Consider $$\partial_{t}u -y\partial_{x}u +x\partial_{y}u = 0$$ in $t>0, (x,y) \in \mathbb{R}^{2}$, with initial condition $$ u(0,x,y) = u_{0}(x,y) $$ for $(x,y) \in \mathbb{R}^{2}$. Find the stationary solutions (no dependence on $t$) and discuss their physical intepretation.
My attempt: Suppose a solution $u$ is stationary. Then $\partial_{t}u = 0$, so the PDE becomes $$-y\partial_{x}u +x\partial_{y}u = 0.$$ We can solve by using the method of characteristics which gives us the equations:
\begin{align} \frac{dx}{ds} &= -y, \\ \frac{dy}{ds} &= x, \\ \frac{du}{ds} &= 0. \end{align}
From here I am unsure. The first two equations can be solved to give $x=A\cos(s) +B\sin(s)$ and $y=C\cos(s) +D\sin(s)$, and we also get that $u(s) = \rm constant$. How can we proceed?
Note that $y=-x'$ gives $$ y=A\sin(s)-B\cos(s), $$ so that in fact there are only two coefficients involved.
It means that $u$ is constant on circles $x^2+y^2=r^2=A^2+B^2$. Or $u_0(x,y)=f_0(r)$. The only condition that I see is that $f_0\in C^1$ and $f_0'(0)=0$.