 Solution

Choose depth $z$ positive downwards, the free surface at $z=0$ and the interface at $z=h$. A Love wave has only transverse displacement field $u_y=U(z)e^{i(\kappa x-\omega t)}$. Take $\omega >0$, $\kappa =\omega /c_L$, and defineA trapped mode requires ${\beta }_0<c_L<\beta$, so both $q,p$ are positive. The SH equations are $U_0^{\prime \prime }+q^2{U}_0=0$ in the layer and $U^{\prime \prime }-p^{2}U=0$ in the half-space. Zero shear traction at the free surface and decay at large depth selectContinuity of displacement field and shear traction at $h$ gives $C=A\cos (qh)$ and $-{\mu }_{0}qA\sin (qh)=-\mu pC$. ConsequentlySubstituting the definitions of $q,p$ yields the requested speed-frequency dispersion relation. For negative frequency use $|\omega |$ in both vertical wavenumbers; the real displacement field has conjugate frequency partners. Neither equality in the trapped-speed interval gives a strictly localized finite-frequency Love mode.