unique solution of pde using method of characteristics I'm solving the pde: $\begin{cases}
xu_x + yu_y = 1 + y^2 \\
u(x, 1) = x + 1
\end{cases}$
Make sure to include all pictures of characteristics and justify whether or not you have found the unique solution.
I've found a solution using the method of characteristics: $u = \ln(y)+\frac{y^2}{2}+\frac{x}{y}+\frac{1}{2}$. How can I argue using the picture of projections on $(x,y)$-plane that the solution is not uniquely determine $u(x, y)$ for $y < 0$?
 A: The projections of the characteristics are lines through the origin $x/y=c_1$ (btw, this is the picture you need). From the boundary conditions, at first sight it seems that every characteristic curve has a initial value assigned as we are given the value for $u$ along a line cutting every characteristic curve (the line $y=1$ or $(x,1)$ ) e. g. To pick a value for $c_1$ unambiguously determines (apparently) the value for $u$ along the line $x=c_1y$, so is,  $u_0=u(c_1,1)=1+c_1$ then $u(c_1y,y)=\ln |y|+y^2/2+c_1+1/2$
But I said "apparently" because in fact the projections of the characteristics are rays from the origin, not lines through the origin. The formula for the general solution, $u = \ln|y|+\frac{y^2}{2}+f\left(\frac{x}{y}\right)$ shows that in any case is defined the value for $y=0$. So is, the characteristics are cut in two for each value of $c_1$ We have $u(c_1y,y)=\ln |y|+y^2/2+c_1+1/2$ for $y>0$, but for $y<0$ we can have the perfectly compatible equation $u(c_1y,y)=\ln |y|+y^2/2+g(c_1)$, being $g$ any function, because the given boundary conditions don't cut the rays laying into the half plane with $y<0$: is not given any value for $u$ at some point of these rays.
