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Wave Equation Representation Formulas

1 · Prerequisites

2 · Summary

This page develops the classical representation formulas for the Cauchy problem of the wave equation $□_{c} u = \partial_{t}^{2} u - c^{2} Δ u = 0$ in $R^{n}$ . After fixing the vocabulary of wave operator, Cauchy data and wave speed, together with the unit-speed rescaling convention that transfers the unit-speed formulas of the sources to general speed, the page builds the one-dimensional theory: factorisation of the operator, the general solution as a sum of a right- and a left-travelling wave, d'Alembert's formula with uniqueness and data attainment, the domain of dependence, the forced formula over the characteristic triangle. The companion examples page contains a counterexample showing that data on one characteristic line do not determine the solution. In parallel it develops spherical means: their smooth signed-radius extension and parity, the vanishing first moment of the sphere measure, the ball–sphere radial identity, the Euler–Poisson–Darboux equation, the projection of sphere integrals of cylindrical functions onto weighted ball integrals, and the two iterated radial-derivative lemmas behind the odd-dimensional reduction. These tools yield Kirchhoff's formula in three dimensions, Poisson's formula in two dimensions by descent, the odd-dimensional formula by iterated spherical means, the even-dimensional formula by descent, the attainment of the Cauchy data, the support dichotomy between sphere-supported odd-dimensional kernels and interior even-dimensional kernels, time-reversal invariance, Duhamel's principle with the forced three-dimensional retarded potential, local determination by the Cauchy data, and a remark separating the wave Poisson formula from the harmonic Poisson kernel.

All statements of the page are read under the Axiom of Countable Choice, which supplies the polar surface measure and the sphere integrals; the one-dimensional construction is choice-free. Uniqueness of classical solutions in dimensions $n \geq 2$ is deliberately deferred to the energy page that consumes this one.

For odd dimensions, sphere dependence means dependence on the data and finitely many transverse derivatives along the sphere; agreement on an open neighbourhood suffices, while bare equality of values on the sphere generally does not. Duhamel's principle assumes joint continuity of the source's spatial derivatives through order $⌊ n / 2 ⌋ + 1$ , without requiring time derivatives.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6.1-sol)Open item page →

Wave equation, Cauchy data and wave speed

Definition

Let $n \geq 1$ , $T > 0$ and $c > 0$ , let $\partial_{t}, \partial_{j}$ be the partial derivatives of Directional derivatives and partial derivatives of a map $U \subseteq R^{m} \to R^{n}$ and let $Δ = \sum_{j = 1}^{n} \partial_{j}^{2}$ be the Laplacian of The Laplacian of a $C^{2}$ function and of a $C^{2}$ vector field. The wave operator of speed $c$ is $□_{c} : = \partial_{t}^{2} - c^{2} Δ,$ and the wave equation with source $f$ is the equation $□_{c} u = f$ on a slab $R^{n} \times (0, T)$ ; the equation is homogeneous when $f = 0$ . A classical solution on the time slab $R^{n} \times [0, T)$ is a function $u$ of class $C^{2}$ on $R^{n} \times (0, T)$ in the sense of $C^{k}$ maps and multi-index derivative notation in Euclidean space which satisfies the equation at every point of $R^{n} \times (0, T)$ . This is the classical-solution and Cauchy-data vocabulary of Scalar partial differential equations, order, and classical solutions, and $□_{c}$ is a linear second-order operator in the sense of Linear, semilinear, quasilinear, and fully nonlinear partial differential equations.

The Cauchy problem for $□_{c} u = f$ prescribes a displacement $u_{0} : R^{n} \to R$ and a velocity $u_{1} : R^{n} \to R$ and asks for a classical solution attaining them as $t ↓ 0$ in the pointwise sense: $u (x, t) \to u_{0} (x)$ and $\partial_{t} u (x, t) \to u_{1} (x)$ for every $x \in R^{n}$ as $t ↓ 0$ . Only the limits are part of the data, so $u$ need not be defined at $t = 0$ a priori.

Unit-speed rescaling convention. A function $u$ solves $□_{c} u = 0$ on $R^{n} \times [0, T)$ with data $(u_{0}, u_{1})$ if and only if $v (x, s) : = u (x, s / c)$ solves $\partial_{s}^{2} v = Δ v$ on $R^{n} \times [0, c T)$ with data $(u_{0}, u_{1} / c)$ ; equivalently $u (x, t) = v (x, c t)$ . Indeed continuous coordinate partial derivatives imply total differentiability (If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative), so the chain rule (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ ) gives $\partial_{t} u (x, t) = c \partial_{s} v (x, c t)$ and $\partial_{t}^{2} u (x, t) = c^{2} \partial_{s}^{2} v (x, c t)$ while spatial derivatives are unchanged, so $□_{c} u = c^{2} (\partial_{s}^{2} v - Δ v)$ evaluated at $s = c t$ , and the initial displacement limits agree, while the initial velocity limit of $u$ is $c$ times that of $v$ . The cited treatments state their formulas at unit speed, and this convention is the only place where the general speed enters those formulas.

Sphere normalisation. Let $σ$ be the polar surface measure on the unit sphere (The polar surface set function on the unit sphere). Write $V_{m} : = ∣ B_{1}^{m} ∣$ for the unit ball and $ω_{m} : = σ (S^{m}) = (m + 1) V_{m + 1} > 0$ for the total polar measure of the unit sphere $S^{m} \subseteq R^{m + 1}$ (Sphere and ball measures scale in Rn); thus the boundary sphere of a ball in $R^{n}$ has total measure $ω_{n - 1} = n V_{n}$ . For an integer $m \geq 0$ put $(2 m + 1)!! : = (2 m + 1) (2 m - 1) \dots 3 \cdot 1$ and $(2 m)!! : = (2 m) (2 m - 2) \dots 2$ , so that $0!! = 1!! = 1$ and $(n - 2)!!$ is defined for every odd $n \geq 3$ . All sphere and weighted-ball integrals on this page are read under the Axiom of Countable Choice of The Axiom of Countable Choice ( ${A C}_{ω}$ ), under which the polar measure and its integrals are supplied.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

The iterated radial-derivative identity behind the odd-dimensional reduction

Statement

Let $k \geq 1$ and let $D_{r} : = r^{- 1} \partial_{r}$ be the radial derivative on $(0, \infty)$ . For every $f \in C^{k + 1} ((0, \infty))$ , $\frac{\partial^{2}}{\partial r^{2}} D_{r}^{k - 1} (r^{2 k - 1} f (r)) = D_{r}^{k - 1} [r^{2 k - 1} \frac{1}{r^{2 k}} \frac{\partial}{\partial r} (r^{2 k} \partial_{r} f (r))]$ on $(0, \infty)$ . Both sides are finite combinations of derivatives of $f$ computed by the product and quotient rules; no integral and no differential equation for $f$ is used.

Facts & Assumptions

Given: an integer $k \geq 1$ , a function $f \in C^{k + 1} ((0, \infty))$ , and the operator $D_{r} = r^{- 1} \partial_{r}$ acting on functions on $(0, \infty)$ .

[F1]

Sums, products, quotients of differentiable functions are differentiable, with the usual sum, product, quotient rules; nonnegative integer powers are differentiated by repeated product rules, and reciprocals by the quotient rule on nonzero domains (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 $g (c) \neq 0$ ).

Proof

1.1F1given

Put $g : = r^{2 k - 1} f$ , so that $g \in C^{k + 1} ((0, \infty))$ and $f = r^{1 - 2 k} g$ . The left-hand side is $\partial_{r}^{2} D_{r}^{k - 1} g$ , and for the right-hand side the product rule gives $r^{2 k} \partial_{r} f = (1 - 2 k) g + r g^{'}$ , hence $r^{- 1} \partial_{r} (r^{2 k} \partial_{r} f) = g^{''} + (2 - 2 k) r^{- 1} g^{'}$ ; since $r^{2 k - 1} r^{- 2 k} = r^{- 1}$ , the right-hand side is $D_{r}^{k - 1} (g^{''} + (2 - 2 k) r^{- 1} g^{'})$ . It therefore suffices to prove $\partial_{r}^{2} D_{r}^{k - 1} g = D_{r}^{k - 1} (g^{''} + (2 - 2 k) r^{- 1} g^{'})$ for every $g \in C^{k + 1} ((0, \infty))$ , which is what the following steps do.

1.2F1algebra

On $(0, \infty)$ the operators satisfy $\partial_{r} = r D_{r}$ , hence $\partial_{r}^{2} = \partial_{r} (r D_{r}) = D_{r} + r^{2} D_{r}^{2}$ ; and for every $j \geq 1$ and every $v \in C^{j} ((0, \infty))$ one has $D_{r}^{j} (r^{2} v) = r^{2} D_{r}^{j} v + 2 j D_{r}^{j - 1} v$ . The last identity is proved by induction on $j$ : for $j = 1$ , $D_{r} (r^{2} v) = r^{- 1} (2 r v + r^{2} v^{'}) = 2 v + r^{2} D_{r} v$ ; and if it holds for $j$ , then $D_{r}^{j + 1} (r^{2} v) = D_{r} (r^{2} D_{r}^{j} v + 2 j D_{r}^{j - 1} v) = r^{2} D_{r}^{j + 1} v + 2 D_{r}^{j} v + 2 j D_{r}^{j} v = r^{2} D_{r}^{j + 1} v + 2 (j + 1) D_{r}^{j} v$ .

1.3F1algebra

Fix $g \in C^{k + 1} ((0, \infty))$ . If $k = 1$ , then $D_{r}^{k - 1} (r^{2} D_{r}^{2} g) = r^{2} D_{r}^{2} g = r^{2} D_{r}^{k + 1} g + 2 (k - 1) D_{r}^{k} g$ directly; if $k \geq 2$ , the second identity of the previous step with $j = k - 1$ and $v = D_{r}^{2} g$ gives the same equality. Also $g^{''} = \partial_{r}^{2} g = D_{r} g + r^{2} D_{r}^{2} g$ by the first identity of the previous step, so $D_{r}^{k - 1} g^{''} = D_{r}^{k} g + D_{r}^{k - 1} (r^{2} D_{r}^{2} g) = r^{2} D_{r}^{k + 1} g + (2 k - 1) D_{r}^{k} g$ . Moreover $r^{- 1} g^{'} = D_{r} g$ , so $D_{r}^{k - 1} ((2 - 2 k) r^{- 1} g^{'}) = (2 - 2 k) D_{r}^{k} g$ .

2.1F1algebra∎

Adding the two pieces of the previous step gives $D_{r}^{k - 1} (g^{''} + (2 - 2 k) r^{- 1} g^{'}) = r^{2} D_{r}^{k + 1} g + (2 k - 1 + 2 - 2 k) D_{r}^{k} g = r^{2} D_{r}^{k + 1} g + D_{r}^{k} g$ , and by the first identity of the second step this last quantity is $\partial_{r}^{2} D_{r}^{k - 1} g$ . This proves the equivalent identity for $g$ and hence, by the substitution of the first step, the identity of the statement.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

Radial-derivative expansion of the Euler–Poisson–Darboux transform and its zero-radius limit

Statement

Let $k \geq 1$ and let $D_{r} = r^{- 1} \partial_{r}$ on $(0, \infty)$ . For every $f \in C^{k - 1} ((0, \infty))$ there are real constants $α_{k, j}$ , $0 \leq j \leq k - 1$ , such that $D_{r}^{k - 1} (r^{2 k - 1} f (r)) = \sum_{j = 0}^{k - 1} α_{k, j} r^{j + 1} f^{(j)} (r), α_{k, 0} = (2 k - 1)!! .$ Consequently, if the finite limit $f (0) : = \lim_{r ↓ 0} f (r)$ exists and each derivative $f^{(j)}$ with $1 \leq j \leq k - 1$ is bounded on $(0, 1]$ (an empty condition when $k = 1$ ) — in particular if $f$ extends to a $C^{k - 1}$ function on $[0, \infty)$ — then $r^{- 1} D_{r}^{k - 1} (r^{2 k - 1} f) (r) ⟶ (2 k - 1)!! f (0) (r ↓ 0) .$ Every coefficient $α_{k, 0} = (2 k - 1)!!$ is nonzero, so the transformed average recovers the value at $r = 0$ with the exact dimensional constant. The limit assumption is needed even for $k = 1$ , when $f (r) = \sin (1 / r)$ is bounded but has no limit. Boundedness of the higher derivatives is also substantive: for $k = 2$ the function $f (r) = r \sin (r^{- 2})$ , which is continuous at $0$ , has $r f^{'} (r) = r \sin (r^{- 2}) - 2 r^{- 1} \cos (r^{- 2})$ unbounded, and then $r^{- 1} D_{r} (r^{3} f)$ does not tend to $3 f (0)$ .

Facts & Assumptions

Given: an integer $k \geq 1$ , a function $f \in C^{k - 1} ((0, \infty))$ , and the operator $D_{r} = r^{- 1} \partial_{r}$ on $(0, \infty)$ .

[F1]

Sums, products, quotients of differentiable functions are differentiable with the usual rules, and polynomial and reciprocal functions are differentiable on their domains (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 $g (c) \neq 0$ ).

Proof

1.1F1given

Base case. For $k = 1$ and $f \in C^{0} ((0, \infty))$ one has $D_{r}^{0} (r^{2 k - 1} f) = r f = α_{1, 0} r^{1} f^{(0)}$ with $α_{1, 0} = 1 = (2 \cdot 1 - 1)!!$ ; under the limit hypothesis, $r^{- 1} D_{r}^{0} (r f) = f (r) \to f (0)$ as $r ↓ 0$ .

1.2given

Induction hypothesis. Fix $k \geq 1$ and assume that for every $h \in C^{k - 1} ((0, \infty))$ there are real constants $α_{k, 0}, \dots, α_{k, k - 1}$ such that $D_{r}^{k - 1} (r^{2 k - 1} h) = \sum_{j = 0}^{k - 1} α_{k, j} r^{j + 1} h^{(j)}$ and $α_{k, 0} = (2 k - 1)!!$ .

1.3F1algebra

Shape of the $(k + 1)$ -st transform. Let $f \in C^{k} ((0, \infty))$ and put $h : = r^{2} f$ , so that $h \in C^{k} ((0, \infty)) \subseteq C^{k - 1} ((0, \infty))$ and $r^{2 k + 1} f = r^{2 k - 1} h$ . The induction hypothesis gives $D_{r}^{k - 1} (r^{2 k + 1} f) = \sum_{j = 0}^{k - 1} α_{k, j} r^{j + 1} h^{(j)}$ , and the product rule gives $h^{(j)} = r^{2} f^{(j)} + 2 j r f^{(j - 1)} + j (j - 1) f^{(j - 2)}$ for $0 \leq j \leq k - 1$ , where the last two terms are read as $0$ for $j = 0$ and $j = 1$ .

1.4F1algebra

Applying $D_{r}$ once to the finitely many resulting terms, and using $D_{r} (r^{m} v) = m r^{m - 2} v + r^{m - 1} v^{'}$ , produces a finite sum $\sum_{i = 0}^{k} β_{i} r^{i + 1} f^{(i)}$ with constants $β_{i}$ independent of $f$ : the term $α_{k, j} r^{j + 1} \cdot r^{2} f^{(j)}$ contributes $(j + 3) α_{k, j} r^{j + 1} f^{(j)}$ and $α_{k, j} r^{j + 2} f^{(j + 1)}$ ; the term $2 j α_{k, j} r^{j + 2} f^{(j - 1)}$ contributes $2 j (j + 2) α_{k, j} r^{j} f^{(j - 1)}$ and $2 j α_{k, j} r^{j + 1} f^{(j)}$ ; and the term $j (j - 1) α_{k, j} r^{j + 1} f^{(j - 2)}$ contributes $j (j - 1) (j + 1) α_{k, j} r^{j - 1} f^{(j - 2)}$ and $j (j - 1) α_{k, j} r^{j} f^{(j - 1)}$ . Every displayed monomial $r^{m} f^{(i)}$ has $m = i + 1$ with $0 \leq i \leq k$ , and negative powers do not occur because the terms with $j = 0, 1$ have one or both of the last two summands read as zero.

1.5F1algebra

Leading coefficient. Evaluating the identity of the previous step at the constant function $f \equiv 1$ , which lies in $C^{k}$ , gives $β_{0} r = D_{r}^{k} (r^{2 k + 1}) = [(2 k + 1) (2 k - 1) \dots 3] r = (2 k + 1)!! r$ : indeed $D_{r} (r^{2 k + 1}) = (2 k + 1) r^{2 k - 1}$ , each further application of $D_{r}$ lowers the exponent by $2$ and multiplies the coefficient by the previous exponent, and $k$ applications leave the exponent $1$ . Hence $β_{0} = (2 k + 1)!!$ , and setting $α_{k + 1, i} : = β_{i}$ for $0 \leq i \leq k$ extends the conclusion of the induction hypothesis from $k$ to $k + 1$ .

2.1givenalgebra∎

Conclusion. By the base case and the induction step, the expansion $D_{r}^{k - 1} (r^{2 k - 1} f) = \sum_{j = 0}^{k - 1} α_{k, j} r^{j + 1} f^{(j)}$ with $α_{k, 0} = (2 k - 1)!!$ holds for every $k \geq 1$ and every $f \in C^{k - 1} ((0, \infty))$ . If $f (0) : = \lim_{r ↓ 0} f (r)$ is finite and each $f^{(j)}$ with $1 \leq j \leq k - 1$ is bounded on $(0, 1]$ , then dividing by $r$ gives $r^{- 1} D_{r}^{k - 1} (r^{2 k - 1} f) (r) = \sum_{j = 0}^{k - 1} α_{k, j} r^{j} f^{(j)} (r) \to α_{k, 0} f (0) = (2 k - 1)!! f (0)$ as $r ↓ 0$ : the $j = 0$ term converges by the assumed limit, and each term with $j \geq 1$ is bounded by $∣ α_{k, j} ∣ r^{j} \sup_{(0, 1]} ∣ f^{(j)} ∣ \to 0$ .

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

Reflection invariance and vanishing first moment of the sphere measure

Statement

Assume the Axiom of Countable Choice. Let $n \geq 1$ and let $σ$ be the polar surface measure on the unit sphere $S^{n - 1} \subseteq R^{n}$ (The polar surface set function on the unit sphere). Then the reflection $R ω = - ω$ preserves $σ$ , and the first moment vanishes: $\int_{S^{n - 1}} ω d σ (ω) = 0 \in R^{n}, so \int_{S^{n - 1}} a \cdot ω d σ (ω) = 0 (a \in R^{n}) .$ The integrals are finite because $σ$ is a finite Borel measure.

Facts & Assumptions

Given: the Axiom of Countable Choice, an integer $n \geq 1$ and the polar surface measure $σ$ on $S^{n - 1}$ .

[F1]

$σ (E) = n λ_{n} {r ω : ω \in E, 0 < r \leq 1}$ for Borel $E \subseteq S^{n - 1}$ (The polar surface set function on the unit sphere).

[F2]

For an invertible linear $T : R^{n} \to R^{n}$ with matrix $A$ and every Lebesgue measurable $E$ , $λ_{n} (T [E]) = ∣ \det A ∣ λ_{n} (E)$ ; in particular $λ_{n} (- E) = λ_{n} (E)$ (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).

[F3]

Under Countable Choice, $σ$ is a finite Borel measure on $S^{n - 1}$ (Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma).

Proof

1.1F1F2algebra

Reflection invariance. For Borel $E \subseteq S^{n - 1}$ the set $- E$ is Borel and ${r θ : θ \in - E, 0 < r \leq 1} = - {r ω : ω \in E, 0 < r \leq 1}$ , so [F1] and [F2] give $σ (- E) = n λ_{n} (- {r ω : ω \in E, 0 < r \leq 1}) = n λ_{n} {r ω : ω \in E, 0 < r \leq 1} = σ (E)$ .

1.2F3algebra

Vanishing of the moment. Each coordinate function $g_{i} (ω) = ω_{i}$ is Borel and bounded by $1$ on $S^{n - 1}$ , so $\int g_{i} d σ$ is a finite real number by [F3]. Because $σ$ is invariant under the bijection $ω \mapsto - ω$ , the substitution formula for a measure-preserving bijection — valid for indicators by definition, for simple functions by linearity, and for bounded Borel functions by the supremum definition of the integral — gives $\int_{S^{n - 1}} g_{i} d σ = \int_{S^{n - 1}} g_{i} (- ω) d σ (ω) = - \int_{S^{n - 1}} g_{i} d σ$ ; hence $\int g_{i} d σ = 0$ for every $i$ , that is $\int_{S^{n - 1}} ω d σ (ω) = 0$ .

2.1algebra∎

For $a \in R^{n}$ , linearity of the integral in the integrand gives $\int_{S^{n - 1}} a \cdot ω d σ (ω) = a \cdot \int_{S^{n - 1}} ω d σ (ω) = 0$ .

LemmaStatement: Literature-sourcedProof: AI-adaptedOpen item page →

Differentiating an integral with moving endpoints

Statement

Let $I \subseteq R$ be an open interval, let $α, β \in C^{1} (I)$ with $α (t) < β (t)$ for $t \in I$ , let $J \subseteq R$ be an open interval containing the closure of the union of the intervals $[α (t), β (t)]$ over $t \in I$ , and let $F : I \times J \to R$ be continuous with continuous partial derivative $\partial_{t} F$ . Then $G (t) : = \int_{α (t)}^{β (t)} F (t, y) d y$ is $C^{1}$ on $I$ and $G^{'} (t) = F (t, β (t)) β^{'} (t) - F (t, α (t)) α^{'} (t) + \int_{α (t)}^{β (t)} \partial_{t} F (t, y) d y .$

Facts & Assumptions

Given: open intervals $I, J$ , functions $α, β \in C^{1} (I)$ with $α < β$ , and a continuous $F : I \times J \to R$ whose partial derivative $\partial_{t} F$ exists and is continuous on $I \times J$ , with $J$ containing the closure of the union of the intervals $[α (t), β (t)]$ .

[F1]

If $I_{0} \subseteq R$ is order-convex with at least two elements and $f : I_{0} \to R$ is continuous, then for $c_{0} \in I_{0}$ , $F (x) = \int_{c_{0}}^{x} f$ is a primitive of $f$ and $\int_{a}^{b} f = G (b) - G (a)$ for every primitive $G$ of $f$ and $a < b$ in $I_{0}$ (Every continuous function on an interval has a primitive; two primitives differ by a constant; and $\int_{a}^{b} f = G (b) - G (a)$ for any primitive $G$ ).

[F2]

Let $a < b$ , $c < d$ and let $g, h : [a, b] \times [c, d] \to R$ be continuous with $x \mapsto g (x, t)$ differentiable on $(a, b)$ and derivative $h (x, t)$ for every fixed $t \in [c, d]$ . Then $G (x) = \int_{c}^{d} g (x, t) d t$ is differentiable on $[a, b]$ with $G^{'} (x) = \int_{c}^{d} h (x, t) d t$ (Leibniz's rule on a compact rectangle: an interior parameter derivative with a continuous extension may be passed through a Riemann integral).

[F3]

If $f : U \to V \subseteq R^{n}$ is totally differentiable at $a$ and $g : V \to R^{p}$ is totally differentiable at $f (a)$ , then $g \circ f$ is totally differentiable at $a$ with $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ ). The required total differentiability follows from continuous coordinate partial derivatives (If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative).

[F4]

Sums and differences of differentiable functions are differentiable, with derivative the sum respectively difference of the derivatives (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 $g (c) \neq 0$ ).

Proof

1.1F1given

Localisation. Fix $t_{0} \in I$ . Choose a compact interval $K \subset I$ with $t_{0}$ in its interior $I_{0}$ . The endpoint functions are bounded on $K$ by A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value, so choose $a < b$ in $J$ with $a < α (t) < β (t) < b$ for $t \in K$ , and choose $r_{*} \in J$ with $r_{*} < a$ . Define $Φ (u, t) : = \int_{r_{*}}^{u} F (t, y) d y$ on $(a, b) \times I_{0}$ . By [F1], $G (t) = Φ (β (t), t) - Φ (α (t), t)$ .

1.2F1F2algebra

The primitive is $C^{1}$ on an open neighbourhood of the endpoint curves. By [F1], $\partial_{u} Φ (u, t) = F (t, u)$ ; by [F2] on compact rectangles inside $I \times J$ , $\partial_{t} Φ (u, t) = \int_{r_{*}}^{u} \partial_{t} F (t, y) d y$ . This last expression is jointly continuous: on a fixed compact rectangle, uniform continuity of $\partial_{t} F$ (Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous) bounds the change in $t$ by the interval length times a uniform error, and boundedness bounds the change in $u$ by a constant times $∣ u - u_{0} ∣$ . Thus both partial derivatives are continuous, and If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative makes $Φ$ totally differentiable.

2.1F1F3F4step 1.2algebra∎

Apply [F3] to the curves $t \mapsto (β (t), t)$ and $t \mapsto (α (t), t)$ in the open domain of $Φ$ , and subtract using [F4]. Evaluation of the primitive gives $G^{'} (t) = F (t, β (t)) β^{'} (t) - F (t, α (t)) α^{'} (t) + \int_{α (t)}^{β (t)} \partial_{t} F (t, y) d y$ . The same uniform estimate as in step 1.2 shows this derivative is continuous. Since $t_{0}$ was arbitrary, the formula holds throughout $I$ .

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Spherical means and the weighted ball integral of space-dependent data

Definition

Assume the Axiom of Countable Choice. Let $n \geq 1$ , let $f : R^{n} \to R$ be continuous, let $σ$ be the polar surface measure on $S^{n - 1}$ (The polar surface set function on the unit sphere) and let $ω_{n - 1} = σ (S^{n - 1}) = n V_{n} > 0$ , where $V_{n} = ∣ B_{1}^{n} ∣$ (Sphere and ball measures scale in Rn). The spherical mean of $f$ is $M_{f} (x, r) : = \frac{1}{ω_{n - 1}} \int_{S^{n - 1}} f (x + r ω) d σ (ω) (x \in R^{n}, r > 0),$ the average of $f$ over the sphere of centre $x$ and radius $r$ with respect to the polar measure; this is the mean of Spherical averages and local ball means in Rn with $u = f$ , restricted to continuous data. One sets $M_{f} (x, 0) : = f (x)$ ; that value is a convention whose consistency as a limit is proved later on this page, not assumed here. The unnormalised sphere integral is $S_{f} (x, r) : = \int_{S^{n - 1}} f (x + r ω) d σ (ω) = ω_{n - 1} M_{f} (x, r) (x \in R^{n}, r > 0) .$ For even $n$ and $r > 0$ put $W_{f} (x, r) : = \frac{1}{n!! V_{n}} \int_{B_{r} (x)} \frac{f (y)}{\sqrt{r^{2} - ∣ y - x ∣^{2}}} d y,$ where $n!! = n (n - 2) \dots 2$ and $V_{n} = ∣ B_{1}^{n} ∣$ is the volume of the unit ball (Sphere and ball measures scale in Rn); for $n = 2$ this is the weighted disk integral appearing in the two-dimensional Poisson formula below. The integral defining $W_{f} (x, r)$ is absolutely convergent and hence well defined: the weight $y \mapsto (r^{2} - ∣ y - x ∣^{2})^{- 1 / 2}$ is integrable over $B_{r} (x)$ — by translation invariance (Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation) and the polar formula its integral is $ω_{n - 1} \int_{0}^{r} s^{n - 1} (r^{2} - s^{2})^{- 1 / 2} d s < \infty$ (Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma) — while $f$ is bounded on the closed ball $\overline{B_{r} (x)}$ because that ball is compact (For $n \geq 1$ , every Euclidean closed ball and every Euclidean sphere of positive radius is compact) and a continuous function is bounded on a compact set (A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value); the product of an integrable function and a bounded function is Lebesgue integrable. The ball average used on this page is the normalised mean $A_{g} (x, r)$ of The average of a locally integrable function over a Euclidean ball, and every sphere or ball integral below is read under the Axiom of Countable Choice of The Axiom of Countable Choice ( ${A C}_{ω}$ ), which supplies the polar measure and its integrals.

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Factorisation of the one-dimensional wave operator

Statement

Let $c > 0$ and let $u$ be $C^{2}$ on an open subset of $R^{2}$ (Wave equation, Cauchy data and wave speed). Then $\partial_{t}^{2} u - c^{2} \partial_{x}^{2} u = (\partial_{t} - c \partial_{x}) (\partial_{t} + c \partial_{x}) u = (\partial_{t} + c \partial_{x}) (\partial_{t} - c \partial_{x}) u .$ In the characteristic coordinates $ξ = x - c t$ , $η = x + c t$ one has $\partial_{t}^{2} u - c^{2} \partial_{x}^{2} u = - 4 c^{2} \partial_{ξ} \partial_{η} u,$ so $u$ solves the homogeneous one-dimensional wave equation exactly on the open set where $u_{ξ η} = 0$ .

Facts & Assumptions

Given: a speed $c > 0$ and a $C^{2}$ function $u$ on an open subset of $R^{2}$ , with coordinates $(x, t)$ and characteristic coordinates $ξ = x - c t$ , $η = x + c t$ .

[F1]

If $f$ is $C^{2}$ on an open subset of $R^{m}$ , then $\partial_{i} \partial_{j} f = \partial_{j} \partial_{i} f$ for every pair of coordinate indices (Clairaut--Schwarz theorem for continuous second partial derivatives).

[F2]

[F3]

Sums, constant multiples of differentiable functions are differentiable with the usual rules (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 $g (c) \neq 0$ ).

Proof

1.1F1F3algebra

Expanding the two compositions and using [F1] for the mixed terms, $(\partial_{t} - c \partial_{x}) (\partial_{t} + c \partial_{x}) u = \partial_{t}^{2} u + c \partial_{t} \partial_{x} u - c \partial_{x} \partial_{t} u - c^{2} \partial_{x}^{2} u = \partial_{t}^{2} u - c^{2} \partial_{x}^{2} u$ , and $(\partial_{t} + c \partial_{x}) (\partial_{t} - c \partial_{x}) u = \partial_{t}^{2} u - c \partial_{t} \partial_{x} u + c \partial_{x} \partial_{t} u - c^{2} \partial_{x}^{2} u = \partial_{t}^{2} u - c^{2} \partial_{x}^{2} u$ ; hence both factorisations equal the operator applied to $u$ .

1.2F2F3algebra

Write $U (ξ, η) : = u (x, t)$ with $ξ = x - c t$ , $η = x + c t$ . By [F2] the chain rule for the substitution $(x, t) \mapsto (ξ, η) = (x - c t, x + c t)$ gives $\partial_{x} u = U_{ξ} + U_{η}$ and $\partial_{t} u = - c U_{ξ} + c U_{η}$ , with the right-hand sides evaluated at $(x - c t, x + c t)$ , hence $\partial_{t} - c \partial_{x} = - 2 c \partial_{ξ}$ and $\partial_{t} + c \partial_{x} = 2 c \partial_{η}$ as operators on $U$ ; composing, $\partial_{t}^{2} u - c^{2} \partial_{x}^{2} u = (\partial_{t} - c \partial_{x}) (\partial_{t} + c \partial_{x}) u = (- 2 c \partial_{ξ}) (2 c \partial_{η}) U = - 4 c^{2} U_{ξ η}$ .

2.1algebra∎

Since $c > 0$ the factor $- 4 c^{2}$ is nonzero, so $\partial_{t}^{2} u = c^{2} \partial_{x}^{2} u$ at a point if and only if $U_{ξ η} = 0$ there; this proves the claimed equivalence and completes the factorisation identities.

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The radial recursion between dimensions n and n+2

Statement

Let $c > 0$ , $n \geq 1$ and let $w$ be a $C^{3}$ function of $(r, t)$ on a domain with $r > 0$ . Write $L_{m} w : = w_{t t} - c^{2} (w_{r r} + \frac{m - 1}{r} w_{r})$ for the radial $m$ -dimensional wave operator and $Δ_{m} g : = g_{r r} + \frac{m - 1}{r} g_{r}$ for the radial Laplacian, in the sign convention of the wave operator of Wave equation, Cauchy data and wave speed. Then $L_{n + 2} [\frac{1}{r} \partial_{r} w] = \frac{1}{r} \partial_{r} [L_{n} w] (r > 0) .$ Consequently $w \mapsto r^{- 1} \partial_{r} w$ maps radial classical solutions of the $m$ -dimensional homogeneous wave equation to radial classical solutions of the $(m + 2)$ -dimensional one; at $m = 1$ this is the correspondence between one-dimensional waves and radial three-dimensional waves. It is the mechanism by which the kernels in dimension $n + 2$ are radial derivatives of the kernels in dimension $n$ .

Facts & Assumptions

Given: a speed $c > 0$ , an integer $n \geq 1$ , and a $C^{3}$ function $w$ of $(r, t)$ on a domain with $r > 0$ .

[F1]

Sums, products, quotients of differentiable functions are differentiable with the usual product, quotient rules (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 $g (c) \neq 0$ ).

[F2]

Proof

1.1F1algebra

Put $h : = r^{- 1} w_{r}$ . Differentiating the product and quotient, $h_{r} = r^{- 1} w_{r r} - r^{- 2} w_{r}$ and $h_{r r} = r^{- 1} w_{r r r} - 2 r^{- 2} w_{r r} + 2 r^{- 3} w_{r}$ , so the radial $(n + 2)$ -dimensional Laplacian of $h$ is $Δ_{n + 2} h = h_{r r} + \frac{n + 1}{r} h_{r} = r^{- 1} w_{r r r} + \frac{n - 1}{r^{2}} w_{r r} - \frac{n - 1}{r^{3}} w_{r}$ .

1.2F1algebra

On the other hand $Δ_{n} w = w_{r r} + \frac{n - 1}{r} w_{r}$ , whence $\partial_{r} (Δ_{n} w) = w_{r r r} + \frac{n - 1}{r} w_{r r} - \frac{n - 1}{r^{2}} w_{r}$ and $r^{- 1} \partial_{r} (Δ_{n} w) = r^{- 1} w_{r r r} + \frac{n - 1}{r^{2}} w_{r r} - \frac{n - 1}{r^{3}} w_{r}$ , the same expression as the radial Laplacian of the previous step.

2.1F2algebra∎

Since $w$ is $C^{3}$ , [F2] gives $w_{r t t} = w_{t t r}$ , so time differentiation commutes with $r^{- 1} \partial_{r}$ : $h_{t t} = (r^{- 1} w_{r})_{t t} = r^{- 1} \partial_{r} (w_{t t})$ . Subtracting $c^{2}$ times the identity $Δ_{n + 2} h = r^{- 1} \partial_{r} (Δ_{n} w)$ from this equality gives $L_{n + 2} h = h_{t t} - c^{2} Δ_{n + 2} h = r^{- 1} \partial_{r} (w_{t t}) - c^{2} r^{- 1} \partial_{r} (Δ_{n} w) = r^{- 1} \partial_{r} (w_{t t} - c^{2} Δ_{n} w) = r^{- 1} \partial_{r} (L_{n} w)$ ; if in particular $L_{n} w = 0$ then $L_{n + 2} (r^{- 1} w_{r}) = 0$ , which is the stated mapping of radial solutions.

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Ball means and sphere means are related by a radial derivative

Statement

Assume the Axiom of Countable Choice, let $n \geq 1$ , let $g \in C^{0} (R^{n})$ , and for $x \in R^{n}$ , $r > 0$ let $A_{g} (x, r) : = ∣ B_{r} (x) ∣^{- 1} \int_{B_{r} (x)} g$ be the ball average of The average of a locally integrable function over a Euclidean ball and $M_{g}$ the spherical mean of Spherical means and the weighted ball integral of space-dependent data. Then for every $r > 0$ $A_{g} (x, r) = \frac{n}{r^{n}} \int_{0}^{r} s^{n - 1} M_{g} (x, s) d s, M_{g} (x, r) = A_{g} (x, r) + \frac{r}{n} \partial_{r} A_{g} (x, r) .$

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , $g \in C^{0} (R^{n})$ , and the means $A_{g}$ and $M_{g}$ of the statement.

[F1]

Under Countable Choice, $\int_{R^{n}} f d λ_{n} = \int_{0}^{\infty} \int_{S^{n - 1}} f (s θ) s^{n - 1} d σ (θ) d s$ for every Borel measurable $f : R^{n} \to [0, \infty]$ (Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma).

[F2]

$M_{g} (x, s) = ω_{n - 1}^{- 1} \int_{S^{n - 1}} g (x + s ω) d σ (ω)$ and $ω_{n - 1} = σ (S^{n - 1})$ for $x \in R^{n}$ , $s > 0$ (Spherical means and the weighted ball integral of space-dependent data).

[F3]

$∣ B_{r} (x) ∣ = ω_{n - 1} r^{n} / n$ for $r > 0$ (Sphere and ball measures scale in Rn).

[F4]

A continuous function on a compact metric space is uniformly continuous (Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous); Euclidean closed balls are compact (For $n \geq 1$ , every Euclidean closed ball and every Euclidean sphere of positive radius is compact, in the sense of Open cover, subcover, compact metric space, and compact subset of a metric space).

[F5]

If $I \subseteq R$ is order-convex with at least two elements and $f : I \to R$ is continuous on $I$ , then for $r_{0} \in I$ , $r \mapsto \int_{r_{0}}^{r} f$ is a primitive of $f$ on $I$ (Every continuous function on an interval has a primitive; two primitives differ by a constant; and $\int_{a}^{b} f = G (b) - G (a)$ for any primitive $G$ ).

Proof

1.1F1F2F3algebra

Polar coordinates applied to the positive and negative parts of the function $u \mapsto g (x + u) 1_{B_{r} (0)} (u)$ give, using translation invariance (Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation) to set $y = x + u$ , $∣ B_{r} (x) ∣ A_{g} (x, r) = \int_{B_{r} (x)} g = \int_{0}^{r} \int_{S^{n - 1}} g (x + s ω) s^{n - 1} d σ (ω) d s = ω_{n - 1} \int_{0}^{r} s^{n - 1} M_{g} (x, s) d s$ , where the inner integral is $ω_{n - 1} M_{g} (x, s)$ by the definition of the spherical mean; dividing by $∣ B_{r} (x) ∣ = ω_{n - 1} r^{n} / n$ gives the first identity.

1.2F4algebra

The function $s \mapsto M_{g} (x, s)$ is continuous on $(0, \infty)$ : for $s, s_{0} > 0$ with $∣ s - s_{0} ∣ < 1$ one has $∣ M_{g} (x, s) - M_{g} (x, s_{0}) ∣ \leq \sup_{ω \in S^{n - 1}} ∣ g (x + s ω) - g (x + s_{0} ω) ∣$ , and the two points $x + s ω$ and $x + s_{0} ω$ lie in the compact ball $\overline{B_{s_{0} + 1} (x)}$ at distance $∣ s - s_{0} ∣$ ; since $g$ is uniformly continuous on that ball by [F4], the supremum tends to $0$ as $s \to s_{0}$ .

1.3F5algebra

Hence $s \mapsto s^{n - 1} M_{g} (x, s)$ is continuous on $(0, \infty)$ , so locally at any $r > 0$ split the integral at a fixed $r_{0} \in (0, r)$ ; the first part is constant and [F5] differentiates the second part. Thus $r \mapsto \int_{0}^{r} s^{n - 1} M_{g} (x, s) d s$ is a primitive of $s^{n - 1} M_{g} (x, s)$ , and the first identity gives $\partial_{r} (r^{n} A_{g} (x, r)) = n r^{n - 1} M_{g} (x, r)$ for $r > 0$ .

2.1algebra∎

Differentiating the product gives $\partial_{r} (r^{n} A_{g} (x, r)) = n r^{n - 1} A_{g} (x, r) + r^{n} \partial_{r} A_{g} (x, r)$ , so equating with the previous display and dividing by $n r^{n - 1} > 0$ yields $M_{g} (x, r) = A_{g} (x, r) + \frac{r}{n} \partial_{r} A_{g} (x, r)$ , the second identity.

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Sphere integrals of a cylindrical function project to weighted ball integrals

Statement

Assume the Axiom of Countable Choice. Let $n \geq 1$ and $g \in C^{0} (R^{n})$ , and let $G (ξ, z) : = g (ξ)$ be the extension of $g$ to $R^{n + 1}$ independent of the last coordinate. For $r > 0$ and $x \in R^{n}$ , with $S_{r}^{n} (x) = {(ξ, z) \in R^{n + 1} : ∣ ξ - x ∣^{2} + z^{2} = r^{2}}$ the sphere of radius $r$ in $R^{n + 1}$ and $ω_{n} = ∣ S^{n} ∣$ its total polar measure (The polar surface set function on the unit sphere), $\int_{S_{r}^{n} (x)} G d S = 2 r \int_{B_{r}^{n} (x)} \frac{g (y)}{\sqrt{r^{2} - ∣ y - x ∣^{2}}} d y,$ and the spherical mean of $G$ over $S_{r}^{n} (x)$ equals $\frac{2}{ω_{n} r^{n - 1}} \int_{B_{r}^{n} (x)} \frac{g (y)}{\sqrt{r^{2} - ∣ y - x ∣^{2}}} d y$ . In particular, for even $n = 2 k$ , every $c > 0$ and $t > 0$ , with $W_{g}$ the weighted ball integral of Spherical means and the weighted ball integral of space-dependent data, $t^{n - 1} M_{G} ((x, 0), c t) = \frac{(n - 1)!!}{c^{n - 1}} W_{g} (x, c t),$ where $M_{G}$ is the $(n + 1)$ -dimensional spherical mean of $G$ with centre $(x, 0)$ . The connecting constant identity is $2 n!! V_{n} / ω_{n} = (n - 1)!!$ .

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , $g \in C^{0} (R^{n})$ , the cylindrical extension $G (ξ, z) = g (ξ)$ , and $r > 0$ , $x \in R^{n}$ .

[F1]

Surface measure is chart-independent; in graph coordinates $(y, h (y))$ its density is $\sqrt{1 + ∣ D h (y) ∣^{2}}$ (Surface integration on compact C1 hypersurfaces, Chart and partition independence of surface measure). On spheres it agrees with polar measure and scales by the appropriate radius power (Agreement with the existing polar sphere measure).

[F2]

Under Countable Choice, $\int_{R^{n}} f d λ_{n} = \int_{0}^{\infty} \int_{S^{n - 1}} f (s θ) s^{n - 1} d σ (θ) d s$ for every Borel $f : R^{n} \to [0, \infty]$ (Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma).

[F3]

$∣ \partial B_{r}^{m} ∣ = ω_{m - 1} r^{m - 1}$ and $∣ B_{r}^{m} ∣ = ω_{m - 1} r^{m} / m$ , so in particular $∣ S_{r}^{n} (x) ∣ = ω_{n} r^{n}$ for the sphere in $R^{n + 1}$ and $ω_{m} = (m + 1) V_{m + 1}$ for the unit sphere $S^{m}$ (Sphere and ball measures scale in Rn).

[F4]

$V_{m} = π^{m / 2} / Γ (m / 2 + 1)$ for every $m \geq 1$ (The closed form for the volume of the unit $n$ -ball).

[F5]

$Γ (s + 1) = s Γ (s)$ for $s > 0$ with $Γ (1) = 1$ (The real Gamma functional equation $Γ (s + 1) = s Γ (s)$ ); $Γ (1 / 2) = \sqrt{π}$ ( $Γ (1 / 2) = \sqrt{π}$ from the Gaussian integral); $Γ (m + 1) = m!$ for every integer $m \geq 0$ (Gamma at the positive integers).

Proof

1.1F1algebra

Graph sheets. The upper and lower open hemispheres of $S_{r}^{n} (x)$ are the graphs $y \mapsto (y, \pm \sqrt{r^{2} - ∣ y - x ∣^{2}})$ over $B_{r}^{n} (x)$ . Their graph density is $\sqrt{1 + ∣ y - x ∣^{2} / (r^{2} - ∣ y - x ∣^{2})} = r / \sqrt{r^{2} - ∣ y - x ∣^{2}}$ by [F1]. The equator has zero surface measure: near each of its points choose a sphere graph omitting a nonzero one of the first $n$ coordinates. Its parameter set for the equator lies in the coordinate hyperplane $z = 0$ , which has Lebesgue measure zero (for $n = 1$ , it is a singleton, null because it lies in intervals of arbitrarily small length); Fubini's theorem for L^1 functions on a sigma-finite product applied to its indicator proves nullity, and the continuous graph density preserves it. A finite chart cover suffices by compactness of the sphere (For $n \geq 1$ , every Euclidean closed ball and every Euclidean sphere of positive radius is compact).

1.2F1F2algebra

Integrating the two sheets gives $\int_{S_{r}^{n} (x)} G d S = 2 r \int_{B_{r}^{n} (x)} g (y) (r^{2} - ∣ y - x ∣^{2})^{- 1 / 2} d y$ . These integrals are absolutely convergent: $g$ is bounded on the closed ball by A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value, and [F2] reduces the weight integral to $ω_{n - 1} \int_{0}^{r} s^{n - 1} (r^{2} - s^{2})^{- 1 / 2} d s$ , bounded by $ω_{n - 1} r^{n - 1} \int_{0}^{r} (r^{2} - s^{2})^{- 1 / 2} d s = ω_{n - 1} r^{n - 1} π / 2$ . Thus the graph computation applies separately to positive and negative parts.

1.3F3algebra

The mean. By [F3], $∣ S_{r}^{n} (x) ∣ = ω_{n} r^{n}$ , so the spherical mean of $G$ over $S_{r}^{n} (x)$ is $ω_{n}^{- 1} r^{- n} \int_{S_{r}^{n} (x)} G d S = \frac{2}{ω_{n} r^{n - 1}} \int_{B_{r} (x)} \frac{g (y)}{\sqrt{r^{2} - ∣ y - x ∣^{2}}} d y$ .

1.4F3F4F5algebra

The even-dimensional form. Let $n = 2 k$ be even, $c > 0$ , $t > 0$ and $r = c t$ . Substituting $r = c t$ in the mean identity, $t^{n - 1} M_{G} ((x, 0), c t) = \frac{2}{ω_{n} c^{n - 1}} \int_{B_{c t} (x)} \frac{g (y)}{\sqrt{c^{2} t^{2} - ∣ y - x ∣^{2}}} d y = \frac{2 n!! V_{n}}{ω_{n} c^{n - 1}} W_{g} (x, c t)$ by the definition of $W_{g}$ . For the constant: $V_{n} = π^{n / 2} / Γ (k + 1) = π^{k} / k!$ by [F4]; the functional equation and $Γ (1 / 2) = \sqrt{π}$ give by induction $Γ (k + 1 / 2) = \frac{(2 k - 1) (2 k - 3) \dots 1}{2^{k}} \sqrt{π} = \frac{(2 k)! \sqrt{π}}{4^{k} k!}$ , so with $ω_{n} = ∣ S^{n} ∣ = (n + 1) V_{n + 1} = \frac{2 π^{k + 1 / 2}}{Γ (k + 1 / 2)} = \frac{2^{2 k + 1} π^{k} k!}{(2 k)!}$ one gets $2 n!! V_{n} = 2 \cdot 2^{k} k! \cdot π^{k} / k! = 2^{k + 1} π^{k}$ and $\frac{2 n!! V_{n}}{ω_{n}} = \frac{2^{k + 1} π^{k} (2 k)!}{2^{2 k + 1} π^{k} k!} = \frac{(2 k)!}{2^{k} k!} = (2 k - 1)!! = (n - 1)!!$ .

2.1algebra∎

Substituting the constant gives $t^{n - 1} M_{G} ((x, 0), c t) = \frac{(n - 1)!!}{c^{n - 1}} W_{g} (x, c t)$ , which together with the two integral identities proves all assertions.

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General solution of the one-dimensional wave equation

Statement

Let $c > 0$ and let $R \subseteq R^{2}$ be a nonempty open rectangle. If $u \in C^{2} (R)$ satisfies $u_{t t} = c^{2} u_{x x}$ on $R$ , then there are intervals $I, J \subseteq R$ and functions $F \in C^{2} (I)$ , $G \in C^{2} (J)$ with $u (x, t) = F (x - c t) + G (x + c t) ((x, t) \in R),$ where $I$ and $J$ are the projections of $R$ onto the $ξ$ - and the $η$ -axis under $ξ = x - c t$ , $η = x + c t$ . Conversely, every such sum is a $C^{2}$ solution of $u_{t t} = c^{2} u_{x x}$ on $R$ . The pair $(F, G)$ is unique up to the replacement $F \mapsto F + k$ , $G \mapsto G - k$ with $k \in R$ , and no other freedom remains.

Facts & Assumptions

Given: a speed $c > 0$ , a nonempty open rectangle $R \subseteq R^{2}$ , and a $C^{2}$ function $u$ on $R$ .

[F1]

On the domain of $u$ , $\partial_{t}^{2} u - c^{2} \partial_{x}^{2} u = - 4 c^{2} \partial_{ξ} \partial_{η} U$ where $U (ξ, η) : = u (x, t)$ and $ξ = x - c t$ , $η = x + c t$ (Factorisation of the one-dimensional wave operator).

[F2]

If $f : U \to V \subseteq R^{n}$ is totally differentiable at $a$ and $g : V \to R^{p}$ at $f (a)$ , then $g \circ f$ is totally differentiable at $a$ with $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ ). The required total differentiability follows from continuous coordinate partial derivatives (If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative).

[F3]

Let $I \subseteq R$ be order-convex and $f : I \to R$ continuous and differentiable at every interior point with $f^{'} = 0$ there. Then $f$ is constant; if moreover $f, g$ are continuous with $f^{'} = g^{'}$ at every interior point then $f - g$ 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).

[F4]

Let $I$ be order-convex with at least two elements and $f : I \to R$ continuous. Fix $c_{0} \in I$ ; then $F (x) = \int_{c_{0}}^{x} f$ is a primitive of $f$ on $I$ , and primitives differ by constants (Every continuous function on an interval has a primitive; two primitives differ by a constant; and $\int_{a}^{b} f = G (b) - G (a)$ for any primitive $G$ ).

[F5]

Proof

1.1F1F2algebra

The affine change of variables $(x, t) \mapsto (ξ, η) = (x - c t, x + c t)$ is a bijection of $R^{2}$ with inverse $x = (ξ + η) / 2$ , $t = (η - ξ) / (2 c)$ , and the image $R^{'}$ of the rectangle $R$ is a nonempty open convex affine image of $R$ (a parallelogram when $R$ is bounded); the projections $I$ of $R^{'}$ onto the $ξ$ -axis and $J$ onto the $η$ -axis are nonempty open intervals, and every section ${η : (ξ, η) \in R^{'}}$ and ${ξ : (ξ, η) \in R^{'}}$ is a nonempty open interval. By [F2] the function $U (ξ, η) : = u ((ξ + η) / 2, (η - ξ) / (2 c))$ is $C^{2}$ on $R^{'}$ , and [F1] gives $u_{t t} - c^{2} u_{x x} = - 4 c^{2} U_{ξ η}$ on $R^{'}$ .

1.2F3F4algebra

Suppose $u$ solves $u_{t t} = c^{2} u_{x x}$ on $R$ ; since $c > 0$ , $U_{ξ η} = 0$ on $R^{'}$ . For fixed $ξ \in I$ the section $J_{ξ} : = {η : (ξ, η) \in R^{'}}$ is a nonempty open interval and $\partial_{η} (U_{ξ} (ξ, \cdot)) = U_{ξ η} (ξ, \cdot) = 0$ there, so by [F3] the value $U_{ξ} (ξ, η)$ is independent of $η$ in that section; call it $p (ξ)$ . To see $p$ is $C^{1}$ , fix $ξ_{0} \in I$ and choose $η_{0}$ with $(ξ_{0}, η_{0}) \in R^{'}$ ; openness gives an interval $I_{0}$ about $ξ_{0}$ on which $(ξ, η_{0}) \in R^{'}$ , so $p (ξ) = U_{ξ} (ξ, η_{0})$ on $I_{0}$ . Since $U$ is $C^{2}$ , this local representative is $C^{1}$ , and hence $p \in C^{1} (I)$ . Choose a primitive $P$ of $p$ on $I$ by [F4]; since $p \in C^{1}$ , $P \in C^{2} (I)$ . Then $\partial_{ξ} (U - P) = 0$ on $R^{'}$ , and the same section argument in the $ξ$ -direction gives that $U - P$ is independent of $ξ$ : there is $G : J \to R$ with $U (ξ, η) - P (ξ) = G (η)$ for all $(ξ, η) \in R^{'}$ . To see $G$ is $C^{2}$ , fix $η_{0} \in J$ and choose $ξ_{0}$ with $(ξ_{0}, η_{0}) \in R^{'}$ ; openness gives a neighbourhood $J_{0}$ on which $G (η) = U (ξ_{0}, η) - P (ξ_{0})$ , a $C^{2}$ function. Thus $G \in C^{2} (J)$ and $u (x, t) = P (x - c t) + G (x + c t)$ on $R$ .

1.3F2F5algebra

Conversely, if $F \in C^{2} (I)$ and $G \in C^{2} (J)$ , then $(x, t) \mapsto F (x - c t) + G (x + c t)$ is $C^{2}$ on $R$ by [F2] and [F5], and two applications of the chain rule give $\partial_{t}^{2} [F (x - c t) + G (x + c t)] = c^{2} F^{''} (x - c t) + c^{2} G^{''} (x + c t) = c^{2} \partial_{x}^{2} [F (x - c t) + G (x + c t)]$ ; hence every such sum solves the wave equation.

1.4F3F5algebra

Uniqueness of the pair. If $F_{1} (ξ) + G_{1} (η) = F_{2} (ξ) + G_{2} (η)$ for all $(ξ, η) \in R^{'}$ , then $Φ (ξ, η) : = F_{1} (ξ) - F_{2} (ξ) - G_{2} (η) + G_{1} (η)$ vanishes on $R^{'}$ . For fixed $ξ \in I$ the section in $η$ is a nonempty interval, so $\partial_{η} Φ = G_{1}^{'} - G_{2}^{'} = 0$ on $J$ , and for fixed $η \in J$ similarly $\partial_{ξ} Φ = F_{1}^{'} - F_{2}^{'} = 0$ on $I$ ; hence $F_{1} - F_{2}$ and $G_{2} - G_{1}$ are the same constant $k$ by [F3], that is $F_{1} = F_{2} + k$ and $G_{1} = G_{2} - k$ .

2.1given∎

Therefore every $C^{2}$ solution of the one-dimensional homogeneous wave equation on a nonempty open rectangle has the form $F (x - c t) + G (x + c t)$ with $F, G$ of class $C^{2}$ on the projections, every such sum is a solution, and the decomposition is unique up to the additive shift $F \mapsto F + k$ , $G \mapsto G - k$ .

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Smoothness, parity and zero-radius limits of spherical means

Statement

Assume the Axiom of Countable Choice of Spherical means and the weighted ball integral of space-dependent data. Let $n \geq 1$ , $k \geq 1$ and $f \in C^{k} (R^{n})$ , and let $M_{f}$ be the spherical mean with the convention $M_{f} (x, 0) : = f (x)$ . Then: (i) $(x, r) \mapsto M_{f} (x, r)$ is $C^{k}$ on $R^{n} \times (0, \infty)$ , and every derivative is obtained by differentiating $f$ under the sphere integral: for every multi-index $α$ and every $m \geq 0$ with $∣ α ∣ + m \leq k$ , $D_{x}^{α} \partial_{r}^{m} M_{f} (x, r) = \frac{1}{ω_{n - 1}} \int_{S^{n - 1}} D_{x}^{α} \partial_{r}^{m} [f (x + r ω)] d σ (ω) (x \in R^{n}, r > 0),$ where $\partial_{r}^{m} [f (x + r ω)]$ is the $m$ -th $r$ -derivative of the composed function $r \mapsto f (x + r ω)$ . (ii) $M_{f}$ extends to a continuous function on $R^{n} \times [0, \infty)$ with $M_{f} (x, 0) = f (x)$ , and $\partial_{r} M_{f} (x, r) \to 0$ as $r ↓ 0$ , uniformly for $x$ in compact subsets of $R^{n}$ . (iii) The signed-radius integral $ω_{n - 1}^{- 1} \int_{S^{n - 1}} f (x + r ω) d σ (ω)$ for $r \in R$ is an even $C^{k}$ extension of $M_{f}$ to $R^{n} \times R$ . Its differentiated-integral formula holds also at $r = 0$ ; in particular every available odd-order radial derivative vanishes there. (iv) $∣ \partial_{r} M_{f} (x, r) ∣ \leq \sup_{B_{r} (x)} ∣ D f ∣$ for every $x \in R^{n}$ and $r > 0$ .

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , $k \geq 1$ , $f \in C^{k} (R^{n})$ , and the spherical mean $M_{f}$ with $M_{f} (x, 0) : = f (x)$ .

[F1]

The spherical mean is integration against the finite measure $σ / ω_{n - 1}$ of total mass one (Spherical means and the weighted ball integral of space-dependent data).

[F2]

The mean value theorem bounds a difference quotient by the corresponding derivative on its segment (The mean value theorem, as the case $g (x) = x$ of Cauchy's: for $f$ continuous on $[a, b]$ with $a < b$ and differentiable on $(a, b)$ there is $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$ ). Continuous partial derivatives imply total differentiability (If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative), so the chain rule applies to $C^{1}$ data composed with affine maps (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ ).

[F3]

A continuous map from a compact metric space to a metric space is uniformly continuous (Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous); Euclidean closed balls are compact (For $n \geq 1$ , every Euclidean closed ball and every Euclidean sphere of positive radius is compact, in the sense of Open cover, subcover, compact metric space, and compact subset of a metric space); a continuous real function on a nonempty compact metric space is bounded and attains its bounds (A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value).

[F4]

Under Countable Choice reflection $ω \mapsto - ω$ preserves the polar measure, and its first moment vanishes: $\int_{S^{n - 1}} ω d σ (ω) = 0$ (Reflection invariance and vanishing first moment of the sphere measure).

Proof

1.1F1F2F3algebra

Differentiation, including signed radii. Put $H (x, r) : = ω_{n - 1}^{- 1} \int_{S^{n - 1}} f (x + r ω) d σ (ω)$ for all real $r$ . On a bounded parameter neighbourhood, all points $x + r ω$ lie in a fixed compact ball. For a coordinate parameter $p$ and a continuous derivative integrand $q (p, ω)$ with continuous $\partial_{p} q$ , [F2] gives $∣ (q (p + h, ω) - q (p, ω)) / h - \partial_{p} q (p, ω) ∣ \leq \sup_{∣ s - p ∣ \leq ∣ h ∣, ω} ∣ \partial_{p} q (s, ω) - \partial_{p} q (p, ω) ∣$ . Uniform continuity on the compact ball makes this bound tend to zero uniformly in $ω$ ; integration against the probability measure [F1] preserves the bound. Iterating through total order $k$ therefore gives every stated derivative under the integral, with continuity again following from the same uniform estimate. This works for $n = 1$ as well, since it requires only a finite measure, not positive-dimensional surface charts.

1.2F3algebra

Part (ii), continuity at $r = 0$ . Let $K \subseteq R^{n}$ be compact and choose $R > 0$ with $K \subseteq B_{R} (0)$ ; the ball $\overline{B_{R + 1} (0)}$ is compact, so by [F3] $f$ is bounded there and uniformly continuous on it. Given $ε > 0$ choose $δ \in (0, 1)$ such that $∣ f (u) - f (v) ∣ < ε$ whenever $u, v \in \overline{B_{R + 1} (0)}$ and $∣ u - v ∣ < δ$ . For $x \in K$ , $0 \leq r < δ$ and $ω \in S^{n - 1}$ one has $x + r ω \in \overline{B_{R + 1} (0)}$ and $∣ x + r ω - x ∣ = r < δ$ , so $∣ M_{f} (x, r) - f (x) ∣ \leq \sup_{ω} ∣ f (x + r ω) - f (x) ∣ < ε$ . Hence $M_{f} (x, r) \to f (x)$ as $r ↓ 0$ uniformly on compact subsets, and the extension by $M_{f} (\cdot, 0) = f$ is continuous because $f$ is.

1.3F3F4algebra

Part (ii), the derivative limit. By part (i) with $∣ α ∣ = 0$ , $m = 1$ , $\partial_{r} M_{f} (x, r) = ω_{n - 1}^{- 1} \int_{S^{n - 1}} D f (x + r ω) \cdot ω d σ (ω)$ for $r > 0$ , so adding and subtracting $D f (x)$ inside the integral gives $\partial_{r} M_{f} (x, r) = ω_{n - 1}^{- 1} \int_{S^{n - 1}} [D f (x + r ω) - D f (x)] \cdot ω d σ (ω) + ω_{n - 1}^{- 1} D f (x) \cdot \int_{S^{n - 1}} ω d σ (ω)$ , where the second term vanishes by [F4]; the first is bounded by $\sup_{ω \in S^{n - 1}} ∣ D f (x + r ω) - D f (x) ∣$ , which tends to $0$ as $r ↓ 0$ uniformly for $x$ in a fixed compact set by uniform continuity of the continuous function $D f$ on a large compact ball, again by [F3]. Hence $\partial_{r} M_{f} (x, r) \to 0$ uniformly on compact subsets of $R^{n}$ .

2.1F1F3F4step 1.1algebra

Parity and the bound. Reflection preserves $σ$ by [F4], hence the substitution $ω \mapsto - ω$ gives $H (x, - r) = H (x, r)$ . Since $H$ is $C^{k}$ by step 1.1, its odd-order radial derivatives at zero vanish whenever their orders are at most $k$ . For (iv), $∣ \partial_{r} M_{f} (x, r) ∣ \leq ω_{n - 1}^{- 1} \int ∣ D f (x + r ω) ∣ d σ (ω)$ . By continuity, each boundary value of $∣ D f ∣$ is at most $\sup_{B_{r} (x)} ∣ D f ∣$ , and integration gives the stated bound.

3.1step 1.1step 1.2step 1.3step 2.1∎

Thus $M_{f}$ has the differentiated-integral formula, the stated uniform zero-radius limits and gradient bound, and an even $C^{k}$ signed-radius extension.

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The Euler–Poisson–Darboux equation for spherical means

Statement

Assume the Axiom of Countable Choice, let $n \geq 1$ , let $f \in C^{2} (R^{n})$ and let $M = M_{f}$ be the spherical mean of Spherical means and the weighted ball integral of space-dependent data. Then for every $x \in R^{n}$ and $r > 0$ $Δ_{x} M (x, r) = M_{r r} (x, r) + \frac{n - 1}{r} M_{r} (x, r) .$ The right-hand side extends continuously to $r = 0$ with value $Δ f (x)$ : with $M (x, 0) : = f (x)$ one has $M_{r} (x, r) \to 0$ and $(M_{r r} + \frac{n - 1}{r} M_{r}) (x, r) \to Δ f (x)$ as $r ↓ 0$ . If $u \in C^{2} (R^{n} \times I)$ is a classical solution of $□_{c} u = 0$ on an interval $I$ , then its space-time mean $(x, r, t) \mapsto M_{u} (x, r, t) : = ω_{n - 1}^{- 1} \int_{S^{n - 1}} u (x + r ω, t) d σ (ω)$ satisfies $\partial_{t}^{2} M_{u} = c^{2} (M_{r r} + \frac{n - 1}{r} M_{r})$ . No equation for $f$ is needed for the identity itself.

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , $f \in C^{2} (R^{n})$ , the spherical mean $M = M_{f}$ with $M (x, 0) = f (x)$ , and, when stated, a $C^{2}$ solution $u$ of $□_{c} u = 0$ on $R^{n} \times I$ .

[F1]

For $k \geq 1$ and $h \in C^{k} (R^{n})$ , $(x, r) \mapsto M_{h} (x, r)$ is $C^{k}$ on $R^{n} \times (0, \infty)$ with all derivatives obtained by differentiating $h$ under the sphere integral; it extends continuously to $r = 0$ with $M_{h} (x, 0) = h (x)$ , and its radial derivative tends to $0$ uniformly on compacta (Smoothness, parity and zero-radius limits of spherical means).

[F2]

If $g \in C^{2} (R^{n})$ and $r > 0$ , then $\partial_{r} M_{g} (x, r) = \frac{r}{n} A_{Δ g} (x, r)$ , where $A_{Δ g} (x, r) = ∣ B_{r} (x) ∣^{- 1} \int_{B_{r} (x)} Δ g$ is the ball average (Radial derivative of a spherical average, The average of a locally integrable function over a Euclidean ball).

[F3]

For $h \in C^{0} (R^{n})$ one has $M_{h} (x, r) = A_{h} (x, r) + \frac{r}{n} \partial_{r} A_{h} (x, r)$ for all $r > 0$ (Ball means and sphere means are related by a radial derivative).

[F4]

A continuous real function on a nonempty compact metric space is bounded; Euclidean closed balls are compact (A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value, For $n \geq 1$ , every Euclidean closed ball and every Euclidean sphere of positive radius is compact).

Proof

1.1F1algebra

The spatial Laplacian passes under the sphere integral: by [F1] the map $(x, r) \mapsto M (x, r)$ is $C^{2}$ , and differentiating the defining integral twice in $x$ , $Δ_{x} M (x, r) = ω_{n - 1}^{- 1} \int_{S^{n - 1}} Δ f (x + r ω) d σ (ω) = M_{Δ f} (x, r)$ for every $r > 0$ .

1.2F2F3algebra

Radial identities. Put $A (x, r) : = A_{Δ f} (x, r)$ . Applying [F2] only to $f$ gives $M_{r} (x, r) = \frac{r}{n} A (x, r)$ . Since $Δ f$ is continuous, [F3] applied to $Δ f$ gives $M_{Δ f} (x, r) = A (x, r) + \frac{r}{n} \partial_{r} A (x, r)$ . Differentiating the first identity in $r$ gives $M_{r r} (x, r) = \frac{1}{n} A (x, r) + \frac{r}{n} \partial_{r} A (x, r)$ ; therefore $M_{r r} + \frac{n - 1}{r} M_{r} = \frac{1}{n} A + \frac{r}{n} \partial_{r} A + \frac{n - 1}{n} A = A + \frac{r}{n} \partial_{r} A = M_{Δ f} (x, r) = Δ_{x} M (x, r)$ . The radial-derivative formula is used only with the $C^{2}$ function $f$ ; no differentiability of $Δ f$ is assumed.

1.3F4algebra

The limits at $r ↓ 0$ . For $0 < r \leq 1$ the bound $∣ A_{Δ f} (x, r) ∣ \leq \sup_{B_{1} (x)} ∣ Δ f ∣ < \infty$ holds, the supremum being finite by [F4]; hence $M_{r} (x, r) = \frac{r}{n} A_{Δ f} (x, r) \to 0$ . Moreover the identity just proved gives $M_{r r} (x, r) + \frac{n - 1}{r} M_{r} (x, r) = M_{Δ f} (x, r) = Δ_{x} M (x, r)$ . Since $Δ f$ is continuous at $x$ , $∣ M_{Δ f} (x, r) - Δ f (x) ∣ \leq \sup_{ω \in S^{n - 1}} ∣ Δ f (x + r ω) - Δ f (x) ∣ \to 0$ as $r ↓ 0$ .

1.4F1algebra

The space-time version. If $u \in C^{2}$ solves $□_{c} u = 0$ , applying the differentiation-under-the-sphere-integral computation of [F1] to the $C^{2}$ function $(x, t) \mapsto u (x, t)$ yields $Δ_{x} M_{u} (x, r, t) = M_{Δ_{x} u} (x, r, t)$ and $\partial_{t}^{2} M_{u} (x, r, t) = M_{u_{t t}} (x, r, t)$ for all $r > 0$ and $t \in I$ . Since $u_{t t} = c^{2} Δ u$ , this gives $\partial_{t}^{2} M_{u} = c^{2} M_{Δ u} = c^{2} Δ_{x} M_{u}$ ; and the identity of the first two steps applied to the $C^{2}$ spatial function $u (\cdot, t)$ gives $Δ_{x} M_{u} (x, r, t) = M_{r r} (x, r, t) + \frac{n - 1}{r} M_{r} (x, r, t)$ . Substituting, $\partial_{t}^{2} M_{u} = c^{2} (M_{r r} + \frac{n - 1}{r} M_{r})$ .

2.1given∎

Collecting: the Euler–Poisson–Darboux identity holds for every $C^{2}$ datum, its right-hand side has the stated continuous extension at $r = 0$ with value $Δ f (x)$ , and the space-time means of solutions satisfy the same radial equation with two time derivatives on the left.

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d'Alembert's formula and uniqueness in one dimension

Statement

Let $c > 0$ , $u_{0} \in C^{2} (R)$ and $u_{1} \in C^{1} (R)$ . Then $u (x, t) : = \frac{1}{2} (u_{0} (x - c t) + u_{0} (x + c t)) + \frac{1}{2 c} \int_{x - c t}^{x + c t} u_{1} (y) d y$ is a $C^{2}$ function on $R \times [0, \infty)$ and is the unique classical solution of the homogeneous Cauchy problem $u_{t t} = c^{2} u_{x x}, u (\cdot, 0) = u_{0}, u_{t} (\cdot, 0) = u_{1} .$ Its value depends on $u_{0}$ only through the two endpoints $x \mp c t$ and on $u_{1}$ only through its integral over $[x - c t, x + c t]$ .

Facts & Assumptions

Given: a speed $c > 0$ , data $u_{0} \in C^{2} (R)$ , $u_{1} \in C^{1} (R)$ , and the displayed function $u$ .

[F1]

A continuous function on an interval has the primitive $P (z) = \int_{0}^{z} u_{1} (y) d y$ when the interval is $R$ ; since $u_{1} \in C^{1}$ , this primitive is $C^{2}$ , and $\int_{a}^{b} u_{1} = P (b) - P (a)$ (Every continuous function on an interval has a primitive; two primitives differ by a constant; and $\int_{a}^{b} f = G (b) - G (a)$ for any primitive $G$ ).

[F2]

If $f$ is totally differentiable at $a$ and $g$ at $f (a)$ , then $g \circ f$ is totally differentiable at $a$ with $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ ). The required total differentiability follows from continuous coordinate partial derivatives (If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative).

[F3]

Sums, constant multiples, products of differentiable functions are differentiable with the usual rules (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 $g (c) \neq 0$ ).

[F4]

Every $C^{2}$ solution of $u_{t t} = c^{2} u_{x x}$ on a nonempty open rectangle has the form $F (x - c t) + G (x + c t)$ with $F, G$ of class $C^{2}$ on the projections, and conversely; the pair is unique up to $F \mapsto F + k$ , $G \mapsto G - k$ (General solution of the one-dimensional wave equation).

Proof

1.1F1F2F3algebra

Regularity and the equation. The first summand is $C^{2}$ because $u_{0} \in C^{2}$ , and the integral term is $(P (x + c t) - P (x - c t)) / (2 c)$ with $P \in C^{2}$ by [F1], so it is $C^{2}$ even across $t = 0$ by [F2]: $\partial_{t} u (x, t) = \frac{1}{2} (- c u_{0}^{'} (x - c t) + c u_{0}^{'} (x + c t)) + \frac{1}{2} (u_{1} (x + c t) + u_{1} (x - c t))$ and $\partial_{x} u (x, t) = \frac{1}{2} (u_{0}^{'} (x - c t) + u_{0}^{'} (x + c t)) + \frac{1}{2 c} (u_{1} (x + c t) - u_{1} (x - c t))$ . Differentiating once more by [F2] and [F3], $\partial_{t}^{2} u = \frac{c^{2}}{2} (u_{0}^{''} (x - c t) + u_{0}^{''} (x + c t)) + \frac{c}{2} (u_{1}^{'} (x + c t) - u_{1}^{'} (x - c t))$ and $c^{2} \partial_{x}^{2} u = \frac{c^{2}}{2} (u_{0}^{''} (x - c t) + u_{0}^{''} (x + c t)) + \frac{c}{2} (u_{1}^{'} (x + c t) - u_{1}^{'} (x - c t))$ , so $u_{t t} = c^{2} u_{x x}$ on $R \times [0, \infty)$ .

1.2F1algebra

Both data are attained. At $t = 0$ the displacement terms give $u (x, 0) = \frac{1}{2} (u_{0} (x) + u_{0} (x)) + 0 = u_{0} (x)$ , and the velocity formula of the previous step gives $\partial_{t} u (x, 0) = \frac{1}{2} (- c u_{0}^{'} (x) + c u_{0}^{'} (x)) + \frac{1}{2} (u_{1} (x) + u_{1} (x)) = u_{1} (x)$ .

1.3F2F3F4algebra

Uniqueness. Let $v$ be any classical solution with the same pointwise displacement and velocity limits at zero and put $w = u - v$ . Apply [F4] to the open rectangle $R \times (0, T)$ , where $T > 0$ is arbitrary. Both characteristic projections are all of $R$ , so $w (x, t) = F (x - c t) + G (x + c t)$ with $F, G \in C^{2} (R)$ . At each fixed $x$ , letting $t ↓ 0$ gives $F (x) + G (x) = 0$ and $- c F^{'} (x) + c G^{'} (x) = 0$ . Differentiating the first identity and combining with the second yields $F^{'} = G^{'} = 0$ everywhere; their sum is zero, so $w = 0$ throughout this rectangle. As $T$ is arbitrary, uniqueness holds on the whole time slab.

2.1given∎

The display of $u$ therefore defines the unique classical solution, and its value at $(x, t)$ involves $u_{0}$ only through the endpoint values $u_{0} (x \mp c t)$ and $u_{1}$ only through the integral over $[x - c t, x + c t]$ .

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The d'Alembert expression attains both initial data

Statement

Let $c > 0$ , $u_{0} \in C^{2} (R)$ , $u_{1} \in C^{1} (R)$ and let $A$ be the d'Alembert expression of d'Alembert's formula and uniqueness in one dimension, $A (x, t) : = \frac{1}{2} (u_{0} (x - c t) + u_{0} (x + c t)) + \frac{1}{2 c} \int_{x - c t}^{x + c t} u_{1} (y) d y .$ Then $A \in C^{2} (R \times [0, \infty))$ , $A (x, 0) = u_{0} (x)$ for every $x$ , and $\partial_{t} A (x, t) = \frac{1}{2} (- c u_{0}^{'} (x - c t) + c u_{0}^{'} (x + c t)) + \frac{1}{2} (u_{1} (x + c t) + u_{1} (x - c t)),$ so $\partial_{t} A (x, 0) = u_{1} (x)$ . The orientation of the velocity integral is the $+$ sign in the second bracket: both ends of the characteristic base are traversed with speed $c$ , and the two endpoint contributions add.

Facts & Assumptions

Given: a speed $c > 0$ , data $u_{0} \in C^{2} (R)$ , $u_{1} \in C^{1} (R)$ , and the expression $A$ of the statement.

[F1]

Let $α, β \in C^{1} (I)$ with $α < β$ and let $F$ be continuous on $I \times J$ with continuous $\partial_{t} F$ , where $J$ contains the closure of the union of the intervals $[α (t), β (t)]$ . Then $G (t) = \int_{α (t)}^{β (t)} F (t, y) d y$ is $C^{1}$ with $G^{'} (t) = F (t, β (t)) β^{'} (t) - F (t, α (t)) α^{'} (t) + \int_{α (t)}^{β (t)} \partial_{t} F (t, y) d y$ (Differentiating an integral with moving endpoints).

[F2]

[F3]

If $f$ is totally differentiable at $a$ and $g$ at $f (a)$ , then $g \circ f$ is totally differentiable at $a$ with $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ ). The required total differentiability follows from continuous coordinate partial derivatives (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

1.1algebra

Displacement at $t = 0$ . At $t = 0$ the two displacement terms are both $u_{0} (x)$ and the integral has equal endpoints, so $A (x, 0) = \frac{1}{2} (u_{0} (x) + u_{0} (x)) + 0 = u_{0} (x)$ for every $x \in R$ .

1.2F1F2F3algebra

Velocity. For $t > 0$ , by [F1] applied to the integral term with $α (t) = x - c t$ , $β (t) = x + c t$ , $α^{'} (t) = - c$ , $β^{'} (t) = c$ and inner integrand $u_{1}$ , $\partial_{t} A (x, t) = \frac{1}{2} (- c u_{0}^{'} (x - c t) + c u_{0}^{'} (x + c t)) + \frac{1}{2 c} (c u_{1} (x + c t) + c u_{1} (x - c t)) = \frac{1}{2} (- c u_{0}^{'} (x - c t) + c u_{0}^{'} (x + c t)) + \frac{1}{2} (u_{1} (x + c t) + u_{1} (x - c t))$ ; the displacement term is differentiated by [F2] and [F3]. Continuity of the data and the $C^{2}$ regularity supplied by d'Alembert's formula and uniqueness in one dimension extend this derivative formula to $t = 0$ .

2.1F1algebra∎

Setting $t = 0$ in the velocity formula gives $\partial_{t} A (x, 0) = \frac{1}{2} (- c u_{0}^{'} (x) + c u_{0}^{'} (x)) + \frac{1}{2} (u_{1} (x) + u_{1} (x)) = u_{1} (x)$ ; the regularity $A \in C^{2} (R \times [0, \infty))$ is established in d'Alembert's formula and uniqueness in one dimension. Hence $A$ attains both initial data, and the two endpoint contributions of the velocity integral add with a $+$ sign as displayed.

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The one-dimensional value depends on the characteristic interval

Statement

Let $c > 0$ and let $u$ be the solution of d'Alembert's formula and uniqueness in one dimension for data $u_{0} \in C^{2} (R)$ , $u_{1} \in C^{1} (R)$ . For every $(x, t) \in R \times [0, \infty)$ , the value $u (x, t)$ is determined by the restrictions of $u_{0}$ and $u_{1}$ to the closed interval $[x - c t, x + c t]$ : if $({\tilde{u}}_{0}, {\tilde{u}}_{1})$ are admissible data agreeing with $(u_{0}, u_{1})$ there, then the corresponding solution satisfies $\tilde{u} (x, t) = u (x, t)$ . In particular, changing the data outside $[x - c t, x + c t]$ does not change the value at $(x, t)$ .

Facts & Assumptions

Given: a speed $c > 0$ , data $u_{0} \in C^{2} (R)$ , $u_{1} \in C^{1} (R)$ , a point $(x, t) \in R \times [0, \infty)$ , and admissible data $({\tilde{u}}_{0}, {\tilde{u}}_{1})$ with ${\tilde{u}}_{0} = u_{0}$ and ${\tilde{u}}_{1} = u_{1}$ on $[x - c t, x + c t]$ .

[F1]

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

[F2]

The d'Alembert expression attains both initial data and defines the unique classical solution of the corresponding Cauchy problem (The d'Alembert expression attains both initial data, d'Alembert's formula and uniqueness in one dimension).

Proof

1.1F1algebra

The formula of [F1] evaluates $u_{0}$ only at the two endpoints $x - c t$ and $x + c t$ of the interval and integrates $u_{1}$ only over that interval; hence replacing $(u_{0}, u_{1})$ by any admissible pair with the same restrictions to $[x - c t, x + c t]$ leaves the right-hand side unchanged, so the d'Alembert expression of the new data equals $u (x, t)$ at the given point.

2.1F2algebra∎

By [F2] both expressions are the solutions of their respective Cauchy problems, so the solution $\tilde{u}$ of the data $({\tilde{u}}_{0}, {\tilde{u}}_{1})$ satisfies $\tilde{u} (x, t) = u (x, t)$ ; in particular the value at $(x, t)$ is unchanged by altering the data off $[x - c t, x + c t]$ .

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The forced one-dimensional wave formula over the characteristic triangle

Statement

Let $c > 0$ , $u_{0} \in C^{2} (R)$ , $u_{1} \in C^{1} (R)$ and let $f$ be of class $C^{1}$ on $R \times [0, \infty)$ , so that $f$ , $\partial_{x} f$ and $\partial_{t} f$ are continuous. Then $u (x, t) = \frac{1}{2} (u_{0} (x - c t) + u_{0} (x + c t)) + \frac{1}{2 c} \int_{x - c t}^{x + c t} u_{1} (y) d y + \frac{1}{2 c} \int_{0}^{t} \int_{x - c (t - s)}^{x + c (t - s)} f (y, s) d y d s$ is the unique classical solution of $u_{t t} = c^{2} u_{x x} + f$ with $u (\cdot, 0) = u_{0}$ , $u_{t} (\cdot, 0) = u_{1}$ . The source integral is over the backward characteristic triangle with vertex $(x, t)$ : $0 \leq s \leq t$ , $∣ y - x ∣ \leq c (t - s)$ , and its coefficient is $1 / (2 c)$ .

Facts & Assumptions

Given: a speed $c > 0$ , data $u_{0} \in C^{2} (R)$ , $u_{1} \in C^{1} (R)$ , a source $f \in C^{1} (R \times [0, \infty))$ , and the displayed function $u$ .

[F1]

[F2]

For continuous $φ$ , the function $x \mapsto \int_{a (x)}^{b (x)} φ$ has derivative $φ (b (x)) b^{'} (x) - φ (a (x)) a^{'} (x)$ ; more generally the primitive of a continuous function is recovered by evaluation at the endpoints (Every continuous function on an interval has a primitive; two primitives differ by a constant; and $\int_{a}^{b} f = G (b) - G (a)$ for any primitive $G$ ).

[F3]

With data $u_{0}, u_{1}$ the homogeneous d'Alembert expression is the unique $C^{2}$ solution of $u_{t t} = c^{2} u_{x x}$ with those data (d'Alembert's formula and uniqueness in one dimension).

[F4]

Sums, products, constant multiples of differentiable functions are differentiable with the usual rules (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 $g (c) \neq 0$ , The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ , 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

1.1F1F2F4algebra

The source term. Extend $f (y, s)$ to $s < 0$ by $f (y, 0)$ ; its value and first spatial derivative remain continuous. Define $Φ (x, t, s) : = \int_{x - c (t - s)}^{x + c (t - s)} f (y, s) d y$ as an oriented integral, also when $t < s$ . Primitives [F2] give $Φ_{t} = c [f (x + c (t - s), s) + f (x - c (t - s), s)]$ and $Φ_{x} = f (x + c (t - s), s) - f (x - c (t - s), s)$ on an open rectangle in $(t, s)$ , including $s = t$ . Since $Φ (x, t, t) = 0$ , [F1] gives $D_{t} = \frac{1}{2} \int_{0}^{t} [f (x + c (t - s), s) + f (x - c (t - s), s)] d s$ for $D = (2 c)^{- 1} \int_{0}^{t} Φ d s$ . The compact-rectangle theorem Leibniz's rule on a compact rectangle: an interior parameter derivative with a continuous extension may be passed through a Riemann integral permits spatial differentiation under this fixed $s$ -integral. Thus $D_{x} = (2 c)^{- 1} \int_{0}^{t} [f (x + c (t - s), s) - f (x - c (t - s), s)] d s$ and $D_{x x} = (2 c)^{- 1} \int_{0}^{t} [f_{x} (x + c (t - s), s) - f_{x} (x - c (t - s), s)] d s$ .

2.1F1F4step 1.1algebra

Equation and regularity. Differentiating $D_{t}$ by [F1] gives $D_{t t} = f (x, t) + \frac{c}{2} \int_{0}^{t} [f_{x} (x + c (t - s), s) - f_{x} (x - c (t - s), s)] d s = f + c^{2} D_{x x}$ . The chain rule The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ gives the displayed integrand derivatives; differentiation of $D_{t}$ in $x$ gives $D_{t x} = \frac{1}{2} \int_{0}^{t} [f_{x} (x + c (t - s), s) + f_{x} (x - c (t - s), s)] d s$ , also equal to $D_{x t}$ by differentiating $D_{x}$ . All these derivatives are continuous because their integrands are continuous on local compact rectangles. At zero, $D, D_{x}, D_{t}, D_{x x}, D_{x t}$ tend to zero and $D_{t t} \to f (x, 0)$ locally uniformly, proving the asserted $C^{2}$ regularity up to the initial time.

3.1F3step 1.1step 2.1algebra

Data and uniqueness. At $t = 0$ the source integral vanishes, so $u (\cdot, 0) = u_{0}$ and $u_{t} (\cdot, 0) = u_{1}$ are exactly the statements of [F3] for the homogeneous part. If $v$ is any classical solution of the forced problem with the same data, then $w : = u - v$ is $C^{2}$ with $w_{t t} = c^{2} w_{x x}$ and zero data, so $w = 0$ by the uniqueness clause of [F3]; hence $u$ is the unique classical solution.

4.1given∎

The double integral runs over $0 \leq s \leq t$ and $∣ y - x ∣ \leq c (t - s)$ , the backward characteristic triangle with vertex $(x, t)$ , and its coefficient is $1 / (2 c)$ ; this completes the identification of the displayed solution.

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Kirchhoff's formula in three dimensions

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $c > 0$ , $u_{0} \in C^{3} (R^{3})$ , $u_{1} \in C^{2} (R^{3})$ and let $M$ be the spherical mean of Spherical means and the weighted ball integral of space-dependent data. Then $u (x, t) : = \frac{\partial}{\partial t} [t M_{u_{0}} (x, c t)] + t M_{u_{1}} (x, c t)$ defines a $C^{2}$ function on $R^{3} \times (0, \infty)$ solving $u_{t t} = c^{2} Δ u$ . Equivalently, for $t > 0$ , $u (x, t) = \frac{1}{4 π c^{2} t^{2}} \int_{\partial B_{c t} (x)} (u_{0} (y) + \nabla u_{0} (y) \cdot (y - x) + t u_{1} (y)) d S (y),$ the sphere integral being the unnormalised form of the mean because $ω_{2} = 4 π$ (Sphere and ball measures scale in Rn). The formula uses the values of $u_{0}, u_{1}$ and the normal derivative of $u_{0}$ on $\partial B_{c t} (x)$ ; pointwise agreement of the two data only on that sphere need not give the same solution value.

Facts & Assumptions

Given: Countable Choice, $c > 0$ , $u_{0} \in C^{3} (R^{3})$ , $u_{1} \in C^{2} (R^{3})$ , and the means $A (x, r) : = M_{u_{0}} (x, r)$ , $B (x, r) : = M_{u_{1}} (x, r)$ .

[F1]

For $k \geq 1$ and $h \in C^{k} (R^{3})$ , the spherical mean $M_{h}$ is $C^{k}$ on $R^{3} \times (0, \infty)$ , every derivative being obtained by differentiating $h$ under the sphere integral (Smoothness, parity and zero-radius limits of spherical means).

[F2]

For $h \in C^{2} (R^{3})$ one has $Δ_{x} M_{h} (x, r) = M_{r r} (x, r) + \frac{2}{r} M_{r} (x, r)$ for $r > 0$ (The Euler–Poisson–Darboux equation for spherical means).

[F3]

If $f$ is totally differentiable at $a$ and $g$ at $f (a)$ , then $g \circ f$ is totally differentiable at $a$ with $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ ). The required total differentiability follows from continuous coordinate partial derivatives (If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative).

[F4]

[F5]

If $h$ is $C^{2}$ on an open set of $R^{m}$ , then $\partial_{i} \partial_{j} h = \partial_{j} \partial_{i} h$ (Clairaut--Schwarz theorem for continuous second partial derivatives).

[F6]

$∣ \partial B_{r}^{3} ∣ = ω_{2} r^{2}$ with $ω_{2} = 3 V_{3}$ , and the map $ω \mapsto a + R ω$ multiplies surface measure by $R^{2}$ : $\int_{S^{2}} h (ω) d σ (ω) = R^{- 2} \int_{\partial B_{R} (a)} h ((y - a) / R) d S (y)$ (Sphere and ball measures scale in Rn, Agreement with the existing polar sphere measure).

[F7]

$V_{3} = π^{3 / 2} / Γ (5 / 2) = 4 π / 3$ , so $ω_{2} = 3 V_{3} = 4 π$ (The closed form for the volume of the unit $n$ -ball, The real Gamma functional equation $Γ (s + 1) = s Γ (s)$ , $Γ (1 / 2) = \sqrt{π}$ from the Gaussian integral).

Proof

1.1F1F3F4algebra

Time derivatives of the candidate. Write $r = c t$ . By [F1] the means $A, B$ are $C^{3}$ respectively $C^{2}$ on $R^{3} \times (0, \infty)$ , so [F3] and [F4] give $u = A + c t A_{r} + t B$ , $\partial_{t} u = 2 c A_{r} + c^{2} t A_{r r} + B + c t B_{r}$ and $\partial_{t}^{2} u = 3 c^{2} A_{r r} + c^{3} t A_{r r r} + 2 c B_{r} + c^{2} t B_{r r}$ on $R^{3} \times (0, \infty)$ , all derivatives being evaluated at $(x, c t)$ .

1.2F2F5F4algebra

Spatial derivatives. By [F2] applied to $u_{0}$ and to $u_{1}$ , $Δ_{x} A = A_{r r} + \frac{2}{r} A_{r}$ and $Δ_{x} B = B_{r r} + \frac{2}{r} B_{r}$ ; since $A$ is $C^{3}$ and $\partial_{r}$ commutes with the $x$ -derivatives by [F5], $Δ_{x} A_{r} = \partial_{r} (Δ_{x} A) = A_{r r r} + \frac{2}{r} A_{r r} - \frac{2}{r^{2}} A_{r}$ . Hence $Δ u = Δ_{x} A + c t Δ_{x} A_{r} + t Δ_{x} B = (A_{r r} + \frac{2}{r} A_{r}) + c t (A_{r r r} + \frac{2}{r} A_{r r} - \frac{2}{r^{2}} A_{r}) + t (B_{r r} + \frac{2}{r} B_{r})$ at $(x, c t)$ .

1.3F1F3F6F7algebra

The unnormalised form. Since $∣ S^{2} ∣ = ω_{2} = 4 π$ by [F6] and [F7], $M_{u_{0}} (x, r) = \frac{1}{4 π} \int_{S^{2}} u_{0} (x + r ω) d σ (ω) = \frac{1}{4 π r^{2}} \int_{\partial B_{r} (x)} u_{0} d S$ , and likewise for $u_{1}$ ; also $\partial_{r} M_{u_{0}} (x, r) = \frac{1}{4 π} \int_{S^{2}} \nabla u_{0} (x + r ω) \cdot ω d σ (ω) = \frac{1}{4 π r^{2}} \int_{\partial B_{r} (x)} \nabla u_{0} (y) \cdot \frac{y - x}{r} d S (y)$ by [F1], [F3] and the scaling in [F6]. Substituting $r = c t$ into $u = A + c t A_{r} + t B$ and collecting the common factor $\frac{1}{4 π c^{2} t^{2}}$ gives $u (x, t) = \frac{1}{4 π c^{2} t^{2}} \int_{\partial B_{c t} (x)} (u_{0} (y) + \nabla u_{0} (y) \cdot (y - x) + t u_{1} (y)) d S (y)$ .

2.1F4step 1.1step 1.2algebra

Comparison. Multiplying step 1.2 by $c^{2}$ and substituting $r = c t$ gives $c^{2} Δ u = c^{2} A_{r r} + (2 c / t) A_{r} + c^{3} t A_{r r r} + 2 c^{2} A_{r r} - (2 c / t) A_{r} + c^{2} t B_{r r} + 2 c B_{r} = 3 c^{2} A_{r r} + c^{3} t A_{r r r} + c^{2} t B_{r r} + 2 c B_{r}$ . This equals $u_{t t}$ from step 1.1, proving the equation. The $C^{3}$ and $C^{2}$ regularity of $A$ and $B$ gives $u \in C^{2}$ .

3.1step 2.1step 1.3∎

Both displays define the same $C^{2}$ solution. The unnormalised display uses the data values and the normal derivative of $u_{0}$ on the sphere, or equivalently the data on an open neighbourhood of that sphere.

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Poisson's formula in two dimensions by descent

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $c > 0$ , $u_{0} \in C^{3} (R^{2})$ , $u_{1} \in C^{2} (R^{2})$ and let $W$ be the weighted ball integral of Spherical means and the weighted ball integral of space-dependent data. Then $u (x, t) : = \frac{1}{2 π c} \frac{\partial}{\partial t} \int_{B_{c t} (x)} \frac{u_{0} (y)}{\sqrt{c^{2} t^{2} - ∣ y - x ∣^{2}}} d y + \frac{1}{2 π c} \int_{B_{c t} (x)} \frac{u_{1} (y)}{\sqrt{c^{2} t^{2} - ∣ y - x ∣^{2}}} d y (t > 0)$ defines a $C^{2}$ function on $R^{2} \times (0, \infty)$ solving $u_{t t} = c^{2} Δ u$ ; here the first $\partial_{t}$ differentiates the $C^{1}$ function $t \mapsto \int_{B_{c t} (x)} u_{0} (y) (c^{2} t^{2} - ∣ y - x ∣^{2})^{- 1 / 2} d y$ , which is legitimate by the projection identity, not by termwise differentiation of a singular integrand. Equivalently, for $t > 0$ , $u (x, t) = \frac{1}{2 π c t} \int_{B_{c t} (x)} \frac{u_{0} (y) + \nabla u_{0} (y) \cdot (y - x) + t u_{1} (y)}{\sqrt{c^{2} t^{2} - ∣ y - x ∣^{2}}} d y .$ The extension to the initial time and the attainment of the data are treated later on this page.

Facts & Assumptions

Given: Countable Choice, $c > 0$ , $u_{0} \in C^{3} (R^{2})$ , $u_{1} \in C^{2} (R^{2})$ , the weighted ball integral $W$ of Spherical means and the weighted ball integral of space-dependent data, and the extension $U_{j} (ξ, z) : = u_{j} (ξ)$ of $u_{j}$ to $R^{3}$ .

[F1]

In three dimensions the Kirchhoff expression $\partial_{t} [t M_{U_{0}}^{(3)} ((ξ, z), c t)] + t M_{U_{1}}^{(3)} ((ξ, z), c t)$ is a $C^{2}$ solution of $V_{t t} = c^{2} Δ_{3} V$ (Kirchhoff's formula in three dimensions).

[F2]

For even $n$ , $t^{n - 1} M_{G}^{(n + 1)} (x, 0, c t) = \frac{(n - 1)!!}{c^{n - 1}} W_{g} (x, c t)$ for the cylindrical extension $G (ξ, z) = g (ξ)$ , where $M^{(n + 1)}$ is the $(n + 1)$ -dimensional spherical mean (Sphere integrals of a cylindrical function project to weighted ball integrals).

[F3]

$W_{f} (x, c t) = (n!! V_{n})^{- 1} \int_{B_{c t} (x)} f (y) (c^{2} t^{2} - ∣ y - x ∣^{2})^{- 1 / 2} d y$ for even $n$ , with $n!! = n (n - 2) \dots 2$ and $V_{n}$ the unit-ball volume (Spherical means and the weighted ball integral of space-dependent data); in particular, the weight $w (z) = (1 - ∣ z ∣^{2})^{- 1 / 2}$ is integrable on $B_{1} \subset R^{2}$ by the polar-coordinate integrability statement in that definition.

[F4]

If measurable functions converge pointwise almost everywhere and are dominated by one nonnegative integrable function, their integrals converge (Dominated convergence).

[F5]

If $f$ is totally differentiable at $a$ and $g$ at $f (a)$ then $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ ); for an invertible linear $T$ with matrix $A$ , $λ_{2} (T [E]) = ∣ \det A ∣ λ_{2} (E)$ (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); sums and products are differentiated by the usual rules (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 $g (c) \neq 0$ ). The required total differentiability follows from continuous coordinate partial derivatives (If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative).

[F6]

A continuous function on a compact metric space is bounded, and closed Euclidean balls are compact (For $n \geq 1$ , every Euclidean closed ball and every Euclidean sphere of positive radius is compact, A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value).

[F7]

A real function continuous on a closed interval and differentiable on its interior has a difference quotient equal to a derivative at an interior point (The mean value theorem, as the case $g (x) = x$ of Cauchy's: for $f$ continuous on $[a, b]$ with $a < b$ and differentiable on $(a, b)$ there is $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$ ).

Proof

1.1F1algebra

Descent. Extend $u_{0}, u_{1}$ to $U_{0}, U_{1}$ on $R^{3}$ by $U_{j} (ξ, z) : = u_{j} (ξ)$ ; these are $C^{3}$ respectively $C^{2}$ . By [F1] the function $V (ξ, z, t) : = \partial_{t} [t M_{U_{0}}^{(3)} ((ξ, z), c t)] + t M_{U_{1}}^{(3)} ((ξ, z), c t)$ is $C^{2}$ on $R^{3} \times (0, \infty)$ with $V_{t t} = c^{2} Δ_{3} V$ . Since $U_{j} (ξ + c t ω^{'}, z + c t ω_{3}) = u_{j} (ξ + c t ω^{'})$ does not depend on $z$ , the sphere means of $U_{0}, U_{1}$ at centre $(ξ, z)$ are independent of $z$ ; hence $V$ is independent of $z$ , so $\partial_{z}^{2} V = 0$ , $Δ_{3} V = Δ_{2} V$ , and the restriction $u (x, t) : = V (x, 0, t)$ is a $C^{2}$ function on $R^{2} \times (0, \infty)$ with $u_{t t} = c^{2} Δ_{2} u$ .

1.2F2F3algebra

The weighted-ball form. By [F2] with $n = 2$ and $G = U_{j}$ , $t M_{U_{j}}^{(3)} ((x, 0), c t) = \frac{1!!}{c} W_{u_{j}} (x, c t) = \frac{1}{c} W_{u_{j}} (x, c t)$ ; therefore $u (x, t) = \frac{1}{c} [\partial_{t} W_{u_{0}} (x, c t) + W_{u_{1}} (x, c t)]$ . With $n = 2$ , $n!! V_{n} = 2 π$ , [F3] reads $W_{f} (x, c t) = \frac{1}{2 π} \int_{B_{c t} (x)} f (y) (c^{2} t^{2} - ∣ y - x ∣^{2})^{- 1 / 2} d y$ , so $u$ is the displayed Poisson expression; the derivative $\partial_{t}$ acts on the $C^{1}$ function $t \mapsto W_{u_{0}} (x, c t)$ , since $W_{u_{0}} (x, c t) = c t M_{U_{0}}^{(3)} ((x, 0), c t)$ by [F2], and the spherical mean is $C^{3}$ , and not by differentiating a singular integrand.

1.3F3F4F5F6F7algebra

The integrated equivalent form. For fixed $x$ put $w (z) : = (1 - ∣ z ∣^{2})^{- 1 / 2}$ on $B_{1} \subset R^{2}$ and $A (x, t) : = \int_{B_{1}} u_{0} (x + c t z) w (z) d z$ . By [F3], $w \in L^{1} (B_{1})$ . To differentiate in $t$ , fix compact sets $K \subset R^{2}$ and $J \subset (0, \infty)$ for $x$ and $t$ . Choose a closed ball $Q$ containing $x + c (t + s) z$ for $x \in K$ , $t \in J$ , $∣ s ∣$ sufficiently small and $∣ z ∣ \leq 1$ . By [F6], $C : = \sup_{Q} ∣ \nabla u_{0} ∣ < \infty$ . The difference quotients of $u_{0} (x + c t z)$ in $t$ , for these parameters and sufficiently small increments $h$ , are bounded in absolute value by $c C ∣ z ∣$ by [F7]. After multiplication by $w (z)$ they are dominated by the locally uniform integrable function $c C ∣ z ∣ w (z)$ . They converge pointwise to $c \nabla u_{0} (x + c t z) \cdot z w (z)$ , so [F4] gives $\partial_{t} A (x, t) = c \int_{B_{1}} \nabla u_{0} (x + c t z) \cdot z w (z) d z$ . The same argument for spatial difference quotients, and dominated convergence applied to convergent parameter sequences with the same compact-set majorant, shows that these first derivatives are continuous locally; in particular $A$ is $C^{1}$ in $(x, t)$ . Now [F5] gives $I (x, t) : = \int_{B_{c t} (x)} u_{0} (y) (c^{2} t^{2} - ∣ y - x ∣^{2})^{- 1 / 2} d y = c t A (x, t)$ , hence $\partial_{t} I = c A + c t \partial_{t} A$ . The same change of variables gives $\int_{B_{c t} (x)} \nabla u_{0} (y) \cdot (y - x) (c^{2} t^{2} - ∣ y - x ∣^{2})^{- 1 / 2} d y = (c t)^{2} \int_{B_{1}} \nabla u_{0} (x + c t z) \cdot z w (z) d z = (c t)^{2} \partial_{t} A / c$ . Therefore $\frac{1}{2 π c} [\partial_{t} I + \int_{B_{c t} (x)} u_{1} (y) (c^{2} t^{2} - ∣ y - x ∣^{2})^{- 1 / 2} d y] = \frac{1}{2 π c t} \int_{B_{c t} (x)} \frac{u_{0} (y) + \nabla u_{0} (y) \cdot (y - x) + t u_{1} (y)}{\sqrt{c^{2} t^{2} - ∣ y - x ∣^{2}}} d y$ , which is the equivalent form.

2.1given∎

Both displays therefore define the same $C^{2}$ solution of the two-dimensional homogeneous wave equation on $R^{2} \times (0, \infty)$ ; the limit at $t ↓ 0$ and the attainment of the data are the subject of the data-attainment lemma below.

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The odd-dimensional wave formula by iterated spherical means

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $n = 2 k + 1 \geq 3$ be odd, $c > 0$ , $u_{0} \in C^{k + 2} (R^{n})$ , $u_{1} \in C^{k + 1} (R^{n})$ , put $D_{t} : = t^{- 1} \partial_{t}$ , and let $M$ be the spherical mean of Spherical means and the weighted ball integral of space-dependent data. Then $u (x, t) : = \frac{1}{(n - 2)!!} [\frac{\partial}{\partial t} D_{t}^{k - 1} (t^{n - 2} M_{u_{0}} (x, c t)) + D_{t}^{k - 1} (t^{n - 2} M_{u_{1}} (x, c t))]$ defines a $C^{2}$ function on $R^{n} \times (0, \infty)$ solving $u_{t t} = c^{2} Δ u$ ; each $D_{t}^{j}$ is applied to the $t$ -dependent function $t \mapsto t^{n - 2} M_{f} (x, c t)$ . For $n = 3$ ( $k = 1$ ) this is exactly Kirchhoff's formula Kirchhoff's formula in three dimensions, and the displayed constant $(n - 2)!!$ is the one required by the leading coefficient of Radial-derivative expansion of the Euler–Poisson–Darboux transform and its zero-radius limit.

Facts & Assumptions

Given: Countable Choice, $n = 2 k + 1 \geq 3$ , $c > 0$ , $u_{0} \in C^{k + 2} (R^{n})$ , $u_{1} \in C^{k + 1} (R^{n})$ , and the spherical means $H_{f} (x, r) : = M_{f} (x, r)$ .

[F1]

For $j \geq 1$ and every $φ \in C^{j + 1} ((0, \infty))$ , $\partial_{r}^{2} D_{r}^{j - 1} (r^{2 j - 1} φ (r)) = D_{r}^{j - 1} [r^{2 j - 1} r^{- 2 j} \partial_{r} (r^{2 j} \partial_{r} φ (r))]$ on $(0, \infty)$ , where $D_{r} = r^{- 1} \partial_{r}$ (The iterated radial-derivative identity behind the odd-dimensional reduction).

[F2]

For $m \geq 1$ and $h \in C^{m} (R^{n})$ , $(x, r) \mapsto M_{h} (x, r)$ is $C^{m}$ on $R^{n} \times (0, \infty)$ with all derivatives obtained by differentiating $h$ under the sphere integral (Smoothness, parity and zero-radius limits of spherical means).

[F3]

For $h \in C^{2} (R^{n})$ the Euler–Poisson–Darboux identity $Δ_{x} M_{h} (x, r) = M_{r r} (x, r) + \frac{n - 1}{r} M_{r} (x, r)$ holds for every $r > 0$ (The Euler–Poisson–Darboux equation for spherical means).

[F4]

If $f$ is totally differentiable at $a$ and $g$ at $f (a)$ , then $g \circ f$ is totally differentiable at $a$ with $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ ); sums, products and quotients of differentiable functions are differentiable with the usual rules on their domains (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 $g (c) \neq 0$ ). The required total differentiability follows from continuous coordinate partial derivatives (If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative).

[F5]

If $h$ is $C^{2}$ on an open subset of $R^{m}$ , then $\partial_{i} \partial_{j} h = \partial_{j} \partial_{i} h$ (Clairaut--Schwarz theorem for continuous second partial derivatives).

Proof

1.1F1F2

Fix $f \in C^{k + 1} (R^{n})$ and put $W (x, t) : = D_{t}^{k - 1} (t^{n - 2} H_{f} (x, c t))$ , where $n - 2 = 2 k - 1$ . Applying [F1] with $j = k$ and $φ (t) : = H_{f} (x, c t)$ — admissible for each fixed $x$ because $H_{f} (x, \cdot) \in C^{k + 1}$ by [F2] — gives $\partial_{t}^{2} D_{t}^{k - 1} (t^{2 k - 1} H_{f} (x, c t)) = D_{t}^{k - 1} [t^{2 k - 1} t^{- 2 k} \partial_{t} (t^{2 k} \partial_{t} H_{f} (x, c t))]$ .

1.2F3F4algebra

The inner expression. By [F4], $\partial_{t} H_{f} (x, c t) = c (H_{f})_{r} (x, c t)$ and $\partial_{t}^{2} H_{f} (x, c t) = c^{2} (H_{f})_{r r} (x, c t)$ , so with $r = c t$ and $2 k = n - 1$ , $t^{- 1} \partial_{t} (t^{2 k} \partial_{t} H_{f} (x, c t)) = t^{- 1} \partial_{t} (t^{n - 1} c (H_{f})_{r} (x, c t)) = c t^{n - 3} [(n - 1) (H_{f})_{r} (x, c t) + c t (H_{f})_{r r} (x, c t)] = c t^{n - 3} [(n - 1) (H_{f})_{r} (x, r) + r (H_{f})_{r r} (x, r)] = c^{2} t^{n - 2} Δ_{x} H_{f} (x, c t)$ , the last equality because $r ((H_{f})_{r r} + \frac{n - 1}{r} (H_{f})_{r}) = r Δ_{x} H_{f}$ by [F3].

1.3F5algebra

Hence $\partial_{t}^{2} W (x, t) = c^{2} D_{t}^{k - 1} (t^{n - 2} Δ_{x} H_{f} (x, c t))$ . The operator $Δ_{x}$ acts only on the $x$ -variables while $D_{t}^{k - 1}$ and multiplication by $t^{n - 2}$ act only on $t$ , and the mixed partials involved commute by [F5] since $H_{f}$ is $C^{k + 1}$ ; therefore $D_{t}^{k - 1} (t^{n - 2} Δ_{x} H_{f} (x, c t)) = Δ_{x} D_{t}^{k - 1} (t^{n - 2} H_{f} (x, c t)) = Δ_{x} W (x, t)$ , that is $\partial_{t}^{2} W = c^{2} Δ_{x} W$ on $R^{n} \times (0, \infty)$ .

1.4F2algebra

Regularity and superposition. For $f = u_{0} \in C^{k + 2} (R^{n})$ the function $t \mapsto t^{n - 2} H_{u_{0}} (x, c t)$ is $C^{k + 2}$ by [F2], so $W_{u_{0}} = D_{t}^{k - 1} (\cdot)$ is $C^{3}$ ; for $f = u_{1} \in C^{k + 1} (R^{n})$ similarly $W_{u_{1}}$ is $C^{2}$ . Hence $u = (n - 2)!!^{- 1} [\partial_{t} W_{u_{0}} + W_{u_{1}}]$ is $C^{2}$ on $R^{n} \times (0, \infty)$ and $u_{t t} = (n - 2)!!^{- 1} [\partial_{t} \partial_{t}^{2} W_{u_{0}} + \partial_{t}^{2} W_{u_{1}}] = (n - 2)!!^{- 1} [\partial_{t} (c^{2} Δ_{x} W_{u_{0}}) + c^{2} Δ_{x} W_{u_{1}}] = c^{2} Δ_{x} u$ .

2.1given∎

For $k = 1$ , that is $n = 3$ , the formula reads $\partial_{t} [t M_{u_{0}} (x, c t)] + t M_{u_{1}} (x, c t)$ with $(n - 2)!! = 1$ , which is Kirchhoff's expression of Kirchhoff's formula in three dimensions. The prefactor $(n - 2)!!$ is the leading coefficient of $D_{r}^{k - 1} (r^{2 k - 1} φ)$ by Radial-derivative expansion of the Euler–Poisson–Darboux transform and its zero-radius limit: with $r = c t$ and with the evenness of the means, which kills the first-order term, that expansion makes the normalised combination the data-carrying normalisation; the precise attainment of $u_{0}$ and $u_{1}$ is proved by the data-attainment lemma below.

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The even-dimensional wave formula by descent

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $n = 2 k \geq 2$ be even, $c > 0$ , $u_{0} \in C^{k + 2} (R^{n})$ , $u_{1} \in C^{k + 1} (R^{n})$ , let $W$ be the weighted ball integral of Spherical means and the weighted ball integral of space-dependent data and put $D_{t} = t^{- 1} \partial_{t}$ . Then $u (x, t) : = c^{1 - n} [\frac{\partial}{\partial t} D_{t}^{k - 1} W_{u_{0}} (x, c t) + D_{t}^{k - 1} W_{u_{1}} (x, c t)]$ defines a $C^{2}$ function on $R^{n} \times (0, \infty)$ solving $u_{t t} = c^{2} Δ u$ . For $n = 2$ ( $k = 1$ ) this is exactly Poisson's formula Poisson's formula in two dimensions by descent, and the factor $c^{1 - n}$ is the exact rescaling of the unit-speed formula obtained by substituting $s = c t$ in the $(n + 1)$ -dimensional odd-dimensional formula.

Facts & Assumptions

Given: Countable Choice, $n = 2 k \geq 2$ , $c > 0$ , $u_{0} \in C^{k + 2} (R^{n})$ , $u_{1} \in C^{k + 1} (R^{n})$ , and the cylindrical extensions $U_{j} (ξ, z) : = u_{j} (ξ)$ to $R^{n + 1}$ .

[F1]

For odd $m = 2 k + 1$ and data in $C^{k + 2}$ respectively $C^{k + 1}$ , the formula of The odd-dimensional wave formula by iterated spherical means with $D_{t} = t^{- 1} \partial_{t}$ defines a $C^{2}$ solution of $v_{t t} = c^{2} Δ v$ on $R^{m} \times (0, \infty)$ .

[F2]

For even $n$ , $t^{n - 1} M_{U}^{(n + 1)} ((x, 0), c t) = \frac{(n - 1)!!}{c^{n - 1}} W_{u} (x, c t)$ for the cylindrical extension of $u$ , where $M^{(n + 1)}$ is the spherical mean in $n + 1$ variables (Sphere integrals of a cylindrical function project to weighted ball integrals).

[F3]

For $n = 2$ and $k = 1$ the formula reduces to Poisson's formula Poisson's formula in two dimensions by descent with $W_{f} (x, c t) = \frac{1}{2 π} \int_{B_{c t} (x)} f (y) (c^{2} t^{2} - ∣ y - x ∣^{2})^{- 1 / 2} d y$ (Spherical means and the weighted ball integral of space-dependent data).

[F4]

Sums, products, constant multiples of differentiable functions are differentiable with the usual rules, and constants commute with differentiation (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 $g (c) \neq 0$ ).

Proof

1.1F1algebra

Descent. The number $n + 1 = 2 k + 1$ is odd, and $U_{0} \in C^{k + 2} (R^{n + 1})$ , $U_{1} \in C^{k + 1} (R^{n + 1})$ , so [F1] applies in dimension $n + 1$ with the same $k$ : $V (ξ, z, t) : = \frac{1}{(n - 1)!!} [\partial_{t} D_{t}^{k - 1} (t^{n - 1} M_{U_{0}}^{(n + 1)} ((ξ, z), c t)) + D_{t}^{k - 1} (t^{n - 1} M_{U_{1}}^{(n + 1)} ((ξ, z), c t))]$ is $C^{2}$ on $R^{n + 1} \times (0, \infty)$ with $V_{t t} = c^{2} Δ_{n + 1} V$ . Since $U_{j} (ξ + c t ω, z + c t ω_{n + 1}) = u_{j} (ξ + c t ω)$ is independent of $z$ , the means of $U_{0}, U_{1}$ at centre $(ξ, z)$ do not depend on $z$ ; hence $V$ does not depend on $z$ , so $\partial_{z}^{2} V = 0$ , $Δ_{n + 1} V = Δ_{n} V$ , and the restriction $u (x, t) : = V (x, 0, t)$ is a $C^{2}$ function on $R^{n} \times (0, \infty)$ with $u_{t t} = c^{2} Δ_{n} u$ .

1.2F2F4algebra

Rewriting with the weighted ball integral. By [F2] with the cylindrical extensions, $t^{n - 1} M_{U_{j}}^{(n + 1)} ((x, 0), c t) = \frac{(n - 1)!!}{c^{n - 1}} W_{u_{j}} (x, c t)$ for $j = 0, 1$ ; substituting into the definition of $V$ and using [F4] to move the constant $\frac{(n - 1)!!}{c^{n - 1}}$ and the factor $\frac{1}{(n - 1)!!}$ through the $t$ -derivatives gives $u (x, t) = c^{1 - n} [\partial_{t} D_{t}^{k - 1} W_{u_{0}} (x, c t) + D_{t}^{k - 1} W_{u_{1}} (x, c t)]$ .

1.3F3algebra

The two-dimensional case. For $n = 2$ , $k = 1$ and $c^{1 - n} = c^{- 1}$ , so the formula reads $c^{- 1} [\partial_{t} W_{u_{0}} (x, c t) + W_{u_{1}} (x, c t)]$ , which by [F3] is exactly the Poisson expression of Poisson's formula in two dimensions by descent.

2.1given∎

Therefore the displayed even-dimensional formula defines a $C^{2}$ solution of the homogeneous wave equation, it reduces to Poisson's formula when $n = 2$ , and its prefactor $c^{1 - n}$ is the one produced by substituting $s = c t$ in the $(n + 1)$ -dimensional odd formula.

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The dimension formulas attain the Cauchy data

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $c > 0$ , $n \geq 2$ and let $u$ be one of the functions constructed from data $u_{0}, u_{1}$ of the regularity required by the corresponding formula: Kirchhoff ( $n = 3$ , Kirchhoff's formula in three dimensions), Poisson ( $n = 2$ , Poisson's formula in two dimensions by descent), the odd-dimensional formula (The odd-dimensional wave formula by iterated spherical means) or the even-dimensional formula (The even-dimensional wave formula by descent). Then, as $t ↓ 0$ , for every $x$ $u (x, t) ⟶ u_{0} (x), \partial_{t} u (x, t) ⟶ u_{1} (x),$ and the extension $u (x, 0) : = u_{0} (x)$ is continuous on $R^{n} \times [0, \infty)$ .

Facts & Assumptions

Given: Countable Choice, $c > 0$ , $n \geq 2$ , and one of the four representation formulas with its data classes.

[F1]

For $j \geq 1$ and every $φ \in C^{j + 1} ((0, \infty))$ , $D_{r}^{j - 1} (r^{2 j - 1} φ (r)) = \sum_{i = 0}^{j - 1} α_{j, i} r^{i + 1} φ^{(i)} (r)$ with $α_{j, 0} = (2 j - 1)!!$ (Radial-derivative expansion of the Euler–Poisson–Darboux transform and its zero-radius limit).

[F2]

For $m \geq 1$ and $h \in C^{m} (R^{n})$ , the spherical mean $M_{h}$ is $C^{m}$ on $R^{n} \times (0, \infty)$ with $\partial_{r}^{i} M_{h}$ obtained by differentiating $h$ under the sphere integral, and $r \mapsto M_{h} (x, r)$ is even (Smoothness, parity and zero-radius limits of spherical means).

[F3]

The odd- and even-dimensional formulas define $C^{2}$ solutions of $u_{t t} = c^{2} Δ u$ on $R^{n} \times (0, \infty)$ (The odd-dimensional wave formula by iterated spherical means, The even-dimensional wave formula by descent); for $n = 3$ the odd formula is Kirchhoff's expression and for $n = 2$ the even formula is Poisson's expression (Kirchhoff's formula in three dimensions, Poisson's formula in two dimensions by descent).

[F4]

Sums, products and quotients are differentiated by the usual rules (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 $g (c) \neq 0$ ).

Proof

1.1F1F2F4algebra

Differentiated finite expansion. Suppose $n = 2 k + 1$ and put $h_{f} (x, t) = M_{f} (x, c t)$ , using the signed-radius extension of [F2]. This is $C^{k + 1}$ and even when $f \in C^{k + 1}$ , so $h_{f} (x, 0) = f (x)$ and $\partial_{t} h_{f} (x, 0) = 0$ . By [F1], $T_{f} (x, t) : = D_{t}^{k - 1} (t^{2 k - 1} h_{f} (x, t)) = \sum_{j = 0}^{k - 1} α_{k, j} t^{j + 1} \partial_{t}^{j} h_{f} (x, t)$ , where $a : = α_{k, 0} = (2 k - 1)!!$ . Differentiate this finite sum itself: $T_{f}^{'} = \sum_{j} α_{k, j} ((j + 1) t^{j} h_{f}^{(j)} + t^{j + 1} h_{f}^{(j + 1)})$ and $T_{f}^{''} = \sum_{j} α_{k, j} (j (j + 1) t^{j - 1} h_{f}^{(j)} + 2 (j + 1) t^{j} h_{f}^{(j + 1)} + t^{j + 1} h_{f}^{(j + 2)})$ , with the first summand omitted for $j = 0$ . Every derivative used has order at most $k + 1$ . Continuity and $h_{f}^{'} (x, 0) = 0$ give $T_{f} \to 0$ , $T_{f}^{'} \to a f (x)$ and $T_{f}^{''} \to 0$ , uniformly for $x$ in compact sets. For $T_{f}^{''}$ , the only terms without a positive power of $t$ are constant multiples of $h_{f}^{'}$ , which vanish at zero. These formulas never differentiate an unspecified error term.

2.1F3step 1.1algebra

Odd-dimensional data. The odd formula is $u = a^{- 1} (T_{u_{0}}^{'} + T_{u_{1}})$ and $u_{t} = a^{- 1} (T_{u_{0}}^{''} + T_{u_{1}}^{'})$ . Step 1.1 applies to both data, since their classes are at least $C^{k + 1}$ . It follows that $u \to u_{0}$ and $u_{t} \to u_{1}$ , locally uniformly in $x$ . The even smooth signed-radius means in step 1.1 also show that $T_{f}$ and its first two derivatives extend continuously through zero.

3.1F3step 2.1

Even-dimensional data by descent. For $n = 2 k$ , extend the data cylindrically to $R^{n + 1}$ . Their differentiability classes are exactly those of the odd formula in dimension $n + 1 = 2 k + 1$ . The construction in The even-dimensional wave formula by descent identifies the even solution with the restriction of that odd solution to the last coordinate zero. The limits of step 2.1 therefore apply without differentiating a singular ball weight.

4.1F3step 2.1step 3.1∎

The locally uniform displacement limit and continuity of $u_{0}$ give joint continuity of the extension $u (x, 0) = u_{0} (x)$ . The velocity limit holds as stated. The cases $n = 3$ and $n = 2$ are Kirchhoff and Poisson by [F3].

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Duhamel's principle for the wave equation

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $c > 0$ , $n \geq 2$ , $T > 0$ and let $f : R^{n} \times [0, T] \to R$ be continuous, compactly supported in $x$ for each fixed $s$ , with all spatial derivatives $D_{x}^{α} f$ through order $q : = ⌊ n / 2 ⌋ + 1$ existing and jointly continuous on $R^{n} \times [0, T]$ . Thus every slice is in the velocity-data class, and the additional derivatives needed to differentiate the launched solution twice are jointly continuous. No time derivative of $f$ is required. For an admissible velocity datum $g$ let $W [g] (x, τ)$ be the homogeneous solution constructed from the formulas above with zero displacement and velocity datum $g$ , so that, by The dimension formulas attain the Cauchy data, $W [g] (x, 0) = 0$ and $\partial_{τ} W [g] (x, 0) = g (x)$ . Then $u (x, t) : = \int_{0}^{t} W [f (\cdot, s)] (x, t - s) d s (0 \leq t \leq T)$ is a $C^{2}$ function with $u (\cdot, 0) = u_{t} (\cdot, 0) = 0$ and $u_{t t} = c^{2} Δ u + f$ on $R^{n} \times (0, T)$ .

Facts & Assumptions

Given: Countable Choice, $c > 0$ , $n \geq 2$ , a source $f$ of the stated class, and for each admissible $g$ the launched solution $W [g]$ with $W [g] (x, 0) = 0$ , $\partial_{τ} W [g] (x, 0) = g (x)$ .

[F1]

The formulas of the page define $C^{2}$ solutions of the homogeneous equation on $R^{n} \times (0, \infty)$ for admissible data, and the data are attained in the limit sense (The dimension formulas attain the Cauchy data).

[F2]

[F3]

In odd dimension $n = 2 k + 1$ , a zero-displacement launch is $W [g] (x, τ) = a^{- 1} \sum_{j = 0}^{k - 1} α_{k, j} τ^{j + 1} \partial_{τ}^{j} M_{g} (x, c τ)$ , $a = (2 k - 1)!!$ , by Radial-derivative expansion of the Euler–Poisson–Darboux transform and its zero-radius limit. The signed-radius mean is $C^{k + 1}$ (Smoothness, parity and zero-radius limits of spherical means). Differentiating this finite sum through total order two involves at most $k + 1$ spatial derivatives of $g$ and nonnegative powers of $τ$ . For $g = f (\cdot, s)$ , the uniform integral estimate in the smoothness lemma applies also with the continuous parameter $s$ : the assumed joint continuity on compact spatial-time sets makes $W$ and its first two $(x, τ)$ derivatives jointly continuous, including at $τ = 0$ . In even dimension use the cylindrical launch in dimension $n + 1$ from The even-dimensional wave formula by descent, with the same derivative order $q = k + 1$ .

Proof

1.1F1F2F3

First derivative. The launched solutions vanish at $τ = 0$ and have velocity $g$ there by [F1]: $W [g] (x, 0) = 0$ and $\partial_{τ} W [g] (x, 0) = g (x)$ . Applying [F2] to the moving-endpoint integral $u (x, t) = \int_{0}^{t} W [f (\cdot, s)] (x, t - s) d s$ in the form $G (t) = \int_{0}^{t} F (t, s) d s$ with $F (t, s) = W [f (\cdot, s)] (x, t - s)$ — defined also for negative $t - s$ by the signed-radius finite sum in [F3]; extend the source slices constantly for $s < 0$ and $s > T$ . Then $F$ and its first two $t$ -derivatives are continuous on a rectangular neighbourhood of the integration region by [F3] — gives $u_{t} (x, t) = \int_{0}^{t} \partial_{t} W [f (\cdot, s)] (x, t - s) d s + W [f (\cdot, t)] (x, 0) = \int_{0}^{t} \partial_{t} W [f (\cdot, s)] (x, t - s) d s$ , since $W [\cdot] (x, 0) = 0$ ; in particular $u (\cdot, 0) = 0$ and $u_{t} (\cdot, 0) = 0$ .

1.2F1F2F3algebra

Second derivative. Differentiating once more with [F2], $u_{t t} (x, t) = \partial_{t} W [f (\cdot, t)] (x, 0) + \int_{0}^{t} \partial_{t}^{2} W [f (\cdot, s)] (x, t - s) d s = f (x, t) + \int_{0}^{t} c^{2} Δ_{x} W [f (\cdot, s)] (x, t - s) d s = f (x, t) + c^{2} Δ_{x} u (x, t)$ , where the last equality uses $\partial_{τ} W [f (\cdot, t)] (x, 0) = f (x, t)$ and the homogeneous equation for every launched solution from [F1]. To justify moving $Δ_{x}$ through the integral, fix any compact set of $x$ -values and a compact time interval $[0, T_{0}] \subseteq [0, T]$ . By [F3], the integrand and its first two $x$ -derivatives are jointly continuous on the resulting compact $(x, s, t - s)$ parameter set; applying [F2] twice with the fixed $s$ -interval endpoints therefore permits differentiating under the $s$ -integral locally in $x$ . The mixed derivative $u_{t x}$ is obtained similarly from the integral formula for $u_{t}$ . These derivative integrals and their boundary terms are continuous; their bounds on local compact sets give continuous one-sided derivatives also at $t = 0, T$ . No common compact support of all source slices is needed.

2.1given∎

Hence $u$ is $C^{2}$ with zero Cauchy data and $u_{t t} - c^{2} Δ u = f$ on $R^{n} \times (0, T)$ ; the constructed $u$ is a classical solution of the forced problem.

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The forced three-dimensional version as a retarded potential

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $c > 0$ and let $f$ be continuous on $R^{3} \times [0, \infty)$ with $f (\cdot, t) \in C^{2} (R^{3})$ and compactly supported in $x$ for each $t$ . Assume that $\nabla_{x} f$ and every second spatial partial derivative $\partial_{x_{i}} \partial_{x_{j}} f$ are jointly continuous in $(x, t)$ ; no time derivatives of $f$ are required. Then the Duhamel construction gives a classical solution of $u_{t t} = c^{2} Δ u + f$ with zero Cauchy data, namely $u (x, t) = \frac{1}{4 π c^{2}} \int_{B_{c t} (x)} \frac{f (y, t - ∣ y - x ∣ / c)}{∣ y - x ∣} d y (t > 0),$ the retarded potential over the backward light cone of $(x, t)$ . In particular the value uses $f$ only on ${(y, s) : 0 \leq s \leq t, ∣ y - x ∣ = c (t - s)}$ , and the radius factor is $1 / (4 π c^{2})$ .

Facts & Assumptions

Given: Countable Choice, $c > 0$ , a source $f$ of the stated class, and the launched Kirchhoff solutions with zero displacement.

[F1]

For admissible sources the Duhamel principle gives the forced solution as $u (x, t) = \int_{0}^{t} W [f (\cdot, s)] (x, t - s) d s$ , where $W [g]$ is the homogeneous solution with zero displacement and velocity datum $g$ (Duhamel's principle for the wave equation).

[F2]

In three dimensions $W [g] (x, τ) = τ M_{g} (x, c τ)$ , and the mean is the normalised sphere integral with $∣ S^{2} ∣ = ω_{2} = 4 π$ : $M_{g} (x, ρ) = \frac{1}{ω_{2}} \int_{S^{2}} g (x + ρ z) d σ (z) = \frac{1}{4 π c^{2} τ^{2}} \int_{\partial B_{c τ} (x)} g d S$ , where $ω_{2} = 3 V_{3} = 4 π$ follows from $V_{3} = π^{3 / 2} / Γ (5 / 2)$ and the Gamma values (Kirchhoff's formula in three dimensions, The dimension formulas attain the Cauchy data, Sphere and ball measures scale in Rn, The closed form for the volume of the unit $n$ -ball, The real Gamma functional equation $Γ (s + 1) = s Γ (s)$ , $Γ (1 / 2) = \sqrt{π}$ from the Gaussian integral).

[F3]

Under Countable Choice, $\int_{R^{3}} F d λ_{3} = \int_{0}^{\infty} \int_{S^{2}} F (ρ z) ρ^{2} d σ (z) d ρ$ for every Borel $F \geq 0$ (Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma).

[F4]

Sums, products, constant multiples of differentiable functions are differentiable with the usual rules (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 $g (c) \neq 0$ ); the substitution $ρ = c (t - s)$ is an orientation-reversing change of the integration variable.

Proof

1.1F1F2

Duhamel form. By [F1] the solution is $u (x, t) = \int_{0}^{t} W [f (\cdot, s)] (x, t - s) d s$ , and by [F2] the launched solution is $W [f (\cdot, s)] (x, τ) = τ M_{f (\cdot, s)} (x, c τ)$ ; substituting $τ = t - s$ gives $u (x, t) = \int_{0}^{t} (t - s) M_{f (\cdot, s)} (x, c (t - s)) d s$ .

1.2F2F4algebra

Sphere-integral form. Writing the mean over $S^{2}$ as the normalised integral with $ω_{2} = 4 π$ from [F2], $u (x, t) = \frac{1}{4 π} \int_{0}^{t} (t - s) \int_{S^{2}} f (x + c (t - s) z, s) d σ (z) d s$ ; substituting $ρ = c (t - s)$ , so $s = t - ρ / c$ , $(t - s) = ρ / c$ and $d s = - d ρ / c$ , gives $u (x, t) = \frac{1}{4 π c^{2}} \int_{0}^{c t} ρ \int_{S^{2}} f (x + ρ z, t - ρ / c) d σ (z) d ρ$ .

2.1F3step 1.2algebra

Ball form. The source is bounded on the compact backward cone by A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value and For $n \geq 1$ , every Euclidean closed ball and every Euclidean sphere of positive radius is compact. The weight $∣ y - x ∣^{- 1}$ is integrable on $B_{c t} (x)$ , since [F3] gives its integral as $4 π \int_{0}^{c t} ρ d ρ < \infty$ . Give the integrand any value at $y = x$ , a null singleton. Apply [F3] separately to the positive and negative parts after translating by $x$ (Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation). This gives $\int_{B_{c t} (x)} f (y, t - ∣ y - x ∣ / c) ∣ y - x ∣^{- 1} d y = \int_{0}^{c t} ρ \int_{S^{2}} f (x + ρ z, t - ρ / c) d σ (z) d ρ$ . Multiplication by $1 / (4 π c^{2})$ identifies this with step 1.2.

3.1F1algebra∎

The integrand is evaluated at $∣ y - x ∣ = c (t - s)$ with $s = t - ∣ y - x ∣ / c \in [0, t]$ , that is on the backward light cone of $(x, t)$ , and the coefficient is $1 / (4 π c^{2})$ ; this is the retarded potential.

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Sphere-supported versus interior-supported free wave kernels

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $c > 0$ and let $u$ be the free solution assigned to admissible data $(u_{0}, u_{1})$ by the formulas of the page (Kirchhoff, Poisson, odd- and even-dimensional formulas). (i) If $n \geq 3$ is odd and $t > 0$ , the value $u (x, t)$ depends on the displacement and velocity data only through their restrictions to a neighbourhood of the sphere $\partial B_{c t} (x)$ : if two admissible data pairs agree on such a neighbourhood, their solutions agree at $(x, t)$ . Pointwise agreement only on the sphere is not asserted. (ii) If $n \geq 2$ is even, $t > 0$ and $u_{1}$ is supported in a compact subset of the open ball $B_{c t} (x)$ , then the velocity contribution has the kernel form $c^{1 - n} D_{t}^{k - 1} W_{u_{1}} (x, c t) = \int_{B_{c t} (x)} u_{1} (y) K (∣ y - x ∣, t) d y, K (ρ, t) : = c^{1 - n} D_{t}^{k - 1} [\frac{1}{n!! V_{n}} (c^{2} t^{2} - ρ^{2})^{- 1 / 2}]$ for $0 \leq ρ < c t$ , where $K$ is smooth on that region and $K (0, t) = a t^{- (n - 1)}$ with $a = c^{- n} (- 1)^{k - 1} (2 k - 3)!! / (n!! V_{n}) \neq 0$ (read $(2 k - 3)!! = 1$ for $k = 1$ ); consequently some data supported strictly inside $B_{c t} (x)$ give a nonzero velocity contribution, so the kernel fills the interior of the ball rather than sitting on the sphere. For displacement data also supported in a compact subset of the open ball, the displacement term has the corresponding kernel $\partial_{t} K (ρ, t)$ .

Facts & Assumptions

Given: Countable Choice, $c > 0$ , even $n = 2 k \geq 2$ , and velocity data $u_{1}$ supported in a compact subset of $B_{c t} (x)$ .

[F1]

The odd-dimensional formula expresses $u (x, t)$ as a finite combination of $t$ -derivatives of $t \mapsto t^{n - 2} M_{u_{0}} (x, c t)$ and $t \mapsto t^{n - 2} M_{u_{1}} (x, c t)$ (The odd-dimensional wave formula by iterated spherical means).

[F2]

For $m \geq 1$ and $h \in C^{m} (R^{n})$ , $(x, r) \mapsto M_{h} (x, r)$ is $C^{m}$ with all derivatives obtained by differentiating $h$ under the sphere integral (Smoothness, parity and zero-radius limits of spherical means).

[F3]

The even-dimensional formula reads $u = c^{1 - n} [\partial_{t} D_{t}^{k - 1} W_{u_{0}} (x, c t) + D_{t}^{k - 1} W_{u_{1}} (x, c t)]$ , with $W_{f} (x, c t) = (n!! V_{n})^{- 1} \int_{B_{c t} (x)} f (y) (c^{2} t^{2} - ∣ y - x ∣^{2})^{- 1 / 2} d y$ (The even-dimensional wave formula by descent, Spherical means and the weighted ball integral of space-dependent data).

[F4]

If $G (y, s)$ is continuous with continuous $\partial_{s} G$ on a compact rectangle, then $s \mapsto \int G (y, s) d y$ is $C^{1}$ with derivative $\int \partial_{s} G$ ; iterating gives the higher derivatives when they are continuous (Leibniz's rule on a compact rectangle: an interior parameter derivative with a continuous extension may be passed through a Riemann integral).

[F5]

Proof

1.1F1F2algebra

Part (i). By [F1] the value $u (x, t)$ is computed from the functions $r \mapsto M_{u_{0}} (x, r)$ and $r \mapsto M_{u_{1}} (x, r)$ and their $r$ -derivatives at $r = c t$ , and by [F2] these are obtained by differentiating the defining sphere integrals. For $r$ in a neighbourhood of $c t$ , $M_{f} (x, r)$ is the average of $f$ over $\partial B_{r} (x)$ , a sphere contained in the chosen neighbourhood of $\partial B_{c t} (x)$ ; hence all these quantities depend on $f$ only through its restriction to that neighbourhood, and so does $u (x, t)$ .

1.2F3F4algebra

Part (ii), kernel form. Fix $t > 0$ and let $u_{1}$ vanish on a neighbourhood of $\partial B_{c t} (x)$ ; the assumed compact support lies inside the open ball, so there is $ε \in (0, t)$ with $supp u_{1} \subseteq B_{c (t - ε)} (x)$ . For $∣ s - t ∣ < ε / 2$ the integrand $u_{1} (y) (c^{2} s^{2} - ∣ y - x ∣^{2})^{- 1 / 2}$ and all its $s$ -derivatives are continuous on a fixed box containing the support, with the product integrand extended by zero where $u_{1} = 0$ , for $s \in [t - ε / 2, t + ε / 2]$ , so by [F4] applied successively in each coordinate of the box, with Fubini (Fubini's theorem for L^1 functions on a sigma-finite product), the derivative $D_{t}^{k - 1}$ passes under the integral sign: $c^{1 - n} D_{t}^{k - 1} W_{u_{1}} (x, c t) = \int_{B_{c t} (x)} u_{1} (y) K (∣ y - x ∣, t) d y$ with $K (ρ, t) = c^{1 - n} D_{t}^{k - 1} [(n!! V_{n})^{- 1} (c^{2} t^{2} - ρ^{2})^{- 1 / 2}]$ . For displacement data $u_{0}$ with the same compact-interior support condition, one additional time differentiation under the fixed-box integral gives the kernel $\partial_{t} K (∣ y - x ∣, t)$ .

1.3F5algebra

The interior kernel. The identity $D_{t} (c^{2} t^{2} - ρ^{2})^{- p} = - 2 p c^{2} (c^{2} t^{2} - ρ^{2})^{- p - 1}$ gives by induction $K (ρ, t) = b (c^{2} t^{2} - ρ^{2})^{- (2 k - 1) / 2}$ , where $b = c^{1 - n} (- 1)^{k - 1} (2 k - 3)!! c^{2 k - 2} / (n!! V_{n}) \neq 0$ and $(- 1)!! = 1$ for $k = 1$ . Thus $K$ has a fixed nonzero sign throughout $0 \leq ρ < c t$ . In particular $K (0, t) = c^{- n} (- 1)^{k - 1} (2 k - 3)!! t^{- (n - 1)} / (n!! V_{n})$ , the stated value.

2.1F3step 1.2step 1.3algebra

Admissible interior data. Choose $0 < δ < c t / 2$ and, by A smooth bump between concentric Euclidean balls translated to centre $x$ , choose $u_{1} \in C_{c}^{\infty} (B_{2 δ} (x))$ with $0 \leq u_{1} \leq 1$ and $u_{1} = 1$ on ${\overline{B}}_{δ} (x)$ . This is admissible in every dimension here. The actual contribution is $\int_{B_{2 δ} (x)} u_{1} (y) K (∣ y - x ∣, t) d y$ , including the transition annulus. By step 1.3 its integrand has one sign and is strictly of that sign on the inner ball, of positive volume, so the contribution is nonzero. The same construction around any point strictly inside the ball shows that the interior kernel is nonzero throughout, rather than just at the centre.

3.1given∎

Collecting: in odd dimensions the value depends only on data near the sphere $\partial B_{c t} (x)$ (a statement about open neighbourhoods, not about pointwise traces), while in even dimensions the velocity kernel is the explicitly displayed smooth function of $ρ < c t$ , nonzero at the centre, so interior data contribute.

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The constructed classical solutions are locally determined by the Cauchy data

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $c > 0$ and let $u$ be a free solution constructed by the formulas of Kirchhoff's formula in three dimensions, Poisson's formula in two dimensions by descent, The odd-dimensional wave formula by iterated spherical means or The even-dimensional wave formula by descent from compactly supported admissible data. Then for every $(x, t)$ with $t > 0$ the value $u (x, t)$ is determined by the data restricted to ${\overline{B}}_{c t} (x)$ : two admissible data pairs agreeing there produce the same value at $(x, t)$ . For the forced solution of The forced three-dimensional version as a retarded potential, the value is determined by the source on the backward cone ${(y, s) : 0 \leq s \leq t, ∣ y - x ∣ \leq c (t - s)}$ ; two sources agreeing there produce the same value at $(x, t)$ .

Facts & Assumptions

Given: Countable Choice, $c > 0$ , a point $(x, t)$ with $t > 0$ , and two admissible configurations agreeing on the stated set.

[F1]

The four free formulas express $u (x, t)$ through the spherical means $M_{u_{j}} (x, c t)$ and their $r$ -derivatives (odd case) or through $W_{u_{j}} (x, c t)$ and its $t$ -derivatives with the substitution $y = x + c t z$ (even case), and the forced solution is the retarded potential (The odd-dimensional wave formula by iterated spherical means, The even-dimensional wave formula by descent, Kirchhoff's formula in three dimensions, Poisson's formula in two dimensions by descent, The forced three-dimensional version as a retarded potential).

[F2]

For $m \geq 1$ and $h \in C^{m} (R^{n})$ , the spherical mean $M_{h}$ and its derivatives are obtained by differentiating $h$ under the sphere integral (Smoothness, parity and zero-radius limits of spherical means).

[F3]

For even $n$ , writing $w (z) = (1 - ∣ z ∣^{2})^{- 1 / 2}$ on $B_{1} \subset R^{n}$ gives $W_{h} (x, c t) = (n!! V_{n})^{- 1} (c t)^{n - 1} \int_{B_{1}} h (x + c t z) w (z) d z$ , and $w$ is integrable (Spherical means and the weighted ball integral of space-dependent data, 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).

[F4]

Difference quotients converging pointwise almost everywhere under one integrable majorant have convergent integrals (Dominated convergence).

[F5]

A scalar function continuous on a closed interval and differentiable on its interior has a difference quotient equal to one of its derivatives on the interior (The mean value theorem, as the case $g (x) = x$ of Cauchy's: for $f$ continuous on $[a, b]$ with $a < b$ and differentiable on $(a, b)$ there is $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$ ).

[F6]

Continuous derivatives are bounded on compact Euclidean balls (For $n \geq 1$ , every Euclidean closed ball and every Euclidean sphere of positive radius is compact, A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value).

Proof

1.1F1F2F3F4F5F6algebra

Free case. Let $(u_{0}, u_{1})$ and $({\tilde{u}}_{0}, {\tilde{u}}_{1})$ be admissible data agreeing on ${\overline{B}}_{c t} (x)$ and put $δ_{j} : = u_{j} - {\tilde{u}}_{j}$ . Each $δ_{j}$ vanishes on the open ball, so all its derivatives through the orders in the formulas vanish on the closed ball by continuity. In the odd-dimensional formula, every sphere-average ingredient is an average of a derivative of $δ_{j}$ evaluated on $\partial B_{c t} (x)$ , hence is zero by [F1, F2]. For even $n$ , put $G_{δ_{j}} (t^{'}) : = \int_{B_{1}} δ_{j} (x + c t^{'} z) w (z) d z$ . Fix a compact interval $J$ about $t$ and a closed ball containing all $x + c t^{'} z$ for $t^{'} \in J$ , $∣ z ∣ \leq 1$ . For each derivative order $0 \leq m < k$ , the difference quotients in $t^{'}$ of $c^{m} D^{m} δ_{j} (x + c t^{'} z) [z, \dots, z] w (z)$ are bounded by $c^{m + 1} C_{m + 1} w (z)$ on $J$ , where $C_{m + 1}$ bounds the next derivative on that ball by [F5, F6]. This is an integrable majorant independent of $t^{'}$ ; [F4] therefore justifies differentiating under the integral successively through order $k$ . At $t^{'} = t$ , all integrand derivatives vanish because $x + c t z \in {\overline{B}}_{c t} (x)$ , so $G_{δ_{j}}^{(m)} (t) = 0$ for $m \leq k$ . By [F3] the even-formula terms are finite combinations of these derivatives and hence vanish. Thus replacing the data by $({\tilde{u}}_{0}, {\tilde{u}}_{1})$ changes no term of the formula and leaves $u (x, t)$ unchanged.

2.1F1given∎

Forced case and conclusion. The retarded potential of the forced three-dimensional formula is an integral of the source over the backward cone $∣ y - x ∣ \leq c (t - s)$ , $0 \leq s \leq t$ ; sources agreeing there give equal integrals, hence equal values at $(x, t)$ . This proves the local determination of the constructed solutions by the stated data or source; no uniqueness claim for arbitrary $C^{2}$ solutions is made, that being the energy statement of the wave-energy page.

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Time reversal of the homogeneous wave equation

Statement

Let $c > 0$ , let $I \subseteq R$ be an open interval and let $u \in C^{2} (R^{n} \times I)$ satisfy $□_{c} u = 0$ on $R^{n} \times I$ . For $τ \in I$ define $v (x, t) : = u (x, 2 τ - t)$ for $t \in 2 τ - I$ . Then $□_{c} v = 0$ on $R^{n} \times (2 τ - I)$ , and $v (\cdot, τ) = u (\cdot, τ), \partial_{t} v (\cdot, τ) = - \partial_{t} u (\cdot, τ) .$ Thus the homogeneous wave flow is reversible: the same equation propagates the time-reversed state, and the velocity is negated.

Facts & Assumptions

Given: an open interval $I$ , a parameter $τ \in I$ , a $C^{2}$ function $u$ on $R^{n} \times I$ with $□_{c} u = 0$ , and $v (x, t) = u (x, 2 τ - t)$ on $R^{n} \times (2 τ - I)$ .

[F1]

If $g$ is totally differentiable at $a$ and $f$ at $g (a)$ , then $f \circ g$ is totally differentiable at $a$ with $D (f \circ g) (a) = D f (g (a)) \circ D g (a)$ (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ ). The required total differentiability follows from continuous coordinate partial derivatives (If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative).

[F2]

Sums, products, constant multiples of differentiable functions are differentiable with the usual rules and derivatives; in particular the affine map $t \mapsto 2 τ - t$ has derivative $- 1$ (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 $g (c) \neq 0$ ).

Proof

1.1F1F2algebra

Apply [F1] to the composition $(x, t) \mapsto (x, 2 τ - t) \mapsto u (x, 2 τ - t)$ : the inner map is affine with differential $(h, s) \mapsto (h, - s)$ , so $\partial_{t} v (x, t) = - \partial_{t} u (x, 2 τ - t)$ ; applying the same rule once more, $\partial_{t}^{2} v (x, t) = (- 1)^{2} \partial_{t}^{2} u (x, 2 τ - t) = \partial_{t}^{2} u (x, 2 τ - t)$ . In the spatial directions the inner map is the identity, so $\partial_{x_{j}} v (x, t) = \partial_{x_{j}} u (x, 2 τ - t)$ for each $j$ and hence $Δ v (x, t) = Δ u (x, 2 τ - t)$ .

2.1F2algebra∎

Therefore for every $(x, t)$ with $t \in 2 τ - I$ , $□_{c} v (x, t) = \partial_{t}^{2} u (x, 2 τ - t) - c^{2} Δ u (x, 2 τ - t) = □_{c} u (x, 2 τ - t) = 0$ , while at $t = τ$ the substitutions $2 τ - τ = τ$ give $v (x, τ) = u (x, τ)$ and $\partial_{t} v (x, τ) = - \partial_{t} u (x, τ)$ . This is the reversibility of the homogeneous wave flow.

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Two different objects are called Poisson's formula

Remark

The name "Poisson's formula" denotes two different objects that should not be conflated. The two-dimensional wave formula of Poisson's formula in two dimensions by descent is the weighted disk integral $u (x, t) = \frac{1}{2 π c} [\frac{\partial}{\partial t} \int_{B_{c t} (x)} \frac{u_{0} (y) d y}{\sqrt{c^{2} t^{2} - ∣ y - x ∣^{2}}} + \int_{B_{c t} (x)} \frac{u_{1} (y) d y}{\sqrt{c^{2} t^{2} - ∣ y - x ∣^{2}}}],$ whose kernel $(2 π c)^{- 1} (c^{2} t^{2} - ∣ y - x ∣^{2})^{- 1 / 2}$ is an interior singular weight on the expanding disk, while the harmonic Poisson kernel of Poisson kernel of a Euclidean ball integrates boundary data against a positive density on the sphere. The two solve different problems — an initial-value (Cauchy) problem in space-time versus the Dirichlet boundary-value problem — use respectively a time parameter $t$ and a fixed radius $R$ . The shared name does not identify their kernels or transfer estimates between these problems.

The wave formula is stated for the speed- $c$ convention of Wave equation, Cauchy data and wave speed; the harmonic kernel is the ball boundary-value kernel of the Poisson-problem page. The remark asserts no new mathematics: it isolates the naming collision so that no consumer imports a boundary-value estimate into the wave representation.

5 · Examples, counterexamples and false statements

None yet.