Past exam of the mathematics course of the University of Cambridge 2025 iii Paper 105 3 a Solution Created 2026-09-24 Updated 2026-09-24
For and , define the characteristic curve byThe bounded derivative makes globally Lipschitz, uniformly in . On each finite time interval, , so Gronwall inequality prevents finite-time escape. The Picard-Lindelof theorem therefore gives a unique trajectory for every finite . Differentiation in givesso the characteristic flow map is a increasing diffeomorphism.
Along a characteristic, the chain rule changes the equation intoTracing backward by the flow therefore givesThe regularity of the flow makes this a classical solution. Conversely, every classical solution obeys the same ordinary differential equation along every characteristic, so the formula also proves uniqueness.
Past exam of the mathematics course of the University of Cambridge 2025 iii Paper 105 3 c Solution Created 2026-09-24 Updated 2026-09-24
With , pull the bounded measurable initial value back along the characteristic flow map:The flow is measurable and invertible, so is measurable and . Choose smooth converging to in with uniformly bounded essential suprema, and definePart a makes each a classical, hence weak, solution. On every compact subset of spacetime, the change-of-variables formula for the flow and its locally bounded Jacobian determinant give in . Passing to the limit in the weak identity by dominated convergence proves that is a bounded weak solution with initial datum .
Past exam of the mathematics course of the University of Cambridge 2026 iii Paper 105 3 a Solution Created 2026-09-24 Updated 2026-09-24
The characteristic flow map solves the ordinary differential equationand henceAlong this characteristic curve, the chain rule givesThe value is therefore constant, and tracing back to time zero gives the classical solutionDirect differentiation verifies both the linear transport equation and its initial value.
Past exam of the mathematics course of the University of Cambridge 2026 iii Paper 105 3 c Solution Created 2026-09-24 Updated 2026-09-24
Solve the adjoint transport equationbackward with terminal value zero. Along the characteristic flow map , the required solution isDifferentiation under the integral verifies the equation. If has compact support in , then vanishes for and for , so as required.
When the initial datum is zero, inserting this into the weak formulation givesfor every . Thus almost everywhere. The difference of two bounded weak solutions has zero initial datum, so this proves uniqueness.