Method of Green’s Functions 18.303 Linear Partial Differential Equations Matthew J. Hancock Fall 2006 We introduce another powerful method of solving PDEs. First, we need to consider some preliminary definitions and ideas. 1 Preliminary ideas and motivation 1.1 The delta function Ref: Guenther & Lee §10.5, Myint-U & Debnath §10.1 Definition [Delta Function] The δ-function is defined by the following three properties, ( δ (x) = 0, ∞, x �= 0, x = 0, Z ∞ δ (x) dx = 1 −∞ Z ∞ f (x) δ (x − a) dx = f (a) −∞ where f is continuous at x = a. The last is called the sifting property of the δ-function. To make proofs with the δ-function more rigorous, we consider a δ-sequence, that is, a sequence of functions that converge to the δ-function, at least in a pointwise sense. Consider the sequence δn (x) = n −(nx)2 √πe Note that Z ∞ Z ∞ Z ∞ 2−zδn (x) dx = √2nπ e −(nx)2 dx = √2 π e dz = erf (∞) = 1 −∞ 0 0 1( Z Z Z Z Z Z Z Z Z Z Z Z Z   q Definition [2D Delta Function] The 2D δ-function is defined by the following three properties, 0, (x, y) = 0,δ (x, y) = �∞, (x, y) = 0, δ (x, y) dA = 1, f (x, y) δ (x − a, y − b) dA = f (a, b) . 1.2 Green’s identities Ref: Guenther & Lee §8.3 Recall that we derived the identity D (G∇ · F + F · ∇G) dA = C (GF) · nˆdS (1) for any scalar function G and vector valued function F. Setting F = ∇u gives what is called Green’s First Identity, dA = nˆ) dS (2) D  G∇ 2 u + ∇u · ∇G  C G (∇u · Interchanging G and u and subtracting gives Green’s Second Identity, u∇ 2G − G∇ 2 u dA = (u∇G − G∇u) nˆdS. (3) · D C 2 Solution of Laplace and Poisson equation Ref: Guenther & Lee, §5.3, §8.3, Myint-U & Debnath §10.2 – 10.4 Consider the BVP 2 ∇ u = F in D, (4) u = f on C. Let (x, y) be a fixed arbitrary point in a 2D domain D and let (ξ, η) be a variable point used for integration. Let r be the distance from (x, y) to (ξ, η), r =(ξ − x)2 + (η − y)2 . Considering the Green’s identities above motivates us to write ∇ 2G = δ (ξ −x, η − y) = δ (r) in D, (5) G = 0 on C. 2Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z    The notation δ (r) is short for δ (ξ − x, η − y). Substituting (4) and (5) into Green’s second identity (3) gives u (x, y) − GF dA = f∇G nˆdS · D C Rearranging gives u (x, y) = GF dA + f∇G nˆdS (6) · D C Therefore, if we can find a G that satisfies (5), we can use (6) to find the solution u (x, y) of the BVP (4). The advantage is that finding the Green’s function G depends only on the area D and curve C, not on F and f. Note: this method can be generalized to 3D domains. 2.1 Finding the Green’s function To find the Green’s function for a 2D domain D, we first find the simplest function that satisfies ∇2v = δ (r). Suppose that v (x, y) is axis-symmetric, that is, v = v (r). Then 12 ∇ v = vrr + rvr = δ (r) For r > 0, 1 vr = 0vrr + r Integrating gives v = A ln r + B For simplicity, we set B = 0. To find A, we integrate over a disc of radius ε centered at (x, y), Dε, 1 = δ (r) dA = 2vdA ∇Dε Dε From the Divergence Theorem, we have ∇ 2vdA = ∇v · ndS Dε Cε where Cε is the boundary of Dε, i.e. a circle of circumference 2πε. Combining the previous two equations gives ∂v  A 1 = ndS =∇v · ∂r dS = εdS = 2πA Cε Cε r=εCε Hence 1 v (r) = ln r 2π 3q q q This is called the fundamental solution for the Green’s function of the Laplacian on 2D domains. For 3D domains, the fundamental solution for the Green’s function of the Laplacian is −1/(4πr), where r =(x − ξ)2 + (y − η)2 + (z − ζ)2 . The Green’s function for the Laplacian on 2D domains is defined in terms of the corresponding fundamental solution, 1 G (x, y; ξ, η) = ln r + h, 2π h is regular, ∇ 2h = 0, (ξ, η) ∈ D, G = 0 (ξ, η) ∈ C. The term “regular” means that h is twice continuously d ifferentiable in (ξ, η) on D. Finding the Green’s function G is reduced to finding a C2 function h on D that satisfies ∇ 2h = 0 (ξ, η) ∈ D, 1 h = −2π ln r (ξ, η) ∈ C. The definition of G in terms of h gives the BVP (5) for G. Thus, for 2D regions D, finding the Green’s function for the Laplacian reduces to finding h. 2.2 Examples Ref: Myint-U & Debn ath §10.6 (i) Full plane D = R2 . There are no boundaries so h = 0 will do, and 11  2 2 G = ln r = ln (ξ − x) + (η − y)2π 4π (ii) Half plane D = {(x, y) : y > 0}. We find G by introducing what is called an “image point” (x, −y) corresponding to (x, y). Let r be the distance from (ξ, η) to (x, y) and r ′ the distance from (ξ, η) to the image point (x, −y), r =(ξ − x)2 + (η − y)2 , r ′ = (ξ − x)2 + (η + y)2 We add 1 ′ 1 q 2 2h = ln r = ln (ξ − x) + (η + y)−2π −2π to G to make G = 0 on the boundary. Since the image point (x, −y) is NOT in D, then h is regular for all points (ξ, η) ∈ D, and satisfies Laplace’s equation, ∂2h ∂2h = = 0∇ 2h ∂ξ2 + ∂η2 4   G(21/2,21/2;ξ,η) 0 −0.1 −0.2 −0.3 −0.4 −0.5 −0.6 −5 −0.7 00 ξ2 4 6 η 8 105 Figure 1: Plot of the Green’s function G (x, y; ξ, η) for the Laplacian operator in the upper half plane, for (x, y) = (√2,√2). for (ξ, η) ∈ D. Writing things out fully, we have 2 21 1 1 ′ 1 r 1 (ξ − x) + (η − y)G =2π ln r + h =2π ln r −2π ln r =2π ln r ′ =4π ln (ξ − x)2 + (η + y)2 (7) G (x, y; ξ, η) is plotted in th …