The Allee-type ideal free distribution

Abstract

The ideal free distribution (IFD) in a two-patch environment where individual fitness is positively density dependent at low population densities is studied. The IFD is defined as an evolutionarily stable strategy of the habitat selection game. It is shown that for low and high population densities only one IFD exists, but for intermediate population densities there are up to three IFDs. Population and distributional dynamics described by the replicator dynamics are studied. It is shown that distributional stability (i.e., IFD) does not imply local stability of a population equilibrium. Thus distributional stability is not sufficient for population stability. Results of this article demonstrate that the Allee effect can strongly influence not only population dynamics, but also population distribution in space.

Appendices

Appendix A: The IFD

First, I study under which conditions strategy $u=(1,0)$ or $(0,1)$ is a strict NE at overall population abundance $x$. These two strategies correspond to population distribution $(x_1^1,x_2^1)=(x,0)$ and $(x_1^2,x_2^2)=(0,x)$, respectively. A strict NE $u^*=(u_1^*,u_2^*)$ satisfies $W(u^*,u^*;x)>W(u,u^*;x)$ for any other strategy $u=(u_1,u_2)\ne u^*.$ Thus, $u^*=(1,0)$ is a strict NE provided $V_1(x)=\min \{r_1 (1-x/K_1),a_1 (x-A_1)\}>V_2(0)=\min \{r_2,-a_2 A_2\}=-a_2 A_2$ due to assumption (1). For $x\le x_1^{\max }=\frac{ K_1 (a_1 A_1+r_1)}{a_1 K_1+ r_1},$ $V_1(x)=a_1 (x-A_1)$ which implies that $u^*=(1,0)$ is a strict NE when $x>\frac{a_1 A_1-a_2 A_2}{a_1}.$ For $x>x_1^{\max },$ $V_1(x)=r_1 (1-x/K_1)$ which implies that $u_1^*=1$ is a strict NE when $x<\frac{r_1+a_2 A_2}{r_1} K_1.$ Thus, for population densities satisfying $\frac{a_1 A_1-a_2 A_2}{a_1} <x<\frac{r_1+a_2 A_2}{r_1} K_1$, $u^*=(1,0)$ is a strict NE. Similarly, $u^*=(0,1)$ is a strict NE when $x<\frac{r_2+a_1 A_1}{r_2} K_2.$ It follows that $u^*=(0,1)$ is a strict NE when $\frac{a_2 A_2-a_1 A_1}{a_2} <x<\frac{r_2+a_1 A_1}{r_2} K_2.$

Second, I consider interior ESSs. These occur at distributions $x_1=u_1 x$ and $x_2=u_2 x$ where $V_1(x_1)=V_2(x_2)$. A strategy $u^*=(u_1^*,u_2^*)$ is a (local) ESS, provided

$$\begin{aligned}&W(u^*,u;x)-W(u,u;x)=(u_1^*-u_1) V_1(x_1)+(u_2^*-u_2) V_2(x_2)\\&\quad = (u_1^*-u_1) (V_1(x_1)-V_1(x_1^*))+(u_2^*-u_2) (V_2(x_2)-V_2(x_2^*))>0 \end{aligned}$$

for all perturbed distributions $u=(u_1,u_2)\ne (u_1^*,u_2^*)$ in a neighborhood of $u^*.$ Let $DV_i(x_i,w_i)$ denote directional derivative of $V_i$ at $x_i$ in a direction $w_i.$ Because $V_i$ is piece-wise linear, I get

$$\begin{aligned} W(u^*,u;x)-W(u,u;x)= (u_1^*-u_1) \left( DV_1(x_1^*,x_1 -x_1^*)-DV_2(x_2^* ,x_2-x_2^*) \right) . \end{aligned}$$

In particular, when $x_1^*\ne x_1^{\max }$ and $x_2^*\ne x_2^{\max }$, $DV_i(x_i^*,x_i-x_i^*)=V_i'(x_i^*) (x_i-x_i^*)$, and

$$\begin{aligned} W(u^*,u;x)-W(u,u;x)= -(u_1^*-u_1)^2 \left( V_1'(x_1^*)+V_2'(x_2^*) \right) x. \end{aligned}$$

The distribution $(x_1^*,x_2^*)$ is an ESS provided the sum of slopes $V_1'(x_1^*)+V_2'(x_2^*)$ is negative. In what follows I study evolutionarily stable distributions as a function of the overall population abundance (for definitions of population thresholds $\tilde{x}^i$ see (2)).

First, I consider a distribution that satisfies $0<x_1<x_1^{\max }$ and $0<x_2<x_2^{\max }.$ For such a distribution $V_1(x_1)=V_2(x_2)$ if

$$\begin{aligned} a_1 ( x_1 - A_1) = a_2 ( x_2 - A_2) \end{aligned}$$

which yields

$$\begin{aligned} (x_1,x_2)= \left( \frac{a_1 A_1 - a_2 A_2 + a_2 x}{a_1 + a_2}, \frac{a_2 A_2 - a_1 A_1 + a_1 x}{a_1 + a_2}\right) \!. \end{aligned}$$

This distribution exists for population abundances satisfying

$$\begin{aligned} \max \left\{ \frac{a_1 A_1-a_2 A_2}{a_1},\frac{a_2 A_2-a_1 A_1}{a_2}\right\} < x < \min \{\tilde{x}^2,\tilde{x}^4\}. \end{aligned}$$

However, this distribution cannot be an ESS, because the sum of slopes $V_1'(x_1)+V_2'(x_2)=a_1+a_2$ is positive.

Second, I consider a distribution that satisfies $0<x_1<x_1^{\max }$ and $x_2^{\max }<x_2.$ For such a distribution $V_1(x_1)=V_2(x_2)$ if

$$\begin{aligned} a_1 ( x_1 - A_1) = r_2 (1 - x_2/K_2) \end{aligned}$$

which yields

$$\begin{aligned} (x_1^3,x_2^3)=\left( \frac{r_2 (x -K_2)-a_1 A_1 K_2}{ r_2- a_1 K_2 } ,\frac{K_2 (r_2 + a_1 (A_1 - x))}{r_2-a_1 K_2} \right) . \end{aligned}$$

Such a distribution is an ESS provided $V_1'(x_1^3)+V_2'(x_2^3)=a_1-r_2/K_2<0$, i.e., $a_1<r_2/K_2.$ Under this assumption, this distribution satisfies constraints $0<x_1^3<x_1^{\max }$ and $x_2^{\max }<x_2^3$ provided

$$\begin{aligned} \frac{K_2 (a_1 A_1+r_2)}{r_2}< x < \min \{\tilde{x}^1,\tilde{x}^2\}. \end{aligned}$$

Third, I consider a distribution that satisfies $x_1^{\max }<x_1$ and $x_2<x_2^{\max }.$ For such a distribution $V_1(x_1)=V_2(x_2)$ if

$$\begin{aligned} r_1 (1 - x_1/K_1) = a_2 (x_2 - A_2) \end{aligned}$$

which yields

$$\begin{aligned} (x_1^4,x_2^4)= \left( \frac{K_1 (r_1+a_2 (A_2-x))}{r_1-a_2 K_1}, \frac{r_1 (x-K_1)- a_2 A_2 K_1}{r_1-a_2 K_1} \right) . \end{aligned}$$

This distribution is an ESS provided $V_1'(x_1^4)+V_2'(x_2^4)= a_2-r_1/K_1<0$, i.e., $a_2<r_1/K_1.$ Under this assumption, this distribution satisfies constraints $x_1^{\max }<x_1^4$ and $x_2^4<x_2^{\max }$ provided

$$\begin{aligned} \frac{K_1 (a_2 A_2+r_1)}{r_1}< x< \min \{\tilde{x}^3,\tilde{x}^4\}. \end{aligned}$$

Fourth, I consider a distribution that satisfies $x_1^{\max }<x_1$ and $x_2^{\max }<x_2.$ For such a distribution $V_1(x_1)=V_2(x_2)$ if

$$\begin{aligned} r_1 (1 - x_1/K_1) = r_2 (1 - x_2/K_2) \end{aligned}$$

which yields

$$\begin{aligned} (x_1^5,x_2^5)=\left( \frac{K_1 (K_2 (r_1-r_2)+r_2 x)}{K_2 r_1+ K_1 r_2}, \frac{K_2 (K_1 (r_2-r_1)+r_1 x)}{K_2 r_1+ K_1 r_2} \right) . \end{aligned}$$

Because $V_1'(x_1^5)+V_2'(x_2^5)= -r_1/K_1-r_2/K_2<0,$ this distribution is always an ESS and satisfies constraints $x_1^{\max }<x_1^5$ and $x_2^{\max }<x_2^5$ whenever

$$\begin{aligned} \max \{\tilde{x}^1,\tilde{x}^3\}< x. \end{aligned}$$

It remains to specify under which conditions population distributions such that either $x_1^*= x_1^{\max }$ or $x_2^*= x_2^{\max }$ are IFDs.

Let us assume that $x_1^*= x_1^{\max }.$ Then $x_2^*= x-x_1^{\max }.$ Provided $x>x_1^{\max }$, $x_2^*>0.$ There are two possibilities: Either $x_2^*< x_2^{\max }$ or $x_2^*> x_2^{\max }.$ If $x_2^*< x_2^{\max }$ then from $V_1(x_1^*)=V_2(x_2^*)$ I get that

$$\begin{aligned} x_2^*=A_2+\frac{r_1 a_1 (K_1-A_1)}{a_2 (a_1 K_1 + r_1)}. \end{aligned}$$

For $x_1<x_1^{\max }$

$$\begin{aligned} W(u^*,u;x)-W(u,u;x)&= -(u_1^*-u_1)^2 \left( V_1'(x_{1-}^*)+V_2'(x_2^*) \right) x \\&= -(u_1^*-u_1)^2 (a_1+a_2)x<0, \end{aligned}$$

where $V_1'(x_{1-}^*)$ denotes the left derivative. Thus, such a distribution cannot be an ESS.

Now I consider the situation where $x_2^*>x_2^{\max }$. The corresponding distribution is then

$$\begin{aligned} x_2^*=K_2+\frac{a_1 r_1 K_2(A_1 - K_1) }{(a_1 K_1 + r_1) r_2}. \end{aligned}$$

For $x_1<x_1^{\max }$

$$\begin{aligned} W(u^*,u;x)-W(u,u;x)&= -(u_1^*-u_1)^2 \left( V_1'(x_{1-}^*)+V_2'(x_2^*) \right) \\&= -(u_1^*-u_1)^2 \left( a_1-\frac{r_2}{K_2}\right) x \end{aligned}$$

and for $x_1>x_1^{\max }$

$$\begin{aligned} W(u^*,u;x)-W(u,u;x)&= (u_1^*-u_1)^2 \left( V_1'(x_{1+}^*)+V_2'(x_2^*) \right) \\&= (u_1^*-u_1)^2 \left( \frac{r_1}{K_1}+\frac{r_2}{K_2}\right) x>0, \end{aligned}$$

where $V_1'(x_{1+}^*)$ denotes the right derivative. Thus, distribution $(x_1^{\max },x_2^*)$ where $x_2^*>x_2^{\max }$ is an ESS provided $a_1<r_2/K_2$.

Similar considerations show that distribution $(x_1^*,x_2^{\max })$ is not an ESS when $x_1^*<x_1^{\max }$, while for $x_1^*>x_1^{\max }$, the distribution is an ESS provided that $a_2<r_1/K_1.$

Finally, the distribution $(x_1^{\max },x_2^{\max })$ is an ESS provided $V_1(x_1^{\max })=V_2(x_2^{\max }),$ $a_1<r_2/K_2$ and $a_2<r_1/K_1$.

Appendix B: Population equilibrium

Now I calculate population equilibria of model (6) assuming population distribution is at the IFD at current population density.

• (a) For population densities that satisfy

  $$\begin{aligned} \frac{a_1 A_1-a_2 A_2}{a_1} \le x\le \frac{(r_1+a_2 A_2)K_1}{r_1} \end{aligned}$$

  and the IFD distribution $(x_1^1,x_2^1)=(x,0)$ population dynamics are

  $$\begin{aligned} \displaystyle {\frac{dx}{dt}}= a_1 x (x-A_1). \end{aligned}$$

  For $x(0)>A_1$ the population is increasing and for $x(0)<A_1$ the population will die out. These are the typical bi-stable population dynamics caused by the positive dependent population growth.

• (b) For population densities that satisfy

  $$\begin{aligned} \frac{a_2 A_2-a_1 A_1}{a_2}\le x\le \frac{(r_2+a_1 A_1)K_2}{r_2} \end{aligned}$$

  and the IFD $(x_1^2,x_2^2)=(0,x)$ population dynamics are

  $$\begin{aligned} \displaystyle {\frac{dx}{dt}}= a_2 x (x-A_2). \end{aligned}$$

  Qualitatively these are the same population dynamics as in the case (a).

• (c) For the interior IFD given by $(x_1^3,x_2^3)$ (assuming that $a_1<r_2/K_2$) the corresponding population dynamics (6) are

  $$\begin{aligned} \displaystyle {\frac{dx}{dt}}=x \frac{a_1 r_2 (A_1 + K_2 - x)}{a_1 K_2 - r_2} \end{aligned}$$

  with the interior equilibrium

  $$\begin{aligned} x_1^*=A_1+K_2. \end{aligned}$$

  Because $a_1<r_2/K_2$, this equilibrium is unstable.

• (d) For the IFD given by $(x_1^4,x_2^4)$ (assuming $a_2<r_1/K_1$) the corresponding population dynamics (6) are

  $$\begin{aligned} \displaystyle {\frac{dx}{dt}}=x \frac{a_2 r_1 (A_2 + K_1 - x)}{a_2 K_1 - r_1} \end{aligned}$$

  with the interior equilibrium

  $$\begin{aligned} x_1^*=A_2+K_1. \end{aligned}$$

  As I assume $a_2<r_1/K_1$, this equilibrium is unstable.

• (e) For the IFD given by $(x_1^5,x_2^5)$ the corresponding dynamics (6) are

  $$\begin{aligned} \displaystyle {\frac{dx}{dt}}=x \frac{r_1 r_2 (K_1+K_2 - x)}{K_1 r_2 +K_2 r_1} \end{aligned}$$

  with the interior equilibrium

  $$\begin{aligned} x_1^*=K_1+K_2. \end{aligned}$$

  This equilibrium is locally asymptotically stable.

Appendix C: Stability of population-distributional equilibria

Now I study stability of system (6) and (7). First I consider the Jacobian of this system evaluated at the interior equilibrium ($0<u_1<1$) where $V_1(u_1 x)=V_2 (u_2 x)=0.$ I will assume that payoffs are differentiable at this point. The Jacobian is

$$\begin{aligned} J_1= \begin{pmatrix} u_1^2 x V_1'(u_1 x) + u_2^2 x V'_2(u_2 x),&{} x (u_1 x V_1'(u_1 x) - u_2 x V'_2(u_2 x))\\ u_1 u_2 \delta (u_1 V'_1(u_1 x) - u_2 V'_2(u_2 x)),&{} u_1 u_2 x \delta (V'_1(u_1 x) +V'_2(u_2 x)) \end{pmatrix}. \end{aligned}$$

The trace is

$$\begin{aligned} {{\mathrm{Tr}}}J_1=x (u_1 (u_1+\delta u_2) V_1'(u_1 x) + u_2 (u_2+\delta u_1) V_2'(u_2 x)) \end{aligned}$$

and the determinant is

$$\begin{aligned} \det J_1=u_1 u_2 x^2 \delta V_1'(u_1 x) V_2'(u_2 x). \end{aligned}$$

For local asymptotic stability of the interior equilibrium the trace must be negative and the determinant positive which requires both $V_1'(u_1 x)$ and $V_2'(u_2 x)$ to be negative. The only locally stable interior equilibrium is $x=K_1+K_2$ with the corresponding population distribution $u_1=K_1/(K_1+K_2)$.

Now I calculate the Jacobian at the boundary equilibrium where $u=(1,0)$ and $x=K_1,$

$$\begin{aligned} J_2= \begin{pmatrix} K_1 V_1'(K_1),&{} K_1 (K_1 V_1'(K_1) - V_2(0)\\ 0,&{} \delta V_2(0) \end{pmatrix}. \end{aligned}$$

Since $V_1'(K_1) =-r_1/K_1<0,$ this equilibrium is locally asymptotically stable provided $V_2(0)=-a_2 A_2<0.$ Similar considerations show that the equilibrium where $u=(0,1)$ and $x=K_2$ is locally asymptotically stable provided $V_1(0)=-a_1 A_1<0$.

Finally, I consider the extinction population equilibrium $x=0.$ The corresponding Jacobian matrix is

$$\begin{aligned} J_0= \begin{pmatrix} u_1 V_1(0) +u_2 V_2(0),&{}\quad 0\\ u_1 u_2 \delta (u_1 V_1'(0)-u_2 V_2'(0)),&{}\quad (u_2-u_1)\delta (V_1(0)- V_2(0)) \end{pmatrix}. \end{aligned}$$

The extinction equilibrium is locally asymptotically stable provided $u_2 V_1(0)+u_1 V_2(0)<0$ and $(u_2-u_1)(V_1(0)- V_2(0)) (u_1 V_1(0)+ u_2 V_2(0))>0.$ It follows that when $V_2(0)<V_1(0)<0$, then equilibrium $x=0$, $u_1=1$ is locally asymptotically stable. Similarly, if $V_1(0)<V_2(0)<0$ then equilibrium $x=0$, $u_1=0$ is locally asymptotically stable.