 Solution

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

E (t) = \int_{0}^{1} u (x, t)^{2} d x

u

, followed by integration by parts, gives

\frac{1}{2} E^{'} (t) = [u u_{x}]_{0}^{1} - \int_{0}^{1} u_{x}^{2} d x + a (t) \int_{0}^{1} u sin^{n} u d x .

x = 1

x = 0

u_{x} (0, t) = αu (0, t)

, so subtraction of the lower endpoint gives

- αu (0, t)^{2}

. In particular the sign is dissipative. Thus the exact energy identity is

\frac{1}{2} E^{'} (t) + \int_{0}^{1} u_{x}^{2} d x + αu (0, t)^{2} = a (t) \int_{0}^{1} u sin^{n} u d x,

which implies

\frac{1}{2} \frac{d}{d t} \int_{0}^{1} ∣ u ∣^{2} d x \leq a (t) \int_{0}^{1} u sin^{n} u d x .

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