Article Contents
Article Contents

# Asymptotic population abundance of a two-patch system with asymmetric diffusion

• * Corresponding author: Donald DeAngelis

The second author is supported by NSF grant of China (11571382)

• This paper considers a two-patch system with asymmetric diffusion rates, in which exploitable resources are included. By using dynamical system theory, we exclude periodic solution in the one-patch subsystem and demonstrate its global dynamics. Then we exhibit uniform persistence of the two-patch system and demonstrate uniqueness of the positive equilibrium, which is shown to be asymptotically stable when the diffusion rates are sufficiently large. By a thorough analysis on the asymptotic population abundance, we demonstrate necessary and sufficient conditions under which the asymmetric diffusion rates can lead to the result that total equilibrium population abundance in heterogeneous environments is larger than that in heterogeneous/homogeneous environments with no diffusion, which is not intuitive. Our result extends previous work to the situation of asymmetric diffusion and provides new insights. Numerical simulations confirm and extend our results.

Mathematics Subject Classification: Primary: 34C37, 92D25; Secondary: 37N25.

 Citation:

• Figure 1.  Phase-plane diagram of system (6). Stable equilibrium is displayed by solid circle. Vector fields are shown by gray arrows. Isoclines of nutrient $u_1$ and consumer $v_1$ are represented by red and blue lines, respectively. According to parameter values in experiments by Zhang et al. (2017), let $N_{01} = 0.02, r = 0.1, k_1 = 0.1, \gamma = 0.01, g_1 = 0.0001$. Then $u_{01} = 0.2 < r^2/g_1$. Numerical simulations show that all positive solutions of (6) converge to equilibrium $E_1^+$, which is consistent with Theorem 2.1(ⅱ)

Figure 2.  Numerical simulations for comparison of $T_1$ and $T_0$ when $s$ varies, Let $r = 0.1, u_{01} = 0.06, u_{02} = 0.0002, g_1 = 0.001, g_2 = 0.0005, D = 100$. When $s = 0.1$, we obtain $T_1 = 11.9531 >8.5095 = T_0$ by numerical computations on (4)

Figure 3.  Numerical simulations for comparison of $T_1$ and $T_2$ when $s$ varies, Let $r = 0.1, u_{01} = 0.06, u_{02} = 0.0002, g_1 = 0.001, g_2 = 0.0005, D = 100$. Then $u_{mean} = 0.0301$. When $s = 0.1$, we obtain $T_1 = 13.4364> 13.2452 = T_2$ by numerical computations on (4)

Figure 4.  Numerical simulations for comparison of $T_1$ and $T_2$ when $s$ varies, Let $r = 0.1, u_{01} = 0.06, u_{02} = 0.0002, g_1 = 0.001, g_2 = 0.0005, D = 100$. When the initial values are $(1.4, 1.4, 1.4, 1.4)$, $(3.4, 3.4, 3.4, 3.4)$, $(4, 4, 4, 4)$, $(7, 7, 7, 7)$ and $(8, 8, 8, 8)$, numerical computations on (4) show that all solutions converge to the same equilibrium $(0.1549, 4.4737, 0.0002, 8.9457)$, while the component $v_1(t)$ is displayed in this figure

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