 Solution

Retain $a=p_0^{1/\gamma }/{\rho }_0$ from part (b), and use time translation invariance to write $G(\mathbf{x},t;\mathbf{y},\tau )=g(\mathbf{x},\mathbf{y};t-\tau )$. Define the temporal Fourier transform byMultiplying the transformed Green function equation by $-a(\mathbf{x})$ givesThis is a symmetric divergence-form spatial operator, although the original unweighted operator need not be symmetric in the ordinary volume measure.

Let $G_j(\mathbf{x})=\tilde{G}(\mathbf{x},{\mathbf{y}}_j;\omega )$. The product rule gives the bilinear Green second identityThe frequency terms cancel. Integrating over the domain, the right side becomesThe 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 reciprocityThe factor is frequency independent, so inverse Fourier transform gives the same relation at equal time lag. Both time arguments below have lag $t-{\tau }_2$, henceReciprocity exchanges source and receiver while preserving elapsed time; it does not turn a causal response into an advanced one.