 Solution

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

ζ = z^{3}

c^{3}

there has an opposite image at

\overset{c}{ˉ}^{3}

, so the pulled-back potential is

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

The two boundary rays map to the real axis, where the imaginary part is constant, giving the required impermeability. Factoring

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

gives the printed form up to branch constants.

The fluid vortex moves with the regular part of the complex velocity, excluding its own singular term. Since

w^{'} = u - i v

, its motion is the conjugated derivative. Differentiating the remaining terms and setting

z = c

gives

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

The overbar on the time derivative is essential.

Write

c = r e^{i θ}

. Using

c^{3} - \overset{c}{ˉ}^{3} = 2 i r^{3} sin 3 θ

and multiplying the velocity equation by

e^{i θ}

gives

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

Thus

\overset{r}{˙} = 3Γ cot 3 θ / (4 π r)

and

\dot{θ} = - Γ/ (4 π r^{2})

. Direct differentiation now yields

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

Hence

r sin (3 θ) = constant

along the trajectory while it stays inside the wedge.

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