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?