Mathematical Methods for Physics and Engineering - Matematica.NET

PDES: SEPARATION OF VARIABLES AND OTHER METHODSwritten as∫u(r 0 )=V∫ [G(r, r 0 )ρ(r) dV (r)+ u(r) ∂G(r, r 0)− G(r, r 0 ) ∂u(r) ]dS(r).S ∂n∂n(21.84)Clearly, we can interchange the roles of r and r 0 in (21.84) if we wish. (Rememberalso that, for a real Green’s function, G(r, r 0 )=G(r 0 , r).)Equation (21.84) is central to the extension of the Green’s function methodto problems with inhomogeneous boundary conditions, and we next discuss itsapplication to both Dirichlet and Neumann boundary-value problems. But, beforedoing so, we also note that if the boundary condition on S is in fact homogeneous,so that u(r) =0or∂u(r)/∂n =0onS, then demanding that the Green’s functionG(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 isdiscussed in exercise 21.28.21.5.3 Dirichlet problemsIn a Dirichlet problem we require the solution u(r) of Poisson’s equation (21.80)to take specific values on some surface S that bounds V ,i.e.werequirethatu(r) =f(r) onS where f is a given function.If we seek a Green’s function G(r, r 0 ) for this problem it must clearly satisfy(21.82), but we are free to choose the boundary conditions satisfied by G(r, r 0 )insuch a way as to make the solution (21.84) as simple as possible. From (21.84),we see that by choosingG(r, r 0 )=0 forr on S (21.85)the second term in the surface integral vanishes. Since u(r) =f(r) onS, (21.84)then becomes∫∫u(r 0 )= G(r, r 0 )ρ(r) dV (r)+ f(r) ∂G(r, r 0)dS(r). (21.86)VS ∂nThus we wish to find the Dirichlet Green’s function that(i) satisfies (21.82) and hence is singular at r = r 0 ,and(ii) obeys the boundary condition G(r, r 0 )=0forr on S.In general, it is difficult to obtain this function directly, and so it is useful toseparate these two requirements. We therefore look for a solution of the formG(r, r 0 )=F(r, r 0 )+H(r, r 0 ),where F(r, r 0 ) satisfies (21.82) and has the required singular character at r = r 0 butdoes not necessarily obey the boundary condition on S, whilst H(r, r 0 ) satisfies756