Let $\Omega=B_{\bar{r}}(0)$, where $\bar{r}>0$. We study the equation $\begin{align} \Delta u(x,y) &= 0, & \text{on } \Omega \\ u(x,y) &= f(x), & \text{on }\partial \Omega \end{align}$where $\Delta =\partial_{xx}+\partial_{yy}$. Motivated by the [[Radial Symmetry of Laplace Operator|radial symmetry of Laplace operator]], we rewrite the equation using polar coordinates. Let $w(r, \theta) =u(x,y)$, where $x=r\cos\theta$ and $y=r\sin\theta$. Then, as shown in [[2D Laplace operator in polar coordinates]], the problem now reads $\begin{align} w_{rr} + \frac{1}{r} w_{r} + \frac{1}{r^{2}} w_{\theta\theta} &= 0, & (r, \theta)\in [0,\bar{r}]\times[0,2\pi] \\ w(\bar{r},\theta) &= g(\bar{r}, \theta), & \theta \in [0,2\pi] \end{align}$ Suppose $w(r, \theta)=R(r)\Theta(\theta)$. This yields the following ODEs $\begin{align} \\ r^{2}R''(r) + rR'(r) - \lambda R(r) &=0, \\ \Theta'' (\theta)- \lambda \Theta(\theta) &= 0, & \Theta(0)=\Theta(2\pi), \Theta'(0) = \Theta'(2\pi) \end{align}$for some constant $\lambda\in \mathbb{R}$ (since this quantity neither depends on $r$ nor $\theta$). The conditions $\Theta(0)=\Theta(2\pi), \Theta'(0) = \Theta'(2\pi)$ ensure that the solution is $C^2$. We can't have $\lambda \leq 0$ \[why?\]. We obtain the following solutions $\Theta_{n}(\theta)$ Equation $(1)$ is an example of a [[Cauchy–Euler ordinary differential equation|Cauchy-Euler equation]] which as the general solution