Vector Analysis Problem Involving Divergence

In summary, by building a vector field as a linear combination of the gradient of the scalar functions f and g, it is possible to write the given expression as the divergence of this vector field. This approach simplifies the problem and allows for the identification of the vector function in terms of f and g.
  • #1
MyName
10
0

Homework Statement


[/B]
Let [itex]f[/itex] and [itex]g[/itex] be scalar functions of position. Show that:
[tex]\nabla f \cdot \nabla(\nabla ^2 g)-\nabla g \cdot \nabla(\nabla ^2f)[/tex]
Can be written as the divergence of some vector function given in terms of [itex]f[/itex] and [itex]g[/itex].

Homework Equations


[/B]
All the identities given at https://en.wikipedia.org/wiki/Vector_calculus_identities, I suppose. Especially relevant would be the second derivative and divergence identites. Also, [itex]\nabla ^2 =\nabla \cdot \nabla[/itex]

The Attempt at a Solution


[/B]
After considerable time messing around with various vector identites, I've been able to show the above is equivalent to:

[tex]\nabla \cdot (f \nabla (\nabla ^2 g)-g\nabla (\nabla ^2 f))+g(\nabla \cdot \nabla(\nabla ^2 f))-f(\nabla \cdot \nabla(\nabla ^2 g))[/tex]

This is painfully close to the result I want, but I can't seem to show that the second and third terms either cancel or are themselves a divergence. I'd really like any hints, and can provide more detail as to the specific identities and manipulations I've used thus far if needed, thanks.
 
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  • #2
Have you tried working it from the other end starting with the answer or is that an unknown?
 
  • #3
MyName said:

Homework Statement


[/B]
Let [itex]f[/itex] and [itex]g[/itex] be scalar functions of position. Show that:
[tex]\nabla f \cdot \nabla(\nabla ^2 g)-\nabla g \cdot \nabla(\nabla ^2f)[/tex]
Can be written as the divergence of some vector function given in terms of [itex]f[/itex] and [itex]g[/itex].

Homework Equations


[/B]
All the identities given at https://en.wikipedia.org/wiki/Vector_calculus_identities, I suppose. Especially relevant would be the second derivative and divergence identites. Also, [itex]\nabla ^2 =\nabla \cdot \nabla[/itex]

The Attempt at a Solution


[/B]
After considerable time messing around with various vector identites, I've been able to show the above is equivalent to:

[tex]\nabla \cdot (f \nabla (\nabla ^2 g)-g\nabla (\nabla ^2 f))+g(\nabla \cdot \nabla(\nabla ^2 f))-f(\nabla \cdot \nabla(\nabla ^2 g))[/tex]

This is painfully close to the result I want, but I can't seem to show that the second and third terms either cancel or are themselves a divergence. I'd really like any hints, and can provide more detail as to the specific identities and manipulations I've used thus far if needed, thanks.

Those terms simplify to [itex]g \nabla^4 f - f \nabla^4g[/itex], which doesn't cancel.

Rather than building a vector field as a linear combination of [itex]\nabla(\nabla^2 f)[/itex] and [itex]\nabla(\nabla^2 g)[/itex], I would have started by building one as a linear combination of [itex]\nabla f[/itex] and [itex]\nabla g[/itex].
 
  • #4
Have you tried working it from the other end starting with the answer or is that an unknown?

Reference https://www.physicsforums.com/threads/vector-analysis-problem-involving-divergence.909435/

Unfortuntely the end result is unknown, otherwise that'd be a great suggestion, thanks!

Those terms simplify to g∇4f−f∇4g, which doesn't cancel. Rather than building a vector field as a linear combination of ∇(∇2f) and ∇(∇2g), I would have started by building one as a linear combination of ∇f and ∇g.

Reference https://www.physicsforums.com/threads/vector-analysis-problem-involving-divergence.909435/

Yeah, I managed to get to that simplification, which like you said defnitely doesn't cancel, so I guess I must be able to somehow write the quantity as a divergence. I'm sorry, but I don't really understand what you mean by building a vector field?

Thanks for the help so far, I appreciate it.
 
  • #5
I would write,$$\nabla f \cdot\nabla\left ( \nabla^2 g\right )-\nabla g\cdot \nabla \left ( \nabla^2 f \right ) =\nabla \cdot \vec V $$
Then do the work of expanding the l.h.s. in it's spatial components and compare with the r.h.s.
 
  • #6
MyName said:
Unfortuntely the end result is unknown, otherwise that'd be a great suggestion, thanks!
Yeah, I managed to get to that simplification, which like you said defnitely doesn't cancel, so I guess I must be able to somehow write the quantity as a divergence. I'm sorry, but I don't really understand what you mean by building a vector field?

Thanks for the help so far, I appreciate it.

Consider the divergence of [itex]D(g, \nabla^2 g)\nabla f - D(f, \nabla^2 f) \nabla g[/itex] for some function [itex]D[/itex].
 
  • #7
I would write,∇f⋅∇(∇2g)−∇g⋅∇(∇2f)=∇⋅⃗V Then do the work of expanding the l.h.s. in it's spatial components and compare with the r.h.s.

That is a great idea, thanks! I managed to solve it using this idea.
 

Related to Vector Analysis Problem Involving Divergence

1. What is vector analysis?

Vector analysis is a branch of mathematics that studies vectors, which are quantities that have both magnitude and direction. It involves the use of mathematical operations, such as addition, subtraction, and multiplication, to manipulate and analyze vectors in various contexts.

2. What is divergence in vector analysis?

Divergence is a measure of how much a vector field is spreading out or converging at a particular point. It is represented by the mathematical operator ∇ ·, which is read as "del dot." Divergence is used to analyze the behavior of vector fields in three-dimensional space.

3. How do you solve a vector analysis problem involving divergence?

To solve a vector analysis problem involving divergence, you first need to identify the vector field and the point at which you want to calculate the divergence. Then, use the formula ∇ · F = ∂Fx/∂x + ∂Fy/∂y + ∂Fz/∂z to calculate the divergence at that point. Finally, interpret the result to understand the behavior of the vector field at that point.

4. What are some real-world applications of vector analysis involving divergence?

Vector analysis involving divergence has various applications in physics, engineering, and fluid mechanics. For example, it can be used to calculate fluid flow rates, analyze electromagnetic fields, and model the behavior of air currents in weather forecasting.

5. Are there any common mistakes to avoid when solving vector analysis problems involving divergence?

Yes, some common mistakes to avoid include forgetting to use the ∇ · operator, miscalculating the partial derivatives, and not considering the direction of the vector field in the interpretation of the result. It is important to carefully follow the formula and pay attention to the details to avoid making errors in solving these types of problems.

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