Divergence Theorem

  • Thread starter roldy
  • Start date
  • #1
232
1

Main Question or Discussion Point

I've tried to make sense of this conjecture but I can't wrap my head around it.

We've been learning about the divergence theorem and the Neumann problem.
I came across this question.

Use the divergence theorem and the partial differential equation to show that

[tex]\underbrace{\int\int\int}_{D}f(x,y,z)dxdydz=0[/tex] is a necessary condition for the Neumann problem to have a solution.

Where the Neumann problem is [tex]\Delta=f(x,y,z)[/tex] in D, [tex]\frac{\partial u}{\partial n}=0[/tex] on [tex]\partial D[/tex].
 

Answers and Replies

  • #2
fzero
Science Advisor
Homework Helper
Gold Member
3,119
289
You will want to consider

[tex]\int_D \Delta u d^3x[/tex]

using [tex]\Delta u = \nabla \cdot (\nabla u)[/tex]. Using the divergence theorem you can relate this integral to the boundary condition. On the other hand, you can use the Poisson equation to relate it to the integral of the source function f.
 
  • #3
232
1
Have to be honest here, I'm still lost on this one. I can't remember exactly how to implement the divergence theorem. I have to use only what's provided in the problem.
 
  • #4
fzero
Science Advisor
Homework Helper
Gold Member
3,119
289
Can you state the divergence theorem as it was taught in class?
 
  • #5
232
1
Only the definition, and only because I had to look it up. He didn't cover the divergence theorem.
 
  • #6
fzero
Science Advisor
Homework Helper
Gold Member
3,119
289
Only the definition, and only because I had to look it up. He didn't cover the divergence theorem.
But you said

We've been learning about the divergence theorem
In any case, the divergence theorem says that if [tex]\mathbf{F}[/tex] is a continuously differentiable vector field, then the volume integral

[tex]\int_V \nabla\cdot \mathbf{F} ~dV = \oint_{S=\partial V} \mathbf{n}\cdot \mathbf{F}~dS,[/tex]

where [tex]\mathbf{n}[/tex] is the outward pointing unit normal vector field to the bounding surface [tex]S=\partial V[/tex]. Can you attempt to apply this to the integral in post #2?

You will also want to figure out an expression for [tex]\partial u/\partial n[/tex] in terms of the gradient of [tex]u[/tex] as well.
 
  • #7
232
1
I'll give it a try. What I meant to say in the first post is that he introduced the divergence theorem. Learning was a bad choice of word since he didn't really teach anything about it.
 
  • #8
dextercioby
Science Advisor
Homework Helper
Insights Author
12,985
540
As a side comment, it's interesting that he asked you to solve a problem involving some notions he didn't properly explain in class.

Well, how fortunate of you to find Physicsforums. Some guys here can succesfully replace your math teacher, unfortunately for them without getting his salary as well. :wink:

Leaving jokes aside though, either he didn't ask you to solve the problem and you've seen it just glancing through your textbook, or he really commited some pedagogical mistake.
 
Last edited:
  • #9
232
1
Yeah, it really makes it difficult when he doesn't show examples. The book is worthless too. I've been trying to search on google for examples using the divergence theorem with relation to the Neumann problem.
 
  • #10
232
1
I guess that I really don't fully understand what [tex]\Delta u[/tex] is.
 
  • #12
fzero
Science Advisor
Homework Helper
Gold Member
3,119
289
I guess that I really don't fully understand what [tex]\Delta u[/tex] is.
Well I already told you that [tex]\Delta u = \nabla \cdot (\nabla u)[/tex]. Independently of this vector calc formula, [tex]\Delta[/tex] is a differential operator called the Laplacian. In terms of [tex](x,y,z)[/tex], it is

[tex] \Delta = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}. [/tex]

The equation

[tex]\Delta u=0[/tex]

is known as the Laplace equation, while

[tex]\Delta u=f[/tex]

is Poisson's equation.
 
  • #13
232
1
So all that I have done is this.

[tex]
\underbrace{\int\int\int}_{D} \left(\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2}\right)dxdydz=0
[/tex]

Just basic substitution in for [tex] \Delta u [/tex]

And substitution into the divergence theorem.

[tex]
\underbrace{\int\int\int}_{D} \left(\frac{\partial f}{\partial x}\hat{i} + \frac{\partial f}{\partial y}\hat{k}+ \frac{\partial f}{\partial z}\hat{k}\right)dx=\underbrace{\int\int}_{S}f\cdot n ds
[/tex]
 
  • #14
fzero
Science Advisor
Homework Helper
Gold Member
3,119
289
So all that I have done is this.

[tex]
\underbrace{\int\int\int}_{D} \left(\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2}\right)dxdydz=0
[/tex]
You can't just set this equal to 0. You have to use the divergence theorem and the Neumann boundary condition to show that it vanishes.

Just basic substitution in for [tex] \Delta u [/tex]

And substitution into the divergence theorem.

[tex]
\underbrace{\int\int\int}_{D} \left(\frac{\partial f}{\partial x}\hat{i} + \frac{\partial f}{\partial y}\hat{k}+ \frac{\partial f}{\partial z}\hat{k}\right)dx=\underbrace{\int\int}_{S}f\cdot n ds
[/tex]
Where does [tex]\nabla f[/tex] come from?

Go back and reread post #2 for hints on how to do this properly.
 
  • #15
232
1
I don't see what I did wrong. [tex]\nabla \cdot f =\frac{\partial f}{\partial x}\hat{i} + \frac{\partial f}{\partial y}\hat{k}+ \frac{\partial f}{\partial z}\hat{k}[/tex].

[tex]
\nabla=\frac{\partial }{\partial x}\hat{i} + \frac{\partial}{\partial y}\hat{k}+ \frac{\partial}{\partial z}\hat{k}
[/tex]

And all's that I did for the first part was substitute in what f(x,y,z) is. [tex]f(x,y,z)=\Delta[/tex]

And [tex]\Delta u=\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2}[/tex]
 
Last edited:
  • #16
fzero
Science Advisor
Homework Helper
Gold Member
3,119
289
I don't see what I did wrong. [tex]\nabla \cdot f =\frac{\partial f}{\partial x}\hat{i} + \frac{\partial f}{\partial y}\hat{k}+ \frac{\partial f}{\partial z}\hat{k}[/tex].
[tex]\nabla f[/tex] is the gradient of a scalar [tex]f[/tex]. The gradient of a scalar is a vector field. When we write [tex]\nabla \cdot \mathbf{F}[/tex], this is the divergence of a vector field [tex]\mathbf{F}[/tex]. The divergence is a scalar.

When you write

[tex]

\underbrace{\int\int\int}_{D} \left(\frac{\partial f}{\partial x}\hat{i} + \frac{\partial f}{\partial y}\hat{k}+ \frac{\partial f}{\partial z}\hat{k}\right)dx=\underbrace{\int\int}_{S}f\cdot n ds

[/tex]

you have the volume integral of a vector field on the LHS, while the RHS is the surface integral of something you've written as the dot product of a scalar f with something you call n, presumably meant to be the normal vector. This equation is mathematically inconsistent and is not a valid application of the divergence theorem.

On the other hand, it is perfectly straightforward to apply the divergence theorem to

[tex]\int_V \Delta u ~dV = \int_V \nabla\cdot ( \nabla u) ~dV.[/tex]
 

Related Threads on Divergence Theorem

Replies
0
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
7
Views
4K
Replies
6
Views
3K
  • Last Post
Replies
10
Views
33K
Replies
3
Views
5K
Replies
1
Views
3K
Replies
1
Views
1K
Replies
0
Views
1K
Replies
3
Views
3K
Top