Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 79 / 1 / c / Solution

The rigid bottom reflects the lower-layer wave. Its repeated returns to the interface create an internal-wave cavity response containing both upward and downward components in $-H<z<0$, while the upper layer contains the imposed incident wave and one net reflected wave.

Let the lower downward and upward particle-displacement coefficients be $D$ and $U$, multiplying $e^{i{m}_{2}z}$ and $e^{-i{m}_{2}z}$ respectively. Their vertical displacement has opposite polarization signs:Zero normal velocity at $z=-H$ gives $U=Dq$, where $q=e^{-2i{m}_{2}H}$. Interface continuity consequently becomesFor the specified angles, $\sin {\theta }_2/\sin {\theta }_1=1/\sqrt{3}$ and $\cos {\theta }_2/\cos {\theta }_1=\sqrt{3}$. ThusThe dependence on bottom depth is thereforeIn particular,Here $k=\pi /{\lambda }_i$ and $m_2=k/\sqrt{3}$, so the positive depths giving this maximum areThe lower vertical-displacement standing-wave envelope is $2\cos {\theta }_2|D||\sin [m_2(z+H)]|$, while its horizontal-displacement envelope has the complementary cosine form. These distinguish the total standing disturbance from the amplitude of either traveling component. The traveling amplitude has minimum $|{\eta }_i|/\sqrt{3}$ at $m_{2}H=(n+1/2)\pi$.

Since $|q|=1$, $|-1+2q|=|2-q|$. Hence the net upward-propagating wave in the upper layer has amplitude magnitude $|{\eta }_i|$ for every $H$, with the phase given above. At an enhancement maximum, $q=1$ and ${\eta }_r={\eta }_i$. No energy escapes through the bottom, so complete eventual reflection is consistent with energy conservation.