Past exam of the calculus of variations course of the University of Oxford / 2009 / 2 / b / i / Solution

At zero, the sixth-power term has zero second derivative, soFor $0<\lambda <1$, the Jacobi solution $z(x)=\cos (\lambda (x-\pi /2))$ is strictly positive on $[0,\pi ]$. The preceding identity gives $Q_{\lambda }(u)=\int z^2[(u/z{)}^{\prime }{]}^2>0$ for nonzero zero-trace $u$.

To control the nonlinear remainder, a uniform lower bound follows directly. Let $m=\min z>0$, $K=\max |z^{\prime }|$, and $v=u/z$. Since $\max z=1$ and $\int v^2\le ({\pi }^2/2)\int v^{\prime 2}$ by the elementary endpoint estimate,Thus $Q_{\lambda }(u)\ge c_{\lambda }\int u^{\prime 2}$ with $c_{\lambda }=m^2/(K^2{\pi }^2+2)>0$. Also $\int u^6\le \Vert u{\Vert }_{\infty }^4({\pi }^2/2)\int u^{\prime 2}$, soThis is positive for every nonzero admissible $u$ sufficiently close to zero in $W^{1,\infty }$. Therefore zero has positive second variation and is a strict weak local minimizer. No sharp Poincaré inequality has been assumed in this proof.