= Solution
Use fixed spatial coordinates in a stationary heterogeneous Earth and define the operator
$$
(\mathcal L u)_i=\rho(\mathbf x)\partial_t^2u_i
-\partial_{x_j}[c_{ijkl}(\mathbf x)\partial_{x_l}u_k].
$$
Here the <elastic stiffness tensor> has its minor and major symmetries, and the physical <body force> per unit volume is $q_i=\rho f_i$. The retarded <elastodynamic Green tensor> $G_i^{\ j}(\mathbf x,\boldsymbol\xi,t)$ is the displacement in direction $i$ due to a unit physical force impulse in direction $j$ at $\boldsymbol\xi$:
$$
\mathcal L_{ik}G_k^{\ j}
=\delta_i^{\ j}\delta(\mathbf x-\boldsymbol\xi)\delta(t),\qquad
G_i^{\ j}=0\quad(t<0).
$$
It satisfies the actual Earth's homogeneous <boundary conditions>, including zero <traction> on the free surface and continuity of displacement and <traction> at bonded internal interfaces. Integration through $t=0$ gives $\rho(\mathbf x)\partial_tG_i^{\ j}(0^+)=\delta_i^{\ j}\delta(\mathbf x-\boldsymbol\xi)$, with $G(0^+)=0$. No spherical symmetry is required.
For zero <initial conditions>, form the <convolution>
$$
\boxed{u_i(\mathbf x,t)=\int_0^t\!d\tau\int_V
G_i^{\ j}(\mathbf x,\boldsymbol\xi,t-\tau)\rho(\boldsymbol\xi)f_j(\boldsymbol\xi,\tau)\,dV_\xi.}
$$
Applying $\mathcal L$ to this expression uses the defining <Dirac delta function> to give $\rho(\mathbf x)f_i(\mathbf x,t)$. Causality gives zero <initial conditions>, and each Green column supplies the required <boundary conditions>. The positive-energy uniqueness argument then identifies this expression with the physical response. The source factor is $\rho(\boldsymbol\xi)$ because $f$ is force per unit mass, whereas the Green tensor was normalized to unit force.
A force dipole is a limiting pair of opposite point forces separated by a small vector. Put $+F_j$ at $\boldsymbol\xi+\varepsilon\mathbf d/2$ and $-F_j$ at $\boldsymbol\xi-\varepsilon\mathbf d/2$, and hold $D_{jk}=\varepsilon F_jd_k$ fixed as $\varepsilon\to0$. Expansion of the two <Dirac delta functions> gives
$$
q_j(\boldsymbol\eta,t)=-D_{jk}(t)\partial_{\eta_k}\delta(\boldsymbol\eta-\boldsymbol\xi).
$$
Its resultant force is zero, and <integration by parts> gives its radiated displacement as $\int_0^t D_{jk}(\tau)\partial_{\xi_k}G_i^{\ j}(\mathbf x,\boldsymbol\xi,t-\tau)\,d\tau$. Thus the source derivative of the <elastodynamic Green tensor> represents a dipole with force direction $j$ and separation direction $k$.
For the surface source choose a normal $\mathbf n$ from the minus side to the plus side and define
$$
b_i=[u_i]=u_i^+-u_i^-,\qquad [t_i]=(\sigma_{ij}^+-\sigma_{ij}^-)n_j.
$$
The same normal is used in both tractions. If the elastic coefficients are continuous through the surface, the distributional gradient of a piecewise smooth displacement is $\partial_j u_i=(\partial_j u_i)_{\mathrm{reg}}+b_i n_j\delta_\Sigma$. Its <stress> therefore has a singular part $c_{ijkl}b_k n_l\delta_\Sigma$. Taking the divergence includes the regular <stress> jump, so a field satisfying the source-free equations on both sides has equivalent source
$$
q_i=-[t_i]\delta_\Sigma-\partial_j[c_{ijkl}b_k n_l\delta_\Sigma].
$$
In particular, the signs follow from the chosen jump convention, not from an unspecified orientation. The <elastodynamic surface-jump representation> is consequently
$$
\boxed{u_i=\int_0^t\!d\tau\int_\Sigma
\left[-G_i^{\ j}[t_j]+M_{jk}\partial_{\xi_k}G_i^{\ j}\right]dS_\xi,
\qquad M_{jk}=c_{jk\ell m}b_\ell n_m,}
$$
with the Green tensor evaluated at $(\mathbf x,\boldsymbol\xi,t-\tau)$. The first term is a surface distribution of forces, and the second is a surface distribution of dipoles.
One can derive the same formula without multiplying discontinuous material coefficients by a surface delta. Apply <integration by parts> on the two sides of $\Sigma$ to the field $u$ and the reciprocal Green field $v$. Major symmetry cancels the volume <strain> terms; the spatial identity is
$$
\int (v_i\mathcal L u_i-u_i\mathcal L v_i)\,dV
=\int_{\partial V}[u_i t_i(v)-v_i t_i(u)]\,dS
$$
when the temporal terms are handled by the retarded time <convolution>. More explicitly, take $v_i(\boldsymbol\xi,\tau)=G_i^{\ a}(\boldsymbol\xi,\mathbf x,t-\tau)$, extend the zero-initial-data solution causally to $\tau<0$, and integrate over all source times. Integration by parts cancels the two temporal terms, with zero endpoint terms; the causal supports subsequently reduce the representation to $0\leq\tau\leq t$. The Green impulse contributes $-u_a(\mathbf x,t)$ on the left. The two outward face normals are $\mathbf n$ and $-\mathbf n$, so their sum on the right is $-[u_i]t_i(v)+v_i[t_i]$. Moving it to the other side gives $[u_i]t_i(v)-v_i[t_i]$. The same identity applied to two force Green fields with interchanged source and receiver proves the reciprocal relation $G_i^{\ j}(\mathbf x,\boldsymbol\xi,t)=G_j^{\ i}(\boldsymbol\xi,\mathbf x,t)$; hence the displayed surface formula follows. On a bonded material interface the Green displacement and common Green <traction> are continuous: $[u_i]t_i(v)$ is then the unambiguous version of the dipole term, evaluated with the matching one-sided coefficients. External free-surface terms vanish because both tractions vanish.
For a fault in <isotropic> material, the <Lamé parameters> give
$$
c_{jk\ell m}=\lambda\delta_{jk}\delta_{\ell m}
+\mu(\delta_{j\ell}\delta_{km}+\delta_{jm}\delta_{k\ell}),\qquad
M_{jk}=\lambda(\mathbf b\cdot\mathbf n)\delta_{jk}
+\mu(b_jn_k+n_jb_k).
$$
Tangential slip obeys $\mathbf b\cdot\mathbf n=0$, and ordinary fault <traction> is continuous across the two faces. Therefore \b[the source is a <double-couple fault source>] with
$$
\boxed{M_{jk}=\mu(b_jn_k+n_jb_k),\qquad [t_j]=0.}
$$
For a uniform patch of area $A$ with $\mathbf b=b\mathbf e_1$ and $\mathbf n=\mathbf e_3$, its integrated <seismic moment tensor> is
$$
\mathsf M=\begin{pmatrix}0&0&M_0\\0&0&0\\M_0&0&0\end{pmatrix},\qquad M_0=\mu A b.
$$
The $13$ dipole is a force in the slip direction separated in the normal direction; the $31$ dipole is a normal force separated in the slip direction. Their equal and opposite <torques> cancel. The symmetric trace-free <seismic moment tensor> has <eigenvalues> $M_0,-M_0,0$, with zero resultant force and <torque>. A temporal <stress> drop on a fault is not a jump of <traction> between its two simultaneously matching faces; it does not add a single-force layer to this tangential-slip representation.
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