xn>0

, B_R−(z)= x∈BR(z) :xn<0. (2.1)

We denote byB,B+_,B−_{,Γ, respectively,B1(0), B}+

1(0),B1−(0),Γ1(0). For every setEwe

denote by ¯Eits closure, and by|E|its Lebesgue measure. (f)R=(1/|BR|)

letter c is used throughout to denote a positive constant, not necessarily the same at each occurrence.

SinceΩis compact,∂Ωcan be covered by a finite number of neighbourhoodsVof its points. It is enough to prove the better regularity ofuholds true inV∩Ω. Since∂Ωis a Lipschitz boundary, one can findTwhich 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 mappingT the variational inequality inΩis transformed to a variational inequality of the same form inB+_{, for ¯u=u◦T}−1_{which satisfies}

B+ ¯

A(x,∇u¯)·(∇v− ∇u¯)dx

≥

B+ ¯

H(x, ¯u,∇u¯)(v−u¯)dx+

B+ ¯

F(x, ¯u)·(∇v− ∇u¯)dx, ∀v∈¯_, (2.3)

where ¯ = {v∈W01,p(B+), v≥ψ, a.e. inB+}, ¯A, ¯H, ¯F satisfy assumptions of type (i)–

(vii), (1.2), (1.3) with different constants.

In order to simplify the notations, we still denote ¯u, ¯_A¯_{, ¯H, ¯Fbyu,,A,H,F,} re-spectively.

Since the originalu∈W01,p(Ω), we define then

u(x) := ⎧ ⎪ ⎨ ⎪ ⎩

u(x), ifx∈B+_,

−3ux1,. . .,xn−1,−xn

+ 4u

x1,. . .,xn−1,−xn

2

, ifx∈B−. (2.4)

In light of Extension theorem [5, page 254], we only need to prove a better regularity of uinB+_.

Definition 2.1. The functionu∈ that satisfies (2.3) for allv∈ is called a weak solu-tion to the obstacle problem with obstacleψ.

Definition 2.2. Call f ∈C0,α_{(Γ), if for allx∈Γ, there existsBr(x) (a ball centered atxof} radiusr),r >0, such that f ∈C0,α_(Br(x)).

In the sequel, we will abbreviateB+_∩BR(y

0)=B+_R,B+∩Bρ(y0)=B_ρ+, for 0< ρ < R≤

1, the pointy0∈Γto be understood.

In the following, we will use some lemmas which we state below.

Lemma2.3. Letw∈W1,p_(B+

R)be a solution of the Dirichlet problem

B+ R

A(x,∇w)∇φ dx=0 ∀φ∈W01,p

B+

R

,

w−u∈W01,p

B+

R

.

For0< ρ < R/2,σ∈(0, 1),

B+ R

|∇w|p_dx≤c

B+ R

|∇u|p_{dx, (2.6)}

B+ ρ

|∇w|p_{dx≤c ρ}

R N

B+ R

|∇w|p_{dx, (2.7)}

B+ ρ

_{∇w−(∇w)ρ}p dx≤c

_ρ

R N+σ

B+ R

_{∇w−(∇w)R}p

dx. (2.8)

Proof. We can easily get (2.6) by insertingφ=w−uin (2.5).

An argument similar to the one in [15, Lemma 2.2] shows that (2.7) hold.

The proof of (2.8) is similar to that of [9, Theorem 1.7].

Lemma2.4. Letv∈W1,p_(B+

R)be a solution of the Dirichlet problem

B+ R

A(x,∇v)∇φ dx=

B+ R

A(x,∇ψ)∇φ dx,

w−v∈W01,p

B+R

, ∀φ∈W01,p

B+R

,

(2.9)

then

B+ R

|∇w|p_dx≤c

B+ R

|∇v|p_{dx, (2.10)}

B+ R

|∇v− ∇w|p_dx≤c B+

R

|∇ψ|p_{dx. (2.11)}

Proof. Formula (2.10) follows immediately from takingφ=w−vin (2.5).

Insertingφ=v−win (2.5) and (2.9), by monotone inequality (iii) and H¨older’s in-equality, we have

B+ R

|∇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+ R

|∇ψ|p_dx

(p−1)/ p

B+ R

|∇v− ∇w|p_dx 1/ p

(2.12)

from which we get (2.11).

Lemma2.5. Ifv∈W1,p_(B+

R)is a solution of the Dirichlet problem (2.9), thenv≥ψinB+R. Proof. It follows fromv=uon∂B_R+,u∈ , thatv≥ψon∂B+_R. Letξ=min(v,ψ),ξ=ψ on∂B+R,ξ−ψ∈W

1,p

monotony inequality (iii), we have

0=

B+ R

A(x,∇v)−A(x,∇ψ)· ∇(ξ−ψ)dx

=

B+ R

{x,v(x)≤ψ(x)}A(x,∇v)−A(x,∇ψ)· ∇(v−ψ)dx

≥a

B+ R

{x,v(x)≤ψ(x)}|∇v− ∇ψ|

p_dx

=a

B+ R

|∇ξ− ∇ψ|p_dx

(2.13)

thereforeξ=ψa.e. inB+R, that is,v≥ψa.e. inB+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 inequality.

We extendvtoB+_{by settingv=uonB}+_\B+

R, and hencev∈ . We have the following corollary.

Corollary2.6. Supposeuis a weak solution to the obstacle problem (2.3),v∈W1,p_(B+

R) is a solution of the Dirichlet problem (2.9), thenv∈ satisfies the variational inequality

B+A(x,∇u)·(∇v− ∇u)dx≥

B+H(x,u,∇u)(v−u)dx+

B+F(x,u)·(∇v− ∇u)dx. (2.14)

Lemma2.7. Assumeuis a weak solution to the obstacle problem (2.3), whereH,F verify (1.2), (1.3), respectively,g∈Lt_(B+_{)witht > N/ p,h∈L}s_(B+_{)withs > p,vsatisfies (2.9),}

then

B+ R

|∇v|p_dx≤c B+

R

|∇u|p_dx+

B+ R

|∇ψ|p_{dx, (2.15)}

B+ R

|∇u− ∇v|p_dx≤c

B+ R

|∇u|p_+|u|r_dx 1+δ

+

B+ R

|∇u|p_dxq/ p₊ B+

R

|u|r_dxq/ p

+RN p(1−1/r−1/t)/(p−1)_+RN(1−p/s)_+RN(1−p/m)

,

(2.16)

whereδ=(r−p)/r(p−1)>0.

Proof. By insertingφ=v−uin (2.9), an application of H¨older’s inequality and Young’s inequality yields

B+ R

|∇v|p_dx≤c

B+ R

A(x,∇v)· ∇v dx

=c

B+ R

A(x,∇v)· ∇u dx+

B+ R

≤c

B+ R

|∇v|p−1_|∇u|dx+

B+ R

|∇ψ|p−1_{|∇u− ∇v|dx}

≤c

B+ R

|∇v|p_dx

(p−1)/ p

B+ R

|∇u|p_dx 1/ p

+

B+ R

|∇ψ|p_dx

(p−1)/ p

B+ R

|∇u− ∇v|p_dx

1/ p

≤c1

B+ R

|∇v|p_dx+c1,p B+

R

|∇u|p_dx+c2

B+ R

|∇u− ∇v|p_dx

+c2,p B+

R

|∇ψ|p_dx

≤c1+c2

B+ R

|∇v|p_dx+c1,p+c2 B+

R

|∇u|p_dx

+c2,p B+

R

|∇ψ|p_dx

(2.17)

for (c1+c2) sufficiently small (c1+c2<1), we can get (2.15).

Byψ∈W1,m_{(Ω),m > N, we have}

BR|∇ψ|

p_{dx≤c∇ψmR}p N(1−p/m)_{. (2.18)}

Combining monotone inequality (iii), (2.9), (2.14), and (1.2) and using Poincare’s inequality, H¨older’s inequality, we have

B+ R

|∇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+ R

|∇u|p(1−1/r)_+|u|r−1_{+|g||u−v|dx}

+c

B+ R

|u|q/ p

+h(x)|∇u− ∇v|dx+c

B+ R

|∇ψ|p−1_{· |∇u− ∇v|dx}

≤c

B+ R

|∇u|p_+|u|r_+|g|r/(r−1)_dx

1−1/r

B+ R

|u−v|r_dx 1/r

+

B+ R

|u|q_+|h|p_dx 1/ p

B+ R

|∇u− ∇v|p_dx 1/ p

+

B+ R

≤c

B+ R

|∇u|p_+|u|r_dx1−1/r_+gtRN(1−1/r−1/t)

×R1−N(1/ p−1/r)

B+ R

|∇u− ∇v|p_dx 1/ p

+c

B+ R

|u|p_+|∇u|p_dxq/ pp

+hsRN(1/ p _−1/s)

+|∇ψ|p−1

p/(p−1)

×

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_+|u|r_dx 1+δ

+

B+ R

|∇u|p_dx q/ p

+

B+ R

|u|p_dxq/ p_+gp/(p−1)

t RN p(1−1/r−1/t)/(p−1)

+hsp/(p−1)RN(1−p _/s)

+

B+ R

|∇ψ|p_dx ≤c B+ R

|∇u|p_+|u|r_dx

1+δ +

B+ R

|∇u|p_dxq/ p

+

B+ R

|u|p_dx q/ p

+RN p(1−1/r−1/t)/(p−1)_+RN(1−p/s)_+RN(1−p/m)

≤c B+ R

|∇u|p_+|u|r_dx

1+δ +

B+ R

|∇u|p_dxq/ p

+

B+ R

|u|r_dx q/r

BR

(q/ p)(1−p/r)

+RN p(1−1/r−1/t)/(p−1)_+RN(1−p/s)_+RN(1−p/m)

≤c B+ R

|∇u|p_+|u|r_dx

1+δ +

B+ R

|∇u|p_dxq/ p

+

B+ R

|u|r_dx q/ p

+RN(q/ p)_+RN p(1−1/r−1/t)/(p−1)_+RN(1−p/s)_+RN(1−p/m)

≤c B+ R

|∇u|p_+|u|r_dx

1+δ +

B+ R

|∇u|p_dxq/ p

+

B+ R

|u|r_dx q/ p

+RN p(1−1/r−1/t)/(p−1)_+RN(1−p/s)_+RN(1−p/m)

(2.20)

3.C0,λ_regularity

Theorem3.1. Assume that H(x,u,∇u)satisfies (1.2),g∈Lt_(B+_{)witht > N/ p,F(x,u)}

satisfies (1.3),h∈Ls_(B+_{)withs > N/(p−1), andψ∈W}1,m_(B+_{)withm > N. Ifu∈}

makes (2.3) hold, thenu∈C0,λ_{(Γ)withλ=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 important observation. It is a well-known result.

Proposition3.2. If f ∈W1,p_{(Ω), then for all constantsk∈R}N_,

Bρ

_{∇f −(∇f)ρ,x} 0

p

dx≤C(p)

Bρ|∇f−k|

p_{dx (3.1)}

for everyρfor whichBρ(x0)⊂Ω.

Proof. By elementary inequality, we have

Bρ

_{∇f −(∇f)ρ,x} 0

p

dx≤C(p)

Bρ|∇f −k| p_dx+

Bρ

_{k−(∇f)ρ,x} 0

p dx

. (3.2)

Moreover

Bρ

_{k−(∇f)ρ,x} 0

p

dx=Bρk−(∇f)ρ,x0 p

=Bρk−1

Bρ

Bρ∇f dx p

=Bρ1 Bρ

Bρ(k− ∇f)dx

p=Bρ1−p

Bρ(k− ∇f)dx p

≤Bρ

1−p

Bρ|∇f−k| p_dxB

ρp

(1−1/ p) =

Bρ|∇f −k| p_dx.

(3.3)

Therefore (3.1) holds for anyk∈RN_.

Proof ofTheorem 3.1. To get the regularity, we need to prove the following inequality:

B+ ρ

|∇u|p_dx≤cρN−p+pλ_{. (3.4)}

Let us consider three different situations.

(1) IfB2R(y0)⊂B+, inequality (3.4)–(4.1) has been proved in [13], since it is related

to interior regularity.

(2) IfBR(y0)⊂B−, by Extension theorem [5, page 254], if we can getC1,αregularity

ofuinB+_{, we can deduce the same result foruinB}−_{, so we need not care about}

(3) IfBR(y0)∩B+ =Ø, we also give three different situations as follows:

(a) y0∈Γ,

(b) y0∈B−,

(c) y0∈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].

Assumeh∈Ls_(B+_,RN_{) withs > N/(p−1),ψ∈W}1,m_(B+_{) withm > N, we see that}

B+ R

|h|p/(p−1)_{≤ h}p/(p−1)

s RN[1−p/s(p−1)],

B+ R

|∇ψ|p−2_{∇ψ−|∇ψ|}p−2_∇ψ

R

p/(p−1)dx≤c

B+ R

|∇ψ|p_dx≤c∇ψp_mRN(1−p/m)_.

(3.5)

Combining (2.7), (2.10), (2.11), (2.16), and (3.5), we have

B+ ρ

|∇u|p_dx

≤c

B+ ρ

|∇w|p_dx+c

B+ ρ

|∇u− ∇w|p_dx

≤c _ρ

R N

B+ R

|∇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

+c

B+ R

|∇u|p_+|u|r_dx 1+δ

+

B+ R

|∇u|p_dx q/ p

+

B+ R

|u|r_dx q/ p

+RN p(1−1/r−1/t)/(p−1)+RN[1−p/s(p−1)]+RN(1−p/m)

≤c _ρ

R N

B+ R

|∇u|p_dx+c B+

R

|∇u|p_+|u|r_dx1+δ

+c

B+ R

|∇u|p_dxq/ p_+c B+

R

|u|r_dxq/ p_+cRN−p+pλ_,

(3.6)

whereλ=min{1−N(1/t+ 1/r−1/ p)/(p−1), 1−N/s(p−1), 1−N/m}. Byt > N/ p, we have the following.

(i) If 2≤p < N, then 1/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 thatris a positive number sufficiently large, such that: 1/t+1/r <1, soN(1/t+1/r−1/N)/(N−1)<(N/(N−1))(1−1/N)=1. Hence, if 2≤p≤N, we always haveN(1/t+ 1/r−1/ p)/(p−1)<1.

In the meantime, by Poincare’s inequality and H¨older’s inequality, we also have

B+ ρ

|u|r_dx≤c

B+ ρ

_uRr dx+c

B+ R

_u−uRr dx

≤c _ρ

R N

B+ R

|u|r_dx+cRr[1−N(1/ p−1/r)]

B+ R

|∇u|p_dxr/ p_,

(3.7)

where

r

1−N ₁

p− 1 r

=

⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩

N p N−p

1−N

1 p−

N−p N p

=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)}

we obtain

φ(ρ)≤c

_ρ

R N

+χ(R)

φ(R) +cRN−p+pλ, (3.10)

where

χ(R)=

B+ R

|∇u|p_+|u|r_dxδ

+

B+ R

|∇u|p_dx

(q−p)/ p +

B+ R

|u|r_dx

(q−p)/ p

+ ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩

B+ R

|∇u|p_dx(r−p)/ p_{, if 2≤p < N;}

RN

B+ R

|∇u|p_dx(r−p)/ p_{, ifp=N.}

(3.11)

We can always getχ(R)→0 asR→0+_{. Applying [7, page 86, Lemma 2.1], we deduce}

that forρsufficiently small,

B+ ρ

|∇u|p_{dx≤φ(ρ)≤cρ}N−p+pλ_{. (3.12)}

4.C1,α1_regularity

Theorem 4.1. Assume that H(x,u,∇u) satisfies (1.2), g ∈Lt_(B+_{) with t > N, F} ∈C0,β_(B+_{,R)withβ >0,ψ∈C}1,γ_(B+_{)for some γ >0. If u∈ makes (2.3) hold, then}

u∈C1,α1₍Γ_{)for some0< α}

1< σ/ p.

To get our result, we need to prove the following inequality:

B+ ρ

_{∇u−(∇u)ρ}p

dx≤cρN+pα1_{. (4.1)}

It is easy to see that|∇ψ|p−2_∇ψ∈C0,γ_(B+_{) ifψ∈C}1,γ_(B+_{) and 2≤p≤N.}

Utilizing the conditions ofTheorem 4.1, we see that:

B+ R

_F−FRp/(p−1)

dx≤ F_Cp/0,(βp₍−_B1)+₎RN+βp/(p−1), (4.2) _∇ψ(x)p−2

∇ψ(x)−∇ψ(y)p−2∇ψ(y)

≤|∇ψ|p−2_∇ψ

C0,γ_(B+₎|x−y|γ, ∀x,y∈B+.

(4.3)

Byψ∈C1,γ_(B+_{), we can get|∇ψ|}p−1_{≤c, so combining condition (vi) we have}

_{A(x,∇ψ)−A(y,∇ψ)≤c|x−y|}α0_{. (4.4)}

By condition (vii), (4.3), and (4.4), we see that for allφ∈W01,p(BR+), there holds

B+ R

A(x,∇ψ)· ∇φ dx

=

B+ R

A(x,∇ψ)−A(x,∇ψ)_R· ∇φ dx

=

B+ R

BR

− A(x,∇ψ(x)−Ay,∇ψ(y)· ∇φ(x)d y dx

=

B+ R

BR

− Ax,∇ψ(x)−Ax,∇ψ(y)

+Ax,∇ψ(y)−Ay,∇ψ(y)· ∇φ(x)d y dx

≤c

B+ R

BR

− |x−y|α0₊∇_ψ(x)p−2∇_ψ(x)−∇_ψ(y)p−2∇_ψ(y)∇_{φ(x)d y dx}

≤c

B+ R

Rα0_+Rγ∇_φ(x)dx.

(4.5)

From last formula, we see that

B+ R

A(x,∇ψ)· ∇φ dx≤c

B+ R

Rα0_+Rγ∇_{φ(x)dx. (4.6)}

Lemma4.2. Assume thatψ∈C1,γ_(B+_),w,v∈W1,p_(B+

R)solve the Dirichlet problem (2.5), (2.9), respectively, then there holds

B+ R

|∇v− ∇w|p_dx≤cRN+γ p/(p−1)_+RN+α0p/(p−1)

. (4.7)

Proof. Insertingφ=v−w in (2.5) and (2.9), by (4.6), monotone inequality (iii), and H¨older inequality, we have

B+ R

|∇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

Rα0_+Rγ|∇_v− ∇_w|_dx

≤cRγ+N(p−1)/ p_+Rα0+N(p−1)/ p B+

R

|∇v− ∇w|p_dx

1/ p .

(4.8)

From last formula, we get (4.7).

Lemma4.3. Assume thatA(x,ξ)satisfies condition (i)–(vii),uis a weak solution to obstacle problem (1.1), whereHverifies (1.2),g∈Lt_(B+_{),t > N;F∈C}0,β_(B+_{),β >0;ψ∈C}1,γ_(B+_),

γ >0.v∈W1,p_(B+

R)solves the Dirichlet problem (2.9), then there holds

B+ R

|∇u− ∇v|p_dx≤c

B+ R

|∇u|p_+|u|r_dx 1+δ

+RN p(1−1/r−1/t)/(p−1)

+RN+βp/(p−1)_+RN+γ p/(p−1)_+RN+α0p/(p−1)

,

(4.9)

whereδ=(r−p)/r(p−1)>0.

Proof. Combining monotone inequality (iii), and (1.2), (2.14), (2.18), (4.6), H¨older in-equality, we have

B+ R

|∇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+ R

|∇u|p(1−1/r)_+|u|r−1_{+|g||u−v|dx}

+c

B+ R

_{F−FR|∇u− ∇v|dx+c} B+

R

≤c

B+ R

|∇u|p_+|u|r_+|g|r/(r−1)_dx1−1/r

B+ R

|u−v|r_dx1/r

+

B+ R

_F−FRp dx

1/ p

B+ R

|∇u− ∇v|p_dx 1/ p

+

B+ R

Rα0_+Rγ|∇_u− ∇_v|_dx

≤c

B+ R

|∇u|p_+|u|r_dx

1−1/r

+gtRN(1−1/r−1/t)

×R1−N(1/ p−1/r)

B+ R

|∇u− ∇v|p_dx

1/ p

+c

B+ R

_F−FRp dx

1/ p

+Rγ+N(p−1)/ p_+Rα0+N(p−1)/ p

×

B+ R

|∇u− ∇v|p_dx 1/ p

.

(4.10)

Since 0< R≤1, by (4.2), and last formula, we have

B+ R

|∇u− ∇v|p_dx≤c

B+ R

|∇u|p_+|u|r_dx 1+δ

+RN p(1−1/r−1/t)/(p−1)

+RN+βp/(p−1)+RN+γ p/(p−1)+RN+α0.p/(p−1)

.

(4.11)

Hence (4.9) holds.

Similar to the proof of (3.12), we get for anyμ∈(0, 1)

B+ ρ

|∇u|p_dx,

B+ ρ

|∇v|p_dx,

B+ ρ

|u|r_dx≤cρN−p+pμ_{. (4.12)}

In view oft > N, we have the following.

(i) If 2≤p < N, 1−1/r−1/t >1−1/r−1/N=(p−1)/ p, from which we getN p(1−

1/r−1/t)/(p−1)> N.

(ii) If p=N, we can assume thatris a positive number large such that 1/t+ 1/r < 1/N,N p(1−1/r−1/t)/(p−1)> N2_{(1−1/N)/(N−1)=N.}

By (4.2), (4.3), (4.9), Lemmas2.3,4.2, and4.3, we have

B+ ρ

_{∇u−(∇u)ρ}p dx

≤c

B+ ρ

_{∇u−(∇w)ρ}p dx

≤c

B+ ρ

_{∇w−(∇w)ρ}p dx+c

B+ ρ

|∇u− ∇v|p_dx+c

B+ ρ

|∇v− ∇w|p_dx

≤c

B+ ρ

_{∇w−(∇w)ρ}p dx

+c

B+ R

|∇u|p_+|u|r_dx1+δ

+RN p(1−1/r−1/t)/(p−1)_+RN+βp/(p−1)_+RN+γ p/(p−1)_+RN+α0p/(p−1)

+cRN+γ p/(p−1)+RN+α0p/(p−1)

≤c _ρ

R N+σ

B+ R

_{∇u−(∇u)R}p dx

+c

B+ R

|∇u|p_+|u|r_dx1+δ

+RN p(1−1/r−1/t)/(p−1)_+RN+βp/(p−1)_+RN+γ p/(P−1)_+RN+α0p/(p−1)

≤c _ρ

R N+σ

B+ R

_{∇u−(∇u)R}p dx

+cR(1+δ)(N−p+pμ)_+RN p(1−1/r−1/t)/(p−1)_+RN+βp/(p−1)

+RN+γ p/(P−1)+RN+α0p/(p−1)_.

(4.13)

We can selectμsufficiently close to 1, such that (1 +δ)(N−p+pμ)> N. Hence we get

B+ ρ

_{∇u−(∇u)ρ}p dx≤c

_ρ

R N+σ

BR

_{∇u−(∇u)R}p

dx+cRN+pα1 _(4.14)

for some 0< α1< σ/ p.

Applying [7, page 86, Lemma 2.1], we get

B+ ρ

_{∇u−(∇u)ρ}p

dx≤cρN+pα1 _(4.15)

forρsufficiently small. By [7, page 72, Theorem 1.3], we obtain thatu∈C1,α1

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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Meng Junxia: College of Mathematics and Computer Science, Hebei University, Hebei 071002, China E-mail address:[email protected]