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

For fixed-endpoint variations, the second variation iswith $p=f_{vv}$, $r=f_{uv}$, $s=f_{uu}$ evaluated along $(x,\bar{u},{\bar{u}}^{\prime })$. The Jacobi equation, written without differentiating $r$ separately, isWhen $r$ is differentiable this is $-(p{z}^{\prime }{)}^{\prime }+(s-r^{\prime })z=0$.

Positive definiteness means $q(\phi )>0$ for every nonzero admissible variation. Uniform or coercive positivity of a second variation is the stronger assertion that $q(\phi )\ge c\int {\phi }^{\prime 2}$ for some fixed $c>0$. On a fixed-endpoint interval this is equivalent to a positive lower bound in the $W^{1,2}$ norm.

The strict Legendre implication uses the uniform interpretation of positivity. If $q\ge c\int {\phi }^{\prime 2}$, localized rapidly oscillating variations show $p(x)\ge c$: otherwise support them in an interval where $p-c<0$, so the derivative term of $q-c\int {\phi }^{\prime 2}$ is negative of order frequency squared, overwhelming its mixed and zero-order terms. Continuity extends the bound to the endpoints, giving $p>0$ on $[a,b]$. Under mere positive definiteness this strict conclusion is false: $f=u^2$ at $\bar{u}=0$ gives $q=2\int {\phi }^2>0$ but $p=0$. In that weaker convention only the ordinary Legendre necessary condition $p\ge 0$ follows. This distinction makes the implication precise.

Under the stated regular condition $p>0$, a conjugate point $c\in (a,b]$ to $a$ is an endpoint admitting a nonzero Jacobi solution with $z(a)=z(c)=0$. The disconjugacy form of the Jacobi positivity criterion says that no such point in $(a,b]$ exists if and only if there is a Jacobi solution strictly positive on all of $[a,b]$.

For the required identity, let that positive solution be $z$ and write $\phi =zh$. Set $A=pz{z}^{\prime }+r{z}^2$. The Jacobi equation givesExpanding the second-variation integrand therefore yieldsFor compactly supported $\phi$, the boundary term vanishes, provingThis is the Jacobi identity for the second variation. The divergence form is also appropriate under the original $C^1$ regularity; with $p>0$, the Euler-Lagrange equation gives the usual further regularity along the extremal.

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.

For every $0<\lambda <1$, positivity in the preceding part gives $\int u^{\prime 2}\ge {\lambda }^2\int u^2$ for each fixed admissible $u$. Let $\lambda \uparrow 1$ to obtainThe sharp constant is attained by $u(x)=\sin x$, consistently with the first conjugate point occurring at the endpoint for $\lambda =1$.

Choose any nonzero admissible $u_0$ and set $u=\gamma u_0$. ThenThe coefficient of $-{\gamma }^6$ is strictly positive, so the energy tends to minus infinity as $|\gamma |\to \infty$. Hence