Past exam of the partial differential equations for pure and applied mathematicians and calculus of variations course of the University of Oxford / 2007 / B / 7

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

A strong local minimizer of a variational integral has the same property for all admissible $v$ with $\Vert v-\bar{u}{\Vert }_{\infty }<\rho$; its derivative need not be close. Since the supremum norm is bounded by the Sobolev norm, this neighborhood contains a sufficiently small weak neighborhood. Thus every strong local minimizer is a weak local minimizer of a variational integral.

A global minimizer satisfies $I(\bar{u})\le I(v)$ for every admissible $v$.

A point of convexity $s$ of a differentiable scalar function is one where its tangent supports the entire graph: $g(q)\ge g(s)+g^{\prime }(s)(q-s)$ for every $q\in \mathbb{R}$. The Weierstrass necessary condition for strong minimality, under its differentiability assumptions in the slope variable, isfor almost every $x$. In other words the reference slope is a point of convexity of the slope integrand. The later example is smooth, so these derivative-based conditions apply there.

Write $g(v)=-\cos (\pi v/2)$. For the stipulated $0<\delta <1$, the displayed tent is continuous at its junction, has zero endpoint values, and its slopes are $\delta$ and $\delta -4$. At both slopes $g^{\prime }$ has the same value $c=(\pi /2)\sin (\pi \delta /2)>0$, and $g^{\prime \prime }=({\pi }^2/4)\cos (\pi v/2)$ is strictly positive on common sufficiently small neighborhoods of them. Taylor's theorem gives, for small $\Vert {\varphi }^{\prime }{\Vert }_{\infty }$,Thus $u^{\delta }$ is a strict weak local minimizer of a variational integral. It is not a strong local minimizer of a variational integral: the global supporting-tangent requirement at either slope has nonzero slope $c$, whereas $g$ is bounded. Letting the trial slope tend to positive infinity violates that requirement. The Weierstrass necessary condition therefore fails.

The opening parameter range $\delta \in (0,1)$ is essential. The subsequent phrase “for each $\delta >0$” must be read within it; taken literally beyond that range it is false. At $\delta =2$ the tent slopes are $2,-2$, where $g=1$ and $g^{\prime \prime }<0$, so this is not a weak minimum.

If $u$ is a strong local minimizer of a variational integral, the Weierstrass condition makes $u^{\prime }(x)$ a point of convexity of the bounded function $g$ for almost every $x$. A global supporting line to a bounded function must have zero slope, and its contact value must be a global minimum. Here that value is $-1$. Therefore $g(u^{\prime })=-1$ almost everywhere and $I(u)=-1$, while every admissible energy is at least $-1$. Hence every strong local minimizer is global.

For every positive integer $n$, set $u_n(x)=4n\min \{x,1-x\}$. Its slopes are $4n,-4n$ almost everywhere, both giving $g=-1$. Hence $I(u_n)=-1$, the universal lower bound, and these tents are all distinct. Thus there are infinitely many global minimizers. Zero is another one.