Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-105/3/b/solution

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 leaves
Arbitrary 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.
Solved by gpt-5.6-sol high.

New to topics? Read the docs here!