Past exam of the calculus of variations course of the University of Oxford / 2009 (C5.2b, DMAC 2B65)

A convex function satisfiesThus its value on every line segment lies below the chord joining the endpoint values.

A strictly convex function satisfies the same inequality strictly whenever $x\ne y$ and $0<t<1$.

For the unlabelled Hessian criterion, convexity implies $F(z_0+th)+F(z_0-th)-2F(z_0)\ge 0$. Divide by $t^2$ and let $t\to 0$ to obtain $h^T{D}^{2}F(z_0)h\ge 0$ for every $h$. Thus the Hessian matrix is positive semidefinite.

Conversely, if every Hessian is positive semidefinite, Taylor's integral formula with $h=z-z_0$ givesThe allowed supporting-plane characterization then provesFor the integral inequality put $m=(b-a{)}^{-1}{\int }_a^{b}u$. The supporting tangent at $m$ gives $F(u(x))\ge F(m)+F^{\prime }(m)(u(x)-m)$ almost everywhere. Integrating cancels the linear term, soThis is Jensen's inequality. It remains valid with value $+\infty$ on the right: the affine lower bound has integrable negative part, so the extended integral is well-defined.

An admissible $\bar{u}$ is a global minimizer ifIn particular it attains the infimum; no closeness condition on competing admissible functions is imposed.

The integrand is nonnegative. Define the Lipschitz triangular wavesThey have zero endpoint values, slope $\pm 2$ almost everywhere, and height $1/N$. Hence $(u_N^{\prime 4}-16{)}^4=0$ almost everywhere. On each of the $N$ equal cells, direct integration of the fourth power of the two affine pieces gives $1/(5N^5)$. ThereforeThis is an oscillating minimizing sequence with no minimizer.

If an admissible $u$ attained the infimum, nonnegativity and $I(u)=0$ would force $u^4=0$ almost everywhere, so its continuous Sobolev representative would be identically zero. Its derivative would then vanish almost everywhere, but the other integrand term would equal $16^4>0$. This is impossible. Equivalently, zero energy would simultaneously require $u=0$ and $|u^{\prime }|=2$ almost everywhere. Thus

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

A weak local minimizer of a variational integral is an admissible $\bar{u}$ for which some $\rho >0$ makes $I(v)\ge I(\bar{u})$ whenever $v\in \mathcal{A}$ and $\Vert v-\bar{u}{\Vert }_{W^{1,\infty }}<\rho$. Both the function and its derivative are required to be close.

A strong local minimizer of a variational integral is an admissible $\bar{u}$ for which the same conclusion holds whenever $\Vert v-\bar{u}{\Vert }_{\infty }<\rho$. Every strong local minimizer is weak, because Sobolev closeness implies uniform closeness. A global minimizer is both, and the converse implication from weak to strong is not generally valid.

If the integrand is independent of $u$, a weak local minimizer has vanishing first variation:The bounded function $h(x)=f_v(x,{\bar{u}}^{\prime }(x))$ therefore has zero distributional derivative, so is constant almost everywhere. For an elementary test-function argument, any smooth compactly supported $\psi$ of integral zero is the derivative of a compactly supported smooth $\phi$. Hence $\int h\psi =0$; subtracting a fixed unit-integral test function from arbitrary tests shows that $h$ agrees with one constant against every test. The fundamental lemma then givesThis is the first integral for a variational integrand independent of the function value.

A point of convexity $v_0$ of a differentiable real function $g$ has a globally supporting tangent: $g(v)\ge g(v_0)+g^{\prime }(v_0)(v-v_0)$ for every $v$. The Weierstrass necessary condition for a strong minimum says that ${\bar{u}}^{\prime }(x)$ is such a point for $v\mapsto f(x,\bar{u}(x),v)$ for almost every $x$. Equivalently, the Weierstrass excess function is nonnegative at every trial slope.

Zero is neither a weak nor a strong local minimizer. Take $u_{\epsilon }(x)=-\epsilon \sin (\pi x)$. It tends to zero in $W^{1,\infty }$ and has zero endpoint values. Since $\sqrt{1+t}\le 1+t/2$ for $t\ge 0$,for sufficiently small positive $\epsilon$. Thus even weakly close admissible competitors decrease the energy.

Zero is neither a weak nor a strong local minimizer. For any nonzero smooth zero-trace $\phi$, Taylor expansion givesThe mixed integral $\int \phi {\phi }^{\prime }=0$ by the endpoint values. Writing $D=\int {\phi }^{\prime 2}$ and $U=\int {\phi }^2$, the quadratic coefficient is $-3D/8+U/2$. The elementary estimate $U\le D/2$ on $[0,1]$ makes this at most $-D/8<0$. Thus arbitrarily small variations decrease the energy. The separate coefficient one on $u^2$ is retained in this calculation.

Zero is a strict weak local minimizer, but is not a strong local minimizer. The derivative integrand has expansion $\cosh v\cos (v^2)=1+v^2/2+O(v^4)$, so for small $|v|$ it is at least $1+3v^2/8$. Concavity gives $(1+u^2{)}^{1/4}\le 1+u^2/4$ for every real $u$. Therefore, when $\Vert u^{\prime }{\Vert }_{\infty }$ is small,using $\int u^2\le \frac{1}{2}\int u^{\prime 2}$. This proves strict weak minimality.

For the strong claim, let $v_N(x)=\sqrt{\pi }\mathrm{dist}(x,N^{-1}\mathbb{Z})$. Its amplitude tends to zero and its derivative is $\pm \sqrt{\pi }$ almost everywhere, so the derivative integrand is the constant $-\cosh \sqrt{\pi }$. The other integrand term is at most $-2$. HenceThis disproves strong minimality. The same obstruction is detected by the Weierstrass necessary condition, since the supporting-tangent inequality at slope zero fails at $v=\sqrt{\pi }$.