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From PDE Evans, 2nd edition: Chapter 8, Exercise 12:

Assume $u$ is a smooth minimizer of the area integral $$I[w]=\int_U (1+|Dw|^2)^{1/2} \, dx,$$ subject to given boundary conditions $w=g$ on $\partial U$ and the constraint $$J[w] = \int_U w \, dx = 1.$$ Prove that graph of $u$ is a surface of constant mean curvature.

(Hint: Recall Example 4 in §8.1.2.)

First, I will print Example 4 from the textbook (page 457) below:

Example 4 (Minimal surfaces). Let $$L(p,z,x)=(1+|p|^2)^{1/2},$$ so that $$I[w] = \int_U (1+|Dw|^2)^{1/2} \, dx$$ is the area of the graph of the function $w : U \to \mathbb{R}$. The associated Euler-Lagrange equation is $$\sum_{i=1}^n \left(\frac{u_{x_i}}{(1+|Du|^2)^{1/2}} \right)_{x_i}=0 \quad \text{in }U.$$ $\quad$ This partial differential equation is the minimal surface equation. The expression $\operatorname{div} \left(\frac{Du}{(1+|Du|^2)^{1/2}} \right)$ on the left side of $(10)$ is $n$ times the mean curvature of the graph of $u$. Thus a minimal surface has zero mean curvature.

I don't have much work started on this, but this question looks interesting. (This is not a homework assignment, as with all my other PDE Evans questions.) But I am asking here because I do not understand fundamentally how mean curvature is applied to Euler-Lagrange equations. In particular, this question uses concepts of differential geometry, which I have not taken any courses in yet in my academic career.

Now, what I do know so far, is that maybe I should show that the graph of $u$ is a minimal surface. This would mean $u$ has zero mean curvature, and hence a constant mean curvature (zero is constant, obviously).

How should I start about this problem?

Community
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Cookie
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    The extra constraint $J[w]=1$ will in general prevent the graph from being minimal. The text you quoted gave an expression for $n$ times the mean curvature - you need to find an argument to show this expression is constant. – Anthony Carapetis Jun 12 '15 at 03:48
  • Look up Lagrange Multipliers, apply the result using first order variations. – Jeb Jun 12 '15 at 05:46

2 Answers2

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you don't need to know any differential geometry to confirm the PDE.

The minimal case is this: replace $w$ by $w + t \phi.$ Here $\phi$ refers to a $C^\infty$ function with compact support, the support (closure of the set where $\phi$ is nonzero) being contained in the interior of $\Omega.$ In order to have $I(w)$ a minimizer (or any critical point) it is necessary that $$ \frac{d}{dt} \; I(w+t \phi) $$ be zero when $t=0.$ You write this out carefully. The main lemma you need is this: if a continuous function $F,$ which will be some combination of $w$ and first and second partials of $w,$ satisfies $\int_\Omega F\phi =0$ for all such $\phi,$ then $F$ is constant zero.

For the constant mean curvature case, replace the $\phi$ by functions $\psi$ with compact support in the interior of $\Omega,$ with the additional constraint that $\int_\Omega \psi =0.$ Vary with $w + t \psi.$ This time, the main lemma is that, if continuous $F$ satisfies $\int_\Omega F\psi =0$ for all such $\phi,$ then $F$ is constant, but the constant is allowed to be nonzero if that is how it works out.

The calculations in the above descriptions are simply not difficult; I recommend getting your hands dirty. It will help when later you take differential geometry. See if you can find proofs of the two lemmas. I cannot tell what you know about test functions, so it may be a matter of looking things up, or asking your professor or TA.

Will Jagy
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  • I haven't read your answer thoroughly yet (though, once I do, I will most likely attempt with follow-up questions), but in response to the very last sentence of your answer: For the record, I ask here on Math SE because I'm studying PDE theory completely on my own (not enrolled in a graduate PDE class). – Cookie Jun 12 '15 at 04:51
  • In textbook notation, one can set $i(\tau):=I[u+\tau v]$, then compute $i'(\tau)$. Then the theory in the textbook says that, since $u$ is a minimizer, we have $i'(0)=0$; that is, $I[u+\tau v]$ achieves its minimum at $\tau = 0$. Does this sound to be basically what you said your answer? (This is, by the way, the same procedure I used in confirming that the Lagrangian I found satisfies a certain Euler-Lagrane equation. Examples: http://math.stackexchange.com/questions/1308820/euler-lagrange-equation, http://math.stackexchange.com/questions/1310342/elliptic-regularization-of-the-heat-equation) – Cookie Jun 12 '15 at 05:13
  • @les, this is pretty good, https://en.wikipedia.org/wiki/Bump_function where the example $\psi$ they give will be used around some point $x_0 \in \Omega$ as $\psi ( \epsilon , |x - x_0 |)$ as the more general bump used for minimal surfaces, for constant mean curvature, two points $x_1 \neq x_2$ and $\psi ( \epsilon , |x - x_1 |) -\psi ( \epsilon , |x - x_2 |)$ which integrates over $\Omega$ to $0.$ – Will Jagy Jun 12 '15 at 15:36
  • Let's see, with some continuous function I called $F,$ if, say, $F$ is positive at some point $x_0,$ then there is a small $\epsilon$ ball around $x_0$ where $F$ remains positive, so the integral of $F$ times $\psi(\epsilon |x - x_0|)$ over the ball (and thus over $\Omega$) is positive and nonzero. That is the first lemma I mentioned. – Will Jagy Jun 12 '15 at 15:40
  • Question on notation: As I'm reading Evans' PDE, the author states explicitly that he's using $U$, rather than the usual $\Omega$. I think your "$\Omega$" is the same as the textbook's "$U$". Now, what is the lower-case $\omega$ referring to when you have, for example, $\int_\omega \psi = 0$? – Cookie Jun 12 '15 at 16:55
  • @les, it means the editor did not give me an error flag because \omega and \Omega are both legitimate Latex, and I did not proof-read carefully enough last night. Upper case Omega is correct because that is what I was calling the region. And yes, my $\Omega$ is the same as your $U,$ I was not paying adequate attention. – Will Jagy Jun 12 '15 at 17:40
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May be I am seeing the trees but here the wood also.. by means of these geometry examples underlying the question leading to the same result from same simpler situation. I see this pattern in the ODEs:

  • In Dido's problem ... Maximum area enclosed for given boundary length ( also same as given area of minimum boundary length),the optimal situation and Lagrangian in the plane are

$$\int w \,dx - \lambda \int \sqrt{1+w^2}dx ,\, F = w - \lambda \sqrt{1+w^2} ; $$

Using Euler-Lagrange equation of calculus of variations/ Beltrami special case where Lagrangian appears independent of $x$ explicitly.

We are led to a constant value of curvature:

$$ \dfrac{w^{''}}{(1+w^2)^{\frac32} } = \dfrac{1}{\lambda} $$

  • For the surface of revolution that maximizes volume for given surface area ( or for given volume contained within minimum surface area )

the optimal situation Lagrangian in $\mathbb R^3$are

$$\int \pi w^2 dx - \lambda \int 2 \pi w \sqrt{1+w^2}dx ; \,F = w^2 - 2 \lambda\, w \sqrt{1+w^2}; $$

we are led to another constant mean curvature $H = 1/ \lambda^{'} $

$$ \dfrac{w^{''}}{(1+w^{'2})^{\frac32} } + \dfrac{1} {w\, \sqrt{1+w^{'2}}}= \dfrac{1}{\lambda ^{'}} $$

So the minimal surfaces have $$ \kappa = const. or \, H= const. $$

Narasimham
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