# Mean curvature as an elliptic operator

Hi,

maybe this question will appear a little bit vague. How well is it known that the mean curvature on a hypersurface is a uniformly elliptic operator? For people from differential geometry this may be well-known, but I would like to see a proof of this. Can you please give me some references for good exposition on this matter?

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Hint: the first question you need to ask yourself is "What is the mean curvature operator acting on?" It outputs a number, but what is its input? – Willie Wong Jul 11 '10 at 22:59

## 2 Answers

Whether the mean curvature operator is uniformly elliptic or not seems to me to depend on how you represent the hypersurface. If you represent the hypersurface as the graph of a function, then it is a straightforward computation to derive the formula for mean curvature in terms of the function and its first and second derivatives. Linearizing the expression then shows that if the normal to the hypersurface remains uniformly bounded away from being horizontal, then the linearized second order operator is uniformly elliptic. As a first step, it is useful to observe that at a point where the normal vector is vertical, then the second fundamental form is given by the Hessian matrix of the function and the mean curvature is the trace of the Hessian.

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As you ask for a reference, there is an entire chapter in Gilbarg and Trudinger, "Elliptic Partial Differential Equations of Second Order." Half of the chapter on equations of mean curvature type was written by Leon Simon. I have an early edition, the 2001 edition has 17 chapters. Meanwhile, as you can can see from the answer by Prof. Yang and the comment by Willie Wong, you have not necessarily interpreted things correctly.

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