Mathematical Methods for Physics and Engineering : A Comprehensive Guide

PDES: SEPARATION OF VARIABLES AND OTHER METHODS

written as

u(r 0 )=

∫

V

G(r,r 0 )ρ(r)dV(r)+

∫

S

[

u(r)

∂G(r,r 0 )

∂n

−G(r,r 0 )

∂u(r)

∂n

]

dS(r).

(21.84)

Clearly, we can interchange the roles ofrandr 0 in (21.84) if we wish. (Remember

also that, for a real Green’s function,G(r,r 0 )=G(r 0 ,r).)

Equation (21.84) iscentralto the extension of the Green’s function method

to problems with inhomogeneous boundary conditions, and we next discuss its

application to both Dirichlet and Neumann boundary-value problems. But, before

doing so, we also note that if the boundary condition onSis in fact homogeneous,

so thatu(r)=0or∂u(r)/∂n=0onS, then demanding that the Green’s function

G(r,r 0 ) also obeys the same boundary condition causes the surface integral in

(21.84) to vanish, and we are left with the familiar form of solution given in

(21.78). The extension of (21.84) to a PDE other than Poisson’s equation is

discussed in exercise 21.28.

21.5.3 Dirichlet problems

In a Dirichlet problem we require the solutionu(r) of Poisson’s equation (21.80)

to take specific values on some surfaceSthat boundsV,i.e.werequirethat

u(r)=f(r)onSwherefis a given function.

If we seek a Green’s functionG(r,r 0 ) for this problem it must clearly satisfy

(21.82), but we are free to choose the boundary conditions satisfied byG(r,r 0 )in

such a way as to make the solution (21.84) as simple as possible. From (21.84),

we see that by choosing

G(r,r 0 )=0 forronS (21.85)

the second term in the surface integral vanishes. Sinceu(r)=f(r)onS, (21.84)

then becomes

u(r 0 )=

∫

V

G(r,r 0 )ρ(r)dV(r)+

∫

S

f(r)

∂G(r,r 0 )

∂n

dS(r). (21.86)

Thus we wish to find theDirichlet Green’s functionthat

(i) satisfies (21.82) and hence is singular atr=r 0 ,and

(ii) obeys the boundary conditionG(r,r 0 )=0forronS.

In general, it is difficult to obtain this function directly, and so it is useful to

separate these two requirements. We therefore look for a solution of the form

G(r,r 0 )=F(r,r 0 )+H(r,r 0 ),

whereF(r,r 0 ) satisfies (21.82) and has the required singular character atr=r 0 but

does not necessarily obey the boundary condition onS, whilstH(r,r 0 ) satisfies