Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 50 / 5

Use the conservative continuity equation and momentum balance, with ${\sigma }_{ij}$ the viscous stress:Differentiate the first equation in time and take the divergence of the second. Eliminating momentum givesSubtract $c_0^2\Delta \rho$, and define the Lighthill stress tensor byThe constant reference subtraction has no double divergence. The exact Lighthill acoustic analogy isThis is an exact rearrangement; linearization is not used to eliminate the nonlinear source tensor.

Convolving with the retarded acoustic Green function and transferring the source derivatives by integration by parts givesThe acoustic compact-source approximation requires source size $\ell$ to satisfy $\Omega \ell /c_0\ll 1$ for its characteristic frequency $\Omega$; the acoustic far field additionally requires $r\gg \ell$ and $\Omega r/c_0\gg 1$. Internal retardation is then negligible at leading order, and the radiating parts of the two observer derivatives act on the common retarded time. With $n_i=x_i/r$ and $S_{ij}=\int T_{ij}{d}^{3}y$, the far-field acoustic force and stress moments giveThe two spatial derivatives supply two factors of $1/c_0$, in addition to the $1/c_0^2$ in the Green function.

For a subsonic fluctuating source of velocity scale $V$, size $\ell$ and turnover time $\ell /V$, suppose its stress scale is $T_{ij}=O({\rho }_0{V}^2)$. Then $S_{ij}=O({\rho }_0{V}^2{\ell }^3)$ and ${\ddot{S}}_{ij}=O({\rho }_0{V}^4\ell )$. The compact acoustic quadrupole Mach-number scaling is thereforeThis fourth-power statement concerns the acoustic density amplitude under the specified eddy-time and stress scaling. With an independently imposed frequency, the general estimate instead retains $V^2{\Omega }^2{\ell }^3/c_0^{4}r$.

With a mass source, the continuity equation becomes ${\rho }_t+{\partial }_i(\rho u_i)=M$. If the conservative momentum equation has no added source, repeating the elimination gives the mass-injection term in the acoustic analogy:More generally, injection can carry momentum. If its momentum source is $S_i$, the additional term is ${\partial }_{t}M-{\partial }_i{S}_i$; prescribing $M$ alone does not determine this dipole contribution. The displayed monopole formula assumes no separately imposed momentum source.

Let $\mathcal{M}(t)=\int M(\mathbf{y},t)d^{3}y$ be the total mass rate. Convolution with the retarded acoustic Green function, followed by the acoustic compact-source approximation, gives the extra acoustic monopoleThe stress quadrupole from the preceding part remains present. If required, a compact injected momentum rate ${\mathcal{S}}_i=\int S_i{d}^{3}y$ adds the dipole $n_i{\dot{\mathcal{S}}}_i/(4\pi c_0^{3}r)$.

A small pulsating bubble displaces liquid with volume flux $\dot{\mathcal{V}}(t)$ and acts acoustically like a mass rate $\mathcal{M}={\rho }_0\dot{\mathcal{V}}$. Spherical symmetry makes this a monopole, withFor volume oscillation $\mathcal{V}={\mathcal{V}}_0+\Delta \mathcal{V}\cos \omega t$, the density amplitude is ${\rho }_0{\omega }^2|\Delta \mathcal{V}|/(4\pi c_0^{2}r)$. At fixed volume-displacement amplitude, it scales as ${\omega }^2$. Holding volume-flux amplitude fixed would instead give an $\omega$ scaling. The bubble must remain small compared with the wavelength, and a frequency-dependent dynamical bubble amplitude is a separate effect.

On a region with smooth positive area, set $q=G(Z)Q(\theta ,\zeta )$, $\zeta =h(Z)$. Substitution into the simple wave equation givesRemove the last term by $G^{\prime }/G=-A^{\prime }/(2A)$, and choose $h^{\prime }=G$. Starting at a regular section $Z_0$, one convenient normalized area transformation of a nonlinear acoustic simple wave isIn the notation $g(\zeta )$, set $g(\zeta )=G(h^{-1}(\zeta ))$. The remaining equation is $Q_{\zeta }-Q{Q}_{\theta }=0$. With initial value $f$ at $Z_0$, the method of characteristics givesThe map from the initial label loses invertibility when $1-f^{\prime }(\xi )\zeta =0$. For smooth data with bounded derivative and $M=\sup f^{\prime }>0$, the first forward characteristic crossing is at ${\zeta }_s=1/M$; if $M\le 0$, no forward shock occurs.

For spherical spreading, write physical radius as $R>0$ and specify data at an emitting radius $R_0>0$. Since $A(R)=R^2$,The shock distance for a spherical simple wave launched at finite radius isIf the axial distance is measured from the emitting surface, $Z=R-R_0$, then $A(Z)=(R_0+Z{)}^2$ andNo finite shock distance exists in this model when $M\le 0$.

The printed combination $A=Z^2$ with general bounded data at $Z=0$ is singular and does not define the intended outgoing spherical initial-value problem. Along a spherical characteristic, $d\theta /dR=-q$ and $dq/dR=-q/R$, so $Rq$ is constant. A bounded trace at $R=0$ forces that constant to be zero for every characteristic curve emanating from the origin; arbitrary nonzero bounded $f$ cannot be launched there. For example, the ostensibly admissible data $f\equiv 1$ would demand both $Rq=0$ and a nonzero trace, a contradiction. The logarithmic transformed distance is also divergent at the origin. Thus an unqualified numerical shock formula from those literal data would be unjustified. The finite-radius formulas above give the physical interpretation once the launching radius, or the shifted area law, is supplied.