Surfaces of revolution with constant mean curvature (II)

Following the above entry, the case $H=0$ is known: the plane and the catenoid are the only rotational minimal surfaces. If $H=0$, then we have 

$$\frac{f(z)}{\sqrt{1+f'(z)^2}}=c,\ c>0,$$ that is, $$\frac{f'}{\sqrt{f^2-c^2}}=\frac{1}{c}.$$ Then the solution is $$f(z)=c\cosh(\frac{1}{c}z+d),\ d\in{\mathbb R} (*).$$

Minimal surfaces are models of soap films. In this particular case of the catenoid, the surface is the soap film formed by two coaxial circles, that is, two circles $C_1\cup C_2$ in parallel planes and the straight-line joining their centers is orthogonal to the planes containing the circles. It is natural to ask if there exists a soap film joining two given circles in parallel planes. In order to simplify the arguments, we suppose 

1. the radii of the circles are identical, namely, $r>0$. 
2. the circles $C_1$ and $C_2$  are contained in the planes of equation $z=-h$ and $z=h>0$, respectively. 

Then we pose the next:

Problem. Under what conditions on $r$ and $h$ does exist a catenoid $S$ joining $C_1$ and $C_2$? In such a case, how many catenoids do exist?

By the symmetry of the hyperbolic cosine, and since $f(h)=f(-h)$, we conclude $d=0$ in (*). Thus, the problem reduces to find $c>0$ such that $$c\cosh(\frac{1}{c}h)=r (**).$$

It is natural to think that if the circles lie very close, then there do exists a catenoid, that is, if $h$ is small, then there exists a solution of (**). 

We propose the problem in the next direction. We suppose that the circles are given (the radius $r$, which we suppose $r=1$). If they are close, there exists a catenoid, but if we separate far then the catenoid is destroyed, that is, there do no exist a catenoid between both circles.

In order to simplify the problem, we do a homothety of the ambient space from the origin and we suppose that the value $r$ of the radius is $r=1$. Consider the function $$g(c)=c\cosh(\frac{1}{c}h).$$ Our idea is using the mean value theorem. It is not difficult to see that $$\lim_{c\rightarrow 0}g(c)= \lim_{c\rightarrow \infty}g(c)=\infty,$$

so we have to study carefully the monotonicity intervals of $g$. We calculate the critical points of $g$. We have

$$g'(c)=\cosh(h/c)-\frac{h}{c}\sinh(h/c).$$ By letting $y=h/c$, this is equivalent to find $y>0$ such that $$\frac{1}{y}=\tanh(y).$$ The function $1/y$ is decreasing from $-\infty$ to $0$ and $\tanh(y)$ is increasing from $0$ to $1$, s   there is only one critical point $c_0$ (with $h>c_0$). See the next figure:

Because the limits are $\infty$, then this critical point is a minimum, $c=c_0$. This the graphic of $g$ when $h=0,5$, $h=1$ and $h=4$, and the graphic of $y=1$.

Let us observe that the minimum increases with $h$! so we have to find that it is possible to choose $h$ so the value of this minimum is $1$ at more.

We compute the value of the minimum, that is, $g(c_0)$. We know that $c_0/h=\tanh(h/c_0)$, and numerically we obtain, $h/c_0=1.19968$. Then 

$$g(c_0)=\frac{hc_0}{\sqrt{h^2-c_0^2}}=\frac{h}{\sqrt{(h/c_0)^2-1}}=1,50888 h.$$

When $h$ is close to $0$, $g(c_0)<1$, thus the graphic of $g$ has points under the line $y=1$, proving that there exists two catenoids spanning $C_1\cup C_2$. For a certain height $h=h_0$, this minimum is exactly $1$, so there exists only one catenoid and when $h>h_0$ there do not exist a catenoid joining $C_1$ and $C_2$. The value of $h_0$ is

$$h_0=\frac{1}{1,50888}=0,6627.$$
Thus, and after a homothety, we obtain:

Theorem. Let $d$ be the distance $d$ between two coaxial circles of radii $r>0$.

1. If $d<1,3254 r$, there exists exactly two catenoids spanning $C_1\cup C_2$.
2. If $d=1,3254 r$, there exists exactly two catenoids spanning $C_1\cup C_2$.
3. If $d>1,3254 r$, there do not  exist a catenoid spanning $C_1\cup C_2$.

Now we give one example of two circles that bound two catenoids. Take $r=1$ and we choose $h=0.5$. The solutions of $g(c)=1$ are: $c_1=0,235095$ and $c_2=0,848338$. The picture of the two catenoids is

Finally a remark: when one dips two coaxial circles in a soapy water container, only one catenoid is formed. In the above case, it would be the blue catenoid. Among the two catenoids, the physical systems chooses that catenoid with minimum area (minimum energy) and in this case, is the 'exterior' catenoid.