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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:

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

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?

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    $\begingroup$ 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. $\endgroup$ Commented Oct 11, 2016 at 17:10

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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 $z = d$.

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  • $\begingroup$ 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 $g'(u)=\pm \sqrt{1-f'(u)^2}$) $\endgroup$
    – Masacroso
    Commented Feb 18, 2022 at 5:12

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