Tamtam Proceedings - lamsin

Cauchy problem for Laplace’s equation 4691. IntroductionConsider a bounded and connected domain Ω ∈ IR 2 with Lipschitz boundary ∂Ω, andlet Γ c be an open part of ∂Ω. The classical Cauchy problem is written as∆u(x) = 0, x ∈ Ω (1)u(x) = f(x), x ∈ Γ c (2)∂u∂n (x) = g(x), x ∈ Γ c (3)where ∆ is the Laplacian operator and n is the outer unit normal with respect to ∂Ω. It iswell known that the problem is highly ill-posed since Hadamard (See [1, 2, 8, 9]). In [5]J. Cheng et al. proof that the initial Cauchy problem [1]-[3] is equivalent to the followingmoment problem ∫ Γ ivβds = µ v (f, g) Where β is an unknown function defined on Γ i , vis an harmonic function such that∆v(x) = 0, x ∈ Ω (4)∂v∂n (x) = 0, x ∈ Γ i (5)and ds is the curve element. They take Re((x 1 + i ∗ x 2 ) j ), j = 0, 1, ...where Re denotesthe real part. With their choice, according to Talenti [10] they approach the solution ofthe moment problem. And they use the Gram-Schmidt matrix to found the Legendremoments; who is badly conditioned.We will give a method that permit to reconstruct directly the solution u on Γ i usingthe Legendre polynomials. In addition, we applied this result to some inverses problems,Robin coefficient determination and crack identification.2. Presentation of the methodIn this paper, to overcome the Gram-Schmidt matrix, we propose a method which,not only rebuild the solution u on Γ, but also push even more numerical results fromconvergence. Then our originality is to make appear directly the components accordingto an orthonormal basis. Stability results and regularization method in three-dimensionalcase will appear in a forthcoming works.Let v an harmonic function solution of the equations∆v(x) = 0, x ∈ Ω (6)v = 0, x ∈ Γ i (7)TAMTAM –Tunis– 2005