Question

How do you prove this @rational

Answer 1

@amistre64

Answer 2

\[\int\limits\limits \int\limits\limits_S ( f \nabla g - g \ \nabla f ) n dS = \int\limits\limits \int\limits\limits \int\limits\limits_E (f \nabla ^2 g - g \nabla ^2 f ) dV\]

Answer 3

im betting rational will have a better grasp on it :)

looks gradient tho

Answer 4

a change of dS to dV usually invokes a jacobian if memory serves

Answer 5

@rational ok so the proof shows as such, I thought I understood it but I honestly don't haha but maybe you can help me see what exactly is going on. \[\int\limits \int\limits _S ((f \nabla g) - (g \nabla f)) \cdot n dS\]
\[\int\limits \int\limits _ S ( f \nabla g) \cdot n dS - \int\limits \int\limits _S (g \nabla f) \cdot n dS\]
\[= \left( \int\limits \int\limits \int\limits _E f( \nabla^2 g) + \nabla f \cdot ( \nabla g) dV \right) - \left( \int\limits \int\limits \int\limits_E g( \nabla ^2 f) + \nabla g \cdot ( \nabla f) dV \right)\]
\[\implies \int\limits \int\limits \int\limits _ E f ( \nabla ^2 g) + \nabla \cdot g - g ( \nabla ^2 f) - \nabla g \ \cdot ( \nabla f) dV\]\[\implies \int\limits \int\limits \int\limits _E f ( \nabla ^2 g) - g( \nabla ^2 f ) dV\]

Answer 6

it seems they are using divergene thm at step 3 when changing double integral into triple

Answer 7

Yup, that's what I speculated!

Answer 8

I initially thought this was just proving the divergence theorem

Answer 9

\[\text{div}\left(f\nabla g\right) ~~=~~f( \nabla^2 g) + \nabla f \cdot ( \nabla g) \]


is that true ?

Answer 10

shouldn't be hard to check i guess...

Answer 11

replace \(\nabla g\) with \(\langle g_x, g_y, g_z\rangle \) and we can work the divergence but im pretty sure that result was mentioned somewhere in the textbook in previous pages...

Answer 12

Is that not the laplace operator?

Answer 13

it is a product of f and del g right

Answer 14

Yup

Answer 15

\[\nabla \cdot \nabla g = \nabla^2g\]
\[\nabla \cdot f\nabla g= ?\]
needs some work i think..

Answer 16

Yes, that makes sense, I was just thinking about \[div ( \nabla f ) = \nabla \cdot (\nabla f)\]

Answer 17

\[\nabla ^2 fg ~~or~~ f(\nabla^2 g)~~or~~g( \nabla ^2f)\]