K$^{-1}$ s$^{-1}$ [@P53]. In this paper, we adopt the following fitting formula of the net cooling rate [@KI02], $$\rho{\cal L}(\rho,T) = \frac{\rho}{m_\mathrm{H}}\left( -\Gamma + \frac{\rho}{m_\mathrm{H}} \Lambda(T) \right)\;\;\mathrm{erg\;cm^{-3}\;s^{-1}},
 \label{cooling rate}$$ $$\Gamma = 2\times10^{-26}\;\;\mathrm{erg\;s^{-1}}, \nonumber$$ $$\frac{\Lambda(T)}{\Gamma} = 10^7\exp\left( -\frac{118400}{T+10^3} \right) 
        + 1.4\times10^{-2}\sqrt{T}\exp\left( -\frac{92}{T} \right) \nonumber.$$ Figure \[fig:eq\] shows the thermal equilibrium state of this net cooling rate in the $(n,P)$ plane. The gas is subject to the cooling (heating) above (below) the thermal equilibrium curve. The bistable fluid consisting of the WNM and CNM can exist in the pressure range of $P_\mathrm{min}<P<P_\mathrm{max}$, where $P_\mathrm{min}/k_\mathrm{B}=1596$ K cm$^{-3}$ and $P_\mathrm{max}/k_\mathrm{B}=5012$ K cm$^{-3}$. The saturate pressure where the static front is realized is as large as $P_\mathrm{sat}/k_\mathrm{B}=2823$ K cm$^{-3}$ in the plane-parallel geometry.

![ Thermal equilibrium state of the net cooling rate ${\cal L}(\rho,T)$ in $(P,n)$ plane. []{data-label="fig:eq"}](fig1.eps){width="8.0cm"}

In steady state $(\partial/\partial t=0)$, equations (\[eoc\]) and (\[eom\]) can be integrated with respect to $x$ to give $$\rho v = j
        \label{const}$$ and $$\rho v^2 + P = M,
        \label{const mom}$$ respectively. The energy equation (\[eoe\]) becomes $$c_pj \frac{dT}{dx}
        -v \frac{dP}{dx} - \frac{d}{dx}\left( \kappa\frac{dT}{dx } \right)+\rho{\cal L}(\rho,T)=0,
        \label{integ eq0}$$ where $j$ and $M$ denotes the mass and momentum fluxes, respectively, that are spatially constant, and $c_p=\gamma 