Mathematical Methods for Physics and Engineering - Matematica.NET

21.5 INHOMOGENEOUS PROBLEMS – GREEN’S FUNCTIONSVVSˆnS 1S 2ˆnˆn(a)Figure 21.11regions V .Surfaces used for solving Poisson’s equation in different(b)where on the RHS it is common to write, for example, ∇ψ · ˆn dS as (∂ψ/∂n) dS.The expression ∂ψ/∂n stands for ∇ψ · ˆn, the rate of change of ψ in the directionof the unit outward normal ˆn to the surface S.The Green’s function for Poisson’s equation (21.80) must satisfy∇ 2 G(r, r 0 )=δ(r − r 0 ), (21.82)where r 0 lies in V . (As mentioned above, we may think of G(r, r 0 )asthesolutionto Poisson’s equation for a unit-strength point source located at r = r 0 .) Let usfor the moment impose no boundary conditions on G(r, r 0 ).If we now let φ = u(r) andψ = G(r, r 0 ) in Green’s theorem (21.81) then weobtain∫[u(r)∇ 2 G(r, r 0 ) − G(r, r 0 ) ∇ 2 u(r) ] dV (r)V∫ [= u(r) ∂G(r, r 0)− G(r, r 0 ) ∂u(r) ]dS(r),S ∂n∂nwhere we have made explicit that the volume and surface integrals are withrespect to r. Using (21.80) and (21.82) the LHS can be simplified to give∫[u(r)δ(r − r 0 ) − G(r, r 0 )ρ(r)] dV (r)V∫ [= u(r) ∂G(r, r 0)− G(r, r 0 ) ∂u(r) ]dS(r). (21.83)S ∂n∂nSince r 0 lies within the volume V ,∫u(r)δ(r − r 0 ) dV (r) =u(r 0 ),Vand thus on rearranging (21.83) the solution to Poisson’s equation (21.80) can be755