= Solution
The first two terms are the radiative <P wave> and <S wave> displacements:
$$
\boxed{\mathbf u_P^{\rm rad}=\frac{\lambda+2\mu\gamma^2}{4\pi\rho\alpha^3R}\,\mathbf e\,\dot V(t-R/\alpha),\qquad
\mathbf u_S^{\rm rad}=\frac{2\mu\gamma}{4\pi\rho\beta^3R}(\mathbf n-\gamma\mathbf e)\dot V(t-R/\beta)}.
$$
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 $V_0$ and
$$
I=\frac{V_0R^2}{2}\left(\frac1{\beta^2}-\frac1{\alpha^2}\right).
$$
The remaining terms combine to the static <displacement field>
$$
\boxed{\mathbf u_{\rm stat}=\frac{V_0}{4\pi\rho R^2}
\left\{\left[\frac{\lambda+\mu-3\mu\gamma^2}{\alpha^2}
+\frac{\mu(3\gamma^2-1)}{\beta^2}\right]\mathbf e
+\frac{2\mu\gamma}{\alpha^2}\mathbf n\right\}}.
$$
It 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> becomes
$$
\boxed{\mathbf u_P^{\rm rad}=\frac{\dot V_P}{12\pi\sqrt3\,\beta R}(1+2\cos^2\theta)\mathbf e,\qquad
\mathbf u_S^{\rm rad}=-\frac{\dot V_S}{4\pi\beta R}\sin2\theta\,\mathbf e_\theta}.
$$
At 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 to
$$
\boxed{\mathbf u_{\rm stat}=\frac{V_0}{12\pi R^2}
[(6\cos^2\theta-1)\mathbf e+2\cos\theta\mathbf n]
=\frac{V_0}{12\pi R^2}[(8\cos^2\theta-1)\mathbf e-\sin2\theta\mathbf e_\theta]}.
$$
Its 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.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2003/iii/paper-78-tensile-crack-patterns.png]
{title=Transient P and S displacement patterns and final static displacement for a tensile crack in a Poisson solid}
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.
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