So I have a Hermite-Gaussian profile given by the initial value problem
$$
iu_z + u_{xx}=0,\quad (x,z)\in\mathbb{R}\times\mathbb{R}_+
$$
where the condition initially is given by this
$$
u_0(x) = H_n\left(\sqrt{2b}x\right)e^{-bx^2}
$$
where $H_n(x)$ is the Hermite polynomial of nth order
The problem is to find $u(x,z)$ and compute various moments of $x$ to determine the behavior of the profile with propagation distance.
I know that the Fourier Transform of this Hermite polynomial is given by thanks to this answer
$$
\mathscr{F}(H_n(\sqrt{2b}x)e^{-bx^2}) =(-i)^n\sqrt{\frac{2\pi}{2b}}H_n\left(\frac{\xi}{\sqrt{2b}}\right)e^{-\xi^2/4b}
$$
I took the Fourier Transform of the initial equation and got a new equation with two variables $\xi_1, \xi_2$ , namely
$$
\mathscr{F}(iu_z + u_{xx}) = [i(i\xi_1) + i^2\xi^2_2]\widehat u = -(\xi_1+\xi_2^2)\widehat u = 0
$$
I can't for the life of me understand how to extend this initial condition to two variables (x,z) to find the inverse Fourier Transform.
Any help is appreciated.
Best Answer
(Partial answer)
You only need to take the Fourier transform with respect to $x$, such that your differential equation becomes $i\hat{u}_z-\xi^2\hat{u}=0$, hence $\hat{u}(z,\xi) = A(\xi)e^{-iz\xi^2}$. The initial condition gives then $$ A(\xi) = \hat{u}(0,\xi) = \hat{u}_0(\xi) = (-i)^n\sqrt{\frac{\pi}{b}}H_n\left(\frac{\xi}{\sqrt{2b}}\right)e^{-\xi^2/4b}. $$ I let you taking the inverse Fourier transform $-$ N.B. : the fact that Hermite functions are eigenfunctions of the Fourier transform will help (see here).
Addendum. I don't know in which context your differential equation arises (it looks like Schrödinger equation for a free particle), but it is to be noticed that it can take the form of the standard 1D heat equation, i.e. $u_t = u_{xx}$, after a Wick rotation $z \mapsto -it$.