= Solution
Let $H(t)=\int_0^tE(r)\,dr$. For positive <rainfall> excess it is <strictly increasing>. The <characteristic equations> are
$$
\frac{dh}{dt}=E(t),\qquad\frac{ds}{dt}=\frac53h^{2/3}.
$$
A <characteristic> starting on the initial line at position $s_0\geq0$ has $h=H(t)$ and $s=s_0+S_0(t)$, where $S_0(t)=\frac53\int_0^tH(r)^{2/3}\,dr$. It supplies the downstream region $s\geq S_0(t)$.
A <characteristic> emitted at the upstream boundary at time $\tau$ has
$$
\boxed{h=H(t)-H(\tau),\qquad s=\frac53\int_\tau^t[H(r)-H(\tau)]^{2/3}\,dr.}
$$
For fixed $t$, this position decreases strictly from $S_0(t)$ to zero as $\tau$ increases from zero to $t$. Thus it uniquely determines $\tau=\tau(s,t)$ in the remaining region. The complete solution is
$$
\boxed{h(s,t)=\begin{cases}H(t)-H(\tau(s,t)),&0\leq s\leq S_0(t),\\H(t),&s\geq S_0(t).\end{cases}}
$$
The formulas agree at the separator, satisfy both zero data and cannot form a <characteristic> crossing while $E>0$. At the upstream endpoint $\tau=t$, height is zero.
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