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