Geometrical optics formulaes for Helmholtz equation

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1 Geometrical optics formulaes for Helmholtz equation Francois Cuvelier To cite this version: Francois Cuvelier. Geometrical optics formulaes for Helmholtz equation <hal > HAL Id: hal Submitted on 7 Feb 013 HAL is a multi-disciplinary open access archive for the deposit and dissemination of scientific research documents, whether they are published or not. The documents may come from teaching and research institutions in France or abroad, or from public or private research centers. L archive ouverte pluridisciplinaire HAL, est destinée au dépôt et à la diffusion de documents scientifiques de niveau recherche, publiés ou non, émanant des établissements d enseignement et de recherche français ou étrangers, des laboratoires publics ou privés.

2 GEOMETRICAL OPTICS FORMULAES FOR HELMHOLTZ EQUATION FRANÇOIS CUVELIER Abstract. The present work deals with high frequency Helmholtz equation resolution using geometrical optics. We give formulaes in dimension and 3 for mixed Dirichlet, Neumann and Robin boundaries conditions. 1. Introduction Let K j j1,,n be a set of regular, disjoint and strictly convex compacts in R d d or 3. We suppose that for j l, K j Kl. Let Ω j Kj c and Ω N Ω j. We note Γ j the boundary of K j and Γ N Γ j. j1 Let u inc be the incident wave given by u inc x e ikϕinc x a inc x 1.1 which satisfy Helmholtz equation in R d. We want to solve, at high frequency, Helmholtz equation u+k u 0, in Ω 1. j1 with boundaries conditions uγ u inc γ, γ Γ D, 1.3 u γ uinc n n γ, γ Γ N, 1.4 αk u αk n γ+βkuγ uinc n γ+βkuinc γ, γ Γ R, 1.5 and outgoing Sommerfeld radiation condition u r r iku bound for r x. 1.6 We assume that and αk βk j0 j 1 α j k j, α j C, 1.7 β j k j, β j C, 1.8 β 1 +ıα 0 µ 0 µ [0;1] 1.9 Here we take Γ D,Γ N and Γ R such that Γ Γ D Γ N Γ R, o Γ D o Γ N, o Γ D o Γ R and o Γ N o Γ R. Date: September 1, 01. 1

3 front of the incident wave Σ Γ D Γ D K j Γ R Γ N K i Γ N Γ R Figure 1. Exterior Helmholtz problem We assume that, Σ, the front of the incident wave is given by Σ { x R d ϕ inc x ϕ Σ } is a regular and orientable surface d 3 or curve d. We chose the orientation Nσ ϕinc σ ϕ inc σ. Let Σ + { x R d σ Σ, t > 0,x σ +t ϕ inc σ }, we also made following assumption : i N K j Σ + j1 ii curvature radius of Σ are negatives At high frequency means that wave length is small with respect to the obstacle boundary curvature radius, geometrical optics approximation for this problem is based on k 1 asymptotic expansion of vx ux + u inc x, where u is solution of problem 1.,1.3,1.4,1.5,1.6. The first term of the k 1 expansion is the geometrical optics ansatz.. Notations and results Definition 1 Geometrical optic ray. Let ρ σ,γ 1,...,γ l γ 0,γ 1,...,γ l Σ Γ l, we say that ρ is a geometrical optic ray coming from Σ, going through x γ l+1 and reflected l times if 1 γ j γ j 1, j {1,...,l+1}, l+1 ]γ j 1 ;γ j [ Γ, j1 γ 1 γ 0 3 γ 1 γ 0 ϕinc γ 0, 4 reflections conditions : j {1,...,l} γ j+1 γ j γ j+1 γ j γ j γ j 1 γ j γ j 1 5 non grazing conditions : j {1,...,l} γj γ j 1 γ j γ j 1,nγ j 0, where nγ note the exterior normal at Γ in γ. We note R l x a such set of ρ and Rx l N R l x. γj γ j 1 γ j γ j 1,nγ j nγ j,

4 Definition Grazing ray. Let ρ σ,γ 1,...,γ l γ 0,γ 1,...,γ l Σ Γ l, we say that ρ is a grazing ray coming from Σ, going through x γ l+1 and reflected l times if conditions 1 to 4 of definition 1 are satisfied and not condition 5. We note G l x a such set of ρ and Gx l N G l x. Definition 3 Propagation operator. Let t > 0, we note S d t R d d the set of matrix A such that I+tA is regular. We note S t the following application : S t : S d t R d d A AI+tA 1 Definition 4 Reflection operator. Let B R d d a symmetric matrix. Let η R d and ζ R d such that ζ,η 0. We note TB,η,ζ d : Rd d R d d given by : A R d d, x R d T d B,η,ζ Ax A ζ,η Bx η,x Aη [ +Bζ Aη +Bζ,x η + Aη,η Bζ,ζ ζ,η ] η,x η Definition 5 Reflection coefficient. Let η R d, we note b η the function defined on Γ by 1, if γ Γ D, 1, if γ Γ b η γ N, ıα 0 η,nγ +β 1, if γ Γ R. ıα 0 η,nγ β 1 Theorem 1. Let x Σ +, such that Gx, and ρ γ 0,γ 1,...,γ l R l x, l N. We note Bγ the curvature matrix of Γ at γ and nγ the exterior normal to Γ at γ. Let u G.O. ρ x be the ray contribution : with and for j l to 1, ϕ j u G.O. ρ x a ρ xe ikϕx;ρ ϕx;ρ ϕ l ρ γ l + x γ l a ρ x a l ρ γ l deti+ x γ l A l ρ γ l ρ γ j ϕ j 1 ρ γ j 1 + γ j γ j 1 a j 1 ρ γ j 1 a ρ j γ j b γ j γ j 1 γ j γ j γ j 1 det I+ γ j γ j 1 A j 1 ρ γ j 1 A ρ j γ j T d γ j γ j 1 S γj γ j 1 A j 1 ρ γ j 1 Bγ j,nγ j, γ j γ j 1 where ϕ 0 ρ γ 0 ϕ inc γ 0, A 0 ρ γ 0 Hessϕ inc γ 0 and a 0 ρ γ 0 a inc 0 γ 0. Then, we have and u inc x ρ R 0x ux ρ Rx u G.O. ρ x+o 1 k 1 u G.O. ρ x+o k 3. Proof To obtain theorem 1, we first give geometrical optics formulaes without obstacle, then with a stricly convox compact and finally with an union of stricly convex compacts. 3

5 3.1. Without obstacles. We want to solve, at high frequency, Helmholtz equation using an asymptotic expansion in k 1 of the solution u+k u 0 in R d 3.1 ux;k j N where we suppose that u is known on Σ. By substituting expansion 3. in Helmholtz equation, we obtain a j x k j e ıkϕx 3. k 1 ϕx [ a 0 x +k ı{ ϕx, a 0 x+a 0 x ϕx}+1 ϕx a 1 x + k j ı{ ϕx, a k+1 x+a k+1 x ϕx}+ a k x 0 j N Equating the coefficients of different powers of k, we have : Eikonal equation ϕx Transport equation for a 0 Transport equation for a j in function of a j 1 ϕx, a 0 x + 1 a 0x ϕx ϕx, a j x + 1 a jx ϕx ı a j 1x Solution of Eikonal equation 3.3. Derivating eikonal equation 3.3 give This equation can be solved by the caracteristics method : Let X t ϕxt with X0 σ Σ then we have : Hessϕx ϕx d dt ϕxt HessϕXtX t HessϕXt ϕxt 0 That is to say ϕxt ϕx0 hence Xt X0+t ϕx0. and ϕxt is constant along the line Xt X0+t ϕx0. Owing to the construction of Σ, we have Xt σ +tn σ with σ Σ and N σ ϕσ unitary normal vector to Σ at σ. We also have : d dt ϕxt ϕxt,x t ϕx0, ϕx0 1 Thus ϕxt ϕx0+t. We have proved following lemma : Lemma 1. 1 Caracteristic curves of Eikonal equation are geometrical optic rays, Phase is linear along geometrical optic rays : 3 We have σ Σ and σ Σ, t 0 ϕσ +tn σ ϕσ+t. 3.7 ϕσ N σ 3.8 HessϕσN σ

6 3.1.. Solution of transport equation 3.4. We want to solve { ϕx, a0 x + 1 a 0x ϕx 0 a 0 given on Σ 3.10 In fact, we only have to solve this equation along the ray σ +tn σ where σ Σ and N σ ϕσ. Thus, transport equation for a 0 becomes { d dt a 0σ +tn σ+ 1 ϕσ +tn σa 0 σ +tn σ 0, 3.11 a 0 σ given on Σ. We now have to compute ϕσ +tn σ TrHessϕσt. Letσ Σ,wenoteford 3resp. d B Σ σ {u,v,n}σresp. B Σ σ {u,n}σthedirect orthonormal curvature basis of Σ at σ. Here u and v are the direction of maximum and minimal principal curvature k 0 1 σ and k0 σ resp. u is the tangent vector and k0 σ the curvature. By hypothesis on Σ, we have k 0 σ k0 1 σ 0 resp. k0 σ 0. With these notations, we have in basis B Σ σ Hessϕσ 1 σ k 0 σ 0 resp. Hessϕσ k 0 Lemma. Let σ Σ. We note σt σ +tn σ then k 0 σ t 0, Hessϕσt S t Hessϕσ 3.1 In basis B Σ σ for d 3 resp. d k 0 1 σt 0 0 Hess ϕσt 0 k 0 σt 0 resp. Hessϕσt with Proof. To simplify notations, we note t 0 0 k 0 j σt k0 j σ 1 tk 0 j σ k0 j σ j 1, resp. 0 k 0 σt k0 σ 1 tk 0 σ k0 σ A t σ Hessϕσ +tn σ. Let Vσ R d be a neighborhood of σ Σ. We have, x Vσ k 0 σt ϕx ϕσ+a 0 σx σ+o x σ 3.13 Substituing ϕx by 3.13 in Eikonal equation 3.3 give, x Vσ : 1 ϕσ+a 0 σx σ+o x σ Thus Using A 0 σ symmetry, we obtain ϕσ + A 0 σx σ, ϕσ +O x σ 1+ A 0 σx σ,n σ +O x σ. x Vσ, A 0 σn σ,x σ 0 A 0 σn σ We now want to obtain a similar formula for A t σ. We know that A 0 σ is the curvature matrix of Σ at σ and in basis B Σ σ we have for d 3 resp. d k1 Σ σ 0 0 A 0 σ 0 k Σ σ 0 k resp. A 0 σ Σ σ

7 By hypothesis, curvature radius of Σ are negative and k Σ σ k Σ 1 σ 0 resp. k Σ σ 0 Thus, t 0, I+tA 0 σ is regular. Let x σ +tn σ with t > 0, then, there exists a neighborhood Vx R d of x such that x Vx, σ Vσ and t in a neighborhood of t verifying x σ +t N σ So we obtain by Taylor s developpement ϕx ϕx+a t σx x+o x x 3.15 Substituing ϕx for 3.15 in eikonal equation 3.3 give To prove formula 3.1 we write A t σn σ x x σ +t N σ σ tn σ σ σ +t tn σ+t N σ N σ. But, due to 3.13 N σ N σ A 0 σσ σ+o σ σ 3.17 and, thus x x σ σ +t tn σ+t A 0 σσ σ+o σ σ Substituing x x for 3.18 in 3.15 and using 3.16 give N σ N σ+a t σσ σ+t A t σa 0 σσ σ+o σ σ + t t. Now, with formula 3.17 we found A 0 σσ σ+o σ σ A t σσ σ+t A t σa 0 σσ σ+o σ σ + t t That is to say A 0 σ A t σ+ta t σa 0 σ A t σi+ta 0 σ. But, t 0, the matrix I+tA 0 σ is regular and formula 3.1 is proved. Lemma 3. Let σ Σ. Along the ray σt σ +tn σ, t 0, we have a 0 σt a 0 σ deti+thessϕσ 3.19 and t > 0, t 0 such that t > t a 0 σt a 0 σt. 3.0 The previous inegality becomes strict if, at least, one of the curvature radius of Σ is strictly negative. Proof. By definition of A t σ see proof of lemma Then we solve equation 3.11: a 0 σt a 0 σe 1 ϕσt TrA t σ t 0 TrAτσdτ t 0 Tr[A0σI+τA0σ 1 ]dτ a 0 σe 1 a 0 σe 1 [logdeti+τa0σ]t 0 a 0σ. deti+ta0σ We have seen, in proof of lemma, that eigenvalues of A 0 σ are positive, thus deti+t A 0 σ deti+ta 0 σ 1. and inequality 3.0 is immediatly prooved. We can remark that if, at least, one of the curvature of Σ at σ is strictly negative then, for d 3 resp. d k Σ σ < k Σ 1 σ 0 resp. k Σ σ < 0 and thus inequality 3.0 become strict. 6

8 Conclusion. The knowledge of a 0, ϕ, ϕ and Hessϕ on Σ is sufficient to compute them for all x in Σ + using formulaes along geometrical optic rays comming through x and we have Σ t Σ σ N σ σt σ +tn σ Figure. Wave Propagation Theorem. Let x Σ +, then there exist an unique σ Σ such that x σ + x σ N σ 3.1 and solution of is given by a 0 σ 1 ux;k e ikϕσ+ x σ +O deti+ x σ Hessϕσ k 3. Proof. By definition of Σ +, we have existence of σ Σ satisfying 3.1. For unicity, we suppose there exists σ 1 Σ and σ Σ satisfying 3.1. By convexity hypothesis on Σ, we have σ σ 1,N σ 1 0 and then But so we obtain By convexity hypothesis on Σ, we have σ σ 1,x σ 1 0. σ σ 1,x σ 1 σ σ 1,x σ + σ σ 1 σ σ 1,x σ σ 1 σ,n σ 0 7

9 and then σ 1 σ,x σ Thus, inequalities 3.3 and 3.4 give us σ 1 σ. Formula 3. is just an application of propagation Lemmas lemma 1 and lemma 3 along the ray σ R 0 x. 3.. Outside a strictly convex compact. Let K R d be a regular and stricly convex compact, Ω K c and Γ K be its boundary. Remark 1. Owing to the strict convexity of K and the hypotheses on Σ, we have There is only one reflexion on K. We denote by x Σ + Ω, l > 1, R l x. Γ s {γ Γ R 0 γ } and Γ e Γ s c Let γ Γ e and σ R 0 γ. Using previous formulas give us a 0 0, ϕ0, ϕ 0 and Hessϕ 0 on γ. If we can compute reflected wave on γ i.e. a 1 0, ϕ1, ϕ 1 and Hessϕ 1 on γ then we can use propagation formulaes along the reflected ray given by γ +t ϕ 1 γ, t > 0 a 0 0 σ ϕ Given 0 σ ϕ 0 σ Hessϕ 0 σ a 1 0 σ ϕ? 1 σ ϕ 1 σ Hessϕ 1 σ σ σ N σ nγ Σ 1 Σ γ Γ Figure 3. Wave reflection So, the local problem is only to find how to compute the reflected wave in γ. That is to say : a 0 0 γ ϕ 0 γ ϕ 0 γ Hessϕ 0 γ Computation of ϕ 1 and ϕ 1 in γ. Lemma 4. Let γ Γ, then given, how to compute Proof. Due to boundary condition : If γ Γ D, we have a 1 γe ikϕ1 γ a 0 γe ikϕ0 γ a 1 0 γ ϕ 1 γ ϕ 1 γ Hessϕ 1 γ ϕ 1 γ ϕ 0 γ 3.5 8? 3.6

10 If γ Γ N, we have a 1 n γ ıka1 γ ϕ1 n γ e ıkϕ1 γ a 0 n γ ıka0 γ ϕ0 n γ e ıkϕ0 γ If γ Γ R, we have [ ] a αk 1 n γ ıka1 γ ϕ1 n +βka γ 1 γ [ ] a αk 0 n γ ıka0 γ ϕ0 n +βka γ 0 γ By identification, we immediately obtain 3.5. e ikϕ1 γ e ikϕ0 γ Lemma 5. Let γ Γ, and nγ the exterior normal to Γ at γ. We have ϕ 1 γ ϕ 0 γ ϕ 0 γ,nγ nγ 3.9 Proof. Let B Γ γ {uγ,vγ,nγ} be the direct orthonormal curvature basis of Γ at γ. Then we have the local parametrization of Γ at γ : γu,v γ +u,v,gu,v with gu,v 1 kγ 1γu +k Γ γv +ou +v Taylor s expansion at order 1 of ϕ 0 and ϕ 1 on Γ at point γ are ϕ 0 γu,v ϕ 0 γ+ ϕ 0 γ,uuγ+vvγ ϕ 1 γu,v ϕ 1 γ+ ϕ 1 γ,uuγ+vvγ +Ou +v +Ou +v Due to relation 3.5 we obtain ϕ 1 γ ϕ 0 γ,uuγ+vvγ +Ou +v that is to say So, exists λ R such that ϕ 1 γ ϕ 0 γ,uγ ϕ 1 γ ϕ 0 γ,vγ 0 ϕ 1 γ ϕ 0 γ+λnγ Taking the norm of previous relation and using eikonal equation 3.6 give λ ϕ 0 γ,nγ. A similar proof will act in dimension Computation of a 1 0 in γ. Lemma 6. Let γ Γ e. If we know a 0 0 γ and ϕ0 γ then a 1 where the function b ϕ 0 γ is given in definition 5. Proof. If γ Γ D Γ e, formula 3.6 give 0 γ b ϕ γγa γ a 1 γ a 0 γ Using asymptotic expansions of a 0 and a 1 k j a 1 j γ k j a 0 j γ j N j N With high frequency hypothesis, we obtain the leading term in power of k: a 1 0 γ a0 0 γ 9

11 If γ Γ N Γ e, formula 3.7 give a 1 n γ ıka1 γ ϕ1 n γ Using asymptotic expansions of a 0 and a 1 a1 j j k j N j N k The leading term in power of k is But lemma 5 give ϕ1 n obtain a 1 a 0 n γ ıka0 γ ϕ0 n γ n γ ık j+1 a 1 j γ ϕ1 n γ a0 j j n γ ık j+1 a 0 j γ ϕ0 n γ 0 γ ϕ1 γ a0 0 n γ ϕ0 n γ γ ϕ0 n γ. Like γ Γe, we have ϕ 0 γ,nγ 0. So we a 1 0 γ a0 0 γ If γ Γ R Γ e, formula 3.8 give a 1 αk n γ ıka1 γ ϕ1 [ a 0 αk n γ +βka 1 γ n γ ıka0 γ ϕ0 n γ Using asymptotic expansions of a 0, a 1, αk and βk j Nk j α j j N + k j a1 j n γ ık j+1 a 1 j Nk j+1 β j 1 j Nk j a 1 j γ j γ ϕ1 ] +βka 0 γ n γ j Nk j α j k j a0 j n γ ık j+1 a 0 j γ ϕ0 n γ j N j Nk j+1 β j 1 j Nk j a 0 j γ The leading term in power of k is ıα 0 a 1 0 γ ϕ1 n γ+β 1a 1 0 γ ıα 0 a 0 0 γ ϕ0 n γ+β 1a 0 0 γ But ϕ1 n γ ϕ0 n γ, so a 1 0 γ β 1 +ıα 0 ϕ 0 n with hypothesis β 1 > α 0 we have So we obtain a 1 a 0 0 γ ϕ 0 β 1 ıα 0 n ϕ 0 β 1 +ıα 0 n 0 ϕ 0 γ β 1 ıα 0 0 n β 1 +ıα 0 ϕ 0 n 10 a 0 0 γ

12 We notice that γ Γ e b ϕ γγ This the case when there is no absorption in the boundary condition Calculus of Hessϕ 1 at γ Γ e. Lemma 7. Let γ Γ e. We note Bγ the curvature matrix of Γ at γ. Then, in dimension d or 3, we have Hessϕ 1 γ T d Bγ,nγ, ϕ 0 γ Hessϕ0 γ. 3.3 Proof. We have seen lemma 4 that ϕ 1 γ ϕ 0 γ. Let V Γ γ be a neighboorhood of γ in Γ. Taylor s expansion at order 1 of ϕ 0 and ϕ 1 are, for all γ V Γ γ, ϕ 0 γ ϕ 0 γ+hessϕ 0 γγ γ+o γ γ 3.33 and ϕ 1 γ ϕ 1 γ+hessϕ 1 γγ γ+o γ γ 3.34 Let γu,v be the local parametrization of Γ at γ define in section?? We compute now, Taylor s expansion at order 1 of nγu,v in dimension and 3. For that, we first evaluate γ u γ k1γu Γ u, v k Γ v γv +Ou +v, 1 We obtain B Γγ nγ+bγγu,v γ+ou +v. nγu,v nγ+bγγu,v γ +Ou +v. 1+k1 Γγu +k Γγv 1 But 1+k1γu Γ +kγv Γ 1 1+Ou +v so, we have nγ nγ+bγγ γ+o γ γ 3.35 In similar way, we obtain the same formula in dimension d. Using formulas 3.33 and 3.35 in equation 3.9 gives ϕ 1 γ ϕ 1 γ+hessϕ 0 γ ϕ 0 γ,nγ Bγγ γ ϕ 0 γ,bγγ γ nγ Hessϕ 0 γγ γ,nγ nγ+o γ γ So we obtain with formula 3.34, for all γ V Γ γ, Hessϕ 1 γγ γ Hessϕ 0 γ ϕ 0 γ,nγ Bγγ γ ϕ 0 γ,bγγ γ nγ Hessϕ 0 γγ γ,nγ nγ+o γ γ Now, we want to extend the previous formula for all x in Vγ R d, a neighborhood of γ. So we must add to previous formula a function from R d to R d which vanishes on Γ. That is to say at order, there exists a constant vector Cγ R d such that : for all x Vγ Hessϕ 1 γx γ Hessϕ 0 γ ϕ 0 γ,nγ Bγx γ ϕ 0 γ,bγx γ nγ Hessϕ 0 γx γ,nγ nγ + x γ,nγ Cγ+o x γ 3.36 To compute Cγ, we take x Vγ such that x γ ε ϕ 1 γ with ε > 0 and use 3.6 for ϕ 0 and ϕ 1 Hessϕ 0 x ϕ 0 x 0 and Hessϕ 1 x ϕ 1 x 0 11

13 So, formula 3.36 become Hessϕ 0 γ ϕ 0 γ,nγ Bγ ϕ 1 γ ϕ 0 γ,bγ ϕ 1 γ nγ Hessϕ 0 γ ϕ 1 γ,nγ nγ + ϕ 1 γ,nγ Cγ 0 Using 3.9 and Bγnγ 0 we obtain Bγ ϕ 1 γ Bγ ϕ 0 γ and so [ Bγ ϕ Cγ Hessϕ 0 0 γ, ϕ 0 γ γnγ,nγ ]nγ ϕ 0 γ,nγ [ Hessϕ 0 γnγ+bγ ϕ 0 γ ] Replacing Cγ by previous formula in 3.36 immediately give 3.3. These results are given in [ec89] Properties of Hessϕ 1 γ. Lemma 8. The matrix Hessϕ 1 γ is symmetric and Hessϕ 1 γ ϕ 1 γ In dimension d 3, eigenvalues of Hessϕ 1 γ are k 1 1 γ, k1 γ,0 and k 1 γ k1 1 γ < 0, k1 1 γ+k1 γ < k0 1 γ+k0 γ and k1 1 γk1 In dimension d, eigenvalues of Hessϕ 1 γ are k 1 γ,0 and k 1 γ < k 0 γ 0. γ < k0 1 γk0 γ. Proof. As we have seen in section 3.1., in dimension d 3 resp. d, eigenvalues of Hessϕ 0 γ are k 0 1 γ, k0 γ,0 resp. k0 γ,0 where k 0 γ k0 1 γ 0 resp. k0 γ 0. So we only have to apply lemma 1 in dimension or lemma 13 in dimension 3 to end the proof Conclusion. Outside a stricly convex compact, we proved following result Theorem 3. Let x Σ + Ω, such that Gx. Then #Rx #R 1 x 1, #R 0 x 1, and u inc x 1 u G.O. ρ x+o k and ρ R 0x ux ρ R 1x 1 u G.O. ρ x+o k 3.3. Outside an union of strictly convex compacts. To prove theorem 1, we just have to verify that we can use propagation and reflection lemmas along each ray coming through x. For that, we have Lemma 9. Let x Σ + Ω, such that Gx. Let l N such that R l x and ρ γ 0,γ 1,...,γ l R l x. In dimension d 3 resp. d, if eigenvalues of the symmetric matrix A 0 ρ γ 0 are λ 0 1 γ 0, λ 0 γ 0, 0 resp. λ 0 γ 0,0 where 0 λ 0 1 γ 0 λ 0 γ 0 resp. 0 λ 0 γ 0 then j {1,,l} eigenvalues of A j ρ γ j are λ j 1 γ j,λ j γ j,0 resp. λ j γ j,0 where 0 < λ j 1 γ j λ j γ j resp. 0 < λ j γ j 1

14 Proof. Due to hypothesis we can apply propagation lemma to obtain that A 0 ρ γ 1 S γ1 γ 0 A 0 ρ γ 0 is well defined. and its eigenvalues are λ 0 1 γ 1,λ 0 γ 1,0 where 0 λ 0 1 γ 1 λ 0 γ 1. By hypothesis ρ R l x so γ 1 γ 0,nγ 1 < 0 and we can apply reflection lemma 7 to obtain that A 1 ρ γ 1 T γ Bγ1,nγ 1, 1 γ 0 A 0 ρ γ 1 γ 1 γ 0 is well defined. Furthermore,lemma8giveusthatmatrixA 1 ρ γ 1 issymmetricanditseigenvaluesareλ 1 1 γ 1,λ 1 γ 1,0 where 0 < λ 1 1 γ 1 λ 1 γ 1. Then a simply recurrence proof give us lemma : Let j {1,,l 1} and suppose that A j ρ γ j is symmetric and its eigenvalues are λ j 1 γ j,λ j γ j,0 where 0 < λ j 1 γ j λ j γ j we can apply propagation lemma to obtain that A j ρ γ j+1 S γj+1 γ j A j ρ γ j is well defined. and its eigenvalues are λ j 1 γ j+1,λ j γ j+1,0 where 0 < λ j 1 γ j+1 λ j γ j+1. By hypothesis ρ R l x so γ j+1 γ j,nγ j+1 < 0 and we can apply reflection lemma 7 to obtain that A ρ j+1 γ j+1 T γ Bγj+1,nγ j+1 γ j A j j+1, ρ γ j+1 is well defined. Furthermore,lemma8giveusthatmatrixA ρ j+1 where and γ j+1 γ j 0 < λ j+1 1 γ j+1 λ j+1 γ j+1. A similar proof will act in dimension d. With theorem 1 hypothesis and this lemma, we immediatly have j {1,...,l}, det I+ γ j γ j 1 A j 1 ρ γ j 1 > 0 γ j+1 issymmetricanditseigenvaluesareλ j+1 1 γ j+1,λ j+1 γ j+1,0 det I+ x γ l A l ρ γ l > 0. So along each ray comming throug x we can apply propagation and reflection lemmas. Then, by adding the contribution of each ray we obtain theorem 1 results. Lemma 10. Let x Σ + Ω, such that Gx. Let l > 1 such that R l x and ρ γ 0,γ 1,...,γ l R l x. We have a 0 ρ γ 0 a ρ x < [ 1+dxλ Γ min 1+dmin λ Γ min l 1 ]d where and 0 < dx min x γ, γ Γ min λ 1γ if d 3, 0 < λ Γ γ Γ min min λγ if d,, γ Γ 0 < d min min i j min γ i γ j. γ i Γ i,γ j Γ j 13

15 Proof. Using 3.31 and definition of a j ρ γ j, we obtain j {1,,l}, a j+1 ρ γ j+1 where t j γ j+1 γ j and γ j+1 x. In dimension d, we can use equation 4.1 of lemma 1 to obtain λ j γ j λ j 1 γ j t j 1 λ Γ γ j nγ j,γ j γ j 1 But we have 1 t j 1 nγ j,γ j γ j 1 [ 1;0[ and then From proposition 1 we get and so λ j γ j λ j 1 γ j +λ Γ min > 0. λ j 1 γ j From equation 3.39, we immediately obtain and thus a j ργ j 3.39 det I+ γ j+1 γ j A j ρ γ j λ j 1 γ j 1 1+t j 1 λ j 1 γ j 1 λ j 1 γ j > 0 j {,,l} λ 0 γ 1 0 a 1 ρ γ 1 a 0 ρ γ 0 a ρ j+1 γ j+1 < a ρ j γ j 1+d min λmin Γ 1/ j {1,,l 1} a l ρ x a ρ x < a l ρ γ l 1+dxλ Γ 1/ min a 0 ρ γ 0 a ρ x < [. 1+dxλ ]1 Γ min 1+dmin λmin Γ l 1. In dimension d 3, We have det I+t j A ρ j γ j 1+λ j 1 γ j+λ j γ jt j +λ j 1 γ jλ j γ jt j and we can use equations 4.5 and 4.6 of lemma 13 to obtain λ j 1 γ j+λ j γ j λ j 1 1 γ j +λ j 1 γ j +4λ Γ min > 0, λ j 1 γ jλ j γ j λ j 1 1 γ j λ j 1 γ j +4λ Γ min > 0. So, with these inequalities, we have det I+t j A ρ j γ j 1+4λ Γ mint j +4λ Γ mint j + + λ j 1 1 γ j λ j 1 γ j From proposition 1 we get λ j 1 λ j 1 α γ j 1 α γ j 1+t j 1 λ j 1 α γ j 1, α {1,} t j λ j 1 1 γ j +λ j 1 γ j t j and, as λ j 1 γ j 1 λ j 1 1 γ j 1 > 0 for j {,,l +1} and λ 0 γ 0 λ 0 1 γ 0 0 see lemma 13 we have λ j 1 γ j λ j 1 1 γ j > 0 j {,,l} λ 0 γ 1 λ 0 1 γ. 1 0 With these inequalities, we have det I+t j A ρ j γ j > 1+λ Γ mint j if j > 1 det I+t 1 A 1 ρ γ 1 1+λ Γ mint 1 14

16 From equation 3.39, we immediately obtain and thus a 1 ρ γ 1 a 0 ρ γ 0 a ρ j+1 γ j+1 < a ρ j γ j 1+d min λmin Γ 1 j {1,,l 1} a l ρ x a ρ x < a l ρ γ l 1+dxλ Γ 1 min a ρ x < a 0 ρ γ 0 1+dxλ Γ min 1+dmin λ Γ min l Technical results Proposition 1. Let t > 0 and A S d t. Assume that λ is an eingenvalue of A with corresponding eingenvector ζ then λ 1+tλ is an eingenvalue of S ta with corresponding eingenvector ζ. Proof. By definition S t A AI+tA 1 and by hypothesis I+tAζ 1+tλζ with 1+tλ 0. So 1+tλI+tA 1 ζ ζ and we obtain Thus 1+tλAI+tA 1 ζ Aζ. AI+tA 1 ζ λ 1+tλ ζ. Lemma 11. Let η R d and ζ R d such that η 1 and η,ζ 0 Let A and B symmetric matrices in R d d. Assume Aζ 0 and Bη 0 then Proof. Note ξ ζ ζ,η η then, by definition, T d B,η,ζ Due to hypothesis, we have and So, we obtain T d B,η,ζ T d B,η,ζAζ ζ,η η 0 Aξ Aξ ζ,η Bξ η,ξ Aη [ +Bζ Aη +Bζ,ξ η + Aη,η Bζ,ζ ζ,η Aξ ζ,η Aη, Bξ Bζ, η,ξ ζ,η Aη +Bζ,ξ ζ,η Aη,η + Bζ,ζ. ] η,ξ η Aξ ζ,η Aη +Bζ+ ζ,η Aη [ +Bζ+4 ζ,η Aη,η η ] Bζ,ζ η η,ζ η 0. Aη,η Bζ,ζ ζ,η Lemma 1. Let B u,w and B I u I,w I two direct orthonormal basis of R such that w,w I < 0. Let B diagλ,0 in B basis and A I diagλ I,0 in B I basis. Then A R T B,w,w I A I is a symmetric matrix having for eingenvalues λ R,0 where λ R λ I λ w,w I 4.1 and If λ > 0 and λ I 0 then A R w I w,w I w 0 4. λ R > λ I

17 Proof. In basis B, we note u I u j j1, wi w j j1 and AI a ij i,j1. By hypothesis, AI is symmetric and A I w I 0. As w,w I 0, we can compute A R : x R A R x A I w,w I B x w,x A I w +Bw I A I w +Bw I,x w + [ A I w,w BwI,w I w,w I ] w,x w in basis B, x x j j1 and A R a11 a x 1 x1 λx1 a1 +λw w a 1 a x x 1 0 a [ ] a1 +λw 1 x1 0, + a a x 1 λw 1 0 x w 1 That is to say, A R a11 λw a 1 λw 1 a 1 λw 1 a λ w 1 w So we have A R symmetry. We can remark that formula 4. is a direct application of Lemma 11. Now, we study eingenvalues of A R. Let H be the normal matrix given by u1 w H 1 u w Then HA R H t is a similar to A R. Using A I u I λ I u I and A I w I 0 give a11 λw u 1 +a 1 +λw 1 u a 11 λw w 1 +a 1 +λw 1 w A R H a 1 +λw 1 u 1 a λ w 1 w u a 1 +λw 1 w 1 a λ w 1 w w λ I u 1 +λw 1 u w u 1 0 λ I u +λ w 1 w u w 1 u 1 0 But u I,w I is a direct and orthonormal basis, so Then H t A R H w u 1 w 1 u 1, w 1 u 1 +w u 0 and u 1 +u 1. [ ] λ I u 1 +u +λ w 1 u w u 1 u 1 w 1 w u +w 1 u 1 u λ I u 1 w 1 +u w +λw1u w u 1 w 1 w1u +w u 1 w 1 λ I λ w [w u 1 w 1 u ] λ I λ w Lemma 13. Let B u,v,w and B I u I,v I,w I two direct orthonormal basis of R 3 such that w,w I < 0. Let B diagλ 1,λ,0 in B basis and the symmetric matrix A I diagλ I 1,λI,0 in B I basis. Then A R T 3 A I is a symmetric matrix having for eingenvalues λ R B,w,w I 1,λ R,0 and A R w I w,w I w Under the hypothesis we have 0 0 < λ 1 λ and 0 λ I 1 λ I H λ R 1 +λ R λ I 1 +λ I +4λ 1 > 0, 4.5 λ R 1 λ R λ I 1 λi +4λ 1 >

18 and thus λ R 1 > 0, λ R > Proof. In basis B, u I u j 3 j1, vi v j 3 j1, wi w j 3 j1 and AI a ij 3 i,j1. By hypothesis, A I is symmetric and A I w I 0. As w,w I 0, we can compute A R : x R 3 in basis B, x x j 3 j1 and A R x A I w,w I B x w,x A I w +Bw I A I w +Bw I,x w A R x + [ A I w,w BwI,w I w,w I a 11 a 1 a 13 a 1 a a 3 a 13 a 3 a 33 a 13 +λ 1 w 1 x 3 a 3 +λ w a 33 x 1 x x 3 a 13 +λ 1 w 1 a 3 +λ w, a 33 ] w,x w λ 1 x 1 λ x 0 x 1 x x 3 [ + a 33 λ 1w1 +λ w ] 0 x That is to say, a 11 λ 1 a 1 a 13 λ 1 w 1 A R a 1 a λ a 3 λ w a 13 λ 1 w 1 a 3 λ w a 33 λ 1 w1 +λ w So we have A R symmetry. To proove formula 4.4, we make computation in basis B where w I w,w I w 1 w w So A R w R a 11 w 1 λ 1 w 1 +a 1 w +a 13 +λ 1 w 1 a 1 w 1 +a w λ w +a 3 +λ w a 13 w 1 λ 1 w1 a 3 w λ w a 33 +λ 1 w1 +λ w Then, using A I w I 0 give immediatly formula 4.4. To establish the formulas under the assumptions H, we will study the eigenvalues of matrix A R. For that, let H be the normal matrix given by u 1 v 1 w 1 H u v w u 3 v 3 Then HA R H t is similar to A R. Using that u I,v I,w I is a direct and orthonormal basis, we have u v 3 u 3 v w 1 u I v I w I u 1 v 3 +u 3 v 1 w 4.8 u 1 v u v 1 v I w I u I v v 3 w v 1 +v 3 w 1 u 1 u 4.9 v 1 w v w 1 u 3 17

19 w u 3 u v 1 w I u I v I w 1 u 3 + u 1 v 4.10 w 1 u w u 1 v 3 u I 1 u 1 +u +u v I 1 v 1 +v +v w I 1 w 1 +w +w Using A I u I λ I 1 ui, A I v I λ I 1 vi and A I w I 0 give { A R H λ I 1 u 1 +λ 1 w 1 u 3 u 1 λ I v 1 +λ 1 w 1 v 3 v 1 0 λ I 1 u +λ w u 3 u λ I v +λ w v 3 v 0 λ I 1 u 3 λ 1 w 1 u 1 u3w 1 λ w u u3w } { λ I v 3 λ 1 w 1 v 1 v3w 1 λ w v v3w } 0 To obtain H t A R H, we compute H t A R Hx,y x,y {u,v,w} We can remark that H t A R Hx,w 0 x {u,v,w} Now, we compute the six leading terms of matrix H t A R H Calculus of H t A R Hu,u H t A R Hu,u λ I 1 u 1 +λ 1 w 1 u 3 u 1 u 1 + λ I 1 u +λ w u 3 u u λ I 1 u 3 λ 1 w 1 u 1 u 3w1 λ w u u 3w u 3 By hypothesis u I 1 so H t A R Hu,u Using formula 4.10 give Calculus of H t A R Hu,v λ I 1 + λ 1 w3 w 1 u 3 u 1 w w 3u 1 w1u λ w3 w u u 3 w 3u w u 3 λ I 1 λ 1 u 1 w 1 u 3 λ u w u 3 H t A R Hu,u λ I 1 λ1 v +λ v 1 H t A R Hu,v λ I 1 u 1 +λ 1 w 1 u 3 u 1 v 1 + λ I 1 u +λ w u 3 u v λ I 1 u 3 λ 1 w 1 u 1 u 3w1 λ w u u 3w 18 v 3

20 By hypothesis u I,v I 0 so H t A R Hu,v λ 1 w3 w 1 u 3 v 1 w 3u 1 v 1 + w 1 u 1 v 3 w 1u 3 v 3 + λ w3 w u 3 v w 3u v + w u v 3 w u 3 v 3 λ 1 u 1 w 1 u 3 w 1 v 3 v 1 + λ u w u 3 w v 3 v Using formulas 4.9 and 4.10 give H t A R Hu,v λ 1 u v +λ u 1 v 1 Calculus of H t A R Hu,w H t A R Hu,w λ I 1 u 1 +λ 1 w 1 u 3 u 1 w 1 + λ I 1 u +λ w u 3 u w λ I 1 u 3 λ 1 w 1 u 1 u 3w1 λ w u u 3w By hypothesis u I,w I 0 so H t A R Hu,w 0 Calculus of H t A R Hv,u H t A R Hv,u λ I v 1 +λ 1 w 1 v 3 v 1 u 1 + λ I v +λ w v 3 v u λ I v 3 λ 1 w 1 v 1 v 3w1 λ w v v 3w u 3 By hypothesis u I,v I 0 so H t A R Hv,u λ 1 w1 u 1 v 3 w 3u 1 v 1 +w 1 u 3 v 1 w 1u 3 v 3 + λ w u v 3 w 3u v +w u 3 v w u 3 v 3 λ 1 u 1 w 1 u 3 w 1 v 3 v 1 + λ u w u 3 w v 3 v Using formulas 4.9 and 4.10 give H t A R Hv,u λ 1 u v +λ u 1 v 1 Calculus of H t A R Hv,v H t A R Hv,v λ I v 1 +λ 1 w 1 v 3 v 1 v 1 + λ I v +λ w v 3 v v λ I v 3 λ 1 w 1 v 1 v 3w1 λ w v v 3w 19 v 3

21 By hypothesis v I 1 so H t A R Hv,v λ I + λ 1 w1 v 1 v 3 w w 3v 1 w1v λ w v v 3 w 3v w v 3 λ I λ 1 v 1 w 1 v 3 λ v w v 3 Using formula 4.9 give H t A R Hv,v λ I λ1 u +λ u 1 Calculus of H t A R Hv,w H t A R Hv,w λ I v 1 +λ 1 w 1 v 3 v 1 w 1 + λ I v +λ w v 3 v w λ I v 3 λ 1 w 1 v 1 v 3w1 λ w v v 3w By hypothesis v I,w I 0 so H t A R Hu,w 0 With all these formulas we obtain : λ I H t A R 1 w λ1 3 v +λ v 1 λ 1 u v +λ u 1 v 1 0 H λ 1 u v +λ u 1 v 1 λ I w λ1 3 u +λ u As H t A R H is similar to A R, we refind that 0 is an eigenvalue of A R and the two others eigenvalues λ R 1 and λ R are also eingenvalues of λ I 1 A w λ1 3 v +λ v1 λ 1 u v +λ u 1 v 1 λ 1 u v +λ u 1 v 1 λ I w λ1 3 u +λ u 1 Furthermore, we have that TrA λ R 1 +λ R 4.15 deta λ R 1 λ R 4.16 To establish inequality 4.5, we use equation 4.15 under hypothesis H and < 0 to get We remark that and so, we obtain λ R 1 +λ R λ I 1 +λ I λ 1 u 1 +u +v 1 +v. u 1 +u +v 1 +v u 1 +v +u v 1 u 1 v u v 1 u 1 +v +u v 1 λ R 1 +λ R λ I 1 +λ I λ 1 u1 +v +u v 1 λ I 1 +λ I +4λ 1 +4λ 1 u1 +v +u v 1 λ I 1 +λ I +4λ 1. 0

22 To establish inequality 4.6, we use equation 4.16 to get λ R 1 λ R λ I 1 λ1 v +λ v 1 λ I λ1 u +λ u 1 4 w3 λ 1 u v +λ u 1 v 1 λ I 1 λi + 4 [ λ1 w3 v +λ v1 λ1 u +λ u 1 λ1 u v +λ u 1 v 1 ] λ1 v +λ v I 1 λ + λ 1 u +λ u I 1 λ 1 λ I 1 λi λ1 v +λ v I 1 λ + λ 1 u +λ u I 1 λ [ λ w3 1 u v +λ 1 λ u 1v +λ 1 λ v1u λ 1u v λ 1 λ u 1 v 1 u v λ 1 λ v1u ] λ I 1 λi λ1 v +λ v I 1 λ + λ 1 u +λ u I 1 λ [ w3λ 1 λ u 1 v u 1 v 1 u v +v1u ] λ I 1 λi λ1 v +λ v I 1 λ + λ 1 u +λ u I 1 λ w3 λ 1 λ u 1 v v 1 u From formula 4.8, we have u 1 v v 1 u and so we obtain λ R 1 λ R λ I 1 λi λ1 v +λ v I 1 λ + λ 1 u +λ u I 1 λ 1 +4λ 1 λ Under hypothesis H and < 0, we clearly have and we obtain formula 4.6. λ R 1 λ R λ I 1 λi +4λ 1 > 0 References [ec89] M.Balabane et C.Bardos. Equation des ondes : solutions asymptotiques et singularités. Compte Rendus INRIA de la session électromagnétisme, Université Paris XIII, Institut Galilée, LAGA CNRS UMR 7539,, 99, Av. J.-B. Clément F Villetaneuse, France address: cuvelier@math.univ-paris13.fr 1