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

Parametrize the surface of revolution byThe magnitude of the cross product ${\mathbf{r}}_x\times {\mathbf{r}}_{\theta }$ is $y\sqrt{1+y^{\prime 2}}$. Integrating the surface area element over $\theta$ givesPositivity of $y$ removes an otherwise necessary absolute value in the area element.

For $F(y,y^{\prime })=y\sqrt{1+y^{\prime 2}}$, the Euler-Lagrange equation isSince $F$ has no explicit $x$ dependence, the Beltrami identity givesSolving this first integral, or substituting into the second-order equation, yieldsThe constants satisfy $y_1=C\cosh ((x_1-K)/C)$ and $y_2=C\cosh ((x_2-K)/C)$. There need not be a solution for arbitrary endpoint data, and when there is one these equations can have more than one pair of constants. These are the possible stationary catenoids, so the axisymmetric second variation of a catenoid still matters when deciding whether they minimize surface area.

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.