For simplicity, we first look at the continuum equations of motion for $(\phi, \pi_T)$ without dissipation or noise:
$$\begin{aligned}
\dot{\phi} & = - \nabla_\mu ( \phi , \pi_{T,\mu} ) = - \pi_{T,\mu} \nabla_\mu \phi, \\
\dot{\pi}_{T,i} & = - \hat{{\cal T}}_{ij} \nabla_k ( \pi_{T,k} \pi_{T,j} + \nabla_k \phi \nabla_j \phi ) \\
& = - \hat{{\cal T}}_{ij} ( \nabla_k \left(\pi_{T,k} \pi_{T,j} \right) + \nabla_j \phi \nabla^2 \phi )
\end{aligned}$$
The time derivative of the Hamiltonian is (assumed $\rho = 1$),
$$\dot{H} = \int d^3x [ - \dot{\phi} \nabla^2 \phi + \pi_{T,i} \dot{\pi}_{T,i} + V'(\phi) \dot{\phi} ] $$
Substituting
$$\begin{aligned}
\dot{{\cal H}} = - \int d^3x \ \left[ - \nabla^2\phi \\
\vec{\pi}_T \cdot \vec{\nabla} \phi
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+ \vec{\pi}_T \cdot \left( \vec{\pi}_T \cdot \vec{\nabla} \right) \vec{\pi}_T
%
+ \nabla^2\phi \ \vec{\pi}_T \cdot \vec{\nabla} \phi
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+ \vec{\nabla} \cdot \left( V(\phi) \vec{\pi}_T \right) \right].
\end{aligned}$$
The first and third terms cancel each other. In the continuum, the
second term can be written as a divergence,
$$\begin{aligned}
\vec{\pi}_T \cdot \left( \vec{\pi}_T \cdot \vec{\nabla} \right) \vec{\pi}_T = \vec{\nabla} \cdot \left( \vec{\pi}_T \ \frac{\vec{\pi}_T^2}{2} \right).
\end{aligned}$$
However, this is not guaranteed in the discretised version if we use the
divergence type term
∇k(πT, kπT, j). Morinishi et al suggests we use a
skew-symmetric form to take care of this:
$$\begin{aligned}
\nabla_k \left(\pi_{T,k} \pi_{T,j} \right) = \frac{1}{2} \ \nabla_k \left(\pi_{T,k} \pi_{T,j} \right) + \frac{1}{2} \ \pi_{T,k} \ \nabla_k \pi_{T,j}.
\end{aligned}$$
Also, in the discretised version we expect the first and third terms
to cancel each other if we use the central difference scheme throughout,
i.e, for (∇jϕ∇2ϕ) term and also the kinetic term,
(∇ϕ)2, in the Hamiltonian:
$$\begin{aligned}
\nabla_j \phi &\to \frac{\phi(\vec{x}+\hat{\nu}_j) - \phi(\vec{x} - \hat{\nu}_j)}{2}, \nonumber \\
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\nabla^2\phi &\to \sum_\nu \frac{1}{4} \ \left(\phi(\vec{x} + 2 \hat{\nu}) + \phi(\vec{x} - 2 \hat{\nu}) - 2 \ \phi(\vec{x}) \right)
\end{aligned}$$
We note that the central difference scheme satisfies
∫d3x(∇ϕ)2 → − ∫d3x ϕ∇2ϕ
in Hamiltonian:
$$\begin{aligned}
{\cal H} &= \frac{1}{2} \ \sum_i \ \left(\frac{\phi_{i+1} - \phi_{i-1}}{2}\right)^2 = \frac{1}{2} \ \frac{1}{4} \sum_{i} \left( \phi_{i+1}^2 + \phi_{i-1}^2 - 2 \ \phi_{i+1} \phi_{i-1} \right), \nonumber \\
%
& = \frac{1}{2} \ \frac{1}{4} \ \sum_i \left( 2 \phi_i^2 - \phi_i \ \phi_{i+2} \ - \phi_i \phi_{i-2} \right) = - \frac{1}{2} \ \sum_i \ \phi_i \left( \frac{\phi_{i+2} + \phi_{i-2} - 2 \phi_i}{4} \right)
\end{aligned}$$
If we discretize the spatial derivatives we get (leaving aside the
projector and using the skew symmetric form of Morinishi et al for
∇k(πT, kπT, i)),
$$\begin{aligned}
\dot{\phi} &= - \pi_{T,\mu}(x) \ \nabla^c_{\mu} \phi \\
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\dot{\pi}_{T,\mu} &= - \Bigg[ \frac{1}{2} \nabla^c_{\nu} \left( \pi_{T,\nu} \pi_{T,\mu} \right) + \frac{1}{2} \ \pi_{T,\nu} \ \nabla^c_{\nu} \pi_{T, \mu} \nonumber \\
%
& + \nabla^c_\mu \phi \ \sum_\nu \ \frac{ \left(\phi(\vec{x} + 2 \hat{\nu}) + \phi(\vec{x} - 2 \hat{\nu}) - 2 \ \phi(\vec{x}) \right)}{4} \Bigg],
\end{aligned}$$
where
∇μcψ(x⃗) = (ψ(x⃗+μ̂)−ψ(x⃗−μ̂))/2.
For the projector we use
$$\begin{aligned}
P_{\mu\nu} = \delta_{\mu\nu} - \frac{\tilde{k}_\mu \ \tilde{k}_\nu}{\tilde{k}^2},
\end{aligned}$$
with k̃μ = sin (kμ), and
kμ = (2π/L)n̂k, μ. The time
derivatives can be solved
using Runge-Kutta methods.
Writing the evolution
equations as,
$$\begin{aligned}
\dot{\phi}_\mu = {\cal F}_\mu (\phi_\mu),
\end{aligned}$$
the ‘third-order’ RK scheme gives,
$$\begin{aligned}
\phi^{n+1}_\mu = \phi^n_{\mu} + \Delta t \ \left( \frac{1}{6} \ k_{\mu,1} + \frac{1}{6} \ k_{\mu,2} + \frac{2}{3} \ k_{\mu,3} \right),
\end{aligned}$$
where kμ, i are given by
$$\begin{aligned}
k_{\mu,1} &= {\cal F}_\mu (\phi^n_\mu), \\
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k_{\mu, 2} &= {\cal F}_\mu \left( \phi^n_\mu + \Delta t \ k_{\mu,1} \right) \\
%
k_{\mu,3} &= {\cal F}_\mu \left( \phi^n_\mu + \frac{\Delta t}{4} \ \left( k_{\mu,1} + k_{\mu,2} \right) \right).
\end{aligned}$$