Conservative vector field?

  • Thread starter Damidami
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  • #1
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Main Question or Discussion Point

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]
so
[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?
Thanks!
 

Answers and Replies

  • #2
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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.
 
  • #3
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* I mean the potential function [itex]\phi(x, y)=\arctan(\frac{y}{x})+C[/itex].
 
  • #4
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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?
Thanks.
 

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