Math Help - Locally Exact Differential....

1. Locally Exact Differential....

A differential $pdx+qdy$(in $\mathbb {C}$)is said to be a locally exact differential in a region $\Omega$if it is an exact differential in some neighbourhood of each $x\in\Omega$.
Show that every rectangle with sides parallel to axes $R\subset\Omega$ satifies $\int_{\partial R}pdx+qdy=0$if it is a locally exact differential..

2. Originally Posted by ynj
A differential $pdx+qdy$(in $\mathbb {C}$)is said to be a locally exact differential in a region $\Omega$if it is an exact differential in some neighbourhood of each $x\in\Omega$.
Show that every rectangle with sides parallel to axes $R\subset\Omega$ satifies $\int_{\partial R}pdx+qdy=0$if it is a locally exact differential..
If pdx+ qdy is locally exact, then it is, as you say, an exact differential. That means that there exist some function F(x,y) such that dF= pdx+ qdy, at least inside $\Omega$.

But then $\oint pdx+ qdy= \oint dF$. Let t be any parameter for the curve and we have $\oint pdx+ qdy= \int_a^b \frac{dF}{dt}dt= F(x(b),y(b))- F(x(a),y(a))$ where a and b are the beginning and ending values for t. But since this is a closed curve, t= a and t= b give the same point so x(b)= x(a), y(b)= y(a) and F(x(b),y(b))- F(x(a),y(a))= 0.

3. Originally Posted by HallsofIvy
If pdx+ qdy is locally exact, then it is, as you say, an exact differential. That means that there exist some function F(x,y) such that dF= pdx+ qdy, at least inside $\Omega$.

But then $\oint pdx+ qdy= \oint dF$. Let t be any parameter for the curve and we have $\oint pdx+ qdy= \int_a^b \frac{dF}{dt}dt= F(x(b),y(b))- F(x(a),y(a))$ where a and b are the beginning and ending values for t. But since this is a closed curve, t= a and t= b give the same point so x(b)= x(a), y(b)= y(a) and F(x(b),y(b))- F(x(a),y(a))= 0.
Hmm....you misunderstand the problem..Locally exact means that $\forall x\in\Omega,\exists\delta>0,pdx+qdy$is exact in $B(x,\delta)$.It is weaker than 'exact differential'