Waves produced by a pulsating source travelling beneath a free surface

The propagation of surface waves induced by a pulsating source travelling beneath a free surface is investigated by introducing Rayleigh’s dissipative force. By defining a disturbance velocity potential $R\varphi \left ( {x,y} \right ) exp \left ( {i\omega t} \right )$, it is shown that $\varphi \left ( {x,y} \right )$ has to satisfy throughout the flow field not only the Laplace equation, but also a differential equation resulting from the analytic extension of the free surface boundary condition. The solution is obtained in the form \[ \varphi \left ( {x,y} \right ) = f\left ( z \right ) + g\left ( {\bar z} \right ),\qquad z = x + iy\], i.e., $\varphi \left ( {x,y} \right )$ is a complex harmonic function of $x$, $y$ but is not an analytic function of $z$. The wave propagation is found to depend on a parameter $\tau$, which is the ratio of the pulsation frequency $\omega$ of the source strength to the fundamental frequency ${\omega _0}$ of the surface wave produced by a constant strength source travelling at the same speed. The case $\tau = 0$, corresponding to the travelling source of constant strength, does give a single undamped harmonic wave train on the downstream side, of wave length $2\pi {c^2}/g$, (or frequency ${\omega _0}$), as is expected. For $0 < \tau < 1/4$, there are four, and for $\tau > 1/4$, two undamped harmonic wave trains of different wave lengths on the downstream side. It is further observed that there exists a critical frequency at $\tau = 1/4$, at which frequency resonance phenomena occur. Thus violent disturbance is predicted at $\tau = 1/4$ by the present analysis. No disturbance is found to propagate infinitely upstream. This result evidently justifies the imposition of an asymptotic upstream condition of “vanishing disturbance at infinity” to replace the effect of the dissipative force in rendering the solution unique.$^{1}$