I'm having some issues showing a PDE is locally or globally solvable. I'm familiar with the Local Existence Theorem in Evans' book on PDE, but I don't know how to apply it. Here's an example:
Use the method of characteristics to find a solution of the following boundary value $$xu_x(x,y,z)+2yu_y(x,y,z)+u_z(x,y,z)=3u(x,y,z)$$ $$u(x,y,0)=g(x,y) $$ Is this PDE linear, semilinear or quasilinear? Does there exist a unique local solution of the boundary value problem? Does the solution exist globally?
I managed to solve this problem explicitly using the method of characteristics, namely $u(x,y,z)=g(\frac{x}{e^z},\frac{y}{e^{2z}})e^{3z}$.
I can see this particular boundary value problem is linear, but I don't have a clue on where to start with the other questions. Do I need to show the triplet $(p^0,z^0,x^0)$ is noncharacteristic? If so, how would one show that? I know that it is noncharacteristic when it is admissible and $F_{P_n}(p^0,z^0,x^0)\neq0$, but I dont exactly knows what that means and how to show it. (notation from Evans' book $\S$3.2.3.)
The left-hand side $x u_x + 2y u_y + u_z$ is the directional derivative (or Lie derivative) of $u$ along the vector field $(x,2y,1)$, so what the PDE dictates is how $u$ is supposed to change as you move along the flow lines of that vector field (i.e., along the characteristic curves). Now if you prescribe the values of $u$ along some surface, then the vector field had better not be tangential to that surface, since then your prescribed values on the surface might contradict the values that the PDE will evolve along the surface. But if the vector field is transversal to the surface, then things are fine: the PDE will evolve your prescribed values from the surface out into the surrounding space (rather than along the surface).
In your case, the values are prescribed on the plane $z=0$, and the vector field has a nonzero $z$-component (everywhere, but in particular at every point of that plane; this is what that “$F_{p_n} \neq 0$” condition reduces to in this case, if you decipher the notation). So the vector field is transversal to the plane $z=0$, and you're guaranteed to have a local solution at least.