Catenoid is a minimal surface

Question

i want to show that the catenoid is a minimal surface. I have given

$f:I \times (0,2\pi)\longrightarrow \mathbb{R}^3$ with $f(r,\phi)=\left( \begin{array}{c}\cosh(r) \;\cos(\phi)\\\cosh(r) \;\sin(\phi)\\r\end{array} \right)$.

I know that:

$f$ is minimal surface $\Longleftrightarrow$ $\Delta f=0$.

$f$ is given in polar coordinates so i have to calculate the following:

$\Delta f= \frac{\partial^2f}{\partial r^2}+\frac{1}{r}\frac{\partial f}{\partial r}+\frac{1}{r^2}\frac{\partial^2f}{\partial \phi^2}$

$\frac{\partial f}{\partial r}= \left( \begin{array}{c}\sinh(r) \;\cos(\phi)\\\sinh(r) \;\sin(\phi)\\1\end{array} \right)$ , $\frac{\partial^2f}{\partial r^2}=\left( \begin{array}{c}\cosh(r) \;\cos(\phi)\\\cosh(r) \;\sin(\phi)\\0\end{array} \right)$

$\frac{\partial f}{\partial \phi}=\left( \begin{array}{c}-\cosh(r) \;\sin(\phi)\\\cosh(r) \;\cos(\phi)\\0\end{array} \right)$ , $\frac{\partial^2 f}{\partial \phi^2}=\left( \begin{array}{c}-\cosh(r) \;\cos(\phi)\\-\cosh(r) \;\sin(\phi)\\0\end{array} \right)$.

But when I put all together I can not show that $\Delta f$ is 0. What did I do wrong? Can someone help me please?

Thanks in advance.

Answer 1

I think you are a little bit confused about the harmonic characterization of (conformally immersed) minimal surfaces.

We know that every regular $2$-dimensional surface can be described locally in isothermal coordinates (i.e. for neighborhood of the surface, there's a coordinate map that preserves angles aka is conformal aka has 1st fundamental form satisfying $E=G$, $F=0$).

So we can cover the surface by a family of coordinate maps $$\vec{x_\alpha}(u,v) = \big(x_1(u,v), x_2(u,v), x_3(u,v)\big)$$

with each $\vec{x_\alpha}$ conformally mapping an open subset of $\mathbb{R}^2$ to $\mathbb{R}^3$.

The harmonic characterization says that the surface is minimal iff for each $\vec{x}_\alpha$ in such a family, the coordinates $x_i(u,v)$ are harmonic functions with respect to the coordinates (u,v). I would advise going back to look at the proof of this characterization for clarification, and thinking about geometrically what it means to be conformal (preserve angles).

I think where you have been misled is in thinking of this as a polar parametrization and using the so-called "polar form of the laplacian."

In your case you have a conformal coordinate map describing the entire catenoid in coordinates $r, \varphi$. What does $$\big(\cosh(r)\cos(\varphi)\big)_{rr} + \cosh(r)\cos(\varphi)\big)_{\varphi \varphi}$$ look like?

Answer 2

This not a complete answer but that's too long for me to post it in comment.

Preliminary for differential geometry of surfaces

\begin{align*} \mathbf{x}(u,v) &= \begin{pmatrix} x(u,v) \\ y(u,v) \\ z(u,v) \end{pmatrix} \\ \mathbf{x}_u &= \frac{\partial \mathbf{x}}{\partial u} \\ \mathbf{x}_v &= \frac{\partial \mathbf{x}}{\partial v} \\ \mathbf{N} &= \frac{\mathbf{x}_u \times \mathbf{x}_v}{|\mathbf{x}_u \times \mathbf{x}_v|} \tag{unit normal vector} \\ \end{align*}

First fundamental form $$\mathbb{I}= \begin{pmatrix} E & F \\ F & G \end{pmatrix}= \begin{pmatrix} \mathbf{x}_{u} \\ \mathbf{x}_{v} \end{pmatrix} \begin{pmatrix} \mathbf{x}_{u} & \mathbf{x}_{v} \end{pmatrix} $$

Second fundamental form $$\mathbb{II}= \begin{pmatrix} e & f \\ f & g \end{pmatrix}= -\begin{pmatrix} \mathbf{N}_{u} \\ \mathbf{N}_{v} \end{pmatrix} \begin{pmatrix} \mathbf{x}_{u} & \mathbf{x}_{v} \end{pmatrix}$$

Metric $$ds^2=E\, du^2+2F\, du\, dv+G\, dv^2$$

Element of area $$dA=|\det \mathbb{I}| \, du \, dv =|\mathbf{x}_u \times \mathbf{x}_v| \, du\, dv =\sqrt{EG-F^2} \, du\, dv$$

Principal curvatures

Let $\begin{pmatrix} \mathbf{N}_{u} \\ \mathbf{N}_{v} \end{pmatrix}= \mathbb{A} \begin{pmatrix} \mathbf{x}_{u} \\ \mathbf{x}_{v} \end{pmatrix}$ where $\mathbb{A}= \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}$.

Now \begin{align*} \begin{pmatrix} \mathbf{N}_{u} \\ \mathbf{N}_{v} \end{pmatrix} \begin{pmatrix} \mathbf{x}_{u} & \mathbf{x}_{v} \end{pmatrix} &= \mathbb{A} \begin{pmatrix} \mathbf{x}_{u} \\ \mathbf{x}_{v} \end{pmatrix} \begin{pmatrix} \mathbf{x}_{u} & \mathbf{x}_{v} \end{pmatrix} \\ -\begin{pmatrix} e & f \\ f & g \end{pmatrix} &= \mathbb{A} \begin{pmatrix} E & F \\ F & G \end{pmatrix} \\ \mathbb{A} &= -\begin{pmatrix} e & f \\ f & g \end{pmatrix} \begin{pmatrix} E & F \\ F & G \end{pmatrix}^{-1} \end{align*}

The principal curvatures $k_{1}, k_{2}$ are the eigenvalues of $-\mathbb{A}$.

Mean curvature

$$H=\frac{k_{1}+k_{2}}{2} =-\frac{1}{2} \operatorname{tr} \mathbb{A} =\frac{eG-2fF+gE}{2(EG-F^2)}$$

Gaussian curvature

$$K=k_{1} k_{2} =(-1)^{2} \det \mathbb{A} =\frac{eg-f^2}{EG-F^2}$$

For minimal surface, $$H=0$$

Answer 3

Theorem: If $(u,v) \to f(u,v)$ is an isothermal parametrisation, then $$f_{uu}+f_{vv} = 2 E H\mathbf{N}$$ where $\mathbf{N}$ is the principal normal to the surface. It is clear from this that a) we must check the parametrisation is isothermal, and b) that the 'Laplacian' is not the usual $\Delta u = u_{xx} + u_{yy}$ (and the coordinates you use are not 'polars'. They are just abstract coordinates).

The definition of isothermal is that the first fundamental form takes the form $$\pmatrix{\lambda^{2} & 0 \\ 0 & \lambda^{2}}$$ (I leave it to you to check this). Your coordinates are $(r,\phi)$, so we look at $$f_{rr} = \pmatrix{\cosh(r) \cos (\phi) \\ \cosh(r) \sin(\phi) \\ 0}$$ and $$F_{\phi \phi} = \pmatrix{\cosh(r) (-\cos(\phi)) \\ \cosh(r) (-\sin (\phi)) \\ 0}$$ from which we see that the catenoid is minimal.