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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?
Jens
  • 95

1 Answers1

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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 $t=1$.

Jacky Chong
  • 25,739