#2

Post
by **Grigorios Kostakos** » Sat Aug 11, 2018 10:29 am

First we write down a useful theorem: If a continuously differentiable vector field $\overline{F}:U\subseteq{\mathbb{R}}^n\longrightarrow{\mathbb{R}}^n\,,$ where $U$ is open, is conservative, then, for every $\overline{x}\in U$, the Jacobian matrix ${\bf{D}}\overline{F}(\overline{x})$ of $\overline{F}$ is symmetric.

Note that the condition "${\bf{D}}\overline{F}(\overline{x})$ is symmetric" is necessary, but not sufficient. It becomes sufficient, iff the set $U$ is star shaped domain.

Let's go to the example: The Jacobian matrix of $\overline{F}$ is symmetric, but ${\mathbb{R}}^2\setminus\{(0,0)\}$ is not star shaped domain. So, we can't conclude that $\overline{F}$ is conservative (or nonconservative ).

We suppose that $\overline{F}$ is conservative. Then there exists a continuously differentiable function $\varphi:{\mathbb{R}}^2\setminus\{{(0,0)}\}\longrightarrow{\mathbb{R}}$, such that

\[{\rm{grad}}\,{\varphi}(x,y)=\overline{F}(x,y)\quad\Leftrightarrow\quad \displaystyle\Bigl({\frac{\partial}{\partial x}\varphi(x,y),\,\frac{\partial}{\partial y}\varphi(x,y)}\Bigr)=\Bigl({-\frac{y}{x^2+y^2},\,\frac{x}{x^2+y^2}}\Bigr)\,.\] We have

\begin{align*}

\displaystyle\frac{\partial}{\partial x}\varphi(x,y)=-\frac{y}{x^2+y^2}\quad&\Rightarrow\quad \varphi(x,y)=\int-\frac{y}{x^2+y^2}\,dx\\

&\Rightarrow\quad \varphi(x,y)=-\arctan\big(\tfrac{x}{y}\big)+g(y)\\

\frac{\partial}{\partial y}\varphi(x,y)=\frac{\partial}{\partial y}\Bigl({-\arctan\big(\tfrac{x}{y}\big)+g(y)}\Bigr)\quad&\Rightarrow\quad\frac{x}{x^2+y^2}=\frac{x}{x^2+y^2}+\frac{d}{dy}\,g(y)\nonumber\\

&\Rightarrow\quad g(y)=c\,,

\end{align*} where $c$ constant. So, all the possible functions are the functions

\[\varphi(x,y)=c-\arctan\big(\tfrac{x}{y}\big)\,,\] of which none is continuously differentiable in ${\mathbb{R}}^2\setminus\{{(0,0)}\}$ (in fact, these functions are not defined in $\big\{(x,y)\in{\mathbb{R}}^2\;|\; y=0\big\}$). So, does not exist an antiderivative (potential) of $\overline{F}$ and, therefore, $\overline{F}$ is not conservative.

**edit:12:00, 12/8/2018. **Corrected solution.

Grigorios Kostakos