Past exam of the calculus of variations course of the University of Oxford / 2011 / 3

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.