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✓ 15 results · all verified · 6 also independently AI-judged

Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 9 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Wave Energy, Finite Propagation and Huygens' Principle

1 · Prerequisites

2 · Summary

This page develops the energy method for the wave equation $□_{c} u = \partial_{t}^{2} u - c^{2} Δ u = f$ in $R^{n}$ and uses it to prove uniqueness, continuous dependence, finite speed of propagation and the sharp dimension-dependent support of Huygens' principle. It begins with the energy density $\frac{1}{2} (u_{t}^{2} + c^{2} ∣ D u ∣^{2})$ and flux $- c^{2} u_{t} D u$ , the local balance law $\partial_{t} e + div q = f u_{t}$ , and the three admissible settings in which the total energy is constant: fixed compact spatial support, integrable flux, and homogeneous Dirichlet or Neumann data on a bounded $C^{1}$ domain (with the interval case in one dimension). The stability estimate $\sqrt{2 E (t)} \leq \sqrt{2 E (0)} + \int_{0}^{t} ∥ f (s) ∥_{2} d s$ records continuous dependence with the sharp $L_{t}^{1} L_{x}^{2}$ forcing constant. Energy uniqueness for the Cauchy problem, for the Dirichlet problem on bounded domains, and backwards in time from terminal data follows, with the zero-energy residual constant fixed by the displacement datum.

The causal half of the page defines the forward and backward cones and the domain of influence, proves the truncated-cone geometry and its outward normals $(\hat{x - x_{0}}, c) / \sqrt{1 + c^{2}}$ , and integrates the local law over a space-time frustum to obtain the cone energy identity, whose lateral flux is a sum of squares $\frac{c}{2} [(u_{t} - c \partial_{r} u)^{2} + c^{2} ∣ D_{\tan} u ∣^{2}]$ . Finite propagation speed, the expansion of compact support at speed at most $c$ , local uniqueness on the domain of dependence, and the formal definition of the strong Huygens principle follow. The principle is then proved for odd $n \geq 3$ by the odd-dimensional representation formula of the preceding page, and refuted for $n = 1$ and even $n$ by data supported strictly inside the base ball; a closing remark separates the shell statement of Huygens from the solid-cone bound of finite propagation, which holds in every dimension.

All statements of the page are read under the Axiom of Countable Choice, which supplies the piecewise $C^{1}$ divergence theorem and surface integrals; the pointwise one-dimensional differential computations require no choice, while the stated Lebesgue and surface integration arguments retain this assumption. The proofs use the representation formulas of wave-equation-representation-formulas and the published surface-measure, differentiation and integration results; the design's "depends on" is formalised by the local uniqueness theorem, not used as an undefined physical phrase.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (gpt-6.1-sol)Open item page →

Forward and backward wave cones, domain of dependence and influence

Definition

Let $n \geq 1$ , let $c > 0$ and let $x_{0} \in R^{n}$ , $t_{0} \geq 0$ . Write $∣ z ∣ : = ∥ z ∥_{2}$ for the Euclidean norm and $B_{r} (x_{0}) = {x \in R^{n} : ∣ x - x_{0} ∣ < r}$ , ${\overline{B}}_{r} (x_{0}) = {x \in R^{n} : ∣ x - x_{0} ∣ \leq r}$ for the open and closed balls, $r \geq 0$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ). Space-time points are written $(x, t) \in R^{n} \times R$ , with $t$ the time coordinate.

The (closed) backward cone with vertex $(x_{0}, t_{0})$ and speed $c$ is

$K^{-} (x_{0}, t_{0}) : = {(x, t) : 0 \leq t \leq t_{0}, ∣ x - x_{0} ∣ \leq c (t_{0} - t)} .$

Its base ball is $B_{c t_{0}} (x_{0}) \times {0}$ , and its lateral boundary, including the base rim and excluding the vertex, is ${(x, t) : 0 \leq t < t_{0}, ∣ x - x_{0} ∣ = c (t_{0} - t)}$ . The (closed) forward cone with vertex $(x_{0}, t_{0})$ is

$K^{+} (x_{0}, t_{0}) : = {(x, t) : t \geq t_{0}, ∣ x - x_{0} ∣ \leq c (t - t_{0})} .$

For data given on a set $S \subseteq R^{n}$ at time $0$ , the domain of influence of $S$ at time $t \geq 0$ is

$S + {\overline{B}}_{c t} (0) = {x + y : x \in S, ∣ y ∣ \leq c t},$

the set of points that data on $S$ can reach by time $t$ at speed at most $c$ .

The domain of dependence of $(x_{0}, t_{0})$ is not left as an undefined physical phrase: the causal content of the name is formalised by the local uniqueness theorem Domain of dependence and local uniqueness, which says, under Countable Choice and for $t_{0} > 0$ , that two $C^{2}$ solutions on a neighbourhood of the closed cone with the same Cauchy data on $B_{c t_{0}} (x_{0})$ and the same source on the cone $K^{-} (x_{0}, t_{0})$ agree at every point of $K^{-} (x_{0}, t_{0})$ (in particular at the vertex). Truncated cones $K (t_{1}, t_{2}) = {(x, t) : t_{1} < t < t_{2}, ∣ x - x_{0} ∣ < c (t_{0} - t)}$ with $0 < t_{1} < t_{2} < t_{0}$ , their piecewise $C^{1}$ presentation and their outward normals are those of Truncated wave cones: convexity, piecewise C1 presentation and outward normals.

The slope $c$ is part of the definition and not decoration: at $n = 1$ the backward cone is the characteristic triangle with the two characteristic lines $x = x_{0} \pm c (t_{0} - t)$ , and the open base ball is the interval $(x_{0} - c t_{0}, x_{0} + c t_{0})$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length). No propagation, uniqueness or support claim is asserted by this item; those are Finite propagation speed for the wave equation, Domain of dependence and local uniqueness and Compact support expands at speed at most c.

DefinitionDefinition: Literature-sourcedProof: Not applicableOpen item page →

Wave energy density, energy flux and total energy

Definition

Let $n \geq 1$ , $c > 0$ , let $U \subseteq R^{n}$ be open, let $I \subseteq R$ be an open interval and let $u : U \times I \to R$ be $C^{2}$ ( $C^{k}$ maps and multi-index derivative notation in Euclidean space). Write $u_{t} : = \partial_{t} u$ and let

$D u : = (\partial_{0} u, \dots, \partial_{n - 1} u)$

be the spatial gradient of The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case, a $C^{1}$ field $U \times I \to R^{n}$ whose Euclidean norm is $∣ D u ∣$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

The kinetic density, potential density and energy density of $u$ are the continuous functions

$e_{k i n} : = \frac{1}{2} u_{t}^{2}, e_{p o t} : = \frac{c^{2}}{2} ∣ D u ∣^{2}, e : = e_{k i n} + e_{p o t} = \frac{1}{2} (u_{t}^{2} + c^{2} ∣ D u ∣^{2}),$

and the energy flux is the $C^{1}$ vector field

$q : = - c^{2} u_{t} D u$

(Divergence and curl of a $C^{1}$ vector field). For a Lebesgue-measurable $Ω \subseteq U$ and a time $t \in I$ with $\int_{Ω} e (x, t) d x < \infty$ , the total energy in $Ω$ is the real number

$E_{Ω} (t) : = \int_{Ω} e (x, t) d x$

(The nonnegative Lebesgue integral); when the integral is infinite the total energy is $+ \infty$ and no real value is assigned.

Sign convention. If $Ω$ is admissible for the divergence theorem with outward unit normal $ν$ , then $\int_{\partial Ω} q \cdot ν d S$ is the energy leaving $Ω$ across $\partial Ω$ per unit time. This is the convention under which the pointwise identity $\partial_{t} e + div q = u_{t} □_{c} u$ of The local wave-energy conservation law holds with $□_{c} = \partial_{t}^{2} - c^{2} Δ$ (Wave equation, Cauchy data and wave speed): the flux vector $q = - c^{2} u_{t} D u$ points in the direction of energy transport, and $div q$ is the local rate at which energy leaves a point.

Speed convention. The gradient term carries the exact factor $c^{2}$ , so at unit speed the energy density is the familiar $\frac{1}{2} u_{t}^{2} + \frac{1}{2} ∣ D u ∣^{2}$ and the flux is $- u_{t} D u$ . The general-speed statements of this page are rendered at $c > 0$ throughout.

No equation for $u$ , no finiteness of a particular integral and no regularity of any boundary are asserted here: each statement that uses these objects states its own hypotheses, and sufficient settings in which $E_{Ω}$ is finite and constant are supplied by Conservation of total wave energy in three admissible settings.

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)Open item page →

The integral of the divergence of an integrable C1 field vanishes

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $n \geq 1$ and let $F \in C^{1} (R^{n}; R^{n})$ (Divergence and curl of a $C^{1}$ vector field) satisfy $∣ F ∣ \in L^{1} (λ_{n})$ (the meaning of vector integrability $F \in L^{1}$ ) and $div F \in L^{1} (λ_{n})$ , where $λ_{n}$ is $n$ -dimensional Lebesgue measure (Lebesgue measurable sets, the family $L (R^{n})$ , and the restricted set function $λ_{n}$ , The class $L^{1} (μ)$ of integrable functions). Then

$\int_{R^{n}} div F d λ_{n} = 0.$

Both integrability hypotheses are used: $F \in L^{1}$ is what dominates the cutoff term $⟨ D χ_{R}, F ⟩$ , and $div F \in L^{1}$ is both the integrand whose integral is computed and its own dominating function. The conclusion fails if $F \in L^{1}$ is dropped (a compactly supported $C^{1}$ function $h$ with $\int h = 1$ has $F (x) = \int_{- \infty}^{x} h$ , so $F \in C^{1}$ has $F^{'} = h \in L^{1}$ but $F \notin L^{1}$ and $\int_{R} F^{'} = 1$ ), and without $div F \in L^{1}$ the displayed integral need not be defined. No decay of the flux is asserted beyond the two stated integrabilities.

Facts & Assumptions

Given: The Axiom of Countable Choice ${A C}_{ω}$ ; an integer $n \geq 1$ ; a field $F \in C^{1} (R^{n}; R^{n})$ with $∣ F ∣ \in L^{1} (λ_{n})$ (the meaning of vector integrability $F \in L^{1}$ ) and $div F \in L^{1} (λ_{n})$ .

[F1]

Divergence theorem: for $n \geq 2$ , a bounded $C^{1}$ domain $Ω$ and $G \in C^{1} (\overline{Ω}; R^{n})$ , $\int_{Ω} div G d λ_{n} = \int_{\partial Ω} G \cdot ν d S$ , under ${A C}_{ω}$ . (Divergence on a bounded C1 Euclidean domain)

[F2]

For every $0 < r < R$ there is a smooth bump $ρ : R^{n} \to [0, 1]$ with $ρ = 1$ on ${\overline{B}}_{r} (0)$ and $supp ρ \subseteq B_{R} (0)$ . (A smooth bump between concentric Euclidean balls)

[F3]

For $C^{1}$ fields $F$ and a $C^{1}$ scalar $φ$ , the product rule $div (φ F) = ⟨ \nabla φ, F ⟩ + φ div F$ holds. (Divergence and curl are linear and satisfy the scalar product rules)

[F4]

Chain rule: $D (G \circ φ) (a) = D G (φ (a)) \circ D φ (a)$ for totally differentiable composites. (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ )

[F5]

Dominated convergence: if $f_{k} \to f$ almost everywhere and $∣ f_{k} ∣ \leq g$ almost everywhere with $\int g d μ < + \infty$ , then $\int f_{k} d μ \to \int f d μ$ . (Dominated convergence)

[F6]

The second fundamental theorem: if $G$ is differentiable at every point of $[a, b]$ and $G^{'}$ is integrable, then $\int_{a}^{b} G^{'} = G (b) - G (a)$ (Darboux integral). (The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ )

[F7]

On a closed bounded interval a bounded function is Darboux integrable exactly when it is Riemann integrable, with the same value. (The Darboux and Riemann definitions agree: a bounded $f$ on $[a, b]$ is Darboux integrable with integral $I$ if and only if for every real $ε > 0$ there is a real $δ > 0$ such that $∣ S (f, P, ξ) - I ∣ < ε$ for every tagged partition of mesh below $δ$ )

[F8]

A bounded Borel Riemann integrable $f$ on a closed interval $Q$ lies in $L^{1} (λ_{1} ∣_{Q})$ and its Lebesgue and Riemann integrals over $Q$ agree (under ${A C}_{ω}$ ). (A bounded Riemann integrable function on a closed bounded interval is Lebesgue measurable and has the same integral)

[F9]

The nonnegative Lebesgue integral is additive over a measurable decomposition of the domain. (Additivity of the nonnegative Lebesgue integral)

[F10]

Closed bounded Euclidean sets are compact, and continuous real functions on nonempty compact sets are bounded. (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line, A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value)

[F11]

A continuous real function on a closed bounded interval is bounded and Darboux integrable. (A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion)

Proof

1.1chooseconstructF2F4F10

Insert the cutoff: by [F2] with $r = 1$ , $R = 2$ choose $χ : R^{n} \to [0, 1]$ smooth with $χ = 1$ on ${\overline{B}}_{1} (0)$ and $χ = 0$ outside $B_{2} (0)$ ; for $R > 0$ set $χ_{R} (x) : = χ (x / R)$ , so that $χ_{R}$ is $C^{1}$ with $0 \leq χ_{R} \leq 1$ , $χ_{R} = 1$ on ${\overline{B}}_{R} (0)$ and $χ_{R} = 0$ outside $B_{2 R} (0)$ ; the chain rule [F4] applied to the map $x \mapsto x / R$ gives $D χ_{R} (x) = R^{- 1} (D χ) (x / R)$ , so with $C_{χ} : = \sup_{R^{n}} ∣ D χ ∣ < \infty$ by [F10] on ${\overline{B}}_{2} (0)$ , since $D χ = 0$ off that ball one has $∣ D χ_{R} (x) ∣ \leq C_{χ} / R$ for every $x$ and every $R > 0$ .

2.1F1F6F7F8F9step 1.1F11

The cutoff divergence integrates to zero: fix $R > 0$ ; the field $χ_{R} F$ is $C^{1}$ on $R^{n}$ and vanishes outside the compact set ${\overline{B}}_{2 R} (0)$ ; if $n \geq 2$ , apply [F1] on the open ball $Ω = B_{3 R} (0)$ , a bounded $C^{1}$ domain whose closure contains $supp (χ_{R} F)$ in its interior, where the boundary term vanishes because $χ_{R} F = 0$ on $\partial Ω$ , so that $\int_{Ω} div (χ_{R} F) d λ_{n} = 0$ and hence, the integrand vanishing off $Ω$ , also $\int_{R^{n}} div (χ_{R} F) d λ_{n} = 0$ ; if $n = 1$ , put $g : = χ_{R} F \in C^{1} (R)$ , so $g$ vanishes outside $(- 2 R, 2 R)$ , and with $a : = - 3 R$ , $b : = 3 R$ the derivative $g^{'}$ is continuous on $[a, b]$ , hence bounded and Darboux integrable, [F6] gives $\int_{a}^{b} g^{'} = g (b) - g (a) = 0$ , [F7] makes this Darboux integral equal to the Riemann integral of $g^{'}$ over $[a, b]$ , [F8] applied on the box $Q = [a, b]$ makes that Riemann integral equal to the Lebesgue integral $\int_{[a, b]} g^{'} d λ_{1}$ , and $g^{'} = 0$ on $R ∖ [a, b]$ , so [F9] applied to the positive and negative parts gives $\int_{R} g^{'} d λ_{1} = \int_{[a, b]} g^{'} d λ_{1} = 0$ ; in both cases $\int_{R^{n}} div (χ_{R} F) d λ_{n} = 0$ .

3.1step 1.1step 2.1F3F5∎

Expand and let $R \to \infty$ : by [F3] one has pointwise $div (χ_{R} F) = ⟨ D χ_{R}, F ⟩ + χ_{R} div F$ , and each of the three functions lies in $L^{1} (λ_{n})$ , the left side because it is continuous with compact support, $χ_{R} div F$ because $∣ χ_{R} div F ∣ \leq ∣ div F ∣$ , and $⟨ D χ_{R}, F ⟩$ because step 1.1 gives $∣ ⟨ D χ_{R}, F ⟩ ∣ \leq (C_{χ} / R) ∣ F ∣$ ; by linearity of the Lebesgue integral on $L^{1} (λ_{n})$ , $\int_{R^{n}} div (χ_{R} F) d λ_{n} = \int_{R^{n}} χ_{R} div F d λ_{n} + \int_{R^{n}} ⟨ D χ_{R}, F ⟩ d λ_{n}$ for every $R > 0$ , with left-hand side $0$ by step 2.1; as $R \to \infty$ through positive integers (so $R \geq 1$ ), the functions $χ_{R} div F$ converge pointwise to $div F$ and are dominated by $∣ div F ∣ \in L^{1} (λ_{n})$ , while $⟨ D χ_{R}, F ⟩$ converge pointwise to $0$ and are dominated by $C_{χ} ∣ F ∣ \in L^{1} (λ_{n})$ , so [F5] gives $\int_{R^{n}} χ_{R} div F d λ_{n} \to \int_{R^{n}} div F d λ_{n}$ and $\int_{R^{n}} ⟨ D χ_{R}, F ⟩ d λ_{n} \to 0$ ; passing to the limit gives $0 = \int_{R^{n}} div F d λ_{n} + 0$ , which is the claim.

Remarks

The lemma supplies the vanishing flux integral in Conservation of total wave energy in three admissible settings(b).

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)Open item page →

Truncated wave cones: convexity, piecewise C1 presentation and outward normals

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $n \geq 1$ , $c > 0$ , $x_{0} \in R^{n}$ , $t_{0} > 0$ and $0 < t_{1} < t_{2} < t_{0}$ . In space-time $R^{n + 1}$ with coordinates $(x, t)$ (Euclidean, so $(0, - 1)$ and $(0, 1)$ below have zero space part) put

$K (t_{1}, t_{2}) : = {(x, t) : t_{1} < t < t_{2}, ∣ x - x_{0} ∣ < c (t_{0} - t)} .$

Then:

(i) $K (t_{1}, t_{2})$ is a nonempty open bounded convex set (A convex subset of $R^{m}$ contains every line segment between two of its points);

(ii) it has the finite piecewise $C^{1}$ presentation of Specified finite piecewise C1 boundary presentations whose non-edge faces are the bottom disk ${\overline{B}}_{c (t_{0} - t_{1})} (x_{0}) \times {t_{1}}$ , the top disk ${\overline{B}}_{c (t_{0} - t_{2})} (x_{0}) \times {t_{2}}$ and the lateral frustum $L : = {(x, t) : ∣ x - x_{0} ∣ = c (t_{0} - t), t_{1} \leq t \leq t_{2}},$ the edge set being the two boundary circles $S_{c (t_{0} - t_{1})} (x_{0}) \times {t_{1}}$ and $S_{c (t_{0} - t_{2})} (x_{0}) \times {t_{2}}$ ;

(iii) the corresponding outward unit normals are $(0, - 1)$ on the bottom disk, $(0, 1)$ on the top disk, and $ν_{l a t} (x, t) = (\frac{x - x_{0}}{∣ x - x_{0} ∣}, c) / \sqrt{1 + c^{2}}$ at points of $L$ ;

(iv) the closed backward cone $K^{-} (x_{0}, t_{0}) = {(x, t) : 0 \leq t \leq t_{0}, ∣ x - x_{0} ∣ \leq c (t_{0} - t)}$ is compact and convex, and $K (t_{1}, t_{2})$ is its interior intersected with the slab ${t_{1} < t < t_{2}}$ .

All statements are also true at $c = 1$ , and $n = 1$ is the characteristic trapezium.

Facts & Assumptions

Given: ${A C}_{ω}$ ; $n \geq 1$ , $c > 0$ , $x_{0} \in R^{n}$ , $t_{0} > 0$ and $0 < t_{1} < t_{2} < t_{0}$ ; the Euclidean structure of The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ on $R^{n + 1}$ ; the function $φ : R^{n + 1} \to R$ , $φ (x, t) : = ∣ x - x_{0} ∣ + c t$ .

[F1]

A finite piecewise $C^{1}$ presentation of a nonempty bounded open $Ω$ consists of compact faces covering its boundary, each a compact Borel subset of a regular $C^{1}$ hypersurface patch, together with a compact edge set $E$ , and it requires: $S_{j} \cap E$ surface-null in each face, the edge set to contain the relative face boundaries and all overlaps, the boundary to be locally a single $C^{1}$ graph with $Ω$ on one side off $E$ , and each face to carry its actual outward unit normal off $E$ . (Specified finite piecewise C1 boundary presentations)

[F2]

On a compact embedded $C^{1}$ hypersurface the chart integral is a finite Borel measure independent of the charts; in graph coordinates $X (y) = (y, h (y))$ its density is $\sqrt{1 + ∣ D h (y) ∣^{2}}$ , and on a one-sided domain boundary the outward unit normal agrees on chart overlaps. (Chart and partition independence of surface measure)

[F3]

$∥ \cdot ∥_{2}$ is a norm on $R^{n}$ for every $n$ , so it is subadditive and absolutely homogeneous. (Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation)

[F4]

For every real $α$ the function $s \mapsto s^{α}$ is differentiable on $(0, \infty)$ with derivative $α s^{α - 1}$ . (Continuity and derivatives of positive-base real powers)

[F5]

Chain rule: $D (G \circ H) (a) = D G (H (a)) \circ D H (a)$ for composable totally differentiable maps. (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ )

[F6]

Finite sums and products of $C^{k}$ Euclidean maps are $C^{k}$ , and composites of composable $C^{k}$ maps are $C^{k}$ . ( $C^{k}$ Euclidean maps are closed under componentwise algebra and composition)

[F7]

For $n \geq 1$ and $r > 0$ , $∣ B_{r} ∣ = ω_{n - 1} r^{n} / n$ , with $0 < ω_{n - 1} < \infty$ (Sphere and ball measures scale in Rn). Consequently every positive-radius sphere has Lebesgue measure zero: for $0 < ε < r$ , it lies in $B_{r + ε} ∖ B_{r - ε}$ , whose measure is $ω_{n - 1} ((r + ε)^{n} - (r - ε)^{n}) / n$ ; monotonicity and finite additivity bound its measure by this quantity, and letting $ε ↓ 0$ gives zero.

[F8]

A Lipschitz self-map of $R^{n}$ carries $λ_{n}$ -null sets to $λ_{n}$ -null sets, under ${A C}_{ω}$ . (A Lipschitz self-map of $R^{n}$ carries Lebesgue null sets to Lebesgue null sets)

[F9]

A subset $U \subseteq R^{m}$ is convex when $(1 - t) x + t y \in U$ for all $x, y \in U$ and $t \in [0, 1]$ . (A convex subset of $R^{m}$ contains every line segment between two of its points)

[F10]

A closed bounded subset of Euclidean space is compact. (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line)

Proof

1.1givenF3F9algebra

The function $φ$ is convex: for $p = (x, t)$ , $q = (y, s) \in R^{n + 1}$ and $λ \in [0, 1]$ , writing $u : = (1 - λ) (x - x_{0})$ and $v : = λ (y - x_{0})$ , the triangle inequality and absolute homogeneity of the Euclidean norm [F3] give $∣ u + v ∣ \leq (1 - λ) ∣ x - x_{0} ∣ + λ ∣ y - x_{0} ∣$ , while $c ((1 - λ) t + λ s) = (1 - λ) c t + λ c s$ by linearity, so $φ ((1 - λ) p + λ q) \leq (1 - λ) φ (p) + λ φ (q)$ ; consequently $K (t_{1}, t_{2}) = {φ < c t_{0}} \cap {t_{1} < t < t_{2}}$ is convex, because both sets are convex: for ${φ < c t_{0}}$ this is the inequality just proved applied to two points with values below $c t_{0}$ , and the slab is defined by two affine conditions [F9].

2.1givenstep 1.1F3algebra

Basic topological properties: $K (t_{1}, t_{2})$ is open since [F3] gives $∣ ∣ x - x_{0} ∣ - ∣ y - x_{0} ∣ ∣ \leq ∣ x - y ∣$ , so $φ$ and the coordinate $t$ are continuous and the half-lines $(t_{1}, t_{2})$ and $(- \infty, c t_{0})$ are open; it is nonempty because $(x_{0}, (t_{1} + t_{2}) / 2)$ has $φ = 0 + c (t_{1} + t_{2}) / 2 < c t_{0}$ and lies in the slab; it is bounded because every point has $t_{1} < t < t_{2}$ and $∣ x - x_{0} ∣ < c (t_{0} - t) < c t_{0}$ . This is (i).

3.1givenstep 2.1algebra

The boundary decomposition: $\partial K (t_{1}, t_{2}) = D_{b} \cup D_{t} \cup L$ with $D_{b} : = {\overline{B}}_{c (t_{0} - t_{1})} (x_{0}) \times {t_{1}}$ , $D_{t} : = {\overline{B}}_{c (t_{0} - t_{2})} (x_{0}) \times {t_{2}}$ and $L$ as in the statement; the only overlaps are $D_{b} \cap L = S_{c (t_{0} - t_{1})} (x_{0}) \times {t_{1}}$ and $D_{t} \cap L = S_{c (t_{0} - t_{2})} (x_{0}) \times {t_{2}}$ , and $D_{b} \cap D_{t} = \emptyset$ . Indeed $\partial {φ < c t_{0}} \subseteq {φ = c t_{0}}$ and $\partial {t_{1} < t < t_{2}} \subseteq {t = t_{1}} \cup {t = t_{2}}$ , and $K$ is the intersection of these three open sets, so a boundary point of $K$ lies in one of the three level sets; conversely a point $p = (x, t)$ with $φ (p) = c t_{0}$ and $t \in (t_{1}, t_{2})$ (a point of $L$ ) has $p + δ (x - x_{0}, c) \notin K$ and $p - δ (x - x_{0}, c) \in K$ for small $δ > 0$ , while a point with $t = t_{1}$ and $∣ x - x_{0} ∣ < c (t_{0} - t_{1})$ has $p - δ e_{t} \notin K$ and $p + δ e_{t} \in K$ , and at a rim point $∣ x - x_{0} ∣ = c (t_{0} - t_{1})$ , $t = t_{1}$ , the points $(x_{0} + \frac{t_{0} - t_{1} - 2 ε}{t_{0} - t_{1}} (x - x_{0}), t_{1} + ε)$ lie in $K$ for $0 < ε < \min ((t_{0} - t_{1}) / 2, t_{2} - t_{1})$ , since their spatial radius is $c (t_{0} - t_{1} - 2 ε) < c (t_{0} - t_{1} - ε)$ , and tend to $p$ ; at a top rim point $(x, t_{2})$ the points $(x, t_{2} - ε)$ lie in $K$ for $0 < ε < t_{2} - t_{1}$ and tend to it, while the interior of the top disk is approached vertically.

4.1F4F5F6step 3.1algebra

Each face is a compact Borel subset of a regular $C^{1}$ hypersurface patch: $D_{b}$ and $D_{t}$ are closed balls in the hyperplanes ${t = t_{i}}$ , which are graphs of the constant (hence $C^{1}$ ) functions over $R^{n}$ with nonvanishing gradient of $(x, t) \mapsto t$ ; the lateral face lies in the graphic hypersurface ${(y, h (y)) : y \in O}$ where $O : = R^{n} ∖ {x_{0}}$ and $h (y) : = t_{0} - ∣ y - x_{0} ∣ / c$ , which is $C^{1}$ because $y \mapsto ⟨ y - x_{0}, y - x_{0} ⟩$ is a finite sum of products of the $C^{1}$ coordinate functions [F6], the square root is differentiable on $(0, \infty)$ with derivative $\frac{1}{2} s^{- 1 / 2}$ [F4], and the chain rule [F5] applies on the open set where the inner value is positive, namely $O$ .

5.1F1F2F7F8F10step 3.1step 4.1

The presentation is verified with $E : = (S_{c (t_{0} - t_{1})} (x_{0}) \times {t_{1}}) \cup (S_{c (t_{0} - t_{2})} (x_{0}) \times {t_{2}})$ : the three faces are closed bounded Borel subsets, hence compact by [F10], of regular $C^{1}$ patches by step 4.1 and cover $\partial K$ by step 3.1; $E \subset \partial K$ is compact and contains the relative boundaries of the faces in their patches (the rim circles of the two disks and the two boundary circles of the annulus ${c (t_{0} - t_{2}) \leq ∣ y - x_{0} ∣ \leq c (t_{0} - t_{1})}$ parametrizing $L$ ) and all pairwise overlaps, which by step 3.1 are exactly the two rim circles; off $E$ the boundary is locally a single $C^{1}$ graph with $K$ on one side, namely $t = t_{i}$ over a small ball in the interior of each disk with $K$ on the side $t > t_{1}$ , respectively $t < t_{2}$ , and $t = h (y)$ over a small ball in $O$ for interior points of $L$ , with $K$ locally ${t < h (y)}$ by the definition of $K$ ; and $S_{j} \cap E$ is surface-null in each face: on a disk the surface measure is $n$ -dimensional Lebesgue measure transported by the graph chart [F2], whose rim is a sphere of positive radius, null by [F7] and [F8] applied to the homothety $z \mapsto x_{0} + r z$ (and for $n = 1$ a two-point set), while on $L$ the graph density is the constant $\sqrt{1 + 1 / c^{2}}$ because $∣ D h ∣ = 1 / c$ on $O$ , so a Borel subset of $L$ is surface-null exactly when its $y$ -projection is $λ_{n}$ -null [F2], and the projection of $E \cap L$ is the union of two positive-radius spheres, null by [F7] and [F8]. Thus $K (t_{1}, t_{2})$ has the specified finite piecewise $C^{1}$ presentation with faces $D_{b}, D_{t}, L$ and edge set $E$ : this is (ii).

5.2givenF2step 4.1algebra

The outward normals: on the bottom disk the region lies locally in ${t > t_{1}}$ , so the outward unit normal is $(0, - 1)$ ; on the top disk it is $(0, 1)$ ; on $L$ the field $\nabla φ = ((x - x_{0}) / ∣ x - x_{0} ∣, c)$ is continuous and nonvanishing near $L$ because $∣ x - x_{0} ∣ = c (t_{0} - t) \geq c (t_{0} - t_{2}) > 0$ there, $K$ is locally the side ${φ < c t_{0}}$ and $L \subseteq {φ = c t_{0}}$ , so the outward unit normal is $\nabla φ / ∣ \nabla φ ∣ = ((x - x_{0}) / ∣ x - x_{0} ∣, c) / \sqrt{1 + c^{2}}$ , using the outward-normal convention for one-sided graph boundaries [F2] and the gradient of The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case. This is (iii).

6.1step 1.1step 2.1F9F10algebra∎

The closed cone: $K^{-} = {φ \leq c t_{0}} \cap {t \geq 0} \cap {t \leq t_{0}}$ is closed because $φ$ is continuous, bounded because $0 \leq t \leq t_{0}$ and $∣ x - x_{0} ∣ \leq c t_{0}$ , and thus compact by [F10], and convex because the sublevel set is convex by the inequality of step 1.1 and the two half-spaces are convex [F9]; its interior is ${0 < t < t_{0}, φ < c t_{0}}$ : the inclusion $\supseteq$ is openness of the right-hand set inside $K^{-}$ , and conversely a point with $φ = c t_{0}$ is not interior, since for $z : = (x_{0}, 0)$ with $φ (z) = 0 < c t_{0}$ the points $r (σ) : = z + σ (p - z)$ , $σ > 1$ , satisfy $p = \frac{1}{σ} r (σ) + (1 - \frac{1}{σ}) z$ , so convexity gives $φ (p) \leq \frac{1}{σ} φ (r (σ)) + (1 - \frac{1}{σ}) φ (z)$ and hence $φ (r (σ)) \geq c t_{0} + (σ - 1) (c t_{0} - φ (z)) > c t_{0}$ , with $r (σ) \to p$ as $σ ↓ 1$ ; a point with $t = 0$ or $t = t_{0}$ is not interior because $p - δ e_{t}$ , respectively $p + δ e_{t}$ , lies outside $K^{-}$ for every $δ > 0$ . Intersecting $int K^{-}$ with the slab ${t_{1} < t < t_{2}} \subseteq {0 < t < t_{0}}$ gives exactly $K (t_{1}, t_{2})$ , which is (iv).

Remarks

The outward normals of (iii) supply the geometric data used in The energy identity on a truncated wave cone.

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Vanishing gradient and time derivative force constancy on convex sets

Statement

Let $U \subseteq R^{m}$ be an open convex set (A convex subset of $R^{m}$ contains every line segment between two of its points) and let $w \in C^{1} (U)$ ( $C^{k}$ maps and multi-index derivative notation in Euclidean space) with $D w = 0$ on $U$ , that is $\partial_{i} w = 0$ on $U$ for every coordinate $i$ (Directional derivatives and partial derivatives of a map $U \subseteq R^{m} \to R^{n}$ ). Then $w$ is constant on $U$ .

In particular, if $(x, t) \mapsto u (x, t)$ is $C^{1}$ on an open convex subset $V$ of space-time $R^{n + 1}$ with $\partial_{t} u = 0$ and $D_{x} u = (\partial_{0} u, \dots, \partial_{n - 1} u) = 0$ on $V$ , then $u$ is constant on $V$ ; this is the conclusion used when the energy density of a wave vanishes identically on a cone or a ball and the displacement is recovered from $u_{t} = D u = 0$ .

Facts & Assumptions

Given: An open convex set $U \subseteq R^{m}$ and a $C^{1}$ function $w : U \to R$ whose total derivative vanishes on $U$ ; for the last sentence an open convex subset $V$ of space-time and a $C^{1}$ function $u$ on $V$ with $\partial_{t} u = 0$ and all spatial partial derivatives zero.

[F1]

If $f : U \to R^{n}$ is totally differentiable at $a$ , then $D_{v} f (a)$ exists for every $v$ and equals $D f (a) v$ ; in particular $\partial_{j} f (a) = D f (a) e_{j}$ , and the matrix of $D f (a)$ is the Jacobian $J f (a)$ . (A total derivative computes every directional derivative, and its matrix is the Jacobian)

[F2]

For scalar-valued $f$ its gradient is $\nabla f (a) = (\partial_{0} f (a), \dots, \partial_{m - 1} f (a))$ . (The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case)

[F3]

If $U \subseteq R^{m}$ is convex and open and $f : U \to R^{n}$ is totally differentiable at every point with $D f (z) = 0$ for every $z \in U$ , then $f$ is constant on $U$ . (A totally differentiable map with zero derivative on a convex open set is constant)

[F4]

A subset $U \subseteq R^{m}$ is convex when $(1 - t) x + t y \in U$ for all $x, y \in U$ and $t \in [0, 1]$ . (A convex subset of $R^{m}$ contains every line segment between two of its points)

[F5]

Continuous partial derivatives imply total differentiability, with the Jacobian as its matrix. (If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative)

Proof

1.1givenF1F2F5algebra

The two forms of the hypothesis are equivalent: at every $a \in U$ the map $w$ is totally differentiable by its $C^{1}$ regularity and [F5], so by [F1] the Jacobian $J w (a)$ is the matrix of $D w (a)$ and its entries are exactly $\partial_{i} w (a)$ , the coordinates of $\nabla w (a)$ [F2]; a linear map is zero exactly when its matrix (equivalently, all its partial derivatives) vanishes, so $D w = 0$ on $U$ if and only if $\partial_{i} w = 0$ on $U$ for every $i$ .

2.1step 1.1F3F4

Constancy: if $D w = 0$ on the open convex $U$ , then [F3] applied to $f = w$ gives that $w$ is constant on $U$ ; conversely if all partial derivatives of $w$ vanish, step 1.1 converts this to $D w = 0$ and the same conclusion follows, so the first claim holds under either form of the hypothesis.

3.1givenstep 1.1step 2.1∎

The space-time case: an open convex subset $V$ of $R^{n + 1}$ with its Euclidean coordinates is an instance of the first claim for $m = n + 1$ , and the hypothesis $\partial_{t} u = 0$ together with $D_{x} u = 0$ says precisely that every coordinate partial derivative of $u$ vanishes on $V$ ; by steps 1.1 and 2.1 the function $u$ is constant on $V$ .

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The local wave-energy conservation law

Statement

Let $n \geq 1$ , $c > 0$ , $U \subseteq R^{n}$ open, $I \subseteq R$ an open interval and $u \in C^{2} (U \times I)$ ( $C^{k}$ maps and multi-index derivative notation in Euclidean space). Put $f : = □_{c} u = \partial_{t}^{2} u - c^{2} Δ u$ (Wave equation, Cauchy data and wave speed, The Laplacian of a $C^{2}$ function and of a $C^{2}$ vector field) and let $e, q$ be the energy density and flux of Wave energy density, energy flux and total energy. Then the pointwise identity

$\partial_{t} e + div q = f u_{t}$

holds on $U \times I$ ; in particular $\partial_{t} e + div q = 0$ for every classical solution of the homogeneous equation. For a homogeneous solution at unit speed the identity reads $\partial_{t} [\frac{1}{2} (u_{t}^{2} + ∣ D u ∣^{2})] = div (u_{t} D u)$ : indeed $q = - c^{2} u_{t} D u$ gives $- div (u_{t} D u) = div q$ at $c = 1$ . No integration, integrability or boundary regularity is used or asserted.

Facts & Assumptions

Given: $n \geq 1$ , $c > 0$ , an open set $U \subseteq R^{n}$ , an open interval $I$ and $u \in C^{2} (U \times I)$ ; the fields $e = \frac{1}{2} (u_{t}^{2} + c^{2} ∣ D u ∣^{2})$ and $q = - c^{2} u_{t} D u$ of Wave energy density, energy flux and total energy; write $D u_{t} : = \partial_{t} (D u)$ and $Δ u = div (D u)$ .

[F1]

Clairaut–Schwarz: on an open set where $u$ is $C^{2}$ , $\partial_{i} \partial_{j} u = \partial_{j} \partial_{i} u$ for every pair of coordinate indices; in particular $\partial_{t} \partial_{i} u = \partial_{i} \partial_{t} u$ , so $D u_{t} = D (u_{t})$ and the mixed derivatives of $u$ commute. (Clairaut--Schwarz theorem for continuous second partial derivatives)

[F2]

Product rule for the divergence: $div (φ F) = ⟨ \nabla φ, F ⟩ + φ div F$ for $C^{1}$ scalar $φ$ and $C^{1}$ field $F$ . (Divergence and curl are linear and satisfy the scalar product rules)

[F3]

The Laplacian is the divergence of the gradient: $Δ f = div \nabla f = \sum_{i < n} \partial_{i} \partial_{i} f$ . (The Laplacian of a $C^{2}$ function and of a $C^{2}$ vector field)

[F4]

Product rule in one variable: $(g h)^{'} = g^{'} h + g h^{'}$ for differentiable $g, h : R \to R$ . (Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ )

[F5]

The Euclidean inner product is symmetric, $⟨ x, y ⟩ = ⟨ y, x ⟩$ , and $∣ z ∣^{2} = ⟨ z, z ⟩$ . (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ )

Proof

1.1givenF1F4F5algebra

Time derivative of the energy density: at every point of $U \times I$ the product rule [F4] gives $\partial_{t} (u_{t}^{2}) = 2 u_{t} u_{t t}$ and, since $∣ D u ∣^{2} = ⟨ D u, D u ⟩$ [F5], $\partial_{t} ∣ D u ∣^{2} = 2 ⟨ D u, \partial_{t} (D u) ⟩ = 2 ⟨ D u, D u_{t} ⟩$ , where $\partial_{t} (D u) = D (u_{t})$ by [F1]; hence $\partial_{t} e = \frac{1}{2} (2 u_{t} u_{t t} + c^{2} \cdot 2 ⟨ D u, D u_{t} ⟩) = u_{t} u_{t t} + c^{2} ⟨ D u, D u_{t} ⟩$ .

1.2givenF1F2F3algebra

Divergence of the flux: the scalar $φ : = - c^{2} u_{t}$ and the field $F : = D u$ are $C^{1}$ on $U \times I$ because $u$ is $C^{2}$ , so the product rule [F2] and [F3] give $div q = div (φ F) = ⟨ \nabla φ, F ⟩ + φ div F = - c^{2} ⟨ \nabla u_{t}, D u ⟩ - c^{2} u_{t} Δ u = - c^{2} ⟨ D u_{t}, D u ⟩ - c^{2} u_{t} Δ u$ , using $\nabla u_{t} = D u_{t}$ from [F1] in the last step.

2.1givenstep 1.1step 1.2F5algebra∎

The balance law: adding the identities of steps 1.1 and 1.2, $\partial_{t} e + div q = u_{t} u_{t t} + c^{2} ⟨ D u, D u_{t} ⟩ - c^{2} ⟨ D u_{t}, D u ⟩ - c^{2} u_{t} Δ u = u_{t} u_{t t} - c^{2} u_{t} Δ u = u_{t} (u_{t t} - c^{2} Δ u) = u_{t} □_{c} u = f u_{t}$ , the inner-product terms cancelling by symmetry [F5]; for $f = 0$ this is $\partial_{t} e + div q = 0$ , and at $c = 1$ it is the stated unit-speed form.

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The energy identity on a truncated wave cone

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $n \geq 1$ , $c > 0$ , $x_{0} \in R^{n}$ , $t_{0} > 0$ and $0 < t_{1} < t_{2} < t_{0}$ ; let $K = K (t_{1}, t_{2})$ be the space-time frustum of Truncated wave cones: convexity, piecewise C1 presentation and outward normals and let $u \in C^{2}$ solve $□_{c} u = f$ on a neighbourhood of $\overline{K}$ (Wave equation, Cauchy data and wave speed). With $e, q$ as in Wave energy density, energy flux and total energy, $D u$ the spatial gradient and

$E (t) : = \int_{B_{c (t_{0} - t)} (x_{0})} e (x, t) d x (t_{1} \leq t \leq t_{2}),$

write $\partial_{r} u : = D u \cdot \frac{x - x_{0}}{∣ x - x_{0} ∣}$ and $D_{t a n} u : = D u - (\partial_{r} u) \frac{x - x_{0}}{∣ x - x_{0} ∣}$ for the radial and tangential parts of $D u$ on a sphere centred at $x_{0}$ . Then

$\int_{K} f u_{t} d x d t = E (t_{2}) - E (t_{1}) + \int_{t_{1}}^{t_{2}} ⁣ \int_{\partial B_{c (t_{0} - t)} (x_{0})} ℓ d S d t,$

where the lateral flux density $ℓ : = c e - c^{2} u_{t} \partial_{r} u = q \cdot \frac{x - x_{0}}{∣ x - x_{0} ∣} + c e$ satisfies

$ℓ = \frac{c}{2} ((u_{t} - c \partial_{r} u)^{2} + c^{2} ∣ D_{t a n} u ∣^{2}) \geq 0.$

Thus the lateral term is a sum of squares, vanishing identically exactly when $u_{t} = c \partial_{r} u$ and $D_{t a n} u = 0$ on the lateral surface. In the homogeneous case $f = 0$ , the identity gives $E (t_{2}) \leq E (t_{1})$ for $t_{1} < t_{2}$ . Normalisation note. The density $ℓ$ above is the one for which the sphere surface measure $d S$ makes the displayed identity an identity: the lateral area element of $K$ carries the graph factor $\sqrt{1 + c^{2}}$ , which cancels the $1 / \sqrt{1 + c^{2}}$ in $V \cdot ν$ , where $V = (q, e)$ , when $d S$ is measured on the sphere $\partial B_{c (t_{0} - t)} (x_{0})$ .

Facts & Assumptions

Given: ${A C}_{ω}$ ; the frustum $K = K (t_{1}, t_{2})$ with its faces $D_{b} = {\overline{B}}_{c (t_{0} - t_{1})} (x_{0}) \times {t_{1}}$ , $D_{t} = {\overline{B}}_{c (t_{0} - t_{2})} (x_{0}) \times {t_{2}}$ and lateral frustum $L$ , edge set $E$ the two rim spheres; a $C^{2}$ function $u$ solving $□_{c} u = f$ on a neighbourhood of $\overline{K}$ ; the fields $e = \frac{1}{2} (u_{t}^{2} + c^{2} ∣ D u ∣^{2})$ , $q = - c^{2} u_{t} D u$ of Wave energy density, energy flux and total energy; the space-time field $V : = (q, e)$ on $R^{n} \times R$ .

[F1]

Local conservation: $\partial_{t} e + div q = f u_{t}$ pointwise. (The local wave-energy conservation law)

[F2]

Piecewise divergence theorem: if $Ω$ has a specified finite piecewise $C^{1}$ presentation and $F \in C^{1} (\overline{Ω}; R^{n})$ , then $\int_{Ω} div F = \sum_{j} \int_{S_{j}} F \cdot ν_{j} d S$ , the faces counted once off the edge set $E$ . (Divergence for finite piecewise C1 presentations)

[F3]

The frustum $K$ has the finite piecewise $C^{1}$ presentation with faces $D_{b}, D_{t}, L$ and edge set the two rim spheres, with outward unit normals $(0, - 1)$ on $D_{b}$ , $(0, 1)$ on $D_{t}$ , and $ν = ((x - x_{0}) / ∣ x - x_{0} ∣, c) / \sqrt{1 + c^{2}}$ on $L$ . (Truncated wave cones: convexity, piecewise C1 presentation and outward normals)

[F4]

On a compact embedded $C^{1}$ hypersurface the chart integral is a finite Borel measure independent of charts; in graph coordinates $X (y) = (y, h (y))$ its density is $\sqrt{1 + ∣ D h (y) ∣^{2}}$ , and on a one-sided boundary the outward unit normal agrees on chart overlaps. (Chart and partition independence of surface measure)

[F5]

The Euclidean inner product is symmetric and the orthogonal decomposition $D u = (\partial_{r} u) \hat{x - x_{0}} + D_{t a n} u$ on a sphere centred at $x_{0}$ gives $∣ D u ∣^{2} = (\partial_{r} u)^{2} + ∣ D_{t a n} u ∣^{2}$ . (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ )

[F6]

Polar integration has radial density $r^{n - 1} d r d σ$ ; for $n \geq 2$ the polar measure equals chart surface measure and radius- $r$ sphere integrals have factor $r^{n - 1}$ . In $n = 1$ each point of $S^{0}$ has mass one and each lateral segment has length element $\sqrt{1 + c^{2}} d t$ . (Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma, Agreement with the existing polar sphere measure)

Proof

1.1givenF1F2F3

The divergence theorem applies to $V = (q, e)$ on the frustum: $u$ is $C^{2}$ on a neighbourhood of $\overline{K}$ , so $q$ and $e$ are $C^{1}$ there, and by [F1] the space-time divergence of $V$ is ${div}_{(x, t)} V = div q + \partial_{t} e = f u_{t}$ ; since $K$ has the finite piecewise $C^{1}$ presentation [F3] with edge set of surface measure zero, [F2] gives $\int_{K} f u_{t} d x d t = \int_{D_{b}} V \cdot ν d S + \int_{D_{t}} V \cdot ν d S + \int_{L} V \cdot ν d S$ .

2.1givenF3F4step 1.1algebra

The caps: with the outward normals $(0, - 1)$ on $D_{b}$ and $(0, 1)$ on $D_{t}$ from [F3], the flux density on the bottom face is $V \cdot (0, - 1) = - e$ and on the top face $V \cdot (0, 1) = e$ , and the chart integral on a face contained in a coordinate hyperplane reduces to the $n$ -dimensional Lebesgue integral of the trace by [F4]; hence $\int_{D_{b}} V \cdot ν d S = - \int_{B_{c (t_{0} - t_{1})} (x_{0})} e (x, t_{1}) d x = - E (t_{1})$ and $\int_{D_{t}} V \cdot ν d S = E (t_{2})$ , so the two caps contribute $E (t_{2}) - E (t_{1})$ .

2.2givenF3F4F5F6step 1.1algebra

The lateral face: by [F3] the outward unit normal on $L$ is $ν = (\hat{x - x_{0}}, c) / \sqrt{1 + c^{2}}$ , so $V \cdot ν = (q \cdot \hat{x - x_{0}} + c e) / \sqrt{1 + c^{2}} = ℓ / \sqrt{1 + c^{2}}$ with $ℓ : = q \cdot \hat{x - x_{0}} + c e$ ; parametrizing the lateral frustum by $(ω, t) \mapsto (x_{0} + c (t_{0} - t) ω, t)$ over $S^{n - 1} \times [t_{1}, t_{2}]$ , or equivalently using the graph density $\sqrt{1 + 1 / c^{2}}$ of [F4] for the graph $t = t_{0} - ∣ x - x_{0} ∣ / c$ , the graph density and polar integration [F6], with $r = c (t_{0} - t)$ and $∣ d r ∣ = c d t$ , give area element $\sqrt{1 + c^{2}} [c (t_{0} - t)]^{n - 1} d ω d t$ , so $\int_{L} V \cdot ν d S = \int_{t_{1}}^{t_{2}} \int_{S^{n - 1}} ℓ [c (t_{0} - t)]^{n - 1} d ω d t = \int_{t_{1}}^{t_{2}} \int_{\partial B_{c (t_{0} - t)} (x_{0})} ℓ d S d t$ ; and, since $q \cdot \hat{x - x_{0}} = - c^{2} u_{t} \partial_{r} u$ , the density is $ℓ = c e - c^{2} u_{t} \partial_{r} u = \frac{c}{2} (u_{t}^{2} + c^{2} ∣ D u ∣^{2}) - c^{2} u_{t} \partial_{r} u = \frac{c}{2} ((u_{t} - c \partial_{r} u)^{2} + c^{2} (∣ D u ∣^{2} - (\partial_{r} u)^{2})) = \frac{c}{2} ((u_{t} - c \partial_{r} u)^{2} + c^{2} ∣ D_{t a n} u ∣^{2}) \geq 0$ by [F5], with equality exactly when both squares vanish.

3.1step 1.1step 2.1step 2.2algebra∎

Substituting steps 2.1 and 2.2 into the identity of step 1.1 gives $\int_{K} f u_{t} d x d t = E (t_{2}) - E (t_{1}) + \int_{t_{1}}^{t_{2}} \int_{\partial B_{c (t_{0} - t)} (x_{0})} ℓ d S d t$ with $ℓ = \frac{c}{2} ((u_{t} - c \partial_{r} u)^{2} + c^{2} ∣ D_{t a n} u ∣^{2}) \geq 0$ , $ℓ$ vanishing identically on $L$ exactly when $u_{t} = c \partial_{r} u$ and $D_{t a n} u = 0$ there; this is the displayed identity and the sum-of-squares form.

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Conservation of total wave energy in three admissible settings

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $n \geq 1$ , $c > 0$ , $T > 0$ , let $U \subseteq R^{n}$ be open and let $u \in C^{2} (U \times (0, T))$ solve the homogeneous equation $□_{c} u = 0$ (Wave equation, Cauchy data and wave speed), with $e, q$ as in Wave energy density, energy flux and total energy and The local wave-energy conservation law. Then $E_{Ω} (t)$ is constant in $t$ in each of the following settings, and in each the vanishing boundary term is:

(a) fixed spatial support: $Ω = R^{n}$ , $U = R^{n}$ and there is a compact $K$ with $supp u (\cdot, t) \subseteq K$ for all $t \in (0, T)$ (The support of a function on $R^{n}$ and its compactly supported Riemann integral); the flux term through $\partial B_{R}$ vanishes for a large ball $B_{R} \supset K$ .

(b) integrable flux (sufficient decay): $U = Ω = R^{n}$ , $e (\cdot, t), ∣ q (\cdot, t) ∣, div q (\cdot, t) \in L^{1} (R^{n})$ for every $t$ , and $t \mapsto E_{R^{n}} (t)$ is differentiable with $E_{R^{n}}^{'} (t) = \int_{R^{n}} \partial_{t} e (x, t) d x$ (automatic, for instance, when $\partial_{t} e$ is dominated on compact time intervals by a fixed $L^{1}$ function); then $\int_{R^{n}} div q (\cdot, t) d x = 0$ by The integral of the divergence of an integrable C1 field vanishes; this integrability is exactly the hypothesis a plane wave fails.

(c) bounded domain with homogeneous Dirichlet or homogeneous Neumann data: for $n \geq 2$ , $U$ is a bounded $C^{1}$ domain (Bounded C1 domains and their outward normals); for $n = 1$ , $U$ is a finite union of disjoint bounded open intervals with pairwise disjoint closures, and the outward unit normals at the left and right endpoints are $- 1$ and $+ 1$ . Require $u \in C^{2} (\overline{U} \times (0, T))$ in the interior-up-to-boundary convention, $Ω = U$ , and either $u (x, t) = 0$ on all of $\partial U \times (0, T)$ or $\partial_{ν} u (x, t) : = D u (x, t) \cdot ν (x) = 0$ there. In the Dirichlet case $u_{t} ∣_{\partial U} = 0$ ; in the Neumann case $\partial_{ν} u = 0$ . Thus in both cases the outward flux $q \cdot ν = - c^{2} u_{t} \partial_{ν} u$ vanishes.

Facts & Assumptions

Given: ${A C}_{ω}$ ; $n \geq 1$ , $c > 0$ , $T > 0$ , an open set $U \subseteq R^{n}$ and a $C^{2}$ solution $u$ of $□_{c} u = 0$ on $U \times (0, T)$ ; the energy density $e = \frac{1}{2} (u_{t}^{2} + c^{2} ∣ D u ∣^{2})$ and flux $q = - c^{2} u_{t} D u$ of Wave energy density, energy flux and total energy.

[F1]

Local balance for a homogeneous solution: $\partial_{t} e + div q = 0$ pointwise; equivalently $\partial_{t} e = - div q$ . (The local wave-energy conservation law)

[F2]

Differentiation under the integral sign: if $x \mapsto f (x, t)$ is integrable for every $t$ , $t \mapsto f (x, t)$ is differentiable for almost every $x$ , the $t$ -derivative is measurable and dominated on the time interval by a fixed integrable $g$ , then $F (t) = \int f (x, t) d μ (x)$ is differentiable with $F^{'} (t) = \int \partial_{t} f (x, t) d μ (x)$ . (Differentiation under the integral sign)

[F4]

Divergence theorem on a bounded $C^{1}$ domain $Ω$ : for $G \in C^{1} (\overline{Ω}; R^{n})$ , $\int_{Ω} div G d λ_{n} = \int_{\partial Ω} G \cdot ν d S$ . (Divergence on a bounded C1 Euclidean domain)

[F5]

If $G \in C^{1} (R^{n}; R^{n})$ has $G, div G \in L^{1} (λ_{n})$ , then $\int_{R^{n}} div G d λ_{n} = 0$ . (The integral of the divergence of an integrable C1 field vanishes)

[F6]

A continuous function on an interval whose derivative vanishes at every interior point is constant. (A function continuous on an interval $I$ whose derivative vanishes at every interior point of $I$ is constant on $I$ ; consequently two such functions with the same derivative differ by a constant)

[F7]

Second fundamental theorem: for $G$ differentiable on $[a, b]$ with integrable $G^{'}$ , $\int_{a}^{b} G^{'} = G (b) - G (a)$ (Darboux integral). (The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ )

[F8]

On a closed bounded interval a bounded function is Darboux integrable exactly when it is Riemann integrable, with the same value; a bounded Borel Riemann integrable function on a closed interval lies in $L^{1}$ there and its Lebesgue and Riemann integrals agree. (The Darboux and Riemann definitions agree: a bounded $f$ on $[a, b]$ is Darboux integrable with integral $I$ if and only if for every real $ε > 0$ there is a real $δ > 0$ such that $∣ S (f, P, ξ) - I ∣ < ε$ for every tagged partition of mesh below $δ$ , A bounded Riemann integrable function on a closed bounded interval is Lebesgue measurable and has the same integral)

[F9]

Closed bounded Euclidean sets are compact; continuous functions on nonempty compact sets are bounded and uniformly continuous. (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line, A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value, Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous)

[F10]

A continuous real function on a closed bounded interval is bounded and Darboux integrable. (A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion)

Proof

1.1givenF1

Localisation in time and the common shape of the three cases: fix a nondegenerate compact interval $[a, b] \subseteq (0, T)$ ; it suffices to prove that $t \mapsto E_{Ω} (t)$ is constant on $[a, b]$ , since $[a, b]$ is arbitrary and $(0, T)$ is an interval [F6]; the local balance [F1] gives $\partial_{t} e = - div q$ pointwise, while the differentiation and boundedness arguments needed to integrate this identity are supplied separately under the hypotheses of (a), (b), and (c).

2.1givenstep 1.1F1F2F4F6F7F8F9F10

Case (a): choose $R > 0$ with the compact $K$ of the statement contained in $B_{R} (0)$ ; for $t \in [a, b]$ the support condition gives $e (x, t) = 0$ for $x \notin K$ , so $E_{R^{n}} (t) = \int_{B_{R} (0)} e (x, t) d x$ , and $u$ , $u_{t}$ and $D u$ all vanish identically on the open complement of $K$ , hence on a neighbourhood of $\partial B_{R} (0)$ , so $q = 0$ there; [F2] applied on the fixed ball with the domination constant $\sup_{{\overline{B}}_{R} \times [a, b]} ∣ \partial_{t} e ∣$ gives $E_{R^{n}}^{'} (t) = \int_{B_{R} (0)} \partial_{t} e (x, t) d x = - \int_{B_{R} (0)} div q (x, t) d x$ on $(a, b)$ , for $n \geq 2$ , [F4] on the ball gives $\int_{B_{R} (0)} div q d λ_{n} = \int_{\partial B_{R} (0)} q \cdot ν d S = 0$ because $q$ vanishes on the boundary; for $n = 1$ , [F7] and [F8] instead give $\int_{- R}^{R} \partial_{x} q d x = q (R, t) - q (- R, t) = 0$ ; hence $E_{R^{n}}^{'} = 0$ on $(a, b)$ and $E_{R^{n}}$ is constant on $[a, b]$ by [F6].

2.2givenstep 1.1F1F2F5F6

Case (b): the differentiation hypothesis gives $E_{R^{n}}^{'} (t) = \int_{R^{n}} \partial_{t} e (x, t) d x$ for every $t \in (0, T)$ (the stated sufficient Lebesgue criterion follows from [F2] on an open interval compactly contained in $(0, T)$ , with the fixed dominating $L^{1}$ function), and [F1] makes the integrand $- div q (\cdot, t)$ , which lies in $L^{1} (λ_{n})$ by hypothesis; hence $E_{R^{n}}^{'} (t) = - \int_{R^{n}} div q (x, t) d x = 0$ by [F5], and [F6] makes $E_{R^{n}}$ constant on $[a, b]$ .

2.3givenstep 1.1F1F2F4F6F9

Case (c), dimension $n \geq 2$ : $e$ and $\partial_{t} e$ are continuous on the compact $\overline{U} \times [a, b]$ , so [F2] gives $E_{U}^{'} (t) = \int_{U} \partial_{t} e (x, t) d x = - \int_{U} div q (x, t) d x$ on $(a, b)$ [F1], and [F4] gives $\int_{U} div q d λ_{n} = \int_{\partial U} q \cdot ν d S$ ; the boundary integrand vanishes: in the Dirichlet case the map $t \mapsto u (p, t)$ is identically zero at every $p \in \partial U$ and differentiable with derivative $\partial_{t} u (p, t)$ (the $C^{1}$ extension to $\overline{U}$ makes the difference quotient converge to the continuous extension of $\partial_{t} u$ ), so $\partial_{t} u (p, t) = 0$ and hence $q (p, t) \cdot ν (p) = - c^{2} u_{t} (p, t) \partial_{ν} u (p, t) = 0$ ; in the Neumann case $\partial_{ν} u = 0$ on the boundary by hypothesis; either way $q \cdot ν = 0$ on $\partial U \times (a, b)$ , so $E_{U}^{'} = 0$ on $(a, b)$ and [F6] gives constancy on $[a, b]$ .

2.4givenstep 1.1F1F2F6F7F8F9F10

Case (c), dimension $n = 1$ : write $U$ as the disjoint union of its finitely many bounded open intervals $(α_{j}, β_{j})$ with pairwise disjoint closures; for each $j$ the function $x \mapsto q (x, t)$ is $C^{1}$ on $[α_{j}, β_{j}]$ , so on that interval [F7] gives the Darboux integral $\int_{α_{j}}^{β_{j}} \partial_{x} q (x, t) d x = q (β_{j}, t) - q (α_{j}, t)$ , [F8] converts this Darboux value first to the Riemann and then to the Lebesgue integral of $\partial_{x} q (\cdot, t)$ over the interval, and at each endpoint both boundary conditions kill $q$ : $q (β_{j}, t) = - c^{2} u_{t} (β_{j}, t) u_{x} (β_{j}, t)$ and $q (α_{j}, t) = - c^{2} u_{t} (α_{j}, t) u_{x} (α_{j}, t)$ , and in the Dirichlet case $u_{t} = 0$ at both endpoints while in the Neumann case the outward normal is $+ 1$ at $β_{j}$ and $- 1$ at $α_{j}$ , so $u_{x} (β_{j}, t) = 0 = u_{x} (α_{j}, t)$ ; hence $\int_{U} div q (\cdot, t) d λ_{1} = \sum_{j} (q (β_{j}, t) - q (α_{j}, t)) = 0$ . Since $e$ and $\partial_{t} e$ are continuous on $\overline{U} \times [a, b]$ , [F2] gives $E_{U}^{'} (t) = \int_{U} \partial_{t} e d λ_{1} = - \int_{U} div q d λ_{1} = 0$ on $(a, b)$ , and [F6] gives constancy on $[a, b]$ .

3.1step 2.1step 2.2step 2.3step 2.4F6∎

Completion: in each of the three settings, and in both dimensions of case (c), the energy $E_{Ω}$ has vanishing derivative on every nondegenerate compact subinterval of $(0, T)$ , hence is constant on each such subinterval by [F6]; a function constant on every compact subinterval of an interval is constant on the interval, so $E_{Ω}$ is constant on $(0, T)$ in all three settings.

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Energy continuous dependence for the forced wave equation

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $c > 0$ , $T > 0$ and let $u \in C^{2}$ solve the forced equation $□_{c} u = f$ either on $R^{n} \times [0, T)$ or on $U \times [0, T)$ with $U$ a bounded $C^{1}$ domain and homogeneous Dirichlet boundary data; assume the total energy $E (t) = \int_{Ω} e (x, t) d x$ (with $Ω = R^{n}$ , respectively $Ω = U$ ) is finite and continuous on $[0, T)$ , differentiable on $(0, T)$ with the energy identity

$E^{'} (t) = (f (t), u_{t} (t)) : = \int_{Ω} f (x, t) u_{t} (x, t) d x,$

which is what differentiating the energy and inserting The local wave-energy conservation law with vanishing boundary flux gives, and assume that $t \mapsto ∥ f (t) ∥_{2}$ is continuous on $[0, T)$ (Cauchy-Schwarz inequality for $L^{2}$ ). Then for every $t \in [0, T)$

$\sqrt{2 E (t)} \leq \sqrt{2 E (0)} + \int_{0}^{t} ∥ f (s) ∥_{2} d s, hence E (t)^{1 / 2} \leq E (0)^{1 / 2} + \int_{0}^{t} ∥ f (s) ∥_{2} d s .$

This is stability of the classical solution with the sharp $L_{t}^{1} L_{x}^{2}$ forcing constant, not only conservation; the second display uses $1 / \sqrt{2} \leq 1$ .

Facts & Assumptions

Given: ${A C}_{ω}$ ; a finite continuous energy $E : [0, T) \to [0, \infty)$ with $E^{'} (t) = (f (t), u_{t} (t))$ on $(0, T)$ and $E (t) = \frac{1}{2} ∥ u_{t} (t) ∥_{2}^{2} + \frac{c^{2}}{2} ∥ D u (t) ∥_{2}^{2}$ ; a continuous map $t \mapsto ∥ f (t) ∥_{2}$ . Write $w (t) : = ∥ f (t) ∥_{2}$ .

[F1]

Cauchy–Schwarz in $L^{2}$ : $∣ \int g h d μ ∣ \leq ∥ g ∥_{2} ∥ h ∥_{2}$ for $g, h \in L^{2} (μ)$ . (Cauchy-Schwarz inequality for $L^{2}$ )

[F2]

Monotonicity from the derivative: a function continuous on an interval that is differentiable at every interior point with derivative $\leq 0$ there is nonincreasing. (On an interval $I$ , for $f$ continuous on $I$ and differentiable at every interior point: $f^{'} \geq 0$ throughout gives $f$ nondecreasing, $f^{'} > 0$ gives $f$ increasing, $f^{'} \leq 0$ and $f^{'} < 0$ give the two decreasing forms; conversely a nondecreasing $f$ has $f^{'} \geq 0$ and a nonincreasing $f$ has $f^{'} \leq 0$ wherever it is differentiable, and no strict converse is claimed)

[F3]

Chain rule for composites of totally differentiable maps. (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ )

[F4]

For every real $α$ , $s \mapsto s^{α}$ is continuous on $(0, \infty)$ and differentiable there with derivative $α s^{α - 1}$ . (Continuity and derivatives of positive-base real powers)

[F5]

First fundamental theorem: the integral function of a function continuous at a point has derivative equal to the integrand there. (The first fundamental theorem: if $f$ is integrable on $[a, b]$ and continuous at $c$ , then $F^{'} (c) = f (c)$ ; in particular a continuous $f$ has $F$ as a primitive)

[F6]

Continuous functions on a closed bounded interval are integrable, and on such an interval Darboux, Riemann and Lebesgue integrals of a continuous function agree. (A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion, The Darboux and Riemann definitions agree: a bounded $f$ on $[a, b]$ is Darboux integrable with integral $I$ if and only if for every real $ε > 0$ there is a real $δ > 0$ such that $∣ S (f, P, ξ) - I ∣ < ε$ for every tagged partition of mesh below $δ$ , A bounded Riemann integrable function on a closed bounded interval is Lebesgue measurable and has the same integral)

Proof

1.1givenF3F4algebra

The regularised energy: for $ε > 0$ define $φ_{ε} (t) : = \sqrt{2 E (t) + ε^{2}}$ on $[0, T)$ . The radicand is positive, so by the chain rule [F3] and the power rule [F4] with $α = \frac{1}{2}$ , the function $φ_{ε}$ is continuous on $[0, T)$ and differentiable on $(0, T)$ with $φ_{ε}^{'} (t) = \frac{2 E^{'} (t)}{2 \sqrt{2 E (t) + ε^{2}}} = \frac{(f (t), u_{t} (t))}{φ_{ε} (t)}$ .

2.1givenstep 1.1F1algebra

An upper bound for the derivative: by Cauchy–Schwarz [F1], $(f (t), u_{t} (t)) \leq ∣ (f (t), u_{t} (t)) ∣ \leq ∥ f (t) ∥_{2} ∥ u_{t} (t) ∥_{2}$ , and since $2 E (t) = ∥ u_{t} (t) ∥_{2}^{2} + c^{2} ∥ D u (t) ∥_{2}^{2} \geq ∥ u_{t} (t) ∥_{2}^{2}$ we have $∥ u_{t} (t) ∥_{2} \leq \sqrt{2 E (t)} \leq φ_{ε} (t)$ ; hence $φ_{ε}^{'} (t) \leq ∥ f (t) ∥_{2} = w (t)$ for every $t \in (0, T)$ .

3.1givenstep 2.1F2F5F6algebra

Monotonicity of the defect: the function $W (t) : = \int_{0}^{t} w (s) d s$ is, by the continuity of $w$ and [F6], the integral function of a continuous integrand, so by [F5] it is differentiable with $W^{'} = w$ ; hence $ψ : = φ_{ε} - W$ is continuous on $[0, T)$ , differentiable on $(0, T)$ , and $ψ^{'} = φ_{ε}^{'} - w \leq 0$ there by step 2.1; [F2] makes $ψ$ nonincreasing, so for every $t \in [0, T)$ one has $φ_{ε} (t) \leq φ_{ε} (0) + \int_{0}^{t} w (s) d s = \sqrt{2 E (0) + ε^{2}} + \int_{0}^{t} ∥ f (s) ∥_{2} d s$ .

4.1givenstep 3.1F4algebra∎

Letting $ε ↓ 0$ : by the continuity of the square root [F4] and $0 \leq E (t) < \infty$ , $φ_{ε} (t) \to \sqrt{2 E (t)}$ and $φ_{ε} (0) \to \sqrt{2 E (0)}$ , so the inequality of step 3.1 passes to the limit and gives $\sqrt{2 E (t)} \leq \sqrt{2 E (0)} + \int_{0}^{t} ∥ f (s) ∥_{2} d s$ ; dividing by $\sqrt{2} \geq 1$ gives $E (t)^{1 / 2} \leq E (0)^{1 / 2} + \int_{0}^{t} ∥ f (s) ∥_{2} d s$ .

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Energy uniqueness for the wave Cauchy problem

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $c > 0$ , $T > 0$ and let $u$ be a classical solution of $□_{c} u = 0$ on $R^{n} \times (0, T)$ with Cauchy data $(u_{0}, u_{1})$ and differentiable displacement $u_{0}$ , so $D u_{0}$ in the initial-energy hypothesis is defined, and assume the total energy has a finite initial value and is conserved in the sharp form $E_{R^{n}} (t) = E_{R^{n}} (0)$ for every $t \in (0, T)$ , where $E_{R^{n}} (0) = \frac{1}{2} \int_{R^{n}} (u_{1}^{2} + c^{2} ∣ D u_{0} ∣^{2}) d x$ . This holds, for instance, when the solution has fixed compact spatial support (by Conservation of total wave energy in three admissible settings(a) together with continuity of $e$ up to $t = 0$ on the fixed support with its value given by the data density), or when the integrability hypotheses of that theorem's case (b) hold and $E_{R^{n}}$ has a continuous extension to $0$ with value equal to the displayed data energy.

(i) If the Cauchy data vanish, $u_{0} = u_{1} = 0$ , then $E_{R^{n}} (0) = 0$ , hence $E_{R^{n}} (t) = 0$ for all $t$ , hence $u_{t} (\cdot, t) = D u (\cdot, t) = 0$ and $u (\cdot, t)$ is constant on $R^{n}$ for every $t \in (0, T)$ ; the constant is the common limit of $u (\cdot, t)$ as $t ↓ 0$ , namely $u_{0} = 0$ , so $u \equiv 0$ .

(ii) More generally, if $u_{1} = 0$ and $D u_{0} = 0$ (equivalently $E_{R^{n}} (0) = 0$ under the conserved-solution hypotheses), then $u (\cdot, t) \equiv u_{0}$ for every $t$ , where the displacement datum $u_{0}$ is then constant: the energy sees only $(u_{t}, D u)$ , and the displacement datum fixes the residual spatial constant. Consequently, two classical solutions with equal Cauchy data in a class closed under differences, in which each difference has the stated sharp energy conservation, agree.

Facts & Assumptions

Given: ${A C}_{ω}$ ; a classical solution $u$ of $□_{c} u = 0$ on $R^{n} \times (0, T)$ with Cauchy data $(u_{0}, u_{1})$ and differentiable displacement $u_{0}$ , so $D u_{0}$ in the initial-energy hypothesis is defined, whose total energy $E_{R^{n}} (t) = \int_{R^{n}} e (x, t) d x$ has the finite initial value $E_{R^{n}} (0) = \frac{1}{2} \int (u_{1}^{2} + c^{2} ∣ D u_{0} ∣^{2})$ and is conserved in the sharp form $E_{R^{n}} (t) = E_{R^{n}} (0)$ for $t \in (0, T)$ ; the density $e = \frac{1}{2} (u_{t}^{2} + c^{2} ∣ D u ∣^{2}) \geq 0$ of Wave energy density, energy flux and total energy.

[F1]

In the whole-space settings (a) and (b), $E_{R^{n}}$ is constant on $(0, T)$ ; the hypothesis of this corollary records the sharp form in which that constant is the initial value, $E_{R^{n}} (t) = E_{R^{n}} (0)$ for all $t \in (0, T)$ ; $E_{R^{n}} (0)$ is the displayed data energy, not an assertion about endpoint derivatives. (Conservation of total wave energy in three admissible settings)

[F2]

A nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere. (A nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere)

[F3]

On an open convex set, a $C^{1}$ function with vanishing gradient is constant. (Vanishing gradient and time derivative force constancy on convex sets)

[F4]

$R^{n}$ is convex as a subset of itself. (A convex subset of $R^{m}$ contains every line segment between two of its points)

[F5]

The wave operator is linear in its argument, so a difference of two solutions of $□_{c} u = 0$ is again a solution. (Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , Wave equation, Cauchy data and wave speed)

Proof

1.1givenF1F2algebra

Vanishing of the density: if $E_{R^{n}} (0) = 0$ (in particular if $u_{0} = u_{1} = 0$ ), then [F1] gives $E_{R^{n}} (t) = 0$ for every $t \in (0, T)$ . Since $e (\cdot, t) \geq 0$ is continuous, [F2] makes it zero almost everywhere, hence everywhere: a positive value would persist on a ball of positive measure. The sum of squares $2 e = u_{t}^{2} + c^{2} ∣ D u ∣^{2}$ then gives $u_{t} = D u = 0$ pointwise at every positive time.

2.1givenstep 1.1F3F4

Constancy: the space-time set $R^{n} \times (0, T)$ is open and convex by [F4], so [F3] and step 1.1 make $u$ a single constant there. The displacement limit identifies this constant with $u_{0} (x)$ for every $x$ . When $u_{0} = u_{1} = 0$ , this proves clause (i).

3.1givenstep 1.1step 2.1algebra

General zero-energy data: if $u_{1} = 0$ and $D u_{0} = 0$ , the displayed data energy is zero, so steps 1.1 and 2.1 show that $u$ equals the constant datum $u_{0}$ at every positive time. Conversely, if $E_{R^{n}} (0) = 0$ , those steps make $u$ a single constant with $u_{t} = 0$ ; its Cauchy limits give $u_{0}$ constant and $u_{1} = 0$ , hence $D u_{0} = 0$ . This proves clause (ii) and its equivalence without assuming continuity of $D u_{0}$ or $u_{1}$ .

4.1givenstep 2.1F5∎

Uniqueness: for two solutions $u, v$ in the stated class with equal Cauchy data, $w : = u - v$ is homogeneous by [F5] and has zero Cauchy limits. The class hypothesis supplies sharp conservation for $w$ , so clause (i) gives $u = v$ on $R^{n} \times (0, T)$ . Their Cauchy extensions, defined at $t = 0$ by the common displacement datum, also agree there; independently assigned endpoint values are not constrained by the Cauchy limits.

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Energy uniqueness for homogeneous Dirichlet waves on bounded domains

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $c > 0$ , $T > 0$ , let $n \geq 1$ and let $U \subseteq R^{n}$ be a bounded open interval if $n = 1$ , or a bounded connected $C^{1}$ domain if $n \geq 2$ (Bounded C1 domains and their outward normals) and let $u \in C^{2} (\overline{U} \times [0, T])$ solve $□_{c} u = 0$ on $U \times (0, T)$ with homogeneous Dirichlet boundary data $u (\cdot, t) ∣_{\partial U} = 0$ for every $t \in (0, T)$ and homogeneous initial data $u (\cdot, 0) = u_{t} (\cdot, 0) = 0$ . Then $u \equiv 0$ on $\overline{U} \times [0, T]$ . More generally, two such Dirichlet solutions with equal initial data agree on $\overline{U} \times [0, T]$ .

The trace condition $u ∣_{\partial U} = 0$ is stated explicitly because it is exactly the boundary flux that is being killed. Connectedness is retained so the proof can treat $U$ as one spatial component; the same argument applies componentwise on a disconnected domain with the corresponding boundary regularity.

Facts & Assumptions

Given: ${A C}_{ω}$ ; a bounded interval ( $n = 1$ ) or bounded connected $C^{1}$ domain ( $n \geq 2$ ) $U$ , a $C^{2}$ function $u$ on $\overline{U} \times [0, T]$ solving $□_{c} u = 0$ on $U \times (0, T)$ with $u = 0$ on $\partial U \times (0, T)$ and $u (\cdot, 0) = u_{t} (\cdot, 0) = 0$ ; the energy density $e = \frac{1}{2} (u_{t}^{2} + c^{2} ∣ D u ∣^{2}) \geq 0$ of Wave energy density, energy flux and total energy and $E_{U} (t) = \int_{U} e (x, t) d x$ .

[F1]

Conservation in case (c): for a bounded $C^{1}$ domain with homogeneous Dirichlet data, $E_{U}$ is constant on $(0, T)$ . (Conservation of total wave energy in three admissible settings)

[F2]

A nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere; a continuous function on $U$ that vanishes almost everywhere vanishes identically, because a positive value at one point persists on a ball of positive measure. (A nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere)

[F3]

Every connected component of an open subset of $R^{n}$ is open and polygonally connected, and any two points of a polygonally connected set are joined by a polygonal path inside it. (Every connected component of an open subset of $R^{n}$ is open and polygonally connected, Polygonal paths and polygonally connected subsets of $R^{n}$ )

[F4]

[F5]

Chain rule, and linearity of differentiation: a difference of two solutions of $□_{c} u = 0$ is again a solution. (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ , Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , Wave equation, Cauchy data and wave speed)

[F6]

Proof

1.1givenF1F2algebraF6

Vanishing of the density and of the first derivatives: $E_{U} (0) = 0$ because $u_{t} (\cdot, 0) = D u (\cdot, 0) = 0$ ; since $e$ is continuous on the compact set $\overline{U} \times [0, T]$ , it is uniformly continuous there, and boundedness of $U$ gives $∣ E_{U} (t) - E_{U} (0) ∣ \leq λ_{n} (U) \sup_{x \in \overline{U}} ∣ e (x, t) - e (x, 0) ∣ \to 0$ as $t ↓ 0$ . By [F1], $E_{U}$ is constant on $(0, T)$ , so this continuity identifies that constant with $E_{U} (0) = 0$ ; for each $t \in (0, T)$ , nonnegativity and continuity of $e$ together with [F2] give $e (\cdot, t) = 0$ on $U$ , hence $u_{t} (\cdot, t) = 0$ and $D u (\cdot, t) = 0$ there, and continuity of these derivatives extends their vanishing to $t = 0$ and $t = T$ .

2.1givenstep 1.1F3F4F5algebra

Spatial constancy on the connected domain: fix $t$ and $p, q \in U$ ; since $U$ is open and connected, [F3] supplies a polygonal path in $U$ from $p$ to $q$ , say with successive vertices $p = a_{0}, \dots, a_{m} = q$ ; for each segment put $g (s) : = u (a_{j - 1} + s (a_{j} - a_{j - 1}), t)$ for $s \in [0, 1]$ ; by the chain rule [F5], $g$ is continuous on $[0, 1]$ and differentiable there with $g^{'} (s) = D u (a_{j - 1} + s (a_{j} - a_{j - 1}), t) \cdot (a_{j} - a_{j - 1}) = 0$ , so [F4] makes $g$ constant; chaining over $j = 1, \dots, m$ gives $u (p, t) = u (q, t)$ , so $u (\cdot, t)$ is constant on $U$ .

3.1givenstep 2.1algebra

The constant is zero: $U$ is nonempty, bounded and open, so $\partial U \neq \emptyset$ ; fix $t \in [0, T]$ and $q \in \partial U$ and a sequence $p_{k} \in U$ with $p_{k} \to q$ ; by step 2.1, $u (p_{k}, t) = u (p_{1}, t)$ for all $k$ , while continuity of $u$ on $\overline{U} \times [0, T]$ and the boundary condition give $u (p_{k}, t) \to u (q, t) = 0$ ; hence $u (\cdot, t) \equiv 0$ on $U$ , and by continuity on $\overline{U}$ .

4.1givenstep 3.1F5∎

Uniqueness for two solutions: if $u$ and $v$ are two such Dirichlet solutions with equal initial data, their difference $w : = u - v$ is $C^{2}$ on $\overline{U} \times [0, T]$ , solves $□_{c} w = 0$ there by linearity [F5], vanishes on $\partial U \times (0, T)$ and has $w (\cdot, 0) = w_{t} (\cdot, 0) = 0$ ; steps 1.1–3.1 applied to $w$ give $w \equiv 0$ on $\overline{U} \times [0, T]$ , that is, $u = v$ .

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Finite propagation speed for the wave equation

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $n \geq 1$ , $c > 0$ , $x_{0} \in R^{n}$ , $t_{0} > 0$ and let $u \in C^{2}$ solve $□_{c} u = f$ on a neighbourhood of the closed backward cone $K^{-} (x_{0}, t_{0}) = {(x, t) : 0 \leq t \leq t_{0}, ∣ x - x_{0} ∣ \leq c (t_{0} - t)}$ (Forward and backward wave cones, domain of dependence and influence). If $f = 0$ on $K^{-} (x_{0}, t_{0})$ and $u (\cdot, 0) = u_{t} (\cdot, 0) = 0$ on the base ball $B_{c t_{0}} (x_{0})$ , then $u \equiv 0$ on $K^{-} (x_{0}, t_{0})$ ; in particular $u (x_{0}, t_{0}) = 0$ . Data and source vanishing in a backward cone control the whole cone: the source term is included, in the sharp form of the enrichment row thm-finite-propagation-for-forced-waves.

Facts & Assumptions

Given: ${A C}_{ω}$ ; a $C^{2}$ function $u$ solving $□_{c} u = f$ on a neighbourhood of the closed cone $K^{-} = K^{-} (x_{0}, t_{0})$ , with $f = 0$ on $K^{-}$ and $u (\cdot, 0) = u_{t} (\cdot, 0) = 0$ on $B_{c t_{0}} (x_{0})$ ; the density $e = \frac{1}{2} (u_{t}^{2} + c^{2} ∣ D u ∣^{2}) \geq 0$ of Wave energy density, energy flux and total energy and $E (t) = \int_{B_{c (t_{0} - t)} (x_{0})} e (x, t) d x$ for $0 < t < t_{0}$ .

[F1]

Cone energy identity: for $0 < t_{1} < t_{2} < t_{0}$ , with $ℓ \geq 0$ on the lateral surface, $\int_{K (t_{1}, t_{2})} f u_{t} = E (t_{2}) - E (t_{1}) + \int_{t_{1}}^{t_{2}} \int_{\partial B_{c (t_{0} - t)} (x_{0})} ℓ d S d t$ . (The energy identity on a truncated wave cone)

[F2]

Dominated convergence: if $f_{k} \to f$ pointwise and $∣ f_{k} ∣ \leq g$ with $\int g < \infty$ , then $\int f_{k} \to \int f$ . (Dominated convergence)

[F3]

A nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere; a continuous nonnegative function with vanishing integral on an open ball vanishes identically there. (A nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere)

[F4]

On an open convex set, a $C^{1}$ function with vanishing gradient is constant; the open cone $int K^{-} = {0 < t < t_{0}, ∣ x - x_{0} ∣ < c (t_{0} - t)}$ is convex, being the increasing union of the convex frusta $K (t_{1}, t_{2})$ . (Vanishing gradient and time derivative force constancy on convex sets, Truncated wave cones: convexity, piecewise C1 presentation and outward normals)

[F5]

Proof

1.1givenF2algebraF5

The energy tends to zero at the base: for every $t \in (0, t_{0})$ the integral $E (t)$ over the ball $B_{c (t_{0} - t)} (x_{0})$ is finite and $E (t) \geq 0$ because $e \geq 0$ is continuous on the compact set $K^{-}$ and hence bounded there; as $t ↓ 0$ , the functions $x \mapsto 1_{B_{c (t_{0} - t)} (x_{0})} (x) e (x, t)$ converge pointwise on $B_{c t_{0}} (x_{0})$ to $1_{B_{c t_{0}} (x_{0})} (x) e (x, 0)$ by continuity of $e$ up to $t = 0$ , and they are dominated by the constant $\sup_{K^{-}} e < \infty$ , so [F2] gives $E (t) \to \int_{B_{c t_{0}} (x_{0})} e (x, 0) d x = \frac{1}{2} \int_{B_{c t_{0}} (x_{0})} (u_{t} (x, 0)^{2} + c^{2} ∣ D u (x, 0) ∣^{2}) d x = 0$ , the last equality because both Cauchy data vanish on the base ball.

2.1givenstep 1.1F1algebra

Monotonicity and vanishing of the energy: for $0 < t_{1} < t_{2} < t_{0}$ the frustum $K (t_{1}, t_{2})$ lies in $K^{-}$ , where $f = 0$ , so [F1] gives $E (t_{2}) - E (t_{1}) = - \int_{t_{1}}^{t_{2}} \int_{\partial B_{c (t_{0} - t)} (x_{0})} ℓ d S d t \leq 0$ because $ℓ \geq 0$ ; thus $E$ is nonincreasing on $(0, t_{0})$ , with $E \geq 0$ and $E (t) \to 0$ as $t ↓ 0$ by step 1.1, so $E (t) = 0$ for every $t \in (0, t_{0})$ .

3.1givenstep 2.1F3algebra

Vanishing of the derivatives on the open cone: fix $t \in (0, t_{0})$ ; by step 2.1 $e (\cdot, t) \geq 0$ has vanishing integral over the open ball $B_{c (t_{0} - t)} (x_{0})$ , so [F3] and continuity give $e (x, t) = 0$ for every $x$ in that ball, and hence $u_{t} (x, t) = 0$ and $D u (x, t) = 0$ there; letting $t$ vary gives $u_{t} = D u = 0$ on the open cone $int K^{-}$ .

4.1givenstep 3.1F4algebra∎

Constancy and conclusion: the open cone $int K^{-}$ is convex [F4], so the vanishing-gradient lemma [F4] makes $u$ constant on it; the constant is $0$ because $u$ is continuous on a neighbourhood of the closed cone and $u (\cdot, 0) = 0$ on the base ball, so evaluating along points of the open cone tending to a base point gives $u \equiv 0$ on $int K^{-}$ ; finally $K^{-}$ is the closure of $int K^{-}$ (each point of the base, of the lateral surface or the vertex is a limit of interior points), so continuity gives $u \equiv 0$ on $K^{-}$ , and in particular $u (x_{0}, t_{0}) = 0$ .

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Compact support expands at speed at most c

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $c > 0$ , $T > 0$ , let $K \subseteq R^{n}$ be compact and let $u$ be a classical solution of $□_{c} u = f$ on a neighbourhood of $R^{n} \times [0, T)$ with Cauchy data $(u_{0}, u_{1})$ satisfying $supp u_{0} \cup supp u_{1} \subseteq K$ (The support of a function on $R^{n}$ and its compactly supported Riemann integral) and $supp f \subseteq {(x, t) : 0 \leq t < T, dist (x, K) \leq c t}$ (support relative to this time slab). For $K = \emptyset$ , use $dist (x, K) = + \infty$ , so the source is zero and the asserted support is empty. Then for every $t \in [0, T)$

$supp u (\cdot, t) \subseteq K + {\overline{B}}_{c t} (0) = {x : dist (x, K) \leq c t},$

the time- $t$ domain of influence of the data support (Forward and backward wave cones, domain of dependence and influence).

Facts & Assumptions

Given: ${A C}_{ω}$ ; a compact $K$ , a $C^{2}$ solution $u$ , defined near the closed initial slab, of $□_{c} u = f$ with data supported in $K$ and source supported in ${(y, s) : dist (y, K) \leq c s}$ .

[F1]

Finite propagation: for $t > 0$ , if $w$ solves $□_{c} w = g$ on a neighbourhood of the closed backward cone $K^{-} (x, t)$ with $g = 0$ there and $w (\cdot, 0) = w_{t} (\cdot, 0) = 0$ on the base ball $B_{c t} (x)$ , then $w (x, t) = 0$ . (Finite propagation speed for the wave equation)

[F2]

$supp u_{0} = \overline{{x : u_{0} (x) \neq 0}}$ and likewise for $u_{1}$ ; a point outside $supp f$ has $f = 0$ there. (The support of a function on $R^{n}$ and its compactly supported Riemann integral)

[F3]

The base ball of $K^{-} (x, t)$ is the open ball $B_{c t} (x) = x + B_{c t} (0)$ . For nonempty compact $K$ , the continuous function $z \mapsto ∣ x - z ∣$ attains a minimum on $K$ , so $K + {\overline{B}}_{c t} (0) = {x : dist (x, K) \leq c t}$ is closed. Also $∣ x - z ∣ \leq ∣ x - y ∣ + ∣ y - z ∣$ for every $z \in K$ ; taking infima gives $dist (x, K) \leq ∣ x - y ∣ + dist (y, K)$ . (A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value, Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation) (Forward and backward wave cones, domain of dependence and influence)

Proof

1.1givenF1F2F3algebra

Reduction to a point outside the domain of influence: if $K = \emptyset$ , all data and the source vanish, so [F1] applied at every $(x, t)$ with $t \in (0, T)$ gives $u = 0$ there; at $t = 0$ , continuity and the Cauchy displacement limit give $u (\cdot, 0) = u_{0} = 0$ , so the support inclusion holds throughout $[0, T)$ . Otherwise let $t \in (0, T)$ and $x \notin K + {\overline{B}}_{c t} (0)$ , so $dist (x, K) > c t$ ; then the base ball of $K^{-} (x, t)$ is $B_{c t} (x)$ by [F3], and $B_{c t} (x) \cap K = \emptyset$ : if $z \in B_{c t} (x) \cap K$ then $dist (x, K) \leq ∣ x - z ∣ < c t$ , contradicting $dist (x, K) > c t$ ; hence the initial data vanish on $B_{c t} (x)$ by [F2]: $u_{0} = u_{1} = 0$ there; and the source vanishes on the cone: if $(y, s) \in K^{-} (x, t)$ had $dist (y, K) \leq c s$ , then $dist (x, K) \leq ∣ x - y ∣ + dist (y, K) \leq c (t - s) + c s = c t$ by [F3], a contradiction, so $dist (y, K) > c s$ and $(y, s) \notin supp f$ , i.e. $f (y, s) = 0$ .

2.1givenstep 1.1F1F3∎

Conclusion: by step 1.1 the data and source of $□_{c} u = f$ vanish in the cone $K^{-} (x, t)$ , so [F1] applied to $u$ gives $u (x, t) = 0$ ; as $x \notin K + {\overline{B}}_{c t} (0)$ was arbitrary, every point outside $K + {\overline{B}}_{c t} (0)$ has $u (\cdot, t) = 0$ , and the containing set is closed by [F3], whence $supp u (\cdot, t) \subseteq K + {\overline{B}}_{c t} (0)$ for every $t \in (0, T)$ ; at $t = 0$ , continuity and the Cauchy displacement limit give $u (\cdot, 0) = u_{0}$ , so the inclusion is exactly the support hypothesis on $u_{0}$ .

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Time-reversed energy uniqueness from final data

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $c > 0$ , $T > 0$ and let $u, v \in C^{2} (R^{n} \times [0, T])$ solve $□_{c} w = 0$ , with $u$ and $v$ lying in a class of homogeneous solutions that is closed under taking differences and in which the total energy is conserved on $[0, T]$ in one of the whole-space senses (a) or (b) of Conservation of total wave energy in three admissible settings (for instance both have fixed compact spatial support inside a common compact set). If $u (\cdot, T) = v (\cdot, T)$ and $u_{t} (\cdot, T) = v_{t} (\cdot, T)$ , then $u = v$ on $R^{n} \times [0, T]$ .

Thus equal terminal displacement and velocity determine the same finite-energy homogeneous solution backward in time; reversibility is a consequence of energy uniqueness together with time-reversal symmetry, not of any representation formula. The time-reversal item Time reversal of the homogeneous wave equation itself belongs to the preceding pair and is consumed here, not reproved.

Facts & Assumptions

Given: ${A C}_{ω}$ ; two homogeneous solutions $u, v$ in the stated class with equal Cauchy data at time $T$ ; the difference $w = u - v$ lies in the class, so its energy is conserved on $[0, T]$ .

[F1]

Apply the time-reversal theorem to $w$ on the open interval $I = (0, T)$ with $τ = T / 2$ : $\tilde{w} (x, s) : = w (x, T - s)$ is $C^{2}$ and homogeneous in the interior. Continuity of $w$ and its derivatives to the endpoints gives the reflected extension on $[0, T]$ , with $\tilde{w} (\cdot, 0) = w (\cdot, T)$ and $\partial_{s} \tilde{w} (\cdot, 0) = - w_{t} (\cdot, T)$ . (Time reversal of the homogeneous wave equation)

[F2]

Energy uniqueness for the Cauchy problem: a homogeneous solution whose total energy is conserved on the time interval and whose Cauchy data vanish at the initial time is identically zero; two conserved solutions with equal initial data agree. (Energy uniqueness for the wave Cauchy problem)

[F3]

The energy density of the time-reversed solution at time $s$ equals the energy density of the original solution at time $T - s$ : the spatial gradient is unchanged and $\partial_{s} \tilde{w} = - \partial_{t} w$ . (Wave energy density, energy flux and total energy)

Proof

1.1givenF1

The difference and its reversal: $w : = u - v$ is $C^{2}$ and homogeneous by linearity, and it lies in the stated class, so its total energy is conserved on $[0, T]$ ; by hypothesis $w (\cdot, T) = 0$ and $w_{t} (\cdot, T) = 0$ ; define $\tilde{w} (x, s) : = w (x, T - s)$ for $s \in [0, T]$ .

2.1givenstep 1.1F1F3algebra

The reversal is a conserved homogeneous solution with zero initial data: by [F1], $\tilde{w}$ is a $C^{2}$ homogeneous solution on $R^{n} \times [0, T]$ with $\tilde{w} (\cdot, 0) = w (\cdot, T) = 0$ and $\partial_{s} \tilde{w} (\cdot, 0) = - w_{t} (\cdot, T) = 0$ ; by [F3] the energy density of $\tilde{w}$ at time $s$ equals that of $w$ at time $T - s$ , so $E_{\tilde{w}} (s) = E_{w} (T - s)$ and conservation of $E_{w}$ transfers to $\tilde{w}$ .

3.1step 2.1F2∎

Conclusion: [F2] applied to the conserved solution $\tilde{w}$ with vanishing Cauchy data gives $\tilde{w} \equiv 0$ ; hence $w \equiv 0$ , that is, $u = v$ on $R^{n} \times [0, T]$ .

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The strong Huygens principle in the homogeneous Cauchy setting

Definition

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $n \geq 1$ and $c > 0$ , and fix the homogeneous Cauchy setting of this page. Admissible data for dimension $n$ is a pair $(u_{0}, u_{1})$ in the regularity class of the dimension- $n$ representation formula of the preceding pair wave-equation-representation-formulas (Wave equation, Cauchy data and wave speed, Spherical means and the weighted ball integral of space-dependent data), and the solution is the $C^{2}$ solution on $R^{n} \times (0, \infty)$ defined by that formula and attaining the data at $t = 0$ (The dimension formulas attain the Cauchy data for $n \geq 2$ and d'Alembert's formula and uniqueness in one dimension for $n = 1$ ); in a class closed under differences with sharp energy conservation it is the unique such solution by Energy uniqueness for the wave Cauchy problem.

Fix $x_{0} \in R^{n}$ , $t_{0} > 0$ and put

$S : = \partial B_{c t_{0}} (x_{0}) = {y \in R^{n} : ∣ y - x_{0} ∣ = c t_{0}} .$

The homogeneous Cauchy problem with speed $c$ satisfies the strong Huygens principle in dimension $n$ when, for every $(x_{0}, t_{0})$ and every admissible data pair $(u_{0}, u_{1})$ , the value $u (x_{0}, t_{0})$ is unchanged when $(u_{0}, u_{1})$ is replaced by an admissible pair that agrees with $(u_{0}, u_{1})$ on a neighbourhood of $S$ ; equivalently, the data-to-value functional is carried by the sphere $S$ : admissible perturbations of the data that vanish on a neighbourhood of $S$ do not change the value $u (x_{0}, t_{0})$ .

The following formulations are equivalent for these linear representation formulas, which have base-ball locality: data vanishing on a neighbourhood of the closed base ball contribute zero. To check equivalence, choose a smooth radial cutoff equal to one on the closed base ball and supported in a slightly larger ball (A smooth bump between concentric Euclidean balls). Multiplying a perturbation by this cutoff does not change its value in the formula, since the data and all derivatives read at radius $c t_{0}$ are unchanged; for the even formula this follows by writing its weighted ball integral on the fixed unit ball and differentiating the smooth data there. A compactly supported perturbation vanishing near $S$ splits into an interior part, supported compactly in $B_{c t_{0}} (x_{0})$ , and an exterior part, vanishing near the closed base ball; the split is smooth because the perturbation is zero on a collar of $S$ . Thus (i) implies neighbourhood invariance, while the reverse follows because a closed support contained in the open base ball is compact and separated from $S$ . Neighbourhood invariance implies (ii) by comparing with zero data; conversely (ii), applied to the cutoff perturbation whose compact support misses $S$ , implies neighbourhood invariance. This proves the equivalence of (i), (ii) and (iii).

(i) Strictly-inside form. The value does not depend on the data at points $y$ with $∣ y - x_{0} ∣ < c t_{0}$ : changes supported in the open base ball $B_{c t_{0}} (x_{0})$ do not affect $u (x_{0}, t_{0})$ .

(ii) Shell form for compactly supported data. If the data are supported in the compact set $K$ , then $u (x_{0}, t_{0}) = 0$ whenever $S \cap K = \emptyset$ , that is, the value at $(x_{0}, t_{0})$ is carried by the $c t_{0}$ -sphere shell of $K$ ; this is the classical sharp-Huygens picture.

(iii) Germ form. Data agreeing on a neighbourhood of $S$ give the same value. In odd dimensions, the finite radial jets that suffice are identified in The strong Huygens principle in odd spatial dimensions.

Caution. The value may involve finitely many transverse (radial) derivatives of the data, so agreement of the bare restrictions to $S$ is not in general enough. For $n = 3$ , $c = 1$ , the radial datum $u_{0} (y) = (1 - ∣ y ∣) ψ (∣ y ∣)$ with $ψ$ smooth, $ψ \equiv 1$ near $∣ y ∣ = 1$ and $ψ \equiv 0$ near $0$ , and $u_{1} = 0$ vanishes on $S = \partial B_{1} (0)$ , yet Kirchhoff's formula (Kirchhoff's formula in three dimensions) gives $u (0, 1) = u_{0} (1) + u_{0}^{'} (1) = 0 - 1 = - 1 \neq 0$ ; the data-to-value functional reads the first radial derivative of $u_{0}$ along $S$ , as well as its values, and cannot be represented by integrating only the bare restriction. The precise positive statement is The strong Huygens principle in odd spatial dimensions and the failures are Wave tails in one and even spatial dimensions: strong Huygens fails. This definition asserts no existence or uniqueness beyond the classical class above, and it does not imply that the principle holds; whether it holds is exactly the content of those theorems.

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Domain of dependence and local uniqueness

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $n \geq 1$ , $c > 0$ , $x_{0} \in R^{n}$ , $t_{0} > 0$ , and let $u, v \in C^{2}$ solve $□_{c} u = f$ and $□_{c} v = g$ on a neighbourhood of the closed backward cone $K^{-} (x_{0}, t_{0})$ (Forward and backward wave cones, domain of dependence and influence). If $f = g$ on $K^{-} (x_{0}, t_{0})$ and $u (\cdot, 0) = v (\cdot, 0)$ , $u_{t} (\cdot, 0) = v_{t} (\cdot, 0)$ on the base ball $B_{c t_{0}} (x_{0})$ , then $u = v$ on $K^{-} (x_{0}, t_{0})$ .

This is the formal content, required by the design's well-definedness note, of the phrase "the value at $(x_{0}, t_{0})$ depends only on the Cauchy data on the base ball and the source on the cone".

Facts & Assumptions

Given: ${A C}_{ω}$ ; two $C^{2}$ functions $u, v$ solving $□_{c} u = f$ , $□_{c} v = g$ near the cone $K^{-} (x_{0}, t_{0})$ , with $f = g$ on the cone and equal Cauchy data on the base ball $B_{c t_{0}} (x_{0})$ .

[F1]

Finite propagation: if $w$ solves $□_{c} w = h$ near $K^{-} (x^{'}, t^{'})$ with $h = 0$ on the cone and zero Cauchy data on its base ball, then $w = 0$ on $K^{-} (x^{'}, t^{'})$ . (Finite propagation speed for the wave equation)

[F2]

The wave operator is linear, so $u - v$ solves $□_{c} (u - v) = f - g$ and the Cauchy data of the difference are the differences of the data. (Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , Wave equation, Cauchy data and wave speed)

Proof

1.1givenF2

The difference solves a homogeneous equation on the cone: $w : = u - v$ is $C^{2}$ on a neighbourhood of $K^{-} (x_{0}, t_{0})$ and, by linearity of differentiation [F2], $□_{c} w = f - g$ , which vanishes on the cone because $f = g$ there; moreover $w (\cdot, 0) = 0$ and $w_{t} (\cdot, 0) = 0$ on the base ball $B_{c t_{0}} (x_{0})$ because the Cauchy data of $u$ and $v$ agree there.

2.1step 1.1F1∎

Conclusion: the hypotheses of [F1] are met by $w$ on the cone $K^{-} (x_{0}, t_{0})$ , so $w = 0$ there, that is, $u = v$ on $K^{-} (x_{0}, t_{0})$ ; in particular the value at the vertex is determined by the base data and the source on the cone.

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The strong Huygens principle in odd spatial dimensions

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $n = 2 k + 1 \geq 3$ be odd, $c > 0$ , and let $u$ be the $C^{2}$ solution given by the odd-dimensional formula The odd-dimensional wave formula by iterated spherical means for admissible data $(u_{0}, u_{1})$ with $u_{0} \in C^{k + 2} (R^{n})$ , $u_{1} \in C^{k + 1} (R^{n})$ (Spherical means and the weighted ball integral of space-dependent data). Fix $x_{0} \in R^{n}$ , $t_{0} > 0$ and $S = \partial B_{c t_{0}} (x_{0})$ . Then:

(a) (germ form) $u (x_{0}, t_{0})$ is a finite linear combination $\sum_{j \leq k} α_{j} \partial_{r}^{j} M_{u_{0}} (x_{0}, c t_{0}) + \sum_{j \leq k - 1} β_{j} \partial_{r}^{j} M_{u_{1}} (x_{0}, c t_{0})$ of radial derivatives of spherical means, and $\partial_{r}^{j} M_{f} (x_{0}, r) = \frac{1}{ω_{n - 1}} \int_{S^{n - 1}} \partial_{r}^{j} [f (x_{0} + r ω)] d σ (ω) (r = c t_{0}),$ so the value is determined by the jet of $(u_{0}, u_{1})$ on $S$ : admissible data agreeing on a neighbourhood of $S$ give the same value at $(x_{0}, t_{0})$ ;

(b) (shell form) if the data are supported in the compact $K$ , then for $t > 0$ , $u (x, t) = 0$ whenever $\partial B_{c t} (x) \cap K = \emptyset$ ; in particular for $K = {\overline{B}}_{r} (x_{0})$ and $c t > r$ the ball $B_{c t - r} (x_{0})$ is quiet, and its support is contained in the annulus $c t - r \leq ∣ x - x_{0} ∣ \leq c t + r$ (this does not assert that both bounding spheres are occupied).

Hence the strong Huygens principle of The strong Huygens principle in the homogeneous Cauchy setting holds in dimension $n$ (germ form (iii), and with it the strictly-inside form (i) and the shell form (ii)); bare agreement of the restrictions to $S$ is not sufficient in general, exactly because the transverse derivatives displayed above may differ (the caution in The strong Huygens principle in the homogeneous Cauchy setting).

Facts & Assumptions

Given: ${A C}_{ω}$ ; odd $n = 2 k + 1 \geq 3$ , $c > 0$ , admissible data $(u_{0}, u_{1})$ in the stated classes; the solution $u (x, t) = \frac{1}{(n - 2)!!} [\partial_{t} D_{t}^{k - 1} (t^{n - 2} M_{u_{0}} (x, c t)) + D_{t}^{k - 1} (t^{n - 2} M_{u_{1}} (x, c t))]$ of The odd-dimensional wave formula by iterated spherical means, with $D_{t} = t^{- 1} \partial_{t}$ and $M_{f}$ the spherical mean of Spherical means and the weighted ball integral of space-dependent data.

[F1]

For $C^{m}$ data, $r \mapsto M_{f} (x, r)$ is $C^{m}$ for $r > 0$ and the derivatives may be taken under the compact sphere integral. (Smoothness, parity and zero-radius limits of spherical means)

[F2]

Chain rule: $\partial_{t}^{j} M_{f} (x, c t) = c^{j} \partial_{r}^{j} M_{f} (x, c t)$ for the one-variable composition. (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ )

[F3]

Differentiation under an integral over the compact sphere with continuous integrand. (Differentiation under the integral sign)

[F4]

Product rule and linearity of differentiation for the expansion of $D_{t}^{k - 1} [t^{n - 2} g (t)]$ . (Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ )

[F5]

Nonempty disjoint compact subsets of $R^{n}$ have a positive distance gap. (A compact set and a disjoint closed set have a positive norm-distance gap, For $n \geq 1$ , every Euclidean closed ball and every Euclidean sphere of positive radius is compact)

[F6]

The support of a function is the closure of its nonzero set; a function vanishing on a neighbourhood of $S (x, c t)$ has zero data there. (The support of a function on $R^{n}$ and its compactly supported Riemann integral)

Proof

1.1givenF1F2F3F4algebra

The value is a finite functional of radial spherical-mean derivatives: expanding $D_{t}^{k - 1} [t^{n - 2} g (t)]$ by the product rule [F4] gives $\sum_{j \leq k - 1} p_{j} (t) g^{(j)} (t)$ with coefficients $p_{j} (t) = a_{j} t^{1 + j}$ , where the constants $a_{j}$ depend only on $n, k$ ; this follows by induction because $D_{t} (t^{a} g^{(j)}) = a t^{a - 2} g^{(j)} + t^{a - 1} g^{(j + 1)}$ ; applying this to $g (t) = M_{u_{0}} (x_{0}, c t)$ and to $g (t) = M_{u_{1}} (x_{0}, c t)$ in the odd-dimensional formula, and applying one further $\partial_{t}$ to the first bracket, exhibits $u (x_{0}, t_{0})$ as $\sum_{j \leq k} α_{j} \partial_{t}^{j} M_{u_{0}} (x_{0}, c t_{0}) + \sum_{j \leq k - 1} β_{j} \partial_{t}^{j} M_{u_{1}} (x_{0}, c t_{0})$ with finite coefficients $α_{j}, β_{j}$ ; the chain rule [F2] converts $\partial_{t}^{j} M_{f} (x_{0}, c t_{0})$ into $c^{j} \partial_{r}^{j} M_{f} (x_{0}, c t_{0})$ , and [F1] with [F3] gives the displayed sphere-integral formula $\partial_{r}^{j} M_{f} (x_{0}, r) = \frac{1}{ω_{n - 1}} \int_{S^{n - 1}} \partial_{r}^{j} [f (x_{0} + r ω)] d σ (ω)$ .

2.1givenstep 1.1F1algebra

The jet along $S$ determines the value: each derivative $\partial_{r}^{j} [f (x_{0} + r ω)]$ at $r = c t_{0}$ is the $j$ -th radial derivative of $f$ at the point $x_{0} + c t_{0} ω \in S$ , hence is determined by the values of $f$ on any neighbourhood of that point; consequently, if two admissible data pairs agree on a neighbourhood of $S$ , their difference $f$ vanishes on that neighbourhood and all derivatives of $f$ vanish along $S$ , so every integral in step 1.1 vanishes for the difference and the two data pairs give the same value $u (x_{0}, t_{0})$ ; this proves (a) and the germ form (iii).

3.1givenstep 2.1F5F6algebra

The shell form: suppose the data are supported in the compact $K$ and $\partial B_{c t} (x) \cap K = \emptyset$ ; if $K = \emptyset$ the data are zero and the conclusion is immediate; otherwise the two compact sets $\partial B_{c t} (x)$ and $K$ have a positive distance gap [F5], so the data vanish on a neighbourhood of $\partial B_{c t} (x)$ by [F6]; comparing the given data with the zero data — which agree on that neighbourhood and give the solution $0$ with value $0$ — step 2.1 gives $u (x, t) = 0$ ; hence the value is carried by the $c t$ -sphere shell of $K$ , and for $K = {\overline{B}}_{r} (x_{0})$ and $c t > r$ every $x$ with $∣ x - x_{0} ∣ < c t - r$ has $\partial B_{c t} (x) \cap K = \emptyset$ , so the interior ball is quiet, giving (b).

4.1step 2.1step 3.1∎

Conclusion: the germ form and the shell form established above are exactly the forms (iii) and (ii) of The strong Huygens principle in the homogeneous Cauchy setting, and the strictly-inside form (i) follows because a perturbation supported in the open base ball is supported away from $S$ ; hence the strong Huygens principle holds in every odd dimension $n \geq 3$ .

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Wave tails in one and even spatial dimensions: strong Huygens fails

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Strong Huygens fails in dimension $1$ and in every even dimension $n \geq 2$ , in the sharp sense that there are admissible compactly supported data whose difference from the zero data is supported in a compact subset of the open base ball $B_{c t_{0}} (x_{0})$ (hence vanishes in a neighbourhood of the sphere $S (x_{0}, t_{0})$ , The strong Huygens principle in the homogeneous Cauchy setting) yet whose values at $(x_{0}, t_{0})$ differ.

(i) $n = 1$ : for every $x_{0} \in R$ , $t_{0} > 0$ and every $u_{1} \in C_{c}^{1} (R)$ supported in $(x_{0} - c t_{0}, x_{0} + c t_{0})$ with $\int_{R} u_{1} \neq 0$ , the pair $(0, u_{1})$ has $u (x_{0}, t_{0}) = \frac{1}{2 c} \int_{x_{0} - c t_{0}}^{x_{0} + c t_{0}} u_{1} \neq 0$ by d'Alembert's formula d'Alembert's formula and uniqueness in one dimension, while the zero data give $u \equiv 0$ .

(ii) $n = 2 k$ even: for every $x_{0}, t_{0}$ and all sufficiently small $ε > 0$ there are $u_{1} \in C_{c}^{k + 1} (B_{ε} (x_{0}))$ , $u_{1} \geq 0$ , $u_{1} \equiv̸ 0$ , with $u (x_{0}, t_{0}) = \int_{B_{ε} (x_{0})} u_{1} (y) K (∣ y - x_{0} ∣, t_{0}) d y \neq 0,$ where, with $D_{t} = t^{- 1} \partial_{t}$ , the kernel of the even-dimensional formula The even-dimensional wave formula by descent is $K (ρ, t) = c^{1 - n} D_{t}^{k - 1} [\frac{(c^{2} t^{2} - ρ^{2})^{- 1 / 2}}{n!! V_{n}}], K (0, t_{0}) = c^{- n} (n!! V_{n})^{- 1} (- 1)^{k - 1} (2 k - 3)!! t_{0}^{- (n - 1)} \neq 0$ (with $(2 k - 3)!! = 1$ for $k = 1$ ), so $K$ has a constant sign on a small ball and the integral is nonzero; the $n = 2$ instance is Poisson's formula Poisson's formula in two dimensions by descent.

Thus data supported strictly inside the base ball affect the value: failure is proved by interior data, not merely by a formula contrast.

Facts & Assumptions

Given: ${A C}_{ω}$ ; the two families of admissible data of (i) and (ii); for (ii) the descended even-dimensional formula with $D_{t} = t^{- 1} \partial_{t}$ , $V_{n} = ∣ B_{1}^{n} ∣$ , $ω_{n - 1} = n V_{n}$ (Sphere and ball measures scale in Rn).

[F1]

Even-dimensional formula by descent: for admissible data $(u_{0}, u_{1})$ in dimension $n = 2 k$ the solution is given by the descended spherical-mean formula, whose velocity term is $c^{1 - n} D_{t}^{k - 1} W_{u_{1}} (x, t)$ with $W_{u_{1}} (x, t) = \int_{B_{c t} (x)} u_{1} (y) (c^{2} t^{2} - ∣ y - x ∣^{2})^{- 1 / 2} (n!! V_{n})^{- 1} d y$ . (The even-dimensional wave formula by descent, Spherical means and the weighted ball integral of space-dependent data)

[F2]

Poisson's formula is the case $n = 2$ of the same family. (Poisson's formula in two dimensions by descent)

[F3]

D'Alembert's formula: $u (x, t) = \frac{1}{2} [u_{0} (x + c t) + u_{0} (x - c t)] + \frac{1}{2 c} \int_{x - c t}^{x + c t} u_{1}$ . (d'Alembert's formula and uniqueness in one dimension)

[F4]

Differentiation under the integral sign, applied on the compact ball where the integrand and all its $t$ -derivatives are smooth. (Differentiation under the integral sign)

[F5]

The nonnegative integral is monotone and positively homogeneous; a nonnegative function has integral zero exactly when it vanishes almost everywhere. A continuous nonzero function of constant sign on an open ball therefore has nonzero integral, since its absolute value is positive on a smaller ball of positive measure. (Monotonicity and nonnegative homogeneity of the nonnegative integral, A nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere, Sphere and ball measures scale in Rn)

[F6]

For $D_{t} = t^{- 1} \partial_{t}$ one has $D_{t} [t^{α}] = α t^{α - 2}$ , hence $D_{t}^{k - 1} [t^{- 1}] = (- 1)^{k - 1} (2 k - 3)!! t^{- (2 k - 1)}$ . (Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ )

[F7]

The strong Huygens principle holds in every odd spatial dimension $n \geq 3$ (The strong Huygens principle in odd spatial dimensions).

[F8]

Translated smooth bumps exist: for $0 < a < b$ choose the supplied $0 \leq ρ \leq 1$ , equal to one on ${\overline{B}}_{a} (0)$ and supported in $B_{b} (0)$ , and use $y \mapsto ρ (y - x_{0})$ . (A smooth bump between concentric Euclidean balls)

Proof

1.1givenF3algebra

The one-dimensional tail: with $u_{0} = 0$ , [F3] gives $u (x_{0}, t_{0}) = \frac{1}{2 c} \int_{x_{0} - c t_{0}}^{x_{0} + c t_{0}} u_{1}$ for every admissible $u_{1} \in C_{c}^{1} (R)$ , and if $u_{1}$ is supported in the open interval $(x_{0} - c t_{0}, x_{0} + c t_{0})$ and has nonzero integral then this value is nonzero, while the zero data give $u \equiv 0$ ; the difference $(0, u_{1})$ is supported strictly inside the base ball $(x_{0} - c t_{0}, x_{0} + c t_{0})$ , so this is a genuine failure of strong Huygens in dimension $1$ .

1.2givenF1F2F4F6algebra

The descended kernel and its value at the centre: for even $n = 2 k$ and data $(0, u_{1})$ , [F1] reads $u (x, t) = c^{1 - n} D_{t}^{k - 1} W_{u_{1}} (x, t)$ with the displayed $W_{u_{1}}$ ; for any $u_{1}$ supported in $B_{ε} (x_{0})$ with $0 < ε < c t_{0}$ , choose $δ > 0$ with $c (t_{0} - δ) > ε$ ; then $(c^{2} t^{2} - ρ^{2})^{- 1 / 2}$ is $C^{\infty}$ on a neighbourhood of $[t_{0} - δ, t_{0} + δ] \times {\overline{B}}_{ε} (x_{0})$ on this compact parameter set, its derivatives times the bounded compactly supported $u_{1}$ have a constant integrable majorant, so [F4] lets $D_{t}^{k - 1}$ be taken under the integral, giving $u (x_{0}, t_{0}) = \int_{B_{ε} (x_{0})} u_{1} (y) K (∣ y - x_{0} ∣, t_{0}) d y$ with $K$ as displayed; at $ρ = 0$ , [F6] with $α = - 1$ gives $K (0, t_{0}) = c^{1 - n} (n!! V_{n})^{- 1} c^{- 1} D_{t}^{k - 1} [t^{- 1}] = c^{- n} (n!! V_{n})^{- 1} (- 1)^{k - 1} (2 k - 3)!! t_{0}^{- (2 k - 1)} \neq 0$ , and for $k = 1$ the empty product is $1$ , recovering Poisson's kernel at the centre [F2].

2.1givenstep 1.2F5F8algebra

Nonzero value from interior data: $K (ρ, t_{0})$ is continuous in $ρ$ on a neighbourhood of $0$ because the differentiated expression is smooth there, and $K (0, t_{0}) \neq 0$ , so there is $0 < ε_{0} < c t_{0}$ with $K$ of one constant sign on $[0, ε_{0}]$ ; choosing now $0 < ε < ε_{0}$ and any $u_{1} \in C_{c}^{k + 1} (B_{ε} (x_{0}))$ with $u_{1} \geq 0$ , $u_{1} \equiv̸ 0$ (take the translated smooth bump of [F8] with inner radius $ε / 2$ and outer radius $ε$ ), the product $u_{1} (y) K (∣ y - x_{0} ∣, t_{0})$ is continuous on the ball, of constant sign and nonzero somewhere, so by [F5] its integral $u (x_{0}, t_{0})$ is nonzero; the data $(0, u_{1})$ are admissible and supported in a compact subset of the open base ball, while the zero data give the value $0$ , so strong Huygens fails in dimension $n$ .

3.1step 1.1step 2.1F7∎

Conclusion: dimensions $1$ and every even $n \geq 2$ admit admissible data that vanish on a neighbourhood of the sphere $S (x_{0}, t_{0})$ yet change the value $u (x_{0}, t_{0})$ , by [step 1.1] and [step 2.1]; by [F7] the strong Huygens principle holds in every odd dimension $n \geq 3$ . Thus it fails in exactly dimensions $1$ and even $n \geq 2$ , and the failures are witnessed by data supported strictly inside the base ball.

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Finite propagation is not the Huygens principle

Remark

Assume the Axiom of Countable Choice. For homogeneous waves with compactly supported initial data, the strong Huygens principle implies finite propagation, but the converse is false. The interior tails in dimension $1$ and every even dimension give witnesses to this failure of the converse.

Finite propagation (Compact support expands at speed at most c) holds in every spatial dimension $n \geq 1$ : data supported in a compact $K$ (with a source supported in the corresponding cone) have $supp u (\cdot, t) \subseteq K + {\overline{B}}_{c t} (0)$ at time $t$ , the full solid $c t$ -neighbourhood of the data support. It bounds the outer front and nothing more.

The strong Huygens principle (The strong Huygens principle in the homogeneous Cauchy setting) is the stronger shell statement: the value at $(x_{0}, t_{0})$ is carried by the sphere $S (x_{0}, t_{0}) = \partial B_{c t_{0}} (x_{0})$ , so data supported strictly inside the base ball have no effect and, for compactly supported data, the disturbance is carried by the shell rather than by the solid cone. It holds for odd $n \geq 3$ (The strong Huygens principle in odd spatial dimensions) and fails for $n = 1$ and every even $n$ (Wave tails in one and even spatial dimensions: strong Huygens fails), where a lasting interior tail remains after the front has passed; the failure witnesses are compactly supported and obey the finite-speed bound.

Two consequences deserve care. First, an interior tail is not a violation of finite speed: the tail stays inside the cone $∣ x - x_{0} ∣ \leq c t$ — it is the interior of that cone, not its exterior, that remains affected. Second, the informal shorthand "the value depends only on the values of the initial data on $\partial B_{t} (x)$ " is not a correct reading of strong Huygens: the functional is a finite linear combination of sphere integrals of radial derivatives, as The strong Huygens principle in odd spatial dimensions proves; neighbourhood agreement preserves those derivatives, whereas bare restriction agreement need not. This is consistent with the germ form (iii) of The strong Huygens principle in the homogeneous Cauchy setting. The positive theorem above is the sharp statement, and this remark records the exact distinction fixed by the drift review of this page.

Remarks

Explicit compactly supported failure witnesses are recorded in Finite speed of propagation does not imply strong Huygens ↗.

5 · Examples, counterexamples and false statements

None yet.

Wave Energy, Finite Propagation and Huygens' Principle

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

Forward and backward wave cones, domain of dependence and influence

Definition

Wave energy density, energy flux and total energy

Definition

The integral of the divergence of an integrable C1 field vanishes

Statement

Facts & Assumptions

Proof

Remarks

Truncated wave cones: convexity, piecewise C1 presentation and outward normals

Statement

Facts & Assumptions

Proof

Remarks

Vanishing gradient and time derivative force constancy on convex sets

Statement

Facts & Assumptions

Proof

The local wave-energy conservation law

Statement

Facts & Assumptions

Proof

The energy identity on a truncated wave cone

Statement

Facts & Assumptions

Proof

Conservation of total wave energy in three admissible settings

Statement

Facts & Assumptions

Proof

Energy continuous dependence for the forced wave equation

Statement

Facts & Assumptions

Proof

Energy uniqueness for the wave Cauchy problem

Statement

Facts & Assumptions

Proof

Energy uniqueness for homogeneous Dirichlet waves on bounded domains

Statement

Facts & Assumptions

Proof

Finite propagation speed for the wave equation

Statement

Facts & Assumptions

Proof

Compact support expands at speed at most c

Statement

Facts & Assumptions

Proof

Time-reversed energy uniqueness from final data

Statement

Facts & Assumptions

Proof

The strong Huygens principle in the homogeneous Cauchy setting

Definition

Domain of dependence and local uniqueness

Statement

Facts & Assumptions

Proof

The strong Huygens principle in odd spatial dimensions

Statement

Facts & Assumptions

Proof

Wave tails in one and even spatial dimensions: strong Huygens fails

Statement

Facts & Assumptions

Proof

Finite propagation is not the Huygens principle

Remark

Remarks

5 · Examples, counterexamples and false statements

Sources