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The local wave energy estimate is, for $T\ge 0$, $r>0$, and any center $x_0$,To prove it, let $R=r+T$ and integrate the local energy estimate identity ${\partial }_{t}e=\mathrm{div}(u_t\nabla u)$, where $e=(u_t^2+|\nabla u{|}^2)/2$, over the shrinking ball $B_{R-t}(x_0)$. Differentiation of this moving-domain integral yieldsIntegrating in time proves the local wave energy estimate, without assumptions at spatial infinity.

Let the union of the initial supports be a compact set $K$. If $\mathrm{dist}(x_0,K)>T$, choose $r>0$ with $r+T<\mathrm{dist}(x_0,K)$. The initial energy on $B_{r+T}(x_0)$ vanishes. Applying the shrinking-ball identity up to every intermediate time shows $u_t$ and $\nabla u$ vanish throughout that cone. In particular, along the vertical segment through $x_0$, $u_t=0$; its initial value is also zero, so $u(T,x_0)=0$. This last value check removes the constant ambiguity invisible to gradient energy.

Consequently the finite propagation speed conclusion isThis set is compact for each finite $t$. The speed is at most one in these units, or $c$ for $u_{tt}-c^2\Delta u=0$. Time reversal gives the same conclusion for negative time.

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.

We prove finite propagation speed using a shrinking cone energy argument. Fix $T>0$ and $x_{\ast }$, and suppose the Cauchy data vanish on $B(x_{\ast },T)$. For $0\le s<T$, define the local wave energyThe homogeneous wave equation gives the local conservation lawDifferentiate the integral over the moving ball. Its boundary moves inward with speed one, so the divergence theorem givesHere $n$ is the outward unit normal, ${\partial }_n$ the normal derivative, and ${\nabla }_{\tan }$ the component of the gradient tangent to the boundary. Since $E(0)=0$ and $e\ge 0$, we have $E(s)=0$. Hence both ${\varphi }_t$ and $\nabla \varphi$ vanish inside the backward light cone. Integrating ${\varphi }_t(s,x_{\ast })=0$ from the zero initial displacement gives $\varphi (s,x_{\ast })=0$; continuity then gives $\varphi (T,x_{\ast })=0$.

Applying the same energy estimate to the difference of two solutions proves the domain of dependence assertion. If $\mathrm{dist}(x_{\ast },K)>T$, the initial ball misses $K$, and the preceding argument proves the stated support bound. No disturbance propagates faster than one.

Use the radial reduction of the three-dimensional wave equation $w=r\varphi$ and extend $w$ oddly in $r$. Write $w=w_h+w_F$, where $w_h$ is the homogeneous wave equation solution with the prescribed Cauchy data, and $w_F$ has zero Cauchy data. With $w_0(r)=r{\varphi }_0(r)$ and $w_1(r)=r{\varphi }_1(r)$, the D'Alembert formula givesBoth odd initial profiles have compact support, soThe Duhamel principle, applied to $w_{tt}-w_{rr}=-rF$, givesNow write $r=t+a$ with $1/2\le a\le 1$. The lower integration limit is $s+a>0$, so this domain of dependence never crosses the axis. Intersecting its integration interval with the support of the source leaves a subset of $[s+a,s+1]$, of length at most one. On this interval,The outgoing shell source estimate for a radial wave therefore yieldsFinally $r\ge t+1/2\ge (1+t)/2$, and thusThis constant depends only on the Cauchy data and the fixed unit bound for the source. The logarithm comes from integrating the shell's accumulated forcing, whereas the factor $(1+t{)}^{-1}$ comes from division by $r$.

We establish a quantitative almost global existence for wave equations estimate. Take $0<\epsilon \le 1$, fix an integer $m\ge 6$, and use the commutation vector fields for the wave equation from the preceding part. Put $\partial =({\partial }_t,\nabla )$ and define the commuted wave energyAll these L2 norms are finite on any smooth existence interval by finite propagation speed. At $t=0$, the polynomial coefficients of the vector fields are bounded on the fixed compact support of the Cauchy data. Whenever a higher time derivative occurs, use ${\varphi }_{tt}=\Delta \varphi -{\varphi }_t^2$ and its differentiated versions to express it in terms of initial spatial derivatives. Every term contains at least one factor of $\epsilon$; consequentlyfor a constant depending only on finitely many derivatives and the support radius of ${\varphi }_0,{\varphi }_1$.

The commutators $[Z,{\partial }_{\mu }]$ are constant linear combinations of translations. Together with $[\square ,S]=2\square$ and the Leibniz rule, this shows that each commuted source is a finite linear combination of productsThis statement includes the extra copies of the original source produced by the scaling vector field. In each product put the factor with fewer commutations in the Lp norm $L^{\infty }$ and the other in the L2 norm. The lower order is at most $\lfloor m/2\rfloor$. Applying the Klainerman-Sobolev inequality to ${\partial }_{\mu }{Z}^J\varphi$ costs at most two additional commutations; commuting those past ${\partial }_{\mu }$ introduces only lower-order translations. Since $\lfloor m/2\rfloor +2\le m$,The inhomogeneous wave energy estimate now givesLet $T_{\epsilon }={\epsilon }^{-N}$. Use a bootstrap argument with $A_m(t)\le 2C_0\epsilon$ up to the smaller of $T_{\epsilon }$ and the maximal existence time. The energy estimate improves this toFor each fixed $N$,Choose ${\epsilon }_N\le 1$ so that $4C_m{C}_0\epsilon \log (1+{\epsilon }^{-N})\le 1/2$ for every $0<\epsilon \le {\epsilon }_N$. Then $A_m(t)\le (3/2)C_0\epsilon$, a strict improvement. A continuity argument closes the bootstrap argument.

Finally, the translation terms in $A_m$ control ordinary spatial Sobolev norms of $\partial \varphi$. The missing L2 norm of $\varphi$ satisfiesThus the full local-existence Sobolev norms remain bounded on this finite interval. The smooth continuation criterion for semilinear wave equations extends the solution past any finite endpoint before $T_{\epsilon }$. To see smooth persistence explicitly, the tame Sobolev product estimate gives $\Vert {\varphi }_t^2{\Vert }_{H^k}\le C_k\Vert {\varphi }_t{\Vert }_{\infty }\Vert {\varphi }_t{\Vert }_{H^k}$. Ordinary differentiated wave energy estimates therefore bound each higher derivative energy by its initial value times $\exp (C_k{\int }_0^t\Vert {\varphi }_t(s){\Vert }_{\infty }ds)$. This is finite on the interval already controlled by the base commuted wave energy; no separate ${\epsilon }_N$ is needed for each derivative order. ThereforeFor $\epsilon =0$ the zero solution is global. The same energy estimate in fact permits an exponential lower bound for the lifespan, which is stronger than any fixed inverse power.

Assume $\omega \ne 0$. With $h=0$, the method of characteristics gives $f(t,\Phi (t,0,z))=f_0(z)$. HenceLet $K=\mathrm{supp}{f}_0$ and $E_{\ast }={\max }_K{H}_{\omega }$; if $f_0=0$, the conclusion is immediate. By the Hamiltonian energy balance, every image point of $K$ remains in the energy sublevelThis is a fixed compact ellipsoid in phase space. Therefore the support bound is uniform in time:This uniform support bound from a coercive conserved energy requires no explicit solution of the curves.

The nonzero-frequency qualification is necessary. At $\omega =0$ the energy does not control $x$, and the equation is free transport equation. Choose a smooth compactly supported $f_0$ that is nonzero at $(x_0,v_0)$ with $v_0\ne 0$. Its transported value at $(x_0+t{v}_0,v_0)$ stays nonzero, so the union of the supports is unbounded. Each individual support is compact, but there is no fixed compact support for all times.

For a sequence of functions $f_n:X\to \mathbb{R}$, pointwise convergence to a function $f:X\to \mathbb{R}$ meansIt is the ordinary limit of a sequence at each fixed input. The threshold $N=N(x,\epsilon )$ may depend on the input. Uniform convergence instead requires one threshold for the entire domain of a function.^[ref]

Consider the polynomial functions $f_n(x)=x^n$, for positive integers $n$, on the interval $[0,1]$. For $0\le x<1$ the geometric sequence $(x^n)$ tends to zero, while $f_n(1)=1$. ThusEach $f_n$ is a continuous function, but its pointwise limit is discontinuous at one. Holding an input still hides what happens near that endpoint. The analogous even-power example on $[-1,1]$ appears in the source.^[ref]

For $0<x<1$ and $0<\epsilon <1$, taking the logarithm of $x^n<\epsilon$ and dividing by the negative real number $\log x$ reverses the inequality:The floor function followed by adding one handles the strict inequality even when the ratio is an integer. For tolerance $\epsilon =0.01$, the first successful indices at $x=0.5$, $0.8$, and $0.99$ are respectively $7$, $21$, and $459$. At $x=0.5$, $2^{-6}=0.015625$ fails and $2^{-7}=0.0078125$ succeeds. At $x=0$ and $x=1$, the error from this pointwise limit is already zero, so $N=1$ works. Nevertheless, $N(x,\epsilon )$ grows without bound as $x\to 1$ from below. A finite list of accurate test inputs cannot establish uniform convergence.

An even sharper warning concerns integration. On $[0,1]$, define the continuous functionsTheir supports are $[0,1/n]$, with height $n$ at $1/(2n)$ and value zero at both ends. Every fixed $x>0$ eventually lies beyond the support, and $g_n(0)=0$ always, so pointwise convergence gives the zero function. Nevertheless, each integral is the area of a triangle: ${\int }_0^1{g}_n(x)dx=\frac{1}{2}(1/n)n=\frac{1}{2}$. Shrinking width compensates increasing height. Hence the limit of a sequence of these integrals differs from the integral of the pointwise limit; interchanging them requires additional conditions, such as uniform convergence preserves integrals on a compact interval.