Past exam of the mathematics course of the University of Cambridge / 2007 / ib / Paper 3 / 15E / Solution

Seek a power series $y(x)={\sum }_{k\ge 0}{a}_k{x}^k$ in the Legendre differential equation. Equating coefficients of $x^k$ givessoThe even and odd coefficients form separate chains. Choose a nonzero seed of the same parity as $n$ and set the other seed to zero. In the chosen chain, none of the factors vanishes before $k=n$, and the factor $k-n$ vanishes at $k=n$. Therefore this chain terminates with a nonzero $a_n$: it gives a polynomial of degree exactly $n$.

To impose the normalization, its value at $x=1$ must be nonzero. Suppose a nonzero polynomial solution were $y=(x-1{)}^{m}q(x)$ with $m\ge 1$ and $q(1)\ne 0$. The lowest-order coefficient in $(1-x^2)y^{\prime \prime }-2x{y}^{\prime }$ would be $-2m^{2}q(1)(x-1{)}^{m-1}$, whereas $n(n+1)y$ has order $m$. This cannot satisfy the equation. Thus divide by $y(1)$ to obtain the Legendre polynomial $P_n$ with $P_n(1)=1$. Substitution or the coefficient recurrence givesFor an axisymmetric solution of the Laplace equation regular at the two polar axes, the separated modes and their superposition in spherical polar coordinates areThis is the usual separation of variables expansion; the singular angular solutions are excluded by regularity at the axes.

Here ${\sin }^2\theta =\frac{2}{3}(P_0(\cos \theta )-P_2(\cos \theta ))$, so only degrees $0$ and $2$ are required. The degree-zero radial function $A_0+B_0/r$ takes the same value $2/3$ at $r=a,b$; since $a<b$, this forces $B_0=0$, $A_0=2/3$. Write the degree-two coefficient as $-(2/3)T(r)$, where $T=A{r}^2+B{r}^{-3}$ and $T(a)=T(b)=1$. Multiplying these boundary equations by $a^3,b^3$ and subtracting givesConsequentlyEach term is harmonic in the shell, and the radial factor is $1$ at both boundaries, so this satisfies both Dirichlet boundary conditions. The maximum principle for harmonic functions, applied to the difference of two solutions, proves uniqueness among solutions continuous on the closed shell.