Nonhomogeneous Neumann problem for the Poisson equation in domains with an external cusp

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J. Math. Anal. Appl. 31 (5) 397 411 www.elsevier.com/locate/jmaa Nonhomogeneous Neumann problem for the Poisson equation in domains with an external cusp Gabriel Acosta a,1, María G.

Gutierrez 115, Los Polvorines, B1613GSX Provincia de Buenos Aires, Argentina b Departamento de Matemática, Facultad de Ciencias Exactas y Naturales, Universidad de Buenos Aires, 148 Buenos Aires,

1 J. Math. Anal. Appl. 31 (5) Nonhomogeneous Neumann problem for the Poisson equation in domains with an external cusp Gabriel Acosta a,1, María G. Armentano b,,, Ricardo G. Durán b,, Ariel L. Lombardi b a Instituto de Ciencias, Universidad Nacional de General Sarmiento, J.M. Gutierrez 115, Los Polvorines, B1613GSX Provincia de Buenos Aires, Argentina b Departamento de Matemática, Facultad de Ciencias Exactas y Naturales, Universidad de Buenos Aires, 148 Buenos Aires, Argentina Received 1 May 4 Available online 3 March 5 Submitted by G. Komatsu Abstract In this work we analyze the existence and regularity of the solution of a nonhomogeneous Neumann problem for the Poisson equation in a plane domain with an external cusp. In order to prove that there exists a unique solution in H 1 () using the Lax Milgram theorem we need to apply a trace theorem. Since is not a Lipschitz domain, the standard trace theorem for H 1 () does not apply, in fact the restriction of H 1 () functions is not necessarily in L ( ). So, we introduce a trace theorem by using weighted Sobolev norms in. Under appropriate assumptions we prove that the solution of our problem is in H () and we obtain an a priori estimate for the second derivatives of the solution. 5 Elsevier Inc. All rights reserved. Supported by ANPCyT under grant PICT 3-59, by CONICET under grant PIP 66/98 and Fundación Antorchas. * Corresponding author. addresses: gacosta@ungs.edu.ar (G. Acosta), garmenta@dm.uba.ar (M.G. Armentano), rduran@dm.uba.ar (R.G. Durán), aldoc7@dm.uba.ar (A.L. Lombardi). 1 The first author is partially supported by PICT The second and third authors are members of CONICET, Argentina. -47X/$ see front matter 5 Elsevier Inc. All rights reserved. doi:1.116/j.jmaa

2 398 G. Acosta et al. / J. Math. Anal. Appl. 31 (5) Keywords: Cuspidal domains; Traces; Neumann problem; Regularity 1. Introduction This paper deals with an elliptic equation in a domain with an external cusp. Since this kind of domains are not Lipschitz, the standard arguments to prove existence cannot be applied when nonhomogeneous Neumann boundary conditions are imposed on some part of the boundary. Indeed, to apply the Lax Milgram theorem in this case one needs to use some trace theorem for Sobolev spaces. However, simple examples show that, for some cusps, there are functions in H 1 () such that their restriction to the boundary are not in L ( ). Therefore the classic trace theorems for Lipschitz domains are not valid in this case. We consider the following model problem: let be the plane domain defined by = { (x, y): <x<1, <y<ϕ(x) }, with ϕ C (, 1), ϕ,ϕ,ϕ > on(, 1), ϕ() = ϕ () = (a typical example is ϕ(x) = x α, α>1), and Γ = Γ 1 Γ Γ 3 the boundary of, where Γ 1 ={ x 1, y= }, Γ ={x = 1, y 1} and Γ 3 = { x 1, y= ϕ(x) } (see Fig. 1). We seek u such that u = f in, u ν = onγ 1, u = onγ, u ν = g on Γ 3, where ν denotes the outside normal to. (1.1) Fig. 1. Cuspidal domain.

3 G. Acosta et al. / J. Math. Anal. Appl. 31 (5) In [6] the authors characterize the traces of the Sobolev spaces W 1,p (), 1 p<, for domains of the class considered here by using some weighted norm on the boundary. Existence of solutions of (1.1) can be derived from their results under certain hypothesis on the data. In order to obtain existence results for more general data we present a different kind of trace results by introducing a weighted Sobolev space in such that the restriction to the boundary of functions in that space are in L p (Γ ). Once the existence of a solution is known, the question about its regularity arises naturally. For the Poisson problem with homogeneous boundary conditions on cuspidal domains it is known that, if the right-hand side of the equation is in L (), then the solution belongs to H () (see [,5]). We show that the technique introduced by Khelif in [5] can be extended to treat nonhomogeneous Neumann type boundary conditions. In this way we prove that the solution of our model problem belongs to the space H ().. Existence and uniqueness of solution In this section we prove some trace results and apply them to obtain existence and uniqueness of solution of our model problem using the Lax Milgram theorem. Let V ={v H 1 (): v Γ = }. The variational problem associated with (1.1) is given by: Find u V such that a(u,v) = L 1 (v) + L (v) v V, where a(u,v) = u v, L 1 (v) = fv and L (v) = Using the Poincaré inequality, it is easy to see that the bilinear form a(, ) is coercive and continuous on V. Therefore, in order to prove that there exists a unique solution in V using the Lax Milgram theorem, we need to impose conditions on the data f and g which guarantee that the linear operators L 1 and L are continuous on V. For the continuity of L 1 it is enough to assume that f L (). On the other hand, the continuity of L when g L (Γ 3 ), in the case of a Lipschitz domain, is proved by using well-known results on restrictions of H 1 () to the boundary. However, since our domain is not Lipschitz, the standard trace theorem for H 1 () does not apply, in fact, the following example shows that for some cusps the restriction of H 1 () functions is not necessarily in L (Ɣ). Example.1. Consider ϕ(x) = x α, α>1, and the function u(x,y) = x γ. Then, an easy computation shows that u H 1 () iff γ< α 1. However, u L (Γ ) iff γ< 1. So, for α>, taking 1 γ<α 1, we have examples of functions which are in H 1 () and such that their restrictions to the boundary are not in L (Γ ). In [6], Mazya et al. characterize the space of traces of W 1,p (), for non-lipschitz domains of the type considered here, by using some weighted norms on the boundary. In particular, it follows from their results that there exists a constant C such that uϕ 1 L (Γ ) C u H 1 (). (.1) Γ 3 gv.

4 4 G. Acosta et al. / J. Math. Anal. Appl. 31 (5) Indeed, the left-hand side agrees with the first term in the norm TW 1 p (), with p =, introduced in [6, p. 18] which, as proved in that paper, is bounded by the H 1 norm. The inequality (.1) can be used to prove the continuity of L under the assumption that gϕ 1 L (Γ 3 ), in fact we have L (u) = Γ 3 gϕ 1 1 uϕ gϕ 1 C gϕ 1 L (Γ 3 ) u H 1 (). L (Γ 3 ) uϕ 1 L (Γ 3 ) Let us observe that assuming continuity of g the condition gϕ 1 L (Γ 3 ) implies that g has to vanish at the origin, which does not seem to be a natural condition for the existence of a solution. Therefore, our goal is to relax the assumption on g by introducing a trace result of a different nature of those in [6]. More precisely, we want to give sufficient conditions to have traces in L p of the boundary. In order to do that we introduce the weighted Sobolev space Wϕ 1,p () as the closure of C ( ) in the norm u p := 1 uϕ Wϕ 1,p p p () L p () + ( uϕ p 1 ) p p L p (). In what follows we use the letter C to denote a generic constant which depends only on p. Lemma.1. There exists a constant C such that for any u Wϕ 1,p () with 1 p<, u L p (Γ ) C ( uϕ p 1 L p () + ( uϕ p 1 ) p L ()). p Proof. We will use the following change of variables which is a generalization of that introduced by Grisvard [3] for power type cusps. Let ξ = 1 ϕ (x) and η = y ϕ(x) then, is transformed in given by { = (ξ, η): ξ> 1 } ϕ (1), <η<1, see Fig.. We denote by Γ 1 ={(ξ, η): ξ 1 ϕ (1),η= }, Γ ={(ξ, η): ξ = 1 ϕ (1), η 1} and Γ 3 ={(ξ, η): ξ 1 ϕ (1),η= 1}. First we give the proof for the case p = 1. Writing v(ξ,η) = u(x,y) we have Γ 3 u = 1 = C ( ) u x,ϕ(x) 1 + ϕ (x) dx C 1 ϕ (1) 1 ( ) u x,ϕ(x) dx v(ξ,1) J(ξ)dξ, (.)

5 G. Acosta et al. / J. Math. Anal. Appl. 31 (5) Fig.. where J(ξ)= ϕ (x) ϕ (x). Applying the following standard trace inequality in : ( ) w L 1 ( Γ 3 ) C w L 1 ( ) + w η, L 1 ( ) to the function w(ξ,η) = v(ξ,η)j(ξ), we get v(ξ,1) ( J(ξ)dξ C v(ξ,η) J(ξ)dξ dη + 1 ϕ (1) v(ξ,η) η and therefore, changing variables and using (.) and (.3), we have ( ) u C u ϕ(x) 1 dxdy + u y dxdy. Γ 3 ) J(ξ)dξ dη Applying the same argument on Γ 1 and a standard trace theorem on Γ, we obtain u L 1 (Γ ) C( uϕ 1 L 1 () + u ) L 1 () concluding the proof for the case p = 1. Now, for any p such that 1 <p<,weuse(.4)foru p to obtain ( ) u p C u p ϕ 1 + p u p 1 u Γ ( ) = C u p ϕ 1 + p u p 1 ϕ q 1 u ϕ q 1, p (.3) (.4) where q = p 1, and therefore, the proof concludes by using the inequality ab q 1 aq + 1 p bp in the last term on the right-hand side.

6 4 G. Acosta et al. / J. Math. Anal. Appl. 31 (5) Remark.1. With an argument analogous to that used in the previous lemma one can prove the following result, which is stronger than (.1): uϕ p 1 L p (Γ ) C( u L p () + uϕ L ()) p. Existence results for more general data g can be obtained from the previous lemma and embedding theorems. During the rest of this section we will restrict ourselves to the case of power type cusps, for which embedding theorems are well known. Let ϕ(x) = x α with α>1. In the next theorem we prove that the restriction of H 1 () functions are in L p (Γ ) under appropriate assumptions on the values of α and p. Inthe proof we will make use of the inclusion H 1 () L r (α + 1) () for r (.5) α 1 which is a particular case of the results given in [1]. Theorem.1. Let u H 1 () and 1 p.ifα<1 + p then u Lp (Γ ) and u L p (Γ ) C u H 1 (). (.6) Proof. From Lemma.1 we know that u L p (Γ ) C ( ux p α L p () + p 1 α( ux ) ) p L p () C ( ux p α L p () + u L ()) p. (.7) To bound the first term on the right-hand side of (.7) we use the Hölder inequality with an exponent q to be chosen below. Then, ( ) 1 ( q ) q 1 u p x α u pq x α q q q 1. From (.5), if p (1+α) q (α 1)p we have ( ) 1 q u pq C u p H 1 (). On the other hand, ( q x α q 1 ) q 1 q is bounded if q>1 + α. So, if α<1 + p we can take q such that 1 + α<q (1+α) (α 1)p and we obtain (.6). Remark.. In particular, it follows from the previous theorem that for α< the functions in H 1 () have traces in L (Γ ), while from Example.1 we know that this is not true for α>. Therefore our result is almost optimal. Now we can give an existence result for problem (1.1) under appropriate assumptions on g and α.

7 G. Acosta et al. / J. Math. Anal. Appl. 31 (5) Theorem.. Let 1 p, g L q (Γ 3 ) with q = p p 1, and f L (). Ifα<1 + p then there exists a unique solution u V of problem (1.1). Proof. Since the bilinear form a(, ) is coercive and continuous on V, the existence of a unique solution will be a consequence of the Lax Milgram theorem if we show that the linear functional L := L 1 + L is continuous on V. Since f L (), L 1 is continuous and therefore it only remains to prove the continuity of L. From Theorem.1 we know that u L p (Γ ) C u H 1 () and so, L (u) = gu g L q (Γ 3 ) u L p (Γ 3 ) C g L q (Γ 3 ) u H 1 () Γ 3 and the theorem is proved. 3. Regularity of the solution In this section we analyze the regularity of the solution u of problem (1.1). Under appropriate conditions on g we prove, in the next theorem, that u H (). In order to obtain this result we will apply the method introduced by Khelif [,5] which is based in approximating the domain by a sequence of Lipschitz domains. Theorem 3.1. Let f L (), and g such that, if h(t) := g(t,ϕ(t)), hϕ 1 L (, 1) and h (ϕ ) 1 L (, 1). Assume also that ϕ ϕ (ϕ ) L (,1) < 1. Then the problem (1.1) has a unique solution u belonging to H (), and there exists a constant C such that u H () C{ f L () + hϕ 1 L (,1) + h (ϕ ) 1 L (,1)}. (3.1) Proof. The existence of a unique solution u H 1 () follows from the results of Section. Then it only remains to show that u H (). Let = 1/n and define and n = { (x, y) : <x<1 }, Γ1 n = { (x, ): x 1 }, Γ = { (1,y): y 1 }, Γ3 n = {( x,ϕ(x) ) : x 1 }, Γ4 n = { (,y): y ϕ( ) }, see Fig. 3.

8 44 G. Acosta et al. / J. Math. Anal. Appl. 31 (5) Fig. 3. We consider the following problem in n : u n = f in n, u n = onγ, u n ν = g on Γ 3 n, (3.) u n ν = onγ 1 n Γ 4 n. In what follows the letter C will denote a constant which may depend on ϕ. Observe first that the solution u n satisfies u n H 1 ( n ) C{ f L () + hϕ 1 } L (3.3) (,1) with C independent of n. Indeed, this estimate follows by standard arguments using a trace theorem as that given in Remark.1 applied on n. Note that the argument of Lemma.1 can be applied to n providing a constant independent of n. It is known that the solution of problem (3.) belongs to H +ε ( n ) [,4], for some positive ε, in particular its first derivatives are continuous. Our goal is to obtain an estimate for u n H ( n ) valid uniformly in n. Using a method introduced by Khelif [,5] we will show that u n H ( n ) C{ f L () + hϕ 1 L (,1) + h (ϕ ) 1 } L (3.4) (,1) with C independent of n. For any ρ and ψ in H 1 ( n ) we have ρ ρ x ψ y = ρ y ψ x + τ, n n n ψ where τ is the unit tangent vector oriented clockwise. Note that the right-hand side has to be understood in a weak sense, i.e., ρ τ H 1/ ( n ). Taking ρ = u n x and ψ = u n y in the equation given above we obtain f = ( u n ) = (ρ x + ψ y ) = ρx + ρ x ψ y + n n n n n n ψ y

9 G. Acosta et al. / J. Math. Anal. Appl. 31 (5) = ρx + ρ y ψ x + ψy + ρ ψ τ n n n n = u n H ( n ) + ρ τ, (3.5) n ψ where u n H ( n ) denotes the seminorm of u n in H ( n ). To simplify notation we introduce the one variable functions v(t) := u n ( ) t,ϕ(t) x and w(t) := u n ( ) t,ϕ(t). y Then, the boundary conditions imply u n y = on Γ 1 n Γ, w = vϕ + h 1 + (ϕ ) on Γ3 n, u n x = onγ 4 n. Therefore, (3.5) becomes u n H ( n ) = n f 1 w(t)v (t) dt, (3.6) and so, we have to bound the last term on the right-hand side. From the boundary condition on Γ3 n we have 1 w(t)v (t) dt = 1 For the first term we have I = 1 = 1 ( ) w(t) w(t) ϕ dt (t) 1 ( h(t) 1 + (ϕ w(t) (t)) ) ϕ dt (t) = I + II. (3.7) w(t)w 1 (t) ϕ (t) dt 1 ( w(t) ) 1 ϕ (t) dt 1 1 w(t) ϕ (t) ϕ (t) dt w(t) ϕ (t) ϕ (t) dt. Now, since u n y is continuous, it follows from the boundary condition on Γ that w(1) =. Therefore, integrating by parts, we obtain for the first term in the right-hand side of the last equation, 1 1 ( w(t) ) 1 ϕ (t) dt = 1 w() 1 ϕ ( ) w(t) ϕ (t) ϕ (t) dt

10 46 G. Acosta et al. / J. Math. Anal. Appl. 31 (5) and then, I = 1 w() 1 ϕ ( ) 1 1 w(t) ϕ (t) ϕ dt. (3.8) (t) Using now the boundary condition on Γ4 n and the fact that u n x is continuous, it follows that v( ) = and so, from the boundary condition on Γ3 n we obtain w( ) = h( ) 1 + ϕ ( ). Therefore, replacing in (3.8) we have I = 1 h ( ) ( 1 + ϕ ( ) ) 1 ϕ ( ) 1 1 w ϕ (t) (t) ϕ dt. (3.9) (t) To bound the first term on the right-hand side we observe that, for any s (, 1), h(s) h() s = h (t) ( s ) 1 h (ϕ ) 1 L ϕ (t) dt (,1) = h (ϕ ) 1 L (,1) ϕ (s) 1. In particular h is continuous at and consequently, since hϕ 1 L (, 1), it follows that h() = (recall that <ϕ(t)<t for all t small enough). Moreover, h ( ) ϕ ( ) h (ϕ ) 1 L (,1) and so, we obtain from (3.9), I C h (ϕ ) 1 L (,1) w ϕ (t) (t) ϕ dt. (3.1) (t) Let us now estimate the second term on the right-hand side of (3.7). A simple computation shows that 1 w(t)h(t)ϕ (t) 1 II = ϕ (t) 1 + ϕ (t) dt + w(t)h (t) 1 + ϕ (t) ϕ dt = III + IV. (t) Using the arithmetic geometric inequality ab 1 ɛ a + ɛ b valid for all ɛ>, we have III ɛ 1 w(t) ϕ (t) ϕ (t) dt + 1 ɛ 1 h(t) ϕ (t) ϕ (t) (1 + ϕ (t) ) dt

11 ɛ 1 G. Acosta et al. / J. Math. Anal. Appl. 31 (5) w(t) ϕ (t) ϕ (t) while, on the other hand, we have IV ɛ ɛ 1 1 w(t) ϕ (t) ϕ (t) w(t) ϕ (t) ϕ (t) dt + 1 hϕ 1 ɛ dt + 1 ɛ 1 L (,1) ϕ ϕ (ϕ ) h (t) (1 + ϕ (t) ) ϕ (t) dt + C h (ϕ ) 1 ɛ L (,1). L (,1), So 1 w(t) ϕ (t) II ɛ ϕ (t) dt + 1 hϕ 1 ϕ ϕ ɛ L (,1) (ϕ ) L p (,1) n + C h (ϕ ) 1 ɛ L (,1). (3.11) Therefore, using the estimates (3.1) and (3.11), we obtain from (3.7), 1 ( ) 1 w(t)v 1 (t) dt + ɛ w(t) ϕ (t) ϕ (t) dt + 1 ϕ ϕ ɛ (ϕ ) hϕ 1 L L (,1) (,1) ( + C ) h (ϕ ) 1 ɛ L (,1). (3.1) But, from the boundary condition on Γ1 n we know that u n (t, ) = and therefore, w (t) = u n ( ) t,ϕ(t) = y and consequently, 1 w (t) ϕ(t) dt u n y (t, y) ( ϕ(t) L ( n ) Therefore, replacing in (3.1) we obtain 1 w(t)v (t) dt (1 + ɛ) ϕ ϕ (ϕ ) y dt ) u n (t, y) dy ϕ(t) y. L (,1) ϕ(t) u n (t, y) y u n (t, y) y L ( n ) dy

12 48 G. Acosta et al. / J. Math. Anal. Appl. 31 (5) ϕ ϕ ɛ (ϕ ) hϕ 1 L L (,1) (,1) ( + C + 1 ) h (ϕ ) 1 ɛ L (,1). (3.13) Hence, using this estimate in (3.6), we conclude that ( u n H ( n ) f L () + C + 1 ) { hϕ 1 ɛ L (,1) + h (ϕ ) 1 } L (,1) + (1 + ɛ) ϕ ϕ u n (ϕ ) (t, y) L (,1) y, (3.14) L ( n ) where we have used that ϕ ϕ (ϕ ) < 1. L (,1) From this fact, we also observe that ɛ>may be chosen in such a way that (1 + ɛ) ϕ ϕ (ϕ ) < 1. L (,1) So, recalling now (3.3), we obtain (3.4). Now, using a standard argument and the Rellich theorem, one can show that there is a subsequence, that for simplicity we continue calling u n, such that, for each k, u n is defined on k for n large enough and converges weakly in H ( k ) and strongly in H 1 ( k ). Moreover, if we call u the limit function, it follows from (3.4) and the weak convergence in H, that u satisfies the estimate (3.1). So, it remains only to show that u is the solution of (1.1). Therefore we have to see that u v = fv+ gv v V. Γ 3 It is enough to show that, given v V, u v fv gv k Γ 3 when k. Moreover, by density, we can assume that v W 1, () V.Forn k, we have u v fv gv k Γ 3 = ( u u n ) v + u n v fv gv k k Γ 3 = ( u u n ) v + u n v u n v fv gv k n n \ k Γ 3

13 = G. Acosta et al. / J. Math. Anal. Appl. 31 (5) k ( u u n ) v \ n fv Γ 3 \Γ n 3 gv where we have used that u n is the solution of problem (3.). But, u n v u n H 1 ( n ) v W 1, () n \ k 1 n \ k n \ k u n v, (3.15) and, since u n H 1 ( n ) are uniformly bounded, the last term on the right-hand side of (3.15) can be made smaller than any positive constant by taking k large enough. Then, the proof concludes by using that, for k fixed, ( u u n ) v k when n. Observe that the domains with power type cusps, i.e., ϕ(t) = t α, α>1, are in the class considered here. In fact, ϕ ϕ (ϕ ) = α 1 α. In what follows we will show that the hypothesis hϕ 1 L (, 1) assumed in the previous theorem is not too restrictive and cannot be substantially relaxed. With this goal we consider ϕ(t) = t α, α>1. In this case, the hypothesis is ht α L (, 1) and we will prove that, if the solution of problem (1.1) belongs to H () then, ht α(r 1) r L r (, 1) for any r<. In particular, if h is continuous at t =, it follows that h() =. We will show in the next lemma that, for u H (), u ν is the restriction to Γ 3 of a function in W 1,r (), forr<. Then, the result will follow by using again the results of [6]. Lemma 3.1. Let u H (), and consider v = η u, where 1 η(x,y) := ( αy, x). x + α y Then, (i) v = u ν on Γ 3, (ii) v = u ν on Γ 1, (iii) v W 1,r () for r<. Proof. The first two assertions follow immediately from the fact that η(x,y) agrees with the outward normal on Γ 3 and with the inward normal on Γ 1. To prove (iii), let us call a(x,y) := αy x + α y and b(x,y) := x x + α y.

14 41 G. Acosta et al. / J. Math. Anal. Appl. 31 (5) Then, we have v = a(x,y) u x + b(x,y) u y. (3.16) Since a and b are bounded functions, we have that v L (). Therefore we have to show that the first derivatives of v are in L r () for any r<. Now, a straightforward computation yields and a y = αx, (x + α y ) 3 b x = α y (x + α y ) 3 a x = αxy b, (x + α y ) 3 y = α xy. (x + α y ) 3 Integrating these expressions over one can easily check that a s 1 y C x α s dx, and b s x C a s x C b s y C x sα 3s+α dx, x s(α+1) 3s+α dx, x s(α+1) 3s+α dx. Therefore, a y Ls () if s<α+ 1, (3.17) { b x s if α 3 Ls (),, s< 3 α 1+α if α< 3, (3.18) b y, a { s if α, x Ls (), s< 1+α α if α<. (3.19) Now, let w be any of the first derivatives of u. Then, in view of (3.16), in order to prove (iii) it is enough to see that, for r<, x a a b b w, y w, xw, y w Lr (), and this is the aim of the rest of the proof. We will make use of the imbedding theorem (.5). (α+1) First choose p = (α+1) r(α 1). Since w H 1 (), it follows from (.5) that w L rq (), where q = (α+1) r(α 1) is the dual exponent of p. On the other hand, since r<,

15 G. Acosta et al. / J. Math. Anal. Appl. 31 (5) we have rp < α + 1 and so, we obtain from (3.17) that a y Lrp (). Then, applying the Hölder inequality we obtain that a y w Lr (). In a similar way, using (3.18), (3.19), and again (.5), we can prove that a x b x w L () choosing now p = (α+1) and q = α+1 α 1. Therefore, taking derivatives in the expression (3.16) we obtain v L r (), forr<, concluding the proof. w, b y w, x L () and v In [6], the authors characterize the traces of W 1,r for general cuspidal domains. Applying their results for our case it follows in particular that for v W 1,r () (see [6, p. 18]), 1 v(t,t α ) v(t,) r y t α(r 1) dt C v W 1,r (). (3.) From this estimate and our previous lemma we can easily obtain the following corollary. Corollary 3.1. Let u be the solution of problem (1.1) and h(t) := g(t,t α ).Ifu H () then, 1 h(t) r < tα(r 1) for any r<. (3.1) Proof. Let v defined from u as in Lemma 3.1. Then, we know from that lemma that v W 1,r (). Therefore, (3.1) follows immediately from (3.) and the fact that v = on Γ 1. Acknowledgments We thank Julián Fernández Bonder and Noemí Wolanski for helpful comments. References [1] R.A. Adams, Sobolev Spaces, Academic Press, New York, [] P. Grisvard, Elliptic Problems in Nonsmooth Domains, Pitman, Boston, [3] P. Grisvard, Problemes aux limites dans des domaines avec points de rebroussement, Ann. Fac. Sci. Toulouse 4 (1995) [4] P. Grisvard, Problemes aux limites dans les polygones Mode d emploi (Boundary value problems in plane polygons. Instructions for use), EDF Bull. Direction Études Rech. Sér. C Math. Inform. 1 (1986) [5] A. Khelif, Equations aux derivees partiellles, C. R. Acad. Sci. Paris 87 (1978) [6] V.G. Mazya, Yu.V. Netrusov, V. Poborchi, Boundary values of functions in Sobolev spaces on certain non- Lipschitzian domains, St. Petersburg Math. J. 11 ()

Error estimates for moving least square approximations

Applied Numerical Mathematics 37 (2001) 397 416 Error estimates for moving least square approximations María G. Armentano, Ricardo G. Durán 1 Departamento de Matemática, Facultad de Ciencias Exactas y