# Catenary curve for minimum surface of revolution

Given two fixed points in plane, we can hang catenaries of different length. However maybe just one of them satisfies the minimum surface of revolution condition. Which one is that?

• Interesting question. A good candidate would be the (false) catenary which is the line segment joining the two points, generating a cone frustrum...Have you had a look at the wonderful article (en.wikipedia.org/wiki/Catenary) ? Commented Jan 26, 2017 at 6:56
• I think the answer is the catenary of the form $a\cosh(x/a)$. With this form we may have two curves or one curve or no curve passing through specified points. If we add a vertical scaling factor $b$, we have all possible catenaries for that pair of points. Commented Jan 26, 2017 at 7:59
• You know the "soap film" physical analogy for minimal surfaces ? Have a look to (en.wikipedia.org/wiki/Minimal_surface_of_revolution) Commented Jan 26, 2017 at 8:24
• All catenaries have an equation of the form $y=a cosh(x/a)$. Commented Jan 26, 2017 at 17:08
• @JeanMarie: If that is so, then IMO the length of the catenary cannot be an additional degree of freedom, apart from the fixed points in the plane. And then there is no difference too between the catenary and the catenary curve determining the catenoid. See my answer below. Am I right or wrong? Commented Feb 9, 2017 at 14:57

Relevant references:

According to the first reference, the area of a surface of revolution is given by: $$A_x=2\pi \int _{a}^{b} y(x) \sqrt {1+\left({\frac{dy}{dx}}\right)^2}\,dx$$ According to the second reference, the Euler-Lagrange equations, resulting from minimizing this area, are given by: $$\frac{\partial L}{\partial q_k} - \frac{d}{dt} \left(\frac{\partial L}{\partial \dot{q}_k}\right) = 0$$ In our case there is only one such equation: $$\frac{\partial L}{\partial y} - \frac{d}{dx} \left(\frac{\partial L}{\partial y'}\right) = 0 \qquad \mbox{with} \qquad L(y,y') = y \sqrt {1+\left(y'\right)^2}$$ Here goes: $$\frac{\partial L}{\partial y} = \sqrt{1+\left(y'\right)^2} \quad ; \quad \frac{\partial L}{\partial y'} = \frac{y\, y'}{\sqrt{1+\left(y'\right)^2}}\\ \frac{\partial L}{\partial y} - \frac{d}{dx}\frac{\partial L}{\partial y'} = \frac{\left[1+\left(y'\right)^2\right]^2}{\left[1+\left(y'\right)^2\right]^{3/2}} - \frac{\left[\left(y'\right)^2+y\,y''\right]\left[1+\left(y'\right)^2\right] - y\,\left(y'\right)^2\,y''} {\left[1+\left(y'\right)^2\right]^{3/2}} = 0$$ Simplification results in the following ODE: $$1 + (y')^2 - y\cdot y'' = 0 \qquad \mbox{or} \qquad y\cdot y'' - (y')^2 = 1$$ The Ansatz $\;y(x) = a\cdot \cosh(x/a+b)\;$ with $(a,b)$ real constants results in the same structure: $$\left[a\cdot \cosh(x/a+b)\right]\left[a\cdot \cosh''(x/a+b)\right] - \left[a\cdot \cosh'(x/a+b)\right]^2 = \\ = \cosh^2(x/a+b) - \sinh^2(x/a+b) = 1$$ We conclude that the catenoid with $\;y(x) = a\cdot \cosh(x/a+b)\;$ is a solution.
(The present approach does not reveal if that solution is the only one, though)

Boundary conditions. Indeed, given two fixed points in plane, we can hang catenaries of different length. Mathematically, this is expressed by $\;y(x) = A\cosh(Bx+C)\;$ with three constants $\,(A,B,C)$ . Apart from the two fixed points, this enables the length to become an additional degree of freedom.
The situation is different with the catenoid. Instead of three constants $\,(A,B,C)\,$ we have only two of these : $(a,b)$ . A physical consequence is that there is only one soap film between two rings:

Mathematics please. So let's try to solve for $\,(a,b)$ , given the fixed points $\,(x_1,y_1),(x_2,y_2)$ : $$\begin{cases}y_1 = a\,\cosh(x_1/a+b) \\ y_2 = a\,\cosh(x_2/a+b) \end{cases}$$ Two equations with two unknowns. Doing it by hand seems to be hopeless. Feeding it into my favorite computer algebra system (MAPLE) results in a two page long expression that the OP doesn't want to know, I think.
Anyway, the length $L$ of the soapfilm catenary can be expressed in the solutions $\,(a,b)\,$ and the abscissae of the end-points, therefore it is no longer a degree of freedom: $$L = \int_{x_1}^{x_2} \sqrt{1+\left[y'(x)\right]^2}\,dx = \int_{x_1}^{x_2} a\sqrt{1+\left[\sinh(x/a+b)\right]^2}\,d(x/a+b) \\ = \int_{x_1/a+b}^{x_2/a+b} a\cosh(u)\,du = a\sinh(x_2/a+b)-a\sinh(x_1/a+b)$$ Brute force. Numerical methods may offer a panacea where analytical methods fail. For ease of the calculations, replace $\,b\,$ by $\,(-p/a)\,$ - should have done this from the beginning - and consider instead: $$\begin{cases}y_1 = a\,\cosh((x_1-p)/a) \\ y_2 = a\,\cosh((x_2-p)/a) \end{cases}$$ Then $\,x = p\,$ is the place where the catenary has its the minimum $\,a$ . It follows that, without loss of generality (??) : $x_1 \le p \le x_2\,$ and $\,0 < a \le \min(y_1,y_2)\,$ . With the home-made tools I have at my disposal, I can easily make an isoline chart / contour plot of the place where the following functions are zero: $$\begin{cases} \color{red}{f(p,a) = a\,\cosh((x_1-p)/a)-y_1 = 0} \\ \color{green}{g(p,a) = a\,\cosh((x_2-p)/a) - y_2 = 0} \end{cases}$$ Now determine numerically the intersection point of the green and red isolines and you're done.

An example with $(x_1,y_1) = (1,1)$ and $(x_2,y_2) = (2,2)$ results in the following values of $(p,a)$:

1.43600867678959E+0000 1.82608695652174E-0001

And the catenary of the catenoid can be sketched with these values, at last:

• Unfortunately, I can't find confirmation of some of the above at Wikipedia and other websites, where they say that the general equation of the catenary is $\,y=a\cosh(x/a)\,$ : same as with the catenoid. Which is contrary to my findings so far. Am I goofing somewhere? Commented Feb 9, 2017 at 14:53
• My suggestion is try simplifying the problem. For instance take $y_1=y_2$. Commented Feb 9, 2017 at 15:10
• I think we should discard the $b$ constant since the absolute location of two points is not important. What is important for us is the relative position of points wrt each other. In other words, adding a horizontal scaling does not affect the surface created. Commented Feb 9, 2017 at 17:05
• @AhmedBilâl: It's a horizontal translation BTW, but I get your point. (Oh, well, it's actually a bad mixture of both) Commented Feb 9, 2017 at 18:27
• I describe the cable hanging case to myself by imagining a line segment which describes the end points of the cable. Then I take it near to the coshx function in my mind. The segment fits just one location without changing the orientation. Then I zoom in and out and again move my segment without rotate and fit it again. Commented Feb 9, 2017 at 18:40