Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 345 4 d Solution Created 2026-10-03 Updated 2026-10-05
Eliminate the transport amplitudes in the linear stability analysis. For a general bed shear response , the dispersion relation isThis is the most specific result available without adding the missing hydrodynamic closure. With , setMultiplication by gives the growth rate and migration velocityThe packed-bed volume fraction cancels. For a real physical closure the negative wavenumber response is the complex conjugate; the displayed formulas cover the independent normal modes.
If , are treated as constants, the upstream shear stress phase lead promotes growth, while the inclined-bed sediment threshold reduces it and the finite saturation length delays sediment transport. There is an unstable band precisely when : . Within this band , so the growing bedforms migrate downstream. The fastest normal mode satisfiesThis last maximization assumes constant ; if the hydrodynamic coefficients vary with wavenumber, their derivatives also enter the selection condition. When and , all nonzero normal modes decay.
Just above the sediment entrainment threshold,Thus the common regime is stable near onset: the slope penalty overwhelms the hydrodynamic phase lead. This conclusion is conditional on that inequality, not universal. For constant , instability begins only when as well as . The transport sensitivity vanishes at onset for , is finite for , and is singular for . Exactly at threshold, the positive-part transport law needs separate treatment; the above linearization with perturbations small compared with is unavailable.
Far above the sediment entrainment threshold, . If , the unstable cutoff tends to , and the dominant wavelength is set by the saturation length. Both growth rate and migration velocity scale with , apart from their length factors. Their numerical values and any detailed dependence on wavenumber still require a flow closure.
For example, prescribing uniform shear stress independently of the bed gives andThis equally admissible closure illustrates why instability cannot be asserted from the printed sediment equations alone.
The plots use constant and three illustrative ratios . They show how the sign of the shear phase lead minus the slope correction determines stability; they are not numerical predictions for unspecified flow conditions.
Past exam of the mathematics course of the University of Cambridge 2018 ii Paper 2 38C Solution Created 2026-09-24 Updated 2026-10-05
Take ; reflection in supplies the result for negative wavenumber. Since each region has uniform base velocity, the perturbation is an incompressible, irrotational flow with a velocity potential satisfying the Laplace equation. Its decaying outer forms areFor a varicose mode of a planar jet, reflection symmetry makes the inner potential proportional to ; for a sinuous mode of a planar jet, it is proportional to .
At the upper interface, the linearized kinematic boundary condition isThe first equation gives . The second givesThe linearized Unsteady Bernoulli equation gives . Applying pressure continuity therefore produces the equal-density top-hat planar-jet dispersion relationSolving the quadratic equation givesEquivalently,The plus sign is the unstable Kelvin-Helmholtz instability branch. In particular, the two modes have exactly the same growth rate,Its limiting forms areThus neither type grows faster in this equal-density idealization, although their phase speeds differ. The unbounded large- growth is the familiar short-wave pathology of an inviscid vortex sheet.
Growth rate of varicose and sinuous planar-jet disturbances
. The two modes have the same dimensionless growth curve, with the long-wave and short-wave asymptotes shown. Past exam of the mathematics course of the University of Cambridge 2018 ii Paper 3 41E b Solution Created 2026-09-24 Updated 2026-10-03
For a discrete Fourier mode , putThe recurrence becomesso its amplification polynomial of a multilevel finite difference scheme isIts roots are real. Moreover,The root-location criterion in the question therefore puts both roots in exactly when . Fourier stability analysis for every wavenumber is equivalent to , so
Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 332 3 b Solution Created 2026-10-03 Updated 2026-10-05
Put ahead of the front and behind. With viscosity , Darcy's law gives the planar pressure gradients and . Write the displacement as with , and take the unperturbed interface to be in its translating frame.
The liquid flux is divergence-free away from the interface, so the pressure perturbations are harmonic. The decaying normal modes areExpanding pressure continuity at the displaced interface and matching normal Darcy flux givesHenceThe common normal-flux perturbation is . The local enthalpy balance converts it into a front-speed perturbation, . Thus the melting-front instability due to permeability contrast hasFor a signed transverse wavenumber, replace by .
Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 332 3 c Solution Created 2026-10-03 Updated 2026-10-05
Melting raises permeability of a porous medium. An advancing protrusion therefore offers a less resistive path to hot liquid, attracting larger Darcy flux and receiving more heat. The resulting faster melting reinforces the protrusion: this is a reactive infiltration instability. A permeability decrease would reverse this feedback, while zero contrast gives no growth in this ideal model.
For positive contrast, the growth rate is proportional to transverse wavenumber, so the model has no finite fastest-growing wavelength. The latent-to-sensible heat ratio slows the front but supplies no short-wave cutoff. Finite thermal transport and a resolved melting zone can introduce a length scale; pore geometry eventually limits the continuum description. If interface curvature changes the equilibrium melting temperature through the Gibbs--Thomson relation, that can also oppose short-wave corrugations. The modified dispersion relation must include such effects before a preferred wavelength can be calculated; a length scale alone does not guarantee that every added mechanism selects one.
Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 336 3 a Solution Created 2026-10-03 Updated 2026-10-05
The linear dispersion relation for a mode is . The proposed modes share phase velocity , so . A quadratic wave interaction generates the second harmonic and the difference harmonic. With only two positive wavenumbers, phase matching for a quadratic wave interaction requires . Equating their phase velocities givesand thereforeThis is a two-to-one resonance of dispersive waves. Its amplitude changes accumulate on . Choose the slow time ; this harmless constant rescaling makes the later amplitude formulas simple. With , the leading real field is .
For the fundamental, the order- cross derivative in is , while the fundamental coefficient in is . For the second harmonic the corresponding terms are and . The solvability condition in the method of multiple scales removes the resonant forcing and givesThe nonresonant third and fourth harmonics enter the correction. They do not change these leading amplitude equations.
Write , and . Taking real and imaginary parts gives the explosive two-to-one amplitude system in polar form:The polar phases are used where their corresponding amplitudes are nonzero. These equations implyFor the second identity, differentiate using : the terms proportional to cancel. Thereforeare first integrals, and in particular
For constant phases with , these real equations reduce to and . The initial data give , so the Riccati equation for is . Integrating and using givesThe tangent and secant function have a pole at , corresponding to . Both modes grow through their locked resonant interaction. This is formal finite-time blowup of the reduced amplitude equations; the weakly nonlinear expansion loses validity as the amplitudes become large, so it cannot establish a singularity of the full partial differential equation.
Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 345 1 a Solution Created 2026-10-03 Updated 2026-10-05
Put , write for perturbation pressure divided by , and set . The linearized Boussinesq approximation, incompressibility and material derivative giveTaking the curl eliminates pressure; differentiating the resulting vorticity equation in and using incompressibility gives . A further application of therefore yieldsFor a stationary nonzero horizontal Fourier mode, , so cancellation of gives the stationary Taylor–Goldstein equationThis division requires on the interval considered: a zero of is a critical level of an internal gravity wave. The mean profiles must be sufficiently smooth for the displayed derivatives, with background hydrostatic pressure and stable density stratification, , for propagating internal gravity waves. The horizontally uniform component is excluded from the cancellation. For a horizontal wavenumber , local vertical propagation additionally requires ; this is distinct from the smoothness restrictions. If stability of the background against other disturbances is needed, the Miles–Howard theorem supplies the sufficient condition , rather than a necessary condition for deriving the equation.
Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 345 1 b Solution Created 2026-10-03 Updated 2026-10-06
Take and assume an initially propagating mode, . Put and . Neglecting as stipulated, the Scorer parameter and stationary internal-wave WKB solution areThe positive root selects upward group velocity for the negative intrinsic frequency branch . The WKB approximation needs a slowly varying background relative to the local vertical wavelength, in particular and , away from a turning level or singular mean profile.
The local dispersion relation is . Its intrinsic and laboratory group velocities are, with ,Thus intrinsic energy propagation is upstream and upward, while laboratory energy propagation is downstream and upward, parallel to . Constant-phase lines of an internal gravity wave have slope and normal spacing ; the horizontal spacing remains .
If , grows with height. The wavevector and laboratory energy ray become more vertical; phase lines become flatter and closer together. For finite nonzero , the heightis a critical level of an internal gravity wave: and the vertical wavenumber diverges. The laboratory group velocity tends to zero. Importantly, , so the formal WKB condition need not deteriorate before this level if that ratio is small. Nevertheless the stationary equation is singular at the level, the phase has no finite limit, and the horizontal perturbation velocity grows as ; the inviscid linear wave cannot be continued uniformly through it without additional physics.
If , decreases to zero atThe wavevector and laboratory ray become horizontal, the phase lines become vertical, and the normal spacing increases. Above , and the disturbance is an evanescent wave. Here diverges, so the WKB approximation fails in a turning region and must be replaced by a local connection solution.
Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 345 1 c Solution Created 2026-10-03 Updated 2026-10-05
Use a common factor , with and . For the upward propagating branches with , defineThese are the inclinations of upward-sloping constant-phase lines of an internal gravity wave to the horizontal. Write the incident, reflected and transmitted complex velocity amplitudes as . Incompressibility gives , , and .
The jump conditions for stratified inviscid shear flow require a common interface displacement, not a common vertical velocity: on each side. Since the background mass density is continuous, pressure is continuous without a hydrostatic jump. The horizontal momentum equation gives . HenceThese two matching equations determine internal-wave transmission across a velocity jump:The amplitude formula printed in the PDF is inconsistent with these material-interface matching conditions. In particular, identical layers have and , whereas its expression labelled “reflected” equals one. The displayed results above distinguish reflection from transmission and retain the intrinsic-frequency factors required by the kinematic boundary condition and pressure balance.
For equal buoyancy frequencies and the specified opposing current, , , andThe incident phase lines have slope , the reflected lines slope , and the transmitted lines slope : they steepen above the interface. For the other specified current, . The transmitted vertical wavenumber then diverges and there is no regular propagating upper-layer wave of that frequency: this is the critical level of an internal gravity wave limit. Approaching it from gives and vanishing transmitted vertical energy flux; setting the intrinsic frequency to zero directly is outside the regular matching calculation.
Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 346 1 a Solution Created 2026-10-03 Updated 2026-10-05
Write for a comoving coordinate, for the peculiar velocity, for the Hubble parameter, and for the density contrast. For a homogeneous self-gravitating fluid, the linearized continuity equation, Euler equations for an inviscid fluid, and cosmological Poisson equation areHere is the physical adiabatic sound speed, gradients refer to comoving coordinates, and the homogeneous background force has already been subtracted. Differentiating the first equation and eliminating and givesA Fourier mode with comoving wavenumber therefore obeysThe Jeans wavenumber occurs where the coefficient in parentheses vanishes. Since a comoving wavelength is , the result isBelow this comoving Jeans length, pressure restores a displaced fluid element: there are acoustic waves rather than growing Jeans instability. In a static background they oscillate; expansion changes their amplitude and frequency, and dissipation can damp them. Pressure support alone does not erase them. In the pre-recombination photon-baryon fluid, Silk damping supplies a separate erasure mechanism.
For collisionless matter, replace the sound speed by a characteristic velocity dispersion. For an isotropic Maxwell-Boltzmann velocity distribution, defining , the static collisionless marginal-stability calculation givesThe coefficient depends on the distribution and the convention for velocity dispersion. Small-scale suppression is now collisionless free streaming and phase mixing, rather than collisional acoustic waves. An instantaneous collisionless Jeans length must be distinguished from the accumulated comoving distance traveled since decoupling.
For the requested thermal histories, an Einstein-de Sitter universe describes the matter-era limit: it cannot literally also have a radiation-dominated epoch. Interpret the question as a spatially flat universe with negligible cosmological constant, passing from radiation domination, , to matter domination, . For the usual schematic dynamical Jeans scale use in the estimate, soNear the particle horizon a relativistic perturbation calculation replaces the Newtonian derivation; the scaling estimate still identifies the relevant sound-crossing or streaming scale.
Before cosmological recombination, adiabatic initial conditions keep the entropy per baryon fixed in the tightly coupled photon-baryon fluid. With , its photon-baryon sound speed isFor example, this follows from and . In the radiation-dominated limit , ; in the strongly baryon-loaded matter-era limit , . Thus the baryon Jeans length across recombination has the asymptotic historyThe transition to the plateau is smooth: retaining gives in matter domination. If baryon loading is still small, this intermediate segment instead rises as until loading becomes important.
At photon decoupling, radiation ceases to provide pressure support to the baryons, so the sound speed drops from the coupled-fluid value to that of a nonrelativistic ideal gas, , with . The drop in comoving Jeans length at fixed density is the same ratio of sound speeds, typically several orders of magnitude. Subsequently an isentropic process gives , so and . Residual energy exchange through Compton scattering can initially keep , producing a short approximately flat gas segment before the decreasing segment. The idealized sketch assumes immediate adiabatic cooling:
log(comoving baryon Jeans scale)
^
| __________________
| / |
| / | loss of photon pressure
| / |
| / +---\
| / \
+-------------------|-----------------|----------> log(time)
t_eq t_rec
slope 1/2 slope ~0 slope -1/3For the cold-dark-matter kinetic Jeans scale, take the stated toy history to mean that collisions maintain a particle temperature equal to the photon temperature until kinetic decoupling. This is an assumption about thermal coupling; a negligible collision rate after decoupling makes the subsequent treatment collisionless. Assume the ordering . While relativistic, the characteristic particle speed is constant; while nonrelativistic but thermally coupled, ; after kinetic decoupling, particle momentum redshifts as and . Combining these with the background dynamical density givesThere is no sudden change at cosmological recombination: cold dark matter has already decoupled. Its small initial velocity dispersion makes its suppression scale much smaller than the pre-recombination baryon scale.
log(comoving CDM dynamical Jeans scale)
^
| _______________
| ___/ \
| ___/ \ no jump
| / \ at t_rec
| / \
+-----------|------------|--------------|----|------> log(time)
t_NR t_dec t_eq t_rec
slope 1/2 slope 1/4 slope 0 slope -1/3The density convention matters. If one instead defines a species' self-gravitating collisionless Jeans length using only its own nonrelativistic density , its thermally coupled segment is constant and its decoupled segment decreases as even during radiation domination. The constant radiation-era segment in the second sketch is the usual instantaneous background dynamical streaming estimate, not an exact single-species self-gravity threshold. A complete multicomponent calculation uses the separate perturbation equations and their combined cosmological Poisson equation.
Past exam of the mathematics course of the University of Cambridge 2021 iii Paper 333 1 i Solution 2026-09-29
The basic state has velocity fieldand buoyancy . For disturbances independent of , the linearized equations areSubstituting a plane wave proportional to and eliminating , , and gives the dispersion relationThus the requested coefficients are
An instability exists precisely when some wavenumber pair makes . Since the Brunt–Väisälä frequency satisfies , this is possible exactly whenThe basic relative vorticity is , so its absolute vorticity is . Its Ertel potential vorticity is thereforeThe instability criterion can consequently be written as : the vertical absolute vorticity has the opposite sign to the planetary vorticity. This is inertial instability, approached most directly by disturbances with .
Past exam of the mathematics course of the University of Cambridge 2021 ii Paper 1 15B d Solution Created 2026-09-24 Updated 2026-09-25
In the far past, , the term is negligible and the rescaled mode is a quantum harmonic oscillator of angular frequency . The stated choice of its ground state is the Bunch-Davies vacuum.
At late times, after the mode is far outside the Hubble scale, . Its variance becomesSince has dimensions of length cubed, the dimensionless power is proportional to . More precisely, with the standard isotropic Fourier convention,which is independent of the wavenumber . Thus equal logarithmic intervals in carry equal late-time power. This is the sense in which inflation naturally generates a scale-invariant inflationary power spectrum.
Past exam of the mathematics course of the University of Cambridge 2022 iii Paper 345 1 e Solution 2026-09-28
The incident internal-wave ray has equation . Intersecting it with the bottom givesIts distance to the slope is consequentlyThe subcritically reflected ray rises through the same vertical distance at angle , so its next free-surface reflection is at
On the first leg, viscous attenuation of an internal-wave beam multiplies the energy density by . The bottom reflection multiplies it by and changes the wavenumber to . Since the corresponding Reynolds number is , attenuation on the second leg contributes . Ignoring boundary-layer enhancement as requested,
Past exam of the mathematics course of the University of Cambridge 2023 iii Paper 321 2 f Solution 2026-09-28
Take the Fourier transform in , with convention . The differentiation rules and turn the forced equation intoFor the unforced equation, impliesIt is therefore exactly the shearing-wave oscillator found in part c, now parametrized by radial wavenumber rather than time.
Past exam of the mathematics course of the University of Cambridge 2023 ii Paper 4 38C d Solution Created 2026-09-24 Updated 2026-09-25
The linearized equations are homogeneous and have constant coefficients in and . Consequently, spatial Fourier modes evolve independently, and the superposition principle reconstructs arbitrary sufficiently regular initial disturbances from them. It is therefore enough to seek the normal modesHere the real is the wavenumber, is the possibly complex temporal exponent, and taking the real part produces a physical displacement.
Past exam of the mathematics course of the University of Cambridge 2024 iii Paper 332 3 iii Solution 2026-09-28
Under the quasi-stationary approximation, and with advective heat transport neglected, each temperature perturbation satisfies the Laplace equation. The decaying normal-mode solutions areKeeping the displaced ice interface at omits the curvature correction described by the Gibbs--Thomson relation.
The fixed melting temperature at the displaced ice interface givesExpanding temperature and conductive heat flux continuity at the displaced outer interface gives
The fixed melting temperature at the displaced ice interface givesExpanding temperature and conductive heat flux continuity at the displaced outer interface gives
Put . Since , the temperature condition in the limits and givesAlthough is small, the product is therefore order one and cannot be discarded. The flux condition then givesThis is why the stated asymptotic warning matters.
The perturbed Stefan condition at isUsing and yieldsSubstitution of the interface-amplitude ratio from part i produces the thin-film icicle-ripple instability dispersion relation
Set . Rationalizing this complex number givesso its growth rate and imaginary part areDifferentiating the growth rate with respect to the wavenumber shows that its positive stationary point satisfiesIt is the unique global maximum, and henceAt this wavenumber . With the convention , a constant phase travels with phase velocityThereforeBecause increases downward, the negative sign means that the icicle ripples migrate upward with speed .
Past exam of the mathematics course of the University of Cambridge 2024 iii Paper 333 3 e Solution 2026-09-28
Multiplyby and integrate from the rigid bottom to an upper endpoint at which . The bottom term also vanishes because . Integration by parts givesTherefore the horizontal wavenumber has the Rayleigh quotientFor a trapped wave the upper endpoint may be taken to infinity because the eigenfunction decays.
For a face of signed slope , conservation of frequency and tangential wavenumber gives, on the branch relevant to an incident downward-right ray,At , the reflected wavenumber diverges and the reflected group velocity becomes tangent to the face. On a supercritical face the denominator changes sign: horizontal propagation reverses and rays from the two faces are directed towards a sawtooth corner. Successive reflections therefore focus energy and shorten the wavelength.
The inviscid ray pattern cannot persist indefinitely. Near-critical focusing amplifies gradients until kinematic viscosity, mass diffusivity, nonlinear wave steepening, and wave breaking matter; a real corner is also rounded on some finite scale. These effects replace the singular ray construction by dissipative boundary layers, mixing, and a finite-width reflected beam.


