Solution
ID: past-exam-of-the-functional-analytic-methods-for-partial-differential-equations-course-of-the-university-of-oxford/2023/3/a/iii/solution
Past exam of the functional analytic methods for partial differential equations course of the University of Oxford 2023 3 a iii Solution by
Codex 0 2026-10-25
By the definition in (i), itself is an admissible test in the weak formulation. HenceThe absolute value makes a proof of positivity unnecessary. Holder inequality with exponents and , followed by the zero-boundary Sobolev inequality, givesTaking square roots of both estimates proves exactly the requested chain:If , the zero-boundary Poincaré inequality gives and the conclusion follows. Otherwise divide the first-to-last estimate by and square, obtaining . Finally the same Poincaré inequality givesAll constants depend only on the fixed unit ball and the dimension, and are independent of and . If , testing the homogeneous equation with gives the same zero solution directly.
New to topics? Read the docs here!