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✓ 9 results · all verified · 2 also independently AI-judged

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Wave Energy, Finite Propagation and Huygens' Principle — Examples

1 · Prerequisites

2 · Summary

These companions illustrate and test the energy and propagation theory of the main page. The conserved energy of a one-dimensional travelling packet is computed explicitly, with its equal kinetic and potential split and with the caution that in dimensions $n \geq 2$ the same profile has infinite total energy; the plane-wave family shows that the characteristic speed $c$ is attained by the moving support, not merely an upper bound. Two counterexamples probe the hypotheses of the conservation theorem: a plane-wave profile whose local conservation law holds while the total energy is infinite, and a compactly supported packet leaving an interval, where the interior energy decays exactly by the flux through the open boundary. Odd reflection realises the Dirichlet half-line problem, with the reflected wave re-entering with reversed sign and the half-line energy equal to half of the conserved whole-line energy. The zero-energy example isolates the residual spatial constant that the displacement datum fixes. The final comparison exhibits the dimensional dichotomy at the heart of Huygens' principle: a three-dimensional spherical pulse leaves a quiet interior behind its expanding front, while the two-dimensional pulse keeps a positive tail at the centre after the front has passed; the last counterexample uses one- and two-dimensional interior-data witnesses to show that finite propagation does not imply strong Huygens.

All statements are read under the Axiom of Countable Choice; the travelling-packet, plane-wave and open-boundary computations use only Lebesgue integration, differentiation under the integral and the fundamental theorem of calculus.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

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Conserved energy of a travelling wave packet

Example

Assume Countable Choice for the Lebesgue measure and multidimensional volume assertions below (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $c > 0$ and let $F \in C_{c}^{2} (R)$ be compactly supported, and put

$u (x, t) : = F (x - c t) (x, t \in R),$

a right-moving travelling packet. Then $u$ is a classical solution of $□_{c} u = 0$ (Wave equation, Cauchy data and wave speed) and for every $t \in R$ its total energy (Wave energy density, energy flux and total energy) is finite, independent of $t$ , and splits equally between its kinetic and potential parts:

$E (t) = \int_{R} \frac{1}{2} (u_{t}^{2} + c^{2} u_{x}^{2}) d x = c^{2} \int_{R} F^{'} (s)^{2} d s, E_{k i n} (t) = E_{p o t} (t) = \frac{1}{2} c^{2} \int_{R} F^{'} (s)^{2} d s .$

The equal split is the signature of a nondispersive packet: pointwise $u_{t} = - c F^{'} (x - c t)$ and $∣ D u ∣ = ∣ F^{'} (x - c t) ∣$ , so both densities equal $\frac{1}{2} c^{2} F^{'} (x - c t)^{2}$ and the energy density is $c^{2} F^{'} (x - c t)^{2}$ .

The finite-energy statement is genuinely one-dimensional. In dimension $n \geq 2$ the profile $u (x, t) = F (ω \cdot x - c t)$ with a unit vector $ω$ is still a classical solution and still satisfies $u_{t} = - c F^{'}$ and $D u = F^{'} ω$ pointwise, so the two densities still split equally; but the density $c^{2} F^{'} (ω \cdot x - c t)^{2}$ then depends on $x$ only through the single variable $ω \cdot x - c t$ , and whenever $F^{'} \equiv̸ 0$ it is bounded below by a positive constant on a slab of infinite $n$ -dimensional measure, so the total energy over $R^{n}$ is $+ \infty$ . The conserved finite total energy computed here is therefore the energy of a one-dimensional packet.

Facts & Assumptions

Given: Countable Choice; $c > 0$ and $F \in C_{c}^{2} (R)$ , and $u (x, t) = F (x - c t)$ on $R \times R$ ; write $F_{2} (s) : = F^{'} (s)^{2}$ for the continuous compactly supported density profile.

[F1]

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: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ )

[F2]

Change of variables: for a $C^{1}$ diffeomorphism $T : U \to V$ of open sets and a continuous compactly supported $f : V \to R$ , $\int_{V} f d λ_{n} = \int_{U} (f \circ T) ∣ \det D T ∣ d λ_{n}$ . (The published Riemann change-of-variables theorem already gives the Lebesgue formula for continuous compactly supported integrands)

[F3]

The energy density and flux of a $C^{2}$ function are $e = \frac{1}{2} (u_{t}^{2} + c^{2} ∣ D u ∣^{2})$ and $q = - c^{2} u_{t} D u$ ; the wave operator is $□_{c} = \partial_{t}^{2} - c^{2} Δ$ . (Wave energy density, energy flux and total energy, Wave equation, Cauchy data and wave speed)

[F4]

The nonnegative Lebesgue integral is monotone and positively homogeneous; a continuous compactly supported function is bounded and supported in a set of finite measure. (Monotonicity and nonnegative homogeneity of the nonnegative integral, The support of a function on $R^{n}$ and its compactly supported Riemann integral)

[F5]

An orthogonal linear map preserves Lebesgue measure, because its determinant has absolute value one; translations also preserve it. (A linear map $T$ of $R^{n}$ sends Lebesgue measurable sets to Lebesgue measurable sets, with $λ_{n} (T [E]) = ∣ \det T ∣ λ_{n} (E)$ when $T$ is invertible and $T [E]$ Lebesgue null when it is not, Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation)

Verification

1.1givenF1F3algebra

Derivatives and the pointwise split: by [F1], $u_{t} = - c F^{'} (x - c t)$ , $u_{t t} = c^{2} F^{''} (x - c t)$ , $u_{x} = F^{'} (x - c t)$ and $u_{x x} = F^{''} (x - c t)$ , so $□_{c} u = c^{2} F^{''} (x - c t) - c^{2} F^{''} (x - c t) = 0$ and $u$ is a classical solution with continuous second derivatives [F3]; moreover $u_{t}^{2} = c^{2} F^{'} (x - c t)^{2}$ and $∣ D u ∣^{2} = F^{'} (x - c t)^{2}$ , so $e_{k i n} = e_{p o t} = \frac{1}{2} c^{2} F_{2} (x - c t)$ and $e = c^{2} F_{2} (x - c t)$ .

2.1givenstep 1.1F2F4algebra

The total energy: for each $t$ , $E (t) = \int_{R} c^{2} F_{2} (x - c t) d x = c^{2} \int_{R} F_{2} (y) d y$ by [F2] applied to the diffeomorphism $T (x) = x + c t$ of $R$ , whose derivative is $1$ ; the value is finite because $F_{2} = F^{' 2}$ is continuous with compact support, so it is bounded by a constant and vanishes outside a bounded interval, and [F4] bounds its integral by the constant times the finite length of that interval; similarly $E_{k i n} (t) = E_{p o t} (t) = \frac{1}{2} c^{2} \int_{R} F_{2} (y) d y$ by [F4]. Hence $E (t)$ is finite, independent of $t$ , and equals $c^{2} \int_{R} F^{'} (s)^{2} d s$ , with the equal split $E_{k i n} = E_{p o t} = \frac{1}{2} c^{2} \int_{R} F^{'} (s)^{2} d s$ .

3.1givenstep 1.1F4F5algebra∎

The multidimensional caution: for $n \geq 2$ , given a unit vector $ω$ and $u (x, t) = F (ω \cdot x - c t)$ , the same chain rule gives $u_{t} = - c F^{'} (ω \cdot x - c t)$ , $D u = F^{'} (ω \cdot x - c t) ω$ and $u_{t t} = c^{2} F^{''}$ , $Δ u = F^{''} ∣ ω ∣^{2} = F^{''}$ , so $u$ is again a classical solution and the densities again satisfy $e_{k i n} = e_{p o t} = \frac{1}{2} c^{2} F^{'} (ω \cdot x - c t)^{2}$ ; but if $F^{'} \equiv̸ 0$ then $F^{' 2} \geq m > 0$ on some nondegenerate interval $I = (a, b)$ by continuity. Choose an orthogonal $O$ with $O e_{0} = ω$ : take $O = I$ if $ω = e_{0}$ , and otherwise set $v = (e_{0} - ω) / ∣ e_{0} - ω ∣$ and $O = I - 2 v v^{T}$ ; direct multiplication gives $O^{T} O = I$ and $O e_{0} = ω$ . For each $R > 0$ , the rotated box $c t ω + O (I \times (- R, R)^{n - 1})$ lies in the slab ${x : ω \cdot x - c t \in I}$ . By [F5] its measure is $(b - a) (2 R)^{n - 1}$ . The density is at least $c^{2} m$ throughout it, so [F4] gives $E_{R^{n}} (t) \geq c^{2} m (b - a) (2 R)^{n - 1}$ for every $R$ ; letting $R \to \infty$ proves $E_{R^{n}} (t) = + \infty$ . Thus the finite total energy of the Example cannot be extended beyond $n = 1$ .

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Plane-wave support translates at the characteristic speed

Example

Let $n \geq 1$ , $c > 0$ , let $ω \in R^{n}$ be a unit vector and let $F \in C^{2} (R)$ have compact nonempty support. Put

$u (x, t) : = F (ω \cdot x - c t) (x \in R^{n}, t \in R) .$

Then $u$ is a classical solution of $□_{c} u = 0$ (Wave equation, Cauchy data and wave speed), and for every $t$ its support is exactly the closed set

$supp u (\cdot, t) = {x \in R^{n} : ω \cdot x - c t \in supp F}$

(The support of a function on $R^{n}$ and its compactly supported Riemann integral), which is the translate $supp u (\cdot, 0) + c t ω$ of the initial support. Consequently the disturbance travels with velocity $c ω$ : its two bounding hyperplanes (the front and the back of the plane wave) advance with speed exactly $c$ along $ω$ . This exhibits the characteristic speed $c$ as an attained speed and not merely an upper bound, in contrast with the general estimate of Finite propagation speed for the wave equation. The support need not be compact when $n \geq 2$ : the support is unbounded in transverse directions; for $n = 1$ it is compact and may be disconnected. The support equals its enclosing slab only if $supp F$ is an interval.

Facts & Assumptions

Given: $n \geq 1$ , $c > 0$ , a unit vector $ω \in R^{n}$ , a function $F \in C^{2} (R)$ with compact nonempty support, and $u (x, t) = F (ω \cdot x - c t)$ ; write $ℓ (x, t) : = ω \cdot x - c t$ and $A : = {s \in R : F (s) \neq 0}$ , so that $supp F = \overline{A}$ .

[F1]

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: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ )

[F2]

The support of $f : R^{n} \to R$ is $supp f = \overline{{x : f (x) \neq 0}}$ , and $f$ is compactly supported when that closure is compact. (The support of a function on $R^{n}$ and its compactly supported Riemann integral)

[F3]

The wave operator of speed $c$ is $□_{c} = \partial_{t}^{2} - c^{2} Δ$ , with $Δ = div \nabla$ the Laplacian. (Wave equation, Cauchy data and wave speed, The Laplacian of a $C^{2}$ function and of a $C^{2}$ vector field)

[F4]

Partial and directional derivatives are the ordinary one-variable derivatives of the line maps $t \mapsto f (a + t v)$ ; the Euclidean gradient is $(\partial_{0} f, \dots, \partial_{n - 1} f)$ . (Directional derivatives and partial derivatives of a map $U \subseteq R^{m} \to R^{n}$ , The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case)

[F5]

A continuous real function on a nonempty compact metric space attains its maximum and minimum. (A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value)

Verification

1.1givenF1F3F4algebra

The profile solves the wave equation: by [F1] and [F4], $u_{t} = - c F^{'} (ℓ)$ , $u_{t t} = c^{2} F^{''} (ℓ)$ , $\partial_{i} u = ω_{i} F^{'} (ℓ)$ and $\partial_{i} \partial_{i} u = ω_{i}^{2} F^{''} (ℓ)$ for every spatial index $i$ , so $Δ u = (\sum_{i} ω_{i}^{2}) F^{''} (ℓ) = F^{''} (ℓ)$ because $∣ ω ∣ = 1$ ; hence $□_{c} u = u_{t t} - c^{2} Δ u = c^{2} F^{''} (ℓ) - c^{2} F^{''} (ℓ) = 0$ on $R^{n} \times R$ [F3], and $u$ is a classical solution with all second derivatives continuous.

2.1step 1.1F2algebra

The support identity: for every $x$ one has $u (x, t) \neq 0$ exactly when $ℓ (x, t) \in A$ , so $supp u (\cdot, t) = \overline{{x : ℓ (x, t) \in A}}$ [F2]; since $x \mapsto ℓ (x, t)$ is continuous, the set ${x : ℓ (x, t) \in supp F} = {x : ℓ (x, t) \in \overline{A}}$ is closed and contains ${x : ℓ (x, t) \in A}$ , whence the closure is contained in it; conversely, if $ℓ (x, t) \in supp F$ , then for every $ε > 0$ closure supplies $s \in A$ with $∣ s - ℓ (x, t) ∣ < ε$ ; the point $x_{s} : = x + (s - ℓ (x, t)) ω$ satisfies $∣ x_{s} - x ∣ < ε$ and $u (x_{s}, t) = F (s) \neq 0$ . Every neighbourhood of $x$ therefore meets the nonzero set, so $x$ is in its closure. Therefore $supp u (\cdot, t) = {x : ω \cdot x - c t \in supp F}$ .

3.1givenstep 2.1algebraF5∎

Translation and speed: the identity $ω \cdot (x + c t ω) = ω \cdot x + c t$ gives $supp u (\cdot, t) = supp u (\cdot, 0) + c t ω$ directly from step 2.1; writing $a : = \min supp F$ and $b : = \max supp F$ , both attained by [F5] applied to the identity on the nonempty compact $supp F$ , the support is contained in the closed slab ${a + c t \leq ω \cdot x \leq b + c t}$ with $a < b$ (continuity and a nonzero value imply that the support contains an interval); the two bounding hyperplanes therefore translate by $c t ω$ , so each moves with velocity $c ω$ and speed exactly $c$ along the direction $ω$ , and the support meets both bounding hyperplanes because $a, b \in supp F$ .

For $n = 1$ the enclosing slab is the interval described by $a + c t \leq ω x \leq b + c t$ , and its front and back endpoints move with velocity $c ω$ ; for $n \geq 2$ the same hyperplanes bound the unbounded slab, and the statement is about the direction of propagation $ω$ and not about compact support at time $t$ . The statement of Finite propagation speed for the wave equation only gives the upper bound, which this family attains.

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The local conservation law need not integrate to a finite conserved energy

Statement refuted

The claim refuted is that the pointwise local conservation law $\partial_{t} e + div q = 0$ by itself integrates to a finite conserved total energy on all of $R^{n}$ . Explicitly: not every classical solution of $□_{c} u = 0$ has finite total energy in the sense of Wave energy density, energy flux and total energy, and for such a solution the local law supplies no finite conserved energy; the integrability hypotheses of Conservation of total wave energy in three admissible settings(b) are therefore not redundant.

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )) for the Lebesgue product measure used below. Witness. Let $n \geq 2$ , $c > 0$ , let $e_{0}$ be the first standard basis vector, let $F \in C^{2} (R)$ with $F^{'} \equiv̸ 0$ (for instance $F (s) = \cos s$ ), and put

$u (x, t) : = F (x_{0} - c t) (x = (x_{0}, \dots, x_{n - 1}) \in R^{n}) .$

Then $u$ is a classical solution of $□_{c} u = 0$ on $R^{n} \times R$ and the local law The local wave-energy conservation law holds pointwise, the energy density is $e (x, t) = c^{2} F^{'} (x_{0} - c t)^{2} \geq 0$ , and $\int_{R^{n}} e (x, t) d x = + \infty$ for every $t$ , although the one-dimensional profile is $C^{2}$ . The same divergence occurs for $n = 1$ with $F (s) = s$ , since its energy density is the positive constant $c^{2}$ .

Facts & Assumptions

Given: Countable Choice; $n \geq 2$ , $c > 0$ , $F \in C^{2} (R)$ with $F^{'}$ not identically zero, $u (x, t) = F (x_{0} - c t)$ and $e = \frac{1}{2} (u_{t}^{2} + c^{2} ∣ D u ∣^{2})$ , $q = - c^{2} u_{t} D u$ as in Wave energy density, energy flux and total energy; write $λ_{n}$ for $n$ -dimensional Lebesgue measure.

[F1]

The local balance law: $\partial_{t} e + div q = f u_{t}$ with $f = □_{c} u$ ; for a homogeneous classical solution, $\partial_{t} e + div q = 0$ . (The local wave-energy conservation law)

[F2]

Chain rule for totally differentiable composites, applied to $x \mapsto x_{0} - c t$ and $F$ . (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ )

[F3]

Tonelli's theorem: for a product-measurable $f \geq 0$ , the double integral is the iterated integral. (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product)

[F4]

The nonnegative Lebesgue integral is monotone: if $0 \leq f \leq g$ then $\int f \leq \int g$ . (Monotonicity and nonnegative homogeneity of the nonnegative integral)

[F6]

$λ_{n}$ is the $n$ -dimensional Lebesgue measure on the Lebesgue measurable sets. (Lebesgue measurable sets, the family $L (R^{n})$ , and the restricted set function $λ_{n}$ )

Counterexample

1.1givenF1F2algebra

The profile solves the equation and its density is a one-variable function: by [F2], $u_{t} (x, t) = - c F^{'} (x_{0} - c t)$ , $\partial_{0} u (x, t) = F^{'} (x_{0} - c t)$ and $\partial_{i} u (x, t) = 0$ for $i \neq 0$ , so $u_{t t} = c^{2} F^{''} (x_{0} - c t)$ , $Δ u = F^{''} (x_{0} - c t)$ and $□_{c} u = 0$ ; hence [F1] holds, $u_{t}^{2} = c^{2} F^{'} (x_{0} - c t)^{2}$ , $∣ D u ∣^{2} = F^{'} (x_{0} - c t)^{2}$ and $e (x, t) = c^{2} F^{'} (x_{0} - c t)^{2}$ , a nonnegative function of the single variable $x_{0} - c t$ .

2.1givenstep 1.1algebra

A positive lower bound on a slab: since $F^{'}$ is continuous and not identically zero, there are $a < b$ and $m > 0$ with $F^{'} (s)^{2} \geq m$ for $s \in [a, b]$ ; consequently, for every $t$ and every $x^{'} = (x_{1}, \dots, x_{n - 1})$ , the density satisfies $e (x_{0}, x^{'}, t) = c^{2} F^{'} (x_{0} - c t)^{2} \geq c^{2} m$ whenever $x_{0} \in [a + c t, b + c t]$ .

3.1givenstep 1.1step 2.1F3F4algebra

The total energy diverges: fix $t$ and $R > 0$ large enough that $[a + c t, b + c t] \subseteq [- R, R]$ ; Tonelli [F3] applied to the nonnegative product-measurable function $e$ (written in the product coordinates $(x_{0}, x^{'})$ , with $d λ_{n}$ the measure of [F6]; Borel product measurability and agreement of the product with Euclidean Lebesgue measure follow from The Borel product of R^m and R^n is the Borel sigma-algebra of R^{m+n} and On Borel subsets of R^{m+n}, the product lambda_m times lambda_n agrees with lambda_{m+n}) gives $\int_{(- R, R)^{n}} e d λ_{n} = \int_{(- R, R)^{n - 1}} (\int_{- R}^{R} c^{2} F^{'} (x_{0} - c t)^{2} d x_{0}) d x^{'}$ , and by step 2.1 the inner integral is at least $c^{2} m (b - a) > 0$ for every $x^{'}$ ; since the box $(- R, R)^{n - 1}$ has measure $(2 R)^{n - 1}$ in each coordinate, this integral is at least $c^{2} m (b - a) (2 R)^{n - 1}$ ; as $R \to \infty$ the right-hand side tends to $+ \infty$ , while for each fixed $R$ monotonicity [F4] bounds $\int_{(- R, R)^{n}} e d λ_{n} \leq \int_{R^{n}} e d λ_{n}$ ; hence $\int_{R^{n}} e d λ_{n} = + \infty$ . (Equivalently, the iterated integral in the $x_{0}$ -variable alone has value $c^{2} \int_{R} F^{'} (s)^{2} d s > 0$ and the remaining $(n - 1)$ -fold integral of the constant $1$ is $+ \infty$ , which [F3] turns into the same conclusion.)

4.1givenstep 3.1F4∎

Failure of a finite-energy conclusion: since $\int_{R^{n}} e (x, t) d x = + \infty$ for every $t$ , the total energy of Wave energy density, energy flux and total energy is not a finite conserved quantity for this solution, so the local differential law [F1] holds pointwise while no finite global identity follows; in dimension $n = 1$ and for $F (s) = s$ one has $e = c^{2}$ , whose integral over $[- R, R]$ is $2 R c^{2} \to + \infty$ as $R \to \infty$ , so the whole-line integral is infinite by [F4]; thus finite energy is not implied by the local identity. This witness does not show that every sufficient hypothesis of the conservation theorem is necessary, and its extended total energy is the constant $+ \infty$ .

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Wave energy need not be conserved through an open boundary

Statement refuted

Assume Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )) for the Lebesgue and Riemann integral bridge used below. The claim refuted is that the total energy of a classical wave solution is automatically constant whenever the domain is bounded, without any hypothesis on the boundary flux. Witness: let $c > 0$ , let $Ω = (0, 1) \subseteq R$ and let $F \in C_{c}^{2} (R)$ have $F^{'}$ nonzero somewhere in $(0, 1)$ ; put

$u (x, t) : = F (x - c t),$

the right-moving packet. Then $u$ solves $□_{c} u = 0$ on $R \times R$ , but its energy in the fixed interval,

$E_{Ω} (t) = c^{2} \int_{0}^{1} F^{'} (x - c t)^{2} d x,$

equals $c^{2} \int_{0}^{1} F^{'} (x)^{2} d x > 0$ at $t = 0$ and is $0$ for all sufficiently large $t$ , once the packet has left the interval. The decrease is exactly the boundary flux: with $q = - c^{2} u_{t} u_{x} = c^{3} F^{' 2}$ ,

$\frac{d}{d t} E_{Ω} (t) = q (0, t) - q (1, t),$

so the energy lost through the right endpoint is accounted for, and Conservation of total wave energy in three admissible settings may not be invoked on a domain with an open boundary without the vanishing-flux hypothesis.

Homogeneous-boundary comparison on the half-line. If $v \in C^{2} ([0, \infty) \times I)$ solves $v_{t t} - c^{2} v_{x x} = 0$ on an open time interval $I$ , and for each compact $J \subseteq I$ there is $R_{J} < \infty$ with $v_{x} = v_{t} = 0$ for $x \geq R_{J}$ and $t \in J$ , then $E_{+} (t) : = \frac{1}{2} \int_{0}^{\infty} (v_{t}^{2} + c^{2} v_{x}^{2}) d x$ is constant under either $v (0, t) = 0$ for every $t \in I$ or $v_{x} (0, t) = 0$ for every $t \in I$ (the homogeneous Dirichlet and Neumann comparisons for the linear case of the cited problem). No nonlinear potential term is asserted.

Facts & Assumptions

Given: Countable Choice; $c > 0$ , $Ω = (0, 1)$ , $F \in C_{c}^{2} (R)$ with $F^{'}$ nonzero somewhere in $(0, 1)$ , $u (x, t) = F (x - c t)$ , and the fields $e = \frac{1}{2} (u_{t}^{2} + c^{2} u_{x}^{2})$ , $q = - c^{2} u_{t} u_{x}$ of Wave energy density, energy flux and total energy; for the last part a solution $v$ on $[0, \infty) \times I$ with the stated support hypothesis.

[F1]

The local balance: $\partial_{t} e + \partial_{x} q = f u_{t}$ , hence $\partial_{t} e = - \partial_{x} q$ for a classical solution. (The local wave-energy conservation law)

[F2]

Chain rule, scalar product rule, and equality of mixed second partials for C2 functions. (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 $g (c) \neq 0$ , Clairaut--Schwarz theorem for continuous second partial derivatives)

[F3]

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 $x$ , and the $t$ -derivative is dominated on the time interval by a fixed integrable function, then $F (t) = \int f (x, t) d x$ is differentiable with $F^{'} (t) = \int \partial_{t} f (x, t) d x$ . (Differentiation under the integral sign)

[F4]

Second fundamental theorem: if $G$ is differentiable on $[a, b]$ with integrable derivative, then $\int_{a}^{b} G^{'} = G (b) - G (a)$ (Darboux integral); on a closed bounded interval continuous functions are integrable, Darboux and Riemann integrals agree, and a bounded Borel Riemann integrable function on a closed interval has the same Lebesgue integral. (The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ , 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)

[F5]

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)

[F6]

The support of $f$ is the closure of ${f \neq 0}$ ; for the translate, $supp F (\cdot - c t) = c t + supp F$ . (The support of a function on $R^{n}$ and its compactly supported Riemann integral)

Proof

1.1givenF1F2algebra

The packet and its flux: by the chain rule [F2], $u_{t} = - c F^{'} (x - c t)$ and $u_{x} = F^{'} (x - c t)$ , so $u_{t t} = c^{2} F^{''} (x - c t)$ , $u_{x x} = F^{''} (x - c t)$ and $□_{c} u = 0$ ; hence [F1] holds and the energy density and flux are $e = c^{2} F^{'} (x - c t)^{2}$ and $q = - c^{2} u_{t} u_{x} = c^{3} F^{'} (x - c t)^{2}$ .

2.1givenstep 1.1F6algebra

Positive initial energy and late vanishing: since $F^{'}$ is continuous and nonzero somewhere in $(0, 1)$ , there are a subinterval of $(0, 1)$ on which $F^{' 2} \geq m > 0$ and hence $E_{Ω} (0) = c^{2} \int_{0}^{1} F^{'} (x)^{2} d x > 0$ ; and by [F6] the support of $x \mapsto u (x, t)$ is $c t + supp F$ , so for every $t > (1 - \min supp F) / c$ the packet is disjoint from $[0, 1]$ and $E_{Ω} (t) = 0$ .

2.2givenstep 1.1F3F4algebra

The flux identity: $e$ and $\partial_{t} e$ are continuous on $[0, 1]$ and bounded on compact time intervals, so [F3] gives $E_{Ω}^{'} (t) = \int_{0}^{1} \partial_{t} e (x, t) d x = - c^{3} \int_{0}^{1} (F^{' 2})^{'} (x - c t) d x$ , and by [F4] the Darboux fundamental theorem applies to the continuous function $x \mapsto F^{' 2} (x - c t)$ on $[0, 1]$ , whose Lebesgue integral equals that Darboux integral, giving $\int_{0}^{1} (F^{' 2})^{'} (x - c t) d x = F^{' 2} (1 - c t) - F^{' 2} (- c t)$ ; therefore $E_{Ω}^{'} (t) = c^{3} (F^{' 2} (- c t) - F^{' 2} (1 - c t)) = q (0, t) - q (1, t)$ .

3.1step 2.1step 2.2algebra

Conclusion for the open boundary: by steps 2.1 and 2.2 the energy $E_{Ω}$ is positive at $t = 0$ , zero for all large $t$ , and its rate of change is exactly the difference of the outward fluxes at the two endpoints; so $E_{Ω}$ is not constant, the drop is accounted for by the flux through the right endpoint, and automatic conservation cannot be inferred without controlling the boundary flux.

4.1givenstep 1.1F2F3F4F5∎

The half-line comparison: fix a compact $J \subseteq I$ and $R \geq R_{J}$ ; for $t \in J$ the integrand of $E_{+}$ vanishes for $x \geq R_{J}$ , so $E_{+} (t) = \frac{1}{2} \int_{0}^{R} (v_{t}^{2} + c^{2} v_{x}^{2}) d x$ , and [F3] with the domination constant $\sup_{[0, R] \times J} ∣ v_{t} v_{t t} + c^{2} v_{x} v_{x t} ∣$ gives $E_{+}^{'} (t) = \int_{0}^{R} (v_{t} v_{t t} + c^{2} v_{x} v_{x t}) d x = c^{2} \int_{0}^{R} (v_{t} v_{x x} + v_{x} v_{x t}) d x = c^{2} \int_{0}^{R} \partial_{x} (v_{t} v_{x}) d x$ , using $v_{t t} = c^{2} v_{x x}$ and the product rule [F2]; by [F4], $\int_{0}^{R} \partial_{x} (v_{t} v_{x}) d x = v_{t} (R, t) v_{x} (R, t) - v_{t} (0, t) v_{x} (0, t)$ ; the upper endpoint term is zero by the support hypothesis, and the lower endpoint term is zero because in the Dirichlet case the trace $t \mapsto v (0, t)$ is identically zero and differentiable with derivative $v_{t} (0, t)$ , while in the Neumann case $v_{x} (0, t) = 0$ directly; hence $E_{+}^{'} (t) = 0$ for every interior $t \in J$ and, $E_{+}$ being continuous on $J$ with vanishing derivative there, [F5] makes $E_{+}$ constant on $J$ ; as $J$ is an arbitrary compact subinterval of $I$ , $E_{+}$ is constant on $I$ .

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Odd reflection at a Dirichlet endpoint

Example

Assume the Axiom of Countable Choice. Let $c > 0$ , let $u_{0} \in C_{c}^{2} (R)$ and $u_{1} \in C_{c}^{1} (R)$ be odd — equivalently, data on the half-line $x > 0$ extended oddly — and let $U$ be the d'Alembert solution of the whole-line problem with data $(u_{0}, u_{1})$ (d'Alembert's formula and uniqueness in one dimension). Then:

(i) $U (\cdot, t)$ is odd for every $t$ , so $u : = U ∣_{x > 0}$ solves the Dirichlet half-line problem $u_{t t} = c^{2} u_{x x}$ on $x > 0$ with $u (0, t) = 0$ and data $u_{0} ∣_{x > 0}$ , $u_{1} ∣_{x > 0}$ ;

(ii) for data obtained by oddly extending $u_{0} = ϕ$ , $u_{1} = c ϕ^{'}$ from $x > 0$ , where $ϕ \in C_{c}^{2} ((0, \infty))$ , the reflected part re-enters with reversed sign:

$u (x, t) = ϕ (x + c t) - ϕ (c t - x) (0 < x < c t);$

(iii) the half-line energy $E_{(0, \infty)} (t) = \frac{1}{2} \int_{0}^{\infty} (u_{t}^{2} + c^{2} u_{x}^{2}) d x$ equals half the whole-line energy of $U$ and is constant in $t$ (Ivrii's Dirichlet case of the half-line energy problem).

Facts & Assumptions

Given: ${A C}_{ω}$ ; $c > 0$ ; odd compactly supported data $u_{0} \in C_{c}^{2} (R)$ , $u_{1} \in C_{c}^{1} (R)$ ; the d'Alembert solution $U$ of the whole-line problem, and for part (ii) a fixed $ϕ \in C_{c}^{2} ((0, \infty))$ with $u_{0} = ϕ$ , $u_{1} = c ϕ^{'}$ on $x > 0$ .

[F1]

D'Alembert's formula: for $U_{0} \in C^{2} (R)$ , $U_{1} \in C^{1} (R)$ the whole-line solution is $U (x, t) = \frac{1}{2} [U_{0} (x + c t) + U_{0} (x - c t)] + \frac{1}{2 c} \int_{x - c t}^{x + c t} U_{1} (s) d s$ , a $C^{2}$ classical solution of $u_{t t} = c^{2} u_{x x}$ . (d'Alembert's formula and uniqueness in one dimension)

[F2]

Conservation in case (a): a homogeneous solution whose spatial support is contained in a fixed compact set throughout a time interval has constant total energy on that interval. (Conservation of total wave energy in three admissible settings)

[F3]

The energy density is $e = \frac{1}{2} (U_{t}^{2} + c^{2} U_{x}^{2})$ . By [F1] and the support definition, the spatial support of $U (\cdot, t)$ is contained in $(supp U_{0} \cup supp U_{1}) + [- c t, c t]$ . (Wave energy density, energy flux and total energy, The support of a function on $R^{n}$ and its compactly supported Riemann integral)

[F4]

The integral of a continuous derivative on a closed interval equals the endpoint difference; its Darboux, Riemann and Lebesgue integrals agree. (The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ , 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)

[F5]

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)

Verification

1.1givenF1algebra

Oddness is preserved and the half-line problem is solved: if $U_{0} = u_{0}$ and $U_{1} = u_{1}$ are odd, then each term of [F1] is odd in $x$ : for the first term, replacing $x$ by $- x$ interchanges the two arguments of the odd function $U_{0}$ and changes the sign, and for the integral term the substitution $s \mapsto - s$ together with oddness of $U_{1}$ reverses the orientation of the interval and the sign of the integrand, leaving the integral odd in $x$ ; hence $U (\cdot, t)$ is odd for every $t$ , so $U (0, t) = 0$ ; therefore $u : = U ∣_{x > 0}$ is a $C^{2}$ solution of $u_{t t} = c^{2} u_{x x}$ on $x > 0$ with trace $u (0, t) = 0$ and the prescribed initial data $u_{0} ∣_{x > 0}$ , $u_{1} ∣_{x > 0}$ , which is (i).

2.1givenstep 1.1F1F4algebra

Reflection with reversed sign: take $u_{0} = ϕ$ and $u_{1} = c ϕ^{'}$ with $ϕ$ compactly supported in $(0, \infty)$ , extended oddly, and $0 < x < c t$ ; in [F1] the first term is $\frac{1}{2} [ϕ (x + c t) - ϕ (c t - x)]$ because $x - c t < 0 < x + c t$ and $U_{0} (- ξ) = - ϕ (ξ)$ for $ξ > 0$ ; the integral term is $\frac{1}{2 c} [c ϕ (x + c t) - c ϕ (c t - x)]$ by [F4] applied to the odd extension of $c ϕ^{'}$ on the two subintervals cut by $0$ ; adding, $u (x, t) = ϕ (x + c t) - ϕ (c t - x)$ as claimed; for $x > c t$ the same computation gives $u (x, t) = ϕ (x + c t)$ , the incoming left-moving profile, so the second term is precisely the reflection.

3.1givenstep 1.1F1F2F3algebraF5∎

Half-line energy: for each $t$ the density $e (U) (x, t) = \frac{1}{2} (U_{t}^{2} + c^{2} U_{x}^{2})$ is even in $x$ , because $U (\cdot, t)$ odd makes $U_{t} (\cdot, t)$ odd and $U_{x} (\cdot, t)$ even; hence $\int_{R} e d x = 2 \int_{0}^{\infty} e d x$ , that is, $E_{(0, \infty)} (t) = \frac{1}{2} E_{R} (t)$ ; fix $T_{0} > 0$ ; by [F3] the support of $U (\cdot, t)$ is contained in the fixed compact set $K : = (supp u_{0} \cup supp u_{1}) + [- c T_{0}, c T_{0}]$ for every $t \in [0, T_{0}]$ , so [F2] makes $E_{R}$ constant on $(0, T_{0})$ ; moreover $E_{R} (t) = \int_{K} e (U) (x, t) d x$ there and at the endpoints, and uniform continuity of $e (U)$ on $K \times [0, T_{0}]$ together with the finite measure of $K$ makes this energy continuous on $[0, T_{0}]$ , so the constancy extends to both endpoints; hence $E_{(0, \infty)} = \frac{1}{2} E_{R}$ is constant on $[0, T_{0}]$ , and since $T_{0}$ was arbitrary it is constant on $[0, \infty)$ , which is (iii).

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Zero wave energy means a spatial constant, fixed by the displacement datum

Example

Assume the Axiom of Countable Choice. Let $c > 0$ and let $u$ be a classical solution of the homogeneous equation $□_{c} u = 0$ 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, with $E (t) = E (0)$ for every $t \in (0, T)$ — the sharp form of conservation in the senses of Conservation of total wave energy in three admissible settings — and with $E (0) = \frac{1}{2} \int_{R^{n}} (u_{1}^{2} + c^{2} ∣ D u_{0} ∣^{2}) d x = 0$ . Then $u_{1} = 0$ and $D u_{0} = 0$ , so the displacement datum $u_{0}$ is constant on $R^{n}$ ; the energy seminorm sees only $(u_{t}, D u)$ and cannot fix that constant, and the evolution keeps it: $u (\cdot, t) \equiv u_{0}$ for every $t$ .

In particular every constant displacement with zero initial velocity, $u (x, t) \equiv k$ for a fixed $k \in R$ , is a genuine classical solution of zero energy. Thus "zero energy" is strictly weaker than "zero solution": the displacement datum is what fixes the residual constant (Energy uniqueness for the wave Cauchy problem(ii)).

Facts & Assumptions

Given: ${A C}_{ω}$ ; a classical homogeneous solution $u$ with Cauchy data $(u_{0}, u_{1})$ and differentiable displacement $u_{0}$ , so $D u_{0}$ in the initial-energy hypothesis is defined, conserved total energy in the sharp form $E (t) = E (0)$ for $t \in (0, T)$ and $E (0) = \frac{1}{2} \int (u_{1}^{2} + c^{2} ∣ D u_{0} ∣^{2}) = 0$ ; 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.

[F1]

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

[F2]

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

[F3]

In each setting of the conservation theorem the energy is constant on the interval; the hypothesis of this Example records the sharp form $E (t) = E (0)$ for all $t \in (0, T)$ . (Conservation of total wave energy in three admissible settings)

[F4]

A continuous function on an interval with vanishing derivative 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)

Verification

1.1givenF1F2F3algebra

Vanishing at positive times: sharp conservation gives $E (t) = E (0) = 0$ for each $t \in (0, T)$ . The nonnegative continuous density therefore vanishes everywhere by [F1], so $u_{t} (\cdot, t) = D u (\cdot, t) = 0$ . By [F2], each spatial slice is constant.

2.1givenstep 1.1F4algebra

Time constancy and the data: for each fixed $x$ , the function $t \mapsto u (x, t)$ has derivative zero on $(0, T)$ , so [F4] makes it constant there. Together with step 1.1 this gives one constant $k$ on all space-time. The Cauchy limits then give $u_{0} (x) = k$ and $u_{1} (x) = 0$ at every $x$ , hence $D u_{0} = 0$ and $u (\cdot, t) \equiv u_{0}$ . This derives pointwise data vanishing without inferring it from an almost-everywhere statement at $t = 0$ .

3.1givenstep 2.1algebra∎

The converse check: the constant displacement $u (x, t) \equiv k$ has $u_{t} = D u = 0$ , so $e \equiv 0$ and $E \equiv 0$ , and $□_{c} u = 0$ trivially; hence the zero-energy solutions are exactly the constant displacements, and that constant is precisely the initial displacement datum, which the energy cannot see.

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A three-dimensional spherical pulse leaves a quiet interior

Example

Assume the Axiom of Countable Choice. Let $c > 0$ , $x_{0} \in R^{3}$ , $r > 0$ , and let $(u_{0}, u_{1}) \in C_{c}^{3} \times C_{c}^{2}$ be supported in ${\overline{B}}_{r} (x_{0})$ (The support of a function on $R^{n}$ and its compactly supported Riemann integral). Then the Kirchhoff solution $u$ of Kirchhoff's formula in three dimensions satisfies $u (x, t) = 0$ whenever $∣ x - x_{0} ∣ > c t + r$ or (for $c t > r$ ) $∣ x - x_{0} ∣ < c t - r$ ; at time $t$ the pulse is carried by the spherical shell

$c t - r \leq ∣ x - x_{0} ∣ \leq c t + r,$

and the interior $B_{c t - r} (x_{0})$ behind the front is quiet. This is the concrete illustration of the strong Huygens principle in three dimensions (The strong Huygens principle in odd spatial dimensions(b)), and it is the three-dimensional side of the contrast with A two-dimensional pulse has a tail inside the cone.

Facts & Assumptions

Given: ${A C}_{ω}$ ; $c > 0$ , $x_{0} \in R^{3}$ , $r > 0$ , compactly supported data $(u_{0}, u_{1})$ with support in ${\overline{B}}_{r} (x_{0})$ ; the Kirchhoff solution $u$ .

[F1]

Kirchhoff's formula defines the solution and evaluates it from $u_{0}$ , its radial derivative, and $u_{1}$ on the sphere $\partial B_{c t} (x)$ , for $t > 0$ . (Kirchhoff's formula in three dimensions)

[F2]

Shell form of strong Huygens in odd dimensions: if the data are supported in a compact $K$ and $\partial B_{c t} (x) \cap K = \emptyset$ , then $u (x, t) = 0$ . (The strong Huygens principle in odd spatial dimensions, The strong Huygens principle in the homogeneous Cauchy setting)

Verification

1.1givenalgebra

The support ball lies inside the sphere: if $∣ x - x_{0} ∣ < c t - r$ (with $c t > r$ ) and $∣ y - x_{0} ∣ \leq r$ , then $∣ y - x ∣ \leq ∣ y - x_{0} ∣ + ∣ x_{0} - x ∣ < r + (c t - r) = c t$ , so ${\overline{B}}_{r} (x_{0})$ is disjoint from $\partial B_{c t} (x)$ with positive distance; if $∣ x - x_{0} ∣ > c t + r$ and $∣ y - x_{0} ∣ \leq r$ , then $∣ y - x ∣ \geq ∣ x - x_{0} ∣ - ∣ y - x_{0} ∣ > c t + r - r = c t$ , so again the support ball is disjoint from the sphere with positive distance.

2.1givenstep 1.1F1F2algebra∎

Vanishing: in either case of step 1.1 the data are supported in a compact set disjoint from $\partial B_{c t} (x)$ , so [F2] gives $u (x, t) = 0$ ; hence $u (\cdot, t)$ vanishes both outside the outer sphere $∣ x - x_{0} ∣ = c t + r$ and inside the inner sphere $∣ x - x_{0} ∣ = c t - r$ , so its support is contained in the closed shell $c t - r \leq ∣ x - x_{0} ∣ \leq c t + r$ ; the quiet interior behind the front is the case $∣ x - x_{0} ∣ < c t - r$ , and the statement is exactly the three-dimensional instance of the shell form [F2], evaluated from the data on $\partial B_{c t} (x)$ as [F1] prescribes.

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A two-dimensional pulse has a tail inside the cone

Example

Assume the Axiom of Countable Choice. Let $c > 0$ , $x_{0} \in R^{2}$ , $r > 0$ , and let $u_{1} \in C_{c}^{2} (R^{2})$ be nonnegative and not identically zero with support in $B_{r} (x_{0})$ ; let $u$ be the Poisson solution with data $(0, u_{1})$ (Poisson's formula in two dimensions by descent). Then for every $t > r / c$

$u (x_{0}, t) = \frac{1}{2 π c} \int_{B_{c t} (x_{0})} \frac{u_{1} (y)}{\sqrt{c^{2} t^{2} - ∣ y - x_{0} ∣^{2}}} d y > 0.$

The centre of the forward cone keeps seeing the pulse after the front has passed: the two-dimensional pulse has a tail inside the cone, in contrast with the quiet three-dimensional interior of A three-dimensional spherical pulse leaves a quiet interior.

Facts & Assumptions

Given: ${A C}_{ω}$ ; $c > 0$ , $x_{0} \in R^{2}$ , $r > 0$ , and a nonnegative $u_{1} \in C_{c}^{2} (R^{2})$ , $u_{1} \equiv̸ 0$ , supported in $B_{r} (x_{0})$ ; the Poisson solution $u$ with data $(0, u_{1})$ .

[F1]

Poisson's formula for data $(0, u_{1})$ : $u (x, t) = \frac{1}{2 π c} \int_{B_{c t} (x)} (c^{2} t^{2} - ∣ y - x ∣^{2})^{- 1 / 2} u_{1} (y) d y$ for $t > 0$ . (Poisson's formula in two dimensions by descent)

[F2]

In dimension two, strong Huygens fails: admissible data supported strictly inside the base disk $B_{c t} (x)$ can affect the value $u (x, t)$ at its vertex. (Wave tails in one and even spatial dimensions: strong Huygens fails)

[F3]

The nonnegative integral is monotone and positively homogeneous, and has value zero exactly for a function vanishing almost everywhere. A continuous function positive at a point is bounded below by a positive constant on a smaller ball, whose measure is positive. (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)

Verification

1.1givenF1algebra

The value at the centre: setting $x = x_{0}$ in [F1] gives the displayed formula $u (x_{0}, t) = \frac{1}{2 π c} \int_{B_{c t} (x_{0})} (c^{2} t^{2} - ∣ y - x_{0} ∣^{2})^{- 1 / 2} u_{1} (y) d y$ , the integrand being defined and continuous on the open disk because $∣ y - x_{0} ∣ < c t$ there.

2.1givenstep 1.1F3algebra

Positivity: if $t > r / c$ then ${\overline{B}}_{r} (x_{0}) \subseteq B_{c t} (x_{0})$ , and on that closed support ball $∣ y - x_{0} ∣ \leq r < c t$ gives $\sqrt{c^{2} t^{2} - ∣ y - x_{0} ∣^{2}} \leq c t$ , hence the weight $(c^{2} t^{2} - ∣ y - x_{0} ∣^{2})^{- 1 / 2} \geq 1 / (c t) > 0$ . Since $u_{1} \geq 0$ is continuous and nonzero, it is positive on a nonempty open subset of $B_{r} (x_{0})$ , so [F3] gives $\int_{B_{c t} (x_{0})} u_{1} (y) d y > 0$ and the displayed integral is at least $(2 π c)^{- 1} (c t)^{- 1} \int_{B_{c t} (x_{0})} u_{1} (y) d y > 0$ ; this shows that the centre still sees a positive displacement at every time after the front $∣ x - x_{0} ∣ = c t$ has passed beyond the support, that is, for $t > r / c$ .

3.1givenstep 2.1F2∎

The tail is carried by the interior: for $t > r / c$ the data are supported strictly inside $B_{c t} (x_{0})$ and vanish near its boundary, yet step 2.1 gives $u (x_{0}, t) > 0$ . This is the interior tail and illustrates the failure of sphere-only dependence in [F2].

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Finite speed of propagation does not imply strong Huygens

Statement refuted

The finite-speed property (Compact support expands at speed at most c) coexists with three distinct wave phenomena: the strong Huygens principle can fail (The strong Huygens principle in the homogeneous Cauchy setting), regularity need not improve, and nonnegative displacement data need not produce a nonnegative solution. The following compactly supported witnesses make these distinctions explicit.

Failure of strong Huygens. Let $c > 0$ , $x_{0} \in R^{n}$ , $t_{0} > 0$ and $n \in {1, 2}$ . In dimension $1$ , choose $u_{0} = 0$ and a nonnegative nonzero $u_{1} \in C_{c}^{1} (R)$ supported in $(x_{0} - c t_{0}, x_{0} + c t_{0})$ . Then the d'Alembert formula gives $u (x_{0}, t_{0}) = \frac{1}{2 c} \int u_{1} > 0$ . In dimension $2$ , choose $u_{0} = 0$ and a nonnegative nonzero $u_{1} \in C_{c}^{2} (R^{2})$ supported in $B_{r} (x_{0})$ for some $r < c t_{0}$ ; the Poisson formula gives $u (x_{0}, t_{0}) > 0$ . In both cases the data vanish on a neighbourhood of $S (x_{0}, c t_{0})$ , yet the value is nonzero, while finite speed still holds.
No smoothing. There is a compactly supported $F \in C^{2} (R)$ that is not $C^{3}$ . The traveling wave $u (x, t) = F (x - c t)$ solves the one-dimensional equation and remains $C^{2}$ but not $C^{3}$ for every $t \geq 0$ . Its support translates at speed $c$ .
No maximum principle. In dimension $2$ , take a nonnegative nonzero $u_{0} \in C_{c}^{\infty} (B_{r} (x_{0}))$ and $u_{1} = 0$ . At any time $t_{0} > r / c$ , the displacement term of Poisson's formula gives $u (x_{0}, t_{0}) < 0$ . Thus the solution can change sign although the initial displacement is nonnegative and the initial velocity is zero.

All four data pairs are compactly supported. The d’Alembert witnesses extend $C^{2}$ through time zero; for the smooth Poisson witnesses the descended sphere expression $\partial_{t} [t M_{U_{0}}^{(3)} ((x, 0), c t)] + t M_{U_{1}}^{(3)} ((x, 0), c t)$ extends smoothly through zero, since the signed-radius means are integrals of smooth data over the fixed compact sphere and may be differentiated there on any compact parameter set. Thus the initial regularity hypothesis is met and Compact support expands at speed at most c supplies the finite-speed bound. The no-smoothing and sign-change examples are independent of the Huygens witnesses; they show why the qualitative properties listed by Hunter require separate arguments.

Facts & Assumptions

Given: ${A C}_{ω}$ ; $c > 0$ ; the compactly supported smooth data chosen in the proof; and a compactly supported $C^{2}$ profile.

[F1]

For $u_{0} \in C^{2} (R)$ and $u_{1} \in C^{1} (R)$ , the unique classical solution is $u (x, t) = \frac{1}{2} (u_{0} (x - c t) + u_{0} (x + c t)) + \frac{1}{2 c} \int_{x - c t}^{x + c t} u_{1} (y) d y .$ (d'Alembert's formula and uniqueness in one dimension)

[F2]

For $u_{0} \in C^{3} (R^{2})$ and $u_{1} \in C^{2} (R^{2})$ , the two-dimensional solution is the Poisson expression $u (x, t) = \frac{1}{2 π c} \frac{\partial}{\partial t} \int_{B_{c t} (x)} \frac{u_{0} (y)}{\sqrt{c^{2} t^{2} - ∣ y - x ∣^{2}}} d y + \frac{1}{2 π c} \int_{B_{c t} (x)} \frac{u_{1} (y)}{\sqrt{c^{2} t^{2} - ∣ y - x ∣^{2}}} d y .$ (Poisson's formula in two dimensions by descent)

[F3]

Finite propagation: for a $C^{2}$ solution defined on a neighbourhood of the initial slab, with data supported in a compact $K$ and a source supported in ${(x, t) : dist (x, K) \leq c t}$ (in particular for a zero source), $supp u (\cdot, t) \subseteq K + {\overline{B}}_{c t} (0)$ for every $t$ . (Compact support expands at speed at most c)

[F4]

The strong Huygens principle in the homogeneous Cauchy setting is the statement that the value is carried by the sphere $S (x_{0}, c t_{0}) = \partial B_{c t_{0}} (x_{0})$ : admissible perturbations vanishing on a neighbourhood of $S$ do not change $u (x_{0}, t_{0})$ . (The strong Huygens principle in the homogeneous Cauchy setting)

[F5]

For any centre and radius there is a smooth nonnegative bump equal to one on the concentric half-radius ball and supported inside the full ball, by translating A smooth bump between concentric Euclidean balls.

[F6]

A nonnegative integral is monotone and positively homogeneous; it vanishes exactly for functions zero almost everywhere. A continuous function positive somewhere is bounded below by a positive constant 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)

[F7]

A parameter derivative may pass under the integral when dominated by a fixed integrable function on the parameter interval. (Differentiation under the integral sign)

[F8]

Chain, product and real-power rules apply to the explicit profile off its join points; the integral of a continuous derivative is its endpoint difference, and Darboux, Riemann and Lebesgue interval integrals agree. (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 $g (c) \neq 0$ , Continuity and derivatives of positive-base real powers, The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ , 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

1.1givenF1F3F4F5F6

The one-dimensional Huygens witness: choose $u_{0} = 0$ and a smooth nonnegative nonzero bump $u_{1}$ as in [F5] supported in $(x_{0} - c t_{0}, x_{0} + c t_{0})$ . Formula [F1] gives $u (x_{0}, t_{0}) = \frac{1}{2 c} \int_{x_{0} - c t_{0}}^{x_{0} + c t_{0}} u_{1} (y) d y > 0.$ The data are compactly supported inside the open base interval, hence vanish on a neighbourhood of its boundary sphere; [F3] supplies finite speed.

1.2givenF2F3F4F5F6

The two-dimensional Huygens witness: choose $0 < r < c t_{0}$ and the nonnegative nonzero smooth bump $u_{1}$ of [F5] supported in $B_{r} (x_{0})$ , with $u_{0} = 0$ . Formula [F2] gives $u (x_{0}, t_{0}) = \frac{1}{2 π c} \int_{B_{r} (x_{0})} \frac{u_{1} (y)}{\sqrt{c^{2} t_{0}^{2} - ∣ y - x_{0} ∣^{2}}} d y > 0.$ Again the data vanish on a neighbourhood of $S (x_{0}, c t_{0})$ , and [F3] applies.

1.3givenF1F8

Compact traveling wave with no smoothing: define $F (s) : = {\begin{cases} (1 - s^{2})^{3} ∣ s ∣^{5 / 2}, & ∣ s ∣ < 1, \\ 0, & ∣ s ∣ \geq 1. \end{cases}$ At $s = \pm 1$ the factor $(1 - s^{2})^{3}$ makes $F, F^{'}, F^{''}$ tend to zero, so the extension by zero is compactly supported and $C^{2}$ . The power and product rules [F8] give $F \to 0$ and $F^{'} (s) = \frac{5}{2} sgn (s) ∣ s ∣^{3 / 2} + O (∣ s ∣^{7 / 2}) \to 0$ as $s \to 0$ . The difference quotients of $F$ and $F^{'}$ at zero tend to zero, so $F^{'} (0) = F^{''} (0) = 0$ . Near $s = 0$ , $F^{''} (s) = \frac{15}{4} ∣ s ∣^{1 / 2} + O (∣ s ∣^{5 / 2}),$ so $F^{''} (h) / h \to + \infty$ as $h ↓ 0$ ; hence $F$ is not $C^{3}$ . Take $u_{0} = F$ , $u_{1} = - c F^{'}$ . In [F1], by the FTC in [F8], the integral term equals $- \frac{1}{2} (F (x + c t) - F (x - c t))$ , so the solution simplifies to $u (x, t) = F (x - c t)$ . It is $C^{2}$ and not $C^{3}$ at $x = c t$ , and its compact support is translated exactly at speed $c$ .

1.4givenF2F3F5F6F7

Nonnegative displacement becomes negative: take the nonnegative nonzero smooth bump $u_{0}$ of [F5] supported in $B_{r} (x_{0})$ , $u_{1} = 0$ , and $t_{0} > r / c$ . Since the support is strictly inside $B_{c t} (x_{0})$ for $t$ near $t_{0}$ , on a small closed time interval about $t_{0}$ the quantity $c^{2} t^{2} - r^{2}$ has a positive lower bound. Thus the integrand and its time derivative are uniformly bounded on the compact support, providing a constant integrable majorant; [F7] permits differentiating the displacement integral in [F2] over the fixed support: $u (x_{0}, t_{0}) = - \frac{c t_{0}}{2 π} \int_{B_{r} (x_{0})} \frac{u_{0} (y)}{(c^{2} t_{0}^{2} - ∣ y - x_{0} ∣^{2})^{3 / 2}} d y < 0.$ The strict inequality follows by [F6] because the kernel is positive and $u_{0}$ is nonnegative and nonzero. Thus positivity is not preserved, despite nonnegative displacement and zero initial velocity; [F3] still gives finite speed.

2.1step 1.1step 1.2F3F4

Huygens conclusion: in steps 1.1 and 1.2 each data pair is supported strictly inside the relevant base ball, so it agrees with the zero pair on a neighbourhood of the sphere but gives a nonzero value at the vertex. This contradicts the defining data-insensitivity in [F4]. Finite propagation [F3] remains true; therefore finite speed does not imply strong Huygens, as recorded in Finite propagation is not the Huygens principle.

3.1step 1.3step 1.4∎

Regularity and positivity conclusions: step 1.3 retains a second-derivative cusp under translation, so the wave flow has no smoothing; step 1.4 gives an explicit failure of positivity preservation, hence of a maximum principle.

Wave Energy, Finite Propagation and Huygens' Principle — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

Conserved energy of a travelling wave packet

Example

Facts & Assumptions

Verification

Plane-wave support translates at the characteristic speed

Example

Facts & Assumptions

Verification

The local conservation law need not integrate to a finite conserved energy

Statement refuted

Facts & Assumptions

Counterexample

Wave energy need not be conserved through an open boundary

Statement refuted

Facts & Assumptions

Proof

Odd reflection at a Dirichlet endpoint

Example

Facts & Assumptions

Verification

Zero wave energy means a spatial constant, fixed by the displacement datum

Example

Facts & Assumptions

Verification

A three-dimensional spherical pulse leaves a quiet interior

Example

Facts & Assumptions

Verification

A two-dimensional pulse has a tail inside the cone

Example

Facts & Assumptions

Verification

Finite speed of propagation does not imply strong Huygens

Statement refuted

Facts & Assumptions

Proof

Sources