 Solution

ID: past-exam-of-the-fluids-and-waves-course-of-the-university-of-oxford/2023/1/c/ii/solution

Write

P = p_{\infty} / ρ > 0

2 R^{2} \dot{R}

gives an exact derivative:

\frac{d}{d t} (R^{3} \dot{R}^{2}) = 2 R^{2} \dot{R} (R \ddot{R} + \frac{3}{2} \dot{R}^{2}) = - 2 P R^{2} \dot{R} = - \frac{2 P}{3} \frac{d}{d t} (R^{3}) .

R^{3} \dot{R}^{2} = \frac{2 P}{3} (a^{3} - R^{3}), \dot{R} = - \frac{2 p _{\infty}}{3 ρ} (\frac{a ^{3}}{R ^{3}} - 1) .

Choose the negative square root because the cavity starts accelerating inward. It cannot turn around while

0 < R < a

, since the squared speed is then strictly positive. Separating variables, with

s = R / a

t_{collapse} = \frac{3 ρ}{2 p _{\infty}} \int_{0}^{a} \frac{R ^{3/2}}{a ^{3} - R ^{3}} d R = a \frac{ρ}{p _{\infty}} \frac{3}{2} \int_{0}^{1} \frac{s ^{3/2}}{1 - s ^{3}} d s .

Hence

t_{collapse} = k \frac{ρ a ^{2}}{p _{\infty}}, k = \frac{3}{2} \int_{0}^{1} \frac{s ^{3/2}}{1 - s ^{3}} d s .

s^{3/2}

near zero and as

[3 (1 - s)]^{- 1/2}

near one. The divergent inward speed as

R \to 0

signals a singular endpoint of this ideal incompressible flow model; the finite collapse time follows within that model.

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