- #1
Dustinsfl
- 2,281
- 5
Consider Laplace's equation on a sphere of unit radius with the boundary condition
$$
u(1,\theta,\varphi) = f(\theta,\varphi)\begin{cases}
100 & -\pi/4 < \varphi < \pi/4\\
0 & \text{otherwise}
\end{cases}
$$
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 $m = -\ell,\ldots,\ell$.
$$
u(1,\theta,\varphi) = f(\theta,\varphi)\begin{cases}
100 & -\pi/4 < \varphi < \pi/4\\
0 & \text{otherwise}
\end{cases}
$$
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 $m = -\ell,\ldots,\ell$.