• PDF
• Cite
• Share
Article Contents  Article Contents

# Asymmetric diffusion in a two-patch mutualism system characterizing exchange of resource for resource

The author is supported by NSF of P.R. China (11571382)

• This paper considers a two-patch mutualism system derived from exchange of resource for resource, where the obligate mutualist can diffuse asymmetrically between patches. First, we give a complete analysis on dynamics of the system without diffusion, which exhibit how resource production of the obligate mutualist leads to its survival/extinction. Using monotone dynamics theory, we show global stability of a positive equilibrium in the three-dimensional system with diffusion. A novel finding of this work is that the obligate species' final abundance is explicitly expressed as a function of the diffusion rate and asymmetry, which demonstrates precise mechanisms by which the diffusion and asymmetry lead to the abundance higher than if non-diffusing, even though the facultative species declines. It is shown that for a fixed diffusion rate, intermediate asymmetry is favorable while extremely large asymmetry is unfavorable; For a fixed asymmetry, small diffusion is favorable while extremely large asymmetry is unfavorable. Initial densities of the species are also shown to be important in species' persistence and abundance. Numerical simulations confirm and extend our results.

Mathematics Subject Classification: 34C12, 37N25, 34C28, 37G20.

 Citation: • • Figure 1.  Phase-plane diagram of system (5). Stable and unstable equilibria are identified by solid and open circles, respectively. Vector fields are shown by green arrows. Isoclines of species $u,v_1$ are represented by red and blue lines, respectively. Fix $r = r_1 = c = 1$. (a) Let $a_{12} = 2.5, a_{21} = 1.3$. Equilibrium $E^+(2.7,2.5)$ is globally asymptotically stable. (b) Let $a_{12} = 4.5, a_{21} = 0.9$. There are two positive equilibria $E^-(1.25,0.14)$ and $E^+(2.79,1.5)$. (c) Let $a_{12} = 3.5, a_{21} = 0.9$. The equilibria $E^-$ and $E^+$ coincide and form a saddle-node point $E^\pm(1.6,0.45)$. In the cases of (b-c), the separatrices (the black line) of $E^-$ subdivide the first quadrant into two regions: one is the basin of attraction of $E_1$ while the other is that of $E^+$. (d) Let $a_{12} = 2.5, a_{21} = 0.5$. All positive solutions converge to equilibrium $E_1(1,0)$

Figure 4.  Comparison of $T_1(s, 0)$ and $T_2(s, D)$ when there is diffusion $D$ as shown in Theorem 4.1(ⅰ). The solid blue line represents $T_2(s, D)$, while the dash-dot red line represents $T_1(s, 0)$. Let $r = 0.2, c = 0.05, a_{12} = 0.1, b = b_1 = 1, s = 1$, $a_{21} = 0.5, r_1 = 0.8, r_2 = 0.1$. (a) When $s = 1$, we have $T_2>T_1$ for $D>0$. (b) When $s = 10$, we have $T_2>T_1$ as $0<D< 0.0135$ while $T_2<T_1$ as $D> 0.0135$

Figure 5.  Comparison of $T_1(s, 0)$ and $T_2(s,100)$ when there is asymmetry $s$ in large diffusion as shown in Theorem 4.2(ⅰ). The solid blue line represents $T_2(s,100)$, while the dash-dot red line represents $T_1(s, 0)$. Let $r = 0.2, c = 0.05, a_{12} = 0.1, b = b_1 = 1, D = 100$, $a_{21} = 0.5, r_1 = 0.8, r_2 = 0.1$. Then we have $T_1(s, 0) = 1.9233$ and $T_2(s,100)>T_1(s, 0)$ as $0.1935< s< 9.3201$. Numerical computation shows that the function $T_2 = T_2(s,100)$ is convex upward with $T_2(s,100) = 0$ as $s\ge 12.1537$

Figure 2.  Comparison of $T_1(s, D)$ and $T_1(s, 0), T_2(s, D)$ and $T_2(s, 0)$ when there is a small diffusion rate $D$, as shown in Theorem 4.1(ⅰ). The solid red and blue lines represent $T_1(s, D)$ and $T_2(s, D)$, while the dash-dot red and blue lines represent $T_1(s, 0)$ and $T_2(s, 0)$, respectively. Let $r = 0.2, c = 0.05, a_{12} = 0.1, b = b_1 = 1, D = 0.1, s = 1$, $a_{21} = 0.5, r_1 = 0.8, r_2 = 0.1$. Then we have $T_1(1, 0.1)<T_1(1,0)$ but $T_2(1,0.1)>T_2(1,0)$

Figure 3.  Comparison of $T_1(s, D)$ and $T_1(s, 0), T_2(s, D)$ and $T_2(s, 0)$ when there is a large diffusion rate $D$, as shown in Theorem 4.2(ⅰ). The solid red and blue lines represent $T_1(s, D)$ and $T_2(s, D)$, while the dash-dot red and blue lines represent $T_1(s, 0)$ and $T_2(s, 0)$, respectively. Let $r = 0.2, c = 0.05, a_{12} = 0.1, b = b_1 = 1, D = 100, s = 0.1$, $a_{21} = 0.5, r_1 = 0.8, r_2 = 0.1$. Then we have $T_1(0.1,100)<T_1(0.1,0)$ but $T_2(0.1,100)>T_2(0.1,0)$

Figure 6.  The surface of $T_2 = T_2(s,D)$ when both of $s$ and $D$ varies. Let $r = 0.2, c = 0.05, a_{12} = 0.1, b = 4, b_1 = 1$, $a_{21} = 0.5, r_1 = 0.8, r_2 = 0.1, 0<s<6, 0<D<6$. Numerical computation shows that for fixed $s$, the surface decreases monotonically, which is consistent with Fig. 4. For fixed $D$, the surface is convex upward, which is consistent with Fig. 5

• ## Article Metrics  DownLoad:  Full-Size Img  PowerPoint