Proof on a uniqueness theorem in electrostatics

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The discussion centers on proving the uniqueness theorem in electrostatics, which states that the electric field is uniquely determined by a given charge density and either the potential or its normal derivative specified on boundary surfaces. The user initially explores the implications of having two electric fields satisfying the conditions, leading to the conclusion that their difference must satisfy Laplace's equation. They express uncertainty about how to proceed, particularly regarding the normal derivative condition when the potential is not specified. Ultimately, the user discovers that setting the normal derivative to zero simplifies the problem, allowing them to complete the proof. The discussion highlights the importance of boundary conditions in determining the uniqueness of electrostatic fields.
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Homework Statement


Prove that the field is uniquely determined when the charge density ##\rho## is given and either ##V## or the normal derivative ##\partial V/\partial n## is specified on each boundary surface. Do not assume the boundaries are conductors, or that V is constant over any given surface.

Homework Equations


Poisson's equation which the potential ##V## must satisfy:
\nabla^2V = -\frac{\rho}{\epsilon_0}
Gauss' law in differential form:
\nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0}
The definition of the potential:
\vec{E} = -\nabla V
Perhaps various product rules of vector calculus will also be useful.

The Attempt at a Solution


I must admit that I'm not sure about the strategy I should use in this problem so I just played around with the equations trying to get some insight and using the proofs given for the two classical uniqueness theorems as models.
First, suppose there are two fields that satisfy the conditions of the problem:
\vec{E_1} = \frac{\rho}{\epsilon_0} ; \vec{E_2} = \frac{\rho}{\epsilon_0}
Looking at their difference ##\vec{E_3} = \vec{E_1} - \vec{E_2}##, it must satisfy ##\nabla \cdot \vec{E_3} = 0## or in integral form: ##\oint \vec{E_3}\cdot d\vec{a} = 0## over each boundary surface. Looking at the potential difference ##V_3 = V_1 - V_2##, it must satisfy ##V_3 = 0## on the boundary or ##\partial V_3/\partial n = 0## on the boundary.
In addition, ##\nabla^2V = 0## thus ##V_3## satisfies Laplace's equation and must have it's minima or maxima on the boundary.
Now I don't have any idea on how to proceed from here. In particular, I feel that the normal derivative can be exploited somehow but since I can't assume that the surfaces are conductors, I don't know how. I would highly appreciate any insight on this problem as well as general tips for going about such a problem in the future.

Thanks very much in advance.

EDIT: After thinking a bit more, if ##V## is given then it follows from the above that ##V_3 = 0## in the region (since it is zero on the boundary and has both maxima and minima zero there). If this reasoning is correct, the only thing left is to prove the case where ##\partial V/\partial n## is specified but ##V## is not.
 
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The usual proof goes by noting that ## \Delta \cdot \left(V \Delta V\right) = (\Delta V)^2 + V \Delta^2 V = (\Delta V)^2 ## and then taking the integral of the equation over the entire region within the surface, and making use of the divergence theorem to obtain a condition on the boundary.
 
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voko said:
The usual proof goes by noting that ## \Delta \cdot \left(V \Delta V\right) = (\Delta V)^2 + V \Delta^2 V = (\Delta V)^2 ## and then taking the integral of the equation over the entire region within the surface, and making use of the divergence theorem to obtain a condition on the boundary.
Thank you very much!
I figured out how to use ##\partial V/\partial n = 0##. This turns the surface integral that follows from the divergence theorem into zero leading to the standard proof. So the question is solved.

Note to self: write ##\hat{n}da## instead of ##d\vec{a}##. The former gives much more insight.
 
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