Maxwell equations in absence of magnetic field I always thought that a change in electric field induces a magnetic field and vice-versa. Moreover, I imagined that any current distribution will give rise to a magnetic field. But then I wrote this down: Maxwell's equations in absence of magnetic field.
\begin{align}
\nabla \cdot \mathbf{E}(\mathbf{x},t) &= \frac {\rho(\mathbf{x},t)} {\varepsilon_0}\\
\nabla \times \mathbf{E}(\mathbf{x},t) &= 0\\
\frac{\partial \mathbf{E}(\mathbf{x},t)} {\partial t}&= -\frac{\mathbf{j}(\mathbf{x},t)}{\varepsilon_0} 
\end{align}
the second equation gives
 \begin{align}
\mathbf{E}(\mathbf{x},t)=-\nabla \phi(\mathbf{x},t)    
\end{align}
so that the rest becomes
\begin{align}
\nabla^2 \phi(\mathbf{x},t) &= -\frac {\rho(\mathbf{x},t)} {\varepsilon_0}\\
\nabla \frac{\partial\phi(\mathbf{x},t)} {\partial t}&= \frac{\mathbf{j}(\mathbf{x},t)}{\varepsilon_0} 
\end{align}
(Note: already we can see charge conservation, i.e. $\partial_t \rho+\nabla\cdot\mathbf{j}=0$)
Then we get the usual solution of the first equation
\begin{align}
\phi(\mathbf{x},t) =  \iiint  \frac{\rho(\mathbf{x}',t)}{4\pi\epsilon_0 |\mathbf{x} - \mathbf{x}'|}\, \mathrm{d}^3\! x',
\end{align}
which can be written as 
\begin{align}
\partial_t\phi(\mathbf{x},t) =  -\iiint  \frac{\nabla_{\mathbf{x}'}\cdot\mathbf{j}(\mathbf{x}',t)}{4\pi\epsilon_0 |\mathbf{x} - \mathbf{x}'|}\, \mathrm{d}^3\! x',
\end{align}
and the second equation, due to gradient theorem, becomes
\begin{align}
\partial_t\phi(\mathbf{x},t)&=  \phi(\mathbf{0},t) + \frac{1}{\varepsilon_0}\int_0^1  \mathbf{j}(\lambda\mathbf{x},t)\cdot\mathbf{x}\,\mathrm d\lambda
\end{align}
so that
\begin{align}
 \phi(\mathbf{0},t) + \frac{1}{\varepsilon_0}\int_0^1  \mathbf{j}(\lambda\mathbf{x},t)\cdot\mathbf{x}\,\mathrm d\lambda=-\iiint  \frac{\nabla_{\mathbf{x}'}\cdot\mathbf{j}(\mathbf{x}',t)}{4\pi\epsilon_0 |\mathbf{x} - \mathbf{x}'|}\, \mathrm{d}^3\! x'
\end{align}
I guess what I'm trying to show here is $\mathbf{E}(\mathbf{x},t)=\mathbf{E}(\mathbf{x})$ so that $\mathbf{j}(\mathbf{x},t)=0$, but I might be wrong. Also I don't see an obvious way to continue with this derivation.
EDIT:
A current density doesn't  always produce a magnetic field, so that configurations with $\mathbf{E}(\mathbf{r},t)$ and $\mathbf{B}=0$ do exist. The final question I'm wondering about is whether the last equation is some sort of a constraint on $\mathbf{j}(\mathbf{x},t)$ or is it an equality true in general?
 A: I think you are looking at this the right way but it is probably easier to think in terms of fields than potentials. Plugging $\newcommand{b}{\mathbf{B}}\renewcommand{e}{\mathbf{E}}\renewcommand{ed}{\dot{\e}}\newcommand{j}{\mathbf{j}}\renewcommand{z}{\mathbf{0}} \b=\z$ into$ \nabla \times \b = \ed + \j$ we get $\ed=-\j$. Then $\e = \e_0+\int_{t_0}^t -\j dt'$. We can now check if this definition of $\e$, together with $\b=\z$, satisfies Maxwell's equations. The ones concerning $\b$ are satisfied by construction. Checking Gauss's law, we find $$\nabla \cdot \e = \nabla \cdot \e_0 + \int_{t_0}^t -\nabla \cdot \j\, dt'=\rho_0 + \int_{t_0}^t \dot{\rho}\, dt'=\rho.$$
So Gauss's law checks out.
Now lets check the last equation.
$$\z=-\dot{\b}=\nabla \times \e = \nabla \times \e_0 + \int_{t_0}^t -\nabla \times \j\, dt'.$$
If the rightmost side is to be zero for all $t$, then we must have that $\nabla \times \e_0=\z$ and then for all $t$, $\nabla \times \j=0$. The first equation tells us that $\e_0$ must be conservative, and the second tells us that $\j$ must be irrotational for all time.
In summary, we have found that $\b$ is zero then it is necessary to have $\ed = -\j$, so that $\e=-\int \j\, dt$, and then we found it is also necessary for $\j$ to be irrotational. Moreover, these two conditions are sufficient since you can construct a solution. 
So in conclusion you can find a $\b=\z$ solution precisely when $\j$ is irrotational, in which case the solution is $\e=-\int \j\, dt$.
A: As shown here section 18.2, it is possible to have configurations in which current density is non-zero but the magnetic field is zero. My understanding is that it is perfectly legitimate to have a time-varying Electric field and null magnetic field at all times. The simplest case is a variable current source $j(r,t)$ eminating radially from a source. Since $j(r,t)$ has spherical symmetry, $B=0$, however $E(r,t)$ varies in both space and time. 
A: I think you have overlooked that a sentence like

a change in electric field induces a magnetic field and vice-versa.

is true in the vacuum, i.e. it is not valid in a region where non-zero charge density and current exist.
A: Magnetic fields are a consequence of special relativity as follows:
Given two charges A and B, and looking at the effect of A on B, you will get the correct result without any thought for magnetic fields unless both charges have a velocity. In any frame where one of the charges is motionless, magnetic field has no effect.
There are further restrictions. A must have a velocity that is perpendicular to the vector from A to B. B must have a velocity that is on the plane made by the AB line and A's velocity vector.
Magnetic fields provide the fudge factor required to correct for frames that display a relativity effect.
