A multi-time-scale analysis of chemical reaction networks: II. Stochastic systems

Abstract

We consider stochastic descriptions of chemical reaction networks in which there are both fast and slow reactions, and for which the time scales are widely separated. We develop a computational algorithm that produces the generator of the full chemical master equation for arbitrary systems, and show how to obtain a reduced equation that governs the evolution on the slow time scale. This is done by applying a state space decomposition to the full equation that leads to the reduced dynamics in terms of certain projections and the invariant distributions of the fast system. The rates or propensities of the reduced system are shown to be the rates of the slow reactions conditioned on the expectations of fast steps. We also show that the generator of the reduced system is a Markov generator, and we present an efficient stochastic simulation algorithm for the slow time scale dynamics. We illustrate the numerical accuracy of the approximation by simulating several examples. Graph-theoretic techniques are used throughout to describe the structure of the reaction network and the state-space transitions accessible under the dynamics.

Appendix

1.1 The general case in Example 10

When there are $N_0$ molecules, there are $N_0+1$ fast components and the transition matrix K is block tridiagonal, i.e.

$$\begin{aligned} K = \left[ \begin{array}{cccccc} \dfrac{1}{\epsilon }K_1^f+K_1^s &{} K_{1,2}^s &{} &{} &{} &{}\\ K_{2,1}^s &{} \dfrac{1}{\epsilon }K_2^f+K_2^s &{} K_{2,3}^s &{} &{} &{}\\ &{} K_{3,2}^s &{} \ddots &{} \ddots &{} &{} \\ &{} &{} \ddots &{} \ddots &{} \ddots &{} \\ &{} &{} &{} \ddots &{} \ddots &{} K_{N_0,N_0+1}^s\\ &{} &{} &{} &{} K_{N_0+1,N_0}^s &{} \dfrac{1}{\epsilon }K_{N_0+1}^f+K_{N_0+1}^s \end{array} \right] , \end{aligned}$$

where the fast blocks are given by

$$\begin{aligned} K_1^f= & {} \left[ \begin{array}{cccccc} -N_0k_1 &{} k_2 &{} &{} &{} &{}\\ N_0k_1 &{} -[(N_0-1)k_1+k_2] &{} 2k_2 &{} &{} &{}\\ &{} (N_0-1)k_1 &{} -[(N_0-2)k_1+2k_2] &{} 3k_2 &{} &{} \\ &{} &{} \ddots &{} \ddots &{} \ddots &{} \\ &{} &{} &{} \ddots &{} \ddots &{} N_0k_2\\ &{} &{} &{} &{} k_1 &{} -N_0k_2 \end{array} \right] _ {(N_0+1)\times (N_0+1)}\\ K_2^f= & {} \left[ \begin{array}{cccccc} -(N_0-1)k_1 &{} k_2 &{} &{} &{} &{}\\ (N_0-1)k_1 &{} -[(N_0-2)k_1+k_2] &{} 2k_2 &{} &{} &{}\\ &{} (N_0-2)k_1 &{} -[(N_0-3)k_1+2k_2] &{} 3k_2 &{} &{} \\ &{} &{} \ddots &{} \ddots &{} \ddots &{} \\ &{} &{} &{} \ddots &{} \ddots &{} (N_0-1)k_2\\ &{} &{} &{} &{} k_1 &{} -(N_0-1)k_2 \end{array} \right] _ {N_0\times N_0} \cdots \\ K_{N_0}^f= & {} \left[ \begin{array}{cc} -k_1 &{} k_2\\ k_1 &{} -k_2 \end{array} \right] , K_{N_{0}+1}^f = 0. \end{aligned}$$

The slow off-diagonal blocks are given by

$$\begin{aligned}&K_{1,2}^s=\left[ \begin{array}{cccc} k_5 &{} &{} &{} \\ k_4 &{} k_5 &{} &{} \\ &{} \ddots &{} \ddots &{} \\ &{} &{} \ddots &{} k_5\\ &{} &{} &{} k_4 \end{array}\right] _{(N_0+1)\times N_0}, K_{2,3}^s=\left[ \begin{array}{cccc} 2k_5 &{} &{} &{} \\ 2k_4 &{} 2k_5 &{} &{} \\ &{} \ddots &{} \ddots &{} \\ &{} &{} \ddots &{} 2k_5\\ &{} &{} &{} 2k_4 \end{array}\right] _{N_0\times (N_0-1)}, \cdots ,\\&K_{N_{0},N_{0}+1}^s = \left[ \begin{array}{c}N_0k_5\\ N_0k_4 \end{array}\right] . \end{aligned}$$

For the lower diagonal blocks,

$$\begin{aligned} K_{2,1}^s= & {} \left[ \begin{array}{ccccc} N_0k_6 &{} k_3 &{} &{} &{} \\ &{} (N_0-1)k_6 &{} 2k_3 &{} &{}\\ &{} &{} \ddots &{} \ddots &{}\\ &{} &{} &{} k_6 &{} N_0k_3 \end{array}\right] _{N_0\times (N_0+1)} \\ K_{3,2}^s= & {} \left[ \begin{array}{ccccc} (N_0-1)k_6 &{} k_3 &{} &{} &{} \\ &{} (N_0-2)k_6 &{} 2k_3 &{} &{}\\ &{} &{} \ddots &{} \ddots &{}\\ &{} &{} &{} k_6 &{} (N_0-1)k_3 \end{array}\right] _{(N_0-1)\times N_0}, \end{aligned}$$

$\cdots , \qquad K_{N_0+1,N_0}^s = \left[ \begin{array}{cc}k_6&k_3 \end{array}\right]$. Finally, the $K_i^s$ along the diagonal are diagonal matrices of the same dimension as $K_i^f$, and the (j, j)-th entry of $K_i^s$ is the negative sum of rates leaving j-th node of the i-th fast component.

In this case the transition rate is given by

$$\begin{aligned} \tilde{k}_{i,j}^s=L_iK_{i,j}^s\varPi _j. \end{aligned}$$

1.2 Moment equations of the invariant distributions

The low-order moments of the distributions for the fast systems play a role in the QSS reduction in Sect. 4, and here we consider the low-order moment equations.

Theorem 20

Let r be the total number of the reactions in the system. Then the invariant (steady-state) distribution of P(n, t), which we denote P(n), satisfies

$$\begin{aligned} \sum _{\ell =1}^{r} \mathcal{R}_{\ell }(n-\nu \mathcal{E}_{(\ell )} P(n-\nu \mathcal{E}_{(\ell )}) = \sum _{\ell =1}^{r} \mathcal{R}_{\ell }(n) P(n) \end{aligned}$$

(8.1)

and the first two moment equations lead to

$$\begin{aligned}&E[ \nu \mathcal{E}\mathcal{R}(n)]=0, \end{aligned}$$

(8.2)

$$\begin{aligned}&\sum _{i=1}^{r}\big [ \nu \mathcal{E}_{(i)} \otimes E[n \mathcal{R}_i(n)] + E[n \mathcal{R}_i(n)] \otimes \nu \mathcal{E}_{(i)} + \nu \mathcal{E}_{(i)} \otimes \nu \mathcal{E}_{(i)} E[ \mathcal{R}_i(n)] \big ] = 0. \nonumber \\ \end{aligned}$$

(8.3)

Proof

At the steady-state

$$\begin{aligned} \sum _{\ell =1}^{r} \mathcal{R}_{\ell }(n-\nu \mathcal{E}_{(\ell )}) P(n-\nu \mathcal{E}_{(\ell )}) = \sum _{\ell =1}^{r} \mathcal{R}_{\ell }(n) P(n) \end{aligned}$$

(8.4)

By multiplying by n and summing over all the values of $n\in \mathcal{L}(n_0)$, we obtain

$$\begin{aligned} \sum _{n}\sum _{\ell =1}^{r} n \mathcal{R}_{\ell }(n-\nu \mathcal{E}_{(\ell )}) P(n-\nu \mathcal{E}_{(\ell )}) = \sum _{n}\sum _{\ell =1}^{r} n \mathcal{R}_{\ell }(n) P(n). \end{aligned}$$

Using the transformation $n-\nu \mathcal{E}_{(\ell )} \rightarrow n$ on the left side, we obtain

$$\begin{aligned} \sum _{n}\sum _{\ell =1}^{r} (n+\nu \mathcal{E}_{(\ell )}) \mathcal{R}_{\ell }(n) P(n) = \sum _{n}\sum _{\ell =1}^{r} n \mathcal{R}_{\ell }(n) P(n). \end{aligned}$$

By subtracting the right side from the left one,

$$\begin{aligned} \sum _{\ell =1}^{r} \nu \mathcal{E}_{(\ell )} \sum _{n} \mathcal{R}_{\ell }(n) P(n) = \sum _{\ell =1}^{r} \nu \mathcal{E}_{(\ell )} E[\mathcal{R}_{\ell }(n)]= 0. \end{aligned}$$

Thus we conclude that

$$\begin{aligned} \nu \mathcal{E}E[ \mathcal{R}(n)]= 0. \end{aligned}$$

If the deficiency $\delta \equiv \rho (\mathcal{E}) - \rho (\nu \mathcal{E})$ is zero, then $E[ \mathcal{R}(n)]$ is a cycle in the graph (Othmer 1979).⁷

At the next order we multiply Eq. (8.4) by a tensor product $n\otimes n$ and sum over n. Then by a similar argument, we obtain

$$\begin{aligned}&\sum _{i=1}^{r}\big [ \nu \mathcal{E}_{(i)} \otimes \sum _{n} n \mathcal{R}_i(n) p(n) + \sum _{n} n \mathcal{R}_i(n) p(n)\otimes \nu \mathcal{E}_{(i)}\\&\qquad +\, \nu \mathcal{E}_{(i)} \otimes \nu \mathcal{E}_{(i)} \sum _{n} \mathcal{R}_i(n) p(n) \big ]\\&\quad = \sum _{i=1}^{r}\big [ \nu \mathcal{E}_{(i)} \otimes E[n \mathcal{R}_i(n)] + E[n \mathcal{R}_i(n)] \otimes \nu \mathcal{E}_{(i)} + \nu \mathcal{E}_{(i)} \otimes \nu \mathcal{E}_{(i)} E[ \mathcal{R}_i(n)] \big ]\\&\quad = 0. \end{aligned}$$

If $\delta =0$, the two lowest moment equations can be simplified to

$$\begin{aligned}&E[ \mathcal{R}(n)]=0 \end{aligned}$$

(8.5)

$$\begin{aligned}&\quad \sum _{i=1}^{r}\big [ \nu \mathcal{E}_{(i)} \otimes E[n \mathcal{R}_i(n)] + E[n \mathcal{R}_i(n)] \otimes \nu \mathcal{E}_{(i)} \big ] = 0. \end{aligned}$$

(8.6)

When all reactions are linear the problem is much simpler, and the evolution equations for the first and second moments can be written explicitly in terms of those moments (Gadgil et al. 2005).

As a consequence of Theorem 20, similar equations can be obtained for the quasi-steady-state of the probability distribution for the fast subsystem in a two-time scale stochastic network. We first define the expectation of a function f(n) over a discrete reaction simplex $\mathcal{L}_f$ for the fast subsystem as follows.

$$\begin{aligned} E_{\mathcal{L}_f}[f(n)] \equiv \sum _{n\in \mathcal{L}_f} f(n) p(n). \end{aligned}$$

Corollary 21

Let $r_f$ be the total number of fast reactions. Then at the steady-state of the fast subsystem, the governing equation is given by

$$\begin{aligned} \sum _{i=1}^{r_f} \mathcal{R}^f_i(n-\nu \mathcal{E}^f_{(i)}) P(n-\nu \mathcal{E}^f_{(i)}) = \sum _{i=1}^{r_f} \mathcal{R}^f_i(n) P(n) \end{aligned}$$

(8.7)

and for each discrete reaction simplex $\mathcal{L}_f$,

$$\begin{aligned}&E_{\mathcal{L}_f} [ \mathcal{R}^f(n)]=0, \end{aligned}$$

(8.8)

$$\begin{aligned}&\sum _{i=1}^{r_f}\big [ \nu \mathcal{E}^f_{(i)} \otimes E_{\mathcal{L}_f}[n \mathcal{R}^f_{i}(n)] + E_{\mathcal{L}_f}[n \mathcal{R}^f_i(n)] \otimes \nu \mathcal{E}^f_{(i)} \big ] = 0. \end{aligned}$$

(8.9)

Proof

The ‘state-wise’ form of the master Eq. (3.1) can be written

$$\begin{aligned} \dfrac{d}{dt}P(n,t)= & {} \sum _{\ell }\dfrac{1}{{\epsilon }} \big [\mathcal{R}^f_{\ell }(n-\nu \mathcal{E}^f_{(\ell )})\cdot P(n-\nu \mathcal{E}^f_{(\ell )},t)-\mathcal{R}^f_{\ell }(n)\cdot P(n,t)\big ] \nonumber \\ \nonumber \\&+ \sum _{k} \big [\mathcal{R}^s_{k}(n-\nu \mathcal{E}^s_{(k)})\cdot P(n-\nu \mathcal{E}^s_{(k)},t)- \mathcal{R}^s_{k}(n)\cdot P(n,t)\big ], \end{aligned}$$

(8.10)

where $\mathcal{R}^f$ and $\mathcal{R}^s$ are the transition rates of fast and slow reactions, respectively and $\mathcal{E}^f$ and $\mathcal{E}^s$ are incidence matrices for fast and slow reactions, respectively.

In the previous theorem, substitute $\mathcal{E}^f, \mathcal{R}^f$ and $E_{\mathcal{L}_f}[\cdot ]$ into $\mathcal{E}, \mathcal{R}$ and $E[\cdot ]$ and use the full rank assumption on $\nu \mathcal{E}^f$.