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.
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.