= Solution
Substituting both profiles into the shear <traction> condition gives
$$
2b_1A_1+\frac{b_2^2+k^2}{b_2}A_2=0.
$$
For the normal <traction>, use $Mb_1^2=Mk^2-\rho k^2c^2$ and $b_2^2+k^2=k^2(2-c^2/c_T^2)$. Removing the common nonzero factors yields the second equation. Thus
$$
\boxed{\begin{pmatrix}
2b_1&(b_2^2+k^2)/b_2\\
(b_2^2+k^2)/k&2k
\end{pmatrix}\binom{A_1}{A_2}=0.}
$$
A nonzero <displacement field> requires the <determinant> to vanish:
$$
\boxed{(b_2^2+k^2)^2=4k^2b_1b_2.}
$$
Let $s=c^2/c_T^2$ and $\eta=c_T^2/c_L^2$. Cancelling $k^4$ gives the <Rayleigh secular equation>
$$
\boxed{(2-s)^2=4\sqrt{1-s}\sqrt{1-\eta s},\qquad0<s<1.}
$$
Squaring and excluding the zero root gives $s^3-8s^2+(24-16\eta)s-16(1-\eta)=0$; any candidate must still satisfy the unsquared equation and the decay inequalities. For stable isotropic elasticity $0<\eta<3/4$, this cubic has one root in $(0,1)$: its values at zero and one have opposite signs, and its strictly negative second <derivative> on this interval prevents a second crossing back to a positive final value. Both sides of the unsquared equation are positive there, so this root is valid.
The root $s=0$ of the squared equation is not a propagating surface <wave>: at $c=0$, the two roots coincide, and the null combination $A_2=-A_1$ in this representation produces zero <displacement field>. As a numerical example, $\lambda=\mu$ gives $\eta=1/3$ and $c\simeq0.9194c_T$. The surface-wave speed is independent of $k$ in this homogeneous half-space model.
Back to article page
