Let $\Omega = \{(x,y) \in \mathbb{R}^2 \mid x^2 + y^2< 1\} = \{(r,\theta) \mid 0 \leq r < 1, \; 0 \leq \theta < 2\pi\}$. Use seperation of variables $(r,\theta)$ to solve the Dirichlet problem $$\begin{cases} \vartriangle u = 0 \text{ in }\Omega \\ u(1,\theta) = g(\theta) \end{cases}$$
Assuming $u(r,\theta) = X(r)Y(\theta)$ then we get $X''(r)Y(\theta) = -X(r)Y''(\theta)$. Dividing both sides of the equation by $X(r)Y(\theta)$, we obtain $$\frac{X''(r)}{X(r)} = \frac{Y''(\theta)}{Y(\theta)} = - \lambda$$.
Since $u(1,\theta) = g(\theta)$, then $X(1)Y(\theta) = g(\theta)$.
But, I do not know what to do from here. Can anyone provide some help?
The Laplacian in polar coordinates is $$ \Delta f(r,\theta)= \frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial}{\partial r}f\right)+\frac{1}{r^2}\frac{\partial^2 f}{\partial\theta^2}. $$ Let $f(r,\theta)=R(r)\Theta(\theta)$, divide $\Delta f=0$ by $f$ to obtain the separated equations: $$ \frac{r}{R}\frac{d}{dr}r\frac{dR}{dr}=\lambda=-\frac{\Theta''}{\Theta} \\ r(rR')'-\lambda R=0,\;\;\Theta''+\lambda\Theta =0. $$ Periodicity in $\theta$ requires $\lambda=n^2,\;\;n=0,1,2,3,\cdots$, and $\Theta_n(\theta)=A_n\cos n\theta + B_n\sin n\theta$, which reduces to $A_n$ for $n=0$. The corresponding solutions in $r$ are $C_n r^n+Dr^{-n}$ for $n\ne 0$ and $C_0 + D_0 \ln(r)$ for $n=0$. The solution that is non-singular at $r=0$ has the series form $$ f(r,\theta) = E_0+\sum_{n=1}^{\infty}E_nr^n\cos n\theta+F_nr^{n}\sin n\theta. $$ The condition $f(1,\theta)=g(\theta)$ determines the coefficients $E_n,F_n$ through the Fourier series $$ g(\theta)=E_0+\sum_{n=1}^{\infty}E_n\cos n\theta+F_n\sin n\theta. $$