Application of an exponential whose power is a second derivative

Question

Specifically I need to apply the exponential operator: $$\exp\left[{\alpha\dfrac{\partial^2}{\partial x^2}}\right]$$ (where $\alpha$ is a purely complex number, so $\alpha = b i$, with $b$ real) to a Gaussian function of the form $$\exp\left[ \frac{-(2\sigma x + C)^2}{4\sigma} \right]$$ where $\sigma$ and C are a complex numbers of the form $\beta= C_{1} + C_{2} i$, with $C_{1}$ and $C_{2}$ reals.

Any idea to correctly apply this operator?

Answer 1

Hint: express the operator as an infinite sum of derivatives, then work in Fourier space.

Answer 2

Method 1. Under the unitary Fourier transform $ \mathcal{F}[f](\xi) = \int_{\mathbb{R}} f(x)e^{-2\pi i \xi x} \, \mathrm{d}x $ defined on the Schwarz space $\mathcal{S}(\mathbb{R})$,

$$ \mathcal{F}\big[e^{\alpha \partial_x^2} f(x)\big](\xi) = \sum_{n=0}^{\infty} \frac{\alpha^n}{n!} \mathcal{F}[f^{(2n)}](\xi) = \sum_{n=0}^{\infty} \frac{\alpha^n}{n!} (2\pi i \xi)^{2n} \mathcal{F}[f](\xi) = e^{-4\pi^2 \alpha \xi^2} \mathcal{F}[f](\xi). $$

Now recall that, if $\operatorname{Re}(\sigma) > 0$ and $f(x) = \exp\left\{ -\frac{(2\sigma x+C)^2}{4\sigma} \right\}$, then

$$ \mathcal{F}\left[ f \right](\xi) = \sqrt{\frac{\pi}{\sigma}} \exp\left\{ \frac{-\pi^2\xi^2 + iC\pi\xi}{\sigma }\right\}. $$

Plugging this back,

$$ \mathcal{F}\big[e^{\alpha \partial_x^2} f(x)\big](\xi) = \sqrt{\frac{\pi}{\sigma}} \exp\left\{\frac{ -(1+4\alpha\sigma)\pi^2\xi^2 + iC\pi\xi}{\sigma }\right\}. $$

Taking inverse Fourier transform,

$$ e^{\alpha \partial_x^2} f(x) = \frac{1}{\sqrt{4\alpha\sigma+1}} \exp\left\{ -\frac{(2\sigma x+C)^2}{4\sigma(1+4\alpha \sigma)} \right\}. $$

---

Method 2. Assume for a moment that $\alpha$ is real. Since $\partial_x^2$ is the infinitesimal generator of the heat equation, we know that $u(t, x) = e^{t\partial_x^2} f(x)$ solves the equation

$$ \partial_t u = \partial_x^2 u \qquad \text{and} \qquad u(0, x) = f(x) $$

So the solution can be written as the convolution with the fundamental solution of the heat equation:

$$ u(t, x) = \int_{\mathbb{R}} f(y) \cdot \frac{1}{\sqrt{4\pi t}} e^{-(x-y)^2/4t} \, \mathrm{d}y. $$

Plugging $f(x) = \exp\Big\{ -\frac{(2\sigma x+C)^2}{4\sigma} \Big\}$ and computing the resulting gaussian integral, we obtain the same as before. Finally, we can check that the map $\alpha \mapsto e^{\alpha \partial_x^2}f(x)$ is analytic, and so, the same answer applies to complex $\alpha$ by the principle of analytic continuation.