Consider the problem $$xu_t+u_x = 0, \quad u(0,x) = \sin x.$$ We're asked to prove that the problem doesn't have a solution defined in all of $\Bbb R^2$, and to give a possible open set in $\Bbb R^2$ on which we actually have a solution.
Solving ${\rm d}t/{\rm d}x = x$, I ultimately get that we'd have solutions of the form $u(t,x) = f(2t-x^2)$, for differentiable functions $f$. The initial condition forces $f(-x^2) = \sin x$, and if this goes for all $x$, using $-x$ instead along with the fact that $\sin$ is an odd function, we obtain $\sin x = -\sin x$ for all $x$.
I'm having trouble giving an example of region where the solution does exist. Can someone give me a push?
You have found the solution is constant along $x^2 - 2t = C$. So we are looking $f(x^2) = sin x$ at $t = 0$. So $f(x) = sin\sqrt{x}$.
Hence, $$u(x, t) = sin \sqrt{x^2 - 2t}$$.
From here, yuo can see where the region is for the solution defined.