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Theorem 15.7 in Larson's Calculus 9th Edition describes conditions for conservative vector fields over a region R. In the Theorem it is stated that R must be "open connected" but does not state that the region must be "simply" connected. In attempting to understand why the classic example $F=\left<\frac{y}{x^2+y^2},\frac{-x}{x^2+y^2}\right>$ for a loop around the origin doesn't satisfy the theorem, it seems to me that the "hole" at the origin is the problem, but I don't see why a donut shape in $R^2$ isn't connected. I only see that it isn't "simply connected", but this isn't stated in the theorem. Do I misunderstand the meaning of "open connected", or should the theorem state "simply connected? Theorem as stated in Larson, 9th, p1088

Tim
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  • Which of the three conditions do you think this function satisfies? All the theorem says is that a function either satisfies all $3$ conditions or none of them. (I'm not making an assertion, just trying to understand the question.) – saulspatz Apr 10 '20 at 22:37
  • Thank you for your follow-up. I believe that the function has continuous first partial derivatives in a donut shaped open region which omits the origin. If this is correct, the theorem implies that the integral should yield 0 for any closed loop. However, a closed loop containing the origin does not yield 0, implying that one of the theorem's hypotheses is not satisfied. I'm trying to figure out why the theorem does not apply. Thank you again. – Tim Apr 10 '20 at 22:54
  • It's been a long time since I did this stuff, and I may be wrong, but I think you're missing something. You need $F=\nabla f$ for some $f$. That is, an exact form is closed. If the domain is simply-connected, then you have that a closed form is exact. (I think. It's been many, many years.) See https://math.stackexchange.com/questions/1072852/proof-that-this-differential-form-is-not-exact – saulspatz Apr 10 '20 at 23:01
  • I believe that $f(x,y)=arctan(\frac{x}{y})$ is the needed $f$, It, too, fails at the origin but (I just noticed) it fails along the axes as well. Hmm.... – Tim Apr 10 '20 at 23:32
  • https://math.berkeley.edu/~ehallman/spring-2016/worksheets/manyTrueFalse-sols.pdf. #37 here seems to suggest that the requirement is simply connected, not just connected. – Tim Apr 11 '20 at 02:59
  • This, too, suggests the region must be simply connected. Is Larson wrong? https://opentextbc.ca/calculusv3openstax/chapter/conservative-vector-fields/ – Tim Apr 11 '20 at 03:08
  • The last link you give states, "a vector field F on an open and connected domain is conservative if and only if it is independent of path," which is just what Larson says. Simple connectedness is required in the other direction: one a simply-connected domain, a closed one-form is exact. See the first sentence of https://www.csun.edu/~vcmth02i/Forms.pdf – saulspatz Apr 11 '20 at 03:23

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The responses by saulspatz clarified my misunderstanding. The theorem I quoted provides sufficient conditions for the three statements to be equivalent but not (as I mistakenly read) sufficient to claim any one of these to be true. A correct reading would be that, if the conditions are met, either all three are false, or all three are true.

The key step is to determine if one of them is true, or false. In most cases, I would imagine, the one to consider is #1 (conservative). The definition of a conservative field F is one for which there exists a differentiable function whose gradient is F. The $f(x,y)$ I proposed is not differentiable at the origin, and indeed does not exist along the x-axis. If one chooses a region in the plane where it is differentiable, then F has the properties of a conservative field.

For the particular exercise, it is shown that despite curl(F) being zero wherever defined, the line integral for a closed loop containing the origin and passing through the x-axis does not yield 0 as might be expected.

Tim
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