3
$\begingroup$

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 $n \geq 2$.

$\endgroup$
3
  • $\begingroup$ You need to make some assumptions about $\phi$ for that evolution equation to be well posed. The first constraint is $\phi>0$. Next you a need to make some assumptions about the behavior of $\phi$ at $\infty$ so the solutions of the evolution equation do not blow up before $t=1$. $\endgroup$ Sep 15, 2022 at 11:42
  • $\begingroup$ Can you please elaborate on $\phi(x) > 0$ constraint? In case of $n=1$, e.g., $e^{-2x\frac{d}{dx}}f(x)=f(e^{-2}x)$. Is $\phi(x) > 0$ constraint necessary for $n \geq 2$? $\endgroup$
    – Mirar
    Sep 15, 2022 at 12:25
  • $\begingroup$ Have you tried $e^{-2xD^2_x} f$? The condition $\phi>0$ guarantees the equation $ D_t u =\phi(x) D^2_xu$, $t>0$, is parabolic and at leas has some local existence. If $\phi<0$ the equation may not even have unique solution. $\endgroup$ Sep 16, 2022 at 12:41

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.