 Solution

ID: past-exam-of-the-calculus-of-variations-course-of-the-university-of-oxford/2009/2/b/ii/solution

For every

0 < λ < 1

, positivity in the preceding part gives

\int u^{'2} \geq λ^{2} \int u^{2}

for each fixed admissible

u

. Let

λ ↑ 1

to obtain

\int_{0}^{π} u^{'2} d x \geq \int_{0}^{π} u^{2} d x .

The sharp constant is attained by

u (x) = sin x

, consistently with the first conjugate point occurring at the endpoint for

λ = 1

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