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Use the complex amplitude and take its real part at the end; the equations are linear with real coefficients. Let $M=\lambda +2\mu$, $\omega =kc$, and $v=(A,B{)}^T$. The differential operators on this exponential function are ${\partial }_X=ik$, ${\partial }_Y=-b$, and ${\partial }_t=-i\omega$. Substitution givesThe determinant factors asConsequently choose the positive decay rootsThe Rayleigh wave decay exponents must both give strict decay into the half-space. A genuine nonzero Rayleigh wave combines both, as the free-surface equations below also show. Thus $b_1,b_2>0$ requires $\boxed{0<c<\min (c_T,c_L)}$. In a stable ordinary isotropic solid $c_T<c_L$, so it is slower even than the transverse wave. The roots depend on the candidate speed as well as the material constants and wavenumber; the speed is fixed by the boundary condition.

For a Rayleigh wave at a free surface of a homogeneous isotropic elastic half-space, define $s=c^2/c_T^2$ and $\eta =c_T^2/c_L^2$. Cancelling the two surface tractions between the constituents with Rayleigh wave decay exponents gives the displayed equation with $0<s<1$. Equivalently $(b_T^2+k^2{)}^2=4k^2{b}_L{b}_T$. Squaring gives a factor $s$ times $P(s)=s^3-8s^2+(24-16\eta )s-16(1-\eta )$. For stable Lamé parameters, $0<\eta <3/4$, so $P(0)<0$, $P(1)=1$, and $P^{\prime \prime }(s)=6s-16<0$ on $(0,1)$. Its concavity ensures exactly one crossing in this interval. Both sides of the unsquared equation are positive, so this root gives the physical speed. The zero root is a degenerate static cancellation, not a propagating Rayleigh wave.