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The problem is "Calculating $\oint_{L} \frac{xdy - ydx}{x^2 + y^2}$, where L is a smooth, simple closed, and postively oriented curve that does not pass through the orgin".

Here is my solution:
Let $$P(x,y) = \frac{-y}{x^2 + y^2}, Q(x,y) = \frac{x}{x^2 + y^2}$$ Get $$\frac{\partial{P}}{\partial{y}} = \frac{y^2 - x^2}{(x^2 + y^2)^2} = \frac{\partial{Q}}{\partial{x}}$$ Acorrding to the Green formula: $$\oint_{L} \frac{xdy - ydx}{x^2 + y^2} = \iint (\frac{\partial{Q}}{\partial{x}} - \frac{\partial{P}}{\partial{y}})dxdy = 0$$

What's wrong with my solution?

Jichao
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  • Requires that $ L $ is smooth, $oriented, closed$ curve to used the Theorem . – Bryan Yocks Nov 27 '10 at 05:22
  • @Bryan: Simple, too, and as Timothy Wagner pointed out, hypotheses on $P$ and $Q$ are key. – Jonas Meyer Nov 27 '10 at 05:30
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    The reason why you cannot apply Green's theorem if the curve $L$ encloses the origin is that the functions $\frac{x}{x^2+y^2}$ and $\frac{y}{x^2+y^2}$ are discontinuous at the origin. So the assumption which fails here is that the functions $\frac{x}{x^2+y^2}$ and $\frac{y}{x^2+y^2}$ are not in $C^{(1)}$. (In fact there are not even in $C^{(0)}$). –  Nov 27 '10 at 07:33

6 Answers6

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That is valid as long as $L$ is a simple closed curve and the origin is not "inside" $L$, i.e., in the bounded region determined by $L$. For example, this would work if $L$ is a unit circle centered at $(2,0)$, but not if it is the unit circle centered at the origin, and not if it is not a simple closed curve.

(As Timothy Wagner already pointed out before I finished writing, you need to check the hypotheses on the theorem you invoked to see when it works.)

If $L$ is closed, your integral should be $2\pi$ times the number of times $L$ winds around the origin.

Jonas Meyer
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  • What if $L$ include the orgin or $L$ is not a simple closed curve? – Jichao Nov 27 '10 at 05:20
  • @Jichao: If $L$ includes the origin, it is a problem in this case, since $P$ and $Q$ are not differentiable at the origin. – Timothy Wagner Nov 27 '10 at 05:27
  • Then it doesn't work, so the integral will have to be determined by other means. I just added the remark that if $L$ is closed, you are trying to find $2\pi$ times the winding number, and sometimes this may be determined geometrically. Depending on the given parametrization of $L$, you may be able to directly compute the line integral. – Jonas Meyer Nov 27 '10 at 05:27
  • @Jonas Meyer: Is there any other simple method to get $2\pi$? I did not know differntial geometry. – Jichao Nov 27 '10 at 05:33
  • @Jichao: Again, it will depend on the given parametrization. If you took the unit circle parametrized by $(\cos(t),\sin(t))$ on $[0,2\pi]$, for example, then it is easy to compute that the integral is $2\pi$. For other curves it may not be easy to evaluate analytically. If you use the winding number interpretation, then you may be able to determine geometrically the net number of times the curve winds around the origin counterclockwise, then multiply by $2\pi$. Or you could use numerical methods to approximate the integral, knowing that you only need to find the nearest multiple of $2\pi$. – Jonas Meyer Nov 27 '10 at 05:37
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    @Jichao: You can prove this where $L$ is a circle of any radius centered at the origin. Then for any arbitrary $L$ that is sufficiently nice, consider the region between $L$ and a circle centered at the origin which completely contains $L$ and apply Green's theorem to this region. – Timothy Wagner Nov 27 '10 at 05:39
  • @Jichao: Now that I noticed you updated your hypotheses on $L$, the response that Timothy Wagner has just given is the best way as long as the origin is contained in the bounded region determined by $L$. (And of course if it is not, you can just apply Green's theorem directly.) – Jonas Meyer Nov 27 '10 at 05:44
  • @Timothy: +1, thanks it works and not only circle but square would be ok :) – Jichao Nov 27 '10 at 07:11
  • @Jonas: Thanks for your continous and kindly replies. – Jichao Nov 27 '10 at 07:13
  • @Jichao: Yes, any nice curve would work so long as you can parametrize it and are able to evaluate the resulting one dimensional integral. – Timothy Wagner Nov 27 '10 at 07:18
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Hint: What does the hypothesis of Green's theorem say about the requirements on functions $P$ and $Q$?

Timothy Wagner
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Just to add that I think that the value of the integral (i.e., 2Pi times the winding number) depends only on the homotopy-type of the curve, i.e., if a curve C is homotopic to the circle L that winds around the origin, then the integral over C will have the same value as the integral over L.

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The answer should be $2\pi$ if the origin lies "inside" $L$, otherwise it should be 0.

TCL
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One way of understanding why Green's theorem doesn't necessarily work in this context is to view Green's theorem as two separate equations added together: $\oint_{L} P dx = \int\int-{\partial P \over \partial y} dy dx$ and $\oint_{L} Q dy = \int\int{\partial Q \over \partial x} dx dy$.

Basically, Green's theorem holds because you can prove each of these equations by integrating by parts. In the first equation you integrate by parts in $y$ for a fixed $x$ and then integrate the result in $x$. In the second, you integrate by parts in $x$ for a fixed $y$ and then integrate the result in $y$. Since in your case both $-{\partial P \over \partial y}$ and ${\partial Q \over \partial x}$ blow up at the origin, if the origin is contained in the interior of your curve these integrations by parts won't work... and you'll get the wrong answer.

(The curve is assumed to be simple and closed in the above.)

Zarrax
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Notice that $\frac {xdt-ydx}{ x^2+y^2}$ is the total derivative of $arctan(x/y)$, so when you integrate about a closed curve, by the fundamental theorem of contour integrals, you get (after parametrizing) $$\int_{C} \frac {xdy-ydx}{x^2+y^2} = arctan(x/y) /_{(c,d)}^{(a,b)}$$ with $(a,b), (c,d)$ points on your curve. If $C$ is a circle, then these points will be of the form $(r, \sqrt {1-r^2)}$ or $(rcos\theta,rsin\theta)$ , so that, with the latter you get $$ tan^{-1}(sin \theta/ cos \theta)=tan^{-1}(tan\theta) = \theta$$ so that the integral equals $$ \theta_1 - \theta_2 $$ though you have to take the periodicity of $tan \theta$ into account.

MSIS
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