Past exam of the mathematics course of the University of Cambridge / 2005 / ib / Paper 2 / 15E / Solution

Fixed endpoint variations yield the Euler–Lagrange equationWhen $F$ is independent of $z$, differentiation givesThis is the Beltrami identity. Denote the constant by $1/k$ when nonzero; for a general Lagrangian a zero constant is also possible.

The film area is $S=2\pi {\int }_{-H}^{H}r\sqrt{1+r^{\prime 2}}dz$. Suppressing its harmless positive factor $2\pi$, take $F=r\sqrt{1+r^{\prime 2}}$. The Euler–Lagrange equation reduces toThe first integral is $r/\sqrt{1+r^{\prime 2}}=1/k$ with $k>0$. Hence $1+r^{\prime 2}=k^2{r}^2$ and, using the differential equation without dividing by $r^{\prime }$, $r^{\prime \prime }=k^{2}r$. The positive solution obeying that first integral is $r=k^{-1}\cosh (k(z-z_0))$. Equal endpoint radii force $z_0=0$, givingThus every smooth positive stationary film in this axisymmetric class is a catenoid.

Put $u=kH>0$. The boundary equation is $R/H=f(u)=\cosh u/u$. The function tends to infinity at both endpoints of $(0,\infty )$, andIts minimum occurs when $u=\coth u$. The function $u-\coth u$ is strictly increasing, with derivative $1+{\mathrm{csch}}^{2}u>0$, and has limits $-\infty$ and $+\infty$. It therefore has exactly one positive zero $A$. At this point $1/A=\tanh A$, so $f(A)=\sinh A$. This proves the catenoid existence threshold for equal rings:At equality there is one stationary solution and above it two. The equation describes candidates for an area minimum, not a guarantee that both branches minimize. The thinner-neck branch is unstable to suitable axisymmetric variations; topology-changing collapse can also compete with connected films. In particular two separate planar disks have area $2\pi R^2$. The derived threshold concerns existence of the smooth annular stationary film; it does not exclude such disconnected competitors.