Estimates for the Stokes Operator in Lipschitz Domains

Transcription

1 age 1183) Estimates for the Stokes Operator in Lipschitz omains Russell M. Brown & Zhongwei Shen Abstract. We study the Stokes operator A in a threedimensional Lipschitz domain. Our main result asserts that the domain of A is contained in W 1,p 0 () W 3/, () for some p>3. Certain L -estimates are also established. Our results may be used to improve the regularity of strong solutions of Navier-Stokes equations in nonsmooth domains. In the appendix we provide a simple proof of area integral estimates for solutions of Stokes equations. Introduction. In a recent interesting paper, euring and von Wahl [W] consider strong solutions of the nonstationary Navier-Stokes equations in (0,T): u = u (u )u π + f, div u =0, with the initial-irichlet condition { u(x,t) =0for(X,t) (0,T), u(x,0) = u 0 (X), X, where is a bounded Lipschitz domain in R 3. Based on the functional analytical approach of Fujita and Kato [FK] and the Rellich estimates of Shen [S1], they show that, if u 0 (A 1/4+ε )forsomeε (0, 1 )andf is bounded and locally Hölder continuous, then a solution (u,π) exists for some T>0and u C ( (0,T],(A) ) C ( [0,T],(A 1/4+ε ) ), where A = P denotes the Stokes operator. The purpose of this note is to describe (A), the domain of A, intermsof Sobolev s spaces. In the case of smooth domains, it is well known that (A) =W, () W 1, 0 () L σ() 1183 Indiana University Mathematics Journal c, Vol. 44, No. 4 (1995)

2 1184 R. M. Brown & Z. Shen where L σ() denotes the space of solenoidal functions in L () (e.g., see [CF]). One can not expect such results in Lipschitz domains, as the W, -estimate, in general, fails in nonsmooth domains. Our main results in this paper assert that (0.1) (A) W 1,p 0 () W 3/, () for some p = p() > 3 (Theorem.17). In particular, it follows from Sobolev s imbedding that for every t (0,T], u(t) C α ( ) for some α = α() > 0, i.e., the strong solution of the Navier-Stokes equations is Hölder continuous up to the boundary as a function of X. We also obtain the following L estimates: (0.) u L () C u 1/ u L () C u 1/4 L () L () Au 1/ Au 3/4 L (), L () for u (A). To establish (0.), we use the reverse Hölder inequality, (0.1) and some localization techniques. See Theorem 3.1 and Corollary 3.. Estimates like (0.) are very useful in the study of Navier-Stokes equations. See [CF] and [H] in the case of smooth domains. To prove (0.1), we shall study the irichlet problem for the Stokes equations with a forcing term, and interpolate between the L estimates in [FKV] and the Hölder estimates in [S]. The following area integral estimate, u(x) dist(x, ) dx C u dσ, for solutions of Stokes equations u = π, divu =0in,playsanimportant role. This estimate is due to E. Fabes, C. Kenig and G. Verchota, but no proof has appeared in the literature. In the appendix of this paper, which is due to Z. Shen, we will provide a simple proof of the area integral estimates for solutions of Stokes equations. The proof given here is based on the idea of a recent paper by B. ahlberg, C. Kenig, J. Pipher, and G. Verchota [KPV] for higher order equations and systems, together with some observations on the pressure term π. Finally, the second author would like to thank C. Kenig for many helpful conversations. 1. Notation and definitions. In this section we collect the definitions for Lipschitz domains, the nontangential maximal function, the Sobolevand Besov spaces we will use, and the Stokes operator. We shall also recall a few elementary facts regarding complex interpolation.

3 Estimates for the Stokes Operator in Lipschitz omains 1185 Lipschitz domains. Let be a bounded, open, connected set in R n.we say that is a Lipschitz domain if for each P, there exists a coordinate system (x,x n ), which is isometric to the usual coordinates, and a Lipschitz function ψ : R n 1 R, a radius r>0sothat B(P,r) ={(x,x n ) R n : x n >ψ(x ) B(P,r), B(P,r) ={(x,x n ) R n : x n = ψ(x ) B(P,r). The nontangential maximal function. For a function u on, we define its nontangential maximal function (u) by (1.1) (u) (P ) = sup{ u(x) : X, X P dist(x, ), P. We now give the definition of the function spaces we will use. Sobolev and Besov spaces. For R n, p [1, ) andk =1,,..., we let W k,p () denote the space of functions u on such that the norm u W k,p () α u Lp () α k is finite. For 1 p< and 0 <α<1, we define W α,p () to be the collection of functions u on with the norm ( u(x) u(y ) p ) 1/p (1.) u W α,p () u Lp () + X Y n+αp dx dy <. For 1 <α<, we define W α,p () to be the set of functions on such that u W α,p () u L p () + u W α 1,p () <. We may define W α,p ( ) for 0 <α<1and1 p< in a similar manner with the integral over. For a region above the graph of a Lipschitz function ϕ, i.e., = {(x,x n ) R n : x n > ϕ(x ), we define the space W 1, ( ) as the space of functions f ( x,ϕ(x ) ) = g(x )whereg W 1, (R n 1 ). Using a partition of unity for, we may extend this definition to the boundary of a bounded Lipschitz domain forw 1, ( ). We remark that W α,p ( ) may also be defined in this manner. See [Gr, p. 0]. We will use W k,p 0 () to denote the closure of C0 () in the norm of W k,p (). If k is a nonnegative integer, we use W k,p () to denote the dual of W k,p 0 (). We will use the same notation L p (), W α,p (), W k,p (), etc., for vectorial counterparts.

4 1186 R. M. Brown & Z. Shen Complex interpolation. We will need the following results on complex interpolation: (1.3) [ L ( ),W 1, ( ) ] ϑ = W ϑ, ( ) and (1.4) [ W α, ( ),C γ ( ) ] ϑ = W t,p ( ) where 1 p = ϑ, and t = αϑ + γ(1 ϑ). When is replaced by R n, (1.3) and (1.4) are well known (e.g., see [BL]). To extend this result to boundaries of Lipschitz domains, we use the following easy proposition, whose proof is omitted. Proposition 1.5. Suppose A i, B i (i =0,1) are Banach spaces and {C(ϑ) : ϑ [0,1] is a family of Banach spaces such that C(0) = A 0, C(1) = A 1 and C(ϑ) A 0 + A 1. Also assume that there exist linear operators I : A 0 + A 1 B 0 + B 1 and P : B 0 + B 1 A 0 + A 1 such that I : C(ϑ) [B 0,B 1 ] ϑ, P : [B 0,B 1 ] ϑ C(ϑ) are bounded for each ϑ [0,1] and P I is the identity on A 0 + A 1.Then [A 0,A 1 ] ϑ = C(ϑ). To see (1.3), let {B j = B(P j,r) : j = 1,,...,N be a covering of by balls as in the definition of Lipschitz domain, and η j C0 (R n ) such that supp η j B j,0 η j 1, j η j =1on. Let ϕ j be an isometry of R n such that ϕ j (B j ) = {(x,x n ):x n >ψ j (x ), (x,x n ) r. We now choose A 0 = L ( ), A 1 = W 1, ( ), B 0 = L (R (n 1),R N ), B 1 = W 1, (R (n 1),R N ), and C(ϑ) =W ϑ, ( ). If g is a function of, we define Ig by letting the j th component be given by Ig(x ) j =(η j g) ( ϕ 1 ( j x,ψ j (x ) )). ( We let η j C0 B(Pj,r) ), j =1,...,N, be functions satisfying η j =1on B(P j,r), we let π n (x,x n )=x be projection on the first n 1 coordinates and define P ( ) N ( (f j ) 1 j N (Q) = η j (Q)f j πn ϕ j (Q) ) for Q. j=1 With these definitions, (1.3) then follows easily from Proposition 1.5. The statement (1.4) may be proved in the same manner.

5 Estimates for the Stokes Operator in Lipschitz omains 1187 The Stokes operator. To introduce the Stokes operator A, let C 0,σ() = {ϕ C 0 () : div ϕ =0, and L σ() be the closure of C 0,σ() in L (). We first define a quadratic form Q on C 0,σ(): Q(u,v) = u vdx = n j=1 u j v j dx. We then extend this form to (Q), the closure of C 0,σ() in the norm u Q = u L () + {Q(u,u) 1/. It is known that ([CF]) (Q) ={u W 1, 0 () : div u =0. We now define the Stokes operator A : (A) L σ() by Au ϕ = Q(u,ϕ) for all ϕ C0,σ(), where the domain of A, (A), is the collection of u in (Q) such that v Q(u,v) can be extended to a bounded linear functional on L σ(). It is well known that A gives a self-adjoint operator on L σ(). Also, it is not hard to see that and (A) ={u (Q) : u + π L σ() for some π L () Au = u + π, for u (A). This is the definition of the Stokes operator given in [W]. Finally, we will follow the standard practice of letting C denote a constant which varies. Throughout this paper C will depend at most on the dimension n and the Lipschitz domain through the collection of balls used to cover the boundary and the maximum of ψ i where ψ i are the functions whose graphs define.

6 1188 R. M. Brown & Z. Shen. The Imbedding of (A). In this section we will establish the imbedding estimate (0.1) for the domain of A. We will need to consider both the irichlet problem with nonzero forcing term f: u + π = f in, (P) div u =0 in, u =0 on, and the problem with inhomogeneous boundary data: u + π =0 in, (BVP) div u =0 in, u = g on. Obviously, since div u = 0 in, the boundary data g should verify the necessary condition (.1) g,n dσ =0, where N denotes the outward unit normal to and, the scalar product on R n. The following result is due to Fabes, Kenig and Verchota [FKV]. The estimates for the nontangential maximal functions, the existence and the uniqueness may be found in their paper. The estimates for the solution in Sobolevspaces were announced in that paper, but no proof has appeared. The proof may be obtained by combining the area integral estimates in the appendix of this paper, with the argument given by E. Fabes in [F] to establish the corresponding Sobolevestimates for harmonic functions. We remark that in the first inequality on the top of page 69 of [F], the integration on R n 1 R n 1 should only range over {(x,y) : x y >t. Theorem. (Fabes, Kenig and Verchota [FKV]). Let be a bounded Lipschitz domain in R n, n 3. Suppose g L ( ) and verifies the condition (.1). Then there exists a solution u to (BVP) which satisfies (u) dσ C g dσ. This solution is unique in the class of u satisfying (u) L ( ). In addition, we have the estimate u W 1/, () C g dσ. Furthermore, if g W 1, ( ), then we also have ( u) L ( ) + u W 3/, () C g W 1, ( ). We shall also need the Hölder estimates established in [S, Theorem 0., p. 801] for the three-dimensional Lipschitz domains.

7 Estimates for the Stokes Operator in Lipschitz omains 1189 Theorem.3 (Z. Shen [S]). Let be a bounded Lipschitz domain in R 3. There exists α 0 > 0 so that if 0 <α<α 0,andg C α ( ) verifies the condition (.1), then the solution to (P) lies in C α ( ) and satisfies where δ(x) = dist(x, ). sup δ(x) 1 α u(x) C g C α ( ), X Our estimates for the Stokes operator A will be obtained by interpolating between the estimates of Theorem. and.3. We begin with the following lemma. Lemma.4. Let be a bounded Lipschitz domain in R 3. There exists ε>0 so that if p 3+ε, andg W 1 1/p,p ( ) satisfies the condition (.1), then the solution u of (BVP) for Stokes equations with boundary data g satisfies ( 1/p u(x) dx) p C g W ( ). 1 1/p,p Proof. Let η be a smooth vector field on R 3 such that η,n c 0 > 0a.e. on. Recall that N is the outward unit normal to. For g W α,p ( ), we define Sg by Sg(P )=g(p) g,n dσ η(p ) for P. η,n dσ It is easy to see that Sg,N dσ =0. Now let u be the solution of (BVP) with boundary data Sg. We observ e that Theorem., the area integral estimate in Appendix A (Theorem A.1) and interior estimates imply that ( 1/ δ(x) 3 u(x) dx) C Sg L ( ) C g L ( ), ( 1/ δ(x) u(x) dx) C Sg W 1, ( ) C g W 1, ( ).

8 1190 R. M. Brown & Z. Shen It then follows from the complex interpolation that ( ( δ(x) 3/ α u(x) ) ) 1/ (.5) dx C g W α, ( ), 0 <α<1. wherewehaveused(1.3). Now, since R 3, we may apply Theorem.3 and the interior estimates to obtain (.6) sup {δ(x) γ u(x) C Sg C γ ( ) C g C γ ( ) X for 0 <γ<α 0. If we interpolate between (.5) and (.6) and use (1.4), we obtain that ( { (.7) δ(x) u(x) ) 1/p p dx C g W 1 1/p,p ( ) for <p<3+ε. To see this, set ϑ =/p < 1, and choose α and γ so that ( ) 3 1= α ϑ +( γ)(1 ϑ). Thus, t = αϑ + γ(1 ϑ) =1 1 p. We use (1.4) to identify the interpolation space [W α, ( ),C γ ( )] ϑ = W t,p ( ). Note that /p < 1 implies that p> and the restrictions that α<1and γ<α 0 imply that ϑ = 1 γ α + 1 > (1 α 0) γ 3 α 0 or p = ϑ < 3 α 0 3+ε, ε = α 0 > 0. 1 α 0 1 α 0 Finally, by the Hardy inequality [St, p. 7], u(x) p dx C δ(x) u(x) p dx +sup u(x) p X K C g p W 1 1/p,p ( ), where K is a compact subset of. The proof is complete.

9 Estimates for the Stokes Operator in Lipschitz omains 1191 Let Γ(X) = ( Γ ij (X) ) be the matrix of fundamental solutions and 1 i,j n q(x) = ( q j (X) ) be the corresponding pressure vector where 1 j n (.8) Γ ij (X) = 1 ( 1 δ ij ω n (n ) X n + x ) ix j X n, q j (X) = 1 x j ω n X n. Theorem.9. Let be a bounded Lipschitz domain in R 3.Supposef W 1,p () where (3 + ε)/( + ε) <p<3+ε and ε is the same as in Lemma.4. Then there exist u W 1,p 0 () and π L p () so that { u + π = f, (.10) div u =0 in and u L p () +inf π c L c R p () C f W 1,p (). The solution u is unique and π is unique up to a constant. Proof. Let <p<3+ε and f W 1,p (). We may extend f to lie in W 1,p (R 3 ). Let v =Γ f W 1,p (R 3 ). By the trace theorem [Gr, p. 33], v W 1 1/p,p ( ) and we obtain (.11) v W 1 1/p,p ( ) C v W 1,p () C f W 1,p (). Now let u = v w where w is the solution of (BVP) for the Stokes equations with boundary data v.thenusatisfies (.10) and u L p () v L p () + w L p () C { f W 1,p () + v W 1 1/p,p ( ) C f W 1,p () by Lemma.4 and (.11). We may obtain the existence of a solution and the estimates of u for (3 + ε)/( + ε) <p< by duality. For the pressure term π, wehave inf π c L c R p () π W 1,p () u W 1,p () + f W 1,p () C u Lp () + f W 1,p () C f W 1,p (). Finally, the uniqueness for p follows by energy estimates. If p<, let u + π = 0 and u W 1,p 0 (). Choose f L p (). We may solve v + q = f, divv =0in,v W 1,p (), and thus obtain that f,u dx = v udx =0, for any f L p (), or that u =0.

10 119 R. M. Brown & Z. Shen Our next result, which is valid for Lipschitz domains in R n (n 3), gives a sharper estimate for solutions of the irichlet problem. It will yield the best embedding of (A) in the scale of Sobolevspaces W s, (). Theorem.1. Let be a bounded Lipschitz domain in R n (n 3) and f L (). Suppose that (u,π) is the solution of (P) with data f, then for q 0 =n/(n +1), we have the estimate u W 3/, () +inf π c W c R 1/, () C f L q 0 (). Proof. We construct the solution of (P) as the sum of the free space solution (v, π) =(Γ f,q f) and the solution of the boundary value problem w + ψ =0 in, div w =0 in, w = v on. By the Calderón-Zygmund estimates [St], we have (.13) v L p (R n ) + π L p (R n ) C f L p (R n ) for 1 < p <. This estimate with p = q 0 = n/(n + 1) and the Sobolev embedding theorem imply that v W 3/, () + π W 1/, () C f L q 0 (). Now consider (w,ψ). We claim that the boundary values of w, Γ f, satisfy (.14) w W 1, ( ) C f L q 0 (). Then the desired estimate, w W 3/, () + ψ W 1/, () C f L q 0 (), will follow from Theorem.. To establish (.14), we observe that (.15) v ( dσ C v + v v ) dx. This follows by applying the divergence theorem to η v where η is a smooth vector field on R n with the property η,n c 0 > 0a.e.on. By (.13), we have v L q0 (R n ), while v(x) C f(y ) dy X Y n 1 satisfies (.16) v q C f L 0 (R n ) L q 0 (R n ), q 0 = q 0 q 0 1. Since q 0 >, the estimate for w on follows from (.15), (.16) and (.13). The estimate for w is easier and we omit the details.

11 Estimates for the Stokes Operator in Lipschitz omains 1193 Theorem.17. Suppose is a bounded Lipschitz domain in R 3.Then (A) W 1,p 0 () W 3/, () for some p = p() > 3. Moreover, for u (A), Proof. Let u (A). Then where f = Au L σ(). Since u W 1,p 0 () + u W 3/, () C Au L (). u + π = f in, div u =0 in, u W 1, 0 (), L () W 1,p () for p 6 in R 3, the theorem follows immediately from Theorems.9 and Some L estimates. In this section we will give the proof of the estimate (0.). We are only able to prove this estimate in three dimensions and thus throughout this section we will assume that is a bounded Lipschitz domain in R 3. Theorem 3.1. There exists a constant C>0 such that, for u (A), u L () C u 1/ L () Au 1/ L (). By the definition of A and a limiting argument, Au udx = u udx. It then follows from the Cauchy inequality that u L () Au L () u L (). Thus, we have:

12 1194 R. M. Brown & Z. Shen Corollary 3.. There exists a constant C>0 such that, for u (A), u L () C u 1/4 L () Au 3/4 L (). As in the case of smooth domains, A 1 : L σ() L σ() is a compact operator. Hence, there exists a sequence of positive numbers {λ k and an orthonormal basis {ω k of L σ() such that Aω k = λ k ω k, ω k (A) and lim k λ k =. Corollary 3.3. exists c>0 such that Let be a bounded Lipschitz domain in R 3. Then there ( ) /3 k λ k c. With Corollary 3., the proof is exactly the same as in the smooth case. See [CF, p ]. To prove Theorem 3.1, we start with a reverse Hölder inequality. Lemma 3.4. Let X 0, R>0 besmall,and(x 0,R)=B(X 0,R). Assume (u,π) satisfies the Stokes equations { u + π =0, div u =0, in (X 0,8R) and u =0on. Then ( 1 R 3 (X 0,R) where p 0 > depends only on. 1/p0 ( 1 u dx) p0 C R 3 (X 0,4R) u dx) 1/, Proof. We begin with a Caccioppoli type inequality for the Stokes equations (see [S, Lemma 1.5, p. 804]). We consider two cases. If B(X 0,R) Ø, we have u dx C R u dx. (X 0,R) On the other hand, if B(X 0,R), we get u dx C R (X 0,R) (X 0,3R) (X 0,3R) u u R dx where u R = 1 u(x) dx. B(X 0,R) B(X 0,R)

13 Estimates for the Stokes Operator in Lipschitz omains 1195 In both cases, by Sobolev-Poincaré inequality, we obtain ( ) 1/ 1 u dx (X 0,R) ( 1 C (X 0,3R) (X 0,R) (X 0,3R) u 6/5 dx) 5/6. The lemma then follows from the usual reverse Hölder inequality (e.g., see [Gi, Proposition 1.1, p. 1]). We now give the: Proof of Theorem 3.1. Let u (A) andf = Au L σ(). Fix X 0. Let t = B(X 0,t) wheret (0,1) is to be chosen later. Let f 1 = f in t and 0 otherwise. Let f = f f 1. We use (u i,π i ), i =1,, to denote solutions of (P) for the Stokes equations with data f i. By uniqueness, u = u 1 + u. First we estimate u 1 (X 0 ). We claim that u 1 (X 0 ) Ct 1/ f L (). Indeed, for any X t, by the imbedding theorem of Morrey [GT, Theorem 7.17, p. 163], u 1 (X) u 1 (X 0 ) Ct γ u 1 Lp (), where p>3andγ =1 3/p. It then follows from Theorem.9 that u 1 (X) u 1 (X 0 ) Ct γ f 1 W 1,p () Ct γ f 1 L q () t 1/ f L (), where 1/q =1/p +1/3 and we also used the Sobolevimbedding theorem [GT, Theorem 7.6, p. 171] and Hölder inequality (we also assume that p<6). Now we integrate the inequality above in X over t. This, together with the Sobolevimbedding, gives ( ) 1/6 1 u 1 (X 0 ) u 1 6 dx + Ct 1/ f L() t t Ct 1/ u 1 L () + Ct 1/ f L () Ct 1/ f 1 W 1, () + Ct 1/ f L () Ct 1/ f 1 L 6/5 () + Ct 1/ f L () Ct 1/ f L ().

14 1196 R. M. Brown & Z. Shen To estimate u (X 0 ), let s [t/8,t/4] and X s.notethat u + π = 0 and div u =0in s. We may apply Theorem.3 on s to obtain u (X) u (X 0 ) Ct α u (P 1 ) u (P ) sup P 1,P s P 1 P α P 1 P Ct α ( s u p0 dσ) 1/p0, where p 0 =/(1 α) > and we have used the Sobolev imbedding on the set s.byintegrationinsover [t/8,t/4], we get u (X) u (X 0 ) Ct α 1/p0 ( t/4 u p0 dx ( 1 Ct u p0 dx t/4 t/4 ( 1 Ct t u dx t { ( 1 Ct u dx t t ) 1/ ) 1/p0 ) 1/p0 Ct 1/ u L () + Ct 1/ u 1 L () Ct 1/ u L () + Ct 1/ f L (), ) 1/ ( ) 1/ 1 + u 1 dx t t whereweusedthereversehölder inequality (Lemma 3.4) in the third inequality. Thus, ( ) 1/6 1 u (X 0 ) u 6 dx + Ct 1/ u L t () + Ct 1/ f L () t Ct 1/ u L6 () + Ct 1/ u L () + Ct 1/ f L () Ct 1/ u L () + Ct 1/ u L () + Ct 1/ f L () Ct 1/ u L () + Ct 1/ u 1 L () + Ct 1/ f L () Ct 1/ u L () + Ct 1/ f L (). To summarize, we have proved that u(x 0 ) Ct 1/ u L () + Ct 1/ f L ()

15 Estimates for the Stokes Operator in Lipschitz omains 1197 for any t (0,c 0 ), where c 0 depends only on. Finally, by the energy estimate, u L () C 0 f L (), sowemaychoose t = c 0 u L () C 0 f L () <c 0 to obtain u(x 0 ) C u 1/ L () Au 1/ L (). The proof is finished. Appendix A. In this appendix, we present a simple proof of the area integral estimates for solutions of Stokes equations. Theorem A.1. Let be a bounded Lipschitz domain in R n, n 3. Suppose u = π, div u =0in, and(u) L ( ). Then u(x) δ(x) dx C u dσ and { u dσ C u(x) δ(x) dx + u(x) dx, where δ(x) = dist(x, ). Theorem A.1 is due to E. Fabes, C. Kenig and G. Verchota (unpublished). The proof given here is based on an idea of B. ahlberg, C. Kenig, J. Pipher and G. Verchota [KPV] developed for elliptic systems, and some observations on the pressure term π. The following lemma is due to C. Kenig and E. Stein. Lemma A.. Suppose ψ : R n 1 R is a Lipschitz function. Let ϑ C0 (R n 1 ) be radial, and ϑ 0, ϑdx=1.then,ifλ C(n, ψ L ), (x,t) (x,y) = ( x,ϕ(x,t) ) is a bi-lipschitz map from to = {(x,y) :y>ψ(x) where x R n 1,t,y R and ϕ(x,t) =λt +(ϑ t ψ)(x), ϑ t (x) = 1 ( x ) t n 1 ϑ. t Moreover, ϕ(x,t) tdxdt is a Carleson measure on.

16 1198 R. M. Brown & Z. Shen Lemma A.3. Let = {(x,y) R n : y>ψ(x) where ψ : R n 1 R is a Lipschitz function. Suppose that { u + π = f, div u = g, in. Also assume that u, π have compact supports and (u) L ( ). Then { u δ(x) dx C (u) dσ + π δ(x) dx + π δ 3 (X) dx and + g +divf u δ (X) dx + f u δ(x) dx { u dσ ε (u) dσ + C ε u δ(x) dx + π δ(x) dx for any ε>0. + π δ 3 (X) dx + Proof. Let ρ : be defined by g +divf u δ (X) dx + ρ(x,t) =(x,y) = ( x,ϕ(x,t) ), f u δ(x) dx where ϕ(x,t) =λt + ϑ t ψ(x) is given in Lemma A.. Let v = u ρ, q = π ρ be defined on.notethatv and q have compact supports in and satisfy (A.4) v = u y ρ ϕ, v = u ρ + u y ρ ϕ, i =1,,...,n 1. It follows from integration by parts that (A.5) u ρ dx = v dx = R n 1 R n 1 = t v R n + = t ϕ u R n y ρ + dx dt + t v R n vdxdt + dx dt + t ϕ v vdxdt u ρ u ρdxdt+ G, y

17 Estimates for the Stokes Operator in Lipschitz omains 1199 where G denotes terms which are bounded in absolute value by (A.6) C t ϕ u ρ u ρ dx dt + C t ϕ π ρ u ρ dx dt + C + C + C t ϕ π ρ u ρ dx dt t π ρ u ρ dx dt t π ρ u ρ dx dt. By the Stokes equations u = π f, wehave u y ρ = u x i = = ρ + π ρ f ρ { u ρ + u ρ ϕ + π ρ f ρ y { u ρ + { u y ρ ϕ u y ρ xϕ + π ρ f ρ, where the repeated index i is summed over {1,,...,n 1 and x = ( ),..., x n 1. x 1 It follows that { u y ρ = 1 { { u ρ + u 1+ xϕ y ρ ϕ + π ρ f ρ.

18 100 R. M. Brown & Z. Shen Thus, we may use integration by parts to obtain t ϕ u y ρ u ρdxdt = ( ) ϕ t 1+ x ϕ u ρ {u ρ dx dt ( ) ϕ ϕ t R n 1+ + x ϕ u y ρ {u ρ dx dt = + ( ) ϕ t { π ρ f ρ u ρdxdt+ G 1+ x ϕ ( ) ϕ t 1+ x ϕ u ρ dx dt ( ) ϕ t x ϕ R n 1+ + x ϕ u y ρ dx dt ( ) ϕ t + { π ρ f ρ u ρdxdt+ G. R n 1+ + x ϕ This, together with (A.5), implies that ( ) ϕ t (A.7) u ρ dx = R 1+ n 1 x ϕ u ρ dx dt ( ) ϕ t + { π ρ f ρ u ρdxdt+ G. R n 1+ + x ϕ It remains to estimate ( ) ϕ t π ρ u ρdxdt. 1+ x ϕ

19 Estimates for the Stokes Operator in Lipschitz omains 101 To this end, we let u =(u 1,u,...,u n ) and note that π ρ u ρ = π ρ u i ρ + π y ρ u n ρ = {π ρ u i ρ + π y ρ Thus, the integration by parts yields { u n ρ u i ρ ϕ. (A.8) ( ) ϕ t π ρ u ρdxdt 1+ x ϕ = ( ) ϕ t π 1+ x ϕ { y ρ u n ρ u i ρ ϕ dx dt + G. Now, we write t = /. Using integration by parts, we obtain = ( ) ϕ t 1+ x ϕ π ) 3 ( ϕ t 1+ x ϕ π { y ρ { y ρ u n ρ u i ρ ϕ u n ρ u i ρ ϕ dx dt dx dt + G. Since u = π f, wehave π = (divu)+divf = g +divf. Thus, π y ρ = π It follows that x i = ρ + g ρ +(divf) ρ { π ρ + { π y ρ ϕ π y ρ xϕ + g ρ +(divf) ρ. { π y ρ = 1 { { π ρ + π 1+ xϕ y ρ ϕ + g ρ +(divf) ρ.

20 10 R. M. Brown & Z. Shen Hence, the integration by parts again yields = ( ) ϕ t 1+ x ϕ π ) 3 ( ϕ t (1 + x ϕ ) { y ρ u n ρ u i ρ ϕ { { g +divf ρ dx dt u n ρ u i ρ ϕ dx dt + G. Thus, putting together (A.7), (A.8) and the estimate above, we have proved that u ρ dx R n 1 ( ) ϕ t = R n 1+ + x ϕ u ρ dx dt ) 3 ( ϕ t (1 + x ϕ ) { { g +divf ρ u n ρ u i ρ ϕ dx dt This implies that ( ) ϕ t f ρ u ρdxdt+ G. 1+ x ϕ u dσ C u δ(x) dx + C + C f u δ(x) dx + G g +divf u δ (X) dx and u δ(x) dx C u dσ + C g +divf u δ (X) dx + C f u δ(x) dx + G.

21 Estimates for the Stokes Operator in Lipschitz omains 103 Since ϕ tdxdt is a Carleson measure, by the Cauchy-Schwartz inequality and (A.6), we have ( ) 1/ ( ) 1/ G C u δ(x) dx (u) dσ ( ) 1/ ( + C ( + C π δ(x) dx π δ 3 (X) dx ( + C π δ(x) dx ( + C π δ 3 (X) dx (u) dσ ) 1/ ( ) 1/ ( ) 1/ ( ) 1/ (u) dσ u δ(x) dx The lemma now follows easily from the Cauchy inequality. ) 1/ ) 1/ 1/ u δ(x) dx). Lemma A.9. Let be a bounded Lipschitz domain in R n, n 3. Suppose that u = π, div u =0in, and(u) L ( ). Then there exists a function π such that π = π + c and π(x) δ(x) dx + π(x) δ 3 (X) dx C u dσ. Proof. By the L estimates in [FKV], u can be represented in terms of a double layer potential: { u i (X) = {Γ ij (X Y )N k (Y ) q i (X Y )N j (Y ) h j (Y ) dσ(y ), y k where h L ( ) C u L ( ), ( Γ ij (X) ) is the matrix of fundamental solutions and ( q i (X) ) is the corresponding pressure vector given in (.8). Note that Γ ij (X) = 1 { xi x j ω n X n = 1 ( ) xi ω n X n x j = 1 { xj ω n X n. Thus, u i (X) = = {Γ ij (X Y )N k (Y )h j (Y ) dσ(y ) y k x k x j y j ω n X Y n N k(y )h j (Y ) dσ(y )= π,

22 104 R. M. Brown & Z. Shen where π(x) = w k(x) x j y j and w k (X) = x k po ω n X Y n N k(y )h j (Y ) dσ(y ). Clearly, π = π + c since π = π. Note that w k is harmonic in and (w k ) L ( ). By the area integral estimates for the harmonic function [], π δ(x) dx n k=1 C w k δ(x) dx C h dσ C u dσ. n k=1 (w k ) dσ Also, since π = 0 in, by interior estimates, π(x) δ 3 (X) dx C π(x) δ(x) dx C u dσ. This completes the proof. We are now in a position to give the: Proof of Theorem A.1. Fix P andr > 0 small. Using linear transformations both in the variable X and functions u, π (see [S1, p. 347]), we may assume that B(P,3r) ={(x,y) R n : y>ψ(x) B(P,3r). ( ) Let η C0 B(P,r) such that η 1onB(P,r). Since u = π, div π =0in,wehave (uη) =( u)η + u η + u η =( π)η + u η + u η = (πη) π η + u η + u η = (πη) f where f = π η u η u η. Note that f C{ π + u + u. Also, div (uη) =u η g and g C u. Moreover, it is not hard to see that g +divf C{ u + π + π + u.

23 Estimates for the Stokes Operator in Lipschitz omains 105 We now apply Lemma A.3 to the equations (uη)+ (πη) =f, div(uη) =g in = {(x,y) R n : y>ψ(x). Let r = B(P,r). We obtain u δ(x) dx C u dσ + C π δ(x) dx + C π δ 3 (X) dx r + C u u δ (X) dx + C u dx C u dσ, where we have used Lemma A.9 in the second inequality. It follows that u δ(x) dx C u dσ. To finish the proof, note that Lemma A.3 also gives { u dσ ε (u) dσ + C ε u δ(x) dx + B(P,r) π δ(x) dx + u dx π δ 3 (X) dx It then follows from the L -estimates [FKV] that { u dσ C u δ(x) dx + π δ(x) dx + u dx. This estimate holds for any pressure term π. In particular, we may choose π such that π(x 0 )=0forsomeX 0. Then,bytheHardyinequality[St,p. 7], π δ(x) dx C π δ 3 (X) dx C u δ 3 (X) dx C u δ(x) dx, where the last inequality follows from the interior estimates. Thus, { u dσ C u δ(x) dx + u dx. The proof is complete..

24 106 R. M. Brown & Z. Shen References [BL] J. Bergh & J. Löfström, Interpolation Spaces, An Introduction, Springer- Verlag, Berlin, [CF] P. Constantin & C. Foias, Navier-Stokes Equations, Univ. of Chicago Press, [] B. ahlberg, Weighted norm inequalities for the Lusin area integral and the nontangential maximal functions for harmonic functions in a Lipschitzdomain, Studia Math. 67 (1980), [KPV], C. Kenig, J. Pipher & G. Verchota, Area integral estimates and maximum principles for higher order elliptic equations and systems on Lipschitz domains, (Preprint). [W] P. euring & W. von Wahl, Strong solutions of the Navier-Stokes system on Lipschitzbounded domains, Math. Nachr. 171 (1995), [F] E. Fabes, Layer potential methods for boundary value problems on Lipschitzdomains, Lecture Notes in Math (1988), [FKV],C.Kenig& G. Verchota, The irichlet problem for the Stokes system on Lipschitzdomains, uke Math. J. 57 (1988), [FK] H. Fujita & T. Kato, On the Navier-Stokes initial value problem, Arch. Rat. Mech. Anal. 16 (1964), [Gi] M. Giaquinta, Multiple Integrals in the Calculus of Variation and Nonlinear Elliptic Systems, Princeton Univ. Press, [GT]. Gilbarg & N. S. Trudinger, Elliptic Partial ifferential Equations of Second Order, nd Edition, Springer-Verlag, [Gr] P. Grisvard, Elliptic Problems in Nonsmooth omains, Pitman, [H] J. G. Heywood, The Navier-Stokes equations: on the existence, regularity and decayofsolutions, Indiana Univ. Math. J. 9 (1980), [JK]. Jerison & C. Kenig, The inhomogeneous irichlet problem in Lipschitzdomains, J. Functional Analysis 130 (1995), [S1] Z. Shen, Boundary value problems for parabolic Lamé systems and a nonstationary linearized system of Navier-Stokes equations in Lipschitz cylinders, Amer. J. Math. 113 (1991), [S], A note on the irichlet problem for the Stokes system in Lipschitzdomains, Proc. Amer. Math. Soc. 13(3) (1995), [St] E. Stein, Singular integrals and differentiability properties of functions, Princeton Univ. Press, The authors were supported in part by the NSF. The first named author also received support from the Commonwealth of Kentucky through the NSF-EPSCoR program. epartment of Mathematics University of Kentucky Lexington, KY Received: July 3rd, 1995; revised: ecember 7th, 1995.