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 Incoming links: Weighted acoustic Green-function reciprocity

Adjoint reciprocity for a weighted Helmholtz operator Created 2026-10-14 Updated 2026-11-05

For smooth positive

f

, let

L = f^{- 1} \nabla \cdot (f \nabla) + k^{2}

. Its unweighted bilinear transpose is the displayed operator. If

v

satisfies

αv + β \partial_{n} v = 0

, its transpose boundary condition is

αu + β (\partial_{n} u - u \partial_{n} lo g f) = 0

. At parameters for which the corresponding boundary-value inverses exist, their Green kernels satisfy

G_{L} (r_{2}, r_{1}) = G_{L^{T}} (r_{1}, r_{2})

. Equivalently

G_{L^{T}} (r, r_{2}) = f (r) G_{L} (r, r_{2}) / f (r_{2})

, exhibiting weighted acoustic Green-function reciprocity.

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Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 88 / 2 / b / Solution 2026-10-14

Let

G_{1} = ϕ (r ∣ r_{1})

and

G_{2} = ϕ (r ∣ r_{2})

. Green's second identity gives

\int_{Ω} [G_{1} (Δ + k_{0}^{2}) G_{2} - G_{2} (Δ + k_{0}^{2}) G_{1}], d V = \int_{S} (G_{1} \partial_{n} G_{2} - G_{2} \partial_{n} G_{1}), d S .

The common Robin boundary condition makes the surface integrand vanish. Where

β \neq = 0

, substitute

\partial_{n} G_{i} = - α G_{i} / β

; a Dirichlet portion also contributes zero. The delta sources therefore give the Helmholtz Green-function reciprocity

ϕ (r_{1} ∣ r_{2}) = ϕ (r_{2} ∣ r_{1}) .

This is a bilinear identity, with no complex conjugation. Assume the boundary-value problem has a unique Green kernel; in an exterior problem use the same outgoing radiation condition for both kernels.

For the weighted operator, assume smooth positive

f

and set

b = \nabla lo g f

. Then

L = \frac{1}{f} \nabla \cdot (f \nabla) + k_{0}^{2} = Δ + b \cdot \nabla + k_{0}^{2} .

Its unweighted transpose, obtained by integration by parts, is

L^{T} u = Δ u - \nabla \cdot (b u) + k_{0}^{2} u = \nabla \cdot [f \nabla (u / f)] + k_{0}^{2} u .

The associated boundary identity is

\int_{Ω} (uLv - v L^{T} u), d V = \int_{S} [u \partial_{n} v - v \partial_{n} u + (\partial_{n} lo g f) uv], d S .

To cancel it when

v

satisfies the given Robin condition, the required adjoint kernel satisfies

\nabla \cdot [f \nabla (\frac{ψ}{f})] + k_{0}^{2} ψ = δ (r - r_{2}),

α ψ + β (\partial_{n} ψ - ψ \partial_{n} lo g f) = 0.

Using these two delta sources in the boundary identity proves

ψ (r_{2} ∣ r_{1}) = ψ (r_{1} ∣ r_{2})

, the adjoint reciprocity for a weighted Helmholtz operator. As a check,

L^{T} (f v) = f Lv

, so

ψ (r ∣ r_{2}) = \frac{f ( r )}{f ( r _{2} )} ψ (r ∣ r_{2}) .

This is equivalent to weighted acoustic Green-function reciprocity,

f (r_{2}) ψ (r_{2} ∣ r_{1}) = f (r_{1}) ψ (r_{1} ∣ r_{2})

. Unit delta normalization is essential here. A Hermitian adjoint would instead conjugate coefficients and change the radiation convention; the requested equality uses the transpose.

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Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 75 / 1 / c / Solution Created 2026-10-03 Updated 2026-10-06

Retain

a = p_{0}^{1/ γ} / ρ_{0}

from part (b), and use time translation invariance to write

G (x, t; y, τ) = g (x, y; t - τ)

. Define the temporal Fourier transform by

G (x, y; ω) = \int_{R} g (x, y; s) e^{- iω s} d s .

Multiplying the transformed Green function equation by

- a (x)

gives

\nabla \cdot (a \nabla G) + \frac{a ω ^{2}}{c _{0}^{2}} G = - a (y) δ (x - y) .

This is a symmetric divergence-form spatial operator, although the original unweighted operator need not be symmetric in the ordinary volume measure.

Let

G_{j} (x) = G (x, y_{j}; ω)

. The product rule gives the bilinear Green second identity

\nabla \cdot {a (G_{1} \nabla G_{2} - G_{2} \nabla G_{1})} = G_{1} \nabla \cdot (a \nabla G_{2}) - G_{2} \nabla \cdot (a \nabla G_{1}) .

The frequency terms cancel. Integrating over the domain, the right side becomes

- a (y_{2}) G (y_{2}, y_{1}; ω) + a (y_{1}) G (y_{1}, y_{2}; ω) .

The boundary integral is zero for common homogeneous Dirichlet boundary condition, Neumann boundary condition or reciprocal Robin boundary condition conditions. In an unbounded domain, use the same outgoing limiting-absorption prescription for both Green functions; it gives the corresponding vanishing boundary pairing. This identity has no complex conjugation: it proves transpose wave reciprocity, not a Hermitian or time-reversal identity. We obtain the weighted acoustic Green-function reciprocity

G (y_{1}, y_{2}; ω) = \frac{a ( y _{2} )}{a ( y _{1} )} G (y_{2}, y_{1}; ω) .

The factor is frequency independent, so inverse Fourier transform gives the same relation at equal time lag. Both time arguments below have lag

t - τ_{2}

, hence

G (y_{1}, t; y_{2}, τ_{2}) = G (y_{2}, t + τ_{1} - τ_{2}; y_{1}, τ_{1}) \frac{[ p _{0} ( y _{2} ) ] ^{1/ γ} ρ _{0} ( y _{1} )}{[ p _{0} ( y _{1} ) ] ^{1/ γ} ρ _{0} ( y _{2} )} .

Reciprocity exchanges source and receiver while preserving elapsed time; it does not turn a causal response into an advanced one.

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Stratified acoustic pressure equation Created 2026-10-06 Updated 2026-11-05

Consider adiabatic perturbations of a static ideal gas, with background pressure

p_{0} (x)

ρ_{0} (x)

and constant heat-capacity ratio

γ

. Assume a prescribed force density

- \nabla χ

that does not change when the fluid is perturbed. Static balance is

\nabla (p_{0} + χ) = 0

. Linearized Euler equations and adiabatic compression then give

ρ_{0} u_{t} = - \nabla p^{'}, p_{t}^{'} = - u \cdot \nabla p_{0} - γ p_{0} \nabla \cdot u .

Differentiating the second equation and substituting the first yields

c_{0}^{- 2} p_{tt}^{'} - a^{- 1} \nabla \cdot (a \nabla p^{'}) = 0, c_{0}^{2} = γ p_{0} / ρ_{0}, a = p_{0}^{1/ γ} / ρ_{0} .

The divergence form gives weighted acoustic Green-function reciprocity. Neither uniform mass density nor uniform entropy is required. A prescribed gravitational acceleration instead produces a force density proportional to the perturbed mass density; its extra buoyancy term cannot be omitted under this derivation. The homogeneous limit is the acoustic pressure wave equation.

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