= Solution
Put $\Delta=fb$ and $v=u_1+\Delta/2$, the mean of the two wall velocities. The <zero-potential-vorticity rotating channel flow> has $h_2=h_1-\Delta v/g$. Substitution into the <volume flux> formula gives
$$
Q=b h_1v-\frac{b\Delta}{2g}v^2.
$$
The resulting quadratic equation has two roots, $v=gh_1/\Delta\pm\sqrt{g^2h_1^2/\Delta^2-2gQ/(b^2f)}$. The positive-depth condition $h_2>0$ selects the minus sign. Thus
$$
\boxed{u_1=-\frac{bf}{2}+\frac{gh_1}{bf}-\sqrt{\frac{g^2h_1^2}{b^2f^2}-\frac{2gQ}{b^2f}}.}
$$
The discarded sign would make $h_2<0$. On the retained branch $h_2=\sqrt{h_1^2-2fQ/g}$, which also shows the admissibility condition $h_1^2\ge2fQ/g$.
The depth-based specific energy at the first wall is
$$
\boxed{E(h_1;Q,b)=h_1+\frac{1}{2g}\left[-\frac{bf}{2}+\frac{gh_1}{bf}-\sqrt{\frac{g^2h_1^2}{b^2f^2}-\frac{2gQ}{b^2f}}\right]^2.}
$$
Because $\partial_y(h+u^2/(2g))=0$, this same specific energy holds everywhere across the channel at fixed $x$. Its sum with the bed elevation is the <Bernoulli function> divided by $g$. Zero <shallow-water potential vorticity> makes that <Bernoulli function> constant throughout the connected steady channel flow, so
$$
\boxed{E(x)+H(x)=B_0/g.}
$$
Consequently the depth-based specific energy decreases as the bed rises and increases as it falls. It is constant along a level-bed channel, even if the width varies; that width variation instead changes the wall depths and velocities compatible with the conserved <volume flux> and <Bernoulli function>.
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