| Cauchy problem arises naturally in many scientific fields and engineering appli-cations, such as plasma physics, geophysics, nondestructive testing and cardiology.It is well-known that such a problem is ill-posed, i.e., the solution does not depend con-tinuously on Cauchy data and a small perturbation in the data may cause an enormous deviation in the solution. In recent years, some numerical methods have been proposed to solve such an inverse problem, such as quasi-reversibility method, conjugate gradient method, moment method, fundamental solution method, and etc.The main purpose of this paper is to provide a simple and effective numerical method for solving Cauchy problem connected with the Laplace equation. The main idea is to approximate the solution to the Cauchy problem by harmonic polynomials, which are solutions of the Laplace equation.Let D c R2be a bounded and simply connected domain with a regular boundary (?)D,and Γ be a portion of the boundary (?)D. Consider the following Cauchy problem: Given Cauchy data f and g on Γ,find u on∑=(?)D\Γ such that u satisfies wherev is the unit normal to the boundary (?)D directed into the exterior of D. Without loss of generality, we make the assumption on the measured data that∫∈H1(Γ),and g∈L2(Γ), and suppose that the Cauchy problem has a unique solution u∈H3/2(D). Main idea:The main idea is to approximate the solution to the Cauchy problem by harmonic polynomials,which are solutions of the Laplace equationâ–³uN=0.in R2of the form Here φk,1(x),φk,2(x)(k=1,2,…)are some harmonic polynomials of degree k,and φ0(x)is a harmonic polynomial of degree0. With this idea in mind,we derive two equations for the coefficients on the specified boundary,which can be solved by regularization methods.Then the data for the solution on the unspecified boundary∑can be obtained by simple calculation of the harmonic polynomials.Before proceeding the approxim ation,it is necessary to recall that for k=0,1,,the harmonic polynomials are φk,1(x)=Re {(x1+ix2)k}, φk,2(x)=Im {(x1+ix2)k} where x=(x1,x2)∈R2.Define the harmonic polynomials φ0(x),φk,J(x)(j=1,2)by where constants0<λ<1and Mk≥‖(?)k,j‖L2((?)D)+‖((?)(?)k,j)/(?)v‖L2((?)D),j=1,2Now we turn to introducing the harmonic polynomial approximation. The ap-proximation proceeds in two steps:First we approximate the solution by the sum of a single-layer potential and a constant.Second we approximate this potential by some harmonic polynomial.To this end,we introduce the single-layer potential Vφ,which is a solution of the Laplace equation,of the form where B is a Lipschitz domain,φ∈L2((?)B),andΦ(x,y)=-(1/2Ï€)ln|x-y|is the funda-mental solution to the Laplace equation.Then we have the following result. theorem1. Let u∈H(3/2)(D) satisfy the Laplace equation. Let B be a bounded and simply connected domain with (?)B∈C2such that D cc B. Then for every ε>0, there exist a constant s0∈R, and a single-layer potential vφ of the form for some (?)∈L2((?)B), such that and especiallyTo approximate the potential v(?) in D, we need the classical approximation theory in L∞with harmonic polynomials.lemma1. Let Br={x∈R2:|x|<r}, BR={x∈R2:|x|<R} andBp={x∈R2:|x|<p} with r<R<p. Assume that v is harmonic on Bp. Then there is a sequence{Pn}N∞=of harmonic polynomials of degree N such that where Ï„=(R/r)and σ=(P/R)-1.By using Lemmal, we obtain the following main result in this section.theorem2. Let u∈H(2/3)(D) satisfy the Laplace equation. Then for every ε>0, there exists a harmonic polynomial uN of the form such that and especially From Theorem2, we see that ifuN is approximated, the approximation of a can be obtained. Our aim here is to get the numerical approximation of the coefficients of uN.To this end, define the trace operator An:R2n+1â†'L2(Γ)×L2(Γ)by where cn=(c0, c1.1, c1.2,...,cn1,cn2)T∈R2n-1. Then, the following property of the operator An holds. theorem3. The operator An: R2n+1â†'L2(Γ)×L2(Γ) is compact and injective.Now, we turn to introducing our numerical algorithm. First, we approximate the coefficients by solving the following equations:(Ancn)(x)=b(x), x∈Γ, where b=(f,g)T∈H1(Γ) x L2(Γ). And then the harmonic polynomial defined by will be the approximation of uN, and therefore of the solution.In general, equations (Ancn)(x)=b(x), x∈Γ, are not solvable since we cannot assume that the Cauchy data b, especially the measured noisy data bδ, are in the range An(R2n+1) of An. Therefore, we will solve (Ancn)(x)=b(x), x∈Γ, by a regularization method in the next section, and then give the error estimates.A regularization method for solving the equationsIn this section, we will use Tikhonov regularization method with Morozov’s dis-crepancy principle to solve the equations (Ancn)(x)=b(x), x∈Γ, and then give the error estimates and convergence results.Due to the ill-posedness,we need to consider the perturbed equations Ancnδ=bδ Here bδ∈L2(Γ)×L2(Γ)are measured noisy data satisfying‖b-bδ‖L2(Γ)×L2(Γ)≤δ. Tikhonov regularization of (Ancn)(x)=b(x), x∈Γ, is to solve the following equation αcnαδ+An*Ancna,δ=An*bδ. By introducing the regularization operator Ra:=(αI+An*An)1An*, for α>0, we get the regularized solution cnα,δ=Rαbδ of Ancnδ=bδ, which is the unique minimum of the Tikhonov functional Jα,δn(cn):=‖AnCn-bδ‖L2(Γ)×L2(Γ)2+α|cn|2Since the measured noisy data bδ may not be in the range An(R2n+1)of An,we will introduce a method to choose the regularization parameter α. theorem4.Let ε,μ be positive constabt,n≥N and δ+ε+μ<‖bδ‖L2(Γ)×L2(Γ).Let the (a) There exists a function h∈L2(Γ) x L2(Γ) such that|cN*-AN*h|<ε;(b) Let‖h‖L2(Γ)xL2(Γ)≤E, thenFrom Theorem4, the approximation of the coefficients is obtained. Define the harmonic polynomial una(δ),δ in the form Then, we have the following main result in this paper.theorem5. Let the assumptions in Theorem4hold. Then the following estimate on boundary∑holds Moreover,At the end of this section, we describe the algorithm for determining the threshold of the discretization parameter N.The parameter,N can be obtained by the following:· Choose r, R and p such that r<R<p and D cc Br;· Compute Ï„=(R/r) and σ=(P/R)-1;· Compute N by solving the inequality:... |
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