Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 341 7 Solution Created 2026-10-03 Updated 2026-10-06
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 iii Paper 343 1 iii Solution Created 2026-10-03 Updated 2026-10-06
An impenetrable hard wall imposes the Dirichlet boundary condition . Choose the bulk complex argument to be zero and write a stationary solution as with as . The dimensional Gross–Pitaevskii equation becomesMultiplication by and use of the bulk limit yield the first integralfor the increasing wall profile. Integrating gives the hard-wall condensate healing profileThe PDF's expression without the factor is correct for the normalized number density , not for the dimensional number density defined at the start of the paper. For example, a bulk number density must approach , whereas the literal printed expression approaches . If both number density and coordinates in this part are understood to have already been normalized, the same result reads and the dimensionless width is . These are two descriptions of the same profile, not different physical healing lengths.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 105 3 a Solution Created 2026-10-03 Updated 2026-10-05
On the zero-boundary Sobolev space , the associated bilinear form isThe homogeneous Dirichlet boundary condition removes the boundary term, so for sufficiently regular and .
Formal self-adjointness means that equals its formal adjoint, or equivalently for compactly supported smooth functions. HereWith the symmetric real principal coefficients in the question, is equivalent to . It makes a symmetric bilinear form.
For the assertions in part (b), positivity must mean strict Dirichlet positivity:This is the positive-definite operator convention. The weaker positive semidefinite operator convention does not suffice: has zero energy on a first Dirichlet eigenfunction. Under strict positivity, the compactness argument in part (b)(i) gives a uniform lower bound as well.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 105 3 c Solution Created 2026-10-03 Updated 2026-10-05
Take , the natural initial-data assumption needed for the requested convergence but not explicitly stated in this subpart. Use the genuine Dirichlet eigensystem from part (b)(iii) and put . The spectral construction of a parabolic solution isEach finite sum has zero Dirichlet boundary condition and satisfies .
To justify all derivatives, fix . For any integers , the squared norm of a tail of the th time derivative isuniformly for . Use the corrected Sobolev domains of powers of an elliptic Dirichlet operator, not the inaccurate unqualified assertion in the PDF. Their norm equivalence gives convergence in every ordinary . The Sobolev embedding theorem then gives convergence of every desired spatial derivative by choosing large enough. Thus termwise differentiation is justified, , and for every . The trace remains zero, and holds pointwise.
Finally Parseval identity givesby the dominated convergence theorem, since each factor tends to zero and is bounded by one. No boundary compatibility of the initial data is required for positive-time smoothness.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 107 3 b Solution Created 2026-10-03 Updated 2026-10-05
For zero Dirichlet boundary condition, the global Schauder estimate isThe constant depends on , domain regularity, the ellipticity lower bound, and coefficient norms. No sign restriction on is needed. Nonzero boundary data contribute the norm of an extension.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 107 5 b Solution Created 2026-10-03 Updated 2026-10-05
Failure of the bound would give normalized with , zero Dirichlet boundary condition, and . The global Schauder estimate bounds them in . By compact embedding of Hölder spaces, a subsequence converges in to with , zero boundary data and . The uniform Hölder seminorm bounds pass to the limit, so . This contradicts the trivial null space, provingThe global Schauder estimate then gives , the bounded-inverse estimate used next.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 107 5 d Solution Created 2026-10-03 Updated 2026-10-05
Let , so has zero Dirichlet boundary condition. Using the supplied form of the minimal surface equation for a graph, setFor and , the difference bound impliesandThe shifted arguments stay in the permitted radius-two ball, and . The Laplacian with zero Dirichlet boundary condition has trivial null space by the maximum principle for harmonic functions. Choose small enough that and , for the constants in part (c) with . Its small-data existence for a nonlinear elliptic Dirichlet problem supplies , and solves the minimal surface equation for a graph with the required boundary values.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 306 1 iii Solution Created 2026-10-03 Updated 2026-10-05
In conformal gauge, . The Weyl invariance of the Polyakov action removes , leavingIts complete variation isFix the string embedding map on the initial and final time slices. The interior Euler-Lagrange field equation is the wave equationAt each spatial endpoint the remaining variation vanishes under either Dirichlet boundary conditions, , or Neumann boundary conditions, , independently for each target-space component. Thus a fixed endpoint or a free endpoint makes the boundary contribution vanish. Fixing a spatial point throughout time imposes Dirichlet boundary conditions on its spatial coordinates, while the time coordinate may retain Neumann boundary conditions.
The independent metric tensor equation must also be retained after gauge fixing. Its Virasoro constraints areTogether with the wave equation, these ensure equivalence to the Nambu–Goto action; varying only the already fixed metric would lose these Virasoro constraints.
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 341 4 a Solution Created 2026-10-03 Updated 2026-10-05
For a sufficiently regular solution with zero Dirichlet boundary conditions, differentiate its squared L2 norm. Since ,Integration by parts and the zero boundary trace give . The potential contribution is real because is real. The quantity in parentheses is consequently real, and its product with has zero real part. HenceThis is the energy method for a skew-adjoint evolution. For general L2 space data, the same result follows from the strongly continuous unitary group generated by a self-adjoint realization of . A sufficient assumption is a bounded real potential with the Dirichlet Laplacian; arbitrary real potentials require appropriate domain and self-adjointness assumptions. The differentiated proof presumes the regularity needed for the displayed integrals, rather than asserting it for every pointwise real function .
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 341 6 Solution Created 2026-10-03 Updated 2026-10-05
The starting point of a conforming finite element method is a weak formulation in a Hilbert space . For example, a homogeneous Dirichlet boundary condition on a bounded Lipschitz domain gives . One seekswhere is a bounded bilinear form, , and is a bounded linear functional. A conforming finite element space is finite dimensional and is typically built from functions that are polynomial on each element of a finite element mesh. With a basis , the Galerkin method imposes the same identity only for , givingThe stiffness matrix is sparse when basis supports overlap only locally. Local element integrals assemble its entries; the mass matrix represents an term. This separates the choice of trial functions from the variational principle determining their coefficients.
The fundamental well-posedness theorem is the Lax-Milgram theorem: if for some , there is a unique solution for each , with . The same theorem on gives a unique discrete solution. The constants must be independent of the mesh for uniform stability of a numerical method. Boundary conditions are built into the trial space when they are essential constraints; flux conditions arise through boundary terms in integration by parts and are natural constraints in the weak formulation.
When is symmetric and a coercive bilinear form, the Ritz method minimizesover . Its first variation is exactly the Galerkin method, so Ritz minimization and Galerkin testing coincide for symmetric coercive problems. The energy norm turns the approximation into an orthogonal projection. Indeed, subtracting the exact and discrete weak equations gives Galerkin orthogonality, . For every the Pythagorean identity becomessoThis exact best-approximation property is the central advantage of the Ritz method.
Symmetry is unnecessary for the Galerkin method. For any bounded, coercive form, the Céa lemma givesTo prove it, put . Galerkin orthogonality yields , so coercivity and boundedness give . Divide when and take the infimum. Thus approximation capability plus uniform coercivity gives convergence. In contrast, minimizing for a nonsymmetric form differentiates its symmetric part and generally does not solve the original weak equation.
A concrete symmetric example is the one-dimensional reaction-diffusion problem from Question 5. Continuous piecewise-linear hat functions give the diffusion stiffness matrix with diagonal and neighbouring entries , and the mass matrix with diagonal and neighbouring entries . Therefore and . Its positive-definite matrix is the coordinate form of the continuous energy. The exact smooth solution makes this an explicit test of the method, not just an abstract existence result.
For a nonsymmetric example, take positive , constant real , andIts form is . Since , it has coercivity constant in the full norm, although it is not symmetric when . The Galerkin method remains well posed. On the uniform hat basis, , where , and . The skew-symmetric matrix cancels from , proving nonsingularity. A Ritz energy would omit precisely this advection contribution. Small may nevertheless make the approximation constant large and produce poorly resolved layers; algebraic solvability is not an accuracy guarantee.
Quantitative convergence uses a finite element interpolation estimate. On a shape-regular mesh, for piecewise-linear conforming elements and in the usual low-dimensional setting,The Céa lemma therefore gives an energy error. The Aubin–Nitsche duality argument improves the L2 norm error when the dual elliptic problem has regularity: solve and useConsequentlywith constants depending on the solution and regularity bounds. The second estimate is conditional on dual regularity; corners or insufficient data regularity can reduce the rate.
The same Galerkin method handles evolution equations. For the homogeneous heat equation, the semidiscrete finite element heat equation is . Because is a positive-definite matrix and is a positive semidefinite matrix,The decreasing quadratic form is exactly the L2 norm squared of the finite-element function. This separates spatial stability of a numerical method from the subsequent time integrator, just as in the Schrodinger equation example. Conformity, coercivity and approximation estimates together explain convergence.
Past exam of the mathematics course of the University of Cambridge 2019 ib Paper 2 5B Solution Created 2026-09-24 Updated 2026-09-29
For an axisymmetric separated solution of the Laplace equation whose angular factor is the Legendre polynomial , the radial ordinary differential equation has the two powers and . Thusfor constants .
Write . Since , only the spherical harmonics of degrees zero and two occur. The degree-zero radial coefficient satisfying its two Dirichlet boundary conditions is . For degree two, write ; thenso and . By linearity and uniqueness of this boundary value problem,Direct substitution at verifies both boundary values.
Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 105 2 b i Solution 2026-10-03
The Dirichlet boundary condition is built into the first component's space, while the Neumann boundary condition is natural. Thus a weak solution is a pairsuch that for every ,If are up to the boundary, taking compactly supported test functions and applying the fundamental lemma of the calculus of variations gives both differential equations pointwise in . Membership of gives on . Applying integration by parts to the second identity and using its differential equation leavesfor every smooth boundary trace . Hence on , so the equations and both boundary conditions hold classically.
Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 341 3 a Solution Created 2026-10-03 Updated 2026-10-05
Write and set . In the discrete L2 norm , the energy method givesTo obtain this identity, shift the indices in the centered second-difference sum. The centered first-difference matrix is skew-symmetric, so its contribution has zero real part even for complex grid values. The homogeneous Dirichlet boundary conditions remove the endpoint terms.
Therefore the semidiscretization is stable for every real and every positive , with the mesh-independent bound . No restriction on is needed for this energy estimate.
Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 341 7 Solution Created 2026-10-03 Updated 2026-10-05
Consider a bounded Lipschitz domain and homogeneous Dirichlet boundary conditions forAssume is a bounded real symmetric matrix field with uniformly, is bounded, and defines a bounded linear functional on , the zero-boundary Sobolev space. The weak formulation isThe Cauchy-Schwarz inequality gives continuity . Uniform ellipticity and the Poincare inequality give a coercive bilinear form, . The Lax-Milgram theorem therefore supplies a unique weak solution, with . Its proof represents by a bounded linear operator using the Riesz representation theorem; coercivity gives an injective operator with closed image, and a zero orthogonal complement makes that image all of . This yields existence, uniqueness and the displayed estimate.
For a conforming finite element space , the Galerkin method finds such that for every . With a basis , this is the linear system , where the stiffness matrix is and . Its positive-definite matrix property proves unique discrete solvability. Local support gives a sparse matrix through elementwise assembly.
For the symmetric problem, the Ritz method minimizes over . Differentiating in every direction gives precisely the Galerkin method equations. Conversely, if solves them, , proving it is the unique minimum. The same argument over gives , where is the energy norm. Thus the Ritz method chooses the best trial function in that norm.
Subtracting the continuous and discrete equations gives Galerkin orthogonality, . For arbitrary and , it follows that . Coercivity and continuity prove the Céa lemma:In the symmetric energy norm, the Cauchy-Schwarz inequality improves the constant to one. Equivalently, Galerkin orthogonality gives the Pythagorean identity . These estimates establish stability and show that approximation properties of the trial space determine convergence.
For continuous piecewise-linear elements on a shape-regular mesh in dimensions at most three, a finite element interpolation estimate gives when . Combining it with the Céa lemma yields energy norm error. More generally, degree- elements give error in under the corresponding regularity and approximation assumptions. Mere mesh refinement cannot supply an order whose required solution regularity is absent.
An additional order in the L2 norm follows under a dual elliptic regularity assumption. Solve and assume . The Aubin–Nitsche duality argument uses Galerkin orthogonality to obtainTherefore , giving for the piecewise-linear case with . This improvement explicitly requires regularity of the dual problem.
As a concrete example, solve on with zero endpoint values. On a uniform mesh of spacing , the piecewise-linear hat functions give , and . The discrete equations are , whose solution is . The exact solution is , so is its piecewise-linear nodal interpolant. On each interval ,Integrating the squared error and squared derivative error gives the explicit rateswhere the norm here is .
For nonsymmetric coercive problems, the Galerkin method, Galerkin orthogonality and the Céa lemma still apply, but the symmetric quadratic minimization interpretation of the Ritz method generally does not. Nonconforming trial spaces, inexact integration and noncoercive equations need additional arguments beyond the conforming coercive theory proved here.
Past exam of the mathematics course of the University of Cambridge 2020 ia Paper 2 10B c Solution Created 2026-09-24 Updated 2026-09-29
Suppose are solutions and put . Then and satisfies the homogeneous Robin boundary conditionFor , Green's first identity givesThe left side is nonnegative, so both sides vanish and . For , the Dirichlet boundary condition on and the same identity again imply . The solution is therefore unique for every .
For a regular spherically symmetric solution, the Laplacian equation isRegularity at the origin gives , hence . Applying the boundary condition at yieldsAt the Neumann boundary condition would require , which is impossible for . Equivalently, its compatibility condition would require .
Past exam of the mathematics course of the University of Cambridge 2021 ib Paper 1 14C i Solution Created 2026-09-24 Updated 2026-09-25
For a normalized wavefunction obeying the infinite-wall Dirichlet boundary condition, integration by parts gives the energy expectationThus every energy eigenvalue is nonnegative.
For , the Time-independent Schrodinger equation and the wall conditions givewhere and . Continuity of and at the finite potential step givesDividing and rearranging yields the bound-state quantization condition