Using the Fourier transform, $\mathscr{F}\{\cdot\} = \hat{f}(\cdot)$, may I transform the $u_{xy}$ where $u=u(x,y,t)$, $x,y \in (-\infty, \infty)$ and $t \in (0,1]$? The initial condition is $u(x,y,t)=\delta(x) \delta(y)$ for $t=0$.
I tried the following:
$ \begin{align} \mathscr{F}\{u_{xy}\}=\hat{f}(\xi_1, \xi_2)& =\int_{\mathbb{R}}\int_{\mathbb{R}}\frac{\partial^{2}{u}} {\partial{x}\partial{y}} e^{-2 \pi i (\xi_1 x + \xi_2 y)} dx dy \tag{1} \\ & = \int_{\mathbb{R}} \frac{\partial}{\partial{y}} e^{-2 \pi i \xi_2 y} \left(\int_{\mathbb{R}} \frac{\partial{u}}{\partial{x}} e^{-2 \pi i \xi_1 x} dx\right) dy \tag{2} \\ & = 2 \pi i \xi_1 \hat{f}(\xi_1) \int_{\mathbb{R}} -2 \pi i \xi_2 e^{-2 \pi i \xi_2 y} dy \tag{3} \\ & = 4 \pi^{2} \xi_1 \hat{f}(\xi_1) \xi_2 \delta(\xi_2) \tag{4} \end{align} $
where in (2) I used Leibniz rule, in (3) differentiated the integrand w.r.t. $y$ and in (4) used the Fourier transform of $1$ referenced here.
If this is so, then I can take the inverse of the transform to get the original function.
Edit: The PDE is the following:
$ \begin{cases} u_{t} = u_{xx} + 2 \rho u_{xy} + u_{yy}, \ & \text{for} \ t \in (0,1],\\ u(x,y,t) = \delta(x) \delta(y), \ & \text{for} \ t = 0 \end{cases} $
where $\rho \in (-1, 1)$.
Thank you.
Let $\widehat u(\xi,\eta,t)$ be the Fourier transform (in the spatial variables) of $u(x,y,t)$. Each time you derivate $u$ with respect to $x$ you multiply $\widehat u$ by $-2\,\pi\,\xi\,i$; each time you derivate $u$ with respect to $y$ you multiply $\widehat u$ by $-2\,\pi\,\eta\,i$. Taking Fourier transforms in the equation we get then $$ \frac{d\widehat u}{dt}=-4\,\pi^2\bigl(\xi^2+2\,\rho\,\xi\,\eta+\eta^2\bigr)\widehat u. $$ This is first order linear ODE, which must be complemented with the initial condition $$ \widehat u(\xi,\eta,0)=\mathcal{F}(\delta(x)\,\delta(y))=1. $$