Past exam of the Probability, statistics and options II course of the University of Oxford / 2007 / G / 2 / iii / Solution

The functional is unbounded above. Let $\ell$ be the positive affine interpolant of the endpoint values and choose a nonnegative, nonzero, continuously differentiable $h$ vanishing at both endpoints. The admissible functions $y_M=\ell +Mh$ satisfyThis excludes a global maximum, but does not prove that a stationary curve is a minimum. The concluding assertion is false without additional conditions.

An explicit counterexample uses $x_1=-2$, $x_2=2$, $y_1=y_2=\cosh 2$ and the stationary function $y=\cosh x$. Its second variation, for a perturbation $h$ vanishing at the endpoints, isChoose $h=1-x^2/4$. Since ${\mathrm{sech}}^{2}x\le 4e^{-2|x|}$, the positive contribution is at most ${\int }_{\mathbb{R}}{x}^2{e}^{-2|x|}dx=1/2$. On $[-1,1]$, $h^2\ge 9/16$, so the magnitude of the negative contribution is at least $(9/8)\tanh 1>9/16>1/2$. Thus ${\delta }^{2}J[h]<0$. Small perturbations in this direction preserve positivity and the endpoint data while decreasing surface area. Consequently this catenoid is not a local minimum. The valid conclusion from part (ii) is a necessary stationary form; minimization requires a separate stability and existence analysis.