Tamtam Proceedings - lamsin

470 Ben Abda et al.Denoted by H 1 = {v | v satisfies (6)-(7)} and H 2 = {v | v satisfies (4)-(5)}. Let {P j } j∈INbe a set of functions such that the following conditions are fulfilled{ Span{pj }(A)∞ j=0 = L2 (Γ i ).{P j } j∈IN is an orthogonal basis in L2 (Γ i ).In all this work, we take for the functions filled the condition (A) the Legendre-Fourierpolynomials. Let L 0 (x), L 1 (x), . . . , be a shifted Legendre polynomials, normalized by∫ 10 (L j(x)) 2 dx = 1. This Polynomials are defined for all j = 0, 1, . . . by L i (x) =i∑C ij x j , where C 00 = 1, C j,0 = (2j + 1) 1 2 , C j,k = −C j,k−1 ( j kj=0+ 1)(j+1k − 1)j = 0, 1, 2, 3, ... k = 1, 2, 3, ..., j. For illustration, let Ω be a bounded, simply Lipschitzconnected domain of IR 2 defined by Ω = { (x, y) ∈ IR 2 / 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 } withΓ i = { (x, y) ∈ IR 2 / y = 0, 0 ≤ x ≤ 1 } and Γ c is a Lipschitz curve which connects thetwo points (0, 0) and (1, 0) such that { (x, y) ∈ IR 2 /y = 0, 0 ≤ x ≤ 1 } ∪ Γ i = ∂Ω. Tofound the moment of the solution directly on the orthogonal basis {L j } j∈IN is equivalentto determine an harmonic function in H 1 such that ∂vn∂y (x, 0) = L n(x), n = 0, 1, . . . .For that it is sufficient to choose v n (x, y) = Im(Q n (x + iy)), n = 0, 1, . . .where Q jindicates a primitive of P j and Im is the imaginary part. For all x ∈ IR, we defined thesite of maps {q j } j∈IN such that q j(x) = ∫ x1 L j(t)dt, for Z = (x, y) in IR 2 we deduce thesite of maps Q j (x, y) = Im(q j (x + iy)). We can proof easily that for all j in IN, Q j is inH 1 , by using Green’s formula we find∫Γ c( ∂Qj∂n u − ∂u∂n Q j) ∫ ( ∂Qjds =Γ i∂n u − ∂u )∂n Q j ds (8)on Γ i we have ∂Qj∂n (x, 0) = L j(x, 0) and Q j (x, 0) = 0, ∀j ∈ IN. Then∫ ( ∂QjΓ c∂n u − ∂u ) ∫∂n Q j ds = L j uds. (9)Γ iWhat gives us the Legendre moments of the solution of problem (1) − (3) on Γ i . Toapply this method for the identification of Robin coefficient and the cracks, we mustcomplete the function ∂u∂n on Γ i. For that reason we defined the site of maps {d j } j∈INsuch that d j (x) = dLjdx (x); for Z = (x, y) in IR2 we deduce the site of maps D j (x, y) =Re(d j (x + iy)). Take v j = D j in (8), then on Γ i we have∂D j∂y (x, 0) = 0 and D j(x, 0) = Re(L j (x, 0)), ∀j ∈ Nhence v j ∈ H 2 . Hence the equation (8) becomes,∫ ( ∂DjΓ c∂n u − ∂u ) ∫∂n D ∂uj ds = L j ds (10)Γ i∂nTAMTAM –Tunis– 2005