Exercise 5.5.4: A field given an initial kick

The two initial conditions are $f(z)+g(z)=0$ and $-v{f}^{\prime }(z)+v{g}^{\prime }(z)=G(z)$. The first gives $f=-g$, then the second $2v{g}^{\prime }=G$, so $g(s)=\frac{1}{2v}{\int }_0^{s}G(s^{\prime })d{s}^{\prime }$ (any constant cancels) and $f=-g$:

$E_x$ is $G_0/(2v)$ times the length of the overlap of the interval $[z-vt,z+vt]$ with the slab $[-a,a]$.

At $t=a/v$ the interval has the half-width $a$. At $z=0$ it coincides with the slab, the overlap is $2a$ and $E_x=G_{0}a/v$. The overlap shrinks linearly to zero at $|z|=2a$: a triangle of height $G_{0}a/v$ and base $4a$. At $t=3a/v$ the interval has the half-width $3a$ and contains the whole slab for $|z|\le 2a$: a plateau $E_x=G_{0}a/v$. The overlap then falls linearly to zero at $|z|=4a$: a trapezoid (Figure 5.5.20).

For $vt>a$ the point $z=0$ always sees the whole slab inside its interval, so

At $t=0$ the interval has zero length and $E_x=0$. For small $t$ and $|z|<a$, $E_x=G_{0}2vt/(2v)=G_{0}t$, so $\partial E_x/\partial t=G_0$. The two ramps of the trapezoid move outwards at $\pm v$, as d'Alembert requires.