Past exam of the calculus of variations course of the University of Oxford / 2010 / 1

A weak local minimizer of a variational integral is an admissible $\bar{u}$ for which there is $\rho >0$ such thatThe Sobolev norm controls both the function and its first weak derivative. The boundary values remain fixed because both functions are admissible.

A strong local minimizer of a variational integral is an admissible $\bar{u}$ for which some $\rho >0$ satisfiesOnly the function values must be close; admissible derivatives may be large. Every strong local minimizer is therefore also a weak local minimizer of a variational integral.

A point of convexity of $g\in C^1(\mathbb{R})$ is a point $v_0$ where its tangent line supports the entire graph from below:The Weierstrass necessary condition says that, at almost every $x$, ${\bar{u}}^{\prime }(x)$ is a point of convexity of $v\mapsto f(x,\bar{u}(x),v)$. In terms of the Weierstrass excess function,

Now use the stated derivative coercivity. If $\bar{u}$ were not a local minimizer in $L^1$, there would be $v_n\in \mathcal{A}$ with $\Vert v_n-\bar{u}{\Vert }_1\to 0$ and $I(v_n)<I(\bar{u})$. The energy bound givesThe common value $v_n(a)=\alpha$ controls their function norms as well, by $|v_n(x)|\le |\alpha |+{\int }_a^b|v_n^{\prime }|$. Thus $(v_n)$ is bounded in $W^{1,2010}(a,b)$. The one-dimensional compact Sobolev embedding into $C([a,b])$ supplies a uniformly convergent subsequence. Its limit must be $\bar{u}$, since the same subsequence converges to $\bar{u}$ in $L^1$. Eventually $\Vert v_n-\bar{u}{\Vert }_{\infty }<\rho$, contradicting strong local minimality. Hence $\bar{u}$ is also a local minimizer in $L^1$. This is coercivity transfers strong minimality to L1 minimality.

Zero is a strict weak local minimizer, but is not a strong local minimizer. Put $D={\int }_0^1|u^{\prime }{|}^{2}dx$. Since $u(0)=0$, the Cauchy-Schwarz inequality gives $|u(x){|}^2\le xD$, and thereforeFor sufficiently small $\Vert u^{\prime }{\Vert }_{\infty }$, $\sin (u^{\prime 2})\ge \frac{1}{2}{u}^{\prime 2}$ pointwise. Hencewith equality only when $u=0$. Thus zero is a strict weak local minimizer of a variational integral.

For the strong claim, define a Lipschitz triangular waveIt has zero endpoint values, $\Vert u_N{\Vert }_{\infty }=\sqrt{\pi }/(2N)\to 0$, and $u_N^{\prime }=\pm \sqrt{\pi }$ almost everywhere. Thus $\sin (u_N^{\prime 2})=\sin \pi =0$, while $u_N$ is nonzero on a set of positive measure. ConsequentlyThese uniformly small competitors disprove the strong local minimizer of a variational integral property.

Zero is a strict strong local minimizer, and therefore also a strict weak local minimizer. Here $I(0)=-2/2010$, andFor $|u|\le 1$, Taylor's theorem gives $\cosh u-1\le \frac{1}{2}\cosh (1)u^2$. Also $u(0)=0$ and the Cauchy-Schwarz inequality give ${\int }_0^1{u}^{2}dx\le \frac{1}{2}{\int }_0^1{u}^{\prime 2}dx$. ThereforeThe coefficient is positive. Thus every nonzero admissible $u$ with $\Vert u{\Vert }_{\infty }\le 1$ has strictly greater energy, proving that zero is a strict strong local minimizer of a variational integral.