Dynamics of Lotka–Volterra Competition Patch Models in Streams with Two Branches

Abstract

Streams may have many branches and form complex river networks. We investigate two competition patch models associated with two different river network modules, where one is a distributary stream with two branches at the downstream end, and the other is a tributary stream with two branches at the upstream end. Treating one species as resident species and the other one as mutant species, it is shown that, for each model, there exists a invasion curve such that the mutant species can invade when rare if and only if its dispersal strategy is below this curve, but the shapes of the invasion curves are different. Moreover, we show that the global dynamics of the two models can be similar or different depending on river networks. Especially, if the drift rates of the two species are equal, then the global dynamics are similar for small drift rate and different for large drift rate. Our results also confirm a conjecture in Jiang et al. (Bull Math Biol 82:131, 2020).

Appendix

In this part, we first show the monotonicity of the principle eigenvalue. The following result follows directly from the Perron–Frobenius Theorem.

Lemma A.1

Suppose that A is an irreducible and essentially nonnegative matrix. Then s(A) is the unique principal eigenvalue of A.

Then We cite the following result from Altenberg (2012), Chen et al. (2022c) on the monotonicity of spectral bound.

Lemma A.2

Assume that $A=(a_{ij})_{n\times n}$ is an irreducible and essentially nonnegative matrix with $\sum _{i=1}^na_{ij}=0$ for each $j=1,\dots ,n$, and $M=\text {diag}(m_i)$ is a real diagonal matrix. Let $\lambda _1(\rho )$ be the spectral bound (resp. principal eigenvalue) of $\rho A+M$ for $\rho >0$. Then $\lambda _1'(\rho )\le 0$ for $\rho \in (0, \infty )$ and the inequality is strict except for the case $m_1=\cdots =m_n$. Moreover,

$$\begin{aligned} \lim _{\rho \rightarrow 0}\lambda _1(\rho )=\max _{1\le i\le n}\{m_i\} \;\; \text {and} \;\; \lim _{\rho \rightarrow \infty }\lambda _1(\rho )=\sum _{i=1}^n{\theta _im_i}, \end{aligned}$$

where ${\varvec{\theta }}=(\theta _1,\dots ,\theta _n)^T\gg 0$ satisfies $\sum _{i=1}^n{\theta _i}=1$ and is the eigenvector of A corresponding to eigenvalue 0.

Then we show that each of (i)–(iv) in Theorem 2.1 (resp. (i)–(ii) in Theorem 2.2) can occur varying $(d_1,q_1)$.

Proposition A.3

Suppose that $\mathrm ({\textbf{H}})$ holds, and $q_1\ge 0$. Then there exists $d^*(q_1)>0$ such that the following statements hold for $0<d_1<d^*(q_1)$:

• (i) If $0<q_1\ll r_1/2$ (i.e., $q_1$ is sufficiently small), then (i) of Theorem 2.1 holds when $k_1/k_3+k_1/k_2>2$, and (ii) of Theorem 2.1 holds when $k_1/k_3+k_1/k_2<2$;

• (ii) If $r_1/2<q_1< r_1k_2/(k_2-k_3)$, then (ii) of Theorem 2.1 holds;

• (iii) If $q_1>r_1k_2/(k_2-k_3)$, then (iv) of Theorem 2.1 holds.

Proof

By (60), we see that ${\varvec{u}}^*$ is bounded for $d_1\in (0,+\infty )$. Then, up to a subsequence,

$$\begin{aligned} \displaystyle \lim _{d_1\rightarrow 0} u^*_1=u_1^0,\;\;\lim _{d_1\rightarrow 0} u^*_2=u_2^0,\;\;\lim _{d_1\rightarrow 0} u^*_3=u_3^0. \end{aligned}$$

Taking $d_1\rightarrow 0$ for each equation of (13), we have

$$\begin{aligned}&\displaystyle r_1u^0_1\left( 1-\frac{2q_1}{r_1}-\frac{u^0_1}{k_1}\right) =0, \end{aligned}$$

(1a)

$$\begin{aligned}&\displaystyle q_1u^0_1+r_2u^0_2\left( 1-\frac{u^0_2}{k_2}\right) =0, \end{aligned}$$

(1b)

$$\begin{aligned}&\displaystyle q_1u^0_1+r_3u^0_3\left( 1-\frac{u^0_3}{k_3}\right) =0. \end{aligned}$$

(1c)

If $0<q_1<r_1/2$, by a tedious computation, we deduce that $u^0_1=k_1(1-2q_1/r_1)$, and $u^0_2,u^0_3>0$ can be uniquely determined by (1b) and (1c) with $u^0_1=k_1(1-2q_1/r_1)$, respectively. Then, by (1b), we have $1-u^0_2/k_2<0$, and hence $\lim _{d_1\rightarrow 0}1-u^*_2/k_2<0$. In addition, we compute from (1) that

$$\begin{aligned} \displaystyle \lim _{q_1\rightarrow 0}\sum ^3_{i=1}\frac{1}{q_1}r_i\left( 1-\frac{u^0_i}{k_i}\right) =\lim _{q_1\rightarrow 0}\left( 2-\frac{u^0_1}{u^0_2} -\frac{u^0_1}{u^0_3}\right) =2-\frac{k_1}{k_2}-\frac{k_1}{k_3}. \end{aligned}$$

Therefore, for $0<q_1\ll r_1/2$, we have $\lim _{d_1\rightarrow 0}\sum _{i=1}^3r_i(1-u^*_i/k_i)<0$ when $k_1/k_3+k_1/k_2>2$ and $\lim _{d_1\rightarrow 0}\sum _{i=1}^3r_i(1-u^*_i/k_i)>0$ when $k_1/k_3+k_1/k_2<2$. Therefore, (i) holds.

If $q_1>r_1/2$, we compute that $(u^0_1,u^0_2,u^0_3)=(0,k_2,k_3)$, and hence

$$\begin{aligned} \lim _{d_1\rightarrow 0}\sum ^3_{i=1}r_i(1-u^*_i/k_i)=r_1>0. \end{aligned}$$

Taking the derivative of (13a)–(13b) with respect to $d_1$ at $d_1=0$, we have

$$\begin{aligned} \begin{aligned}&u_2^0+u_3^0-2u_1^0+(r_1-2q_1)\left. \frac{\partial u^*_1}{\partial d_1}\right| _{d_1=0}=0,\\&u_1^0-u_2^0+q_1\left. \frac{\partial u^*_1}{\partial d_1}\right| _{d_1=0}-r_2\left. \frac{\partial u^*_2}{\partial d_1}\right| _{d_1=0}=0, \end{aligned} \end{aligned}$$

(2)

which yields

$$\begin{aligned} \displaystyle \left. \frac{\partial u^*_2}{\partial d_1}\right| _{d_1=0}=\frac{1}{r_2}\left[ q_1\frac{k_2+k_3}{2q_1-r_1}-k_2\right] . \end{aligned}$$

(3)

Noticing that $\lim _{d_1\rightarrow 0}u_2^*=k_2$, we see from (2)–(3) that (ii) and (iii) hold. $\square$

Proposition A.4

Suppose that $\mathrm ({\textbf{H}})$ holds, and $d_1>0$. Then there exists $q^*(d_1)>0$ such that the following statements hold for $q_1>q^*(d_1)$:

• (i) If $r_1<-\sum _{i=2}^3r_i(1-{\hat{u}}_i/k_i)$, then (iii) of Theorem 2.1 holds;

• (ii) If $r_1>-\sum _{i=2}^3r_i(1-{\hat{u}}_i/k_i)$, then (iv) of Theorem 2.1 holds.

Here $({\hat{u}}_2,{\hat{u}}_3)\gg {\varvec{0}}$ is the unique positive solution of (62) with $k_2>{\hat{u}}_2>{\hat{u}}_3>k_3$.

Proof

By the proof of Theorem 2.9 for model (I), we see that $\lim _{q_1\rightarrow \infty } u^*_1=0$, $\lim _{q_1\rightarrow \infty }u^*_2={\hat{u}}_2$ and $\lim _{q_1\rightarrow \infty } u^*_3={\hat{u}}_3$. Then $\lim _{q_1\rightarrow \infty }(1-u^*_2/k_2)>0$. By (62), we compute that

$$\begin{aligned} \displaystyle \sum _{i=2}^3r_i\left( 1-\frac{{\hat{u}}_i}{k_i}\right) =-\frac{d_1}{2}\left( \frac{{\hat{u}}_3}{{\hat{u}}_2}+\frac{{\hat{u}}_2}{{\hat{u}}_3} -2\right) <0. \end{aligned}$$

In addition, we have

$$\begin{aligned} \displaystyle \lim _{q_1\rightarrow \infty }\sum _{i=1}^3r_i\left( 1-\frac{u^*_i}{k_i}\right) =r_1+\sum _{i=2}^3r_i\left( 1-\frac{{\hat{u}}_i}{k_i}\right) . \end{aligned}$$

It follows that (i) and (ii) holds. $\square$

Proposition A.5

Suppose that $\mathrm ({\textbf{H}})$ holds, and $d_1>0$. Then there exists ${\overline{q}}^{(2)}(d)> \underline{q}^{(2)}(d_1)>0$ such that (i) of Theorem 2.2 holds for $q_1<\underline{q}^{(2)}(d_1)$ and (ii) of Theorem 2.2 holds for $q_1>{\overline{q}}^{(2)}(d)$.

Proof

Clearly, $\lim _{q_1\rightarrow 0}(u^*_1,u^*_2,u^*_3)=(u^0_1,u^0_2,u^0_3)$, where $(u^0_1,u^0_2,u^0_3)\gg {\varvec{0}}$ satisfies

$$\begin{aligned}&\displaystyle d_1u_1-d_1u_3=r_1u_1\left( 1-\frac{u_1}{k_1}\right) , \end{aligned}$$

(4a)

$$\begin{aligned}&\displaystyle d_1u_2-d_1u_3=r_2u_2\left( 1-\frac{u_2}{k_2}\right) , \end{aligned}$$

(4b)

$$\begin{aligned}&\displaystyle 2d_1u_3-d_1u_1-d_1u_2=r_3u_3\left( 1-\frac{u_3}{k_3}\right) . \end{aligned}$$

(4c)

If $u^0_1=u^0_3$, we see from (4b) and (4c) that $k_2>u^0_2>u^0_3>k_3$; and if $u^0_2=u^0_3$, we see from (4a) and (4c) that $k_1>u^0_1>u^0_3>k_3$. Then we see from (4) that

$$\begin{aligned} \displaystyle \sum _{i=1}^3r_i\left( 1-\frac{u^0_i}{k_i}\right) =-d_1\left( \frac{u^0_1}{u^0_3}+\frac{u^0_3}{u^0_1}+\frac{u^0_2}{u^0_3}+\frac{u^0_3}{u^0_2}-4\right) <0, \end{aligned}$$

which implies that (i) holds. By the proof of Theorem 2.9 for model (II), we have

$$\begin{aligned} \displaystyle \lim _{q_1\rightarrow \infty }u_1^*=0,\;\;\lim _{q_1\rightarrow \infty }u_2^*=0\;\;\text {and}\;\;\lim _{q_1\rightarrow \infty }u_3^*=k_3. \end{aligned}$$

It follows that $\lim _{q_1\rightarrow \infty }\sum _{i=1}^3r_i(1-u^*_i/k_i)=r_1+r_2>0$, and hence (ii) holds. $\square$

Proposition A.6

Suppose that $k_1>k_2=k_3$ (resp. $k_1=k_2<k_3$) holds, $d_1,d_2>0$, and $q_1=q_2=q\ge 0$, and denote

$$\begin{aligned} {\bar{q}}^{(1)}= & {} \max \left\{ \frac{r_2}{k_2}\left( k_1-k_2\right) ,\;\; \frac{r_3}{k_3}\left( k_1-k_3\right) \right\} , \\ {\bar{q}}^{(2)}= & {} \max \left\{ \frac{r_1}{k_1}\left( k_1-k_3\right) ,\;\; \frac{r_2}{k_2}\left( k_2-k_3\right) \right\} . \end{aligned}$$

Then the following statements hold for model (I) (resp. (II)):

• (i) If $d_1>d_2>0$, then $({\varvec{u}}^*, {\varvec{0}})$ is globally asymptotically stable for $q>{\bar{q}}^{(1)}$(resp. $q>{\bar{q}}^{(2)}$);

• (ii) If $0<d_1<d_2$, then $({\varvec{0}},{\varvec{v}}^*)$ is globally asymptotically stable for $q>{\bar{q}}^{(1)}$(resp. $q>{\bar{q}}^{(2)}$).

Proof

We only prove that (i) and (ii) hold for model (I), and model (II) can be proved similarly. We divide the rest of the proof into steps.

Step 1. We show that $f_1^{(1)},f_2^{(1)}<0$, where $f_1^{(1)}$ and $f_2^{(1)}$ are defined in (11).

By Lemma 3.1, $f_2^{(1)}<0$ and $u^*_1<k_1$. Suppose to the contrary that $f^{(1)}_1\ge 0$. Then we see from (11), (13b) and (13c) that $k_2\ge u_2^*>u_3^*>k_3$, which contradicts $k_2=k_3$.

Step 2. We show that $T_1^{(1)},T_2^{(1)}>0$, where $T_1^{(1)}$ and $T_2^{(1)}>0$ are defined in (10).

We only prove that $T_1^{(1)}>0$, and $T_2^{(1)}>0$ can be proved similarly. Suppose to the contrary that $T_1^{(1)}= u^*_2-u^*_1\le 0$. This combined with (12b) and $q>{\bar{q}}^{(1)}$ implies that

$$\begin{aligned} \displaystyle u^*_1\ge u^*_2 \ge k_2\left( 1+\frac{q}{r_2}\right) >k_1, \end{aligned}$$

which contradicts $u_1^*<k_1$ (by Lemma 3.1).

Step 3. We claim $\lambda _1^{(1)}\left( d_2,q,{\varvec{1}}-{{\varvec{u}}^*}/{{\varvec{k}}}\right) <0$ for $d_2<d_1$ and $\lambda _1^{(1)}\left( d_2,q,{\varvec{1}}-{{\varvec{u}}^*}/{{\varvec{k}}}\right) >0$ for $d_2>d_1$.

We first show $\lambda _1^{(1)}\left( d_2,q,{\varvec{1}}-{{\varvec{u}}^*}/{{\varvec{k}}}\right) \ne 0$ for $d_2\ne d_1$. Suppose to the contrary that $\lambda _1^{(1)}\left( d_2,q,{\varvec{1}}-{{\varvec{u}}^*}/{{\varvec{k}}}\right) =0$ with the corresponding eigenvector ${\varvec{\phi }}=(\phi _1,\phi _2,\phi _3)^T\gg {\varvec{0}}$. Note that $f_i^{(1)}$ and $g_i^{(1)}$ have the same signs for $i=1,2$, where $g_1^{(1)}$ and $g_2^{(1)}$ are defined in (42). Then we have $g_1^{(1)}<0$ and $g_2^{(1)}<0$. By Steps 1–2 and (44b), we get

$$\begin{aligned} 0=\displaystyle \sum ^2_{j=1}T^{(1)}_{j}g^{(1)}_{j}<0, \end{aligned}$$

which is the contradiction. In addition, by Steps 1–2 and Lemma 3.6,

$$\begin{aligned} \displaystyle \frac{\partial \lambda _1^{(1)}}{\partial d}\left( d_1,q,{\varvec{1}}-\frac{{\varvec{u}}^*}{{\varvec{k}}}\right) =\frac{-\left( T_1^{(1)}f_1^{(1)}+T_2^{(1)}f_2^{(1)}\right) }{(d_1+q_1)\left( u^*_1\right) ^2+d_1\left( u^*_2\right) ^2+d_1\left( u^*_3\right) ^2}>0. \end{aligned}$$

Then we obtain the desired results.

Step 4. For $d_1\ne d_2$, model (I) admits no positive equilibrium.

Suppose to the contrary that model (I) has a positive equilibrium, denoted by $({\varvec{u}},{\varvec{v}})$. Using the similar arguments as in the proof of Steps 1–2, we see that ${{\tilde{f}}}_1^{(1)},{{\tilde{f}}}_2^{(1)},{{\tilde{g}}}_1^{(1)},{{\tilde{g}}}_2^{(1)}<0$ and ${\mathcal {T}}_1^{(1)},{\mathcal {T}}_2^{(1)},{\mathcal {S}}_1^{(1)},{\mathcal {S}}_2^{(1)}>0$, where ${{\tilde{f}}}^{(1)}_i,{{\tilde{g}}}^{(1)}_i,{\mathcal {T}}^{(1)}_i,{\mathcal {S}}^{(1)}_i (i=1,2)$ are defined in (54) and (55). By (58a), we have

$$\begin{aligned} 0=\displaystyle \sum ^2_{j=1}{\mathcal {S}}_j^{(1)}{\tilde{f}}_{j}^{(1)}<0, \end{aligned}$$

which is the contradiction.

By Step 3–4, we see that, for $d_2>d_1$, $({\varvec{u}}^*, {\varvec{0}})$ is unstable and model (I) admits no positive equilibrium. Then, by Lam and Munther (2016, Theorem 1.3), (ii) holds for model (I). Since the nonlinear terms of model (I) are symmetric, we obtain that (i) also holds. $\square$