= Solution
A <causal Green function> vanishes before the impulse. For $t>\tau$ it solves the homogeneous <linear ordinary differential equation>, whose characteristic roots are $-q\pm2i$. Thus it has the form
$$
G(t,\tau)=e^{-q(t-\tau)}[A\cos2(t-\tau)+B\sin2(t-\tau)].
$$
Continuity at $t=\tau$ is necessary to avoid a delta derivative in $G_{tt}$, so $A=0$. Integrating its forced equation across the impulse gives the <jump condition for an impulse>, $G_t(\tau^+,\tau)-G_t(\tau^-,\tau)=1$. The derivative from below is zero and the derivative from above is $2B$, giving
$$
\boxed{G(t,\tau)=\frac12\Theta(t-\tau)e^{-q(t-\tau)}\sin[2(t-\tau)].}
$$
Here $\Theta$ is the <Heaviside step function>. The zero initial data exclude any additional homogeneous response. By the <superposition principle>, the output for a general locally integrable input is
$$
\boxed{x(t)=\int_0^tG(t,\tau)f(\tau)\,d\tau
=\frac12\int_0^te^{-q(t-\tau)}\sin[2(t-\tau)]f(\tau)\,d\tau.}
$$
The upper endpoint causes no unwanted term in the first derivative because $G(t,t)=0$; in the second derivative its unit derivative jump reproduces the input.
Back to article page
