= Solution
At a fixed time, the complete spatial derivative is
$$
\boxed{\frac{dS}{dx}=\left(\frac{\partial S}{\partial x}\right)_h+\frac{\partial S}{\partial h}h_x.}
$$
The first term describes geometric variation along the channel at a fixed surface elevation. The second describes the extra wetted area produced by a varying water level; $S_h$ is the surface width.
Assume a fixed channel, negligible lateral inflow, uniform cross-sectional velocity, incompressible water and the hydrostatic long-wave approximation. Integrating <volume conservation> over a short channel segment gives
$$
\boxed{S_t+(Su)_x=0,\qquad
S_h(h_t+uh_x)+Su_x+uS_x=0.}
$$
Inviscid horizontal momentum, with $h$ measured from the fixed datum, is $u_t+uu_x+gh_x=0$. The coefficient matrix acting on $(h_x,u_x)$ has eigenvalues
$$
\boxed{\frac{dx}{dt}=u\pm c,\qquad c^2=\frac{gS}{S_h}.}
$$
These are the <variable-section channel characteristics>. Put $D_\pm=\partial_t+(u\pm c)\partial_x$. Adding the corresponding linear combinations of the two equations gives
$$
\boxed{\frac gcD_\pm h\pm D_\pm u=-\frac{uc}{S}S_x.}
$$
The opposite signs in the two combinations cancel both derivative terms in the source calculation.
If $c=c(h)$, then $D_\pm c=c_hD_\pm h$ and $dc^2/dh=2cc_h$. Taking $\xi=t$ along a characteristic gives the requested form
$$
\boxed{\frac{2g}{dc^2/dh}\frac{dc}{d\xi}\pm\frac{du}{d\xi}
=-\frac{uc}{S}\left(\frac{\partial S}{\partial x}\right)_h.}
$$
A sufficient geometry assumption is $S(x,h)=b(x)A(h)$, for which $S/S_h=A/A'$ is independent of $x$ at fixed height.
For a completely general $S(x,h)$, the printed form omits an explicit wave-speed geometry term. The correct conversion is instead
$$
\frac{2g}{\partial_h(c^2)}\left[D_\pm c-(u\pm c)(\partial_xc)_h\right]\pm D_\pm u
=-\frac{uc}{S}S_x.
$$
For example, a lake at rest with constant $h$ but position-dependent $c$ has $D_\pm c=\pm c c_x$. The printed left side is then nonzero while its right side is zero. The $h$-based compatibility equation remains valid, including when $\partial_h(c^2)=0$ makes the speed-based expression singular.
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