Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 1 / 37E / Solution

The isotropic elastic displacement equation is $\rho \ddot{u}=(\lambda +\mu )\nabla (\nabla \cdot u)+\mu {\nabla }^{2}u$. With $u=\nabla \varphi +\nabla \times \mathbf{\psi }$, its dilational and shear potentials satisfyFor the specified two-dimensional potential, $u_x={\varphi }_x+{\psi }_y$, $u_y={\varphi }_y-{\psi }_x$.

Choose a real wave number $k>0$ and put $\omega =kc$. Decay into $y<0$ requires the combinationswith $0<c<c_S$. Real parts give the physical fields. Then $u_x=ikA{e}^{py}+qB{e}^{qy}$ and $u_y=pA{e}^{py}-ikB{e}^{qy}$, 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 becomeA nonzero pair exists precisely whenFor such a root, one can choose $A\ne 0$ and $B=-2ikpA/(k^2+q^2)$, providing the requested wave explicitly. Squaring and excluding the static zero root gives the equivalent Rayleigh polynomialThe physical positive root is unique as allowed in the question. All factors of $k$ cancel in the speed equation, so the Rayleigh surface wave is nondispersive, with $c$ determined only by the material constants. The potentials decay exponentially over depths of order $k^{-1}$.