Opinion dynamics in multi-agent systems: selected analytic models and verifying simulations

Abstract

In this paper opinion dynamics in multi-agent systems is investigated analytically using a kinetic approach. Interactions among agents are interpreted as collisions among molecules in gases and opinion dynamics is described according to the Boltzmann equation. Starting from a microscopic description of single interactions, global properties of the opinion distribution are derived analytically. The proposed analytic model is general enough to allow reproducing features of real societies of agents, such as positive and negative influences and bounded confidence, which are typically used to study opinion distribution models. Analytic results relative to emergent and global characteristics of considered multi-agent systems are verified by simulations obtained via direct implementation of the proposed microscopic interactions rules. Simulations confirm analytic results.

Appendix A

This appendix details how the stationary solutions $g_{\infty }$ shown in (37) and (48) are derived. First, let us consider the results in Subsect. 3.2 obtained with the diffusion function $D_1(\cdot )$. The function $g_{\infty }$ is found by solving (35) with $C=0$. Let us start by considering $v>0$, so that (35) can be written as

$$\frac{\lambda }{2}(1-v)^2\frac{\partial g}{\partial v} + \left[ (v-u) + \lambda (v-1)\right] g = 0.$$

(62)

Dividing both sides by g one obtains

$$\frac{g'}{g} = \frac{2}{(1-v)}+ \frac{2(u-v)}{\lambda (1-v)^2}$$

(63)

and observing that

$$\frac{2(u-v)}{\lambda (1-v)^2} = \frac{{\mathrm {d}}}{{\mathrm {d}}v} \left( -\frac{2}{\lambda } \log (1-v) + \frac{2(u-1)}{\lambda (1-v)}\right)$$

(64)

Eq. (63) can be written as

$$(\log g(v))' = \left( \log (1-v)^{-2-\frac{2}{\lambda }} + \frac{2(u-1)}{\lambda (1-v)} \right) '$$

(65)

where we used the fact that

$$\frac{2}{(1-v)} = -2 (\log (1-v))'.$$

(66)

Integrating (65), one finally obtains the following expression for the stationary profile

$$g_{\infty }(v) = \tilde{c}_{u, \lambda } (1-|v|)^{-2 - \frac{2}{\lambda }} \exp \left( \frac{2(u-1)}{\lambda (1-|v|)}\right)$$

(67)

where $\tilde{c}_{u, \lambda }$ is a normalization constant that depends on the average opinion u and on $\lambda$.

Let us now consider $v<0$ so that Eq. (35) becomes

$$\frac{\lambda }{2}(1+v)^2\frac{\partial g}{\partial v} + \left[ (v-u) + \lambda (v+1)\right] g = 0.$$

(68)

Dividing both sides by g leads to

$$\frac{g'}{g} = -\frac{2}{(1+v)}+ \frac{2(u-v)}{\lambda (1+v)^2}$$

(69)

and applying analogous manipulations to the case with $v>0$ one obtains

$$(\log g(v))' = \left( \log (1+v)^{-2-\frac{2}{\lambda }} - \frac{2(u+1)}{\lambda (1+v)} \right) '.$$

(70)

Integrating (70) leads to the following formula for the stationary profile

$$g_{\infty }(v) = \hat{c}_{u, \lambda } (1-|v|)^{-2 - \frac{2}{\lambda }} \exp \left( \frac{-2(u+1)}{\lambda (1-|v|)}\right)$$

(71)

where v has been substituted by $-|v|$ and $\hat{c}_{u, \lambda }$ is a normalization constant.

Since $g_{\infty }$ is the solution of a differential equation it must be continuous. From (67) and (71) it is evident that $g_{\infty }$ is continuous for $v>0$ and $v<0$. By imposing that $g_{\infty }$ is also continuous in $v=0$, the following equality needs to hold

$$\tilde{c}_{u, \lambda } \exp \left( \frac{2u}{\lambda }\right) = \hat{c}_{u, \lambda } \exp \left( \frac{-2u}{\lambda }\right) .$$

(72)

Finally, the solution of (35) is

$$g_{\infty }(v)= c_{u, \lambda } (1-|v|)^{-2 - \frac{2}{\lambda }} \exp \left[ -\frac{2(1-uv)}{\lambda (1-|v|)}\right]$$

(73)

as stated in (37).

Let us now consider the results in Subsect. 3.3 obtained with the diffusion function $D_2(\cdot )$. The function $g_{\infty }$ is found by solving (46) with $C=0$. Using standard techniques, from (46), one obtains

$$\frac{g'}{g} = \frac{4v}{1-v^2} + \frac{2(u-v)}{\lambda (1-v^2)^2}.$$

(74)

Integrating the right hand side of (74) leads to an explicit expression of $\log g$, and, therefore, of g. The first addend on the right hand side of (74) can be written as

$$\frac{{\mathrm {d}}}{{\mathrm {d}}v} \left( -2\log (1-v^2) \right) .$$

(75)

Concerning the remaining terms in (74), first observe that

$$\frac{2u}{\lambda }\frac{1}{(1-v^2)^2} = \frac{{\mathrm {d}}}{{\mathrm {d}}v}\left( \frac{u}{2\lambda } \log \left( \frac{1+v}{1-v}\right) + \frac{uv}{\lambda (1-v^2)}\right) .$$

(76)

Moreover one can compute

$$\begin{aligned} -\frac{2v}{\lambda (1-v^2)^2}&= \frac{1}{2\lambda } \left( \frac{1}{(1+v)^2} - \frac{1}{(1 - v)^2} \right) = -\frac{1}{2\lambda }\frac{{\mathrm {d}}}{{\mathrm {d}}v} \left( \frac{1}{1+v} + \frac{1}{1-v}\right) \\&= -\frac{1}{\lambda }\frac{{\mathrm {d}}}{{\mathrm {d}}v} \frac{1}{1-v^2}. \end{aligned}$$

(77)

Finally, using (75), (76), and (77), Eq. (74) can be written as

$$\frac{{\mathrm {d}}}{{\mathrm {d}}v}\log g(v)=\frac{{\mathrm {d}}}{{\mathrm {d}}v} \left[ -2\log (1-v^2) + \frac{u}{2\lambda } \log \left( \frac{1+v}{1-v}\right) + \frac{uv-1}{\lambda (1-v^2)} \right]$$

and, therefore,

$$\log g(v)= \log (1-v^2)^{-2} + \log \left( \frac{1+v}{1-v}\right) ^{\frac{u}{2\lambda }} + \frac{uv-1}{\lambda (1-v^2)} + \alpha _{u, \lambda }$$

(78)

where $\alpha _{u, \lambda }$ is a constant which depends on the average opinion u and on the value of $\lambda$. Taking the exponential of (78), the following expression for the stationary solution is derived

$$g_{\infty }(v) = c_{u, \lambda } (1+v)^{-2+\frac{u}{2\lambda }} (1-v)^{-2-\frac{u}{2\lambda }} \exp \left( \frac{uv-1}{\lambda (1-v^2)} \right)$$

(79)

which coincides with that shown in (48).