Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 50 (Acoustics and Stability)

Take $\omega >0$, with time convention $e^{-i\omega t}$, and use the Fourier transform pair $\hat{h}(k)=\int h(x)e^{-ikx}dx$, $h(x)=(2\pi {)}^{-1}\int \hat{h}(k)e^{ikx}dk$. For an acoustic velocity potential, $\mathbf{u}=\nabla \varphi$ and $p=-{\rho }_0{\varphi }_t$. Its harmonic amplitude solves the Helmholtz equation. The outgoing Fourier component has the formThe outgoing acoustic square-root branch is fixed by decay for evanescent components and upward radiation for propagating components:A precise causal prescription is to replace $\omega$ by $\omega +i\sigma$, $\sigma >0$, select $\mathrm{Re}\gamma >0$ for real $k$, invert on the real $k$ axis, and take $\sigma \downarrow 0$. It corresponds to switching on the forcing from the remote past. The limiting inversion contour $C$ passes below the positive outgoing surface-wave pole and above the negative one. The branch points are approached with the same causal prescription; analytic continuation during contour deformation must remain on this sheet. A principal real square root with positive imaginary part inside the acoustic interval would produce incoming sound and the wrong solution.

The elastic membrane's normal velocity equals the fluid velocity at the boundary, so $-\gamma A=-i\omega \hat{\eta }$. ConsequentlyThe elastic membrane balance therefore givesWriting $D=(T{k}^2-m{\omega }^2)\gamma -{\rho }_0{\omega }^2$, the one-sided fluid-loaded membrane radiation formulas areThe sign of the fluid term describes positive added mass of an evanescent fluid layer, not negative inertia.

For $0<\theta <\pi$, the acoustic saddle point is $k_s=k_0\cos \theta$, with ${\gamma }_s=-i{k}_0\sin \theta$. Applying the stated method of steepest descent result givesSince $p=i{\rho }_0\omega \varphi$ in amplitude notation, the angular acoustic directivity of the pressure, apart from angle-independent factors, isThe intensity directivity is proportional to ${\mathcal{D}}^2$. This cylindrical radiating field has amplitude of order $r^{-1/2}$. The formula is for fixed interior angles; grazing-angle limits require a separate uniform approximation.

For real $k>k_0$, $\gamma$ is positive, $D\to -{\rho }_0{\omega }^2$ as $k\downarrow k_0$, and $D\to +\infty$ as $k\to \infty$. A root exists. A root also requires $T{k}^2-m{\omega }^2>0$. In that range,so the root is unique. Evenness supplies the second root, $-k_{\ast }$. There are no real roots inside $|k|<k_0$: the fluid-loading term is real and nonzero while the other term is purely imaginary. Thus there are exactly two real physical-sheet roots, $\pm k_{\ast }$, for nonzero forcing frequency, positive elastic-sheet tension and positive fluid mass density. The zero-frequency static limit is not a two-distinct-root statement.

These roots are counterpropagating evanescent acoustic surface waves coupled to the membrane:Their normal profiles decay as $e^{-{\gamma }_{\ast }y}$, and their phase speed is below both $c_0$ and the vacuum membrane speed. They matter for the elastic membrane motion and for observations near its surface, but are not automatically part of the leading far-field radiation. A residue is picked up only when the causal contour deformation crosses its pole; at fixed $y/r>0$ its magnitude is exponentially small, $e^{-{\gamma }_{\ast }r\sin \theta }$.

For example, in the right half-plane put $k=k_0\cos \zeta$, $\gamma =-i{k}_0\sin \zeta$ and $k_{\ast }=k_0\cosh a$. The pole is at $\zeta =ia$ and the saddle at $\zeta =\theta$. The steepest-descent path has $\mathrm{Re}\cos (\zeta -\theta )=1$. At the pole height its left branch has real coordinate $\theta -\arccos (\mathrm{sech}a)$; comparing it with the original contour just to the right of the pole shows that its residue is crossed forThe analogous sector lies near $\theta =\pi$ for the negative pole. This is why the existence of real dispersion zeros alone does not make their residues relevant in every observation direction.

Let the perturbation velocity and pressure be $(u,w,p)(z)e^{ik(x-ct)}$, with real $k>0$. Linearizing the Euler equations about the inviscid parallel shear flow givesUse $u={\varphi }^{\prime }$, $w=-ik\varphi$ to satisfy incompressibility. The first momentum equation gives $p/{\rho }_0=-(U-c){\varphi }^{\prime }+U^{\prime }\varphi$; differentiate it and use the second equation. The Rayleigh equation for inviscid shear flow follows:A growing temporal normal mode has $c_i=\mathrm{Im}c>0$ and growth rate $k{c}_i$.

Rayleigh's inflection-point theorem states that such an instability of a smooth parallel profile requires $U^{\prime \prime }$ to change sign somewhere. To prove it, divide by $U-c$, multiply by ${\varphi }^{\ast }$, integrate over $z$ and integrate the second derivative by parts. Decay removes the endpoint terms, givingTaking the imaginary part yieldsThe weighting is nonnegative. A one-signed $U^{\prime \prime }$ that is not identically zero cannot satisfy this identity for a nontrivial mode; otherwise the mode vanishes on an interval and uniqueness of the differential equation forces it to vanish throughout. If $U^{\prime \prime }\equiv 0$, the real part of the identity is already a strictly positive integral for a nontrivial decaying mode. This proves the necessary sign-change condition.

Fjørtoft's theorem strengthens this: if $U_s$ is the velocity at an inflection point, instability requires $U^{\prime \prime }(U-U_s)<0$ somewhere. This is a necessary condition, not a sufficient criterion. For $U=\tanh z$, $U_s=0$ andThe possibility of instability is therefore consistent with Fjørtoft's theorem.

For the piecewise-linear approximation, $U^{\prime \prime }=0$ within each of the three regions. The decaying solutions outside and the two independent interior solutions can be writtenAt each corner, the normal velocity is continuous, so $\varphi$ is continuous. Pressure matching at a piecewise-linear shear interface additionally gives $[(U-c){\varphi }^{\prime }-U^{\prime }\varphi ]=0$, hence $(U-c)[{\varphi }^{\prime }]=[U^{\prime }]\varphi$. At $z=1$ the slope jumps from $1$ to $0$; at $z=-1$ it jumps from $0$ to $1$. Substituting the decaying outer solutions givesThus the inviscid instability of a piecewise-linear mixing layer has dispersion relationAt $k=1/2$, $c=\pm i/e$, so the root with positive imaginary part is an explicitly growing mode, with growth rate $1/(2e)$. More generally the unstable interval is $0<k<k_c$, where the unique $k_c>1/2$ solves $2k_c-1=e^{-2k_c}$. This direct calculation establishes instability of the approximation; passing a necessary inflection test alone would not have established it.

A singular perturbation is one for which setting the parameter to zero does not give a regular approximation uniformly over the region of interest. The limiting equation may lose a derivative or boundary condition, or a formally small correction may accumulate over a long time. Here the last mechanism occurs.

At fixed $t$, substitute a regular asymptotic expansion. The leading equations and initial conditions give $y_0=3$ and $x_0=\sin t$. At first order,with zero initial corrections. ThereforeThe secular terms become comparable with the leading terms at $t=O({ϵ}^{-1})$. The fixed-time expansion is not uniform on the slow relaxation time.

Introduce $\tau =ϵt$ through the method of multiple scales, and write $x=A(\tau )\sin t+O(ϵ)$, $y=Y(\tau )+O(ϵ)$. Average the second equation over the rapid oscillation. Since the sine has zero mean, $Y^{\prime }=1-Y$, where the prime now denotes $d/d\tau$. With $Y(0)=3$, $Y=1+2e^{-\tau }$.

In the oscillator equation, $2(y^{\prime }{)}^2$ is of order ${ϵ}^2$. The order-$ϵ$ resonant cosine forcing is $-(2A^{\prime }+3YA)\cos t$. A bounded first correction requires its coefficient to vanish, soIntegration gives the slow oscillator damping by a relaxing auxiliary variable:Hence the uniform leading solution isTo check the uniform ordering, the instantaneous sine forcing in the auxiliary equation is absorbed by a bounded first correction $-ϵA(\tau )\cos t$, plus a bounded slow correction chosen to match the initial data. The resonant first-order oscillator forcing has already been removed. Remaining order-${ϵ}^2$ terms accumulate by at most $O(ϵ)$ over $0\le t\le K/ϵ$ for fixed $K$. The slow coefficients stay bounded on this interval, and variation of constants for the oscillator and the auxiliary first-order equation gives the displayed uniform errors. Expanding the amplitude near $\tau =0$ gives $A=1-9\tau /2+\cdots$, recovering the secular term of the failed regular expansion.

For the positive-drift equation, the outer expansion obeys $(1+x)y_0^{\prime }+y_0=0$. Select the boundary condition at the right endpoint, obtaining $y_0=2/(1+x)$. The first correction satisfiesIntegrating $[(1+x)y_1{]}^{\prime }=-4/(1+x{)}^3$ givesThe boundary condition mismatch is at $x=0$. Set $X=x/ϵ$ and use the inner expansion $Y=Y_0+ϵY_1+\cdots$. The inner equation is $Y_{XX}+(1+ϵX)Y_X+ϵY=0$. At leading order, matching to $2$ and imposing $Y(0)=1$ give $Y_0=2-e^{-X}$. At first order,The outer expansion in the overlap is $2+ϵ(3/2-2X)$. Matching its constant and linear terms givesThe additive composite expansion adds inner and outer approximations and subtracts the common overlap:It meets the left condition exactly and the right condition up to exponentially small terms, with an $O({ϵ}^2)$ uniform asymptotic error. This is the first-order composite expansion for a positive drift specialized to the present coefficients.

Reversing the drift moves the decaying boundary layer to $x=1$; the outer solution is selected at $x=0$. For the second equation, $y_{\mathrm{out}}=1+x$ is in fact an exact solution, but predicts $2$ at the right boundary instead of $1+ϵ$. Put $X=(1-x)/ϵ$. The inner equation becomesMatching to $2-ϵX$ givesIndeed, $Y_1^{\prime \prime }+2Y_1^{\prime }=-2+(1+2X)e^{-2X}$ and $Y_1(0)=1$, as required by the order-$ϵ$ boundary value. Since $d/dx=-{ϵ}^{-1}d/dX$, $Y_0^{\prime }(0)=2$ and $Y_1^{\prime }(0)=-4$ give the endpoint derivative in a negative-drift boundary layer:The order-one term cannot be obtained by differentiating the leading inner solution alone.

Write $L=\log \psi$, with $\psi >0$. Under the Cole-Hopf transformation $q=2ϵL_{\theta }$, direct differentiation givesSince $L_{\theta \theta }+L_{\theta }^2={\psi }_{\theta \theta }/\psi$, Burgers' equation is equivalent toThe ratio is therefore a function $a(Z)$ alone. Multiplying $\psi$ by $\exp [-\int a(Z)dZ]$ leaves $q$ unchanged and removes this scalar freedom. Thus one may choose the normalization for whichThis is the heat equation. The positive sign in $q=2ϵ{\partial }_{\theta }\log \psi$ corresponds to the negative conservative flux $-q^2/2$ used here; the more usual positive-flux convention has the opposite sign.

Integrate the initial Cole-Hopf transformation and choose a continuous positive normalization:Split the heat kernel convolution at zero, and complete the square in the positive half-line contribution. DefineThe resulting heat equation solution isDifferentiating the moving endpoints gives ${\partial }_{\theta }{I}_{-}=-e^{-{\theta }^2/(4ϵZ)}$ and $E{\partial }_{\theta }{I}_{+}=e^{-{\theta }^2/(4ϵZ)}$. These terms cancel. Hence ${\psi }_{\theta }=(4\pi ϵZ{)}^{-1/2}UE{I}_{+}/(2ϵ)$, and the viscous Burgers step solution with negative flux isThe lower limit of $I_{-}$ is $\theta$, as printed in the original PDF. The converted TeX's $-\theta$ does not give the correct convolution or endpoint cancellation.

For the vanishing-viscosity limit, the conservative flux is $F(q)=-q^2/2$ and the characteristic speed is $F^{\prime }(q)=-q$. If $U>0$, the characteristics from the right move left and collide with those on the left. The Rankine-Hugoniot condition gives shock speed $[F]/[q]=-U/2$. ThusThe viscous transition has width $O(ϵ/U)$, and its centre value is exactly $U/2$ by the midpoint symmetry of a viscous Burgers step.

If $U<0$, the right characteristics move right and separate from the left ones. The limit is a rarefaction wave,This follows either from the Burgers Riemann problem with negative flux or the large-argument Gaussian-tail asymptotics of the exact formula. The two signs therefore give a shock and a spreading rarefaction respectively, not two shocks.

Set $s=Z+c\theta$ and write $q=Q(s)$. Substitution into the modified Burgers equation with cubic flux givesAfter one integration,A regular constant limiting state has $Q^{\prime }\to 0$. The state $Q=0$ at the negative end sets $C=0$, and the nonzero state $Q=\beta$ at the positive end givesThus the first-order equation is $ϵc^2{Q}^{\prime }=Q(1-Q^2/{\beta }^2)$. Put $W=Q^2$. Its equation is logistic,The heteroclinic travelling wave connecting the prescribed endpoints is consequentlywhere $s_0$ is an arbitrary translation. Taking the sign from $\beta$ is essential; squaring the equation alone loses the negative-front solution. A fixed level propagates with $d\theta /dZ=-1/c=-{\beta }^2/3$, also obtained from the Rankine-Hugoniot condition for flux $-q^3/3$.

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.