$u_{tt}-16u_{xx}=0, \quad 0
Determine $u(x,t)$ in terms of x and t for $(x,t)$ in regions 1, 2 , and 3 determined by the characteristics.
So I know for region 1 $u(x,t)$ can simply be found using d'Alembert's solution. What I am not sure of is for regions 2 and 3. Suppose $P: (x,t)$ is in region 2. You form a characteristic quadrilateral having one vertex on the line $x=0$ and two vertices on the piece of the characteristic from the origin bounding region 1. You can find this by $u(P)=u(A)+u(B)-u(C)$ where A is on x=0 and C and B are on the piece of the characteristic from the origin bounding region 1. I have how you find u(A), u(B), and u(C) in my notes, but I do not really understand it.
$\hskip1.5in$