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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.

The Gauss map sends $p$ to the chosen unit normal $N(p)$. Differentiating $⟨N,N⟩=1$ shows that $d{N}_p$ maps $T_{p}S$ into itself. The shape operator $S_p=-d{N}_p$ is self-adjoint because its associated second fundamental form is symmetric. Its eigenvalues $k_1,k_2$ are the principal curvatures. A point is umbilical when $k_1=k_2$, and the surface is minimal when its mean curvature $H=(k_1+k_2)/2$ vanishes everywhere.

In an orthonormal principal basis, $d{N}_p$ has diagonal matrix $\mathrm{diag}(-k_1,-k_2)$. It is conformal exactly whenwhich is equivalent to $k_1^2=k_2^2\ne 0$. At a nonumbilical point this says $k_1=-k_2$, exactly the condition $H=0$. Hence, when there are no umbilical points, the surface is minimal if and only if its Gauss map is conformal.

A plane has both principal curvatures zero, hence $K=H=0$. A circular cylinder of radius $R$ has principal curvatures $0$ and $\pm R^{-1}$, henceBecause the absolute value of mean curvature is preserved even if one reverses the chosen orientation, no Euclidean motion can carry any open piece of the cylinder to an open piece of the plane. Thus the condition $K\equiv 0$ does not characterize planar pieces.

There are analogous examples for both signs of constant Gaussian curvature. Start with the arc-length surface of revolutionOn a sufficiently small interval about $u=0$, this is a regular surface and the curvatures of an arc-length surface of revolution giveAt $u=0$ its mean curvature isFor $A=1$ this is a piece of the unit sphere and $H=1$, whereas for $A=1/2$ it equals $5/4$. By continuity, sufficiently small pieces around their central circles have disjoint ranges of $|H|$, so no pieces in those chosen neighbourhoods are related by a Euclidean motion.

For curvature minus one, useagain on a sufficiently small interval. ThenThe choices $A=1$ and $A=2$ give $H(0)=0$ and $H(0)=-3/4$. Restricting to small enough pieces again separates the ranges of $|H|$. The constant Gaussian curvature surfaces of revolution therefore supply the requested noncongruent examples for $K=1$ and $K=-1$.

For an axisymmetric surface $r=a(z,t)$, twice the mean curvature with the outward-normal convention isWriting $a=a_0+\eta$ and retaining linear terms gives

At the unperturbed boundary $r=a_0$, the linearized kinematic boundary condition, tangential-stress condition, and Young–Laplace equation areFor a normal mode $e^{ikz+st}$ these become $u(a_0)=s\eta$, ${\sigma }_{rz}(a_0)=0$, and

The Papkovich–Neuber representation of body-force-free Stokes flow iswhere $\mathbf{\Phi }$ and $\chi$ are harmonic.

Let $x=kr$ and $E=e^{ikz+st}$. Substitution of the given radial vector potential and scalar potential givesUsing the modified Bessel function identity $(x{I}_1{)}^{\prime }=x{I}_0$, the tangential stress simplifies toThe stress-free condition at $r=a$ therefore yields

For a cylinder of radius $r$ and length $L$, the mean curvature is $H=1/(2r)$ and the area is $2\pi rL$. Its Helfrich membrane energy per unit length is thereforeThe first term favors a narrow tube and the second penalizes its curvature. Setting the derivative with respect to $r$ to zero givesThe positive second derivative confirms a minimum.

The spatial metric is stationary and only ${\beta }^r=b(r)=Mr/(r+M{)}^2$ is nonzero. The evolution equation therefore saysthe Lie derivative of the spatial metric along the shift vector. For the radial component,soFor the angular components,and spherical symmetry suppliesAll off-diagonal components vanish.

Contracting with the inverse spatial metric gives the mean curvatureThus

Choose a smooth unit normal $N$. For $v,w\in T_p\Sigma$, the first fundamental form and second fundamental form arewhere the shape operator is $S_p=-d{N}_p$. The principal curvatures are the eigenvalues ${\kappa }_1,{\kappa }_2$ of $S_p$; the mean curvature and Gaussian curvature areThe surface is minimal when $H=0$ everywhere.

For a local parametrization $X(u^1,u^2)$, the first fundamental form is the Euclidean inner product restricted to each tangent plane:Writing $u^1=u,u^2=v$, this is

After choosing an orientation, the Gauss map isThe second fundamental form is $II(v,w)=⟨S(v),w⟩$, where the shape operator is $S=-dN$. In coordinates,The Gaussian curvature and mean curvature are respectively the determinant and half the trace of $S$:Changing the chosen normal reverses $II$ and $H$ but leaves $K$ unchanged.