Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 64 3 Solution Created 2026-10-03 Updated 2026-10-07
The bounded classical solution is the heat-kernel convolutionThe two-dimensional heat kernel has integral one. Boundedness of permits differentiation under the integral for , proving and . Its approximate identity property gives , locally uniformly since is continuous. This bounded solution satisfies the required Gaussian-growth bound.
For heat-equation uniqueness under Gaussian growth, let be the difference of two solutions, so and on a finite time interval. Choose and a slab . The positive comparison solutionsatisfies . For any , its faster spatial growth makes on a sufficiently large lateral cylinder boundary. At the initial boundary, . The heat equation maximum principle applied to proves the same inequality inside. First allow the cylinder radius to increase, then let . Repeating these slabs proves uniqueness on every finite interval with the stated uniform growth bound. No spatial integrability of is required.
At time , this is Gaussian filtering with covariance , hence per coordinate. With the angular-frequency Fourier transform convention , the Gaussian filtering Fourier multiplier isFor merely bounded , this identity is understood through tempered distributions. The multiplier leaves the zero frequency unchanged and attenuates large frequencies exponentially: it suppresses rapid noise fluctuations but blurs image edges. In cycles-per-length frequency , the multiplier is .
For the printed Perona-Malik equation, retain the supplied conductance , which includes an extra factor . In one dimension write and . Then , withThe forward-backward threshold for gradient-weighted exponential diffusion is thereforeAt and the coefficient vanishes, giving degenerate diffusion. Increasing increases the forward-diffusion range. Small nonzero slopes smooth, while large slopes formally sharpen; a negative coefficient produces short-wave growth and ill-posedness, so this is a dynamics explanation, not a general existence theorem for arbitrary data. Replacing the printed conductance by would give threshold , but that is a different equation.
The four-neighbour mean expansion follows by Taylor expansion: opposing first and third derivatives cancel, soIf for every sufficiently small , division by and passage to the limit give . This is a pointwise conclusion; it does not assert harmonicity in a whole neighbourhood.
For the area median over the disk, the disk-median curvature expansion, at a point where , isConsequently the analogous median fixed-point condition yieldsThis is the vanishing of the level-line curvature at : to second order the level line is straight there. It need not be a straight segment. For example has zero disk median at the origin for every radius, by odd symmetry, but its zero level line is the cubic . Vanishing curvature on an entire connected regular level arc would force that arc to be straight. The factor is for the disk's uniform area measure; a circle-boundary median has a different scale factor, while producing the same zero-curvature equation.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 64 4 Solution Created 2026-10-03 Updated 2026-10-14
An image segmentation separates an observed image signal into regions that are smooth within themselves and separated by meaningful image edges. Pure image smoothing blurs the very transitions that ought to define these regions. The Mumford–Shah segmentation model instead chooses the reconstruction and its discontinuity set together. For a bounded planar Lipschitz domain and bounded grey-value data , one standard normalization isHere is the relatively closed edge set and can have different traces on its two sides. The first term is quadratic fidelity, the second penalizes variation within regions, and the Hausdorff measure term charges total edge length. It balances fitting, denoising and economical region geometry. The model was developed by David Mumford and Jayant Shah; their 1989 paper formulates this joint variational approach.
All three terms matter. Without fidelity, a constant reconstruction with no edge has zero energy. Without the gradient term, smooth data can be fitted exactly with no edge penalty. Without the length term, fine partitions with nearly constant region means can drive fidelity arbitrarily low while creating excessive boundaries. This is segmentation overfitting without an edge penalty. Increasing favors flatter regions; increasing makes extra boundaries more expensive and can remove small features. These parameter effects describe a balance, not a guaranteed monotone evolution of every individual boundary.
For fixed , varying gives the fixed-edge Euler-Lagrange equation for Mumford–Shah:The boundary condition is a separate one-sided Neumann boundary condition on each side of an edge, not continuity of across it. The fidelity makes this fixed-edge problem strictly convex, so its weak solution is unique by the Lax-Milgram theorem. Optimizing the edge set remains a different geometric problem. The nonconvexity of Mumford–Shah segmentation prevents a general uniqueness assertion or a guarantee that a numerical stationary point is globally optimal.
The piecewise-constant Mumford–Shah problem imposes on regions . It minimizesThe factor counts each shared internal boundary once. The region means in piecewise-constant segmentation give , provided . Thus the remaining optimization concerns the partition. For a smooth interface between and , moving it in the normal pointing out of changes fidelity by per unit displacement, while length changes by its curvature . The resulting segmentation interface curvature balance isWith equal isotropic interface costs, three freely meeting smooth edges satisfy the triple-junction angle in isotropic segmentation: force balance of their unit tangents gives angles of degrees. These are local stationarity conditions on regular interfaces, not a description of every singular edge configuration.
A simple piecewise-constant segmentation contrast threshold explains why small objects can disappear. On a domain of area , suppose two constant intensities differ by , occupying areas and with internal boundary length . Keeping that boundary fits the data exactly and costs . Merging both regions costs the within-region squared error . Among these two candidates, splitting wins precisely when . Other partitions may beat either candidate, so this is not a universal global segmentation formula.
This original synthetic example compares those two candidate geometries using their least-squares means. It illustrates the edge cost; it does not claim to compute the globally optimal Mumford–Shah segmentation.
Existence is cleanly stated in the relaxed SBV space formulation:where is the jump set of a bounded-variation function and the Cantor part of a bounded-variation derivative is absent. Clip a minimizing sequence to the bounded data range: this cannot increase fidelity, gradient energy or jump length. Its values are uniformly bounded, its gradients bounded in , and its jump lengths bounded. The SBV compactness theorem supplies an -convergent subsequence staying in the special class; bounded values also give strong convergence. The fidelity then converges, while gradient energy and jump length are lower semicontinuous. The direct method in the calculus of variations produces a minimizer. The essential closedness of Mumford–Shah jump sets is the additional regularity result connecting this relaxed minimizer to a closed-edge formulation; existence in alone does not assert that every edge set is smooth.
A practical continuous approximation is the Ambrosio–Tortorelli approximation, introduced in Ambrosio and Tortorelli's 1990 paper. An auxiliary field is near one in regions and near zero at edges. One normalized energy isThe edge field weakens smoothing across a narrow transition, and its own energy approximates interface length. The Gamma-convergence result, together with the required compactness, relates global minimizing sequences to the limiting segmentation energy; it does not make the finite-parameter problem jointly convex. Alternating the two fields solves quadratic elliptic subproblems, but initialization and stopping can affect which local stationary configuration is found. The strengths of the model are joint denoising and segmentation, sharp transitions and a geometric cost. Limitations include competing local minima, sensitivity to scale parameters, loss of fine texture, and boundaries driven by intensity rather than semantic object identity.
Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 68 4 Solution Created 2026-10-03 Updated 2026-10-06
In diffusion image processing, the observed grey-value image signal is the initial condition for an evolution ; time controls the image smoothing scale. A useful model must suppress image noise while respecting the image signal's geometric boundaries. Unless exterior values are intended, use periodic boundaries or Neumann boundary conditions, so the filter does not lose intensity through the image signal border.
The basic linear model is the heat equation, , . On the whole plane,It is Gaussian image smoothing with variance in each coordinate. In Fourier variables, : high spatial frequencies are damped most strongly. Under zero-flux or periodic boundaries the mean is conserved, the maximum principle keeps values within the initial range, andThese give stable image noise suppression, but a sharp step also contains high frequencies and is blurred across a width of order . Constant diffusivity has no mechanism for distinguishing image noise from an image edge.
Linear sharpening by the backward heat equation would multiply Fourier modes by and amplify arbitrarily fine image noise without bound. It is an ill-posed initial-value problem. A controlled unsharp masking step instead uses , with multiplier . This amplifies high frequencies by at most ; it can improve apparent contrast but also amplifies image noise and cannot reliably restore information already lost by image smoothing.
Nonlinear models use image signal structure to select the diffusion. A gradient-based energy and its formal gradient flow areFor a smooth solution with the corresponding zero-flux condition, . The fidelity term prevents indefinite drift towards a constant reconstruction; pure diffusion is usually stopped at a selected finite time.
The local principal diffusion coefficients distinguish suppression of diffusion from backward diffusion. Where , the flux derivative isIts coefficient along an image signal level curve is ; across that curve it is . Forward parabolic behavior requires both coefficients nonnegative, with strict positive lower bounds giving uniform parabolicity. Merely choosing and decreasing does not establish well-posedness.
For a convex area-type penalty , the coefficients areDiffusion across steep image edges is weak but remains forward. As , the total variation flow formally becomes . Its convex subgradient formulation handles flat regions and discontinuities. It favors piecewise constant image signals and preserves sharp transitions better than Gaussian image smoothing, but can produce staircasing in total variation denoising, shrink small objects and move boundaries by curvature. image edge preservation does not mean exact preservation of all image edge locations or amplitudes.
The Perona-Malik equation takes a decreasing diffusivity such as . Small gradients are smoothed strongly, while large gradients have weak flux. More precisely,For , diffusion in the gradient direction is backward, so a strong transition can steepen. This gives formal image edge enhancement but also causes instability and continuum ill-posedness. The exponential choice similarly has a negative normal coefficient for . Discrete results depend on the stencil, step size and implicit regularization; appealing visual results are not a proof of a well-posed PDE.
A regularized Perona-Malik diffusion computes the conductance from a smoothed image signal, for exampleThe flux still acts on , but the image edge detector is less sensitive to raw image noise. With a smooth kernel and a positive conductance bounded away from zero on the attained range, the local principal diffusion is forward; appropriate regularity and boundary assumptions give a well-posed nonlocal parabolic model. The exact regularization and fixed image smoothing scale are part of the model.
A genuinely directional filter uses a diffusion tensor for image filtering:Here estimates the image edge normal and its tangent. A structure tensor provides robust directions at a second averaging scale. Strong tangent diffusion smooths image noise along an image edge, while weak normal diffusion reduces mixing across it. Related coherence-enhancing choices connect elongated features along their dominant orientation. Positive tensor eigenvalues preserve forward parabolicity; this type of enhancement must be distinguished from Perona–Malik's backward normal diffusion.
For more direct sharpening, a shock filter for image enhancement formally evolves . It is a Hamilton–Jacobi-type transport mechanism, not a positive diffusion operator, and steepens transitions around inflection boundaries. In practice it can be combined with regularized forward diffusion to control image noise. Any enhancement method needs a scale or stopping rule and an honest treatment of image noise amplification.
Linear heat flow offers predictable image smoothing but blurs image edges; convex nonlinear diffusion can preserve them; backward or shock mechanisms sharpen them at a greater stability cost. Gradient thresholds, conductance regularization and positive tensor directions determine which of these behaviors a proposed filter actually has.
Past exam of the mathematics course of the University of Cambridge 2016 iii Paper 329 2 Solution Created 2026-10-03 Updated 2026-10-06
In the surfactant transport with exchange relaxation, is the material derivative following an interfacial material element. The term accounts for dilution by tangential stretching, while accounts for changing area through normal motion and curvature. The surface diffusion term redistributes surfactant down concentration gradients. The exchange term relaxes its concentration towards , through adsorption or desorption from a surrounding reservoir, on time scale .
In the bubble frame the spherical interface is stationary and impermeable. A steady material derivative is . For a linear response in the rising speed this and are second order, whereas stretching of is first order. ThusRotational symmetry, linear response to , and zero normal velocity allow only , where is the surface tangent projector. This is the dipolar, axisymmetric sector; axisymmetry alone, without the linear-response assumption, would allow other angular dependence.
On a sphere and . Applying the surface tangent projector to the derivative gives the sphere surface derivative identitiesFor example, taking a second surface divergence of the first identity yields the third. Equivalently, the components of are degree-one spherical harmonics. Hence a dipolar surfactant distribution solves the linear equation:The required smallness condition is . Surface diffusion or exchange must smooth the concentration faster than it is redistributed by the actual interfacial velocity. We require , and ; with neither process there is no such steady linear balance at nonzero . A small capillary number also justifies the spherical approximation.
Take the normal from the inner bubble to the outer liquid, define , and set . The interfacial stress balance with variable surface tension isBecause and , its tangential part becomesThis is a Marangoni stress: surface-tension gradients oppose the clean-bubble circulation.
Use the Unscaled Papkovich–Neuber representation, and . The constant vector potential gives . In the decaying, linear axisymmetric sector, a vector monopole and scalar dipole provide the two constants needed at the sphere; other equivalent potential choices amount to a representation freedom. The PDF potential contains the factor omitted in the TeX aid. With that factor restored, the exterior velocity isAt the interface, no penetration gives , and the tangential velocity gives . Using the traction coefficient printed in equation (3), the tangential stress condition gives . The requested algebraic answers are therefore
There is a factor-of-two error in the PDF traction. Direct differentiation of the preceding velocity with the Newtonian fluid stress tensor givesrather than its printed coefficient . In particular the tangential traction is . Keeping the same definition of , the physically consistent translating surfactant-coated bubble instead has andThe two sets must not be silently conflated. As a useful check, the no-slip limit has the usual total Stokes flow drag with the corrected traction; the printed traction would double it.
Both versions give , , when : the interface is tangentially stress-free and circulates as a clean inviscid bubble. For , , , : the interface is immobile in the bubble frame and behaves as a no-slip sphere. The Marangoni stress remains finite because has a nonzero limit. The concentration smallness condition still has to hold; large is not by itself a complete linearization criterion.
The normal component is balanced by the inner gas pressure and the normal capillary stress , including the variation of , together with the hydrostatic pressure difference responsible for buoyancy. A spatially uniform inner pressure alone cannot balance the dipolar traction. To make the normal stress balance on a translating bubble explicit, let be the outer-minus-inner mass density. After removing the equilibrium Laplace pressure, the corrected outer normal traction has dynamic part . Its degree-one balance isThus, if a terminal speed is desired, , interpolating between the clean-bubble and rigid-sphere speeds. A degree-one shape displacement is merely a translation of the sphere, so the balance determines its rise speed, rather than a dipolar shape distortion.
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 309 2 c Solution Created 2026-10-03 Updated 2026-10-06
Fix the Riemann curvature tensor conventionFor the limiting affine connection, every product of two nonzero connection coefficients is zero, because its only nonzero coefficients have spatial upper index and both lower indices . Its only possibly nonzero curvature components, apart from antisymmetry in the last two indices, areThe Ricci tensor is thereforeAll other components vanish, so the limiting connection is a connection with vanishing Ricci tensor exactly when satisfies the Laplace equation on spatial Euclidean space. This is Ricci-flatness, not necessarily vanishing curvature: is a harmonic function but has nonzero Hessian matrix and hence a nonflat limiting connection. Reversing the curvature sign convention reverses the displayed Ricci sign but leaves the equivalence unchanged.
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 309 4 e Solution Created 2026-10-03 Updated 2026-10-06
Use linearization around Minkowski spacetime after constant rescaling of : set , , . To first order,Hence is affine in , while is the freely propagating profile. The affine transverse trace in a diagonal linearized plane wave is pure gauge: it has zero linearized curvature and is locally a coordinate artifact. To see this without imposing extra boundary conditions, use a first-order linearized coordinate gauge transformation , withThe changes cancel and the remaining time/longitudinal changes are proportional to . The transverse perturbation becomeswith all time and longitudinal components zero. It is a transverse-traceless gauge plane gravitational wave travelling in the positive -direction, with plus polarization and no cross polarization in these transverse axes.
For freely falling nearby particles, geodesic deviation gives opposite transverse tidal accelerations, since and in vacuum. Thus the transverse geodesic deviation eigenvalues have opposite signs; the stretching and compressing roles interchange when the curvature profile changes sign. If as well, there is no linearized tidal wave: that special profile is also locally flat to this order. The exact diagonal metric describes a fixed transverse polarization through its opposite vacuum curvature eigenvalues; the linearized calculation identifies it as plus relative to the chosen axes.
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 342 1 a Solution Created 2026-10-03 Updated 2026-10-06
Take positive downwards and write for the filament bending modulus. Under the material-modulus interpretation of the printed as Young's modulus, the circular cross-section gives the second moment of area and bending modulusIndeed a small curvature gives axial strain , so integrating over the section gives bending energy per unit length. Keeping this distinction is essential to the powers of radius in the answer.
After allowing for buoyancy, the effective downward force per unit length isThe filament loses potential energy as it moves down; the displaced fluid's hydrostatic contribution is included through the density difference. For an inextensible filament with arc-length coordinate , the geometric bending and gravitational energy is , up to a constant. For the small-slope Monge representation this reduces toThe upward-positive convention reverses the sign of and the gravitational term together. There is no stretching contribution at this order when bending a slender filament without imposed axial tension. The wording does not explicitly distinguish a material elastic modulus from a flexural modulus: if instead denotes the filament bending modulus in the intended convention, set throughout; then the factor is already included in .
Past exam of the mathematics course of the University of Cambridge 2017 ii Paper 2 23I Solution Created 2026-09-24 Updated 2026-10-05
For a regular parameter , the curvature and torsion of a space curve areThe torsion of a space curve requires . Reparameterize by arc length and set , , . Differentiating the pairwise inner products of this orthonormal Frenet frame shows that its derivative matrix is skew symmetric. Since , defining gives the Frenet-Serret formulasThe fundamental theorem of regular space curves says that prescribed smooth and smooth on an interval determine a unit-speed curve uniquely up to an orientation-preserving Euclidean rigid motion. Solve this frame's linear differential equation with an initial oriented orthonormal frame, then integrate ; uniqueness proves the assertion.
Under the intended convention that the shared parameter is arc length, equality on an open interval gives equal position and equal Frenet frames at one point. Equal give identical differential equations, so uniqueness gives . The printed “regular smooth” hypothesis alone does not establish this convention. Without common arc length, an orientation-preserving smooth change of speed that is the identity on but not elsewhere reparameterizes a circle into a counterexample: both curves have , and agree on , but differ elsewhere as parameterized curves.
For an oriented surface with unit normal , define normal curvature and geodesic curvature . ThusA unit-speed curve is a geodesic exactly when . If also , then , so it is a straight line. A vertical ruling of a circular cylinder is such a geodesic, although the cylinder contains no open piece of a plane. A nonconstant regular straight line cannot be a simple closed curve. Hence no planar patch is required, and such a geodesic cannot be simple and closed.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 332 1 Solution Created 2026-10-03 Updated 2026-10-05
Take in the injection direction and along the flat interface. The specified must be a Darcy velocity, so the pore-fluid velocity and unperturbed interface speed are . Write the perturbed interface as , with , and choose the normal from fluid 1 into fluid 2. The background pressure gradients from Darcy law are .
By incompressibility and uniform permeability of a porous medium, pressure perturbations are harmonic functions. A transverse Fourier mode with positive wavenumber has the decaying formsThe kinematic boundary condition and equal normal Darcy flux implyWithout capillarity the pressure is continuous at the displaced interface. Its linear perturbation is thereforeConsequently the planar viscous-fingering dispersion relation reduces toThis is the Saffman–Taylor instability: less viscous fluid advances preferentially through protrusions. Without a short-wave regularizing mechanism the idealized rate has no finite maximum.
For the prescribed apparent capillary pressure, the chosen normal is , so . Subtract the flat-front pressure jump . Linearization of the jump at givesThere is no first-order product of the surface-tension perturbation and the perturbed macroscopic curvature. ThusUsing the capillary number convention and , this isIf is measured from the unperturbed front and denotes its local apparent tension, set to obtain the usual fixed-coefficient expression. For a physically fixed spatial gradient, and the displayed rate is an instantaneous, generally time-dependent growth rate; the amplitude solves , rather than one constant exponential law for the entire displacement. The parameter has dimensions of inverse length; the absolute tension gradient is .
Assume , as required for positive apparent surface tension, and define . If , unstable modes satisfy . Differentiating the cubic dispersion relation gives the fastest-growing modeWhen , every nonzero mode decays, with a neutral zero-wavenumber translation in the infinite-domain limit. For , surface-tension-gradient stabilization of viscous fingering suppresses the instability entirely at positive injection speeds satisfyingIf , no positive injection speed eliminates every long-wave unstable mode in an unbounded interface. The prescribed wetting variation is treated as an effective normal pressure law for porous-media flow.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 332 3 Solution Created 2026-10-03 Updated 2026-10-05
For an axisymmetric thin elastic plate, define . Uniform excess pressure and bending stiffness give . The roof is attached to undeformed rock at the front, giving clamped boundary conditions , and regularity excludes singular displacement or curvature at . A regular particular solution is ; adding a constant and a multiple of and enforcing the two edge conditions gives the axisymmetric clamped-plate deflectionIntegrating the magma intrusion volume,therefore yieldsHere the axisymmetric phrase requires the radial plate biharmonic operator; a Cartesian one-dimensional beam equation would not give the same pressure coefficient. Hydrostatic contributions are excluded as instructed.
For the early front, let be the leading propagation speed. The fluid and fracture radii differ only within a local region of length . In lubrication theory, the elastic pressure gradient near the front is of order , and the volume flux through the gap is of order . In a frame following the fluid front that flux is of order . The viscous peeling of an elastic plate law is thereforeMatch the peeling region to the interior curvature: . The stated vapour-gap relation , with interpreted as the relevant pressure magnitude, then givesSubstitution gives the vapour-tip peeling of a magma intrusion speedWith constant supplied volume flux and negligible initial intrusion volume, . Integrating gives , orOrder-one coefficients require the full local peeling profile. This early regime is a front-controlled asymptotic regime, not an assertion that a continuum peeling zone remains thin at arbitrarily small .
When fracture toughness instead imposes fixed , equating it to the outer edge curvature gives the toughness-controlled magma intrusion lawIts pressure can be written in either useful formThe pressure decreases as the intrusion grows on this advancing branch.
If is the reservoir volume before any magma was transferred to the initially negligible intrusion, volume conservation and negligible conduit storage give . With the supplied pressure-volume law and conduit conductance, the reservoir-fed magma intrusion obeysIn this law is an effective reservoir pressure-volume stiffness, with dimensions of pressure per volume. If the stated initial reservoir volume refers instead to the start of the late regime, when the intrusion already has volume , replace by the total throughout. An independent value of would then be needed.
For an advancing solution, inflow ceases when the reservoir and intrusion pressures equalize. Its final volume is the larger positive root in ofEquivalently, solves . The larger root is essential: rises from zero to a maximum at and then decreases. A nonempty interval of positive inflow exists only ifThere are then two positive equilibria . At equilibrium the derivative of the volume ordinary differential equation is , so is unstable and has asymptotic stability. An initial advancing intrusion with approaches . The equality case has a double equilibrium and no positive-inflow interval; above the threshold there is no advancing equilibrium branch of this model.
The late formula is singular at and cannot nucleate the intrusion: the earlier propagation regime must supply its finite seed. Nor does this fixed-edge-curvature propagation law prescribe how an irreversible fracture closes under negative inflow. The final-size answer is conditional on reaching the advancing late branch. In particular the reservoir is generally not emptied, and controls the approach rate rather than the equilibrium size.
Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 329 1 Solution Created 2026-10-03 Updated 2026-10-05
The material derivative follows the interfacial surfactant. The term describes dilution by tangential expansion and concentration by compression. The term accounts for changing surface area due to normal motion of a curved interface. Finally, describes surface diffusion. Together these give conservation of insoluble surfactant on a moving interface.
The relevant surface Péclet number isWrite and . Since has zero trace and , the surface divergence isThere is no normal motion. To first order in , the steady surfactant balance becomes . The traceless quadratic is a degree-two spherical harmonic, so . The mean of is zero by total surfactant conservation; consequentlyThe discarded advective term is smaller by . This is the quadrupolar surfactant distribution on a spherical interface.
With the unit normal directed from the bubble into the exterior, the interfacial stress balance with variable surface tension isUsing the specified first-order curvature and , together with , givesThe first term is the spherical capillary pressure; the last contains the normal tension correction and the tangential Marangoni stress.
The ambient Stokes flow has the symmetry of the symmetric traceless tensor . In the Unscaled Papkovich–Neuber representation, is harmonic and yields . The decaying vector potential must have the form , and the scalar disturbance must have the degree-two form . They are harmonic outside the bubble and have precisely the required rotational covariance. Dimensionless amplitudes may therefore be writtenFor a steady bubble, the no-penetration boundary condition gives . Its tangential velocity gives . The tangential stress boundary condition then yieldsHence , and . The inviscid interior supplies only the constant pressure balancing . Matching the remaining normal stress givesEliminating givesAs , : the Marangoni stress suppresses tangential motion and effectively immobilizes the interface. This limit must retain the small surface Péclet number and small-deformation assumptions; can grow through increasing without invalidating the linear concentration approximation.
Past exam of the mathematics course of the University of Cambridge 2021 iii Paper 329 1 Solution 2026-09-29
In a cross-section normal to the cylinder axes, measure from the line of contact. The two circular boundaries have the parabolic approximationfor their separation. The contact point of the meniscus is at , so the leading cross-sectional area isThe area of the small meniscus cap is and is lower order. At its upper end the gap has width . A tangent semicircle therefore has radius and curvature
The Young–Laplace equation makes the liquid pressure, relative to the nearly uniform gas pressure,At a fixed , lubrication theory gives a planar Poiseuille flow through a gap of width , with axial flux per unit Integrating across the cusp,The continuity equation now gives
Set . Since , this is the porous medium equationConservation of the fixed volumeand dimensional analysis give the self-similar solution . One integration of the resulting ordinary differential equation yields the compactly supported Barenblatt solutionEquivalently,Its tip is . Usingin the volume constraint gives . Therefore
For vertical cylinders at equilibrium, hydrostatic pressure gives . Balancing this with the capillary pressure gives the large-height profile
The signed curvature isAn orientation-preserving isometry sends to and leaves unchanged, soAn orientation-reversing isometry reverses the chosen normal and hence the sign convention for signed curvature, while the geometric curvature remains invariant.
Under the same independence assumption, inclusion-exclusion givesOn ,There is one interior inflection point, where the curvature changes from positive to negative:At the graph reaches one with zero slope and then stays constant. There is no further change from one sign of curvature to the opposite sign there.
Without independence, still holds, but only is forced by the marginals. Its inflection behaviour is then not determined. The shared- construction in (e), for example, gives before the plateau.
The endpoint constraints are . The constants in the three costs contribute only , and respectively, so they do not affect the minimizing shapes. For companies B and C, the fixed arc length isThe original PDF has as company A's first variable term. The local TeX transcription has lost that derivative dot and incorrectly repeats ; the calculation below follows the PDF.
Company A. The relevant functional is . The Euler-Lagrange equation givesIts general solution is . Adding and subtracting the endpoint equations gives and . ThusThis curve is odd, negative on the left half and positive on the right half; company A has no lower-half-plane restriction. It is the unique global minimum. Indeed, for any endpoint-zero variation , integration by parts and the differential equation cancel all linear terms, leavingunless .
Company B. Minimize under the length constraint. Introduce a constant Lagrange multiplier and apply the Euler-Lagrange equation to :The curvature is constant, so the required lower arc is a circular arc. Writing its radius as and imposing the endpoints givesSince , its length determines the radius:Equivalently, introduce with . Then , which increases strictly from to . The stipulated range of therefore selects one minor arc. At it becomes the lower semicircle, with vertical endpoint tangents; it is still a single-valued finite-length graph.
Global minimality follows from the convexity of the augmented slope cost with . For a competitor ,Consequently the augmented cost difference is at least . Competitors have the same length, so the multiplier contribution cancels and the original cost is minimized. This proves that a circular arc maximizes area below a fixed chord: minimizing negative signed height maximizes the area between the fence and the chord.
Company C. The variable cost is . The augmented integrand is . Since it has no explicit dependence, the Beltrami identity givesThe hanging branch with integrates to . Equal endpoint heights force , and the endpoint value fixes . ThusIts length isWriting , the equation has one positive solution for every : its derivative has numerator . The curve is a catenary, is even, and lies strictly below the chord except at the endpoints.
A direct argument proves this is a global minimum, rather than merely an extremum of an augmented functional. Parametrize any competitor by arc length , from its left endpoint to its right endpoint. Then , , and integration by parts givesThe Cauchy-Schwarz inequality gives the last step. Equality requires and , whose integrals are precisely the displayed catenary with the stated length. Thus it attains a lower bound valid for every competitor. This proves that a catenary minimizes the integral of height along a fixed-length curve.
Past exam of the particle Astrophysics course of the University of Cambridge 2012 Paper 1 1 Solution 2026-10-20
The metric source is a gravitational frequency shift. photons following null geodesics through a time-dependent spatial metric change momentum relative to the background comoving frame. The directional projection of that metric change gives the factor. It is not a collision term or continuing photon scattering after recombination. The normalization and sign belong to the specified definition of and ; a distribution perturbation should not silently be identified with temperature without fixing that normalization.
Define , where dots here mean conformal-time derivatives. In an Einstein-de Sitter background,A power law therefore satisfies , or . Thuswith the required growing mode . This is a solution in the perturbation convention used in the question, rather than a statement that every gauge-invariant gravitational potential grows as .
Use an integrating factor for the photon equation:For the growing mode . For the polynomial-source photon transport solution, a particular solution is : direct substitution gives . The initial-value solution is consequentlyThis also follows by integrating the source twice by parts. The apparent inverse powers of cancel in the complete long-wavelength limit; the integral form remains regular. At , the source vanishes and the complete expression reduces to .
Physically, the exponential transports the recombination pattern along photon trajectories; its direction-dependent phase projects a Fourier perturbation onto different angular multipoles. The extra terms are the accumulated gravitational contribution, here expressible as endpoint terms because the source has this special time dependence. For endpoint monopole and dipole in photon transport, at the observer the last two terms are respectively a directional dipole and an angular monopole for each mode; they do not supply a new sky pattern at multipoles . The propagating recombination combination contains the intrinsic perturbation and gravitational/velocity endpoint contributions. Individual terms depend on the perturbation gauge; observable temperature anisotropy requires the consistent emitter and observer contributions.
In pure matter domination the gauge-invariant scalar potentials are constant for a growing linear mode: the matter contrast grows as while the potential scales as its contrast divided by . Thus growth of this is compatible with no late Integrated Sachs-Wolfe effect in Einstein-de Sitter. When dark energy suppresses matter growth, the potentials generally evolve; photons gain and lose unequal energies entering and leaving a potential well. In a convention where are the dimensionless Newtonian-gauge potentials, the integrated temperature source is along the light path.
Large-angle cosmic microwave background data can constrain that late contribution, but low-multipole power alone has large cosmic variance and depends on the primordial radiation spectrum, geometry and foreground treatment. Cross-correlating the temperature map with low-redshift galaxies or other matter tracers isolates a late-time connection; decaying wells typically correlate overdensities with hotter cosmic microwave background directions. Boughn and Crittenden compared WMAP temperature data with the HEAO1 hard X-ray background and examined possible foreground origins of the correlation. This tests potential evolution under a cosmological model, not dark energy independently of all gravity and curvature assumptions.[ref]
A cosmological constant has fixed density and equation of state . A separately conserved dynamical component obeysIf departs from , its history changes the Hubble parameter, luminosity distances and angular-diameter distances. Type Ia supernovae probe a relative distance-redshift relation, baryon acoustic oscillations provide standard-ruler distances and expansion information, and the cosmic microwave background fixes early-time scales and geometric combinations. Combining these breaks some degeneracies with matter density, curvature and calibration. Distance observables integrate the expansion history, so they smooth rapid variations in and cannot determine an arbitrary function exactly.
Structure-growth measurements add a dynamical test. Galaxy clustering, redshift-space distortions and weak gravitational lensing constrain the growth of matter perturbations and metric potentials. The late Integrated Sachs-Wolfe effect probes potential evolution through CMB-matter correlations. A scalar-field dark-energy model can have perturbations and an effective sound speed different from other models with a similar background expansion. Comparisons of geometry and growth can therefore distinguish some models and test gravity assumptions.
A statistically significant, robust departure from or from constant dark-energy density would disfavor a pure cosmological constant within the stated framework. Consistency with over a finite redshift range does not prove exact constancy: a slowly rolling field can mimic it. Supernova calibration/evolution, galaxy bias, photometric redshift errors, intrinsic galaxy alignments and foregrounds must be controlled; modified gravity can also change growth. Parameterizations such as are useful finite-dimensional tests rather than a complete classification of all dark-energy histories.
Past exam of the physics A course of the University of Cambridge 2022 Paper 1 Section A 1 Solution Created 2026-10-18 Updated 2026-10-20
Let the infinite outer walls be at , the finite steps at , and the three potential energy values be . The original drawing specifies the orderingThe energy eigenstates can be chosen real. The Dirichlet boundary conditions are ; wavefunction matching at a finite potential step requires both and to be continuous at . In particular, neither finite step acts as an additional hard wall.
The time-independent Schrödinger equation givesThus a wavefunction is sinusoidal in a classically allowed region, with wavenumber , and a linear combination of exponential functions in a classically forbidden region, with inverse decay length . Its curvature changes sign relative to at a step crossing ; its slope does not jump.
By the Sturm oscillation theorem, the three energy eigenstates have exactly zero, one, and two interior nodes, respectively. Choose their global phases so that their slopes at the left wall are positive:
- : a positive lobe in the left-hand potential well, followed by a nonoscillatory tail through the barrier and right-hand region. The tail reaches zero at the right wall. There are no interior nodes.
- : two lobes of opposite sign in the left-hand potential well, with one interior node, followed by a nonoscillatory tail of the sign of the second lobe. Both the middle and right regions are classically forbidden regions.
- : sinusoidal behavior in both outer regions, joined through the nonoscillatory barrier. There are two interior nodes in total. The right-hand wavelength is longer than the left-hand wavelength, because .
For , the right-hand piece is proportional to , rather than a pure decaying exponential: this enforces the finite-distance hard wall exactly. The barrier pieces likewise contain both exponential solutions. Their small tails are an instance of quantum tunnelling.
This is an original illustrative calculation, with vertical lines marking only the region boundaries. It uses , , , , , , and in arbitrary energy units. Its lowest energy eigenvalues are approximately , so it has exactly the required ordering. Here both nodes of the third wavefunction happen to lie towards the left; its dominant lobe is on the right. These precise positions and relative amplitudes are illustrative: the original drawing supplies no numerical lengths, barrier heights, or mass, so it does not determine a unique quantitative wavefunction. The required sketch features are the matching, curvature, hard-wall zeros, and ordered node counts.
Past exam of the physics B course of the University of Cambridge 2011 Paper 1 Section A A5 Solution 2026-10-20
Use a local horizontal plane, constant gravity and first order in Earth's angular speed. Let point along the initial horizontal shot, to the gunner's left and upward. At the North pole, . Maximum unperturbed range at equal launch and landing heights requires a elevation, givingThe leading Coriolis acceleration is , while vertical motion has no first-order Coriolis correction. With zero initial lateral displacement and velocity,Here , and . The miss is to the right of the aiming direction, independent of its azimuth at the pole. The flight time is about , so . This polar projectile Coriolis deflection uses the usual flat, uniform-gravity approximation; planetary curvature and variation of gravity are beyond its accuracy. Using the polar gravity rather than changes the estimate by less than one percent. Earth's sidereal rotation period, rather than the solar day, sets .[ref]
Earth Fact Sheet by National Aeronautics and Space Administration. Orbital parameters: sidereal rotation period 23.9345 hours; bulk parameters: polar surface acceleration
Polar projectile deflection
. Details. Past exam of the physics B course of the University of Cambridge 2011 Paper 2 Section C C11 Solution 2026-10-20
Let denote distance from the neutral axis within a cross section. Bending to radius gives longitudinal strain , so Hooke's law gives stress . Its moment isHere is Young's modulus and is the second moment of area, not a mass moment of inertia. For the stated square orientation, .
For self-weight deflection of a square cantilever, the downward load per unit length is . A free-end segment from to has shear-force magnitude and bending-moment magnitude . Thus, with the given sign conventions,Using the small-slope curvature for downward deflection,The clamp fixes position and slope, and the unloaded free end has zero moment and shear:Four integrations yieldThe linear beam model requires small slopes and strains; the strong dependence eventually invalidates that approximation if other dimensions and material properties are fixed.




