Domain for PDE Solution I'm working on PDE exercises for solutions via characteristics. The problems ask for the solution and its domain. 
For $$y^{-1}u_x + u_y = u^2, \quad u(x,1) = x^2$$
I found the solution to be 
$$ u(x,y) = \left( \frac{1}{(x- \ln |y|)^2} - y +1 \right) ^{-1}$$.
So far, so good, but I can't find a clear way to express the domain. The best I've come up with is that it's "the union of open connected sets $U \subset \mathbb{R}^2$ such that $x \neq \ln |y|$ and $(x - \ln |y|)^2(y-1) \neq 1$ for all $(x,y) \in U$ and there exists a point $(x_0, 1) \in U$.    I.e., nothing in the solution breaks, and part of the initial data is in the set. 
For another, 
$$ u_x + u^{1/2} u_y = 0, \quad u(x,0) = x^2+1,$$
I get the implicit formula $u(x,y) = (x-\frac{y}{u(x,y)^{1/2}})^2 +1$, and here again, I don't see a good way to describe the domain.
Is there some other way to express the solutions, or another way to think about them, which would make the domain clearer? I'm running into this problem over and over, and would appreciate any guidance. Thanks. 
 A: For the  second IVP, the parametric equations describing your solution are $x(t,s)=t+s$, $y(t,s)=t \sqrt{s^2+1}$ and $u(x,y)=s^2+1$ - this is consistent w/ your solution. Clearly, $(t,s)$ is allow to span the entire $\mathbf{R}^2$ and hence so does $(x,y)$ - done; no restrictions.
For the first case, sometimes (=when no IVs are specified) the problem can lie with the change of coordinates one selects - there's no unique coordinate system that will reduce the PDE to a parametric ODE.  Clearly, this is not the case here: the solution is unique & you found it. (Have you checked that it's the solution, btw?)
Edit. For that first problem, $x\ne{\rm  ln}|y|$ is not needed - you do of course need that $y\ne0$, but that's not strange given the PDE coefficients & the way the characteristics look. That is because turning your composite fraction (I'm counting the negative power here) into a regular one puts $(x-{\rm  ln}|y|)^2$ in the numerator's place. There only remains the second condition in that problem, the important thing about which - IMO - is why it arises, not how it can be rewritten. (Surely it's somewhat surprising that such a condition is even needed? After all, the problem "seems" to be well-posed & all that.) The reason this occurs is two-fold: first, recall that every such 1st order problem is effectively an ODE problem, b/c the first order terms correspond to a directional derivative; second, the ODE $\dot{u}=u^2$ blows up in finite time. Not so surprising after all, then, that this would happen.
