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)
 A: 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.
