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Conservative vector field?

  1. Dec 23, 2011 #1
    Let [itex] f(x,y) = ( \frac{-y}{x^2 + y^2}, \frac{x}{x^2 + y^2}) [/itex] with [itex] f : D \subset \mathbb{R}^2 \to \mathbb{R}^2 [/itex]

    I know if I take [itex] D = D_1 = \mathbb{R}^2 - \{ (0,0) \} [/itex] the vector field is not conservative, for the circulation over a circunference centered at the origin does not equal zero.

    But let's see what happens if I take [itex] D = D_2 = \{(x,y) \in \mathbb{R}^2 : (x,y) \neq (0,a), a \geq 0\} [/itex], that is [itex] \mathbb{R}^2 [/itex] without the ray starting at the origin and going in the positive-y axis.

    With [itex] D_2 [/itex] one has [itex] f [/itex] has to be conservative, because the domain is simply connected, and it satisfices the necesary condition (jacobian continuous and symmetric). So together they satisfty a sufficient condition.

    The problem arises when I want to calculate the potential function. I get

    [itex] \phi'_x = \frac{-y}{x^2 + y^2}[/itex]
    [itex] \phi'_y = \frac{x}{x^2 + y^2}[/itex]
    [itex] \phi \approx - \arctan(\frac{x}{y}) + c(y) [/itex]
    [itex] \phi \approx \arctan(\frac{y}{x}) + c(x) [/itex]

    And can't see how to get the expression for the potential function.
    Is it [itex] \phi(x,y) = \arctan(\frac{y}{x}) - \arctan(\frac{x}{y}) + c [/itex] ?

    Any idea what is happening here?
  2. jcsd
  3. Dec 23, 2011 #2
    Do you remember the trig identity [itex] \arctan(x)+\arctan(\frac{1}{x})=\frac{\pi}{2}[/itex]? Your expressions are in fact consistent :P The potential function [itex]\arctan(x)+C[/itex] works fine.
  4. Dec 23, 2011 #3
    * I mean the potential function [itex]\phi(x, y)=\arctan(\frac{y}{x})+C[/itex].
  5. Dec 24, 2011 #4
    Hi identity1,
    Thanks for your reply.
    Some things still I don't understand.
    That equation [itex] \arctan(x) + \arctan(\frac{1}{x}) = \frac{\pi}{2} [/itex] seems to work only for positive real values of x, since for example
    [itex] \arctan(-2) + \arctan(\frac{-1}{2}) = \frac{-\pi}{2} [/itex]

    Say, for example, I want to calculate the circulation of [itex] f [/itex] with domain [itex]D_2 [/itex] from a curve that starts at [itex]A = (-1,1)[/itex] and ends at [itex]B=(1,1)[/itex]

    If I use [itex]\phi_1(x,y) = -\arctan(\frac{x}{y}) + C[/itex]
    [itex]\phi_1(B) - \phi_1(A) = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2} [/itex]

    Instead if I use [itex]\phi_2(x,y) = \arctan(\frac{y}{x}) + C[/itex]
    [itex]\phi_2(B) - \phi_2(A) = -\frac{\pi}{4} - \frac{\pi}{4} = \frac{-\pi}{2} [/itex]

    So it changes the sign for the same curve oriented in the same direction from A to B.

    Something seems to be wrong, the sign shouldn't change, doesn't it?
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