Boundary problems for the Ginzburg-Landau equation

Transcription

1 Boundary problems for the Ginzburg-Landau equation David CHIRON Laboratoire Jacques-Louis LIONS, Université Pierre et Marie Curie Paris VI, 4, place Jussieu BC 87, 755 Paris, France chiron@ann.jussieu.fr Abstract We provide a study at the boundary for a class of equation including the Ginzburg- Landau equation as well as the equation of travelling waves for the Gross-Pitaevskii model. We prove Clearing-Out results and an orthogonal anchoring condition of the vortex on the boundary for the Ginzburg-Landau equation with magnetic field. Introduction This paper is devoted to the study at the boundary for the equation for the complex-valued function u in a bounded regular domain Ω R N, N, i log ε cx u = u + ε u u log ε dxu, where c : Ω R N is a bounded lipschitz vector field, d : Ω R + is a lipschitz non negative bounded function and ε > 0 is a small parameter. For instance, the Ginzburg-Landau equation with magnetic field i A/ u = ε u u is of the type considered. Another problem that can be written like equation is the equation for the travelling waves for the Gross-Pitaevskii equation. This equation writes i ψ t + ψ + ψ ψ = 0, 3 where ψ : R R N C. Travelling waves solutions to this equation are solutions of the form possibly rotating the axis ψt, x = Ux Ct, x,..., x N. Equation 3 reads now on U ic U x = U + U U.

2 In dimension N 3, if the propagation speed is small, it is convenient to perform the scaling x ux := U, c := C ε ε log ε in dimension N =, the scaling for the speed is C = ε, and the equation becomes then ic log ε u x = u + ε u u and we expect c to be of order one. This equation is of the type with d 0 and c = c e. If N =, the equation is i u x = u + ε u u, which is also of the considered type with d 0 and c = e log ε. We will be interested in in the asymptotic ε 0 with and we supplement this equation with either the Dirichlet condition div c = 0, 4 u = g ε on Ω, 5 either the Coulomb gauge and the homogeneous Neumann condition u = 0 and c n = 0 on Ω. 6 Furthermore, we will assume that there exists a constant Λ 0 > 0 independent of ε such that c L Ω + c L Ω + d L Ω + d L Ω Λ 0. 7 Finally, we may assume 0 < ε ε 0 Λ 0 / small enough so that To this problem, is associated the energy E ε u := u + a εx u = ε where Ω Λ 0 ε / log ε. 8 a ε x := dxε log ε. Ω e ε u,

3 . Anchoring condition at the boundary Our first result is about the anchoring condition of the vortex on the boundary for the Ginzburg- Landau equation with Neumann condition. Assuming the upper bound E ε u M log ε, for the function u, we expect that the energy of u concentrates at its vortices, which are curves Γ in dimension N = 3. We therefore introduce the measure µ ε := e εu log ε dx, the mass of which is bounded by M by hypothesis. We may then assume, up to a subsequence, that as ε 0, µ ε µ weakly as measures. Moreover, we define the N -dimensional density of µ Θ x := lim inf r 0 µ B r x r N and the geometrical support of µ Σ µ := {x Ω, Θ x > 0}. From Theorem 3 in [BOS], we know that Σ µ is closed in Ω and countably N -rectifiable. Let us assume that the magnetic field H = log ε curl c obeys the London equation H + H = π δ Γ. We may then describe further Σ µ near the boundary. In this regime of energy, Γ consists in a finite number of curves of finite length. Therefore, from London equation, we expect H to be of order one, that is log ε curl c = H, and thus, since c n = 0 on Ω, c 0 if ε 0. Our result is concerned with the anchoring of Σ µ at the boundary, under the only hypothesis c ε 0 in C 0 Ω as ε 0. 9 We note that by hypothesis, c ε is bounded in C 0, Ω, thus we may assume for a subsequence that c ε c in C 0 Ω. We then only assume c = 0. In the case of the Neumann boundary condition 6, we will use the reflection principle. There exists δ > 0 such that the nearest point projection map Π : Ω δ Ω is well-defined in the δ-neighborhood Ω δ of Ω and a smooth fibration. A point x Ω δ may therefore be described by the couple y, t, where y = Πx is its projection on Ω and 3

4 t = ±distx, Ω = ± x Πx, the sign ± being + if x is inside Ω and otherwise. We then define the reflection map φ : W := Ω c Ω δ V := Ω Ω δ, where φx is the point described by the couple y, t if x is described by y, t. We define the varifold Ṽ by Ṽ := V in Ω and Ṽ := φ V in W, that is Ṽ consists in V union its reflection with respect to the boundary Ω. We then consider the manifold M := Ω δ endowed with the smooth riemannian metric g defined by g = g 0 in Ω and g = φ g 0 in W, where g 0 is the euclidian metric on Ω. Theorem. Assume 4 and 7. Let u ε be a family of solutions of -6 satisfying the energy bound E ε u M log ε for a vector field c ε satisfying c ε 0 in C 0 Ω as ε 0. Then, the varifold VΣ µ, Θ is stationary in Ω. Moreover, Ṽ is a stationary varifold in M, g. Remark. In the case where Ω is locally the half-plane R N + = R + R N, then the theorem states that Ṽ is a stationary varifold in locally RN for the usual metric. This Theorem says that, in some weak sense, the union of the varifold V and its symmetric with respect to the boundary is smooth, that is V must meet the boundary Ω orthogonally. Since V is not in general a smooth curve, we may only use a weak formulation of this orthogonality. However, if V is a smooth curve up to the boundary, then Theorem states that, denoting τ the tangent unit vector to V, τ = ±n on Ω. The fact that the vortex must meet the boundary orthogonally can be found in the literature. For instance, in [CH], Chapman and Heron considered a domain which is the half-plane in R 3 {z < 0} and a straight line vortex Γ, defined by y = 0, x = mz 0 for a 0 m < +, meeting the boundary {z = 0} at 0. Using the London equation and the boundary conditions for the magnetic field, they proved, computing the propagation speed of the vortex at 0, that the coefficient m must be zero, for otherwise, the propagation speed would be infinite. However, their computation does not exclude the case of two vortices, defined by y = 0, x = mz 0 and y = 0, x = mz 0, since in that case, the propagation is, due to the symmetry, zero. Our Theorem states that there can not be another possibility involving two such coplanar straight lines vortices, that is vortices defined by y = 0, x = mz 0 and y = 0, x = m z 0 with 0 m, m < + and m m can not hold. Our Theorem even states that if we have two straight line vortices in the half plane {z < 0} meeting at 0, then they must be in a plane orthogonal to {z = 0}. At the opposite of [CH], our approach is based on equation only, whereas the London equation is the limit equation for the current see below, which is the second equation of the Ginzburg-Landau equation with magnetic field. Remark. In the case c ε c 0 as ε 0, by Theorem 3 in [BOS], we know that the varifold V satisfies inside the domain Ω the curvature equation H = c dj, 0 dµ 4

5 where H is the generalized mean curvature of V and, up to a subsequence, J is a weak limit of the jacobian Ju ε and refers to Hodge duality. Theorem generalizes then in the form see Section 5 H = c d J, d µ where c, J and µ are the extensions of c, J and µ by reflection, and H the generalized mean curvature of Ṽ in M, g. Equation also implies in somme weak sense that the vortex must be orthogonal to the boundary. We show in the figure below some non-admissible and admissible configurations for a vortex we assume regular for instance d J dµ =, in the case where c can be non zero. In Theorem, the stationarity of V thus of Ṽ inside the domain Ω is a direct consequence of 0 with c = 0. Ω Ω non admissible configuration admissible configuration non admissible configuration admissible configuration Non-admissible and admissible configurations a one vortex b two vortices.. Monotonicity and Clearing-Out Theorems The second result is a Clearing-Out theorem for this equation. This result is also called η- compactness in [R], [LR] and η-ellipticity Lemma in [BBO]. We recall the definition of the scaled energy, for a map u : Ω C, Ẽ ε u, x 0, r := r N E εu, Ω B r x 0 = = r N Ω B rx 0 u r N Ω B rx 0 + a εx u 4ε e ε u and finally set ˇB r x 0 := Ω B r x 0 and r ε := ε µ log ε /N, where µ 0, is a constant depending only on N. We can now state our Clearing-Out result for the Dirichlet boundary condition 5. First, we make the following standard hypothesis for the boundary datum g ε C and g ε C ε. 5

6 We also introduce the following quantity, for 0 r r and 0 < ν, r Tε ν x 0, r, r := e r r N +ν ε g ε dr, Ω B rx 0 where e ε g ε := g ε + a εx g ε. ε Theorem. Assume 4 and 7. Let u be a solution of -5 on Ω, with g ε satisfying. Let x 0 Ω, 0 < ν, rε / r minr, + Λ 0 /ν, where R > 0 depends only on Ω, and σ > 0 be given. Then, there exist constants η > 0 and ε 0 > 0 depending on σ, ν, N, Ω, Λ 0 and the constant C in but independent of u and g ε such that, for ε ε 0, if and then T ν ε x 0, ε, r / ε η, 3 T ν ε x 0, r ε, r η log ε, 4 Ẽ ε u, x 0, r η log ε, 5 ux 0 σ. Note that one may take different ν s for 3 and 4, but we can always assume they are equal. Remark 3. We emphasize that the quantity involved in Tε ν the scaled energy for g ε, namely e r N 3 ε g ε. Ω B rx 0 is related to the decay as r 0 of We make an hypothesis at small scales r rε / for 3, which is the suitable assumption for g ε is smooth enough and of modulus one, and an hypothesis at large scales r can be of order one for 4, which is an hypothesis on g ε similar to the one made on u for 5. Remark 4. If there exist δ 0, ] and a constant M > 0 such that, for 0 < ρ r, ρ N 3 g ε + a εx g ε Mρ δ, 6 ε Ω B ρx 0 then, for 0 < r r r and 0 < ν < minδ,, T ν ε x 0, r, r M r r r δ ν dr M rδ ν δ ν. Therefore, the hypothesis 3 is verified for ε sufficiently small depending on ν, δ and M, and 4 is verified for r and ε sufficiently small depending on M. One may even consider for 6 a constant M log ε. In particular, if g ε is uniformly M-lipschitzian of modulus on Ω B r x 0, then by 8 ρ N 3 Ω B ρx 0 g ε + a εx g ε ε CM ρ + Cρ Λ 0ε log ε 4 CM + 4 ρ, where C depends only on Ω, thus 6 is satisfied with δ =. 6

7 Remark 5. We would like to emphasize that we do not impose g ε near the point x 0. The condition 3 for r ε however implies g ε x 0 if x 0 is at distance less than ε from the boundary. We enlarge the conditions on the boundary datum already used in [LR] and [BBO]. In this case, g ε is a suitable smooth approximation of a map of modulus smooth outside a finite union of smooth submanifolds of Ω of dimension N 3. The Clearing-Out Theorem is then stated far away from these submanifolds. Our Clearing-Out result for the the Neumann boundary condition 6 is the following. Theorem 3. Assume 4 and 7. Let u be a solution of -6 on Ω, x 0 Ω and σ > 0 be given, and let rε / r minr, / + Λ 0, where R > 0 depends only on Ω. There exist constants η > 0 and ε 0 > 0, depending on N, Ω, σ and Λ 0 but independent of u, such that, for 0 < ε < ε 0, if Ẽ ε u, x 0, r η log ε, then ux 0 σ. Remark 6. These theorems do not give compactness on the solution u as ε 0. For the Dirichlet problem, the compactness properties follow from hypothesis on the whole boundary see for instance [BBBO] for compactness in W,p, p < N/N. These results rely strongly on monotonicity formulas of the scaled energy of solutions of. For the Dirichlet problem, the result is the following. Proposition. Assume 4 and 7. Let ν 0, ] and g ε satisfying. There exist R > 0, depending only on Ω, C > 0, depending on Ω, ν and the constant C in only, and β > 0 depending on N only such that, if u is a solution of -5, 0 < r minr, + Λ 0 /ν and x 0 Ω, then for any 0 < θ < /, we have Ẽ ε x 0, θr C Ẽε r + Tε ν x 0, θr, r + Λ 0 ε. β 7 For the Neumann problem, the result is the following. Proposition. Assume 4 and 7. There exist β > 0, R > 0 and C > 0 depending on Ω and N only such that, if u is a solution of -6, x 0 Ω and 0 < r minr, / + Λ 0, then for any 0 < θ < /, Ẽ ε x 0, θr C Ẽε r + Λ 0 ε β. 8.3 Models involving equation We would like to discuss some models involving equation, as well as the boundary conditions. Note that can be rewritten as i log ε cx u = u + ε ua εx u, 9 where a ε x := dxε log ε. 7

8 When div c = 0, it is also equivalent to where i log ε c u + ε ub εx u = 0, 0 b ε x := a ε x + ε log ε cx. 4 If c log ε = A and d = c /4 with div A = 0, then this equation is the first equation in the Ginzburg-Landau system of superconductivity, namely i A/ u = ε u u. The second equation for the induced magnetic field H := log ε curl c in dimension N = 3 is iu, A u = curl H + curl H ex = log ε curl c + curl H ex, where H ex is the imposed magnetic field and A = i log ε c is the covariant derivative. Equations - are the Euler-Lagrange equations of the Ginzurg-Landau functional Ju, c = u i log ε cu + u + H Hex. ε In this case, the natural boundary condition is Ω The functional J is gauge-invariant, that is, if ψ H Ω, then n iu, A u = 0. 3 Jue iψ, c + ψ = Ju, c. We can freeze the gauge-invariance by choosing, for instance, the Coulomb gauge { div c = 0 in Ω, c n = 0 on Ω. In this case, the boundary condition 3 becomes with the Coulomb gauge u = 0 on Ω. This justifies the study at the boundary with the homogeneous Neumann condition 6. Writing in the form has the advantage to include in the same analysis the equation already mentioned in dimension N 3 ic log ε u = u + ε u u related to the travelling waves for the Gross-Pitaevskii equation with small speeds. This equation is used as a model for superfluidity, nonlinear optics and Bose-Einstein condensates. It is close to the Ginzburg-Landau equation and a similar asymptotic analysis as ε 0 can be carried out for this equation. 8

9 We would like to mention that in [C], we have been interested in travelling vortex helices to the Gross-Pitaevskii equation. In this case, we approximate the problem on cylinders of axis x, and we impose the Dirichlet boundary condition u = e iθ on the lateral surface of the cylinder that forces the solution u to have a degree one in the plane orthogonal to x. Therefore, this study required a Dirichlet boundary condition, whereas the Neumann condition is the natural one for the gauge-invariant functional J. We refer to [BOS] for the generalization of the analysis of equation inside the domain see Theorems and also 3 there. The proofs of Theorems and 3 will follow the same lines as in appendix A of [BOS]. We also mention the study of minimizers in dimension 3 for the U-Higgs model in [R]. For the study near the boundary for the Ginzburg-Landau functional without magnetic field d = c 0 and Dirichlet datum smooth outside a finite union of smooth N -dimensional submanifolds of Ω, we refer to [LR] for minimizers in dimension N 3 and [BBO] for the general case. The paper is organized as follows. In Section, we state and prove two lemmas concerning basic L bounds for u and u. Section 3 is devoted to the monotonicity formulas and the proof of Propositions and. In Section 4, we prove the Clearing-Out Theorems and 3, while the result about the orthogonal anchoring of the vortex on the boundary of Theorem is given in Section 5. Basic L bounds We first state two lemmas related to L bounds for u and u. Dirichlet problem. The first one concerns the Lemma. Assume 4 and 7. Let u be a solution of -5, with g ε satisfying. Then, u max g ε, b ε C and u C ε for a constant C depending on Ω, Λ 0 and the constant C in only. The second one is for the Neumann problem. Lemma. Assume 4 and 7. Let u H L 4 Ω be a solution of -6. Then, u C,α Ω for some α > 0 and where K depends on Ω and Λ 0. u b ε and u K ε, In particular, for the Ginzburg-Landau functional with magnetic field where d c /4, thus b ε, Lemma states that u. 9

10 . Proof of Lemma It is close to the proof of Lemma 3 in [BOS]. From, we deduce u = u, u + u = ε u a ε u + log ε u, i c u + u ε u a ε u c log ε u u + u = ε u a ε u c c + u u log ε u log ε ε u b ε u. Therefore, the function w := max g ε, b ε u satisfies and by the maximum principle, we deduce w + ε u w 0 in Ω, w 0 w 0 in Ω. on Ω, Concerning the bound on the gradient, we consider the scaled map ûx := uεx, which satisfies û + ûâ ε û = iε log ε ĉ û in Ω ε, û = g ε εx on Ω ε, where ĉx := cεx and ˆdx := dεx. By standard elliptic estimates see [GT], since g ε εx C by hypothesis, û L C for a constant C depending on Ω and Λ 0, and the estimate for u is obtained by scaling back.. Proof of Lemma The proof of the C,α regularity of u uses a standard bootstrap argument and the fact that the coefficients c,d are lipschitzian. Concerning the L bounds, as in the proof of Lemma, we find that w := b ε u satisfies w + ε u w 0 in Ω, w = 0 on Ω, since w u = u, = 0 by 6. We then adapt an idea of [F], used in the proof of universal bounds for travelling waves for the Gross-Pitaevskii equation. For f : Ω R, we decompose f = f + f in its nonpositive and nonnegative part f +, f 0, f + f = 0. Since w and w are Hölder continuous, we have by Kato s inequality see [B], [K] w = w + sign + w w u ε sign+ w w. 4 0

11 Therefore if w 0, the right-hand side of 4 is zero, and if w < 0, then u > b ε 0, w b ε ε w 0. 5 From 5, it is clear that we can not have w cte > 0, since b ε > 0. As a consequence, in view of 5, we deduce by the strong maximum principle Ω is connected that either w = cte, and then this constant must be zero, either w achieves its maximum only on the boundary, for instance at x 0 Ω. Assuming w 0, we have w x 0 > 0. In particular, since w C,α Ω, in a neighborhood of x 0 in Ω, w = w > 0 is C,α. It is then well-known that in this case, since w > 0 in this neighborhood, we have by 5, w = w > 0. This contradicts the boundary condition w = 0. Therefore, w 0 and w = w + 0, that is u b ε, which finishes the proof for the L bound. For the estimate on the gradient, we consider the scaled map ûx := uεx, which satisfies By standard elliptic estimates, we have û + ûâ ε û = iε log ε ĉ û in Ω ε, û = 0 on Ω ε. û L C and we conclude by scaling back. 3 Monotonicity formulas at the boundary As already mentioned, we follow the lines of the proof of Theorem of [BOS] given in appendix A there. When this will not lead to a confusion, we will denote Ẽεu, x 0, r and ˇB r x 0 by Ẽεx 0, r, or even Ẽεr, and ˇB r. We first recall the Pohozaev identity. Lemma 3.. Let u be a solution of on Ω, then for any z 0 R N and ω Ω, N u + N a ω 4ε ε x u N log ε ω ω Ju, c i xξ i x z 0 i [ = x z 0 ne ε u u ] ω, x z log ε 0 u + a ε x u x z 0 dx. ω Here, ξ i stands for the -form ξ i := N x j dx i dx j. j i

12 3. The Dirichlet problem In this subsection, we assume that u is a solution to the Dirichlet problem -5. We will denote, for r > 0, G ε x 0, r := r N g ε + a εx g ε = e Ω B rx 0 ε r N ε g ε. Ω B rx 0 Note that G ε is not the scaled energy for g ε. We fix 0 < ν. In the sequel, C denotes a constant depending on N, Ω and ν only. Lemma 3.. Let u be a solution of -5 on Ω, then for r > 0 and x 0 Ω d dr Ẽεx 0, r = u + r N Ω B rx 0 r N B rx 0 Ω x x 0 n u + a ε x u r N ˇB rx 0 ε N log ε Ju, c rn i xξ i x x 0 6 ˇB rx 0 i log ε a r N ε x u x x 0 dx ˇB rx 0 x x r N 0 ne ε g ε u, x x 0 g ε, B rx 0 Ω where x x 0 is the orthogonal projection of x x 0 on the tangent hyperplane to Ω at x. Proof. Up to a translation, we may assume x 0 = 0. One has dẽε dr = N r N E εr + = N r N + r N r N Ω Br u u + N ˇB r 4ε a ε x u + ˇBr ε r N + a εx u 4ε a ε x u ˇBr Ω B r e ε u. We use Lemma 3. with ω = ˇB r x 0 and z 0 = x 0 for the first term and then split ˇB r into ˇB r = B r Ω Ω B r

13 to obtain, since x n = r on Ω B r, dẽε dr = a ε x u + e r N ˇBr ε r N ε u Ω B r N log ε Ju, c rn i xξ i x ˇB r i log ε a r N ε x u x dx ˇB r x ne r N ε u u ˇB r, x u, = a ε x u + u r N ˇBr ε r N Ω B r N log ε Ju, c rn i xξ i x ˇB r i log ε a r N ε x u x dx 7 ˇB r x ne r N ε u u, x u. B r Ω It suffices then to write, on B r Ω, e ε u = u + g ε + a εx g ε x = u + e 4ε ε g ε and x = x nn + x, thus u, x u = x n u + u, x g ε, to finally deduce x ne ε u u, x u = x ne ε g ε x n u u, x g ε. Inserting this in the last integral yields 6. We note in equality 6 the last term involving the normal derivative of u. The next lemma provides an estimate for this term. Lemma 3.3. Control of the normal derivative. Let u be a solution of -5. There exist C and R depending only on Ω such that, for all x 0 Ω and 0 < r < R, there exists z 0 ˇB r x 0 such that r u C e ε u + r e ε Ω B rx 0 g ε + log ε ˇBrx0 Ω B rx 0 ω Ju, c i xξ i x z 0 i log ε + a ε u x z 0 d, 8 where ω ˇB r x 0 depends on u, x 0 and r. ω 3

14 The proof is, as in [LR], based on a Pohozaev identity at a point z 0 around which ˇB r x 0 is strictly starshaped. However, we will not use a good extension of g inside the domain Ω as in [LR] see Lemma II.5 there, since it requires a strong regularity hypothesis for instance, g ε bounded in C, around x 0 and will not enable us to treat the case of the monotonicity at large scale see Remark 3. below. Proof. For simplicity, assume first that Ω is locally the half-plane R N + = R N {0}. We also assume up to a translation x 0 = 0,..., 0, a. We assume first that 0 a r/4, that is x 0 is close to the boundary Ω. We define y := 0,..., 0, b for b a and ρ := r a + b /. The intersection of R N + and the balls B ρy and B r x 0 is the ball in R N + = RN {0} centered at 0 and of radius r a /. By averaging, there exists r r, 9r/8 such that, for ρ = r and b = r r + a /, e ε u C e ε u. 9 R N + B r y r ˇB rx 0 ρ b a z 0 X N r Moreover, since r 9r/8 and 0 a < r/4, b r /64, then r + b r r/8 r/3. We then set z 0 := 0,..., 0, r +b ˇB r x 0 and easily see that, since a r/4, ω := ˇB r x 0 B r y is strictly starshaped around z 0, that is there exists α > 0 such that x z 0 n αr. 30 Next, we apply the Pohozaev identity of Lemma 3. with z 0, x 0 and ω to obtain N u + N a ω 4ε ε x u N log ε ω ω Ju, c i xξ i x z 0 i [ = x z 0 ne ε u u ] ω, x z log ε 0 u + a ε x u x z 0 dx. ω As in the proof of Lemma 3., we write x z 0 = x z 0 nn + x z 0 to deduce 4

15 C e ε u + C log ε ω ω ω Ju, i c i xξ i x z 0 + log ε a ε x u x z 0 dx x z 0 n u x z 0 n u + u, x z 0 u x z 0 n a ε u. ε In the last integral, in view of the starshapedness assumption 30, the third term has an absolute value x z 0 n u + Cr u. Thus, using the starshapedness assumption 30 and splitting ω into Ω B r x 0 and Ω B r y, C e ε u + C log ε ω ω Ju, c i xξ i x z 0 + log ε a ε x u x z 0 dx i ω x z 0 n u Cr u + a ε u ω ε αr u Cr e ε g ε Cr e ε u. ω Ω B rx 0 ω Ω B r y We conclude estimating the last term by 9. We assume now that a r/4, that is x 0 is far enough from the boundary. Then, we have n = e N and, if x R N +, then x N = 0 and x x 0 n = a x N = a r 4. 3 In other words, ˇBr x 0 is strictly starshaped around x 0. We conclude then as in the previous case. This concludes the proof in the case Ω is locally an half-plane. For the general case, we use local charts and note that the starshapedness assumptions 30 or 3 will still be true at least for r < R depending on Ω. We then prove a first monotonicity formula useful for small scales, which is the boundary version of Lemma 4 in [BOS]. Note that the presence of the term r ν in front of Λ and r ν in front of G ε r is specific to the Dirichlet condition. Lemma 3.4. Monotonicity at small scales. There exist C and 0 < R <, depending only on ν and Ω, such that for any solution u of -5 on Ω, denoting Λ := C + Λ 0 log ε and Q := CΛ 0 ε log ε, and any x 0 Ω and 0 < r minr, Λ /ν, we have d expλr ν Ẽεr + Q dr Λ a ε x u + r N Ω B rx 0 ε + u C G εr. r N r ν B rx 0 Ω r N B rx 0 Ω x n Proof. First, we note that, if 0 < r RΩ is sufficiently small, then for all x ˇB r x 0, u x x 0 n x x 0 n Cr, 3 5

16 where C depends on Ω only. This fact was already used in [LR] Lemma II.5. We recall the argument. One may assume that, for r < R sufficiently small, ˇBr is the uppergraph of ψ : B 0 R N R and that ψ0 = ψ0 = 0, so that the tangent hyperplane at Ω at ψ0 = 0 is R N {0}. Therefore, the outward normal writes n = + ψ / e N N i= i ψ e i, where e i i N is the canonical basis of R N. In order to prove 3, it suffices then to prove x x 0 e N Cr, since ψ0 = 0, so ψ Cr. This last inequality is a direct consequence of the fact that T 0 Ω = R N {0} and Ω is locally the uppergraph of ψ. We now turn to the proof of Lemma 3.4. Once more, we assume x 0 = 0. We have to estimate each term on the right hand side of 6. For the fourth one, we use the rough estimate for the jacobian Jux C ux and ξ i x Cr for all x B r, which yields N r log ε N Ju, ˇB rx 0 i c i xξ i x C r c log ε ux N ˇB r CΛ 0 log ε Ẽεr. 33 For the fifth one, by Cauchy-Schwarz, log ε a r N ε x u x dx ˇB r Concerning the second term, we have by 3 r N x n u r N B r Ω C r Λ 0ε log ε a ε x u N ˇB r ε C r Λ 0ε log ε N/ a ε x u / r N ˇB r ε CrΛ 0 ε log ε Ẽ ε r / Ẽεr + Cr Λ 0 ε log ε B r Ω x n u C u. 35 r N 3 B r Ω We use 8 of Lemma 3.3 to estimate r u C e ε u + log ε Ω B r ˇBr ω Ju, c i xξ i x z 0 i + r e Ω Br ε g log ε ε + a ε u x z 0 d. 36 In 36, we estimate the second term as in 33 and the fourth one as in 34 since ω ˇB r to obtain u C Ẽε r + rg r N 3 ε r + Λ 0 log ε Ẽεr + Cr Λ 0 Ω B r ε log ε ω

17 Inserting 37 in 35 yields x n u r N B r Ω r N B r Ω x n u where Λ = C + Λ 0 log ε. For the last term in 6, we have first x ne r N ε g ε G ε r and since x r, r N B r Ω B r Ω u, x g ε r N r ν νr N ΛẼεr CrG ε r Cr Λ 0ε log ε 4, 38 B r Ω B r Ω u g ε e ε g ε + ν rν r N 3 B r Ω u which yields, using 37, the estimate of the last term in 6 x ne r N ε g ε u Ω B r, x g ε νλ r Ẽεr + C G εr + Cr ν Λ ν r ν 0 ε log ε Inserting estimates 33, 34, 35, 38 and 39 into 6 gives dẽε dr r N + r N B r Ω Ω B r x n u + a ε x u r N ˇBr ε u νλ r Ẽεr C G εr Cr ν Λ ν r ν 0 ε log ε 4, from which we infer for r Λ /ν note that r ν r ν since 0 < ν d expλr ν Ẽε dr = expλr ν dẽε dr + νλ r ν expλrν Ẽεr x n u + a ε x u r N B r Ω r N ˇBr ε + u C expλr ν G εr C expλr ν r ν Λ r N Ω B r r ν 0 ε log ε 4 x n u + a ε x u r N B r Ω r N ˇBr ε + u C G εr d Q r N r ν dr Λ expλrν Ω B r and the proof is complete. The previous monotonicity formula is useful for r C + Λ 0 log ε /ν = Λ /ν, that is r small if Λ 0 > 0. As in [BOS], the monotonicity formula for large scales will be a consequence of the refined estimates on jacobians as in [JS]. 7

18 Lemma 3.5. Jerrard & Soner. Assume u Hloc Ω, C, ϕ C0, c Ω, Λ R N. There exist K and α 0,, depending only on N and Ω, such that, denoting K := Suppϕ, Ju, ϕ K log ε ϕ E ε u, K + Kε α dϕ + K + E ε u, K. 40 Ω The advantage of this estimate is the factor log ε dividing the energy. Note that this lemma is stated with the energy E ε and not the usual Ginzburg-Landau energy used in [JS] corresponding to d 0, but these two energies are close with our hypothesis, since one may infer from 0 d Λ 0 that u a ε u Ω Λ ε ε 0 ε log ε 4 u + ε. Ω ε Ω Ω Remark 3.. We emphasize that this is the Pohozaev identity we used for Lemma 3.3 which provides the control of the normal derivative using the estimate of Jerrard and Soner of Lemma 3.5. The extension procedure of [LR] would have led to a term log ε ˇB r N ic k x k u, x l l gx, k,l= where g is a good extension inside Ω of g, and this term would be difficult to handle since it is not a jacobian if g u, thus we do not expect a compensation property. For our purpose, we will need for our study a boundary version of this result, in order to have an estimate close to 40 for a ϕ having a support intersecting Ω. This will be done by a standard extension of g in a neighborhood of Ω as in [BO]. Nevertheless, in order to apply Lemma 3.5, we need a map ϕ which has compact support say in ˇB r x 0, hence, as in [BOS], we adapt the definition of the energy temporarily. We define a cut-off function f : R + R + R + For x 0 Ω and r > 0, we then set fa, b := Ē ε x 0, r := r N if b a, b/a if a b a, 0 if a b. ˇB r x 0 e ε ufr, x x 0 dx. An integration by parts shows that, for any F 0 measurable, F xfr, x = F x dt. 4 ˇB r ˇB rt This formula is the link between the usual scaled energy Ẽε and Ēε. 8

19 Lemma 3.6. Assume u satisfies -5, x 0 Ω and r > 0. Then, d dr Ēεx 0, r = e r N ε ufr, x x 0 + t u Ω B r x 0 r N Ω B trx 0 + r N B trx 0 Ω x x 0 n u dt + a ε x u fr, x x r N ˇB r x 0 ε 0 N log ε Ju, c rn i xξ i x x 0 fr, x x 0 4 ˇB r x 0 i log ε a r N ε x u x x 0 dxfr, x x 0 ˇB r x 0 x x r N 0 ne ε g ε fr, x x 0 + r N B r x 0 Ω B r x 0 Ω u, x x 0 g ε fr, x x 0. Proof. We still assume x 0 = 0. First, one has dēε dr = N e r N ε ufr, x + e ˇB r r N ε ufr, x Ω B r + e r N ε u r fr, x. 43 ˇB r In 43, we have then for the first term by 4 N e r N ε ufr, x = N ˇB r r N and for the third term e r N ε u r fr, x = ˇB r r N thus dēε dr = r N + r r e ε ufr, x Ω ˇB r t N N tr N ˇB rt e ε u dt Ω B ρ e ε u ρ r dρ = r N ˇB rt e ε u + e rt N ε u dt. Ω B rt t e ε u, Ω B tr The term between parenthesis is Ẽ εrt, hence inserting formula 6 for rt and using formula 4 we are led to the conclusion. Lemma 3.7. Monotonicity at large scales. There exist constants R > 0, depending only on Ω, and C, depending only on Ω and ν, such that, for any x 0 Ω, r ε r minr, + Λ 0 /ν and u solution to -5, we have for every r ε s < r, Ē ε s CĒεr + T ν ε s, r + Λ 0 ε β. 9

20 Proof. We assume x 0 = 0 and we estimate each term on the right-hand side of 4. For the sixth term, we have as for 34 log ε a r N ε x u x dxfr, x Ẽεr + Cr Λ 0ε log ε ˇB r Concerning the fifth one, we proceed as in [BO] Proposition. there. First, we extend u outside the domain. There exists δ 0 > 0 such that the nearest point projection Π is well-defined and is a smooth fibration from the δ 0 -neighborhood Ω δ0 of Ω onto Ω, inducing smooth diffeomorphisms Π t : Ω t Ω 0 t δ 0. We extend u in a map ũ in Ω δ0 by setting ũ = u Π on Ω δ0 \ Ω. We extend in the same way the c i s i N and d on Ω δ0. Finally, we extend the ξ i s as in [BO], that is we write on Ω ξ i = ξ i + ξ i N, where ξ i and ξ i N are respectively the tangential and the normal components of ξ i on Ω see the Appendix of [BBO] for notations, and then we set, if dx, Ω = t, where Π t N r N log ε ξ i x = Π t ξ i x + ξ i N Πx, denotes the inverse of the diffeomorphism Π t. Next, we write Ju, c i xξ i xfr, x C ˇB r r log ε Jũ, ϕ + N i B r B r \Ω Jũ, ϕ, where ϕx := i c i x ξ i xfr, x. The first integral is estimated with Lemma 3.5. Since ϕ Cc 0, B r, Λ R N and ϕ CΛ 0 r, dϕ CΛ 0, 45 we obtain log ε Jũ, ϕ CΛ r N 0 Ẽ ε r + C B r r Λ 0ε α log ε + E N ε r. For the second integral, we have as in [BO], using the coarea formula and with d = dist., Ω verifying d =, Jũ, ϕ = d Π Jg ε, ϕ B r \Ω = B r \Ω r 0 dt CΩr d t B r Ω fr, x c i x Π t Jg ε, Π t ξ i i Jg ε, ϕ. If N 3 if N =, Jg ε 0, to estimate the last integral, we also invoke Jerrard-Soner s result of Lemma 3.5 with this time the smooth manifold B r Ω of dimension N and ϕ Cc 0, B r Ω, Λ R N satisfying also 45, thus log ε Jũ, ϕ CΛ r N 0 rg ε r + C r Λ 0ε α log ε + r N G N ε r. B r \Ω 0

21 We therefore deduce the estimate for the fifth term in 4 r r log ε Ju, c N i xξ i xfr, x ˇB r i CΛ 0 Ẽ ε r + C r Λ 0ε α log ε + E N ε r + CΛ 0 rg ε r + C r Λ 0ε α log ε + r N G N ε r CΛ 0 + εα log ε r Ēε4r + CΛ 0 rg ε r + εα log ε ε α log ε + CΛ r r N For the seventh term, we have clearly x ne r N ε g εfr, x N G ε r. 47 B r Ω We estimate also the normal derivative as for 37, using estimates similar to 44 and 46, u C + Λ r N εα log ε Ēε 4r 48 Ω B r r + C + Λ 0 r + εα log ε G ε r + CΛ 0 r r ε log ε 4 ε α log ε + CΛ 0. r N We infer from 48 the estimate for the third term in 4 as for 38 using 3 r N C B trx 0 Ω x n u dt r N Ēε 4r C + Λ 0 + εα log ε r B r Ω x n u + rλ 0 + εα log ε r G ε r Λ 0ε log ε 4 Λ 0 ε α log ε r N. 49 For the last term in 4, we obtain as for 39 and using 48 u r N B r Ω, x g ε fr, x C + Λ 0 + εα log ε ν Ē ε 4r r 4r + C + Λ ν 0 r + εα log ε Gε r 50 r r ν + CΛ 0 ε log ε 4 r ν ε α log ε + CΛ 0 r. N +ν Combining estimates 44, 46, 47, 49 and 50 with 4 yields dēε dr t u + r N Ω B tr r N + a ε x u fr, x C r N ˇBr ε C + Λ 0 r + εα log ε r B tr Ω x n u dt + Λ 0 + εα log ε r ν Ē ε 4r 5 4r ν Gε r CΛ r ν 0 ε log ε 4 r ν ε α log ε CΛ 0 r. N +ν

22 Now, we assume r large enough so that r ν N ε α log ε r N ε α log ε hence r ε α log ε tends to zero with ε. We therefore assume r ε β ε α log ε /N =: r ε, 5 where 0 < β < α is fixed for instance, β = α/, and set µ := α β > 0, so that 5 implies, for r ε r minr, + Λ 0 /ν with dēε dr Ar C + Λ 0 Ēε4r 4r ν + G εr r ν + Λ 0 ε β, 53 Ar := r N t u + Ω B tr a ε x u ˇBr + r N In particular, since Ar 0, for r r ε, r N B tr Ω x n u dt ε fr, x. 54 dēε dr C + Λ 0 Ēε4r 4r ν + G εr r ν + Λ 0 ε β. 55 To conclude the proof, we will need the following discrete Gronwall inequality. Lemma 3.8. Discrete Gronwall inequality. Let 0 < s < 4s < s, f : [s, s ] R + be continuous and assume h : [s, s ] R + is continuously differentiable and satisfies { hs θ N hθs if θ [, s /s ], s [s, s ] and θs [s, s ], h s C h4s 56 fs for all s [s 4s ν, s /4], for constants C > 0 and ν 0, ]. Then, for all s s < t s, where t/4 hs 4 N expcλt ν ht + λ := s 4 ν 4 ν. fr dr expcλt ν, 57 Proof. We proceed by induction. Let s s < t < s. Assume t/4 s t. Then, by 56, Assume that for some k N, it holds k hs 4 N ht + 3Ctν 4 + νi+ i= hs 4 N ht. t/4 k s k fρ dρ + j= t/4 j t/4 j fρ dρ for all t 4 k s t 4 k. If t 4 k+ s t 4 k, then, by 56 and using the fact that k i=j + 3Ctν 4 +νi t/4 k s 4r ν dr t 4 k s t/4 k ν 3t 4 k+ 4 νk t ν = 3tν 4 kν+,

23 we obtain hs ht/4 k + C t/4 k s k + j= t/4 k s k fρ dρ + 4 N ht t/4 j t/4 j fρ dρ h4r t/4 k dr + fρ dρ 4r ν s i= + 3Ctν 4 νi+ + k + 3Ctν 4 +νi i=j + 3Ctν t/4k fρ dρ + 4νk+ t/4 k 4 N ht k + j= 4 N ht k i= t/4 j + 3Ctν 4 νi+ + k j= t/4 k s t/4 j fρ dρ + 3Ctν k t/4 4 j kν+ k i= + 3Ctν 4 νi+ + t/4 k s t/4 k t/4 k fρ dρ + 3Ctν k 4 νk+ 4N ht t/4 j fρ dρ k fρ dρ + + 3Ctν 4 kν+ i=j fρ dρ + + 3Ctν 4 +νi k j= i= + 3Ctν 4 +νi i=j t/4 k t/4 j + 3Ctν 4 νi+ t/4 k fρ dρ t/4 j fρ dρ The conclusion follows then from the inequality, valid for all m N, m i= by definition of λ. Indeed, we have + 3Ctν 4 νi+ exp 3 4 Ctν 4 νi expcλt ν, i= k i=j + 3Ctν 4. +νi hs 4 N ht expcλt ν + and the proof is complete. t/4 k s fρ dρ + expcλt ν t/4 4 N ht expcλt ν + expcλt ν fρ dρ s k j= t/4 j t/4 j fρ dρ To conclude the proof of Lemma 3.7, we apply Lemma 3.8 with s = r ε, s = r, h = Ēε, fr = Cr ν G ε r + Λ 0 ε β, C = C + Λ 0. Note that λs ν = + Λ 0 r ν. The first hypothesis in 56 is easily verified for the modified scaled energy Ēε and the second one is 55. We then infer that, for every r ε s < r minr, + Λ 0 /ν, Ē ε s CĒεr + T ν ε s, r + Λ 0ε β. This finishes the proof of Lemma The Neumann problem In this subsection, we point out the modifications to make in order to handle the Neumann case, that is for solutions u to -6. In the sequel, C is a constant depending only on Ω and N. 3

24 Lemma 3.9. Let u be a solution of -6 in Ω, then for r > 0 and x 0 Ω d dr Ẽεx 0, r = u + a ε x u r N Ω B rx 0 r N ˇB rx 0 ε N log ε Ju, c rn i xξ i x x 0 58 ˇB rx 0 i log ε a r N ε x u x x 0 dx ˇB rx 0 x x r N 0 ne ε u. B rx 0 Ω Proof. Assuming x 0 = 0, we still have formula 7. It suffices to use the Neumann condition 6 to obtain 58, since the last term in the last integral in 7 is 0. This time, the last term in equality 58 involves the energy on the boundary of u. Note that this term is not so bad since the term x x 0 n is expected to be of order r if x 0 is close enough to the boundary. The next lemma, analoguous to Lemma 3.3, provides an estimate for this term. Lemma 3.0. Control of the boundary energy. Let u be a solution of -6. There exist C and 0 < R depending only on Ω such that, for all x 0 Ω and 0 < r < R, there exists z 0 ˇB r x 0 such that r Ω B rx 0 e ε u C e ε u + log ε ˇBrx0 where ω ˇB r x 0 depends on u, x 0 and r. + log ε ω Ju, c i xξ i x z 0 i a ε u x z 0 d, 59 Proof. The proof begins as for Lemma 3.3, that is assuming first that Ω is locally the half-plane R N + = R N {0}, that a r/4 and exhibiting by averaging y and r r, 9r/8 such that e ε u C e ε u. 60 R N + B r y r ˇB rx 0 We also have for an α > 0, ω x z 0 n αr. 6 We also apply the Pohozaev identity of Lemma 3. with z 0, x 0 and ω and use the Neumann condition 6 to obtain C e ε u + C log ε ˇB rx 0 ω Ju, c i xξ i x z 0 i + log ε a ε x u x z 0 dx x z 0 ne ε u Cr e ε u. Ω B r ω Ω B r y 4

25 The last integral is estimated by 60 and for the before last integral, we use the starshapedness assumption 6 to obtain C e ε u + C log ε ˇB rx 0 ω Ju, c i xξ i x z 0 i + C log ε a ε x u x z 0 dx αr e ε u ω Ω B rx 0 and the conclusion follows. If a r/4 or for a general domain Ω, the proof is the same as for Lemma 3.3. The monotonicity formula for small scales is then given in the following lemma, where χ stands for the characteristic function. Lemma 3.. Monotonicity at small scales. There exist C and 0 < R, depending only on N and Ω, such that for any solution u of -6 in Ω any x 0 Ω and 0 < r minr, Λ, with d 0 := distx 0, Ω, Λ := C + Λ 0 log ε, Q := CΛ 0 ε log ε, we have, with the convention d 0 χ {r d0 } d 0 r = 0 if d 0 = 0, d exp [Λr+Cd 0 χ {r d0 } ]Ẽε dr d 0 r + Q r N Ω B rx 0 Λ a ε x u + ε r N B rx 0 Ω u 0. In particular, exp[λr + Cd 0 χ {r d0 } d 0 r ]Ẽε + Q is a nondecreasing function on 0, R. Λ Proof. First, we note that, if 0 < r RΩ sufficiently small, for all x ˇB r x 0, x x 0 n Cd 0 + r, 6 where C depends on Ω only. This is a basic difference with Lemma 3.4. Arguing as in Lemma 3.4, that is assuming that for r < R sufficiently small, ˇBr x 0 is the uppergraph of a map ψ : B 0 R N R and that ψ0 = ψ0 = 0, so that the tangent hyperplane at Ω at ψ0 = 0 is R N {0}, we are led to prove, as for the proof of 3, that for x ˇB r x 0, It is clear that N x x 0 n = x x 0 e N i ψ e i Cd 0 + r. i= x x 0 e N Cd 0 + r 63 since either d 0 is of greater than or of the order of r and then inequality 63 is true, either d 0 r R and then inequality 63 is also true. Since ψ Cr and x x 0 r, the second term in 6 is Cr. Therefore, 6 holds. We now turn to the proof of Lemma 3.. We proceed as in Lemma 3.4 and estimate each term on the right hand side of 58. The third 5

26 one and the fourth one are treated as in 33 using the rough estimate for the jacobian and 34, which yields respectively N log ε Ju, c rn i xξ i x CΛ 0 log ε Ẽεr 64 ˇB rx 0 i and log ε r N a ε x u x dx Ẽεr + CΛ 0 ε log ε ˇB r We use 59 of Lemma 3.0 to estimate the last term in 58 C r e ε u e ε u + log ε Ω B r ˇB r ω Ju, c i xξ i x z 0 i log ε + a ε u x z 0 d. 66 ω Note that we do not need to estimate the last term in 58 if r < d 0, since in this case, ˇBr =. Therefore, 6 and 66 imply d x ne r N ε u Cχ 0 + r {r d0 } e Ω B r r N ε u + log ε ˇB r ω Ju, c i xξ i x z 0 i log ε + a ε u x z 0 d. We estimate the two last terms as in 64 and 65 to infer + rλ r N 0 log ε Ẽεr + Λ 0 ε log ε Ω B r x ne ε u Cχ {r d0 } + d 0 r Inserting estimates 64, 65 and 67 into 58 gives, with Λ = C+Λ 0 log ε and for r Λ, dẽε dr u + a ε x u r N r N ˇBr ε Ω B r Λ + Cχ {r d0 } To conclude, we introduce the primitive r 0 d 0 r Ẽεr C + χ {r d0 } r Λ 0ε log ε 4. ω d 0 χ {ρ d0 } d 0 ρ dρ = d 0χ {r d0 } d 0 r 0. Consequently, d exp [Λr + Cd 0 χ {r d0 } ]Ẽε dr d 0 r [ = exp Λr + Cd 0 χ {r d0 } ] d d 0 r Ẽ ε dr + Λ + Cd 0χ {r d0 } d 0 r Ẽεr u + a ε x u r N Ω B r r N ˇBr ε d [ C + χ 0 {r d0 } r exp Λr + Cd 0 χ {r d0 } ]Λ d 0 r 0 ε log ε 4. 6

27 To conclude the proof, just note that the last term is Λ C d dr Q Λ exp [ Λr + Cd 0 χ {r d0 } d 0 r ], where Q = CΛ 0 ε log ε. In the next lemma, we compute the derivative of the modified scaled energy Ēεx 0, r in the same way as for Lemma 3.6. Lemma 3.. Assume u satisfies -6, x 0 Ω and r > 0. Then, d dr Ēεx 0, r = r N + r N Ω B r x 0 ˇB r x 0 N log ε rn log ε r N r N e ε ufr, x x 0 + t u r N Ω B trx 0 a ε x u fr, x x ε 0 Ju, c i xξ i x x 0 fr, x x 0 68 ˇB r x 0 i a ε x u x x 0 dxfr, x x 0 ˇB r x 0 B r x 0 Ω x x 0 ne ε ufr, x x 0. Lemma 3.3. Monotonicity at large scales. There exist constants C and 0 < R, depending only on Ω such that, if x 0 Ω, r ε r minr, + Λ 0 and u is a solution to -6, then for every r ε s < r, Ē ε s C exp C + Λ 0 t 4 + d 0χ {t d0 } d 0 t Ēεr + Λ 0 ε β. Proof. We assume x 0 = 0 and we estimate each term on the right-hand side of 68. For the fifth term, we have as for 65 log ε a r N ε x u x dxfr, x Ẽεr + CΛ 0 ε log ε ˇB r Concerning the fourth one, we may use a reflection with respect to the boundary. We assume r δ 0. We extend u in a map ũ defined on U := ˇB r φ ˇB r by setting for x = φ y φ ˇB r y ˇB r, ũx := uy = u φx. It is then clear that on φ ˇB r, and that Jũ = φ Ju E ε ũ, U CE ε u, ˇB r. We also extend the -form i c ixξ i x by this way setting ϕ := φ i c ixξ i x in φ ˇB r. This -form is in C 0, 0 U, Λ R N and satisfies ϕ CΛ 0 r and dϕ CΛ

28 Moreover, Jũ, ϕx = Ju, c i xξ i xfr, x. U ˇB r i We apply the result of Jerrard and Soner of Lemma 3.5 for the first integral to infer Ju, c i xξ i xfr, x CΛ 0 r E εr ˇB r log ε + CΛ 0ε α + E ε r. i Consequently, we have the following estimate for the fourth term in 68 log ε Ju, c r N i xξ i xfr, x CΛ 0 + εα log ε Ẽεr + CΛ 0ε α log ε. 7 ˇB r r r N i We estimate also the boundary energy as for 67. We first apply Lemma 3.0 to obtain log ε e r N 3 ε u C Ẽε r + Ω B r r ω Ju, c N i xξ i xfr, x i log ε + a r N ε x u x dxfr, x. Using then 6, Ẽ ε r Ēε4r, r and estimates similar to 69 and 7, we infer for r minr, + Λ 0 the estimate of the last term in 68 as in 67 x ne r N ε ufr, x Ω B r d C + χ 0 {r d0 } Ēε 4r + Λ 4r 0 ε log ε 4 ε α log ε + Λ 0. 7 r N Combining estimates 69, 7 and 7 with 68 yields dēε dr t u + a ε x u fr, x 73 r N Ω B tr r N ˇBr ε d C + χ 0 {r d0 } 4r + + Λ 0 + εα log ε Ēε4r + Λ 0 r ε log ε 4 ε α log ε + Λ 0. r N As in Lemma 3.7, we assume ω r ε β ε α log ε /N = r ε, 74 where 0 < β < α is fixed and take µ = α β > 0 so that r ε α log ε r N ε α log ε ε β. Hence, with Br := t u + a ε x u fr, x, 75 r N Ω B tr r N ˇBr ε 73 implies dēε dr Br C + Λ 0 + χ {r d0 } d 0 Ēε 4r + Λ 4r 0 ε β. 76 In particular, since Br 0, for r ε r minr, + Λ 0, dēε dr C d + Λ 0 + χ 0 {r d0 } Ēε 4r + Λ 4r 0 ε β. 77 To conclude the proof, we also make use of a discrete Gronwall inequality. 8

29 Lemma 3.4. Discrete Gronwall inequality. Let 0 < s < 4s < s and h : [s, s ] R + be continuously differentiable and such that { hs θ N hθs if θ [, s /s ], s [s, s ] and θs [s, s ], h s C + Λ 0 + χ {r d0 } d 0 h4s + D for all s [s 4r, s /4], where C, D and Λ 0 are positive constants. Then, for all s s < t s, hs 4 N exp C + Λ 0 t/4 + d 0 χ {t d0 } d 0 t ht + D. 78 Proof. We reduce the proof to the case D = 0 considering gs := hs + D. We have gs = hs + D θ N hθs + θ N D = θ N gθs 79 if θ [, s /s ], s [s, s ] and θs [s, s ] and g s = h s C + Λ 0 + χ {s d0 } d 0 4s h4s + D = C for all s [s, s /4]. Let s s < t < s. Assume t/4 s t. Then, by 79, gs 4 N gt. By induction, assume that for some k N it holds for all t 4 k s t 4 k, where we have set If α i t := t s t, then, by 80, 4 k+ 4 k t/4 k gs 4 N gt t/4 i k + Cα i t i= d 0 + Λ 0 + χ {r d0 } t/4 4r dr. i gs gt/4 k + C + Λ 0 + χ {r d0 } g4r dr s 4r k k 4 N gt + Cα i t + 4 N Cgt + Cα i t i= k+ = 4 N gt + Cα i t. i= d 0 i= d + Λ 0 + χ 0 {s d0 } g4s 80 4s t/4 k d 0 + Λ 0 + χ {r d0 } t/4 4r dr k+ The conclusion then follows from the definition of α i t and the inequality, valid for all m N, m + Cα i t exp C i= i= α i t = exp C t/4 0 d + Λ 0 + χ 0 {r d0 } 4r dr = exp C + Λ 0 t 4 + d 0χ {t d0 } d 0 t. 9

30 Coming back to h, we deduce hs gs 4 N exp C + Λ 0 t 4 + d 0χ {t d0 } d 0 t ht + D, and the proof is complete. To conclude the proof of Lemma 3.3, we apply Lemma 3.4 with s = r ε, s = r, h = Ēε, C = C and D = Λ 0 ε β. The first hypothesis needed is still verified for the modified scaled energy Ē ε and the second one is 77. We then infer that, for every r ε s < r, Ē ε s C exp C + Λ 0 t 4 + d 0χ {t d0 } d 0 t Ēεr + Λ 0 ε β. This finishes the proof of Lemma Proofs of Propositions and Before giving the proof, we notice that for any d 0 0 and t 0, 0 d 0 χ {t d0 } d 0 t, since, for t d 0 > 0, d 0 /d 0 /t = d 0 /t [0, ]. Therefore, in the Neumann case, this extra term is less than a constant. We assume x 0 = 0 and first consider the case θr ρ := + Λ 0 log ε r/. 8 By Lemma 3.4 resp. Lemma 3. in the Dirichlet case resp. the Neumann case, we deduce Ẽ ε θr CẼερ + T ν ε θr, ρ + Λ 0 ε log ε 4 8 resp. Ẽ ε θr CẼερ + Λ 0ε log ε Next, by Lemma 3.7 resp. Lemma 3.3, recalling r ε ρ for 0 < ε < ε 0 sufficiently small, applied with s = ρ and r/, Ẽ ε ρ Ēερ CẼεr + T ν ε ρ, r/ + Λ 0ε β 84 resp. Ẽ ε ρ Ēερ CẼεr + Λ 0 ε β. 85 Combining 8 and 84 resp. 83 and 85 yields 7 resp. 8 if 8 holds. If θr r/ ρ, we only use Lemma 3.4 resp. Lemma 3. as for 8 resp. 83, and if ρ θr r/, we only use Lemma 3.7 resp. Lemma 3.3 as for 84 resp. 85. The proof is complete. 30

31 4 Proof of Theorems and 3 We follow step by step the lines of [BBO] Theorem bis and [BOS] Theorem. The proof is divided in three parts. Let 0 < δ < /3 be a constant to be determined later, depending only on N and Ω, and, in the Dirichet case, on ν and the constant C in. 4. Proof of Theorem Part A: Choosing a good radius. Lemma 4.. Assume 0 < ε < δ /β, that u is a solution of -5 and that Ẽ ε r η log ε and T ν ε x 0, r ε, r η log ε 86 holds for a rε / r minr, + Λ 0 /ν. Then, there exists a radius r 0 r ε, rε / such that a ε u Cη + Λ r0 N ˇBr0 ε 0 ε β log ε log δ, Ẽεr 0 N Ẽ ε δr 0 Cη + Λ 0 ε β log ε log δ. Proof. From 53, we have for r ε r minr, + Λ 0 dēε dr Ar C + Λ 0 Ēε4r 4r ν + G εr r ν + Λ 0 ε β, 87 where Ar is defined in 54. Let k be the greatest integer such that r ε δ 4 k r/8 and define the intervals I j := r ε δ 4 j+, r ε δ 4 j, j k. These intervals are clearly disjoint and k j=i j r ε, r/8. From r r / ε and log r ε C log ε, we infer log ε k C log δ. 88 We integrate 87 over each I j, j k, and use the monotonicity formula of Proposition k j= I j Ar C + Λ 0 Ēε4r 4r ν + G εr r ν + Λ 0 ε β dr k Ē ε r ε δ 4 j+ Ēεr ε δ 4 j j= CẼε r + T ν ε r ε, r + Λ 0 ε β Cη log ε + Λ 0 ε β 89 3

32 by hypothesis. Moreover, still with the monotonicity formula of Proposition, we have k j= I j + Λ 0 Ēε4r 4r ν +G εr r ν + Λ 0 ε β dr C + Λ 0 Ẽε r + T ν ε r ε, r + Λ 0 ε β r r ε dr r ν + CT ν ε r ε, r + CΛ 0 ε β Cη log ε + Λ 0 ε β, 90 by 86 and the hypothesis r ν + Λ 0. We deduce from 88, 89 and 90 the existence of some j 0 {,..., k} such that I j0 Ar Ēεr ε δ 4 j 0+ Ēεr ε δ 4 j 0 Cη + Λ 0 ε β log ε log δ. 9 In particular, by the mean value formula, there exists some rε r 0 δ 4 j 0, r ε δ 4 j 0 I j0 such that ˇBr0 a ε u r0 N ε Cη + Λ 0 ε β log ε log δ, which is the first assertion of the Lemma. Noticing that δ r 0 I j0, we infer from 9 Ẽ ε r 0 N Ẽ ε δr 0 Ēεr 0 Ēε δ r 0 Cη + Λ 0 ε β log ε log δ, where we have used once more Lemma 3.7. This is the second assertion of the Lemma. Part B: δ-energy decay. Lemma 4.. There exist constants C and ε 0, depending on N, ν and Ω, such that, if u is a solution of -5 with g ε satisfying and T ν ε x 0, ε, r / ε η, with ε < ε 0 and ε < r rε /, then E ε δr C γ + δ N + γ 4 a ε u E r N ˇBr ε ε r + Cγ a ε u ˇB r ε + Cr N + γ 4 a ε u η + Λ r N ˇBr ε 0 ε β + C ηr N. For the ease of presentation, we will assume that Ω is locally the half plane R N + = R N R +. By the mean-value inequality, there exists r/3 r r/6 such that r u 96 u, 9 Ω B r x 0 r a ε u 96 Ω B r x 0 ˇB rx 0 a ε u, 93 ˇB rx 0 r N G ε r 96ηr N. 94 3

33 To clarify the last one, if ρ N G ε ρ 96ηr N, then ρ ν G ε ρ 96ηρ ν 96ηε. The proof is divided in four steps. Step : Hodge-de Rham decomposition of u u. Since u is a solution of and div c = 0, d u du = u u = u, c u log ε = d log ε u c, 95 where c := N i= c ixdx i. We consider the solution of the auxiliary problem ξ = 0 in ˇB r x 0, ξ c n ξ = 0 on R N + B r x 0, on R N + B r x 0, which exists and is unique. By 9 and 93, we have ξ Cr u + Λ ˇB r R N + Br x 0 0 ε log ε from which we infer by standard estimates ξ Cδ N ˇB δr R N + Br x 0 u ε CE ε r + r N Λ 0 ε β, 96 ˇB r ξ Cδ N E ε r + r N Λ 0 ε β. 97 We turn now to the Hodge-de Rham decomposition of u du. By construction of ξ and from 95, we have in D R N + [ d u du log ε u ] c dξ χ = By classical Hodge theory see for instance the Appendix of [BBO], Proposition A.8, there exists some -form ϕ on R N + such that d ϕ = u du log ε u c dξ χ in D R N +, 99 dϕ = 0 in D R N +, 00 ϕ L R N + CE εr + ξ L ˇB r + rn Λ 0 ε β, 0 ϕ = 0 on R N +, 0 ϕx x N 0 as x Step : Improved estimates for ϕ on ˇB δr x 0. Let f : R + R + be any smooth function such that ft = if γ t + γ, t ft = if t γ or t + γ, f t 4 for any t 0. 33

34 We consider the function on R N + defined by { f τx := ux in ˇB r, outside, so that, by construction, 0 τ 4γ in R N Note that thus, in ˇB r, f u u du = f u u df u u, dτu du = d[f u u df u u] = i<j i f u u j f u u dx i dx j. Turning now to ϕ, we apply the d operator to 99 to deduce that, in D R N +, ϕ = dd u ϕ = dχ τu du d χ dξ d χ log ε c + dχ τu du = ω + ω + ω 3 + ω 4 + ω 5, where, χ standing for the characteristic function of ˇBr x 0, ω := χ dτu du = χ i<j i f u u j f u u dx i dx j, ω := σ Br R dr f u u du σ N + Br R dx N N f g + ε g ε dg ε r = x x 0, ω 3 := dχ dξ = σ Br R N dr dξ, + u ω 4 := d χ log ε c, ω 5 := dχ τu du and σ stands for surface measure. We denote also the -forms on R N + A = A = A 3 = 0, A 4 := χ log ε g ε c and A 5 := χ f g ε g ε dg ε. From the Appendix of [BBO], we know that the solutions of the problems on R N + ϕ i = ω i in R N +, ϕ i = 0 on R N + = RN {0}, d ϕ i = A i on R N + = RN {0}, for i =,, 3, 4 and 5 exist in H loc RN +, but are not unique for i = 4 and 5. We will consider in this case the solutions given by convolutions the ω i s and A i s have compact support. Note that this prevents us from imposing condition at infinity since, a priori, the integrals over R N + of the components of A 4 and A 5 are not zero. Concerning ϕ, we note that the second measure in ω 34

35 involves a measure supported on R N +, but the weak formulation of the equation has a meaning for test functions in C,c Λ RN + see Lemma A.4 in [BBO], namely, for all ζ C,c Λ RN +, dϕ, dζ + d ϕ, d ζ = ω, ζ = R N + Br dr f u u du ζ f g ε g ε dg ε ζ. R N + Let φ := ϕ 5 i= ϕ i. Then, by 99 and the condition ξ = 0 on B r x 0 R N +, d φ = d ϕ A 4 A 5 = χ f g ε g ε dg ε =: A. Consequently, φ is the solution given by convolution of φ = 0 in R N +, φ = 0 on R N + = RN {0}, d φ = A on R N + = RN {0}, 05 and We turn now to estimate φ and the ϕ i s. ϕ = φ + 5 ϕ i. i= Estimate for φ. We have φ Cr ˇB r x 0 B rx 0 R N + g ε Cr N G ε r Cηr N. 06 This is a direct consequence of standard estimates see the Appendix of [BBO] and a scaling argument for the equation 05 combined with the bound A L R N + C g ε. B r x 0 R N + Estimate for ϕ 5. We claim that ϕ 5 Cγ u Cγ E ε r. 07 ˇB r R N + Indeed, since ϕ 5 is a solution of ϕ 5 = ω 5 = dχ τu du, we obtain, multiplying by ϕ 5 and integrating see Lemma A.4 in [BBO] ϕ 5 L R N + = dχ τu du, ϕ5 A 5 ϕ 5. Moreover, integration by parts once more yields dχ τu du, ϕ5 = χ τu du, d ϕ 5 + R N + R N + R N + 35 R N + R N + χ f g ε g ε dg ε ϕ 5.