 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-331/1/ii/solution

The centrifugal criterion concerns axisymmetric disturbances, so set

m = 0

. The azimuthal equation gives

v^{'} = - \frac{2Ω + r Ω ^{'}}{σ} u^{'} .

Eliminating

w^{'}

with incompressibility and then

p^{'}

with the axial equation yields

\frac{d}{d r} [\frac{1}{r} \frac{d}{d r} (r u^{'})] - k^{2} (1 + \frac{Φ ( r )}{σ ^{2}}) u^{'} = 0,

Φ (r) = 2Ω (2Ω + r Ω^{'}) = \frac{1}{r ^{3}} \frac{d}{d r} (r^{4} Ω^{2}) .

Impermeability at the two solid walls gives

u^{'} (r_{i}) = u^{'} (r_{o}) = 0

Multiply the equation by

r \overline{u^{'}}

, integrate between the walls, and use integration by parts. The boundary terms vanish and one obtains

σ^{2} = - \frac{k ^{2} \int _{r_{i}}^{r_{o}} r Φ∣ u ^{'} ∣ ^{2} d r}{\int _{r_{i}}^{r_{o}} \frac{∣ ( r u ^{'} ) ^{'} ∣ ^{2}}{r} d r + k ^{2} \int _{r_{i}}^{r_{o}} r ∣ u ^{'} ∣ ^{2} d r} .

The denominator is positive. Therefore

Φ \geq 0

throughout the annulus excludes positive real

σ^{2}

and gives centrifugal stability. Since

r^{4} Ω^{2} = (r^{2} Ω)^{2}

\frac{d}{d r} (r^{2} Ω)^{2} \geq 0 for centrifugal stability .

An outward decrease of squared specific angular momentum permits an axisymmetric centrifugal instability.

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