 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2002/iii/paper-52/4/b/solution

V = 0

, the particle budget gives

ϕ = 1

, so

g^{'} = g_{0}^{'}

is constant. The primitive momentum conservation equation integrates to

\frac{1}{2} u^{2} + g_{0}^{'} h = \frac{1}{2} u_{0}^{2} + g_{0}^{'} h_{0}, u_{0} = Q_{0} / h_{0} .

Introduce

q = Q / Q_{0} = 1 - W x / Q_{0}

y = h / h_{0}

and

ε = Fr_{0}^{2} = Q_{0}^{2} / (g_{0}^{'} h_{0}^{3}) ≪ 1

. The exact relation on the subcritical flow branch is

y + \frac{ε q ^{2}}{2 y ^{2}} = 1 + \frac{ε}{2} .

Expanding uniformly for

0 \leq q \leq 1

gives

\frac{h}{h _{0}} = 1 + \frac{Fr _{0}^{2}}{2} (1 - q^{2}) + O (Fr_{0}^{4}), h \sim h_{0} .

The depth increases slightly because horizontal kinetic energy decreases as fluid is removed. In fact the depth equation has

h^{'} \geq 0

on this branch, and

Fr^{2} = ε \frac{q ^{2}}{y ^{3}} \leq ε ≪ 1.

Thus the small-Froude number approximation becomes better downstream, including as the flux vanishes. At leading order

u = (Q_{0} - W x) / h_{0}

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