Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 329 / 1

Table of contents
- a 1
  - Solution a
- b 1
  - Solution b
- c 1
  - Solution c

a

Solution

The Papkovich–Neuber representation writes a homogeneous Stokes flow in terms of a harmonic vector field

Φ

and a harmonic scalar

χ

μ u = Φ - \frac{1}{2} \nabla (x \cdot Φ) + \nabla χ, p = - \nabla \cdot Φ .

For translation, rotational symmetry and decay at infinity restrict the trial harmonic fields to the fundamental harmonic

1/ r

and its directional derivatives contracted with

U

. For rotation, the only decaying isotropic axial-vector field with the required boundary value is proportional to

Ω \times x / r^{3}

. Matching the no-slip boundary condition

u = U + Ω \times x

at

r = a

gives the superposition of the translating sphere in Stokes flow and the rotating sphere in Stokes flow:

u = \frac{3 a}{4 r} (I + \frac{xx}{r ^{2}}) U + \frac{a ^{3}}{4 r ^{3}} (I - 3 \frac{xx}{r ^{2}}) U + \frac{a ^{3}}{r ^{3}} Ω \times x, p = \frac{3 μ a}{2 r ^{3}} U \cdot x .

Each term decays at infinity, and direct substitution at

r = a

gives the prescribed rigid velocity.

When

U = 0

, the pressure is constant and may be set to zero. Differentiating the rotational velocity and using the Newtonian fluid stress tensor gives

σ = - \frac{3 μ a ^{3}}{r ^{5}} [(Ω \times x) x + x (Ω \times x)] .

b

Solution

For two zero-body-force Stokes flows

(u, σ)

and

(u, σ)

in the same domain, the Lorentz reciprocal theorem for Stokes flow states

\int_{\partial D} u \cdot σ n d S = \int_{\partial D} u \cdot σ n d S .

Apply it first with the auxiliary translating-sphere solution and then with the auxiliary rotating-sphere solution. The swimmer is force-free and torque-free, while the auxiliary surface tractions are known. The resulting surface slip velocity formulas are

V = - \frac{1}{4 π a ^{2}} \int_{r = a} u_{s} d S, ω = - \frac{3}{8 π a ^{4}} \int_{r = a} x \times u_{s} d S .

On

r = a

, the first part of the prescribed slip is

\frac{( A \times x ) \times x}{a ^{2}} = \frac{x ( A \cdot x )}{a ^{2}} - A .

Its surface average is

- 2 A /3

, whereas the

B

term has zero average by oddness. Thus

V = \frac{2}{3} A .

The

A

term contributes no rotation. For the other term, the isotropic second and fourth surface moments give

\int_{r = a} x \times u_{s} d S = \frac{8 π a ^{3}}{15} ∣ B ∣^{2} B,

and hence

ω = - \frac{∣ B ∣ ^{2}}{5 a} B .

c

Solution

When

B = 0

, part (b) gives

V = 2 A /3

and

ω = 0

. The total boundary velocity is therefore

V + u_{s} = - \frac{1}{3} A + \frac{x ( A \cdot x )}{a ^{2}} .

The decaying velocity potential

ϕ = - \frac{a ^{3}}{3} \frac{A \cdot x}{r ^{3}}

is harmonic for

r > a

, and

u = \nabla ϕ = - \frac{a ^{3}}{3} \nabla (\frac{A \cdot x}{r ^{3}})

has exactly this value at

r = a

. It is a force-free potential-dipole field and decays as

r^{- 3}

.

When

A = 0

, define the degree-three harmonic polynomial

H_{3} (x) = (B \cdot x)^{3} - \frac{3}{5} ∣ B ∣^{2} r^{2} (B \cdot x) .

An appropriate decaying harmonic potential is

Ψ = C \frac{H _{3} ( x )}{r ^{7}}, u = \nablaΨ \times x .

Indeed, the tangential boundary value is proportional to

[\frac{( B \cdot x ) ^{2}}{a ^{3}} - \frac{∣ B ∣ ^{2}}{5 a}] B \times x,

which combines the prescribed slip with the rigid rotation found in part (b). Since

H_{3} / r^{7} = O (r^{- 4})

and multiplication of its gradient by

x

preserves that order, the exterior velocity decays as

∣ u ∣ = O (r^{- 4}) .

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