Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 78 / 3 / i / Solution

Let the layer occupy $0<z<h$, with a free surface at $z=0$ and rigid base at $z=h$. Write an SH-wave as $u_y=Y(z)e^{i(kx-\omega t)}$. The equation and boundary conditions giveThus $Y=A\cos (qz)$ and the guided SH modes between rigid and free planes haveThe propagating branch has $\omega >{\omega }_n$ and $k=\sqrt{{\omega }^2-{\omega }_n^2}/{\beta }^{\prime }$. Hence the rigid-base approximation to Love waves predictsAt each threshold $c_n\to \infty$ and $U_n\to 0$. As frequency rises, the phase velocity decreases towards ${\beta }^{\prime }$ and the group velocity increases towards ${\beta }^{\prime }$. Frequencies below ${\omega }_n$ have imaginary horizontal wavenumber and do not give propagating guided modes. The infinite threshold phase speed does not violate causality: the limiting group speed is zero. These quarter-wave thresholds differ from the actual Love-wave cutoff frequencies of an elastic substrate.