= Solution
Taking the <divergence> and <curl> of the elastic equation gives
$$
(\nabla\cdot u)_{tt}=c_P^2\Delta(\nabla\cdot u),\qquad
(\nabla\times u)_{tt}=c_S^2\Delta(\nabla\times u),
\qquad
\boxed{c_P^2=(\lambda+2\mu)/\rho,\quad c_S^2=\mu/\rho}.
$$
Substitution of the <plane wave> gives $\omega^2A=c_P^2k(k\cdot A)-c_S^2k\times(k\times A)$. Resolve $A$ parallel and perpendicular to $k$. A <P wave> has longitudinal polarization and $\omega^2=c_P^2|k|^2$; an <S wave> has transverse polarization and $\omega^2=c_S^2|k|^2$, with two transverse polarizations.
For the <Rayleigh wave>, take displacement potentials in the $x,z$ plane,
$$
u_x=\phi_x-\psi_z,\quad u_z=\phi_z+\psi_x,\qquad
\phi=Ae^{k\alpha z}e^{ik(x-ct)},\quad
\psi=Be^{k\beta z}e^{ik(x-ct)},
$$
where $\alpha=\sqrt{1-c^2/c_P^2}$, $\beta=\sqrt{1-c^2/c_S^2}$. For $0<c<c_S<c_P$, both waves decay into $z<0$. These are an evanescent longitudinal wave and an evanescent vertically polarized shear wave. At $z=0$, the <Cauchy stress tensor> $\sigma_{ij}=\lambda(\nabla\cdot u)\delta_{ij}+\mu(u_{i,j}+u_{j,i})$ gives
$$
\sigma_{xz}=\mu k^2[2i\alpha A-(1+\beta^2)B]e^{ik(x-ct)},\qquad
\sigma_{zz}=\mu k^2[(1+\beta^2)A+2i\beta B]e^{ik(x-ct)}.
$$
The third traction component is already zero. The two remaining stress-free conditions have a nonzero solution $(A,B)$ exactly when their determinant vanishes:
$$
\boxed{(1+\beta^2)^2=4\alpha\beta}.
$$
Substituting $\alpha,\beta$ gives the required Rayleigh wave equation. The nonzero subsonic root supplies a self-sustained surface wave; no existence proof for that root is required. The formal zero-speed root is degenerate and does not describe a propagating wave.
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