 Solution

ID: past-exam-of-the-calculus-of-variations-course-of-the-university-of-oxford/2011/2/b/i/solution

For any zero-endpoint admissible

u

, use

∣ sin u ∣ \leq ∣ u ∣

(0, 1)

I (u) \geq \frac{1}{2} \int_{0}^{1} (u^{'})^{2} - \int_{0}^{1} u^{2} \geq (\frac{1}{2} - \frac{1}{π ^{2}}) \int_{0}^{1} (u^{'})^{2} \geq 0.

The coefficient is positive, and equality forces

u^{'} = 0

, hence

u = 0

by its endpoint values. Thus

u = 0 is the unique strict global minimizer .

It is therefore both a strict strong and a strict weak local minimizer.

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