Solution

ID: past-exam-of-the-probability-statistics-and-options-ii-course-of-the-university-of-oxford/2011/k/k3/solution

On the circle put . The Joukowski map gives , so its image is the slit . Away from that circle, the inverse equation is . For , neither root has modulus , and their product is , so exactly one root has modulus greater than . Selecting that root defines the exterior inverse.
More explicitly, define to be the branch analytic on normalized by at infinity. It is positive for real and negative for real . Then
Existence and analyticity of this branch also follow from the exterior inverse: is nonzero for , so the selected simple root varies analytically; has the stated normalization. The exterior map is injective because with distinct roots would imply , impossible for two exterior points. Thus is a conformal bijection throughout .
The proposed potential is a uniform stream plus its cylinder doublet and the circulation term. On , the first two terms sum to the real quantity , while the imaginary part of the logarithmic term is the constant . Thus the cylinder is a streamline. At infinity the complex velocity tends to and the circulation is , verifying both far-field conditions. Logarithms may be chosen locally; their derivative gives a single-valued velocity.
Composition therefore gives the plate potential
At the edge , corresponding to , the physical complex velocity is . Since the denominator has a simple zero, finiteness requires
which is exactly
With this cancellation, the limiting ratio is . The Kutta value of the circulation gives and . Hence the finite edge velocity is

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