= Hydrostatic lower bound for a travelling-pressure response
{title2=$a\geq\frac{p_1}{\rho g}$}
For the subcritical <travelling-pressure response of finite-depth surface gravity waves>, put $c_*^2=g\tanh(kh)/k$. When $0\leq c<c_*$, the <amplitude> is
$$
a(c)=\frac{p_1}{\rho g}\frac1{1-(c/c_*)^2}\geq\frac{p_1}{\rho g}.
$$
The inequality follows immediately because the denominator belongs to $(0,1]$. Equality holds precisely at $c=0$, where <hydrostatic equilibrium> gives $\rho g\eta=-p_1\cos(kx)$. If only strictly positive $c$ is allowed, $p_1/(\rho g)$ is the <infimum>, approached as $c\downarrow0$, and is not an attained <minimum>. This bound applies to the travelling forced component, rather than an arbitrary superposition with free <surface gravity waves>.
Back to article page
