characteristic quadrilateral in region II: wave equation

Question

Solve the following problem: $$ \begin{aligned} u_{t t} &=u_{x x} \text { for } 0<x<2, t>0 \\ u(x, 0) &=x(2-x), u_{t}(x, 0)=0 \text { for } 0<x<2 \\ u(0, t) &=-t^{2}=u(2, t) \text { for } t>0 \end{aligned} $$

Characteristics through the origin and $(2,0)$ divide the strip $0\leq x\leq 2, T\geq 0$ into regions, labeled I, II, III, IV, $\cdots$

I know how to solve it for region I by using d'Alembert's formula. For region II by parallelogram rule we have $$ u(x, t)=u(A)+u(B)-u(C) $$ where $A$ is on the left boundary of the strip, $u(B)$ and $u(C)$ are on the boundary of region I. But I couldn't understand how they get coordinates for $A(0,t_A),B(x_B,t_B)\:\&\:C(x_C,t_C)$ of those points? It will be a great help if anyone help me to figure out this.

Thanks in advance.

Answer 1

A picture is worth a thousand words

Can you find out those coordinate now?

Hint: Use the coordinate $(x,t)$ of $P$ to find out the constant $\alpha,\beta$ and $\gamma$.