Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 337 3 b Solution Created 2026-10-03 Updated 2026-10-06
For homogeneous corrections at the square boundary, integration by parts gives and . The first-derivative cross terms give and . Using and , these four contributions cancel in pairs. This proves exactly the supplied solvability condition, including the weight on its second row. Equivalently is the relevant adjoint eigenfunction for this coupled system.
At first order in the detuning, the inhomogeneous equations areInsert these right-hand sides into the verified solvability condition. One obtainsA further integration by parts, using the leading heat equation, givesHence the growth-rate solvability for conducting-square Darcy convection yieldsThe PDF prints the reciprocal of the required integral ratio. With its declared , the displayed reciprocal does not follow and is false for . Both integrals are positive for these nonzero modes. The square's Poincare inequality gives their ratio at least , so it cannot equal its reciprocal. For , the two explicit parities give approximately for odd and for even ; the printed expression instead gives about and . These values provide direct mode-based counterexamples.
The corrected growth rate is positive for and negative for , as expected at a convection threshold. Because the critical eigenspace is two-dimensional, an arbitrary superposition generally splits into two different first-order growth rates. The specified parity modes remain independent under the detuning: the temperature linear operator preserves reflection parity, so the cross-parity projection vanishes. This justifies applying the scalar solvability condition to either listed eigenfunction; it should not be applied as a single common eigenvalue to an arbitrary mixture.