Past exam of the partial differential equations for pure and applied mathematicians and calculus of variations course of the University of Oxford / 2008 / B / 6

A weak local minimizer of a variational integral is an admissible $\bar{u}$ for which some $\rho >0$ gives $I(v)\ge I(\bar{u})$ for all admissible $v$ with $\Vert v-\bar{u}{\Vert }_{W^{1,\infty }}<\rho$.

A strong local minimizer of a variational integral obeys the same inequality for all admissible $v$ with $\Vert v-\bar{u}{\Vert }_{\infty }<\rho$. The derivative of a competitor need not be close in this definition.

A global minimizer has $I(\bar{u})\le I(v)$ for every admissible $v$. Global minimality implies strong local minimality, which implies weak local minimality. The reverse implications are not valid in general.

Where $f_v$ exists, the strengthened Weierstrass condition says that for some $\delta >0$, the excessis nonnegative for every $x\in [a,b]$, every trial slope $q\in \mathbb{R}$, and all $y,p$ with $|y-\bar{u}(x)|<\delta$, $|p-{\bar{u}}^{\prime }(x)|<\delta$. This is a neighborhood condition in height and reference slope, not just an inequality on the reference curve.

The classical Weierstrass sufficient condition, with the usual smooth integrand regularity in addition to continuity, states that an admissible $C^1$ curve satisfying the Euler-Lagrange equation, a coercively positive second variation, and this strengthened condition is a strong local minimizer. A strict excess away from the reference slope gives the corresponding strict conclusion. The differentiability assumptions are part of that classical sufficient theorem; continuity alone does not define its first and second variations.

Zero is a strict weak local minimizer and is not a strong local minimizer. Write $D={\int }_0^1{u}^{\prime 2}$. For zero endpoint values, Cauchy-Schwarz inequality applied togives $|u(x){|}^2\le x(1-x)D$, hence ${\int }_0^1{x}^2{u}^2\le D/20$. For sufficiently small slopes, Taylor expansion gives $\frac{1}{2}[\sin (1+u^{\prime 2})-\sin 1]\ge (\frac{1}{2}\cos 1-\epsilon )u^{\prime 2}$. Thereforefor a suitably small fixed $\epsilon$ and every nonzero sufficiently weakly close competitor.

For strong minimality take $a=\sqrt{3\pi /2-1}$ and $u_N(x)=a\mathrm{dist}(x,N^{-1}\mathbb{Z})$. Their amplitudes tend to zero but slopes are $\pm a$, so $\sin (1+u_N^{\prime 2})=-1$. Thus $I(u_N)\le -1/2<I(0)=\sin 1/2$. Uniform closeness therefore does not prevent a decrease in energy.

The inequality $\cos t\ge 1-t^2/2$ givesThe elementary endpoint estimate $|u(x){|}^2\le x{\int }_0^1{u}^{\prime 2}$ implies $\int u^2\le \frac{1}{2}\int u^{\prime 2}$. ConsequentlyThus zero is the unique global minimizer, and in particular a strict strong and strict weak local minimizer.