In one of my Calculus III classes the professor presented the following vector field, defined on the set $S$ of all points $(x,y) \neq(0,0)$ : $$ \bbox[6px,border:1px solid black] { \vec{F}=\frac{-y}{x^2+y^2}\vec{i} + \frac{x}{x^2+y^2}\vec{j} } $$

Although $\vec{F}$ is not a gradient on $S$, it is a gradient on the set $\Omega=\mathbb{R}^2-\left\{(x,y)\,|\,y=0, \, x\geq0\right\}$, i.e, all points in the xy-plane except those on the positive x-axis. In the class, my professor wrote a potential function for $\vec{F}$ on $\Omega$ using the following functions:

$$ \bbox[6px,border:1px solid black] { \begin{alignat}{0} \Psi_1(x,y)=-\arctan\left(\frac x y \right), \,y\neq0 &\text{and} &\Psi_2(x,y)=\arctan\left(\frac y x \right), \,x\neq0 \end{alignat} } $$

The first is a potential of $\vec F$ on $\Omega^+=\{(x,y)\in\mathbb{R}^2\,|\,y>0\}$ and on $\Omega^-=\{(x,y)\in\mathbb{R}^2\,|\,y<0\}$.

The last one is a potential of $\vec F$ too, but on $\Omega_+=\{(x,y)\in\mathbb{R}^2\,|\,x>0\}$ and on $\Omega_-=\{(x,y)\in\mathbb{R}^2\,|\,x<0\}$

The potential function wrote by my professor: Class Notes $$ \bbox[8px,border:1px solid black] { \phi (x,y) = \begin{cases} \Psi_1(x,y) & \text{if $(x,y) \in R_1$} \\[2ex] \Psi_2(x,y)+\frac\pi 2 & \text{if $(x,y) \in R_2$} \\[2ex] \Psi_1(x,y)+2\pi & \text{if $(x,y) \in R_3$} \\ \end{cases} } $$ (Note: I remember that my professor started with the argument that since $\nabla\Psi_1=\vec F = \nabla\Psi_2$, then $\Psi_1-\Psi_2=k, \,k \in \mathbb{R}$)

My doubts:

  • I've tried a lot, but still have not figured out how to build $\phi (x,y)$ using $\Psi_1$ and $\Psi_2$. So, how to find this specific potential function of $\vec F$ ?
  • Why can't I simply write potential functions of $\vec{F}$ in the following ways $$ \phi_A (x,y) = \begin{alignat}{0} \begin{cases} \Psi_1(x,y) & \text{if $(x,y) \in R_1$} \\[2ex] \Psi_2(x,y) & \text{if $(x,y) \in R_2$} \\[2ex] \Psi_1(x,y) & \text{if $(x,y) \in R_3$} \\ \end{cases} &\text{or} &\phi_B(x,y)= \begin{cases} \Psi_1(x,y) & \text{if $y>0$} \\[2ex] 0 & \text{if $y=0$ and $x<0$} \\[2ex] \Psi_1(x,y) & \text{if $y<0$} \\ \end{cases} \end{alignat} $$ since $\nabla\phi_A=\nabla\phi_B=\vec F$ on $\Omega$ ?
  • 1
    $\begingroup$ Exactly how are $R_1$, $R_2$ and $R_3$ defined? $\endgroup$
    – md2perpe
    May 12, 2019 at 14:37
  • 1
    $\begingroup$ Hint: What values do $\Psi_1$ and $\Psi_2$ have on the four halfaxes? Are $\phi_A$ and $\phi_B$ continuous? $\endgroup$
    – md2perpe
    May 12, 2019 at 14:39
  • 1
    $\begingroup$ Using polar coordinates $(r,\alpha)$, where $\alpha$ is the counterclockwise angle from the positive x-axis, we have $\phi=\alpha - \pi/2$ on all of $\Omega.$ The need to split $\Omega$ into regions occurs when we want to express $\phi$ in terms of $x$ and $y$. $\endgroup$
    – md2perpe
    May 12, 2019 at 15:32
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    $\begingroup$ Since they aren't continuous, they are not potentials for all of $\Omega$. $\endgroup$
    – md2perpe
    May 12, 2019 at 15:38
  • 1
    $\begingroup$ In that case you have one patch covering all of the domain, which excludes origin. $\endgroup$
    – md2perpe
    May 12, 2019 at 16:17


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