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 Post subject: $\nabla\times\nabla f$Posted: Sat Nov 19, 2016 6:04 pm

Joined: Sat Nov 14, 2015 6:32 am
Posts: 159
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 $f$ .

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

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 Post subject: Re: $\nabla\times\nabla f$Posted: Sun Nov 20, 2016 2:16 pm
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Joined: Mon Nov 09, 2015 1:36 am
Posts: 460
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. ${\rm{div}} \,({\rm{curl }}\,\overline{f}\,)\equiv 0$.

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Grigorios Kostakos

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 Post subject: Re: $\nabla\times\nabla f$Posted: Wed Dec 07, 2016 12:55 am

Joined: Sat Nov 14, 2015 6:32 am
Posts: 159
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.

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

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