functions. To find the ADM Hamiltonian corresponding to the action (\[action1\]), we proceed as in [@Nar72] and [@MatRyaTot73]. We rewrite the action as follows: $$\label{action2}
S=(16\pi)^{-1}\int(\Pi^{ij}\frac{\partial{g_{ij}}}{\partial{t}}+\Pi^{\phi}\frac{\partial{\phi}}{\partial{t}}-NC^0-N_iC^i)d^4x$$ The $\Pi_{ij}$ and $\Pi_\phi$ are respectively the conjugate momenta of the metric functions and scalar field. The lapse and shift functions now play the role of Lagrange multipliers. By varying (\[action2\]) with respect to $N$ and $N_i$, we get the constraints $C^0=0$ and $C^i=0$ with: $$\label{C0}
C^0 =-\sqrt{^{(3)}g}^{(3)}R-\frac{1}{\sqrt{^{(3)}g}}(\frac{1}{2}(\Pi^k _k )^2 -\Pi^{ij}\Pi_{ij})+\frac{1}{\sqrt{^{(3)}g}}\frac{\Pi_\phi ^2 \phi^2 }{6+4\omega}+\sqrt{^{(3)}g}U(\phi)$$ $$C^i =\Pi^{ij}_{\mid j}$$ We rewrite the metric functions $g_{ij}$ as $e^{-2\Omega+2\beta_{ij}}$. It means that $\Omega$ stands for the isotropic part of the metric and $\beta_{ij}$ for the anisotropic parts. Then, using Misner parameterisation [@Mis62]: $$p_k^i=2\pi\Pi_k^i-2/3\pi\delta_k^i\Pi_l^l$$ $$6p_{ij}=diag(p_++\sqrt{3}p_-,p_+-\sqrt{3}p_-,-2p_+)$$ $$\beta_{ij}=diag(\beta_++\sqrt{3}\beta_-,\beta_+-\sqrt{3}\beta_-,-2\beta_+)$$ and rewriting the Hamiltonian as $H=2\pi\Pi_k^k$, from the expression (\[C0\]) and the constraint $C^0=0$, we get: $$\label{hamiltonien}
H^2 = p_+ ^2 +p_- ^2 +12\frac{p_\phi ^2 \phi^2}{3+2\omega}+24\pi^2 R_0 ^6 e^{-6\Omega}U+V(\Omega,\beta_+,\beta_-)$$ with $p_\phi=\pi \Pi_\phi$. The form of $V(\Omega,\beta_+,\beta_-)$ specifies each Bianchi model and is given in table 1. From (\[hamiltonien\]), we derive the Hamiltonian equations: $$\label{betapm}
\dot{\beta}_ \pm = \frac{\partial H}{\partial p_ \pm}=\frac{p_\pm}{H}$$ $$\label{phip}
\dot{\phi}=\frac{\partial H}{\partial p_\phi}=\frac{12\phi^2 p_\phi }{(3+2\omega)H}$$ $$\label{ppp}
\dot{p}_+=-\frac{\partial H}{\partial \beta_ \pm}=-\frac{\partial V(\Omega,\beta_+,\beta_-)/\partial \beta_+}{2H}$$ $$\label{pmp}
\dot{p}_-=-\frac{\partial H}{\partial \beta_ \pm}=-\frac{\partial V(\Omega,\beta_+,\beta_-)/\partial \beta_-}{2H}$$ $$\label{pphip}
\dot{p}_\phi=-\frac{\partial 