Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 82 / 1

Put $r=|\mathbf{x}|$, $\mathbf{n}=\mathbf{x}/r$, and take a compactly supported Lighthill stress tensor. The isotropic term in that tensor is $(p-c_0^2\rho ){\delta }_{ij}$: the Kronecker delta is implicit in the printed shorthand. Convolving the Lighthill acoustic analogy with the retarded acoustic Green function and integrating twice by parts in the source coordinates givesThe boundary terms vanish because of compact support. For source size $\ell$, the acoustic compact-source approximation requires $\omega \ell /c_0\ll 1$. At $r\gg \ell$, we may therefore replace the denominator by $r$ and the retarded argument by $t-r/c_0$ in the leading source integral. Define $S_{ij}(t)=\int T_{ij}(\mathbf{y},t)d^{3}y$. In the radiation region $\omega r/c_0\gg 1$, the leading two spatial derivatives act on the retarded time rather than on $1/r$ or $\mathbf{n}$:Thus the leading acoustic quadrupole density isFor the compact acoustic quadrupole Mach-number scaling, let $U$ be the source fluctuation speed, $m=U/c_0\ll 1$, and use the advective source time $\ell /U$. With $T_{ij}=O({\rho }_0{U}^2)$, one has $S_{ij}=O({\rho }_0{U}^2{\ell }^3)$ and ${\ddot{S}}_{ij}=O({\rho }_0{U}^4\ell )$. ConsequentlyThe fourth power refers to the radiation coefficient with geometric spreading separated. It assumes that the source frequency scales as $U/\ell$; an independently imposed frequency or an independently strong thermal source would change that Mach number estimate.

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.

Let $D_{ij}=T_{ij}^{+}-T_{ij}^{-}$, so the Lighthill stress tensor is $T_{ij}^{-}+D_{ij}H(S)$. Write $S_i={\partial }_{i}S$ and $S_{ij}={\partial }_i{\partial }_{j}S$. The distributional derivative of the Heaviside step function gives ${\partial }_{i}H(S)=S_i\delta (S)$. Applying the product rule a second time yields the surface sources of a discontinuous Lighthill stress tensor:Repeated indices are summed. The first line gives the two bulk acoustic quadrupole distributions. The second line consists of additional surface delta distributions and their derivatives, all supported on the shock wave $S=0$. For a symmetric Lighthill stress tensor, the first two coefficients of $\delta (S)$ combine to $2D_{ij,i}{S}_j$ after relabelling indices.

A discontinuity therefore adds both a surface-delta source and a surface-delta-derivative source. The level function need not be a signed distance: retaining its derivatives makes the expression invariant under a smooth reparametrization of the same surface. If the two bulk fields are only continuously differentiable, their second derivatives in the first line are read weakly; no classical second derivative is being assumed. The coefficients multiplying ${\delta }^{\prime }(S)$ are kept as extensions, not replaced prematurely by traces.

Use the jump convention $[f]=f_1-f_2$, consistent with region 1 occupying $x>Vt$. The one-dimensional Lighthill stress tensor is $T=\rho u^2+p-c_0^2\rho$, so $T=T_2+[T]H(x-Vt)$. Both bulk stresses are constant. The distributional jump formula for a Heaviside product therefore givesThe Rankine-Hugoniot condition for conservation of mass is $[\rho u]=V[\rho ]$, while conservation of momentum gives $[\rho u^2+p]=V[\rho u]$. Hence a physical moving shock wave also satisfiesThis sign can be checked directly. Subtracting a constant reference density does not affect derivatives, and ${\rho }^{\prime }=\text{constant}+[\rho ]H(x-Vt)$. Thus ${\rho }_{tt}^{\prime }=V^2[\rho ]{\delta }^{\prime }(x-Vt)$ and ${\rho }_{xx}^{\prime }=[\rho ]{\delta }^{\prime }(x-Vt)$, exactly reproducing the coefficient above.