On the electric field taken in the proof of the second uniqueness theorem

Question

In the proof of uniqueness theorem, we consider $$vec{E}_3 = vec{E}_2 - vec{E_1}$$ where $vec{E_2}$ and $vec{E_1}$ are electric fields which satisfy all the boundary condition required.
Now, it maybe noted that due to both $vec{E}_1$ and $vec{E}_2$ satisfying all the conditions that the difference $vec{E}_3$ doesn't really satisfy any of the conditions.
Since $nabla cdot vec{E}_3 = 0$ everywhere and $oint vec{E_3} cdot dA= 0$ over every boundary surface.
This leads me to wonder, is the $vec{E}_3$ we define meant to be a physical field or just an 'auxiliary' mathematical function used in proof? I ask this because from my understanding if an electric field is physical then it must equal the charge density by $epsilon_o$ wherever there is charge.
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Michael Seifert · Accepted Answer

It's more or less just a mathematical construct.  If you want to solve the problem of the electric field created by a charge density $rho$, then you have $vec{nabla} cdot vec{E} = rho$.  If we assume that both $vec{E}_1$ and $vec{E}_2$ satisfy this equation, then their difference satisfies
$$
vec{nabla} cdot (vec{E}_2 - vec{E}_1) = vec{nabla} cdot vec{E}_2 - vec{nabla} cdot vec{E}_1 = rho - rho = 0.
$$
So it wouldn't be a solution to the "real problem" you're trying to solve.
You could, however, interpret it to be a solution to a different problem, one in which there are no charges present ($rho = 0$) in the volume of interest.  In other words, it's not entirely unphysical, it's just the solution to a different physical situation than the original one.

On the electric field taken in the proof of the second uniqueness theorem

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