On the Robin Boundary Condition for Laplace s Equation in Lipschitz Domains

Transcription

1 COMMUNICATIONS IN PARTIAL DIFFERENTIAL EQUATIONS Vol. 29, Nos. 1& 2, pp , 2004 On the Robin Boundary Condition for Laplace s Equation in Lipschitz Domains Loredana Lanzani 1 * and Zhongwei Shen 2 1 Department of Mathematics, University of Arkansas, Fayetteville, Arkansas, USA 2 Department of Mathematics, University of Kentucky, Lexington, Kentucky, USA ABSTRACT Let be a bounded Lipschitz domain in n n 3 with connected boundary. We study the Robin boundary condition u/n + bu = f L p on for Laplace s equation u = 0in, where b is a non-negative function on. For 1 <p<2 +, under suitable compatibility conditions on b, we obtain existence and uniqueness results with non-tangential maximal function estimate u p Cf p, as well as a pointwise estimate for the associated Robin function. Moreover, the solution u is represented by a single layer potential. Key Words: Robin boundary condition; Lipschitz domains; Laplace s equation. 1. INTRODUCTION Let be a bounded Lipschitz domain in n n 3. For Laplace s equation u = 0in, the Dirichlet and Neumann problems with boundary data in L p are well understood. We refer the reader to Kenig (1994) for the references. In this Correspondence: Loredana Lanzani, Department of Mathematics, University of Arkansas, Fayetteville, AR 72701, USA; lanzani@uark.edu. 91 DOI: /PDE Copyright 2004 by Marcel Dekker, Inc (Print); (Online)

2 92 Lanzani and Shen paper we will be concerned with the Robin boundary condition: u + bu = f on (1.1) N for Laplace s equation on. Here N denotes the outer unit normal to, and bu denotes the pointwise multiplication of u by the so-called Robin coefficient b, a given function defined on. The Robin condition arises naturally in heat conduction problems as well as in Physical Geodesy (see e.g., Heiskanen and Moritz, 1967; Kellogg, 1954; Otero, 1998). Under suitable compatibility conditions on b (see (1.7) and (1.8) below), we will show that, given any f L p with 1 <p 2, there exists a unique solution u satisfying: u = 0 in u L p (1.2) u/n + bu = f on (see Theorem 4.3). Our method also yields a pointwise uniform estimate for the Robin function associated to (1.2). Here u denotes the non-tangential maximal function of u, namely: u Q = supux X Q for Q (1.3) where Q denotes the interior non-tangential approach region Q = X X Q < 2 distx (1.4) Although we will never again explicitly state so, all boundary relations should be interpreted as holding non-tangentially almost everywhere with respect to the surface measure on, that is, for instance, the Robin condition (1.1) should read: lim ux NQ+bQuX = fq for a.e. Q (1.5) X Q X Q where denotes the inner product in n. It is well known (see Hunt and Wheeden, 1968) that the existence of the limit above is granted a.e. on by the harmonicity of u and u, and by the conditions: u < u < a.e. on (for the latter, see Lemma 2.6 below). Throughout this paper we shall assume that b 0 and b 0on (1.6) Clearly if b 0, (1.1) reduces to the Neumann condition. It is known that the Neumann problem on Lipschitz domains with boundary data in L p is uniquely solvable (up to a constant) for 1 <p<2 + where = > 0 (see Jerison and Kenig, 1981, for p = 2, and Dahlberg and Kenig, 1987, for 1 <p<2 + ). Moreover, given any p>2, one may find a bounded Lipschitz domain and a

3 Laplace s Equation in Lipschitz Domains 93 harmonic function u on such that u L 2 and u u/n L, but u L p (see Kenig, 1994). By the uniqueness of the Robin problem (1.2) with p = 2, this implies that the Robin problem (1.2) on Lipschitz domains for p>2 is not solvable in general even with b L. However, like the Neumann problem, for a given Lipschitz domain, the Robin problem is solvable for 2 <p<2 + under suitable conditions on b (see Theorem 2.8). Note that if 1 <p<n 1 and u L p, then u L q where 1/q = 1/p 1/n 1. Also u L n 1 implies that u L q for any q< (see Lemma 2.6). Since bu = f u/n L p, in view of Hölder s inequality, we assume that and b L n 1 if 1 <p 2 and n 4 or 1 <p<2 and n = 3 (1.7) b L s for some s>2ifp = 2 and n = 3 (1.8) In the context of Lebesgue spaces, we believe that conditions (1.7) (1.8) are optimal. Our solution of the Robin problem is based on the method of layer potentials. Let gx = 1 gq w n 2 n X Q dq X n 2 n (1.9) denote the single layer potential with density g, where w n denotes the surface measure of the unit sphere in n. If ux = SgX, then u = 0 in u L p Cg L p for 1 <p<, and u ( N + bu = 12 ) I + K g + bg on (1.10) where K gp = pv 1 n P Q NP Q P n gqdq (1.11) is bounded on L p for 1 <p< (see Colfman et al., 1982; Fabes et al., 1978; Verchota, 1984). Let = ( 12 I + K ) + bs (1.12) It is not hard to show that under conditions (1.7) (1.8), is bounded on L p for 1 <p 2. One of the main goals of this paper is to establish the invertibility of on L p for I<p 2. Once this is done, the solution u of problem (1.2) may be represented by ux = 1 f X X (1.13)

4 94 Lanzani and Shen To this end, we first note that operator K is not compact for a general Lipschitz domain. However, it has been established that 1 I + 2 K is a Fredholm operator with index zero on L p for 1 <p<2 + (see Verchota, 1984, for p = 2, and Dahlberg and Kenig, 1987 or Mitrea and Taylor, 2001, for 1 <p<2 + ). Under conditions (1.7) (1.8), we will show that the operator b is in fact compact on L p for 1 <p 2. It follows that is also a Fredholm operator with index zero. Consequently the invertibility of as well as the existence of solutions follow once we establish the uniqueness of the solution for the Robin problem (1.2). See Sec. 2 for details. The uniqueness for the case p = 2 is easily obtained via integration by parts. The case 1 <p<2 is more involved. To deal with it, we first use the L 2 solvability to construct the Robin function, the analogue to the Green function for the Dirichlet problem. We then establish a pointwise uniform estimate on the Robin functions for a sequence of smooth domains which approximate from inside. This uniform estimate, together with an integral representation of u by the Robin functions, yields the uniqueness for 1 <p<2. See Secs. 3 and 4 for details. In Sec. 5, we consider the case where is a C 1 domain in n n 3. We show that the Robin problem over a C 1 domain with boundary data in L p is uniquely solvable for 1 <p< and b in L s where n 1 if 1 <p<n 1 s = n 1 + if p = n 1 (1.14) p if p>n 1 Finally in the appendix, we give a proof of the Friedrichs inequality on that will be needed in this paper. Although the inequality may be proved easily by a compactness argument (see Necas, 1967, p. 20), our proof shows that the constants in the Friedrichs inequalities on remain bounded if is approximated by in a certain manner. This is important in our estimate of the Robin functions on. We will use p to denote the norm in L p. Throughout Secs. 2 4, will be a bounded Lipschitz domain in n n 3 with connected boundary. In Sec. 5, will denote a bounded C 1 -domain. Acknowledgments. The results in this paper were announced in Lanzani (2001). The authors wish to thank M. Mitrea for recently pointing out a simpler method to obtain the uniqueness for 1 <p<2, based on the following considerations from the theory of Fredholm operators. Since is invertible on L 2 L 2 is dense in L p for p<2 and is a Fredholm operator on L p L p L p must be onto. It follows that is invertible on L p since the Fredholm index of on L p is zero. This gives the existence of the solution. The uniqueness also follows as every harmonic function in with u L p may be represented by a single layer potential (Dahlberg and Kenig, 1987). This approach was used in Mitrea and Taylor (2002, Theorem 4.2) to solve the Robin problem on Lipschitz domains in Riemmanian manifolds with b L. We remark that this method works equally well for b L n 1 and n 4. It also works in the case n = 3 once we realize that the integration by parts argument actually yields the uniqueness for 4/3 p 2. However, due to its nature, the

5 Laplace s Equation in Lipschitz Domains 95 approach in Mitrea and Taylor (2002) cannot yield any explicit information on the Robin function and its growth rate, which is one of the main goals of this paper. The uniform pointwise estimate of the Robin function we obtain in Sec. 3 as well as the uniform Friedrichs inequalities in Sec. 6 are of independent interest. 2. GENERAL REMARKS AND THE CASE: p = 2 We begin by recalling two important results concerning the single layer potential on a Lipschitz domain that will be essential to this paper. Let L p 1 denote the space of functions in L p whose first derivatives are also in L p. Theorem 2.1. For 1 <p<, the operator L p L p 1 is bounded. If 1 <p 2 L p L p 1 is also invertible. The boundedness of L p L p 1 follows from a deep theorem of Coifman-McIntosh-Meyer (1982). The invertibility for 1 <p 2 is established by Verchota (1984). In fact L p L p 1 is invertible for 1 <p<2 + (see Dahlberg and Kenig, 1987). Theorem 2.2 (Dahlberg and Kenig, 1987; Verchota, 1984). There exists = > 0 such that if 1 <p<2 +, 1 2 I + K L p L p is a Fredholm operator with index zero. Recall that we are interested in the invertibility of = 1 2 I + K + b on L p where b denotes the pointwise multiplication of the single layer potential on by the Robin coefficient b. Lemma 2.3. Suppose b L S for some s>n 1. Then the operator b L p L p is compact for any p 1s. If b L n 1 b L p L p is compact for any p 1n 1. Proof. The proof rests on Hölder inequality: bg p b s g q q = qp s (2.1) where qp s = ps/s p for 1 <p<sand qs s =, as well as on the Sobolev imbedding theorem on. First assume that b L s for some s>n 1. Let 1 <p s. Since L p 1 is compactly embedded in L q for q = qp s (see e.g., Brezis, 1983), the compactness of b on L p is an immediate consequence of Theorem 2.1 and inequality (2.1). Let 1 <p<n 1. Note that the embedding: L p 1 Lqpn 1 is bounded but not compact. To treat the case s = n 1, we resort to approximation. We choose a sequence of bounded functions b on such that b b n 1 0 as. Observe that Hölder inequality, Theorem 2.1 and the Sobolev embedding yield: b bf p Cb b n 1 f p (2.2)

6 96 Lanzani and Shen This, together with the fact that each b is compact, implies that b is compact. The proof is complete. Corollary 2.4. Let 1 <p 2. Suppose b satisfies (1.7) (1.8). Then L p L p is a Fredholm operator with index zero. Proof. Suppose b satisfies (1.7) (1.8). By Lemma 2.3, b is compact on L p. In view of Theorem 2.2, this implies that = 1 I + 2 K + b L p L p is a Fredholm operator with index zero. Lemma 2.5. Let 1 <p 2. Suppose b satisfies (1.7) (1.8). If the solution of the Robin problem (1.2) is unique (if it exists), then L p L p is one-to-one. Consequently L p L p is invertible. Proof. Suppose g = 0 for some g L p. Let ux = gx. Then u satisfies (1.2) with boundary datum f = 0. By the uniqueness assumption, u 0 in. It follows that g = 0on. Hence g = 0 by Theorem 2.1. Lemma 2.6. Let u C 1. Suppose u L p for some p>1. (a) (b) (c) If 1 <p<n 1, then u L q where 1/q = 1/p 1/n 1. If p = n 1, then u L q for any q<. If p>n 1, then u L. Proof. Let K = X distx where >0. Using the Fundamental Theorem of Calculus and a simple geometric observation, one may show that for small, u u P Q sup ux+c dp (2.3) X K c Q pn 1 Lemma 2.6 then follows from (2.3) and the fractional integral Theorem (Stein, 1970). Lemma 2.7. If p 2, the solution of the Robin problem (1.2) is unique. Proof. We may assume p = 2 since L p L 2 for p>2. Suppose u is a harmonic function in such that u L 2 and u/n + bu = 0on. Note that u L 2 implies that u L 2. Using integration by parts, we have u 2 dx = u u N d = bu 2 d 0 (2.4) where we also used the assumption b 0. We should point out that the integration by parts used in (2.4) may be ustified by approximating by a sequence of smooth domains from inside in the manner described in Verchota (1984, Theorem 1.12).

7 Laplace s Equation in Lipschitz Domains 97 The condition u u L 1 is needed to apply the Lebesgue dominated convergence theorem. It follows from (2.4) that u is constant in, and bu = 0on. Since b 0 and b 0on (see (1.6)), we must have u 0in. Note that if b takes negative values, the solution of (1.2) may not be unique. As a simple example, each u x 1 x n = x = 1 2n satisfy (1.2) on the unit ball in n with b = 1 and f = 0. This is the main reason why we have introduced condition (1.6). Theorem 2.8. Let be a bounded Lipschitz domain in n n 3 with connected boundary. Assume the Robin coefficient b satisfies (1.6). Assume that b L n 1 if n 4, and b L s for some s>2 if n = 3. Then there exists 1 = 1 > 0 such that, for any f L p with 2 p<2 + 1, the Robin problem (1.2) has a unique solution u. Moreover, u p Cf p and u admits the representation (1.13). Proof. The uniqueness was proved in Lemma 2.7. To finish the proof, we only need to show that is invertible on L p. Clearly the case p = 2 follows from Lemmas 2.5 and 2.7. For p>2, we note that by Lemma 2.3, b is compact on L p where 2 <p<n 1ifn 4, and 2 <p<sif n = 3. Let 1 = { min n 3 if n 4 min s 2 if n = 3 (2.5) where is the same as in Theorem 2.2. In view of Theorem 2.2, is a Fredholm operator on L p if 2 <p< The invertibility of follows from the uniqueness property as in the case p = 2. The proof is complete. We remark that, if 1 <p<2, the argument in the proof of Lemma 2.7 do not apply because u u may not be in L 1. Instead in the next sections we use the L 2 solvability to construct the Robin functions R X Y, the analogue of the Green functions, for a sequence of smooth domains which approximate from inside. In Sec. 3 we will show that R X Y CX Y 2 n for X Y (2.6) where C is a constant independent of. The desired uniqueness property follows from (2.6) and an integral representation formula of u by the Robin functions. 3. A POINTWISE ESTIMATE OF ROBIN FUNCTIONS This section is devoted to the proof of (2.6). We begin with a local L estimate for harmonic functions with Robin boundary condition. Let, DX r = BX r where BX r denotes the ball in n centered at X with radius r.

8 98 Lanzani and Shen Lemma 3.1. Let be a bounded, connected Lipschitz domain in n n 3. There exists r 0 > 0 such that, if u satisfies u = 0 in DX 0 r u L 2 BX 0 r u N + bu = 0 on BX (3.1) 0 r for some r<r 0 and X 0, then ( 1/2 sup ux C u dx) 2 (3.2) X DX 0 r/2 DX 0 r where C is a positive constant which depends only on n and the Lipschitz character of, and denotes the integral mean. Proof. We use the Moser method. The proof is similar to that in the case of Dirichlet boundary condition (see e.g., Gilbarg and Trudinger, 1983). We begin by choosing r 0 > 0 so small that for any Q BQ r 0 is given by the intersection of BQ r 0 and the region above a Lipschitz graph after a possible rotation of the coordinate system. Next we fix X 0 and r 0r 0. We may assume that X 0 = Q. For M>0, we let { ux if 0 <ux<m v M X = (3.3) 0 otherwise Then v M X = { ux 0 if 0 <ux<m otherwise (3.4) Let C0 BQ r and D = DQ r. Note that, for 1, u v u M 2 dx = D D N v M 2 d = bu v M 2 d BQr = bv +1 M 2 d 0 (3.5) BQr since b 0on. It follows that v M 2 v 1 M 2 dx + 2 v M v MdX 0 (3.6) D D Letting = 1 + /2 in the inequality above, we obtain: v M D 2 2 dx + 2 v 2 M vmdx 0 (3.7) D

9 Laplace s Equation in Lipschitz Domains 99 Using a few elementary identities, we can re-write inequality (3.7) as: D v M 2 dx Schwartz inequality now yields: D v M 2 dx ( 1 1 ) M Dv vmdx + 1 v M 2 dx (3.8) D ( ) Dv M 2 dx (3.9) This, together with the Friedrichs inequality (see Theorem 6.1) and the Sobolev embedding: L 2 1 Lp 1/p = 1/2 1/n, gives ( ) 2 vm p ( p dx C ) D M Dv 2 dx 2 1 p = (3.10) n for all 1 = 1 + /2, and for any C function with support in BQ r. Here C is a positive constant that depends only on n and the Lipschitz character of. Next, given any two positive numbers t and s with 0 <t<s<r, we select C0 BQ r such that 0 1 1onBQ t 0 outside BQ s, and C/s t. Applying (3.10) to yields: Let ( v p M dx DQt ) 1 p [ ( C ) 1 2 ] 1 1 ( ) 1 v M dx (3.11) s t DQs ( ) 1 q t = vmdx q q and = n DQt n 2 > 1 (3.12) We have proved that q t [ ] 2 Cq q q s for any q 2 0 <t<s<r (3.13) s t Applying inequality (3.13) to q = m q>q 2m= 1 2 3, yields: m+1 q t ( ) C m 2 q m q m q s (3.14) s t By iterating (3.14) m times with t = r/2 + r/2 m+1 and s = r/2 + r/2 m, we conclude ( m+1 q r 2 + r 2 m+1 ) ( ) 2 m=0 1 2Cq q m=0 q 3 q r (3.15) r

10 100 Lanzani and Shen Finally we take q = 2 and let m in (3.15). Note that =0 1/3 = / 1 = n/2. We may conclude that ( ) 1/2 sup v M C v 2 M dx (3.16) DQr/2 DQr Let M in (3.16), we obtain ( 1/2 sup u + C u dx) 2 (3.17) DQr/2 DQr where u + denotes the positive part of u. By applying (3.17) to u, we obtain a similar inequality for the negative part of u. Proof of Proposition 3.1 is then complete. Next we choose an increasing sequence of smooth domains which approximate in a uniform, non-tangential fashion, as described in Theorem 1.12 of Verchota (1984). In particular, we have (a) There are homeomorphisms which satisfy sup Q Q Q 0as, and Q Q for all Q. (b) There are positive functions h on such that h and h 1 are uniformly bounded, h d = d (3.18) (c) E E for any Borel set E, and h 1 a.e. on as. The outer unit normal N to satisfies N Q NQ a.e. on as. We now proceed to construct the Robin function on each. For Y n, let Y X denote the fundamental solution for the Laplacian, namely: Y 1 1 X = (3.19) 2 n n X Y n 2 Let b Q = b 1 Q for n 4. If n = 3, we choose b Q = b 1 Q where is the characteristic function over the set P bp. Note that b L n 1 for n 4, and b L for n = 3. Thus Y /N + b Y L 2 if Y. It follows from Theorem 2.8 that for each Y, there exists a harmonic function W Y on satisfying W Y L 2 W Y + b N W Y = Y (3.20) + b N Y on We now define the Robin function R X Y = R Y X = Y X W Y X for X Y (3.21)

11 Laplace s Equation in Lipschitz Domains 101 Proposition 3.2. Let X Y and X Y. Then R X Y = R Y X (3.22) Proof. It suffices to show W Y X = W X Y. To this end, we apply Green s formula on to the harmonic function W Y X. We obtain [ W Y X = W Y Q X Q X Q W Y ] Q dq N N ( W X ) = W Y Q Q + b N QW X Q [ X Q ] Y Q + b N Q X Q Y Q dq (3.23) A similar identity holds for W X Y. By combining these two identities and applying the divergence theorem to the integral we obtain { Y W X QW N Q W Y W X Y Y W X = X } QW Q N dq (3.24) ( X Q Y Q Y Q ) X Q dq (3.25) N N It is standard to show that the right hand-side of (3.25) vanishes for any X Y X Y. The proof of Proposition 3.2 is concluded. For f C0, let v X = R X Y fy dy (3.26) Let K f be the support of f. Since distk f >0, by Theorem 2.8, W Y 2 is uniformly bounded for Y K f and a fixed. Using this fact, it is not hard to verify that v L 2 1 and is a weak solution of v = f in v (3.27) + b N v = 0 on Lemma 3.3. There exists a constant C independent of such that, if v is given by (3.26) with f C 0, then { } n 2 v n 2 2n 2n dx { } n+2 C f n+2 2n 2n dx (3.28) If n = 3, we also require to be sufficiently large.

12 102 Lanzani and Shen Proof. Since v L 2 1 is a weak solution of (3.27), we have v 2 dx + b v 2 d = v f dx (3.29) By Hölder s inequality, the right hand side of (3.29) is bounded by v L 2n n 2 f L 2n n+2 Thus it suffices to show that v 2 is bounded by a constant C times the left L n 2 2n hand side of (3.29). Furthermore, by the Sobolev imbedding theorem, we only need to show that, for any u L 2 1 { } u 2 dx C u 2 dx + b u 2 d (3.30) with constant C independent of. To this end, we choose >0 so that E >0 where E = Q bq 1/. Let E = E. By the Friedrichs inequality (6.2), { } u 2 dx C u 2 dx + u 2 d E { } C u 2 dx + b u 2 d (3.31) for n 4 and 1. In the case n = 3, (3.31) holds for all 1/. This completes the proof. We are now in a position to estimate the growth of the Robin function. Theorem 3.4. R X Y There exists a constant C > 0 independent of such that C X Y n 2 for any X Y (3.32) Proof. Fix X 0 Y 0. We let r = minr 0 X 0 Y 0 /8 where r 0 is given in Lemma 3.1. Note that R Y 0 = R Y 0 satisfies R Y 0 = 0 in BX 0 r R Y 0 L 2 BX 0 r (3.33) R Y 0 + b N R Y 0 = 0 on BX 0 r By Lemma 3.1, we have R Y 0 X 0 sup BX 0 r/2 R Y 0 C r n 2 2 R Y 0 L 2n n 2 BX 0 r (3.34)

13 Laplace s Equation in Lipschitz Domains 103 where C is a positive constant which depends only on n and the Lipschitz character of. We claim that R Y 0 L n 2 2n BX 0 r C r n 2 2 (3.35) Clearly this inequality, together with (3.34), gives the desired estimate in Theorem 3.1. To this end, given f C0 BX 0 r, welet vx = v X = R X Y fy dy (3.36) Then v = 0 in BY 0 r v/n + b v = 0 on, and v L 2. It follows from Lemma 3.1 that vy 0 sup v BY 0 r/2 C r n 2 2 f L 2n n+2 C r n 2 2 v L 2n n 2 (3.37) where we have used Lemma 3.3 in the last inequality. In view of (3.36), this implies the Claim (3.35) by duality. The proof is complete. 4. THE ROBIN PROBLEM IN L p In this section we state and prove the main result (Theorem 4.3) of the paper. We begin with a representation formulas for harmonic functions in by the Robin functions R constructed in the last section. Lemma 4.1. Let u be a harmonic function in. Then, for any X, ( ) u ux = R X Q Q + b N QuQ dq (4.1) Proof. By the definition of R, the integral in the right hand side of (4.1) equals I 1 I 2 where ( ) u I 1 = X Q Q + b N QuQ dq (4.2) ( ) u I 2 = W X Q Q + b N QuQ dq (4.3) Since u = 0in and u C 2, we may use the Green s formula to obtain ( ) I 1 = X Q + b N Q X Q uqdq + ux (4.4)

14 104 Lanzani and Shen Similarly, by applying Green s identity to the harmonic functions u and W X,weget ( W X I 2 = Q + b N QW X )uqdq Q ( ) = X Q + b N Q X Q uqdq (4.5) The proof is concluded, as we have ux = I 1 I 2. Next we use Lemma 4.1 and Theorem 3.4 to establish the uniqueness property for the Robin problem (1.2) in the case 1 <p<2. Lemma 4.2. Let 1 <p<2 and b L n 1. Suppose that u is a harmonic function in such that u L p and u/n + bu = 0 on. Then u 0 in. Proof. Fix X. There exists an integer J = JX > 0 such that X for all J. Using Lemma 4.1 and the homeomorphisms, we may write ux = R X Qu Q N Q +b Qu Qh QdQ (4.6) for J, where h is given in (3.18). It then follows from Theorem 3.4 that ux C X u Q N Q+b Qu Qh QdQ (4.7) Let F Q = u Q N Q+b Qu Qh Q. Recall that b Q = bq if n 4, and b Q = bq if n = 3 where is the characteristic function over the set P bp. Also recall that Q Q and N Q NQ h Q 1 a.e. on as. It follows that F u/n + bu = 0 a.e. on as. Since sup J F Q u Q + bqu Q sup h (4.8) and u + bu L 1, we may apply the Lebesgue dominated convergence theorem to conclude that F QdQ 0 as (4.9) By (4.7), this implies that ux = 0. Thus u 0in. The proof is finished. The following is the main theorem of the paper. Theorem 4.3. Let be a bounded Lipschitz domain in n n 3 with connected boundary. Let 1 <p 2. Suppose that b satisfies the conditions (1.6) (1.8). Then, given

15 Laplace s Equation in Lipschitz Domains 105 any f L p, there exists a unique harmonic function u in such that u L p and u/n + bu = f on. Moreover, u p Cf p and u admits the representation (1.13). Proof. The case p = 2 was dealt with in Theorem 2.8. For 1 < p < 2, the uniqueness is proved in Lemma 4.2. By Lemma 2.5, this implies that the operator is invertible on L p. The existence and the representation formula (1.13) now follow. 5. THE ROBIN PROBLEM IN C 1 -DOMAINS If is a C 1 -domain, as in the cases of Dirichlet and Neumann problems, the Robin problem (1.2) is uniquely solvable for any 1 <p<. Theorem 5.1. Let be a bounded domain of class C 1 in n n 3 with connected boundary. Let 1 <p<. Suppose b satisfies (1.6) and (1.14). Then, given any f L p, there exists a unique harmonic function u such that u L p and u/n + bu = f on. Moreover, u p Cf p and u admits the representation (1.13). Proof. The uniqueness is contained in Lemma 4.2 since the uniqueness in L p implies the uniqueness in L q for any q>p. To prove the existence, we note that for a C 1 domain K is in fact compact on L p for any 1 <p< (see Fabes et al., 1978). By Lemma 2.3, if b satisfies (1.14), the operator b is also compact on L p. It follows that = 1/2 + K + b is a Fredholm operator on L p with index zero. Since is one-to-one on L p for 1 <p 2 by Theorem 4.3, is also one-to-one on L p for 2 <p<. This implies that is invertible for any 1 <p<. The existence as well as the representation formula (1.13) follows as before. 6. APPENDIX: THE FRIEDRICHS INEQUALITY The purpose of this appendix is to give a proof of the following Friedrichs inequality (see Necas, 1967, p. 20). Theorem 6.1. Let be a bounded, connected Lipschitz domain in n n 2. Then, given any set E with E > 0, there exists a constant C which depends on n E and such that { } u 2 dx C u 2 dx + u 2 d for any u L 2 1 (6.1) E It follows from our proof that if we approximate by a sequence of smooth domains with homeomorphisms in the manner described in Theorem 1.12 of Verchota (1984), then { } u 2 dx C u 2 dx + u 2 d u H 1 (6.2) E where E = E and constant C is independent of.

16 106 Lanzani and Shen We begin with an elementary observation. Lemma 6.2. Let V be a bounded measurable set in n. Then for any function u C 1 n and any point Y n, we have V where ux uy 2 dx 1 n ( ) n+1 sup X Y X V F ux 2 dx (6.3) X Y n 1 F = FV Y = x n X= Y + sz Y for some Z V and some s 0 1 (6.4) Proof. Given any point X V, the Fundamental Theorem of Calculus and Hölder inequality yield: ( ) X Y ( ) ux uy 2 sup Z Y u X Y 2 Y + t dt (6.5) Z V 0 X Y We now let: l= sup Z Y Z V vy = { uz 2 0 Z F otherwise (6.6) By integrating (6.5) over V and using polar coordinates for the right-hand side, we obtain: V ux uy 2 dx ln+1 n = ln+1 n Sn 1 n 0 vy + t dt d vx dx (6.7) X Y n 1 The proof of Lemma 6.2 is concluded. For a measurable set V in, we define V = Y FV Y (6.8) Lemma 6.3. Let be a bounded domain in n n 2. Let V be a measurable set in. Then for any u C 1 n, we have uy 2 dy C { } ux 2 dx + ux 2 dx (6.9) V V V where C depends only on n and the diameter of.

17 Laplace s Equation in Lipschitz Domains 107 Proof. Note that by Lemma 6.2, we have uy 2 V 2 ux 2 dx + 2 ux 2 V n diamn+1 dx (6.10) X Y n 1 for any Y V. Integrating the above inequality over V, we obtain V uy 2 dy V 2 V ux 2 dx + 2 n diamn+1 sup Y dx X Y ux 2 dx n 1 (6.11) From this, inequality (6.9) follows easily. We are now in a position to give The Proof of Theorem 6.1. It suffices to prove the Friedrichs inequality for u C 1 n. Since E > 0, there exists a point P E such that lim r 0 E BP r = 1 (6.12) BP r i.e., P is a regular point of E. Without loss of generality, we may assume that P = 0. Since is a Lipschitz domain, we may also assume that there exist r 0 > 0 and a Lipschitz function n 1 such that 0 = 0 0 = 0, and B0r 0 = B0r 0 X x n n 1 x n > X (6.13) It is not difficult to show that there is a small positive constant c 0, which depends only on n and, with the following property: X c 0 r 0 0 <t c 0 r 0 X X + t B0r 0 (6.14) Thus, for X c 0 r 0 and 0 <t c 0 r 0 we may write: ux X + t ux X = Hölder inequality now yields: t ux X + t 2 2uX X 2 + 2t Integrating both sides of (6.16) over the set: 0 u x n X X + s ds (6.15) t 0 ux X + s 2 ds (6.16) E = X n 1 X c 0 r 0 X X E (6.17)

18 108 Lanzani and Shen We obtain: { ux X + t 2 dx C E uq 2 d + E } ux 2 dx (6.18) It follows by integrating both sides of (6.18) over t 0c 0 t 0 that { } ux 2 dx C ux 2 dx + uq 2 d 0 E (6.19) where A 0 = X = X X + t X c 0 r 0 0 <t<c 0 r 0 (6.20) In view of (6.12), it is easy to see that A 0 > 0. Let A = A 1 for 1 (see (6.8)). Since X 0 = 0c 0 r 0 /2 A 0 A 1 contains the ball BX 0 c 1 r 0 where c 1 = c 1 >0. It follows that A BX 0 c 1 r 0 > 0 for all 1. In view of Lemma 6.3, this implies that { } ux 2 dx C ux 2 dx + uq 2 d (6.21) E Finally we observe that, since satisfies the interior cone condition, there exists an integer M which only depends on the Lipschitz character of, such that any point X in may be connected to X 0 by a continuous path consisting of m line segments with m M. It follows that we must have A M =. This, together with (6.21), completes the proof of Theorem 6.1. ACKNOWLEDGMENTS First author was supported by NSF Grant No. DMS Second author was supported by NSF Grant No. DMS REFERENCES Brezis, H. (1983). Analyse Fonctionnelle. Paris: Masson. Colfman, R., McIntosh, A., Meyer, Y. (1982). L intégrale de cauchy définit un opérateur borné sur L 2 pour les courbes Lipschitziennes. Ann. Math. 116: Dahlberg, B., Kenig, C. (1987). Hardy spaces and the Neumann problem in L p for Laplace s equation in Lipschitz domains. Ann. Math. 125: Fabes, E., Jodeit, M. Jr., Riviére, N. (1978). Potential techniques for boundary value problems on C 1 -domains. Acta Math. 141: Gilbarg, D., Trudinger, N. S. (1983). Elliptic Partial Differential Equations of Second Order. 2nd ed. Springer-Verlag. Heiskanen, W., Moritz, H. (1967). Physical Geodesy. San Francisco/Londres: W. H. Freeman and Co.

19 Laplace s Equation in Lipschitz Domains 109 Hunt, R., Wheeden, R. (1968). On the boundary values of harmonic functions. Trans. Amer. Math. Soc. 132: Jerison, D., Kenig, C. (1981). The Neumann problem on Lipschitz domains. Bull. Amer. Math. Soc. 4: Kellogg, O. (1954). Foundations of Potential Theory. Dover. Kenig, C. (1986). Elliptic boundary value problems on Lipschitz domains. Beiing Lectures on Harmonic Analysis. Annals of Math. Studies 112: Kenig, C. (1994). Harmonic Analysis Techniques for Second Order Elliptic Boundary Value Problems. Regional Conference Series in Math. 83, AMS. Lanzani, L. (2001). The Cl n -valued Robin boundary value problem on Lipschitz domains in n. In: Clifford Analysis and Its Applications (Prague, 2000). NATO Sci. Ser. II Math. Phys. Chem. 25, Dordrecht: Kluwer Acad. Publ. Medkóva, D. (1997). The boundary value problems for Laplace equation on domains with Non-Smooth boundary. Arch. Math. (Brno) 34: Medkóva, D. (1998). Solution of the Robin problem for the Laplace equation. Appl. Math. 43: Medkóva, D. (2000). Solution of the Robin and Dirichlet problem for the Laplace equation. In: Direct and Inverse Problems of Mathematical Physics (Newark, DE, 1997). Int. Soc. Anal. Appl. Comput. 5, Dordecht: Kluwer Acad. Publ., pp Mitrea, M., Taylor, M. (2001). The method of layer potentials on Lipschitz domains in Riemannian manifolds. Comm. Anal. Geom. 9(2): Mitrea, M., Taylor, M. (2002). Potential theory on Lipschitz domains in Riemannian manifolds: the case of Dini metric tensors. Trans. Amer. Math. Soc. 355(5): Necas, J. (1967). Les Methods Directes en Théorie des Elliptiques. Prague: Academia. Otero, J. (1998). A uniqueness theorem for a Robin boundary value problem of Physical Geodesy. Quart. Appl. Math. 56: Stein, E. M. (1970). Singular Integrals and Differentiability Properties of Functions. Princeton University Press. Taylor, A. E. and Lay, D. C. (1980). Introduction to Functional Analysis. 2nd ed. John Wiley & Sons. Verchota, G. (1984). Layer potentials and regularity for the Dirichlet problem for Laplace s equation on Lipschitz domains. J. Funct. Anal. 59: Received October 2002 Accepted July 2003

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