When I'm learning laplace's equation with the case is dirichlet problem for a circle, i know that naturally i have to separate variables in polar coordinate.
And i derive from rectangle to polar coordinat, then i separate the variable until i have two ODE as follows
$\Theta''+\lambda\Theta=0 \\ r^2R''+rR-\lambda R=0$
Solve the ODE and Plug in to the solution $u(r,\theta)=R(r)\cdot \Theta(\theta)$ and i have:
$u=\left(Cr^n+Dr^{-n}\right)(A\cos n\theta+B\sin n\theta)$
But i still don't know, in the textbooks that i've read told me that D is vanish and summing the remaining solution, it become full Fourier Series as follows:
$u=\frac{1}{2} A_0+\displaystyle\sum_{n=1}^{\infty}r^n(A_n\cos n\theta+B_n\sin n\theta)$
Please, explain to me about the last part. Thanks.
There is a boundary condition in the disk given which makes the other part go away.
$$ | u(0,\theta)| \leq \infty $$
This conditions enforces that
$$ | R(0) | \leq \infty $$
For this to work you note that in
$$ R(r) = Cr^{n} + Dr^{-n} $$
when $ r \to 0$ this will blow up. So for it to work $ D = 0$.
You'll be left with
$$ u(r,\theta) = \sum_{n=0}^{\infty} A_{n}r^{n} \cos(n\theta) + \sum_{n=1}^{\infty} B_{n} r^{n} \sin(n\theta) $$
Then we take out $A_{0}$ since
$$ A_{0} r^{0} \cos(0 \cdot \theta) = A_{0} $$
So we can change $n$ to $n=1$
$$ u(r,\theta) = A_{0} + \sum_{n=1}^{\infty} A_{n} r^{n} \cos(n \theta) + \sum_{n=1}^{\infty} B_{n} r^{n} \sin(n \theta) = A_{0} + \sum_{n=1}^{\infty} r^{n} (A_{n} \cos(n \theta) + B_{n} \sin(n \theta) ) $$
I believe $\frac{A_{0}}{2} $ comes from the interval.