 Solution

ID: past-exam-of-the-calculus-of-variations-course-of-the-university-of-oxford/2009/3/b/i/solution

Zero is neither a weak nor a strong local minimizer. Take

u_{ε} (x) = - ε sin (π x)

. It tends to zero in

W^{1, \infty}

and has zero endpoint values. Since

1 + t \leq 1 + t /2

for

t \geq 0

I (u_{ε}) - I (0) \leq \frac{ε ^{2} π ^{2}}{4} - \frac{4018}{π} ε < 0

for sufficiently small positive

ε

. Thus even weakly close admissible competitors decrease the energy.

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