 Solution

ID: past-exam-of-the-viscous-flow-course-of-the-university-of-oxford/2023/1/a/iii/solution

V (t)

\frac{d}{d t} \int_{V (t)} ρ u_{i} d V = \int_{\partial V (t)} t_{i} (n) d S = \int_{\partial V (t)} σ_{ij} n_{j} d S .

Use the Reynolds transport theorem on the left. Expanding the product and using the continuity equation leaves

\partial_{t} (ρ u_{i}) + \partial_{j} (ρ u_{i} u_{j}) = ρ (\partial_{t} u_{i} + u_{j} \partial_{j} u_{i}) + u_{i} [\partial_{t} ρ + \partial_{j} (ρ u_{j})] = ρ \frac{D u _{i}}{D t} .

On the right, the divergence theorem applied to each row of the Cauchy stress tensor gives

\int_{\partial V (t)} σ_{ij} n_{j} d S = \int_{V (t)} \partial_{j} σ_{ij} d V .

Consequently the volume integral of

ρ D u_{i} / D t - \partial_{j} σ_{ij}

vanishes on every material volume. Localizing as in part (i) gives the Cauchy momentum equation:

ρ (\frac{\partial u}{\partial t} + (u \cdot \nabla) u) = e_{i} \frac{\partial σ _{ij}}{\partial x _{j}} .

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