for this specific choice form of $F(R)$ has the form, $$\begin{aligned}
 V(\Phi)&=&\frac{1}{8\kappa^2\alpha} \exp{(-\frac{5}{2\sqrt{3}}\kappa\Phi)}\big[\exp{(\frac{3}{2\sqrt{3}}\kappa\Phi)} - 1\big]^2\nonumber\\
 &-&\Lambda \bigg[\exp{(-\frac{5}{2\sqrt{3}}\kappa\Phi)}-1\bigg]
 \label{scalar_potential}\end{aligned}$$

One can check that the above potential (in eqn.(\[scalar\_potential\])) is stable for the parametric regime $\alpha > 0$. The stable value ($<\Phi>$) as well as the mass squared ($m_{\Phi}^2$) of the scalar field ($\Phi$) are given by the following two equations $$\exp{\bigg(\frac{3}{2\sqrt{3}}\kappa<\Phi>\bigg)} = \bigg[\sqrt{9 - 40\kappa^2\alpha\Lambda} - 2\bigg]
 \label{vev_phi}$$ and $$m_{\Phi}^2 = \frac{1}{8\alpha} \bigg[\sqrt{9 - 40\kappa^2\alpha\Lambda}\bigg] \bigg[\sqrt{9 - 40\kappa^2\alpha\Lambda} - 2\bigg]^{-\frac{2}{3}}
 \label{mass_phi}$$ Furthermore, the minimum value of the potential i.e. $V(<\Phi>)$ is non zero and serves as a cosmological constant. Thus the effective cosmological constant in scalar-tensor theory is $\Lambda_{eff} = \Lambda - V(<\Phi>)$ where $V(<\Phi>)$ is, $$\begin{aligned}
 V(<\Phi>)&=&\Lambda + \bigg[\sqrt{9 - 40\kappa^2\alpha\Lambda} - 2\bigg]^{-\frac{5}{3}}\nonumber\\
 &\bigg[&-\Lambda + (1/8\kappa^2\alpha)\bigg[\sqrt{9 - 40\kappa^2\alpha\Lambda}-3\bigg]^2\bigg]
 \nonumber\\\end{aligned}$$ Above form of $V(<\Phi>)$ with $\Lambda < 0$ clearly indicates that $\Lambda_{eff}$ is also negative.\

Solutions in scalar-tensor and in the corresponding F(R) theory
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Considering $\xi$ as the fluctuation of the scalar field over its vev, the final form of action for the scalar-tensor theory in the bulk can be written as, $$\begin{aligned}
 S = \int d^4x dy \sqrt{\tilde{G}} [\frac{\tilde{R}}{2\kappa^2} + \frac{1}{2}\tilde{G}^{MN}\partial_M\xi 
 \partial_N\xi - (1/2)m_{\Phi}^2\xi^2 + \Lambda_{eff}]
 \label{action2ST}\end{aligned}$$ where the 