Past exam of the options II course of the University of Oxford / 2007 / G / G2 / iii / Solution

The functional is unbounded above. Let $y_0$ be the positive affine interpolant between the two endpoint values, and choose a smooth nonnegative function $\varphi$ that vanishes at the endpoints and is positive somewhere inside. The admissible curves $y_M=y_0+M\varphi$ satisfyThe claimed minimum conclusion does not follow and is false in general. A stationary point may have a negative second variation, even though the functional is unbounded above. Here is an explicit example within the allowed positive-graph class.

Take the stationary profile $y(x)=\cosh x$ on $[-2,2]$, and the endpoint-zero variation $\eta (x)=\cosh x-\cosh 2$. For sufficiently small perturbation parameter $s$, $y+s\eta$ remains positive and preserves both endpoint values. Put $C=\cosh 2$. Differentiating the area twice and integrating its mixed term by parts givesThis is strictly negative: the first term is less than $2\pi C$, while $(C+C^{-1})\tanh 2>\pi$. For example $e^2>7$ gives $C>7/2$ and $\tanh 2>24/25$, whose product already exceeds $\pi$. Thus nearby positive profiles have smaller area. This catenoid is not a local minimum.

The corrected conclusion is that any smooth positive minimizer, if attained, must have the catenoid form. That necessary condition is not sufficient, and existence or stability requires a separate argument.