Solving Laplace's Equation on a Disk: Question About Logic of Assumptions When Solving Radial ODE. I am told that Laplace's equation on a disk of radius $\rho$ is $\dfrac{\partial^2{u}}{\partial{r}^2} + \dfrac{1}{r} \dfrac{\partial{u}}{\partial{r}} + \dfrac{1}{r^2} \dfrac{\partial^2{u}}{\partial{\theta}^2}$.
We assume a separation of variables solution: $u(r, \theta) = R(r)\Theta(\theta)$.
Therefore, the PDE gives us $R''(r)\Theta(\theta) + \dfrac{R'(r)\Theta(\theta)}{r} + \dfrac{R(r)\Theta''(\theta)}{r^2} = 0$
$\implies r^2 \dfrac{R''}{R} + r \dfrac{R'}{R} = -\dfrac{\Theta''}{\Theta} = \lambda$
This gives us the to ODEs
$r^2 R'' + r R' - \lambda R = 0$
and
$\Theta'' + \lambda \Theta = 0$
My confusion is when $\lambda = 0$.
If $\lambda = 0$, then we get that
$\Theta(\theta) = A + B \theta$.
We know that we require periodic solutions, so assume that we have $\Theta(0) = \Theta(2\pi)$.
$\therefore B = 0$ $\implies \Theta(\theta) = A$.
The corresponding radial ODE is $r^2 R'' + r R' = 0$.
Here is my confusion.
We solve this by assuming that $r \not= 0$ and dividing by $r$:
$r^2 R'' + r R' = 0$
$\implies r R'' + R' = 0$
$\implies [rR']' = 0$ ($\because$ By the product rule)
$\implies rR'(r) = D$
$\implies R(r) = C + D \ln(r)$ ($\because$ By using separation of variables)
But here I am told that, since we're dealing with a physical problem, we require the solution to be finite everywhere in the disk, so we need $D = 0$.
I understand that, if $r = 0$, then we get $\ln(0) = -\infty$. However, we JUST made the assumption that $r \not= 0$ to allow us to divide by $r$ and get to $R(r) = C + D \ln(r)$! Doesn't this make it so that we've already dealt with $r = 0$ by assuming that $r \not= 0$? So that begs the question, why do we need to still have $D = 0$? Why can't we just have $R(r) = C + D \ln(r) \forall r \not= 0$? From a logical standpoint, this part doesn't make any sense to me.
I would greatly appreciate it if people could please take the time to clarify this.
 A: By assuming that $r > 0$ you found an ODE for $R$ with solutions $C+D\ln r$, which is a perfectly valid solution of the Laplace equation in $0 < r < \infty$. It is the most general solution that depends on $r$ only. However, if you want to solve the Laplace equation in a disk $x^2 + y^2 < R_0$ then $D=0$ is required.
If you have a solution $\psi(x,y)$ of the Laplace equation in an open region $\Omega$ of the plane, and if $(x_0,y_0)\in\Omega$, then the following is another solution of the Laplace equation in $\Omega\setminus\{ (x_0,y_0)\}$ for any constant $E$:
$$
               \varphi(x,y) = \psi(x,y)+E\ln((x-x_0)^2+(y-y_0)^2).
$$
The reason for this is that the solution multiplying $E$ is a translate of the simple radial solution $\ln(r)$. You would not accept $\varphi$ as a solution in the open region $\Omega$, but it is an acceptable solution in $\Omega\setminus\{(x_0,y_0)\}$. The author has short-cut this discussion by saying that, based on Physical grounds, you exclude unbounded solutions, which is not the best way of saying it. Physically, the solution $\ln((x-x_0)^2+(y-y_0)^2)$ is the solution of Laplace's equation in the presence of a point charge at $(x_0,y_0)$. So the Physical principle at work is that you are solving the potential equation in a region where there are no charges in the interior.
