A Global Picture of the Gamma-Ricker Map: A Flexible Discrete-Time Model with Factors of Positive and Negative Density Dependence

Abstract

The gamma-Ricker model is one of the more flexible and general discrete-time population models. It is defined on the basis of the Ricker model, introducing an additional parameter $\gamma >0$. For some values of this parameter ($\gamma \le 1)$, population is overcompensatory, and the introduction of an additional parameter gives more flexibility to fit the stock–recruitment curve to field data. For other parameter values ($\gamma >1$), the gamma-Ricker model represents populations whose per-capita growth rate combines both negative density dependence and positive density dependence. The former can lead to overcompensation and dynamic instability, and the latter can lead to a strong Allee effect. We study the impact of the cooperation factor in the dynamics and provide rigorous conditions under which increasing the Allee effect strength stabilizes or destabilizes population dynamics, promotes or prevents population extinction, and increases or decreases population size. Our theoretical results also include new global stability criteria and a description of the possible bifurcations.

Appendix

Proof of Proposition 2

Computing the Schwarzian derivative of f and simplifying, we get

$$\begin{aligned} (Sf)(x)=\frac{-q(x)}{2x^2(\gamma -\delta x)^2},\quad \forall \; x\ne \gamma /\delta , \end{aligned}$$

(8)

where

$$\begin{aligned} q(x)=\delta ^4x^4-4\gamma \delta ^3x^3+6\gamma ^2\delta ^2x^2-4(\gamma ^3-\gamma )\delta x+\gamma ^4-\gamma ^2. \end{aligned}$$

Since $q''(x)=12\delta ^2(\gamma -\delta x)^2$, it follows that q is convex. On the other hand, equation $q'(x)=0$ has a unique real root $x_0=(\gamma -\gamma ^{1/3})/\delta$, and

$$\begin{aligned} m=q(x_0)=3\gamma \left( \gamma -\gamma ^{1/3}\right) \end{aligned}$$

is a global minimum of q.

It is clear that $m>0$ if $\gamma >1$, and therefore, $q(x)>0$ for all $x\in {\mathbb {R}}$. If $\gamma =1$, then $x_0=m=0$, and hence, $q(x)>0$ for all $x>0$. In both cases, it follows from (8) that $(Sf)(x)<0$ for all $x\in (0,\infty ), x\ne \gamma /\delta .$

If $0<\gamma <1$, then q(x) has two real roots $x_1,x_2$, with $x_1<0<x_2$. Moreover, $q(x)>0$ for all $x>x_2$. Since $q(\gamma /\delta )=3\gamma ^2>0$, it follows that $\gamma /\delta >x_2$, and therefore, $q(x)>0$ for all $x>\gamma /\delta$. Again (8) ensures that $(Sf)(x)<0$ for all $x\in (\gamma /\delta ,\infty )$. $\square$

Proof of Theorem 1

We need two auxiliary results that we include here for completeness of the proof. The first one is a consequence of Corollary 2.10 in El-Morshedy and Jiménez-López (2008):

Proposition 5

Let $f:(0,\infty )\rightarrow (0,\infty )$ be a ${\mathcal {C}}^3$ map with a unique fixed point p and a unique critical point c, which is a local maximum. If $-1\le f'(p)<1$ and $(Sf)(x)<0$ for all $x\in (c,\infty )$, then p is globally asymptotically stable.

The second one is taken from Proposition 1 in Liz and Ruiz-Herrera (2015):

Proposition 6

Consider the difference equation

$$\begin{aligned} x_{n+1}=f(x_n), \end{aligned}$$

where f satisfies the following conditions:

• (H1) $f:[0,\infty )\rightarrow [0,\infty )$ is a ${\mathcal {C}}^1$-unimodal map, with a unique critical point $c>0$, such that $f'(x)>0$ for all $x\in (0,c)$ and $f'(x)<0$ for all $x\in (c,\infty )$.

• (H2) $f(0)=f'(0)=0$, and $\lim _{x\rightarrow \infty }f(x)=0$.

• (H3) f has three fixed points $0<x_1<x_2$, so that $f(x)<x$ for all $x\in (0,x_1)\cup (x_2,\infty )$ and $f(x)>x$ for all $x\in (x_1,x_2)$.

• (H4) f is three times differentiable, and $(Sf)(x)<0$ whenever $f'(x)\ne 0$.

If $f^2(c)>x_1$ and $f'(x_2)\ge -1$, then $x_2$ is asymptotically stable and its immediate basin of attraction is $(x_1,f^{-1}(x_1)$.

We proceed with the proof of Theorem 1 (A). First, we observe that condition (6) is equivalent to $f'(p)\ge -1$. Indeed,

$$\begin{aligned} f'(p)= \beta p^{\gamma -1}e^{-\delta p} (\gamma -\delta p)=\frac{f(p)}{p} (\gamma -\delta p)=\gamma -\delta p. \end{aligned}$$

Hence,

$$\begin{aligned} f'(p)\ge -1\Longleftrightarrow p\le \frac{\gamma +1}{\delta } \Longleftrightarrow f\left( \frac{\gamma +1}{\delta } \right) \le \frac{\gamma +1}{\delta }. \end{aligned}$$

It is clear that the last inequality is equivalent to (6).

Now, the result follows easily from Propositions 1, 2 and 5. Notice that the inequality $f'(p)<1$ always holds because

$$\begin{aligned} f'(p)=\gamma -\delta p<1 \Longleftrightarrow \delta p>\gamma -1. \end{aligned}$$

The last inequality trivially holds because $\gamma -1\le 0$.

If (6) does not hold, then $f'(p)<-1$, and therefore, p is unstable.

Fig. 5

(Color Figure Online) Graph of the map $\beta =T_{\delta }(\gamma )$ showing the stability switches for the largest positive equilibrium of (1). a $T_{\delta }$ is decreasing for $\delta =0.1<e^{-2}$, b $T_{\delta }$ is unimodal for $\delta =1>e^{-2}$

The proof of Theorem 1 (B) follows from Proposition 6. Propositions 1 and 2 ensure that assumptions (H1)–(H4) hold.

Finally, the convergence to the equilibrium p is eventually monotone if and only if $f'(p)\ge 0$, which is equivalent to say that $f(c)\le c$ for the unique critical point $c=\gamma /\delta$ of f. It is clear that condition $f(\gamma /\delta )\le \gamma /\delta$ is equivalent to (7). $\square$

Proof of Proposition 3

The equilibrium is asymptotically stable if $\beta \le T_{\delta }(\gamma )$. It is easily seen that $T_{\delta }(0)=e/\delta$ and $\lim _{x\rightarrow \infty }T_{\delta }(x)=0$. Moreover, $T_{\delta }'(x)=0$ if and only if

$$\begin{aligned} (1+x)e^{-2/(1+x)}=\delta . \end{aligned}$$

(9)

It is clear that Eq. (9) has a unique positive solution ${\bar{\gamma }}$ if and only if $\delta >e^{-2}$.

Thus, if $\delta \le e^{-2}$ then $T_{\delta }$ is decreasing on $(0,\infty )$, and therefore, increasing $\gamma$ is destabilizing if $\beta <e/\delta$, and the fixed point p is unstable for all values of $\gamma >0$ if $\beta \ge e/\delta$. See Fig. 5a.

If $\delta >e^{-2}$, then $T_{\delta }$ attains a local maximum at ${\bar{\gamma }}>0$. Thus, there are three possibilities (see Fig. 5b):

• If $\beta \le e/\delta$, the increasing $\gamma$ is destabilizing.

• If $e/\delta<\beta <T_{\delta }(\gamma )$, then there are two stability switches.

• If $\beta >T_{\delta }(\gamma )$, then p is unstable for all $\gamma >0$.

A singular case is $\beta =T_{\delta }(\gamma )$, for which p is only asymptotically stable if $\gamma ={\bar{\gamma }}$.

Proof of Proposition 4

The proof of Proposition 4 is very similar to the proof of Proposition 3, using the map $G_{\delta }$ that defines the extinction boundary:

$$\begin{aligned} G_{\delta }(x):=e^{x-1}\left( \frac{x-1}{\delta }\right) ^{1-x}=\left( \frac{\delta e}{x-1}\right) ^{x-1},\quad x>1. \end{aligned}$$

In this case, $G_{\delta }$ is unimodal, with a global maximum $G_{\delta }(1+\delta )=e^{\delta }$. Moreover, $\lim _{x\rightarrow 1^+}G_{\delta }(x)=1, \lim _{x\rightarrow \infty }G_{\delta }(x)=0.$ See Fig. 6. We leave the details to the reader.

Fig. 6

(Color Figure Online) Graph of the map $\beta =G_{\delta }(\gamma )$ showing the survival/extinction switches for (1)

Proof of Theorem 2

The equation that defines a positive equilibrium of (1) is $\beta p^{\gamma -1}e^{-\delta p}=1$, or, equivalently, $p^{1-\gamma }e^{\delta p}-\beta =0$. For a fixed value of $\beta >0$, define the map

$$\begin{aligned} F(p,\gamma ):=p^{1-\gamma }e^{\delta p}-\beta ,\quad p>0,\gamma >0. \end{aligned}$$

By the implicit function theorem, equation $F(p,\gamma )=0$ defines a function $p=p(\gamma )$ if $\partial F/\partial p\ne 0$. By Proposition 4, we know that $p(\gamma )$ exists for $\gamma \in (0,\infty )$ if $\beta \ge e^{\delta }$, and otherwise, $p(\gamma )$ exists in two open intervals $(0,\gamma _1], [\gamma _2,\infty )$, with $0<1\le \gamma _1<\gamma _2$.

Moreover, we can compute

$$\begin{aligned} \frac{\partial p}{\partial \gamma }=\frac{-\partial F/\partial \gamma }{\partial F/\partial p}=\frac{p \ln (p)}{1-\gamma +\delta p}. \end{aligned}$$

First we show that the denominator is always positive. Indeed, if $\gamma \le 1$, then $1-\gamma +\delta p\ge \delta p>0$. If $\gamma >1$, then it is clear that the largest equilibrium point p satisfies $p>p^*$, where $p^*$ is the fixed point at which $f'(p^*)=1$. Now, it is easy to check that $p^*=(\gamma -1)/\delta$, and therefore,

$$\begin{aligned} p>\frac{\gamma -1}{\delta }\Longrightarrow 1-\gamma +\delta p>0. \end{aligned}$$

Thus, $\partial p/\partial \gamma >0$ if $p>1$, and $\partial p/\partial \gamma <0$ if $p<1$.

Now, if $\gamma \le 1$, then it is clear that $p=1$ if and only if $\beta =e^{\delta }$. Moreover, since $f(x)>x$ for $x<p$ and $f(x)<x$ for $x>p$, it follows that $p>1$ if $\beta >e^{\delta }$, and $p<1$ if $\beta <e^{\delta }$.

If $\gamma >1$ and $\beta >e^{\delta }$, a simple graphical analysis also shows that $p>1$. Hence, $p(\gamma )$ is an increasing function of $\gamma$ for $\gamma \in (0,\infty )$ if $\beta >e^{\delta }$.

If $\gamma >1$ and $\beta <e^{\delta }$, then two extinction switches occur at $\gamma _1, \gamma _2$, with $1\le \gamma _1<\gamma _2$. Moreover, since $\gamma _1<1+\delta <\gamma _2$, we get that $p(\gamma _1)=(\gamma _1-1)/\delta<1<(\gamma _2-1)/\delta =p(\gamma _2)$. Thus, $p(\gamma )$ is a decreasing function of $\gamma$ in $(0,\gamma _1)$ and an increasing function of $\gamma$ in $(\gamma _2,\infty )$.

Finally, if $\beta =e^{\delta }$, then $x=1$ is an equilibrium of (1) for all $\gamma >0$. Moreover, $p=1$ is the largest positive equilibrium if $\gamma \le 1+\delta$; for $\gamma > 1+\delta , x=1$ is unstable, and the largest positive equilibrium is $p>1$. Thus, $p(\gamma )$ is an increasing function of $\gamma$ in $(1+\delta ,\infty )$. $\square$