Nontrivial invariant transformations for heat equations

Question

It is well known that if $u$ is a harmonic function on $\mathbb R^2$ then its Kelvin transform defined by $$ v(r,\theta) = u(\frac{1}{r},\theta)$$ is also harmonic for $r>0$. Note that the Kelvin transform is defined in all dimensions but I stated it in dimension two for simplicity.

My question is whether something similar can be done for the heat equation $$ \partial_t u - \Delta u =0,$$ that is to say, find a transformation $\psi: \mathbb R^{1+n} \to \mathbb R^{1+n}$ so that if $u$ solves heat equation then so does $u\circ \psi$. Of course, I don't want the trivial transformations like the parabolic scaling $\psi(t,x) = (\alpha^2 t, \alpha x)$.

Answer 1

At least when $n=1$, there are no nontrivial transformations of this kind.

Indeed, suppose that $v(t,x)=u(\tau(t,x),\xi(t,x))$, where $u_t=u_{xx}$. Then $u_{tx}=u_{xxx}$ and $u_{tt}=u_{xxxx}$, so that the equation \begin{equation*} v_t=v_{xx} \tag{1}\label{1} \end{equation*} can be rewritten as \begin{equation*} \xi_x(u_{xxx}\tau_x+u_{xx}\xi_x) +\tau_x\,(u_{xxxx}\tau_x+u_{xxx}\xi_x) \\ +u_{xx}\tau_{xx}+u_x\xi_{xx}-u_{xx}\tau_t-u_x\xi_t=0. \tag{2}\label{2} \end{equation*} Since at any point $(t,x)$ the partial derivatives $u_x,u_{xx},u_{xx},u_{xxx},u_{xxxx}$ can take arbitrary real values, the coefficients of $u_x,u_{xx},u_{xx},u_{xxx},u_{xxxx}$ in \eqref{2} must all be zero. In particular, the coefficient $\tau_x^2$ of $u_{xxxx}$ in \eqref{2} must be $0$, so that $\tau_x=\tau_{xx}=0$, and we can write $\tau(t,x)=T(t)$ for some function $T$.

So, \eqref{2} implies \begin{equation*} u_{xx}(\xi_x^2-T'(t))+u_x(\xi_{xx}-\xi_t)=0, \end{equation*} so that \begin{equation*} \xi_x^2=T'(t),\quad\xi_{xx}=\xi_t. \end{equation*} Hence, $\xi_x\xi_{xx}=0$ and, if $\xi$ is real analytic, then \begin{equation*} \xi_{xx}=0, \quad\text{whence}\quad\xi_t=0, \end{equation*} whence for some real $a,b,c$ we have $\xi(t,x)=a+bx$, $T'(t)=\xi_x^2=b^2$, $\tau(t,x)=T(t)=c+b^2t$. Thus, \begin{equation*} \xi(t,x)=a+bx,\quad\tau(t,x)=c+b^2t. \end{equation*}

The case $n\ge2$ should be similar.

Answer 2

Yes, how about the Appell transform, see here.

$$v(t,x)\mapsto\Gamma(t,x)v(-\frac1t,\frac xt)$$ where $\Gamma$ is the heat kernel.

Truly nontrivial if you ask me. Of course, this inverts time and not space!