# Proof of Green's theorem in Apostol book [closed]

Hello,

I got a question about one point in the Green's theorem proof that appears in Apostol's Mathematical Analysis, first edition, Ed. Addison Wesley. More exactly, in the theorem 10-42, immediately previous (and essential) to Green's (p. 287-289). Please, if you do not have time for helping me, just tell me where I can look for assistance.

(Sorry for the clumsy redaction, I am not the better in english).

Well, Apostol has a contour denoted by the greek letter "gamma", described by a function "alpha", which has its component functions, "gamma 1" and "gamma 2", all of them defined in [a,b] as parametric interval.

At certain point (which is irrelevant for the purpose of my question) Apostol crosses horizontally the contour with a line segment L. This of course produces two new contours, "gamma one" and "gamma two". Be aware that each of these new contours is NOT only an arc of the mother-contour "gamma", because they have a line side that the mother-contour does not: it is L.

In p. 289 Apostol puts this equation:

Total variation of real function "alpha 2" on the parametric interval of gamma (which as I said is [a, b]) =

Total variation or real function "alpha 2" in the parametric interval of gamma 1, plus Total variation or real function "alpha 2" in the parametric interval of gamma 2."

This is my problem: I can not find a way to justify this equation. (For Apostol it may be obvious, because he doesnt prove it). It is not easy to find the parametric intervals corresponding to contours gamma 1 and gamma 2. Remember that these two are not merely arcs of "mother-contour" gamma: gamma 1 and gamma 2 include also a line segment. For that reason I can not simply take a number from [a, b] (say "c"), state that [a, c] corresponds to contour gamma 1, [c. b] to contour gamma 2, and apply theorem 8.11 (sum of total variations).

What I am trying to do? Well, I am parametrizing contour gamma 1 this way: a function for its curvilinear section (which is part of contour gamma), another function for its line segment section, and then I concatenate (join) both of them in a single parametrization. I am doing the same for contour gamma 2.

But I have doubts about this tactic because I have to assume that certain points in the mother contour gamma corresponds to several images of the range of function alpha 2. An undesirable restriction.

Please, if you have not enough time to answer, just tell me where I can find assistance. I am studying analysis by myself (no college, no university). I am blocked in this at least 2 weeks ago. Thanks.

pedro

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## closed as too localized by Ben Webster♦Jun 13 '10 at 20:11

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If I understand you correctly, I think what you are doing is okay. I don't have Apostol 1st ed. handy, so let me try to summarize what you wrote: $\gamma$ is a closed contour (let's say a circle). $\gamma_1,\gamma_2$ are two closed contours, they are two half-circles with the diameter included. What you do is to apply Theorem 8.11 to $\gamma_1$ and $\gamma_2$ as the Arc part + the diameter part. Then Apostol's claim follows if you observe the diameter parts of $\gamma_{1,2}$ cancel each other. The "double points" contribute infinitesimally to the integral, so you don't need to worry there. – Willie Wong Apr 26 '10 at 18:22
Also, your question is not quite at the right level for this website. It will fit in better at Art of Problem Solving, or NRICH, links to which are available here: mathoverflow.net/faq#whatnot Best of luck on your studies! – Willie Wong Apr 26 '10 at 18:25

Uh, you are talking about Green's theorem in a plane, right? Then your contour integral is in fact a line integral. Recall that in the statement of Green's theorem the curve is required usually to be positively oriented in the sense that it goes counter-clockwise around the closed region. To be more precise, say $\gamma$ is parametrized by $[0,1]$ and the top semicircle is $\gamma([0,1/2])$. Let $\tau:[0,1]\to L$ parametrize the diameter. Then we can parametrize $\gamma_1$ over $[0,1/2]$ by $\gamma_1(t) = \gamma(t/2)$ if $t < 1/4$, and $\gamma_1(t) =$\tau((t+1)/4)$when$1/4 < t < 1/2$. – Willie Wong Apr 27 '10 at 9:29$\gamma_2$then, over$[1/2,1]$can be$\gamma_2(t) = \tau((3-t)/4)$if$1/2 < t < 3/4$and so on. This way I hope it is sort of clear that the contours for that two$L\$ segments, traversed in opposite directions, cancel out. – Willie Wong Apr 27 '10 at 9:32