Consider the rate of transmission of energy due to a single sinusoidal component of wave number \( k \). The energy flux across the vertical plane \( x = 0 \) is the pressure work done by the fluid in the region \( x < 0 \) on the fluid in the region \( x > 0 \). Writing \( p \) as the sum of a perturbation \( p' \) and a background pressure \( = -\rho g z \), the time average energy flux \( EF \) per unit length of crest is

\[ EF = \frac{\omega}{2\pi} \int_0^{2\pi/\omega} \int_{-H}^0 p u \, dz \, dt = \frac{\omega}{2\pi} \int_0^{2\pi/\omega} \int_{-H}^0 (p' - \rho g z) u \, dz \, dt \\ = \frac{\omega}{2\pi} \int_0^{2\pi/\omega} \int_{-H}^0 p' u \, dz \, dt + \frac{\omega}{2\pi} \frac{\rho g H^2}{2} \int_0^{2\pi/\omega} u \, dt \]

The wave period is \( 2\pi / \omega \). The final integral \(\boxed{\int_0^{2\pi/\omega} u \, dt}\) is zero because the time average of \( u \) over one wave period is zero. Substituting for \( p' \) from \(p' = -\rho \frac{\partial \phi}{\partial t} = \rho \frac{a \omega^2}{k} \frac{\cosh(k(z + H))}{\sinh(kH)} \cos(kx - \omega t) = \rho g a \frac{\cosh(k(z + H))}{\cosh(kH)} \cos(kx - \omega t)\) and \( u \) from \(u = a \omega \frac{\cosh(k(z + H))}{\sinh(kH)} \cos(kx - \omega t)\), the equation for \(EF\) becomes

\[ EF = \frac{\omega}{2\pi} \int_0^{2\pi/\omega} \cos^2(kx - \omega t) \, dx \cdot \frac{\rho a^2 \omega^3}{k \sinh^2 kH} \int_{-H}^0 \cosh^2[k(z + h)] \, dz = \frac{\rho g a^2}{2} \left[ \frac{c}{2} \left( 1 + \frac{2kH}{\sinh(2kH)} \right) \right] \]

The first factor of the final form for the energy flux is the wave energy per unit area given in the total wave energy in the water column per unit horizontal area \(E = E_k + E_p = \rho g \overline{\eta^2} = \frac{1}{2} \rho g a^2\). Therefore, the second factor must be the speed of propagation of the wave energy of component \( k \). This energy propagation speed is called the group speed

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