## PDE Heat Equation

I'm having trouble with differentiating this properly...

Let be a solution of ,

Show that for constants a, $x_{0},t_{0}$, the function

$v(x,t)=u(ax-x_{0}, a^{2}t-t_{0})$, satisfies $v_{t}=kv_{xx}$.

I know I have to use chain rule, but I'm having trouble with the notation.
This is what I thought I should do:
Let $\alpha =ax-x_{0}$, and $\beta=a^{2}t-t_{0}$
$v(x,t) = u(\alpha, \beta)$
$v_{x} = \frac{\delta u}{\delta \alpha}\frac{\delta \alpha}{\delta x} + \frac{\delta u}{\delta \beta}\frac{\delta \beta}{\delta x} = au_{\alpha}$
$v_{xx} = a^{2}\frac{\delta^{2}u}{\delta \alpha^{2}} = a^{2}u_{\alpha \alpha}$
$v_{t}=\frac{\delta u}{\delta \alpha}\frac{\delta \alpha}{\delta t} + \frac{\delta u}{\delta \beta}\frac{\delta \beta}{\delta t}= a^{2}u_{\beta}$

But for $v_{t}=kv_{xx}$, I end up with $a^{2}u_{\beta}=ka^{2}u_{\alpha \alpha}$
If the $u_{\beta}$ and $u_{\alpha \alpha}$ weren't there, I could say it works for k = 1, but I don't think I can just make those notation just "disappear."

How do I write this out properly, so that the solution is verified?