I need to understand the proof of the following statement:
if $u(x,y)$ is a function that's $C^1(\Bbb R)$ and the graph of $u(x,y)$ is on $\Bbb R^3 $is a union of the characteristic lines of the given equation. So $u(x,y)$ is solution of the equlation.
So far I know that I suppose to determine a fixed $(x_0,y_0)$ point and argue that the point $u(x_0,y_0,u(x_0,y_0))$ is on the surface of the solution. I suppose we need to prove that the point $(x_0,y_0)$ satisfy the equation at that point.
stuck from here, and don't fully understand the whole concept of the "characteristic lines" idea(I'm familiar with the solution techniques itself), so please explain me the solution with as many details as you can.