mathimatikoi.org https://www.mathimatikoi.org/forum/ 

Show that a vector field is not conservative (example) https://www.mathimatikoi.org/forum/viewtopic.php?f=23&t=1322 
Page 1 of 1 
Author:  andrew.tzeva [ Thu Aug 09, 2018 11:19 am ] 
Post subject:  Show that a vector field is not conservative (example) 
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 $$\vec{\nabla}\times \vec{F}=0\,\vec{i}+0\,\vec{j}+\bigg(\frac{y^2x^2}{(x^2+y^2)^2}\frac{y^2x^2}{(x^2+y^2)^2}\bigg)\vec{k}=\vec{0}$$ 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? 
Author:  Grigorios Kostakos [ Sat Aug 11, 2018 10:29 am ] 
Post subject:  Re: Show that a vector field is not conservative (example) 
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. 
Author:  Grigorios Kostakos [ Sun Aug 12, 2018 11:01 am ] 
Post subject:  Re: Show that a vector field is not conservative (example) 
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. 
Author:  andrew.tzeva [ Mon Aug 13, 2018 3:01 pm ] 
Post subject:  Re: Show that a vector field is not conservative (example) 
Thank you. The 2nd solution (with the direct counterexample) is much more helpful. 
Author:  Grigorios Kostakos [ Tue Aug 14, 2018 6:41 am ] 
Post subject:  Re: Show that a vector field is not conservative (example) 
andrew.tzeva wrote: Thank you. The 2nd solution (with the direct counterexample) is much more helpful. Sure, in this case! But in general, to find a suitable curve isn't easy.

Page 1 of 1  All times are UTC [ DST ] 
Powered by phpBB® Forum Software © phpBB Group https://www.phpbb.com/ 