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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.

Depends on

Used by

The constructed classical solutions are locally determined by the Cauchy data Corollary

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Sources

Victor Ivrii, Partial Differential Equations (University of Toronto, 2018, CC BY-SA) (standard reference, not scraped)
Gerald Teschl, Partial Differential Equations: From Classical to Modern (2025 archived author manuscript, AMS Graduate Studies in Mathematics) (standard reference, not scraped)
John K. Hunter, Notes on Partial Differential Equations (revised 18 June 2014, UC Davis) (standard reference, not scraped)