I am stuck with the following problem :
Let $F_1,F_2:\bf R^2 \to R$ be functions such that $$\begin{align*} F_1(x_1,x_2)=\frac{-x_2}{x_1^2+x_2^2} \\ F_1(x_1,x_2)=\frac{x_1}{x_1^2+x_2^2}\end{align*}$$ Then which of the following ARE correct?
$\displaystyle \frac{\partial{F_1}}{\partial{x_2}}=\frac{\partial{F_2}}{\partial{x_1}}$
$\exists\,\,$ a function $f \colon \bf R^2 \backslash \{(0,0)\} \to R$ such that $\displaystyle \frac{\partial{f}}{\partial{x_1}}=F_1$ and $\displaystyle \frac{\partial{f}}{\partial{x_2}}=F_2$
$\nexists \,\,$ a function $f \colon \bf R^2 \backslash \{(0,0)\} \to R$ such that $\displaystyle \frac{\partial{f}}{\partial{x_1}}=F_1$ and $\displaystyle \frac{\partial{f}}{\partial{x_2}}=F_2$
$\exists$ a function $f \colon \bf D \to R$ where $\bf D$ is the open disc of radius $1$ centered at $(2,0)$ which satisfies $\displaystyle \frac{\partial{f}}{\partial{x_1}}=F_1$ and $\displaystyle \frac{\partial{f}}{\partial{x_2}}=F_2$
I have only been able to show that option 1 is true and $\displaystyle \frac{\partial{F_1}}{\partial{x_2}}= \frac{\partial{F_2}}{\partial{x_1}}=\frac{-x_1^2+x_2^2}{(x_1^2+x_2^2)^2}$ .
But I am not sure about other options .Clearly, only one of the options 2,3 can hold (I do not know how to prove it.) Abou option 3, I have no idea.
Can someone please clarify by giving a detailed explanations . Thank you very much.
From the condition $\frac{\partial F_1}{\partial y}=\frac{\partial F_2}{\partial x}$ we know that $(F_1,F_2)$ has an antiderivative on every simply connected domain. This answers question 4.
The contour integral on a circle around the origin (say $x=\cos t$, $y=\sin t$) does not vanish, so there is no global antiderivative.
The local antiderivative function is $angle(x,y)+C$, the direction of the point from the origin. (Try to differentiate $\arctan\frac{y}{x}$.) This "angle" can be continued continously to angle domains but cannot be continued continously to the entire domain.