Past exam of the mathematics course of the University of Cambridge 2011 ii Paper 1 14E ii Solution Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2012 ia Paper 3 12C ii a Solution 2026-10-07
The Dirichlet boundary condition gives , so the unique radial solution isIts radial Laplacian is one, it is smooth at the origin, and its boundary value is zero. The uniqueness result in part (i) shows that this is also the only solution among all classical functions on the ball, not merely among spherically symmetric ones.
Past exam of the mathematics course of the University of Cambridge 2012 ia Paper 3 12C i Solution Created 2026-09-24 Updated 2026-10-07
For two solutions of the Poisson equation with the same Dirichlet boundary condition, let . Then in and on its boundary. By Green's first identity,Thus is constant on each connected component, and the zero boundary values make each constant zero. The Dirichlet solution is unique whenever it exists. This argument requires the regularity needed for the displayed integration by parts; it does not assert existence for arbitrary rough data.
With a Neumann boundary condition, integrating the Poisson equation and using the divergence theorem instead gives the necessary compatibility conditionOn a disconnected region this condition must hold on each connected component separately. If two solutions have the same normal derivative, their difference is harmonic with . The same Green's first identity again forces . Thus Neumann solutions are unique only up to an additive constant on each connected component. Adding a constant preserves both the equation and normal derivative, so uniqueness without a normalization cannot hold. On a connected region, prescribing the mean value removes this one-dimensional freedom.
Past exam of the mathematics course of the University of Cambridge 2012 iii Paper 12 2 Solution Created 2026-10-03 Updated 2026-10-07
Consider two finite assemblies of the same number of congruent bounded Euclidean tiles. Label their matching faces so the same prescribed face identifications are used in both assemblies. For each face type , encode the first assembly by a symmetric involution : a paired face exchanges its two tile indices, an exterior Dirichlet boundary condition contributes diagonal , and an exterior Neumann boundary condition contributes diagonal . Let encode the second assembly. Face coordinates are pulled back to the same reference face; if identifications have additional face isometries, these pullbacks must be included in the intertwining operators.
The transplantation theorem says that a constant matrix with for every face type sends the vector of tile restrictions of a Laplacian eigenfunction on the first assembly to one on the second, by . If is invertible, it is a bijection of each eigenspace and the assemblies are isospectral with multiplicities.
To prove it, each component solves the same interior equation , and a constant linear combination solves that equation too. On a face, let be the vector of boundary values and the vector of outward normal derivatives. All matching and exterior conditions are exactlyFor an internal face these equations say that values agree and the two outward derivatives sum to zero. For diagonal they impose zero value, and for diagonal zero normal derivative. The intertwining identity gives and , so the transplanted functions match and satisfy the correct boundary conditions. Applying proves the eigenspace bijection. At corners the same reasoning is interpreted in the finite-energy weak domain: no value jump and cancellation of normal flux prevent an extra distributional source. It gives the usual self-adjoint Dirichlet/mixed Laplacians on polygonal assemblies.
Here is an explicit transplantation by reflection parity. Take a rectangle of width and height , with the Dirichlet boundary condition on every exterior side. Cut it at its vertical midline into two congruent half-rectangles, pulling the right-hand half back by reflection. The midline gluing matrix isReplace the connected assembly by the disjoint union of two width-, height- rectangles: the first has Dirichlet conditions on all sides; the second has Dirichlet on three sides and Neumann on the side representing the cut. Its corresponding matrix is . The invertible, indeed orthogonal, matrixintertwines the midline conditions. Every other face has matrix on both assemblies, which automatically intertwines. The transplantation theorem proves equality of the complete spectra, although one surface is connected and the other has two components with the required uniform and mixed conditions.
Equivalently, odd reflection eigenfunctions vanish on the cut, while even reflection eigenfunctions have zero normal derivative there. A direct separation-of-variables check gives the connected rectangle's eigenvaluesEven gives the fully Dirichlet half-rectangle spectrum, and odd , , gives the mixed half-rectangle spectrum. The split preserves every multiplicity, including coincidences between different pairs of indices.
Past exam of the mathematics course of the University of Cambridge 2012 iii Paper 50 1 Solution Created 2026-10-03 Updated 2026-10-07
Set , , and . The Trotter product formula separates the kinetic and potential energy evolution to first order in . Inserting resolutions of the identity in the position operator basis and using the free short-time kernel givesThe exponent converges formally to the action along the path. The prefactors and intermediate integrations define the time-sliced configuration-space path integral measure, soA real-time path integral is an oscillatory limit with the Feynman i-epsilon prescription, rather than an ordinary probability integral. The operator derivation assumes the usual self-adjointness and product-formula hypotheses for the Hamiltonian.
For a quadratic potential energy write , with and . The first variation vanishes by the Euler-Lagrange equation and the fixed endpoints. Because is quadratic, the remaining expansion is exact:Here integration by parts has no boundary contribution. Translation of the integration variables leaves the time-sliced measure unchanged. Thus the Gaussian path integral over is independent of the endpoint values, and the entire endpoint dependence lies in .
Expanding in eigenfunctions satisfying Dirichlet boundary conditions, each real mode contributes an inverse square root of its quadratic eigenvalue, with the oscillatory phase fixed by continuation. Normalizing against gives a precise determinant-ratio form:The determinant ratio is meaningful after a common regulator; the phase is part of the prescription. It is equivalent to writing with an endpoint-independent normalization.
For example, if , the Dirichlet oscillator determinant ratio isA linear or constant term in changes the action but not this prefactor. The formula holds away from conjugate times, when the boundary problem and determinant are nonsingular. At , it must be continued as a distribution with its Maslov index phase; the assertion of an ordinary finite prefactor cannot be used literally there. For the unshifted oscillator, at the kernel is .
For the full-line source integral, impose vacuum boundary conditions by adiabatic damping. Take the source initially as a test function, so its pairings with the Green function are defined. After integration by parts, writeWith the Fourier transform convention , the inverse isSet . Completing the square givesand consequentlyThe Gaussian functional integral is source independent and contains the regulated determinant. This is most naturally read as a normalized vacuum generating functional. For , closing the frequency contour gives ; the time-ordered oscillator two-point function is . This checks the sign of the source exponent and identifies the required vacuum prescription.
Past exam of the mathematics course of the University of Cambridge 2012 iii Paper 51 2 Solution Created 2026-10-03 Updated 2026-10-07
Let be the induced worldvolume metric and write . Varying the embedding in the auxiliary-metric brane action and integrating by parts givesThe boundary contribution, including its relative factor, isAt initial and final times one fixes the endpoint configurations or uses variations of compact support. At a spatial boundary, free target directions require Neumann boundary conditions ; fixed directions require Dirichlet boundary conditions . Mixed conditions must make this boundary pairing vanish. A closed brane has no spatial boundary. Since the auxiliary worldvolume metric enters without derivatives, its variation produces no extra boundary term.
Using , variation of the inverse worldvolume metric givesTaking the trace yields . For , this forces , and substitution gives . For a nondegenerate timelike embedding, eliminating the auxiliary worldvolume metric therefore gives twice the worldvolume area, with the overall physical brane tension supplied by the action normalization.
For , the traced metric equation is an identity, and the remaining equation only says . It fixes the worldsheet metric up to a Weyl transformation, rather than determining it uniquely. Indeed is Weyl invariant precisely in two worldvolume dimensions, and the constant term vanishes precisely at . This is the Weyl-invariance exception for the string among branes. In a nondegenerate interior it gives the familiar classical equivalence of Polyakov and Nambu–Goto actions; degeneracies at a free string endpoint must be treated through the original equations.
For the open string, choose conformal gauge on . The embedding equation is the wave equation, and NN means at both spatial endpoints. For these operator formulas restore the conventional overall normalization , with ; the overall factor does not change the preceding classical equations. Its open-string mode expansion isReality requires . Canonical quantization givesThe metric equation is the vanishing of the worldsheet stress tensor: . The two endpoint-compatible mode expansions contain the same string oscillator family. Their quadratic coefficients are the classical Virasoro constraints.
For the operators, use normal ordering with positive-index string oscillators as annihilators:Thus has modes before the quantum ordering correction. The physical-state Virasoro conditions for an open string areOne imposes only the positive modes on kets, with the adjoint conditions on bras, as in Gupta-Bleuler quantization. Requiring every positive and negative mode to annihilate the same state would conflict with the Virasoro central extension. The intercept is the zero-mode ordering constant. The standard critical bosonic string theory has and ; in light-cone gauge in string theory the transverse zero-point energy gives , while full anomaly-free Lorentz or BRST quantization fixes the critical values. The mass constraint is then , with .
To compute the Virasoro algebra, commute a quadratic generator with one string oscillator:These identities and the Jacobi identity imply that commutes with every string oscillator and with the center-of-mass coordinates and momenta. Mode number permits a scalar term only for . Its coefficient follows from the formal zero-momentum Fock vacuum, on which . For ,The two possible string oscillator contractions giveThe timelike target coordinate still contributes one to this central charge: its two metric signs cancel in . ConsequentlyThis is the free-boson Virasoro central term. It vanishes for the three global conformal modes .
To keep the intercept convention separate, define . The Virasoro zero-mode shift changes the displayed central term toThe matter central charge here is , not . Covariant worldsheet ghost fields contribute ; their inclusion cancels the anomaly at , while the intercept is handled by the appropriate zero-mode and physical-state convention.
Past exam of the mathematics course of the University of Cambridge 2013 ib Paper 2 16B Solution Created 2026-09-24 Updated 2026-10-07
For the Dirichlet Green's function, solve with zero endpoint values. Away from , is linear. Continuity at and the derivative jump giveFor nonzero Dirichlet boundary conditions, subtract their linear interpolant. The resulting boundary value problem has solutionIn the final heat equation, a steady state has zero time derivative. Here , , , . Integrating twice and imposing the endpoint values givesIts second derivative and both boundary conditions verify the answer directly.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 44 1 Solution Created 2026-10-03 Updated 2026-10-07
Write , with . Independent variations of and give the Euler-Lagrange equation . Away from its unique endpoint solution isIntegration by parts cancels the linear fluctuation terms and gives . On the classical solution, the action is the boundary term . Substituting the endpoint derivatives therefore givesThe remaining Gaussian path integral contains two real fluctuation coordinates per mode, so it contributes an inverse functional determinant, rather than its inverse square root. Thus , with the measure normalization fixing the otherwise arbitrary constant.
For Dirichlet boundary conditions, the normalized sine modes have eigenvalues , . Their determinant ratio is the convergent Dirichlet oscillator determinant ratioThe last equality is the sine infinite product. With the prescribed free determinant this givesThe kernel uses the Feynman i-epsilon prescription; at a caustic it is a distributional limit, not an ordinary finite function. Its limit is .
There is a sign error in the printed complex-integral hint. For a positive damping parameter , polar integration gives the regulated complex Fresnel integralThis regulated value, together with the stated free-kernel normalization, fixes the phase consistently.
For and positive imaginary-time length , Wick rotation givesPut with and use the convergent real Gaussian integral. Writing givesThis is the parity-twisted oscillator thermal trace. The complex coordinate describes two independent real quantum harmonic oscillators, each with mass two in these units. Their total energy is ; level has degeneracy , and spatial inversion has parity operator eigenvalue . The plus sign is ; the minus sign is . The alternating trace inserts parity into a bosonic system; it does not change the oscillators into fermions.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 46 1 Solution Created 2026-10-03 Updated 2026-10-07
Actions and equations. Use a mostly-plus Minkowski metric and define the induced worldsheet metric . For a nondegenerate timelike worldsheet, the Nambu–Goto action isVarying the independent metric in the Polyakov action sets its worldsheet stress tensor to zero:In two dimensions this says for a positive local factor. The factor drops out of , and substitution gives the Nambu–Goto action. Conversely, any nondegenerate induced metric solves the auxiliary-metric equation up to a Weyl transformation. This classical equivalence of Polyakov and Nambu–Goto actions is a statement about classical embeddings; quantum equivalence additionally requires treatment of the metric measure and anomaly.
Variation of in the metric action, followed by elimination of , gives the Nambu–Goto equations of motionA closed string has periodic and no spatial endpoint variation.
For a background metric , Kalb–Ramond field , and dilaton , one consistent Lorentzian convention isThe orientation fixes the two-form sign; here the Lorentzian curvature convention is chosen so Wick rotation gives the positive Euclidean dilaton term . This avoids hiding the convention in the topology argument. For constant vacuum value , the Gauss-Bonnet theorem gives . A connected closed oriented surface of genus has Euler characteristic , soThe connected vacuum amplitude has a string genus expansion ; disconnected vacuum diagrams exponentiate the connected sum. Each additional handle supplies a factor . This is dilaton Euler-characteristic weighting.
The rotating circle. For the specified embedding, the squared spatial speed and tangent length are both , and their spatial inner product is zero. Including therefore givesThe metric is constant. Its equation reduces to , satisfied by the left- and right-moving trigonometric components and the linear time component. On a constant-time slice with , the proper length of a string isIt is time-independent.
Let be an ordinary two-dimensional rotation matrix. Rotate the first coordinate pair by and the second by . In these time-dependent Cartesian coordinates both pairs become . A further fixed orthogonal change of basis givesThus every spatial slice is a planar circle of radius ; its plane rotates in the ambient four-space. This is a rigid circular string in four spatial dimensions, not a pulsating circle. The rotating axes establish its spatial shape, not a transformation to an inertial spacetime frame.
The momentum density obtained from the Nambu–Goto action is . Here , hence the conserved target-space energy isThe material velocity is transverse to the tangent and has magnitude . Its Lorentz factor is , so the excess over is kinetic energy. The spatial circle being stationary in rotating axes does not eliminate this energy.
Constraints and endpoints. Varying the multipliers in the Nambu-Goto phase-space action imposesThese are first-class constraints generating normal and tangential worldsheet diffeomorphisms. They remove the two longitudinal embedding degrees of freedom; they are not extra physical force laws. Hamilton's equations areFor an open string, integration by parts leaves the spatial boundary variationAllowed variational open-string boundary conditions must make this vanish for every permitted endpoint displacement, and be preserved by the evolution. The bracketed expression is the open-string endpoint momentum flux. For free displacements, set that flux to zero. In a boundary-preserving gauge with and finite nonzero , this is the Neumann boundary condition at each end. This shows that free-end string boundary conditions are consistent.
At such an end, gives , and gives . For a nontrivial endpoint trajectory with nonzero time velocity, its spatial speed is therefore one. This null motion of a free string endpoint follows from the boundary condition together with the constraints, not merely from the bulk wave equation. The induced metric may degenerate right at a free end; the result is understood as the endpoint limit or in the Polyakov description.
For a fixed spatial -plane, split coordinates into along its -dimensional worldvolume, including time, and transverse to it. Set and at the ends. Tangential variations are free and their momentum flux vanishes; normal variations vanish, so their flux need not vanish. Evolution preserves the fixed normal values with at the boundary in the same gauge. These mixed Neumann boundary conditions and Dirichlet boundary conditions consistently restrict the endpoint worldlines to the plane.
The massless open-string excitations then separate into a gauge vector along the worldvolume and transverse scalar fields. A scalar displacement changes the corresponding ; its interpretation is a fluctuation of the plane's embedding. These worldvolume fields from open-string massless states support the interpretation as a dynamical planar D-brane. In the bosonic theory the still lower ground state is a tachyon, indicating an unstable brane; the interpretation does not require pretending that this tachyon is a stable massless field. Appropriate superstring sectors can remove that instability.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 63 1 2 Solution 2026-10-07
The zero Dirichlet boundary conditions permit an exact finite-interval calculation. Let on the interior grid points. The Dirichlet discrete Laplacian has an orthogonal eigenvector basisConsequently the update matrix has modal amplification factorsFor every physical , all denominators are at least one. Orthogonal diagonalization therefore provesuniformly in , and . The same bound controls initial perturbations; a forcing increment is propagated by contraction, so successive increments accumulate at most by their sum. The highest-order consistent method is unconditionally stable for all . This proof incorporates the boundary conditions, whereas a periodic Fourier mode calculation alone would not do so.
For completeness, if “range” is interpreted algebraically to include negative on a fixed grid, the exact power-stable range isIndeed, for every denominator is less than one, so requires for every . The strongest restriction comes from . Equality gives a simple amplification factor and is allowed because the update is orthogonally diagonalizable. Other negative values either amplify a mode or make the update singular. This extra branch describes backward time stepping on a fixed spatial grid; it is not a positive-time diffusion discretization, and no fixed negative remains in that branch as .
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 63 3 3 Solution Created 2026-10-03 Updated 2026-10-07
The appropriate bounded-operator formulation uses the zero-boundary Sobolev spaceThe Poincare inequality makes this a Hilbert space norm equivalent to the usual norm, and the zero endpoint traces encode the Dirichlet boundary conditions. Under the usual regular-coefficient assumptions, for example continuous on the closed interval with and , setIf , then Cauchy-Schwarz inequality and the Poincare inequality giveBy the Riesz representation theorem, a unique bounded linear operator satisfies . Symmetry of makes self-adjoint, and the lower bound makes it elliptic and uniformly positive definite. The differential expression in the question is represented weakly by : for smooth functions, integration by parts giveswith the boundary term zero. A forcing becomes the bounded functional , or its Riesz representative in .
Equivalently, for sufficiently smooth coefficients one may realize the differential operator itself on with domain . Its regular Sturm-Liouville operator realization is self-adjoint andThis realization is an unbounded operator, so it should not be confused with the bounded weak operator used above.
The PDF states sign conditions without coefficient regularity. If only almost everywhere and are bounded, the same form is still bounded and strictly positive: a zero energy would force almost everywhere and the zero traces force . Uniform ellipticity, however, requires a positive lower bound. For instance for , , and satisfy literal pointwise positivity. Derivatives of unit norm supported in and of integral zero define functions in , with energy at most . Thus pointwise positivity without regularity does not imply coercivity in this norm. These distinctions supply the precise regularity and operator domain behind the intended positive-definiteness proof.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 63 7 Solution Created 2026-10-03 Updated 2026-10-07
A stability analysis asks whether perturbations in an evolution remain bounded on a fixed physical time interval by a constant independent of the mesh. For a linear update , the basic requirement isThis permits physical growth bounded by ; it need not require every step to be a contraction. Consistency of a numerical method measures how well its stencil reproduces the differential equation, whereas stability of a numerical method controls the accumulated defects. For a well-posed linear problem, the Lax equivalence theorem connects stability and convergence for a consistent approximation in the specified norm.
On the whole line or a periodic grid, constant-coefficient stencils commute with translations. The Fourier modes diagonalize such spatial operators, turning a scalar one-step scheme intoThe Fourier symbol is found simply by replacing a shift by . The Parseval identity makes this modal calculation a norm estimate: if at every resolvable frequency, then the discrete norm cannot increase. More generally, a uniform gives and a fixed-time stability bound. All frequencies matter, including those near the grid scale; a small-wavenumber expansion establishes consistency but cannot test the entire stability range.
For the heat equation, put and . The Forward Euler diffusion scheme hasThe condition for every is exactly . This illustrates a parabolic time-step restriction . The Backward Euler diffusion scheme instead haswhich is contractive for every . The Crank-Nicolson diffusion scheme hasand is likewise unconditionally stable. However, its high-frequency amplification approaches for large , so it need not damp rapidly decaying modes effectively and may produce alternating transients from nonsmooth data. An A-stable time integrator can stabilize a diffusion semidiscretization with left-half-plane spectrum, but damping and accuracy remain separate questions.
For transport , define the hyperbolic Courant number . Forward Euler with a centered spatial difference gives and . It is unstable under a fixed nonzero CFL ratio. For , the upwind finite difference scheme hasThus is the stable range. Upwinding for negative reverses the spatial difference and gives . The Lax-Wendroff advection scheme gives a second-order example:It is stable for , but that result does not imply monotonicity or exclude oscillations at shocks. The Courant–Friedrichs–Lewy condition has a domain-of-dependence interpretation for explicit hyperbolic schemes; the Fourier calculation supplies the sharper algebraic stability criterion for the particular stencil.
For multilevel methods, each frequency gives an amplification polynomial and a companion matrix. Every amplification root must lie in the closed unit disk, with unit-modulus roots simple. Uniformity across frequencies and meshes is essential: one also needs control of the associated matrix powers, for example through a uniformly bounded diagonalization or the uniform power bound from separated amplification roots. A repeated unit root generates a Jordan block and polynomial growth in the number of steps. The same issue occurs for systems: checking only eigenvalue moduli of a nonnormal Fourier amplification matrix is insufficient. For instancehas unit eigenvalues but no fixed-time mesh-uniform bound as . For a hyperbolic system with a constant coefficient matrix admitting a uniformly bounded characteristic diagonalization, scalar modal estimates for each characteristic speed do give a system estimate. A positive energy symmetrizer provides another route. Frozen-coefficient Fourier tests for variable coefficients are useful local diagnostics, but a global estimate must also account for coefficient variation and boundary terms.
Boundary conditions change the allowable modes and can change the conclusion. A periodic calculation is exact only for a periodic closure. On a finite interval with zero Dirichlet boundary conditions, the centered diffusion matrix is diagonalized instead by the discrete sine transform. On interior points its eigenvalues are before scaling by . Thus the exact fixed-grid Forward Euler restriction iswhich tends to the whole-line threshold as the grid is refined. Backward Euler and Crank-Nicolson remain stable for all nonnegative . An appropriate Neumann boundary condition uses cosine-type modes and includes a constant mode with eigenvalue zero. The constant mode must be preserved, rather than forced to decay; depending on the boundary stencil, the natural discrete inner product may have endpoint weights.
For hyperbolic transport, the continuous and discrete boundary closures must respect the direction of propagation. If on , prescribe the left inflow value; prescribing an independent outflow value can overdetermine the problem. The energy method exposes this directly:Homogeneous inflow is dissipative, while forced inflow contributes data to the estimate. A compatible upwind boundary update has the same one-sided information flow. For diffusion,The last term vanishes for homogeneous Dirichlet or Neumann conditions. With Robin boundary conditions of dissipative sign it contributes a further nonpositive term. A non-dissipative boundary can produce genuine PDE growth, which a valid stability estimate should represent rather than incorrectly suppress.
Even a perfectly stable interior symbol cannot detect every defective boundary closure of a difference scheme. As a concrete counterexample, leave a stable explicit heat stencil on all grid points away from the left endpoint, but replace its first interior row by , still keeping . The interior Fourier symbol is unchanged, yet makes the finite-interval update unstable. The faulty boundary-adjacent row creates a mode absent from the periodic interior calculation. Thus the entire finite update matrix, or an energy estimate including its boundary rows, must be examined.
A more systematic boundary stability of a finite-difference method analysis transforms time and any tangential spatial variables, then looks for modes with and on a half-line. The interior dispersion relation selects decaying spatial modes; the boundary equations must determine their coefficients without admitting a nonzero growing homogeneous mode. A uniform inverse estimate for the boundary system, expressed by the Uniform Kreiss--Lopatinskii condition, also rules out arbitrarily weak boundary control near limiting frequencies. Merely excluding one obvious unstable mode is weaker than proving a mesh-uniform estimate. Discrete summation by parts and suitable boundary penalties offer an energy-based alternative.
Finally, nonlinear scalar conservation laws can develop discontinuities, so linearized Fourier stability is only one part of the analysis. A monotone conservative scheme, such as the Engquist-Osher method under its CFL restriction, has an invariant state range and L1 contraction of a monotone conservative scheme, with a discrete entropy inequality selecting admissible shocks. These complement the Fourier and boundary estimates. For an evolution problem, the meaningful outcome is a stability bound in a specified norm, uniform over the chosen mesh family and compatible with the actual boundary and initial data.
Past exam of the mathematics course of the University of Cambridge 2014 ib Paper 1 16A ii Solution Created 2026-09-24 Updated 2026-10-06
The method of images places an image charge at the reflected position below the plane. The candidate potential in the physical upper half-space isAt the two terms cancel, so and . Direct differentiation givesThus the upper-side field satisfies exactly the conductor boundary conditions from part (i). The fictitious pair does have a field below the plane, but that part of its solution is not used for the physical conductor.
In , the potential has the correct real point source, the correct zero Dirichlet boundary condition, and the required decay at infinity. By Uniqueness of the Dirichlet problem, it is the physical potential there. This is why the artificial charge can represent the induced conductor response without being a second physical charge.
Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 46 1 Solution Created 2026-10-03 Updated 2026-10-06
Use the mostly-plus Minkowski metric, and write for equality on the constraint surface. Assume that the mechanical constraints are locally independent. They are first-class constraints whenThus their Poisson brackets vanish on the constraint surface, and their Hamiltonian flows preserve that surface. The structure functions of a constraint algebra may depend on the phase space point. The finite real span of the constraints is a Lie algebra if it closes with constant structure coefficients, in a suitable choice of generators. The Jacobi identity then gives the usual conditions on the structure constants of a Lie algebra. With general structure functions the finite real span need not close, even though the Poisson bracket of all smooth functions is itself a Lie bracket.
To see the gauge invariance directly, let generate a canonical gauge transformation:The variation of the phase-space action integrand isThe second term cancels without using the equations of motion. Taking to vanish at the temporal boundaries leaves the action invariant. Arbitrary functions therefore relate different descriptions of the same physical motion. This reasoning also works with structure functions; constant structure coefficients are only needed for the finite-dimensional Lie algebra claim.
For a closed string, choose and periodic fields. A convenient Nambu-Goto phase-space action isHere and are Lagrange multipliers. The Nambu–Goto phase-space constraints are and , with canonical Poisson bracketsLet . Differentiating the periodic Dirac delta function givesThe opposite signs in and cancel these terms, so . Replace the original constraints by the equivalent chiral densitiesTheir mixed Poisson brackets vanish. Choose opposite Fourier orientations for the two sectors:The chiral constraint algebra of a closed string isEach is the Witt algebra: the vector fields on a circle satisfy . Fourier expansion identifies each real algebra, with , with the Lie algebra of vector fields on the circle. The two commuting copies give , not a quantum central extension.
For an open string, allowed boundary conditions must remove the endpoint term in the variation of the action, consistently with the allowed endpoint variations. The spatial boundary term isIt expresses the open-string endpoint momentum flux. In the temporal gauge for a string , take a boundary-adapted parametrization with at the ends. Fixing gives , a Dirichlet boundary condition. At the other end allow arbitrary spatial variations; for nonzero these require , a Neumann boundary condition. Also in this temporal gauge for a string, so . The constraint at this free-end string boundary condition reduces to . Hamilton's equation consequently gives there. Since , the free endpoint has spatial speed one. This is the null motion of a free string endpoint.
A straight rotating string with one fixed endpoint supplies the required solution in at least two spatial dimensions. Set , , andTake . The Hamilton's equations become , which holds because both second derivatives give . The Nambu–Goto phase-space constraints are satisfied byThe endpoint at stays at the origin, while at and the endpoint moves around a circle of radius with angular speed . At each time the whole string lies on a straight radial segment. Its spatial proper length isThe velocity is everywhere perpendicular to the segment, so this also equals the sum of local rest-frame lengths. The induced worldsheet metric becomes degenerate at the null free endpoint, as expected for the limiting free-end solution.
Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 66 1 a Solution Created 2026-10-03 Updated 2026-10-06
Take real and use the L2 norm on the spatial interval. Existence, uniqueness and continuous dependence are the three requirements of Hadamard well-posedness. An energy method supplies the decisive estimate. For a smooth solution with homogeneous Dirichlet boundary conditions, integration by parts givesThe drift contributes only a boundary term, which vanishes. The Poincare inequality further givesApply the same argument to the difference of two solutions to obtain uniqueness and continuous dependence on the initial data.
For existence, use the Dirichlet gauge transform for constant drift: satisfies with zero boundary values. Expanding in its Fourier sine series givesFor the series defines a solution continuous in down to and smooth for positive time; multiplication by the fixed bounded exponentials preserves this interpretation. Its energy estimate follows by approximation with smooth initial data. For a classical solution at the initial corners, require the usual smoothness and boundary compatibility instead. The problem is well posed in , with a contraction estimate independent of the initial data.
Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 66 7 Solution Created 2026-10-03 Updated 2026-10-06
The finite element method replaces an infinite-dimensional variational problem by a finite-dimensional one, usually using functions that are polynomial on small mesh elements. Ritz method and Galerkin method describe how the discrete equations are selected; neither intrinsically requires piecewise polynomials, though those spaces make assembly local and sparse.
For a concrete realization of the two-point problem, choose homogeneous Dirichlet boundary conditions at both endpoints of . The differential expression alone is not a complete boundary value problem, so this boundary choice is an explicit assumption. Let with , with , and . Use the zero-boundary Sobolev space with norm . By the Poincare inequality this is equivalent to its full first-order Sobolev norm. Multiplying by a test function and applying integration by parts givesThe endpoint terms vanish by the essential boundary condition. The form is a symmetric bilinear form, bounded byand it is a coercive bilinear form:The load is a bounded linear functional. The Lax-Milgram theorem gives a unique weak solution and a bound . Under the stated coefficient regularity it solves the differential equation in the usual weak sense.
The Ritz method minimizes the energyover , or over a conforming finite-dimensional subspace . Its first variation is ; symmetry and coercivity make the functional strictly convex. In fact, if is the weak solution,Thus the energy minimizer is exactly the weak solution. The Galerkin method instead directly asks for with for all . For this symmetric coercive problem, Ritz-Galerkin equivalence for a symmetric coercive form shows that the discrete minimizer and Galerkin solution coincide. For a nonsymmetric problem, a Galerkin formulation can still apply but the same quadratic minimization generally cannot represent its full bilinear form; appropriate coercivity or inf-sup conditions must then justify the chosen spaces.
Subtract the continuous and discrete weak equations to obtain Galerkin orthogonality . Since defines the energy norm , the Pythagorean identity givesIn a general reference norm, the Céa lemma yieldsThe proof uses together with coercivity and continuity. Dense approximation spaces therefore imply convergence; the estimate reduces a numerical error problem to an approximation problem.
For implementation, partition the interval by nodes . Choose continuous piecewise-linear functions and the interior piecewise-linear hat functions as a basis, with the two boundary coefficients prescribed. Write . The coefficients satisfyThe stiffness matrix is symmetric positive definite: for nonzero interior coefficients. The local supports make it tridiagonal in this one-dimensional linear-element case.
On an element of length , for constant element coefficients , the two-node matrix and constant-load vector areFor variable coefficients, integrate against the same shape functions rather than silently replacing them by constants. Element contributions are added at shared nodes, and boundary values are eliminated or lifted into the right side. A sparse direct solve or a suitable iterative method gives the nodal coefficients. Numerical quadrature should preserve the required accuracy and, with positive coefficients and weights, the positive energy structure.
A finite element interpolation estimate gives for these elements, where is the largest element size and the solution has the stated regularity. The Céa lemma then gives first-order energy error. If the dual elliptic problem has regularity, the Aubin–Nitsche duality argument gives second-order error: for , solve , then use to gain one further factor of . Higher-order elements improve rates when the solution is sufficiently smooth.
Nonzero Dirichlet data are handled by a boundary lifting and an affine trial space. Neumann boundary conditions enter naturally through the integrated boundary term; Robin terms modify both the form and the load. With pure Neumann conditions and , constants lie in the kernel: existence needs the load/flux compatibility condition, uniqueness needs a mean constraint, and the preceding Dirichlet coercivity argument cannot simply be reused. Conforming approximation, boundary conditions and coercivity are the ingredients connecting the finite-element construction to a justified Ritz or Galerkin solution.
Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 69 1 iv Solution Created 2026-10-03 Updated 2026-10-06
The partial differential equation. Each term in is a translated heat kernel and satisfies for . The same holds for , either by direct differentiation or because it is . For , and its derivatives vanish faster than any power as . Therefore differentiating the boundary convolution creates no extra upper-endpoint term. Gaussian domination justifies differentiating both integrals on compact subsets of , proving the advection-diffusion equation.
The initial condition. For fixed , the first Gaussian in is an approximate identity centered at . Its integral tends to . The reflected Gaussian is exponentially small as , because its center lies outside the half-line. The boundary convolution also tends to zero for . Thus .
The Dirichlet boundary condition. The identityshows that , so the initial-data contribution vanishes at zero. The boundary contribution must be evaluated as a limit, not by substituting inside its singular integral. The positive kernel satisfiesfor every . Hence it is a one-sided approximate identity at zero time, and continuity of givesThe mass formula follows from the Gaussian Laplace integral, or from the decaying solution of the corresponding constant-coefficient ordinary differential equation. Compatibility gives the continuous corner value. These arguments also verify the equivalent contour solution in (iii). In the decaying energy class the solution is unique: the difference of two solutions has zero data and for real solutions, with the analogous modulus identity for complex solutions.
The unheaded sine transform question. The direct classical Fourier sine transform does not close on . Ifthen integration by parts givesThe drift introduces an unknown cosine transform; the usual scalar sine-transform solution of the heat equation is therefore unavailable directly.
A Dirichlet gauge transform for constant drift does provide a qualified alternative. Set . Then , with and . If these weighted data have the decay needed for an ordinary Fourier sine transform, it solves the transformed problem and produces exactly the kernel above. Mere decay of does not ensure this weighted integrability. Thus not directly by the classical sine transform of ; yes after a gauge transformation when the required weighted-transform hypotheses hold, or after a justified cutoff/limit argument.
