Q:

Problem verifying Gauss's law

I am trying to verify Gauss's law, differential form, SI units, non-special relativity regime, ignoring time retardation, by performing the differentiation on the left-side of the equations to see if the results match the right side of the equations.  I can't get Gauss's law (electric) nor Ampere's law to match their left and right sides.
For example, I am using
$$ \vec{E} = -\nabla \Phi -\partial \vec{A} / \partial t $$
Taking the divergence of both sides I get
$$ \nabla \bullet \vec{E} = \dfrac{1}{\epsilon_0}\rho - \nabla \bullet \partial \vec{A} / \partial t $$
Some people have suggested that I now need to apply Coulomb gauge condition
$$\nabla \bullet \vec{A} = 0$$
but 
1) That is not indicated explicitly in Maxwell's equations.  I only see
$$ \nabla \bullet \vec{E} = \dfrac{1}{\epsilon_0}\rho . $$  Where is the Coulomb gauge requirement indicated?
And 2) doesn't selecting the Coulomb gauge fix the gauge, removing the gauge freedom from Maxwell's system of equations? 
Does this mean that Maxwell's equations are only valid (mathematically consistent) in the Coulomb gauge?  I thought Maxwell's equations were mathematically valid in general and for any gauge.  But Gauss's Law doesn't seem to be valid for any gauge.
If you look at Jackson's book, for example, Gauss's Law is only derived in his Chapter 1 on Electrostatics.
Any comments on the following line of reasoning would be appreciated....
Let us do some expansion (also called a decomposition) of the electrodynamic field $\vec{E}$
$$ \vec{E} = \vec{E}_1 + \vec{E}_2 + \vec{E}_3 + ... $$
The decomposition could be a Helmholtz decomposition (2 terms only) of the vector field.  But in this case, I'll take the decomposition to be
$$ \vec{E} = \vec{E}_1 + \vec{E}_2 = -\nabla \Phi -\partial \vec{A} / \partial t $$ where
$$ \vec{E}_1 = -\nabla \Phi $$ and
$$ \vec{E}_2 =  -\partial \vec{A} / \partial t $$
Is there a mathematical problem in doing this?  I can then write
$$ \vec{E} = \vec{E}_1 -\partial \vec{A} / \partial t $$
If I understand what some people are telling me, maybe I misunderstand what they are saying, I must now relabel $ \vec{E}_1 \rightarrow \vec{E}$ (because in Jackson's book Chapter 1 $ \vec{E} = -\nabla \Phi$) and write
$$ \vec{E} = \vec{E} -\partial \vec{A} / \partial t $$
I don't think this is mathematically correct.
Why isn't it valid to write Gauss's Law as
$$ \nabla \bullet \vec{E}_1 = \dfrac{1}{\epsilon_0}\rho $$
And so, with this, Maxwell's Gauss's law is 
$$ \nabla \bullet \vec{E}_1 = \dfrac{1}{\epsilon_0}\rho $$
without any gauge fixing.  Maxwell's system for electrodynamics then continues to have gauge freedom.
If this is not mathematically correct, can you point out what mathematically went wrong?  
Again, I am trying to verify mathematically all of Maxwell's equations as a single system of differential equations and running into some issues....  Any help would be appreciated.

A:

Your mistake is simply that $\nabla^2 \phi = -\rho/\epsilon_0$ only in Coulomb gauge. Otherwise you just have
$$\nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon_0} = -\nabla^2 \phi - \frac{\partial \nabla\cdot\mathbf{A}}{\partial t}.$$
Maxwell's equations in terms of the fields are valid always, and so are the relations between the fields and potentials, $\mathbf{E} = - \nabla \phi - \partial \mathbf{A} / \partial t$ and $\mathbf{B} = \nabla \times \mathbf{A}$ together with the equations for the potentials:
$$\nabla^2 \phi + \frac{\partial \nabla \cdot \mathbf{A}}{\partial t} = - \frac{\rho}{\epsilon_0}$$
and
$$\frac{1}{c^2} \frac{\partial^2 \mathbf{A}}{\partial t^2} - \nabla^2 \mathbf{A} + \nabla(\nabla \cdot \mathbf{A}) + \frac{1}{c^2} \frac{\partial \nabla \phi}{\partial t} = \mu_0 \mathbf{J}.$$
Simplified versions of these only hold in certain gauges. For example, in Coulomb gauge, where $\nabla \cdot \mathbf{A} = 0$, we have $\nabla^2 \phi = - \rho/\epsilon_0$, but not in other gauges.

