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 Incoming links: Helmholtz Green-function reciprocity

Let

G_{1} = ϕ (r ∣ r_{1})

and

G_{2} = ϕ (r ∣ r_{2})

\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 .

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})

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

, so

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

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.