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For the wave equation $u_{tt}=c^2{u}_{xx}$ with zero initial displacement and initial velocity $g$, D'Alembert's formula becomesFor $c=1$ this is the original speed-one form $u(t,x)=\frac{1}{2}{\int }_{x-t}^{x+t}f(s)ds$ when the initial velocity is named $f$. The integral records the total initial kick inside the domain of dependence.

For a computed example, set $g(x)=V$ on $[-a,a]$ and zero elsewhere, with $V>0$. This discontinuous datum gives a weak solution:At the centre, $u(0,t)=V\min (t,a/c)$. For $a=1\mathrm{m}$, $c=2\mathrm{m}{\mathrm{s}}^{-1}$ and $V=0.4\mathrm{m}{\mathrm{s}}^{-1}$, this is $0.1\mathrm{m}$ at $t=0.25\mathrm{s}$ and $0.2\mathrm{m}$ at $t=1\mathrm{s}$. Once $ct>a$, a central plateau $u=Va/c$ occupies $|x|\le ct-a$. In contrast, central zero region for odd compactly supported initial velocity follows when the full integral of $g$ vanishes. Finite propagation speed permits an interior tail; it does not imply the Strong Huygens principle.

For the unforced, constant-$c$ one-dimensional wave equation, finite propagation speed says that changing Cauchy data outside $[x_{\ast }-c{t}_{\ast },x_{\ast }+c{t}_{\ast }]$ cannot change the solution at $(x_{\ast },t_{\ast })$, where $c>0$ and $t_{\ast }\ge 0$. This interval is the initial-data domain of dependence. Its endpoints are found by following the two characteristic lines for the one-dimensional wave equation backwards from the observation. D'Alembert's formula makes the dependence sharper: initial displacement is sampled at the endpoints, while initial velocity is integrated over the entire interval.^[ref]

For the synthetic choice $c=1$, $(x_{\ast },t_{\ast })=(0.5,1.5)$ has initial-data domain of dependence $[-1,2]$. A change confined near $x=3$ is irrelevant to that observation, however large it is. If both Cauchy data vanish outside $[-a,a]$, their compact support can influence only $[-a-ct,a+ct]$ at $t\ge 0$. Thus an observer at $x>a$ sees exactly zero displacement for $0\le t<(x-a)/c$. There is no nonzero precursor beyond this causal cone for the constant-speed wave equation in the ideal model. This is an exact statement about the partial differential equation, stronger than an experimental failure to detect a small signal.

Support widening for the wave equation is a bound on possible influence, not a claim that every point inside the expanding interval moves. A bump function released with zero initial velocity splits into two separated travelling waves; the middle can become empty. An initial kick can instead leave an interior tail, as in the D'Alembert formula with initial velocity. Consequently finite propagation speed does not imply the Strong Huygens principle. A finite string needs boundary conditions, and the whole-line formula applies locally only until its backward causal cone for the constant-speed wave equation encounters an endpoint.

For the three-dimensional wave equation $u_{tt}=c^2\Delta u$ on all of ${\mathbb{R}}^3$, with $c>0$, smooth initial displacement $f$ and initial velocity $g$, define the normalized spherical meanFor $t>0$, the Kirchhoff formula isThe factor $t$ and the normalization of the mean are essential. Equivalently, $t{M}_{ct}g=(4\pi c^{2}t{)}^{-1}{\int }_{|y-x|=ct}g(y)d{S}_y$. At $t=0$, the limiting value and time derivative recover $f,g$. For instance constant data give $u=f+tg$, which directly checks the coefficients.^[ref]

The solution uses spherical data rather than the full interior ball, giving the Strong Huygens principle. The differentiated displacement term also samples its normal variation, so this is not merely an average of two lists of pointwise boundary values. The method of descent for the wave equation changes the formula in two spatial dimensions to a disk integral with an interior tail.

For $t>0$, the three-dimensional Kirchhoff formula with zero displacement isThe Strong Huygens principle says that the value depends only on the initial velocity on the sphere of radius $t$, not on points strictly inside it. In particular a compactly supported disturbance leaves no tail at a fixed point after the sphere has passed the support.

For method of descent for the wave equation, extend the two-dimensional initial velocity independently of the third coordinate. Projecting the three-dimensional sphere onto its planar disk gives two sheets, each with area element $tdy/\sqrt{t^2-|y-x{|}^2}$. ThereforeThe local argument can use compact cutoffs in the extra coordinate, so global Schwartz decay there is unnecessary. The weak Huygens principle is finite propagation speed: only data in the closed disk of radius $t$ can influence the point. Unlike the three-dimensional free equation, the full interior contributes and a tail can persist.

For the equation with potential $|x{|}^2$, multiply by $u_t$ and use the product rule:This is the claimed local energy conservation with flux $p=-u_t\nabla u$.

Fix $(t_0,x_0)$ and let $E(t)$ be the integral of $e$ over the shrinking ball $|x-x_0|\le t_0-t$, $0\le t\le t_0-a$ with $a>0$. Its boundary moves inward at speed one. The divergence theorem givesbecause the negative integrand isAll initial data are zero, so $E(0)=0$. Nonnegativity then forces $E(t)=0$. Letting the small terminal ball approach the cone tip proves $u=0$ there; alternatively $u_t=0$ throughout the cone and the initial value fixes $u$. The same argument applied to a difference of solutions whose data agree on the initial ball proves local domain of dependence. Thus the potential equation has propagation speed at most one and satisfies the weak Huygens principle, without requiring finite total energy on all space. It does not assert the strong no-tail property.

For the wave equation with d'Alembert operator $\square =-{\partial }_t^2+\Delta$, the wave speed is one. The domain of dependence of $(t,x)$ is the initial ball $\overline{B}(x,t)$. More precisely, two solutions whose Cauchy data agree on a neighbourhood of that ball have the same value at $(t,x)$. Thus, if both Cauchy data have support in $K$, finite propagation speed givesThe Strong Huygens principle is sharper in three spatial dimensions: only the initial data on the sphere $|y-x|=t$ contribute, including the first spatial derivatives of the initial displacement there. The Kirchhoff formula makes this precise:In particular, if the Cauchy data have compact support in $\overline{B}(0,R)$, thenFor $t>R$, the solution therefore vanishes behind the inward edge $|x|=t-R$ as well as outside $|x|=t+R$. The absence of an interior tail is the additional content of the Strong Huygens principle.

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 $c$.

The Strong Huygens principle in three spatial dimensions says that the solution at $(t,x)$ depends only on the initial data on the sphererather than on the full ball bounded by that sphere. This follows immediately from the Kirchhoff formula and its time derivative. Consequently a disturbance has no tail inside the light cone: if the initial data are supported in a compact set $K$, then $u(t,x)=0$ whenever that sphere misses $K$. This is sharper than finite propagation speed, which only excludes influence from outside the ball.

The weak Huygens principle is the domain-of-dependence statement that initial data outside a backward light cone cannot influence its tip. It permits an interior tail and is weaker than the Strong Huygens principle. For $u_{tt}-\Delta u+V(x)u=0$ with $V\ge 0$, integrating the local energy over shrinking balls proves this statement: the boundary contribution is $u_t{\partial }_{\nu }u-\frac{1}{2}(u_t^2+|\nabla u{|}^2+V{u}^2)\le 0$.