Say $F(x)=v(x)=0$ when $|x| \geq a$ whilst $F(x)=h(x)$ and $v(x)=c \ h'(x)$, both when $|x|<a$. As we know, the wave equation in D'Alembert's form is:
$$
y(x,t)=\frac{1}{2}\left[F(x-ct)+F(x+ct)\right] + \frac{1}{2 c} \int_{x-ct}^{x+ct} v(s) \,ds.
$$
I calculated that $y(a,t)=\frac{h(a)}{2}$ in $t \in (0, \frac{2a}{c})$ but I'm curious how to calculate $y(a,t)$ for $t > \frac{2a}{c}$?
I think the answer is $0$ because the wave moves outwards from the center at speed $c$ so eventually it will pass the point $a$. However, if time is on the $y$ axis, does this mean it stays at position $a$ and just increases on $y$?
Can someone please clarify my intuition of the characteristic diagram involving the time parameter?
When dealing with piecewise defined initial conditions, it helps to draw the regions of dependency of each piece. To do this, draw the line $x=x_0\pm ct$ for each point $x_0$ separating two pieces:
I put big fat $0$s in the regions which do not depend on $[-a,a]$. The rest of space-time is divided into three parts.
In part (iii), $$y(x,t)=\frac{1}{2}\left[h(x+ct) + h(x-ct)\right] + \frac{1}{2 c} \int_{x-ct}^{x+ct} ch' (s) \,ds = h(x+ct) $$ This describes the wave moving to the left keeping its profile.
In Part (i), $$y(x,t)=\frac{1}{2}\left[h(x+ct)\right] + \frac{1}{2 c} \int_{-a}^{x+ct} ch' (s) \,ds = h(x+ct) - \frac12 h(-a) $$ This is also the wave moving to the left in its original form, but with the added shock from the discontinuity at $-a$ (which doesn't exist if $h(-a)=0$).
Part (ii) is similar and is left as an exercise. It will have $h(x-ct)$, plus the shock from the discontinuity at $a$.