21
$\begingroup$

Is there a good measure theoretic definition of curl?

To give an idea of the sort of equation that I'm looking for, here's now I define grad and div. For the gradient, say we are given a Fréchet differentiable function $f:X\rightarrow\mathbb{R}$, then we can define $\nabla f(x)$ to be the element in $X$ such that $$ \langle \nabla f(x),\eta \rangle = f^\prime(x)\eta. $$ Hence, $\nabla f(x)$ is the Riesz representative of the Fréchet derivative (Note, we've assumed a Hilbert space.) For the divergence, we have $f:X\rightarrow X$ and can define $$ \nabla \cdot f(x) = \lim_{\Omega\rightarrow \{x\}}\frac{1}{\mu(\Omega)} \int_{\Omega} \nabla\cdot f(x) d\mu =\lim_{\Omega\rightarrow \{x\}}\frac{1}{\mu(\Omega)} \int_{\partial \Omega} f(x)\cdot n d\mu. $$ Here, $\lim_{\Omega\rightarrow \{x\}} 1/\mu(\Omega)$ is shorthand for $\lim_{k\rightarrow \infty} 1/\mu(\Omega_k)$ where $\Omega_{k+1}\subseteq\Omega_k$ and $\lim_{k\rightarrow\infty}\Omega_k = \{x\}$. In addition, $\mu$ denotes some measure. In any case, the first equality follows from the Lebesgue differentiation theorem and the second follows from integration by parts. In this way, we require that $X$ be finite dimensional.

Now, I'd like to get something along these lines for curl. More specifically, I'm interested in a definition of curl that does not use differential forms or the exterior calculus.

I haven't checked this closely, but if it helps, I believe that for $f:X\rightarrow X$, we have that $$ (\nabla \times f(x))\times \delta x = f^\prime(x)\delta x-f^\prime(x)^*\delta x $$ where $f^\prime(x)^*$ denotes the adjoint of the Fréchet derivative of $f$ at $x$. In other words, the cross product between the curl of $f$ and $\delta x$ is the antisymmetric part of the Fréchet derivative.

$\endgroup$
3
  • 2
    $\begingroup$ What about the definition on the Wikipedia page? en.wikipedia.org/wiki/Curl_%28mathematics%29 $\endgroup$
    – Thomas
    Nov 26, 2012 at 16:06
  • 1
    $\begingroup$ Not sure if this might be useful (probably not), but you should get the curl if you replace $\cdot$ with $\times$ in the definition of divergence. This holds in classical vector calculus (cfr. Schey, Div grad curl and all of that, Ex. III-29), and should extend to general measure-theoretic setting. $\endgroup$ Jan 7, 2013 at 2:16
  • 1
    $\begingroup$ Maybe this answer helps: mathoverflow.net/questions/52829/… $\endgroup$ Sep 10, 2013 at 15:44

1 Answer 1

3
$\begingroup$

Let $v=(v_i)$ be a vector field. If $c=(c_i)$ is the curl of $v$, as you did with the divergence, you can write:

$$ c_i(x) = \lim_{A_i\to \{x\}}\frac{1}{\mu(A_i)} \int_{A_i}(\nabla\times v)_i\,d\mu = \lim_{A_i\to \{x\}}\frac{1}{\mu(A_i)} \int_{\partial A_i} (v\times dx)_i, $$

where: $$ (v\times dx)_i = \sum_{j,k}\epsilon_{ijk} v_j\,dx_k $$

are the components of the usual cross product, and $A_i$ are flat surfaces with smooth boundary, orthogonal to $dx_i$. For example, $A_1$ is parallel to the $yz$ plane, and oriented in that way.

Note: integration by parts (componentwise) is exactly equivalent to exterior differentiation and Stokes theorem.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.