Assume $u: \mathbb{R^{2} \times \mathbb{R}_{\geq 0}} \to \mathbb{R}$ satisfy the following PDE
$$ \nabla u \cdot \langle 1, y, -x \rangle =u$$
where $\nabla u = \langle \partial_{t} u, \partial_{x} u, \partial_{y}u \rangle$ is the "time-space" gradient and $u(0, x, y)= xy.$ This seems like a simple problem, but I spent a couple of hours to solve this PDE. Still, I could not solve it.
I was able to find a curve that has the tangent vector (field) $\langle 1, y, -x \rangle$ in $\mathbb{R^3},$ then using the initial condition, finally, I found the following solution:
$$ u(t, x, y)= xye^{t},$$ but sadly this not satisfy my PDE. Maybe I made some mistakes. Any suggestions/hints/help would be appreciated. Thanks so much.
$\textbf{Hint:}$ Switch to polar coordinates
$$\begin{cases}x = r\cos\theta \\ y = r\sin\theta \end{cases}$$
The differential equation becomes
$$\partial_t u - \partial_\theta u = u$$