GREEN S IDENTITIES AND GREEN S FUNCTIONS

Transcription

1 GREEN S IENTITIES AN GREEN S FUNCTIONS Green s first identity First, recall the following theorem. Theorem: (ivergence Theorem) Let be a bounded solid region with a piecewise C 1 boundary surface. Let n be the unit outward normal vector on. Let f be any C 1 vector field on =. Then f dv = f n ds where dv is the volume element in and ds is the surface element on. By integrating the identity (v u) = v u + v u over and applying the divergence theorem, one gets v u ds = v udv + v u dv (1) where u/ = n u is the directional derivative in the outward normal direction. This is Green s first identity. 1

2 Constraint on Neumann problems Let v = 1. Green s first identity becomes u ds = udv. (2) Consider the Neumann problem in any domain : u = f(x) in u = h(x) on. Applying the above identity, one gets h ds = f dv. Therefore h and f cannot be freely selected, else the problem may have no solution. related to Gauss Law Mean value property In three dimensions, the average value of any harmonic function over any sphere equals its value at the center. Applying the identity Eq.(2) to a sphere with radius a in spherical coordinates, one gets 2π π = a 2 2π u r (a, θ, φ)a 2 sinθdθdφ = π u r (a, θ, φ) sin θdθdφ 2

3 This gives [ d 1 dr 4πr 2 = 1 4π 2π π 2π π ] u(r, θ, φ)r 2 sin θdθdφ u r (r, θ, φ) sin θdθdφ = Thus the mean value inside the rectangular bracket is independent of r and 1 2π π u(r,θ, φ)r 2 sin θdθdφ = u(). 4πr 2 Maximum Principle Theorem: If is a connected solid region, a non-constant harmonic function in cannot take its maximum value inside, but only on. This can be shown in the same way as the two dimensional case. Uniqueness of irichlet s problem Applying the Maximum Principle is one way to show uniqueness, but there is an alternative. By substituting v = u, a harmonic function, in Green s first identity, one gets 3

4 u u ds = u 2 dv. Let u 1 and u 2 be two solutions of the same irichlet problem, then their difference u = u 1 u 2 is a harmonic function satisfying the zero boundary irichlet problem, then u 2 dv = u 2 = u = constant =. For Neumann problem, the solution is unique up to a constant on. irichlet s (energy) Principle Theorem: Among all the functions w(x) continuous on and C 1 on that satisfy the irichlet boundary condition w = h(x) on, the harmonic function u that satisfies the boundary condition minimize the energy: E[w] = 1 w 2 dx. 2 Proof: Let w = u +v where both w and u satisfy the irichlet boundary condition, but u is a harmonic function. Then v is a function that has zero value on. E[w] = 1 (u 2 + v) 2 dx. 4

5 = E[u] + u v dx + E[v] As v = on and u = in, Green s first identity gives u v dx =. Therefore, E[w] = E[u] + E[v] E[u]. This means that the energy is smallest when w = u. Green s second identity Switch u and v in Green s first identity, then subtract it from the original form of the identity. The result is ( (u v v u)dv = u v ) v u ds. (3) This is Green s second identity. It is valid for any general (need not be harmonic) pair of functions u and v. Representation formula Let K(x,x ) 1/(4π x x ). Any harmonic function u in an open solid region can be express as an integral over the boundary as [ u(x ) = u(x) K(x,x ) K(x,x ) u ] ds where x and x is on. 5

6 Proof: First, note that K(x,x ) = except at x = x. Therefore, both u and K are harmonic in the domain = B ɛ (x ) where B ɛ (x ) a ball in, centered at x, and with radius ɛ >. Applying Green s second identity to u and K in, one gets [ u(x) K(x,x ) K(x,x ) u ] ds =. ( B ɛ ) With x put at the origin of a spherical coordinate system, the unit normal for the directional derivative at the surface of B ɛ is pointing towards the origin. Flipping the unit normal outward introduces a minus sign in front of the flux integral over B ɛ. The equation can be rewritten as: [ u(x) K(x,x ) K(x,x ) u ] ds ( ) ( ] 1 1 = 1 4π = 1 4π B ɛ r=ɛ [ u(x) r ) u r ds r r [ u + ɛ u ] sin θdθdφ u() as ɛ +. r The formula is thus obtained. This formula is not useful for finding a solution. For a irichlet problem, u is uniquely determined by its value on. There is no freedom in choosing u/. However, this 6

7 formula is a step towards Green s function, the use of which eliminates the u/ term. Green s Function It is possible to derive a formula that expresses a harmonic function u in terms of its value on only. efinition: Let x be an interior point of. The Green s function G(x,x ) for the operator and the domain is a function defined for x such that: (i) Let K(x,x ) = 1/(4π x x ). The function H(x) G(x,x ) K(x,x ) has continuous second derivatives and is harmonic in (including the point x ). (ii) G(x,x ) = for x. Requirement (i) implies that as a function of x, G(x,x ) possesses continuous second derivatives and G = in, except at the point x = x. Requirement (ii) implies that the value of H on is given by H(x) = K(x,x ) where x is on. If the solution of a irichlet problem with arbitrary boundary value (described by a continuous function on ) exists, then H exists (and so does G). However, this important existence theorem is not to be proven here. 7

8 Theorem: If G(x,x ) is the Green s function, then the solution of the irichlet problem is given by the formula u(x ) = u(x) G(x,x ) ds. Proof: Recall that the representation formula is ( u(x ) = u K ) K u ds. The result of applying Green s second identity to the pair of harmonic functions u and H is ( u H ) H u ds =. Adding the two equations, the result becomes ( u(x ) = u G ) G u ds = u G ds. It is the formula needed. Principle of reciprocity The Green s function G(x,x ) of a region is symmetric, i.e. G(x,x ) = G(x,x) for x x. 8

9 This relation ensures the C 2 and harmonic properties of G as a function of x (as long as x x). Proof: Apply Green s second identity to the pair of functions u(x) G(x, a), v(x) G(x, b) in the region = B ɛ (a) B ɛ (b) in which u, v are harmonic. The result is (u v v u)dv = where I ɛ (a) = and I ɛ (b) = ( u v ) v u ds I ɛ (a) I ɛ (b) B ɛ (a) B ɛ (b) ( u v ) v u ds ( u v ) v u ds. Note that the unit normals for the directional derivatives in these two flux integrals are outward from the centers of the balls. Since u, v = in B ɛ (a) B ɛ (b) and u, v vanish on I ɛ (a) + I ɛ (b) = for any ɛ >. To evaluate I ɛ (a), put a at the origin of a spherical coordinate system, so that x a = r and = r. 9

10 Then, take the limit lim I ɛ(a) = ɛ + lim ɛ + r=ɛ = lim ɛ + [( 1 r=ɛ ) v 4πr + H v r v r 1 4πr 2 r2 sin θdθdφ = v(a). ( 1 )] 4πr + H r 2 sinθdθdφ Similarly, lim ɛ + I ɛ(a) = u(b). Therefore, v(a) + u(b) =. G(a,b) = G(b,a). The Green s function can also be applied to solve the Poisson equation. Theorem: The solution of the problem u = f in is given by u(x ) = u(x) G(x,x ) u = h on ds + f(x)g(x,x ) dv. 1

11 Green s function in special geometries Green s functions can be found explicitly for certain special cases. The half-space For the half-space with z >, the Green s function is 1 G(x,x ) = 4π x x + 1 4π x x. where x = (x,y, z ) for x = (x, y, z ). As n is pointing downward at the boundary, G/ = G/ z z= = 1 z 2π x x 3. Therefore, u(x ) = z h(x) 2π x x ds. 3 The Ball The Green s function for the ball = { x < a} is 1 G(x,x ) = 4π x x + q 4π x x. where q = a/ x (> 1) and x = (a/ x ) 2 x = q 2 x. For any x on the surface of the ball, 11

12 x x = x a x a x x = x a x x = (q ) 1 x x. Therefore, G(x,x ) =. Furthermore, as G = x x 4π x x x x 3 q 4π x x = 1 (q ) 2 x 3 4π x x, 3 G = x a G = a2 x 2 4πa x x 3. Thus, u(x ) = a2 x 2 4πa x =a h(x) x x 3 ds. Problem Set 1 12