Question

Help with multivariable calculus please.

Answer 1

Being F a vector field with continous partial derivatives, If rot(F)=0 and Div(F)=0 is F constant?

Answer 2

I tried using Gauss and stokes theorem but i dont see anything

Answer 3

rot(F) = curl(F)?

Answer 4

yep, sry xD here in spanish we call it rot

Answer 5

i think the solution should be find using those theorems

Answer 6

if the curl(F)=0 => F is a conservative vector field

Answer 7

yep, and if Div(F)=0 => F=curl(G) being G another vector field

Answer 8

it's been a while since i've dealt with this stuff so i'm a bit rusty. sorry

Answer 9

1) curl F = 0 is not the definitive test for a conservative field: http://mathinsight.org/path_dependent_zero_curl



2) i've had a look about for examples and \((2x,2y,−4z)\) has zero curl and divergence

Answer 10

hint:
if we apply this vector identity:
\[\Large \nabla \times \left( {\nabla \times {\mathbf{F}}} \right) = \left( {\nabla \cdot {\mathbf{F}}} \right)\nabla - {\nabla ^2}{\mathbf{F}}\]
using your condition, we have:
\[\Large {\mathbf{0}} = {\nabla ^2}{\mathbf{F}}\]

Answer 11

curl is tendancy to rotate and divergence is compressibility. if they're both 0, doesn't that just mean that an incompressible fluid is flowing in a way that it tends not to rotate?

Answer 12

@pgpilot326 +1

or its just not actually compressing

Answer 13

and does F is constant mean?

like 1,2,3 ?

Answer 14

yep

Answer 15

well we have an example so that is not true

Answer 16

yep thank you very much guys , i really appreciate the help

Answer 17

or think gravity!

zero curl, zero divergence.

Answer 18

we can show, that, from the condition:
\[\Large {\mathbf{0}} = {\nabla ^2}{\mathbf{F}}\]
the general field \( \large {\mathbf{F}}\) depends linearly on the cartesian coordinates \( \large x, \; y, \; z \)

Answer 19

yes and it doesnt have to be necessarily constant, thanks! :)

Answer 20

:)

Answer 21

@Michele_Laino
interesting

why did you set the triple product to zero ?

and what does \(\nabla \times ( \nabla \times\vec A)\) actually mean if we already know that \( \nabla \times\vec A = 0\)

Answer 22

because that nails it :-)

Answer 23

since:
\[\Large \nabla \times {\mathbf{F}} = {\mathbf{0}}\]

Answer 24

indeed. the curl is zero so can we use the bac cab rule for anything?

like 2 x 0 = 3 x 0
2 = 3

Answer 25

yes! I think so, since the vector equation above, it is an identity

Answer 26

furthermore, also this condition:
\[\Large \nabla \cdot {\mathbf{F}} = 0\]
holds

Answer 27

aaahhh! i think i am getting you.

so, ok, ....the gravitational field, inverse square but no curl no divergence....

i hate to think how you fit that into cartesian to get a linear.

Answer 28

I have developed, component by component the cndition:
\[\Large {\mathbf{0}} = {\nabla ^2}{\mathbf{F}}\]
using this other condition:
\[\Large \nabla \times {\mathbf{F}} = {\mathbf{0}}\]

Answer 29

condition*

Answer 30

yep, i get that bit

the triple product is zero because AxB is zero and .....

if div F is zero, the laplacian must also be zero

so we have a linear in x,y,z

i get the reasoning, sure. it's very good.

Answer 31

thanks!

Answer 32

you need a medal :-)

Answer 33

or 2

Answer 34

lol!

Answer 35

no worries! I'm happy so! thanks! @IrishBoy123

Answer 36

good!