How do you explain that Poynting's theorem $P=E\times H$ is still valid for DC fields? Poynting's theorem P = ExH is quite valid for DC circuits with stationary electromagnetic fields as well as for AC circuits and time varying EM fields.  The question arises: Do Maxwell's 4 equations have any significance in stationary electromagnetic fields? i.e. when the time derivative of both E&H is zero?We assume that the proof(s) of mathematical physics exist but should be based on groundbreaking assumptions.
NB: There is no relationship between Maxwell's 4 equations and the Lorentz force F=q(E+vxB) which are two different and totally independent phenomena equations.
 A: 
Do Maxwell's 4 equations have any significance in stationary electromagnetic fields? i.e. when the time derivative of both E&H is zero?

Yes. Maxwell’s equations and the Lorentz force law describe all classical behavior of the electric and magnetic fields and their interactions with charges. They apply for all frequencies, including zero.
Their only limitation is that they are classical, they require large enough pieces of matter that things can be treated classically. Although, for the fields the Schrodinger’s equation reproduces Maxwell’s equations anyway. So Maxwell’s equations do play a non-trivial role even quantum mechanically.
So, the short answer is, yes. Poynting’s theorem does apply at DC and even at DC the Poynting vector represents energy flux. The flow of energy from a battery to a resistive load can be analyzed with it.
To see this, first consider the DC Maxwell's equations $$ \nabla \cdot \vec D = \rho $$ $$ \nabla \cdot \vec B = 0 $$ $$ \nabla \times \vec E = 0 $$ $$ \nabla \times \vec H = \vec J $$ where $\rho$ and $\vec J$ are specifically the free charge and current densities. Now, we can combine the third and fourth equations as follows: $$ \vec H \cdot \left(\nabla \times \vec E \right) + \vec E \cdot \left(\nabla \times \vec H \right) = \vec E \cdot \vec J $$ which simplifies to $$\nabla \cdot \left( \vec E \times \vec H \right) + \vec E \cdot \vec J = 0$$ which applies for time-independent scenarios. Unsurprisingly, this is what we would have obtained using Poynting's theorem directly and removing all of the time-dependent terms.
A: Not sure what you mean by "groundbreaking assumptions" but I think you mean "experimental evidence"?
Poyntings theorem is derived directly from maxwells equations, which is experimentally verified.
$$\vec{S} = \frac{1}{\mu_0}\vec{E} × \vec{B}$$
Is the poynting vector, not poyntings theorem.
