Mathematical Methods for Physics and Engineering - Matematica.NET

PDES: SEPARATION OF VARIABLES AND OTHER METHODSTherefore, in order that G =0at|r| = a, the strength of the image charge must be−a/|r 0 |. Consequently, the Dirichlet Green’s function for the exterior of the sphere is1G(r, r 0 )=−4π|r − r 0 | + a/|r 0 |4π |r − (a 2 /|r 0 | 2 )r 0 | .For a less formal treatment of the same problem see exercise 21.22. ◭If we seek solutions to Poisson’s equation in the interior of a sphere then theabove analysis still holds, but r and r 0 are now inside the sphere and the imager 1 lies outside it.For two-dimensional Dirichlet problems outside the circle |r| = a, we are ledby arguments similar to those employed previously to use the same image pointas in the three-dimensional case, namelyr 1 =a2|r 0 | 2 r 0. (21.100)As illustrated below, however, it is usually necessary to take the image strengthas −1 in two-dimensional problems.◮Solve Laplace’s equation in the two-dimensional region |r| ≤a, subject to the boundarycondition u = f(φ) on |r| = a.In this case we wish to find the Dirichlet Green’s function in the interior of a disc ofradius a, so the image charge must lie outside the disc. Taking the strength of the imageto be −1, we haveG(r, r 0 )= 12π ln |r − r 0|− 12π ln |r − r 1| + c,where r 1 =(a 2 /|r 0 | 2 )r 0 lies outside the disc, and c is a constant that includes the strengthof the image charge and does not necessarily equal zero.Since we require G(r, r 0 )=0when|r| = a, the value of the constant c is determined,and the Dirichlet Green’s function for this problem is given byG(r, r 0 )= 1 (ln |r − r 0 |−ln2π∣ r −a2|r 0 | r 2 0∣ − ln |r )0|. (21.101)aUsing plane polar coordinates, the solution to the boundary-value problem can be writtenas a line integral around the circle ρ = a:∫u(r 0 )= f(r) ∂G(r, r 0)dlC ∂n∫ 2π= f(r) ∂G(r, r 0)0 ∂ρ ∣ adφ. (21.102)ρ=aThe normal derivative of the Green’s function (21.101) is given by∂G(r, r 0 )∂ρ= r|r| · ∇G(r, r 0)= r2π|r| ·( r − r0|r − r 0 | − r − r )1. (21.103)2 |r − r 1 | 2764