1 Introduction

In this paper we derive interior and global Hölder estimates for solutions of quasilinear elliptic equations of the form

divA(x,u,u)=f(x),
(1.1)

where Ω is a bounded open subset in R n , n2, f(x) L q p 1 (Ω) for some q>n. Throughout the paper, the exponent p is denoted as the Hölder conjugate of p, 1<pn.

Suppose that the operator A:Ω×R× R n R n is a Carathéodory mapping satisfying the following assumptions:

  1. (a)

    for a.e. xΩ and all sR, ξ R n ,

    A(x,s,ξ)ξα | ξ | p , | A ( x , s , ξ ) | b ( | s | ) ( k ( x ) + | ξ | p 1 ) ;
  2. (b)

    for a.e. xΩ and all sR, ξ 1 , ξ 2 R n , ξ 1 ξ 2 ,

    ( A ( x , s , ξ 1 ) A ( x , s , ξ 2 ) ) ( ξ 1 ξ 2 )>0;
  3. (c)

    for a.e. xΩ, all sR, ξ R n and all λR,

    A(x,s,λξ)=λ | λ | p 2 A(x,s,ξ),

where α is a positive real constant, k(x) belongs to L q p 1 (Ω) for some q>n and b:[0,+)(0,+) is a continuous function.

Definition 1.1 A function u W loc 1 , p (Ω) is called a weak solution of (1.1) if

Ω A(x,u,u)ϕdx= Ω f(x)ϕdx
(1.2)

for every ϕ W 1 , p (Ω) with compact support.

The Hölder continuity estimate for solutions of nonlinear elliptic equations has always been an important subject in the theory of differential equations and dynamical systems, see, e.g., [14]. Numerous results on the Hölder regularity of elliptic equations with various conditions have been obtained, see [512] and references therein. In this paper, we derive interior and global Hölder estimates for solutions of a class of quasilinear elliptic equations. The key points are the choice of appropriate test functions, the method of iteration and the integral tests developed by Ladyzhenskaya and Ural’tseva, see [1]. Note that we restrict the exponent 1<pn since for the case p>n, the Hölder estimate can be directly obtained by the Sobolev embedding theorem.

2 Preliminary results

The theorems of the following section require some preparatory results which we group together here.

Lemma 2.1 [9]

Let f(τ) be a non-negative bounded function defined for 0 R 0 τ R 1 . Suppose that for R 0 τ<t R 1 we have

f(τ)A ( t τ ) γ +B+θf(t),

where A, B, γ, θ are non-negative constants and θ<1. Then there exists a constant c depending only on γ and θ such that for every ρ, R, R 0 ρ<R R 1 , we have

f(ρ)c [ A ( R ρ ) γ + B ] .

Now we present a very useful lemma which is fundamental in the proof of our theorems, it appears in [1] as follows.

Lemma 2.2 [[1], Lemma 4.8, p.66]

Suppose that the function u(x) is measurable and bounded in some ball B ρ 0 or in a portion of it Ω ρ 0 = B ρ 0 Ω. Consider the balls B ρ and B b ρ , where b is a fixed constant greater than 1, which are concentric with B ρ 0 . Suppose that for arbitrary ρ b 1 ρ 0 , at least one of the following inequalities regarding u(x) is valid:

osc Ω ρ u c 1 ρ ε , osc Ω ρ u ϑ osc Ω b ρ u

for certain positive constants c 1 ,ε1 and ϑ<1. Then, for ρ ρ 0 ,

osc Ω ρ uc ρ 0 m ρ m ,

where

m=min{ log b ϑ,ε},c= b m max { ω 0 , c 1 ρ 0 ε } ,

and

ω 0 = osc Ω ρ 0 u.

3 Main results

3.1 Interior Hölder estimate

Let x 0 Ω and t>0, we denote by B t the ball of radius t centered at x 0 . For k>0, write

A k = { x Ω : u ( x ) > k } , A k , t = A k B t

and denote by B p (Ω,M,γ,δ,1/q) the class of functions u(x) in W 1 , p (Ω) with essential max Ω |u|M such that for u(x) and u(x), the following inequalities are valid in an arbitrary sphere B ρ Ω for arbitrary σ(0,1)

A k , ρ σ ρ | u | p dxγ [ 1 σ p ρ p ( 1 n q ) max A k , ρ ( u k ) p + 1 ] | A k , ρ | 1 p q
(3.1)

for k max B ρ u(x)δ, where the parameters of the class M, γ and δ are arbitrary positive numbers, 1<pn and q>n2. Note that we do not exclude the case q=.

Lemma 3.1 [[1], Lemma 6.2, p.85]

There exists a positive number s such that for an arbitrary ball B ρ belonging to Ω together with the ball B 4 ρ concentric with it and for an arbitrary function u(x) in B p (Ω,M,γ,δ,1/q), at least one of the following two inequalities holds:

osc B ρ u 2 s ρ 1 n q , osc B ρ u ( 1 1 2 s 1 ) osc B 4 ρ u .

To prove the interior Hölder continuity, firstly, by choosing an appropriate test function and making full use of fundamental inequalities, together with Lemma 2.1, we can obtain the following result.

Theorem 3.2 Let 1<pn and f(x) L q p 1 (Ω) for some q>n. Suppose that u is a bounded solution of (1.1), then u B p (Ω,M,γ,δ,1/q).

Proof Let B 1 Ω and 0 R 0 τ<t R 1 be arbitrarily fixed. Let η be a cutoff function such that

η C 0 ( B t ),0η1,η | B τ 1,|η| 2 t τ .

Set M>0 satisfies |u|M. For every k>0, let h>M+k and take

ϕ= T h ( η ( u k ) + ) =max { h , min { h , η ( u k ) + } }

as a test function in (1.2) and obtain

A k , t f ( x ) T h ( η ( u k ) ) d x = Ω f ( x ) T h ( η ( u k ) + ) d x = Ω A ( x , u , u ) T h ( η ( u k ) + ) d x = A k , t A ( x , u , u ) T h ( η ( u k ) ) d x = A k , t A ( x , u , u ) ( η ( u k ) ) d x = A k , τ A ( x , u , u ) u d x + A k , t A k , τ A ( x , u , u ) ( ( u k ) η + η u ) d x .
(3.2)

Then, by applying the structure conditions of mapping A, (3.2) yields

A k , t f ( x ) T h ( η ( u k ) ) d x α A k , τ | u | p d x b A k , t A k , τ ( k ( x ) + | u | p 1 ) | ( u k ) η + η u | d x ,

thus

α A k , τ | u | p d x A k , t | f ( x ) | | η ( u k ) | d x + b A k , t A k , τ ( k ( x ) + | u | p 1 ) | ( u k ) η + η u | d x ,
(3.3)

where b= max [ 0 , M ] b(|s|). Hence it follows from (3.3) and Young’s inequality that

α A k , τ | u | p d x ε A k , t ( u k ) p d x + c ( p , ε ) A k , t | f | p d x + b ε A k , t A k , τ ( k ( x ) + | u | p 1 ) p d x + b c ( n , ε ) A k , t A k , τ | ( u k ) η + η u | p d x c ( b , ε , p ) A k , t A k , τ | u | p d x + c ( b , p ) ε A k , t A k , τ | u | p d x + c ( n , ε , b , p ) ( t τ ) p A k , t A k , τ ( u k ) p d x + ε A k , t ( u k ) p d x + c ( p , ε ) A k , t | f | p d x + c ( b , ε , p ) A k , t A k , τ | k ( x ) | p d x .
(3.4)

Since p < q p 1 for some q>np>1, by applying Hölder’s inequality, we have estimates for the last two integrals on the right-hand side of (3.4)

A k , t | f | p dx | A k , t | 1 p q f q p 1 ; A k , t p p 1 f q p 1 ; Ω p p 1 | A k , t | 1 p q
(3.5)

and

A k , t A k , τ | k | p d x A k , t | k | p d x | A k , t | 1 p q k q p 1 ; A k , t p p 1 k q p 1 ; Ω p p 1 | A k , t | 1 p q .
(3.6)

Since |supp ( u k ) + |=| A k |< 1 k p u p ; Ω p , then there is k 0 >0 such that for k> k 0 , we have | A k | 1 2 | B t |, then we get ( u k ) + W 1 , p ( B t ) and |supp ( u k ) + | 1 2 | B t |.

Take n ˆ =n when p<n, and n ˆ to be any real number satisfying n< n ˆ <q when p=n. Let p ˜ = p n ˆ n ˆ p , then

p ˜ = p n ˆ n ˆ p = { n p n p , p < n , n n ˆ n 2 n ˆ n n n ˆ n 2 n ˆ n ( 1 < n ˆ n 2 n ˆ n < n = p ) , p = n .

According to the Sobolev imbedding inequality, we obtain

( B t | ( u k ) + | p ˜ d x ) 1 p ˜ c ( n , p ) ( B t | ( ( u k ) + ) | p d x ) 1 p = c ( n , p ) ( A k , t | u | p d x ) 1 p .

It then follows from Hölder’s inequality that

A k , t ( u k ) p d x = B t | ( u k ) + | p ˜ p p ˜ 1 d x ( B t ( u k ) p ˜ d x ) p p ˜ | B t | 1 p p ˜ c ( n , p ) | B t | 1 p p ˜ A k , t | u | p d x .
(3.7)

By substituting (3.5)-(3.7) into (3.4), we see that for k> k 0 ,

α A k , τ | u | p d x c 1 A k , t A k , τ | u | p d x + c ( b , p ) ε A k , t A k , τ | u | p d x + c ( n , ε , b , p ) ( t τ ) p A k , t A k , τ ( u k ) p d x + ε c ( n , p ) | B t | 1 p p ˜ × A k , t | u | p d x + c ( p , ε ) f q p 1 ; Ω p p 1 | A k , t | 1 p q + c ( b , ε , p ) k q p 1 ; Ω p p 1 | A k , t | 1 p q ,
(3.8)

where c 1 =c(b,ε,p). Adding to (3.8) both sides

c 1 A k , τ | u | p dx,

we obtain

A k , τ | u | p d x c 1 α + c 1 A k , t | u | p d x + c ( b , p ) ε α + c 1 A k , t A k , τ | u | p d x + c ( n , ε , b , p ) α + c 1 ( t τ ) p max B t ( u k ) p | A k , t | + ε c ( n , p ) α + c 1 | B t | 1 p p ˜ A k , t | u | p d x + c ( b , ε , p ) α + c 1 ( f q p 1 ; Ω p p 1 + k q p 1 ; Ω p p 1 ) | A k , t | 1 p q .
(3.9)

For k> k 0 , we can choose R 1 and ε sufficiently small such that for t R 1 , we get

c ( b , p ) ε α + c 1 + c ( n , p ) ε α + c 1 | B t | 1 p p ˜ 1 2 α α + c 1
(3.10)

and

| A k , t |= | A k , t | 1 p q | A k , t | p q | A k , t | 1 p q | B t | p q = | A k , t | 1 p q t n p q .
(3.11)

By substituting (3.10)-(3.11) into (3.9), we obtain that

A k , τ | u | p d x θ A k , t | u | p d x + γ [ ( t τ ) p t n p q max B t ( u k ) p + 1 ] | A k , t | 1 p q ,
(3.12)

where θ=( 1 2 α+ c 1 )/(α+ c 1 )<1. Thus, let ρ, R be arbitrarily fixed with R 0 ρ<R R 1 , we obtain

A k , τ | u | p d x θ A k , t | u | p d x + γ [ ( t τ ) p R n p q max B R ( u k ) p + 1 ] | A k , R | 1 p q .
(3.13)

Therefore we have deduced that for every t and τ such that R 0 ρτ<tR R 1 , inequality (3.13) holds. Therefore we have from Lemma 2.1 that

A k , ρ | u | p dxC [ ( R ρ ) p R n p q max B R ( u k ) p + 1 ] | A k , R | 1 p q .
(3.14)

Since u is a solution to equation (1.1), we have that −u is a solution to the equation

div A ˜ (x,v,v)=f(x),

where A ˜ (x,v,v)=A(x,v,v). And the operator A ˜ satisfies the same structure conditions (a)-(c), hence the same inequality (3.14) holds with u replaced by −u. Therefore we get that the function u B p (Ω,M,γ,δ,1/q). □

Remark Especially, if q=, i.e., f(x),k(x) L (Ω), then condition (a) of A simplifies into |A(x,u,u)|b(M) | u | p 1 +b(M). Proceeding the process of the proof in Theorem 3.2, we finally have

A k , τ | u | p dxθ A k , t | u | p dx+ ( t τ ) p A k , R ( u k ) p dx+| A k , R |.

Therefore we have from Lemma 2.1 that

A k , ρ | u | p dxC ( ( R ρ ) p A k , R ( u k ) p d x + | A k , R | ) .

Similarly, we get that the same inequality holds with u replaced by −u, hence the function u B p (Ω,M,γ,δ,0).

Then, by applying Lemma 3.1 and Lemma 2.2, we obtain, for arbitrary ρ ρ 0 , that

osc B ρ u 4 m ( ρ ρ 0 ) m ( osc B ρ 0 u + 2 s ρ 0 1 n q ) ,

where m=min{ log 4 (1 1 2 s 1 ),1 n q }. Choosing ρ= ρ 0 /5, we have the following oscillation estimate which is important and fundamental in our main results.

Proposition 3.3 Suppose that u(x) W loc 1 , p (Ω) is a bounded weak solution of (1.1), then

osc B R 5 uγ osc B R u+C R 1 n q
(3.15)

holds for any ball B R Ω, where γ=γ(n,q,s)(0,1), and C is a positive constant depending on n, q and s.

The oscillation estimate Proposition 3.3 can be used to obtain the following interior Hölder estimate by the method of iteration.

Theorem 3.4 Suppose that u(x) W loc 1 , p (Ω) is a bounded weak solution of (1.1), then

osc B r u 5 κ ( r R ) κ osc B R u+C r κ
(3.16)

for any ball B R Ω and 0<r<R<, where κ(0,1) depends on n, q and s, and C=C(n,q,s,Ω).

Proof Let 0<ρR, B R Ω, then Proposition 3.3 yields

osc B ρ 5 uγ osc B ρ u+C ρ 1 n q .
(3.17)

Let ρ 0 =R, ρ ι = 5 ι ρ 0 for ι=0,1,2, and take b= γ + 1 2 (0,1). Let ω ι = osc B ρ ι u, where spheres B ρ ι are concentric with B ρ 0 , and

κ=min { log 5 b γ , 1 n q } .

Then we have 0<κ<1, 5 κ γb. Moreover, denote c 1 = 5 κ C R 1 n q .

Observe that (3.17) implies that

osc B ρ ι 1 5 uγ osc B ρ ι 1 u+C ρ ι 1 1 n q .
(3.18)

From the notations above, we can get from (3.18) that

ω ι γ ω ι 1 +C ρ ι 1 n q .
(3.19)

Write y ι = 5 ι κ ω ι , then we have from (3.19) that

y ι b y ι 1 + c 1 5 κ .
(3.20)

We obtain

y ι b y ι 1 + c 1 5 κ b ( b y ι 2 + c 1 5 κ ) + c 1 5 κ = b 2 y ι 2 + b c 1 5 κ + c 1 5 κ b ι y 0 + ( b ι 1 + + b + 1 ) c 1 5 κ y 0 + 1 1 b c 1 5 κ .

Thus we have

ω ι = 5 ι κ y ι 5 ι κ y 0 + 1 1 b c 1 5 κ 5 ι κ = y 0 ( ρ ι ρ 0 ) κ + 1 1 b c 1 5 κ ( ρ ι ρ 0 ) κ .
(3.21)

For arbitrary 0<r<R= ρ 0 <, there exists ι 0 1 such that ρ ι 0 r ρ ι 0 1 . We obtain

osc B r u osc B ρ ι 0 1 u y 0 5 κ ρ 0 κ ρ ι 0 κ + 1 1 b c 1 ρ 0 κ ρ ι 0 κ 5 κ y 0 ρ 0 κ r κ + 1 1 b c 1 ρ 0 κ r κ = 5 κ ( r R ) κ osc B R u + 1 1 b 5 κ C R 1 n q R κ r κ 5 κ ( r R ) κ osc B R u + C R ( 1 n q ) κ r κ 5 κ ( r R ) κ osc B R u + C r κ ,

where C=C(n,q,s,Ω). □

As an important application of Theorem 3.4, we investigate the global Hölder continuity of weak solutions of (1.1), which is the main result of the paper.

Theorem 3.5 Suppose that u(x)C( Ω ¯ ) W loc 1 , p (Ω) is a weak solution of (1.1). If there are constants L0 and 0<δ1 such that

| u ( x ) u ( x 0 ) | L | x x 0 | δ
(3.22)

for all xΩ and x 0 Ω, then there exist constants L 1 0 and 0< δ 1 1 such that

| u ( x ) u ( y ) | L 1 | x y | δ 1
(3.23)

for all x,y Ω ¯ .

Proof For arbitrary x,y Ω ¯ , we discuss the following three cases:

  1. (A)

    xΩ, yΩ or xΩ, yΩ:

We have by (3.22) that (3.23) holds with δ 1 =δ, and L 1 =L.

  1. (B)

    x,yΩ:

For xΩ, there exists { x k }Ω such that x k x as k. Since uC( Ω ¯ ), we have u( x k )u(x) and | x k y | δ | x y | δ as k. Then we get

| u ( x ) u ( y ) | | u ( x ) u ( x k ) | + | u ( x k ) u ( y ) | | u ( x ) u ( x k ) | + L | x k y | δ .

Let k, then (3.23) is obtained with δ 1 =δ and L 1 =L.

  1. (C)

    x,yΩ:

For case (C), we consider two cases (I) and (II):

  1. (I)

    |xy| 1 2 dist(x,Ω).

Choose x 0 Ω such that | x 0 x|=dist(x,Ω)=r. Then, for arbitrary z B r 2 (x),

| u ( z ) u ( x ) | | u ( z ) u ( x 0 ) | + | u ( x 0 ) u ( x ) | L | z x 0 | δ + L | x 0 x | δ L ( | z x | + | x x 0 | ) δ + L | x 0 x | δ L ( 3 2 r ) δ + L r δ = L r δ ( 1 + ( 3 2 ) δ ) 5 2 L r δ ;

therefore we have, for all z 1 , z 2 B r 2 (x),

| u ( z 1 ) u ( z 2 ) | | u ( z 1 ) u ( x ) | + | u ( x ) u ( z 2 ) | 5 L r δ .

Therefore we obtain osc B r 2 ( x ) u5L r δ . Since x,y B ¯ (x,|xy|) B r 2 (x)Ω, it follows from (3.16) that

| u ( x ) u ( y ) | osc B ¯ ( x , | x y | ) u = osc B ( x , | x y | ) u 5 κ ( | x y | r / 2 ) κ osc B r 2 ( x ) u + C | x y | κ 50 L | x y | κ r δ κ + C | x y | κ .
(3.24)

To estimate (3.24), we consider two cases (i) and (ii).

  1. (i)

    If δ<κ, then δ=min{δ,κ} δ 1 . Since κδ>0 and |xy|<r/2<r<diamΩ, thus | x y | κ δ < r κ δ and | x y | κ δ < ( diam Ω ) κ δ . Then we obtain

    | u ( x ) u ( y ) | 50L | x y | δ 1 +C ( diam Ω ) κ δ 1 | x y | δ 1 .
    (3.25)
  2. (ii)

    If δκ, then κ=min{δ,κ} δ 1 . Since δκ0 and r=dist(x,Ω)diamΩ, thus r δ κ ( diam Ω ) δ κ . Then we obtain

    | u ( x ) u ( y ) | 50L ( diam Ω ) δ δ 1 | x y | δ 1 +C | x y | δ 1 .
    (3.26)

Therefore we have the estimate for case (I) by substituting (3.25) and (3.26) into (3.24) so that

| u ( x ) u ( y ) | ( 50 L max ( 1 , ( diam Ω ) δ δ 1 ) + C max ( 1 , ( diam Ω ) κ δ 1 ) ) | x y | δ 1 .
(3.27)

Next, we estimate case (II) of case (C).

  1. (II)

    |xy| 1 2 dist(x,Ω).

Choose x 0 Ω such that | x 0 x|=dist(x,Ω)=r. Then we have

| u ( x ) u ( y ) | | u ( x ) u ( x 0 ) | + | u ( x 0 ) u ( y ) | L | x x 0 | δ + L | x 0 y | δ L | x x 0 | δ + L ( | x 0 x | + | x y | ) δ L 2 δ | x y | δ + L 3 δ | x y | δ 5 L | x y | δ .
(3.28)

Similarly, to estimate (3.28), we consider two cases (iii) and (iv).

  1. (iii)

    If δ<κ, then δ=min{δ,κ}= δ 1 . Then we obtain

    | u ( x ) u ( y ) | 5L | x y | δ 1 .
    (3.29)
  2. (iv)

    If δκ, then δ 1 =κ, thus we have | x y | δ = | x y | δ 1 | x y | δ δ 1 ( diam Ω ) δ δ 1 | x y | δ 1 . Then we obtain

    | u ( x ) u ( y ) | 5L ( diam Ω ) δ δ 1 | x y | δ 1 .
    (3.30)

Therefore we have the estimate for case (II) by substituting (3.29) and (3.30) into (3.28) so that

| u ( x ) u ( y ) | 5Lmax ( 1 , ( diam Ω ) δ δ 1 ) | x y | δ 1 .
(3.31)

Finally, combined with (3.27) and (3.31), we have the estimate for case (C)

| u ( x ) u ( y ) | ( 50 L max ( 1 , ( diam Ω ) δ δ 1 ) + C max ( 1 , ( diam Ω ) κ δ 1 ) ) | x y | δ 1 .
(3.32)

Therefore the theorem follows with

δ 1 =min{δ,κ}(0,1]

and

L 1 =50Lmax ( 1 , ( diam Ω ) δ δ 1 ) +Cmax ( 1 , ( diam Ω ) κ δ 1 ) .

 □

3.2 Global Hölder estimate

In order to extend the above results to a global Hölder estimate, we need to place an additional constraint on Ω.

Definition 3.6 [1]

We shall say that the boundary Ω of Ω satisfies condition (A) if there exist two positive numbers a 0 and θ 0 such that for an arbitrary ball B ρ with center on Ω of radius ρ a 0 and for an arbitrary component Ω ˆ ρ of B ρ Ω, the inequality

| Ω ˆ ρ |(1 θ 0 )| B ρ |

holds.

Now let

Ω t =Ω B t , Ω k , t = A k Ω t ,

and let B p ( Ω ¯ ,M,γ,δ,1/q) be the class of functions u(x) in B p (Ω,M,γ,δ,1/q) that, together with their negatives, satisfy inequality (3.1) for the balls B ρ with B ρ Ω, the integration region Ω k , ρ , and for k max Ω ρ u(x)δ and k max B ρ Ω u(x).

Lemma 3.7 [[1], Lemma 7.1, p.92]

If Ω satisfies condition (A) and if the function u(x) in B p ( Ω ¯ ,M,γ,δ,1/q) satisfies on Ω a Hölder condition, more precisely, if

osc Ω B ρ uL ρ ϵ ,ϵ>0,
(3.33)

for balls B ρ (where ρ a 0 ) with centers on Ω, then there exists a positive number s such that for an arbitrary ball B ρ , for a ball B 4 ρ (where 4ρ 1 4 min{ a 0 ,1}) with center on Ω that is concentric with it, at least one of the following inequalities holds:

osc Ω ρ u 2 s ρ ϵ 1 , ϵ 1 = min { 1 n q , ϵ } , osc Ω ρ u ( 1 1 2 s 1 ) osc Ω 4 ρ u .

The number s is determined by the parameters of the class B p , by the numbers ϵ and L in (3.33), and by the numbers a 0 and θ 0 in condition (A).

Analogously, we proceed the proof basically the same [10] as Theorem 3.2, and we can prove that u B p ( Ω ¯ ,M,γ,δ,1/q). By applying Lemma 3.7 and Lemma 2.2, for

ρ ρ 0 = 1 4 min{ a 0 ,1},

we obtain that

osc Ω ρ u 4 m ( ρ ρ 0 ) m ( osc Ω ρ 0 u + 2 s ρ 0 ϵ 1 ) ,

where ϵ 1 =min{1 n q ,ϵ}, m=min{ log 4 (1 1 2 s 1 ), ϵ 1 } and ϵ is in (3.33). Choosing ρ= ρ 0 /5, we have the following oscillation estimate.

Proposition 3.8 Suppose that u(x) W loc 1 , p (Ω) is a bounded weak solution of (1.1), then

osc Ω R 5 uγ osc Ω R u+C R ϵ 1 ,

where γ(0,1) and C are positive constants depending on n, q, s and ϵ in (3.33), holds.

Proceeding completely analogously to the proof of Theorem 3.4, we obtain the following.

Theorem 3.9 Suppose that u(x) W loc 1 , p (Ω) is a bounded weak solution of (1.1), then

osc Ω r u 5 κ ( r R ) κ osc Ω R u+C r κ
(3.34)

for any ball B R and 0<r<R< ρ 0 , where κ(0,1) and C>0 depends on n, q, s and ϵ.

We now have the following global Hölder estimate based on the boundary Hölder continuity.

Theorem 3.10 Suppose that Ω is bounded and satisfies condition (A). Let u(x)C( Ω ¯ ) W 1 , p (Ω) is a weak solution of (1.1). If there are constants M0 and 0<γ1 such that

| u ( x ) u ( y ) | M | x y | γ
(3.35)

for all x,yΩ, then there exist constants M 1 0 and γ 1 >0 such that

| u ( x ) u ( y ) | M 1 | x y | γ 1
(3.36)

for all x,y Ω ¯ .

Proof It is clear that we just need to prove (3.22) according to Theorem 3.5. For all xΩ and x 0 Ω, we consider the following two cases:

  1. (i)

    If r=|x x 0 |<1, then there exists r 0 such that r=|x x 0 |< r 0 1. The boundary Hölder estimate (3.34) with R= r 1 2 yields

    | u ( x ) u ( x 0 ) | osc Ω B r ( x 0 ) u C r κ 2 osc Ω B R ( x 0 ) u + C r κ .
    (3.37)

Since u(x)C( Ω ¯ ) and Ω is bounded, we have

sup Ω |u| C 1 <.
(3.38)

Therefore we get by substituting (3.38) into (3.37) that

| u ( x ) u ( x 0 ) | osc Ω B r ( x 0 ) u C 2 C 1 | x x 0 | κ 2 + C | x x 0 | κ C ( 2 C 1 + 1 ) | x x 0 | κ 2 .
(3.39)
  1. (ii)

    If r=|x x 0 |1, then

    | u ( x ) u ( x 0 ) | 2 sup Ω |u|2 C 1 | x x 0 | κ 2 .
    (3.40)

Combining (3.39) and (3.40), we have

| u ( x ) u ( x 0 ) | M 1 | x x 0 | κ 2

with

M 1 =max(C2 C 1 +C,2 C 1 ).

Therefore the theorem follows from Theorem 3.5. □

4 Application

We conclude this paper with an application of Theorem 3.10 in a simple case of (1.1). We consider the following equation:

D i ( a i j ( x ) D j u ) =f(x),
(4.1)

where f(x) L q 2 (Ω) for some q>n, and the coefficients a i j (i,j=1,,n) are assumed to be measurable functions on Ω, and there exist positive constants λ and Λ such that

a i j (x) ξ i ξ j λ | ξ | 2 ,xΩ,ξ R n
(4.2)

and

Σ | a i j ( x ) | 2 Λ 2 .
(4.3)

Let

A i (x,u)= a i j (x) D j u,

then we can easily prove that the operator A satisfies the structural assumption (a)-(c).

To apply Theorem 3.10 for (4.1), we put a more general constraint on Ω to obtain the Hölder continuity up to boundary.

Definition 4.1 [13]

We shall say that Ω satisfies an exterior cone condition at a point x 0 Ω if there exists a finite right circular cone V= V x 0 with vertex x 0 such that Ω ¯ V x 0 = x 0 .

Definition 4.2 [13]

Let us say that Ω satisfies a uniform exterior cone condition on Ω if Ω satisfies an exterior cone condition at every x 0 Ω and the cones V x 0 are all congruent to some fixed cone V.

We now have the following Hölder estimate at the boundary.

Theorem 4.3 [13]

Suppose that u(x) is a W 1 , 2 (Ω) solution of (4.1) in Ω and Ω satisfies an exterior cone condition at a point x 0 Ω. We have, for any 0<rρ,

osc Ω B r ( x 0 ) uC [ ( r ρ ) κ sup Ω B ρ ( x 0 ) | u | + r κ λ 1 f L q 2 ( Ω ) + osc Ω B r ρ ( x 0 ) u ] ,
(4.4)

where C=C(n,λ,Λ,q,ρ, V x 0 ), κ=κ(n,λ,Λ,q, V x 0 ) are positive constants.

Theorem 4.4 Suppose that Ω is bounded and satisfies the uniform exterior cone condition. Let u(x)C( Ω ¯ ) W 1 , 2 (Ω) be a weak solution of (4.1). If there are constants M0 and 0<γ1 such that

| u ( x ) u ( y ) | M | x y | γ
(4.5)

for all x,yΩ, then there exist constants M 1 0 and γ 1 >0 such that

| u ( x ) u ( y ) | M 1 | x y | γ 1
(4.6)

for all x,y Ω ¯ .

Proof It is clear that we just need to prove (3.22) according to Theorem 3.5. For all xΩ and x 0 Ω, we consider the following two cases:

  1. (i)

    If r=|x x 0 |<1, then there exists r 0 such that r=|x x 0 |< r 0 1. The boundary Hölder estimate (4.4) with ρ= r 1 2 yields

    | u ( x ) u ( x 0 ) | osc Ω B r ( x 0 ) u C [ r κ 2 sup Ω B ρ ( x 0 ) | u | + r κ λ 1 f L q 2 ( Ω ) + osc Ω B r 3 4 ( x 0 ) u ] .
    (4.7)

Since u(x)C( Ω ¯ ) and Ω is bounded, we have

sup Ω |u| C 1 <,
(4.8)

and for all z 1 , z 2 Ω B r 3 4 ( x 0 ) , we have from (4.5) that

| u ( z 1 ) u ( z 2 ) | M | z 1 z 2 | γ M ( | z 1 x 0 | + | z 2 x 0 | ) γ 2 M r 3 4 γ .
(4.9)

Thus (4.9) yields that

osc Ω B r 3 4 ( x 0 ) u2M r 3 4 γ .
(4.10)

Therefore we get by substituting (4.8) and (4.10) into (4.7) that

| u ( x ) u ( x 0 ) | osc Ω B r ( x 0 ) u C C 1 | x x 0 | κ 2 + C f L q 2 ( Ω ) | x x 0 | κ + C M | x x 0 | 3 4 γ C max ( C 1 , f L q 2 ( Ω ) , M ) | x x 0 | γ 1 ,
(4.11)

where γ 1 =min( κ 2 , 3 4 γ).

  1. (ii)

    If r=|x x 0 |1, then

    | u ( x ) u ( x 0 ) | 2 sup Ω |u|2 C 1 | x x 0 | γ 1 .
    (4.12)

Combining (4.11) and (4.12), we have

| u ( x ) u ( x 0 ) | M 1 | x x 0 | γ 1

with

M 1 =max ( C C 1 , C f L q 2 ( Ω ) , C M , 2 C 1 )

and

γ 1 =min ( κ 2 , 3 4 γ ) .

Therefore the theorem follows from Theorem 3.5. □