Showing that a given PDE is a solution to harmonic motion via transform OK, here's an problem that should, by all accounts, be pretty simple, but I want to make sure I am approaching this correctly. 
Given: $$\frac{\partial ^2u}{\partial t^2}=c^2\frac{\partial ^2u}{\partial x^2}$$
use the change of variables $\alpha = x+ct$, $\beta = x-ct$ to transform the above equation into $$\frac{\partial ^2u}{\partial \alpha \partial \beta}=0$$
This should be straightforward enough. So, using the chain rule: 
$$ \frac{\partial u}{\partial t}= \frac{\partial u}{\partial \alpha}\frac{\partial \alpha}{\partial t}+ \frac{\partial u}{\partial \beta}\frac{\partial \beta}{\partial t} \text{and } \frac{\partial u}{\partial x}= \frac{\partial u}{\partial \alpha}\frac{\partial \alpha}{\partial x}+ \frac{\partial u}{\partial \beta}\frac{\partial \beta}{\partial x}$$
we take the partial derivatives with respect to t and x in the alpha and beta expressions: 
$$\frac{\partial \alpha}{\partial t}=c, \frac{\partial \alpha}{\partial x} = 1,\frac{\partial \beta}{\partial t}=-c,\frac{\partial \beta}{\partial x}=1$$ 
Plug these back into the above PDEs: 
$$ \frac{\partial u}{\partial t}=c \frac{\partial u}{\partial \alpha}-c \frac{\partial u}{\partial \beta} \text{and } \frac{\partial u}{\partial x}= \frac{\partial u}{\partial \alpha}+ \frac{\partial u}{\partial \beta}$$
taking another derivative with respect to beta: 
$$ \frac{\partial^2 u}{\partial t \partial \beta}=c \frac{\partial^2 u}{\partial \alpha \partial  \beta}-c \frac{\partial^2 u}{\partial \beta^2} \text{and } \frac{\partial^2 u}{\partial x \partial \beta}= \frac{\partial^2 u}{\partial \alpha \partial \beta}+ \frac{\partial^2 u}{\partial \beta^2}$$
I saw there are two terms that are $\frac{\partial^2 u}{\partial \alpha \partial \beta}$ and rearranging things a bit, and noting they are equal: 
$$ c \frac{\partial^2 u}{\partial \alpha \partial  \beta}=c \frac{\partial^2 u}{\partial \beta^2}+\frac{\partial^2 u}{\partial t \partial \beta} \text{and } \frac{\partial^2 u}{\partial \alpha \partial \beta}=\frac{\partial^2 u}{\partial \beta^2}-\frac{\partial^2 u}{\partial x \partial \beta}  $$
The problem I am having is that I feel I messed up a substitution someplace. Because I end up with $$-c\frac{\partial^2 u}{\partial x \partial \beta}=\frac{\partial^2 u}{\partial t \partial \beta}$$
and the terms aren't going to zero. Or there's some stupidly simple step I missed. I guess what I am asking is if I am right so far. 
 A: I prefer to take $α=x+ct, β=x−ct$ and solve $x=\dfrac12(\alpha+\beta),t=\dfrac1{2c}(\alpha-\beta)$ hence $$\frac{\partial x}{\partial\alpha}=\frac12,\frac{\partial x}{\partial \beta}=\frac12,\frac{\partial t}{\partial \alpha}=\frac1{2c},\frac{\partial t}{\partial\beta}=-\frac1{2c}$$hence we have:$$\frac{\partial u}{\partial \alpha}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial\alpha}+\frac{\partial u}{\partial t}\frac{\partial t}{\partial\alpha}=\frac1{2c}\left(c\frac{\partial u}{\partial x}+\frac{\partial u}{\partial t}\right)\\\frac{\partial^2u}{\partial\alpha\partial\beta}=\frac{\partial^2u}{\partial\beta\partial\alpha}=\frac1{2c}\left(c\frac{\partial^2u}{\partial x^2}\frac{\partial x}{\partial\beta}+\frac{\partial^2u}{\partial t^2}\frac{\partial t}{\partial\beta}\right)=\frac1{4c^2}\left(c^2\frac{\partial^2u}{\partial x^2}-\frac{\partial ^2u}{\partial t^2}\right)$$Manipulating our equation a little we find:$$\frac{\partial ^2u}{\partial t^2}=c^2\frac{\partial ^2u}{\partial x^2}\\c^2\frac{\partial ^2u}{\partial x^2}-\frac{\partial ^2u}{\partial t^2}=0$$Hence $$\frac{\partial^2u}{\partial\alpha\partial\beta}=0$$
