麻省理工学院：《偏微分方程式数字方法》（英文版）Lecture 24 notes.pdf_大学文库

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Laplace Problems Exterior Radiation Condition Green’s function Ansatz or Indirect Approach Single and Double Layer Potentials First and Second Kind Equations Greens Theorem Approach First and Second Kind Equations

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Numerical methods for pdes Integral Equation Methods, Lecture 4 Formulating Boundary Integral equat Notes by Suvranu De and J. White April 30, 2003

Numerical Methods for PDEs Integral Equation Methods, Lecture 4 Formulating Boundary Integral Equations Notes by Suvranu De and J. White April 30, 2003

1 Outline Laplace problems Exterior Radiation Condition Ansatz or Indirect Approach Single and Double layer Potentials and Second Kind Eq Greens Theorem Approach First and Second Kind equations 2 3-D Laplace problems 2.1 Differential equatior SLIDE 2 Laplace's equation in 3-D 02u(),02u(2),02u() ay 2 where ∈g2 and s is bounded by t. 2.2 Boundary Conditions sLide 3 Dirichlet Condition (x)x∈r Neumann Condition du(a) dur(a) x∈T PLUS 2.2.1 Radiation Condition SLIDE 4 lim u(x)→0 1im→x3(→O(‖-1)

1 Outline Slide 1 Laplace Problems Exterior Radiation Condition Green’s function Ansatz or Indirect Approach Single and Double Layer Potentials First and Second Kind Equations Greens Theorem Approach First and Second Kind Equations 2 3-D Laplace Problems 2.1 Differential Equation Slide 2 Laplace’s equation in 3-D ∇2u(x) = ∂2u(x) ∂x2 + ∂2u(x) ∂y2 + ∂2u(x) ∂z2 = 0 where x = x, y, z ∈ Ω and Ω is bounded by Γ. 2.2 Boundary Conditions Slide 3 Dirichlet Condition u(x) = uΓ(x) x ∈ Γ OR Neumann Condition ∂u(x) ∂nx = ∂uΓ(x) ∂nx x ∈ Γ PLUS A Radiation Condition 2.2.1 Radiation Condition Slide 4 The Radiation Condition limx→∞u(x) → 0 not specific enough! Need limx→∞u(x) → O(x−1) 1

li ()→O(|-2) 2.3 Greens Function SLIDE 5 Laplace's Equation Greens Functie V2G(动)=4丌() 6(x)≡ impulse in3 Defined by its behavior in an integral 6(r)f(x)ds2=f(0) Not too hard to show ‖ 3 Ansatz(Indirect) Formulations 3.1 Single Layer Potential SLIDE 6 u(i) o(r)dr u(a)automatically satisfies V2u=0 on 12 Must now enforce boundary condition 3.1.1 Boundary Conditions SLIDE 7 Dirichlet Problem r(r) a(x)dr'z∈r Neumann Problem dur(r) a (x)drz∈r

OR limx→∞u(x) → O(x−2) 2.3 Greens Function Slide 5 Laplace’s Equation Greens Function ∇2G(x)=4πδ(x) δ(x) ≡ impulse in 3-D Defined by its behavior in an integral δ(x )f(x )dΩ = f(0) Not too hard to show G(x) = 1 x 3 Ansatz (Indirect) Formulations 3.1 Single Layer Potential Slide 6 Consider u(x) = Γ 1 x − x σ(x )dΓ u(x) automatically satisfies ∇2u = 0 on Ω. Must now enforce boundary conditions 3.1.1 Boundary Conditions Slide 7 Dirichlet Problem uΓ(x) = Γ 1 x − x σ(x )dΓ x ∈ Γ Neumann Problem ∂uΓ(x) ∂nx = ∂ ∂nx Γ 1 x − x σ(x )dΓ x ∈ Γ 2

3.1.2 Care Evaluating Integrals slide 8 On a smooth surface anil-‖ (E )dr 3.1.3 Neumann Problem 2nd Kind! slide 9 2TO()+ani l 3.1.4 Radiation Condition SLIDE 10 limrl-au(G=A=-2zTo( )ar-O() Unless o(adr=0 Then lim→∞(→O(-) 3.2 Double Layer Potential Consider SLIDE 11 闭= anp li-r/(2)dr u(a)automatically satisfies V2u=0 on 12 Must now enforce boundary conditions

3.1.2 Care Evaluating Integrals Slide 8 On a smooth surface: lim x→Γ ∂ ∂nx Γ 1 x − x σ(x )dΓ = 2πσ(x ) + Γ ∂ ∂nx 1 x − x σ(x )dΓ 3.1.3 Neumann Problem 2nd Kind! Slide 9 ∂uΓ(x) ∂nx = 2πσ(x ) + Γ ∂ ∂nx 1 x − x σ(x )dΓ 3.1.4 Radiation Condition Slide 10 limx→∞u(x) = Γ 1 x − x σ(x )dΓ → O(x−1) Unless Γ σ(x )dΓ = 0 Then limx→∞u(x) → O(x−2) 3.2 Double Layer Potential Slide 11 Consider u(x) = Γ ∂ ∂nx 1 x − x µ(x )dΓ u(x) automatically satisfies ∇2u = 0 on Ω. Must now enforce boundary conditions 3

3.2.1 Boundary Conditions Dirichlet Problem ur()=on =- Fo(arier Neumann Problem r(a) a mz=m1如mF|- a(x)drx∈r Neumann Problem generates Hypersingular Integral 3.2.2 Dirichlet Problem 2nd Kind SLIDE 13 dur(E) (x) 3.2.3 Radiation Condition SLIDE 14 nr‖-‖ d→O(引-2) Add Extra Term to slow decay 0 4 Greens Theorem Approach 4.1 Green's Second Identity SLIDE 15 Aluon-uomarl Easy to implement any boundary conditions

3.2.1 Boundary Conditions Slide 12 Dirichlet Problem uΓ(x) = Γ ∂ ∂nx 1 x − x σ(x )dΓ x ∈ Γ Neumann Problem ∂uΓ(x) ∂nx = ∂ ∂nx Γ ∂ ∂nx 1 x − x σ(x )dΓ x ∈ Γ Neumann Problem generates Hypersingular Integral 3.2.2 Dirichlet Problem 2nd Kind! Slide 13 ∂uΓ(x) ∂nx = 2πσ(x ) + Γ ∂ ∂nx 1 x − x σ(x )dΓ 3.2.3 Radiation Condition Slide 14 limx→∞u(x) = Γ ∂ ∂nx 1 x − x σ(x )dΓ → O(x−2) Add Extra Term to slow decay u(x) = Γ ∂ ∂nx 1 x − x σ(x )dΓ + αG(x∗) x∗ Ω 4 Green’s Theorem Approach 4.1 Green’s Second Identity Slide 15 Ω u∇2w − w∇2u dΩ = Γ w ∂u ∂n − u ∂w ∂n dΓ Now let w = 1 x−x 2πu(x) = Γ 1 x − x ∂u ∂n − u ∂ ∂nx 1 x − x dΓ Easy to implement any boundary conditions! 4

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