First order PDE's. Determining when IVP solution is unique

The initial value problem\begin{cases} yu_x+xu_y=0 \\ \\ u(0,y)=e^{-{y^2}} \end{cases}

has for solution $$u(x,y) = e^{x^2-y^2}$$

My question is, where in the $xy$-plane is the solution unique? The initial data asserts, that the solution must be a decreasing exponential on the characteristic curves. It somehow makes sense then, to claim that the region of interest is where $x^2 \leq y^2$.

How does one reach this conclusion rigorously?

Solution 1:

The PDE implies that $u$ is constant, along the characteristics $$ y^2-x^2=c, \quad c\in\mathbb R. $$ In fact, every $(x,y)\in\mathbb R^2$, with the exception of $(0,0)$ lies in a unique characteristic. The origin lies in $y=x$ and $y=-x$, where $u(0,0)=1$.

Nevertheless, only the characteristics with $c\ge 0$ intersect with the $y-$axis, where the initial data are prescribed.

Hence, the IVP enjoys uniqueness only in $$ \{(x,y): |y|\ge |x|\}. $$