# minimizing the norm of a curl over a domain

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 $$\int_\Omega h\cdot\operatorname{curl}(\operatorname{curl}f) dx = 0\quad\forall h,$$ 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
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

OP is mostly perfect until some later parts.

For the curl operator we have a similar integration by parts formula to the divergence theorem: $$\int_{\Omega}\nabla \cdot (u\nabla v) = \int_{\Omega}(u\Delta v + \nabla u\cdot \nabla v) = \int_{\partial \Omega} u\nabla v\cdot \nu \,dS,$$ where $\nu$ is the outward unit normal to the boundary. The curl integration by parts formula is: $$\int_{\Omega}\nabla \cdot (\phi\times \psi) = \int_{\Omega}(\psi\cdot\nabla \times \phi - \phi\cdot \nabla\times \psi) = \int_{\partial \Omega} \phi\times \psi\cdot \nu \,dS = \int_{\partial \Omega} \nu\times \phi\cdot \psi \,dS.\tag{1}$$

Say now you wanna minimize (here I use $u$ and $v$ instead of $f$ and $h$): $$\mathcal{G}(v) = \int_{\Omega}|\nabla\times v|^2\,dx,$$ if the minimizer is $u$ and the variational limit is also no problem: $$0= \frac{d}{d\epsilon}\mathcal{G}(u+\epsilon v)\Bigg|_{\epsilon= 0} = 2\int_{\Omega} \nabla \times u\cdot \nabla \times v.$$

Now we use formula (1), letting $\phi = \nabla \times u$, and $\psi = v$: \int_{\Omega}(v\cdot\nabla \times (\nabla \times u)- \nabla \times u\cdot \nabla\times v) =\left\{\begin{aligned} \int_{\partial \Omega} \big(\nu\times (\nabla \times u)\big)\cdot v\,dS \\ \\ \int_{\partial \Omega} (v\times \nu) \cdot (\nabla \times u)\,dS \end{aligned}\right. There are two choices of way of writing boundary terms to be zero:

1. You can set for the minimizer $\color{blue}{\nu\times (\nabla \times u)|_{\partial \Omega} = 0}$, and only tangential part suffices. $\nabla \times u|_{\partial \Omega = 0}$ is too strong.

2. You can set for the test function spaces $\color{red}{\nu\times v = 0}$, then $\nu \times u$ can be prescribed as any permissible boundary data $g$ (those who have 0 surface divergence).

Choosing blue, you have to modular a gradient in a proper function space $H$ to set up a well-defined problem: \left\{\begin{aligned} \nabla\times (\nabla \times u) &=0 \quad \text{ in } \Omega, \\ \nu \times (\nabla \times u) &= 0\quad \text{ on } \partial \Omega. \end{aligned}\right.\tag{2} For $u \in H/\nabla p$ the quotient space, which means we don't tell the difference between $u$ and $u+\nabla p$, above problem has a unique solution. The only solution for (2) is zero, a.k.a. $u = 0$ or say $u = \nabla p$. Once we add a gauge condition $\nabla \cdot u = f$, then (2) is equivalent to the Neumann problem for $p$: \left\{\begin{aligned} \Delta p &= f \quad \text{ in } \Omega, \\ \nabla p\cdot \nu &= 0\quad \text{ on } \partial \Omega. \end{aligned}\right.

Choosing red, the problem is: \left\{\begin{aligned} \nabla\times (\nabla \times u) &=0 \quad \text{ in } \Omega, \\ \nu \times u &= g\quad \text{ on } \partial \Omega. \end{aligned}\right.\tag{3} Again this problem has a unique solution in $H/\nabla p$, except this time $p$ is constant on the boundary which implies $\nabla p\times \nu =0$. If we set $g$ to be 0, then above problem again has a zero solution, which means $u = \nabla p$. Now a gauge condition $\nabla \cdot u = f$, then (3) can be transformed to a Dirichlet boundary problem for $p$: \left\{\begin{aligned} \Delta p &= f \quad \text{ in } \Omega, \\ p &= \mathrm{Constant}\quad \text{ on } \partial \Omega. \end{aligned}\right.

Summary: the minimizer $u = \mathrm{arg}\min\mathcal{G}(v)$ is a gradient field $\nabla p$, the boundary condition of $p$ relies on the boundary condition of $u$. Both problem (2) and (3) don't require the normal component of $\nabla \times u$ or $u$ to be imposed. If you prescribe $\nabla\times u = 0$ on the boundary, which implies $\nabla\times u\cdot \nu = 0$, the problem may be over-determined.

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thanks a lot for the insight! :) – WhitAngl Jun 25 '13 at 19:10
@WhitAngl Hey, while I was cooking up the answer I didn't realize it was a 2-year old question, have you got any trouble with this anymore recently? – Shuhao Cao Jun 25 '13 at 19:14
I was partly asking out of curiosity, and also due to a project I tried and didn't work out well... this could as well be to the way I formulated it (rather than your simpler Laplace equation), or a bug or anything. I might take another look at some point :) The answer was interesting anyway. Thanks! – WhitAngl Jun 25 '13 at 19:32

(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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