1. Problem 5Solve 2 u = 0 r B4 u = cos2 r B4 .This is a lot of work, rst let u(r, , ) = F (r ) G ( ) H () so we may seperate the DE.In spherical coordinates 2111 u=r r 2 r u + 2 (sin u) + 2 u = 0.r2r sin r 2 sin2 1 ddF 1d dG 1 d2 H0 = G ( ) H () r2+ F (r ) H ( ) sin + F (r ) G ( ) r2 drdr r2 sin dd r2 sin2 d2Dividing by u,1 1 d dF11 ddG 1 1 d2 H 0=r2 + sin + .r2 F (r ) drdrr2 sin G ( ) dd r2 sin2 H () d2With some rearrangement, 1 d2 H 1 1 ddF 11 ddG = r2 sin2 r2+ sin . H () d2r2 F (r ) dr dr r2 sin G ( ) ddNotice that when is xed and r and vary the above equation holds even though the RHS is solely afunction of r, and the LHS is solely a function of . Hence, 1 d2 H = 2 H () d2for some constant . We chose 2 (the RHS is always negative then) to ensure that the solutions to theabove ODE are periodic, and the boundary conditions do not imply that H 0. Note, H () = c1 exp(i),or if one desires; H () = c1 sin() + c2 cos(). 1 d2 HReturning to the larger equation, we substituteH () d2 = 2 so that 1 1 ddF11 d dG 2 0=r2+sin . r2 F (r ) dr dr r 2 sin G ( ) ddr2 sin2 Multiplying through by r2 and rearranging some more,1 ddF21 1 d dGr2 =2 sin .F (r ) drdrsin sin G ( ) d dWe chose another seperation constant to convert the above equation into two ODEs. ddF r2 F (r ) = 0(i) dr dr d dG2sin +G ( ) sin G ( ) = 0 (ii) dd sin 1

2. Equation (i) looks like an ODE for spherical Bessel functions at rst glance, but is infact the so-called Eulerdifferential equation if we let = l (l + 1). To solve, we apply the Frobenius method. Make the ansatz,F (r ) = an r nn =0with xed.d dFd2 FdF r2 = r2 + 2rdrdrdr2 drand we compute the derivativesd2 F = n ( n 1 ) a n r n 2dr2n =0 dF= nan r n1 . dr n =0Plugging into the ODE, n ( n 1) a n r n + 2 nan r n l (l + 1) an r n = 0n =0n =0n =0 = an {n(n 1) + 2n l (l + 1)} rn . n =0 = ann2 + n l ( l + 1) r nn =0 = an {n(n + 1) l (l + 1)} rnn =0Now we make the observation that each term in this series is unique. As such, for such a series to terminateto zero either an = 0 n Z+ (which leads to obvious trivialities) OR n(n + 1) l (l + 1) = 0.n(n + 1) = l (l + 1) and n = (l + 1), l.If n = (l + 1), l then the only possible way to rectify the situation is to let an = 0, hence all an forn = l, (l + 1) must be zero. Only two terms survive, 1Fl (r ) = Al r l + Bl , r l +1where l will be summed over in our nal solution.Equation (ii) is the associated Legendre DE which has solutions Pl (cos ) for = 0, ..., l. I think solving(ii) is another application of the Frobenius method, I omit the derivation to save time; everyone knows thisODE anyways.Finally, our general solution is l 1 u(r, , ) = Al r l + Blr l +1 (, ) ll =0 =lwhere (, ) = ei Pl (cos ) are the spherical harmonics familiar to anyone with a background inlquantum mechanics. We now impose our boundary condition. First note, there is an implied conditionthat r 0 causes singularities in our general solution. Thus, we take Bl = 0 l. l u(r = 4, ) = Al 4l Pl (cos ) l =0 =lWhere we have absorbed the exponentials in into the Al s.2