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If you consider the homogenous wave-equation: $$ -\Delta_x u(x,t) + \frac{\partial^2 u}{\partial t^2} (x,t) \ = \ 0 \ \ \mathrm{in} \ \Omega \times (0, \infty), $$ with some Neumann boundary conditions: $$ \frac{\partial u}{\partial \overrightarrow{n}} (x,t) \ = \ 0 \ \ \mathrm{on} \ \partial \Omega \times (0, \infty). $$ And now, by separation of the variables, you make the ansatz: $$ u(x,t) = v(x) w(t), $$ to get the standing wave solutions of the system. Inserting this special type of solution into the wave-equation, leads to the equation: $$ w(t)(-\Delta v(x)) + v(x) w''(t) \ = \ 0 \ \ \mathrm{in} \ \Omega \times (0, \infty). $$

And now, all literatur i've found so far is saying (just by assuming $u \neq 0$) the above equation is equivalent to: $$ \frac{w''(t)}{w(t)} \ = \ \frac{\Delta v(x)}{v(x)} \ \equiv \ \underbrace{\mathrm{const}}_{:= \ -\lambda}. $$ And therefore equivalent into solving ,,$w''(t) + \lambda w(t) = 0$'' and ,,$-\Delta v(x) = \lambda v(x)$'', respectively.

But what if there are some $x^* \in \Omega$ or $t^* \in (0, \infty)$ with $w(t^*) = 0$ or $v(x^*) = 0$? Then the above equation with the fractions wouldn't make sense anymore. I don't see how $u \neq 0$ is excluding these cases. Am i overseeing something super trivial here or why does noone even mention that case? Thanks in advanve!

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The quick answer is simply that it gives an answer that works.

It's better to view the separation of variables as a way to make an educated guess at solutions (to check later on), rather than a rigorous way to find a solution, and it just so happens that having the additional (unreasonable) assumption that $v$ and $w$ are nonzero everywhere allows you to get something that's valid even when dropping this requirement. Indeed, if $w$ and $v$ do solve them ODEs, you get $$\begin{align}\frac{\partial^2 u}{\partial^2 t}&=v(x)w''(t) = -\lambda v(x) w(t) \text{ and} \\ \Delta_xu(x,t)& = \Delta v(x) w(t) = -\lambda v(x) w(t), \end{align}$$ then subtracting the two equations verifies that it works.

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