For smooth non-increasing and an ordered subsolution and supersolution, begin with and solve successive Poisson equations with forcing and the fixed boundary data. The weak maximum principle for elliptic operators applied to the negatives of consecutive differences givesThe global Schauder estimate and the Hölder interpolation inequality bound the iterates uniformly in . Their bounded monotone sequence converges, and compact embedding of Hölder spaces permits passing to the equation. The uniform Hölder seminorm bounds retain regularity of the limit.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 107 6 a Solution Created 2026-10-03 Updated 2026-10-05
As printed, the hypotheses are insufficient. They order the barriers only on the boundary; the intended monotone iteration for a semilinear elliptic equation needs an ordered subsolution and supersolution throughout the domain. On , take a positive first Dirichlet Laplacian eigenfunction with , and setBoth barriers satisfy and vanish on the boundary, but inside. No function lies between them. This also disproves the unconditional conclusions in (b) and (d).
With the intended extra hypothesis on , solvability of the Poisson equation and the Schauder estimate give a unique with forcing and boundary data . Smooth composed with the barrier has the required forcing regularity. The barrier inequalities giveBoth differences have nonnegative boundary values. The weak maximum principle for elliptic operators applied to their negatives gives under the corrected hypothesis.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 107 6 b Solution Created 2026-10-03 Updated 2026-10-05
Use the interior ordered subsolution and supersolution correction from part (a). Recursively solve the Poisson equation with boundary value , starting from . Part (a) gives . If , thenThe first difference has zero boundary values, the second nonnegative values. The weak maximum principle for elliptic operators applied to their negatives gives . Induction proves the bounded monotone sequenceWithout interior ordering, the counterexample in part (a) disproves the claim.