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 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 23-DLaplaceProblems 2.1 Di?erential Equation Slide 2 Laplace’s equation in 3-D ? 2 u(vectorx)= ? 2 u(vectorx) ?x 2 + ? 2 u(vectorx) ?y 2 + ? 2 u(vectorx) ?z 2 =0 where vectorx = x,y,z ∈ ? and ? is bounded by Γ. 2.2 Boundary Conditions Slide 3 Dirichlet Condition u(vectorx)=u Γ (vectorx) vectorx ∈ Γ OR Neumann Condition ?u(vectorx) ?n vectorx = ?u Γ (vectorx) ?n vectorx vectorx ∈ Γ PLUS A Radiation Condition 2.2.1 Radiation Condition Slide 4 The Radiation Condition lim bardblvectorxbardbl→∞ u(vectorx) → 0 not specific enough! Need lim bardblvectorxbardbl→∞ u(vectorx) → O(bardblvectorxbardbl ?1 ) 1 OR lim bardblvectorxbardbl→∞ u(vectorx) → O(bardblvectorxbardbl ?2 ) 2.3 Greens Function Slide 5 Laplace’s Equation Greens Function ? 2 G(vectorx)=4πδ(vectorx) δ(vectorx) ≡ impulse in 3-D Defined by its behavior in an integral integraldisplay δ(vectorx prime )f(vectorx prime )d? prime = f(0) Not too hard to show G(vectorx)= 1 bardblvectorxbardbl 3 Ansatz (Indirect) Formulations 3.1 Single Layer Potential Slide 6 Consider u(vectorx)= integraldisplay Γ 1 bardblvectorx?vectorx prime bardbl σ(vectorx prime )dΓ prime u(vectorx) automatically satisfies ? 2 u =0on ?. Must now enforce boundary conditions 3.1.1 Boundary Conditions Slide 7 Dirichlet Problem u Γ (vectorx)= integraldisplay Γ 1 bardblvectorx?vectorx prime bardbl σ(vectorx prime )dΓ prime vectorx ∈ Γ Neumann Problem ?u Γ (vectorx) ?n vectorx = ? ?n vectorx integraldisplay Γ 1 bardblvectorx?vectorx prime bardbl σ(vectorx prime )dΓ prime vectorx ∈ Γ 2 3.1.2 Care Evaluating Integrals Slide 8 On a smooth surface: lim x→Γ ? ?n vectorx integraldisplay Γ 1 bardblvectorx?vectorx prime bardbl σ(vectorx prime )dΓ prime =2πσ(vectorx prime )+ integraldisplay Γ ? ?n vectorx 1 bardblvectorx?vectorx prime bardbl σ(vectorx prime )dΓ prime 3.1.3 Neumann Problem 2nd Kind! Slide 9 ?u Γ (vectorx) ?n vectorx =2πσ(vectorx prime )+ integraldisplay Γ ? ?n vectorx 1 bardblvectorx?vectorx prime bardbl σ(vectorx prime )dΓ prime 3.1.4 Radiation Condition Slide 10 lim bardblvectorxbardbl→∞ u(vectorx)= integraldisplay Γ 1 bardblvectorx?vectorx prime bardbl σ(vectorx prime )dΓ prime → O(bardblvectorxbardbl ?1 ) Unless integraldisplay Γ σ(vectorx prime )dΓ prime =0 Then lim bardblvectorxbardbl→∞ u(vectorx) → O(bardblvectorxbardbl ?2 ) 3.2 Double Layer Potential Slide 11 Consider u(vectorx)= integraldisplay Γ ? ?n vectorx prime 1 bardblvectorx?vectorx prime bardbl μ(vectorx prime )dΓ prime u(vectorx) automatically satisfies ? 2 u =0on ?. Must now enforce boundary conditions 3 3.2.1 Boundary Conditions Slide 12 Dirichlet Problem u Γ (vectorx)= integraldisplay Γ ? ?n vectorx prime 1 bardblvectorx?vectorx prime bardbl σ(vectorx prime )dΓ prime vectorx ∈ Γ Neumann Problem ?u Γ (vectorx) ?n vectorx = ? ?n vectorx integraldisplay Γ ? ?n vectorx prime 1 bardblvectorx?vectorx prime bardbl σ(vectorx prime )dΓ prime vectorx ∈ Γ Neumann Problem generates Hypersingular Integral 3.2.2 Dirichlet Problem 2nd Kind! Slide 13 ?u Γ (vectorx) ?n vectorx =2πσ(vectorx prime )+ integraldisplay Γ ? ?n vectorx prime 1 bardblvectorx?vectorx prime bardbl σ(vectorx prime )dΓ prime 3.2.3 Radiation Condition Slide 14 lim bardblvectorxbardbl→∞ u(vectorx)= integraldisplay Γ ? ?n vectorx prime 1 bardblvectorx?vectorx prime bardbl σ(vectorx prime )dΓ prime → O(bardblvectorxbardbl ?2 ) Add Extra Term to slow decay u(vectorx)= integraldisplay Γ ? ?n vectorx prime 1 bardblvectorx?vectorx prime bardbl σ(vectorx prime )dΓ prime + αG(vectorx ? ) vectorx ? owner ? 4 Green’s Theorem Approach 4.1 Green’s Second Identity Slide 15 integraldisplay ? bracketleftbig u? 2 w?w? 2 u bracketrightbig d?= integraldisplay Γ bracketleftbigg w ?u ?n ?u ?w ?n dΓ bracketrightbigg Now let w = 1 bardblvectorx?vectorx prime bardbl 2πu(vectorx)= integraldisplay Γ bracketleftbigg 1 bardblvectorx?vectorx prime bardbl ?u ?n ?u ? ?n vectorx prime 1 bardblvectorx?vectorx prime bardbl dΓ bracketrightbigg Easy to implement any boundary conditions! 4