# Surface of revolution with zero mean curvature

I want to show that the only surfaces of revolution with zero mean curvature are the plane and the catenoid (revolving $x = \frac{1}{a}\cosh(ay+b)$ about $y$-axis).

Aiming to do this, I have calculated the first and the second fundamental form for the general surface of revolution $r(\rho, \phi) = (x(u), \rho(u)\cos\phi, \rho(u)\sin\phi), ~~\rho(u) > 0$, which I am not sure if is correct. The result is \begin{align} b_{11} &= \frac{x''\rho' - x'\rho''}{\sqrt{\rho'^2+x'^2}} \\ b_{12} &= 0 = b_{21} \\ b_{22} &= \frac{x'\rho'}{\sqrt{\rho'^2+x'^2}}, \end{align}

\begin{align} g_{11} &= x'^2 + \rho'^2 \\ g_{12} &= 0 = g_{21} \\ g_{22} &= \rho^2. \end{align}

Denote $G = (g_{ij})$ and $Q = (b_{ij})$. Therefore, the mean curvature is, by definition, the trace of the matrix $G^{-1}Q$, and then I have ended up with some nasty differential equation which hardly seems possible to solve:

$$\frac{x''\rho' - x'\rho''}{(\rho'^2+x'^2)^{3/2}} + \frac{x'\rho'}{\rho^2\sqrt{\rho'^2+x'^2}} = 0.$$

To make things better, I have tried parametrising the surface by different parameters, e.g. the arc length of the curve, which reduces $\rho'^2+x'^2$ to $1$. But so far my efforts are in vain. Is there any errors in my calculations or any way can I take to bypass such difficulties?

• I would recommend that you start by assuming you have a parametrization of the form $(u,h(u))$ for the curve and then you'll end up with a relatively straightforward differential equation. Commented Oct 11, 2016 at 17:10

Consider your surface of revolution with the following parametrization, taking $$z$$ as the axis where the curve revolves, $$\pmb{\mathrm{x}}(u,v)=(f(u)\cos v,f(u)\sin v,g(u)).$$

As you pointed out, for a such a parametrization we can assume that $$f(u) > 0$$, and also we can assume that the profile curve $$u\mapsto(f(u), 0, g(u))$$ is unit-speed, that is, $$f'(u)^2 + g'(u)^2 = 1$$.

Then, $$\pmb{\mathrm{x}}_u=(f'(u)\cos v, f'(u) \sin v, g(u)),\;\pmb{\mathrm{x}}_v=(−f(u) \sin v, f(u) \cos v, 0).$$

Hence, \begin{align} &E=\|\pmb{\mathrm{x}}_u\|^2= f'(u)^2 + g'(u)^2 = 1,\\ &F=\langle\pmb{\mathrm{x}}_u,\pmb{\mathrm{x}}_v\rangle= 0,\\ &G =\|\pmb{\mathrm{x}}_v\|^2 = f(u)^2. \end{align}

So the first fundamental form is given by $$du^2 + f(u)^2dv^2.$$

We also have that \begin{align} \pmb{\mathrm{x}}_u\times\pmb{\mathrm{x}}_v&= (−f(u)g'(u) \cos v,−f(u)g'(u) \sin v, f(u)f'(u)),\\ \|\pmb{\mathrm{x}}_u\times\pmb{\mathrm{x}}_v\|&= f(u),\\ \pmb{N}&= {\pmb{\mathrm{x}}_u\times\pmb{\mathrm{x}}_v\over\|\pmb{\mathrm{x}}_u\times\pmb{\mathrm{x}}_v\|} = (−g'(u) \cos v,−g'(u) \sin v, f'(u)),\\ \pmb{\mathrm{x}}_{uu}&= (f''(u) \cos v, f''(u) \sin v, g''(u)),\\ \pmb{\mathrm{x}}_{uv}&= (−f'(u) \sin v, f'(u) \cos v, 0),\\ \pmb{\mathrm{x}}_{vv}&= (−f'(u) \cos v,−f(u) \sin v, 0),\\ L&= \langle\pmb{\mathrm{x}}_{uu},\pmb{N}\rangle = f'(u)g''(u) − f''(u)g'(u),\\ M&= \langle\pmb{\mathrm{x}}_{uv},\pmb{N}\rangle = 0,\\ N& = \langle\pmb{\mathrm{x}}_{vv},\pmb{N}\rangle = f(u)g'(u), \end{align} so the second fundamental form is $$(f'(u)g''(u) − f''(u)g'(u))\;du^2 + f(u)g'(u)\;dv^2.$$

Now, we use the well-known expression for $$H$$ given by $$H={1\over2}{EN-2FM+GL\over EG-F^2},$$ so we get $$H={1\over2}\left(f'(u)g''(u)-f''(u)g'(u)+{g'(u)\over f(u)}\right).$$

So far, so good.

Suppose now that for some value of $$u$$, say $$u = u_0$$, we have $$g'(u_0)\neq 0$$. Since $$g'(u)$$ is continuous, we have that $$g'(u) \neq 0$$ for $$u$$ in some open interval containing $$u_0$$. Let $$(\alpha, \beta)$$ be the largest such interval. Supposing also that $$u\in(\alpha, \beta)$$, the condition $$f'(u)^2 + g'(u)^2 = 1$$ gives (by differentiating with respect to $$u$$) $$f'(u)f''(u)+g'(u)g''(u)=0$$, and then \begin{align} (f'(u)g''(u)-g'(u)f''(u))g'(u)&=\\ &=-f'(u)^2f''(u)-f''(u)g'(u)^2\\ &=-f''(u)(f'(u)^2+g'(u)^2)=-f''(u), \end{align} so $$f'(u)g''(u)-g'(u)f''(u)={-f''(u)\over g'(u)}$$ and we get $$H={1\over 2}\left({g'(u)\over f(u)}-{f''(u)\over g'(u)}\right).$$

Again by $$g'(u)^2=1-f'(u)^2$$, the surface is minimal if and only if $$f(u)f''(u)=1-f'(u)^2.$$

This is an ODE that can be solved by taking $$h=f'$$ (I will stop writing the dependance on $$u$$) and noticing that $$f''(u)={dh\over dt}={dh\over df}{df\over dt}=h{dh\over df}.$$

Hence, the ODE is now $$fh{dh\over df}=1-h^2.$$

We assumed that $$g'(u)\neq 0$$, so $$h^2\neq 1$$ (as $$h^2+g'(u)^2=1$$). This allows us to rearrange and integrate the equation $$\int {h\over 1-h^2}\;dh\;=\;\int {1\over f}\;df,$$ which gives us $${1\over \sqrt{1-h^2}}\;=\;af\;,$$ (where $$a>0$$ is a constant) and then $$h\;=\;{\sqrt{a^2f^2-1}\over af}.$$

With this, since $$h=f'$$, integrating again $$\int {af\over\sqrt{a^2f^2-1}}\;df\;=\;\int\;du.$$

If we solve this integral equation for $$f$$ we obtain $$f={1\over a}\sqrt{1+a^2(u+b)^2},$$ where $$b$$ is a constant that we can assume that is zero by taking the change of parameter $$u\mapsto u+b$$, so we have $$f={1\over a}\sqrt{1+a^2u^2}.$$

With this expression for $$f$$, we can compute $$g$$ as follows \begin{align} &g'^2=1-f'^2=1-h^2={1\over a^2f^2}\;,\\ &g'=\pm{1\over\sqrt{1+a^2u^2}}\;,\\ &g\;=\;\pm{1\over a}\sinh^{-1}(au)+c, \end{align} for some constant $$c$$.

We will consider the term $$au$$ so we can write $$f$$ in terms of $$g$$: $$au=\pm\sinh(a(g-c))\;,$$ so $$f={1\over a}\cosh(a(g-c)).$$

By this, the profile curve of the surface is $$x={1\over a}\cosh(a(z-c))$$ where, by a translation along the revolving axis, we can assume that $$c=0$$, obtaining the equation of a catenary.

So far, we have shown that the open subset of our surface corresponding to $$u\in(\alpha,\beta)$$ is part of the catenoid, for in the proof we used in an essential way that $$g'(u)\neq 0$$. This is why the proof has so far excluded the possibility that we are in the case of a plane.

Now, suppose that $$\beta < \infty$$. Then, if the curve is defined for $$u\geq\beta$$, we would get $$g'(\beta)=0$$. Otherwise $$g'(u)$$ would be non-zero on an open interval containing $$\beta$$, which would contradict the assumption that $$(\alpha,\beta)$$ is the largest open interval containing $$u_0$$ on which $$g'(u)\neq 0$$. Right, but the formulas above show that $$g''^2={1\over a^2f(u)^2}={1\over 1+a^2u^2}$$ for $$u\in(\alpha,\beta)$$, and since $$g'(u)$$ is continuous, $$g'(\beta)=\pm(1+a^2\beta^2)^{-{1\over 2}}\neq 0$$.

This is a contradiction and then the profile curve can not be defined for $$u\geq\beta$$. In the case $$\beta=\infty$$, we obtain the same.

We should now proceed with a similar argument for $$\alpha$$, showing that $$(\alpha,\beta)$$ is the entire domain of definition of the profile curve. Hence, the whole surface of revolution is an open subset of a catenoid.

The remaining case is that in which $$g'(u) = 0$$. But then $$g(u)$$ is a constant, say $$d$$, and the surface would be an open subset of the plane $$z = d$$.

• alternatively, to solve $f(u)f''(u)=1-f'(u)^2$ faster you can note that $\frac{d}{d u}(f(u)f'(u))=f(u)f''(u)+(f'(u))^2$, so you have the ODE $f(u)f'(u)=u+K_1$, what is immediate to solve, with solution given by $f(u)^2=u^2+2K_1u+K_2$, so the general solution is $f(u)=\pm \sqrt{(u+K_1)^2+K_3}$, what correspond to a plane or a catenary when its parametrized by arc-length (that is, from $f$ now you can find easily $g$ due to the relation $g'(u)=\pm \sqrt{1-f'(u)^2}$) Commented Feb 18, 2022 at 5:12