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Assume a differential of the form $df(x,y) = X(x,y) \,dx + Y(x,y) \,dy$. If $\oint\ df=0$, it's easy to see that $\frac{\partial X}{\partial y} = \frac{\partial Y}{\partial x}$, which can be seen using Greene's theorem.

However, I want to show the stronger case that the line integral being $0$ implies $X = \frac{\partial f}{\partial x}$, and $Y = \frac{\partial f}{\partial y}$, making $df$ a total differential. Is this true, and how do you prove this?

Michael Hardy
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Mondo Duke
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  • Do you mean line integral equals 0 around any closed curve? If you take $\vec{F}(x,y)=\big<1,x^2\big>$ then $\int_{{r=1}}\vec{F}\cdot d\vec{r}=0$ but clearly $\vec{F}$ isn't conservative and hence has no potential function. – Matthew H. Sep 06 '20 at 04:58
  • Also, if you assume $df(x,y)=X(x,y)dx+Y(x,y)dy$ then $X(x,y)=f_x(x,y)$ and $f_y(x,y)=Y(x,y)$. We immediately get $$f_{xy}=\frac{\partial X}{\partial y}=\frac{\partial Y}{\partial x}=f_{yx}$$ by equating mixed partials. No need to use Green's Theorem. It can be shown that if $\int_{C}\vec{F}\cdot d\vec{r}=0$ around any closed curve $C$ then $\vec{F}=\vec{\nabla}f$ for some scalar valued $f$. – Matthew H. Sep 06 '20 at 05:14
  • Crossposted from https://physics.stackexchange.com/q/577833/2451 – Qmechanic Sep 06 '20 at 05:16
  • Thanks! If I didn't assume $df(x,y)$ was a total differential though, would Green's Theorem be the best way to prove it though? – Mondo Duke Sep 06 '20 at 05:18
  • By definition, a vector field $\vec{F}(x,y)$ is conservative if and only if $\vec{F}=\vec{\nabla}f$ for some scalar valued function $f$. This immediately implies the partial derivative condition you're looking for. So, if we can show the line integral of $\vec{F}$ around any closed curve $C$ equals zero implies that $\vec{F}$ is conservative, then we're done. You can see this construction here: https://math.stackexchange.com/questions/1316837/if-the-integral-of-a-vector-field-over-a-closed-curve-equals-zero-is-the-field – Matthew H. Sep 06 '20 at 05:31

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Maybe we can use Green's Theorem to prove the partial derivative condition you're looking for. Let $\vec{F}(x,y)=\big<M(x,y),N(x,y)\big>$ be vector field such that $$l(x,y):=N_x(x,y)-M_y(x,y)$$ defines a continuous map on $\mathbb{R}^2$. Assume that $\int_{C}\vec{F}\cdot d\vec{r}=0$ for any closed curve $C$. Without loss of generaliy assume there exists $(x_0,y_0)\in\mathbb{R}^2$ such that $l(x_0,y_0)>0$ for some $(x_0,y_0)\in \mathbb{R}^2$. By continuity there must exist an open set $O$ containing $(x_0,y_0)$ such that $l(x,y)>0$ for any $(x,y)\in O$. Take $Q$ to be a closed curve that's contained in $O$, is oriented counter$-$clockwise, and encloses some region $R$. Then by Green's Theorem $$\int_{Q}\vec{F}\cdot d\vec{r}=\int\int_{R}l(x,y)dA>0$$ which is a contradiction, so $l(x,y)=0$ everywhere.

Matthew H.
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