Mathematical Methods for Physics and Engineering - Matematica.NET

PDES: SEPARATION OF VARIABLES AND OTHER METHODSand, in a similar way, we might wish to choose ∂G(r, r 0 )/∂n = 0 in the solution ofNeumann problems. However, in general this is not permitted since the Green’sfunction must obey the consistency condition∫∫∂G(r, r 0 )dS = ∇G(r, r 0 ) · ˆn dS = ∇ 2 G(r, r 0 ) dV =1.S ∂nSVThe simplest permitted boundary condition is therefore∂G(r, r 0 )= 1 for r on S,∂n Awhere A is the area of the surface S; this defines a Neumann Green’s function.If we require ∂u(r)/∂n = f(r) onS, the solution to Poisson’s equation is given by∫u(r 0 )= G(r, r 0 )ρ(r) dV (r)+ 1 ∫∫u(r) dS(r) − G(r, r 0 )f(r) dS(r)V A SS∫∫= G(r, r 0 )ρ(r) dV (r)+〈u(r)〉 S − G(r, r 0 )f(r) dS(r), (21.106)Vwhere 〈u(r)〉 S is the average of u over the surface S and is a freely specifiable constant.For Neumann problems in which the volume V is bounded by a surface S atinfinity, we do not need the 〈u(r)〉 S term. For example, if we wish to solve a Neumannproblem outside the unit sphere centred at the origin then r>ais the regionV throughout which we require the solution; this region may be considered as beingbounded by two disconnected surfaces, the surface of the sphere and a surfaceat infinity. By requiring that u(r) → 0as|r| →∞,theterm〈u(r)〉 S becomes zero.As mentioned above, much of our discussion of Dirichlet problems can betaken over into the solution of Neumann problems. In particular, we may use themethod of images to find the appropriate Neumann Green’s function.◮Solve Laplace’s equation in the two-dimensional region |r| ≤a subject to the boundarycondition ∂u/∂n = f(φ) on |r| = a, with ∫ 2πf(φ) dφ =0as required by the consistency0condition (21.105).Let us assume, as in Dirichlet problems with this geometry, that a single image charge isplaced outside the circle atr 1 =a2|r 0 | r 0,2where r 0 is the position of the source inside the circle (see equation (21.100)). Then, from(21.99), we have the useful geometrical result|r − r 1 | = a|r 0 | |r − r 0| for |r| = a. (21.107)Leaving the strength q of the image as a parameter, the Green’s function has the formG(r, r 0 )= 1 (ln |r − r0 | + q ln |r − r 1 | + c ) . (21.108)2π766S∫