Given wave equation: $ u_{tt}-c^2u_{xx}=0 $, let u be a solution.
Points A, B, C, D are vertices of parallelogram of two pairs of characteristic lines: $ x-ct=c1, x-ct=c2, x+ct=d1, x+ct=d2 $
Use parallelogram rule to find u that satisfies this:
$ u_{tt}-u_{xx}=0, u(-a,a)=a,$ $ $ $ u(a,a)=a^2 $ for $ a>0 $
Guys, how can I use parallelogram rule here? I know that if we draw the characteristic lines we will get picture like that.
You are given data on the "angle" $y=-x,\, x\le 0$ and $y=x,\, x\ge 0$. If you take any point $P(x_0,y_0)$ inside this angle (above the given lines) and you draw characteristic lines parallel to the sides of the angle, the value of $u$ there is the sum of the values carried by the characteristics (this is what the general solution of the wave equation is). The intersections with the sides of the angle are $(\frac{x_0+y_0}{2}, \frac{x_0+y_0}{2})$ (characteristic with slope $-1$) and $(\frac{x_0-y_0}{2}, \frac{y_0-x_0}{2})$ (characteristic with slope $+1$). Then, $$ u(x_0,y_0)=\frac{y_0-x_0}{2}+\left(\frac{x_0+y_0}{2}\right)^2 $$ The solution is not determined uniquely outside of the angle.
The parallelogram rule just tells you that the sums of the values of the solution at opposite vertices of a parallelogram are equal. In your case the parallelogram is formed by $(x_0,y_0)$, $(0,0)$ and the intersections of the characteristics with the initial angle I mentioned before. You get the same answer of course.
Hope this helps.