Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 324 3 b i Solution Created 2026-10-03 Updated 2026-10-06
Let denote the coherent exact quantum phase estimation unitary operator from the preceding part, with the data quantum register written first. For every eigenvector of ,because and lies in , with no phase-aliasing ambiguity. Apply coherently to the input quantum state and a clean quantum register; do not measure the eigenvalue label. By linearity it produces .
Append a quantum ancilla in , and use the provided quantum variable rotation with , choosing the nonnegative cosine. The resulting normalized quantum state isThe sum covers both terms; this fixes the potentially ambiguous sum placement in the abbreviated display. The orthonormal basis of eigenvectors and the unit length of each rotated quantum ancilla show that its norm is one. This is quantum spectral filtering by multiplication, not the reciprocal rotation used in the HHL algorithm.
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 339 1 a Solution Created 2026-10-03 Updated 2026-10-06
For a vector in the nonnegative orthant, the PDF's matrix gives the quadratic formThus it is a copositive matrix. On the other hand, with ,so it is not a positive semidefinite matrix. Equivalently, the eigenvalues are and , with eigenvectors and .
The local TeX has corrupted the matrix entries. The PDF has diagonal entries one and off-diagonal entries two. The all-twos matrix in the TeX would instead have quadratic form and would be a positive semidefinite matrix; it cannot demonstrate the requested distinction.
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 341 6 Solution Created 2026-10-03 Updated 2026-10-06
A Runge-Kutta method advances through stagesThe Butcher tableau records , normally with . Linear absolute stability tests the amplification of a linear decay mode; nonlinear B-stability compares distances between two solutions of a dissipative vector field. Both describe propagation of perturbations, but they impose different conditions.
For the Dahlquist test equation, elimination of stages givesThe linear stability domain consists of for which the step is defined and . Poles or singular stage systems must be excluded, even if an output formula has a formal cancellation. For a scalar test equation, the proof is : the powers are bounded exactly when . For , the update is . If is a normal matrix, unitary diagonalization of a normal matrix makes the operator norm equal to the largest scalar amplification modulus. For a diagonalizable matrix, powers are bounded by the condition number of the eigenvector matrix times the maximal scalar power. These constants must remain uniform for mesh-dependent systems. A defective unit-modulus eigenvalue can instead produce a growing Jordan block; the scalar spectral test alone does not control a general matrix.
An A-stable method accepts the entire closed left half-plane. The Explicit Euler method has and a disk of stability , so it is not A-stable. No nonconstant polynomial stability function can be A-stable, since it is unbounded on the negative real axis; this excludes all consistent explicit Runge-Kutta methods. The Backward Euler method has , and when , proving A-stability. It is also L-stable, because for large left-half-plane .
The implicit midpoint rule and the trapezoidal rule have the same scalar stability function,The identity proves A-stability. On the imaginary axis their amplification has unit modulus, and along the negative real axis it tends to . Thus neither is L-stable; very stiff decay can persist as alternating numerical values. In contrast, the two-stage Radau IIA method has , with the explicit nonnegative modulus-gap proof in question 1, and is L-stable. This illustrates why L-stability matters beyond A-stability for a stiff differential equation.
For nonlinear problems, assume a real or complex Euclidean inner product and the dissipativity hypothesisfor all relevant states at each common time. Exact solutions satisfyA B-stable Runge-Kutta method reproduces this nonexpansion of distances whenever its stages exist. Put , , and . Then . Expansion of the output squared norm, followed by substituting in its linear terms, proves the Runge-Kutta contractivity identityIf and is a positive semidefinite matrix, the first sum is nonpositive by dissipativity, and the second is nonnegative: expand each coordinate of the stage vectors to express it as a sum of nonnegative quadratic forms. These are exactly the conditions for algebraic stability of a Runge-Kutta method. Hence algebraic stability implies B-stability. Applying B-stability to the linear dissipative problem , or its two-dimensional real form, also proves A-stability whenever stages are well defined.
The Backward Euler method has , hence ; the implicit midpoint rule has , , hence . Both are algebraically stable and B-stable. For the two-stage Radau IIA method,a positive semidefinite matrix, while both weights are positive. It therefore combines third-order accuracy, B-stability, and L-stability. The distinction between stability and solvability is essential: for Lipschitz continuous , the stage map is a contraction when , so the contraction mapping theorem proves small-step unique solvability. The preceding algebraic stability proof is an estimate for existing stages; it is not a theorem asserting stage existence for arbitrary step sizes and domains.
Linear A-stability is insufficient for nonlinear B-stability. The trapezoidal rule fails B-stability, even though its scalar stability function is identical to that of the B-stable implicit midpoint rule. Its Runge-Kutta method coefficients areFailure of this sufficient algebraic stability test alone would not prove failure of B-stability. An actual counterexample does: take and . Dissipativity follows fromThe unique step map obeys , because the left-hand side is strictly increasing and onto. At , , and implicit differentiation givesNearby initial states expand in distance. This proves a true nonlinear failure rather than merely failure of a coefficient criterion.
Finally, stability of a numerical method on finite intervals connects these contractivity properties to accuracy. If a one-step map is nonexpansive and its exact one-step defect has norm at most , then the error satisfies , hence for . A Lipschitz step bound gives the same order using the discrete Gronwall inequality. For a stiff differential equation, the value of B-stability is that the propagation estimate need not grow with a large negative dissipative rate. The accuracy constant still needs appropriate smoothness and uniform derivative bounds; stability alone does not establish a mesh-uniform error order.
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 341 7 Solution Created 2026-10-03 Updated 2026-10-14
For a fully discrete linear partial differential equation evolution, write in a specified discrete norm. Finite-time stability of a numerical method meanswith independent of the allowed meshes. For a multistep method, augment the state with its time-history values and include a stable starter. Bounds may grow as when the continuous problem has growth; demanding decay for all time would be a stronger assertion. The treatment of initial data, forcing, Dirichlet boundary conditions, and the norm is part of the hypothesis, not a detail supplied by an interior calculation.
For constant coefficients on an infinite or periodic uniform grid, von Neumann stability analysis substitutes the Fourier mode . The discrete Fourier transform and the Parseval identity turn a uniform multiplier estimate into a discrete L2 norm estimate. In a one-step scalar scheme, proves contractivity, while gives finite-time stability. In a multilevel scheme, every root of a polynomial in its amplification polynomial of a multilevel finite difference scheme matters, as does uniform control of the associated companion matrix.
For the Forward Euler diffusion scheme with ,Thus for all frequencies exactly whenSufficiency follows since ; necessity follows from the highest-frequency mode on even periodic grids. This gives the usual parabolic mesh restriction . For the Backward Euler diffusion scheme,so every is stable. The Crank-Nicolson method hasagain of modulus at most one for all , but stiff modes approach rather than zero. These are PDE manifestations of the A-stability and L-stability distinctions for time integration.
For advection , assume and set . The upwind finite difference scheme is , withIt is contractive when . This also has a direct maximum-norm proof: each newest value is a convex combination of two old values. The resulting Courant–Friedrichs–Lewy condition expresses that the numerical domain of dependence covers the physical one. For negative , the upwind direction must be reversed. By contrast, Forward Euler method time stepping with the centered first finite difference has and for nonzero . Under fixed nonzero refinement it is unstable, since a fixed nontrivial mode grows geometrically over steps. This is not a claim of instability under every imaginable mesh coupling: instead bounds its spurious finite-time growth, since .
The leapfrog advection scheme illustrates a multilevel subtlety. Its amplification polynomial of a multilevel finite difference scheme isFor , both roots of a polynomial have unit modulus and separation at least . A uniformly conditioned eigenbasis of the companion matrix then proves power boundedness of a two-level Fourier scheme for arbitrary history data. At and a grid admitting , the roots coincide on the unit circle, and a Jordan block produces growth. Thus the endpoint fails the ordinary arbitrary-history root condition for a multistep method; checking only that both moduli equal one misses the instability. Restricting the starter or filtering the parasitic mode is an additional hypothesis.
The energy method handles variable coefficients and finite boundaries for which Fourier stability analysis may be unavailable. For the Backward Euler diffusion scheme with homogeneous Dirichlet boundary conditions, define . The discrete summation by parts identity isTake the inner product of with . The elementary identity givesThis proves unconditional contractivity including the actual boundary treatment. More generally, a dissipative operator gives of operator norm at most one: with , dissipativity implies , and the Cauchy-Schwarz inequality gives . In finite dimensions this also proves invertibility. For the second-order backward differentiation formula, the BDF2 discrete energy identity used in question 4 controls both time levels and proves unconditional diffusion stability. A repeated root strictly inside the disk is harmless here; the energy argument supplies the uniform bound without a singular eigenvector formula.
A second technique is eigenvalue stability analysis of a finite difference method. Under the method of lines, a time integrator advances by . If is a normal matrix, the scalar linear stability domain criterion on all controls its powers exactly in the corresponding L2 norm. For the centered Dirichlet discrete Laplacian, its eigenvalues lie between and zero; intersecting this interval with a time integrator's stability interval determines its mesh restriction. A uniform bound on the diagonalization of a matrix is needed for nonnormal diagonalizable systems. Eigenvalues alone can be misleading. For example,has both eigenvalues inside the disk for , but the upper-right entry of is . Taking , , this grows like . Hence stable scalar eigenvalues do not give mesh-uniform stability. Boundary closures can introduce precisely the extra growth that an interior Fourier symbol overlooks, so boundary stability of a finite-difference method must be checked separately.
For nonlinear spatial discretizations, a useful replacement for Fourier stability analysis is a convexity argument. If the Forward Euler method map is nonexpansive in a chosen norm for , the second-order strong stability preserving Runge-Kutta methodis also nonexpansive under the same restriction. Indeed, for two inputs, the triangle inequality gives . Its Taylor expansion is , proving order two. Such strong stability preserving Runge-Kutta methods transfer suitable forward-step bounds without relying on a linear spectral calculation.
Finally, consistency of a numerical method explains what stability buys. For a linear Hadamard well-posed problem, the Lax equivalence theorem equates convergence of a consistent discretization with its stability, under the stated approximation-space and norm hypotheses. Directly, if the error satisfies , iteration gives the discrete Duhamel principleThus vanishing normalized local truncation error and initial error yield convergence. With a nonlinear Lipschitz step estimate, the discrete Gronwall inequality plays the same role. A stable but inconsistent stencil, such as the literal defective formulas in questions 3 and 4 under their usual refinement, is not rescued by any amplification bound. The decisive checks are a mesh-uniform evolution bound, a valid boundary closure, and consistency with the actual PDE.
Past exam of the mathematics course of the University of Cambridge 2017 ii Paper 4 16H i Solution Created 2026-09-24 Updated 2026-10-05
Let be the adjacency matrix of a graph and with . Choose a vertex maximizing . ThenDividing givesThis argument works even with a complex eigenvector; for a finite undirected graph the matrix is real symmetric and all eigenvalues are real.
Past exam of the mathematics course of the University of Cambridge 2017 ii Paper 4 16H iv Solution Created 2026-09-24 Updated 2026-10-05
If a regular graph has connected components, the indicator vector of each component is a -eigenvector. They are linearly independent, and the Laplacian argument shows that every such eigenvector is constant on each component. Therefore its multiplicity is exactly , in particular greater than one.
Past exam of the mathematics course of the University of Cambridge 2017 ii Paper 4 16H Solution 2026-10-05
For the Petersen graph, each vertex has three neighbours, adjacent vertices have no common neighbour, and distinct nonadjacent vertices have exactly one common neighbour. The entries of count length-two walks, so diagonal entries are three, adjacent entries zero, and other off-diagonal entries one. HenceThe connected graph has the simple eigenvalue three with eigenvector . On its orthogonal complement , so the remaining eigenvalues satisfy and are one or minus two. If their multiplicities are , then and . Thus
Past exam of the mathematics course of the University of Cambridge 2017 ii Paper 4 21F iii Solution Created 2026-09-24 Updated 2026-10-05
Apply the spectral theorem for compact Hermitian operators to positive . Its positive eigenvalues have orthonormal eigenvectors spanning , with a finite or countable list. In the infinite case . DefineThen , so the are orthonormal. Also , because . Decompose with and apply bounded to the norm-convergent expansion. This gives the singular value decompositionThe series converges in norm; its squared tail norm is . No separability of the whole Hilbert space is necessary: the nonzero spectral subspace of a compact operator is separable. For use the empty list .
Past exam of the mathematics course of the University of Cambridge 2017 ii Paper 4 39A a Solution Created 2026-09-24 Updated 2026-10-05
Let . The zero Dirichlet boundary values introduce no forcing terms, so the scheme isThe common orthogonal sine eigenvectors have -eigenvalues . Thus is invertible for every , and each amplification eigenvalue isThe matrix is orthogonally diagonalizable, so its powers have Euclidean norm at most one, uniformly in the mesh and time step. Therefore Crank-Nicolson diffusion scheme is stable for every . Negative amplification factors at large can create temporal sign oscillations, which do not contradict this norm stability.
Past exam of the mathematics course of the University of Cambridge 2018 ib Paper 2 14G a Solution Created 2026-09-24 Updated 2026-10-05
The characteristic polynomial of isWhen , its discriminant is positive, so has two distinct real eigenvalues and two independent real eigenvectors. Let have those eigenvectors as columns. Rescaling one column makes without changing the diagonalization. Hence andThis is the normal form of a hyperbolic element of PSL2(R).
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 331 4 b v Solution Created 2026-10-03 Updated 2026-10-05
At the formal value , and is a skew-Hermitian matrix. Put . Its two eigenvalues and normalized eigenvectors areSubstitution uses and verifies . Their Hermitian inner product issince . This provides the requested orthogonality in the formal continuation. There is no corresponding physical for the printed system.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 335 2 iii Solution Created 2026-10-03 Updated 2026-10-05
Let be the measured wave scattering response> linear operator, with direct background propagation removed if present. Its time reversal operator is the positive operator . For a normalized emitted signal ,The maximizing signal is a right singular vector, equivalently an eigenvector of with its largest eigenvalue. This is a precise intensity statement independent of a scatterer model.
Under the Born approximation for scalar wave scattering, neglect interactions between point scatterers and writewhere is the receiver response to scatterer , is its illumination by the source array, and is its scattering wave amplitude. Well-resolved point scatterers have approximately orthogonal vectors and . In the ideal orthogonal limit,Thus the most reflective, geometrically weighted scatterer gives the largest eigenvalue, and its normalized steering vector gives the corresponding wave focusing signal. If illumination and reception factors are equal for all scatterers, this is precisely the scatterer with largest . Geometric size by itself is not the quantity being ranked.
Repeated adjoint time reversal acoustics is power iteration on . If its top eigenvalue is simple and the initial signal has a nonzero component along its eigenvector, thenA tied largest eigenvalue leaves a combination in the leading eigenspace, and an initial signal orthogonal to that eigenspace cannot excite it. This is the basis of the DORT method. Mere physical separation is insufficient if the array cannot resolve the scatterers; coherent steering-vector overlap or multiple scattering can mix the modes. Also, if includes unrestricted homogeneous transmission, its largest eigenvalue need not identify any individual scatterer. The correspondence concerns the resolved wave scattering response>. The ideal correspondence and selective wave focusing are analyzed by Prada and Fink.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 341 3 b Solution Created 2026-10-03 Updated 2026-10-05
Let be the tridiagonal matrix with diagonal two and adjacent entries minus one, incorporating zero Dirichlet boundary conditions. The actual PDF scheme isAn orthonormal basis of eigenvectors of has components , and substitution givesThey form an orthonormal basis by the spectral theorem for real symmetric matrices. Hence is invertible for every , and the amplification eigenvalues are . For the discrete L2 norm ,This bound has constant one independent of time step and mesh, which is the required stability of a numerical method. Therefore every positive Courant number is stable. If a forcing or local-defect sequence is added, the discrete variation-of-constants formula gives a bound by the initial norm plus the sum of those perturbation norms; this also justifies the global error conclusion in part (a).
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 344 3 c Solution Created 2026-10-03 Updated 2026-10-05
For the stated uniaxial nematic order normalization, the nematic director is an eigenvector of with eigenvalue , and the two perpendicular eigenvalues are . ThereforeUsing precisely the Landau-de Gennes free energy convention in the preceding solution givesIf the quartic invariant were instead normalized as , its coefficient would be ; the physical predictions are unchanged after redefining .
For a nonzero global minimizer of the free energy, compare opposite values of the scalar nematic order parameter:The lower one has , soThe claim concerns the stable ordered phase, not every metastable stationary point. For the two signs are degenerate. For , equality of the ordered and isotropic free energies, together with stationarity, gives and . The finite jump exhibits the first-order phase transition driven by the cubic invariant.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 345 2 b Solution Created 2026-10-03 Updated 2026-10-05
Use the shallow-water approximation, a hydrostatic vertically mixed layer, a deep motionless ambient, and no ambient fluid entrainment. Particle loss changes reduced gravity but changes total layer volume only at the discarded dilute-particle order. Likewise thermal expansion changes the equation of state while the leading Boussinesq approximation retains volume conservation. DefineDepth integration of mass conservation, horizontal momentum balance, particle transport, and the heating law givesHere the integrated excess hydrostatic pressure is . There is no particle source from the bed. Bed drag and the particle contribution to inertia are neglected at this order. The equivalent variable-buoyancy shallow water equations areThe term is necessary when heating or particle loss makes the reduced gravity vary horizontally.
For , the quasilinear system hasIts characteristic polynomial factors asThus the characteristic curves have slopesOn the two repeated contact characteristic curves, the particle and heating laws are and . Writing and , the gravity-wave compatibility relations areFor and , gravity-wave right eigenvectors can be taken as . The contact eigenspace has dimension two: and , with independent . Therefore has a full set of real eigenvectors.
The system is hyperbolic throughout strict static stability, but not strictly hyperbolic because the contact speed is repeated. At the eigenvalues coalesce and this diagonalizability is lost; when the gravity-wave speeds become complex, reflecting the loss of a statically stable lower layer.
Past exam of the mathematics course of the University of Cambridge 2018 ii Paper 2 40E b Solution Created 2026-09-24 Updated 2026-10-03
Subtracting the fixed point from the weighted Jacobi method givesPut . Acting on the supplied sine eigenvector givesThereforeand the corresponding eigenvalues of the iteration matrix are