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For the catenoid profile $y(x)=E\cosh (x/E)$ spanning fixed-endpoint rings, the second variation of $F[y]=2\pi \int y\sqrt{1+y^{\prime 2}}dx$ isThis refers to axisymmetric graph variations, not all possible changes in surface topology. Strict positivity for every nonzero admissible variation gives a strict local minimum in the $C^1$ topology: the positive lowest Dirichlet eigenvalue of the regular Sturm-Liouville problem gives coercivity, and sufficiently small coefficient perturbations preserve it. The zero variation must be excluded from a strict-positivity hypothesis.

Along a sufficiently smooth stationary curve, differentiation and the Euler-Lagrange equation give the Beltrami identity:The cancellation uses the absence of explicit $x$ dependence.

For an axisymmetric surface of revolution, its area is $I[y]=2\pi {\int }_{-l}^{l}y\sqrt{1+y^{\prime 2}}dx$. A smooth positive-radius minimizer is stationary, so the Beltrami identity, with the constant factor omitted, givesEquivalently the Euler-Lagrange equation gives $y{y}^{\prime \prime }=1+y^{\prime 2}$, whence $y^{\prime \prime }=y/k^2$. The first integral selectsEqual radii at $\pm l$ imply $x_{\ast }=0$ (subtract the two cosh values), givingThis is a catenoid. The conclusion is a necessary form for a smooth connected minimizer; it does not declare every stationary catenoid a minimum.

Put $a=l/k>0$. The boundary equation is $r/l=\cosh a/a$. Its derivative is $(a\sinh a-\cosh a)/a^2$, so its unique critical point solves $a\tanh a=1$. The left side increases strictly from zero to infinity. Thus $\cosh a/a$ decreases to a unique minimum and then increases, diverging as $a\downarrow 0$ and $a\to \infty$. There are exactly two positive values of $k$ whenAt equality there is one value, and below it there is none.

To match the printed second variation convention, write $I[y+\epsilon \eta ]=I[y]+\epsilon \delta I+{\epsilon }^2{\delta }^{2}I+o({\epsilon }^2)$; its ${\delta }^{2}I$ is the quadratic coefficient, half the second derivative with respect to $\epsilon$. Expanding the area integrand givesOn the catenoid these coefficients are $k{\mathrm{sech}}^2(x/k)$ and $2\tanh (x/k)$. Integrate the mixed term by parts; the boundary term is zero because $\eta (\pm l)=0$. Since $(\tanh (x/k){)}^{\prime }={\mathrm{sech}}^2(x/k)/k$, this yieldsThis is the axisymmetric second variation of a catenoid, with the explicitly stated quadratic-coefficient convention.

An immersion $\varphi :U\to {\mathbb{R}}^3$ has differential of rank $2$, equivalently linearly independent ${\varphi }_u,{\varphi }_v$. It has isothermal coordinates when its first fundamental form has $E=G>0$, $F=0$. A minimal surface has zero mean curvature.

For an isothermal immersion, differentiation of $E=G$ and $F=0$ givesand similarly its scalar product with ${\varphi }_v$ is zero. Its normal component is $e+g$. The given mean curvature formula reduces to $H=(e+g)/(2E)$, soThus an isothermal immersion is minimal exactly when each coordinate is a harmonic function.

An explicit example is the Enneper surface, parametrized on ${\mathbb{R}}^2$ byEvery coordinate has zero Laplacian. Direct differentiation gives ${\varphi }_u\cdot {\varphi }_v=0$ and $|{\varphi }_u{|}^2=|{\varphi }_v{|}^2=(1+u^2+v^2{)}^2>0$. Thus it is an isothermal immersion and is minimal everywhere.

It is not an open subset of any of the three listed surfaces. The parameter points $(\sqrt{3},0)$ and $(-\sqrt{3},0)$ both map to $(0,0,3)$, but their tangent planes differ: ${\varphi }_u=(-2,0,\pm 2\sqrt{3})$ while ${\varphi }_v=(0,4,0)$. A plane, catenoid, or helicoid is an embedded smooth surface with a unique tangent plane at each point. If our whole image lay in any one of them, both rank-$2$ tangent planes at this point would have to equal its tangent plane, a contradiction.

Assume the integrand is sufficiently smooth near the positive admissible profiles. Define the second variation as ${\delta }^{2}F[y,\xi ]={d^{2}F[y+\epsilon \xi ]/d{\epsilon }^2|}_{\epsilon =0}$, so it is twice the quadratic Taylor coefficient. Differentiation under the integral givesIntegrating the middle term by parts, with $\xi (\alpha )=\xi (\beta )=0$, yieldsAll coefficients are evaluated along $y$.

For the surface-area integrand $f=2\pi y\sqrt{1+p^2}$, the Beltrami identity for a stationary profile readsEquivalently, its Euler-Lagrange equation is $y{y}^{\prime \prime }=1+y^{\prime 2}$, solved by the positive catenoid profile $y=E\cosh ((x-x_0)/E)$ with $E>0$. Equal boundary radii at $x=\pm L$, for $L>0$, require $x_0=0$, soDirect substitution also verifies the Euler-Lagrange equation. This is a stationary profile whenever a positive $E$ satisfying the boundary relation exists; such an $E$ does not exist for arbitrary ring separations and radii.

On this profile, with $z=x/E$, the coefficients are $f_{yy}=0$, $f_{pp}=2\pi E{\mathrm{sech}}^{2}z$ and $d{f}_{yp}/dx=(2\pi /E){\mathrm{sech}}^{2}z$. Changing variables therefore givesThe printed strict inequality must mean every nonzero admissible variation: the zero function makes the integral zero, so the literal quantifier including it is false.

For completeness, the strict positivity here is sufficient for a strict local minimum in the fixed-endpoint $C^1$ topology, among positive axisymmetric graph profiles. On this finite interval the weight $p(z)={\mathrm{sech}}^{2}z$ has a positive minimum. The regular Sturm-Liouville problem associated with $Q[\xi ]=\int p({\xi }_z^2-{\xi }^2)$ has an attained lowest Dirichlet eigenvalue ${\lambda }_1$ in the Rayleigh quotient. Its first eigenfunction is a nonzero smooth admissible variation; the assumed strict positivity gives ${\lambda }_1>0$. Hence $Q\ge {\lambda }_1\Vert \xi {\Vert }_{L^2}^2$. Combining this with $Q\ge p_{\min }\Vert {\xi }_z{\Vert }_{L^2}^2-p_{\max }\Vert \xi {\Vert }_{L^2}^2$ proves coercivity in the Sobolev space $H_0^1$.

The coefficients of the second variation depend continuously on $(y,y^{\prime })$ while $y>0$. For a sufficiently small $C^1$ perturbation, the quadratic forms along the segment from the stationary profile to the perturbed one retain a uniform positive coercivity bound. Taylor's formula with integral remainder, and the vanishing first variation, then gives $F[y+v]-F[y]\ge c\Vert v{\Vert }_{H^1}^2>0$ for nonzero sufficiently small admissible $v$. The stationary area is therefore a strict local minimum under the stated nonzero-variation condition. This supplies the uniform bound needed in an infinite-dimensional problem, rather than relying on pointwise positivity alone in an arbitrary function space.

A regular surface is a minimal surface when its mean curvature vanishes identically, equivalently when its two principal curvatures sum to zero everywhere.

The catenoidis an example. In part (a), take $f(v)=\cosh v$ and $g(v)=v$, so $q=\cosh v$. The formula there givesA plane and a circular cylinder have zero Gaussian curvature, as does a flat cone away from its vertex. Because a local isometry preserves Gaussian curvature, the catenoid is minimal and is not locally isometric to any of those surfaces; this is the catenoid is a minimal surface.

The map $\varphi$ is an immersion when $d\varphi$ has rank two, equivalently ${\varphi }_u$ and ${\varphi }_v$ are linearly independent. It is isothermal when its first fundamental form hasIn general the mean curvature isThe surface is minimal when $H=0$. In isothermal coordinates the Laplace-Beltrami operator identity isThus $H=0$ exactly when each coordinate function $x,y,z$ is harmonic.

Forthe isothermal conditions areand harmonicity givesConsequentlyand the isothermal equation reduces toThese are all the solutions for which $f$ never vanishes. When $AB>0$ they parameterize a catenoid, up to translations, reflection and scaling. The limiting cases $AB=0$, $c=0$, with exactly one of $A,B$ nonzero, parameterize a punctured plane.

Withthe coordinate tangent vectors satisfyThus the first fundamental form isGeometrically, $S$ is the half of a helicoid with positive radius, winding upwards indefinitely.

Consider the upper half of the catenoid, parametrized as the surface of revolutionIt has the same first fundamental form, so $(u,v)\mapsto Y(u,v)$ defines a local isometry. By the invariance of Gaussian curvature under local isometry, the curvatures agree; directly, the conformal metric above gives