Three-patch Models for the Evolution of Dispersal in Advective Environments: Varying Drift and Network Topology

Abstract

We study the evolution of dispersal in advective three-patch models with distinct network topologies. Organisms can move between connected patches freely and they are also subject to the passive, directed drift. The carrying capacity is assumed to be the same in all patches, while the drift rates could vary. We first show that if all drift rates are the same, the faster dispersal rate is selected for all three models. For general drift rates, we show that the infinite diffusion rate is a local Convergence Stable Strategy (CvSS) for all three models. However, there are notable differences for three models: For Model I, the faster dispersal is always favored, irrespective of the drift rates, and thus the infinity dispersal rate is a global CvSS. In contrast, for Models II and III a singular strategy will exist for some ranges of drift rates and bi-stability phenomenon happens, i.e., both infinity and zero diffusion rates are local CvSSs. Furthermore, for both Models II and III, it is possible for two competing populations to coexist by varying drift and diffusion rates. Some predictions on the dynamics of n-patch models in advective environments are given along with some numerical evidence.

Appendices

Appendix A: The Global Dynamics of Model I

In this section, we mainly study the global dynamics of Model I. By the monotone dynamical system theory (Hess and Lazer 1991, Theorem 1.5) (see also Hsu et al. 1996; Lam and Munther 2016; Smith 1995), in order to show the global stability of the semi-trivial steady state $(U^*,0)$, we need to show the linear instability of the other semi-trivial steady state $(0,V^*)$ and the non-existence of positive steady state of (1).

By replacing $u_i$ and $v_i$ by $ku_i$ and $kv_i$, for all i, we may assume without loss of generality that $k=1$. Henceforth, we will prove our theorems concerning Models I, II and III only for the case $k=1$.

1.1 A.1. Preliminary Estimates on Nonnegative, Non-trivial Steady States

In this subsection, we consider nonnegative and non-trivial solutions of Model I. After setting $k=1$, they satisfy the following system:

$$\begin{aligned} \left\{ \begin{array}{l} d(U_3-U_1)-q_1U_1+U_1(1-U_1-V_1)=0\\ d(U_3-U_2)-q_2U_2+U_2(1-U_2-V_2)=0\\ d(U_1+U_2-2U_3)+q_1U_1+q_2U_2+U_3(1-U_3-V_3)=0\\ D(V_3-V_1)-q_1V_1+V_1(1-U_1-V_1)=0\\ D(V_3-V_2)-q_2V_2+V_2(1-U_2-V_2)=0\\ D(V_1+V_2-2V_3)+q_1V_1+q_2V_2+V_3(1-U_3-V_3)=0\\ \end{array} \right. \end{aligned}$$

(6)

When $d,D>0$, it follows from irreducibility of system (6) that there are at most three types of nonnegative and non-trivial solutions, namely: semi-trivial equilibria $(U^*,0)$, $(0,V^*)$ and positive solutions for which $U_i>0$, $V_i>0$, $i=1,2,3$. Hence, for the simplicity of notation, in this subsection we may denote all of these different types of solutions as (U, V), with the understanding that there are only three possibilities for all $i=1,2,3$: $U_i>0$ and $V_i=0$, or $U_i=0$ and $V_i>0$, or $U_i>0$ and $V_i>0$. We shall establish some a prior estimates of (U, V).

Lemma 1

Assume $q_1,q_2\ge 0$, $(q_1,q_2)\not = (0,0)$ and $d,D>0$.

• (i) If $q_1\ge q_2$, then $U_1\le U_2$ and  $V_1\le V_2$.

• (ii) If $q_1\le q_2$, then $U_1\ge U_2$ and  $V_1\ge V_2$.

In particular, if $q_1=q_2$, then $U_1=U_2$ and  $V_1=V_2$.

Proof

For part (i) we shall prove $U_1\le U_2$ only, as $V_1\le V_2$ follows from a similar argument. We argue by contradiction: if not, assume that there exist some $q_1 \ge q_2$ and a nonnegative, non-trivial solution of (6) such that $U_1> U_2$. By the first and second equations of (6), we get

$$\begin{aligned} (-d-q_1+1-U_1-V_1)U_1=(-d-q_2+1-U_2-V_2)U_2=-dU_3<0, \end{aligned}$$

(7)

so that $-d-q_i+1-U_i-V_i<0$, $i=1,2$. Due to $U_1> U_2$, (7) implies

$$\begin{aligned} (-d-q_1+1-U_1-V_1)U_1> (-d-q_2+1-U_2-V_2)U_1, \end{aligned}$$

(8)

and thus

$$\begin{aligned} q_2-q_1> (U_1+V_1)-(U_2+V_2), \end{aligned}$$

(9)

which together with $q_1\ge q_2$ implies that $U_1+V_1< U_2+V_2$. This implies $(V_1,V_2)\ne (0,0)$ and $V_2>V_1>0$.

Therefore, similar to (8), the equations of $V_1$ and $V_2$ from (6) imply

$$\begin{aligned} (-D-q_1+1-U_1-V_1)V_1<(-D-q_2+1-U_2-V_2)V_1, \end{aligned}$$

which implies $q_2-q_1<(U_1+V_1)-(U_2+V_2)$. This, however, contradicts (9). Therefore, $U_1\le U_2$ holds. This proves (i). The conclusion in (ii) follows by exchanging patches 1 and 2. $\square$

Lemma 2

Assume $q_1,q_2\ge 0$ and $(q_1,q_2)\not = (0,0)$. For any $d,D>0$,

• (i) if $q_1\ge q_2$, then $U_1+V_1<1<U_3+V_3$;

• (ii) if $q_1\le q_2$, then $U_2+V_2<1<U_3+V_3$.

In particular, when $q_1=q_2$, $U_1+V_1=U_2+V_2<1<U_3+V_3$.

Proof

As the proofs of (i) and (ii) are similar, we only prove (i). Firstly, we show

$$\begin{aligned} U_1+V_1<1. \end{aligned}$$

(10)

We argue by contradiction: If not, there exist some $q_1\ge q_2$ and a nonnegative non-trivial solution such that $U_1+V_1 \ge 1$; i.e., $1-U_1-V_1 \le 0$. We claim that

$$\begin{aligned} U_3+V_3>U_1+V_1 \ge 1. \end{aligned}$$

(11)

Without loss of generality, we may assume $(U_i)$ is non-trivial, so that the first equation of (6) implies that $U_3 \ge \frac{d+q_1}{d}U_1>U_1$. If $(V_i)$ is trivial then (11) holds. If not, then applying similar reasoning to the fourth equation of (6), we also get $V_3 \!\ge \! \frac{D+q_1}{D}V_1\!>\!V_1$. This proves (11) for any nonnegative solutions. Due to $q_1\ge q_2$, we get $U_2+V_2\ge U_1+V_1 \ge 1$ by Lemma 1, thus

$$\begin{aligned} U_1(1-U_1-V_1)+U_2(1-U_2-V_2)+U_3(1-U_3-V_3)<0. \end{aligned}$$

However, adding the equations of $U_1, U_2, U_3$ in (6), we get

$$\begin{aligned} U_1(1-U_1-V_1)+U_2(1-U_2-V_2)+U_3(1-U_3-V_3)=0. \end{aligned}$$

(12)

This is a contradiction. This proves (10).

Next, we claim

$$\begin{aligned} U_3+V_3>1. \end{aligned}$$

(13)

We again argue by contradiction and assume that there exist some $q_1 \ge q_2$ and a nonnegative non-trivial solution such that $(U_i)$ is non-trivial and $U_3+V_3 \le 1$. From the third equation of (6), we obtain

$d(U_1+U_2-2U_3)+q_1U_1+q_2U_2 \le 0$,

which together with $U_1\le U_2$ (Lemma 1) implies

$$\begin{aligned} 2d(U_1-U_3)+q_1U_1+q_2U_2 \le 0. \end{aligned}$$

Hence $U_1 < U_3$. If $(V_i)$ is non-trivial, then we can get $V_1<V_3$ by the same method. Therefore, $U_1+V_1<U_3+V_3 \le 1$. In view of (12), we have also

$$\begin{aligned} U_2+V_2>1 \ge U_3+V_3. \end{aligned}$$

Using the second equation of (6), we get $U_3>U_2$, which implies $V_3<V_2$. Hence, by the equation of $V_2$ in (6), we get $1-U_2-V_2>0$, i.e., $U_2+V_2<1$, which is again impossible. Hence, we proved (13). The proof of (i) is completed. $\square$

Lemma 3

Assume $q_1,q_2\ge 0$, $(q_1,q_2)\not = (0,0)$, and $d,D>0$.

• (i) If $q_1\ge q_2$, and $(U_1,U_2,U_3)\ne (0,0,0)$, then $U_1\le U_2<U_3$.

• (ii) If $q_1\ge q_2$, and $(V_1,V_2,V_3)\ne (0,0,0)$, then $V_1\le V_2<V_3$.

• (iii) If $q_1\le q_2$, and $(U_1,U_2,U_3)\ne (0,0,0)$, then $U_2\le U_1<U_3$.

• (iv) If $q_1\le q_2$, and $(V_1,V_2,V_3)\ne (0,0,0)$, then $V_2\le V_1 < V_3$.

In particular, if $q_1=q_2$ then every positive equilibrium satisfies $U_1=U_2<U_3$, $V_1=V_2<V_3$.

Proof

We only prove (i) as (ii)-(iv) follow from a similar argument. To this end, we assume $(U_1,U_2,U_3)\ne (0,0,0)$ and prove $U_1\le U_2<U_3$. From Lemma 1, it suffices to prove $U_3>U_2$. Suppose to the contrary that $q_1\ge q_2$ and there is a nonnegative solution such that $(U_1,U_2,U_3)\ne (0,0,0)$ and $U_3 \le U_2$. We claim that $U_3+V_3 \le U_2 + V_2$. This is immediate if $(V_i)$ is trivial. If $(V_i)$ is non-trivial, then the second equation of (6) implies

$$\begin{aligned} -q_2+1-U_2-V_2\ge 0. \end{aligned}$$

By way of the fifth equation of (6), we obtain $V_3\le V_2$, which again implies $U_3+V_3 \le U_2 + V_2$.

By Lemma 2, we have $1-U_2-V_2\le 1-U_3-V_3<0$. Again using the second equation of (6), we get $U_3>U_2$, a contradiction. The assertions (ii)-(iv) are analogous, by exchanging the role of U and V or the patches one and two. $\square$

Lemma 4

Assume $q_1,q_2\ge 0$, $(q_1,q_2)\not = (0,0)$, and $d,D>0$. Then we have

$$\begin{aligned} 3-\sum _{i=1}^3 (U_i+V_i)>0. \end{aligned}$$

(14)

Proof

By exchanging the two species if necessary, we may assume without loss of generality that $(U_i)$ is non-trivial. Adding the equations of $U_i$, $i=1,2,3$, in (6), we get

$$\begin{aligned} U_1(1-U_1-V_1)+U_2(1-U_2-V_2)+U_3(1-U_3-V_3)=0. \end{aligned}$$

(15)

If $q_1\ge q_2$, applying (15), $U_1+V_1<1<U_3+V_3$ (Lemma 2) and $U_1\le U_2<U_3$ (Lemma 3). Hence

$$\begin{aligned} U_2(1-U_1-V_1)+U_2(1-U_2-V_2)+U_2(1-U_3-V_3)>0, \end{aligned}$$

that is, (14) holds.

The proof of the case $q_1<q_2$ is similar and thus omitted. $\square$

Lemma 5

Assume $q_1,q_2\ge 0$, $(q_1,q_2)\not = (0,0)$ and $d,D>0$.

• (i) If $(U_i)$ is non-trivial, then $-q_1+1-U_1-V_1<0$.

• (ii) If $(V_i)$ is non-trivial, then $-q_2+1-U_2-V_2<0$.

Proof

By Lemma 3 and the first and second equation of (6), we get

$$\begin{aligned} (-q_i+1-U_i-V_i)U_i=d(U_i-U_3)<0, \quad i=1,2. \end{aligned}$$

This proves (i). The proof of (ii) is omitted. $\square$

From the above results, we can establish the non-existence of positive solution of (6).

Lemma 6

Assume $q_1,q_2\ge 0$, $(q_1,q_2)\not = (0,0)$, and $d,D>0$ satisfy $d\not = D$. Then (6) has no positive solution.

Proof

If there exists a positive solution (U, V) with $U_i>0$ and $V_i>0$, then we can rewrite (6) as $E_0(U_1,U_2,U_3)^T=(0,0,0)^T$ and $F_0(V_1,V_2,V_3)^T=(0,0,0)^T$, where the matrices $E_0$ and $F_0$ are defined as

$$\begin{aligned} \begin{aligned} E_0&= \begin{pmatrix}-d-q_1+1-U_1-V_1 &{} 0 &{} d\\ 0 &{} -d-q_2+1-U_2-V_2 &{} d\\ d+q_1 &{} d+q_2 &{} -2d+1-U_3-V_3\end{pmatrix}\\ F_0&= \begin{pmatrix}-D-q_1+1-U_1-V_1 &{} 0 &{} D\\ 0 &{} -D-q_2+1-U_2-V_2 &{} D\\ D+q_1 &{} D+q_2 &{} -2D+1-U_3-V_3\end{pmatrix}. \end{aligned}\nonumber \\ \end{aligned}$$

(16)

Direct calculation gives

$$\begin{aligned} \begin{aligned} 0=\mathrm{det}(E_0)&=d^2 \left( 3-\sum _{i=1}^3(U_i+V_i)\right) +d {P} \\&\quad + (-q_1 + 1-U_1-V_1)(-q_2 + 1 -U_2-V_2)(1 -U_3-V_3), \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} 0=\mathrm{det}(F_0)&=D^2\left( 3-\sum _{i=1}^3(U_i+V_i)\right) +D{P} \\&\qquad + (-q_1 + 1-U_1-V_1)(-q_2 + 1 -U_2-V_2)(1 -U_3-V_3), \end{aligned} \end{aligned}$$

for some constant P depending only on $U_i, V_i$ ($i=1,2,3$) and $q_j$ ($j=1,2$). Multiplying the above two equations by D, d, respectively, and subtracting the resulting equations, in view of $D \ne d$, we obtain

$$\begin{aligned} \begin{aligned}&Dd\left( 3-\sum _{i=1}^3(U_i+V_i)\right) \\&= (-q_1 + 1-U_1-V_1)(-q_2 + 1 -U_2-V_2)(1 -U_3-V_3). \end{aligned} \end{aligned}$$

(17)

From Lemmas 2 and 5, it follows that the right-hand side of (17) is negative. However, the left-hand side of (17) is positive, as implied by Lemma 4. This contradiction finishes the proof. $\square$

1.2 A.2. The Global Stability of Semi-trivial Steady State

In this subsection, we mainly establish Theorem 1. We first study the linear instability of the semi-trivial steady state $(0,V^*):=(0,0,0,V^*_1,V^*_2,V^*_3)$ for Model I, where $V^*$ satisfies

$$\begin{aligned} F_1(V_1^*, V_2^*, V_3^*)^T=(0,0,0)^T, \end{aligned}$$

(18)

with matrix $F_1$ given by

$$\begin{aligned} F_1= \begin{pmatrix}-D-q_1+1-V_1^* &{} 0 &{} D\\ 0 &{} -D-q_2+1-V_2^* &{} D\\ D+q_1 &{} D+q_2 &{} -2D+1-V_3^*\end{pmatrix}. \end{aligned}$$

The linear instability of $(0,V^*)$ is determined by the sign of the principal eigenvalue, denoted as $\varLambda _1$, of the eigenvalue problem

$$\begin{aligned} E_1\begin{pmatrix}\varphi _1\\ \varphi _2\\ \varphi _3\end{pmatrix} + \varLambda \begin{pmatrix}\varphi _1\\ \varphi _2\\ \varphi _3\end{pmatrix} =\begin{pmatrix} 0\\ 0\\ 0\end{pmatrix}, \end{aligned}$$

(19)

where matrix $E_1$ is given by

$$\begin{aligned} E_1= \begin{pmatrix}-d-q_1+1-V_1^* &{} 0 &{} d\\ 0 &{} -d-q_2+1-V_2^* &{} d\\ d+q_1 &{} d+q_2 &{} -2d+1-V_3^*\end{pmatrix}. \end{aligned}$$

Note that $\varLambda _1=\varLambda _1(d,D)$ depends on D by way of $V^*_i$. We first study the sign of $\varLambda _1$ for the case $q_1=q_2=q$. From Jiang et al. (2020), we recall the following two results concerning $\varLambda _1$:

Proposition 1

(Jiang et al. 2020, Proposition 1) Suppose $q_1=q_2=q>0$. Then the derivative of $\varLambda _1$ with respect to d, at d=D, is given by

$$\begin{aligned} \frac{\partial \varLambda _1}{\partial d}{\Big |_{d=D}} =-\frac{(V_1^*-\frac{D}{D+q}V_3^*)(V_3^*-V_1^*)+(V_2^*-\frac{D}{D+q}V_3^*)(V_3^*-V_2^*)}{(V_1^*)^2+(V_2^*)^2+\frac{D}{D+q}(V_3^*)^2}. \end{aligned}$$

(20)

Proposition 2

(Jiang et al. 2020, Proposition 2) Assume $q_1=q_2=q>0$ and $V_1^*+V_2^*+V_3^*\ne 3$. Then $\mathrm{det}(E_1)=0$ if and only if either $d=D$, or $d=F(D)$, where function F is given by

$$\begin{aligned} F(D): = \frac{(-q + 1 - {V_1^*})(-q + 1 - {V_2^*})(1 - {V_3^*})}{D(3-V_1^*-V_2^*-V_3^*)}, \quad D>0. \end{aligned}$$

(21)

Corollary 1

Assume $q_1=q_2=q>0$. For any $d, D>0$, we have $\frac{\partial \varLambda _1}{\partial d}{\big |_{d=D}} < 0$.

Proof

By Lemma 3, we have $V_3^*-V_1^*>0$ and $V_3^*-V_2^*>0$. Using (i) of Lemma 1 and Lemma 2, we get $V_2^*=V_1^*<1$, which together with the equations of $V_1^*$ and $V_2^*$ in (18) yields

$$\begin{aligned} V_1^*-\frac{D}{D+q}V_3^*> 0 \quad and \quad V_2^*-\frac{D}{D+q}V_3^* > 0. \end{aligned}$$

Therefore, the right-hand side of (20) is negative. $\square$

Corollary 2

Assume $q_1=q_2=q>0$. Then for any $d, D>0$, the right-hand side of (21) is strictly negative.

Proof

This lemma is a direct consequence of Lemmas 2, 4 and 5. $\square$

Theorem 8

Assume $q_1=q_2=q>0$. Then for any $d, D>0$, we have

$$\begin{aligned} \varLambda _1(d,D)=\left\{ \begin{aligned} + \quad \quad&D > d; \\ - \quad \quad&D < d. \\ \end{aligned} \right. \end{aligned}$$

Proof

Since the right-hand side of (21) is strictly negative (by Corollary 2), Proposition 2 says that $\varLambda _1(d,D)=0$ if and only if $d=D$. Therefore, by Corollary 1 and the continuity of $\varLambda _1$, $\varLambda _1(d,D)>0$ holds for $D>d>0$ and $\varLambda _1(d,D)<0$ holds for $0<D<d$. $\square$

Next, we consider the sign of $\varLambda _1$ for any $q_1,q_2\ge 0$ and $(q_1,q_2)\not = (0,0)$.

Lemma 7

For any $q_1,q_2\ge 0$ and $(q_1,q_2)\not = (0,0)$, we have $\varLambda _1(d,D)<0$ for $d>D>0$.

Proof

Fix $d> D >0$. By Theorem 8, if $q_1=q_2=q$, then $\varLambda _1(d,D)<0$. Then by the continuity of $\varLambda _1$ in $q_1,q_2$, it is sufficient to show $\varLambda _1 \ne 0$ for any $q_1 \ne q_2$. If not, we assume there exist some $q_1\ne q_2$ such that $\varLambda _1 = 0$. By direct calculation, we get

$$\begin{aligned} \begin{aligned} 0=\mathrm{det}(E_1)&=d^2 \left( 3-V_1^*-V_2^*-V_3^*\right) +MD + (-q_1 + 1 \\&\quad - {V_1^*})(-q_2 + 1 - {V_2^*})(1 - {V_3^*}). \end{aligned} \end{aligned}$$

By (18), we also get

$$\begin{aligned} \begin{aligned} 0=\mathrm{det}(F_1)&=D^2 \left( 3-V_1^*-V_2^*-V_3^*\right) +Md \\&\quad + (-q_1 + 1 - {V_1^*})(-q_2 + 1 - {V_2^*})(1 - {V_3^*}). \end{aligned} \end{aligned}$$

Here M depends on $V_i^* (i=1,2,3), q_1$ and $q_2$. Multiplying the above two equations by d, D, respectively, and subtracting them, we obtain

$$\begin{aligned} (D-d)[(3-V_1^*-V_2^*-V_3^*)Dd-(-q_1+1-V_1^*)(-q_2+1-V_2^*)(1-V_3^*)]=0. \end{aligned}$$

Due to $d>D$, we have

$$\begin{aligned} (3-V_1^*-V_2^*-V_3^*)Dd=(-q_1+1-V_1^*)(-q_2+1-V_2^*)(1-V_3^*). \end{aligned}$$

(22)

Lemmas 2 and 5 imply that the right-hand side of (22) is negative. However, the left-hand side of (22) is positive, as implied by Lemma 4. This contradiction finishes the proof. $\square$

Proof of Theorem 1

Fix $d>D$. By Lemmas 6 and 7, $(0,V^*)$ is linearly unstable, and Model I has no positive equilibria. By the theory of monotone dynamical systems (Hess and Lazer 1991, Theorem 1.5), $(U^*,0)$ is globally asymptotically stable among all nonnegative, non-trivial solutions of (1).$\square$

Appendix B: The Dynamics of Model II

In this section, we mainly study the dynamics of Model II and establish Theorems 2 to 4.

1.1 B.1. Preliminary Estimates on Nonnegative, Non-trivial Steady States

In this subsection, we study the nonnegative and non-trivial solutions of the system

$$\begin{aligned} \left\{ \begin{array}{l} d(U_2-U_1)-q_1U_1+U_1(1-U_1-V_1)=0\\ d(U_1+U_3-2U_2)+q_1U_1-q_2U_2+U_2(1-U_2-V_2)=0\\ d(U_2-U_3)+q_2U_2+U_3(1-U_3-V_3)=0\\ D(V_2-V_1)-q_1V_1+V_1(1-U_1-V_1)=0\\ D(V_1+V_3-2V_2)+q_1V_1-q_2V_2+V_2(1-U_2-V_2)=0\\ D(V_2-V_3)+q_2V_2+V_3(1-U_3-V_3)=0.\\ \end{array} \right. \end{aligned}$$

(23)

Once again, we set $k=1$ and observe that system (23) has at most three types of nonnegative and non-trivial solutions, that is, semi-trivial solution $(U^*,0),(0,V^*)$ and positive solutions for which $U_i>0, V_i>0$ ($i=1,2,3$). In the following, we denote by (U, V) a nonnegative non-trivial solution of (23), in which $U_i>0$ and $V_i=0$, or $U_i=0$ and $V_i>0$, or $U_i>0$ and $V_i>0$ for all $i=1,2,3$. We shall establish a priori estimates of the nonnegative and non-trivial solutions (U, V).

Lemma 8

For any $d,D>0$ and $q_1,q_2>0$, we have $U_1+V_1<1<U_3+V_3$.

Proof

We will prove this conclusion for the case $U_i>0$. The case $U_i \equiv 0$ and $V_i>0$ can be proved by a similar argument.

Step 1: We prove $U_1+V_1<1$. We argue by contradiction and assume that there exist some $q_1,q_2$ such that $U_1+V_1\ge 1$. Then by the first equation of (23), we have $d(U_2-U_1)-q_1U_1 \ge 0$, i.e., $U_2 \ge \frac{d+q_1}{d}U_1 >U_1$. Following from the similar argument, we also get $V_i=0$ for all i, or $V_2>V_1$. Hence $U_2+V_2>U_1+V_1 \ge 1$.

Clearly, we have $U_3+V_3 \ge 1$. If not, assume that $U_3+V_3 <1$ for some $q_1,q_2$. By the equations of $U_3$ and $V_3$, we have $U_3>\frac{d+q_2}{d}U_2>U_2$ and $V_3 \ge \frac{D+q_2}{D}V_2 \ge V_2$ (where equality holds in case $V_i=0$ for all i), which imply $U_3+V_3>U_2+V_2 >1$, a contradiction.

Therefore, we get

$$\begin{aligned} U_1(1-U_1-V_1)+U_2(1-U_2-V_2)+U_3(1-U_3-V_3)<0. \end{aligned}$$

However, adding the equations of $U_i$, $i=1,2,3$, in (23), we find that the left-hand side of the above inequality is equal to zero. This is a contradiction. Hence, $U_1+V_1<1$ holds.

Step 2: We show $U_3+V_3>1$. If not, we assume $U_3+V_3 \le 1$ for some $q_1, q_2$. By the equations of $U_3$ and $V_3$ in (23), we deduce that $U_2<U_3$ and $V_2\le V_3$ (with equality holds if $V_i =0$). Thus $U_2+V_2<U_3+V_3 \le 1$, which together with $U_1+V_1<1$ implies

$$\begin{aligned} U_1(1-U_1-V_1)+U_2(1-U_2-V_2)+U_3(1-U_3-V_3)>0. \end{aligned}$$

Similarly, adding the equations of $U_i$, $i=1,2,3$, in (23), we find that the left-hand side of the above inequality is equal to zero. This is a contradiction. $\square$

Lemma 9

For any $d,D>0$, $q_1, q_2>0$, it holds that

$$\begin{aligned} -q_2+1-U_2-V_2<0. \end{aligned}$$

(24)

Proof

We consider two cases:

Case I. Either $U_3> U_2$ or $V_3> V_2$. Without loss of generality, assume $U_3> U_2$, so that $U_i >0$ for all i. Adding equations of $U_1$ and $U_2$ in (23), we have

$$\begin{aligned} d(U_3-U_2) +U_1(1-U_1-V_1) +U_2(-q_2+1-U_2-V_2)=0. \end{aligned}$$

(25)

It is easy to see that (24) follows from (25), $U_3\ge U_2$ and $U_1+V_1<1$ (Lemma 8).

Case II. $U_3\le U_2$ and $V_3\le V_2$. For this case,

$$\begin{aligned} -q_2+1-U_2-V_2 \le -q_2+1-U_3-V_3<0. \end{aligned}$$

The last inequality follows from Lemma 8. This completes the proof. $\square$

Lemma 10

Let $d,D>0$, and either $q_1\ge 1$ or $\frac{q_2}{2}<q_1<1$.

• (i) Either $U_i \equiv 0$ or $U_1 < U_2$.

• (ii) Either $V_i \equiv 0$ or $V_1 < V_2$.

Proof

We only prove (i), as (ii) follows in a completely analogous manner. Assume $U_i >0$ for all i, we need to show $U_1 < U_2$. Obviously, for $q_1 \ge 1$, this conclusion is true from the first equation of (23).

Next, assume to the contrary that there exist some $\frac{q_2}{2}<q_1<1$ such that $U_1 \ge U_2$. By the first equation of (23), $-q_1+1-U_1-V_1\ge 0$, i.e., $U_1+V_1 \le 1-q_1$. Hence $U_1+V_1<1-\frac{q_2}{2}$. Using the 4th equation of (23), we get $V_2 \le V_1$. So we have

$$\begin{aligned} U_2+V_2 \le U_1+V_1 <1-\frac{q_2}{2}. \end{aligned}$$

(26)

From the second equation of (23), we get

$$\begin{aligned} d(U_1+U_3-2U_2)+q_1U_1-q_2U_2+\frac{q_2}{2}U_2 <0, \end{aligned}$$

which together with $U_1 \ge U_2$ indicates $d(U_3-U_2)+(q_1- \frac{q_2}{2})U_2<0$. Since $q_1>\frac{q_2}{2}$, we have $U_3<U_2$.

We claim that $U_2 + V_2 > U_3 + V_3$. If $V_i \equiv 0$, then it follows from $U_3<U_2$ and we are done. If $V_i >0$ for all i, then we can repeat the above argument to show that $V_3 < V_2$. Using Lemma 8, we have $U_2 + V_2> U_3 + V_3 >1$. This is in contradiction with (26). $\square$

Lemma 11

For any $d,D>0$, if $q_1 \ge 1$ or $\frac{q_2}{2}<q_1<1$, then

$$\begin{aligned} 3-\sum _{i=1}^3 (U_i+V_i)>0. \end{aligned}$$

(27)

Proof

Adding all six equations of (23), we have

$$\begin{aligned}&(U_1+V_1)(1-U_1-V_1) +(U_2+V_2)(1-U_2-V_2) \nonumber \\&\qquad +(U_3+V_3)(1-U_3-V_3) =0. \end{aligned}$$

(28)

We consider two cases:

Case I. $U_2+V_2\le U_3+V_3$. For this case, by $U_3+V_3>1$ we have

$$\begin{aligned} (U_3+V_3)(1-U_3-V_3) \le (U_2+V_2)(1-U_3-V_3). \end{aligned}$$

(29)

By Lemma 10, we have $U_1+V_1<U_2+V_2$. This together with $U_1+V_1<1$ implies

$$\begin{aligned} (U_1+V_1)(1-U_1-V_1) <(U_2+V_2)(1-U_1-V_1). \end{aligned}$$

(30)

It is easy to see that (27) follows directly from (28), (29) and (30).

Case II. $U_2+V_2\ge U_3+V_3$. For this case, by $U_3+V_3>1>U_1+V_1$ (by Lemma 8) we have

$$\begin{aligned} (U_3+V_3)(1-U_3-V_3) <(U_1+V_1)(1-U_3-V_3). \end{aligned}$$

(31)

Since $U_2 + V_2 \ge U_3 + V_3>1 > U_1 + V_1$, we can similarly derive

$$\begin{aligned} (U_2+V_2)(1-U_2-V_2) <(U_1+V_1)(1-U_2-V_2). \end{aligned}$$

(32)

It is easy to see that (27) follows directly from (28), (31) and (32). Note that the above reasoning is valid also when $U_1=U_2=U_3=0$ or $V_1=V_2=V_3=0$. $\square$

Note that the above results are valid when $q_1=q_2>0$. The following result implies that (23) has no positive solution when $q_1 \ge 1$ or $\frac{q_2}{2}<q_1<1$.

Corollary 3

If $q_1 \ge 1$ or $\frac{q_2}{2}<q_1<1$, then system (23) has no positive solutions for $d\not =D$.

Proof

We argue by contradiction. If there exists some positive solution, denoted by (U, V), for (23). Direct calculation, as in the proof of Lemma 6, gives

$$\begin{aligned} \begin{aligned}&Dd(3-\sum _{i=1}^3(U_i+V_i))\\&\quad =(-q_1+1-U_1-V_1)(-q_2+1-U_2-V_2)(1-U_3-V_3). \end{aligned} \end{aligned}$$

(33)

Due to $U_1<U_2$, the equation of $U_1$ implies $-q_1+1-(U_1+V_1)<0$. By Lemma 8, $1-(U_3+V_3)<0.$ By Lemma 9, we see $-q_2+1-(U_2+V_2)<0.$ Hence, the right-hand side of (33) is negative. However, the left-hand side of (33) is positive from Lemma 11, which is a contradiction. $\square$

1.2 B.2. The Global Dynamics of Model II When $q_1 \ge 1$ or $\frac{q_2}{2}<q_1<1$

In this subsection, we shall show that the faster diffuser always wins when $q_1 \ge 1$ or $\frac{q_2}{2}<q_1<1$. We first study the local instability of $(0,V^*):=(0,0,0,V_1^*,V_2^*,V_3^*)$, as determined by the sign of the principal eigenvalue $\varLambda _2$ of the eigenvalue problem

$$\begin{aligned} E_2\begin{pmatrix}\varphi _1\\ \varphi _2\\ \varphi _3\end{pmatrix} + \varLambda \begin{pmatrix}\varphi _1\\ \varphi _2\\ \varphi _3\end{pmatrix} =\begin{pmatrix} 0\\ 0\\ 0\end{pmatrix}, \end{aligned}$$

(34)

where matrix $E_2$ is given by

$$\begin{aligned} E_2= \begin{pmatrix}-d-q_1+1-V_1^* &{} d &{} 0\\ d+q_1 &{} -2d-q_2+1-V_2^* &{} d\\ 0 &{} d+q_2 &{} -d+1-V_3^*\end{pmatrix}. \end{aligned}$$

Proposition 3

When $d=D$, the derivative of $\varLambda _2$ with respect to d satisfies

$$\begin{aligned} \frac{\partial \varLambda _2}{\partial d}\Big |_{d=D}=-\frac{(D+q_1)V_1^*(V_2^*-V_1^*) + DV_2^*(V_1^*+V_3^*-2V_2^*) + \frac{D^2}{D+q_2}V_3^*(V_2^*-V_3^*)}{(D+q_1)(V_1^*)^2 + D(V_2^*)^2 + \frac{D^2}{D+q_2}(V_3^*)^2}.\nonumber \\ \end{aligned}$$

(35)

Proof

Differentiate (34) with respect to d, we get

$$\begin{aligned} \begin{pmatrix} \varphi _2-\varphi _1\\ \varphi _1+\varphi _3-2\varphi _2\\ \varphi _2-\varphi _3\end{pmatrix} + E_2\begin{pmatrix}\varphi _1'\\ \varphi _2'\\ \varphi _3'\end{pmatrix} + \frac{\partial \varLambda _2}{\partial d}\begin{pmatrix}\varphi _1\\ \varphi _2\\ \varphi _3\end{pmatrix} + \varLambda _2 \begin{pmatrix}\varphi _1'\\ \varphi _2'\\ \varphi _3'\end{pmatrix} = \begin{pmatrix}0\\ 0\\ 0\end{pmatrix}, \end{aligned}$$

(36)

where $\varphi _i'=\frac{\partial \varphi _i}{\partial d}$, $i=1,2,3$. Note that when $d=D$,

$$\begin{aligned} \left. E_2\right| _{d=D}\begin{pmatrix}V_1^*\\ V_2^*\\ V_3^*\end{pmatrix}=\begin{pmatrix}0\\ 0\\ 0\end{pmatrix} , \end{aligned}$$

$$\begin{aligned} \left( \left. {E_2}\right| _{d=D}\right) ^T\begin{pmatrix} (D+q_1)V_1^*\\ DV_2^*\\ \frac{D^2}{D+q_2}V_3^*\end{pmatrix}=\begin{pmatrix}0\\ 0\\ 0\end{pmatrix}, \end{aligned}$$

(37)

and when $d=D$, we may choose

$$\begin{aligned} \begin{pmatrix}\varphi _1\\ \varphi _2\\ \varphi _3\end{pmatrix} =\begin{pmatrix}V_1^*\\ V_2^*\\ V_3^*\end{pmatrix}. \end{aligned}$$

(38)

Set $d=D$ in (36) and multiplying it by $\begin{pmatrix}(D+q_1)V_1^*, DV_2^*, \frac{D^2}{D+q_2}V_3^*\end{pmatrix}$, using (37), (38) and $\varLambda _2(D,D)= 0$, we obtain (35). This completes the proof. $\square$

1.2.1 B.2.1. The Sign of $\varLambda _2$ When $q_1=q_2$

Our goal in this subsection is to determine the sign of $\varLambda _2$ when $q_1=q_2$. We first recall the following result:

Proposition 4

(Jiang et al. 2020, Proposition 4) Assume $q_1=q_2=q$ and ${V_1^*+V_2^*+V_3^*}\ne 3$. Then $\mathrm{det}(E_2)=0$ if and only if either $d=D$, or $d=F(D)$, where F(D) is given by (21).

Lemma 12

Suppose $d,D>0$ and $q_1=q_2=q>0$, then $V_2^*<V_3^*$.

Proof

We argue by contradiction. If not, we assume there exist some $d,D,q>0$ such that $V_i >0$ for all i and $V_2\ge V_3$. By the sixth equation of (23), we get

$$\begin{aligned} qV_3^*+V_3^*(1-V_3^*)\le qV_2^*+V_3^*(1-V_3^*) \le 0, \end{aligned}$$

which implies that $V_3^* \ge 1+q >1$.

Therefore, $V_2^*\ge V^*_3 >1$. Hence,

$$\begin{aligned}&d(V_1^*+V_3^*-2V_2^*)+qV_1^*-qV_2^* + V_2^*\left( 1-V_2^*\right) \\&\quad = (d+q)(V_1^*-V_2^*) + d(V_3^*-V_2^*) + V_2^*\left( 1-V_2^*\right) <0, \end{aligned}$$

where we also used the assumption $V_2^*\ge V_3^*$ and $V_1^*<V_2^*$ (Lemma 10). This is in contradiction with the fifth equation of (23). $\square$

Corollary 4

Suppose $q_1=q_2=q>0$, then for any $d,D>0$, the quantity F(D) given in (21) is strictly negative.

Proof

By Lemmas 8 and 9, we have

$$\begin{aligned} V_3^*>1 \quad \text { and }\quad -q+1-V_2^*<0. \end{aligned}$$

By Lemma 10 and the first equation of (23), we get $-q+1-V_1^*<0.$ Using also Lemma 11, the quantity F(D), given in (21), is strictly negative. $\square$

Lemma 13

Suppose $q_1=q_2=q>0$, then for any $d,D>0$, we have $\frac{D+q}{D}V_1^*>V_2^*>\frac{D}{D+q}V_3^*$.

Proof

By the fourth equation of (23) and Lemma 8, we get $d(V_2^*-V_1^*)-qV_1^*<0$, which implies $V_2^*<\frac{D+q}{D}V_1^*$. Similarly, by the sixth equation of (23) and Lemma 8, we have $d(V_2^*-V_3^*)+qV_2^*>0$, i.e., $V_2^*>\frac{D}{D+q}V_3^*$. $\square$

Corollary 5

Suppose $q_1=q_2=q>0$, then for any $d,D>0$, we have $\frac{\partial \varLambda _2}{\partial d}\Big |_{d=D}<0$.

Proof

If $q_1=q_2=q$, (35) can be rewritten as

$$\begin{aligned} \frac{\partial \varLambda _2}{\partial d}\Big |_{d=D}=-\frac{\frac{D+q}{D}V_1^*(V_2^*-V_1^*) + V_2^*(V_1^*+V_3^*-2V_2^*) + \frac{D}{D+q}V_3^*(V_2^*-V_3^*)}{\frac{D+q}{D}(V_1^*)^2 + (V_2^*)^2 + \frac{D}{D+q}(V_3^*)^2}. \end{aligned}$$

Note that

$$\begin{aligned} \begin{aligned}&\frac{D+q}{D}V_1^*(V_2^*-V_1^*)+V_2^*(V_1^*+V_3^*-2V_2^*)+\frac{D}{D+q}V_3^*(V_2^*-V_3^*)\\&\quad =(V_2^*-V_1^*)(\frac{D+q}{D}V_1^*-V_2^*)+(V_3^*-V_2^*)(V_2^*-\frac{D}{D+q}V_3^*), \end{aligned} \end{aligned}$$

which together with Lemmas 10, 12 and 13 yields the conclusion. $\square$

Theorem 9

Assume $q_1=q_2=q>0$. Then for any $d,D>0$, we have

$$\begin{aligned} \varLambda _2(d,D)=\left\{ \begin{aligned} + \quad \quad&D > d; \\ - \quad \quad&D < d.\\ \end{aligned} \right. \end{aligned}$$

Proof

Since the quantity F(D), which is given in (21), is strictly negative (by Corollary 4), Proposition 4 says that $\varLambda _2(d,D)=0$ if and only if $d=D$. Therefore, by Corollary 5 and the continuity of $\varLambda _2$, $\varLambda _2(d,D)>0$ holds for $D>d>0$ and $\varLambda _2(d,D)<0$ holds for $0<D<d$. $\square$

1.2.2 B.2.2. The Proof of Theorem 2

In this subsection, we will study the global dynamics of Model II for $q_1 \ge 1$ or $\frac{q_2}{2}<q_1<1$.

Lemma 14

If $q_1 \ge 1$ or $\frac{q_2}{2}<q_1<1$, $\varLambda _2(d,D)<0$ for $d>D$.

Proof

The proof is similar as that of Lemma 7. It follows from Theorem 9 that, if $q_1=q_2$, then $\varLambda _2(d,D)<0$ for $d>D$. Since $\varLambda _2$ is continuous with respect to parameters $q_1,q_2$, it suffices to show that $\varLambda _2 \ne 0$ for any $q_1\not =q_2$ and $q_1 \ge 1$ or $\frac{q_2}{2}<q_1<1$. We argue by contradiction and assume that there exists some q satisfying the assumptions such that $\varLambda _2 = 0$. By proceeding similarly as in Lemma 7, we derive (22) again. Note that $3-V_1^*-V_2^*-V_3^*>0$ holds, which implies the left-hand side of (22) is positive. Using $V_1^*<V_2^*$ and the first equation of (23), we deduce $-q_1+1-V_1^*<0$. By Lemma 9, $-q_2+1-V_2^*<0$. These together with $V_3^*>1$ imply the right-hand side of (22) is negative, which is a contradiction. $\square$

Proof of Theorem 2

By Corollary 3 and Lemma 14, the equilibrium $(0,V^*)$ is linearly unstable and Model II has no positive equilibria. By the theory of monotone dynamical systems (Hess and Lazer 1991, Theorem 1.5), the equilibrium $(U^*,0)$ is globally asymptotically stable. $\square$

1.3 B.3. Existence of Evolutionarily Singular Strategy

In this subsection, we consider the existence of evolutionarily singular strategy and establish Theorem 3. The linear stability of the semi-trivial steady state, $(U^*,0):=(U_1^*,U_2^*,U_3^*,0,0,0)$, is determined by the sign of the principal eigenvalue ${\tilde{\varLambda }}_2$ of the eigenvalue problem

$$\begin{aligned} F_2\begin{pmatrix}\varphi _1\\ \varphi _2\\ \varphi _3\end{pmatrix} + \varLambda \begin{pmatrix}\varphi _1\\ \varphi _2\\ \varphi _3\end{pmatrix} =\begin{pmatrix} 0\\ 0\\ 0\end{pmatrix}, \end{aligned}$$

(39)

where matrix $F_2$ is given by

$$\begin{aligned} F_2= \begin{pmatrix}-D-q_1+1-{U_1^*} &{} D &{} 0\\ D+q_1 &{} -2D-q_2+1-{U_2^*} &{} D\\ 0 &{} D+q_2 &{} -D+1-{U_3^*}\end{pmatrix}. \end{aligned}$$

Note that $U_i^*$, $i=1,2,3,$ satisfy

$$\begin{aligned} F_2|_{D=d}\begin{pmatrix}U^*_1\\ U^*_2\\ U^*_3\end{pmatrix} =\begin{pmatrix} 0\\ 0\\ 0\end{pmatrix}. \end{aligned}$$

(40)

By exchanging the role of the two species, we can rewrite Proposition 3 as follows:

Proposition 5

When $D=d$, the derivative of ${\tilde{\varLambda }}_2$ with respect to D satisfies

$$\begin{aligned} \frac{\partial {\tilde{\varLambda }}_2}{\partial D}\Big |_{D=d}=-\frac{ (d+q_1)U_1^*(U_2^*-U_1^*) + dU_2^*(U_1^*+U_3^*-2U_2^*) + \frac{d^2}{d+q_2}U_3^*(U_2^*-U_3^*)}{(d+q_1)(U_1^*)^2 + d(U_2^*)^2 + \frac{d^2}{d+q_2}(U_3^*)^2}.\nonumber \\ \end{aligned}$$

(41)

Lemma 15

For any $q_1,q_2>0$, we have $\frac{\partial {\tilde{\varLambda }}_2}{\partial D}(d,d)<0$ for sufficiently large d.

Proof

Set

$$\begin{aligned} M:=(d+q_1)U_1^*(U_2^*-U_1^*) + dU_2^*(U_1^*+U_3^*-2U_2^*) + \frac{d^2}{d+q_2}U_3^*(U_2^*-U_3^*).\nonumber \\ \end{aligned}$$

(42)

By (23), we can rewrite (42) as

$$\begin{aligned} \frac{M}{d}&=(U_2^*-U_1^*)\left[ \frac{d+q_1}{d}U_1^* - U_2^*\right] + (U_2^*-U_3^*)\left[ \frac{d}{d+q_2}U_3^* - U_2^*\right] \nonumber \\&= (U_2^*-U_1^*)\frac{U^*_1(1-U_1^*)}{d} + (U_2^*-U_3^*)\frac{U^*_3(1-U_3^*)}{d+q_2} \end{aligned}$$

(43)

Note that $(U_1^*,U_2^*,U_3^*) \rightarrow (1,1,1)$ as $d \rightarrow \infty$. As $(U_1^*, U_2^*, U_3^*)$ is the unique stable positive solution of (40), it can be shown that it is smooth at $d=\infty$ so that we can expand $U_i$ as

$$\begin{aligned} U_i^* = 1 + \frac{{\tilde{U}}_i}{d} + O(\frac{1}{d^2}) \quad \text { for }i=1,2,3. \end{aligned}$$

To determine ${\tilde{U}}_i$, we substitute the above expansion of $U^*_i$ into the first and third equation in (40) to get

$$\begin{aligned} {\tilde{U}}_1-{\tilde{U}}_2 = { -q_1}\quad \text { and }\quad {\tilde{U}}_2 - {\tilde{U}}_3 = -q_2. \end{aligned}$$

(44)

By adding the first three equations of (40), we obtain $\sum _{i=1}^3 U^*_i(1-U^*_i)=0$, from which we deduce $\sum _{i=1}^3 {\tilde{U}}_i = 0$. Combining this with (44), we obtain

$$\begin{aligned} {\tilde{U}}_1 = -\frac{2q_1+q_2}{3},\quad {\tilde{U}}_2 = \frac{q_1-q_2}{3},\quad {\tilde{U}}_3 = \frac{q_1+2q_2}{3}. \end{aligned}$$

(45)

Having determined ${\tilde{U}}_i$, we may substitute $U^*_i = 1 + {\tilde{U}}_i/d + O({1}/{d^2})$ into (43) to get

$$\begin{aligned} {d^2} M&= ({\tilde{U}}_2-{\tilde{U}}_1)(-{\tilde{U}}_1)+ ({\tilde{U}}_2 - {\tilde{U}}_3)(-{\tilde{U}}_3) + o(1)\\&= \frac{2}{3}(q_1^2 + q_1q_2 + q_2^3) + o(1) >0 \quad \text { for }d \gg 1. \end{aligned}$$

Therefore, by Proposition 5, we have $\frac{\partial {\tilde{\varLambda }}_2}{\partial D}(d,d)<0$ for $d \gg 1$. $\square$

Lemma 16

If $0<q_1<1$ and $q_2>2q_1$, we have $\frac{\partial {\tilde{\varLambda }}_2}{\partial D}(d,d)>0$ for sufficiently small d.

Proof

When $d \rightarrow 0$, we have $U_1^* \rightarrow {\overline{U}}_1:= 1-q_1$ and, passing to a subsequence if necessary, $U_2^*\rightarrow {\overline{U}}_2$ for some nonnegative ${\overline{U}}_2$. We claim that if $2q_1<q_2$, then ${\overline{U}}_2<{\overline{U}}_1$. If not, we assume for some $2q_1<q_2$, ${\overline{U}}_2\ge {\overline{U}}_1$. By the equation of $U_2^*$ and let $d \rightarrow 0$,

$$\begin{aligned} q_1{\overline{U}}_1-q_2{\overline{U}}_2+{\overline{U}}_2(1-{{\overline{U}}_2})=0. \end{aligned}$$

Then we have $q_1{\overline{U}}_2-q_2{\overline{U}}_2+{\overline{U}}_2(1-{{\overline{U}}_2})\ge 0$, which implies that ${\overline{U}}_2 \le 1+q_1-q_2$. Therefore, $1+q_1-q_2 \ge 1-q_1$, i.e., $2q_1 \ge q_2$. This contradiction shows that ${\overline{U}}_2<{\overline{U}}_1$.

Note that $M \rightarrow q_1{\overline{U}}_1({\overline{U}}_2-{\overline{U}}_1)<0$ as $d\rightarrow 0$, where M is given by (42). Note that $0<q_1<1$. Hence, for sufficiently small d, we have $\frac{\partial {\tilde{\varLambda }}_2}{\partial D}(d,d)>0$. $\square$

Proof of Theorem 3

Since $d\mapsto \frac{\partial {\tilde{\varLambda }}_2}{\partial D}(d, d)$ is analytic, all the roots are discrete. By Lemmas 15 and 16, $\frac{\partial {\tilde{\varLambda }}_2}{\partial D}(d, d)<0$ for $d\gg 1$ and $\frac{\partial {\tilde{\varLambda }}_2}{\partial D}(d, d)>0$ for $0<d\ll 1$. This says that the infinity and zero diffusion rates are local CvSSs. Furthermore, there exists at least one $d^*=d^*(q_1,q_2)$ such that $\frac{\partial {\tilde{\varLambda }}_2}{\partial D}(d^*, d^*)=0$, and $\frac{\partial {\tilde{\varLambda }}_2}{\partial D}(d, d)$ changes sign from positive to negative in a neighborhood of $d^*$; i.e., $d^*$ is an evolutionary singular strategy which is not a CvSS. $\square$

1.4 B.4. The Proof of Theorem 4

The proof of Theorem 4 is divided into a series of lemmas. First, we recall that $({\hat{V}}_2,{\hat{V}}_3)$ is the unique positive solution of (4) with $k=1$.

Lemma 17

Let $d=q_1=0$ and $D,d_2>0$ and let $({\hat{V}}_2,{\hat{V}}_3)$ be the unique positive solution of (4). Then ${\hat{V}}_2<1 < {\hat{V}}_3$ and $(1-{\hat{V}}_2,0,0,{\hat{V}}_2,{\hat{V}}_2,{\hat{V}}_3)$ is a nonnegative solution of system (23).

Proof

It is clear that $(1-{\hat{V}}_2,0,0,{\hat{V}}_2,{\hat{V}}_2,{\hat{V}}_3)$ is a nonnegative solution of system (23) when $d=q_1=0$.

Adding the equations of (4), we have

$$\begin{aligned} {\hat{V}}_2(1-{\hat{V}}_2) + {\hat{V}}_3(1-{\hat{V}}_3)=0. \end{aligned}$$

(46)

In view of (46), it is enough to show ${\hat{V}}_3>1$. Suppose not, then ${\hat{V}}_3\le 1$ and the 2nd equation of (4) implies ${\hat{V}}_2 < {\hat{V}}_3 \le 1$, which contradicts (46). $\square$

Lemma 18

The matrix

$$\begin{aligned} {\hat{E}}_1=\begin{pmatrix} -D-q_2+1-2{\hat{V}}_2 &{} D \\ D &{} -D+q_2+1-2{\hat{V}}_3 \end{pmatrix} \end{aligned}$$

is invertible.

Proof

Observe that zero is an eigenvalue of the cooperative matrix

$$\begin{aligned} {\hat{E}}_2:= {\hat{E}}_1+ \begin{pmatrix} {\hat{V}}_2 &{} 0 \\ 0 &{} {\hat{V}}_3 \end{pmatrix} \end{aligned}$$

with a strictly positive eigenvector $({\hat{V}}_2,{\hat{V}}_3)$. Hence, zero is the principal eigenvalue of ${\hat{E}}_2$. Since the principal eigenvalue is strictly monotone with respect to the diagonal entries, we deduce that zero is not an eigenvalue of ${\hat{E}}_1$. $\square$

Set $U=(U_1, U_2, U_3)$ and $V=(V_1, V_2, V_3)$. Define map $F(d,q_1,U, V): {\mathbb {R}}^8\rightarrow {\mathbb {R}}^6$ by

$$\begin{aligned} F(d,q_1, U,V)= \begin{pmatrix} d(U_2-U_1)-q_1U_1+U_1(1-U_1-V_1)\\ d(U_1+U_3-2U_2)+q_1U_1-q_2U_2+U_2(1-U_2-V_2)\\ d(U_2-U_3)+q_2U_2+U_3(1-U_3-V_3)\\ D(V_2-V_1)-q_1V_1+V_1(1-U_1-V_1)\\ D(V_1+V_3-2V_2)+q_1V_1-q_2V_2+V_2(1-U_2-V_2)\\ D(V_2-V_3)+q_2V_2+V_3(1-U_3-V_3) \end{pmatrix}\nonumber \\ \end{aligned}$$

(47)

It is clear that $(U,V)=(U_1,U_2,U_3,V_1,V_2,V_3)$ is a steady state of Model II if and only if $F(d,q_1, U, V)=0$. Now, observe that $F(0,0,{{\hat{U}}}, {{\hat{V}}})=0$. One can further compute

$$\begin{aligned} \begin{aligned}&D_{(U, V)}F(0, 0, {\hat{U}}, {\hat{V}}) \\&= \begin{pmatrix} -{(1-{\hat{V}}_2)} &{} 0 &{} 0 &{}{-(1-{\hat{V}}_2)} &{} 0 &{} 0\\ 0 &{} -q_2+1-{\hat{V}}_2 &{} 0 &{} 0 &{} 0 &{} 0\\ 0 &{} q_2 &{} 1-{\hat{V}}_3 &{} 0 &{} 0 &{} 0\\ -{{\hat{V}}_2} &{} 0 &{} 0 &{} -D-{{\hat{V}}_2} &{} D &{} 0\\ 0 &{} -{{\hat{V}}_2} &{} 0 &{} D &{} -2D-q_2+1-{2{\hat{V}}_2} &{} D\\ 0 &{} 0 &{} -{\hat{V}}_3 &{} 0 &{} D+q_2 &{} -D+1-{2{\hat{V}}_3}\\ \end{pmatrix}. \end{aligned} \end{aligned}$$

Lemma 19

Let $D>0$ and ${q_2\ge 1}$ and consider the eigenvalue problem

$$\begin{aligned} D_{(U, V)}F(0, 0, {\hat{U}}, {\hat{V}})\begin{pmatrix} \varphi \\ \psi \end{pmatrix} = \lambda \begin{pmatrix} \varphi \\ \psi \end{pmatrix} \quad \text { for }\varphi ,\psi \in {\mathbb {R}}^3. \end{aligned}$$

(48)

Then every eigenvalue of (48) lies in $\{z \in {\mathbb {C}}:\, Re \, z <0\}$. In particular, $D_{(U, V)}F(0, 0, {\hat{U}}, {\hat{V}})$ is invertible.

Proof

First, note that system (4) implies

$$\begin{aligned} \frac{D}{D+q_2-1+{\hat{V}}_2} = \frac{{\hat{V}}_2}{{\hat{V}}_3} = \frac{D-1+{\hat{V}}_3}{D+q_2}. \end{aligned}$$

(49)

It suffices to show that the principal eigenvalue of (48), denoted as $\lambda _1^*$, is strictly negative. Suppose to the contrary that (48) holds for some $\varphi ,\psi \in {\mathbb {R}}^3$ and $\lambda ^*_1 \in {\mathbb {R}}$ such that

$$\begin{aligned} \sum _{i=1}^3(|\varphi _i|+|\psi _i|)=1,\quad \varphi _i \le 0, \quad \psi _i \ge 0,\,\,\, \text { for }i=1,2,3, \quad \text { and }\quad \lambda ^*_1 \ge 0. \end{aligned}$$

(50)

We will show that $D=0$, which gives a contradiction.

Now, multiply both sides of equation (48) with $\lambda =\lambda _1^*$ on the left by the row vector

$$\begin{aligned} \mathbf {r}_1:=\left( -\frac{{\hat{V}}_2}{1-{\hat{V}}_2}\big (D-1+{\hat{V}}_3\big ),-M^2,-M, D-1+{\hat{V}}_3,D-1+{\hat{V}}_3, D\right) , \end{aligned}$$

with $M >0$ to be chosen later. We obtain

$$\begin{aligned} \mathbf {R}^T \begin{pmatrix} \varphi \\ \psi \end{pmatrix} = \lambda ^*_1 \mathbf {r}_1\begin{pmatrix} \varphi \\ \psi \end{pmatrix} \quad \text { and }\quad \mathbf {r}_1\begin{pmatrix}\varphi \\ \psi \end{pmatrix} >0, \end{aligned}$$

(51)

where the strict inequality follows from (50), and $\mathbf {R}$ can be computed (using (49) for $R_5$) as

$$\begin{aligned} \mathbf {R} = \begin{pmatrix} R_1\\ R_2\\ R_3\\ R_4\\ R_5\\ R_6 \end{pmatrix} = \begin{pmatrix} 0 \\ M^2(q_2-1+{{\hat{V}}_2}) -M q_2 { -{\hat{V}}_2(D-1+{\hat{V}}_3)} \\ M {({\hat{V}}_3-1)} - {\hat{V}}_3D\\ 0\\ -{{\hat{V}}_2}(D-1+ {\hat{V}}_3)\\ - {{\hat{V}}_3} D \end{pmatrix} \end{aligned}$$

Next choose $M \gg 1$ so that $R_2,R_3<0$ and $R_5,R_6<0$. By inspecting (51) in conjunction with (50), we deduce that

$$\begin{aligned} \varphi _1=-\frac{1}{2},\quad \varphi _2=\varphi _3=\psi _2=\psi _3 = 0, \quad \psi _1= \frac{1}{2} \quad \text { and }\quad \lambda ^*_1=0. \end{aligned}$$

But if we substitute this into the 5th component of (48), we have $D/2=0$. This is a contradiction. $\square$

Lemma 20

Fix any $D>0$ and ${q_2\ge 1}$. Then there exists some $\delta >0$ such that for any $d\in (0, \delta )$, $q_1\in (-\delta , \delta )$ and $d+q_1>0$, Model II has a positive steady state, denoted by $(U^\delta , V^\delta )$, which satisfies $(U^\delta ,V^\delta )\rightarrow ({\hat{U}}, {\hat{V}})$ as $(d,q_1)\rightarrow (0,0)$, where $({\hat{U}}, {\hat{V}})$ is given in (3) and (4) with $k=1$.

Proof

It is easy to check that $F(0,0, {\hat{U}}, {\hat{V}})=0$. Moreover, we have shown in Lemma 19 that $D_{(U, V)}F (0, 0, {\hat{U}}, {\hat{V}})$ is invertible.

By the implicit function theorem, there exists some $\delta >0$ such that for $|d|, |q_1|\le \delta$, there exists $(U^\delta , V^\delta )\in {\mathbb {R}}^6$ such that

$$\begin{aligned} F(d, q_1, U^\delta , V^\delta )=(0,0,0,0,0,0)^T, \end{aligned}$$

and $(U^\delta , V^\delta )\rightarrow ({\hat{U}}, {\hat{V}})$ as $d, q_1\rightarrow 0.$ Finally we show that for all $d,q_1$ small such that $d>0$ and $d+q_1>0$, each component of $(U^\delta , V^\delta )$ is also positive. Since ${\hat{U}}_1, {\hat{V}}_2,{\hat{V}}_3>0$, it suffices to show that $U_2^\delta >0$ and $U_3^\delta >0$. Recall from Lemma 17 that ${\hat{V}}_2< 1 < {\hat{V}}_3$. By setting the second component of (47) to zero we have

$$\begin{aligned} \left( 2d+q_2-1+ {U^\delta _2 + V^\delta _2}\right) U^\delta _2= & {} (d+q_1) U^\delta _1 + dU_3 \\= & {} (d+q_1)(1-{\hat{V}}_2 + o(1)) + o(d)>0. \end{aligned}$$

Using $q_2 \ge 1$, we deduce that $U^\delta _2 >0$. Next, we set the third component of (47) to get

$$\begin{aligned} \left( d -1 + {{\hat{V}}_3+ o(1)}\right) U^\delta _3=\left( d -1 + {U^\delta _3 + V^\delta _3}\right) U^\delta _3 = (d+q_2)U^\delta _2>0. \end{aligned}$$

Since ${\hat{V}}_3 >1$, we deduce that $U^\delta _3 >0$. In summary, we have proved that $U_2^\delta >0$ and $U_3^\delta >0$ for $d\in (0, \delta )$, $q_1\in (-\delta , \delta )$ and $d+q_1>0$. $\square$

Lemma 21

Suppose that $q_2\ge 1$. Let (U, V) denote any positive solution of Model II. Then as $d\rightarrow 0$ and $q_1\rightarrow 0$, $(U, V)\rightarrow ({\hat{U}}, {\hat{V}})$.

Proof

First it is easy to see that $U_i, V_i$, $i=1,2,3,$ are uniformly bounded with respect to small $d, q_1$. Hence, passing to a sub-sequence if necessary we may assume $U_i\rightarrow {\bar{U}}_i$ and $V_i\rightarrow {\bar{V}}_i$ as $d, q_1\rightarrow 0$, where ${\bar{U}}_i, {\bar{V}}_i\ge 0$ satisfy $F(0, 0, {\bar{U}}, {\bar{V}})=0$, with F defined in (47).

Step 1: ${\bar{U}}_2=0.$

This is a consequence of assumption $q_2\ge 1$ and

$$\begin{aligned} -q_2{\bar{U}}_2+{\bar{U}}_2 (1-{\bar{U}}_2-{\bar{V}}_2)=0. \end{aligned}$$

Step 2: If ${\bar{U}}_3>0$, then ${\bar{U}}_1>0$.

Suppose to the contrary that ${\bar{U}}_3>0$ and ${\bar{U}}_1=0$. The first component of $F(0,0,{{\bar{U}}},{{\bar{V}}})=0$ yields ${{\bar{U}}_3+{\bar{V}}_3}=1.$ Therefore, the 4th to 6th component of $F(0,0,{{\bar{U}}},{{\bar{V}}})=0$ can be rewritten as

$$\begin{aligned} \left\{ \begin{aligned}&D({\bar{V}}_2-{\bar{V}}_1)+{\bar{V}}_1(1-{{\bar{V}}_1})=0 \\&D({\bar{V}}_1+{\bar{V}}_3-2{\bar{V}}_2)-q_2{\bar{V}}_2+{\bar{V}}_2(1-{{\bar{V}}_2})=0 \\&D({\bar{V}}_2-{\bar{V}}_3)+q_2{\bar{V}}_2=0. \end{aligned} \right. \end{aligned}$$

By ${\bar{U}}_3+{\bar{V}}_3=1$ and ${\bar{U}}_3>0$, we have ${\bar{V}}_3<1.$ By the third equation above, ${\bar{V}}_2=D/(D+q_2) {\bar{V}}_3<1$. Adding three equations we find

$$\begin{aligned} {\bar{V}}_1(1-{{\bar{V}}_1})+{\bar{V}}_2(1-{{\bar{V}}_2})=0, \end{aligned}$$

which together with ${\bar{V}}_2<1$ implies ${\bar{V}}_1>1$. By the first equation we then obtain ${\bar{V}}_2>{\bar{V}}_1$, which is a contradiction. This completes Step 2.

Step 3: If ${\bar{U}}_3 =0$, then ${\bar{U}}_1>0$.

Suppose to the contrary that ${\bar{U}}_1={\bar{U}}_3=0.$ Then ${\bar{U}}_3$ and ${\bar{V}}_i$ satisfy

$$\begin{aligned} \left\{ \begin{aligned}&{\bar{U}}_3(1-{\bar{U}}_3-{\bar{V}}_3)=0 \\&D({\bar{V}}_2-{\bar{V}}_1)+{\bar{V}}_1(1-{{\bar{V}}_1})=0 \\&D({\bar{V}}_1+{\bar{V}}_3-2{\bar{V}}_2)-q_2{\bar{V}}_2+{\bar{V}}_2(1-{{\bar{V}}_2})=0 \\&D({\bar{V}}_2-{\bar{V}}_3)+q_2{\bar{V}}_2+{\bar{V}}_3(1-{\bar{U}}_3-{\bar{V}}_3)=0 \end{aligned} \right. \end{aligned}$$

If ${\bar{U}}_3=0$, then either ${\bar{V}}_i=0$ for $i=1,2,3$, or ${\bar{V}}_i={V}^*_i>0$ for $i=1,2,3$, where $(0, V^*)$ is one of the semi-trivial steady states of Model II. We rule out both cases as follows:

• 1. For the case $({\bar{U}}, {\bar{V}})=(0,0,0,0,0,0)$, we have $(U,V)\rightarrow (0,0,0,0,0,0)$ as $d, q_1\rightarrow 0$, which implies that $1-(U_i+V_i)>0$ for small $d, q_1$. Adding the equations of $U_i$ for $i=1,2,3$, we have

  $$\begin{aligned} U_1(1-U_1-V_1)+U_2(1-U_2-V_2)+U_3(1-U_3-V_3)=0, \end{aligned}$$

  which is a contradiction as each term in the left-hand side is positive.

• 2. For the case $({\bar{U}}, {\bar{V}})=(0, V^*)$, we normalize $U_i$ by setting ${\tilde{U}}_i\!=\!U_i/(U_1\!+\!U_2\!+\!U_3)$. Then by similar argument we have, by passing to a subsequence if necessary, ${\tilde{U}}_i\rightarrow {\check{U}}_i\ge 0$ as $d, q_1\rightarrow 0$, and ${\check{U}}_i$ satisfy ${\check{U}}_1+{\check{U}}_2+{\check{U}}_3=1$ and

  $$\begin{aligned} {\check{U}}_1(1-V_1^*) =-q_2 {\check{U}}_2+{\check{U}}_2(1-{V_2^*})=q_2{\check{U}}_2+{\check{U}}_3(1-{V_3^*})=0 \end{aligned}$$

  Since $V_1^*<1$, so ${\check{U}}_1=0$. It follows from $q_2\ge 1$ that ${\check{U}}_2=0$. This together with the last equation and $V_3^*>1$ implies that ${\check{U}}_3=0.$ This contradicts ${\check{U}}_1+{\check{U}}_2+{\check{U}}_3=1$.

Having ruled out both cases above, we proved Step 3.

Step 4: ${\bar{U}}_1>0$.

This is a consequence of Steps 2 and 3.

Step 5: ${\bar{U}}_3=0$.

If not, then ${\bar{U}}_3>0$, which leads to ${\bar{U}}_3+{\bar{V}}_3=1$. Therefore,

$$\begin{aligned} \left\{ \begin{aligned}&D({\bar{V}}_3-{\bar{V}}_2)-q_2{\bar{V}}_2+{\bar{V}}_2(1-{{\bar{V}}_2})=0 \\&D({\bar{V}}_2-{\bar{V}}_3)+q_2{\bar{V}}_2=0. \end{aligned} \right. \end{aligned}$$

Adding the above two equations we find ${\bar{V}}_2(1-{{\bar{V}}_2})=0$. Since ${\bar{V}}_2<1$, the only possibility is ${\bar{V}}_2=0$, from which we have ${\bar{V}}_i=0$ for $i=1,2,3$ and ${\bar{U}}_1={\bar{U}}_3=1.$ That is, $(U, V)\rightarrow (1, 0, 1, 0, 0, 0)$ as $d, q_1\rightarrow 0$. We normalize $V_i$ by setting ${\tilde{V}}_i=V_i/(V_1+V_2+V_3)$. Then by passing to a subsequence if necessary, ${\tilde{V}}_i\rightarrow {\check{V}}_i\ge 0$ as $d, q_1\rightarrow 0$, and ${\check{V}}_i$ satisfy ${\check{V}}_1+{\check{V}}_2+{\check{V}}_3=1$, and

$$\begin{aligned} \left\{ \begin{aligned}&{\check{V}}_1={\check{V}}_2\\&D({\check{V}}_1+{\check{V}}_3-2{\check{V}}_2)-q_2{\check{V}}_2 +{\check{V}}_2=0\\&D({\check{V}}_3-{\check{V}}_2)+q_2{\check{V}}_2=0,\\ \end{aligned} \right. \end{aligned}$$

from which we conclude that ${\check{V}}_i=0$ for all i, which is a contradiction. This completes Step 5.

Step 6: ${\bar{V}}_1={\hat{V}}_2$ and ${\bar{U}}_1=1-{\hat{V}}_2.$

As ${\bar{U}}_1>0$ and ${\bar{U}}_2={\bar{U}}_3=0$, we have ${\bar{U}}_1+{\bar{V}}_1=1$, ${\bar{V}}_1={\bar{V}}_2$, and

$$\begin{aligned} \left\{ \begin{aligned}&D({\bar{V}}_3-{\bar{V}}_2)-q_2{\bar{V}}_2+{\bar{V}}_2(1-{{\bar{V}}_2})=0 \\&D({\bar{V}}_2-{\bar{V}}_3)+q_2{\bar{V}}_2+{\bar{V}}_3(1-{{\bar{V}}_3})=0. \end{aligned} \right. \end{aligned}$$

By similar normalization argument we can show that ${\bar{V}}_i>0$ for all i. Hence, ${\bar{V}}_i={\hat{V}}_i$ for $i=2,3.$ Thus ${\bar{V}}_1={\hat{V}}_2$ and ${\bar{U}}_1=1-{\hat{V}}_2.$ This completes the proof. $\square$

Lemma 22

Fix any $D, q_2>0$. Then there exists some $\delta >0$ such that for any $d\in (0, \delta )$ and $q_1\in (-d, \delta )$, the positive steady state $(U^\delta , V^\delta )$, which is given by Lemma 20, is locally stable.

Proof

By previous result, there exists some $\delta >0$ such that for $|d|, |q_1|\le \delta$, there exist $(U^\delta , V^\delta )\in {\mathbb {R}}_+^6$ such that $F(d, q_1, U^\delta , V^\delta )=(0,0,0,0,0,0)^T.$ Since the two-species competition models II and III are strongly monotone, the linearized system at $(U^\delta , V^\delta )$ has a principal eigenvalue (it is real, simple and has the largest real part among all eigenvalues), which we denote as $\lambda ^\delta _1$; i.e.,

$$\begin{aligned} D_{(U,V)}F(d,q_1, U^\delta , V^\delta ) \begin{pmatrix} \varphi ^\delta \\ \phi ^\delta \end{pmatrix}=\lambda _1^\delta \begin{pmatrix} \varphi ^\delta \\ \phi ^\delta \end{pmatrix}, \end{aligned}$$

where $\varphi ^\delta :=(\varphi _1^\delta , \varphi _2^\delta , \varphi _3^\delta )^T$ and $\phi ^\delta :=(\phi _1^\delta , \phi _2^\delta , \phi _3^\delta )^T$. Furthermore, we may choose $\varphi _i^\delta <0$ and $\phi _i^\delta >0$ for $i=1,2,3$, and normalize by

$$\begin{aligned} \sum _{i=1}^3 (|\varphi _i^\delta |+ |\phi _i^\delta |)=1. \end{aligned}$$

We proceed to show that $(U^\delta , V^\delta )$ is stable, that is, $\lambda _1^\delta <0$. To this end, we argue by contradiction and assume $\lambda _1^\delta \ge 0$. Let $\delta \rightarrow 0$ (so that $d\rightarrow 0, q_1\rightarrow 0)$, by passing to a subsequence, we may assume that $\lambda _1^\delta \rightarrow \lambda _1^* \ge 0$, so that $D_{(U,V)}F(0,0,{\hat{U}},{\hat{V}})$ has at least one nonnegative eigenvalue, i.e., that (48) holds for some non-trivial eigenvector $(\varphi ,\psi )$ and nonnegative eigenvalue $\lambda ^*_1$. However, Lemma 19 asserts that $\lambda ^*_1 <0$. This is a contradiction.

Therefore, $\lambda _1^\delta <0$ for sufficiently small $\delta >0$, i.e., $(U^\delta , V^\delta )$ is stable. $\square$

In the following two results, we consider the linear instability of the semi-trivial steady states of (2), denoted by $(U^*,0):=(U_1^*,U_2^*,U_3^*,0,0,0)$ and $(0,V^*):=(0,0,0,V_1^*,V_2^*,V_3^*)$.

Lemma 23

If $q_2\ge 1$, then there exists $\delta >0$ such that $(U^*,0)$ is linearly unstable for every $0 \le d , q_1 \le \delta$.

Proof

Setting $d= 0, q_1= 0$, $(U_1^*,U_2^*,U_3^*)$ is the unique solution of

$$\begin{aligned} \left\{ \begin{array}{l} U_1^*(1-U_1^*)=0\\ -q_2U_2^*+U_2^*(1-U_2^*)=0\\ q_2U_2^*+U_3^*(1-U_3^*)=0.\\ \end{array} \right. \end{aligned}$$

By direct calculations and using $q_2\ge 1$, we get

$$\begin{aligned} (U_1^*,U_2^*,U_3^*)=(1, \max (1-q_2,0), \frac{1+\sqrt{1+4q_2U_2^*}}{2}). \end{aligned}$$

For $q_2\ge 1$, we have $(U_1^*,U_2^*,U_3^*)=(1,0,1)$, and its linear stability of $(U^*,0)$ is determined by eigenvalue problem

$$\begin{aligned} {\tilde{E}}_2 \varphi + \varLambda \varphi =0, \end{aligned}$$

where $\varphi =(\varphi _1,\varphi _2,\varphi _3)^{T}$ and

$$\begin{aligned} {\tilde{E}}_2= \begin{pmatrix}-D &{} D &{} 0\\ D &{} {-2D -q_2 + 1} &{} D\\ 0 &{} D+q_2 &{} -D\end{pmatrix}. \end{aligned}$$

We will test ${\tilde{E}}_2$ by multiplying on the right with the vector $({\hat{V}}_2-\epsilon , {\hat{V}}_2, {\hat{V}}_3)^T$, where ${\hat{V}}_2, {\hat{V}}_3$ is given in Theorem 4 and $\epsilon$ is a small positive constant.

$$\begin{aligned} {\tilde{E}}_2 \begin{pmatrix} {\hat{V}}_2-\epsilon \\ {\hat{V}}_2 \\ {\hat{V}}_3 \end{pmatrix}= \begin{pmatrix}\epsilon D\\ ({\hat{V}}_2)^2-\epsilon D\\ {\hat{V}}_3({\hat{V}}_3-1)\end{pmatrix}. \end{aligned}$$

Since all of the entries of the right-hand side are positive, we can apply the Collatz–Wielandt Formula (Meyer 2000, P. 667) to get

$$\begin{aligned} -\varLambda = \mathop {\max }\limits _{\{ \varphi \ge 0: \varphi \not = 0 \}}\mathop {\min }\limits _{1\le i \le 3, \varphi _i> 0}\frac{[{\tilde{E}}_2 \varphi ]_i}{\varphi _i} \ge \min \left\{ \frac{\epsilon D}{{\hat{V}}_2-\epsilon }, \frac{({\hat{V}}_2)^2-\epsilon D}{{\hat{V}}_2}, \frac{{\hat{V}}_3({\hat{V}}_3-1)}{{\hat{V}}_3} \right\} >0, \end{aligned}$$

that is, $(U^*,0)$ is linearly unstable when $d=q_1=0$. By continuity, it remains linearly unstable for all small d and $q_1$. $\square$

Lemma 24

For each $D, q_2 >0$, there exists $\delta >0$ such that $(0,V^*)$ is linearly unstable for every $0 \le d,q_1 \le \delta$.

Proof

Setting $d=0$ and $q_1=0$, the linear instability of $(0,V^*)$ is determined by the principal eigenvalue of the following problem

$$\begin{aligned} \begin{pmatrix}1-V_1^*&{} 0 &{} 0\\ 0 &{} -q_2+1-V_2^* &{} 0\\ 0 &{} q_2 &{} 1-V_3^*\end{pmatrix}\begin{pmatrix}\varphi _1\\ \varphi _2\\ \varphi _3\end{pmatrix}+ \varLambda \begin{pmatrix}\varphi _1\\ \varphi _2\\ \varphi _3\end{pmatrix}= \begin{pmatrix}0\\ 0\\ 0\end{pmatrix}. \end{aligned}$$

Clearly, $\varLambda =V^*_3-1$ is an eigenvalue with eigenfunction $(0,0,1)^T$. Recalling that $V_1^*<1$ (Lemma 8), we deduce that there is at least one negative eigenvalue. Thus $(0,V^*)$ is linearly unstable. $\square$

Proof of Theorem 4

By Lemma 20, there exists some $\delta >0$ such that for any $d\in (0, \delta )$, $q_1\in (-d, \delta )$, Model II has a unique positive steady state $(U^\delta , V^\delta )$ in a small neighborhood of $({\hat{U}}, {\hat{V}})$. Lemma 21 further ensures that this is the only positive steady state for small positive d and $q_1$. By Lemma 22, $(U^\delta , V^\delta )$ is locally stable. We can then conclude by the theory of monotone dynamical systems (Hess 1991; Hess and Lazer 1991; Smith 1995) and Lemmas 23 and 24 that $(U^\delta , V^\delta )$ is globally stable. $\square$

Appendix C: The Dynamics of Model III

In this section, we mainly consider the dynamics of Model III, i.e., system (5), in homogeneous environments. We consider the nonnegative and non-trivial solutions of

$$\begin{aligned} \left\{ \begin{array}{l} d(U_2+U_3-2U_1)-(q_1+q_2)U_1+U_1(1-U_1-V_1)=0\\ d(U_1-U_2)+q_1U_1+U_2(1-U_2-V_2)=0\\ d(U_1-U_3)+q_2U_1+U_3(1-U_3-V_3)=0\\ D(V_2+V_3-2V_1)-(q_1+q_2)V_1+V_1(1-U_1-V_1)=0\\ D(V_1-V_2)+q_1V_1+V_2(1-U_2-V_2)=0\\ D(V_1-V_3)+q_2V_1+V_3(1-U_3-V_3)=0\\ \end{array} \right. \end{aligned}$$

(52)

There are three types of nonnegative and non-trivial solutions of this system. We denote three different types of solutions as (U, V), in which $U_i>0$ and $V_i=0$, or $U_i=0$ and $V_i>0$, or $U_i>0$ and $V_i>0$ for all $i=1,2,3$. The semi-trivial steady state $(U^*,0):=(U_1^*,U_2^*,U_3^*,0,0,0)$ satisfies

$$\begin{aligned} \left\{ \begin{array}{l} d(U_2^*+U_3^*-2U_1^*)-(q_1+q_2)U_1^*+U_1^*(1-{U_1^*})=0\\ d(U_1^*-U_2^*)+q_1U_1^*+U_2^*(1-{U_2^*})=0\\ d(U_1^*-U_3^*)+q_2U_1^*+U_3^*(1-{U_3^*})=0\\ \end{array} \right. \end{aligned}$$

(53)

The linear stability of $(U_1^*,U_2^*,U_3^*,0,0,0)$ is determined by the sign of the principal eigenvalue ${\tilde{\varLambda }}_3$ of the eigenvalue problem

$$\begin{aligned} F_3\begin{pmatrix}\varphi _1\\ \varphi _2\\ \varphi _3\end{pmatrix} + \varLambda \begin{pmatrix}\varphi _1\\ \varphi _2\\ \varphi _3\end{pmatrix} =\begin{pmatrix} 0\\ 0\\ 0\end{pmatrix}, \end{aligned}$$

(54)

where matrix $F_3$ is given by

$$\begin{aligned} F_3= \begin{pmatrix}-2D-(q_1+q_2)+1-{U_1^*} &{} D &{} D\\ D+q_1 &{} -D+1-{U_2^*} &{} 0\\ D+q_2 &{} 0 &{} -D+1-{U_3^*}\end{pmatrix}. \end{aligned}$$

Proposition 6

When $D=d$, the derivative of ${\tilde{\varLambda }}_3$ with respect to D satisfies

$$\begin{aligned} \frac{\partial {\tilde{\varLambda }}_3}{\partial D}\Big |_{D=d}=-\frac{U_1^*(U_2^*+U_3^*-2U_1^*) + \frac{d}{d+q_1}U_2^*(U_1^*-U_2^*) + \frac{d}{d+q_2}U_3^*(U_1^*-U_3^*)}{(U_1^*)^2 + \frac{d}{d+q_1}(U_2^*)^2 + \frac{d}{d+q_2}(U_3^*)^2}.\nonumber \\ \end{aligned}$$

(55)

Proof

Differentiate (54) with respect to D, we get

$$\begin{aligned} \begin{pmatrix} \varphi _2+\varphi _3-2\varphi _1\\ \varphi _1-\varphi _2\\ \varphi _1-\varphi _3\end{pmatrix} + F_3\begin{pmatrix}\varphi _1'\\ \varphi _2'\\ \varphi _3'\end{pmatrix} + \frac{\partial {\tilde{\varLambda }}_3}{\partial D}\begin{pmatrix}\varphi _1\\ \varphi _2\\ \varphi _3\end{pmatrix} + {\tilde{\varLambda }}_3 \begin{pmatrix}\varphi _1'\\ \varphi _2'\\ \varphi _3'\end{pmatrix} = \begin{pmatrix}0\\ 0\\ 0\end{pmatrix}, \end{aligned}$$

(56)

where $\varphi _i'=\frac{\partial \varphi _i}{\partial D}$, $i=1,2,3$. Note that when $D=d$,

$$\begin{aligned} \left. F_3\right| _{D=d}\begin{pmatrix}U_1^*\\ U_2^*\\ U_3^*\end{pmatrix}=\begin{pmatrix}0\\ 0\\ 0\end{pmatrix} , \end{aligned}$$

$$\begin{aligned} \left. {F_3^T}\right| _{D=d}\begin{pmatrix} U_1^*\\ \frac{d}{d+q_1}U_2^*\\ \frac{d}{d+q_2}U_3^*\end{pmatrix}=\begin{pmatrix}0\\ 0\\ 0\end{pmatrix}, \end{aligned}$$

(57)

and when $D=d$, we may choose

$$\begin{aligned} \begin{pmatrix}\varphi _1\\ \varphi _2\\ \varphi _3\end{pmatrix} =\begin{pmatrix}U_1^*\\ U_2^*\\ U_3^*\end{pmatrix}. \end{aligned}$$

(58)

Set $D=d$ in (56) and multiplying it by $\begin{pmatrix}U_1^*, \frac{d}{d+q_1}U_2^*, \frac{d}{d+q_2}U_3^*\end{pmatrix}$, using (57), (58) and ${\tilde{\varLambda }}_3(d,d)= 0$, we obtain (55). This completes the proof. $\square$

Next, we establish some a prior estimates of $U_i^*$, $i=1,2,3$.

Lemma 25

For any $d,D>0$ and $q_1,q_2>0$, the following results hold:

• (i) If $q_1 \ge q_2$, then $U_2^* \ge U_3^*$;

• (ii) If $q_1 \le q_2$, then $U_2^* \le U_3^*$.

In particular, if $q_1=q_2$, $U_2^*=U_3^*$ holds.

Proof

We prove (i) only and (ii) can be shown similarly. We will assume $U^*_2 < U^*_3$ and deduce $q_1 < q_2$. By the second and third equation of (53),

$$\begin{aligned} \left\{ \begin{aligned}&(-d+1-{U_2^*})U_2^*=-(d+q_1)U_1^*<0,\\&(-d+1-{U_3^*})U_3^*=-(d+q_2)U_1^*<0. \end{aligned} \right. \end{aligned}$$

This implies

$$\begin{aligned} 0<U_2^* < U^*_3 \quad \text { and }\quad 0>-d+1-U_2^* > -d+1-U_3^*. \end{aligned}$$

Combining the above, we have

$$\begin{aligned} -(d+q_2)U_1^* = (-d+1-{U_2^*})U_2^* > (-d+1-{U_3^*})U_3^*=-(d+q_2)U_1^*. \end{aligned}$$

This implies $q_2 > q_1$. This proves (i). $\square$

Lemma 26

For any $d,D>0$ and $q_1,q_2>0$, $U_1^*<1$ always holds.

Proof

By exchanging patches 2 and 3 if necessary (the river network is symmetric), we may assume without loss of generality that $q_1\ge q_2$.

We argue by contradiction and assume that $U_1^* \ge 1$ for some $q_1 \ge q_2>0$. By the first equation of (53), we get

$$\begin{aligned} d(U_2^*+U_3^*-2U_1^*)-(q_1+q_2)U_1^* \ge 0. \end{aligned}$$

Using $U_2^* \ge U_3^*$ (from Lemma 25), we have

$$\begin{aligned} 2d(U_2^*-U_1^*) \ge (q_1+q_2)U_1^* > 0, \end{aligned}$$

so $U_2^*>U_1^*\ge 1$.

In view of $\sum _{i=1}^3 U_i^*(1-U_i^*)=0$ (upon summing (53)), we must have $U_3^*<1$. By the third equation of (53), we get $U_3^*>\frac{d+q_2}{d}U_1^*>U_1^*\ge 1$. This is a contradiction. This finishes the proof for the case $q_1\ge q_2$, and the case $q_1\le q_2$ can be treated similarly. $\square$

1.1 C.1. The Global Dynamics of Model III When $q_1=q_2$

In this part, we shall show Theorem 5. We first establish some a priori estimates of nonnegative and non-trivial steady state of (52) as $q_1=q_2$.

1.1.1 C.1.1. Preliminary Results on Nonnegative, Non-trivial Steady States

Lemma 27

Suppose $q_1=q_2>0$, then for any $d, D>0$, we have $U_2=U_3$ and $V_2=V_3$.

Proof

By the similar argument in the proof of Lemma 1, we can obtain this lemma. $\square$

Lemma 28

Suppose $q_1=q_2:=q>0$, then for any $d, D >0$, we have $U_3+V_3>1$.

Proof

Assume that $(U_i)$ is non-trivial and is thus positive for all i. The same argument applies to the case $U_1=U_2=U_3=0$. Assume to the contrary that $U_3+V_3 \le 1$ for some $d,D,q>0$, then by the third equation of (52), we get $d(U_1-U_3)+qU_1 \le 0$. Thus $U_3 \ge \frac{d+q}{d}U_1>U_1$.

Similarly, we can show $V_3>V_1$, if $V_i>0$ for all i. Hence $1\ge U_3+V_3>U_1+V_1$ holds. By the first equation of (52) and Lemma 27, we obtain

$$\begin{aligned} 2d(U_3-U_1)-2qU_1 = d(U_2+U_3-2U_1)-2qU_1<0, \end{aligned}$$

i.e., $d(U_3-U_1)-qU_1<0$. This together with the third equation of (52) implies that $U_3+V_3>1$, which is a contradiction with our assumption. $\square$

Lemma 29

Suppose $q_1=q_2=q>0$, then for any $d, D >0$, we have $U_1+V_1<1$.

Proof

We argue by contradiction. If $U_1+V_1 \ge 1$ for some $d,D,q>0$, then by the first equation of (52) and Lemma 27, we have

$$\begin{aligned} 2d(U_2-U_1)-2qU_1 = d(U_2+U_3-2U_1)-2qU_1 \ge 0, \end{aligned}$$

so that $d(U_2-U_1)-qU_1 \ge 0$, which together with the second equation of (52) implies $U_2+V_2 \le 1$. Using Lemma 27, we have $U_3+V_3 = U_2+V_2 \le 1$. But this contradicts Lemma 28. $\square$

Lemma 30

Suppose $q_1=q_2:=q>0$ and $d, D >0$.

• (i) If $U_i>0$ for all i, then $U_1 < U_3$.

• (i) If $V_i>0$ for all i, then $V_1 < V_3$.

Proof

In case of $(U^*,0)$ and $(0,V^*)$, the lemma follows from Lemmas 28 and 29. It therefore suffices to consider positive equilibria (U, V). We will prove (i), as (ii) follows from a similar argument.

If $U_1 \ge U_3$ for some $d,D,q>0$, by Lemmas 28 and 29, we have $V_3>V_1$, which together with the 6th equation of (52) implies

$$\begin{aligned} (q+1-U_3-V_3)V_3>qV_1+V_3(1-U_3-V_3)=D(V_3-V_1)>0; \end{aligned}$$

i.e., $q+1-U_3-V_3>0$. However, by the third equation of (52) and $U_1 \ge U_3$, we get

$$\begin{aligned} (q+1-U_3-V_3)U_3 \le qU_1+U_3(1-U_3-V_3)=d(U_3-U_1)\le 0; \end{aligned}$$

i.e., $q+1-U_3-V_3\le 0$. This is a contradiction. $\square$

The following result is a direct consequence of Lemmas 27 and 30, and it provides some insight for the biological interpretation of Theorem 5.

Corollary 6

Assume $q_1=q_2>0$ and $d, D >0$. Then we have

$$\begin{aligned} U_1<1<U_2 = U_3\quad \text { and }\quad V_1<1<V_2 = V_3. \end{aligned}$$

Lemma 31

Suppose $q_1=q_2>0$, then for any $d,D>0$, we have

$$\begin{aligned} 3-\sum _{i=1}^3 (U_i+V_i)>0. \end{aligned}$$

(59)

Proof

By possibly exchanging the role of U and V, we may assume $U_i>0$ for all i. Adding the equations of $U_i \ (i=1,2,3)$ in (52) and using $U_2+V_2=U_3+V_3>1$ (Lemmas 27 and 28), $U_1+V_1<1$ (Lemma 29) and $U_1<U_3=U_2$ (Corollary 6), we obtain

$$\begin{aligned} \begin{aligned}&U_3(1-U_1-V_1)+U_3(1-U_2-V_2)+U_3(1-U_3-V_3)\\&\quad > U_1(1-U_1-V_1)+U_2(1-U_2-V_2)+U_3(1-U_3-V_3)= 0, \end{aligned} \end{aligned}$$

which establishes (59). $\square$

Lemma 32

Suppose $q_1=q_2:=q>0$ and $d,D>0$. Then we have $-2q+1-U_1-V_1<0.$

Proof

Corollary 6 and the first equation of (52) indicate that $-2q+1-U_1-V_1<0$. $\square$

Theorem 10

If $q_1=q_2>0$ and $d, D > 0$, then system (52) has no positive solution.

Proof

We argue by contradiction. If there exists a positive solution (U, V) for (52), by direct calculation, we obtain

$$\begin{aligned} Dd(3-\sum _{i=1}^3(U_i+V_i))=(-2q+1-U_1-V_1)(1-U_2-V_2)(1-U_3-V_3).\qquad \end{aligned}$$

(60)

Lemma 31 shows that the left-hand side of (60) is positive, but the right-hand side of (60) is negative, due to Lemmas 27, 28 and 32. This completes the proof. $\square$

1.1.2 C.1.2. Global Stability of $(U^*,0)$ When $q_1=q_2$

We assume $q_1=q_2:=q$ throughout this subsection. The local instability of $(0,V^*)$ is determined by the sign of the principal eigenvalue, denoted as $\varLambda _3$, of the system

$$\begin{aligned} E_3 \begin{pmatrix}\varphi _1\\ \varphi _2\\ \varphi _3\end{pmatrix} + \varLambda \begin{pmatrix}\varphi _1\\ \varphi _2\\ \varphi _3\end{pmatrix} =\begin{pmatrix} 0\\ 0\\ 0\end{pmatrix}, \end{aligned}$$

where $E_3$ is rewritten as

$$\begin{aligned} E_3= \begin{pmatrix}-2d-2q+1-V_1^* &{} d &{} d\\ d+q &{} -d+1-V_2^* &{} 0\\ d+q &{} 0 &{} -d+1-V_3^*\end{pmatrix}. \end{aligned}$$

Setting $q_1=q_2:=q$ and exchanging the role of the two species, (55) can be rewritten as

$$\begin{aligned} \frac{\partial \varLambda _3}{\partial d}\Big |_{d=D}=-\frac{(V_2^*-\frac{D+q}{D}V_1^*)(V_1^*-V_2^*)+(V_3^*-\frac{D+q}{d}V_1^*)(V_1^*-V_3^*)}{\frac{D+q}{D}(V_1^*)^2+(V_2^*)^2+(V_3^*)^2}. \end{aligned}$$

(61)

We can obtain the following result by direct calculations:

Proposition 7

(Jiang et al. 2020, Proposition 6) Assume $q_1=q_2:=q>0$ and ${V_1^*+V_2^*+V_3^*}\ne 3$. Then $\mathrm{det}(E_3)=0$ if and only if either $d=D$, or

$$\begin{aligned} d = \frac{(-2q + 1 - {V_1^*})(1 - {V_2^*})(1 - {V_3^*})}{D(3-V_1^* - {V_2^*} - {V_3^*})}. \end{aligned}$$

(62)

Corollary 7

Suppose $q_1=q_2:=q>0$, then for any $d,D>0$, $\frac{\partial \varLambda _3}{\partial d}\big |_{d=D}<0.$

Proof

By Corollary 6, we have $V_1^*-V_2^*<0, V_1^*-V_3^*<0$. Using $V_2^*>1$ and the fifth equation of (52), we get

$$\begin{aligned} \frac{D+q}{D}V_1^*-V_2^*>0. \end{aligned}$$

Similarly, by $V_3^*>1$ and the sixth equation of (52), we get

$$\begin{aligned} \frac{D+q}{D}V_1^*-V_3^*>0. \end{aligned}$$

Therefore, the right-hand side of (61) is strictly negative. $\square$

Lemma 33

Suppose $q_1=q_2>0$, then for any $d,D>0$, the right-hand side of (62) is negative.

Proof

Using Lemmas 27 and 28, we get $V_2^*=V_3^*>1$, hence $(1-V_2^*)(1-V_3^*)>0$, which together with Lemmas 31 and 32 shows that the right-hand side of (62) is strictly negative. $\square$

Theorem 11

Suppose $q_1=q_2>0$, then for any $d,D>0$, we have

$$\begin{aligned} \varLambda _3(d,D)=\left\{ \begin{aligned} + \quad \quad&D> d; \\ - \quad \quad&D< d; \\ \end{aligned} \right. \quad \text { and }\quad {\tilde{\varLambda }}_3(d,D)=\left\{ \begin{aligned} - \quad \quad&D > d; \\ + \quad \quad&D < d. \\ \end{aligned} \right. \end{aligned}$$

Proof

Equation (62) cannot hold since the right-hand side is strictly negative, by Lemma 33. Hence, Proposition 7 says that $\varLambda _3(d,D)=0$ if and only if $d=D$. Therefore, by Corollary 7 and the continuity of $\varLambda _3$, $\varLambda _3(d,D)>0$ holds for $D>d>0$ and $\varLambda _3(d,D)<0$ holds for $0<D<d$. The result for ${\tilde{\varLambda }}_3$ follows from the identity ${\tilde{\varLambda }}_3(d,D)=\varLambda _3(D,d)$ for all d, D. $\square$

Proof of Theorem 5

For $d>D$, Theorems 11 and 10 say that $(0,V^*)$ is linearly unstable, and that Model III has no positive equilibria. It follows from the theory of monotone dynamical systems (Hess and Lazer 1991, Theorem 1.5) that the equilibrium $(U^*,0)$ is globally asymptotically stable.$\square$

1.2 C.2. The Local Stability of $(U^*,0)$

In this subsection, we determine the local stability of the semi-trivial steady sate $(U^*,0)$ for more general $q_1, q_2$.

Lemma 34

Suppose $0<q_2\le q_1+\frac{1}{2}$, $0<\frac{q_2}{q_1}\le \sqrt{2}$. Then $U_2^*>1$ holds for all $d>0$.

Proof

Since we have shown $U_2^*>1$ for $q_1=q_2$, it is sufficient to show $U_2^*\not =1$ for any $q_1, q_2$ satisfying the assumptions. We argue by contradiction: Suppose that $U_2^*=1$ for some $q_1, q_2$. By the second equation of (53), we have

$$\begin{aligned} U_1^*=\frac{d}{d+q_1}. \end{aligned}$$

(63)

Adding the equations of $U_1^*, U_2^*, U_3^*$ in (53) and using $U_2^*=1$, we get

$$\begin{aligned} U_1^*(1-{U_1^*})+U_3^*(1-{U_3^*})=0. \end{aligned}$$

(64)

Substituting (63) and (64) into the third equation of (53), we get

$$\begin{aligned} U_3^*=\frac{1}{d+q_1}(d+q_2-\frac{q_1}{d+q_1}). \end{aligned}$$

(65)

By $U_1^*<1$ (from (63)) and (64), we see that $U_3^*>1$. This, together with (65), implies that

$$\begin{aligned} \frac{q_1}{d+q_1} < q_2-q_1. \end{aligned}$$

(66)

Hence, $q_2-q_1>0$. Therefore, we can rewrite (66) to get

$$\begin{aligned} d>\frac{q_1(1-q_2+q_1)}{q_2-q_1}>0, \end{aligned}$$

(67)

which the last inequality follows from $0<q_2-q_1\le \frac{1}{2}$. By (63), (64) and (65), after simplifications, we have

$$\begin{aligned} dq_1=(q_2-q_1-\frac{q_1}{d+q_1})(d+q_2-\frac{q_1}{d+q_1}). \end{aligned}$$

(68)

It follows from (68) that $dq_1<(q_2-q_1)(d+q_2)$, which can be rewritten as

$$\begin{aligned} d(2q_1-q_2)<q_2(q_2-q_1). \end{aligned}$$

(69)

By (67) and (69), note that $2q_1-q_2>0$ by assumption, we have

$$\begin{aligned} 0< \left( \frac{q_1}{q_2-q_1}-q_1\right) (2q_1-q_2)<q_2(q_2-q_1). \end{aligned}$$

(70)

By assumption $0<q_2-q_1 \le \frac{1}{2}$, we have $\frac{q_1}{q_2-q_1} \ge 2q_1$. Hence, by (70)

$$\begin{aligned} q_1(2q_1-q_2)<q_2(q_2-q_1), \end{aligned}$$

which is equivalent to $q_2>\sqrt{2}q_1$, a contradiction to assumption $q_2 \le \sqrt{2}q_1$. $\square$

Lemma 35

Suppose that $0<q_1\le q_2+ \frac{1}{2}$ and $\frac{q_2}{q_1}\ge \frac{1}{\sqrt{2}}$, then $U_3^*>1$ holds for all $d>0$.

Proof

This proof is similar to Lemma 34, by exchanging the role of patches 2 and 3. We omit the proof. $\square$

Directly by Lemmas 34 and 35, we obtain the following result, which also provides some insight for the biological interpretation of Theorem 6.

Corollary 8

Let $q_1,q_2,d,D$ be positive. If $| q_2-q_1 | \le \frac{1}{2}$ and $\frac{1}{\sqrt{2}}\le \frac{q_2}{q_1}\le \sqrt{2}$, then $U_2^*>1$ and $U_3^*>1$ hold.

Proof of Theorem 6

Fix $d > D$. We have shown that if $q_1=q_2$, ${\tilde{\varLambda }}_3(d,D)>0$ in Theorem 11. By the continuity of ${\tilde{\varLambda }}_3$ in $q_1,q_2$, we just need to prove ${\tilde{\varLambda }}_3 \not =0$. By contradiction, we assume that there exist some $q_1,q_2$ such that ${\tilde{\varLambda }}_3 =0$. Then by direct calculation, we get

$$\begin{aligned} Dd(3-U_1^*-U_2^*-U_3^*)=[-(q_1+q_2)+(1-{U_1^*})](1-{U_2^*})(1-{U_3^*}). \end{aligned}$$

(71)

Adding the equations of (53), we get

$$\begin{aligned} U_1^*(1-{U_1^*})+U_2^*(1-{U_2^*})+U_3^*(1-{U_3^*})=0. \end{aligned}$$

(72)

Due to $U_2^*>1$ and $U_3^*>1$, we have $U_1^*<1$, so

$$\begin{aligned} U_i^*(1-U_i^*)<(1-U_i^*), \quad \text { for }i=1,2,3. \end{aligned}$$

Substituting this into (72), we obtain $3-U_1^*-U_2^*-U_3^*>0$. Again using $U_1^*<U_2^*, U_1^*<U_3^*$ and the first equation of (53), $-(q_1+q_2)+(1-{U_1^*})<0$. This together with $U_2^*>1$ and $U_3^*>1$ (Corollary 8) yields the right-hand side of (71) is negative. This contradiction finishes the proof. $\square$

1.3 C.3. Existence of Evolutionarily Singular Strategy

The goal of this subsection is to establish Theorem 7.

Lemma 36

For any $q_1,q_2>0$, we have $\frac{\partial {\tilde{\varLambda }}_3}{\partial D}(d,d)<0$ for sufficiently large d.

Proof

By Proposition 6, the sign of $\frac{\partial {\tilde{\varLambda }}_3}{\partial D}(d,d)$ is the opposite of that of N, where

$$\begin{aligned} N:=U_1^*(U_2^*+U_3^*-2U_1^*)+\frac{d}{d+q_1}U_2^*(U_1^*-U_2^*)+\frac{d}{d+q_2}U_3^*(U_1^*-U_3^*).\qquad \end{aligned}$$

(73)

By (53), we can rewrite (73) as

$$\begin{aligned} N&=(U_2^*-U_1^*)(U_1^*-\frac{d}{d+q_1}U_2^*) +(U_3^*-U_1^*)(U_1^*-\frac{d}{d+q_2}U_3^*)\nonumber \\&=\frac{1}{d+q_1} (U^*_2-U^*_1)U^*_2\left( {U_2^*}-1\right) + \frac{1}{d+q_2} (U^*_3-U^*_1)U^*_3\left( {U_3^*}-1\right) . \end{aligned}$$

(74)

Note that $(U_1^*,U_2^*,U_3^*) \rightarrow (1,1,1)$ as $d \rightarrow \infty$. As $(U_1^*, U_2^*, U_3^*)$ is the unique stable positive solution of (53), it can be shown that it is smooth at $d=\infty$ so that we can expand $U_i^*$ as $U_i^*=1+{\tilde{U}}_i/d+O(1/d^2)$, $i=1,2,3$, for sufficiently large d. From the second and third equation of (53) we have

$$\begin{aligned} {\tilde{U}}_2={\tilde{U}}_1+q_1, \quad {\tilde{U}}_3={\tilde{U}}_1+q_2. \end{aligned}$$

(75)

Recall (72) (resulting from adding three equations in (53)), it follows that

$$\begin{aligned} {\tilde{U}}_1+{\tilde{U}}_2+{\tilde{U}}_3=0. \end{aligned}$$

(76)

By solving (75) and (76), ${\tilde{U}}_1=-(q_1+q_2)/3$, ${\tilde{U}}_2=(2q_1-q_2)/3$, ${\tilde{U}}_3=(2q_2-q_1)/3$. Hence, for large d it holds that

$$\begin{aligned} \left\{ \begin{aligned} U_1^*&=1-\frac{q_1+q_2}{3d} +O(1/d^2),\\ U_2^*&=1+\frac{2q_1-q_2}{3d} +O(1/d^2),\\ U_3^*&=1+\frac{2q_2-q_1}{3d} +O(1/d^2). \end{aligned} \right. \end{aligned}$$

Substituting into (74), we obtain

$$\begin{aligned} d^3 N \rightarrow \frac{2}{3} (q_1^2+q_2^2-q_1q_2)>0 \quad \text { as }d \rightarrow \infty , \end{aligned}$$

(77)

provided that $(q_1, q_2)\not =(0,0)$. Therefore, we conclude by Proposition 6 that $\frac{\partial {\tilde{\varLambda }}_3}{\partial D}(d,d)<0$ for sufficiently large d. $\square$

Lemma 37

Let $q_1,q_2>0$. For sufficiently small d, we have

$$\begin{aligned} \frac{\partial {\tilde{\varLambda }}_3}{\partial D}(d,d)=\left\{ \begin{aligned} -\quad \quad&{\text { if }q_1+q_2 \le 1 \,\text { or }\, q_1+q_2>(q_1-q_2)^2;}\\ + \quad \quad&{\text { if } 1<q_1+q_2<(q_1-q_2)^2.}\\ \end{aligned} \right. \end{aligned}$$

(78)

Proof

We consider three cases:

Case I. $q_1+q_2< 1$. By the first equation of (53), we see that $U_1^* \rightarrow {\overline{U}}_1:= 1-(q_1+q_2)>0$ as $d \rightarrow 0$. By the second and third equation of (53), $U_2^*\rightarrow {\overline{U}}_2\ge 1, U_3^*\rightarrow {\overline{U}}_3 \ge 1$. Thus $N \rightarrow {\overline{U}}_1({\overline{U}}_2+{\overline{U}}_3-2{\overline{U}}_1)>0$ as $d\rightarrow 0$, where N is given by (73). Here we used $q_1>0$ and $q_2>0$. By Proposition 6, we have $\frac{\partial {\tilde{\varLambda }}_3}{\partial D}(d,d)<0$ for sufficiently small d when $q_1+q_2<1$.

Case II. $q_1+q_2=1$. For this case, we have $U_1^* \rightarrow 0$ and $U_i^* \rightarrow 1$ $(i=2,3)$ as $d\rightarrow 0$. By the first equation of (53), we get

$$\begin{aligned} \frac{U_1^*}{\sqrt{d}} \rightarrow \sqrt{2} \quad \quad as \quad d\rightarrow 0. \end{aligned}$$

(79)

Thus $N=2\sqrt{2} d^{1/2}+ o(1)$ is positive for sufficiently small d. Therefore, $\frac{\partial {\tilde{\varLambda }}_3}{\partial D}(d,d)<0$ when $q_1+q_2=1$.

Case III. $q_1+q_2>1$. For this case, we have $U_1^* \rightarrow 0$ and $U_i^* \rightarrow 1$ $(i=2,3)$ as $d\rightarrow 0$. By the first equation of (53), we get

$$\begin{aligned} \frac{U_1^*}{d} \rightarrow \frac{2}{q_1+q_2-1}, \quad \quad as \quad d\rightarrow 0. \end{aligned}$$

(80)

Substituting into (73), we get

$$\begin{aligned} \frac{N}{d}= & {} \frac{2}{q_1+q_2-1}(1 + 1 -o(1)) + \frac{1}{d+q_1} \cdot 1 \cdot (o(1)-1) \\&+ \frac{1}{d+q_2} \cdot 1 \cdot (o(1)-1) + o(1). \end{aligned}$$

Hence,

$$\begin{aligned} \lim _{d\rightarrow 0+} \frac{N}{d }= \frac{(q_1+q_2)-(q_1-q_2)^2}{(q_1q_2)(q_1+q_2-1)}. \end{aligned}$$

Having determined the sign of N for d sufficiently small, (78) follows from Proposition 6. $\square$

Proof of Theorem 7

Since $d\mapsto \frac{\partial {\tilde{\varLambda }}_3}{\partial D}(d, d)$ is analytic, all the roots are discrete. By Lemmas 36 and 37, $\frac{\partial {\tilde{\varLambda }}_3}{\partial D}(d, d)>0$ for d small and $\frac{\partial {\tilde{\varLambda }}_3}{\partial D}(d, d)<0$ for $d\gg 1$. This says that the infinity and zero diffusion rates are local CvSSs. Furthermore, there exists at least one $d^*=d^*(q_1,q_2)$ such that $\frac{\partial {\tilde{\varLambda }}_3}{\partial D}(d^*, d^*)=0$, and $\frac{\partial {\tilde{\varLambda }}_3}{\partial D}(d, d)$ changes sign from positive to negative in a neighborhood of $d^*$; i.e., $d^*$ is an evolutionary singular strategy which is not a CvSS. $\square$