1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Rigorous Divergence Theorem Proof

  1. Oct 4, 2006 #1
    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]​
     
    Last edited: Oct 4, 2006
  2. jcsd
  3. Oct 4, 2006 #2

    StatusX

    User Avatar
    Homework Helper

    It depends what you mean by rigorous. To be completely rigorous, you would need to use the Riemann integral (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.
     
  4. Oct 5, 2006 #3

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Prove it when your region is a box.
     
  5. Oct 6, 2006 #4
    I have done this, but I want a more general proof.

    I was looking at this 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.
     
  6. Oct 6, 2006 #5

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Use the change of variables theorem, then, in conjunction with your box result.
     
  7. Oct 6, 2006 #6
    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.
     
    Last edited: Oct 6, 2006
  8. Oct 6, 2006 #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?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Rigorous Divergence Theorem Proof
  1. Divergence theorem (Replies: 2)

  2. Divergence theorem (Replies: 5)

  3. Divergence Theorem (Replies: 5)

  4. Divergence theorem (Replies: 1)

  5. Divergence theorem (Replies: 2)

Loading...