It is currently Mon Jan 21, 2019 9:16 am


All times are UTC [ DST ]




Post new topic Reply to topic  [ 3 posts ] 
Author Message
 Post subject: $\nabla\times\nabla f$
PostPosted: Sat Nov 19, 2016 6:04 pm 

Joined: Sat Nov 14, 2015 6:32 am
Posts: 156
Location: Melbourne, Australia
Here is something that caught my attention the other day.

Suppose you have in hand the function

$$f(x,y,z)=\frac{\sin^7\left(\frac{x}{y^2+z^2+1}\right)+e^{2xz^2}}{x^2+y^2+z^2+3}$$

Compute $\nabla\times\nabla f$ which is to say, the curl of the gradient of $f$ .

_________________
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$


Top
Offline Profile  
Reply with quote  

PostPosted: Sun Nov 20, 2016 2:16 pm 
Team Member
User avatar

Joined: Mon Nov 09, 2015 1:36 am
Posts: 454
Location: Ioannina, Greece
We'll prove something more general: If $f:U\subseteq{\mathbb{R}}^3\longrightarrow{\mathbb{R}}$ is a twice differentiable function in some open set $U\subseteq{\mathbb{R}}^{3}$, then $\nabla\times\big(\nabla {f}\big)\equiv \overline{0}$ in $U$.

Proof:
\begin{align*}
\nabla{f}&= \bigg(\frac{\partial f}{\partial x}\,,\; \frac{\partial f}{\partial y}\,, \;\frac{\partial f}{\partial z} \bigg)\,,\\
\nabla\times\big(\nabla {f}\big) &=\left|\begin{array}{ccc}
\overline{e}_x & \overline{e}_y &\overline{e}_z \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} &\frac{\partial}{\partial z} \\
\frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} & \frac{\partial f}{\partial z}\end{array}\right|\\
&=\bigg(\frac{\partial^2 f}{\partial y\partial z}-\frac{\partial^2 f}{\partial z\partial y}\,, \;\frac{\partial^2 f}{\partial z\partial x}-\frac{\partial^2 f}{\partial x\partial z}\,, \; \frac{\partial^2 f}{\partial x\partial y}-\frac{\partial^2 f}{\partial y\partial x}\bigg)\\
&=\bigg(\frac{\partial^2 f}{\partial y\partial z}-\frac{\partial^2 f}{\partial y\partial z}\,, \;\frac{\partial^2 f}{\partial z\partial x}-\frac{\partial^2 f}{\partial z\partial x}\,, \; \frac{\partial^2 f}{\partial x\partial y}-\frac{\partial^2 f}{\partial x\partial y}\bigg)\\
&=(0,0,0)\,. \end{align*}


P.S. By the same pattern it can be proved that if $\overline{f}:U\subseteq{\mathbb{R}}^3\longrightarrow{\mathbb{R}}^3$ is a twice differentiable function in some open set $U\subseteq{\mathbb{R}}^{3}$, then $\nabla\cdot\big(\nabla \times{\overline{f}}\big)\equiv 0$ in $U$, i.e. ${\rm{div}} \,({\rm{curl }}\,\overline{f}\,)\equiv 0$.

_________________
Grigorios Kostakos


Top
Offline Profile  
Reply with quote  

PostPosted: Wed Dec 07, 2016 12:55 am 

Joined: Sat Nov 14, 2015 6:32 am
Posts: 156
Location: Melbourne, Australia
Good evening,

well, since the denominator has no root we can conclude that $f$ is a vector field. That the curl is zero. :clap2:

_________________
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$


Top
Offline Profile  
Reply with quote  

Display posts from previous:  Sort by  
Post new topic Reply to topic  [ 3 posts ] 

All times are UTC [ DST ]


Mathimatikoi Online

Users browsing this forum: No registered users and 1 guest


You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot post attachments in this forum

Search for:
Jump to:  
Powered by phpBB® Forum Software © phpBB Group Color scheme created with Colorize It.
Theme created StylerBB.net