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

Substituting the sinusoidal plane wave into the Navier-Cauchy equation gives $\Delta u=-u$, $u_{tt}=-c^{2}u$, and $\nabla \mathrm{div}u=-k(k\cdot a)\sin (k\cdot X-ct)$ because $|k|=1$. Equality of the amplitudes is thereforeThis is an eigenvalue problem for the acoustic tensor $\mu I+(\lambda +\mu )k\otimes k$. Choosing a nonzero amplitude parallel to $k$ gives $\rho c^2=\lambda +2\mu$; choosing any nonzero amplitude perpendicular to $k$ gives $\rho c^2=\mu$. For positive mass density and stable Lamé parameters, both squared speeds are positive. These choices construct nonzero solutions explicitly, proving existence.

Resolve each amplitude by the orthogonal projectionsThe acoustic tensor acts by $\lambda +2\mu$ on the longitudinal line and by $\mu$ on its perpendicular plane. ThusThe longitudinal component is a longitudinal wave, while the two independent transverse polarizations are transverse waves. Each projection obeys its own wave equation, and superposition adds the corresponding solutions.

A general planar disturbance is therefore a sum of the two types. For a single sinusoid with a common speed, both projections cannot be nonzero unless the two speeds coincide: projecting the eigenvalue equation gives $(\rho c^2-\lambda -2\mu )a_L=0$ and $(\rho c^2-\mu )a_T=0$. This distinguishes superposition of waves with different speeds from a mixed polarization traveling rigidly at one speed.

Divide the Navier-Cauchy equation by mass density. Since $\mu /\rho =c_T^2$ and $(\lambda +\mu )/\rho =c_L^2-c_T^2$, it becomesEquivalently, using the vector calculus identity $\Delta u=\nabla \mathrm{div}u-\nabla \times (\nabla \times u)$,The divergence and curl consequently propagate at $c_L$ and $c_T$, respectively.

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 the $b_1$ root the elastic wave polarization is parallel to the complex derivative vector $(ik,-b_1)$, givingFor the $b_2$ root the field has zero divergence, $ik{A}_2-b_2{B}_2=0$, givingThese relations also follow by inserting each decay root into either row of the matrix. The complete decaying displacement field has complex componentsmultiplied by $e^{ik(X-ct)}$, followed by the real part. The first constituent is irrotational and the second is divergence-free; both are needed to cancel the two surface tractions.

The material occupies $Y\ge 0$, so its outward unit normal at the surface is $n=-e_Y$. A free surface has zero traction, $\tau n=0$. ThereforeThe third condition is automatic for the two-dimensional displacement field with zero third component and no dependence on the third coordinate. The first two impose zero shear and normal stress at the free surface.

Write the in-plane displacement field as $(u,v)$. The infinitesimal strain tensor gives $2e_{XY}=u_Y+v_X$, $e_{YY}=v_Y$, and $\mathrm{tr}e=u_X+v_Y$. The two nontrivial zero-traction conditions becomeEquivalently for the complex amplitudes they are $U^{\prime }(0)+ikV(0)=0$ and $ik\lambda U(0)+(\lambda +2\mu )V^{\prime }(0)=0$. Because the constitutive equation is real and linear, vanishing of these complex amplitudes enforces the conditions at every position and time.

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.