Del inverses

[paddo]

G'day,
If you're given a vector q and have that del x p=q (i.e curl(p)=q), how would you find p?
Also for divergences.
cheers

[HallsofIvy]

Right down the equations for the individual components:
\nabla\times p= q
means that
\frac{\partial p_z}{\partial y}- \frac{\partial p_y}{\partial z}= q_x
\frac{\partial p_x}{\partial z}- \frac{\partial p_z}{\partial x}= q_y
\frac{\partial p_y}{\partial x}- \frac{\partial p_x}{\partial y}= q_z
Solve those equations.

[Hurkyl]

You might get some mileage out of Stokes' theorem. (And some real analysis to figure out how to extract information about p) Keep mind mind that p is not uniquely determined, not even up to a constant. You can add any irrotational field to p and get a new solution. I'm pretty sure there's some ugly integral you can write down that gives you a particular solution -- hopefully someone will remember it and post it here.

[atyy]

I'm pretty sure there's some ugly integral you can write down that gives you a particular solution -- hopefully someone will remember it and post it here.

Hurkly, that's funny!

paddo, you may like to compare Ampere's circuital law and the Biot Savart law:
http://en.wikipedia.org/wiki/Maxwell%27s_equations
http://en.wikipedia.org/wiki/Biot-Savart_law

You may also like to compare Gauss's law with Coulomb's law for a continuous charge distribution:
http://en.wikipedia.org/wiki/Coulomb%27s_law

[Anthony]

You can use homotopy operators to do this type of thing.

[Mute]

Do you happen to know the divergence of p, and the projection of p onto the outward pointing normal vector of the boundary of the volume you're solving for p in?

In general, if you know

\nabla \times \mathbf{u} = \mathbf{C}(\mathbf{r})

\nabla \cdot \mathbf{u} = s(\mathbf{r})

and the value of \mathbf{\hat{n}}\cdot \mathbf{u} on the boundary of the volume you're solving in, \partial V, then there is a unique solution for \mathbf{u}. Writing

\mathbf{u} = -\nabla \phi + \nabla \times \mahtbf{A},

then

\phi(\mathbf{r}) = \frac{1}{4\pi}\int d^3\mathbf{r'} \frac{s(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|} + \mbox{const.}

\mathbf{A}(\mathbf{r}) = \frac{1}{4\pi}\int d^3\mathbf{r'} \frac{\mathbf{C}(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|} + \nabla f(\mathbf{r})

[Anthony]

I should probably elaborate on my previous post. You want a linear homotopy operator h:\Lambda^k \rightarrow \Lambda^{k-1} to be such that:

\theta = \mathrm{d}h (\theta) + h (\mathrm{d}\theta)

for a k-form \theta. Clearly if \theta is closed you have \theta = \mathrm{d}\eta with \eta = h(\theta). If we work on star-shaped domains, then the following holds:

h(\theta) = \int_0^1 (\iota_X \theta )[\lambda x] \frac{\mathrm{d}\lambda}{\lambda}

where X= x^i \partial /\partial x^i is the scaling vector field. You might like to try it out with some examples.