= Solution
The isotropic elastic displacement equation is $\rho\ddot u=(\lambda+\mu)\nabla(\nabla\cdot u)+\mu\nabla^2u$. With $u=\nabla\phi+\nabla\times\boldsymbol\psi$, its dilational and shear potentials satisfy
$$
\boxed{\phi_{tt}=c_P^2\nabla^2\phi,\quad\boldsymbol\psi_{tt}=c_S^2\nabla^2\boldsymbol\psi,\qquad c_P^2=(\lambda+2\mu)/\rho,\quad c_S^2=\mu/\rho.}
$$
For the specified two-dimensional potential, $u_x=\phi_x+\psi_y$, $u_y=\phi_y-\psi_x$.
Choose a real wave number $k>0$ and put $\omega=kc$. Decay into $y<0$ requires the combinations
$$
\phi=Ae^{py}e^{i(kx-\omega t)},\qquad\psi=Be^{qy}e^{i(kx-\omega t)},\quad p=k\sqrt{1-c^2/c_P^2},\quad q=k\sqrt{1-c^2/c_S^2},
$$
with $0<c<c_S$. Real parts give the physical fields. Then $u_x=ikAe^{py}+qBe^{qy}$ and $u_y=pAe^{py}-ikBe^{qy}$, times the common propagating factor. The free surface requires $\sigma_{xy}=\mu(u_{x,y}+u_{y,x})=0$ and $\sigma_{yy}=\lambda\nabla\cdot u+2\mu u_{y,y}=0$. At zero these become
$$
2ikpA+(k^2+q^2)B=0,\qquad(k^2+q^2)A-2ikqB=0.
$$
A nonzero pair exists precisely when
$$
\boxed{(k^2+q^2)^2=4k^2pq,\qquad(2-s^2)^2=4\sqrt{1-s^2}\sqrt{1-\kappa s^2},\quad s=c/c_S,\ \kappa=c_S^2/c_P^2.}
$$
For such a root, one can choose $A\ne0$ and $B=-2ikpA/(k^2+q^2)$, providing the requested wave explicitly. Squaring and excluding the static zero root gives the equivalent Rayleigh <polynomial>
$$
s^6-8s^4+(24-16\kappa)s^2-16(1-\kappa)=0,\qquad0<s<1.
$$
The physical positive root is unique as allowed in the question. All factors of $k$ cancel in the speed equation, so \b[the Rayleigh surface wave is nondispersive], with $c$ determined only by the material constants. The potentials decay exponentially over depths of order $k^{-1}$.
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