let k be the total number of nodes distributed on a field. each node performs a truncated Lévy Walk, with the following characteristics:
- Motion speed: Motion speed remains constant, in our case it is the motion of pedestrians.
- Time interval of each displacement: The time interval of each displacement, $t_{fi}$ is directly related to the velocity and length of the displacement. $\xi_i$ is the length of the displacement and v=1 and $t_{fi} = \xi_i$.
- Direction of the displacement: the direction is uniformly distribution and the direction angle $\theta_i \in [0, 2\pi)$
- Length of the displacement: In Truncated Lévy Walk, the length of the displacement is assumed to have a Levy distribution. For $\alpha \le 2$, the density function $f_l(x)$ can be approximated by $\frac{1}{|x|^{1+\alpha}}$. Since it is a truncated Levy Walk, the range of the displacement length does not vary between ($- \infty, + \infty$) but between ($0, \tau_\xi$). $\tau_\xi$ is the maximum allowed displacement length. The value of $\alpha$ is assumed to be between 0.7 and 1.7
- Pause interval: TLW includes a pause time, that takes place after the end of each displacement. The pause time $T_p$ has a levy distribution, which, similarly to the displacement length, can be approximated to $\frac{1}{T_p^{1 + \gamma}}$ with ${T_p} \in (0, \tau_p)$. $\tau_p$ is the maximum allowed pause interval. $\gamma$ is assumed to be equal to 0.7
- Complete displacement time: This time, $\Delta t_s$ is the sum of the displacement time and the pause time.
What i am trying to find, is the resulting distribution of the locations of the k nodes at time t after performing a truncated levy walk for time interval equal to t. Does anyone have any recommendation as to how i can find the answer?
