$D$ and $H$ in macroscopic Maxwell's equation: auxiliary or constitutive? I'm not a physicist.  I want to understand the macroscopic Maxwell's equations. But after reading Wikipedia and other Googled stuffs, I got very confused. In particular, $D$ and $H$ have two different equations, respectively. One group is called auxiliary fields:
$$B = \mu_0(H+M),$$
$$D = \epsilon_0E+P.$$
The other group is called constitutive relations:
$$B = \mu H,$$
$$D = \epsilon E.$$   
Which group of equations are relevant to the macroscopic Maxwell's equation if we consider the case with polarization ($P$ and $M$ present). My aim is to learn computational electromagnetic.
 A: Both pairs of equations are equivalent. In the first equation have the magnetic flux density $B$, which results from the applied magnetic field $H$, and the magnetization $M$. You can think of the magnetization as the response of the material to the applied field. The magnetization is related to the field strength by
$$M = \chi_m H \,.$$
where $\chi_m$ is the magnetic susceptibility. If $\chi_m > 0$, the material is paramagnetic, meaning the response has the same sign as the applied field (thus the total flux density grows). If it is $< 0$, it is diamagnetic, and the resulting flux density is smaller.
There is a different way of writing the same physics:
$$B =  \mu_0( H + M) = \mu_0( H + \chi_m H)  =  \mu_0 (\underbrace{1 + \chi_m}_{ =\; \mu_r}) H$$
$$ \Rightarrow \quad B = \mu_0\mu_r H = \mu H$$
So the same information given by the magnetic susceptibility $\chi_m$ is also in the relative permeability $\mu_r$.
The situation is very similar for the electric field. An applied electric field $E$ moves the electrons in the material, leading to polarization $P$. Together both form the displacement field $D$. There is a slight difference in notation here, though. In the case of the magnetic field, $\mu_0$ is multiplied by $(H+M)$. For the electric field, only $E$ is multiplied by $\varepsilon_0$, not $P$. So  the equation for $P$ needs a factor of $\varepsilon_0$:
$$P = \varepsilon_0 \chi_e E \quad \Rightarrow \quad 1 + \chi_e = \varepsilon_r$$
$$ \Rightarrow \quad D = \varepsilon_0\varepsilon_r E = \varepsilon E$$
