Non-equilibrium processes in an unconserved network model with limited resources

Abstract

We present a study of a peculiar form of network topology comprising of lanes connected via a junction, competing for particles in a reservoir of limited capacity with non-conserving dynamics. We exploit mean-field approximation to thoroughly analyze stationary dynamic properties such density profiles, phase boundaries, and phase diagrams. The steady-state properties have been studied by taking into account the time evolution of particle density. It is found that the ratio of the number of incoming and outgoing lanes from the network junction as well as that of the kinetic rates substantially influences the number of stationary phases and the complexity of the phase diagram concerning the increasing number of particles in the system. The maximal current phase can persist in a fragment or the complete lanes, when the number of incoming and outgoing lanes is equal, as opposed to unequal counterparts. For lower values of the total number of particles, only low density phase is achieved in the left subsystem. All the theoretical findings are supported by extensive Monte Carlo simulations and explained using simple physical arguments.

Appendices

Appendix A

Here, we investigate the transient solution, i.e., the evolution in particle density by analyzing the master equation as stated in Eq. (10) and further utilize it to acquire the corresponding steady-state solution. To completely understand the first-order differential equation in Eq. (10), one needs to study its characteristics which are given by

$$\begin{aligned} \frac{d\rho }{dt}= & {} \Omega _D[Kz-(1+Kz)\rho ] , \end{aligned}$$

(38)

$$\begin{aligned} \frac{d\rho }{dx}= & {} \frac{\Omega _D[Kz-(1+Kz)\rho ]}{1-2\rho }, \end{aligned}$$

(39)

where $Kz=\Omega _A^*/\Omega _D$. We consider the density profile $\rho (x,t)$ that develops for an initial density step

$$\begin{aligned} \rho _0(x,0)={\left\{ \begin{array}{ll} \alpha ^*, &\quad 0\le x<0.5\\ 1-\beta , &\quad 0.5< x \le 1 \end{array}\right. } \end{aligned}$$

(40)

where the left ($x=0$) and the right $(x=1)$ boundaries are fixed at $\alpha ^*$ and $1-\beta$, respectively. Integrating Eq. (38) and utilizing the initial density profile given by Eq. (40), we obtain the solution $\rho _I(x,t)$ given by

$$\begin{aligned} \displaystyle \rho _I(x,t)=\frac{Kz-\bigg ( \Big (Kz-(1+Kz)\rho _0(x,0)\Big )e^{-\Omega _D(1+Kz)t}\bigg )}{1+Kz}. \end{aligned}$$

(41)

For the study of Eq. (39), we introduce a rescaled density of the form

$$\begin{aligned} \sigma =\frac{(Kz+1)(2\rho -1)}{Kz-1}-1, \end{aligned}$$

(42)

where Langmuir isotherm $\rho _l=Kz/(Kz+1)$ eventuate for $\sigma =0$ and is similar to that in reference [20] . It is evident that the above equation is not defined for $Kz=1$. Depending upon the values of Kz, different scenarios are possible. Therefore, we categorize our analysis into two cases, i) $Kz=1$, and ii) $Kz\ne 1$.

1.1 A1.$\ Kz=1$

Let us first discuss the case when $Kz=1$, for which Eq. (38) and (39) simplify considerably and lead to

$$\begin{aligned} \frac{d\rho }{dt}=\Omega _D(1-2\rho )\quad \textrm{and} \quad \frac{d\rho }{dx}=\Omega _D. \end{aligned}$$

(43)

Solving the above equations, we obtain the general solution of Eq. (10) as

$$\begin{aligned} (1-2\rho )e^{2\Omega _D t}=f(\rho -\Omega _D x), \end{aligned}$$

(44)

where f is an arbitrary function to be calculated. Utilizing the initial and boundary conditions, we obtain three solutions:

$$\begin{aligned} \rho _{\alpha }(x)&= \Omega _D x +\alpha ^*, \end{aligned}$$

(45)

$$\begin{aligned} \rho _I(x,t)&= \frac{1-(1-2\rho _0)e^{-2\Omega _D t}}{2}, \end{aligned}$$

(46)

$$\begin{aligned} \rho _{\beta }(x)&= \Omega _D (x-1)+1-\beta , \end{aligned}$$

(47)

where $\rho _{\alpha }(x)$ and $\rho _\beta (x)$ denote the solutions satisfying the left and the right boundary conditions, respectively. Depending upon how $\rho _{\alpha }(x)$, $\rho _{I}(x,t)$ and $\rho _{\beta }(x)$ can be matched, different scenarios for the density profile occur [20]. The density profile $\rho _{\alpha }(x)$ is separated from $\rho _{I}(x,t)$ at $x_{\alpha }$ and $\rho _{I}(x,t)$ is separated from $\rho _{\beta }(x)$ at $x_{\beta }$ whose values are given by:

$$\begin{aligned} x_{\alpha }&=\frac{1-(1-2\rho _0)e^{-2\Omega _D t}-2\alpha ^*}{2\Omega _D}, \end{aligned}$$

(48)

$$\begin{aligned} x_{\beta }&=\frac{1-(1-2\rho _0)e^{-2\Omega _D t}-2(1-\beta )}{2\Omega _D}+1. \end{aligned}$$

(49)

According to the relative positions of $x_{\alpha }$ and $x_{\beta }$, the density profile is obtained as:

• 1. When $x_{\alpha }\le x_{\beta }$, the density profile is given by

  $$\begin{aligned} \rho (x,t)= {\left\{ \begin{array}{ll} \Omega _D x +\alpha ^*, &\quad 0\le x \le x_{\alpha }\\ \frac{1-(1-2\rho _0)e^{-2\Omega _D t}}{2}, &\quad x_{\alpha }\le x \le x_{\beta }\\ \Omega _D (x-1)+1-\beta , &\quad x_{\beta }\le x \le 1. \end{array}\right. } \end{aligned}$$

  (50)

• 2. When $x_{\alpha }> x_{\beta }$, there is necessarily a density discontinued at the point $x_w$ where the currents corresponding to $\rho _{\alpha }(x)$ and $\rho _{\beta }(x)$ matches. The density profile is expressed as

  $$\begin{aligned} \rho (x,t)= {\left\{ \begin{array}{ll} \Omega _D x +\alpha ^*, &\quad 0\le x \le x_{w}\\ \Omega _D (x-1)+1-\beta , &\quad x_{w} \le x \le 1 \end{array}\right. } \end{aligned}$$

  (51)

  where $x_{w}=(\Omega _D-\gamma +\beta )/(2\Omega _D).$

1.2 A2. $Kz\ne 1$

In this case, the transformation given by Eq. (42) is well defined and Eq. (39) in the rescaled form reduces to

$$\begin{aligned} \left( \frac{\sigma +1}{\sigma }\right) \frac{\partial \sigma }{\partial x}=\frac{\Omega _D (Kz+1)^2}{Kz-1}. \end{aligned}$$

(52)

Integrating the above equation yields

$$\begin{aligned} |\sigma (x)|\text {exp}(\sigma (x))=Y(x), \end{aligned}$$

(53)

where Y(x) is

$$\begin{aligned} Y(x)=|\sigma (x_0)|\text { exp}\left\{ \Omega _D \frac{(Kz+1)^2}{Kz-1}(x-x_0)+\sigma (x_0)\right\} , \end{aligned}$$

(54)

and $x_0$ is the reference point. In particular, at the boundaries $x_0$ is equal to 0 or 1, which gives

$$\begin{aligned} Y_{\alpha }(x)&=|\sigma (0)|\text { exp}\left\{ \Omega _D \frac{(Kz+1)^2}{Kz-1}x+\sigma (0)\right\} , \\ Y_{\beta }(x)&=|\sigma (1)|\text { exp}\left\{ \Omega _D \frac{(Kz+1)^2}{Kz-1}(x-1)+\sigma (1)\right\} . \end{aligned}$$

The rescaled Eq. (53) is known to have an explicit solution in the form of a special function known as Lambert W function [19, 20, 48] and can be written as

$$\begin{aligned} \sigma (x)={\left\{ \begin{array}{ll} W_{-1}(-Y(x)), &\quad \sigma (x)<-1 \\ W_0(-Y(x)), &\quad -1\le \sigma (x)<0\\ W_0(Y(x)), &\quad 0< \sigma (x). \end{array}\right. } \end{aligned}$$

(55)

Using the properties of the Lambert W function and the values of $\alpha ^*$ and $\beta$, a specific branch of the Lambert W function can be chosen to obtain the solution to Eq. (53). The rescaled solution satisfying the left boundary condition is the left rescaled solution denoted by $\sigma _{\alpha }(x)$, while the one obeying the right boundary condition is represented by $\sigma _{\beta }(x)$ called as the right rescaled solution. Therefore, the particle densities $\rho _{\alpha }(x)$ and $\rho _{\beta }(x)$ can be obtained by substituting back $\sigma _{\alpha }(x)$ and $\sigma _{\beta }(x)$ in Eq. (42). Employing the suitable solution to $\sigma _\alpha (x)$ and $\sigma _\beta (x)$, the complete solution is constructed from the possible combination of the three solutions, i.e.,

$$\begin{aligned} \rho (x,t)= {\left\{ \begin{array}{ll} \rho _{\alpha }(x), &{} 0\le x \le x_{\alpha }\\ \rho _I(x,t), &{} x_{\alpha }\le x \le x_{\beta }\\ \rho _{\beta }(x), &{} x_{\beta }\le x \le 1 \end{array}\right. } \end{aligned}$$

(56)

where $\rho _I(x,t)$ is given by Eq. (41). The two solutions $\rho _{\alpha }(x)$ and $\rho _I(x,t)$ are matched at the position $x_{\alpha }$ for which the current for both the solutions is equal. Similarly, the currents for $\rho _{I}(x,t)$ and $\rho _{\beta }(x)$ are equated to calculate the value of $x_\beta$.

1.3 A2.1 Explicit solution

Analogous to the homogeneous single-lattice TASEP model with LK coupled to an infinite reservoir [19, 20], the solution $\rho _\alpha (x)$ corresponding to the left boundary condition is stable only if $\alpha ^*\le 0.5$. The entry rate $0\le \alpha ^*\le 0.5$ implies that $0\le \rho _\alpha (x)\le 0.5$. Utilizing rescaled density Eq. (42), one has $\sigma _\alpha \in [-2Kz/(Kz-1),-1]$ for $Kz>1$ and $\sigma _\alpha \in [-1,-2Kz/(Kz-1)]$ for $Kz<1$. Hence, the left rescaled solution is given as follows:

• (a) If $Kz>1$, then

  $$\begin{aligned} \sigma _\alpha (x)=W_{-1}(-Y_\alpha (x)). \end{aligned}$$

  (57)

• (b) If $Kz<1$, then

  $$\begin{aligned} \sigma _\alpha (x)={\left\{ \begin{array}{ll} W_0(Y_\alpha (x)),&\quad 0\le \alpha \le \rho _l\\ 0, &\quad \alpha =\rho _l\\ W_0{(-Y_\alpha (x))},&\quad \rho _l\le \alpha \le 0.5.\\ \end{array}\right. } \end{aligned}$$

  (58)

Similarly, the solution $\rho _\beta (x)$ matching the right boundary is stable only for $\beta \le 0.5$ and is always in high density regime ($\rho _\beta (x)\ge 0.5$). For the right rescaled solution $\sigma _\beta (x)$, employing $\beta \le 0.5$ and $\rho _\beta (x)\ge 0.5$ transforms to $\sigma _\beta (x) \in [-1, 2/(Kz-1)]$ if $Kz>1$ and $\sigma _\beta (x) \in [ 2/(Kz-1),-1]$ for $Kz<1$. Hence, by choosing the suitable branch of the Lambert W function, one obtains

• (a) If $Kz>1$, then

  $$\begin{aligned} \sigma _\beta (x)={\left\{ \begin{array}{ll} W_0(Y_\beta (x)), &\quad 0\le \beta \le 1-\rho _l\\ 0, &\quad \beta = 1-\rho _l\\ W_0(-Y_\beta (x)), &\quad 1-\rho _l\le \beta \le 0.5\\ \end{array}\right. } \end{aligned}$$

  (59)

  where $\rho _l$ is the Langmuir isotherm.

• (b) If $Kz<1$, then

  $$\begin{aligned} \sigma _\beta (x)=W_{-1}(-Y_\beta (x)). \end{aligned}$$

  (60)

For thorough exploration, it is essential to perceive the phase boundaries separating the different phases in the phase diagram, in addition to the particle densities. The following section supplies a different method to calculate the density profiles as well as the phase boundaries.

1.4 A2.2 Implicit solution

An alternative method for obtaining the particle densities $\rho _{\alpha }(x)$ and $\rho _\beta (x)$ is through direct integration of Eq. (39). This equation can be rewritten as

$$\begin{aligned} {dx}=\frac{1-2\rho }{\Omega _D[Kz-(1+Kz)\rho ]} d\rho . \end{aligned}$$

(61)

Upon integration, we have

$$\begin{aligned} \frac{1}{\Omega _D(1+Kz)}\left( 2\rho +\frac{Kz-1}{Kz+1} \ln \left| Kz-(1+Kz)\rho \right| \right) =x+const. \end{aligned}$$

(62)

In the low density situation, the density of the left end $\alpha ^*$ is employed, to find the profile $\rho _\alpha (x)$ through

$$\begin{aligned} \frac{1}{\Omega _D(1+Kz)}\Bigg (2(\rho _\alpha -\overline{\alpha })+\frac{Kz-1}{Kz+1} ln\left| \frac{ Kz-(1+Kz)\rho _\alpha }{ Kz-(1+Kz)\overline{\alpha } } \right| \Bigg )= x \end{aligned}$$

(63)

where $\overline{\alpha }=\text {min}(\alpha ^*,0.5)$. The high density solution $\rho _\beta (x)$ corresponding to the right boundary condition is obtained from

$$\begin{aligned} \frac{1}{\Omega _D(1+Kz)}\Bigg (2(1-\bar{\beta }-\rho _\beta )+\frac{Kz-1}{Kz+1}\ln \left| \frac{ Kz-(1+Kz)(1-\bar{\beta })}{ Kz-(1+Kz)\rho _\beta } \right| \Bigg )=1-x \end{aligned}$$

(64)

for $\bar{\beta }=\text {min}(\beta ,0.5)$.

This implicit solution also retrieves the density solutions $\rho _{\alpha }(x)$ and $\rho _\beta (x)$ obtained for $Kz=1$. Now, the solutions procured for the densities $\rho _{\alpha }(x)$ and $\rho _\beta (x)$ in the subsection A2 can be deployed in Eq. (56) to finally compute the overall density of the lattice. Both the implicit and the explicit solutions are equivalent and can be used interchangeably. For the sake of completeness, we will be utilizing the explicit solutions to calculate the density profiles while the implicit solution will provide the phase boundaries.

Appendix B

Density profiles corresponding to different phases of the phase diagrams are depicted in Fig. 5.

Fig. 5

Typical density profiles for $m=n,\ K=1,\ \Omega _D=0.1$ (a-c with $\mu =2.5$ and d-j with $\mu =200$), $m=2,\ n=1,\ K=1,\ \Omega _D=0.1$ (k-m with $\mu =1$), and $m=2,\ n=1,\ K=3,\ \Omega _D=0.01$ (n-o with $\mu =1$). Solid blue curves and red symbols denote mean-field and Monte Carlo simulations, respectively