in the bulk scalar potential. Here we find the solutions of the field equations when $V(\Phi)$ is retained up to cubic term in $\Phi$. In this scenario, the five dimensional action in ST theory turns out to be, $$\begin{aligned}
 S = \int d^4x dy \sqrt{\tilde{G}} \bigg[\frac{\tilde{R}}{2\kappa^2} + \frac{1}{2}\tilde{G}^{MN}\partial_M\xi 
 \partial_N\xi - (1/2)m_{\Phi}^2\xi^2 + \frac{g}{3}\xi^3 + \Lambda_{eff}\bigg]
 \label{action2ST_final}\end{aligned}$$ where $g$ is the self cubic coupling of $\Phi(\phi)$ and can be easily determined from the form of $V(\Phi)$ presented in eqn.(\[scalar\_potential\]) as, $$\begin{aligned}
 g = -\frac{\sqrt{3}\kappa}{16\alpha}\frac{[2\sqrt{9 - 40\kappa^2\alpha\Lambda} + 3]}{[\sqrt{9 - 40\kappa^2\alpha\Lambda} - 2]^{\frac{2}{3}}}
 \label{g}\end{aligned}$$ Considering the metric ansatz as, $$d\tilde{s}^2 = e^{- 2 A(\phi)} \eta_{\mu\nu} dx^{\mu} dx^{\nu} - dy^2
 \label{grav.sol1.ST_final}$$ the gravitational as well as the scalar field equations of motion take the following form, $$\begin{aligned}
 4 A'^2(y) - A''(y) = -(2\kappa^2/3)(\frac{1}{2}m_{\Phi}^2\xi^2 - \frac{g}{3}\xi^3)
 \label{equation1_final}\end{aligned}$$

$$\begin{aligned}
  A'^2(y)= \frac{\kappa^2}{12}\xi'^2 - (\kappa^2/6)(\frac{1}{2}m_{\Phi}^2\xi^2 - \frac{g}{3}\xi^3)
 \label{equation2_final}\end{aligned}$$

$$\begin{aligned}
 \xi''(y) = 4A'\xi' + m_{\Phi}^2\xi(\phi) - g\xi^2(\phi)
 \label{equation3_final}\end{aligned}$$

To determine the solutions of the above differential equations, we apply the iterative method by considering the form of metric determined in eqn. (\[grav.sol1.ST\]) as the zeroth order solution. In the leading order correction of $\kappa v_h$, $\xi(\phi)$ and $A(\phi)$ turn out to be $$\begin{aligned}
 \xi(y)&=&\bigg[Ae^{(2+\nu)ky} + Be^{(2-\nu)ky}\bigg]\ - \frac{\kappa 