Graphs and the minimal-surface equation

Try

The graph f(x,y)=xy has a saddle at the origin. Its gradient vanishes there and its Hessian has determinant minus one, so K=−1. Its mean curvature is zero at that point, but not throughout the graph. A pointwise cancellation of principal curvatures does not automatically make an entire surface minimal.

Use the upward normal N=(−f_x,−f_y,1)/W, where W=√(1+f_x²+f_y²). This fixes the signs of the second form and mean curvature under S=−dN.

Predict

Predict whether every saddle graph has H=0. Opposite principal signs only imply K<0; their magnitudes need not match.

Predict whether stationary area necessarily means a global minimum among all possible spanning surfaces. It does not. For graphs, convexity of the graph-area integrand gives a useful stronger conclusion within the same graph competitor class, but that scope must be stated.

Experiment

At the origin compare three graphs.

Graph	Principal values with upward normal	K	H
(x²+y²)/2	1,1	1	1
xy	1,−1	−1	0
x²−y²/2	2,−1	−2	1/2

Each gradient vanishes at the origin, so the Hessian is the shape matrix there. Move away from the origin on the graph xy and recompute rather than reusing the origin shortcut.

Understand

For a graph, the metric determinant is W² and b=(Hess f)/W. Therefore K=(f_xx f_yy−f_xy²)/W⁴. Computing half the trace of g⁻¹b gives 2H=((1+f_y²)f_xx−2f_x f_y f_xy+(1+f_x²)f_yy)/W³. Equivalently, 2H=div(∇f/W).

Now vary the graph to f+εh with h smooth and compactly supported. Its area is ∫√(1+||∇f+ε∇h||²). Differentiating at zero yields ∫(∇f/W)·∇h. Integration by parts gives −∫div(∇f/W)h. Stationarity for every such h is equivalent to H=0, the minimal graph equation.

For f=xy, the numerator of 2H is −2xy and W³=(1+x²+y²)^(3/2). Thus H vanishes on the coordinate axes but not where xy≠0. The zero at one point did not prove the equation on a neighborhood.

Apply

The function p↦√(1+||p||²) is convex. Its tangent inequality shows a competitor f+h with the same boundary has area at least that of f plus ∫(∇f/W)·∇h. For a smooth minimal graph and an admissible zero-boundary variation, this last integral vanishes. Hence the graph minimizes area among the stated graph competitors. This proof does not automatically compare arbitrary non-graph surfaces or singular competitors.

Repair “Minimal means all curvatures are zero.” H=0 means the two principal values sum to zero. A catenoid has nonzero opposite values. A plane is only the simplest minimal example.

For optional depth, existence and regularity of minimal graphs with prescribed boundary data are PDE problems. The first-variation calculation characterizes a sufficiently smooth solution but does not guarantee one for every domain and boundary.

Continue

Separate pointwise curvature, the PDE H=0, stationarity, and the competitor class for a minimum claim. Continue to surfaces of revolution to construct and verify a nonplanar minimal surface explicitly.