Past exam of the calculus of variations course of the University of Oxford / 2010 / 2 / b

Here $P=2$, $Q=-2{\lambda }^2$ and the Jacobi equation is $w^{\prime \prime }+{\lambda }^{2}w=0$. With $w(0)=0$ and $w^{\prime }(0)=1$, its solution is $w(x)=\sin (|\lambda |x)/|\lambda |$ for $\lambda \ne 0$, and $w(x)=x$ for $\lambda =0$. Its first positive zero is $\pi /|\lambda |$ when $\lambda \ne 0$.

If ${\lambda }^2<{\pi }^2/b^2$, no conjugate point occurs in $(0,b]$. The Jacobi positivity criterion givesSince $I_{\lambda }$ itself is this quadratic form divided by two, $I_{\lambda }(u)>0=I_{\lambda }(0)$ for every nonzero admissible $u$. Thus zero is the unique global minimizer. This argument establishes the result before using the sharp Poincaré inequality.

Fix an admissible $u$. The preceding result gives ${\int }_0^b{u}^{\prime 2}dx\ge {\lambda }^2{\int }_0^b{u}^{2}dx$ for every ${\lambda }^2<{\pi }^2/b^2$. Letting ${\lambda }^2$ increase to ${\pi }^2/b^2$ provesEquality is attained byIndeed the derivative and function squared integrals are $A^2{\pi }^2/(2b)$ and $A^{2}b/2$. These are all equality cases: an equality function minimizes the nonnegative threshold quadratic form, so its first variation vanishes. The resulting weak equation $u^{\prime \prime }+(\pi /b{)}^{2}u=0$ makes $u$ smooth and, with the endpoint conditions, yields precisely these multiples. This is the Wirtinger inequality.

Use $u_A(x)=A\sin (\pi x/b)$, which is admissible for every real $A$. Direct integration givesThe coefficient is strictly negative, so the energy tends to minus infinity as $|A|\to \infty$. Therefore