Minimal surface

A minimal surface has vanishing mean curvature vector. For a smooth surface in three-dimensional Euclidean space, this means $H=0$, equivalently $k_1+k_2=0$. Thus Gaussian curvature is $K=-k_1^2\le 0$, but neither principal curvature needs to vanish. A plane and a catenoid both satisfy the condition, with very different Gaussian curvatures.

By the first variation of area formula, a minimal surface is a stationary point of surface area for every smooth variation that fixes its boundary, or has compact support in its interior. The word “minimal” specifies this differential equation; it does not guarantee a global minimum of surface area, or even a local minimum. A stable minimal surface imposes a further condition on the second variation.

Let a boundaryless minimal surface be closed in ${\mathbb{R}}^3$, and let its distance to a plane tend to infinity along every sequence escaping bounded sets. If it has no planar point, it must meet the plane. Otherwise absolute signed height has a positive attained minimum by properness. Its Hessian there is semidefinite and equals the second fundamental form up to sign because the plane is tangent. Minimality makes its trace zero, so the entire form vanishes, contradicting the absence of planar points. This argument uses curvature and attained extrema, rather than an unproved contact comparison for Gaussian curvature.

An outer area-minimizing surface is a closed surface in a Riemannian initial-data slice whose area is no greater than that of any enclosing competitor. This is a variational property, not a consequence of containment alone. It is important when choosing the horizon-area quantity in the Riemannian Penrose inequality and when examining an apparent-horizon area comparison.

The isothermal coordinates $(u,v)$ give the immersion $(u-u^3/3+u{v}^2,v-v^3/3+u^{2}v,u^2-v^2)$. Its coordinate functions are harmonic and its metric is $(1+u^2+v^2{)}^2(d{u}^2+d{v}^2)$, so it is a complete immersed minimal surface with self-intersections.

With $A(X,Y)=({\overline{\nabla }}_{X}Y{)}^{\perp }$ and $\mathbf{H}=\mathrm{tr}A$, a variation with velocity $V$ satisfiesThus a boundaryless submanifold is stationary for every compactly supported variation exactly when it is minimal.

If $M\subseteq {\mathbb{R}}^{n+m}$ and $f$ is smooth in the ambient space, then

For the squared distance $r^2(x)=|x{|}^2$, a minimal $n$-submanifold satisfies ${\Delta }_M{r}^2=2n$. On a compact boundaryless manifold, $r^2$ attains a maximum where the Laplacian is nonpositive, a contradiction for $n>0$.

A stable minimal surface has nonnegative second variation of surface area for every admissible normal variation $fN$ fixing the boundary. For a two-sided minimal surface in three-dimensional Euclidean space, the quadratic form isHere ${\nabla }_{M}f$ is the gradient for the induced Riemannian metric. The gradient term penalizes rapid changes in the displacement; the principal curvature term can make the second variation negative. A plane has only the nonnegative gradient term. Nonnegativity alone, especially with zero modes, does not prove a strict local minimum or a global minimum.

For a finite catenoid between fixed rings, the answer depends on the ring separation, waist radius, and allowed variations. The axisymmetric second variation of a catenoid tests a specified restricted class. Comparing with two disconnected disks tests a different class of competitors, including a change in topology. Consequently, a stationary catenoid can have greater surface area than two disks without immediately collapsing under every small perturbation.