$V(\chi,F^2) \equiv {\rm Tr}(V_X(X,F^2))$. The equation of motion can be easily derived as $$\begin{aligned}
\chi^{\prime\prime}(z) + (-\frac{3}{z} + 3A^\prime_s(z) - \phi^\prime(z) + \frac{f^\prime(z)}{f(z)})\chi^\prime(z)+ \frac{e^{2A_s(z)}}{z^2f(z)}\partial_\chi V(\chi(z),F^2) = 0.\label{eq-chi-1}\end{aligned}$$

Generally, the potential $V_X$ could be of any form satisfying the $SU(2)_L\times SU(2)_R$ symmetry. In previous studies [@Chelabi:2015cwn; @Chelabi:2015gpc], it is taken as a simple form $V_X(X,F^2)=M_5^2 X^+ X +\alpha (X^+X)^2$ with $M_5^2=-3$ from the AdS/CFT prescription, or equivalently $$\begin{aligned}
V(\chi, F^2)=-\frac{3}{2}\chi^2+v_4 \chi^4.\end{aligned}$$ It is shown that $v_4$ should be positive in order to have chiral symmetry breaking at low temperature in chiral limit. Therefore, firstly, we would have a test on this simple form of potential. Inserting this potential into Eq.(\[eq-chi-1\]), one can obtain the leading UV($z\rightarrow 0$) expansion of $\chi(z)$ as $$\begin{aligned}
\chi(z) = m_q\zeta z +...+ \frac{\sigma}{\zeta}z^3 + ...,\end{aligned}$$ where $\sigma$ is the chiral condensate(order parameter) and $\zeta = \frac{\sqrt{3}}{2\pi}$ is the normalization constant obtain by matching correlation function$\langle\bar{q}q(p)\bar{q}q(0)\rangle$ with 4D calculation(for details, please refer to [@Cherman:2008eh]). Near the horizon, the regular condition of $\chi$ requires $\frac{1}{f(z)}(f^\prime\chi^\prime + e^2A_s\partial_\chi V(\chi)/z^2)$ to be finite at $z=z_h$. Taking the metric solved in previous sections and imposing the above UV and IR boundary conditions, one can solve $\sigma$ with respect to $m_q$, $T$ 