= Solution
{c}
For $0<c<c^*$, the nonnegative <amplitude> is
$$
a(c)=\frac{p_1\tanh h}{\rho[(c^*)^2-c^2]}
=\frac{p_1}{\rho g}\frac1{1-(c/c^*)^2}.
$$
Its first two <derivatives> are positive in this interval:
$$
\frac{da}{dc}=\frac{2p_1\tanh h\,c}
{\rho[(c^*)^2-c^2]^2}>0,\qquad
\frac{d^2a}{dc^2}=\frac{2p_1\tanh h\,[(c^*)^2+3c^2]}
{\rho[(c^*)^2-c^2]^3}>0.
$$
The graph is increasing and convex, starts with a horizontal tangent at the limiting value $a(0)=p_1/(\rho g)$, and has a vertical <asymptote> at $c=c^*$. Therefore the <hydrostatic lower bound for a travelling-pressure response> is
$$
\boxed{a_{\rm crit}=\frac{p_1}{\rho g},\qquad a(c)>a_{\rm crit}\quad(0<c<c^*)}.
$$
Because the paper assumes $c>0$, this is an <infimum>, approached as $c\downarrow0$, rather than an attained <minimum>. If the stationary limit $c=0$ is admitted, equality holds: the <hydrostatic equilibrium> condition is $\rho g\eta=-p_1\cos x$. Near resonance,
$$
a(c)\sim\frac{p_1}{\rho g}\frac{c^*}{2(c^*-c)}.
$$
The divergence describes the linear model; sufficiently close to $c^*$ the small-disturbance assumptions fail. The lower bound and sketch refer to the travelling forced component, since adding free oscillations can change the instantaneous <surface elevation>.
\Image[/past-exam-of-the-fluids-and-waves-course-of-the-university-of-oxford/2021-forced-wave-amplitude.png]
{title=Subcritical travelling-wave amplitude rises from its hydrostatic lower bound and diverges at resonance}
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