Rigorous Divergence Theorem Proof

In summary, the conversation is about trying to construct a rigorous proof for the divergence theorem by using a basic proof that is filled with errors and assumptions. The question is whether this line of reasoning can be used to prove the theorem for all cases. The proof involves splitting up a closed surface into small rectangular parallelipipeds and approximating the flux of a vector function over the surface using these boxes. The conversation then discusses different approaches to making the proof more rigorous, including using the Riemann integral and the change of variables theorem. The conversation also considers the limitations of the current approach, such as not accounting for source points in the field. Finally, a simplified approach is suggested where the divergence theorem can be rewritten and integrated to prove its validity
  • #1
Saketh
261
2
The Background:
I'm trying to construct a rigorous proof for the divergence theorem, but I'm far from my goal. So far, I have constructed a basic proof, but it is filled with errors, assumptions, non-rigorousness, etc.

I want to make it rigorous; in so doing, I will learn how to construct rigorous proofs.

The Question:
My question is - can I use this line of reasoning to rigorously prove the divergence theorem for all cases? I realize that this is a time-consuming request, but I will greatly appreciate any assistance. I want to finally prove something rigorously, unlike my regular high school "let's prove that [tex]\vec{F} \equiv q\vec{E}[/tex]!" type of proofs.

The "Proof"

Given: Vector function [itex]\bf{F}(x, y, z)[/itex], a closed surface [itex]S[/itex], and an enclosed volume [itex]V[/itex].

We can split up [itex]S[/itex] into many small rectangular parallelipipeds (henceforth ``boxes'') as an approximation. We can now approximate the flux of the vector function over [itex]S[/itex] using these boxes.

[tex]
\Phi \approx \sum \bf{F} \cdot \bf{\hat{n}} \Delta S
[/tex]​

We divide and multiply the flux expression by [itex]\Delta V[/itex].

[tex]
\Delta V \left [ \sum \bf{F} \cdot \bf{\hat{n}} \Delta S \right ] = \sum \bf{F} \cdot \bf{\hat{n}} \Delta S
[/tex]​

Now we limit the expression as [itex]\Delta[/itex] approaches zero. This means that the width, height, and length of the boxes will approach infinitesimal values. By definition, the bracketed quantity will equal the divergence of [itex]\bf{F}[/itex]. In addition, the area for the flux will approach the infinitesimal value [itex]dS[/itex].

[tex]
\lim_{\Delta \to 0} \Delta V \left [ \sum \bf{F} \cdot \bf{\hat{n}} \Delta S \right ] = \nabla \cdot \bf{F} \,dV = \sum \bf{F} \cdot \bf{\hat{n}} \,dS
[/tex]​

If we integrate both sides (one side as a volume, the other as a surface), we get the following expression:

[tex]
\int \!\!\! \int \!\!\! \int_V \nabla \cdot \bf{F} \,dV = \int \!\!\! \int_S \bf{F} \cdot \bf{\hat{n}} \,dS
[/tex]​

If [itex]\bf{F}[/itex] is not a continuous function on [itex]V[/itex] and [itex]S[/itex], then the infinitesimal form cannot be integrated. Therefore, [itex]\bf{F}[/itex] must be continuous on [itex]V[/itex] and [itex]S[/itex].
If \[itex]\bf{F}[/itex] is not differentiable at any point on S, then the dot product of [itex]\bf{F}[/itex] and the unit normal vector is indeterminate. Therefore, the flux would be indeterminate. For the same reason, it follows that the spatial derivative of [itex]\bf{F}[/itex] must be continuous on S.
Thus, we have the divergence theorem: If [itex]\bf{F}(x, y, z)[/itex] is a vector function differentiable on closed surface [itex]S[/itex] and continuous on enclosed volume [itex]V[/itex], then

[tex]
\int \!\!\! \int \!\!\! \int_V \nabla \cdot \bf{F} \,dV = \int \!\!\! \int_S \bf{F} \cdot \bf{\hat{n}} \,dS
[/tex]​
 
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  • #2
It depends what you mean by rigorous. To be completely rigorous, you would need to use the http://en.wikipedia.org/wiki/Riemann_integral" [Broken] (since this is how integration of continuous functions is defined). Have you ever worked with this before? If so, try applying some ideas from there (define a partition of V, etc). If not, I can suggest a few holes in your current argument you could patch up, but I would call what you'd be left with then more of a motivating argument than a proof.
 
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  • #3
Prove it when your region is a box.
 
  • #4
arildno said:
Prove it when your region is a box.
I have done this, but I want a more general proof.

I was looking at http://www.math.hawaii.edu/~lee/calculus/diverg.pdf#search="divergence theorem proof"" earlier today, but I was confused by one step - the one where he splits the Divergence Theorem into three separate identities on the last page. Can someone explain this to me, please? After I understand this, I will be able to formulate a better argument.

Although I have worked with Riemann integration, it was in an informal setting without the definition of partitions, etc.
 
Last edited by a moderator:
  • #5
Saketh said:
I have done this, but I want a more general proof.

I was looking at http://www.math.hawaii.edu/~lee/calculus/diverg.pdf#search="divergence theorem proof"" earlier today, but I was confused by one step - the one where he splits the Divergence Theorem into three separate identities on the last page. Can someone explain this to me, please? After I understand this, I will be able to formulate a better argument.

Although I have worked with Riemann integration, it was in an informal setting without the definition of partitions, etc.
Use the change of variables theorem, then, in conjunction with your box result.
 
Last edited by a moderator:
  • #6
arildno said:
Use the change of variables theorem, then, in conjunction with your box result.
How can I do this?

I derived my box result by finding fluxes through the six faces of the box, and then limiting the distance between the fluxes to get the divergence in terms of flux.

EDIT: In addition, I believe that the box result fails to account for cases where there are source points of field lines e.g. point charges.
 
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  • #7
Something just struck me - we can write an expression for a differential flux element as follows:
[tex]
lim_{\Delta \to 0}\frac{\Delta V \int \!\!\! \int_S \bf{F} \cdot \hat{\bf{n}} \,dS}{\Delta V}
[/tex]
If I'm not wrong, this can be rewritten as
[tex]
\nabla \cdot \bf{F} \,dV
[/tex]
Which is mathematically equivalent to a differential flux element expression
[tex]
\bf{F} \cdot \hat{\bf{n}} \,dS
[/tex]
So we can write this as
[tex]
\nabla \cdot \bf{F} \,dV = \bf{F} \cdot \hat{\bf{n}} \,dS
[/tex]
Integrating both sides
[tex]
\int \!\!\! \int \!\!\! \int_V \nabla \cdot \bf{F} \,dV = \int \!\!\! \int \bf{F} \cdot \hat{\bf{n}} \,dS
[/tex].

This seems mathematically sound to me, but too simple to be correct. Is this a proper way of "proving" the divergence theorem?
 

1. What is the Rigorous Divergence Theorem Proof?

The Rigorous Divergence Theorem Proof is a mathematical proof that is used in vector calculus to relate a surface integral over a closed surface to a triple integral over the volume enclosed by the surface. It is also known as Gauss's Theorem or Green's Theorem in three dimensions.

2. Why is the Rigorous Divergence Theorem Proof important?

The Rigorous Divergence Theorem Proof is important because it is a fundamental tool in vector calculus and is used to solve a wide range of problems in physics, engineering, and other fields. It allows for the conversion of a difficult surface integral into a simpler triple integral, making calculations more manageable.

3. How is the Rigorous Divergence Theorem Proof derived?

The Rigorous Divergence Theorem Proof is derived from the fundamental theorem of calculus and the definition of a divergence in vector calculus. It involves breaking down a vector field into its component functions and using properties of integration to manipulate the resulting equations.

4. What are the assumptions made in the Rigorous Divergence Theorem Proof?

The main assumption made in the Rigorous Divergence Theorem Proof is that the vector field is continuous and differentiable in the region of interest. Additionally, the surface and volume must be well-behaved and satisfy certain conditions, such as being closed and having a smooth boundary.

5. How is the Rigorous Divergence Theorem Proof used in real-world applications?

The Rigorous Divergence Theorem Proof is used in a variety of real-world applications, such as fluid mechanics, electromagnetism, and heat transfer. It allows for the calculation of important physical quantities, such as flow rate and flux, and is also used in the formulation of governing equations in these fields.

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