$$\frac {\partial^2 u}{\partial t^2}=c^2 \space \nabla^2 u$$ $u(x,t)$ is the amplitude of the wave at position $x$ and time $t$. I want to show by dimension analysis that $c$ has to be of the dimension velocity.
My thoughts:
Wave amplitude $u$ is of dimension length, L.
Position $x$ is of dimension length, L.
Time $t$ is of dimension time, T.
$c^2\nabla ^2u=c^2\nabla \cdot \nabla u=c^2\nabla\cdot[\frac {\partial u}{\partial x},\frac {\partial u}{\partial t}]=c^2(\frac {\partial^2 u}{\partial x^2}+\frac {\partial^2 u}{\partial t^2}) $
Substituting the variables with their corresponding unit yields $$L/T^2=[c]^2((L/L^2)+(L/T^2))=[c]^2((1/L)+(L/T^2))$$ $[c]$ means dimension of $c$. Given that I know that $c$ is of dimension velocity, L/T. $$(L^2/T^2)((1/L)+(L/T^2))=L/T^2+L^3/T^4$$ Which makes no sense.
$\nabla^2$ is $\frac{\partial^2}{\partial x^2}$, with no partial w/r to time. Thus: \begin{align} \frac{1}{c^2}\frac{\partial^2 u}{\partial t^2} &=\frac{\partial^2 u}{\partial x^2}\, ,\\ &= \frac{\partial^2 u}{\partial (ct)^2} \end{align} so that $ct$ must have the same dimensions as $x$. The correct dimensionality of $c$ follows immediately.