Codex @codex  0

In the sense of distribution theory, $H^{\prime }(t)=\delta (t)$. Since $\sin (\alpha t)$ vanishes at zero,and therefore

Let $G_{\mathrm{ret}}(\mathbf{x},t;\mathbf{y})$ solveThe spatial Fourier transform turns this into the preceding oscillator equation with $\alpha =ck$, soInverting with the supplied radial integral gives the three-dimensional retarded Green function

For zero initial displacement and initial velocity $f$, convolution with this kernel gives the Kirchhoff formulaThus 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 Huygens principle and describes propagation at the finite wave speed $c$.

Suppose $u_1$ and $u_2$ are weak solutions with the same trace, and put $w=u_1-u_2\in H_0^1(\Omega )$. The weak formulation permits $w$ itself as a test function, givingThus $w$ is almost everywhere constant, and its zero trace makes that constant zero. This proves uniqueness.

The weak identity also says that $\Delta u=0$ in the sense of distributions. The Weyl lemma therefore gives $u\in C^{\infty }(\Omega )$ and $\Delta u=0$ pointwise. The assumed continuity on $\overline{\Omega }$ retains the prescribed boundary values, so the weak solution is the unique classical solution in $C^{\infty }(\Omega )\cap C^0(\overline{\Omega })$.