1
$\begingroup$

I can prove that $\exp(a\partial_x)f(x) = f(x+a)$, but what happens for second derivatives? To be more precise, what is the right-hand side of $\exp(a^2\partial_x^2)f(x)$?

The above operator has an integral representation $$\exp(a^2\partial_x^2)f(x) = \int\limits_{-\infty}^\infty \text{d}y K(x-y)f(y) \, ,$$ and I think that it must be $K(x-y) \propto \frac{1}{\sqrt{a}}\exp(-|x-y|^2/a^2)$. The reason is that in the limit $a\rightarrow 0$ I want to obtain $K(x-y) = \delta(x-y)$ such that $f(x)$ gets mapped to itself.

My questions:

  1. How can I derive a formula for $K(x-y)$?
  2. Why is the expression $\exp(a\partial_x)f(x) = f(x+a)$ so different when I put a second derivative in the exponent instead of a first derivative?
$\endgroup$
4
$\begingroup$

Let us introduce the function \begin{align} u(t, x) = \exp\left(t\alpha^2\partial^2_x \right)f(x) \end{align} then we see that $u$ satisfies the Cauchy problem \begin{align} \partial_t u - \alpha^2\partial^2_x u =0, \ \ u(0, x) = f(x), \end{align} which is just the heat equation. By fundamental PDE, we see that \begin{align} u(t, x) = \frac{1}{\sqrt{4\pi \alpha^2t}} \int^\infty_{-\infty} \exp\left(-\frac{(x-y)^2}{4\alpha^2 t}\right) f(y)\ \text{d}y. \end{align} Finally, set $t=1$ yields \begin{align} \exp\left(\alpha^2\partial^2_x\right)f(x) = \frac{1}{\sqrt{4\pi \alpha^2}}\int^\infty_{-\infty} \exp\left(-\frac{(x-y)^2}{4\alpha^2 }\right) f(y)\ \text{d}y. \end{align}

Note that $u(t, x)= \exp(t\alpha \partial_x)f(x)$ satisfies a transport equation, i.e. \begin{align} \partial_t u-\alpha \partial_x u =0, \ \ u(0, x) = f(x). \end{align} Solving the pde yields $u(t, x) = f(x+\alpha t)$. Set $t=1$.

$\endgroup$
  • $\begingroup$ Thanks, that answers my question 1! Any ideas about question 2? $\endgroup$ – Jens Feb 9 at 5:09
  • 1
    $\begingroup$ I already did. See last sentence. $\endgroup$ – Jacky Chong Feb 9 at 5:09
  • $\begingroup$ What is this transport equation? I am not familiar with this term. $\endgroup$ – Jens Feb 9 at 5:10
  • 1
    $\begingroup$ Okay. I updated the post. $\endgroup$ – Jacky Chong Feb 9 at 5:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.