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Math Help - Locally Exact Differential....

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    ynj
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    Locally Exact Differential....

    A differential pdx+qdy(in \mathbb {C})is said to be a locally exact differential in a region \Omegaif 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=0if it is a locally exact differential..
    Last edited by ynj; February 8th 2010 at 08:17 AM.
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    Quote Originally Posted by ynj View Post
    A differential pdx+qdy(in \mathbb {C})is said to be a locally exact differential in a region \Omegaif 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=0if 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.
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    ynj
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    Quote Originally Posted by HallsofIvy View Post
    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+qdyis exact in B(x,\delta).It is weaker than 'exact differential'
    Last edited by ynj; February 9th 2010 at 12:16 AM.
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