= Solution
In the sense of <distribution theory>, $H'(t)=\delta(t)$. Since $\sin(\alpha t)$ vanishes at zero,
$$
\psi'(t)=H(t)\cos(\alpha t),
\qquad
\psi''(t)=\delta(t)-\alpha H(t)\sin(\alpha t),
$$
and therefore
$$
\boxed{\psi''+\alpha^2\psi=\delta(t)}.
$$
Let $G_{\rm ret}(\mathbf x,t;\mathbf y)$ solve
$$
(\partial_t^2-c^2\nabla^2)G_{\rm ret}
=\delta(t)\delta^{(3)}(\mathbf x-\mathbf y),
\qquad G_{\rm ret}=0\quad(t<0).
$$
The spatial <Fourier transform> turns this into the preceding oscillator equation with $\alpha=ck$, so
$$
\widehat G(\mathbf k,t)=H(t)\frac{\sin(ckt)}{ck}.
$$
Inverting with the supplied radial integral gives the three-dimensional <retarded Green function>
$$
\boxed{G_{\rm ret}(\mathbf x,t;\mathbf y)
=\frac{H(t)}{4\pi c|\mathbf x-\mathbf y|}
\delta(|\mathbf x-\mathbf y|-ct)
=\frac{\delta(t-|\mathbf x-\mathbf y|/c)}{4\pi c^2|\mathbf x-\mathbf y|}}.
$$
For zero initial displacement and initial velocity $f$, convolution with this kernel gives the <Kirchhoff formula>
$$
\begin{aligned}
u(\mathbf x,t)
&=\int_{\mathbb R^3}G_{\rm ret}(\mathbf x,t;\mathbf y)f(\mathbf y)\,d^3y\\
&=\frac{t}{4\pi}\int_{S^2}f(\mathbf x+ct\,\boldsymbol\omega)\,d\Omega
=t\langle f\rangle_t.
\end{aligned}
$$
\b[Thus the value at $(\mathbf x,t)$ depends only on the initial data on the sphere of radius $ct$, not on its interior. This is the <Strong Huygens principle>[Huygens principle] and describes propagation at the finite wave speed $c$.]
Back to article page
