 Solution

Use time dependence $e^{i\omega t}$ and write the outgoing two-dimensional Helmholtz equation kernel as $G_2\sim C{e}^{-i{k}_{0}r}/\sqrt{k_{0}r}$, where $C$ is independent of the Mach number. The harmonic Lighthill acoustic analogy givesFor a compact acoustic quadrupole, the source integral is $O({\rho }_0{U}^2{\ell }^2)$. In the radiation region each derivative of the outgoing exponential supplies a factor of $k_0$, soKeeping the same advective source-frequency scaling as in (i), $\omega =O(U/\ell )$ and hence $k_0\ell =O(m)$. Therefore the two-dimensional compact quadrupole scaling isThe changed exponent comes from cylindrical spreading $(k_{0}r{)}^{-1/2}$. Compactness $k_0\ell \ll 1$ and the far-field condition $k_{0}r\gg 1$ are separate assumptions. At fixed imposed $\omega$ and fixed $\ell$, with only $U$ varied, the stress instead gives an $O(m^2)$ dependence; monochromaticity alone does not supply the additional source-frequency assumption.