Math 2.0

Differential geometry Minimal and constant mean curvature surfaces Problem

Bernstein's problem

Builds on Minimal surface equationMinimal surface

Problem

Let m≥1m \ge 1 and let u ⁣:ℝm→ℝu \colon \R^m \to \R be a twice continuously differentiable function that solves the minimal surface equation div⁡(∇u1+∣∇u∣2)=0\operatorname{div} \left( \frac{\nabla u}{\sqrt{1 + \abs{\nabla u}^2}} \right) = 0 on all of ℝm\R^m. Must uu be an affine function, so that its graph is a hyperplane in ℝm+1\R^{m+1}?

Remarks

The answer is yes for m≤7m \le 7 and no for m≥8m \ge 8. Sergei Bernstein proved the case m=2m = 2: an entire minimal surface graph over the plane is a plane. Wendell Fleming gave a new proof for m=2m = 2 based on area-minimizing cones. This approach led to the cases m=3m = 3 by Ennio De Giorgi, m=4m = 4 by Frederick Almgren and m≤7m \le 7 by James Simons. Enrico Bombieri, Ennio De Giorgi and Enrico Giusti found a non-affine entire solution for m=8m = 8, and so for every m≥8m \ge 8. The dividing line comes from Simons' cone {x∈ℝ8:x12+x22+x32+x42=x52+x62+x72+x82}\set{x \in \R^8 : x_1^2 + x_2^2 + x_3^2 + x_4^2 = x_5^2 + x_6^2 + x_7^2 + x_8^2}, which is area-minimizing. For m=1m = 1 the statement is elementary: the equation says that u′/1+u′2u'/\sqrt{1 + u'^2} is constant, so u′u' is constant.

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  1. Minimal surface concept
  2. Minimal surface equation concept
  3. Bernstein's problem problem

Further reading

  1. Web https://en.wikipedia.org/wiki/Bernstein%27s_problem
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Changed2 November 2026

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