【作者】 李媛；

【导师】 马富明；

【作者基本信息】 吉林大学 ， 计算数学， 2009， 博士

【摘要】 本文研究了定态Schr(?)dinger方程的某些正反散射问题的数值计算方法.文章中对于这些算法给出了一些理论分析,并通过数值实验检验了算法的可行性.全文的具体内容如下:第一章将简单介绍定态Schr(?)dinger方程正反散射问题的研究背景以及研究现状.第二章给出本文研究需要的某些预备知识,对于PML方法及因子分解法的研究背景及进展给予介绍,并分别以一种典型的问题为例来阐述各方法的基本原理.之后的两章是本文的主要工作.第三章研究具有短程位势的二维散射问题的数值计算.针对一类特殊的短程位势,提出了一种PML方法.首先根据复化极径的思想,得到了极坐标下的PML方程,然后在关于吸收参数所做的假设下,证明了PML问题解的存在唯一性,并通过数值实验表明该方法具有一定的可行性.此外,本章还求出了具一类特殊位势的Schr(?)dinger方程的散射解在圆域外的级数表达式.第四章考虑具紧支集位势的Schr(?)dinger方程反散射问题,利用因子分解法来重构位势的支集.首先推导出了散射振幅算子的分解公式;之后考虑内传播问题,研究了这一算子为单射的条件;根据其为单射、正规算子,可以通过散射振幅算子的谱数据来判断抽样点是否位于位势支集内.本章对于三维情况做了详细的理论分析,最后说明该方法可以用于二维问题,并针对二维情况给出了数值算例.更多还原

【Abstract】 In recent years, the field of differential equations has come to distinguish between two different types of problems: the direct and the inverse. The main difference beween these two problems is that an inverse problem is often ill-posed and thus more complicated than the direct one. Scattering describing the processes of collision among particles is a class of basic problems in quantum mechanics. One can detect the structures of particles and promote the developements of theories by the analysis for results of scattering, so it’s very necessary to study the direct and inverse scattering problems for Schr（?）dinger equations in mathematics. In this thesis, we consider the model independent of time, i.e., in stationary state.Nowadays, the numerical computation has been the most important means for solving the Schr（?）dinger equations. Solving numerically a direct scattering problem is related to the computation in unbounded domain, so we have to truncate it into a bounded one, and present an absorbing boundary condition or set an absorbing layer out of the domain such that the error between the numerical and the exact solution isn’t large. The studies on the inverse scattering problems for Schrodinger equations have a long history, including the work on the existence and uniqueness of the potential when scattering data known, the reconstruction of the potential from exact or noisy data and the stability analysis, etc., but there are still many problems open. In this thesis, we study the numerical computation on the direct and inverse scattering problems for stationary Schr（?）dinger equations as follows:Ⅰ. The PML method for the 2D scattering problem with short range potential1. The mathematical desciption for scattering problemWe consider the following Schr（?）dinger equation（?）. （1） Given a plane wave uⁱ as the incident wave, i.e., uⁱ（x） = e^ikω·x, where i = （?）, k∈R is the wave number,ω∈S¹ is the direction of incidence, S¹ denotes the unit circle, V（x） denotes the potential. The incidence of uⁱ gives rise to the scattered wave u^s such that the total wave u =uⁱ + u^s satisfies the Schrodinger equation （1）, and for the require of physical background, u^s must satisfies the Sommerfeld boundness and radiation conditions at infinity:（?）. （2）By the stationary phase lemma, u^s satisfies the asymptotic behaviour uniformly with respect to （?）, whereα（（?）,ω,κ） is called scattering amplitude. The scattering problem can be stated as follows: Given the potential V（x） and the incident wave uⁱ, find the total wave u = uⁱ + u^s satisfying the Schrodinger equation （1） such that the scattered wave u^s satisfies the condition （2）.Let r> 0 and the potential V（x） satisfy the following conditions: V（x）∈L^∞(B_r0) and V（x） > 0 when r < r₀ and V（x） = b/r^δwhen r≥r₀, where B_r0 denotes the circle centered at origin and of radius r₀, both b > 0 andδ> 1 are constants, r= |x|. Since we consider the problem in unbounded domain, the assumption on the potential when r≥r₀ is essential and the refinement on V（x） when r < r₀ can be relaxed and even allowed not to be continuous on r = r₀.Substitute uⁱ = e^ikω·x into （1）, then u^s satisfies the equation（?）. （3）A solution of equation （3） satisfying the condition （2） is called outgoing solution, the existence and uniqueness of such solutions are given by the following theorem:Theorem 1.1 Except for a possibly discrete set composed of countable number of k, the equation （3） has a unique outgoing solution for every k∈R.2. The presentation of PML methodWe set a perfectly matched layer B^PML= {x∈R²,R < |x| <ρ} outside B_R, where R > r₀. The fictitious medium in this layer can absorb the scattered wave very well such that the reflection on the artificial boundary very small or vanishing at all, so the solution in B_R can not be " polluted " heavily. We set up a polar coordinates system by taking the origin and the positive direction of x-axis in rectangular coordinates system as the origin and the direction of polar-axis respectively. Assume that the argument of the incident directionωisθ₁, then the equation （3） can be written in polar form as（?）. （4）By the idea of radius complexed, letα（r） = 1+ iσ（r）, whereσ（r） is the absorbing factor satisfying thatσ∈C（R）,σ≥0 andσ= 0 when r≤R. Let （?） denote the complex radius defined as follows:then（?）. （5） Replace the r in （4） by （?） and using （5） again, we get（?）. （6） where （?）（r,θ） = V（（?）,θ）, The matrix A = A（x） satisfyingand its exact form is:By fomula （6）, the PML problem corresponding to the problem （1）-（2） can be defined as the following boundary value problem: find （?）, such that（?）, in B_ρ, （7）（?）= 0, onΓ_ρ. （8）The aim of the PML method is to solve numerically the problem （7）-（8）.Let （?）: H¹（B_ρ）×H¹（B_ρ）→C be a sesquilinear form: the weak formulation of the problem （7）-（8） is: findφ∈H¹（B_ρ）, such that（?）. （9）where （·,·）_Bρ denotes the L² inner product in B_ρ.For the absorbing factorσ, we assume as follows:（H） For a constantσ₀ > 0 and an integer m≥1,σ=σ₀ （?） ,R≤r≤ρ. Theorem 1.2 Except for a possibly discrete set composed of countable number of k, the equation （9） has a unique solution for every k∈R. In the numerical experiments, we give several examples to observe the effects of the position of absorbing layer placed、the thickness of the layer and the damping rate of potential to the scattered wave. The results show that solving 2D Schrodinger equation with short range potential by PML method is feasible to a certain extent.3. The series expression of scattered wave outside some cicular domains Let V（r） = b/r² oustide B_r0, r₁ > r₀, and the incident wave uⁱ = e^iκω·x we want to derive the expression of the scattered wave u⁸ in r≥r₁. Assume that the incident directionωis the negtive direction of y-axis（The results have no essential differences when the directionωchanges）, we take the origin and the positive direction of x-axis in rectangular coordinates system as origin and the direction of polar-axis in polar coordinates frame. Using the expression e^iκω·x=（?） in 2D and the geometrical relations betweenφand the argumentθ, we know that（?）, （10） whereφis the angle between x andω, J_m denotes the mth order Bessel function. By the symmetry of the potential, we consider the variable seperated solution u（r,θ） =（?） of the equation （1） in polar form, then R_m（kr） satisfies theνth order Bessel equation:（?） （11）whereν² = m²+ b. Since the fundamental solution of Bessel function are J_ν（kr） and H _ν^（1）（kr）, the solution of equation （11） can be written as（?）. （12） By the asymptotic behaviour of J_ν（kr） and H_ν^（1）（kr） when r→∞and the elastic scattering doesn’t change the amplitude of each partial wave（Assume that only elastic scattering occurs）, we getα_ν= e^2iη_ν- 1 = 2isinη_νe^iη_ν, whereη_νis called phase shift in physics. The radiation condition for outgoing solution leads to b_m =（?）. Substitute a_νand b_m into fomula （12）, thenso the scattered wave u^s（r,θ） has the following series expression in r≥r₁:（?）. （13）Ⅱ. The factorization method for the inverse scattering problem with compact support potentialWe extend the factorization method to the inverse scattering problem for Schr（?）dinger equation to reconstruct the support of the potential. Similar conclusions as the cases of inverse obstacle and medium scattering problems in acoustics can be obtained. We make the analysis carefully for the three-dimensioal case and state at last that this method can also be applied to the two-dimensional case.1. The existence and uniqueness of the solutions of direct potential scattering problemsLet D （?） R³ be a bounded domain with C² boundaryαD,R³ \ （?） connected and the potential q（x） satisfyq≥0, supp q = （?）, there exists some a∈（0,1）, such that q∈C^0,α（R³）, （14）where C^0,α（R³） denotes the Holder continuous function space with H（?）lder indexα. Let uⁱ = e^ik（?）·x Denotesαplane wave in R³ of direction （?）∈S²:where S² denotes the unit sphere in R³ and k > 0 denotes the wave number, we consider the following scattering problem for Schrodinger equation:（?）. （15）（?）, （16）（?）, （17） where u^s denotes the scattering wave.We denote byΦ{x,y) the fundamental solution of the Helmholtz equation in i.e.In fact,Φ（x, y） is also dependent of k and thus should be written asΦ_k（x, y）, in this thesis we omit the index k most of the time except for several places. The main properties of the volume potential are summarized in the following lemma.Lemma 2.1 We define the volume potentialωwith densityφ∈L²（D） bythen the following holds:（a）ωis continuous in R³, analytic in the exterior R³ \ （?） of D, and satisfies the Helmholtz equation△ω+ k²ω= 0 in R³\ （?） and the Sommerfeld radiation condition （17）.（b） Ifφ∈C{?), thenω∈C^1,α（D） for everyα∈（0,1）.（c） If evenφ∈C^0,α（D） for someα∈(0,1], thenω∈C^2,α（D） and△ω+ k²ω= -φin D. Furthemore, ifφ∈C^0,α（D） is of compact support, thenω∈C^2,α（R³）. and△ω+ k²ω= -φin R³; whereφis extended by zero into the whole of R³.In order to obtain the existence and uniquness of the solutions of scattering problem （15）-（17）, we need the following several lemmas.Lemma 2.2（Unique Continuation Principle） Let q（x） satisfy （14） and u∈C²（R³） be a solution of Schr（?）dinger equation△u + （k² - q（x））u = 0 in R³ such that u（x） = 0 for all b≥αfor some |x|≥b, where （?） {x : |x|≤α}, then u has to vanish in all of R³.Lemma 2.3（Uniqueness） Let q（x） satisfy （14）. The problem （15）-（17） has at most one solution, i.e., if u is a solution corresponding to uⁱ = 0, then u = 0.Lemma 2.4（Equivalence） Let q（x） satisfy （14）, then（a） If u∈C²（R³） is a solution of the scattering problem （15）-（17）, then u∈C（?） solves the Lippmann-Schwinger integral equation（?）, x∈R³. （18）（b） If, on the other hand, u∈C（?） is a solution of the integral equation （18）, then u can be extended by the right-hand side of （18） to a solution u∈C²（R³） of the scattering problem （15）-（17）. We define the operator T : L²（D）→L²（D） by（?）, x∈D. Equation （18） is then written in short form as（?）. （19）Theorem 2.5 Under the assumptions （14） on q（x） , there exists a unique solution u of the scattering problem （15）-（17） or, equivalently, the Lippmann-Schwinger equation （18）.Lemma 2.6 If q（x） satisfies （14）, the operator I + T is an isomorphism from L²（D） onto itself.2. The factorization of the scattering amplitude operatorThe Lippmann-Schwinger equation （19） leads to a concept of weak solution as follows.Definition 2.7 For h∈L²（D）, the solutionν∈L²（D） of the integral equationν+ Tν= Th （20）is called the weak solution of the direct potential scattering problem.Under the assumption （14） on q（x）, the scattering waveν^s of the scattering problem （15）-（17） satisfies （?）, （21）uniformly with respect to （?）, whereν_α（?） is called scattering amplitude,（?） denotes the direction in which the incident wave is scattered. The inverse potential scattering problem is to reconstruct the potential q（x） by the given dataν_α（?）. The uniqueness of this problem for fixed k² > 0 and all （?）, （?）∈S² has been proved. For weak solutionν∈L²（D） of the problem （20） we conclude from the asymptoticbehavioruniformly with respect toν∈（?） and （?）, that the scattering amplitude pattern ofν^s is given by（?）. （21） As a special case, the scattering amplitude of the scattering problem （15）—（17） takes the form（?）. （22）Now we define the scattering amplitude operator A : L²（S²）→L²（S²）, i.e.（?）. （23）If q（x） is of compact support, then the operator A is normal, and the scattering operator S = （?）A is unitary in L²（S²）, i.e., SS^* = S^*S = I.We define the operator G : L²（D）→L²（S²） which maps h∈L²（D） into the scattering amplitudeν_αof the weak solution v of the problem （20）, i.e., Gh =ν_α. （24）We will work with the weighted space L²（D, q） which is defined with respect to the inner product （?）. （25）In the following lemma we collect properties of the operators T and G in L²（D, q）.Lemma 2.8 （a） The operator T is well defined and compact from L²（D, q） into inself. （b） The adjoint T^# of T in L²（D,q） with respect to〈·,·〉_q is given by（c） Both I+ T and I + T^# are isomorphisms from L²（D,q） onto itself.（d） The operator G is also well defined and compact from L²（D,q） into L²（S²）. Theorem 2.9 The scattering amplitude operator A = -4πG（I + T^#）G^#, whereG^# and T^# are the adjoints of G and T respectively.3. An interior transmission problem First, we formulate the interior transmssion problem in spaces of smooth functions. Given z∈D andα∈C, findν,ω∈C²（D）∩C¹（?） such that（?）, （26）（?）. （27） We will show at the end of this section that there exist at most a countable number of k∈E₊ for which the homogeneous problem （26）-（27）, i.e. forα= 0, admits nontrivial solutions under more assumptions on q（x）. We define two closed subspaces of L²（D, q）:（?）, （28）（?）. （29） Lemma 2.10 If q（x） satisfies the condition （14）, then the follwing conclusions hold:（a） Both I+ T and I + T^# are isomorphisms from H_q onto H₁.（b） The space （?） of Herglotz wavefunctions is dense in H₁. Next we introduce the concept of weak solutions of the interior transmission problem （26）-（27）.Definition 2.11 We fixed z∈D andα∈R,（a ） A pair （ν,ω）∈L²（D,q）×L²（D,q） is called the weak solution of （26）-（27）, if（?）. （30）（b） k² is called an interior transmission eigenvalue if there exists a weak solution （ν,ω）≠（0,0） of the homogeneous interior transmission problem （30）, i.e. forα= 0.The equivalence of （26）-（27） and （30） is shown in the following lemma.Lemma 2.12 Fix z∈D andα∈C, letν,ω∈C²（?） satisfy△ν+ k²ν= 0 and△ω+ （k² - q（x））ω= 0 in D. The boundary conditionsω-ν=αΦ（·,z） and （?） onαD （31）hold if and only ifν=ω+Tωand （?）. （32）We construct a projection operator P from L²（D, q） into the subspace:（?）. （33） Note that H is the orthogonal complement of H₁ with respect to the sesquilinear form By Lax-Milgram theorem, for anyφ∈L²（D,q） there exists a unique （?）∈H with （?） for all （?）. We define P : L²（D,q）→H by （?） and note that P is linear and bounded.We will transform the interior transmission problem into a equivalent equation in the following.Lemma 2.13 Fix z∈D andα∈C, we define d := inf{|x - z| : x∈αD}and chooseφ∈C^∞（R） withφ（t） = 0 for t≤（?） andφ（t） = 1 for t≥（?）. Set（?） and （?）, Note that （?） coincide withΦ（·, z）in a neighbourhood ofαD and therefore f_z is well defined in D, then the pair （ν,ω）∈H₁×H_q is a weak solution of （30） if and only ifω∈L²（D, q） satisfies（?）. （34）In order to show that A is injective, we need the following lemma.Lemma 2.14 LetΩ（?） R³ be a domain that decomposed into two disjoint sub-domains: （?） such that （?）. Let the boundariesαΩ₁ andαΩ₂ be smooth（i.e.,C²）. Let u_j∈C²（Ω_j）∩C¹（?） for j = 1,2 be solutions of the Schr（?）dinger equation△u₁ + （k² - q（x））u₁ = 0 inΩ₁ and the Helmholtz equation△u₂ + k²u₂ = 0 inΩ₂; where q（x） satisfies （14）. Furthermore, let u₁ = u₂ and （?） onΓ, whereΓdenotes the common boundaryΓ:=αΩ₁∩αΩ₂. Then the function u, defined bycan be extended to a function in C²（Ω） that satisfies the equation△u+（k² - q（x））u = 0By the Lemma 2.14, we can prove the following lemma.Lemma 2.15 g∈L²（S²） is a solution of the homogeneous integral equation（?）, （35）if and only if there existsω∈C²（D）∩C¹（?） such that （ν,ω） solve （26）-（27） forα= 0, whereνis the Herglotz function defined by（?）. （36） Next we give the most important theorem of this section.Theorem 2.16 If k² is not an interior transmission eigenvalue then for every z∈D andα∈C, there exists a unique weak solution of the interior transmission problem （30）. In this case, A is injective.Now we give further assumptions on q（x） as follows:（?）, （37）and there exist some k₀∈R₊ such that（?）. （38）whereΦ_k0 （x, y） denotes the fundamental solution of Helmholtz equation for k = k₀. The following conclusion holds:Theorem 2.17 If q（x） satisfies the conditions （14）, （37） and （38）, then except for at most a countable number of k, k² is not an interior transmission eigenvalue for every k∈R₊.Remark 2.18 Assume that D contains in a sphere with radiusα, V and M denote the volume of D and upper bound of q（x） respectively, if（?）. （39） then the condition （38） hold. The condition （39） can be checked easterly compared with （38）.4. The range of GWe denote byλ_j∈C andφ_j∈L²（S²） the eigenvalues and eigenfunctions. respectively, of A. Their existence is assured by the general spectral theory for compact normal operators. Furthermore, {φ_j ,j∈N} is a complete orthogonal system in L²（S²） provided A is injective. Letσ_j = |λ_j|, （?） , then Aφ_j = （?）, （?）, thus（?） is a singular system for A.Defineφ_j∈L²（D,q） by（?）, （40）where we take the branch of the square root with Im（?） > 0. Note that from theunitary of S = （?）A the eigenvaluesλ_j lie on the circle of radius （?） with center （?）, i.e. in particular Im（λ_j） > 0. Then, since A = -4πG（I + T^#）G^#,（?）. （41）We will prove that {φ_j : j∈N} forms a Riesz basis in H_q = R（G^#） = N（G）^? （?） L²（D,q）.Theorem 2.19 Assume that k² is not an interior transmission eigenvalue. Then the functions {φ_j : j∈N} from （40） formαRiesz basis in H_q = N（G）^? （?） L²（D, q）,i.e. every elementφ∈H_q has a unique representation in the form （?） with（?）, and every sequence {α_j} （?） C with （?） generates an element （?）, Furthermore, there exists c > 1 with （?）, and every sequence {α_j} （?） C with （?） generates an element （?）. （42）Theorem 2.20 Assume that k² is not an interior transmission eigenvalue, then we have（?）, （43）whereλ_j,φ_j are the eigenvalues and eigenfunctions, respectively, of A. Here G|_H1 is the restriction of G to the closed subspace H₁ of L²（D,q）.As an application we give a simple characterization of the support of q（x） as follows.Theorem 2.21 Assume that k² is not an interior transmission eigenvalue. Define the function r_z∈L²（S²） by r_z（?） := （?）, （?）∈S². Then z∈D if and only if r_z∈R(G|_H1), i.e. by the previous theorem（?） , （44）whereρ_j^（z）∈C are the expansion coefficients of r_z with respect toφ_j.We make some detailed analysis for three-dimensional case, but all of the results also hold for the two-dimensional case with possibly different constants, the numerical experiments can show this very well.更多还原