1
$\begingroup$

To show that a function $$F(x,y,z)=(f_1,f_2,f_3)$$ is a gradient/conservative field, is enough to show that $$\frac{\partial f_1}{\partial z} = \frac{\partial f_3}{\partial x} \ and \ \ \frac{\partial f_1}{\partial y} = \frac{\partial f_2}{\partial x} \ \ and \ \ \frac{\partial f_2}{\partial z} = \frac{\partial f_3}{\partial y} \ \ ????$$ To show that is NOT it's enough to show that one of this conditions are not true. But my question is: if these 3 conditions are true can i conclue that the function F is a gradient field?

$\endgroup$
1
  • $\begingroup$ if those three conditions are true then $\nabla \times F = 0$ $\endgroup$
    – user317176
    Commented Dec 27, 2016 at 16:06

2 Answers 2

3
$\begingroup$

It's a bit complicated/delicate:

  1. If the domain $D$ of the vector field $F$ is a non-empty, simply connected open set, then every $C^{1}$ vector field whose curl vanishes throughout $D$ is a gradient field in $D$. Particularly, if $F$ is defined in some rectangular box, or on a ball, or on all of $\mathbf{R}^{n}$, then the following are equivalent:

    (i) $\nabla \times F = 0$ throughout $D$.

    (ii) There exists a $C^{2}$ function $f$ in $D$ such that $\nabla f = F$.

  2. The vector field $$ F = \frac{(-y, x, 0)}{x^{2} + y^{2}} $$ is curl-free on the complement $D$ of the $z$-axis in $\mathbf{R}^{3}$. Because the integral of $F$ around the unit circle is $2\pi$, however, $F$ is not a gradient field on $D$.

    If $D'$ denotes the complement of an arbitrary closed half-plane containing the $z$-axis, then $F$ is a gradient field in $D'$. (!) For example, if $D'$ is the complement of the half-plane $x \leq 0$, $y = 0$, and if $\theta:D' \to (-\pi, \pi)$ denotes the branch of the cylindrical angle function, then $F = \nabla \theta$ in $D'$.

In summary, whether or not a particular curl-free vector field $F$ is gradient in a domain $D$ depends not merely on the component functions of $F$, but on the topology of $D$.

$\endgroup$
0
$\begingroup$

If the domain $D$ is simply connected, then the answer to your question is YES.

Added:- Since $\nabla ×F=\begin{vmatrix}\hat i&\hat j&\hat k\\\delta/\delta x&\delta/\delta y&\delta/\delta z\\f_1&f_2&f_3\end{vmatrix}=0$

$\implies \hat i(\delta f_3/\delta y-\delta f_2/\delta z)-\hat j(\delta f_3/\delta x-\delta f_1/\delta z)+\hat k(\delta f_2/\delta x-\delta f_1/\delta y)=0$

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .