= Solution
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 $c_{ijkl}b_kn_l$. For the isotropic <elastic stiffness tensor>, contraction gives
$$
m_{ij}(\mathbf y,t)=d(\mathbf y,t)\left[\lambda\delta_{ij}+2\mu n_i(\mathbf y)n_j(\mathbf y)\right].
$$
The equivalent distributional <body force> is the negative divergence of this prescribed moment density:
$$
\boxed{f_i(\mathbf x,t)=-\partial_{x_j}\int_S
d(\mathbf y,t)[\lambda\delta_{ij}+2\mu n_i(\mathbf y)n_j(\mathbf y)]
\delta^{(3)}(\mathbf x-\mathbf y)\,dS_{\mathbf y}}.
$$
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 $V(t)=S_0d(t)$. Its <seismic moment tensor> and point-source force are
$$
\boxed{M_{ij}(t)=V(t)(\lambda\delta_{ij}+2\mu n_in_j),\qquad
f_i=-M_{ij}(t)\partial_{x_j}\delta^{(3)}(\mathbf x-\mathbf y_0)}.
$$
The 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.
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