= Solution
For the state vector $q=(h,u,g')^T$, the primitive equations are $q_t+\mathsf A q_x=S$, with
$$
\mathsf A=\begin{pmatrix}u&h/2&0\\g'&u&h/3\\0&0&u\end{pmatrix},\qquad
S=\begin{pmatrix}w_e\\-2uw_e/h\\-2g'w_e/h\end{pmatrix}.
$$
The <characteristic polynomial> is
$$
\det(\mathsf A-\lambda I)=(u-\lambda)\left[(u-\lambda)^2-\frac{g'h}{2}\right].
$$
Consequently the three <characteristic speeds> are
$$
\boxed{\frac{dx}{dt}=u,\quad u+c,\quad u-c,\qquad c=\sqrt{\frac{g'h}{2}}.}
$$
For $h>0$ and $g'>0$, the <hyperbolic system> has three distinct real <eigenvalues>. The long-wave speed relative to the current is $c$, while $u$ transports a <buoyancy>/contact disturbance.
A left <eigenvector> for the material characteristic is $(0,0,1)$, giving
$$
\boxed{\frac{dg'}{dt}=-\frac{2g'w_e}{h}\quad\text{along}\quad\frac{dx}{dt}=u.}
$$
For $u\pm c$, left <eigenvectors> can be chosen as
$$
\ell_\pm=\left(\pm\frac{2c}{h},1,\pm\frac{h}{3c}\right).
$$
Multiplying $q_t+\mathsf A q_x=S$ by these rows gives the <characteristic compatibility for an entraining triangular-channel current>. With all differentials evaluated along the indicated characteristic,
$$
\boxed{du\pm\frac{2c}{h}\,dh\pm\frac{h}{3c}\,dg'
=-\frac{2w_e}{h}\left(u\mp\frac c3\right)dt,
\qquad \frac{dx}{dt}=u\pm c.}
$$
The source follows directly from $\ell_\pm S=-2uw_e/h\pm2cw_e/h\mp2g'w_e/(3c)$ and $g'/c=2c/h$.
Since $4\,dc=(2c/h)dh+(h/c)dg'$, an equivalent form is
$$
d(u\pm4c)\mp\frac{4c}{3g'}\,dg'
=-\frac{2w_e}{h}\left(u\mp\frac c3\right)dt.
$$
Combining it with the material <buoyancy> equation also gives
$$
D_\pm(u\pm4c)=\frac{2h}{3}g'_x-\frac{2w_e}{h}(u\pm c),\qquad
D_\pm=\partial_t+(u\pm c)\partial_x.
$$
Only when <entrainment> vanishes and $g'$ is spatially constant do $u\pm4c$ become constant <Riemann invariants> on these <characteristic curves>. Dropping the <buoyancy> differential in the general problem would lose one of the required transport couplings. The characteristic formulas with $1/c$ apply to the nondegenerate region $g'h>0$; dry or neutrally buoyant limits require their conservative form instead.
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