As stated in this article, the Weierstrass transform of $f(x)$ is defined as:
\begin{equation} W[f](x)=\frac{1}{4\pi}\int_{-\infty}^{\infty}f(y)e^{-\frac{(x-y)^{2}}{4}}dy \end{equation}
which can be re-written as $W[f](x)=e^{D^2}f(x)$ where $D=\frac{d}{dx}$.
Question: Can one generalize the above definition to define $e^{\phi(x)D^2} f(x)$ in a similar integral form?
My understanding is that $e^{\phi(x)D^2} f(x)=u(x,1)$ is the solution of $\frac{\partial u(x,t)}{\partial t}=\phi(x)\frac{\partial ^2}{\partial x^2}u(x,t)$ at $t=1$ with $u(x,0)=f(x)$. Assuming that $\phi(x)$ is a sufficiently smooth function, what boundary conditions have to be imposed in order to find a solution in the form of the above integral for this PDE? and how one can find it?
Motivation: It is well-known that $e^{aD}f(x)=f(x+a)$ and $e^{\phi(x)D}f(x)=f(e^{\phi(x)D}x)$. I also know that, similar to the Weierstrass transform, $e^{aD^n}f(x)$ can be expressed as an integral transform (see https://math.stackexchange.com/a/2108889/1027701). I however was wondering about $e^{\phi(x)D^n}f(x)$ for $n \geq 2$.