Past exam of the elasticity and plasticity course of the University of Oxford / 2011 / 2 / h / Solution

Substituting both profiles into the shear traction condition givesFor the normal traction, use $M{b}_1^2=M{k}^2-\rho k^2{c}^2$ and $b_2^2+k^2=k^2(2-c^2/c_T^2)$. Removing the common nonzero factors yields the second equation. ThusA nonzero displacement field requires the determinant to vanish:Let $s=c^2/c_T^2$ and $\eta =c_T^2/c_L^2$. Cancelling $k^4$ gives the Rayleigh secular equationSquaring 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 $c\simeq 0.9194c_T$. The surface-wave speed is independent of $k$ in this homogeneous half-space model.