Interpret the stated ansatz as
$$
a(z,t)=f(t)+\sqrt2\,g(t)\sin(kz).
$$
Multiplying the pressure relation by $a$ gives $2\mu a_t=Pa-\gamma$. Equating its constant and sinusoidal parts yields
$$
f'=\frac{Pf-\gamma}{2\mu},
\qquad
g'=\frac{Pg}{2\mu}.
$$
The volume per wavelength is proportional to the mean of $a^2$, namely $f^2+g^2$. Hence
$$
\boxed{\alpha_0=f(0)^2+g(0)^2=f(t)^2+g(t)^2}.
$$
Choosing
$$
P=\frac{\gamma f}{\alpha_0}
$$
then gives exactly
$$
\boxed{f'=-\frac\gamma{2\mu}\frac{g^2}{\alpha_0}},
\qquad
\boxed{g'=\frac\gamma{2\mu}\frac{fg}{\alpha_0}}.
$$

For $g\ll f\simeq a_0$, the disturbance amplitude obeys
$$
g'\simeq\frac\gamma{2\mu a_0}g,
$$
so its long-wave linear growth rate is
$$
\boxed{s=\frac\gamma{2\mu a_0}}.
$$
The minimum radius is $a_{\min}=f-\sqrt2g$, and
$$
\frac{d a_{\min}}{dt}
=-\frac\gamma{2\mu\alpha_0}(g^2+\sqrt2fg)<0.
$$
Since the trajectory follows the circle $f^2+g^2=\alpha_0$ toward increasing $g$, it reaches $f=\sqrt2g$ and hence $a_{\min}=0$. Thus the nonlinear disturbance narrows monotonically to pinch-off within this approximation.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 329 / 2 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-329/2/a/solution