Andrew Kennedy Atkins
Posted on Thursday, 13 May, 2004 - 09:01 pm:

Points in a punctured plane, R2(0,0) are specified by their polar coordinates (r,q). What are the necessary conditions for the differential w=A(r,q)dr+B(r,q)dq to be exact? Thanks
Matthew Buckley
Posted on Friday, 14 May, 2004 - 08:52 am:

Andrew,

Have a look at this .

Matt.
Kerwin Hui
Posted on Friday, 14 May, 2004 - 08:57 am:

Assume w is smooth (you can probably get away with C1 ). If you just want a necessary condition - that is easy. The exterior derivative of the 1-form w must be zero, i.e. w is closed. More explicitly,
LaTeX Image.

If you want the necessary and sufficient condition, you need to work harder (check - dq is not exact. Why?). Heuristically, if w=df and g is a closed curve, then by Stoke's theorem we expect
LaTeX Image
and this, together with the requirement w is exact, is the integrability criterion.

There are various ways of proving this. You can try to write
LaTeX Image
and prove the integrability criterion with a lot of (heavy) machineary from Fourier analysis. An alternative, which comes with a health warning [useful prerequisite: a course on algebraic topology] is the following:

The fundamental group of X=2 -(0,0) is , the free group generated by the homotopy class of the usual embedding of S1 as r=1,q goes from 0 to 2p [Note that there is a smooth retract of X to the circle r=1]. Hence the first singular homology group is
LaTeX Image.
Once again the generator can be thought of as the usual S1 (choose a suitable triangulation). The first de Rham cohomology group, defined as
LaTeX Image
is, by de Rham's theorem, isomorphic to LaTeX Image, the singular cohomology group with real coefficients, and the isomorphism is given by the de Rham map
LaTeX Image
Thus the necessary and sufficient condition for w to be exact is that dw=0 and the integral along S1 of the 1-form w is zero (by de Rham theorem), i.e.
LaTeX Image
and
LaTeX Image.


Kerwin
Andrew Kennedy Atkins
Posted on Friday, 14 May, 2004 - 02:30 pm:

Ok, thank you to both of you. I was wondering if it would make any difference if the plane is punctured. So, it doesn't make any difference for necessary condition, but it does for sufficient condition. Thanks a lot!