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According to my computations, the function which minimizes $\int_\Omega \|\operatorname{curl} f\|^2~dx$ should satisfy $\operatorname{curl}(\operatorname{curl}(f)) = 0$ everywhere on $\Omega$, provided $\operatorname{curl} f = 0$ on $\partial \Omega$.

I followed the same kind of computation that the one demonstrating the the argmin of $\int_\Omega \|\nabla f\|^2~dx$ should satisfy $\Delta f = 0$. However, I am not sure whether my computations are right... May anyone check that please ?

1) We first start with a functional $G(f) = \int_\Omega \|\operatorname{curl} f\|^2~dx$.
2) We compute $V(f,h) = lim_{\epsilon\rightarrow 0} \frac{G(f+\epsilon h)-G(f)}{\epsilon} = 2\int_\Omega \operatorname{curl} f\cdot\operatorname{curl} h~dx$
3) We use the identity : $\operatorname{div}(A\times B) = -A\cdot\operatorname{curl} B + B\cdot\operatorname{curl} A$, with $A=\operatorname{curl} f$ and $B=h$
4) We obtain $\int_\Omega \operatorname{curl} f\cdot \operatorname{curl} h\:dx = -\int_\Omega \operatorname{div}(\operatorname{curl} f\times h)~dx + \int_\Omega h\cdot \operatorname{curl}(\operatorname{curl}(f))~dx$
5) We use the divergence theorem to obtain : $\int_\Omega \operatorname{curl} f\cdot \operatorname{curl} h~dx = -\int_{\partial\Omega} \operatorname{curl} f\times h~ds + \int_\Omega h\cdot\operatorname{curl}(\operatorname{curl}(f))~dx$
6) We assumed $\operatorname{curl} f = 0$ on $\partial\Omega$, so the first term is zero.
7) $V(f,h)$ should equal zero for all $h$ for the function to be minimized, so $\int_\Omega h\cdot\operatorname{curl}(\operatorname{curl}(f)) dx = 0~~\forall h$, which implies $\operatorname{curl}(\operatorname{curl}(f)) = 0$ locally.

I guess this reasonning may be wrong in several places..... or is it right?? Thanks!

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Your reasoning looks correct to me. I think you don't even need $ \nabla \times F = 0 $ on the boundary because $ h $ is supposed to vanish there anyway, thereby fulfilling (6) without need for the extra information. I may be a bit shaky on my vector calculus, but I think that given $ \nabla \times ( \nabla \times F) = 0 $ we can deduce that $ \nabla \times F = \nabla g $ for some $ g $, which is impossible unless $ g $ is constant and therefore $ \nabla \times F = 0 $ which finally means that $ F = \nabla \phi $ for some $ \phi $. I think. – anon Jun 8 '11 at 19:12
great, thanks :) I'm mainly worried about point 7 : I know that for scalar valued functions, if $\int f g=0~~\forall g$ then $f=0$ (almost everywhere!), but I don't know if this applies to vector valued functions, by replacing the product by a scalar product between them... – WhitAngl Jun 8 '11 at 20:07
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Since it must be true for all $ h(x) = u(x) e_i $ for scalar functions $ u $ and $ \{ e_1, \dots e_n \} $ the vector space basis, we can still make the conclusion with sound logic. – anon Jun 8 '11 at 21:00
@anon: if $\Omega$ is a simply connected domain then $\nabla\times W = 0 \implies W = \nabla g$. Else it is better to just use the double curl identiy to get $\nabla\times(\nabla\times A) = \nabla (\nabla\cdot A) - \nabla^2 A$. – Willie Wong♦ Apr 4 '12 at 13:03
Also note that the minimizers are in general not unique: if you replace $f \to f + \nabla h$ for some scalar function $h$, necessarily that $\nabla \times f = \nabla\times f + \nabla\times(\nabla h)$ so if $f$ is a minimizer and $h$ any function compactly supported in $\Omega$, $f + \nabla h$ is another minimizer. Generally this degree of freedom is gotten rid of by prescribing the divergence of $f$ (the constraint $\nabla\cdot f = k$ for some function $k$), in which case the double curl equation reduces to a Poisson equation. – Willie Wong♦ Apr 4 '12 at 13:07

1 Answer

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(This is supposed to be a shorter comment but I find myself powerless).

I haven't checked you reasoning, but I wanted to offer a simple (standard) proof. Some authors call this Dirichlet's principle.

Define $E(f) := \int_{\Omega} \| \nabla f \|^2 dx$. It's clear that $E(f) \geq 0$.

Assume $f, g$ are equal on $\partial \Omega$ Let's see that, if $f$ is harmonic then $E(f) \leq E(g)$.

For that, let $u := f - g$. Now calculate $E(g) = E(f - u)$ using Green's first identity (here you'll use that $u = 0$ on $\partial \Omega$ and that $\Delta f = 0$).

You'll get immediately that $E(g) = E(f) + E(u)$ , and using that $E(u) \geq 0$ that $E(f) \leq E(g)$.

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