# Same Div, different Curl, Equal on the boundary of some region?

"Find a pair of fields having equal divergences in some region, having the same values on the boundary of that region, and yet having different curls"

Can anyone name a pair of fields that fit this requirement (It would be awesome if the region was the unit sphere)

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Your gravatar is very fitting for this question :-) – joriki Sep 13 '11 at 9:17
Haha, Its a fluke though – uguhu Sep 13 '11 at 9:19
Note that the difference $\mathbf{h}=\mathbf{F-G}$ will (1) be divergence-free, (2) vanish on $\partial D$, and (3) be a nonzero field (i.e. not identically the zero vector). It suffices to find such an $\mathbf{h}$, because then $\mathbf{G}$ can be chosen arbitrarily and $\mathbf{F}$ is determined. Helmholtz decomposition sounds like it might be relevant in some fashion for a general approach. – anon Sep 13 '11 at 9:35
I'm confused by that, can you continue with how exactly, one would use that to find a pair of fields that match the conditions – uguhu Sep 13 '11 at 9:42
@uguhu: Which point of mine are you referring to with the word "that"? I reduced the problem to a simpler one of finding a single vector field satisfying some properties and also mentioned Helmholtz decomposition because it just might be helpful; I don't actually have a solution in mind. Also, if you want to alert a user to a response you need to put "@anon" or equivalent to your comment so that the system pings them. – anon Sep 13 '11 at 10:04

Building on anon's comment (the background to which you can read up on at Wikipedia), we need to find a field that has zero divergence and non-zero curl and vanishes on the boundary of a region, preferably on the unit sphere. (In mathematics, as opposed to physics, "sphere" usually refers to the surface, not to the body, and the body is called a "ball".)

Consider a field of the form $f(r)(\mathbf r\times\mathbf v)$, where $r=|\mathbf r|$ is the distance from the origin and $\mathbf v$ is an arbitrary constant vector. Its divergence vanishes:

$$\nabla\cdot(f(r)(\mathbf r\times\mathbf v))=f'(r)\frac{\mathbf r}{r}\cdot(\mathbf r\times\mathbf v)+f(r)\mathbf v\cdot(\nabla\times\mathbf r)=\mathbf 0\;.$$

Its curl is

$$\begin{eqnarray} \nabla\times(f(r)(\mathbf r\times\mathbf v)) &=& (\mathbf v\cdot\nabla)(f(r)\mathbf r)-\mathbf v(\nabla\cdot(f(r)\mathbf r)) \\ &=& f(r)\mathbf v+ (\mathbf v\cdot\frac{\mathbf r}r)f'(r)\mathbf r-3f(r)\mathbf v-\mathbf vf'(r)\frac{\mathbf r}{r}\cdot\mathbf r \\ &=& \mathbf v(-2f(r)-f'(r)r)+(\mathbf v\cdot\mathbf r)f'(r)\frac{\mathbf r}r\;, \end{eqnarray}$$

which is non-zero for non-zero $f$ and $\mathbf v$. Thus you can choose some function $f$ that vanishes on the unit sphere, e.g. $f(r)=r-1$, or something that yields a smooth field at $r=0$ in case you care about that.

[Edit in response to the comment:]

I didn't understand the requirement to mean that the curls have to be different everywhere. In fact this is impossible if the fields are sufficiently differentiable. By Stokes' theorem, the curl of the difference must be tangent to the sphere, since we can integrate its normal component over an arbitrarily small portion of the surface and this must be equal to the integral of the difference along the boundary curve, which is zero since the fields are equal on the sphere. So the curl of the difference forms a continuous tangent vector field on a $2$-sphere, which must vanish somewhere on the sphere by the hairy ball theorem.

[Edit in response to the other comment:]

You can get a field with zero curl, zero boundary values and non-zero divergence in a similar way. Consider a field of the form $f(r) \mathbf r$. Its curl vanishes:

$$\begin{eqnarray} \nabla\times(f(r) \mathbf r) &=& f(r)\nabla\times\mathbf r+(\nabla f(r))\times\mathbf r \\ &=& f(r)\nabla\times\mathbf r+f'(r)\frac{\mathbf r}r\times\mathbf r \\ &=& \mathbf 0\;. \end{eqnarray}$$

Its divergence is

$$\begin{eqnarray} \nabla\cdot(f(r)\mathbf r) &=& f(r)\nabla\cdot\mathbf r+(\nabla f(r))\cdot\mathbf r \\ &=& f(r)\nabla\cdot\mathbf r+f'(r)\frac{\mathbf r}r\cdot\mathbf r \\ &=& 3f(r)+f'(r)r\;, \end{eqnarray}$$

which is not identically zero unless $f(r)\propto r^{-3}$. Thus, as in the other case, you can choose some function $f$ that vanishes on the unit sphere to obtain a suitable difference between the two fields.

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+1. I was looking for a $(1-r^2)\mathbf{w}$ but didn't think to make $\mathbf{w}=\mathbf{r}\times\mathbf{v}$. – anon Sep 13 '11 at 10:52
Sorry, a term in the curl was missing; I've added it. – joriki Sep 13 '11 at 11:16
amazing using your and @anon 's advice I was able to find the pair of fields that fit the conditions (as far as I can tell) Any tips on how you would do it reverse: "Find a pair of fields having equal curls in some region, having the same values on the boundary of that region, and yet having different divergences." – uguhu Sep 13 '11 at 12:02
OH WAIT, @joriki I finding that there are lots of areas within the uni sphere where the curls end up equal (referring to original question) – uguhu Sep 13 '11 at 13:16
@uguhu: This is inevitable; I edited the answer in response. – joriki Sep 13 '11 at 18:21