Stochastic Simulation of Continuous Time Random Walks: Minimizing Error in Time-Dependent Rate Coefficients for Diffusion-Limited Reactions

Abstract

A reaction limited by standard diffusion is simulated stochastically to illustrate how the continuous time random walk (CTRW) formalism can be implemented with minimum statistical error. A step-by-step simulation of the diffusive random walk in one dimension reveals the fraction of surviving reactants P(t) as a function of time, and the time-dependent unimolecular reaction rate coefficient K(t). Accuracy is confirmed by comparing the time-dependent simulation to results from the analytical master equation, and the asymptotic solution to that of Fickian diffusion. An early transient feature is shown to arise from higher spatial harmonics in the Fourier distribution of walkers between reaction sites. Statistical ‘shot’ noise in the simulation is quantified along with the offset error due to the discrete time derivative, and an optimal simulation time interval $\Delta t_0$ is derived to achieve minimal error in the finite time-difference estimation of the reaction rate. The number of walkers necessary to achieve a given error tolerance is derived, and $W = 10^7$ walkers is shown to achieve an accuracy of $\pm 0.2\%$ when the survival probability reaches $P(t) \sim \frac{1}{3}$. The stochastic method presented here serves as an intuitive basis for understanding the CTRW formalism, and can be generalized to model anomalous diffusion-limited reactions to prespecified precision in regimes where the governing wait-time distributions have no analytical solution.

Appendix: Continuum Limit for Survival Probability Distribution

The continuum expressions for the diffusion-limited reaction are detailed below. Using the following definition of the Fourier transform,

$$\begin{aligned} \widetilde{p}(k,t) = \frac{1}{\sqrt{2\pi }}\int {p}(x,t)e^{-ikx} dx \end{aligned}$$

(53)

Eq. (33) can be expressed as

$$\begin{aligned} \widetilde{p}(k,t) = \sum _{j = -\infty }^{\infty } (-1)^j \left[ \frac{1}{\sqrt{2\pi }}\int G\left( x-j {x_\textrm{tr}},t\right) e^{-ikx} dx\right] \end{aligned}$$

(54)

Taking advantage of the translation property of Fourier transforms, this can be written as

$$\begin{aligned} \begin{aligned} \widetilde{p}(k,t)&= \sum _{j = -\infty }^{\infty } (-1)^j e^{-ij{x_\textrm{tr}}k}\widetilde{G}\left( k,t\right) \\&= \widetilde{G}\left( k,t\right) \left[ \sum _{j = -\infty }^{\infty } e^{i{2jx_\textrm{tr}}k} - \sum _{j = -\infty }^{\infty } e^{i{(2j-1)x_\textrm{tr}}k}\right] \\&= \widetilde{G}\left( k,t\right) \left( 1-e^{-i{x_\textrm{tr}}k}\right) \sum _{j = -\infty }^{\infty } e^{i2j{x_\textrm{tr}}k}~. \end{aligned} \end{aligned}$$

(55)

where the Fourier transform of G(x, t) is:

$$\begin{aligned} \begin{aligned} \widetilde{G}(k,t)&= \frac{1}{\sqrt{2\pi }}\int _{-\infty }^{\infty }(4 \pi D t)^{-\frac{1}{2}} \exp {\left( -\frac{x^2}{4 D t}\right) }e^{-i k x}dx \\&= \frac{1}{\sqrt{2\pi }}e^{-k^2 D t}~. \end{aligned} \end{aligned}$$

(56)

Applying the Poisson summation formula, yields

$$\begin{aligned} \sum _{j = -\infty }^{\infty } e^{i{2jx_\textrm{tr}}k} = \frac{\pi }{{x_\textrm{tr}}}\sum _{\nu = -\infty }^{\infty } \delta \left( k-\frac{\pi \nu }{{x_\textrm{tr}}}\right) . \end{aligned}$$

(57)

Inverting the Fourier transform and using Eqs. (55) and (57), the final expression for the time-dependent probability distribution becomes,

$$\begin{aligned} {p^\dagger }(x,t)&=\frac{1}{\sqrt{2\pi }}\int \frac{1}{\sqrt{2\pi }}e^{-k^2 D t}e^{ikx} \left( 1-e^{-i{x_\textrm{tr}}k}\right) \nonumber \\&~~~~~\times \frac{\pi }{{x_\textrm{tr}}}\sum _{\nu = -\infty }^{\infty } \delta \left( k-\frac{\pi \nu }{{x_\textrm{tr}}}\right) dk \end{aligned}$$

(58)

$$\begin{aligned}&=\frac{1}{{2x_\textrm{tr}}}\sum _{\nu =-\infty }^{\infty } \exp {\left[ -\nu ^2\left( \frac{\pi }{{x_\textrm{tr}}}\right) ^2Dt\right] }\nonumber \\&~~~~~\times \exp {\left[ i\nu \left( \frac{\pi }{{x_\textrm{tr}}}\right) x\right] }\left[ 1-\exp {\left( -i\nu \pi \right) }\right] \end{aligned}$$

(59)

$$\begin{aligned}&=\frac{1}{{x_\textrm{tr}}}\sum _{\nu \text { odd}}\exp {\left[ -\nu ^2\left( \frac{\pi }{{x_\textrm{tr}}}\right) ^2Dt\right] }\nonumber \\&~~~~~\times \exp {\left[ i\nu \left( \frac{\pi }{{x_\textrm{tr}}}\right) x\right] } \end{aligned}$$

(60)

$$\begin{aligned}&=\frac{{2}}{{x_\textrm{tr}}}\sum _{\nu =0}^{\infty } \exp {\bigg \{-\left[ \frac{\pi (2\nu +1)}{{x_\textrm{tr}}}\right] ^2Dt\bigg \}}\nonumber \\&~~~~~\times \cos {\bigg \{\left[ \frac{\pi (2\nu +1)}{{x_\textrm{tr}}}\right] x\bigg \}}. \end{aligned}$$

(61)

Spatial integration will then determine the survival probability as a function of time,

$$\begin{aligned} {P^\dagger }(t)&= \int _{-x_\textrm{tr}/2}^{x_\textrm{tr}/2} {p^\dagger }(x,t) dx\end{aligned}$$

(62)

$$\begin{aligned}&=\frac{{2}}{{x_\textrm{tr}}}\sum _{\nu =0}^{\infty }\exp {\bigg \{-\left[ \frac{\pi (2\nu +1)}{{x_\textrm{tr}}}\right] ^2Dt\bigg \}}\nonumber \\&~~~~~\times \int _{-x_\textrm{tr}/2}^{x_\textrm{tr}/2} \cos {\bigg \{\left[ \frac{\pi (2\nu +1)}{{x_\textrm{tr}}}\right] x\bigg \}}dx&\end{aligned}$$

(63)

$$\begin{aligned}&=\frac{{2}}{{x_\textrm{tr}}}\sum _{\nu =0}^{\infty }\exp {\bigg \{-\left[ \frac{\pi (2\nu +1)}{{x_\textrm{tr}}}\right] ^2Dt\bigg \}}\nonumber \\&~~~~~\times \left[ \frac{{2x_\textrm{tr}}}{\pi (2\nu +1)}(-1)^{\nu }\right]&\end{aligned}$$

(64)

$$\begin{aligned}&=\frac{4}{\pi }\sum _{\nu =0}^{\infty }\frac{(-1)^{\nu }}{2\nu +1} \exp {\bigg \{-\left[ \frac{\pi (2\nu +1)}{{x_\textrm{tr}}}\right] ^2Dt\bigg \}} . \end{aligned}$$

(65)