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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/iii/paper-65/3/d/solution

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Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 65 / 3 / d / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-14

Use mean incompressibility to convert advection into momentum-flux divergence:

U U_{x} + V U_{y} = \partial_{x} (U^{2}) + \partial_{y} (U V) .

Therefore the reduced Reynolds-averaged momentum equation gives

\partial_{x} (U^{2}) + \partial_{y} {U V + \overline{\overset{u}{^} \overset{v}{^}}} = 0.

For a localized free jet,

U V

and the Reynolds stress tend to zero in the quiescent ambient as

y \to \pm \infty

. Assume sufficient decay to differentiate its momentum integral. Integrating across the jet gives

\frac{d}{d x} \int_{- \infty}^{\infty} U^{2} d y = - [U V + \overline{\overset{u}{^} \overset{v}{^}}]_{- \infty}^{\infty} = 0.

Thus

M_{0} = \int U^{2} d y

is constant. It is a specific momentum flux per unit span: multiplication by density gives the momentum flux. A clipped profile with nonzero stress just inside its artificial edge would fail this boundary-flux assumption unless an additional external momentum source were supplied.

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