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

Parametrize the surface by $r(x,\theta )=(x,y(x)\cos \theta ,y(x)\sin \theta )$, with $0\le \theta <2\pi$. Its tangent vectors are orthogonal, of lengths $\sqrt{1+y^{\prime 2}}$ and $y$. Thus the area element is $y\sqrt{1+y^{\prime 2}}dxd\theta$. Integrating over the angle gives

Discard the constant factor $2\pi$ when forming the Euler-Lagrange equation. For $F=y\sqrt{1+y^{\prime 2}}$ it becomesThe Beltrami identity, since $F$ has no explicit $x$ dependence, gives $F-y^{\prime }{F}_{y^{\prime }}=y/\sqrt{1+y^{\prime 2}}=C>0$. Combining with the differential equation yields $y^{\prime \prime }=y/C^2$. Its exponential solution has coefficients whose product is $C^2/4$, by $y^{\prime 2}=y^2/C^2-1$. Positivity therefore puts it in the formIts constants must satisfy $y_1=C\cosh ((x_1-K)/C)$ and $y_2=C\cosh ((x_2-K)/C)$. Depending on the endpoint data, these equations may have no solution or more than one. A constant function can satisfy a formal first-integral equation but fails the full Euler-Lagrange equation; it is not an extra extremal.

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.