Past exam of the calculus of variations course of the University of Oxford / 2011 (C5.2b, DMAC 2B65)

Original solutions to the paper released at the source request. The complete editable transcription is retained in the private examination archive.

For the supporting inequality, assume first that $F$ is convex and fix $z,z_0$. For $0<t\le 1$,Subtract $F(z_0)$, divide by $t$ and let $t\downarrow 0$. Differentiability givesConversely assume this inequality for every pair. Put $w=(1-t)z_0+tz$ and apply it with base point $w$ to both $z_0$ and $z$. Multiply those two inequalities by $1-t$ and $t$ and add. The gradient terms cancel, proving $F(w)\le (1-t)F(z_0)+tF(z)$. Thus the supporting inequality characterizes a differentiable convex function.

For the integral inequality let $\bar{z}=|\Omega {|}^{-1}{\int }_{\Omega }u$. Apply the supporting inequality at $\bar{z}$ to $u(x)$ and integrate. Its affine term has integral $DF(\bar{z})\cdot ({\int }_{\Omega }u-|\Omega |\bar{z})=0$, yieldingThis is Jensen's inequality. Since $u$ is integrable, the supporting affine lower bound has integrable negative part, so the right side is well-defined as an extended real integral. If its positive part is infinite, the inequality remains valid.

A convex function satisfiesIts value on a line segment never lies above the chord joining its endpoint values.

A strictly convex function satisfies the strict inequalityEndpoint and coincident-point cases are excluded because they necessarily give equality.

A global minimizer is an admissible function $\bar{u}$ for whichEquivalently it attains the infimum over the whole admissible class. Its endpoint traces must equal the prescribed boundary values; comparing only with nearby functions would define a local rather than a global notion.

The integrand is nonnegative, so $\inf I\ge 0$. For each positive integer $n$, setThis is a continuous piecewise affine sawtooth with zero endpoints, $0\le u_n\le 1/n$, and $u_n^{\prime }=\pm 2$ almost everywhere. It belongs to $W^{1,\infty }$, and $(u_n^{\prime }{)}^6=64$ almost everywhere. ThereforeTogether with nonnegativity this proves $\boxed{\underset{\mathcal{A}}{\inf }I=0}$. The functions converge uniformly to zero while their derivatives retain magnitude two; small amplitude does not force small gradient.

If an admissible $u$ attained the infimum zero, each nonnegative term of its integrand would vanish almost everywhere. In particular $|u{|}^{2011}=0$ almost everywhere, so $u=0$ in its Sobolev space and its weak derivative is zero. But then the remaining term is $(-64{)}^2=4096$ almost everywhere, giving $I(u)=4096$, a contradiction. Thus the infimum is zero but no global minimizer exists. The oscillatory sequence has no convergence of derivatives compatible with substitution into this nonconvex energy.

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.

A scalar Tonelli existence theorem states: if $f(x,y,p)$ is continuous, convex in $p$, and has a uniform superlinear lower bound $f(x,y,p)\ge \Theta (|p|)-C$ with $\Theta (r)/r\to \infty$, and at least one fixed-endpoint admissible absolutely continuous function has finite energy, then the energy attains its infimum in the fixed-endpoint $W^{1,1}$ class. One may take $\Theta$ nonnegative and convex. The lower bound makes derivatives uniformly integrable; endpoint control then gives uniform compactness of a minimizing sequence, and convexity supplies the required lower semicontinuity.

For the specified functional, the integrand is continuous, nonnegative, and convex in the derivative, withThe affine function $u_0(x)=2x-1$ has the required endpoints and finite energy. Thus a global minimizer exists. In fact any finite-energy minimizer belongs to $W^{1,52}$, since its derivative's 52nd power is integrable.

It is unique. The admissible class is convex, and the integrand is strictly convex in $(y,p)$: $|p{|}^{52}$ and $|y{|}^{2011}$ are strictly convex, with the latter multiplied by $e^x>0$. If two distinct admissible finite-energy functions minimized the energy, their midpoint would have strictly smaller energy. Distinct continuous representatives differ on a set of positive measure, so the strict integral inequality really applies. This contradiction proves uniqueness.

In one dimension, $u\in W^{1,2}$ has an absolutely continuous bounded representative. For any smooth zero-endpoint test function $\eta$, the functions $u+t\eta$ are admissible and arbitrarily close to $u$ for small $t$. Local minimality forces the first variation to vanish. Differentiation under the integral is valid because $u,\eta$ are bounded and the derivative terms are integrable, giving the weak Euler-Lagrange equationIn distributional form this says $(2u^{\prime }{)}^{\prime }=4x{u}^3$. The fundamental theorem for weak derivatives, or the fundamental lemma applied after subtracting an antiderivative, gives a constant $C$ withThis is the integrated Euler-Lagrange equation. The boundary difference $u(1)-u(0)=2$ determinesThe weak identity extends to all zero-trace $W^{1,2}$ test functions by density.

The continuous representative of $u$ makes $x\mapsto 2xu(x{)}^3$ continuous on the closed interval. The integrated formula therefore gives a continuously differentiable representative of $u^{\prime }$, and hence $u\in C^2([0,1])$ withDifferentiate once more. The right side is continuously differentiable because $u\in C^2$, andThus $\boxed{u\in C^3([0,1])}$. These formulas give continuous one-sided derivatives at the endpoints as well. Repeated bootstrapping would actually give smoothness of every order.