 Solution

ID: past-exam-of-the-flows-fluctuations-and-complexity-course-of-the-university-of-oxford/2016/1/d/solution

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Past exam of the flows, fluctuations and complexity course of the University of Oxford / 2016 / 1 / d / Solution by

Codex  0 2026-10-25

u_{r} = 0

and

u_{θ} = (3 v /2) sin θ

. The only surface stagnation points are therefore

(0, 0, R) and (0, 0, - R) .

(1)

There are no other stagnation points in the exterior: away from the axis the polar component is nonzero, and on the axis the radial component vanishes only at

r = R

Near the front point put

(X, Y, Z) = (x, y, z - R)

. Expanding the potential from solution gives

ϕ = - \frac{3 v R}{2} + \frac{3 v}{4 R} (X^{2} + Y^{2} - 2 Z^{2}) + O (∣ X ∣^{3} / R^{2}),

(2)

so the linear velocity field is

(3 v X / (2 R), 3 v Y / (2 R), - 3 v Z / R)

. This agrees with solution for

W = 3 v / (4 R)

At the rear point, using

Z = z + R

instead, the quadratic term has the opposite sign:

W = - 3 v / (4 R)

. Incoming axial flow and tangential outflow at the front become tangential inflow and axial outflow at the back. The local streamline geometry is unchanged while its direction reverses.

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