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Today, in the lecture, we covered an example of a vector field which suffices the necessary condition for integrability, yet is not integrable. The following field also known as the angular form is an example of such a case.

$$\omega = -\frac{y}{r^2}dx + \frac{x}{r^2}dy$$ Where $r = \sqrt{x^2+y^2}$ and the domain is $\Omega = \mathbb{R}^2/${(0,0)} Clearly, we can check that the mixed partials are equal and that the necessary condition holds. But if we compute the circulation about the origin say a unit circle, we notice that it is not zero.

I am aware that the necessary condition is a sufficient condition once we have a simply connected domain $\Omega$. By slicing the plane of the above example we obtain a simply connected domain without the origin and have an integrable vector field.

My question is the following: When we have holes in the domain of the vector field, do we have to check the circulation around each hole to make sure it is zero and then we can find its potential and state that it is integrable? If it is non-zero do we slice the domain and state that it is integrable on the restricted domain? I hope my question makes sense, thanks for any help, ideas or clarification!

Edit: Furthermore after the slicing (removing any smooth curve that connects 0 to $\infty$ ) I claim that $\phi(x,y) = \arctan(\frac{y}{x})$ But it is discontinuous along the y-axis, so then it cannot be the potential function because it is not differentiable there. Am I making a mistake in this logic? Thanks!

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    $\begingroup$ You are basically correct, but this is the start of a much deeper topic known as de rham cohomology, which you can read about in just about any smooth manifolds text. $\endgroup$ Apr 2 at 18:33
  • $\begingroup$ @CharlesHudgins Interesting! Thank you! Do you have any text on manifolds in mind that you would recommend for an undergraduate student? Just curious. $\endgroup$ Apr 2 at 18:37
  • $\begingroup$ "When we have holes in the domain of the vector field, do we have to check the circulation around each hole to make sure it is zero and then we can find its potential and state that it is integrable?" Not quite. The circulation (curl) of this example $\omega$ is zero in all of $\Omega$ but still it is not integrable because $\Omega$ is not simply connected. "If it is non-zero do we slice the domain and state that it is integrable on the restricted domain?" No. If the circulation is not zero slicing of the domain will never ever make the field integrable. $\endgroup$
    – Kurt G.
    Apr 2 at 19:14
  • $\begingroup$ @KurtG. The first part I explained myself, but having a slit plane i.e. in the above example makes it integrable. So, not "never ever". $\endgroup$ Apr 2 at 19:54
  • $\begingroup$ I have a physics background, so I like either Nakahara's classic text or The Geometey of Physics. $\endgroup$ Apr 2 at 19:56

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Edit: Furthermore after the slicing (removing any smooth curve that connects 0 to $\infty$ ) I claim that $\phi(x,y) = \arctan(\frac{y}{x})$ But it is discontinuous along the y-axis, so then it cannot be the potential function because it is not differentiable there. Am I making a mistake in this logic? Thanks!

Too long for a comment:

  • The function $\phi(x,y)=\arctan(\frac yx)$ is a bad choice for a potential function of $$\omega=\frac{-y\,dx+x\,dy}{x^2+y^2}$$ because this $\phi$ is undefined on the entire $y$-axis.

  • A better choice is the function $$ \psi(x,y)=\operatorname{sgn}(y)\arccos\left(\tfrac{x}{\sqrt{x^2+y^2}}\right) $$ which is defined on $\mathbb R^2\setminus\{(0,0)\}$ and discontinuous only on the negative $x$-axis. See Desmos.

  • This $\psi$ is differentiable on $\mathbb R^2$ without the negative $x$-axis.

  • The two argument arcus tangens $\arctan2(x,y)$ was also created to fix those problems.

  • Now think about what the properties of $\psi(x,y)$ have to do with the slicing of the domain $\mathbb R^2\setminus\{(0,0)\}$ on which the above $\omega$ was originally defined.

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  • $\begingroup$ So we have to simply remove 0>=x>-$\infty$ and y$\ne$0 from the domain of $\omega$, Do I understand this correctly? As then we have a simply connected domain $\endgroup$ Apr 5 at 8:53
  • $\begingroup$ @TeodorasPaura Right. Removing that ray prevents closed paths from crossing the line where $\psi$ is discontinuous. All closed line integrals in that sliced domain are zero. $\endgroup$
    – Kurt G.
    Apr 5 at 10:36

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