- #1
Dustinsfl
- 2,281
- 5
Laplace axisymmetric
$u(a,\theta) = f(\theta)$ and $u(b,\theta) = 0$ where $a<\theta<b$.
The general soln is
$$
u(r,\theta) = \sum_{n=0}^{\infty}A_n r^n P_n(\cos\theta) + B_n\frac{1}{r^{n+1}}P_n(\cos\theta)
$$
I am supposed to obtain
$$
u(r,\theta) = \sum_{n = 0}^{\infty}A_n\left[\left(\frac{r}{b}\right)^n - \frac{b}{r}^{n + 1}\right]P_n(\cos\theta)
$$
with
$$
A_nb^n\left[\left(\frac{r}{b}\right)^n - \frac{b}{r}^{n + 1}\right] = \frac{2n + 1}{2}\int_0^{\pi}f(\theta)P_n(\cos\theta) \sin \theta d\theta.
$$
Using the BC I can't obtain that.
$u(a,\theta) = f(\theta)$ and $u(b,\theta) = 0$ where $a<\theta<b$.
The general soln is
$$
u(r,\theta) = \sum_{n=0}^{\infty}A_n r^n P_n(\cos\theta) + B_n\frac{1}{r^{n+1}}P_n(\cos\theta)
$$
I am supposed to obtain
$$
u(r,\theta) = \sum_{n = 0}^{\infty}A_n\left[\left(\frac{r}{b}\right)^n - \frac{b}{r}^{n + 1}\right]P_n(\cos\theta)
$$
with
$$
A_nb^n\left[\left(\frac{r}{b}\right)^n - \frac{b}{r}^{n + 1}\right] = \frac{2n + 1}{2}\int_0^{\pi}f(\theta)P_n(\cos\theta) \sin \theta d\theta.
$$
Using the BC I can't obtain that.