Past exam of the mathematics course of the University of Cambridge 2020 ii Paper 2 38B iii Solution Created 2026-09-24 Updated 2026-09-29
Past exam of the mathematics course of the University of Cambridge 2020 ii Paper 2 38B ii Solution Created 2026-09-24 Updated 2026-09-29
WriteConservation of the two-dimensional volume suggestsSubstitution into the scalar conservation law givesThe integration constant is zero at the dry rear, soThusThe jump from to the dry region is a shock. Its Rankine-Hugoniot condition gives , which is also satisfied by the similarity profile.
Volume conservation determines the front:Therefore the finite-volume gravity-driven thin-film current hasand its head height is
Past exam of the mathematics course of the University of Cambridge 2020 ii Paper 2 38B i Solution Created 2026-09-24 Updated 2026-09-29
Let point down the plane and normal to it, with velocity . The two-dimensional incompressible Navier-Stokes equation isThe lubrication theory limit has film thickness much smaller than its streamwise length. Transverse viscous derivatives dominate, inertia is negligible at late times, and the leading equations areAt the wall, by the no-slip boundary condition. At the free surface , take and , neglecting surface tension. The normal balance gives the hydrostatic pressureIf the current has thickness scale and length scale , thenThus the streamwise pressure gradient is asymptotically smaller than the downslope body force for a fixed nonzero inclination. The streamwise equation becomesUsing and givesThe volume flux isIntegrating incompressibility across the film and using the kinematic boundary condition gives , hence
Past exam of the mathematics course of the University of Cambridge 2020 ii Paper 2 37D iii Solution Created 2026-09-24 Updated 2026-09-29
Since ,Relative to the distant observer, the static clock suffers gravitational time dilation. The orbiting clock suffers the same gravitational effect together with the kinematic time dilation caused by its orbital motion, so it accumulates the least proper time.
Past exam of the mathematics course of the University of Cambridge 2020 ii Paper 2 37D ii Solution Created 2026-09-24 Updated 2026-09-29
Past exam of the mathematics course of the University of Cambridge 2020 ii Paper 2 37D i Solution Created 2026-09-24 Updated 2026-09-29
By spherical symmetry, place the geodesic in the equatorial plane . Withthe geodesic Lagrangian iswhere is the constant squared norm of the tangent: for a unit-speed timelike geodesic and for a null geodesic. The cyclic coordinates and give the conserved quantitiesSubstitution into the normalization givesMultiplying by and rearranging producesThis is the timelike geodesic effective potential form, with the same derivation applying to any affinely parametrized causal geodesic.
Past exam of the mathematics course of the University of Cambridge 2020 ii Paper 2 34A b ii Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2020 ii Paper 2 34A b i Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2020 ii Paper 2 34A a Solution Created 2026-09-24 Updated 2026-09-29
Let . In the interaction picture,and the state obeysThe first term of the Dyson series, or equivalently first-order time-dependent perturbation theory, givesThus, if ,
Past exam of the mathematics course of the University of Cambridge 2020 ii Paper 2 33C iii Solution Created 2026-09-24 Updated 2026-10-03
The first Born approximation for a one-dimensional reflection coefficient follows by replacing the Jost solution inside its Volterra scattering equation by the free wave :
To check the inverse problem, any shallow discrete state contributes only at pointwise, so to first order the Marchenko input isDifferentiating and inserting the Born approximation giveswhere the second line is the Fourier inversion theorem. At first order the quadratic integral term in the Marchenko equation may be dropped, soThe reconstruction formula therefore returnsThus direct and inverse scattering agree to first order.
Past exam of the mathematics course of the University of Cambridge 2020 ii Paper 2 33C ii Solution Created 2026-09-24 Updated 2026-09-29
For one bound state at and no reflection, writeThe rank-one Gelfand-Levitan-Marchenko kernel has the form . Substitution into the integral equation givesand henceChoose by . ThenDifferentiating the reconstruction formula gives the one-soliton solutionIt is a negative soliton of amplitude travelling to the right with speed , as required by this sign convention for KdV.
Past exam of the mathematics course of the University of Cambridge 2020 ii Paper 2 33C i Solution Created 2026-09-24 Updated 2026-09-29
The Korteweg-De Vries equation is the compatibility condition for the Lax pairFor rapidly decaying initial data, first solve the direct KdV Schrodinger spectral problemComparing its Jost solutions at the two spatial infinities produces the KdV scattering data: the reflection coefficient on the continuous spectrum, discrete eigenvalues , and norming amplitudes for the bound states.
The Lax evolution is isospectral. More precisely, the Time evolution of KdV scattering data isThis is the simple linear evolution that makes the nonlinear initial-value problem tractable.
To invert the data, formSolve the Gelfand-Levitan-Marchenko equationand reconstructDirect scattering at , linear evolution of the data, and this inverse step constitute the inverse scattering transform.
Past exam of the mathematics course of the University of Cambridge 2020 ii Paper 2 32E b iii Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2020 ii Paper 2 32E b ii Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2020 ii Paper 2 32E b i Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2020 ii Paper 2 32E a Solution Created 2026-09-24 Updated 2026-10-14
The Bendixson-Dulac criterion says the following. Let be a simply connected domain, let be continuously differentiable, and suppose there is a continuously differentiable function such thathas one sign throughout and does not vanish identically on any open subset. Then has no periodic orbit contained in .
Indeed, if a periodic orbit existed, simple connectivity would put its interior inside . The vector field , and hence , is tangent to , so its outward flux is zero. The divergence theorem would givecontradicting the sign hypothesis.
The Poincaré-Bendixson theorem states that a nonempty compact omega-limit set of a continuously differentiable planar flow that contains no equilibrium point is a periodic orbit. More generally, when the omega-limit set contains only finitely many equilibria, it is either an equilibrium, a periodic orbit, or a union of equilibria and connecting trajectories.
Past exam of the mathematics course of the University of Cambridge 2020 ii Paper 2 31D b ii Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2020 ii Paper 2 31D b i Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2020 ii Paper 2 31D a ii Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2020 ii Paper 2 31D a i Created 2026-09-24 Updated 2026-09-24
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