Past exam of the calculus of variations course of the University of Oxford / 2011 / 2

The variational terminology refers to the size of a neighborhood, rather than weak versus strong convergence in a Banach space. The definitions in the two numbered children giveThe second implication holds because smallness in $W^{1,\infty }$ implies smallness in the uniform norm. Reverse implications need not hold.

For differentiable $f$ in its slope variable, define the Weierstrass excess functionThe strengthened Weierstrass condition means that some $\delta >0$ givesIt requires support at nearby heights and nearby reference slopes, while the trial slope $q$ is arbitrary. Requiring this only exactly on the curve is the unstrengthened necessary condition.

A standard Weierstrass sufficient condition assumes the usual $C^2$ smoothness of $f$, $u\in \mathcal{A}\cap C^1$, the weak Euler-Lagrange equation, a coercively positive second variation, and the strengthened condition above. Explicitly, for some $c>0$,for zero-endpoint variations. Then $u$ is a strong local minimizer. If the excess is strictly positive for $q\ne p$ throughout the tube, the strong minimum is strict. The second-variation hypothesis supplies a neighboring field of extremals. Its slope $p(x,y)$ gives the comparison identityfor uniformly close admissible $v$. This explains how the condition controls competitors with large derivatives.

A weak local minimizer of a variational integral is an admissible $\bar{u}$ for which there is $\epsilon >0$ such thatBoth the function and its first derivative must remain uniformly close, up to the almost-everywhere convention for the derivative.

A strong local minimizer of a variational integral is an admissible $\bar{u}$ for which there is $\epsilon >0$ such thatThe competitor's derivative need not be close. This compares with a larger family of competitors, which is why it is the stronger minimality property.

For any zero-endpoint admissible $u$, use $|\sin u|\le |u|$ and the sharp Poincaré inequality on $(0,1)$:The coefficient is positive, and equality forces $u^{\prime }=0$, hence $u=0$ by its endpoint values. ThusIt is therefore both a strict strong and a strict weak local minimizer.

Here $I(0)=1/2$. First restrict $\Vert u^{\prime }{\Vert }_{\infty }\le \pi /12$. On that slope interval, the second derivative of $h(p)=1-\cos (\pi /3+p)$ is $\cos (\pi /3+p)\ge \cos (5\pi /12)>1/4$. Taylor's theorem givesThe linear term integrates to zero because ${\int }_0^1{u}^{\prime }=0$. Since $x^3\le 1$, the Poincaré inequality yieldsThe coefficient is positive. Thus zero is a strict weak local minimizer.

For failure of strong minimality, take a triangular bump supported on $[1/2-\epsilon ,1/2+\epsilon ]$, with slopes $+2\pi$ and $-2\pi$, and zero outside. Its height is $2\pi \epsilon \to 0$. Almost everywhere its slope is $0$ or $\pm 2\pi$, so the cosine term has exactly its zero-function value everywhere. The potential term is strictly negative on the bump. ThereforeHence it is not a strong local minimizer. The derivatives of these bumps stay large, so they do not contradict weak local minimality.