It is useful to define the new variable $z$ as $$\begin{aligned}
U_z=(U_{KK}^5+U_{KK}^3\, z^2)^{1/5}.\end{aligned}$$ Then by neglecting the higher order of $F^2$ in the expansion, the $D4$ brane action can be written as \[49\] $$\begin{aligned}
S_{D4}= -\widetilde \mu_4(2\pi\alpha')^2 \int d^4x dz\,[\,
\frac{R^4}{4U_z^{5/2}} \eta^{\mu\nu}\eta^{\rho\sigma} F_{\mu
\rho}F_{\nu \sigma}+\frac{25}{8}\frac{U_z^{9/2}}{U_{kk}^3}
\eta^{\mu\nu} F_{\mu z}F_{\nu z}\,] +\mathcal{O}(F^3) \, ,\end{aligned}$$ where $\widetilde \mu_4$ is $$\begin{aligned}
\widetilde \mu_4=\sqrt{\frac{3}{2}} \, \frac{N_c U_{KK}^{3/2}}{5
R^3}\, \mu_4.\end{aligned}$$ The gauge fields $A_\mu$ ($\mu=0,1,2,3$) and $A_z$ have a mode expansion in terms of complete sets $\{\psi_n(z)\}$ and $\{\phi_n(z)\}$ as $$\begin{aligned}
A_\mu(x^\mu,z)&=&\sum_{n} B_\mu^{(n)}(x^\mu) \psi_n(z) \ ,
\label{expand;Av}\\
A_z(x^\mu,z)&=&\sum_{n} \varphi^{(n)}(x^\mu) \phi_n(z) \ .
\label{expand;A}\end{aligned}$$ After calculating the field strengths, the action (14) is rewritten as $$\begin{aligned}
S_{D4}&=&-\widetilde \mu_4 (2\pi\alpha')^2 \int d^4x dz \sum_{m,n}
\bigg[ \frac{R^4}{4U_z^{5/2}} F_{\mu\nu}^{(m)}F^{\mu\nu
(n)}\psi_m\psi_n +\nonumber \\ &&~~
\frac{25}{8}\frac{U_z^{9/2}}{U_{kk}^3} (
\partial_\mu \varphi^{(m)}\partial^\mu \varphi^{(n)}\phi_m\phi_n
~+B_\mu^{(m)}B^{\mu(n)}\dot\psi_m\dot\psi_n
-2\partial_\mu\varphi^{(m)}B^{\mu(n)}\phi_m\dot\psi_n )
\,\bigg],\nonumber\\
 \label{phiphiBB}\end{aligned}$$ where the over dot denotes the derivative respect to the $z$ coordinate.

We introduce the following dimensionless parameters, $$\widetilde z \equiv\frac{z}{U_{KK}} \ , ~~~ K(\widetilde z)\equiv
1+\widetilde z^2=\left(\frac{U_z}{U_{KK}}\right)^5 \ ,$$ and find that the functions $\psi_n$ ($n\ge 1$) satisfy the normalization condition as $$\widetilde \mu_4 (2\pi\alpha')^2\frac{R^4}{U_{KK}^{3/2}} \int
d\widetilde z\,  K^{-1/2}\, \psi_n\psi_m =\delta_{nm} \ .$$ Also, we suppose the functions $\psi_n$ ($n\ge 1$) satisfy the following condition $$\begin{aligned}
