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I would like to check if the differential form $w = y^2xdx + ydy$ is exact.

If $w$ would be exact we would have a function $f$ with $\partial f/\partial y =y$, thus $f = 1/2 y^2 +c$. $f$ would need to satisfy $\partial f/\partial x = y^2x$ as well, so $c = 1/2 y^2 x^2 + c'$, where $c'$ may depend on $y$. Now differentiating our found expression for $f$ with respect to $y$ again we could say how $c'$ must look like, since we know that the differentiative has to be $y$ once again, and so on.

I am struggling to get this argument to an end and to conclude, that $w$ cannot be exact. Any ideas to finish this?

As an alternative I think the following should work:

$w$ is not exact, since $w$ is not closed:

$\partial w_1 /\partial y = 2yx \neq 0 = \partial w_2 / \partial x$ (for $x,y \neq 0$).

Is this correct?

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    $\begingroup$ The second approach looks perfect to me. $\endgroup$
    – lisyarus
    Aug 1, 2016 at 11:56

3 Answers 3

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Your second approach is correct. Combining with the first: If $w= a(x,y)dx + b(x,y) dy$ is exact, then there is $f$ so that $\partial_x f =a$ and $\partial_y f =b$. When $a$ and $b$ are $C^1$ we may take a cross derivative (which commute): $$ \partial_x(\partial_y f) = \partial_y(\partial_x f) \Leftrightarrow \partial_x b = \partial_y a $$ The argument works in any dimension.

Note that: Your domain is ${\Bbb R}^2$ which is simply connected and the condition on equal partial derivates is equivalent to being exact. When the domain is not simply connected this is no longer so (used e.g. in the case of the Cauchy formula in complex analysis).

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  • $\begingroup$ I know that exact always implies closed. So you are saying that if the domain of the differential form is simply connected, then we can conclude exact from closed as well? (Just want to reassure myself) $\endgroup$
    – CHwC
    Aug 1, 2016 at 15:20
  • $\begingroup$ Yes, indeed. The point is that you define $f$ by integrating along a path, and simple connectivity implies that the definition does not depend upon the path! So it gives a consistent global definition of $f$. $\endgroup$
    – H. H. Rugh
    Aug 1, 2016 at 15:31
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For a form to be exact, it is necessary to be closed, which is often easy to check, as you did correctly: it is not closed, and hence not exact, except in $0$.

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After $$f(x,y)=\frac{y^2}2+c(x)$$ you can write

$$\frac{\partial}{\partial x}(x,y)=\frac d{dx}c(x)=d(x)=y^2x$$ which is inconsistent.

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  • $\begingroup$ What do you mean with "inconsistent"? I think this is exactly the part which I am missing in my thoughts. $\endgroup$
    – CHwC
    Aug 1, 2016 at 15:25

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