Show that if $u(x,t)$ is a solution of the heat equation \begin{equation*} u_t - k u_{xx} = 0 \,,\qquad (x\in R,\ t\geq0)\,, \end{equation*} then so is \begin{equation*} v(x,t) = \frac{1}{\sqrt{kt}} \exp\left( \frac{ -x^2}{4kt} \right) \ u( \frac{x}{kt}, - \frac{1}{k^2t}) \,. \end{equation*}
I am using Walter A. Strauss as text, and I try to follow "invariance properties", and I am trying to find a "dilated" function, but got confused on the negative sign before "$\frac{1}{k^2t}$" , I also referring to http://www.math.ucsb.edu/~grigoryan/124A/lecs/lec9.pdf, but it seems not same, so I am not sure where to start, can someone help me with a hint, an example or a better resources?
Thanks in advance!
You are supposed to derive $v$ with the purpose of verifying it is a solution.
You don't need to find any functions; you are given all the functions you need.
To get you started: the product and chain rule yield $$ \begin{split} v_x(x,t) &= \frac{1}{\sqrt{kt}} \frac{-2x}{4kt} \exp\left( \frac{ -x^2}{4kt} \right) \ u\left( \frac{x}{kt}, - \frac{1}{k^2t}\right) \\ &+ \frac{1}{\sqrt{kt}} \exp\left( \frac{ -x^2}{4kt} \right) \frac{1}{kt} \ u_x\left( \frac{x}{kt}, - \frac{1}{k^2t}\right) \end{split}$$
Next, take $x$ derivative again to find $v_{xx}$. (It will be the sum of five terms, some of which will be equal.) Finally, find $v_t$ and check that $v_t=kv_{xx}$.