Past exam of the perturbation methods and applied complex variables course of the University of Oxford / 2009 / B / 4 / iv

Table of contents
- Solution iv

Solution

Separate the real and imaginary parts of the derivative and integrate from

D = (0, 1)

:

x (θ) = - \frac{2 a}{π} \int_{- π /2}^{θ} cos s cot s d s, y (θ) = 1 - \frac{2 a}{π} \int_{- π /2}^{θ} sin s cot s d s .

Because

cos s cot s = csc s - sin s

and

sin s cot s = cos s

, these integrals are

x (θ) = - \frac{2 a}{π} [lo g ∣ tan (θ /2) ∣ + cos θ], y (θ) = 1 - \frac{2 a}{π} (1 + sin θ) .

θ \to 0^{-}

,

x \to + \infty

and

y \to 1 - 2 a / π

. This must equal the outgoing stream depth

a

, so

a = \frac{π}{π + 2} .

Figure 1.
Draining free surface and its potential-strip and hodograph representations
. Details.

The solution has a vertical tangent at

D

and approaches the horizontal line

y = a

downstream, agreeing with both solid-wall and free-streamline conditions.

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