Transformation of corner singularities in presence of small or large parameters

Transformation of corner singularities in presence of small or large parameters Monique Dauge IRMAR, Université de Rennes 1, FRANCE Analysis and Numerics of Acoustic and Electromagnetic Problems October 17-22, 2016, Linz, Austria RICAM Special Semester on Computational Methods in Science and Engineering http://perso.univ-rennes1.fr/monique.dauge 1/23

Outline 1 Joint works and discussions 2 Example 1: Self-similar perturbations of a 2D corner 3 Example 2: Neumann-Robin boundary conditions 4 Example 3: Thin layers 5 Example 4: Conducting material in 2D eddy current formulation 6 Profiles and corner layers 1/23

Joint works M. COSTABEL, M. DAUGE, Example 2 A singularly perturbed mixed boundary value problem, Comm. Partial Differential Equations 21, 11 & 12 (1996), 1919 1949. G. CALOZ, M. COSTABEL, M. DAUGE, G. VIAL Example 3 Asymptotic expansion of the solution of an interface problem in a polygonal domain with thin layer Asymptotic Analysis 50 (1/2) (2006), 121 173. M. DAUGE, S. TORDEUX, AND G. VIAL Example 1 Selfsimilar perturbation near a corner: matching versus multiscale expansions for a model problem, Around the research of Vladimir Maz ya. II, vol. 12 of Int. Math. Ser. Springer (2010), 95 134. M. DAUGE P. DULAR, L. KRÄHENBÜHL, V. PÉRON, R. PERRUSSEL, C. POIGNARD Example 4 Corner asymptotics of the magnetic potential in the eddy-current model Mathematical Methods in the Applied Sciences 37, 13 (2014) 1924 1955 M. COSTABEL, M. DALLA RIVA, M. DAUGE, P. MUSOLINO Example 1 Converging expansions for Lipschitz self-similar perforations of a plane sector, In preparation, (2016). 2D eddy current formulation for a conductor surrounded by a dielectric medium Recent discussions with R. HIPTMAIR, R. CASAGRANDE, K. SCHMIDT, A. SEMIN 1/23

Example 1: Self-similar perturbations of a 2D corner See references 1 Let Γ be the infinite plane sector Γ = {x = (x 1, x 2 ) R 2, r > 0, θ (0, ω)} with polar coordinates (r, θ) and the opening ω (0, π) (π, 2π). 2 Bounded un-perturbed domain Ω Ω 0, with Ω Γ. Assume Ω bounded connected with curvilinear polygonal boundary, r 0 > 0 such that B(0, r 0 ) Ω = B(0, r 0 ) Γ 3 Bounded perturbation pattern P, with P Γ. P connected with curvilinear polygonal boundary, R 0 > 0 such that B(0, R 0 ) P = B(0, R 0 ) Γ 4 Family of perturbed domains, with ε (0, ε 0 ) Ω ε = Ω εp 2/23

Dirichlet Problems for Example 1 Simplifying assumption f L 2 (Ω) with support outside a ball B(0, r 1 ) General assumption f L 2 (Ω) analytic or C inside a ball B(0, r 1 ) We choose the simplifying assumptions for the talk because A zero Taylor expansion of f at the corners allows to discard polynomials from corner asymptotics, so to avoid log terms and convergence issues. The family of problems under consideration is the family of Dirichlet problems for ε [0, ε 0 ) { u ε = f in Ω ε (P ε) u ε = 0 on Ω ε. Unique solution in u ε H 1 0 (Ω ε) 3/23

Two particular cases of Example 1 Question Behavior of u ε close to 0 (in the region B(0, r 0 ) Ω ε) as ε 0. Let us consider two examples 1a A rounded corner: The boundary of P is smooth! 1b A cracked corner: The boundary of P has a crack Σ abutting at 0 (P = Γ \ Σ, with a straight segment Σ). Set α = π ω In both examples, Ω = Ω 0 has a corner of opening ω at 0. Therefore [Kondrat ev 1967] u 0 has an expansion as K u 0 = u0,reg K + γ k r kα sin kαθ, k=1 u K 0,reg H m+1 (Ω B(0, r 0 )), m = [(K + 1)α] 4/23

Paradoxes on singularities for Example 1 On the one hand On the other hand u 0 = γ 1 r α sin αθ + u 0,reg, u 0,reg H 2 (Ω B(0, r 0 )) 1a Rounded corner: u ε is smooth in Ω ε B(0, r 0 ) 1b Thus 1a A cracked corner: The first singularity of u ε behaves like r 1 2 sin θ 2 u ε has no singularity but u 0 has one 1b u ε has a stronger singularity than u 0 How can the singularity r α sin αθ associated with ε = 0 1a Disappear as soon as ε > 0 1b Transform into a stronger singularity as soon as ε > 0? 5/23

Example 2: Neumann-Robin boundary conditions See references Take a polygon Ω with straight sides Choose a side, denoted by Σ R. Denote Ω \ Σ R by Σ N Assume for simplicity that the openings of Ω at the two ends of Σ R are the same, say ω Assume for simplicity that π is not a rational number ω Assume for simplicity that f L 2 (Ω) with support outside neighborhoods of the ends of Σ R Then the Neumann-Robin problem is for ε [0, ε 0 ) u ε = f in Ω (P ε) nu ε = 0 on Σ N ε nu ε + u ε = 0 on Σ R When ε = 0, we obtain a mixed Neumann-Dirichlet problem u 0 = f in Ω (P 0 ) nu 0 = 0 on Σ N u 0 = 0 on Σ R 6/23

Example 2: Variational formulations and singularities Variational spaces: V ε = H 1 (Ω), ε > 0, and V 0 = {u H 1 (Ω), u ΣR = 0}. Variational formulations (P ε) u ε H 1 (Ω), v H 1 (Ω), u ε v dx + 1 u ε v dσ = Ω ε Σ R (P 0 ) u 0 V 0, v V 0, u 0 v dx = f v dx Singularities of u 0 (for a Neumann-Dirichlet corner) recall that α = π ω Ω r α/2 cos 1 2 αθ, r 3α/2 cos 3 2 αθ, r 5α/2 cos 5 2 αθ,... Singularities of u ε (for a Neumann-Robin corner): Ω Ω f v dx r α cos αθ + ε 1 γ 1 r α+1 cos(α + 1)θ + ε 2 γ 2 r α+2 cos(α + 2)θ +... r 2α cos 2αθ + ε 1 γ 1r 2α+1 cos(2α + 1)θ + ε 2 γ 2r 2α+2 cos(2α + 2)θ +... i.e. Neumann singularity plus successive shadows. Negative powers of ε 7/23

Example 3: Thin layer Set Take a polygon Ω with straight sides Consider the union Σ of two consecutive sides σ 1 and σ 2. Let O = σ 1 σ 2 be the corner inside Σ. Assume for simplicity that the opening ω at O is such that π ω Q See references Assume for simplicity that the openings of Ω at the two exterior corners c 1 and c 2 in Σ equal π 2 Define the layer L ε of width ε as the polygon formed by Σ, the segments σ l,ε, l = 1, 2, parallel to σ l at the distance ε outside Ω, and two perpendicular segments of length ε issued from c 1 and c 2. Assume for simplicity that f L 2 (Ω) with support outside a neighb. of O Ω ε = Ω Σ L ε and consider the transmission problem, with a positive number β 1 u ε H0 1 (Ω ε), v H0 1 (Ω ε), u ε v dx + β u ε v dx = Ω L ε Thus u 0 is solution of the Dirichlet pb on Ω = Ω 0. Ω f v dx. 8/23

Example 3: Singularities Singularities of u 0 at 0 (Dirichlet) recall that α = π ω r α sin αθ, r 2α sin 2αθ, r 3α sin 3αθ,... Singularities of u ε at 0: Those of the transmission problem U Ω = F in Γ Ω U L = 0 in Γ L [U] = 0 on Γ Ω Γ L nu Ω + β nu L = 0 on Γ Ω Γ L Here Γ Ω is the sector of opening ω and Γ L is (the interior of) its complement U Ω and U L are the restrictions of U to Γ Ω and Γ L, respectively. n exterior derivative. The singularities of the transmission problems are different from the Dirichlet singularities. 9/23

Example 4: Conducting material in 2D eddy current formulation Take a conductor body M (conductivity σ) surrounded by a dielectric body N. Harmonic Maxwell equations curl E = iκµh and curl H = (iκɛ + σ1 M )E + j 0 The eddy current formulation consists in neglecting κɛ with respect to σ: and add the gauge condition curl E = iκµh and curl H = σ1 M E + j 0 div E = 0 in N. Assume that the domains and the fields do not depend on the variable x 3 and that H = (0, 0, u). With, now, Ω R 2, M Ω, N = Ω \ M the 2D eddy current formulation for a magnetic potential u is (P δ ) u δ H0 1 (Ω), v H0 1 (Ω), u δ v dx + i u δ 2 δ v dx = f v dx with Ω δ 2 = κµσ M N 10/23

Example 4: Singularities The parameter δ is (proportional to) the skin depth. It tends to 0 when σ tends to infinity. Assume that Ω is smooth. Assume for simplicity that the conductor M has only one corner in O. Let 2π ω be its opening, i.e. ω is the opening of N at O. Assume for simplicity that the opening ω is such that π ω Q Assume for simplicity that f L 2 (N) with support outside a neighb. of O In strong form (P δ ) is and (P 0 ) is u δ + i δ 2 1 M u δ = f in Ω, with u δ H 1 0 (Ω) u 0 = f in N, with u 0 H 1 0 (N) Singularities of u 0 at 0 (Dirichlet) recall that α = π ω r α sin αθ, r 2α sin 2αθ, r 3α sin 3αθ,... Singularities of u δ at 0? 11/23

Example 4: Singularities, continued See references Singularities at a corner are determined by the principal part of the operator. The principal part of + i δ 2 1 M is. The corner O is inside Ω. Hence no principal singularity! By the way the solution u belongs to H 2 (Ω). But, still, singularities appear as shadows of harmonic polynomials. Indeed (here * = wildcard) 1 + δ 2 r 2( log 2 r ϕ 2(θ) + log r ϕ 1(θ) + ϕ ) 0 + δ 4 r 4 4 l=0 logl r ϕ l (θ) +... r cos θ + δ 2 r 3 3 l=0 logl r ϕ l (θ) + δ 4 r 5 5 l=0 logl r ϕ l (θ) +... r sin θ + δ 2 r 3 3 l=0 logl r ϕ l (θ) + δ 4 r 5 5 l=0 logl r ϕ l (θ) +... The logarithmic terms of highest degree are present! The strongest singularity of u δ is r 2 log 2 r, Versus the strongest singularity of u 0 that is r π ω 12/23

Two strategies Two strategies We have a family of singularly perturbed problems (P ε) with solutions u ε and want to understand How u ε tends to u 0 How singularities transform as ε 0 2 strategies, in the framework of multiscale analysis: A Convert the singularities s λ (homogeneous of degree λ) of u 0 into profiles K λ solutions of a model problem with ε = 1. Roughly A 0 s λ = 0, A 1 K λ = 0 and K λ s λ = O(r λ ) B Leave the singular part as is, but introduce a corrector Y λ (a corner layer) to reduce the distance to u ε. 13/23

Example 1: Self-similar perturbations of a 2D corner Example 1: Profiles and corner layers u 0 λ Λ c λs λ, with Λ = N π ω and sλ = r λ sin λθ. We have, on the sector Γ: { s λ = 0 in Γ Lemma s λ = 0 on Γ. There exists a unique solution K λ in the perturbation pattern P of the problem K λ = 0 in P K λ = 0 on P K λ = s λ + O(1) as R Proof: Set K 1 1(P) = {u L 2 loc(p), u, r 1 u L 2 (P)}. Define Y λ as the variational solution in K 1 1(P) of the problem { Y λ = 0 in P Y λ = s λ on P \ Γ (note that P \ Γ is compact and s λ = 0 on P Γ). Then set K λ = s λ + Y λ 14/23

Example 1: Self-similar perturbations of a 2D corner Example 1: Profiles and corner layers, asymptotics at infinity Then K λ( x ε ) is harmonic and satisfies the Dirichlet condition of εp. Formally u ε : λ Λ c λ ε λ K λ( x ε ) satisfies the Dirichlet condition of εp. But may differ from u ε on Ω \ Γ. This error is far from O. To analyze it, one needs as negative powers of r. Lemma The complete asymptotic at infinity of K λ There exists a sequence of numbers γ λ,λ so that K λ (X) s λ (X) + λ Λ γ λ,λ R λ sin λ θ, R Then ε λ K λ( x ) s λ (x) + ε λ+λ γ λ,λ r λ sin λ θ, x εr 0 ε λ Λ 15/23

Example 1: Self-similar perturbations of a 2D corner Example 1: Finite expansions A B Choose m N. Using the profiles K λ u ε = u [m] 0 (x) + ε λ u [m] λ (x) + c λ,λ ε λ+λ K λ( x ) + O(ε m ) ε λ m λ+λ m The terms u [m] λ are more or less regular: u[m] 0 H m (in fact in K m). m This is a moving asymptotics. But it allows to capture the transformation of singularities through the profiles K λ Choose m N. Using the corner layers Y λ u ε = u 0 (x) + ε λ u λ (x) + c λ,λ ε λ+λ Y λ( x ) + O(ε m ) ε λ m λ+λ m In certain configurations (essentially when the perturbation consists of small holes dug in Ω), it is possible to prove that the series converge. Here the assumption that f 0 close to 0 is important. When this assumption is relaxed to analyticity, we can only prove conditional convergence. The conditions that Ω Γ and P Γ can be relaxed, with the help of cut-offs: χ 0 localizes near 0 and multiply rapid terms, whereas χ localizes near infinity and χ ( x ε ) multiply slow terms. 16/23

Example 2: Neumann-Robin Example 2: Profiles Denote s λ = r λ cos λθ, for λ Λ = (N 1 ) π, the singularities of the mixed 2 ω Neumann-Dirichlet problem. They satisfies on the sector Γ s λ = 0 in Γ ns λ = 0 if θ = 0, s λ = 0 if θ = ω. Denote by S λ (Γ) the space of homogeneous functions of degree λ on Γ. Lemma There exists a unique solution K λ in the sector Γ to the problem K λ = 0 in Γ nk λ = 0 if θ = 0, nk λ + K λ = 0 if θ = ω, K λ = s λ + O(R λ ) as R that has the form L K λ = s λ + sl λ + O(R 1 2 ), with s λ l S λ l (Γ), L = [λ + 1 ]. 2 l=1 17/23

Example 2: Neumann-Robin Example 2: Profiles Proof. Denote s0 λ := s λ. For l = 1,..., there exists unique sl λ S λ l (Γ) such that sl λ = 0 in Γ nsl λ = 0 if θ = 0, sl λ = nsl 1 λ if θ = ω, The existence as homogeneous solutions and the uniqueness is due to the assumption π Q that implies that λ l does not hit the Mellin spectrum ω Λ = (Z 1 ) π of the mixed Neumann-Dirichlet problem. Introduce the sum 2 ω X λ = χ L l=0 sλ l with a radial smooth cut-off, χ 1 for x 2 and χ 0 for x 1 Then, on the Robin side θ = ω and by construction nx λ + X λ = χ ns λ L The trace ns λ L is homogeneous of degree λ L 1 that is < 1 2. Thus χ ns λ L belongs to L 2 (R +). We enter the variational space V / 18/23

Example 2: Neumann-Robin Example 2: Profiles The variational space for the mixed Neumann-Robin problem on the sector Γ U = F in Γ nu = 0 if θ = 0, nu + U = G if θ = ω, is V = { U L 2 loc(γ), U L 2 (Γ), U θ=ω L 2 (R } +) and the variational formulation is U V, V V, U V dx + U V dr = F V dx + G V dr Γ θ=ω Γ θ=ω Thus we find Y λ V solution of Y λ = 0 in Γ ny λ = 0 if θ = 0, ny λ + Y λ = X λ if θ = ω, and our profile is finally K λ = X λ + Y λ = χ ( s λ + L l=1 sλ l ) + Y λ, with s λ l S λ l, Y λ V 19/23

Example 2: Neumann-Robin Example 2: Full expansion at infinity The constructive approach can be carried on, but will never account for the whole asymptotics at infinity, because part of it is hiding in a variat. solution. Need for an analytic tool to find this asymptotics: The Mellin transform. Since the mixed Robin-Neumann is not an homogeneous operator, this is a non-classical setting. Let us analyze Y λ. Set u = χ Y λ and g = χ 2 nsl, λ so that u = 0 in Γ nu = 0 if θ = 0, nu + u = g if θ = ω, Introduce for µ C and the Mellin transform of u and g: U[µ] = 0 r µ u(r, ) dr r, G[µ] = U[µ] is holomorphic for Re µ > 0 G[µ] is meromorphic with a unique pole of order 1 at µ 0 = λ [λ + 1 2 ] 1 [ µ0 < 1 2 0 r µ g(r) dr r ] 20/23

Example 2: Neumann-Robin Example 2: Full expansion at infinity The mixed Neumann-Robin becomes after Mellin transform ( θ 2 + µ 2 )U[µ] = 0 in (0, ω) ( ) θ U[µ] = 0 if θ = 0, U[µ] = θ U[µ + 1] + G[µ] if θ = ω, The Mellin symbol A[µ] : H 2 (0, ω) L 2 (0, ω) R R of the mixed NR is the operator on the left of ( ). The poles of A[µ] 1 are simple and situated on Λ = (Z 1 ) π. Re-write ( ) as 2 ω U[µ] = A[µ] 1( ) 0, 0, G[µ] θ U[µ + 1](ω) Since U[µ] is holomorphic for Re µ 0, the rhs above is meromorphic for Re µ > 1. Hence meromorphic extension of U[µ] for Re µ > 1. Can be continued step by step up to Re µ. The poles are µ 0 l ( l N) and λ l ( λ Λ R, l N) 21/23

Example 2: Neumann-Robin Example 2: Full expansion at infinity of profiles Recall that the profile is K λ and the corner layer is Y λ. They are associated with the singularity s λ = r λ cos λθ, for λ Λ R + Lemma There exist homogeneous functions s λ λ,l S λ l so that K λ s λ + sl λ + s λ λ,l r l 1 λ Λ, λ<0 l 0 Y λ sl λ + s λ λ,l r l>l λ Λ, λ<0 l 0 The material is ready for the description of the solutions u ε and u 0, along the lines of Example 1. The analysis is more difficult due the singular perturbational character of the operator ε n + I 22/23

Conclusion Conclusion What about examples 3 and 4? The new difficulty there is the presence of a further anisotropic scale (thin layer for Example 3 and boundary layer skin effect for Example 4). This was solved 10 years ago for Example 3, and is still under investigation for Example 4. 23/23