 Solution

Substitution of the proposed normal mode into the wave equation gives the ordinary differential equationsA nonzero exterior displacement field has exponential decay at infinity only when $c<c_2$. Withits wave amplitude is $f_2(y)=D{e}^{-\ell (y-h)}$. At $c=c_2$ the exterior wave amplitude is affine and cannot decay; at $c>c_2$ it is oscillatory and cannot decay. Setting the exterior wave amplitude identically zero would impose both $f_1(h)=0$ and $f_1^{\prime }(h)=0$, forcing the entire normal mode to vanish.

We also need $c>c_1$. Indeed, if $c<c_1$, the rigid boundary condition gives $f_1(y)=A\sinh (qy)$ with $q=k\sqrt{1-c^2/c_1^2}>0$. At the interface,which contradicts continuity of traction because both values of the shear modulus are positive. At $c=c_1$, the layer wave amplitude is $Ay$ and the same contradiction follows from $f_1^{\prime }(h)/f_1(h)=1/h>0$.

Therefore $c_1<c<c_2$. Define the positive transverse wavenumberThe rigid boundary condition now gives $f_1(y)=A\sin (my)$. The interface boundary conditions giveNeither $\sin (mh)$ nor $\cos (mh)$ can vanish for a nontrivial normal mode with $m,\ell >0$. Dividing therefore gives the rigid-backed shear-wave dispersion relation:Conversely, positive $m,\ell$ satisfying these relations generate a nonzero normal mode through the wave amplitudes above. Its physical displacement field is the real part of the complex expression.