therefore we transform the action into Einstein frame by a conformal transformation of metric: $$\begin{aligned}
g^{s}_{\mu\nu}=e^{\frac{4\Phi}{3}}g^{e}_{\mu\nu},\end{aligned}$$ then the action of the dilaton background part becomes: $$\begin{aligned}
S_G^e &&= \frac{1}{16\pi G_5}\int d^5x\sqrt{-g^e}[R^e-\frac{4}{3}\partial_\mu\Phi\partial^\mu\Phi-V^e(\Phi)- \frac{h(\Phi)}{4}F_{\mu\nu}F^{\mu\nu}],
\label{S_G_E}\end{aligned}$$ with $$V^e(\Phi)=e^{4/3\Phi}V^s(\Phi).$$ In the Einstein frame, the metric ansatz becomes: $$\begin{aligned}
ds^2=\frac{e^{2A_e(z)}}{z^2}[-f(z)dt^2+\frac{1}{f(z)}dz^2+d\vec{x}^2].\label{metrice}\end{aligned}$$ Here, the two metric warp factors in two frames follow the relationship of $A_s = A_e + \frac{2\Phi}{3}$. When considering finite chemical potential, the only non-vanishing component of the gauge potential $A_\mu$ of the strength tensor $F_{\mu\nu}$ is the time component $A_t$, i.e. $A=A_t dt$.

Inserting the above ansatz, one can derive the following equations of motion after certain simplifications [@Fang:2015ytf]: $$\begin{aligned}
A^{\prime\prime}_e-A^{\prime2}_e+\dfrac{2}{z}A^{\prime}_e+\dfrac{4\Phi^{\prime2}}{9}
&  =0,\label{eom-A}\\
A_{t}^{\prime\prime}+\left(  \frac{h^{\prime}}{h}+A^{\prime}_e-\dfrac{1}%
{z}\right)  A_{t}^{\prime}  &  =0,\label{eom-At}\\
f^{\prime\prime}+\left(  3A^{\prime}_e-\dfrac{3}{z}\right)  f^{\prime}%
-e^{-2A_e}z^{2}h(\Phi)A_{t}^{\prime2}  &  =0.\label{eom-f}\end{aligned}$$ The Hawking temperature of the black hole solution can be evaluated numerically by $$\begin{aligned}
T = \frac{f'(z_h)}{4\pi}.\end{aligned}$$

Following [@Li:2013oda], we take the dilaton field in the form of $$\Phi(z) =\alpha\ {\rm tanh} (\beta^2z^2+\gamma^4z^4),$$ which tends to $z^2$ power thus goes to the AdS$_5$ limit at the UV boundary and approaches a positive constant for a possible crossover transition at IR. Then one can solve Eqs.(\[eom-A\])-(\[eom-f\]) by imposing the following boundary conditions at the boundary $z=0$ and the horizon 