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Dec 8, 2014 - Keywords: wave guides; transmission; singularities; transparent boundary conditions. Mathematics Subject Classification 2000: 22E46, 53C35, ...
Singularities occuring in multimaterials with transparent boundary conditions Philippe Destuynder, Caroline Fabre

To cite this version: Philippe Destuynder, Caroline Fabre. Singularities occuring in multimaterials with transparent boundary conditions. 2014.

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Preprint

Singularities occuring in multimaterials with transparent boundary conditions

Philippe Destuynder1 and Caroline Fabre2 1 Applied Mathematical Department, CNAM, 292 rue Saint Martin 75015 Paris, France [email protected] 2 Laboratoire de Math´ ematiques d’Orsay UMR 8628 Universit´ e Paris-Sud Bˆ atiment 425 91405 Orsay Cedex, France [email protected]

Computation of waves at the interface between two materials with different wave velocities is an important engineering problem. Transparent boundary conditions on the input and output boundaries are known for a single velocity wave. Adapting them in case of two velocities, singularities appear in the computation of the waves. We precisely exhibit these singularities both with theoritical and numerical points of view. Keywords: wave guides; transmission; singularities; transparent boundary conditions Mathematics Subject Classification 2000: 22E46, 53C35, 57S20

Introduction The theory of transparent boundary conditions is a large and deep domain of mathematical studies. To our knowledge, the first main theoretical paper on the subject concerns simulations of solutions of a linear wave equation in the exterior of a bounded domain and is due to B. Engquist and A. Majda ([5]) : the authors suggest a method, based on a Fourier transform in time and the transverse direction, which leads to exact and non local transparent boundary conditions. They also perform several series developments (with different orders) in order to obtain approximate but local transparent boundary conditions. Let us also notice the works of L. Halpern ([10] [11]) : she (and co-authors) studied well-posedness of different boundary conditions and their numerical schemes. When one studies solutions of linear wave equations with a unique velocity, the boundary conditions on the input and output boundary are transparent boundaries conditions : they ensure that waves should go out of the domain (there is no reflection) and they avoid singularity (see Fig.2). In case of a transmission problem, the difference between wave’velocities involves singularities which are localized at the intersections between the boundary and the interface of transmission (see Fig.3 and Fig.4). We give a precise description of them with a mathematical analysis of 1

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this phenomenon : we prove that the singularities are involved by the gap across the interface on the lateral boundary of the antisymmetric part of the Neumann derivative of the solution. Let us notice that these singularities do not appear for homogeneous Neumann boundary conditions on the whole boundary of Ω. Let us illustrate our claim with the following different situations : figures 1, 2, 3 and 4 below concern solutions of a transmission problem (see system (1.1)) in the square Ω =]0, 1[×]0, 1[ and during the time T = 1.2. The interface is localized at y = 0.2 and the velocities are c = c− if y < 0.2 and c = c+ if y > 0.2 with 0 < c− ≤ c+ . Initial data are null and the right hand side f is localized in the region with velocity c− . Fig. 1 shows the support of the right hand side f . Numerical computations are performed in Matlab with a time step of order 0.001 and space 1 = 0.0083. Figures 2, 3 and 4 are steps (both in the x and y directions) of 120 snap-shots at time T = 0.41294 or T = 0.41367 in the three following situations : in Fig. 2, one has c− = c+ = 1 and the right hand side is a high frequencies time excitation. One can see that there is no singularity. In Fig. 3, one has c+ = 2 and c− = 1 and the right hand side f is a low frequencies time excitation. In Fig. 4, one has c+ = 2 and c− = 0.5 and the right hand side is a high frequency time excitation. In both Fig. 3 and 4, one can see singularities appearing at the intersection between the interface y = 0.2 and the the boundary x = 0. This is what we explain in the following. Others numerical computations are given in order to exhibit more precisely these singularities.

Fig. 1. The right hand side’s support

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Fig. 2. One material

Fig. 3. Two materials and low frequencies time excitation

3

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Fig. 4. Two materials and high frequencies time excitation

In a forthcoming paper, we suggest a way to avoid these singularities introducing efficient new tranparent boundary conditions. Let us notice that our method leads to a non local computation which is classical in the obtention of exact transparent boundary conditions at higher order but new for first order. Our plan is the following one : in a first section, we present the mathematical problem and we focus on the singularity. In a second section, we study existence and regularity results of its solutions. In a third section, we focus on singularities on the boundary of the interface. In this section, we state our main theorem. All along, the paper is completed by numerical simulations which have been performed with Matlab and which illustrate our theoritical results.

1. The wave model used for the discussion Let us consider a two dimensional open set Ω =]0, L[×] − a, a[ (L > 0 and a > 0) of R2 as shown on figure 5. A point x of Ω has coordinates x = (x1 , x2 ). The wave velocity is denoted by c and is piecewise constant. It is c+ in Ω+ = Ω ∩ (x2 > 0) and c− in Ω− = Ω ∩ (x2 < 0) and we assume that 0 < c− < c+ . For any given functions f = f (x, t), u0 = u0 (x) and u1 = u1 (x), let us consider

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5

Fig. 5. The open set Ω and the notations for the theoretical discussion

u = u(x, t) solution of the following mathematical model:  2 ∂ u   − div(c2 ∇u) = f in Q = Ω×]0, T [,  2  ∂t        ∂u     ∂ν = 0 on (Γ+ ∪ Γ− )×]0, T [, (1.1)   ∂u ∂u   +c = 0 on (Γe ∪ Γs )×]0, T [,    ∂t ∂ν         u(x, 0) = u0 (x) ∂u (x, 0) = u1 (x) in Ω. ∂t We write Γe+ = Γe × (x2 > 0), Γe− = Γe × (x2 < 0), Σ+ = Γ+ ×]0, T [, Σ− = Γ− ×]0, T [, Σe = Γe ×]0, T [, and Σs = Γs ×]0, T [. The interface is Γi = Ω ∩ [x2 = 0] and Σi = Γi ×]0, T [. The letter ν denotes the unit outward normal vector on the boundary Γ of Ω. Existence and uniqueness of solutions of system (1.1) are classical results, even if the transparent boundary conditions on Γe ∪ Γs (which imply first order time derivative) require a slightly different strategy as the usual one. Let us summarize them in the following statement. Proposition 1.1. Let us assume that u0 ∈ H 1 (Ω), u1 ∈ L2 (Ω) and f ∈ L2 (Q). Then, there exists a unique solution u to the system (1.1) with : u ∈ L∞ (]0, T [; H 1 (Ω)) ∩ W 1,∞ (]0, T [; L2 (Ω)).

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We don’t give the full proof of proposition 1.1, one can refer to [4] in case of interest : it is based on a Galerkin method, a priori estimate and weak convergence of a subsequence solution of a finite dimensional approximation. Let us recall the variational formulation of (1.1). • The function u is solution of Z 2 Z ∂ u 1 ∀v ∈ H (Ω), (x, t)v(x)dx + c2 ∇u(x, t).∇v(x)dx+ 2 Ω ∂t Z Ω Z (1.2) ∂u c (x, t)v(x) = f (x, t)v(x)dx Γe ∪Γs ∂t Ω • The energy is defined by Z Z 1 ∂u 1 2 E(t) = (x, t) dx + c2 |∇u(x, t)|2 dx 2 Ω ∂t 2 Ω

(1.3)

and one has for f = 0 Z c

E(t) + Σe ∪Σs

∂u (x, t)2 dx = E(0). ∂t

Therefore it decreases with respect to the time variable. In the general case, there exists a constant d > 0, such that for every data (u0 , u1 , f ) ∈ H 1 (Ω) × L2 (Ω) × L2 (Q) : Z E(t) +

c| Σe ∪Σs

∂u (x, t)|2 dx ≤ d[E(0) + ||f ||Q ]2 , ∂t

where || ||Q denotes the L2 (Q)− norm. These results prove that : ∂u ∈ L2 (Σe ∪ Σs ). ∂t ∂u 1 ∂u ∂u Since =− , one has ∈ L2 (Σe ∪ Σs ). ∂ν c ∂t ∂ν ∂u ∈ H −1 (0, T ; H 1 (Ω)), the discontinuity of c implies that Furthermore, even if ∂t ∂u ∂u in general 6∈ D0 (0, T ; H 1/2 (Γe )) and 6∈ D0 (0, T ; H 1/2 (Γs )). ∂ν ∂ν The function u, unique solution of (1.1), satisfies :  2   ∂ u − c2+ ∆u = f in Q+ = Ω+ ×]0, T [,   ∂t2       2 ∂ u (1.4) − c2− ∆u = f in Q− = Ω− ×]0, T [, 2  ∂t        ∂u ∂u   c2+ − c2− = 0 on Σi , ∂νi ∂νi where νi is one of the unit normal vector to Γi and

∂u = ∇u.νi . ∂νi

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The third equation of (1.4) implies that, whatever the smoothness of the data 2 are, there is in general no hope to have u ∈ D0 (0, T ; Hloc (Ω)) if c is not constant across the separation line Γi between the two materials. We now turn to the regularity study of solutions of (1.1). 2. On the smoothness of u in time and space Let us study the regularity of u first with respect to the time variable and then with respect to space variables. Singularities will be discussed in the next section. Based on derivative methods with respect to the time t, one can upgrade the results stated in Proposition 1.1 by assuming that the initial condition u0 doesn’t cross the interface Γi . Theorem 2.1. Let us define by O+, respectively O− , two open subsets whose closures are subsets of Ω+ and Ω− . If the initial conditions are such that u0 ∈ H 2 (Ω) with supp(u0 ) ⊂ O+ ∪ O− , u1 ∈ H 1 (Ω) and if f ∈ H 1 (]0, T [; L2 (Ω)), then the solution u of (1.1) is such that: u ∈ W 1,∞ (]0, T [; H 1 (Ω)) ∩ W 2,∞ (]0, T [; L2 (Ω)). From classical inclusions (see [3]), this implies for instance that u ∈ C 0 ([0, T ]; H 1 (Ω)). ∂u ∂f . The function u˙ is solution of (1.1) with data f˙ = ∈ ∂t ∂t ∂ u ˙ L2 (Q), u(x, ˙ 0) = u1 (x) ∈ H 1 (Ω) and (x, 0) = f (x, 0) + div(c2 ∇u0 ). ∂t The assumption on the initial data u0 ensures that ∇u0 is null on a neighborhood of the interface Γi in Ω and thus div(c2 ∇u0 ) ∈ L2 (Ω). Moreover, f ∈ C 0 ([0, T ]; L2 (Ω)) and f (x, 0) is in L2 (Ω). Applying Proposition 1.1, we easily deduce that u ∈ W 1,∞ (]0, T [; H 1 (Ω)) ∩ W 2,∞ (]0, T [; L2 (Ω)). Proof. Let us set u˙ =

By iterating Theorem 2.1, one easily gets Theorem 2.2. With the same notations as in Theorem 2.1, and if the initial conditions are such that (u0 , u1 ) ∈ H 3 (Ω) × H 2 (Ω) both with compact supports in O+ ∪ O− , and if f ∈ H 2 (0, T ; L2 (Ω)), then the solution u of (1.1) is such that: u ∈ W 2,∞ (]0, T [; H 1 (Ω)) ∩ W 3,∞ (]0, T [; L2 (Ω)) and thus u ∈ C 1 ([0, T ], H 1 (Ω)). Remark 2.1. Let us first recall that, if γ is a non-empty and open part of the whole boundary Γ of Ω, the restriction operator to γ maps H 1/2 (Γ) onto H 1/2 (γ)

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whereas the extension by zero of a function of H 1/2 (γ) is not in general a function of H 1/2 (Γ). In both cases of theorems 2.1 and 2.2, one has u|Γe ∪Γs ∈ C 1 ([0, T ]; H 1/2 (Γe ∪ Γs )) and therefore c

∂u ∈ C 0 ([0, T ]; H 1/2 (Γe ∪ Γs )). ∂ν |Γe ∪Γs 

The function c is discontinuous across Γi and therefore in general (even a piecewise continuous function is not globally H 1/2 of the whole open set) ∂u ∂u 6∈ D0 (0, T ; H 1/2 (Γe )) and 6∈ D0 (0, T ; H 1/2 (Γs )). ∂ν |Γe ∂ν |Γs

(2.1)

We get at least for 0 ≤ s < 1/2, ∂u ∂u ∈ C 0 ([0, T ]; H s (Γe )) and ∈ C 0 ([0, T ]; H s (Γs )). ∂ν |Γe ∂ν |Γs Let us now turn to the smoothness with respect to the space variables and there are several cases. For x ∈ R2 , we denote by Vx a small enough open and non-empty neighborhood of x in R2 . We write 1O the characteristic function of a set O of R2 ◦

and Γi the interior of Γi . Theorem 2.3. Let f ∈ H 1 (Q) and let us assume that the initial data as in Theorem 2.1. Let u be the solution of system 1.1. One has : 2 2 (1) u1Ω+ ∈ L∞ (]0, T [; Hloc (Ω+ )) and u1Ω− ∈ L∞ (]0, T [; Hloc (Ω− )) ∞ 2 ¯ (2) If x ∈ Γ+ ∪ Γ− then u ∈ L (]0, T [; H (Vx ∩ Ω)) ¯ (3) If x ∈ (Γe ∩Ω+ )∪(Γe ∩Ω− )∪(Γs ∩Ω+ )∪(Γs ∩Ω− ) then u ∈ L∞ (]0, T [; H 2 (Vx ∩Ω)) ◦ ∂u (4) If x ∈Γi then ∈ L∞ (]0, T [; H 1 (Vx )). ∂x1

Proof. (1) On Ω+ ×]0, T [, one has div(c2 ∇u) = c2+ ∆(u) =

∂2u ∈ L∞ (]0, T [; L2 (Ω+ )). ∂t2

From a localisation argument, it is classical to prove that u|Ω+ ∈ 2 L∞ (]0, T [; Hloc (Ω+ )). Of course, the same argument can be applied in Ω− . ∂u (2) Let x ∈ Γ+ for example. The boundary conditions = 0 on Σ+ allows us ∂ν to apply a symmetry argument and to consider an extension u ¯ of u across Γ+ which is still solution of 1.1 in a neighborhood Vx of x. The first point of this theorem leads to u ¯ ∈ L∞ (]0, T [; H 2 (Vx )) and thus u ∈ L∞ (]0, T [; H 2 (Vx ∩ Ω)).

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∂u 1 ∂u ∂u ∈ L∞ (]0, T [; H 1 (Ω)), we obtain =− ∈ ∂t ∂ν c ∂t ∞ 1/2 2 L (]0, T [; H (Γe ∩Ω+ )). After localisation, we get ∆u ∈ L ((Vx ∩Ω)×(0, T )) and ∂u ¯ ∈ L∞ (0, T ; H 1/2 (∂(Vx ∩Ω)). Classically, this leads to u ∈ L∞ (0, T ; H 2 (Vx ∩ Ω)). ∂ν ◦ ¯ (4) Let x ∈Γi . In this case, let us notice that we can assume that Vx ⊂ Ω. 0 2 There is no hope (in general) that u ∈ D (0, T ; Hloc (Vx )) excepted for c+ = c− . ∂u Let us consider ρ ∈ D(Ω) with ρ = 1 on Vx and let us write w1 = ρ . Since ∂x1 Γi is parallel to the axis and boundary x1 , the function w1 is still solution of a transmission problem similar to 1.1 with (3) Let x ∈ Γe ∩ Ω+ . Since

 2   ∂ w1 − div(c2 ∇w1 ) ∈ L2 (Q), ∂t2   ∂w1 ∈ L∞ (]0, T [; H 1/2 (∂Vx )), ∂ν and thus w1 ∈ L∞ (]0, T [; H 1 (Vx )). ∂u is L∞ (]0, T [, H 1 (]a, b[×] − a, a[)) in ∂x1 any rectangle ]a, b[×] − a, a[ with 0 < a < b < L.  Remark 2.2. One can easily prove that

We now turn to the main point : the description of the singularities focusing at the point (x1 , x2 ) = (0, 0). 3. Polylog-2 Singularities at the jonction of the interface and the boundary Γe (or Γs ) We focus our study at the point with the coordinates (0, 0) but, of course, an analogous result is valid at the point with coordinates (L, 0). We introduce an even function ρ with ρ = ρ(x2 ) ∈ C ∞ (R)  a a  ρ(x2 ) = 1 for − < x2 < 2 2 (3.1)  ρ(x2 ) = 0 for |x2 | > 3a 4 a a inf(a, L) [×] − , [. We consider a function ρa ∈ C ∞ (Ω) Let us denote W =]0, 2 2 2 such that  ρa = 1 in W, (3.2) ρa = 0 in a neighborhood of Γ+ ∪ Γ− ∪ Γs . Let us set : V = {w ∈ H 1 (Ω), ρw1Ω+ ∈ H 2 (Ω+ ), ρw1Ω− ∈ H 2 (Ω− ); ∂ρw and ∈ H 1 (W )}. ∂x1

(3.3)

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For a given function h, we denote by Ts and Ta the following symmetric and non-symmetric part of h defined by Ts (h)(x1 , x2 ) =

1 [h(x1 , x2 ) + h(x1 , −x2 )], 2

(3.4)

Ta (h)(x1 , x2 ) =

1 [h(x1 , x2 ) − h(x1 , −x2 )]. 2

(3.5)

and

When no mistake can be made, we write hs and ha instead of Ts (h) and Ta (h). Of course, h = hs + ha . We denote by Im(z) the imaginary part of the complex number z and we introduce the two following functions : Li2 is the polylogarithm function of order 2 defined by Li2 (z) =

X zn n2

for |z| ≤ 1.

n≥1

π With z = exp[− (x1 − ix2 )], we consider the function S defined by : a 2a c− c+ S(x1 , x2 ) = 2 2 Im[Li2 (z) − Li2 (−z)] [ 1(x2 >0) + 1(x 0, such that usg (x1 , x2 , t) = (c+ − c− )α0 (t)S(x1 , x2 ) in ]0, ε[×] − ε, ε[. On ]0, ε[×] − ε, ε[, we have x2  2 1 + eiπ a ) − Arg( x2 c− c+  1 − eiπ a  ∇usg (0, x2 , t) = (c+ −c− )[ 1(x2 >0) + 1(x2 0 Ts (g1 , g2 )(x2 ) = 21   [g2 (x2 ) + g1 (−x2 )] if x2 < 0 2

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for g1 ∈ L2 (Γe+ ) and g2 ∈ L2 (Γe− ). We easily get T ∈ L(L2 (Γe+ ) × L2 (Γe− ); L2 (Γe )). If (g1 , g2 ) ∈ H 1 (Γe+ )×H 1 (Γe− ) then Ts (g1 , g2 ) ∈ H 1 (Γe ) (there is no gap across x2 = 0) hence T ∈ L(H 1 (Γe+ ) × H 1 (Γe− ); H 1 (Γe )). By interpolation of order 1/2, we deduce Lemma 3.2.



We now turn to the proof of Theorem 3.1 with the study of u(2) , solution of (3.8) and we first prove the following result concerning the solution u : it defines the gap of u|Γe at the origin. Proposition 3.1. Suppose (u0 , u1 ) ∈ H 1 (Ω) × L2 (Ω). We have u ˜1Ω+ L2 (0, T ; H 3/2 (Ω+ )), u ˜1Ω− ∈ L2 (]0, T [; H 3/2 (Ω− )) and we can write



g˜a = Ta (˜ g ) = α(t) sign(x2 ) + gar where α ∈ H −2 (0, T ) and gar ∈ H −2 (]0, T [; H 1/2 (Γe )). Proof of Proposition 3.1. Let us prove that u ˜1Ω+ ∈ L2 (]0, T [; H 3/2 (Ω+ )). We S A A S introduce u ˜ and u ˜ defined by u ˜ =u ˜−u ˜ and

u ˜S (x1 , x2 , t) =

 2 ˜(x1 , −x2 , t) ˜(x1 , x2 , t) + c2− u c u   + if x2 > 0  2 2  c+ + c−    ˜(x1 , x2 , t) ˜(x1 , −x2 , t) + c2− u c2 u    + if x2 < 0 c2+ + c2−

We have u ˜S ∈ L2 (0, T ; H 1 (Ω) and u ˜S = u ˜ on Γi . Let us prove that −1 2 u ˜ ∈ H (0, T ; H (Ω)). We obtain in Ω+ : S

∂u ˜S 1 ∂u ˜ ∂u ˜ = 2 [c2+ (x1 , x2 , t) − c2− (x1 , −x2 , t)] 2 ∂x2 c+ + c− ∂x2 ∂x2 ∂u ˜S thus (recall that u ˜ satisfies the transmission condition), we get = 0 on ∂ν Γi . An analogous calculus on Ω− proves that there is no gap of the normal derivative of u ˜S through the interface Γi . Since u ˜S ∈ L2 (0, T ; H 1 (Ω)), uS 1Ω− ) ∈ H −1 (0, T ; L2 (Ω− )), with no gap ∆(˜ uS 1Ω+ ) ∈ H −1 (0, T ; L2 (Ω+ )), ∆(˜ of the normal derivative through Γi , we deduce that ∆˜ uS ∈ H −1 (0, T ; L2 (Ω)). On another hand, we have on Ω+ ∂u ˜S 1 ∂u ˜ ∂u ˜ = 2 [c2 (x1 , x2 , t) + c2− (x1 , −x2 , t)] ∂x1 c+ + c2− + ∂x1 ∂x1

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and on Ω− ∂u ˜ ∂u ˜S 1 ∂u ˜ [c2− = 2 (x1 , x2 , t) + c2+ (x1 , −x2 , t)], 2 ∂x1 c+ + c− ∂x1 ∂x1 ∂u ˜S is a symmetric function on Γe and on Γs . We can apply Lemma 3.2 and ∂ν ∂u ˜S we get that ∈ H −1 (0, T ; H 1/2 (Γe ∪ Γs )). We have proved that u ˜S is solution ∂ν of thus

  ∆˜ uS ∈ H −1 (0, T ; L2 (Ω))   S u ˜ = 0 on Γ+ ∪ Γ−  ˜S   ∂u ∈ H −1 (0, T ; H 1/2 (Γe ∪ Γs )). ∂ν Since u ˜S is null in a neighborhood of Γ+ and Γ− , we deduce that u ˜S ∈ −1 2 H (0, T ; H (Ω)). Let us now prove that u ˜A 1Ω+ = (˜ u−u ˜S )1Ω+ ∈ H −1 (0, T ; H 3/2 (Ω+ )). We have A 2 1 S u ˜ ∈ L (0, T ; H (Ω)). Since u ˜=u ˜ on Γi , we get u ˜A |Γi = 0. Moreover, using that the normal direction on Γe (respectively Γs ) is the −x1 ’s one (respectively x1 ’s one), one can easily prove that uA satisfies in Ω+ × (0, T )  ∆˜ uA ∈ H −1 (0, T ; L2 (Ω+ ))        A ∂u ˜ ∂u ˜ ∂u ˜S = − ∈ H −1 (0, T ; H 1/2 (Γe+ ∪ Γs+ ))  ∂ν ∂ν ∂ν       A u = 0 on Γi ∪ Γ+ and therefore u ˜A ∈ H −1 (]0, T [; H 3/2 (Ω+ )) if (u0 , u1 ) ∈ H 1 (Ω) × L2 (Ω). We proved that u ˜ ∈ H −1 (]0, T [; H 3/2 (Ω+ )) + H −1 (]0, T [; H 2 (Ω+ )) and thus u ˜ ∈ −1 3/2 H (]0, T [; H (Ω+ )) for initial data (u0 , u1 ) in H 1 (Ω) × L2 (Ω). Of course, the same is valid in Ω− . ∂u ∂u ˜ |Γe+ and |Γ make sense in H −2 (]0, T [; H 1 (Γe+ )) and We deduce that ∂t ∂t e− H −2 (]0, T [; H 1 (Γe− )) and thus their values at point (0, 0) exist. Furthermore, ∂u ˜ ∈ H −1 (]0, T [; H 1/2 (Γe )), thus these values are equal. Let us write α = g˜a |Γe+ ∂t be the trace of the function g˜a on Γe+ . We have 1 1 ∂u ˜ 1 ∂u α(t) = g˜a (0+, t) = − [ + (0, 0+, t) − − (0, 0−, t)] 2 c ∂t c ∂t =

1 ∂u ˜ (c+ − c− ) (0, 0, t). 2c+ c− ∂t

We write g˜a (x2 , t) = α(t) sign (x2 ) + gar (x2 , t)

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Singularities

Transparent boundary conditions

where sign denotes the sign-function defined by sign (x) = 1 if x > 0 and sign (x) = −1 if x < 0. The function gar is an odd function with respect to x2 , gar 1Γe+ ∈ H −2 (0, T ; H 1 (Γe+ )) and gar 1Γe− ∈ H −2 (0, T ; H 1 (Γe− )) and their values at the point (0, 0) are null therefore gar ∈ H −2 (]0, T [; H 1/2 (Γe )) and proposition 3.1 is proved.  Let us return to the proof of our main theorem. We write u(2) = u(3) + u(4) ∂u(3) where u(3) (respectively u(4) ) is solution of (3.8) with = gar (respectively ∂ν (4) ∂u = α sign (x2 )) on Γe . ∂ν Lemma 3.1 can be apply to u(3) and leads to u(3) ∈ V. We deduce that the singular part of the function u is involved by the lack of continuity at the origin of 1 ∂u the odd part of the function . Let us study w = u(4) . c ∂t We use the same splitting as in proposition 3.1 and we write w = wS + wA . We get wS ∈ H −1 (0, T ; H 2 (Ω)) and thus it is sufficient to study u(5) = wA . We know that ∆u(5) = 0 separately in Ω+ and Ω− and that u(5) = 0 on Γi ∪ Γ+ ∪ Γ− . ∂u(5) ∂u(5) =− on Γe+ and Γe− . We have on Γe+ : Let us compute ∂ν ∂x1 ∂u(5) ∂u(4) ∂wS (x1 , x2 ) = (x1 , x2 ) − (x1 , x2 ) ∂x1 ∂x1 ∂x1 c2+ = −α −

= −α −

∂u(4) ∂u(4) (x1 , x2 ) + c2− (x1 , −x2 ) ∂x1 ∂x1 c2+ + c2−

c2+ (−α) + c2− α 2c2− = −α c2+ + c2− c2+ + c2−

and thus β+ (t) =

∂u(5) c− ∂u ˜ = (c+ − c− ) (0, 0, t) on Γe+ . 2 2 ∂ν c+ (c+ + c− ) ∂t

(3.14)

On Γ− , we get ∂u(4) ∂u(4) c2− (x1 , x2 ) + c2+ (x1 , −x2 ) ∂u(5) ∂ν ∂ν = −α − ∂ν c2+ + c2− = −α −

c2− (−α) + c2+ α 2c2+ = −α c2+ + c2− c2+ + c2−

and thus β− (t) =

∂u(5) c+ ∂u ˜ =− (c+ − c− ) (0, 0, t) on Γe− . ∂ν c− (c2+ + c2− ) ∂t

(3.15)

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Singularities

Polylog singularities

We deduce that the function u(5) is solution of   ∆u(5) = 0 in Ω+ , ∆u(5) = 0 in Ω− ,                  u(5) = 0 on Γ+ ∪ Γi ,  u(5) = 0 on Γ− ∪ Γi , and       ∂u(5) ∂u(5)     = β+ on Γe+ ∪ Γs+ , = β− on Γe− ∪ Γs− ,        ∂ν  ∂ν

19

(3.16)

In order to describe this singularity involved by the function u(5) and du to the fact that the role of the time variable is here a parameter, we begin with Let β ∈ R and let w = w(x) ∈ H 1 (Ω+ ) be solution of :  ∆w = 0 in Ω+ ,        w = 0 on Γ+ ∪ Γi ∪ Γs+ , (3.17)        ∂w = β on Γe+ ∂ν then there exists ε > 0 such that the behavior near the origin is given by

Proposition 3.2.

w(x) =

x1 x2 x1 x2 2a β Im[Li2 (e−2π( L +i a ) ) − Li2 (−e−2π( L +i a ) )] in ]0, ε[2 2 π

Proof. For sake of simplicity we consider another system which leads to the same singularity (the problem is a local one). Let us write Ωa = R+∗ ×]0, a[ and consider the solution v ∈ H 1 (Ωa ) of the system  ∆v = 0 in R+∗ ×]0, a[,          v = 0 on R+∗ × {0, a}, (3.18)    ∂v   = β on {0}×]0, a[.     ∂ν r We write for n ∈ N∗ , wn (x2 ) =

2 nπx2 sin( ) and a a

v(x1 , x2 ) =

X

an wn (x2 )e



nπx1 a

n≥1

the serie being convergent in the spaces H 1 (Ωa ) and H 2 (Ωa ∩ (x2 > 0)). Let us recall that the functions wn (n > 0) represent an orthonormal basis of L2 (]0, a[). We obtain for p ∈ N∗ ,

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Singularities

Transparent boundary conditions

pπx1 a v(x1 , x2 )wp (x2 )dx2 = ap e

a

Z



0

and nπx1 − ∂v πX a an nwn (x2 )e (x1 , x2 ) = − ∂x1 a n thus : Z

a

0

pπ ∂v (0, x2 )wp (x2 )dx2 = − ap ∂x1 a

which leads to : aβ ap = pπ

Z 0

a

a2 β wp (x2 )dx2 = 2 2 p π

r

2 [1 − (−1)p ]. a

We then get a2p = 0 and a2p+1

√ 2a 2a = β (2p + 1)2 π 2

and − 1 4a X (2p + 1)π sin( x2 )e v(x1 , x2 ) = 2 β 2 π (2p + 1) a

(2p + 1)πx1 a .

(3.19)

p≥0

We set πx1 πx2 +i a a z=e −

and we get

v(x1 , x2 ) = Im[

4a X z 2p+1 β ] π2 (2p + 1)2 p≥0

zn n2 ( |z| ≤ 1) and we refer to [14] for very surprising properties of this function. We proved that

Let us recall that the dilogarithm function Li2 is defined by Li2 (z) =

v(x1 , x2 ) =

2a βIm[Li2 (z) − Li2 (−z)]. π2

Let us compute the gradient of v. One obtains with (3.19) :

P

n≥1

December 8, 2014

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Singularities

Polylog singularities

− ∂v 4 X 1 (2p + 1)π =− β sin( x2 )e ∂x1 π (2p + 1) a

21

(2p + 1)πx1 a

p≥0

and − ∂v 4 X 1 (2p + 1)π = β cos( x2 )e ∂x2 π (2p + 1) a

(2p + 1)πx1 a

p≥0

thus (notice that

1+z 6∈ R− ) 1−z

∂v 2 1+z ∂v 4 X z 2p+1 = β log( ) −i = β ∂x2 ∂x1 π (2p + 1) π 1−z p≥0

=

1+z 1+z 2 β [Ln| | + iArg( )] π 1−z 1−z

where the function Arg takes its values in the open set ] − π, +π[. One deduces that πx1 πx1 −2 πx a cos( 2 ) + e a ∂v 1 a = βLn| πx1 πx1 | ∂x2 π − −2 πx2 a 1 − 2e a cos( )+e a 1 + 2e



On Γi , one has πx1 ∂v 2 1+e a 4 = βLn| πx1 | = π β ∂x2 π − 1−e a −

Ln| coth

πx1 |, a

and

lim x1

→0+

∂v = +∞. ∂x2

In a same way, one has on Γe+ , lim +

x2 →0

∂v (0, x2 ) = +∞. ∂x2

Theorem 3.1 is proved. Let us remark that in case of a unique velocity (c+ = c− ), there is no singular part.

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Singularities

Transparent boundary conditions

Conclusion - In this paper, we have discussed the singularity which appears when one uses a transparent boundary condition for a bimaterial. This situation occurs for instance when one tries to simulate wave propagations in an infinite strip replaced by a finite one. The singularity which belongs to the Dilog family is an artefact which doesn’t exist in the physical model. Therefore it seems necessary to eliminate it from the solution. The first step was obviously to make it explicit in order to be able to suggest a method which would improve the boundary condition. For instance, a numerical method could be to compute the contribution of this artificial singularity and to substract it from the global solution. In fact, this is a way for defining an upgrade transparent boundary condition for this kind of problems. This will be discussed in a forthcoming paper. Finally, let us remark that these phenomena appear in the study of the detection of cracks at the interface between two materials, as it is noticed in [7]. References [1] Adams R. 1975 Sobolev spaces, (Academic press, New York) [2] Bui H.D. 1978 M´ecanique de la rupure fragile, (Masson, Paris) [3] Brezis H. 1983 Analyse Fonctionnelle, collection math´ematiques appliqu´ees pour la maˆıtrise (Masson, Paris) [4] Dautray R. and Lions J.-L., 1988 Analyse math´ematique et calcul num´erique T8. Evolution : Semi-groupe, Variationnel, collection enseignement (Masson, Paris) [5] Engquist B. and Majda A., 1977, Absorbing Boundary Conditions for the Numerical Simulation of Waves, Math. of Comp., vol 31(139), p629-651. [6] Destuynder Ph. and Djaoua M., Sur une interpretation math´ematique de l’int´egrale de Rice en m´ecanique de la rupture fragile, Mathematical Methods in the Applied Sciences Volume 3, Issue 1, pages 70-87, (1981) [7] Destuynder Ph. and Fabre C., Few remarks on the use of Love waves in non destructive testing, AIMS Journals, Discrete and continuous dynamical systems, to appear. [8] Fung Y.C. 1965 Foundations of solid mechanics, p.182, (Prentice-Hall Inc., Englewood Cliffs, USA) [9] Grisvard P. 1992 Singularities in boundary value problems, (Masson, Paris) [10] Halpern L., Absorbing Boundary Conditions for the Discretization Schemes of the One-Dimensional Wave Equation, Math. of Comp., vol 38(158), 1982, p. 415-429. [11] Halpern L. and Trefethen L.-N., Well-Posedness of One-Way Wave Equations and Absorbing Boundary Conditions, Math. of Comp., vol 47(176), 1986, p. 421-435. [12] Love A.E.H. 1924 A treatise on the mathematical theory of elasticity, Fourth edition, (Cambridge University press and Dover edition in 1944) [13] Necas J. 1967 Les m´ethodes directes en th´eorie des ´equations elliptiques, (Masson, Paris) [14] Zagier D. The Dilogarithm Function