Control and Cancellation Singularities of Bilaplacian in a Cracked Domains

Abstract We are interested in controlling and removing singularities of the Dirichlet problem involving the bilaplacian operator in a domain with corner. It’s possible of making the solution to the bilaplacian operator regular, through acting on a small part of a cracked domain with corner. Then, the best singularity coefficients can be controlled by simultaneous actions of two controls on a small part of the boundary.


Introduction and Statement of Problem
We consider the Dirichlet problem for the bilaplacian operator in a bounded polygonal domain Ω of R 2 .Since the domain is polygonal, the solution of this problem does not only depend on the regularity of data, but also on the geometry of the domain (Grisvard, P., 1974;Grisvard, P., 1992;Kondratiev, V. A., 1967).This solution is singular in the neighbourhood of non-convex vertices of Ω (see Bayili, G., 2009;Seck, C., Bayili, G., Sène, A., & Niane, M. T., 2011).Niane et al. (2006) proved that it is possible by acting on a small part of the domain or on a small part of the borders, a regular solution of the Laplace equation can be obtained.Let m + 1 the number of non-convex angles of Ω and Ō a non empty open bounded Ω.We will show that there are infinitely differentiable functions with support in Ō and satisfying the following condition if f ∈ L 2 (Ω), (λ i ) 1≤i≤k are the coefficients of the singularities and (g i ) 1≤i≤k the singularities of the problem (2) has an unique solution y ∈ H 4 (Ω).
We will also prove the following result if Γ 1 and Γ 2 are two analytical open sets of Γ whose measure of the intersection is non zero, then there exist k functions (h i , (3) has an unique solution y ∈ H 4 (Ω).Proof.According to the hypothesis and by Gram-Schmidt orthogonalization, there exist v 0 , ..., v m , such that for all i ∈ {0, ..., m}.This implies that v (n)  i , e j H −→ v i , e j H = δ i j as n −→ ∞, and hence the matrix K n = ( v (n)  i , e j H ) 0≤i< j≤m is invertible for n large enough.Fixed this value of n, write The family (ω i| Ō) 1≤i≤k is linearly independent ?
Effectively, assume that there exist real numbers (α i ) 1≤i≤k not all of them zero such that We know that k i=1 α i ω i is an analytical form, according to the unicity theorem of Holmgren's-Kovalevska in L. Hormander (1976), we have k i=1 α i ω i = 0 on Ω, we can deduce by hypothesis that α i = 0, ∀i ∈ {1, ..., k} and consequently (ω i| Ō) 1≤i≤k is linearly independent.
Since D( Ō) is dense in L 2 ( Ō), Niane et al. (2006) and Density Lemma imply that there exists a family (g j ) 1≤ j≤k of functions of D(Ω) with support in Ō such that: Theorem Let Ω be a non-empty bounded open polygon with R n of boundary Γ.Let Γ 1 and Γ 2 be two non-empty analytic open sets of Γ such that mes(Γ 1 ∩ Γ 2 ) 0. Let (ω i ) 1≤i≤k be a linear independent family of biharmonic functions of L 2 (Ω) verifying then there exist k functions Proof.Let the space H = L 2 (Γ 1 ) × L 2 (Γ 2 ) with the following scalar product With this product, H is a Hilbert space.Next we prove that the family is linearly independent.

Assume the existence of real numbers (α
Since ψ = k i=1 α i ω i is an analytical form in virtue of the Holmgren and the Cauchy-Kovalevska theorem (see Hormander, L., 1976).Hence, we can deduce under our hypothesis that According to the Cauchy-Kowalevska Theorem, there exists a non-empty open neighbourhood By Holmgren Theorem (Hormander, L., 1976), we obtain: Consequently, we have: So we can deduce that α i = 0, ∀i and the family

Preliminary Results on Dual Singular Functions
We show that, in a cracked domain, we can obtain a regular solution of the biharmonic problem by acting two simultaneous controls on two small parts of the boundary of intersection not empty and not reduce to a point on the small part O of Ω not intercepting any vertices.
Lemma (P.Grisvard, 1985) . This singular part is described below by its polar coordinates where α i = π ω i is the singularity exponent related to the crack O i and η i is cut-off function equal to 1 on the neighbourhood of vertex of the open O i .
By Grisvard Lemma, in each non-convex vertices, we have finite number of dual singular solutions associated with the domain Ω.
Pose ω * i = r − j S i (r, θ) = r α i − j sin(α i θ)η i (r) f or 1 ≤ i ≤ k and j ∈ {1, 3}.According to Grisvard (1985;1989) and Timouyas (2003), (ω * i ) 1≤i≤k is the family of dual singular solutions associated to m angles of non-convex vertex of domain Ω.This family is linearly independent and verifies are obtained as

Cancellations of Singularities
Theorem It exist k infinitely differentiable functions with compact support contained in Ō such that if f ∈ L 2 (Ω) and (λ i ) 1≤i≤k the singularity coefficients corresponding to the problem then the solution of problem verifies ϕ ∈ H 4 (Ω) ∩ H 2 0 (Ω).Proof.The dual singular solutions of (16) verifies hypotheses of Theorem 2.2.Hence it exist a family (g i ) 1≤i≤k of functions with compact support contained in Ō such that ∀ 0 ≤ i < j ≤ k, Ω ω i g j dx = δ i j (18) Let (λ i ) 1≤i≤k the singularity coefficients associeted with ( 15) and (ζ i ) 1≤i≤k the singularity coefficients of ( 16).So we have: Consequently and the solution is ϕ ∈ H 4 (Ω) ∩ H 2 0 (Ω) Let H be a Hilbert space, D a dense subspace in H and {e 0 , ..., e m } a subset of H.Then, there exist {d 0 , ..., d m } in D such that ∀ 1 ≤ i < j ≤ m, e i , d j H = δ i j .
13) Vol. 4, No. 4; 2012 with V i the i th open neighbourhood of vertex of O i of the domain Ω.The singularity coefficients (λ i ) 1≤i≤k associeted with problem