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In my vector calculus class, when we were introduced to the curl operator the professor gave us this example:

Is it possible to find a vector field $\bf{G}$ such that $$\mathbf{F} = \nabla \times {\mathbf{G}}?$$

As a motivation for the identity that the divergence of a curl of a vector field is $0$.

My question is:

Given that the identity holds true, how do you solve for a general solution for the component functions of $\bf{G}$, given $\bf{F}$?

The best I can do is find one or two solutions. For example, if $\mathbf{F}$ = $\langle-y,-z,-x\rangle$, a solution I find is $\mathbf{G}$ = $\langle xy,0,-\frac{1}{2}y^2+xz\rangle$. I had to go through a system of partial differential equations and made it work, but can the general solution be written explicitly?

Thanks!

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the answer comes from a bit of calculus III which is beyond most of the texts we require students to purchase in America. It's called Helmholtz theorem and it arises in connection with the first, second and third identities of Greene. –  James S. Cook Jul 16 '13 at 3:53
    
Thanks for the reference, I will go look it up. We learn Greene's theorem in a couple days, so it will be a productive read. @JamesS.Cook –  kvmu Jul 16 '13 at 4:00
    
it would be nice if someone wrote a nice complete answer to your question, I couldn't find anything too complete on MSE at present. For now, perhaps pages 72-82 of supermath.info/MultivariateCalculus2011Chapter7.pdf will be helpful. These are my notes based on Susan Colley's excellent vector calculus text. –  James S. Cook Jul 16 '13 at 4:00
    
Hi, kvmu, I removed the differential-equations tag, which is supposed to be for ODE. Also I edited the title to make it more descriptive. Hope you wouldn't mind. :) –  Shuhao Cao Jul 16 '13 at 4:27

2 Answers 2

up vote 3 down vote accepted

$\newcommand{\F}{\mathbf{F}}\newcommand{\p}{\mathbf{p}}$If we consider the unbounded $\mathbb{R}^3$ case, there is a path integral formula to construct the right inverse of the curl operator for divergence free vector field $$ \mathcal{R}(\F) = -(\p - \p_0)\times \int^1_0 \F\Big(\p_0 + t(\p- \p_0)\Big)t\,dt.\tag{1} $$ A proof of this formula can be found here, for more discussion the author pointed to Spivak's book Calculus on Manifolds.

In your case, letting $\p_0 = \langle 0,0,0\rangle $, it is $$ \mathcal{R}(\F) = \langle x,y,z\rangle \times \int^1_0 \langle y,z,x\rangle t^2 dt = -\frac{1}{3}\langle z^2-xy,x^2-yz,y^2-xz\rangle .$$ You can check that this is indeed the right inverse $$ \nabla \times \mathcal{R}(\F) = \F = -\langle y, z, x\rangle. $$ Notice the vector field above is difference from yours, because essentially $\mathcal{R}(\F) + \nabla \phi$ is also the answer for smooth $\phi$.


Further checking: the difference between the right inverse $\mathcal{R}(\F) $ above and your potential field $\langle xy,0,−y^2/2+xz\rangle$ is $$ \mathbf{A} = \langle \frac{2}{3}x y +\frac{1}{3}z^2, \frac{1}{3}(x^2-y z), -\frac{1}{6}y^2+\frac{2}{3}xz\rangle = \nabla \left(\frac{1}{3} x^2y + \frac{1}{3}xz^2 - \frac{1}{6}y^2 z\right), $$ indeed a gradient.

If you wanna eliminate the huge kernel of curl operator, what you can do is (1) choosing a gauge (pinning down the divergence of $\mathbf{G}$), and/or (2) specifying boundary condition restricted on a simply-connected domain $\Omega$ of $\mathbb{R}^3$, by posing the following boundary value problem on some $\Omega$: $$\left\{ \begin{aligned} \nabla \times \mathbf{G} &= \F\quad \text{ in }\Omega, \\ \nabla \cdot \mathbf{G} &= g \quad \text{ in }\Omega, \\ \mathbf{G} \cdot \mathbf{n} &= 0 \quad \text{ on }\Gamma. \end{aligned} \right.$$ You can check the construction of yours $\langle xy,0,−y^2/2+xz \rangle $ is not divergence free, while formula (1) automatically gives you the divergence free potential field. For unbounded $\mathbb{R}^3$, choosing a gauge $g$ will get you the result you want.

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nice. Could you recommend a text where this is further discussed. –  James S. Cook Jul 16 '13 at 4:03
    
@JamesS.Cook I learned it in my diff-geom class a long time ago, I will try to find some references later. –  Shuhao Cao Jul 16 '13 at 4:04

Since we know that the solution to the Maxwell equation

$\nabla \times B$ = j

is given by the Biot-Savart formula

$B(r) = \int \frac{\bf{j(r')} \times {(\bf{r}-\bf{r'})}}{|r-r'|^3} d^3\bf{r'}$

this gives a convenient shortcut (and the derivation/proof is in any E&M book).

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