The problem that I am trying to solve is similar to this exerciese in Evans' book.
We are given a solution $u\in C^\infty(\mathbb{R}^n\times(0,\infty))$ of the equation
\begin{equation} \partial_tu-\Delta u=0\quad\forall (x,t)\in\mathbb{R}^n\times(0,\infty) \end{equation}
that satisfies $u(x,t)=u(ax,a^2t)$ for all positive $a$.
What I now want to show is that there exists a function $v$ on $\mathbb{R}^n$ such that $u(x,t)=v\left(\frac{x}{\sqrt{t}}\right)$.
I also want to find a new differential equation for the function $v$.
For the second part I have tried the following:
I first calculate $(\partial_t-\Delta)v\left(\frac{x}{\sqrt{t}}\right)$.
We have $\partial_tv\left(\frac{x}{t}\right)=-\frac{1}{2}t^{-\frac{3}{2}}x\cdot(\nabla v)\left(\frac{x}{\sqrt{t}}\right)$
and $\partial_{x_i}v\left(\frac{x}{t}\right)=\frac{1}{\sqrt{t}}(\partial_{x_i}v)\left(\frac{x}{\sqrt{t}}\right)$
and $\partial^2_{x_i}v\left(\frac{x}{t}\right)=\frac{1}{t}(\partial_{x_i}v)\left(\frac{x}{\sqrt{t}}\right)$.
Thus
\begin{equation} (\partial_t-\Delta)v\left(\frac{x}{\sqrt{t}}\right)=-\frac{1}{2}t^{-\frac{3}{2}}x\cdot(\nabla v)\left(\frac{x}{\sqrt{t}}\right)-\frac{1}{t}(\Delta v)\left(\frac{x}{\sqrt{t}}\right). \end{equation}
Now since $u$ is a solution to the equation above we get:
\begin{equation} \frac{1}{2\sqrt{t}}x\cdot(\nabla v)\left(\frac{x}{\sqrt{t}}\right)+(\Delta v)\left(\frac{x}{\sqrt{t}}\right)=0 \end{equation}
So my guess would be that the PDE for $v$ is $\frac{1}{2\sqrt{t}}x\cdot\nabla v+\Delta v=0$ but a solution to this would not have the form $v\left(\frac{x}{\sqrt{t}}\right)$ just $v(x)$ right? But I think there should be an easy fix for that that I just do not see.
Regarding the first part I am a little more stuck.
I have tried to use the fact that $u(ax,a^2t)$ is constant in $a$ so the partial derivative $\partial_a u(ax,a^2t)$ has to vanish.
So with this I get:
\begin{align} \partial_a u(a x, a^2 t )&=a\sum_{k=1}^n (\partial_{x_k}u)(ax,a^2t)+a^2(\partial_t u)(ax,a^2t)\\ &=0 \end{align}
Now I don't really see how I can handle this for arbitrary $a>0$ but if I take $a=1$ I get the equation
\begin{equation} \sum_{k=1}^n \partial_{x_k} u+\partial_t u=0. \end{equation}
Evans shows that a solution to the PDE
\begin{equation} \begin{cases} u_t+b\cdot Du=f & \text{ in } \mathbb{R}^n\times(0,\infty)\\ u=g & \text{ on } \mathbb{R}^n\times\{t=0\} \end{cases} \end{equation}
is given by \begin{equation} u(x,t)=g(x-tb)+\int_0^t f(x+(s-t)b,s) \mathrm{d}s \end{equation}
Since in my case $f=0$ and $b=(1,1,...,1)^\top=:I$ I get that $u(x,t)=g(x-tI)$ which seems to go in the right direction but this is firstly not of the form $g(\frac{x}{\sqrt{t}})$ and secondly I do not even have a boundary condition.
I also haven't really worked with the transport equation before so I think there should be a different method that I am not seeing so I would appreciate any help.
Since $u(x,t)=u(ax,a^2t)$ for every $a$, you can just pick $a=1/\sqrt{t}$ thus concluding that $$ u(x,t)=u(x/\sqrt{t},1)=v(x/\sqrt{t}). $$