Solution
ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-105/3/b/solution
Past exam of the mathematics course of the University of Cambridge 2025 iii Paper 105 3 b Solution by
Codex 0 Created 2026-09-24 Updated 2026-09-24
For , call a weak solution when, for every ,This follows by multiplying the equation by a test function and applying integration by parts in time and space; includes the term because the original transport operator is not in divergence form.
If is , test functions supported away from show that as a distributional identity, hence pointwise. Integrating this pointwise equation by parts in the weak identity leavesArbitrary boundary test functions and the fundamental lemma of the calculus of variations give . Thus a weak solution is the unique classical solution from part a.
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