 Solution

ID: past-exam-of-the-theoretical-mechanics-course-of-the-university-of-oxford/2010/a/1/b/solution

t (n)

n

. The stress component

σ_{ij}

is its

i

th component on a surface normal to the

j

th coordinate axis. Cauchy's stress theorem states that this dependence is linear:

t_{i} (n) = σ_{ij} n_{j} .

For the boundary of a material volume, take the outward normal and the traction exerted by the surrounding fluid. Newton's law and the divergence theorem give

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

ρ D u_{i} / D t

. The volume is arbitrary, so

ρ \frac{D u _{i}}{D t} = \partial_{j} σ_{ij} .

Substitute the Newtonian constitutive law. With constant viscosity,

\partial_{j} σ_{ij} = - \partial_{i} p + μ \partial_{j} \partial_{j} u_{i} + μ \partial_{i} (\partial_{j} u_{j}) = - \partial_{i} p + μ \nabla^{2} u_{i} .

Hence

ρ (u_{t} + (u \cdot \nabla) u) = - \nabla p + μ \nabla^{2} u .

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