Ive got the following problem:
Let $\mathbb{D}=\{(x,y) \in \mathbb{R}^2:x^2+y^2<1\}$ be the unit disk.
Let g: $\partial \mathbb{D} \rightarrow \mathbb{R}$ continuously differentiable in Fourier-form: $g(e^{it})=\sum_{k \in \mathbb{Z}}c_ke^{ikt}$
Evaluate u of the Dirichlet-Problem:
$\Delta u=0$ on $\mathbb{D}$
$u=g$ on $\partial \mathbb{D}$
in dependence of $c_k$
Tip: Use the polar form of the laplace-operator and the separation of variables.
What I did so far:
$\Delta=\frac{\partial^2}{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r}+\frac{1}{r^2}\frac{\partial^2}{\partial \phi^2}\\ Let: \ u=R(r)Y(\phi)\\ =>\Delta u=\frac{\partial^2}{\partial r^2}R(r)Y(\phi)+\frac{1}{r}\frac{\partial}{\partial r}R(r)Y(\phi)+\frac{1}{r^2}\frac{\partial^2}{\partial \phi^2}R(r)Y(\phi)$
Separation:
$\frac{r^2}{R(r)}[\frac{\partial^2}{\partial r^2}R(r)+\frac{1}{r}\frac{\partial}{\partial r}R(r)]=-\frac{1}{Y(\phi)}\frac{\partial^2}{\partial \phi^2}Y(\phi)$
Now I dont know how to go further.
Im happy for every answer!
So you start off with separation of variables $u(\varphi, r)=R(r)\Phi(\varphi)$ $$r^{2}\frac{R''}{R}+r\frac{R'}{R}=-\frac{\Phi''}{\Phi}=m^{2}$$ So $$\Phi''+m^{2}\Phi=0$$ $$r^{2}R''+rR'-m^{2}R=0$$ The first equation has a periodic boundary condition $\Phi(\varphi)=\Phi(\varphi+2\pi)$, because $\varphi$ is the angular variable, the most general solution is thus $$\Phi(\varphi)=\sum_{m\in\mathbb{Z}}\Phi_{m}e^{im\varphi}$$ The second equation is the Euler equation and has the solution in the form $$R(r)\propto{r^{k}}$$ This substitution gives $$k=\pm{m}$$ So the solution is $$R(r)=R_{m}r^{m}+R_{-m}r^{-m}$$ So, the solution to the original equation is $$u(\varphi, r)=\sum_{m\in\mathbb{Z}}(R'_{m}r^{m}+R'_{-m}r^{-m})e^{im\varphi}$$ Where we've defined $$R'_{\pm{m}}=R_{\pm{m}}\Phi_{m}$$ The boundary conditions gives $$u(\varphi, 1)=\sum_{m\in\mathbb{Z}}(R'_{m}+R'_{-m})e^{im\varphi}=\sum_{m\in\mathbb{Z}}c_{m}e^{im\varphi}$$ So, $$R'_{-m}=c_{m}-R'_{m}$$ And the solution is $$u(\varphi, r)=\sum_{m\in\mathbb{Z}}(R'_{m}r^{m}+[c_{m}-R'_{m}]r^{-m})e^{im\varphi}$$