# Del inverses

1. Sep 2, 2008

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

2. Sep 2, 2008

### HallsofIvy

Staff Emeritus
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.

3. Sep 2, 2008

### Hurkyl

Staff Emeritus
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.

4. Sep 2, 2008

### atyy

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's_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's_law

5. Sep 3, 2008

### Anthony

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

6. Sep 3, 2008

### 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

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

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

7. Sep 4, 2008

### 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.