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

Let be a solution of ,

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

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

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 $\displaystyle \alpha =ax-x_{0}$, and $\displaystyle \beta=a^{2}t-t_{0}$

$\displaystyle v(x,t) = u(\alpha, \beta)$

$\displaystyle v_{x} = \frac{\delta u}{\delta \alpha}\frac{\delta \alpha}{\delta x} + \frac{\delta u}{\delta \beta}\frac{\delta \beta}{\delta x} = au_{\alpha}$

$\displaystyle v_{xx} = a^{2}\frac{\delta^{2}u}{\delta \alpha^{2}} = a^{2}u_{\alpha \alpha}$

$\displaystyle v_{t}=\frac{\delta u}{\delta \alpha}\frac{\delta \alpha}{\delta t} + \frac{\delta u}{\delta \beta}\frac{\delta \beta}{\delta t}= a^{2}u_{\beta}$

But for $\displaystyle v_{t}=kv_{xx}$, I end up with $\displaystyle a^{2}u_{\beta}=ka^{2}u_{\alpha \alpha}$

If the $\displaystyle u_{\beta}$ and $\displaystyle u_{\alpha \alpha}$ 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?