= Solution
Substitution of the proposed <normal mode> into the <wave equation> gives the <ordinary differential equations>
$$
f_j''+k^2\left(\frac{c^2}{c_j^2}-1\right)f_j=0.
$$
A nonzero exterior <displacement field> has <exponential decay> at infinity only when $c<c_2$. With
$$
\ell=k\sqrt{1-c^2/c_2^2}>0,
$$
its <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 $f_1'(h)=0$, 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,
$$
\frac{f_1'(h)}{f_1(h)}=q\coth(qh)>0,
\qquad
\frac{f_2'(h)}{f_2(h)}=-\ell<0,
$$
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 $f_1'(h)/f_1(h)=1/h>0$.
Therefore $c_1<c<c_2$. Define the positive transverse <wavenumber>
$$
m=k\sqrt{c^2/c_1^2-1}.
$$
The rigid <boundary condition> now gives $f_1(y)=A\sin(my)$. The interface <boundary conditions> give
$$
D=A\sin(mh),\qquad
\mu_1mA\cos(mh)=-\mu_2\ell A\sin(mh).
$$
Neither $\sin(mh)$ nor $\cos(mh)$ can vanish for a nontrivial <normal mode> with $m,\ell>0$. Dividing therefore gives the <rigid-backed shear-wave dispersion> relation:
$$
\boxed{\frac{c^2}{c_1^2}=1+\frac{m^2}{k^2},\qquad
\frac{c^2}{c_2^2}=1-\frac{\ell^2}{k^2},\qquad
\tan(mh)=-\frac{\mu_1m}{\mu_2\ell}.}
$$
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.
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