## Formulation of MZ-guided Markovian/Non-Markovian DPD

The equation of motion (EOM) of the coarse-grained (CG) particles obtained from the Mori-Zwanzig projection is given by

\begin{eqnarray}\label{equ:EoM} \frac{d}{dt}\mathbf{P}_I &=& \frac{1}{\beta} \frac{\partial}{\partial \mathbf{R}_I} {\rm{ln}} \omega(\mathbf{R}) \\ \nonumber &-& {\beta} \sum_{X=1}^{K}\int_{0}^{t} ds \left \langle [\delta\mathbf{F}_I(t-s)] [\delta\mathbf{F}_X(0)]^T \right \rangle \frac{\mathbf{P}_X(s)}{M_X} \\ &+& \delta\mathbf{F}_I(t) \ , \end{eqnarray}

where $\beta = 1/k_BT$ with $T$ the thermodynamic temperature and $k_B$ the Boltzmann constant, $\mathbf{R}=\{\mathbf{R}_1,\mathbf{R}_2,\cdots,\mathbf{R}_K\}$ is a phase point in the CG phase space, and $\omega (\mathbf{R})$ is defined as a normalized partition function of all the microscopic configurations at phase point $\mathbf{R}$ given by

$$\label{equ:omega} \omega (\mathbf{R})=\frac{\int d^N \mathbf{\hat r}\delta(\mathbf{\hat R}-\mathbf{R})e^{-\beta U} } {\int d^N \mathbf{\hat r} e^{-\beta U} } \ ,$$

where $U$ is the potential energy corresponding to the phase point $\mathbf{R}$, and the integrations are performed over all the possible microscopic configurations $\{\mathbf{\hat r}_i\}$.

In the right-hand side of Eq.(\ref{equ:EoM}), the first term represents the conservative force due to the change of microscopic configuration, and it is the ensemble average force on cluster $I$ denoted as $\langle\mathbf{F}_I \rangle$. The last term $\delta\mathbf{F}_I$ is the fluctuating force on cluster $I$ and it is given by $\delta\mathbf{F}_I = \mathbf{F}_I - \langle \mathbf{F}_I \rangle$ in which $\mathbf{F}_I$ is the instantaneous total force acting on the cluster $I$. The second term of Eq. (\ref{equ:EoM}) is the friction force determined by an integral of memory kernel.

${\color{red}{First\ approximation:}}$ Here, we assume that the non-bonded interactions between neighboring clusters in the microscopic system are explicitly pairwise decomposable, and hence the total force consists of pairwise forces, e.g. $\mathbf{F}_I \approx \sum_{J\neq I}\mathbf{F}_{IJ}$ and $\delta\mathbf{F}_I \approx \sum_{J\neq I}\delta\mathbf{F}_{IJ}$.

However, when we consider the force $\mathbf{F}_{I J}$ that a molecule $J$ exerts on another molecule $I$, in principle, $\mathbf{F}_{I J}$ involving many-body effects depends on all the COM coordinates $\mathbf{R}$ as well as their microscopic configurations. Although Eq. (\ref{equ:EoM}) based on the Mori-Zwanzig formalism is accurate, a direct computation of the many-body interactions is very difficult.

${\color{red}{Second\ approximation:}}$ In practice, we neglect the many-body correlations between different pairs, and assume that the force $\mathbf{F}_{I J}$ between two clusters $I$ and $J$ depends only on the relative COM positions $\mathbf{R}_{I}$ and $\mathbf{R}_{J}$ and is independent of the positions of the rest of clusters.

Therefore, the conservative term is in Eq.(\ref{equ:EoM}) approximated by $$\label{equ:FC} \frac{1}{\beta} \frac{\partial}{\partial \mathbf{R}_I} {\rm{ln}} \omega(\mathbf{R}) = \langle \mathbf{F}_I \rangle \approx \sum_{J\neq I}\langle \mathbf{F}_{IJ} \rangle = \sum_{J\neq I}{F_{IJ}^C}(R_{IJ})\mathbf{e}_{IJ}$$

where $\mathbf{e}_{IJ}$ is the unit vector from CG particle $J$ to $I$ given by $\mathbf{e}_{I J}=(\mathbf{R}_I-\mathbf{R}_J)/R_{I J}$ with $R_{I J}=|\mathbf{R}_I-\mathbf{R}_J|$. The rotational symmetry of the CG pairs about the $\mathbf{e}_{IJ}$ suggests that, on average, $\mathbf{F}_{IJ}$ has no preference on the plane perpendicular to $\mathbf{e}_{IJ}$ and remains only nonzero component along $\mathbf{e}_{IJ}$. Here, $F_{IJ}^C(R_{I J})$ represents the magnitude of conservative force $\mathbf{F}_{IJ}^C$, which is time independent but distance dependent.

Similarly, the fluctuating force defined as the deviation from the mean force is also decomposed into pairwise forces \begin{eqnarray}\label{equ:FR} \delta\mathbf{F}_I &=& \mathbf{F}_I - \langle \mathbf{F}_I \rangle \approx \sum_{J\neq I}\delta\mathbf{F}_{IJ} \\ \delta\mathbf{F}_{IJ} &=& \mathbf{F}_{IJ}- \langle \mathbf{F}_{IJ} \rangle = \mathbf{F}_{IJ} -{F_{IJ}^C}(R_{IJ})\mathbf{e}_{IJ} \end{eqnarray}

where $\mathbf{F}_{IJ}$ is the instantaneous force exerted by cluster $J$ on cluster $I$, and $\langle \mathbf{F}_{IJ}\rangle$ is the ensemble average of $\mathbf{F}_{IJ}$ obtained by Eq. (\ref{equ:FC}).

Now we consider the memory kernel in Eq. (\ref{equ:EoM}). Based on the second approximation, the correlation of fluctuating forces between different pairs is ignored. Thus, we have \begin{eqnarray}\label{equ:no_corelation} &&\sum_{J\neq I}\sum_{Y\neq X}\left \langle [\delta\mathbf{F}_{IJ}(t-s)] [\delta\mathbf{F}_{XY}(0)]^T \right \rangle \mathbf{V}_X(s) \nonumber\\ &=&\left \langle [\delta\mathbf{F}_{IJ}(t-s)] [\delta\mathbf{F}_{IJ}(0)]^T \right \rangle \mathbf{V}_I(s)|_{X=I,Y=J} + \nonumber\\ &&\left \langle [\delta\mathbf{F}_{IJ}(t-s)] [\delta\mathbf{F}_{JI}(0)]^T \right \rangle \mathbf{V}_J(s)|_{X=J,Y=I} \nonumber\\ &=&\left \langle [\delta\mathbf{F}_{IJ}(t-s)] [\delta\mathbf{F}_{IJ}(0)]^T \right \rangle \mathbf{V}_{IJ}(s) \end{eqnarray}

where $\mathbf{V}_{IJ}=\mathbf{V}_{J}-\mathbf{V}_{J}$ is the relative velocity of CG particle $I$ to $J$. Moreover, ${\color{red}{Third\ approximation:}}$ we assume that the memory on time is finite, e.g. history length $N\cdot\Delta t$ where $\Delta t$ is the time step of DPD simulations. Therefore, the time correlation between the fluctuating forces is zero when the time interval is larger than $N\Delta t$ $$\label{equ:no_history} \left\langle[\delta\mathbf{F}_I(t)] [\delta\mathbf{F}_X(0)]^T \right \rangle|_{t>N\Delta t} = 0$$

Then, the second term in Eq.(\ref{equ:EoM}) can be expanded as follows:

\begin{eqnarray}\label{equ:FD} &&-{\beta} \sum_{X=1}^{K}\int_{0}^{t} ds \left\langle [\delta\mathbf{F}_I(t-s)] [\delta\mathbf{F}_X(0)]^T \right \rangle \mathbf{V}_X(s) \nonumber\\ &=& -{\beta}\sum_{X=1}^{K}\int_{t-N\Delta t}^{t} ds \left \langle [\delta\mathbf{F}_I(t-s)] [\delta\mathbf{F}_X(0)]^T \right \rangle \mathbf{V}_X(s) \nonumber\\ &=& -{\beta\cdot \Delta t}\sum_{X=1}^{K}\sum_{n=0}^{N}\alpha_n \left \langle [\delta\mathbf{F}_{I}(n\Delta t)] [\delta\mathbf{F}_{X}(0)]^T \right \rangle \mathbf{V}_X(t-n\Delta t) \nonumber\\ &=& -{\beta\cdot \Delta t}\sum_{X=1}^{K}\sum_{J\neq I}\sum_{Y\neq X}\sum_{n=0}^{N}\alpha_n \left \langle [\delta\mathbf{F}_{IJ}(n\Delta t)] [\delta\mathbf{F}_{XY}(0)]^T \right \rangle \mathbf{V}_X(t-n\Delta t) \nonumber\\ &=& -{\beta\cdot \Delta t}\sum_{J\neq I}\sum_{n=0}^{N}\alpha_n \left \langle [\delta\mathbf{F}_{IJ}(n\Delta t)] [\delta\mathbf{F}_{IJ}(0)]^T \right \rangle \mathbf{V}_{IJ}(t-n\Delta t) \end{eqnarray}

We define the friction matrix ${\Gamma}_{IJ,n}$ as $$\label{equ:Gamma} {\Gamma}_{IJ,n}=\beta \left \langle [\delta\mathbf{F}_{IJ}(n\Delta t)] [\delta\mathbf{F}_{IJ}(0)]^T \right \rangle$$

where $\delta\mathbf{F}_{IJ}$ is the fluctuating force. Generally, the fluctuating force $\delta {\bf{F}}_{I J}$ is not parallel to the radial direction $\mathbf{e}_{IJ}$. However, $\delta {\bf{F}}_{I J}$, on average, is transversely isotropic with respect to $\mathbf{e}_{IJ}$ because the instantaneous pairwise force $\mathbf{F}_{IJ}$ has no preference between directions $\perp_1$ and $\perp_2$.

When we calculate the friction matrix, we do not distinguish between the directions $\perp_1$ and $\perp_2$ and decompose $\delta\mathbf{F}_{IJ}$ into two parts

\begin{align} \delta \mathbf{F}_{IJ}&=(\mathbf{e}_{IJ}\mathbf{e}^T_{IJ})\cdot\delta\mathbf{F}_{IJ}+(\mathbf{1}- \mathbf{e}_{IJ}\mathbf{e}^T_{IJ})\cdot\delta\mathbf{F}_{IJ} \nonumber \\ &=\delta\mathbf{F}^{\parallel}_{IJ}+\delta\mathbf{F}^{\perp}_{IJ} \ , \label{equ:randomF} \end{align}

where $\delta\mathbf{F}^{\parallel}_{IJ}$ is the component along vector $\mathbf{e}_{IJ}$ and $\delta\mathbf{F}^{\perp}_{IJ}$ the perpendicular part whose modulus is equally distributed on directions $\perp_1$ and $\perp_2$.

Remark: The memory term given by Eq. (\ref{equ:FD}) can be further simplified with a ${\color{red}{Markovian\ assumption}}$ that the memory of fluctuating force in time is short enough to be approximated by a Dirac delta function

\begin{eqnarray} &\beta \langle [\delta\mathbf{F}_{IJ}(t-s)][\delta\mathbf{F}_{IJ}(0)]^T \rangle = 2 {\gamma}_{IJ} \delta(t-s) \ , \label{equ:Mark_app1} \\ &{\beta}\int_{0}^{t} ds \left \langle [\delta\mathbf{F}_{IJ}(t-s)][\delta\mathbf{F}_{IJ}(0)]^T \right \rangle {\mathbf{V}_{IJ}(s)} = \gamma_{IJ} \cdot {\mathbf{V}_{IJ}(t)} \ , \label{equ:Mark_app2} \end{eqnarray}

where $\gamma_{IJ}$ is the friction tensor defined by $\gamma_{I J} = \beta \int_{0}^{\infty} dt \left \langle [\delta\mathbf{F}_{IJ}(t)][\delta\mathbf{F}_{IJ}(0)]^T \right \rangle$. Then, the equation of motion of DPD particles based on the Markovian approximation can be expressed by

\begin{eqnarray}\label{equ:DPD} \frac{d\mathbf{P}_I}{dt}&=&\sum_{J\neq I}\left\{\right. F^C_{IJ}(R_{IJ})\mathbf{e}_{IJ} - {\gamma}_{IJ}(R_{IJ}) \left( \mathbf{e}_{IJ}\cdot \mathbf{V}_{IJ} \right)\mathbf{e}_{IJ} +\delta\mathbf{F}_{IJ}(t) \left.\right\} \end{eqnarray}