analysis in order to find the degrees of fredoom of our model. To this end, we will use the ADM formalism [@adm], where the line element acquires the form $$\begin{aligned}
 ds^2=g_{\mu\nu}dx^{\mu}dx^{\nu}= -N^2dt^2+\gamma_{ij}(dx^i+N^idt)(dx^j+N^jdt),
 \end{aligned}$$ the matter Lagrangian becomes $$\begin{aligned}
  {\mathcal L}_{matt}
    =\left(\frac{\dot{\phi}^2}{2N^2}-\frac{N^i}{N^2}\dot{\phi}\phi_{,i}-\frac{1}{2}\left(\gamma^{ij}-\frac{N^iN^j}{N^2}\right)\phi_{,i}\phi_{,j}-V(\phi)\right),
  \end{aligned}$$ and the action is given by $$\begin{aligned}
 S=\int_{-\infty}^{\infty}\left\{\int_{\Sigma_t} N\sqrt{\gamma} \left(\frac{1}{2}({\mathcal R}+{\mathcal I})
 +f(-\nabla_{\mu}u^{\mu})+{\mathcal L}_{matt}  \right)d^3x\right\} dt , 
 \end{aligned}$$ where ${\mathcal R}$ is the intrinsic curvature, i.e. the scalar curvature of $\Sigma_t$ , and ${\mathcal I}= K_{ij}K^{ij}-(K_i^i)^2$ is the extrinsic curvature scalar, being $$\begin{aligned}
 K_{ij}=\frac{1}{2N}\left(D_iN_j+D_jN_i-\dot{\gamma}_{ij}     \right)
 \end{aligned}$$ the extrinsic curvature tensor, with $D$ the induced Levi-Civita connection in the slicing $\Sigma_t$.

Introducing a Lagrangian multiplier field, namely $\beta$, the action becomes $$\begin{aligned}
\label{action1}
 S=\int_{-\infty}^{\infty}\left\{\int_{\Sigma_t} N\sqrt{\gamma} \left(\frac{1}{2}({\mathcal R}+{\mathcal I})
 +f(\chi)+ \beta(\chi+ \nabla_{\mu}u^{\mu} )+ {\mathcal L}_{matt}  \right)d^3x\right\} dt=\nonumber\\ 
 \int_{-\infty}^{\infty}\left\{\int_{\Sigma_t} N\sqrt{\gamma} \left(\frac{1}{2}({\mathcal R}+{\mathcal I})
 +f(\chi)+ \beta\chi-  \beta_{, \mu}u^{\mu} + {\mathcal L}_{matt}  \right)d^3x\right\} dt, 
 \end{aligned}$$ with the canonical momenta being $$\begin{aligned}
P^{ij}=\frac{1}{2}\sqrt{\gamma}(K^{ij}-K^l_l\gamma^{ij}), \quad P_{\beta}=\frac{\sqrt{\gamma}}{N}(u_0-N^iu_i), \quad 
P_{\phi}=\frac{\sqrt{\gamma}}{N}(\dot{\phi}-\phi_{,i}N^i)-\frac{N\sqrt{\gamma}}{\sqrt{-\phi_{,\nu}\phi^{,\nu}}}(\beta^{,0}+u^0u^{\mu}\beta_{,\mu}),\end{aligned}$$ and the constraints $$\begin{aligned}
P_{\chi}\approx 0, \quad P_{N}\approx 0, \quad P_{N^i}\approx 0. \end{aligned}$$

Note that, from the equation $P_{\beta}=\frac{\sqrt{\gamma}}{N}(u_0-N^iu_i)$, we obtain $\dot{\phi}$ as a function of $P_{\beta}$. In fact, taking the square of this expression one gets $-\phi_{,\nu}\phi^{,\nu}=\frac{\gamma \gamma^{ij}\phi_{,i}\phi_{,j}}{{P^2_{\beta}-\gamma}}$ and, thus, $$\begin{aligned}
\dot{\phi}=\frac{N\sqrt{\gamma^{ij}\phi_{,i}\phi_{,j}}P_{\beta}}{\sqrt{P_{\beta}^2-\gamma}}+N^i\phi_{,i}.\end{aligned}$$ And, from $P_{\phi}=\frac{\sqrt{\gamma}}{N}(\dot{\phi}-\phi_{,i}N^i)-\frac{N\sqrt{\gamma}}{\sqrt{-\phi_{,\nu}\phi^{,\nu}}}(\beta^{,0}+u^0u^{\mu}\beta_{,\mu})$, we 