Consider a vector field $F: \mathbb{R}^2/\{\vec{0}\} \to \mathbb{R}^2$


$\mathbb{R}^2/\{\vec{0}\}$ is not a simply connected domain . Suppose also that $F$ is irrotational.

Can the following procedure be valid to prove that $F$ is conservative?

  • I find one closed curve $\gamma_1$ that goes around the origin and such that $$\oint_{\gamma_1} F \cdot ds=0$$ (assuming that such curve exists).

  • If I consider any other closed curve $\gamma_2$ that goes around the origin, then I can use Green Theorem and see the union of $\gamma_2 \cup \gamma_1$ as the border of a regular domain in $\mathbb{R}^2$. Let's say this domain is $D$: then its border is $\partial D=\gamma_1 \cup \gamma_2$. I have to choose the positive orientation for $\partial D$ but suppose that $\gamma_1$ and $\gamma_2$ already have the right orientations in order to have $+\partial D$ (I think that this is not restrictive). Therefore I can finally say that:

    $$\oint_{+\partial D} F \cdot ds=\oint_{\gamma_1 \cup \gamma_2 } F \cdot ds=\oint_{\gamma_1 } F \cdot ds+\oint_{\gamma_2 } F \cdot ds= \int \int_{D} \partial_{x} F_2- \partial_{y} F_1=0$$ $$\implies \oint_{\gamma_1 } F \cdot ds=-\oint_{\gamma_2 } F \cdot ds$$ Since $F$ is irrotational. But $\oint_{\gamma_1} F \cdot ds=0$, therefore

$$\oint_{\gamma_2 } F \cdot ds=0$$

This is valid for any curve $\gamma_2$ that goes around the origin. (indipendently from the fact that $\gamma_1$ and $\gamma_2$ are indeed oriented in the proper way to have $+\partial D$).

  • For any curve $\gamma_3$ that does not go around the origin it is possible to find a subset $A \subset \mathbb{R}^2/\{\vec{0}\}$ simply connected such that $\gamma_3 \subset A$, and $F$ is irrotational, $F$ is conservative in $A$ and $$\oint_{\gamma_3} F \cdot ds=0$$
  • So in conclusion we have $$\oint_{\gamma} F \cdot ds=0 \,\,\, \forall \gamma \subset \mathbb{R}^2$$ and $F$ is conservative.

Is the previous proof valid or are there any mistakes?

I think it should be valid, but it looks strange because if it is, I would conclude that $F$ is conservative looking at one only curve $\gamma_1$, which seems a bit reductive I think?


This is a very good question. Your conjecture is correct:

Theorem. If the field $F=(P,Q)$ defined in $\Omega:={\mathbb R}^2\setminus\{0\}$ has vanishing curl: $Q_x-P_y\equiv0$, and if $\int_{\gamma_*}F\cdot dz=0$ for a single generating cycle $\gamma_*$, then $F$ is conservative.

In order to prove this theorem you have to prove that $\int_\gamma F\cdot dz=0$ for all closed curves $\gamma\subset\Omega$. For this you cannot directly appeal to Green's theorem, because such a curve $\gamma$ may have selfintersections, or go around the origin several times, etc. In any case, you cannot assume that $\gamma$ and $\gamma_*$ together form the boundary of a nice domain $D$.

Instead we proceed by constructing first a function $f:\>\Omega\to{\mathbb R}$ with $\nabla f=F$ on $\Omega$, i.e., a potential of $F$. For this construction we shall only need Green's theorem for rectangles. For simplicity assume that you can choose four points $z_i=(x_i,y_i)$ $(i\ {\rm mod}\ 4)$ on your "witness curve" $\gamma_*$, such that $z_i$ is in the open quadrant $Q_i$, and such that the part $\gamma_i$ of $\gamma$ connecting $z_{i-1}$ with $z_i$ completely lies in the halfplane containing $Q_{i-1}\cup Q_i$. Choose an arbitrary $c_0\in{\mathbb R}$ and define a function $f_R$ in the right halfplane $H_R$ by $$f_R(x,y):=c_0+\int_{x_0}^x P(x',y_0)\>dx'+\int_{y_0}^y Q(x,y')\>dy'\ .$$ It follows that ${\partial f_R\over \partial y}=Q$ on $H_R$. Applying Green's theorem to the obvious rectangle we see that we also have $$f_R(x,y)=c_0+\int_{y_0}^y Q(x_0,y')\>dy'+\int_{x_0}^x P(x',y)\>dx'\ ,$$ and this shows that ${\partial f_R\over \partial x}=P$ on $H_R$ as well. It follows that $$\nabla f_R(x,y)=F(x,y)\qquad\bigl((x,y)\in H_R\bigr)\ .$$ We have $f(z_0)=c_0$ and, using a standard fact about potentials, $$f(z_1)=c_0+\int_{\gamma_1} F\cdot dz=:c_1\ .$$ In the same way we construct a potential $f_U$ in the upper halfplane requiring $f_U(z_1)=c_1$. Since $f_U(z_1)=f_R(z_1)$ we can conclude that $f_U$ and $f_R$ coincide in the intersection $Q_1$ of their domains; furthermore $$f_U(z_2)=c_1+\int_{\gamma_2} F\cdot dz=:c_2\ .$$ In the same way we construct a potential $f_L$ in the left halfplane requiring $f_L(z_2)=c_2$, and then a potential $f_B$ in the bottom halfplane requiring $$f_B(z_3)=f_L(z_3)=c_2+\int_{\gamma_3} F\cdot dz=:c_3\ .$$ The function $f_B$ is also defined in the fourth quadrant $Q_4$,where we already have $f_R$. Since $\nabla f_B-\nabla f_R=F-F=0$ in $Q_4$ we know that $f_B$ and $f_R$ differ by a constant there. We therefore compute $$f_B(z_0)=c_3+\int_{\gamma_4}F\cdot dz=\ldots= c_0+\int_{\gamma_*}F\cdot dz=c_0\ .$$ This shows that in fact $f_B=f_R$ in $Q_4$, so that $f_R$, $F_U$, $f_L$, $f_B$ together constitute a globally defined potential function $f:\>\Omega\to{\mathbb R}$ of $F$. Therewith it is proven that $F$ is conservative on $\Omega$.

  • $\begingroup$ Thanks a lot for this great answer! Can I extend this to the three dimensional case? That is: if $F: \Omega \to \mathbb{R}^3$, with $\Omega=\{(x,y,z) \in \mathbb{R}^3 : x^2+y^2\neq0 \}$ has vanishing curl and $\oint_{\gamma^*} F \cdot ds =0$ for a single cycle $\gamma^*$ that goes around the $z$ axis then $F$ is conservative? $\endgroup$ – Gianolepo Nov 16 '16 at 10:48
  • $\begingroup$ In this simple case the same proof works. $\endgroup$ – Christian Blatter Nov 16 '16 at 12:12

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