TITLE: Separation of variables for PDE: dividing by zero?
QUESTION [5 upvotes]: This feels like a question that is both simple and duplicate but I can't find an answer or a previous version of the question. 
Suppose we are given some PDE, for example Laplace's Equation in polar coordinates,
$$ \nabla^2 u = \frac{\partial^2u}{\partial r^2} + \frac{1}{r} \frac{\partial u}{\partial r} + \frac{1}{r^2} \frac{\partial^2u}{\partial \theta^2} = 0.  $$
The usual way to solve this kind of equation is to posit a separated solution $$u(r,\theta) = R(r)\Theta(\theta).$$
We then get something like
$$(R'' + \frac{1}{r} R') \Theta = -\frac{R}{r}\Theta''.$$
At this point we usually divide by $$\frac{\Theta R}{r}$$ to find two ODEs.
Now, I can see that $\Theta R$ can't be the zero function (well, maybe it can be, but we don't like them zero solutions). But what if $\Theta R$ is zero at some specific point? Are we looking for solutions that are everywhere non-zero?
For example, the exercise that I am working on right now is the above problem on the half-sphere $$ H = \{ (x,y)\ | \ x^2 +y^2 \leq a, y \geq 0 \},$$
with boundary conditions $$u(a,\theta) = \sin \theta,$$
$$ u(r,0) = u(r,\pi) = 0.$$
To summarize: I know how to solve the problem, I just don't know how to justify the dividing by $\Theta R$. I can think of one argument: we can try if the answer we get is a solution by substituting in the original equation later. But then it is not clear in general if we have lost any solutions? 
Thanks for any help in advance.

REPLY [5 votes]: At the point "We get something like $\ldots$" we need the following Lemma, written out in terms of variables $x$ and $y$ instead of $r$ and $\theta$:
Lemma. Let $I$ and $J$ be  intervals on the $x$-, resp., the $y$-axis. Assume that
$$a(x)b(y)=c(x)d(y)\qquad\forall x\in I,\quad\forall y\in J\tag{1}$$
with both sides $\ne0$ for at least one point $(x_0,y_0)$. Then there is a constant $\mu\ne0$ with
$$c(x)=\mu a(x)\quad(x\in I),\qquad b(y)=\mu d(y)\quad(y\in J)\ .$$
Proof. Put $x:=x_0$ in $(1)$ and obtain
$$b(y)={c(x_0)d(y)\over a(x_0)}\quad(y\in J)$$
and therefore $b(y)=\mu d(y)$ $(y\in J)$ with $\mu={\displaystyle{c(x_0)\over a(x_0)}}\ne0$.  Now put $y:=y_0$ in $(1)$ and obtain
$$a(x)\>\mu d(y_0)=c(x) d(y_0)\quad(x\in I)\ .$$
This implies $c(x)=\mu a(x)$ $(x\in I)$, as claimed.