not enter the equation of motion of $\chi$. In this situation, only $v_4, v_6$ affect the behavior of $\sigma$. In Fig.\[v4-v6-mu=0\], we show the effects of $v_4, v_6$ on $\sigma$ when $\mu=0$. From the figure, we could see that, since $m_q\neq0$, the phase transition turns from second order to crossover type. Different from chiral limit, $\sigma$ approaches to zero at high temperature instead of being exactly zero at $T_c$. Here, we simply define the pseudo transition temperature $T_c$ as $\frac{\partial^2 \sigma}{\partial^2 T}|_{T_c}=0$. From the figure, we can see that the larger value of $v_4$ or $v_6$ leads to a smaller value of $\sigma$. Also, they would change the pseudo transition temperature. Thus, changing $v_4, v_6$ we can get different values of vacuum condensate $\sigma_0\equiv \sigma(T=0,\mu=0) $ and $T_c$. Therefore, by fitting $\sigma=(0.340\rm{GeV})^3, T_c=0.238GeV$, close to the results with vanishing $\mu$ from previous studies[@Burger2013The; @McLerran:2008ua], we can fix $v_4=2, v_6=100$ firstly. Then, only three free parameters $\lambda_2, \lambda_4, \lambda_6$ are left.

=6.5 cm =4.5 cm =6.5 cm =4.5 cm =6.5 cm =4.5 cm -0.05cm 0.15 cm\

Generally, if one works in the full back-reaction scenario, the parameters $\lambda_2, \lambda_4, \lambda_6$ will back-reacting the background geometric and also the equation of states. 