# PDE, Dirichlet problem for a circle

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:

$$u=\frac{1}{2} A_0+\displaystyle\sum_{n=1}^{\infty}r^n(A_n\cos n\theta+B_n\sin n\theta)$$

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.