Proof of Vector Identity Using Standard Identities | C^2 Scalar Functions

[madachi]

Homework Statement

Let f(x,y,z), g(x,y,z), h(x,y,z) be any C^2 scalar functions. Using the standard identities of vector analysis (provided in section 2 below), prove that

\nabla \cdot ( f \nabla g \times \nabla h ) = \nabla f \cdot ( \nabla g \times \nabla h)

Homework Equations

Note: The identities below require f,g,F,G to be suitable differentiable, either order C^1 or C^2.

1. \nabla (f+g) = \nabla f + \nabla g
2. \nabla (\lambda f) = \lambda \nabla f, where \lambda is a constant
3. \nabla (fg) = f \nabla g + g \nabla f
4. \nabla (\frac{f}{g}) = \frac{g \nabla f - f \nabla g}{g^2}
5. \nabla \cdot (F+G) = \nabla \cdot F + \nabla \cdot G
6. \nabla \times (F+G) = \nabla \times F + \nabla \times G
7. \nabla \cdot (fF) = f \nabla \cdot F + F \cdot \nabla f
8. \nabla \cdot (F \times G) = G \cdot (\nabla \times F ) - F \cdot (\nabla \times G)
9. \nabla \cdot (\nabla \times F) = 0
10. \nabla \times (fF) = f \nabla \times F + \nabla f \times F
11. \nabla \times (\nabla f) = 0
12. {\nabla}^2 (fg) = f{\nabla}^2 g + g{\nabla}^2 f + 2 \nabla f \cdot \nabla g
13. \nabla \cdot (\nabla f \times \nabla g) = 0
14. \nabla (f \nabla g - g \nabla f) = f {\nabla}^2 g - g {\nabla}^2 f

The Attempt at a Solution

Using identity 8,

\nabla \cdot ( f \nabla g \times \nabla h ) = \nabla h \cdot (\nabla \times (f \nabla g)) - (f \nabla g) \cdot (\nabla \times (\nabla h))

One of the terms on RHS, \nabla \times (\nabla h) = 0 by identity 11.

So the equation reduces to

\nabla \cdot ( f \nabla g \times \nabla h ) = \nabla h \cdot (\nabla \times (f \nabla g))

I'm stuck here. There is no identity that I can use to further simplify this to the one required. (from what I can see, or am I wrong?) How do we proceed?

Thanks!

 

[tiny-tim]

Welcome to PF!

Hi madachi! Welcome to PF!

Looking at the RHS of the answer, you need to separate the f from the g and h, and 8. won't do that.

But 7. will, so try 7. ​

 

[Mark44]

Might be a naive question, but is there an f missing on the left side in this equation?
\nabla \cdot ( f \nabla g \times \nabla h ) = \nabla f \cdot ( \nabla g \times \nabla h)

If so, here's the corrected version.
\nabla \cdot ( f \nabla g \times \bold{f} \nabla h ) = \nabla f \cdot ( \nabla g \times \nabla h)

 

[madachi]

Hi madachi! Welcome to PF!

Looking at the RHS of the answer, you need to separate the f from the g and h, and 8. won't do that.

But 7. will, so try 7. ​

How do we use 7 ? Because the original equation doesn't look like "the form" of identity 7. Could you show me the first step?

Thanks!

Might be a naive question, but is there an f missing on the left side in this equation?
\nabla \cdot ( f \nabla g \times \nabla h ) = \nabla f \cdot ( \nabla g \times \nabla h)

If so, here's the corrected version.
\nabla \cdot ( f \nabla g \times \bold{f} \nabla h ) = \nabla f \cdot ( \nabla g \times \nabla h)

No there isn't an f missing on the left side in the equation.

Thanks.

 

[tiny-tim]

Hi madachi!

(have a nabla: ∇ )

f is f, and F is ∇g x ∇h

 

[madachi]

Hi madachi!

(have a nabla: ∇ )

f is f, and F is ∇g x ∇h

Thanks! Using what you suggest,

So

\nabla \cdot ( f \nabla g \times \nabla h)
= f \nabla \cdot (\nabla g \times \nabla h ) + ( \nabla g \times \nabla h ) \cdot \nabla f

Then we use identity 8 to show that the vector field on the first term equal to 0 right? And the second term on RHS is just the identity that we are required to show?

Thanks!

 

[tiny-tim]

Hi madachi!

(just got up :zzz: …)

… Then we use identity 8 to show that the vector field on the first term equal to 0 right?

Not sure how you get it from 8 …

how about 13? ​