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Green's Identities and Green's Functions

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Green's Identities and Green's Functions LECTURE 17 Green's Identities and Green's Functions Let us recall The Divergence Theorem in n-dimensions. Theorem 17.1. Let F : Rn ! Rn be a vector field over Rn that is of class C1 on some closed, connected, simply connected n-dimensional region D ⊂ Rn. Then Z Z r · F dV = F · n dS D @D where @D is the boundary of D and n(r) is the unit vector that is (outward) normal to the surface @D at the point r 2 @D. As a special case of Stokes' theorem, we may set (1) F = rφ with φ a C2 function on D. We then obtain Z Z (2) r2φ dV = rφ · dS : D @D Another special case of Stokes' theorem comes from the choice (3) F = φr : For this case, Stokes' theorem says Z Z (4) r · (φr ) dV = φr · ndS : D @D Using the identity (5) r · (φF) = rφ · F + φr · F we find (4) is equivalent to Z Z Z (6) rφ · r dV + φr2 dV = φr · n dS : D D @D Equation (6) is known as Green's first identity. Reversing the roles of φ and in (6) we obtain Z Z Z (7) r · rφ dV + r2φ dV = rφ · n dS : D D @D Finally, subtracting (7) from (6) we get Z Z (8) φr2 − r2φ dV = (φr − rφ) · n dS : D @D Equation (8) is known as Green's second identity. Now set 1 (r) = jr − roj + 78 17. GREEN'S IDENTITIES AND GREEN'S FUNCTIONS 79 and insert this expression into (8). We then get Z 1 Z 1 φ r2 dV = r2φ dV D jr − roj + " D jr − roj + " Z 1 1 + rφ − φ r · ndS : @D jr − roj + " jr − roj + " Taking the limit ! 0 and using the identities 2 1 n lim r = −4πδ (r − ro) !0 jr − roj + " 1 1 lim = !0 jr − roj + " jr − roj 1 1 lim r = r !0 jr − roj + " jr − roj we obtain R 1 2 −4πφ (ro) = D jr−r j r φ dV (9) o : + R 1 rφ − φ r 1 · ndS @D jr−roj jr−roj Equation (9) is known as Green's third identity. Notice that if φ satisfies Laplace's equation the first term on the right hand side vanishes and so we have −1 R 1 1 φ (ro) = 4π @D jr−r j rφ − φ r jr−r j · ndS (10) o o = 1 R φ @ 1 − 1 @φ dS : 4π @D @n r−ro jr−roj @n @ Here @n is the directional derivative corresponding to the surface normal vector n. Thus, if φ satisfies Laplace's equation in D then its value at any point ro 2 D is completely determined by the values of φ and @φ @n on the boundary of D. 1. GREEN'S FUNCTIONS AND SOLUTIONS OF LAPLACE'S EQUATION, II 80 1. Green's Functions and Solutions of Laplace's Equation, II Recall the fundamental solutions of Laplace's equation in n-dimensions log jrj ; if n = 2 (11) Φn (r; ; θ1; : : : ; θn−2) = 1 : rn−2 ; if n > 2 Each of these solutions really only makes sense in the region Rn − O; for each possesses a singularity at the origin. We studied the case when n = 3, a little more closely and found that we could actually write 1 0 ; if r 6= 0 (12) r2 = −4πδ3 (r) = r 1 ; if r = 0 In fact, using similar arguments one can show that 2 n (13) r Φ(r) = −cnδ (r) n where cn is the surface area of the unit sphere in R . Thus, the fundamental solutions can actually be regarded as solutions of an inhomogeneous Laplace equation where the driving function is concentrated at a single point. Let us now set n = 3 and consider the following PDE/BVP r2Φ(r) = f(r) ; r 2 D (14) Φ(r)j@D = h(r)j@D 3 where D is some closed, connected, simply connected region in R . Let ro be some fixed point in D and set −1 (15) G (r; ro) = + φo(r; ro) 4π jr − roj where φo(r; ro) is some solution of the homogeneous Laplace equation 2 (16) r φo(r; ro) = 0 : Then 2 3 (17) r G (r; ro) = δ (r − ro) : Now recall Green's third identity Z Z (18) Φr2Ψ − Ψr2Φ dV = (ΦrΨ − ΨrΦ) · n dS : D @D If we replace in (18) by G (r; ro) we get R 3 Φ(ro) = D Φ(r)δ (r − ro) dV R 2 = D Φr GdV R 2 R (19) = D Gr Φ dV + @D (ΦrG − GrΦ) · n dS R R @G @Φ = D Gf dV + @D h @n − G @n dS R R @G R @Φ = D Gf dV + @D h @n dS − @D G @n dS : Up to this point we have only required that the function φo satisfies Laplace's equation. We will now make our choice of φo more particular; we shall choose φo(r; ro) to be the unique solution of Laplace's equation in D satisfying the boundary condition 1 (20) = φo(r; ro)j@D 4π jr − roj @D so that G (r; ro)j@D = 0 : 1. GREEN'S FUNCTIONS AND SOLUTIONS OF LAPLACE'S EQUATION, II 81 Then the last integral on the right hand side of (19) vanishes and so we have Z Z @G (21) Φ (ro) = G(r; ro)f(r) dV + h(r) (r; ro) dS : D @D @n Thus, once we find a solution φo (r; ro) to the homogenenous Laplace equation satisfying the boundary condition (21), we have a closed formula for the solution of the PDE/BVP (14) in terms of integrals of @G G (r; ro) times the driving function f(r), and of @n (r; ro) times the function h(r) describing the boundary conditions on Φ. Note that the Green's function G (r; ro) is fixed once we fix φo which in turn depends only on the nature of the boundary of the region D (through condition (20)). Example Let us find the Green's function corresponding to the interior of sphere of radius R centered about the origin. We seek to find a solution of φo of the homogenous Laplace's equation such that (20) is satisfied. This is accomplished by the following trick. Suppose Φ (r; ; θ) is a solution of the homogeneous Laplace equation inside the sphere of radius R centered at the origin. For r > R, we define a function R R2 (22) Φ(~ r; ; θ) = Φ ; ; θ : r r I claim that Φ(~ r; ; θ) so defined also satisfies Laplace's equation in the region exterior to the sphere. To prove this, it suffices to show that 2 2 ~ @ 2 @Φ~ 1 @ @Φ~ 1 @ Φ~ (23) 0 = r rΦ = @r r @r + sin(θ) @θ sin(θ) @θ + sin2(θ) @ 2 or ! ! @ @Φ~ 1 @ @Φ~ 1 @2Φ~ (24) r2 = − sin(θ) − : @r @r sin(θ) @θ @θ sin2(θ) @ 2 Set R2 (25) u = : r so that R2 r = u ~ u (26) Φ(r; ; θ) = R Φ(u; ; θ) @ du @ R2 @ u2 @ @r = − dr @u = − r2 @u = − R2 @u and so @ 2 @Φ~ u2 @ R4 u2 @ u @r r @r = − R2 @u u2 − R2 @u R Φ u2 @ @ = R @u @u (uΦ) u2 @2Φ @Φ = R u @u2 + 2 @u (27) u @ 2 @Φ = R @u u @u u 1 @ @Φ 1 @2Φ = − R sin(θ) @θ sin(θ) @θ + sin2(θ) @ 2 1 @ @Φ~ 1 @2Φ~ = − sin(θ) @θ sin(θ) @θ + sin2(θ) @ 2 Notice that (28) lim Φ(~ r; ; θ) = Φ (r; ; θ) r!R This transform is called Kelvin inversion. 1. GREEN'S FUNCTIONS AND SOLUTIONS OF LAPLACE'S EQUATION, II 82 Now let return to the problem of finding a Green's function for the interior of a sphere of radius. Let R2 R2 (29) ~r = r ; ; θ = r : r r2 In view of the preceding remarks, we know that the functions 1 Φ1 (r) = jr−roj (30) R 1 Φ2 (r) = = Φ~ 1(r) r j~r−roj will satisfy, respectively, 2 3 r Φ1(r) = −4πδ (r − ro) (31) 2 4πR 3 R2r r Φ2 (r) = − r δ r2 − ro : 2 However, notice that the support of r Φ2 (r) lies completely outside the sphere. Therefore, in the interior of the sphere, Φ2 is a solution of the homogenous Laplace equation. We also know that on the boundary of the sphere that we have (32) Φ1(r) = Φ2(r) : Thus, the function R 1 1 G (r; ro) = − r 4π j~r−roj 4πjr−roj (33) = 1 − 1 R r 4πjr−roj 4πj r r− R roj thus satisfies 2 3 (34) rrG (r; ro) = δ (r − ro) for all r inside the sphere and (35) G (r; ro) = 0 or all r on the boundary of the sphere. Thus, the function G (r; ro) defined by (33) is the Green's function for Laplace's equation within the sphere. Now consider the following PDE/BVP r2Φ(r) = f(r) ; r 2 B (36) Φ(R; ; θ) = 0 : where B is a ball of radius R centered about the origin. According to the formula (21) and (33), the solution of (36) is given by Z Z @G Φ(ro) = G(r; ro)f(r) dV + h ( ; θ) (r; ro) dS B @B @n Z = G(r; ro)f(r) dV B To arrive at a more explicit expression, we set ro = (r cos( ) sin(θ); r sin( ) sin(θ); r cos(θ)) r = (ρ cos(α) sin(β); ρ sin(α) sin(β); ρ cos(β)) : Then dV = ρ2 sin2(θ) dρ dα dβ dS = ρ2 sin2(θ) dα dβ 1.
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