Past exam of the mathematics course of the University of Cambridge 2003 iii Paper 59 2 b Solution Created 2026-10-03 Updated 2026-10-14
Let and . The eigenvalues at the conductive equilibrium point are and , both negative since and . The coupled Jacobian matrix has characteristic polynomialwhereA simple zero eigenvalue occurs on . The reflection , with fixed, makes its generic steady-state bifurcation a pitchfork bifurcation.
For a nonzero imaginary pair, substitute into : the real and imaginary parts require and . HenceWith , this Hopf bifurcation curve is present only for . At its intersection with the pitchfork bifurcation curve,and . The zero eigenvalue has geometric multiplicity one: its eigenvector has , . Thus this is a reflection-symmetric Bogdanov–Takens bifurcation, rather than two independent zero modes.
The Routh-Hurwitz stability criterion gives the stable trivial state below for and below for . The rest of the line , beyond , is a stationary bifurcation of an already unstable trivial state. Likewise the extension of below is only an algebraic line, since its putative frequency is not real. The middle panel of the bifurcation diagram in part 1(b) distinguishes the actual curves from this extension.
Past exam of the mathematics course of the University of Cambridge 2003 iii Paper 59 2 c Solution Created 2026-10-03 Updated 2026-10-14
To determine the nonlinear pitchfork bifurcation, solve the four slaved steady equations exactly. SetThe nonzero equilibrium points satisfyThusIf , the coefficient of is positive for every physical : there is no change of pitchfork direction at a nonnegative rotation parameter. If , the change occurs atBelow the nonzero branch lies on ; above it initially lies on . At , , so grows like rather than the ordinary square-root scaling.
The rational equilibrium curve at a degenerate pitchfork also gives the nearby saddle-node bifurcations of the two reflection-related nonzero branches. For ,andThe two saddle-node bifurcations occur on one parameter curve by reflection symmetry; its separation from the pitchfork bifurcation curve is quadratic in .
The stability statement needs a further condition. At , the other coupled eigenvalues have polynomial . They are stable only for . If , a one-dimensional centre manifold gives the local form , where and . The quintic term stabilizes the large branch. The interval then has a stable conductive state and two stable finite-amplitude states, with the two small unstable branches separating their basins of attraction. This is the usual subcritical convection hysteresis. If , the branch still reverses its direction, but the trivial state already has an additional unstable mode; this equilibrium-curve calculation alone does not imply a stable bistable wedge. At equality a two-dimensional centre manifold is essential.
Past exam of the mathematics course of the University of Cambridge 2003 iii Paper 59 2 d Solution Created 2026-10-03 Updated 2026-10-14
A codimension-three degeneracy is possible when the double-zero eigenvalue condition and the vanishing cubic restoring coefficient of the pitchfork bifurcation coincide. Equating and givesEquivalently . For every , this is a physically admissible aspect ratio and . The other two parameters are and .
At this point the fifth-order restoring term is nonzero: the steady relation is with . It is therefore a degenerate reflection-symmetric Bogdanov–Takens bifurcation with three independent conditions: zero determinant, zero frequency, and zero cubic restoring coefficient. Varying the aspect ratio supplies the third parameter, in addition to . At a fixed generic aspect ratio it is not a codimension-two event available by varying only .
It is not a triple-zero eigenvalue: the three-mode characteristic polynomial still has the noncritical factor , while the eigenvalues remain strictly negative. The extra degeneracy is nonlinear. A fifth-order, two-dimensional normal form is needed for its complete unfolding; the one-dimensional quintic pitchfork bifurcation used away from cannot be substituted at the double-zero point.
Past exam of the mathematics course of the University of Cambridge 2003 iii Paper 59 3 a Solution Created 2026-10-03 Updated 2026-10-14
Take small incoming sections and outgoing sections after straightening the stable manifold and unstable manifold of the origin. Its eigenvalues are , so the leading local passage satisfies , . The exit time and transverse exit coordinate areThus the local passage contracts strongly as , while its duration diverges. Smooth global reinjection from the right outgoing section to the upper incoming section has leading transverse coordinate ; the corresponding left-to-lower return has . Choose the parameters as these signed splitting coordinates and absorb into .
After scaling the incoming-section labels to , the leading Poincaré return map isThe second coordinate records which outgoing branch was used, not an independent continuous amplitude. A smooth planar flow preserves the orientation of a transverse-section return. For the right-to-upper pair of sections, the ordered pair consisting of the flow direction and the positive section tangent has the same orientation at both ends; the same is true for the left-to-lower pair. Hence for these sign conventions and a nondegenerate global return.
For general nonlinear , this is an asymptotic return model, with smooth parameter corrections and higher-order local-passage corrections. The hypotheses do not imply an exact power-law map in the original coordinates. Interpreting the printed parameters themselves as linear splitting coordinates also requires a nondegenerate two-parameter unfolding: the two splitting derivatives must be nonzero. For example, composing a generic right-loop splitting with preserves the stated loop at but does not give the printed signed offset. The subsequent region diagram uses the standard generic splitting-coordinate interpretation. Nor is a normal return point: it lies on the stable manifold, whose trajectory takes infinite time to reach the saddle equilibrium. This explains how the map describes nearby recurrent trajectories without extending a finite-time return across the singular separatrix.
Past exam of the mathematics course of the University of Cambridge 2003 iii Paper 59 4 a Solution Created 2026-10-03 Updated 2026-10-14
For , polar coordinates give and ; the remaining amplitude equation is . Interpret in the original complex coordinate, where the phase is undefined.
If , put . The full equilibrium points are . Their eigenvalues areThus is stable for and has a two-dimensional unstable manifold for . The point has a one-dimensional unstable manifold for and is repelling for . On they merge in a saddle-node bifurcation; the simultaneous zero and imaginary eigenvalues at give a fold-Hopf bifurcation.
A steady positive amplitude represents a full periodic orbit, not a full equilibrium. Solving the amplitude equations givesIt exists in the wedge . The transverse amplitude Jacobian matrix has characteristic polynomial . Its two transverse Floquet exponents have negative real part for and positive real part for ; the phase direction supplies the neutral Floquet multiplier one.
The wedge boundaries are Hopf bifurcations at from and from . The left branch is supercritical, producing a stable periodic orbit as loses stability; the right branch is subcritical in the oscillatory subspace, with an unstable periodic orbit on the side where the transverse complex pair of is stable. The real unstable direction of persists throughout. There are no full equilibrium points when , and then excludes a local periodic orbit. The right panel of the bifurcation diagram in part 1(b) shows the primary curves and the perturbed global curve calculated below.
Past exam of the mathematics course of the University of Cambridge 2003 iii Paper 59 4 e Solution Created 2026-10-03 Updated 2026-10-14
The normal form is invariant under . This continuous rotational symmetry makes the axis invariant and reduces the nontrivial geometry to two amplitude variables. The axis connection from to persists without parameter tuning, while rotating the outer connection produces a family of connections from back to . A two-dimensional autonomous amplitude flow cannot by itself produce the proposed saddle-focus return chaos: symmetry reduction hides the phase-dependent reinjection needed for that mechanism.
Generic higher-order terms need not retain this exact rotational symmetry. A normal form can preserve it to a finite truncation order while the full equations have small symmetry-breaking remainders. These unfold the special connection geometry into phase-dependent returns; nearby parameter curves can contain homoclinic orbits to a saddle-focus, and the Shilnikov bifurcation mechanism becomes available. Their exact placement depends on the omitted terms and is not fixed by the displayed truncation alone.
For example, at the perturbed positive-axis saddle, the real unstable eigenvalue is and the stable complex pair has real part . Near the calculated connection curve , the contracting ratio isThis is the positive-saddle-value regime, explaining why the relevant saddle-focus return can be expanding.
In the Shilnikov return map, take a positive return coordinate and , with nondegenerate . At the connection parameter , the fixed point condition isThe right side tends to zero, whereas the phase passes through infinitely many oscillations as . Hence there are infinitely many positive fixed points accumulating at zero, with logarithmically spaced amplitudes. For these points,whose magnitude tends to infinity along the accumulating fixed-point sequence. They represent long-period saddle periodic orbits of the full return, with period growing like , rather than infinitely many stable cycles.
Successive turning branches of this oscillatory return can stretch and fold across suitable intervals. The full saddle-focus Poincaré return map then admits invariant Smale horseshoes and chaotic itineraries near a generic homoclinic orbit at a Shilnikov bifurcation. Varying unfolds accumulating saddle-node and period-doubling events. Whether any resulting invariant set attracts trajectories depends on the other return direction and the omitted terms; the leading scalar map alone does not prove a chaotic attractor for every perturbation. Its domain also excludes negative iterates, which leave the chosen return branch. The degeneracies or would remove the oscillatory mechanism, so the conclusion uses generic, genuinely saddle-focus reinjection.
Past exam of the mathematics course of the University of Cambridge 2003 iii Paper 60 1 ii Solution Created 2026-10-03 Updated 2026-10-14
The third planform is fixed by a diagonal reflection and changes sign under a quarter-turn. Extend the preceding group representation byThe other diagonal reflection changes only the sign of . Consequently the first two components of an equivariant dynamical system have factors and , whereas its third component is even in each of . The quarter-turn exchanges and while negating . Its admissible quadratic coupling is therefore in the third equation and opposite couplings in the first two. At cubic order the only new terms are . The linear action on the two-dimensional irreducible mode has one coefficient , while the distinct scalar representation has its own coefficient . A small splitting of the two marginal linear modes supplies these two unfolding parameters. With real coefficients the resulting amplitude equations areThese equations follow by listing all equivariant monomials through degree three; the displayed minus signs are coefficient conventions, not consequences of symmetry.
In the specified reduced system put , . A nonzero branch exists for , with . Write perturbations as , , and . Linearization givesFor the square-pattern interaction with a sign-changing scalar mode, the remaining eigenvalues have sum and product , where . A real two-dimensional stability matrix has both eigenvalues in the left half-plane exactly when its trace is negative and its determinant is positive. ThereforeIf , the block has negative determinant and one growing direction, explaining the instability at sufficiently small . At the lower endpoint , one eigenvalue is zero and the other is . The symmetry fixes the diagonal state and negates its critical direction. Thus a generic nonlinear unfolding has a pitchfork bifurcation to unequal-amplitude states with nonzero .
At the upper endpoint , the trace vanishes andThe symmetric eigenvalue remains negative. The critical conjugate pair crosses the imaginary axis with derivative . If its cubic Hopf normal form coefficient is nonzero, this is a Hopf bifurcation and nearby periodic solutions exist on one side. Their stability and the side of existence depend on that coefficient; the linear calculation does not determine them. It does establish the generic possibility of oscillating square patterns. The equality is a double-zero degeneracy, outside this nondegenerate Hopf bifurcation argument.
Past exam of the mathematics course of the University of Cambridge 2003 iii Paper 60 1 i Solution Created 2026-10-03 Updated 2026-10-14
Choose amplitude coordinates in which a quarter-turn and a diagonal reflection act as and . These generate the natural two-dimensional group representation of the dihedral group . An equivariant dynamical system satisfies . The two independent sign changes make odd in and even in , and odd in and even in . The quarter-turn equates the two linear coefficients and the two sets of cubic coefficients. No quadratic monomial is allowed. Thus the square-symmetric cubic steady-state normal form isThe bifurcation parameter has been scaled so that the critical linear eigenvalue is . Nondegeneracy at cubic order means , and . The equilibrium equations factor as . When both components are nonzero, subtracting the two bracketed expressions gives . Therefore the complete leading list of small steady branches is:
- The zero state, with two linear eigenvalues .
- Four axial states and , when . Their stability matrix has eigenvalues and .
- Four diagonal states , with independent signs and . Their eigenvalues are and .
Each family is one orbit of the dihedral group. Each nonzero state retains one reflection fixing its amplitude line; its full spatial symmetry depends also on the chosen planforms. Higher-order terms perturb the branch amplitudes but preserve the nondegenerate local classification. For , an axial branch is asymptotically stable precisely when and , while a diagonal branch is asymptotically stable precisely when and . Every cubic branch lying on has a positive radial eigenvalue , so it is unstable. In the exceptional isotropic cubic case , the cubic truncation instead has a circle of equilibria; higher-order anisotropy is then needed, which is why that case was excluded.
Past exam of the mathematics course of the University of Cambridge 2003 iii Paper 60 2 ii Solution 2026-10-14
Let the equilibrium be and write . The linearization of the cubic term gives . Therefore the real stability matrix isIts eigenvalues obeyThe square root may be imaginary; stability uses real parts. The determinant is exactly , so a zero eigenvalue occurs only at the turning points derived above. Away from the cusp those are ordinary saddle-node bifurcations; the parameterization joining point has .
At a candidate Hopf bifurcation, the trace vanishes: . Substitution into the steady cubic gives , which has one positive root. The pair is genuinely imaginary only ifThus and, equivalently,This direct calculation proves the threshold without needing the additional monotonicity assumption. The equality gives a double-zero eigenvalue, not an ordinary Hopf bifurcation. Along a fixed- branch, implicit differentiation gives ; at this crossing . A nonzero nonlinear Hopf coefficient then gives the usual Hopf bifurcation.
The fold and Hopf thresholds of a directly forced cubic amplitude divide the diagrams into three regimes. At noncritical points the two eigenvalues, counted with multiplicity by the signs of their real parts, are classified as follows:
- For , there is one equilibrium for every . It has two stable eigenvalues for and two unstable eigenvalues for .
- For , the largest-amplitude branch has two stable eigenvalues; the intermediate-amplitude branch has one stable and one unstable eigenvalue. The smallest-amplitude branch emerging from the lower fold has two stable eigenvalues up to its Hopf crossing, and two unstable eigenvalues beyond it. The relative location of the Hopf point and the other fold in is not fixed throughout this interval.
- For , the largest-amplitude branch again has two stable eigenvalues, the intermediate branch has one of each, and the smallest-amplitude branch has two unstable eigenvalues. The trace-zero point lies on the saddle segment and is not a Hopf bifurcation because its determinant is negative.
At a fold one eigenvalue is zero; at a Hopf point the pair has zero real part. These nonhyperbolic points are marked separately in the sketches.
Steady amplitudes of the moving forced pattern, with folds, Hopf points and stable/unstable eigenvalue counts in all three regimes
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