= Solution
For a mode of positive <frequency>, differentiation of the causal <convolution> gives
$$
\dot a(t)=\int_0^t\cos[\omega(t-s)]f(s)\,ds,
$$
$$
\ddot a(t)=f(t)-\omega\int_0^t\sin[\omega(t-s)]f(s)\,ds
=f(t)-\omega^2a(t).
$$
The first boundary term vanishes because $\sin0=0$; the second supplies $f(t)$ because $\cos0=1$. Both initial values vanish. This verifies the forced-oscillator equation and its <initial conditions>, including a step source in the piecewise or distributional sense.
The spatial <Dirac delta function> reduces the excitation to
$$
f_\alpha(t)=C_\alpha H(t-t_s),\qquad
C_\alpha=\mathbf M:\mathbf e_\alpha^*(\mathbf x_s).
$$
For an event at $t_s\ge0$, put $\tau=t-t_s$. Integrating the oscillator kernel from $t_s$ to $t$ gives the <normal-mode response to a step seismic moment>:
$$
a_\alpha(t)=\frac{C_\alpha}{\omega_\alpha^2}
[1-\cos(\omega_\alpha\tau)]H(\tau),
$$
$$
\boxed{\mathbf u(\mathbf x,t)=H(\tau)\sum_\alpha
\mathbf u_\alpha(\mathbf x)
\frac{\mathbf M:\mathbf e_\alpha^*(\mathbf x_s)}{\omega_\alpha^2}
[1-\cos(\omega_\alpha\tau)].}
$$
Non-unit mass normalization introduces the same factor $1/N_\alpha$ as above. Complex modes must be combined with their conjugate partners to produce a real displacement. The formal $\omega\to0$ limit is $a=C\tau^2H(\tau)/2$; for rigid-body modes here, $C=0$ because their <strain> is zero.
The step produces a static component plus continuing oscillations in this lossless model. It does not predict physical damping. \b[A point source is appropriate when rupture dimensions are small compared with the <wavelengths> being analysed and rupture duration is short compared with their <periods>.] Long-period observations of a relatively small <earthquake> can meet these conditions; high-frequency finite-fault directivity generally cannot. A finite source-time function replaces the ideal instantaneous step when its duration matters.
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