Laplace equation inside a sphere

Question

Given the boundary value problem: $$ \Delta u(r,\theta,\phi)=0,\quad 0\leq r<2,\quad 0\leq \theta <\pi,\quad 0\leq \phi <2\pi$$ $$u(2,\theta,\phi)=1+\cos(\theta)+7\cos^2(\theta)$$

and the Legendre polynomials: $P_0(x)=1,P_1(x)=x,P_2(x)=\frac12(3x^2-1)$

The Spherical Harmonics formula is: $$ u=\sum_{n=0}^{\infty} \sum_{m=-n}^{n} A_{nm}r^{n}Y_{n}^{m}$$

for $m=0$, the spherical harmonics reduce to the Legendre polynomials in $cos(\theta)$. From there I reckon that: $$A_0^0=1, A_1^0=\frac12$$

but I'm having trouble finding $A_0^2$ to complete the solution. Any help?

Answer 1

We only care about the spherical harmonics with $m=0$ and $n=0,1,2$. So

$$ u(r,\theta,\phi) = A_0^0 + A_1^0r\cos \theta + A_2^0 \frac{r^2}{2}(3\cos^2 \theta - 1) $$

Applying the boundary

\begin{align} u(2,\theta,\phi) &= A_0^0 + 2A_1^0\cos\theta + 2A_2^0(3\cos^2\theta-1) \\ &= (A_0^0 - 2A_2^0) + 2A_1^0\cos\theta + 6A_2^0\cos^2\theta \end{align}

Comparing coefficients

\begin{align} A_0^0 - 2A_2^0 &= 1 \\ 2A_1^0 &= 1 \\ 6A_2^0 &= 7 \end{align}

$\implies A_2^0 = \dfrac76, A_1^0 = \dfrac12, A_0^0 = \dfrac{10}{3} $