Can we find a general formula for the ODEs that govern fundamental solutions of linear evolution equations? A time evolution equation is an equation of the form
$$\partial_tu=\mathscr{F}[u] \\u:(x,t)\mapsto u(x,t)\\(x,t)\in\mathbb {R}\times\mathbb{R}_{\geq 0}$$
Where $\mathscr F[u]$ is some differential operator acting on $u$ involving $u$ and its spacial derivatives only, e.g  the viscous Burgers' equation
$$\partial_t u=\nu\partial_x^2u-u\partial_xu$$
Where in this case $\mathscr F=\nu\partial_x^2-\operatorname{id}\cdot~\partial_x$, where the centered dot means pointwise multiplication.
I'll consider the case where the operator is a polynomial in the spacial derivatives and we couple the equation with some physical initial and boundary conditions:
$$\partial_tu=P(\partial_x)[u]\\u(x,0)=f(x)\\ \int_{\mathbb{R}}u(x,t)^2\mathrm dx~~~\text{converges} ~\forall t\tag{1}$$
The fundamental solution $\Phi$ is one that satisfies $\boldsymbol{(1)}$ with $f(x)=\delta(x)$.
In the case of the heat equation, $P:z\mapsto z^2$ one of course has the classical result
$$\Phi(x,t)=\frac{1}{\sqrt{4\pi t}}\exp\left(\frac{-x^2}{4t}\right)$$
Which satisfies the first order equation
$$\partial_x\Phi(x,t)+\frac{x}{2t}\Phi(x,t)=0$$
Similarly, a result I've recently become familiar with is that in the case $P:z\mapsto z^3$ One has
$$\Phi(x,t)=\frac{1}{(3t)^{1/3}}\operatorname{Ai}\left(\frac{-x}{(3t)^{1/3}}\right)$$
($\operatorname{Ai}$ being the Airy function of the first kind) Which, curiously, satisfies the second order equation
$$\partial_x^2\Phi(x,t)+\frac{x}{3t}\Phi(x,t)=0$$
Which looks quite similar to the form found for the heat equation.
QUESTION: Can anyone find a general pattern in these differential equations to describe the fundamental solutions for general polynomial operators, e.g
$$P(\partial_x)=\sum_{k=0}^n a_k\partial_x^k$$
I conjecture that for $P:z\mapsto z^n$ with $n\geq 2$ that
$$\partial_x^{n-1}\Phi(x,t)+\frac{x}{nt}\Phi(x,t)=0$$
But have not been able to produce a proof yet. Once this formula is found, can we linearly combine different powers to produce a formula for the general polynomial case?
Thanks in advance for any input.
 A: A partial answer: Your conjecture for $P(z)=z^n$ seems to be correct.
(But you won't be able to take linear combinations to get the solution for general $P$; after all, the solution to $y''+y'=0$ is not the sum of the solutions to $y''=0$ and $y'=0$, to take a simple example.)
One way of finding $\Phi$ for the PDE $u_t=\partial_x^n u$ is to first solve it with initial data $u(x,0)=H(x)$ (the Heaviside function) and then let $\Phi = u_x$. Since the PDE and the initial data are invariant under the transformation $(x,t) \mapsto (cx,ct^n)$, it's natural to seek a solution (for $t>0$) of the form $u(x,t)=g(p)$, where $p=x t^{-1/n}$. Inserting this into the PDE gives an ODE for $g$:
$$
- \frac{p}{n} g'(p) = g^{(n)}(p)
.
$$
Here, to satisfy the initial condition for $u$, or more precisely that $u(x,t) = g(x t^{-1/n}) \to H(x)$ as $t \to 0^+$ for each fixed $x$, you want to pick out a solution $g(p)$ which tends to $0$ as $p \to -\infty$ and to $1$ as $p \to +\infty$. I haven't really thought about whether this is possible in general, but at least it works for $n=2$ and $n=3$, where the job is done by a particular antiderivative of the Gaussian and of the Airy function, respectively. Assuming that we can indeed find such a particular solution $g$ to the ODE above, the function
$$
\Phi(x,t) = u_x(x,t) = t^{-1/n} g'(x t^{-1/n})
$$
will be a fundamental solution to the PDE, and the ODE for $g$ implies your conjectured ODE for $\Phi$, as you can easily verify.
