MHB Boundary conditions spherical coordinates

Dustinsfl
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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.
 
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dwsmith said:
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.

I have this solved.
 
I have the equation ##F^x=m\frac {d}{dt}(\gamma v^x)##, where ##\gamma## is the Lorentz factor, and ##x## is a superscript, not an exponent. In my textbook the solution is given as ##\frac {F^x}{m}t=\frac {v^x}{\sqrt {1-v^{x^2}/c^2}}##. What bothers me is, when I separate the variables I get ##\frac {F^x}{m}dt=d(\gamma v^x)##. Can I simply consider ##d(\gamma v^x)## the variable of integration without any further considerations? Can I simply make the substitution ##\gamma v^x = u## and then...

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