 Solution

ID: past-exam-of-the-options-ii-course-of-the-university-of-oxford/2010/k/k4/solution

ζ = z^{3}

sends the wedge

0 < ar g z < π /3

to the upper half-plane. The physical vortex maps to

ζ = c^{3}

; an opposite image vortex at

\overline{c}^{, 3}

makes the real axis a streamline. Pulling this potential back gives

w (z) = - \frac{i Γ}{2 π} lo g (z^{3} - c^{3}) + \frac{i Γ}{2 π} lo g (z^{3} - \overline{c}^{, 3}) .

Since

z^{3} - c^{3} = (z - c) (z^{2} + cz + c^{2})

, splitting the first logarithm, up to an additive constant, gives exactly the required form. The image construction enforces zero normal velocity on both rays.

To find the vortex motion, hold

c

fixed while differentiating with respect to

z

, remove the self-induced singular term

- i Γ/ (2 π (z - c))

, and then evaluate at

z = c

. The remaining complex velocity is

\frac{i Γ}{2 π} [- \frac{2 c + c}{c ^{2} + c ^{2} + c ^{2}} + \frac{3 c ^{2}}{c ^{3} - c ^{, 3}}] .

u - i v

equals

d \overline{c} / d t

, rather than

d c / d t

. Thus

\frac{d c}{d t} = \frac{i Γ}{2 π} (- \frac{1}{c} + \frac{3 c ^{2}}{c ^{3} - c ^{, 3}}) .

The overbar on the source derivative is essential to the trajectory calculation.

Write

c = r e^{i θ}

. Since

c^{3} - \overline{c}^{, 3} = 2 i r^{3} sin (3 θ)

, multiply the equation by

e^{i θ}

to obtain

\overset{r}{˙} - i r \dot{θ} = \frac{3Γ}{4 π r} cot (3 θ) + \frac{i Γ}{4 π r} .

Taking real and imaginary parts gives

\overset{r}{˙} = \frac{3Γ}{4 π r} cot (3 θ), \dot{θ} = - \frac{Γ}{4 π r ^{2}} .

Therefore

\frac{d}{d t} (r sin (3 θ)) = \overset{r}{˙} sin (3 θ) + 3 r cos (3 θ) \dot{θ} = 0,

proving the trajectory invariant

r sin (3 θ) = constant .

The calculation also covers

Γ = 0

, when the vortex position is stationary.

This is the sixty-degree case of a point vortex in a wedge.

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