I am puzzled by the generation of the two ODEs for a Laplace equation in polar coordinates. I have checked around on SE, and found some different and thus confusing answers.
I list my attempt,
The IVBP is:
\begin{equation} \Delta u+u=0\ \ \ \ \ 0<r<1,\ 0<\theta<\pi/4 \\ u(r,0)=0,\ u/r,\pi/4)=0 \ \ \ \ on\ \ 0<r<1\\ u(1,\theta)=1 \ \ \ on \ \ 0<\theta<\pi/4 \end{equation}
Then I set up the operator
\begin{equation} \Delta u =\frac{1}{r}\frac{\partial}{\partial r}r\frac{\partial}{\partial r}+\frac{1}{r^2}\frac{\partial^2}{\partial \theta^2} \end{equation}
$u=R(r)\Theta(\theta)$ gives the PDE
\begin{equation} \frac{1}{r}\frac{\partial}{\partial r}r\frac{\partial}{\partial r}R\Theta+\frac{1}{r^2}\frac{\partial^2}{\partial \theta^2}R\Theta + R\Theta=0 \end{equation}
this gives
\begin{equation} R_{rr}\Theta+\frac{1}{r}R_r\Theta+\frac{1}{r^2}\Theta_{\theta\theta}R+R\Theta=0 \end{equation}
which yields, by division of $R\Theta$:
\begin{equation} \frac{R_{rr}}{R}+\frac{1}{r}\frac{R_r}{R}+1=-\frac{1}{r^2}\frac{\Theta_{\theta\theta}}{\Theta} \end{equation}
Then multiply by $r^2$ and get:
\begin{equation} r^2\frac{R_{rr}}{R}+r\frac{R_r}{R}+r^2=-\frac{\Theta_{\theta\theta}}{\Theta} \end{equation}
These two must equal some common constant, since this is an orthogonal system:
\begin{equation} r^2\frac{R_{rr}}{R}+r\frac{R_r}{R}+r^2=-\frac{\Theta_{\theta\theta}}{\Theta}=\pm k^2 \end{equation}
Thence obtaining two ODEs:
\begin{equation} r^2\frac{R_{rr}}{R}+r\frac{R_r}{R}+r^2=\pm k^2 \\ \frac{\Theta_{\theta\theta}}{\Theta}=\pm k^2 \end{equation}
The second is a simple ODE, while the former is a Bessel equation. Am I right that this form is the correct separated variables form of the Laplace equation in polar coordinates? This is because the Bessel equation seems not easy to solve since it gives
\begin{equation} r^2R_{rr}+rR_r+R(r^2\pm k^2)=0 \end{equation}
Is this a correct outlining of the Laplace equation in polar coordinates for this very case of an IVBP?
Thanks