Uniqueness of solution to 1st order pdes I am given a 1st order partial differential equation $y{\partial \psi\over\partial x}+x{\partial \psi\over\partial y}=0$ subjected to boundary condition $\psi(x,0)=\exp(-x^2)$. I have found that a solution is $\psi(x,y)=\exp(y^2-x^2)$. But I am asked when the solution is unique. Could someone please explain how to answer this? Thanks.
 A: Consider the parametric curves $x = A e^t + B e^{-t}$, $y = A e^t - B e^{-t}$, which satisfy $x' = y$, $y' = x$.  Along such a curve any solution $\psi$ must be constant, according to the chain rule: 
$$ \frac{d}{dt} \psi(x(t),y(t)) = \psi_x \frac{dx}{dt} + \psi_y \frac{dy}{dt} = 0$$
Now the curve intersects $y=0$ if and only if $A$ and $B$ are either both positive (i.e. $x > |y|$), both negative ($x < -|y|$), or both $0$ ($x=y=0$).  So a boundary condition on $y=0$ produces uniqueness only in the regions $|x| \ge |y|$. In the region $|y| > |x|$ the solution  is not unique.   For example, you could add $f(y^2 - x^2)$  to $\psi(x,y)$ where $f$ is differentiable with  $f(s) = 0$ for $s \le 0$.
A: Uniqueness can be addressed in the following way. Let us suppose that exist another solution $\phi(x,y)$ such that $\phi(x,0)=e^{-x^2}$ then, being your equation linear then also $\phi'(x,y)=a\psi(x,y)+b\phi(x,y)$, with two arbitrary coefficients $a$ and $b$ is a solution. The boundary condition will give $a+b=1$. So, unless you do not give another condition on $y$ your solution cannot be unique.
Of course, you have also the other condition in the given problem. From the fact that $\psi(x,0)=e^{-x^2}$ and from the other fundamental result that your equation has the general solution (characteristic method cited in the comments) $\psi(x,y)=\psi(x^2-y^2)$ for you is enough to set $\psi(0,0)=1$ and your solution is unique.
Finally, I would like to point out the simple way the solution OP proposed can be found. One have to search for a solution in the form
$$\psi(x,y)=\phi(y)e^{-x^2}$$
with $\phi(0)=1$ and the solution is immediately obtained, consistent with the characteristic method.
