Past exam of the elasticity and plasticity course of the University of Oxford / 2010 / 2

Assume $\mathbf{k}\ne 0$. Dotting the decomposition with $\mathbf{k}$ givesCrossing it on the left by $\mathbf{k}$ and using $\mathbf{B}\cdot \mathbf{k}=0$ gives $\mathbf{k}\times \mathbf{a}=|\mathbf{k}{|}^2\mathbf{B}$. ThusIndeed $(\mathbf{k}\times \mathbf{a})\times \mathbf{k}/|\mathbf{k}{|}^2=\mathbf{a}-A\mathbf{k}$, verifying the decomposition and the imposed transverse condition.

Apply the wave ansatz to the oscillatory displacement about a static body-force balance, or equivalently to the zero-body-force homogeneous equation. The parallel and perpendicular polarizations have speedsThe first is longitudinal and the second transverse. A wave family is dispersive when its phase speed $\omega /|\mathbf{k}|$ varies with wavenumber. In each of these homogeneous bulk families, $\omega =c|\mathbf{k}|$, so both are nondispersive and group and phase speeds coincide.

The positivity assumptions imply $\lambda >-2\mu /3$, hence $\lambda +\mu >\mu /3>0$. Therefore $c_p^2-c_s^2=(\lambda +\mu )/\rho >0$: longitudinal waves travel faster.

Substitution of $w_i=f_i(y)e^{i(kx-\omega t)}$ givesThe oscillatory seam solution requires $m^2={\omega }^2/c_{s1}^2-k^2$, givingFor an exponentially decaying exterior, $f_2^{\prime \prime }={\ell }^2{f}_2$ with $\ell >0$, so the same differential equation instead requiresThe printed exterior plus sign is inconsistent with its decaying ansatz and must be a sign error. The guided mode has $c_{s1}<c_L<c_{s2}$, so a slower shear-wave seam than its cladding is necessary.

For an even seam mode $f_1=A\cos (my)$, upper-interface displacement and traction continuity giveThe lower interface gives the same equations when $B_2=A_2$. Eliminating the amplitudes yieldsFor an odd seam mode $f_1=B\sin (my)$, the equations areWith $B_2=-A_2$ the lower condition agrees, and elimination givesThese interface conditions together with the two corrected dispersion identities determine the guided-wave branches.

In the formal stiff-cladding limit with a nonvanishing exterior decay rate, the interface displacement tends to zero. The even seam mode then has $\cos (mh)=0$, whereas the odd mode has $\sin (mh)=0$. The positive wavenumbers areThe zero odd index would give an identically zero mode and is excluded. The smallest even transverse wavenumber is $\pi /(2h)$, so its lowest branch satisfiesThis is the symmetric-mode cutoff: the actual limiting branches have the discrete listed values of $m$. If the exterior shear speed remains finite, the separate decay requirement ${\omega }^2<c_{s2}^2{k}^2$ still applies; the cutoff alone does not replace that guided-wave condition.