Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 78 / 4

Let $d(\mathbf{y},t)=[u^{[n]}]$ denote the signed opening, positive when the faces separate, and let $S$ be the crack surface. Its displacement jump is $\mathbf{b}=d\mathbf{n}$. The singular strain associated with this jump has moment density $c_{ijkl}{b}_k{n}_l$. For the isotropic elastic stiffness tensor, contraction givesThe equivalent distributional body force is the negative divergence of this prescribed moment density:The resulting forced elastic wave in an isotropic solid equation is $\rho u_{i,tt}={\partial }_j{\sigma }_{ij}+f_i$. This is a tensile-crack seismic source; it need not be represented as an ordinary smooth volume force. For a small nearly planar patch of area $S_0$, uniform normal $\mathbf{n}$ and opening $d(t)$, define its volume change $V(t)=S_{0}d(t)$. Its seismic moment tensor and point-source force areThe sign convention makes a positive opening a positive extensional moment. Reversing the convention for the displacement jump changes both signs together; the force is independent of reversing the orientation of the crack normal.

Complete the crack normal to an orthonormal basis $({\mathbf{e}}_1,{\mathbf{e}}_2,\mathbf{n})$ with the first two vectors in the crack plane. The seismic moment tensor is diagonal in this basis:Each term gives an elastic force dipole, $f_i=-D{e}_i{e}_j{\partial }_j\delta$: two equal and opposite forces in direction $\mathbf{e}$, separated along $\mathbf{e}$, with moment $D$. Hence there are two identical in-plane dipoles of strength $\lambda V$ and one normal dipole of strength $(\lambda +2\mu )V$. For positive opening and ordinary positive Lamé parameters, all are extensional and the normal one is larger by $2\mu V$.

More generally, stable isotropic linear elasticity requires $\mu >0$ and $3\lambda +2\mu >0$. These imply $\lambda +2\mu >|\lambda |$, so the normal dipole has greater magnitude even if an auxetic material has negative $\lambda$ and compressive in-plane dipoles. The point source has zero net force, because it differentiates a delta function, and zero net torque, because its moment tensor is symmetric. Its normal moment distinguishes opening from a pure shear dislocation.

Treat a small patch as the point tensile-crack seismic source at ${\mathbf{y}}_0$, with $M(t)=KV(t)$ and $K=\lambda I+2\mu \mathbf{n}{\mathbf{n}}^{\mathsf{T}}$. Let $\mathbf{e}=(\mathbf{x}-{\mathbf{y}}_0)/R$, $\gamma =\mathbf{e}\cdot \mathbf{n}$, $\mathbf{a}=K\mathbf{e}$, $m=\mathbf{e}\cdot K\mathbf{e}$ and $s=\mathrm{tr}K$. Source convolution and an integration by parts give the point-moment elastodynamic displacementThe derivatives act on angular factors and on retarded times as well as on powers of $R$; keeping only the explicit $1/R$ Green-function terms would miss the static displacement.

Set $V_P=V(t-R/\alpha )$, $V_S=V(t-R/\beta )$ andConvolving the elastodynamic Green tensor with the opening history givesUse ${\partial }_{j}R=e_j$, ${\partial }_j{e}_i=({\delta }_{ij}-e_i{e}_j)/R$, ${\partial }_j{V}_P=-e_j{\dot{V}}_P/\alpha$, andFor example, $K_{kj}{\partial }_j(e_i{e}_k/R)=[a_i+e_i(s-3m)]/R^2$, while differentiating the near-field angular tensor gives $[6a_i+3e_{i}s-15e_{i}m]/R^4$. Combining these with the endpoint terms in ${\partial }_{R}I$ yieldsFor the normal crack, $\mathbf{a}=\lambda \mathbf{e}+2\mu \gamma \mathbf{n}$, $m=\lambda +2\mu {\gamma }^2$, and $s=3\lambda +2\mu$. Explicitly, the three nonradiative vector coefficients in the last two lines areThus those lines are ${\mathbf{C}}_P{V}_P/({\alpha }^2{R}^2)+{\mathbf{C}}_S{V}_S/({\beta }^2{R}^2)+{\mathbf{C}}_{I}I/R^4$. The small-source approximation assumes the patch size is small relative to both $R$ and the wavelengths of interest; a finite crack is obtained by integrating the same point-patch expression over its surface with its local opening and normal.

The first two terms are the radiative P wave and S wave displacements:They decay as $R^{-1}$ and are purely transient: once the opening rate vanishes, each disappears after its respective arrival time. The P displacement is longitudinal and the S displacement transverse. An abrupt opening history makes these formulas distributions; a smooth finite-duration intrusion gives ordinary pulses.

Suppose the source starts from zero opening and ends with volume change $V_0$. At times later than the end of intrusion plus $R/\beta$, both retarded openings equal $V_0$ andThe remaining terms combine to the static displacement fieldIt decays as $R^{-2}$, so it must not be identified with the transient far-field pulses.

For a Poisson solid, $\lambda =\mu$, ${\alpha }^2=3{\beta }^2$ and $\mu =\rho {\beta }^2$. Let $\theta$ be polar angle from the crack normal, with unit vector ${\mathbf{e}}_{\theta }$ in the increasing-$\theta$ direction. Then $\gamma =\cos \theta$ and $\mathbf{n}-\gamma \mathbf{e}=-\sin \theta {\mathbf{e}}_{\theta }$. The tensile-crack radiation pattern becomesAt fixed $R$ and positive opening rate, the P pattern is outward, has no angular node, and is three times larger along the normal than in the crack plane. The signed S amplitude has four alternating lobes: it vanishes along the normal and in the crack plane, and its magnitude is maximal at $45^{\circ }$ to them. Neither field has an azimuthal SH component about the crack normal.

The final static field simplifies toIts radial part is outward near the poles, changes sign at ${\cos }^2\theta =1/8$, and is inward in the crack plane. The tangential part has the same signed angular factor as the S radiation, but is permanent and has a different distance dependence. Its radial sign changes are not nodes of the whole vector because the tangential component generally remains nonzero there. Rotating a meridional sketch around the crack normal gives the complete three-dimensional axisymmetric pattern.

The arrows show displacement at points on a circle of constant observation radius. Each panel uses its own scale; the separate formulas retain the physical P/S prefactors and arrival times. The normal is vertical and the crack lies horizontally.