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PostPosted: Thu Aug 09, 2018 11:19 am 

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Let $\Omega=\mathbb{R^2}\smallsetminus\{(0,0)\}$ and $$\vec{F}(x,y)=-\frac{y}{x^2+y^2}\vec{i}+\frac{x}{x^2+y^2}\vec{j}$$ First $$\vec{\nabla}\times \vec{F}=0\,\vec{i}+0\,\vec{j}+\bigg(\frac{y^2-x^2}{(x^2+y^2)^2}-\frac{y^2-x^2}{(x^2+y^2)^2}\bigg)\vec{k}=\vec{0}$$ is not a sufficient condition for conservativeness . To show that the vector field $F$ is not conservative, we take the scalar function

$$f(x,y)=-\arctan\bigg(\frac{x}{y}\bigg)$$ which seems to be the potential function of the field $F$. If $$\int_C\vec{F}\cdot\mathrm{d}\vec{r}\neq f\big(x(\beta),y(\beta)\big)-f\big(x(\alpha),y(\alpha)\big)$$ $\big($where $r(t)$ is the parametrization of an arbitrary curve $c$ and $\alpha$ and $\beta$ are its starting and endpoint respectively$\big)$

then the fundumental theorem for line integrals is not satisfied and thus $F$ is not conservative.
I tried using $r(t)=t\vec{i}+t\vec{j}, \space t\in[\alpha,\beta]$, but it didn't work. What curve would be a better choice for $C$ and what's the deal with $\mathrm{rot}\,F$ being zero?

Also, could a quick explanation be that, since there's no function with the same domain as $\vec{F}$ whose gradient is $\vec{F}$ ,$\vec{F}$ is not conservative?


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PostPosted: Sat Aug 11, 2018 10:29 am 
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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.

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PostPosted: Sun Aug 12, 2018 11:01 am 
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andrew.tzeva wrote:
...I tried using $r(t)=t\vec{i}+t\vec{j}, \space t\in[\alpha,\beta]$, but it didn't work. What curve would be a better choice for $C$ and what's the deal with $\mathrm{rot}\,F$ being zero?..

Here is a 2nd solution, choosing an appropriate (closed) curve:

The line integral of $\overline{F}$ over the circle $\overline{c}(t)=(\cos{t},\sin{t})$, $t\in[0,2\pi]$, which is closed curve, is
\begin{align*}
\displaystyle\mathop{\oint}\limits_{C(\overline{0},1)}{\overline{F}\cdot d\overline{s}}&=\int_0^{2\pi}{\bigl(-\sin{t},\cos{t}\bigr)\cdot\Bigl(\frac{d}{dt}\cos{t},\,\frac{d}{dt}\sin{t}\Bigr)\,dt}\\
&=\int_0^{2\pi}{\bigl(-\sin{t},\cos{t}\bigr)\cdot\bigl(-\sin{t},\,\cos{t}\bigr)\,dt}\\
&=\int_0^{2\pi}{1\,dt}\\
&=2\pi\neq 0\,.
\end{align*} Therefore, $\overline{F}$ is not conservative.

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PostPosted: Mon Aug 13, 2018 3:01 pm 

Joined: Wed Nov 15, 2017 12:37 pm
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Thank you. The 2nd solution (with the direct counter-example) is much more helpful.


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PostPosted: Tue Aug 14, 2018 6:41 am 
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andrew.tzeva wrote:
Thank you. The 2nd solution (with the direct counter-example) is much more helpful.
Sure, in this case! But in general, to find a suitable curve isn't easy.

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