Convergence for elliptic equations in periodic perforated domains

J. Differential Equations 255 (2013) 1734–1783

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Journal of Differential Equations www.elsevier.com/locate/jde

Convergence for elliptic equations in periodic perforated domains Li-Ming Yeh Department of Applied Mathematics, National Chiao Tung University, Hsinchu, 30050, Taiwan, ROC

a r t i c l e

i n f o

a b s t r a c t

Article history: Received 20 December 2012 Revised 9 May 2013 Available online 14 June 2013

Convergence for the solutions of elliptic equations in periodic perforated domains is concerned. Let  denote the size ratio of the holes of a periodic perforated domain to the whole domain. It is known that, by energy method, the gradient of the solutions of elliptic equations is bounded uniformly in  in L 2 space. Also, when  approaches 0, the elliptic solutions converge to a solution of some simple homogenized elliptic equation. In this work, above results are extended to general W 1, p space for p > 1. More precisely, a uniform W 1, p estimate in  for p ∈ (1, ∞] and a W 1, p n convergence result for p ∈ ( n− , ∞] for the elliptic solutions in 2 periodic perforated domains are derived. Here n is the dimension of the space domain. One also notes that the L p norm of the second order derivatives of the elliptic solutions in general cannot be bounded uniformly in  . © 2013 Elsevier Inc. All rights reserved.

MSC: 35J05 35J15 35J25 Keywords: Periodic perforated domain Homogenized elliptic equation

1. Introduction Convergence for the solutions of elliptic equations in periodic perforated domains is presented. Let Y ≡ (0, 1)n for n  3 consist of a sub-domain Y m completely surrounded by another connected  ≡ {x | x ∈  (Y − j ) for some j ∈ Zn } with boundary ∂Ω  , sub-domain Y f (≡ Y \ Y m ),  ∈ (0, 1], Ωm m m  be a connected region. The problem that we consider is and Ω f ≡ Rn \ Ωm

⎧ −∇ · (∇ U  + Q  ) = F  in Ω f , ⎪ ⎨ ,  = 0 (∇ U  + Q  ) · n on ∂Ωm ⎪ ⎩ |U  |(x) = o(1) for large |x|, E-mail address: [email protected]. 0022-0396/$ – see front matter © 2013 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.jde.2013.05.023

(1.1)

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  is a unit normal vector on ∂Ωm . |U  |(x) = o(1) for large |x| where Q  , F  are given functions and n means lim|x|→∞ |U  |(x) = 0. If Q  , F  both are bounded with compact support, by energy method, a solution of (1.1) in Hilbert space D 1,2 (Ω f ) (see definition in Section 2) exists uniquely for each  . The L 2 norm of the gradient of the solution of (1.1) in Ω f is bounded uniformly in

 . If, in addition,

Q  = 0 and F  = F in (1.1), by compactness principle [3], there exists a function U 0 ∈ D 1,2 (Rn ) such that the solution U  of (1.1) satisfies

  ∇ U  XΩ f → K∇ U 0 in L 2 Rn weakly as  → 0,

(1.2)

where XΩ  is the characteristic function on Ω f and K is a positive definite matrix depending on Y f . f

Moreover, U 0 in (1.2) satisfies



−∇ · (K∇ U 0 ) = |Y f | F

in Rn ,

|U 0 |(x) = o(1)

for large |x|,

(1.3)

where |Y f | is the volume of Y f . It is interesting to know whether a uniform bound of the solution of (1.1) in  in Ω f can be derived in L p space and whether a convergence rate of U  can be obtained

in L p space for any p ∈ (1, ∞] or not. One example in Remark 2.1 below shows that the L p estimate of the second derivatives of the solution of (1.1) may not be bounded uniformly in  in general. There are some literatures related to this work. Lipschitz estimate and W 2, p estimate for uniform elliptic equations with discontinuous coefficients had been proved in [13,17]. Uniform Hölder, L p , and Lipschitz estimates in  for uniform elliptic equations with periodic smooth oscillatory coefficients were proved in [4,5]. Uniform Lipschitz estimate in  for the Laplace equation in periodic perforated domains was considered in [22]. Uniform Hölder and Lipschitz estimates in  for non-uniform elliptic equations with periodic discontinuous oscillatory coefficients were shown in [24]. By homogenization theory, the solutions of elliptic equations in periodic perforated domains in general converge to a solution of some homogenized elliptic equation with convergence rate  in L 2 norm and with convergence rate  1/2 in H 1 norm as  closes to 0 (see [6,12,20] and references therein). Higher order asymptotic expansion for the solutions of elliptic equations in perforated domains could be found in [7,14]. Rigorous proof of higher order convergence rate for the solution of (1.1) in Hilbert spaces was considered in [6,8,20]. In this work, we shall derive a uniform W 1, p estimate in  for the solution of (1.1) in Ω f and prove a W 1, p convergence result with convergence rate  for the solution of (1.1) for p > 1 case. The approaches to derive above results are similar to those in [4,5] and are based on the following steps: First we prove the existence of the Green’s function of the Laplace operator in perforated domains. Next we find approximations of the Green’s function of the Laplace operator in perforated domains. Then we derive a uniform W 1, p estimate in  for elliptic equations in perforated domains. Finally an asymptotic expansion technique is used to derive the W 1, p convergence rate for the solutions of (1.1) and (1.3). Concerning the approximation of the Green’s function of the Laplace operator in bounded perforated domains, some uniform approximations in  in L ∞ space for the Green’s function can be found in the review paper [18] and references therein. However, they are not enough for our purpose. Uniform approximations in  in L ∞ space for the zero, the first, and the second order derivatives of the Green’s function are needed here, and they are derived by an approach different from [18]. The rest of this manuscript is organized as follows: Notation and main results are stated in Section 2. In Section 3, we present uniform Hölder and uniform Lipschitz estimates in  for the solutions of elliptic equations in periodic perforated domains; prove an L ∞ convergence result for an elliptic equation in perforated domains; derive a priori estimates for some interface problems; show the existence of the Green’s function of the Laplace operator in perforated domains; and give some estimates for the zero order and the first order derivatives of the Green’s function. Approximation of the second order derivatives of the Green’s function of the Laplace operator in perforated domains is derived in Section 4. Main results (that is, a uniform W 1, p estimate in  and a W 1, p convergence result for

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the solution of (1.1)) are shown in Section 5. Finally, uniform Hölder and uniform Lipschitz estimates in  for the solutions of elliptic equations in perforated domains claimed in Section 3 are proved in Section 6. 2. Notation and main result W s, p (D) denotes a Sobolev space with norm  ·  W s, p (D) , C k,α (D) is a Hölder space with s, p

k ,α

norm  · C k,α (D) , W loc (D) ≡ {ζ | ζ ∈ W s, p ( D ) for any compact subset D of D}, C loc (D) ≡ {ζ | ζ ∈

( D ) for any compact subset D of D}, and [ζ ]C 0,α is the Hölder semi-norm of ζ , where s  −1, s ,2 s α ∈ [0, 1] (see [2,11]). H s (D) ≡ W s,2 (D), L p (D) ≡ W 0, p (D), H loc (D) ≡ W loc (D), ∞ 0,0 ∞ C (D) ≡ C (D). C (D) is a space of infinitely differentiable functions in D, C 0 (D) is a subset ∞ (Rn ) is a subset of C ∞ (Rn ) of (0, 1)n -periodic of C ∞ (D) with compact support in D, and C per C

k ,α

p ∈ [1, ∞], k  0,

s ∞ (Rn ) under the H s (resp. C k,α ) norm and (D) (resp. C per (D)) is the closure of C per functions. H per k ,α

s, p

s ζ  H per (D) ≡ ζ  H s (D∩(0,1)n ) (resp. ζ C k,α (D) ≡ ζ C k,α (D∩(0,1)n ) ) for s  1, k  0, α ∈ [0, 1]. W 0 (D) is per

the closure of C 0∞ (D) under the W s, p norm for s  1, p ∈ (1, ∞). supp(ζ ) is the support of ζ . For Banach spaces B1 and B2 , define ζ1 , . . . , ζk B1 ≡ ζ1 B1 + · · · + ζk B1 and ζ B1 ∩B2 ≡ ζ B1 + ζ B2 . Let B r (x) denote a ball centered at x with radius r > 0. For any set D ⊂ Rn , |D| is the volume of D, D is the closure of D, XD is the characteristic function on D, D/r ≡ {x | rx ∈ D} for r > 0, dist(x, ∂D) is the distance between x and ∂D, and

1 − ζ (x) dx ≡ ζ (x) dx if ζ ∈ L 1 (D). |D| D

D



If ζ ∈ L 1 (D), (ζ )x,r ≡ − B (x)∩D ζ ( y ) dy. Let Ωm ≡ {x | x ∈ Y m − j for some j ∈ Zn }, Ω f ≡ Rn \ Ωm , r Ωmω ≡ ωΩm , Ω ωf ≡ ωΩ f , and ∂Ωmω be the boundary of Ωmω for any ω ∈ (0, ∞). D1,2 (Rn ) ≡ 2n

{ζ ∈ L n−2 (Rn ) | ∇ζ ∈ L 2 (Rn )} under the norm ζ D1,2 (Rn ) ≡ ∇ζ  L 2 (Rn ) is a Hilbert space (see p. 168 [15]). D 1,2 (Ω f ) ≡ {ζ |Ω  | for ζ ∈ D 1,2 (Rn )} for any  ∈ (0, 1] with norm ζ D1,2 (Ω  ) ≡ f f

∇ζ L 2 (Ω  ) and D1,2 (Rn \ Y m ) ≡ {ζ |Rn \Y m | for any ζ ∈ D1,2 (Rn )} with norm ζ D1,2 (Rn \Y m ) ≡ f

∇ζ L 2 (Rn \Y m ) . It is easy to see that both D1,2 (Ω f ), D1,2 (Rn \ Y m ) are also Hilbert spaces. Denote by xi for i ∈ {1, . . . , n} the i-th component of x ∈ Rn , Rn+ ≡ {x | xn > 0}, and ∂Rn+ ≡ {x | xn = 0}. Next we present our main results.

Theorem 2.1. Suppose A1. Y m is a smooth simply-connected sub-domain of (0, 1)n for n  3, A2.  ∈ (0, 1], p ∈ (1, ∞), Q  ∈ L p (Ω f ) with support in B t (0) for some t > 0, 1, p

then a W loc (Ω f ) solution of

⎧  ⎪ ⎪ −∇ · (∇ U  + Q  ) = 0 in Ω f , ⎨ ⎪ ⎪ ⎩

 = 0 (∇ U  + Q  ) · n

, on ∂Ωm

|U  |(x) = o(1)

for large |x|

exists uniquely. The solution U  satisfies

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⎧ for p ∈ (1, ∞), ∇ U  L p (Ω f )  c  Q  L p (Ω f ) ⎪ ⎪ ⎪ ⎨ U   nnp  c  Q  L p (Ω f ) for p ∈ (1, n), L − p (Ω f ) ⎪ ⎪ ⎪ ⎩ U  p  L ( B s (0)∩Ω f )  ct ,s  Q   L p (Ω f ) for p ∈ (1, ∞), where s > 0; c is a positive constant independent of  , Q  ; and ct ,s is a positive constant independent of  but dependent on t, s. Theorem 2.2. Suppose A1 and A3.

 ∈ (0, 1], p ∈ (1, ∞), F  ∈ W −1, p (Rn ) with support in B t (0) for t > 0, 1, p

then a W loc (Ω f ) solution of

⎧ in Ω f , −U  = F  ⎪ ⎨   = 0 on ∂Ωm , ∇U · n ⎪ ⎩ |U  |(x) = o(1) for large |x|

(2.1)

U  W 1, p ( B s (0)∩Ω  )  ct ,s  F  X B t (0)∩Ω f W −1, p (Rn ) ,

(2.2)

exists uniquely and satisfies

f

where s > 0 and ct ,s is a constant independent of compact support, the solution of (2.1) satisfies

 but dependent on t, s. If, in addition, F  ∈ L p (Rn ) has

⎧  c  F  L p (Ω f ) for p ∈ (1, n), ∇ U   nnp ⎪ ⎨ L − p (Ω f ) ⎪ ⎩ U  

np L n−2p

(Ω  ) f

 c  F  L p (Ω f )

(2.3)

for p ∈ (1, n/2),

where c is a constant independent of  , F  . Theorem 2.1 and Theorem 2.2 imply that n Corollary 2.1. Suppose A1,  ∈ (0, 1], p ∈ ( n− , ∞), and 2 np

A4. Q  ∈ L p (Ω f ), F  ∈ L n+ p (Rn ) have compact support, then a W 1, p (Ω f ) solution of (1.1) exists uniquely and satisfies

 U  W 1, p (Ω  )  c  Q  L p (Ω f ) +  Q  , F   f

np

L n+ p (Ω f )

+ F 

np L n+2p

(Ω  )

 ,

f

where c is a constant independent of  , Q  , F  . Theorem 2.3. If A1 holds,  ∈ (0, 1], and both Q  , F  ∈ L n+δ (Rn ) for any δ > 0 have compact support, then a D 1,2 (Ω f ) solution of

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L.-M. Yeh / J. Differential Equations 255 (2013) 1734–1783



−∇ · (∇ U  + Q  ) = F  in Ω f ,  = 0 (∇ U  + Q  ) · n

(2.4)

 on ∂Ωm

exists uniquely and satisfies

U  L ∞ (Ω f )  c  Q  , F  Ln+δ (Rn ) ,

(2.5)

where c is a constant independent of  , Q  , F  . If, in addition, Q  = 0 in Rn , then

∇ U  L ∞ (Ω f )  c  F  Ln+δ (Rn ) ,

(2.6)

where c is a constant independent of  , F  . 1 (Ω f ) for i ∈ {1, . . . , n} satisfy, in the cell Y f , Let X(i ) ( y ) ∈ H per

⎧   ⎪ −∇ · ∇X(i ) + ei = 0 in Y f , ⎪ ⎪ ⎪   ⎪ ⎪ ⎨ ∇X(i ) + ei · n  y = 0 on ∂ Y m , ⎪ ⎪ ⎪ X(i ) dy = 0, ⎪ ⎪ ⎪ ⎩Y

(2.7)

f

 y denotes a unit normal vector on ∂ Y m and ei is the unit vector in the i-th coordinate where n direction in Rn . Define

 ⎧ (i ) (i ) y ⎪ ⎪ ( y ) ≡  X for i ∈ {1, . . . , n}, X ⎪  ⎪  ⎨  (1 ) (n)  ⎪ X ≡ X , . . . , X , ⎪ ⎪ ⎪   ⎩ X ≡ X(1) , . . . , X(n) .

(2.8)

Denote by Ξ an n × n matrix function whose (i , j )-component is ∂i X( j ) and define

K≡





I + Ξ ( y ) dy ,

(2.9)

Yf

where I is the identity matrix. By [3], K is a constant symmetric positive definite matrix. For i 1 , i 2 ∈ 1 {1, . . . , n}, find S(i 1 ,i 2 ) ( y ) ∈ H per (Ω f ) satisfying, in the cell Y f ,

⎧ Ki ,i ⎪ ⎪ S(i 1 ,i 2 ) + ∂i 1 X(i 2 ) = −δi 1 ,i 2 − ∂i 1 X(i 2 ) + 1 2 ⎪ ⎪ |Y f | ⎪ ⎪ ⎪ ⎪ ⎨ (i 1 ,i 2 ) ( i2 )  y + X n yi = 0 ∇S ·n 1 ⎪ ⎪ ⎪ ⎪ ⎪ S(i 1 ,i 2 ) dy = 0, ⎪ ⎪ ⎪ ⎩Y f

in Y f , on ∂ Y m ,

(2.10)

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where

δi 1 ,i 2 ≡

1 if i 1 = i 2 , 0

if i 1 = i 2 ,

 y ≡ (n y1 , . . . , n yn ). Ki 1 ,i 2 is the (i 1 , i 2 )-component of K in (2.9), and n y i1 is the i 1 -th component of n y

Let S( y ) ≡ (S(i 1 ,i 2 ) ( y )) be an n × n matrix function and S ( y ) ≡  2 (S(i 1 ,i 2 ) (  )). By energy method and Lax–Milgram Theorem [11], X(i 1 ) in (2.7) and S(i 1 ,i 2 ) in (2.10) for i 1 , i 2 ∈ {1, . . . , n} are solvable 1 (Ω f ). By Theorem 6.30 [11] and tracing the proof of Theorem 8.24 [11], if Y m is a uniquely in H per C 2,α domain for some

α ∈ (0, 1), then

 (i )  X 1 

2,α C per (Ω f )

  + S(i 1 ,i 2 ) C 2,α (Ω )  c for i 1 , i 2 ∈ {1, . . . , n}, per

f

(2.11)

where c is a constant. Remark 2.1. The W 2, p norm of the solution of (1.1) in general is not bounded uniformly in  even if Q  , F  have compact supports and  Q   W 1, p (Ω  ) ,  F   L p (Ω  ) are bounded independent of  . One f f example is as follows. Suppose η is a bell-shaped smooth function satisfying η ∈ C 0∞ ( B 1 (0)), η ∈ [0, 1], and η(x) = 1 in B 1/2 (0). By (2.7) and (2.8), we have the following equation

⎧   (1 )    (1 )  (1 ) ⎪ −∇ · ∇ ηX − X ∇ η + ηe1 = − ∇X + e1 ∇ η in Ω f , ⎪ ⎪ ⎨    (1 )  (1 ) ,  =0 ∇ ηX − X ∇ η + ηe1 · n on ∂Ωm ⎪ ⎪ ⎪ ⎩  (1)  ηX (x) = 0 for large |x|. By (2.11), we see

 (1 )     X ∇ η − ηe1  1, p  +  ∇X(1) + e1 ∇ η p  W (Ω ) L (Ω ) f

f

(1)

is bounded uniformly in  , but ηX  W 2, p (Ω  ) is not bounded uniformly in f

.

Let Π for  ∈ (0, 1] denote an extension operator in Ω f , which maps a function ζ in Ω f to Π ζ in Rn and still keeps the same regularity of ζ (see [1] for the existence of such an operator). By Theorem 4.31 [2] and extension theorem [1], we know, for any ζ ∈ W 1, p (Ω f ), p ∈ (1, n), and  ∈ (0, 1],

ζ 

np

L n− p (Ω f )

 Π ζ 

np

L n− p (Rn )

 c ∇Π ζ L p (Rn )  c ∇ζ L p (Ω f ) ,

(2.12)

where c is independent of  . If F ∈ C 0∞ (Rn ), then the D 1,2 (Ω f ) solution of



−U  = F

in Ω f ,

(2.13)

  = 0 on ∂Ωm ∇U · n exists uniquely by Lax–Milgram Theorem [11], extension theorem [1], and (2.12). Moreover,

U  

2n

L n−2 (Ω f )

+ ∇ U  L 2 (Ω  )  c  F  f

2n

L n+2 (Rn )

,

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where c is independent of  , F . By compactness principle [3], there is a subsequence of the solution of (2.13) satisfying, for any fixed r > 0,

⎧ ⎨ Π U  → U 0





in L 2 B r (0) strongly

as  → 0,

  ⎩ ∇ U  XΩ  → K∇ U 0 in L 2 B r (0) weakly f

where Π is the extension operator used in (2.12), XΩ  is the characteristic function on Ω f , and K is f the one in (2.9). The limit function U 0 ∈ D 1,2 (Rn ) satisfies

−∇ · (K∇ U 0 ) = |Y f | F

in Rn

(2.14)

where |Y f | is the volume of Y f . Moreover, we have the following results: Theorem 2.4. Assume A1 and  ∈ (0, 1]. n (1) If p ∈ ( n− , ∞) and F ∈ W 1, p (Rn ) with compact support, the solutions of (2.13) and (2.14) satisfy 2

⎧ np , ⎨ U  − U 0 L p (Ω f )  c   F W 1, nnp + p (Rn )∩ W 1, n+2p (Rn )   ⎩ ∇ U  − ( I + ∇X )∇ U 0  p   c   F  L (Ω ) 1, p n W

f

(R )∩ W

np 1, n+ p

(Rn )

,

where c is a constant independent of  , F . (2) If δ > 0 and F ∈ W 1,n+δ (Rn ) with compact support, the solutions of (2.13) and (2.14) satisfy

U  − U 0 L ∞ (Ω f )  c   F 

W

1, 2n n +2

(Rn )∩ W 1,n+δ (Rn )

,

where c is a constant independent of  , F . (3) If δ > 0 and F ∈ W 2,n+δ (Rn ) with compact support, the solutions of (2.13) and (2.14) satisfy

  ∇ U  − ( I + ∇X )∇ U 0 

L ∞ (Ω f )

 c  F 

W

1, 2n n +2

(Rn )∩ W 2,n+δ (Rn )

,

where c is a constant independent of  , F . Here I is the identity matrix and X is defined in (2.8). 3. Existence of the Green’s function This section consists of three subsections. The first subsection is to present uniform Hölder and uniform Lipschitz estimates as well as to show a convergence result for elliptic equations in perforated domains Ω f . We also give a local maximum norm estimate for a non-uniform elliptic equation. In the second and the third subsections, we prove the existence and derive some estimates for the Green’s function of the Laplace equation in the perforated domain Ω f and in Rn \ Y m respectively. 3.1. Some auxiliary results Let d0 be a positive constant and let D0 , D1 , D, A, D2 be smooth domains satisfying

⎧ Y ⊂ D 0 ⊂ D 1 ⊂ D ⊂ A ⊂ Y ⊂ D2 , ⎪ ⎨ m dist(Y m , ∂ D0 ), dist(D0 , ∂ D1 ), dist(D1 , ∂ D)  d0 > 0, ⎪ ⎩ dist(D, ∂ A), dist(A, ∂ Y ), dist(Y , ∂ D2 ), dist(∂ D2 , ∂Ωm )  d0 > 0. Next we give a uniform Hölder estimate for elliptic equations in perforated domains.

(3.1)

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δ Lemma 3.1. For any δ > 0 and  ∈ (0, 1], there is a μ ∈ (0, n+δ ) such that any solution of



−∇ · (∇ U  + Q  ) = F  in B 1 (x) ∩ Ω f ,  = 0 (∇ U  + Q  ) · n

(3.2)

 on B 1 (x) ∩ ∂Ωm

satisfies

[U  ]C 0,μ ( B

)

1/2 (x)∩Ω f

   c U  L 2 ( B 1 (x)∩Ω  ) +  Q  , F  Ln+δ ( B 1 (x)∩Ω  ) , f f

where x ∈ Rn and c is a constant independent of  , x. Proof of Lemma 3.1 will be given in Section 6.1. Next we give a pointwise estimate for the solution of (2.13) and the solution of (2.14). Lemma 3.2. If δ > 0,  ∈ (0, 1], and F ∈ W 1,n+δ (Rn ) with compact support, the solution of (2.13) and the solution of (2.14) satisfy

U  − U 0 L ∞ (Ω f )  c   F 

W

1, 2n n +2

(Rn )∩ W 1,n+δ (Rn )

,

where c is a constant independent of  , F . Proof. By Lemma 4.4 [11], (2.14) in [11], and Theorem 3 on p. 39 [23], the solution of (2.14) satisfies

 2  ∇ U 0 

W

1, 2n n +2

(Rn )∩ W 1,n+δ (Rn )

 c F 

W

1, 2n n +2

(Rn )∩ W 1,n+δ (Rn )

,

(3.3)

where c is a constant dependent on n, δ , K but independent of F . Define

ϕ (x) ≡ U  (x) − U 0 (x) − X (x)∇ U 0 (x) − S (x)∇ 2 U 0 (x) in Ω f , where U  , U 0 are the solutions of (2.13), (2.14) respectively, X is defined in (2.8), and S is defined in (2.10). Then function ϕ ∈ D 1,2 (Ω f ) satisfies



  −∇ · ∇ ϕ + S ∇ 3 U 0 = X ∇U 0 + ∇S ∇ 3 U 0 in Ω f ,    .  =0 ∇ ϕ + S ∇ 3 U 0 · n on ∂Ωm

(3.4)

By energy method, (2.11), (2.12), and (3.3), the solution of (3.4) satisfies

ϕ 

2n

L n−2 (Ω f )

+ ∇ ϕ L 2 (Ω  )  c   F  f

W

1, 2n n +2

(Rn )∩ W 1,n+δ (Rn )

where c is independent of  , F . By Lemma 3.1, (2.11), and (3.5), for any

[ϕ ]C 0,μ ( B where

)

1 (x)∩Ω f

 c  F 

W

1, 2n n +2

,

(3.5)

 ∈ (0, 1] and x ∈ Rn ,

(Rn )∩ W 1,n+δ (Rn )

,

δ μ ∈ (0, n+δ ) and c is independent of  , x, F . By Hölder inequality and (3.5)–(3.6),

(3.6)

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    ϕ (x)  ϕ (x) − 

−

 

 

B 1 (x)∩Ω f

 c  F  which implies the lemma.

W

1, 2n n +2

−

ϕ ( y ) dy  + 

 

ϕ ( y ) dy 

B 1 (x)∩Ω f

(Rn )∩ W 1,n+δ (Rn )

,

2

Lemma 3.3. For any p ∈ (1, ∞), s ∈ (n, ∞), and α ∈ (0, 1), there is a constant c such that any solution of



−∇ · (∇ ϕ + Q ) = F

in D2 \ Y m ,

 =0 (∇ ϕ + Q ) · n

on ∂ Y m

(3.7)

satisfies

⎧   ⎨ ϕ W 1, p (Y f )  c ϕ L p (D \Y ) +  Q L p (D \Y ) +  F XD \Y W −1, p (D2 ) , m m 2 2 2  ⎩ ∇ ϕ L ∞ (Y )  c ϕ  ∞ f L (D2 \Y m ) +  Q C 0,α (D2 \Y m ) +  F  L s (D2 \Y m ) ,

(3.8)

 is the unit outward normal vector on ∂ Y m and ∂ D2 . See (3.1) for D2 . where n Proof. Let c denote a constant. Step 1: Claim if Q ∈ C 0∞ (D2 \ Y m ) and F ∈ L p (D2 ) for p ∈ (1, ∞), any solution of

⎧ ⎪ ⎨ −∇ · (∇ψ + Q ) = F  =0 (∇ψ + Q ) · n ⎪ ⎩ ψ =0

in D2 \ Y m , on ∂ Y m ,

(3.9)

on ∂ D2

satisfies

  ψW 1, p (D2 \Y m )  c ψ L p (D2 \Y ) +  Q L p (D2 \Y m ) +  F XD2 \Y m W −1, p (D2 ) .

(3.10)

First we assume ψ L p (D2 \Y ) +  Q  L p (D2 \Y m ) +  F XD2 \Y m  W −1, p (D2 )  1 and let ψˆ be a solution of



−∇ · (∇ ψˆ + Q ) = F

in D2 \ Y m ,

ψˆ = 0

on ∂ Y m ∪ ∂ D2 .

(3.11)

1, p By [19], there exists a unique function ψˆ ∈ W 0 (D2 \ Y m ) satisfying (3.11) and

ˆ ψ W 1, p (D2 \Y m )  c for p ∈ (1, ∞).

(3.12)

Set ψˇ ≡ ψ − ψˆ . (3.9), (3.11), and Q ∈ C 0∞ (D2 \ Y m ) imply

⎧ ˇ in D2 \ Y m , ⎪ ⎨ −ψ = 0 ˇ ˆ  = −∇ ψ · n  on ∂ Y m , ∇ψ · n ⎪ ⎩ ˇ on ∂ D2 . ψ =0

(3.13)

L.-M. Yeh / J. Differential Equations 255 (2013) 1734–1783

1743

By the Green’s formula, (3.13), and Theorem 6.5.1 [9], we see that

ˇ = E∂ Y m (∇ ψˆ · n ˇ 2 − T∂ Y m (ψ)  |∂ Y m ) + E∂ D2 (∇ ψˇ · n  |∂ D2 ) on ∂ Y m , ψ/

(3.14)

where E∂ Y m , E∂ D2 (resp. T∂ Y m ) are the single-layer potentials (resp. double-layer potential) (see (4.1) in [25] for definition). By (3.12), (3.14), and Lemma 4.1 [25],

  ˇ W 1−1/ p, p (∂ Y )  c 1 + ∇ψ · n  W −1/ p, p (∂ D2 ) for p ∈ (1, ∞). ψ m

(3.15)

Let η be a smooth function satisfying η ∈ C ∞ (D2 ), η ∈ [0, 1], η(x) = 1 if dist(x, ∂ D2 ) < d0 /2, η(x) = 0 if x ∈ Y , ∇ η W 1,∞ (D2 )  c. See (3.1) for d0 . For the existence of such a function η , we refer the reader to Lemma 7.2, Chapter 1 [14]. Multiply (3.9) by η to see



  −∇ · ∇(ηψ) − ψ∇ η + η Q = η F − ∇ η(∇ψ + Q ) in D2 \ Y m ,

ηψ = 0

on ∂ Y m ∪ ∂ D2 .

(3.16)

By [19], the solution of (3.16) satisfies

 W −1/ p, p (∂ D2 )  c ηψW 1, p (D2 \Y m )  c . ∇ψ · n

(3.17)

(3.13), (3.15), and (3.17) imply

ˇ ψ W 1, p (D2 \Y m )  c . Together with (3.12), we obtain (3.10). Step 2: Claim if Q ∈ C 0∞ (D2 \ Y m ) and F ∈ L p (D2 ) for p ∈ (1, ∞), any solution of (3.9) satisfies

  ψW 1, p (D2 \Y m )  c  Q L p (D2 \Y m ) +  F XD2 \Y m W −1, p (D2 ) . For any ζ ∈ C 0∞ (D2 \ Y m ) and r ∈ [2, ∞), let

(3.18)

ρ be the solution of

⎧ ⎪ ⎨ −ρ = ζ in D2 \ Y m ,  = 0 on ∂ Y m , ∇ρ · n ⎪ ⎩ ρ =0 on ∂ D2 .

(3.19)

By Lax–Milgram Theorem and Theorem 7.26 [11], the H 1 solution of (3.19) exists uniquely and

ρ 

2n

L n − 2 ( D2 \ Y m )

 c ρ  H 1 (D2 \Y m )  c ζ L 2 (D2 \Y m )  c ζ Lr (D2 \Y m ) .

So, if r ∈ [2, n2n −2 ], then

ρ Lr (D2 \Y m )  c ζ Lr (D2 \Y m ) .

(3.20)

ρ W 1,r (D2 \Y m )  c ζ Lr (D2 \Y m ) ,

(3.21)

By (3.10) and (3.20), if r ∈ [2, n2n −2 ].

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L.-M. Yeh / J. Differential Equations 255 (2013) 1734–1783

If r > n2n −2 , by (3.21),

ρ 

W

1, 2n n −2

( D2 \ Y m )

 c ζ 

2n

L n − 2 ( D2 \ Y m )

 c ζ Lr (D2 \Y m ) .

(3.22)

Theorem 7.26 [11] and (3.22) then imply

⎧ 2n ⎪ ⎪  c ρ  1, 2n if n > 4 and r > , ⎪ ρ  n2n ⎪ − 4 ( D2 \ Y m ) n − 2 ( D2 \ Y m ) n −4 L W ⎪ ⎪ ⎪ ⎧ ⎨ 2n ⎪ ⎪ ⎨ n  4 and r > n − 2 , or ⎪ ⎪ ⎪ ρ Lr (D2 \Y m )  c ρ  1, 2n if ⎪ ⎪ ⎪ ⎪ W n − 2 ( D2 \ Y m ) 2n ⎪ ⎪ ⎩ n > 4 and r  ⎩ . n−4

(3.23)

2n So if n  4 and r > n2n −2 or if n > 4 and r ∈ [2, n−4 ], by (3.10) and (3.22)–(3.23),

ρ W 1,r (D2 \Y m )  c ζ Lr (D2 \Y m ) .

(3.24)

Repeating above argument, we see that the solution of (3.19) satisfies (3.24) for all r  2. Since C 0∞ (D2 \ Y m ) is dense in L r (D2 \ Y m ) for all r  2, we see (3.24) holds for any ζ ∈ L r (D2 \ Y m ). Let ζ ∈ L r (D2 \ Y m ) for r ∈ [2, ∞), multiply (3.9) by ρ obtained from (3.19), and use extension theorem [1] (or see remark above (2.12)), Green Theorem, and Hölder inequality to obtain

−

ψζ dx =

D2 \ Y m

ψρ dx =

D2 \ Y m



=

( Q ∇ ρ − F ρ ) dx

D2 \ Y m

( Q ∇ ρ − F XD2 \Y m Π1 ρ ) dx

D2 \ Y m

   c ζ Lr (D2 \Y m )  Q L p (D2 \Y m ) +  F XD2 \Y m W −1, p (D2 ) ,

(3.25)

for p ∈ (1, 2], 1/ p + 1/r = 1. So

  ψL p (D2 \Y m )  c  Q L p (D2 \Y m ) +  F XD2 \Y m W −1, p (D2 ) for p ∈ (1, 2].

(3.26)

Eqs. (3.10) and (3.26) imply (3.18) for p ∈ (1, 2]. Similarly take ρ to be the solution of (3.19) with ζ ∈ L r (D2 \ Y m ) for r ∈ (1, 2]. Since (3.18) holds for p ∈ (1, 2], the solution of (3.19) satisfies

ρ W 1,r (D2 \Y m )  c ζ Lr (D2 \Y m ) for r ∈ (1, 2].

(3.27)

Again we multiply (3.9) by ρ obtained from (3.19) with ζ ∈ L r (D2 \ Y m ) and r ∈ (1, 2] as well as argue as (3.25) to see that (3.18) holds for p ∈ [2, ∞). So (3.18) holds for p ∈ (1, ∞). Step 3: By Lax–Milgram Theorem [11] and maximal principle (see Theorem 3.1 and Lemma 3.4 [11]), we see that the W 1, p solution of (3.9) exists uniquely and satisfies (3.18) for any Q ∈ C 0∞ (D2 \ Y m ) and F ∈ L p (D2 ) for p ∈ (1, ∞). Since C 0∞ (D2 \ Y m ) is dense in L p (D2 \ Y m ) and L p (D2 ) is dense in W −1, p (D2 ) for p ∈ (1, ∞), the solution of (3.9) exists uniquely and satisfies (3.18) for any Q ∈ L p (D2 \ Y m ), F ∈ W −1, p (D2 ) by using a limiting argument. Step 4: Let ηˆ be another smooth function satisfying ηˆ ∈ C 0∞ (D2 ), ηˆ ∈ [0, 1], ηˆ = 1 in Y , ∇ ηˆ W 1,∞ (D2 )  c. Multiply (3.7) by ηˆ to see

L.-M. Yeh / J. Differential Equations 255 (2013) 1734–1783

1745

⎧   ⎪ ˆ ) − ϕ ∇ ηˆ + ηˆ Q = ηˆ F − ∇ ηˆ (∇ ϕ + Q ) in D2 \ Y m , −∇ · ∇(ηϕ ⎪ ⎨    =0 ˆ ) − ϕ ∇ ηˆ + ηˆ Q · n ∇(ηϕ on ∂ Y m , ⎪ ⎪ ⎩ ˆ =0 ηϕ on ∂ D2 . ˆ and use (3.18) to obtain (3.8)1 . Let ψ = ηϕ Step 5: One can modify the argument from Step 1 to Step 4 and employ Lemma 3.2 [24] and (3.8)1 to obtain (3.8)2 . So we skip it. 2 We also have a uniform Lipschitz estimate for (3.2). Lemma 3.4. If δ > 0,  ∈ (0, 1], and α ∈ (0, 1), any solution U  of (3.2) satisfies

   0,α  ∇ U  L ∞ ( B 1/2 (x)∩Ω f )  c U  L ∞ ( B 1 (x)∩Ω f ) +  −μ/2  Q C (Ω / )   +  −1+μ/2 Q  , F  

f



L n+δ ( B 1 (x)∩Ω f )

,

(3.28)

 (x) = Q  ( x), μ ≡ δ , x ∈ Rn , and c is a constant independent of  , x. where Q n+δ Proof of Lemma 3.4 is in Section 6.2. For any

 Eν ,ω ( y ) ≡

ν ∈ (0, 1] and ω ∈ (0, ∞), we define 1

if y ∈ Ω ω , f

ν 2 if y ∈ Ωmω .

ω , any solution of Lemma 3.5. If ν ∈ (0, 1], ω ∈ [1, ∞), and 0 ∈ ∂Ωm

  ∇ · Eν ,ω ∇ ϕ = 0 in B 1 (0)

(3.29)

satisfies

  ϕ  H k ( B 1/2 (0)∩Ω ω ) + ν ϕ  H k ( B 1/2 (0)∩Ωmω )  c ϕ L 2 ( B 1 (0)∩Ω ω ) + ν ϕ L 2 ( B 1 (0)∩Ωmω ) , f

f

(3.30)

where k ∈ N and c is a constant independent of ν , ω . Proof. Let ∂i denote the partial derivative in the xi direction, ∂it be t times of partial derivative ∂i , and ∂i 1 ,...,ik = ∂i 1 ∂i 2 · · · ∂ik for i , i 1 , . . . , ik ∈ {1, . . . , n}. We consider the following problem:

⎧ −∇ · ( A ∇Φ) + Q ∇Φ = 0 ⎪ ⎪ ⎪ ⎪ ⎨ −ν 2 ∇ · ( A ∇φ) + ν 2 Q ∇φ = 0 ⎪  = ν 2 A ∇φ · n  A ∇Φ · n ⎪ ⎪ ⎪ ⎩ Φ =φ

in B 1 (0) ∩ {x | xn < 0}, in B 1 (0) ∩ {x | xn > 0}, on B 1 (0) ∩ {x | xn = 0},

(3.31)

on B 1 (0) ∩ {x | xn = 0},

 is a normal vector on where A is a positive definite matrix, both A, Q are smooth functions, and n the plane {x | xn = 0}. We claim that there is a constant c independent of ν such that, for any k ∈ N,

Φ H k ( B 1/2 (0)∩{xn <0}) + ν φ H k ( B 1/2 (0)∩{xn >0})    c Φ L 2 ( B 1 (0)∩{xn <0}) + ν φL 2 ( B 1 (0)∩{xn >0}) .

(3.32)

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L.-M. Yeh / J. Differential Equations 255 (2013) 1734–1783

Proof of the claim: This is done by induction on k. Let η be a bell-shaped function satisfying η ∈ C 0∞ ( B 1 (0)), η ∈ [0, 1], and η = 1 in B 3/4 (0). Multiply (3.31) by η(Φ X{xn 0} + φ X{xn 0} ) and do integration by parts to obtain

Φ H 1 ( B 3/4 (0)∩{xn <0}) + ν φ H 1 ( B 3/4 (0)∩{xn >0})    c Φ L 2 ( B 1 (0)∩{xn <0}) + ν φL 2 ( B 1 (0)∩{xn >0}) , where c is independent of ν . So we prove (3.32) for k = 1 case. Let us assume (3.32) for some k ∈ N and k  1. Let ηˆ be a bell-shaped function satisfying ηˆ ∈ C 0∞ ( B 3/4 (0)), ηˆ ∈ [0, 1], and ηˆ = 1 in B 1/2 (0). Take partial derivative ∂i 1 ,...,ik of (3.31) for i 1 , . . . , ik ∈ {1, . . . , n − 1}, multiply the resulting equations by ηˆ (∂i 1 ,...,ik Φ X{xn 0} + ∂i 1 ,...,ik φ X{xn 0} ), and do integration by parts to obtain

∂i 1 ,...,ik Φ H 1 ( B 1/2 (0)∩{xn <0}) + ν ∂i 1 ,...,ik φ H 1 ( B 1/2 (0)∩{xn >0})    c Φ L 2 ( B 1 (0)∩{xn <0}) + ν φL 2 ( B 1 (0)∩{xn >0}) ,

(3.33)

where c is independent of ν . If we take partial derivative ∂n ∂i 1 ,...,ik−1− of (3.31) for i 1 , . . . , ik−1− ∈ {1, . . . , n − 1},  ∈ {0, . . . , k − 1}, and k  1, then

⎧ 2+ ⎪ ⎪ ∂n ∂i 1 ,...,ik−1− Φ ⎪ ⎪ ⎪ ⎪ k+ 1+ 1− s ⎪   ⎪ ⎪ t t ⎪ = C s,t1 ,...,tn−1 ∂ns ∂11 · · · ∂nn−−11 Φ in B 1 (0) ∩ {xn < 0}, ⎪ ⎪ ⎪ ⎨ s=0 t 1 +···+tn−1 =0 ⎪ ⎪ ∂n2+ ∂i 1 ,...,ik−1− φ ⎪ ⎪ ⎪ ⎪ ⎪ k+ 1+ 1− s ⎪   ⎪ ⎪ ⎪ = ⎪ ⎪ ⎩

t

s=0 t 1 +···+tn−1 =0

t

C s,t1 ,...,tn−1 ∂ns ∂11 · · · ∂nn−−11 φ

(3.34)

in B 1 (0) ∩ {xn > 0},

where C s,t1 ,...,tn−1 is smooth in {xn < 0} ∪ {xn > 0}. By (3.33)–(3.34), we obtain (3.32) for k + 1 case and we prove the claim. Because of A1, we can find an open set Oω and a smooth diffeomorphism τω with positive Jaω ) ⊂ {x = 0}, cobian determinant for each ω  1 such that 0 ∈ Oω , τω (Oω ) → B 1 (0), τω (Oω ∩ ∂Ωm n τω (Oω ∩ Ωmω ) ⊂ {xn > 0}, and τω (Oω ∩ Ω ωf ) ⊂ {xn < 0}. After transform by the mapping τω , Eq. (3.29) can be written as (3.31) in the new coordinate system. (3.30) follows from (3.32). 2 Lemma 3.6. If ν ,  ∈ (0, 1] and r > 0, any solution of

  ∇ · Eν , ∇ ϕ = 0 in B r (x)

(3.35)

satisfies

 1/2     ϕ (x)XΩ  + νϕ (x)XΩ    c  − ϕ 2 ( y )XΩ  + ν 2 ϕ 2 ( y )XΩ  dy  ,   m m f f B r (x)

where c is a constant independent of x, ν ,  , r.

(3.36)

L.-M. Yeh / J. Differential Equations 255 (2013) 1734–1783

1747

( y ) ≡ ϕ (r y ), Eν , /r ( y ) ≡ Eν , (r y ). Then (3.35) implies Proof. Assume x = 0 ∈ Ω f and define ϕ

   = 0 in B 1 (0). ∇ · Eν , /r ∇ ϕ

(3.37)

If 1   /r (resp.  /r < 1), Theorems 7.26, 8.24 [11] and Lemma 3.5 for k > n/2 (resp. Lemma 4.3 [24]) imply

[ ϕ ]C 0,μ ( B where

 /r )

1/2 (0)∩Ω f

  L 2 ( B 1 (0)∩Ωm /r ) ,  c  ϕ L 2 ( B 1 (0)∩Ω  /r ) + ν ϕ f

μ > 0 and c is a constant independent of ν ,  , r. So       ϕ (0) = ϕ˜ (0)  ϕ  (0) −

    ( y ) dy  +  ϕ

− B 1/2 (0)∩Ω f /r

ϕ ]C 0,μ ( B  [

1/2

f

  | y |μ dy + 

B 1/2 (0)∩Ω f /r

   c 

− B 1 (0)∩Ω f /r

  = c 

−

 

( y ) dy  ϕ

  2 1/2 ν ϕ  ( y ) dy 

 /r B 1 (0)∩Ωm

1/2    νϕ ( y )2 dy  . 

−

  ϕ ( y )2 dy +

B r (0)∩Ω f

− B 1/2 (0)∩Ω f /r

−

 2 ϕ ( y ) dy +

  ( y ) dy  ϕ

B 1/2 (0)∩Ω f /r

−

(0)∩Ω  /r )

−

(3.38)

 B r (0)∩Ωm

If 0 = x ∈ Ω f , we shift x to 0 and repeat the above argument to see that (3.38) with 0 replaced by x still holds. So (3.36) is proved for x ∈ Ω f .

 and define ϕ ( y ) = ϕ (r y ). So (3.37) still holds. If  /r  1, Next we assume x = 0 ∈  Y m ⊂ Ωm ϕ | in the region r Y m is by maximal principle [11] and 0 ∈ r Y m , we know that maximal value of |  bounded by the maximal value of | ϕ | on the boundary of r Y m . Since (3.36) holds in Ω f ,

      ϕ (0) = ϕ ϕ ( z) (0)  max  z∈∂ r Y m

 1/2     2 2 2      ( y ) XΩ  /r ( y ) + ν ϕ ( y ) XΩm /r ( y ) dy  c − ϕ  max  f z∈∂ Y r

m

B d0 ( z )

 1/2     2 2 ( y ) XΩ  /r ( y ) + ν 2 ϕ ( y ) XΩm /r ( y ) dy   c  − ϕ f B 1 ( 0)

 1/2     2 2  c  − ϕ ( y ) XΩ f ( y ) + ν 2 ϕ ( y ) XΩm ( y ) dy  , B r ( 0)

where d0 is defined in (3.1) and c is independent of Lemma 3.5 with k > n2 imply

ν ,  /r. If  /r > 1, Theorems 7.26, 8.24 [11] and

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L.-M. Yeh / J. Differential Equations 255 (2013) 1734–1783





ν [ ϕ ]C 0,μ ( B 1/4 (0)∩Ωm /r )  c  ϕ L 2 ( B 1 (0)∩Ω f /r ) + ν  ϕ L 2 ( B 1 (0)∩Ωm /r ) , where

μ, c are positive constants independent of ν ,  /r.       ϕ (0) = ϕ (0)  ϕ (0) −

−

 

 

−

( y ) dy  +  ϕ

 /r B 1/4 (0)∩Ωm

 

( y ) dy  ϕ

 /r B 1/4 (0)∩Ωm

  ϕ ]C 0,μ ( B 1/4 (0)∩Ωm /r ) +  ϕ L 2 ( B 1 (0)∩Ωm /r )  c [    c ν −1  ϕ L 2 ( B 1 (0)∩Ω  /r ) + ν  ϕ L 2 ( B 1 (0)∩Ωm /r ) f

 1/2     2 2  c ν −1  − ϕ ( y ) XΩ f ( y ) + ν 2 ϕ ( y ) XΩm ( y ) dy  , B r ( 0)

 case. If x ∈ Ω  and x = 0, we where c is independent of ν ,  /r. So (3.36) holds for x = 0 ∈  Y m ⊂ Ωm m shift the coordinate system such that the origin of the coordinate system is located at x. Then we see . 2 that (3.36) holds for x ∈ Ωm

3.2. The Green’s function in Ω f Let Gν , for

ν ,  ∈ (0, 1] denote the Green’s function of 

  −∇ y · Eν , ( y )∇ y Gν , (x, y ) = δ(x − y ) in Rn ,

(3.39)

Gν , (x, y ) → 0 as |x − y | → ∞. 1, 1

1 By Theorem 5.4 [16] and remark on pp. 62, 67 [16], Gν , (x, ·) ∈ H loc (Rn \ {x}) ∩ W loc (Rn ) exists uniquely when n  3 and

Gν , (x, y ) = Gν , ( y , x) for x = y .

(3.40)

Lemma 3.7. There is a constant c independent of ν ,  ∈ (0, 1] such that

⎧ c |x − y |2−n if x, y ∈ Ω f , ⎪ ⎪ ⎪ ⎪ ⎪   ⎨ c ν −1 |x − y |2−n if x ∈ Ω f , y ∈ Ωm or if x ∈ Ωm , y ∈ Ω f , Gν , (x, y )  ⎪ ⎪ c ν −2 |x − y |2−n if x, y ∈ Ω  , ⎪ m ⎪ ⎪ ⎩ c |x − y |2−n if |x − y | >  ,     ∇ y Gν , (x, y ) + ∇x Gν , (x, y )  c |x − y |1−n if x, y ∈ Ω  . f

(3.41)

(3.42)

For any x ∈ Ωm , there is a number d > 0 such that if 0 < |x − y | < d, then

  ∇ y Gν ,1 (x, y )  c ν −2 |x − y |1−n , where c is a constant independent of ν .

(3.43)

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1749

Proof. Proof of (3.41)1 (that is, for x, y ∈ Ω f case). Set r ≡ |x − y | for x, y ∈ Ω f . Let F ∈ C 0∞ ( B r /3 ( y )) and find

ϕ ∈ D1,2 (Rn ) satisfying   −∇ · Eν , ∇ ϕ = F XΩ f + ν F XΩm .

By Lax–Milgram Theorem [11], extension theorem [1], and (2.12) for p = 2, ϕ is solvable uniquely in D 1,2 (Rn ). By Definition 5.1 and p. 67 [16] and Lemma 3.6, we see, for x ∈ Ω f ,

⎧ ⎪ ϕ ( x ) = Gν , (x, z) F ( z)XΩ  ( z) + Gν , (x, z)ν F ( z)XΩm ( z) dz, ⎪ ⎪ f ⎪ ⎪ ⎪ ⎨ B r /3 ( y )  1/2 ⎪     ⎪ 2 2 2 ⎪   ,     ⎪ ϕ ( x ) c ϕ ( z ) X ( z ) + ν ϕ ( z ) X ( z ) dz −  Ωf Ωm ⎪   ⎪ ⎩

(3.44)

B r /3 (x)

where c is independent of x, y, [11] imply

   

ν ,  , r. (2.12), (3.44), extension theorem [1], and Hölder inequality

  Gν , (x, z) F ( z)XΩ  ( z) + Gν , (x, z)ν F ( z)XΩm ( z) dz f

B r /3 ( y )

 1/2    c  − ϕ 2 ( z)XΩ f ( z) + ν 2 ϕ 2 ( z)XΩm ( z) dz B r /3 (x)

  n −2    2n  2n n −2       c  − ϕ ( z)XΩ f ( z) + νϕ ( z)XΩm ( z) dz B r /3 (x)

c

∇ ϕ XΩ f + ν ∇ ϕ XΩm L 2 (Rn ) r

where c is independent of x, y,

n 2 −1

c

 F L 2 ( B r/3 ( y )) n

r 2 −2

(3.45)

,

ν ,  , r. Multiply (3.45) by r −n to obtain

     − Gν , (x, z) F ( z)XΩ  ( z) + Gν , (x, z)ν F ( z)XΩ  ( z) dz   m f B r /3 ( y )

  n−2  − r c 

1/2  F ( z) dz . 2

(3.46)

B r /3 ( y )

Since y ∈ Ω f , Eqs. (3.39), (3.46) and Lemma 3.6 imply

 1/2         Gν , (x, y )  c  − Gν , (x, z)2 XΩ  ( z) + ν 2 Gν , (x, z)2 XΩ  ( z) dz    m f

c r n −2

.

B r /3 ( y )

So (3.41)1 holds. (3.41)2,3 are proved exactly by following the argument of (3.41)1 . (3.41)4 (that is, |x − y | >  case) follows from Theorem 3.1 [11], (3.41)1 , and (3.1).

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L.-M. Yeh / J. Differential Equations 255 (2013) 1734–1783

Proof of (3.42). Assume x, y ∈ Ω f , x = y, and y = 0. We define r ≡ |x|, then

  −∇z · Eν , ( z)∇z Gν , (x, z) = 0 in B r /2 (0). If

ϕ (z) ≡ Gν , (x, 2r z), then   −∇ · Eν ,2 /r ∇ ϕ ( z) = 0 in B 1 (0).

Suppose 2 /r > 1, by Theorem 2.10 [11], Lemma 3.5 with k >

n 2

+ 1, and (3.41), we get

    ∇ y Gν , (x, 0) = 2 ∇ ϕ (0) 

r c r

ϕ L 2 ( B

2 /r ) 1 (0)∩Ω f

+ ν ϕ L 2 ( B

 2 /r ) 1 (0)∩Ωm

 c |x|1−n ,

(3.47)

where c is independent of x, y, ν ,  /r. Suppose 2 /r  1, by Lemma 5.3 [24] and (3.41), we also get (3.47). Assume x, y ∈ Ω f and x = y. One can shift y to 0, repeat the above process, and obtain

|∇ y Gν , (x, y )|  c |x − y |1−n , where c is independent of ν ,  . By (3.40), we also obtain |∇x Gν , (x, y )|  c |x − y |1−n , where c is independent of ν ,  . So (3.42) is proved. (3.43) can be proved in a similar way as (3.42), so we skip it. 2 Lemma 3.8. There is a unique function G (x, y ) in Ω f × Ω f satisfying, for any x ∈ Ω f ,

⎧ − y G (x, y ) = δ(x − y ) ⎪ ⎪ ⎪ ⎪ ⎪ y = 0 ⎪ ∇ y G (x, y ) · n ⎪ ⎨  G (x, y )  c |x − y |2−n ⎪ ⎪     ⎪ ⎪ ∇ y G (x, y ) + ∇x G (x, y )  c |x − y |1−n ⎪ ⎪ ⎪ ⎩ G (x, y ) = G ( y , x)

in Ω f , on ∂Ωm , if x = y ,

(3.48)

if x = y , if x = y ,

 y is a unit vector normal to ∂Ωm and c is a constant. where n For any x ∈ Ω f , there is a number d > 0 such that if 0 < |x − y | < d, then

  ∇x ∇ y G (x, y )  c |x − y |−n ,

(3.49)

where c is a constant independent of x, y. Proof. If x ∈ Ω f , dist(x, ∂Ωm ) > 0. Assume t i → 0 and si → ∞ as i → ∞. For any si > dist(x, ∂Ωm ) > t i > 0, we define Dt i ,si (x) ≡ { y ∈ Rn | 0 < t i < |x − y | < si }. From (3.39),

  −∇ y · Eν ,1 ∇ y Gν ,1 (x, y ) = 0 in Dt i ,si (x).

(3.50)

Since x ∈ Ω f , by maximal principle (see Theorem 3.1 and Lemma 3.4 [11]) and (3.41)1 , Gν ,1 (x, ·) L ∞ (Dt ,s (x)) with fixed t i , si is bounded independent of ν . If j ∈ Zn and the closure i i

Y − j ⊂ Dt i ,si (x), by Lemma 3.3 [24], we see

  Gν ,1 (x, ·)

C 1,α (D\Y m − j )

  + Gν ,1 (x, ·)C 1,α (Y

m − j)

   c Gν ,1 (x, ·) L ∞ (D

t i ,si (x))

,

L.-M. Yeh / J. Differential Equations 255 (2013) 1734–1783

ν . So there are  ti ,  si satisfying (1)

where Y m ⊂ D ⊂ Y (see (3.1)), α ∈ (0, 1), and c is independent of t i < si < si , (2)  t i → 0,  si → ∞ as i → ∞, and (3) 0 < t i <

  Gν ,1 (x, ·)

C 1,α (Ω f ∩D t ,s (x)) i i

1751

  + Gν ,1 (x, ·)C 1,α (Ω

m ∩D  t i , si (x))

 c,

(3.51)

where α ∈ (0, 1) and c is bounded independent of ν but dependent on t i . Therefore for each fixed α (Ω ∩ D  ti ,  si , there is a convergent subsequence of Gν ,1 (x, ·) in C 1, α < α < 1 as  f t i , si (x)) for some  ν → 0. By a diagonal process, we can even extract a subsequence of Gν ,1 (x, ·) (same notation for subsequence) such that, for all  ti ,  si with 0 < t i < s i < ∞,

 α Ω ∩ D Gν ,1 (x, ·) converges to G (x, ·) in C 1,  f t i , si (x) as ν → 0.

(3.52)

If ζ ∈ C 0∞ (Rn ) and x ∈ / supp(ζ ) (that is, the support of ζ ), there are  ti ,  si such that supp(ζ ) ⊂ Dt i ,si (x). Multiply (3.50) by ζ to see, by (3.51)–(3.52),

0 = lim

ν →0

∇ y Gν ,1 (x, y )∇ζ ( y )XΩ f + ν 2 ∇ y Gν ,1 (x, y )∇ζ ( y )XΩm dy



=

∇ y G (x, y )∇ζ ( y ) dy .

(3.53)

Ωf

(3.40), (3.52)–(3.53), and Lemma 3.7 imply, for all  ti ,  si and for any x ∈ Ω f ,

⎧ − y G (x, y ) = 0 ⎪ ⎪ ⎪ ⎪ ⎪ y = 0 ⎪ ∇ y G (x, y ) · n ⎪ ⎪ ⎨  G (x, y )  c |x − y |2−n ⎪ ⎪   ⎪ ⎪ ∇ y G (x, y )  c |x − y |1−n ⎪ ⎪ ⎪ ⎪ ⎩ G (x, y ) = G ( y , x)

in Ω f ∩ Dt i ,si (x), on ∂Ωm ∩ Dt i ,si (x), for y ∈ Ω f ∩ Dt i ,si (x),

(3.54)

for y ∈ Ω f ∩ Dt i ,si (x), for y ∈ Ω f ∩ Dt i ,si (x),

where c is a constant independent of  ti ,  si . For any F , Q ∈ C (Rn ) with compact support, there is a unique ϕν ,1 ∈ D 1,2 (Rn ) satisfying, by Lax– Milgram Theorem [11], extension theorem [1], and (2.12) for p = 2,

  −∇ · Eν ,1 ∇ ϕν ,1 + Q XΩ f = F XΩ f

in Rn .

By energy method, (2.12), and Lemma 3.3, we see that there is a subsequence of ϕν ,1 (same notation for subsequence) such that (1) ϕν ,1 converges to ϕ in D 1,2 (Ω f ) ∩ W 1, p ( B r (0) ∩ Ω f ) weakly for any r > 0, p ∈ (n, ∞) as ν → 0 and (2) ϕ satisfies

−∇ · (∇ ϕ + Q ) = F

in Ω f ,

 =0 (∇ ϕ + Q ) · n

on ∂Ωm ,

(3.55)

 is a unit vector normal to ∂Ωm . By Definition 5.1 and p. 67 [16], (3.42), and (3.43), where n



ϕν ,1 (x) = Ωf

Gν ,1 (x, y ) F ( y ) dy − Ωf

∇ y Gν ,1 (x, y ) Q ( y ) dy for x ∈ Rn .

(3.56)

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L.-M. Yeh / J. Differential Equations 255 (2013) 1734–1783

By Lemma 3.7, (3.52), and (3.56), for any x ∈ Ω f and d <

  ϕ (x) − 





dist(x,∂Ωm ) , 2



 

G (x, y ) F ( y ) − ∇ y G (x, y ) Q ( y ) dy 

Ω f \ B d (x)

  = lim ϕν ,1 (x) − ν →0





   Gν ,1 (x, y ) F ( y ) − ∇ y Gν ,1 (x, y ) Q ( y ) dy 

Ω f \ B d (x)

       Gν ,1 (x, y ) F ( y ) − ∇ y Gν ,1 (x, y ) Q ( y ) dy  = lim  ν →0 B d (x)

 cd Q , F L ∞ ( B d (x)) , where c is independent of d. So, for any x ∈ Ω f and δ > 0, there is a  d < dist(x, ∂Ωm ) such that if d < d, then cd Q , F  L ∞ ( B d (x)) < δ . So we obtain





ϕ (x) =

G (x, y ) F ( y ) dy −

Ωf

∇ y G (x, y ) Q ( y ) dy .

(3.57)

Ωf

(3.54), (3.55), and (3.57) imply the existence of G (x, y ) and (3.48). The uniqueness of G (x, ·) for any x ∈ Ω f is due to (3.48)3 and maximal principle [11]. Let x ∈ Ω f , d < dist(x, ∂Ωm ), y ∈ B d/4 (x) \ {x}, and r ≡ |x − y |. By (3.48),

−z ∂ y i G ( z, y ) = 0 in B r /2 (x), where ∂ y i is the partial derivative with respect to y i for i ∈ {1, . . . , n}. By Theorem 2.10 [11] and (3.48)4 ,

    ∇x ∂ y G (x, y )  c ∂ y G (·, y ) ∞ i i L (B r

r /2 (x))

where c is a constant independent of x, y. So we prove (3.49).

 c |x − y |−n , 2

Remark 3.1. From (3.55) and (3.57) in the proof of Lemma 3.8, we know that for any F , Q ∈ C (Rn ) with compact support, the D 1,2 (Ω f ) solution of

−∇ · (∇ ϕ + Q ) = F

in Ω f ,

 =0 (∇ ϕ + Q ) · n

on ∂Ωm

(3.58)

exists uniquely and satisfies



ϕ (x) = Ωf

G (x, y ) F ( y ) dy −

∇ y G (x, y ) Q ( y ) dy for x ∈ Ω f .

(3.59)

Ωf

Following the argument of Lemma 4.1 [11] and employing (3.48)4 and (3.59), if F ∈ C (Rn ) with compact support, the D 1,2 (Ω f ) solution of

L.-M. Yeh / J. Differential Equations 255 (2013) 1734–1783

−ϕ = F

1753

in Ω f ,

 = 0 on ∂Ωm ∇ϕ · n satisfies

∇ ϕ (x) =

∇x G (x, y ) F ( y ) dy for x ∈ Ω f .

Ωf

Tracing the proof of Lemma 4.2 [11] as well as employing (3.49), we have Lemma 3.9. If Q ∈ C 0∞ (Rn ) with support in B r (0) for some r > 0, then the D 1,2 (Ω f ) solution of

−∇ · (∇ ϕ + Q ) = 0 in Ω f ,  =0 (∇ ϕ + Q ) · n

on ∂Ωm

satisfies





∂x j ϕ (x) = −

∂x j ∇ y G (x, y ) Q ( y ) dy + Q (x)

B s (0)∩Ω f

∂x j ∇ y Γ (x, y ) dy

B s (0)∩Ω f



+ Q (x)

∇ y Γ (x, y )n j dσ y for any x ∈ Ω f ,

(3.60)

∂( B s (0)∩Ω f )

where s > 1 + r, Γ is the fundamental solution of the Laplace equation in Rn , j ∈ {1, . . . , n}, and n j is the j-th  = (n1 , . . . , nn ) which is a unit outward normal vector on ∂( B s (0) ∩ Ω f ). component of n Moreover, there is a constant c independent of r such that

      ∇ ϕ (x) + ∇x ∇ y G (x, y ) Q ( y ) dy   c  Q (x) for any x ∈ Ω f .  

(3.61)

Ωf

Proof. Define, for any x ∈ Ω f and δ < dist(x, ∂Ωm )/2 < 1,



  ∇ y G (x, y )ηδ |x − y | Q ( y ) dy ,

ϕδ (x) ≡ − B s (0)∩Ω f

where ηδ (x) = η(x/δ) and η ∈ C 0∞ (R) is an even function satisfying η ∈ [0, 1], η(x) = 0 in |x|  1/2, η(x) = 1 in |x|  1, and η (x)  0 for x  0. By (3.48), ϕδ ∈ C 1 (Ω f ) and ϕδ converges to



ϕ (x) ≡ −

∇ y G (x, y ) Q ( y ) dy

B s (0)∩Ω f

in L ∞ (Ω f ) as δ → 0. Define

ρ (x, ·) for x ∈ Ω f as

ρ (x, y ) ≡ G (x, y ) − Γ (x, y ) for y ∈ Ω f .

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L.-M. Yeh / J. Differential Equations 255 (2013) 1734–1783

Then  y ρ (x, ·) = 0 in Ω f

and, by Theorem 6.30 [11] and Lemma 3.8, ρ (x, ·)C 2 (Ω ) and f ∇x ∇ y ρ  are bounded by a constant depending on dist(x, ∂Ωm ). If x = (x1 , . . . , xn ) ∈ Ω f , f ×Ω f ) for any j ∈ {1, . . . , n}, L ∞ (Ω



   ∂x j ∇ y G (x, y )ηδ |x − y | Q ( y ) dy

∂x j ϕδ (x) = − B s (0)∩Ω f



    ∂x j ∇ y G (x, y )ηδ |x − y | Q ( y ) − Q (x) dy

=− B s (0)∩Ω f





      ∂x j ∇ y Γ (x, y )ηδ |x − y | + ∂x j ∇ y ρ (x, y )ηδ |x − y | dy

− Q (x) B s (0)∩Ω f



    ∂x j ∇ y G (x, y )ηδ |x − y | Q ( y ) − Q (x) dy

=− B s (0)∩Ω f





+ Q (x)

   ∂x j ∇ y ρ (x, y )ηδ |x − y | dy .

∇ y Γ (x, y )n j dσ y − Q (x)

∂( B s (0)∩Ω f )

B s (0)∩Ω f

By Lemma 3.8 and arguing as the proof of Lemma 4.2 [11], if δ → 0, then ∂x j ϕδ (x) converges to the right-hand side of (3.60) for any x ∈ Ω f , s > 1 + r, and j ∈ {1, . . . , n}. So we prove (3.60). From (2.13) in [11],

∂ y i ∂ y j Γ (x, y ) dσ y = 0 for any t > 0 and i , j ∈ {1, . . . , n}.

(3.62)

∂ B t (x)

If x ∈ / B r (0) ∩ Ω f , (3.61) is obvious from (3.60). If x ∈ B r (0) ∩ Ω f and s > 1 + r, Divergence Theorem [11] and (3.62) imply

   

  ∇ y Γ (x, y )n j dσ y 



∂x j ∇ y Γ (x, y ) dy +

B s (0)∩Ω f

  = −

∂( B s (0)∩Ω f )



∂ y j ∇ y Γ (x, y ) dy +

B s (0)∩Ω f



  ∇ y Γ (x, y )n j dσ y 

∂( B s (0)∩Ω f )

 2  ∇ Γ (x, y ) dy + c y

 B s+|x|+1 (x)\ B s−|x|−1 (x)



c B s+|x|+1 (x)\ B s−|x|−1 (x)

|x − y |−n dy + c  c ln

s+r +1 s−r −1

+ c,

(3.63)

where c is a constant independent of r, s. We note that (n1 , . . . , nn ) is a unit outward normal vector on ∂( B s (0) ∩ Ω f ) in (3.63). If s is much larger than r, the right-hand side of (3.63) is bounded by a constant independent of r. Together with (3.60), we obtain (3.61). 2

L.-M. Yeh / J. Differential Equations 255 (2013) 1734–1783

1755

If we define

G ω (x, y ) ≡ ω2−n G (x/ω, y /ω)

for any ω ∈ (0, ∞),

(3.64)

then G ω (x, y ) satisfies, by Lemma 3.8,

⎧ ⎪ ⎪ − y G ω (x, y ) = δ(x − y ) ⎪ ⎪ ⎪ ⎪ ω = 0 ∇ G (x, y ) · n ⎪ ⎪ ⎨ y ω  G (x, y )  c |x − y |2−n ⎪ ω ⎪     ⎪ ⎪ ⎪ ∇ y G ω (x, y ) + ∇x G ω (x, y )  c |x − y |1−n ⎪ ⎪ ⎪ ⎩ G ω (x, y ) = G ω ( y , x)

in Ω ω f , ω, on ∂Ωm

for x = y ,

(3.65)

for x = y , for x = y ,

 ω is a unit vector normal to ∂Ωmω and c is a constant independent of where n

ω.

3.3. The Green’s function in Rn \ Y m For

ν ∈ (0, 1] and ω ∈ (0, ∞), let us define  1 ν , ω  E ( y) ≡

if y ∈ Rn \ ω Y m ,

ν 2 if y ∈ ω Y m .

Let Gν∗, for

ν ,  ∈ (0, 1] denote the Green’s function of 

 ν ,  E ( y )∇ y Gν∗, (x, y ) = δ(x − y ) in Rn , −∇ y ·  Gν∗, (x, y ) → 0 as |x − y | → ∞.

(3.66)

1 By Theorem 5.4 [16] and remark on pp. 62, 67 [16], we see that Gν∗, (x, ·) ∈ H loc (Rn \ {x}) ∩ W loc (Rn ) exists uniquely when n  3 and 1, 1

Gν∗, (x, y ) = Gν∗, ( y , x) for x = y .

(3.67)

A modification of the proof of Lemma 4.3 [24], we have: If δ > 0 and ν ,  ∈ (0, 1], any solution ϕ of

  ν , −∇ ·  E ∇ ϕ = 0 in B 1 (x) satisfies

  [ϕ ]C 0,μ ( B 1/2 (x)\ Y m )  c ϕ L 2 ( B 1 (x)\ Y m ) + ν ϕ L 2 ( B 1 (x)∩ Y m ) ,

(3.68)

δ where μ ≡ n+δ , x ∈ Rn , and c is independent of ν ,  , x.

Employing (3.66)–(3.68) and following the proofs of Lemmas 3.6, 3.7, 3.8, 3.9, we see that there is a subsequence of Gν∗,1 (x, ·) converging to G∗0 (x, ·) in C 1,α (Dt i ,si (x) \ Y m ) as ν closes to 0 for all

0 < t i < si < ∞ and x ∈ Rn \ Y m , where α ∈ (0, 1). Here  ti ,  si , Dt i ,si (x) are defined in a same way as those in the proof of Lemma 3.8. Also following the arguments of Lemma 3.8 and Remark 3.1, we know:

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L.-M. Yeh / J. Differential Equations 255 (2013) 1734–1783

G∗0 (x, y ) defined in (Rn \ Y m ) × (Rn \ Y m ) satisfies, for any x ∈ Rn \ Y m ,

⎧ ⎪ − y G∗0 (x, y ) = δ(x − y ) ⎪ ⎪ ⎪ ⎪ ⎪ y = 0 ∇ y G∗0 (x, y ) · n ⎪ ⎪ ⎨  ∗ G (x, y )  c |x − y |2−n ⎪ 0 ⎪  ⎪ ⎪ ⎪ ∇ y G∗0 (x, y )  c |x − y |1−n ⎪ ⎪ ⎪ ⎩ ∗ G0 (x, y ) = G∗0 ( y , x)

in Rn \ Y m , on ∂ Y m , for x = y ,

(3.69)

for x = y , for x = y ,

 y is a unit vector normal to ∂ Y m and c is a constant. where n If F , Q ∈ C 0∞ (Rn ) with support in B r (0) for some r > 0, the D 1,2 (Rn \ Y m ) solution of



−∇ · (∇ ϕ + Q ) = F

in Rn \ Y m ,

 =0 (∇ ϕ + Q ) · n

on ∂ Y m

exists uniquely and satisfies, for x ∈ Rn \ Y m ,

⎧ ∗ ⎪ ϕ ( x ) = G ( x , y ) F ( y ) dy − ∇ y G∗0 (x, y ) Q ( y ) dy , ⎪ 0 ⎪ ⎪ ⎪ ⎪ ⎪ Rn \Y m Rn \Y m ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ∗ ⎪ ⎪ ∂x j ϕ (x) = ∂x j G0 (x, y ) F ( y ) dy − ∂x j ∇ y G∗0 (x, y ) Q ( y ) dy ⎪ ⎪ ⎪ ⎪ ⎨ Rn \Y m B s (0)\Y m ⎪ ⎪ ⎪ + Q (x) ∇ ∂x j ∇ y Γ (x, y ) dy , y Γ (x, y )n j dσ y + Q (x) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ∂( B s (0)\Y m ) B s (0)\Y m ⎪ ⎪ ⎪   ⎪ ⎪       ⎪ ∗ ∗ ⎪  ⎪ ∂x j ϕ (x) + ∂x j ∇ y G0 (x, y ) Q ( y ) − ∂x j G0 (x, y ) F ( y ) dy   c  Q (x), ⎪ ⎪ ⎩

(3.70)

Rn \Y m

where s > 1 + r, c is a constant independent of r, Γ is the fundamental solution of the Laplace equation in Rn ,  = (n1 , . . . , nn ) which is a unit outward normal vector on j ∈ {1, . . . , n}, and n j is the j-th component of n ∂( B s (0) \ Y m ). 4. The second derivatives of the Green’s function G (x, y ) From Lemma 3.8, we know some estimates for the zero order and the first order derivatives of the Green’s function G (x, y ). This section is to find an approximation for the second order derivatives of G (x, y ) and it consists of two subsections. The first subsection is the approximation of G (x, y ) for |x − y |  1. The second subsection is the approximation of G (x, y ) for |x − y |  n + 1. 4.1. Approximation of G (x, y ) for |x − y |  1 Let G 0 (x, y ) denote the Green’s function of



  −∇ y · K∇ y G 0 (x, y ) = δ(x − y ) in Rn , G 0 (x, y ) → 0 as |x − y | → ∞,

(4.1)

L.-M. Yeh / J. Differential Equations 255 (2013) 1734–1783

1757

where K is the constant symmetric positive definite matrix in (2.9). By change of variable, the Green’s function G 0 of (4.1) can be transformed to the fundamental solution of the Laplace equation in some new coordinate system. Together with the results on p. 17 [11], we see that there is a constant c such that

 ⎧  G 0 (x, y )  c |x − y |2−n if x = y , ⎪ ⎪ ⎨  1 − n ∇ y G 0 (x, y )  c |x − y | if x = y , ⎪ ⎪ ⎩ G 0 (x, y ) = ω2−n G 0 (x/ω, y /ω) for any ω > 0.

(4.2)

Lemma 4.1. If ω ∈ (0, 1], |x − y |  12 , and x, y ∈ Ω ω , then f

  G ω (x, y ) − G 0 (x, y )  c ωσ ,

(4.3)

where c , σ > 0 are constants independent of ω . See (3.64) for G ω (x, y ). Proof. We fix x ∈ Ω ω for f

ω ∈ (0, 1] and define c1 ≡

sup

1/5|x− y | y ∈Ω ω f

  G ω (x, y ) − G 0 (x, y ).

By (3.65)3 and (4.2)1 , c 1 is a constant independent of ω , x. (3.65)4 and (4.2)2 imply that G ω (x, y ) (resp. G 0 (x, y )) is a uniformly Lipschitz continuous function (independent of ω ) of y in Ω ω \ B 1/4 (x) f (resp. Rn \ B 1/4 (x)). So there is a positive constant c 2 independent of

    ∇ y G ω (x, z) + ∇ y G 0 (x, y )  c 2 for



ω, x such that

z ∈ Ωω \ B 1/4 (x), f y ∈ Rn \ B 1/4 (x).

(4.4)

Now we define

θω,x ≡

sup

1/2|x− y | y ∈Ω ω f

  G ω (x, y ) − G 0 (x, y ).

By (3.65)3 and (4.2)1 , there is a y ω,x ∈ Ω ω satisfying |x − y ω,x |  f

1 2

and

  θω,x = G ω (x, y ω,x ) − G 0 (x, y ω,x )  c 1 . Pick a number β > n so that

ρω , x ≡

θω,x β c2

 1/4. By (4.4),

  G ω (x, z) − G 0 (x, y )  θω,x /5 for Let F ∈ C 0∞ ( B ρω,x ( y ω,x )) satisfy



z ∈ B ρω,x ( y ω,x ) ∩ Ω ω , f y ∈ B ρω,x ( y ω,x ).

(4.5)

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L.-M. Yeh / J. Differential Equations 255 (2013) 1734–1783

⎧ F ∈ [0, 1], ⎪ ⎨ F ≡ 1 on B ρω,x /2 ( y ω,x ), ⎪ ⎩ ∇ F L ∞ ( B ρω,x ( y ω,x ))  4/ρω,x . Put F in the right-hand side of (2.13) with  = ω and (2.14) to obtain U ω , U 0 . Since is a constant c 3 independent of ω satisfying, by Lemma 3.2,

ω ∈ (0, 1], there

−δ

U ω − U 0 L ∞ (Ω ωf )  c 3 ω|ρω,x | n+δ .

(4.6)

We also note, by (2.13), (2.14), (4.4), (4.5), Remark 3.1, and Taylor expansion,

      U ω (x) − U 0 (x) =  G ω (x, y ) F ( y ) dy − G 0 (x, y )|Y f | F ( y ) dy    Rn

Ωω f

    G 0 (x, y ) F ( y ) dy − G 0 (x, y )|Y f | F ( y ) dy  =  (G ω − G 0 )(x, y ) F ( y ) dy + Ωω f n +1  c 4 ρω ,x − n +1  c4 ω ,x

ρ





Ωω f

Rn





ωn |Y f |G 0 (x, y j ) F ( y j ) − G 0 (x, z j ) F (z j ) for y j , z j ∈ ω(Y − j )

j ∈Zn



−2 1 − c 5 ωρω ,x ,

(4.7) −δ

ω, x. If 1 − c5 ωρω−,2x > 1/2, Eqs. (4.6)–(4.7) imply ρωn+,x1  c ωρωn+δ ,x . √  1/2, then ρω,x  c ω , which also implies (4.3). So this lemma

where c 4 , c 5 are independent of −2 So (4.3) holds. If 1 − c 5 ωρω ,x holds. 2

Lemma 4.2. If ω ∈ (0, 1], |x − y | > 2/3, and x, y ∈ Ω ω , then f

    ∇ y G ω (x, y ) − I + ∇X( y /ω) ∇ y G 0 (x, y )  c ωσ ,

(4.8)

where I is the identity matrix, X is defined in (2.8), and the constants c , σ > 0 are independent of ω . 1 (Ω f ) satisfying, in the cell Y f , Proof. For i 1 , i 2 ∈ {1, . . . , n}, find T(i 1 ,i 2 ) ( y ) ∈ H per

⎧ ⎪ T(i 1 ,i 2 ) + ∂i 1 X(i 2 ) = −δi 1 ,i 2 − ∂i 1 X(i 2 ) in Y f , ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ∇T(i 1 ,i 2 ) · n  y + X( i 2 ) n y i = 0 on ∂ Y m , 1 ⎪ ⎪ ⎪ T(i 1 ,i 2 ) dy = 0. ⎪ ⎪ ⎪ ⎩ Yf

See (2.7) for X(i ) and see (2.10) for δi 1 ,i 2 , n y i . Let T( y ) ≡ (T(i 1 ,i 2 ) ( y )) be an n × n matrix function 1 y and Tω ( y ) ≡ ω2 (T(i 1 ,i 2 ) ( ω )). Note T(i 1 ,i 2 ) is a special case of (2.10). As (2.11),

TC 2,α (Ω )  c for α ∈ (0, 1), per

where c is a constant. Define

f

ϕω in Ω ωf \ B 1/2 (x) as, for any fixed x ∈ Ω ωf ,

(4.9)

L.-M. Yeh / J. Differential Equations 255 (2013) 1734–1783

1759

ϕω ( y ) ≡ G ω (x, y ) − G 0 (x, y ) − Xω ( y )∇ y G 0 (x, y ) − Tω ( y )∇ 2y G 0 (x, y ), where Xω is defined in (2.8). Then function

ϕω satisfies ⎧   ⎨ −∇ · ∇ ϕω + Tω ∇ 3y G 0 = Xω ∇ y  y G 0 + ∇Tω ∇ 3y G 0 in Ω ωf \ B 1/2 (x), ω ω\B ⎩ ∇ ϕ + T ∇ 3 G  · n on ∂Ωm ω ω y 0  =0 1/2 (x).

By Lemma 3.4, Lemma 4.1, (2.11), and (4.9), ∇ ϕω  L ∞ (Ω ω \ B 2/3 (x))  c ωσ . So (4.8) holds. f

2

For any fixed y ∈ Ω ω , we define f



ζω (x) ≡ ∇ y G ω (x, y )   ζ0 (x) ≡ I + ∇X( y /ω) ∇ y G 0 (x, y )

for x ∈ Ω ω f \ B 2/3 ( y ).

By (3.65), we see



−ζω = 0

in Ω ω f \ B 2/3 ( y ),

 ω = 0 on ∂Ωmω \ B 2/3 ( y ). ∇ζω · n By Lemma 4.2, ζω − ζ0  L ∞ (Ω ω \ B 2/3 ( y ))  c ωσ . Tracing the argument from (2.13) to (2.14), we see that f

ζ0 satisfies

−∇ · (K∇ζ0 ) = 0 in Rn \ B 2/3 ( y ), where K is the matrix in (2.9). Following the proof of Lemma 4.2, we obtain , then Lemma 4.3. If ω ∈ (0, 1], |x − y |  3/4, and x, y ∈ Ω ω f

     ∇x ∇ y G ω (x, y ) − I + ∇X(x/ω) I + ∇X( y /ω) ∇x ∇ y G 0 (x, y )  c ωσ where c , σ > 0 are constants independent of ω . Lemma 4.3 then implies Corollary 4.1. There are constants c , σ > 0 so that, for x, y ∈ Ω f and 1  |x − y |,

     ∇x ∇ y G (x, y ) − I + ∇X(x) I + ∇X( y ) ∇x ∇ y G 0 (x, y ) 

ω2−n G ( ωx , ω ) and G 0 (x, y ) = ω2−n G 0 ( ωx , ω ) for η Hence if ω = x, ω = y, and |ξ − η| = 1, then, by Lemma 4.3, Proof. G ω (x, y ) =

y

y

c

|x − y |n+σ

.

ω ∈ (0, 1] by (3.64) and (4.2).

ξ

     ∇x ∇ y G (x, y ) − I + ∇X(x) I + ∇X( y ) ∇x ∇ y G 0 (x, y )      = ∇x ∇ y G (ξ/ω, η/ω) − I + ∇X(ξ/ω) I + ∇X(η/ω) ∇x ∇ y G 0 (ξ/ω, η/ω)      = ∇ξ ∇η G ω (ξ, η) − I + ∇X(ξ/ω) I + ∇X(η/ω) ∇ξ ∇η G 0 (ξ, η)ωn  c ωn+σ . If we take

ω=

1

|x− y | , then

(4.10)

ω ∈ (0, 1] implies |x − y |  1. So we prove (4.10). 2

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4.2. Approximation of G (x, y ) for |x − y |  n + 1 Define γ (x, y ) ≡ G∗0 (x, y ) − Γ (x, y ), where G∗0 is the Green’s function in (3.69) and Γ is the fundamental solution of the Laplace operator in Rn . By (3.69), γ is a function in (Rn \ Y m ) × (Rn \ Y m ) satisfying, for any x ∈ Rn \ Y m ,

⎧ − y γ (x, y ) = 0 ⎪ ⎪ ⎪ ⎪ ⎨ ∇ y γ (x, y ) · n  y = −∇ y Γ (x, y ) · n y   ⎪ γ (x, y )  c |x − y |2−n ⎪ ⎪ ⎪ ⎩ γ (x, y ) = γ ( y , x) where c is a constant. Also note, for each j ∈ Zn , G∗j (x, y ) ≡ Γ (x, y ) +

in Rn \ Y m , on ∂ Y m , for x = y ,

(4.11)

for x = y ,

γ (x + j , y + j ) is defined in (Rn \ Y m − j ) ×

(R \ Y m − j ) and satisfies n

⎧ − y G∗j (x, y ) = δ(x − y ) ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ∇ y G∗j (x, y ) · n y = 0  ∗  ⎪ G (x, y )  c |x − y |2−n ⎪ ⎪ j ⎪ ⎪ ⎩ ∗ G j (x, y ) = G∗j ( y , x)

in Rn \ Y m − j , on ∂ Y m − j , for x = y ,

(4.12)

for x = y .

Let us use notation in (3.1). (4.11)3 implies

⎧    ⎨ γ (x, ·) L ∞ (∂ Y ) + Γ (x, ·)C 2,α (∂ Y )  c if x ∈ Rn \ D0 , m m   ⎩ ∇ y Γ (·, y ) 2,α c if y ∈ Rn \ D0 , C (∂ Y )

(4.13)

m

where

α ∈ (0, 1) and c is a constant independent of x, y. By (4.11), (4.13)1 , and Theorem 3.1 [11], ⎧  ⎨ γ (x, ·) L ∞ (Rn \Y )  c if x ∈ Rn \ D0 , m   ⎩ γ (·, y ) ∞ n  c if y ∈ Rn \ D0 , L (R \Y )

(4.14)

m

where c is independent of x, y. By (4.11), (4.14)1 , and Corollary 6.3 and Theorem 6.30 [11], we obtain

  γ (x, ·) where

C 2,α (Rn \Y m )

 c if x ∈ Rn \ D0 ,

(4.15)

α ∈ (0, 1) and c is independent of x. By (4.11), (4.14)2 , and Corollary 6.3 [11],   γ (x, ·)

C 2,α (Rn \D)

 c if x ∈ D0 \ Y m ,

(4.16)

where c is a constant independent of x. By (4.11), for any y = ( y 1 , . . . , yn ) ∈ Rn \ Y m ,

−x ∂ y i γ (·, y ) = 0 in Rn \ D0 ,

(4.17)

where ∂ y i is the partial derivative with respect to y i for i ∈ {1, . . . , n}. (4.15), (4.17), and Corollary 6.3 [11] imply

L.-M. Yeh / J. Differential Equations 255 (2013) 1734–1783

  ∂ y γ (·, y ) 2,α n  c if y ∈ Rn \ Y m , i C (R \D)

1761

(4.18)

where α ∈ (0, 1) and c is independent of y = ( y 1 , . . . , yn ). ˘ x) = 1 in x ∈ D, Let φ˘ ∈ C 0∞ (Rn ) be a bell-shaped smooth function and satisfy φ˘ ∈ [0, 1], φ(

˘ ⊂ A. Then X˘A (x, y ) ≡ φ( ˘ x)φ( ˘ y ) is a smooth function in Rn × Rn satisfying supp(φ)

⎧ ˘ ˘ ⎪ ⎨ XA (x, y ) = XA ( y , x) ∈ [0, 1],

1 if x, y ∈ D, ⎪ ⎩ X˘A (x, y ) = 0 if x ∈ / A or y ∈ / A.

(4.19)

For any fixed x ∈ Ω f , we define, for y ∈ Ω f \ {x},

G ∗ (x, y ) ≡ Γ (x, y ) +



γ (x + j , y + j )X˘A (x + j , y + j )

j ∈Zn

   ˘ = Γ (x, y ) 1 − XA ( x + j , y + j ) + G∗j (x, y )X˘A (x + j , y + j ). j ∈Zn

(4.20)

j ∈Zn

Because of (4.12)4 and (4.19)1 ,

G ∗ (x, y ) = G ∗ ( y , x) Define, in Ω ω × Ω ωf for f

in Ω f × Ω f \ {x = y }.

(4.21)

ω ∈ [ n+1 1 , ∞), G ∗ω (x, y ) ≡ ω2−n G ∗



x

,

y

ω ω

.

(4.22)

Lemma 4.4. (1) There is a constant c such that

    1 − X˘A (x, y ) γ (x, y )  c for x, y ∈ Rn \ Y m .

(4.23)

1 (2) For ω ∈ [ n+ , ∞), there is a constant c independent of ω such that 1

 ∗  G (x, y )  c |x − y |2−n for x, y ∈ Ω ω and x = y . ω f Proof. Let c denote a constant independent of †

(4.24)

ω ∈ [ n+1 1 , ∞). By (4.14), |γ (x, y )|  c if x ∈/ D or y ∈/ D.

If x, y ∈ D, (4.19)2 implies 1 − XA (x, y ) = 0. So we see that (4.23) holds. (4.24) is from the definition of G ∗ω and (4.11). 2 1 Lemma 4.5. If ω ∈ [ n+ , ∞), |x − y |  12 , and x, y ∈ Ω ωf , then 1

  G ω (x, y ) − G ∗ (x, y )  c ω2−n , ω

where c is a constant independent of ω .

(4.25)

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L.-M. Yeh / J. Differential Equations 255 (2013) 1734–1783

Proof. Fix x ∈ Ω ω and define f

c1 ≡

  G ω (x, y ) − G ∗ (x, y ). ω

sup y ∈Ω ω f

1/5|x− y |

By (3.65)3 and (4.24), we see that c 1 is independent of ω , x. (3.65), (3.69), and (4.22) imply that \ G ω (x, y ) and G ∗ω (x, y ) are uniformly Lipschitz continuous functions (independent of ω ) of y in Ω ω f B 1/4 (x). So there is a positive constant c 2 (independent of ω , x) such that

    ∇ y G ω (x, y ) + ∇ y G ∗ (x, y )  c 2 for y ∈ Ω ω \ B 1/4 (x). ω f

(4.26)

Let us define

θω,x ≡

  G ω (x, y ) − G ∗ (x, y ).

sup

ω

y ∈Ω ω f

1/2|x− y |

Clearly, θω,x  c 1 . By (3.65)3 and (4.24), there is a y ω,x ∈ Ω ω such that f

  θω,x = G ω (x, y ω,x ) − G ∗ω (x, y ω,x ).

1/2  |x − y ω,x | and Take a number β > n so that

ρω,x ≡

θω,x  min{1/4, d0 }. β c2

(4.27)

See (3.1) for d0 . Then we see, by (4.26),

  G ω (x, y ) − G ∗ (x, y )  θω,x /5 for y ∈ B ρ ( y ω,x ) ∩ Ω ω . ω ,x ω f

(4.28)

Because of A1, one can take F ω,x ∈ C 0∞ ( B ρω,x ( y ω,x ) ∩ Ω ω ) such that f



F ω,x ∈ [0, 1], on B ρω,x /β ( y ω,x + yˆ ω,x ).

F ω ,x = 1

(4.29)

The point y ω,x + yˆ ω,x is chosen so that B ρω,x /β ( y ω,x + yˆ ω,x ) ⊂ B ρω,x ( y ω,x ) ∩ Ω ω . f If y ω,x ∈ ω(Y f − j ) for some j ∈ Zn , then we consider the following problems:



−U ω = F ω,x in Ω ωf , ω = 0 ∇Uω · n

ω, on ∂Ωm

 and

− U ω = F ω,x in Rn \ ω(Y m − j ), ω = 0 ∇ Uω · n

on ω(∂ Y m − j ).

(4.30)

By Lax–Milgram Theorem [11], extension theorem [1], and (2.12), both U ω ,  U ω are solvable uniquely in D 1,2 space, and

⎧ −(U ω −  Uω) = 0 ⎪ ⎪ ⎨ ⎪  ω = ⎪ ⎩∇(U ω − U ω ) · n



in Ω ω f ,

0

on ω(∂ Y m − j ),

ω −∇  Uω · n

ω \ ω(∂ Y − j ). on ∂Ωm m

(4.31)

L.-M. Yeh / J. Differential Equations 255 (2013) 1734–1783

We claim that, for any

1763

ω ∈ [ n+1 1 , ∞),

n U ω −  U ω  L ∞ (Ω ω )  c ω2−n  F ω,x  L 1 (Rn )  c ω2−n ρω ,x , f

where c is independent of ω , x. Proof of the claim: Because of y ω,x ∈ ω ∈ [ n+1 1 , ∞),

(4.32)

ω(Y f − j ), there is a smooth domain Bω such that, for

(S1) Bω ⊃ ω(Y m − j ) ∪ supp( F ω,x ), Bω ∩ ω(Y m − i ) = ∅ for i = j, (S2) dist(∂ Bω , supp( F ω,x )) is greater than ωd0 (see (3.1) for d0 ). By (S1) as well as Theorem 3.1 and Lemma 3.4 in [11], we see, from (4.31),

U ω −  U ω  L ∞ (Bω ∩Ω ω )  U ω −  U ω  L ∞ (∂ Bω ) . f

(4.33)

By Remark 3.1, (3.65), (3.70)1 , (4.12), and (4.30), we have, for z ∈ Ω ω \ Bω , f

⎧       c  F ω,x  L 1 (Rn ) ⎪ ⎪ U ω ( z) =  G ω ( z, y ) F ω,x ( y ) dy   ⎪ ⎪  dist( z, supp( F ))n−2 ,  ⎪ ω ,x ⎪ ⎪ ⎨ Ωω f

 ⎪    ⎪ ⎪    ⎪ U ( z ) = ω ⎪  ⎪ ⎪ ⎩



ω2−n G∗j



z

,

y



ω ω

 

F ω,x ( y ) dy  

Rn \ω(Y m − j )

where c is independent of

c  F ω,x  L 1 (Rn ) dist( z, supp( F ω,x ))n−2

,

ω, x, and F ω,x . By (S2),

n U ω −  U ω  L ∞ (Ω ω \Bω )  c ω2−n  F ω,x  L 1 (Rn )  c ω2−n ρω ,x , f

(4.34)

where c is independent of ω , x. (4.33)–(4.34) imply (4.32). By (4.28)–(4.30) and Remark 3.1, we see that, for some j ∈ Zn ,



ρωn+,x1  c

B ρω,x /β ( y ω,x + yˆ ω,x )

θω,x 5

  

dz  



 

G ω (x, y ) − G ∗ω (x, y ) F ω,x ( y ) dy 

Ωω f

  U ω (x)  U ω (x) −         x y x y + j , + j 1 − X˘A + j , + j F ω,x ( y ) dy , + c  ω2−n γ ω ω ω ω Ωω f

where c is independent of

ω, x. By (4.23), (4.32), and (4.35),

ρωn+,x1  c ω2−n ρωn ,x . So (4.25) holds and we prove the lemma.

2

(4.35)

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L.-M. Yeh / J. Differential Equations 255 (2013) 1734–1783

1 Lemma 4.6. If ω ∈ [ n+ , ∞), |x − y | > 2/3, and x, y ∈ Ω ωf , then 1

  ∇ y G ω (x, y ) − ∇ y G ∗ (x, y )  c ω2−n , ω

(4.36)

where c is a constant independent of ω .

\ Proof. If x ∈ Ω ω f



j ∈Zn



ω(A − j ), G ∗ω (x, ·) = Γ (x, ·) in Ω ωf \ {x} by (4.19), (4.20). So

− y (G ω − G ∗ω ) = 0

in Ω ω f \ {x},

 ω = −∇ y Γ (x, ·) · n  ω on ∂Ωmω . ∇ y (G ω − G ∗ω ) · n ω is of order Since the distance from x to ∂Ωm

ω (see (3.1)),

  ∇ y Γ (x, ·)

ω) C 1,α (∂Ωm

 c ω1−n ,

where α ∈ (0, 1) and c is independent of x, ω . By Theorem 6.30 [11] and Lemma 4.5, we obtain (4.36). If x ∈ ω(A \ Y m − j ) for some j ∈ Zn , then





G ∗ω (x, y ) = ω2−n G∗0 (x/ω + j , y /ω + j ) + ω2−n γ (x/ω + j , y /ω + j ) X˘A (x/ω + j , y /ω + j ) − 1 , for y ∈ Ω ω \ {x}. By (4.11), f

⎧ ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪  ⎪   ⎨ ⎪ ∗ · x ⎪ −n = G − G − ⎪ ˘ y ω ω 2 ω (∇ γ ∇ X ) + j , + j ⎪ y y A ⎪ ⎪ ⎪ ⎪ ω ω ⎪ ⎨ ⎪ ⎪  ⎪ ⎪ · x ⎪ −n ⎪ ⎪ ˘ ⎩ + ω (γ  y XA ) ⎪ + j, + j ⎪ ⎪ ω ω ⎪ ⎪ ⎪ ⎧   ⎪ ⎪ ⎪ x ⎪ ⎪ ⎪  ω + j ∇ y Γ (x, ·) · n  ω ⎨ − 1 − φ˘ ⎪ ∗ ⎪ ω ⎪  = ∇y Gω − Gω · n ⎪ ⎪ ⎪ ⎩ ⎩ ω −∇ y Γ (x, ·) · n ω \ ω(∂ Y − j ) is of order Because the distance from x to ∂Ωm m

 in

Ω ωf \ ω(A − j ), or

ω(D \ Y m − j ),

in ω(A \ D − j ), on ω(∂ Y m − j ), ω \ ω(∂ Y − j ). on ∂Ωm m

ω (see (3.1)), by (4.13)1 , (4.15), and (4.16),

⎧     −n ⎪ ⎪ ω (∇ y γ ∇ y X˘A ) x + j , · + j  ⎪ ⎪  0,α ⎪ ω ω ⎪ C (ω(A\D− j )) ⎪ ⎪ ⎪    ⎪ ⎪   −n · x ⎪ ⎪  ˘ +  c ω−n , ⎪ ⎪ ω (γ  y XA ) ω + j , ω + j  0,α ⎨ C (ω(A\D− j ))   ⎪ sup ∇ y Γ (x, ·)C 1,α (ω(∂ Y −i ))  c ω1−n , ⎪ ⎪ m ⎪ ⎪ i ∈Zn ⎪ ⎪ i = j ⎪ ⎪ ⎪    ⎪ ⎪   x ⎪ ⎪   ˘ ⎪ ∇ 1 − φ + j Γ ( x , ·)  c ω1−n , y ⎩  1,α ω C

where

(ω(∂ Y m − j ))

α ∈ (0, 1) and c is independent of x, ω. We get (4.36) by Theorem 6.30 [11] and Lemma 4.5. 2

L.-M. Yeh / J. Differential Equations 255 (2013) 1734–1783

1765

1 Lemma 4.7. If ω ∈ [ n+ , ∞), |x − y | > 3/4, and x, y ∈ Ω ωf , then 1

  ∇x ∇ y G ω (x, y ) − ∇x ∇ y G ∗ (x, y )  c ω2−n ,

(4.37)

ω

where the constant c is independent of ω . Proof. To show (4.37), we modify the argument for (4.36) and use the facts G ω (x, y ) = G ω ( y , x) and ω (x) ≡ ∂ yk G ∗ω (x, y ) for ϕω (x) ≡ ∂ yk G ω (x, y ) and ϕ G ∗ω (x, y ) = G ∗ω ( y , x) (see (3.65)  5 and (4.21)). Define ω ω ∗ k ∈ {1, . . . , n}. If y ∈ Ω f \ j ∈Zn ω (A − j ), then G ω (·, y ) = Γ (·, y ) in Ω f \ { y } and



in Ω ω f \ B 2/3 ( y ),

ω ) = 0 −(ϕω − ϕ

 ω = −∇x ∂ yk Γ (·, y ) · n  ω on ∂Ωmω \ B 2/3 ( y ). ω ) · n ∇(ϕω − ϕ ω is of order Because the distance from y to ∂Ωm

ω (see (3.1)),

  ∇x ∂ y Γ (·, y ) 1,α ω  c ω−n , k C (∂Ω ) m

where α ∈ (0, 1) and c is independent of y, ω , k. By (4.36) and Theorem 6.30 [11], we have (4.37). If y ∈ ω(A \ Y m − j ) for some j ∈ Zn , then

G ∗ω (x, y ) = ω2−n G∗0 (x/ω + j , y /ω + j )

  + ω2−n γ (x/ω + j , y /ω + j ) X˘A (x/ω + j , y /ω + j ) − 1 , y

for x ∈ Ω ω \ { y }. Note (4.11)1,4 imply x γ ( ωx + j , ω + j ) = 0 for x, y ∈ Ω ωf . So f

⎧ ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪  ⎪  ϕ − ϕ ) = −( ⎪ ω ω y · ⎪ −n−1 ⎪ ⎪ ˘ ⎪ ⎪ 2 ω ∂ (∇ γ ∇ X ) + j , + j y x x A ⎪ ⎪ k ⎪ ⎪ ω ω ⎪ ⎪ ⎪ ⎪ ⎪ ⎪  ⎪ ⎨ ⎪ ⎪ y · ⎪ ⎩ + ω−n−1 ∂ yk (γ x X˘A ) + j, + j ⎪ ω ω ⎪ ⎪ ⎪ ⎧  ⎪  ⎪ ⎪ y ⎪ ⎪ ⎪ ⎪ ω + j ∇x ∂ yk Γ (·, y ) · n ⎪ − 1 − φ˘ ⎪ ⎪ ⎪ ⎪ ⎪ ω ⎪ ⎪ ⎨ ⎪ ⎪  ⎪ ⎪ ω = ω ) · n ∇(ϕω − ϕ ⎪ −1 ˘ y ⎪ ⎪ ω ⎪ ∇x Γ (·, y ) · n + ω + j φ ⎪ ⎪ ⎪ ⎪ ω ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎩ ω −∇x ∂ yk Γ (·, y ) · n ω \ ω(∂ Y − j ) is of order Since the distance from y to ∂Ωm m

⎧ ω Ω \ (ω(A − j ) ∪ B 2/3 ( y )), ⎪ ⎨ f in

⎪ ⎩

or

ω(D \ Y m − j ) \ B 2/3 ( y ),

in ω(A \ D − j ) \ B 2/3 ( y ),

on ω(∂ Y m − j ) \ B 2/3 ( y ), ω \ ω(∂ Y − j ). on ∂Ωm m

ω (see (3.1)), by (4.13)2 and (4.18),

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L.-M. Yeh / J. Differential Equations 255 (2013) 1734–1783

  ⎧   −n−1 · y ⎪   ˘ ⎪ ω ∂ (∇ γ ∇ X ) + j , + j ⎪ yk x x A  0,α ⎪ ⎪ ω ω C (ω(A\D− j )) ⎪ ⎪ ⎪    ⎪ ⎪   ⎪ y · ⎪ −n−1 ⎪ + ∂ yk (γ x X˘A ) + j, + j   c ω−n−1 , ⎪  0,α ω ⎪ ω ω ⎪ C ( ω ( A \ D− j )) ⎪ ⎪ ⎪ ⎪ −n ⎨ sup ∇ ∂ Γ (·, y ) 1,α x yk C (ω(∂ Y m −i ))  c ω , i ∈Zn ⎪ i = j ⎪ ⎪ ⎪    ⎪ ⎪   ⎪ y ⎪   ˘ ⎪ ⎪  1 − φ ω + j ∇x ∂ yk Γ (·, y ) 1,α ⎪ ⎪ C (ω(∂ Y m − j )) ⎪ ⎪ ⎪    ⎪ ⎪   ⎪ ⎪ −1 ˘ y  ⎪  c ω−n , ⎩ + ω φ ω + j ∇x Γ (·, y ) 1,α C

where

(ω(∂ Y m − j ))

α ∈ (0, 1) and c is independent of y, ω, k. We get (4.37) by (4.36) and Theorem 6.30 [11]. 2

By Lemma 4.7 and tracing the argument of Corollary 4.1, we have Corollary 4.2. There is a constant c > 0 so that, for x, y ∈ Ω f and |x − y |  n + 1,

  ∇x ∇ y G (x, y ) − ∇x ∇ y G ∗ (x, y )  c |x − y |−2 . 5. Proof of main results In this section, we prove the main results. Theorem 2.1 is in Section 5.1, Theorem 2.2 is proved in Section 5.2, Theorem 2.3 is proved in Section 5.3, and Theorem 2.4 is in Section 5.4. 5.1. Proof of Theorem 2.1 Lemma 5.1. If Q ∈ [C 0∞ (Rn )]n , then

     ∇x ∇ y G (x, y ) Q ( y ) dy    Ωf

L p (Ω f )

 c  Q L p (Ω f ) for any p ∈ (1, ∞),

where c is a constant. Proof. Let η be a bell-shaped function in C 0∞ (R) satisfying η ∈ [0, 1], η( z) = 1 for | z| < n and for | z|  n + 1, and set ∇x ∇ y G (x, y ) = L0 (x, y ) + Lc (x, y ), where

η( z) = 0

       L0 (x, y ) ≡ η |x − y | ∇x ∇ y G ∗ + 1 − η |x − y | I + ∇X(x) I + ∇X( y ) ∇x ∇ y G 0 . By (4.20), we define

∇x ∇ y G (x, y ) Q ( y ) dy = Ωf

where

4  i =1

Fi ( Q )(x),

(5.1)

L.-M. Yeh / J. Differential Equations 255 (2013) 1734–1783







η |x − y | Q ( y )∇x ∇ y

F1 ( Q )(x) ≡

 

G∗0 (x + k, y + k)X˘A (x + k, y + k) dy ,

k∈Zn

Ωf











η |x − y | Q ( y )∇x ∇ y Γ (x, y ) 1 −

F2 ( Q )(x) ≡





X˘A (x + k, y + k) dy ,

k∈Zn

Ωf

F3 ( Q )(x) ≡

1767













1 − η |x − y | Q ( y ) I + ∇X(x) I + ∇X( y ) ∇x ∇ y G 0 (x, y ) dy ,

Ωf



Lc (x, y ) Q ( y ) dy .

F4 ( Q )(x) ≡ Ωf

From (4.19), we know X˘A (x + k, y + k) = 0 for any k ∈ Zn only if |x − y | < x ∈ Y f − j for some j ∈ Zn , by change of variable and the definition of η ,





√

n and x + k, y + k ∈ Y . If



˘ x + j )φ( ˘ y + j ) dy Q ( y )∇x ∇ y G∗0 (x + j , y + j )φ(

F1 ( Q )(x) = Y f −j

˘ x + j) = ∇ φ(







˘ y ) + G∗0 (x + j , y )∇ φ( ˘ y ) dy Q ( y − j ) ∇ y G∗0 (x + j , y )φ(

Yf

˘ x + j) + φ(



˘ y )∇x ∇ y G∗0 (x + j , y ) dy Q ( y − j )φ(

Yf

˘ x + j) + φ(



˘ y ) dy . Q ( y − j )∇x G∗0 (x + j , y )∇ φ(

(5.2)

Yf

(1)

Define V j ( z) ≡

D

1, 2

Yf

˘ y ) dy for z ∈ Rn \ Y m . By (3.70)1 , V (1) is the unique G∗0 (z, y ) Q ( y − j )∇ φ( j

(R \ Y m ) solution of n

⎧ ⎨ − V (j 1) = Q (· − j )∇ φ˘ in Rn \ Y m , ⎩ ∇ V (1 ) · n  =0 j

(2)

Define V j ( z) ≡

Yf

on ∂ Y m .

˘ y ) dy for z ∈ Rn \ Y m . By (3.70)1 , V (2) is the unique ∇ y G∗0 (z, y ) Q ( y − j )φ( j

D1,2 (Rn \ Y m ) solution of

⎧   ⎨ −∇ · ∇ V (j 2) − Q (· − j )∇ φ˘ = 0 in Rn \ Y m , ⎩ ∇ V (2) − Q (· − j )∇ φ˘  · n  =0 on ∂ Y m . j By Lemma 3.3,

 (1 ) (2 )  V , V  j

j

W 1, p (Y f )

   c  V (j 1) , V (j 2)  L p (D

2 \Y )

   +  Q (· − j ) L p (Y ) , f

where p ∈ (1, ∞) and c is a constant independent of j. See (3.1) for D2 . By (3.69),

(5.3)

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L.-M. Yeh / J. Differential Equations 255 (2013) 1734–1783

 (1 )   (2 )   V ( z) +  V ( z)  c j



j

    Q ( y − j ) | z − y |2−n + | z − y |1−n dy for z ∈ D2 \ Y ,

Yf

where c is independent of j. Tracing the proof of Lemma 7.12 [11], we see

 (1 ) (2 )  V , V  j

j

L p ( D2 \ Y )

   c  Q (· − j )L p (Y

f

)

for p ∈ (1, ∞),

(5.4)

where c is a constant independent of j. By (3.70)3 ,

     ∇z ∇ y G∗ (·, y ) Q ( y − j )φ( ˘ y ) dy  0   Yf

L p (Y

f

)

  (2 )  ∇ V j , Q (· − j )L p (Y ) , f

(5.5)

˘ x + j )) ∩ supp(φ( ˘ x + i )) = ∅ if i = j, where p ∈ (1, ∞) and c is independent of j. Since supp(φ( (5.2)–(5.5) imply

  (1) (2)  p   F1 ( Q ) pp  V , V  1, p c j j L (Ω ) W (Y f

j ∈Zn

f)

 p  +  Q (· − j ) L p (Y ) f

p

 c  Q L p (Ω f ) ,

(5.6)

where p ∈ (1, ∞) and c is a constant. So there is a constant c such that

  F1 ( Q ) p  c  Q L p (Ω f ) for p ∈ (1, ∞). L (Ω ) f

(5.7)

By (4.19), F2 ( Q ) can be written as F2 ( Q ) = F21 ( Q ) − F22 ( Q ) − F23 ( Q ), where

F21 ( Q )(x) =

Q ( y )∇x ∇ y Γ (x, y ) dy ,

Ωf



F22 ( Q )(x) = Ωf



F23 ( Q )(x) =







1 − η |x − y | Q ( y )∇x ∇ y Γ (x, y ) dy ,

  ˘ η |x − y | Q ( y )∇x ∇ y Γ (x, y ) XA (x + k, y + k) dy . 



k∈Zn

Ωf

It is not difficult to see, by Theorem 3 on p. 39 [23] and (2.13) in [11],

    F21 ( Q ) p n + F22 ( Q ) p n  c  Q L p (Ω ) for p ∈ (1, ∞), f L (R ) L (R ) where c is a constant. By Lemma 7.12 and Lemma 4.2 [11] and repeating the argument from (5.2) to (5.6), we obtain

  F23 ( Q ) where c is a constant. Therefore,

L p (Ω f )

 c  Q L p (Ω f ) for p ∈ (1, ∞),

L.-M. Yeh / J. Differential Equations 255 (2013) 1734–1783

  F2 ( Q )

L p (Ω f )

 c  Q L p (Ω f ) for p ∈ (1, ∞),

1769

(5.8)

where c is a constant. Since K is a constant symmetric positive definite matrix (see (2.9)), we do change of variable as the remark before Lemma 4.1 as well as use both Theorem 3 on p. 39 [23] and (2.13) in [11] to obtain

  F3 ( Q )

L p (Ω f )

 c  Q L p (Ω f ) for p ∈ (1, ∞),

(5.9)

where c is a constant. Finally,













η |x − y | Lc1 (x, y ) Q ( y ) + 1 − η |x − y | Lc2 (x, y ) Q ( y ) dy ,

F4 ( Q )(x) ≡ Ωf

where

Lc1 (x, y ) = ∇x ∇ y G (x, y ) − ∇x ∇ y G ∗ (x, y ),    Lc2 (x, y ) = ∇x ∇ y G (x, y ) − I + ∇X(x) I + ∇X( y ) ∇x ∇ y G 0 (x, y ). Define



Ωf

P1 ( Q )(x) ≡

η(|x − y |)Lc1 (x, y ) Q ( y ) dy if x ∈ Ω f , if x ∈ Rn \ Ω f .

0

(5.10)

Clearly, P1 is a linear map. For any δ > 0, by Fubini Theorem [21], Corollary 4.2, and change of variable,

    δ  x ∈ Rn  P1 ( Q )(x) > δ         P1 ( Q ) dx  η |x − y | Lc1 (x, y ) Q ( y ) dy dx  Ωf



 Rn

Rn

Ωf Ωf

η(|x − y |) |x −

y |2

  XΩ f (x) Q ( y )XΩ f ( y ) dx dy  c  Q  L 1 (Ω f ) ,

(5.11)

where c is a constant. By (5.10) and Corollary 4.2, if x ∈ Ω f ,

  P1 ( Q )(x) 









η |x − y | Lc1 (x, y ) Q ( y ) dy  c  Q L ∞ (Ω f ) ,

(5.12)

Ωf

where c is a constant. By (5.11)–(5.12) and Theorem 5 on p. 21 [23], we see

  P1 ( Q )

L p (Rn )

 c  Q L p (Ω f ) for p ∈ (1, ∞),

where c is a constant. Similarly, if we define

(5.13)

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L.-M. Yeh / J. Differential Equations 255 (2013) 1734–1783



P2 ( Q )(x) ≡

c Ω f (1 − η(|x − y |))L2 (x, y ) Q ( y ) dy

if x ∈ Ω f , if x ∈ Rn \ Ω f ,

0

as well as employ Theorem 5 on p. 21 [23] and Corollary 4.1, then

  P2 ( Q )

L p (Rn )

 c  Q L p (Ω f ) for p ∈ (1, ∞),

where c is a constant. Together with (5.13), we have

  F4 ( Q )

L p (Ω f )

 c  Q L p (Ω f ) for p ∈ (1, ∞),

where c is a constant. The lemma follows from (5.1), (5.7), (5.8), (5.9), and (5.14).

(5.14)

2 1, p

Lemma 5.2. If p ∈ (1, ∞) and Q ∈ [ L p (Rn )]n with support in B t (0) for some t > 0, then a W loc (Ω f ) solution of

⎧ −∇ · (∇ U + Q ) = 0 in Ω f , ⎪ ⎨  =0 (∇ U + Q ) · n on ∂Ωm , ⎪ ⎩ |U |(x) = o(1) for large |x|

(5.15)

exists uniquely. The solution of (5.15) satisfies

⎧ if p ∈ (1, ∞), ∇ U L p (Ω f )  c  Q L p (Ω f ) ⎪ ⎪ ⎨ U L p ( B s (0)∩Ω f )  ct ,s  Q L p (Ω f ) if p ∈ (1, ∞), ⎪ ⎪ ⎩ U  np  c  Q L p (Ω f ) if p ∈ (1, n), n− p L

(5.16)

(Ω f )

 is a unit vector normal to ∂Ωm , c is a constant independent of t, and ct ,s is a constant dependent where s > 0, n on t, s. Proof. By Remark 3.1, we know that the D 1,2 solution of (5.15) exists uniquely if Q ∈ [C 0∞ (Rn )]n . Lemma 5.1 and (3.61) imply (5.16)1 . By Remark 3.1 and Lemma 3.8, the solution of (5.15) satisfies

    |U |(x) =  ∇ y G (x, y ) Q ( y ) dy   c |x − y |1−n | Q |( y ) dy for x ∈ Ω f . Ωf

(5.17)

Ωf 1, p

By Lemma 7.12 [11] and (5.17), we have (5.16)2 . So a W loc solution of (5.15) for p ∈ (1, ∞) exists. 1, p

By Theorem 1 on p. 119 [23] and (5.17), we obtain (5.16)3 . The uniqueness of the W loc solution of (5.15) for p ∈ (1, ∞) is due to the maximal principle (see Theorem 3.1 and Lemma 3.4 in [11]). So we prove Lemma 5.2 for Q ∈ [C 0∞ (Rn )]n case. If Q ∈ [ L p (Rn )]n with compact support, the lemma can be proved by a limiting argument. Theorem 2.1 is a direct result of Lemma 5.2. 2

L.-M. Yeh / J. Differential Equations 255 (2013) 1734–1783

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5.2. Proof of Theorem 2.2 Assume F  ∈ C 0∞ ( B t (0)) for some t > 0. By Remark 3.1 and (2.12), the D 1,2 (Ω f ) solution of (2.1) exists uniquely and

U   where ct is independent of and Fubini Theorem [21],

2n

L n−2 (Ω f )

 ct  F  X B t (0)∩Ω f  H −1 (Rn ) ,

(5.18)

 but dependent on t. Let ξ ∈ C 0∞ ( B s (0)) and s > 0. By Remark 3.1, (3.64),





∇ U  ξ dx =

B s (0)∩Ω f

∇x G  (x, y ) F  ( y )ξ(x) dy dx B s (0)∩Ω f B t (0)∩Ω f





=

∇x G  (x, y )ξ(x) dx F  ( y ) dy .

(5.19)

B t (0)∩Ω f B s (0)∩Ω f

If we define



ϕ ( y ) ≡

∇x G  (x, y )ξ(x) dx, B s (0)∩Ω f

then, by Remark 3.1 and (3.64)–(3.65),

⎧ −∇ · (∇ ϕ − ξ ) = 0 in Ω f , ⎪ ⎨   = 0 on ∂Ωm , (∇ ϕ − ξ ) · n ⎪ ⎩ |ϕ |( y ) = o(1) for large | y |. By Theorem 2.1, we see

ϕ W 1,r ( B  (0)∩Ω  )  c ,s ξ Lr (Ω f ) for any r ∈ (1, ∞), f

where  > 0 and c ,s is independent of Hölder inequality,



 but dependent on , s. By (5.20), extension theorem [1], and

∇ U  ξ dx = B s (0)∩Ω f

(5.20)



ϕ F  dy = B t (0)∩Ω f

Π ϕ F  X B t (0)∩Ω f dy Rn

 Π ϕ W 1,r ( B t +1 (0))  F  X B t (0)∩Ω f W −1, p (Rn )  ct ,s ξ Lr (Ω f )  F  X B t (0)∩Ω f W −1, p (Rn ) , where

1 r

+

1 p

(5.21)

= 1 and ct ,s is independent of  but dependent on t, s. Since C 0∞ ( B s (0)) is dense in

L r ( B s (0)) for r ∈ (1, ∞) and s > 0, we know

∇ U  L p ( B s (0)∩Ω f )  ct ,s  F  X B t (0)∩Ω f W −1, p (Rn ) for any p ∈ (1, ∞).

(5.22)

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L.-M. Yeh / J. Differential Equations 255 (2013) 1734–1783

If p ∈ [2, ∞), then

 F  X B t (0)∩Ω f  H −1 (Rn )  ct  F  X B t (0)∩Ω f W −1, p (Rn ) ,

(5.23)

1, p

where ct depends on t. (5.18), (5.22), and (5.23) imply that a W loc (Ω f ) solution of (2.1) for p ∈

[2,

2n ] n−2

exists and (2.2) holds. That is,

U  W 1, p ( B s (0)∩Ω  )  ct ,s  F  X B t (0)∩Ω f W −1, p (Rn ) ,

(5.24)

f

where ct ,s is a constant independent of

 F  X B t (0)∩Ω f 

W

2n  but dependent on t, s. If p ∈ ( n2n −2 , n−4 ], we know

−1, n2n −2

(Rn )

 ct  F  X B t (0)∩Ω f W −1, p (Rn ) ,

where ct depends on t. By Theorem 7.26 [11], (5.24), and a similar argument as (2.12), if n2n −2 < n and 2n p ∈ ( n2n , ] , then −2 n−4

U  

2n

L n−4 ( B s (0)∩Ω f )

where ct ,s is independent of

 c s U  

W

1, 2n n −2

( B s (0)∩Ω f )

 ct ,s  F  X B t (0)∩Ω f W −1, p (Rn ) ,

(5.25)

 but dependent on t, s. Theorem 7.26 [11], (5.22), and (5.25) imply 1, p

2n 2n 2n 2n  that, if n2n −2  n and p ∈ ( n−2 , ∞) or if n−2 < n and p ∈ ( n−2 , n−4 ], then a W loc (Ω f ) solution of (2.1) 1, p exists and (2.2) holds. Repeating the same process, we see that a W loc (Ω f ) solution of (2.1) exists and (2.2) holds for p ∈ [2, ∞). Since (2.2) holds for p ∈ [2, ∞), a modification of the argument from (5.19) to (5.21) shows

U  L p ( B s (0)∩Ω f )  ct ,s  F  X B t (0)∩Ω f W −1, p (Rn ) for p ∈ (1, 2], where ct ,s is a positive constant independent of

(5.26)

 but dependent on t, s. (5.22) and (5.26) imply that

1, p

a W loc (Ω f ) solution of (2.1) exists and (2.2) holds for p ∈ (1, 2]. So (2.2) holds for p ∈ (1, ∞). Remark 3.1 and (3.65) imply

⎧ ⎪ U |( x )  c |x − y |2−n | F  |( y ) dy | ⎪  ⎪ ⎪ ⎪ ⎪ ⎨ Ω f ⎪ ⎪ ⎪ |∇ U  |(x)  c |x − y |1−n | F  |( y ) dy ⎪ ⎪ ⎪ ⎩ 

for x ∈ Ω f ,

Ωf

where c is a constant independent of  , F  . By Theorem 1 on p. 119 [23], we obtain (2.3). Uniqueness 1, p of the W loc (Ω f ) solution of (2.1) for p ∈ (1, ∞) is due to Theorem 3.1 and Lemma 3.4 [11]. So Theorem 2.2 holds for F  ∈ C 0∞ (Rn ) case. If F  ∈ W −1, p (Rn ) with compact support, Theorem 2.2 can be proved by a limiting argument.

L.-M. Yeh / J. Differential Equations 255 (2013) 1734–1783

1773

5.3. Proof of Theorem 2.3 By Lax–Milgram Theorem [11], extension theorem [1], and (2.12), we know that the D 1,2 (Ω f ) solution of (2.4) exists uniquely and satisfies

U  

2n L n −2

(Ω  )

  c  Q  L 2 (Rn ) +  F  

f

 2n L n +2

(Rn )

(5.27)

,

where c is independent of  . For any x ∈ Ω f , by Lemma 3.1 and (5.27),

    U  (x)  U  (x) − 

    U  ( y ) dy  + 

− B 1/2 (x)∩Ω f

 [U  ]C 0,μ ( B

B 1/2 (x)∩Ω f

  

−

)

1/2 (x)∩Ω f

  U  ( y ) dy 

−

B 1/2 (x)∩Ω f

  U  ( y ) dy 

−

| y |μ dy + 

B 1/2 (x)∩Ω f

 c  Q  , F  Ln+δ (Rn ) , where μ, δ > 0 and c is independent of Lemma 3.4.

 , x. So we prove (2.5). (2.6) follows from (2.5) and

5.4. Proof of Theorem 2.4 Lemma 5.3. Suppose

⎧   ⎪ ⎨ Q ∈ L p Rn , ⎪ ⎩



| Q |(x) = O |x|

np



F ∈ L n+ p Rn −n−1

 ,



 for p ∈



| F |(x) = O |x|

n n−2

−n−1

,∞ ,



(5.28)

,

then a W 1, p (Ω f ) solution of

⎧ −∇ · (∇ ϕ + Q ) = F ⎪ ⎨  =0 (∇ ϕ + Q ) · n ⎪ ⎩ |ϕ |(x) = o(1)

in Ω f , on ∂Ωm ,

(5.29)

exists uniquely and satisfies

⎧   , +  F  np ⎪ ⎨ ϕ L p (Ω f )  c  Q L nnp + p (Ω ) n + 2p L (Ω f ) f   ⎪ ⎩ ∇ ϕ L p (Ω f )  c  Q L p (Ω f ) +  F  np , n+ p L

(5.30)

(Ω f )

where c is a positive constant independent of Q , F . |ζ |(x) = O (|x|−n−1 ) for ζ ∈ { Q , F } means that when |x| is large, |ζ |(x)  c |x|−n−1 where c is a constant. Proof. The uniqueness of the solution of (5.29) is from maximal principle (see Theorem 3.1 and Lemma 3.4 [11]). If Q , F ∈ C 0∞ (Rn ), by Remark 3.1 and Lemma 3.9, the solution of (5.29) exists uniquely in D 1,2 (Ω f ) and satisfies, for x ∈ Ω f ,

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L.-M. Yeh / J. Differential Equations 255 (2013) 1734–1783



ϕ (x) =

G (x, y ) F ( y ) dy −

Ωf

∇ y G (x, y ) Q ( y ) dy ,

(5.31)

Ωf

      ∇ ϕ (x) − ∇x G (x, y ) F ( y ) dy + ∇x ∇ y G (x, y ) Q ( y ) dy   c  Q (x),   Ωf

(5.32)

Ωf

where c is a constant independent of Q , F . By Theorem 1 on p. 119 [23], (3.48), and (5.31), we get (5.30)1 . By Theorem 1 on p. 119 [23] and (3.48),

     ∇x G (x, y ) F ( y ) dy    Ωf

L p (Ω f )

 c F 

np

L n+ p (Ω f )

,

where c is a constant. Together with Lemma 5.1 and (5.32), we obtain (5.30)2 for Q , F ∈ C 0∞ (Rn ) case. For general case, (5.28) implies Q , F ∈ L 1 (Rn ). We can find Q m , F m ∈ C 0∞ (Rn ) such that

Let

⎧ ⎨ Qm → Q

in L p Rn ∩ L 1 Rn

⎩F → F m

in L n+ p Rn ∩ L 1 Rn



np















as m → ∞.

(5.33)

ϕm be the solution of (5.29) corresponding to Q m and F m . Then (5.30) implies ⎧   , +  F m  np ⎪ ⎨ ϕm L p (Ω f )  c  Q m L nnp + p (Ω ) L n+2p (Ω f ) f   ⎪ ⎩ ∇ ϕm L p (Ω f )  c  Q m L p (Ω f ) +  F m  np , n+ p L

m of where c is independent of Q m , F m . There is a subsequence ϕ and (5.33)–(5.34),

(5.34)

(Ω f )

ϕm such that, by Theorem 7.26 [11]

m → ϕ weakly in W 1, p (Ω f ) and pointwise in Ω f as m → ∞, ϕ

(5.35)

and ϕ satisfies (5.29)1,2 and (5.30). Next we claim ϕ , Q , F satisfy (5.31) almost everywhere in Ω f . For any fixed x ∈ Ω f ,

     G (x, y )( F m − F )( y ) dy    Ωf

   

 



 

G (x, y )( F m − F )( y ) dy  + 

Ω f \ B 1 (x)



 

G (x, y )( F m − F )( y ) dy .

(5.36)

Ω f ∩ B 1 (x)

By (3.48) and (5.33),

  lim  m→∞



Ω f \ B 1 (x)

  G (x, y )( F m − F )( y ) dy   lim c  F m − F  L 1 (Ω f ) = 0. m→∞

By Fubini Theorem [21] and (3.48),

(5.37)

L.-M. Yeh / J. Differential Equations 255 (2013) 1734–1783

 

m→∞

 



lim 

1775

G (x, y )( F m − F )( y ) dy dx

Ω f Ω f ∩ B 1 (x)



 lim c m→∞

Rn B 1 (x)





= lim c m→∞

|x − y |2−n | F m − F |( y ) dy dx | z|2−n | F m − F |(x − z) dz dx  lim c  F m − F L 1 (Ω f ) = 0. m→∞

Rn B 1 (0)

This means that there is a subsequence of



G (x, y ) F m ( y ) dy

converging to

G (x, y ) F ( y ) dy

Ω f ∩ B 1 (x)

(5.38)

Ω f ∩ B 1 (x)

pointwise almost everywhere in Ω f . By (5.36)–(5.38), we prove





Ω f G (x, y ) F m ( y ) dy converges

G (x, y ) F ( y ) dy pointwise almost everywhere in Ω f . By a similar argument, we also have

Ωf

Ω f ∇ y G (x, y ) Q m ( y ) dy converging to Ω f ∇ y G (x, y ) Q ( y ) dy pointwise almost everywhere in Ω f . Together with (5.35), we prove the claim. That is, functions ϕ , Q , F satisfy (5.31) almost everywhere in Ω f . Because (5.28)2 holds and because ϕ , Q , F satisfy (5.31), it is not difficult to see that ϕ satisfies (5.29)3 . Then uniqueness of the solution implies the lemma. 2 to

n If U 0 solves (2.14) and F in (2.14) is in W 1, p (Rn ) with compact support for p ∈ ( n− , ∞), then, 2 by Definition 5.1 and p. 67 [16],

U 0 (x) =

G 0 (x, y )|Y f | F ( y ) dy

for x ∈ Rn ,

Rn

where G 0 is the Green’s function in (4.1). By change of variable, G 0 can be transformed to the fundamental solution of the Laplace equation in some new coordinate system. By Lemma 4.4 [11], (2.14) in [11], and Theorem 3 on p. 39 [23], we know

⎧   ⎪ ⎨ ∇ 2 U 0 

 W i , p (Rn )

 c  F W i, p (Rn ) for p ∈

⎪   ⎩  i  ∇ U 0 (x) = O |x|2−n−i

n n−2

, ∞ and i ∈ {1, 2},

(5.39)

for i ∈ {0, 1, 2, 3},

where c is a constant dependent on n, p, K. Define

ϕ (x) ≡ U  (x) − U 0 (x) − X (x)∇ U 0 (x) − S (x)∇ 2 U 0 (x) for x ∈ Ω f , where U  , U 0 are the solutions of (2.13) and (2.14), X is defined in (2.8), and S is defined in (2.10). As in the proof of Lemma 3.2,

⎧   3 3  ⎪ ⎪ −∇ · ∇ ϕ + S ∇ U 0 = X ∇U 0 + ∇S ∇ U 0 in Ω f , ⎨    ,  =0 ∇ ϕ + S ∇ 3 U 0 · n on ∂Ωm ⎪ ⎪ ⎩ |ϕ |(x) = o(1).

(5.40)

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(5.40)3 is due to (2.11), Remark 3.1, (3.48), and (5.39)2 . By Lemma 5.3, (2.11), and (5.39), the solution of (5.40) satisfies

⎧ np ⎨ ϕ L p (Ω f )  c   F W 1, nnp + p (Rn )∩ W 1, n+2p (Rn ) ⎩ ∇ ϕ L p (Ω  )  c   F  f

W 1, p (Rn )∩ W

np 1, n+ p

 for p ∈

n n−2

(Rn )

,∞ ,

n where c is independent of  . So if p ∈ ( n− , ∞), then 2

⎧ np , ⎨ U  − U 0 L p (Ω f )  c   F W 1, nnp + p (Rn )∩ W 1, n+2p (Rn )   ⎩ ∇ U  − ( I + ∇X )∇ U 0  p   c   F  L (Ω ) 1, p n W

f

np 1, n+ p

(R )∩ W

(Rn )

(5.41)

,

where c is independent of  . Similarly, by (2.11), (5.39)–(5.40), Lemma 3.2, and Lemma 3.4, we have

⎧ , ⎨ U  − U 0 L ∞ (Ω f )  c   F W 1, n2n +2 (Rn )∩ W 1,n+δ (Rn )   ⎩ ∇ U  − ( I + ∇X )∇ U 0  ∞   c   F  1, 2n L (Ω ) n +2

W

f

(Rn )∩ W 2,n+δ (Rn )

,

(5.42)

where c is independent of  . By (5.41) and (5.42), we prove Theorem 2.4. 6. Proof of Lemmas 3.1, 3.4 From Remark 2.1, we know the W 2, p norm of the solutions of (3.2) in B 1/2 (0) ∩ Ω f in general is not bounded uniformly in  even if  Q   W 1, p (Ω  ) ,  F   L p (Ω  ) are bounded independent of  . But f

f

Lemma 3.1 and Lemma 3.4 prove that the Hölder norm and the Lipschitz norm of the solutions of (3.2) in B 1/2 (0) ∩ Ω f are bounded uniformly in  . 6.1. Proof of Lemma 3.1 Lemma 6.1. For any δ > 0, there are θ1 , θ2 ∈ (0, 1) (dependent on δ , Y f ) with θ1 < θ22 and pendent on δ , θ1 , θ2 ) such that if U λ , Q λ , F λ satisfy



−∇ · (∇ U λ + Q λ ) = F λ in B 1 (0) ∩ Ω λf , λ = 0 (∇ U λ + Q λ ) · n

λ on B 1 (0) ∩ ∂Ωm ,

0 ∈ (0, 1) (de-

(6.1)

and





max U λ  L 2 ( B 1 (0)∩Ω λ ) , 0−1  Q λ , F λ  Ln+δ ( B 1 (0)∩Ω λ )  1, f

f

then, for any 0 < λ  0 and θ ∈ [θ1 , θ2 ],

 2 − Πλ U λ − (Πλ U λ )0,θ  dx  θ 2μ

(6.2)

B θ ( 0)

δ  λ is the unit vector normal to ∂Ωmλ , and Πλ is the extension operator in [1] (or see (2.12)). where μ ≡ n+δ ,n

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Proof. Consider the following problem LU ≡ −∇ ·(K∇ U ) = 0, where K is the constant matrix in (2.9). Then U satisfies, by Theorem 9.11 [11],

U W 1,r ( B 1/2 (0))  c U L 2 ( B 3/4 (0)) for any r > n, where c depends on K, r. If

μ satisfies μ < μ < 1, then  2

− U − (U )0,θ  dx  θ 2μ − U 2 dx B θ ( 0)

(6.3)

B 3/4 (0)

for θ sufficiently small (see p. 70 [10]). Fix θ1 , θ2 ∈ (0, 1/2) with θ1 < θ22 such that (6.3) holds for θ ∈ [θ1 , θ2 ]. We claim (6.2). If not, there is a sequence {θλ , U λ , Q λ , F λ } satisfying (6.1) and

⎧ θλ ∈ [θ1 , θ2 ], ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ U λ L 2 ( B 1 (0)∩Ω λf )  1, ⎪ ⎪ ⎨ lim  Q λ , F λ  Ln+δ ( B 1 (0)∩Ω λ ) = 0, f λ→0 ⎪ ⎪ ⎪  2 ⎪ ⎪ 2μ ⎪ − Πλ U λ − (Πλ U λ )0,θλ  dx > θλ . ⎪ ⎪ ⎪ ⎩

(6.4)

B θ λ ( 0)

By compactness principle [3], we can extract a subsequence (same notation for subsequence) such that

⎧ θλ → θ∗ ⎪ ⎪ ⎨ Πλ U λ → U ⎪ ⎪ ⎩ X λ ∇ U λ → K∇ U Ω f









in L 2 B 3/4 (0) strongly

as λ → 0.

(6.5)

in L 2 B 3/4 (0) weakly

Here XΩ λ is the characteristic function on Ω λf and K is the constant positive definite matrix in (2.9). f

Moreover, U satisfies LU = 0 in B 3/4 (0). If θ2 is small enough (dependent on δ , Y f ), Eqs. (6.3)–(6.5) then imply 2μ

2μ

θ∗ = lim θλ λ→0

 2  lim − Πλ U λ − (Πλ U λ )0,θλ  dx λ→0

B θ λ ( 0)

2   2   2μ 2  = − U dx −  − U dx = − U − (U )0,θ∗  dx < θ∗ . B θ ∗ ( 0) 2μ

So we get θ∗

B θ ∗ ( 0)

B θ ∗ ( 0)

2μ

< θ∗ , which is impossible. Therefore we prove (6.2). 2

Lemma 6.2. Let δ , 0 , θ1 , θ2 , μ be same as those in Lemma 6.1. For any  /θ k  0 , if U  , Q  , F  satisfy



  0 , θ ∈ [θ1 , θ2 ], and k with

−∇ · (∇ U  + Q  ) = F  in B 1 (0) ∩ Ω f ,  = 0 (∇ U  + Q  ) · n

, on B 1 (0) ∩ ∂Ωm

(6.6)

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then

 2 − Π U  − (Π U  )0,θ k  dx  θ 2kμ J 2 ,

(6.7)

B θ k ( 0)

where J  ≡ U   L 2 ( B 1 (0)∩Ω  ) + 0−1  Q  , F   Ln+δ ( B 1 (0)∩Ω  ) . f

f

Proof. We only consider J  < ∞ case and this is done by induction on k. For k = 1, we define Uˆ  ≡ UJ  , Qˆ  ≡ QJ  , Fˆ  ≡ FJ  . Then Uˆ  , Qˆ  , Fˆ  satisfy (6.6) and 









max Uˆ   L 2 ( B 1 (0)∩Ω  ) , 0−1  Qˆ  , Fˆ   Ln+δ ( B 1 (0)∩Ω  )  1. f

f

By Lemma 6.1 (in this case λ =  ),

 2 − Π Uˆ  − (Π Uˆ  )0,θ  dx  θ 2μ . B θ ( 0)

This implies (6.7) for k = 1 case. Suppose (6.7) holds for some k satisfying  /θ k  0 , we define

    ⎧ Uˆ  ≡ J −1 θ −kμ U  θ k x − (Π U  )0,θ k ⎪ ⎪ ⎨   Qˆ  ≡ J −1 θ k(1−μ) Q  θ k x ⎪ ⎪   ⎩ Fˆ  ≡ J −1 θ k(2−μ) F  θ k x

in B 1 (0) ∩ Ω f /θ k .

Then they satisfy

⎧ ⎨ −∇ · (∇ Uˆ  + Qˆ  ) = Fˆ  ⎩

in B 1 (0) ∩ Ω f /θ k ,

(6.8)

k   /θ = 0 on B 1 (0) ∩ ∂Ωm /θ k , (∇ Uˆ  + Qˆ  ) · n

k

  /θ is the unit vector normal to ∂Ωm /θ k . By induction, where n





max Uˆ   L 2 ( B 1 (0)∩Ω  /θ k ) , 0−1  Qˆ  , Fˆ   Ln+δ ( B 1 (0)∩Ω  /θ k )  1. f f Since  /θ k  0 , by Lemma 6.1 (in this case λ =  /θ k ), we obtain

 2 − Π /θ k Uˆ  − (Π /θ k Uˆ  )0,θ  dx  θ 2μ .

(6.9)

B θ ( 0)

Note, by [1],

 2 − Π /θ k Uˆ  − (Π /θ k Uˆ  )0,θ  dx = B θ ( 0)

−

|Π U  − (Π U  )0,θ k+1 |2

B θ k +1 ( 0)

Eqs. (6.9)–(6.10) imply the inequality (6.7) for k + 1 case.

2

J 2 θ 2kμ

dx.

(6.10)

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Lemma 6.3. For any δ > 0, there is a μ∗ ∈ (0, μ) such that the solutions of (6.6) satisfy, for any  ∈ (0, 1],

[U  ]C 0,μ∗ ( B

)

1/2 (0)∩Ω f

 c J,

where c is a constant independent of  . μ, J  are same as those in Lemma 6.2. Proof. Let 0 , θ1 , θ2 be same as those in Lemma 6.2, define ∗ ≡ 0 θ2 /2, and let   ∗ . Denote by c a constant independent of  . Because of θ1 < θ22 , for any r ∈ [ /0 , θ2 ], there exist θ ∈ [θ1 , θ2 ] and k ∈ N such that r = θ k . Lemma 6.2 implies that the solutions of (6.6) satisfy, for any r ∈ [ /0 , θ2 ],

 2 − Π U  − (Π U  )0,r  dx  cr 2μ J 2 .

(6.11)

B r ( 0)

 ∈ (0, μ), Since 2 /0 ∈ [ /0 , θ2 ], we define, for any μ

⎧  μ U ( x) − (Π U ) Uˆ  (x) ≡ J −1  − ⎪    0,2 /0 ⎪ ⎨ μ Q ( x) Qˆ  (x) ≡ J −1  1−  ⎪ ⎪ ⎩ μ F ( x) Fˆ  (x) ≡ J −1  2− 

in B 2/0 (0) ∩ Ω f / .

(6.12)

Then they satisfy



−∇ · (∇ Uˆ  + Qˆ  ) = Fˆ  in B 2/0 (0) ∩ Ω f / ,   / = 0 (∇ Uˆ  + Qˆ  ) · n

(6.13)

 / , on B 2/0 (0) ∩ ∂Ωm

  / is the unit vector normal to ∂Ωm / . Take r = 2 /0 in (6.11). We see where n

Uˆ  L 2 ( B 2/

0

(0)∩Ω f / )

+  Qˆ  , Fˆ  Ln+δ ( B 2/

0

(0)∩Ω f / )

 c,

. Tracing the proof of Theorem 8.24 [11], there is a where c is independent of μ

[Uˆ  ]C 0,μ∗ ( B

 / )

1/0 (0)∩Ω f

μ∗ ∈ (0, μ) such that

 c,

(6.14)

. If the μ  in (6.12) is taken to be the μ∗ in (6.14), by Theorem 1.2 on where c is independent of μ p. 70 [10], we see that (6.11) with μ replaced by μ∗ also holds for r   /0 . So (6.11) holds for r ∈ (0, θ2 ]. We then shift the origin to any point x ∈ B 1/2 (0), repeat above argument, and see that (6.11) with 0 (resp. μ) replaced by x (resp. μ∗ ) also holds for r > 0. By Theorem 1.2 on p. 70 [10], we see: For any δ > 0, there is a μ∗ ∈ (0, μ) and an ∗ ∈ (0, 1) (dependent on δ , Y f ) such that if  ∈ (0, ∗ ), then the solutions of (6.6) satisfy

[U  ]C 0,μ∗ ( B

)

1/2 (0)∩Ω f

 c J.

(6.15)

From the proof of Theorem 8.24 [11], we also see: For any δ > 0, there is a μ∗ ∈ (0, μ) such that if  ∈ [∗ , 1], then the solutions of (6.6) satisfy

[U  ]C 0,μ∗ ( B

)

1/2 (0)∩Ω f

Combining (6.15) and (6.16), we prove this lemma.

2

 c J.

(6.16)

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By shifting the coordinate, we see that Lemma 3.1 is a direct consequence of Lemma 6.3. 6.2. Proof of Lemma 3.4 For convenience, let us assume 0 ∈ Ω f . Lemma 6.4. For any δ > 0, there exist a constant θ ∈ (0, 1) (dependent on δ , Y f ) and a constant 0 ∈ (0, 1) (dependent on θ , δ ) so that if





max U λ  L ∞ ( B 1 (0)∩Ω λ ) , 0−1  Q λ , F λ  Ln+δ ( B 1 (0)∩Ω λ )  1, f

f

then the solutions of (6.1) satisfy, for any 0 < λ  0 ,









sup Πλ U λ (x) − Πλ U λ (0) − x + Πλ Xλ (x) bλ   θ 1+μ/2 ,

x∈ B θ (0)

(6.17)

−1

, bλ ≡ | BK (0)| B (0)∩Ω λ ∇ U λ dx, K−1 is the inverse matrix of the positive definite matrix K θ θ f in (2.9), and Xλ is defined in (2.8).

where

μ≡

δ

n+δ

Proof. If U is a solution of −∇ · (K∇ U ) = 0 in B 3/4 (0), then, by Theorem 6.2 [11],

U C 2+μ ( B 1/2 (0))  c U L ∞ ( B 3/4 (0)) . By Taylor’s expansion, if

μ satisfies μ < μ < 1, then, for θ small enough, 



sup U (x) − U (0) − x(∇ U )0,θ   θ 1+

x∈ B θ (0)

μ 2

U L ∞ ( B 3/4 (0)) .

(6.18)

Fix a value θ such that (6.18) holds. We claim (6.17). If not, there is a sequence {U λ , Q λ , F λ } satisfying (6.1) and

⎧ U λ L ∞ ( B 1 (0)∩Ω λ )  1, ⎪ f ⎪ ⎪ ⎪ ⎨ lim  Q λ , F λ  Ln+δ ( B 1 (0)∩Ω λ ) = 0, f λ→0 ⎪     ⎪ ⎪ 1+μ/2  ⎪ . ⎩ sup Πλ U λ (x) − Πλ U λ (0) − x + Πλ Xλ (x) bλ  > θ

(6.19)

x∈ B θ (0)

After extraction of a subsequence (same notation for subsequence), we have, by [1] and Lemma 3.1,

⎧ ⎨ Πλ U λ → U

in L ∞ B 3/4 (0) strongly

⎩ XΩ λ ∇ U λ → K ∇ U

in L 2 B 3/4 (0) weakly

f









as λ → 0.

(6.20)

Here XΩ λ is the characteristic function on Ω λf . Eqs. (6.19) and (6.20) imply that U satisfies −∇ · f

(K∇ U ) = 0 in B 3/4 (0). By (6.18), (6.19)3 , and (6.20), we get

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    θ 1+μ/2  lim sup Πλ U λ (x) − Πλ U λ (0) − x + Πλ Xλ (x) bλ  λ→0 B θ (0)

  μ = sup U (x) − U (0) − x(∇ U )0,θ   θ 1+ 2 U L ∞ ( B 3/4 (0)) . B θ ( 0)

If θ is small enough, the right-hand side of above inequality is less than θ 1+ Then we get contradiction. So (6.17) holds. 2

μ

2

for some

μ

∈ (μ, μ ).

Lemma 6.5. Let δ , θ , 0 , μ be same as those in Lemma 6.4 and let U  , Q  , F  satisfy (6.6). For any   0 and k with  /θ k  0 , there are ak , bk such that

⎧         ⎪ ⎨ ak + bk  c J ,     μ Π U  (x) − Π U  (0) −  a − x + Π X (x) b   θ k(1+ 2 ) J , ⎪ k k ⎩ x∈sup B ( 0)

(6.21)

θk

where J ≡ U   L ∞ ( B 1 (0)∩Ω  ) + 0−1  −1+μ/2 Q  , F   Ln+δ ( B 1 (0)∩Ω  ) and c is a constant independent of  . f

f

−1

Proof. This is done by induction on k. Take a1 = 0, b1 = | BK (0)| θ Q ˆ F Qˆ  ≡ J , F ≡ J . Then Uˆ  , Qˆ  , Fˆ  satisfy (6.1) and  

B θ (0)∩Ω f



∇ U  dx and define Uˆ  ≡

U

 J ,



max Uˆ   L ∞ ( B 1 (0)∩Ω  ) , 0−1  Qˆ  , Fˆ   Ln+δ ( B 1 (0)∩Ω  )  1. f f By Lemma 6.4 (λ =  in this case), we obtain (6.21) for k = 1. If (6.21) holds for some k satisfying  /θ k  0 , then we define, in B 1 (0) ∩ Ω f /θ k ,

      ⎧ Uˆ  (x) ≡ θ −k(1+μ/2) J−1 U  θ k x − Π U  (0) −  ak − θ k x + X (θ k x) bk , ⎪ ⎪ ⎨   Qˆ  (x) ≡ θ −kμ/2 J−1 Q  θ k x , ⎪ ⎪   ⎩ Fˆ  (x) ≡ θ k(1−μ/2) J−1 F  θ k x . Then Uˆ  , Qˆ  , Fˆ  satisfy (6.1) in B 1 (0) ∩ Ω f /θ k (or see (6.8)) by (2.7). By induction,





max Uˆ   L ∞ ( B 1 (0)∩Ω  /θ k ) , 0−1  Qˆ  , Fˆ   Ln+δ ( B 1 (0)∩Ω  /θ k )  1. f f

(6.22)

By Lemma 6.4 (λ =  /θ k  0 in this case), we have









μ

sup Π /θ k Uˆ  (x) − Π /θ k Uˆ  (0) − x + Π /θ k X /θ k (x) b /θ k   θ 1+ 2 ,

(6.23)

B θ ( 0)

−1

where b /θ k ≡ | BK (0)| θ



B θ (0)∩Ω f /θ k





∇ Uˆ  dx. Rewrite (6.23) in terms of U  to obtain, by [1],







sup Π U  θ k x − Π U  (0) +  Π1 X(0)bk − θ k x + Π X θ k x bk

B θ ( 0)

    − J θ k(1+μ/2) x + θ −k Π X θ k x b /θ k   J θ (k+1)(1+μ/2) .

(6.24)

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Define

ak+1 ≡ −Π1 X(0)bk

bk+1 ≡ bk + J θ kμ/2 b /θ k .

and

By (6.22) and (6.23), |b /θ k | is bounded uniformly in

(6.25)

 , k. So we get (6.21)1 . Substituting (6.25) into

(6.24) and making the change of variables θ k x to x, we obtain (6.21)2 .

2

Lemma 6.6. Let 0 , μ, J be same as those in Lemma 6.5. For any   0 , the solutions of (6.6) satisfy

    ∇ U  L ∞ ( B 1/2 (0)∩Ω f )  c J∗, ≡ c J +  −μ/2  Q  ( x)C 0,α (Ω  / ) , f

where α > 0 and the constant c is independent of  . Proof. Let c be a constant independent of as that in Lemma 6.5. By Lemma 6.5,

 . Let k ∈ N satisfy  /θ k  0 <  /θ k+1 , where θ is same









 1+ μ  2  J∗, .  

sup Π U  (x) − Π U  (0) −  ak − x + Π X (x) bk   c 

B  /0 (0)

(6.26)

0

Define, in B 1/0 (0) ∩ Ω f / ,

    ⎧ −1  Uˆ  (x) ≡  −(1+μ/2) J∗, ⎪  U  ( x) − Π U  (0) −  ak −  x + X ( x) bk , ⎪ ⎨ −1 Qˆ  (x) ≡  −μ/2 J∗,  Q  ( x), ⎪ ⎪ ⎩ −1 Fˆ  (x) ≡  1−μ/2 J∗,  F  ( x). Then Uˆ  , Qˆ  , Fˆ  satisfy (6.1) in B 1/0 (0) ∩ Ω f / (or see (6.13)) by (2.7). By (6.26),

Uˆ  L ∞ ( B 1/

0

(0)∩Ω f / )

+  Fˆ  Ln+δ ( B 1/

0

(0)∩Ω f / )

+  Qˆ  C 0,α ( B

 / )

1/0 (0)∩Ω f

 c.

By Theorem 6.30 and Theorem 9.19 [11],

Uˆ  W 1,∞ ( B 1/2

0

Since ∇ Uˆ  (0) =

(0)∩Ω f / )

 c.

∇ U  (0)−( I +∇X(0))bk ,  μ/2 J∗,

     ∇ U  (0) =  I + ∇X(0) b +  μ/2 J∗, ∇ Uˆ  (0)  c J∗, . k

(6.27)

Since one can shift the origin to any point in Ω f and gets the same result as (6.27), we conclude ∇ U   L ∞ ( B 1/2 (0)∩Ω  )  c J∗, . 2 f

By Lemma 6.6, we know that for any δ > 0, there is a constant 0 ∈ (0, 1) such that, for  ∈ (0, 0 ), any solution of (3.2) satisfies (3.28). By (3.8)2 , we also know that for any δ > 0 and  ∈ [0 , 1], any solution of (3.2) satisfies (3.28). So we prove Lemma 3.4.

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