 Solution

ID: past-exam-of-the-differential-equations-1-course-of-the-university-of-oxford/2023/3/a/ii/solution

With zero drift, choose the smooth function

u (x, y, t) = - 1 + \frac{t}{8} (1 - x^{2} - y^{2}) .

It equals

- 1

on the entire parabolic boundary, so

A = - 1

. On the closed cylinder,

- 1 \leq u \leq - 7/8

, while

u_{t} - Δ u = \frac{1 - x ^{2} - y ^{2}}{8} + \frac{t}{2} \leq \frac{5}{8} .

Consequently

u_{t} - Δ u + u^{3} \leq \frac{5}{8} - (\frac{7}{8})^{3} = - \frac{23}{512} < 0.

Nevertheless the maximum at

(0, 0, 1)

\overline{Q} max u = - \frac{7}{8} > - 1 = A .

The negative cubic term permits an interior increase while the boundary stays fixed, explaining why the sign condition in the preceding part matters.

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