Math 2.0

Differential geometry Definition

Mean curvature flow

Builds on Mean curvatureNormal vectorSmooth manifold

Definition

Let MM be a smooth nn-dimensional manifold and let T>0T > 0. A mean curvature flow is a smooth map F ⁣:M×[0,T)→ℝn+1F \colon M \times [0, T) \to \R^{n+1} such that:

  1. for each t∈[0,T)t \in [0, T), the map Ft=F(⋅,t)F_t = F(\cdot, t) is an immersion of MM as a hypersurface;
  2. ∂tF(p,t)=H⃗(p,t)\partial_t F(p, t) = \vec{H}(p, t) for all p∈Mp \in M and t∈[0,T)t \in [0, T).

Here H⃗(p,t)=−Hν\vec{H}(p, t) = -H \nu is the mean curvature vector of FtF_t at pp. The vector ν\nu is a unit normal vector, and HH is the mean curvature, the sum of the principal curvatures with respect to ν\nu.

Remarks

Example: let F(p,t)=R(t) pF(p, t) = R(t)\, p for pp in the unit sphere Sn⊂ℝn+1S^n \subset \R^{n+1}, with R(t)=R02−2ntR(t) = \sqrt{R_0^2 - 2nt}. The sphere of radius RR has mean curvature vector of length n/Rn/R pointing to the centre, so the flow condition is R′(t)=−n/R(t)R'(t) = -n/R(t). This holds since ddtR2=−2n\frac{d}{dt} R^2 = -2n, and the flow exists until T=R02/(2n)T = R_0^2/(2n). Non-example: R(t)=R0+tR(t) = R_0 + t moves the sphere outward, against the mean curvature vector, so it fails clause 2. A plane or any minimal hypersurface, with H⃗=0\vec H = 0, is a stationary flow. Sign conventions for HH and ν\nu vary, but H⃗\vec H always points toward the centre of a round sphere.

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Further reading

  1. Web https://en.wikipedia.org/wiki/Mean_curvature_flow
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