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$$

u(1,\theta,\varphi) = f(\theta,\varphi)\begin{cases}

100 & -\pi/4 < \varphi < \pi/4\\

0 & \text{otherwise}

\end{cases}

$$

Here we will consider a three-term approximation to the solution, i.e., involving the spherical harmonics $Y^m_{\ell}$ for $\ell = 0,1,2$ and $m = -\ell,\ldots,\ell$.

Conclude that the form of the solution will be

$$

u(r,\theta,\varphi) = \sum_{\ell = 0}^2\sum_{m = -\ell}^{\ell}A_{\ell,m}Y^m_{\ell}(\theta,\varphi)

$$

with (shoudn't there be a 100 in front of this integral?)

$$

A_{\ell,m} = \int_{-\pi/4}^{\pi/4}\int_0^{\pi}\bar{Y}^m_{\ell}(\theta,\varphi)\sin\theta d\theta d\varphi.

$$

Let $u(r,\theta,\phi) = G(r,\theta)\Phi(\varphi)$. We have already solved for the axisymmetric case and know the solution is of the form

$$

u(\ell,\theta) = \sum_{\ell = 0}^{\infty}\left(A_{\ell}r^{\ell} + B_{\ell}\frac{1}{r^{\ell + 1}}\right)P_{\ell}(\cos\theta).

$$

Therefore, $\Phi'' = -\lambda^2\Rightarrow n = \pm i\lambda$ (I just put this down since I couldn't derive it. Can someone show me how to get to this part?). That is, $\Phi(\varphi) = e^{\pm i m\varphi}$. So the general solution is

$$

u(r,\theta,\phi) = \sum_{\ell = 0}^{\infty}\sum_{m = -\ell}^{\ell}A_{\ell,m}P^m_{\ell}(\cos\theta)e^{im\varphi}

$$

since $r = 1$.

(There has to be a way to show this two more elogantly than just saying this is this except it) Define $P^{-m}_{\ell}(x) = (-1)^m\frac{(\ell - m)!}{(\ell + m)!}P^m_{\ell}(x)$ and also define

$$

Y^m_{\ell}(\theta,\varphi) = \sqrt{\frac{(2\ell + 1)!(\ell - m)!}{4\pi(\ell + m)!}}P^m_{\ell}(\cos\theta)e^{im\varphi}.

$$

Then we have

$$

u(r,\theta,\varphi) = \sum_{\ell = 0}^2\sum_{m = -\ell}^{\ell}A_{\ell,m}Y^m_{\ell}(\theta,\varphi).

$$

Lastly, using the boundary condition above, we have that

$$

A_{\ell,m} = 100\int_{-\pi/4}^{\pi/4}\int_0^{\pi}\bar{Y}^m_{\ell}(\theta,\varphi)\sin\theta d\theta d\varphi.

$$

How do I do this?

From the definition of the spherical harmonics, write down the explicit expressions for $Y^m_{\ell}$ for $\ell = 0,1,2$ and $m = -\ell,\ldots,\ell$.