Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 79 / 2 / i

Table of contents
- Solution i

Solution

Write

A_{ij} = \partial_{j} u_{i}

and use incompressibility,

A_{ii} = 0

. The displayed vector can be expressed as

D_{i} = A_{ij} A_{jk} u_{k} - \frac{1}{2} u_{i} tr (A^{2}) .

On taking its divergence,

\partial_{i} A_{ij} = 0

. Commuting the remaining partial derivatives gives

A_{ij} (\partial_{i} A_{jk}) u_{k} = u_{k} A_{ij} \partial_{k} A_{ji} = \frac{1}{2} u_{k} \partial_{k} tr (A^{2}),

which cancels the derivative of the second term of

D_{i}

. The derivatives falling on

u_{k}

leave

\partial_{i} D_{i} = A_{ij} A_{jk} A_{ki} = tr (A^{3}) .

In homogeneous turbulence, averaging commutes with differentiation and

⟨ D_{i} ⟩

is position independent, provided these moments exist. Consequently

⟨ tr (A^{3})⟩ = 0

.

Using the strain-rate tensor and vorticity decomposition supplied in the question now gives the Betchov relation:

⟨ S_{ij} S_{jk} S_{ki} ⟩ = - \frac{3}{4} ⟨ ω_{i} ω_{j} S_{ij} ⟩ .

This uses homogeneity and incompressibility; isotropy is unnecessary.

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