Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 4 / 37A / b

Table of contents
- i b
  - Solution i
- ii b
  - Solution ii

i

Solution

The componentwise product rule for the Laplacian and the harmonicity of

Φ

give

\nabla^{2} (x \cdot Φ) = 2\nabla \cdot Φ + x \cdot \nabla^{2} Φ = 2\nabla \cdot Φ .

Hence

\nabla \cdot u = \nabla^{2} (x \cdot Φ + χ) - 2\nabla \cdot Φ = 2\nabla \cdot Φ + 0 - 2\nabla \cdot Φ = 0.

Thus

\nabla \cdot u = 0

.

ii

Solution

Applying the Laplacian to

u

and using

\nabla^{2} Φ = 0

gives

\nabla^{2} u = \nabla [\nabla^{2} (x \cdot Φ + χ)] - 2 \nabla^{2} Φ = 2\nabla (\nabla \cdot Φ) .

For

p = 2 μ \nabla \cdot Φ,

one therefore has

- \nabla p + μ \nabla^{2} u = 0 .

Together with part (i), these are the Stokes equations, so the fields form the Unscaled Papkovich–Neuber representation.

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