Examples of conservative vector fields in the plane whose closed line integral is not zero

Question

In Vector Calculus, the classic example of a conservative (irrotational is more appropriate) vector field in $\mathbb{R}^2\backslash\{(0,0)\}$ whose line integral over a piecewise smooth closed path may not be $0$ is \begin{align} \tag{1} \mathbf F(x,y) = \frac{-y\mathbf i + x\mathbf j}{x^2+y^2}. \end{align}

I was trying to find other interesting examples of conservative irrotational vector fields $\mathbf F:\mathbb{R}^2\backslash\{(0,0)\} \longrightarrow \mathbb{R}^2$ for which that happens.

The best I could do was \begin{align} \tag{2} \mathbf F(x,y) = \frac{-f'(x)g(y)\mathbf i + f(x)g'(y) \mathbf j}{f(x)^2+g(y)^2}, \end{align} where $f,g:\mathbb{R} \longrightarrow \mathbb{R}$ are strictly increasing differentiable functions and $f(0)=g(0)=0$ (perhaps we can be more general).

Then, we can show that the line integral of $\mathbf F$ over any positively oriented, piecewise smooth, simple closed curve that surrounds the origin is $2\pi$.

For example, let $f(x)=x$ and $g(y)=y^3$, then we get $\mathbf F(x,y) = \dfrac{-y^3\mathbf i + 3xy^2 \mathbf j}{x^2+y^6}. \tag{3}$

I would like to know if there exist other simple/interesting examples, i.e., other vector fields that are not linear combinations of the one in the format provided above or other trivial extensions.

Edit: I have found a more general form: \begin{align} \tag{4} \mathbf F = \frac{g \nabla f - f \nabla g}{f^2+g^2}, \end{align} where $f,g:\mathbb{R}^2\longrightarrow \mathbb{R}$ are differentiable functions that only map to zero (simultaneously) at the origin.

Answer 1

The following is a bit informal but can be made rigorous using the pull-back formalism for differential forms. I also hope that, as Matteo's answer did, that his is a motiation to study these things further.

• Your setup $f(0)=0,\,g(0)=0$ (and nonzero elsewhere) leads to a one-form $$\tag{1} \boldsymbol{\omega}=\frac{(g\,\partial_x f-f\,\partial_x g)\,dx-(f\,\partial_y g-g\,\partial_yf)\,dy}{f^2+g^2} $$ on $\mathbb R^2\setminus\{0\}$ which is dual to your vector field $\mathbf{F}$ from (4) and closed, $d\boldsymbol{\omega}=0\,,$ but not exact. Writing $$\tag{2} \boldsymbol{\omega}=\frac{g\,(\partial_x f\,dx+\partial_y f\,dy)-f(\partial_xg\,dx+\partial_yg\,dy)}{f^2+g^2} $$ and using $$\tag{3} df=\partial_x f\,dx+\partial_y f\,dy\,,\quad\quad\quad dg=\partial_xg\,dx+\partial_yg\,dy $$ we get $$\tag{4} \boldsymbol{\omega}=\frac{g\,df-f\,dg}{f^2+g^2}\,. $$ It is now not surprising that $\boldsymbol{\omega}$ is the pull-back of the irrotational vortex $$\tag{5} \boldsymbol{\omega}_0=\frac{y\,dx-x\,dy}{x^2+y^2}\, $$ under the map $\Phi:(x,y)\mapsto (f,g)$ which keeps the origin fixed and therefore maps $\mathbb R^2\setminus\{0\}$ to a subset of it.

• In polar coordinates we know that $\boldsymbol{\omega}_0=d\theta$ and the map becomes $$\tag{6} \Phi:(r,\theta)\mapsto (\rho,\varphi)\,,\quad\rho=\sqrt{f^2+g^2}\,,\quad\varphi=\arctan(g/f)\,,\quad f=\rho\cos\varphi\,,\quad g=\rho\sin\varphi\,. $$ It is not surprising that $\boldsymbol{\omega}=d\varphi\,.$ Since $d\varphi=\partial_r\varphi\,dr+\partial_\theta\varphi\,d\theta$ we can write $$\tag{7} \boldsymbol{\omega}=\partial_r\varphi\,dr+\partial_\theta\varphi\,d\theta $$ which is (1) in polar coordinates.

• We know from $\boldsymbol{\omega}_0=d\theta$ that, for a closed curve in $\mathbb R^2\setminus\{0\}\,,$ $$\tag{8} \int_\gamma\boldsymbol{\omega_0}=2\pi k_0 $$ where $k_0$ is the winding number of $\gamma$ around the origin. When $\gamma$ is closed then $\Phi\circ\gamma$ is closed . Therefore, we know from $\boldsymbol{\omega}=d\varphi$ that $$\tag{9} \int_{\Phi\circ\gamma}\boldsymbol{\omega}=2\pi k $$ where $k$ is the winding number of the curve $\Phi\circ\gamma$ around the origin. See also this post.

• Replacing $\Phi\circ\gamma$ by $\gamma$ it follows now that $$\tag{10} \int_{\color{red}{\gamma}}\boldsymbol{\omega}=2\pi k_{\color{red}{0}} $$ that is: for any closed curve $\gamma$ the integral of $\boldsymbol{\omega}$ is identcial to that of the irrotational vortex $\boldsymbol{\omega}_0\,.$ In other words, in that example, Matteo's constant $c$ is one.

• Since for any closed curve in $\mathbb R^2\setminus\{0\}$ the integral of $\boldsymbol{\omega}-\boldsymbol{\omega}_0$ is zero it follows now from de Rham cohomology that there is a function $h$ defined on all of $\mathbb R^2\setminus\{0\}$ such that $$\tag{11}\boxed{\phantom{\Big|} \boldsymbol{\omega}=\boldsymbol{\omega}_0+dh\,.\quad} $$ That $\boldsymbol{\omega}$ is therefore just a trivial modification of the irrotational vortex $\boldsymbol{\omega}_0\,.$

Answer 2

They are substantially all equal to the first example times a constant (call it $X_0$), in particular any conservative vector field $X=cX_0+\nabla f$. Where $c$ is some constant. Observe that the second vector field has a potential, so it has no effect on the computation on closed line integral. This you can prove yourself by making sure that $X $ and $X_0$ are evaluated equally on the same closed loop, by adjusting the constant.

If instead of $\mathbb{R}^2\setminus 0$ you take the plane minus two points you can prove that a generic conservative vector field can be written as $X=c_1\cdot X_1 + c_2 \cdot X_2+ \nabla f$, where $X_1,X_2$ are conservative and "rotate around" one of the two points. This fact greatly generalize in the De Rham Cohomology, which precisely tells you how many hole you have on a space by counting the conservative forms, which do not differ by a potential.