 Solution

ID: past-exam-of-the-partial-differential-equations-for-pure-and-applied-mathematicians-and-calculus-of-variations-course-of-the-university-of-oxford/2007/b/5/a/ii/solution

Put

w = tϕ

into the differentiability expansion. The remainder is

o (∣ t ∣∥ ϕ ∥)

, so division by

t

and passage to zero gives $g^{'} (0) = D I (u) ϕ$ . For

ϕ = 0

this is immediate as well.

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