Proportional threshold harvesting in discrete-time population models

Abstract

Threshold-based harvesting strategies tend to give high yields while protecting the exploited population. A significant drawback, however, is the possibility of harvesting moratoria with their socio-economic consequences, if the population size falls below the threshold and harvesting is not allowed anymore. Proportional threshold harvesting (PTH) is a strategy, where only a fraction of the population surplus above the threshold is harvested. It has been suggested to overcome the drawbacks of threshold-based strategies. Here, we use discrete-time single-species models and rigorously analyze the impact of PTH on population dynamics and stability. We find that the population response to PTH can be markedly different depending on the specific population model. Reducing the threshold and allowing more harvest can be destabilizing (for the Ricker and Hassell map), stabilizing (for the quadratic map), or both (for the generalized Beverton–Holt map). Similarly, management actions in the form of increasing the threshold do not always improve population stability—this can also be due to bistability. Our results therefore emphasize the importance of a rigorous analysis in investigating the impact of PTH on population stability.

A Appendix

1.1 A.1 Existence and uniqueness of a positive equilibrium of (2.3)

We first prove a technical lemma.

Lemma 1

Let $g:[a,b]\rightarrow {\mathbb {R}}$ be a twice differentiable map such that $g'(x)>0$ and $g''(x)<0$ for all $x\in (a,b)$, and $g(a)>a$. Then g can have at most one fixed point p in (a, b]. Moreover, the fixed point exists if and only if $g(b)\le b$.

Proof

Assume that the set of fixed points of g is not empty, and denote by $p_1$ the infimum of this set. Since $g(a)>a$, it is clear that $g'(p_1)\le 1$. By the hypotheses, $g'(x)<1$ for all $x>p_1$. This excludes the possibility of more fixed points of g, and implies that $g(b)\le b$. Bolzano’s Theorem ensures the existence of a fixed point of g when $g(a)>a$ and $g(b)\le b$. $\square$

Proposition A.1

Assume that f satisifies (H) and $T< K$. Then $F_q$ has a unique positive fixed point $K_1=K_1(q,T)\in (T,K)$ for each $q\in (0,1)$.

Proof

We first prove the uniqueness of the fixed point of $F_q$, dividing the proof into two steps.

Step 1:$F_q$cannot have positive fixed points in$(0,T]\cup [K,\infty )$.

For a fixed $q\in (0,1)$, $F_q$ is a convex combination of f(x) and $F_1(x)$. Actually, $F_q(x)=(1-q)f(x)+cF_1(x)$, for all $x\ge 0$. Hence,

$$\begin{aligned} \left. \begin{array}{c}f(x)\le x,\quad \forall \, x\ge K\\ F_1(x)<x,\quad \forall \, x\ge K\end{array}\right\}&\Longrightarrow F_q(x)<x,\quad \forall \, x\ge K,\\ \left. \begin{array}{c}f(x)> x,\quad \forall \, x\le T\\ F_1(x)\ge x,\quad \forall \, x\le T\end{array}\right\}&\Longrightarrow F_q(x)>x,\quad \forall \, x\le T. \end{aligned}$$

Step 2:$F_q$has a unique positive fixed point$K_1\in (T,K)$.

We notice that $F_q(x)=(1-q)f(x)+qT$ for all $x\in (T,K)$, and therefore $F_q'(x)$ and $F_q''(x)$ have the same sign as $f'(x)$ and $f''(x)$, respectively.

If $d\ge K$, then $F_q'(x)>0$ and $F_q''(x)<0$ for all $x\in (T,K)$. Since $F_q(T)>T$ and $F_q(K)<K$, Lemma 1 ensures that $F_q$ has a unique fixed point $K_1\in (T,K)$.

In the following, we assume that $d<K$ and consider two cases.

Case 1: $F_q(d)<d$.

In this case, $F_q$ cannot have fixed points in (d, K). By Lemma 1, $F_q(T)>T$ and $F_q(d)<d$ imply that $F_q$ has a unique fixed point $K_1\in (T,d)\subset (T,K)$.

Case 2: $F_q(d)\ge d$.

In this case, Lemma 1 guarantees that $F_q$ cannot have any fixed point in (T, d). Since $F_q'(x)<0$ for all $x\in (d,K)$, $F_q(d)\ge d$, and $F_q(K)<K$, it follows from Bolzano’s and Rolle’s theorems that there exists a unique $K_1\in [d,K)\subset (T,K)$ such that $F_q(K_1)=K_1$. $\square$

1.2 A.2 Global stability for (2.3) when $F'(K_1)\ge 0$

If the map f satisfies (H) and F is nondecreasing at the positive equilibrium $K_1$, then $K_1$ attracts all solutions of (2.3) starting at a positive initial condition. This result is a direct consequence of Proposition A.1 and the following auxiliary result:

Lemma 2

(Braverman and Liz 2012, Lemma 1) Let $g:[0,\infty ) \rightarrow [0,\infty )$ be a continuous function such that $g(0)=0$, and g has a unique fixed point p such that $x<g(x)\le p$ for all $x\in (0,p)$, and $0<g(x)<x$ for all $x>p$. Then p is globally attracting for all positive solutions of the equation

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

(A.1)

that is, every solution $\{x_n\}$ of (A.1) with $x_0>0$ converges to p.

1.3 A.3 Local stability for (2.3) when f is concave in (0, K)

Proposition A.2

Assume that f satisifies (H), and $f''(x)<0$ for all $x\in (0,K)$. If K is asymptotically stable for (2.1), then the unique positive equilibrium $K_1=K_1(q,T)$ of (2.3) is asymptotically stable for all $q\in (0,1)$ and $T<K$. Moreover, if $K_1$ is asymptotically stable for given values of $q\in (0,1)$ and $T\in (0,K)$, then decreasing T cannot destabilize the equilibrium.

Proof

Assume first that K is asymptotically stable for (2.1). Since $f''(x)<0$ for all $x\in (T,K)$, $K_1<K$, and $|f'(K)|\le 1$, it follows that

$$\begin{aligned} |F_q'(K_1)|=(1-q)|f'(K_1)|\le (1-q)|f'(K)|<1. \end{aligned}$$

This implies that $K_1$ is asymptotically stable.

If $K_1(q,T_1)$ is asymptotically stable for given values of $q\in (0,1)$ and $T_1\in (0,K)$, then the same argument used above applies to prove that $K_1(q,T_2)$ is asymptotically stable if $T_2<T_1$. We only need to use that $T_2<T_1$ implies $T<K_1(q,T_2)<K_1(q,T_1)<K$ (see Franco and Liz 2013). $\square$

1.4 A.4 Global stability for (2.3) with a stable Ricker map

In this section we prove that if the positive equilibrium $K=1$ of the Ricker map (3.2) is globally asymptotically stable, then the same property holds for all $q\in (0,1]$ and $T>0$.

When $T\ge K$, we prove a general result for maps f satisfying assumption (A). To this end, we shall use the following auxiliary result:

Lemma 3

(El-Morshedy and Jiménez López 2008, Theorem B) Assume that $f:(0,\infty )\rightarrow (0,\infty )$ has a globally attracting equilibrium p. Let $g:(0,\infty )\rightarrow (0,\infty )$ be a continuous map satisfying that $x<g(x)\le \max \{f(x),p\}$ for all $x<p$, and $x>g(x)\ge \min \{f(x),p\}$ for all $x>p$. Then p is a globally attracting equilibrium of g.

Proposition A.3

Assume that f satisifies (A). If $T\ge K$ and K is a global attractor of (2.1), then K is a global attractor of (2.3) for all $q\in (0,1]$.

Proof

If $T\ge K$, then Lemma 3 applies to $g=F_q$ because $x<F_q(x)\le f(x)\le \max \{f(x),K\}$ for all $x<K$, and $x>F_q(x)=f(x)\ge \min \{f(x),K\}$ for all $x>K$ (see Fig. 2a). $\square$

Next we address the case $T<1$ for the Ricker map.

Proposition A.4

If $r\le 2$ and $T<1$, then the unique positive equilibrium $K_1=K_1(q,T)$ of (2.3) with the Ricker map $f(x)=x \,\mathrm {e}^{r(1-x)}$ is globally asymptotically stable.

Proof

It is well known that the equilibrium $K=1$ for the Ricker map $f(x)=x \,\mathrm {e}^{r(1-x)}$ is a global attractor if and only if $r\le 2$ (e.g. Thieme 2003). Moreover, since $f''(x)<0$ for all $x<2/r$, it follows that $f''(x)<0$ for all $x\in (0,1)$ if $r\le 2$. Thus, Propositions A.1 and A.2 ensure that (2.3) has a unique positive equilibrium $K_1$, and it is locally asymptotically stable if $T<1$ and $q\in (0,1)$.

Next we prove that $K_1$ is actually a global attractor. To this end, we use an enveloping theorem due to Cull (2007, Theorem 3). We have to show that $F(x)<2K_1-x$ if $x<K_1$, and $F(x)>2K_1-x$ if $x>K_1$.

Denote by $p_1, p_2$ the points such that $f(p_1)=f(p_2)=T$, $0<p_1<T<K_1<1<p_2$. F is differentiable in $(p_1,p_2)$. We distinguish three cases (see Fig. 11):

• If $x\in (0,p_1)$, then $F(x)=f(x)\le T<2T-x<2K_1-x$.

• If $x\in (p_2,\infty )$, then we use that $f(x)>2-x$ for all $x>1$ (e.g. Cull 2007) to show that $F(x)=f(x)>2-x>2K_1-x$.

• If $x\in (p_1,p_2)$ we have $F'(x)=(1-q)f'(x)>-1$, which makes it impossible that there is a point $q\in (p_1,K_1)\cup (K_1,p_2)$ such that $F(q)=2K_1-q$ (assume that such q exists, and apply Rolle’s Theorem to the map $G(x)=2K_1-x-F(x)$ and the points q and $K_1$ to arrive at a contradiction).

Thus, F must remain below the line $2K_1-x$ for $x<K_1$, and above it for $x>K_1$, and the proof is complete. $\square$

Fig. 11

A diagram showing how the modified map $F(x)=\min \{f(x),(1-q)f(x)+qT\}$ (solid blue line) is enveloped by the line $2K_1-x$. The original Ricker map f(x) is shown in a red dashed line (color figure online)

1.5 A.5 Stability switches for (2.3) with an unstable Ricker map

Here we rigorously prove the results about the Ricker model stated in Sect. 3.2.

Proposition A.5

Assume that $r>2$, $0\le T<1$, and $q\in (0,1)$.

(I) :

The positive equilibrium $K_1$ of (2.3) with $f(x)=x \,\mathrm {e}^{r(1-x)}$ is asymptotically stable if and only if the following inequality holds:

$$\begin{aligned} (1-q)(r g(r,q,T)-1)\le \mathrm {e}^{r(g(r,q,T)-1)}, \end{aligned}$$

(A.2)

where

$$\begin{aligned} g(r,q,T):=\frac{2+rqT+\sqrt{(rqT)^2+4}}{2r} \end{aligned}$$

(A.3)

is the larger root of the quadratic equation $rx^2-(2+rqT)x+qT=0$.

(II) :

Moreover, for each $r>2$ fixed, the boundary of the stability region defined by the implicit equation

$$\begin{aligned} (1-q)(r g(r,q,T)-1)= \mathrm {e}^{r(g(r,q,T)-1)} \end{aligned}$$

defines a decreasing curve $q=q(T)$ in the plane (T, q), $0<T<1$.

Proof

We first prove (I). Since the positive equilibrium $K_1$ of (2.3) belongs to the interval (T, 1), function F is defined as $F(x)=(1-q)f(x)+qT$ in a neighborhood of $K_1$. This map is unimodal, of class $C^3$, and has a negative Schwarzian derivative everywhere. Thus, the condition for local asymptotic stability is $F'(K_1)\ge -1$, which is equivalent to $(1-q)f'(K_1)\ge -1$ (see Liz 2010). In order to get the stability boundary, we solve the system of equations

$$\begin{aligned} x&=(1-q)f(x)+qT \end{aligned}$$

(A.4)

$$\begin{aligned} -1&=(1-q)f'(x) \end{aligned}$$

(A.5)

Since $f(x)=\mathrm {e}^{r(1-x)}$, system (A.4)–(A.5) reads

$$\begin{aligned} (1-q)\mathrm {e}^{r(1-x)}&=\frac{x-qT}{x} \end{aligned}$$

(A.6)

$$\begin{aligned} (1-q)\mathrm {e}^{r(1-x)}&=\frac{-1}{1-rx} \end{aligned}$$

(A.7)

This system leads to

$$\begin{aligned} \frac{x-qT}{x}=\frac{-1}{1-rx}\Longleftrightarrow rx^2-(2+rqT)x+qT=0. \end{aligned}$$

This quadratic equation has two positive roots

$$\begin{aligned} x_{\pm }=\frac{2+rqT\pm \sqrt{(rqT)^2+4}}{2r}. \end{aligned}$$

Since $0<q<1$, we get

$$\begin{aligned} 2-\sqrt{(rqT)^2+4}<0<2rT(1-q), \end{aligned}$$

which implies that $x_{-}<T$.

On the other hand, one can check that $x_{+}=g(r,q,T)\in (T,1)$ for some q such that

$$\begin{aligned} q<\frac{r-2}{(r-1)T}\quad \text{ and }\quad q>\frac{rT-2}{rT-1} \quad \text{ if }~{ rT>2.} \end{aligned}$$

Replacing $x=g(r,q,T)$ in (A.7), we get the boundary of the stability region defined in (A.2).

Next we prove (II). The stability boundary can be written as $H(r,q,T)=0$, where

$$\begin{aligned} H(r,q,T)= \mathrm {e}^{r(g(r,q,T)-1)}-(1-q)(rg(r,q,T)-1). \end{aligned}$$

We show that, for a fixed value of $r>2$, it holds that $\partial H/\partial T$ and $\partial H/\partial q$ are positive at the equilibrium. Thus, the Implicit Function Theorem ensures that $\partial q/\partial T$ is negative, which proves  (II).

It is clear that $\partial g/\partial T>0$ and $\partial g/\partial q>0$, where g is defined in (A.3). On the other hand, it is evident that $rg(r,q,T)>1$. Since

$$\begin{aligned} \frac{\partial H}{\partial T}&=r\frac{\partial g}{\partial T}\left( \mathrm {e}^{r(g(r,q,T)-1)}-(1-q)\right) ,\\ \frac{\partial H}{\partial q}&=r \frac{\partial g}{\partial q}\left( \mathrm {e}^{r(g(r,q,T)-1)}-(1-q)\right) +(rg(r,q,T)-1) \end{aligned}$$

it only remains to prove that $\mathrm {e}^{r(g(r,q,T)-1)}>1-q$.

Replacing $x=g(r,q,T)$ in (A.6), we get

$$\begin{aligned} (1-q)g(r,q,T)\mathrm {e}^{r(1-g(r,q,T))}=g(r,q,T)-qT<g(r,q,T), \end{aligned}$$

from where it follows that $(1-q)\mathrm {e}^{r(1-g(r,q,T))}<1$, as we wanted to prove. $\square$

Corollary A.6

Assume that $r>2$, $0\le T<1$, and $q\in (0,1)$. Denote by $K_1$ the positive equilibrium of (2.3) with $f(x)=x \,\mathrm {e}^{r(1-x)}$.

(i) :

For $T=0$, $K_1$ is asymptotically stable if and only if $q\ge 1-\mathrm {e}^{2-r}$.

(ii) :

As $T\rightarrow 1^-$, the limit form of the asymptotic stability condition for $K_1$ is given by the inequality $q> (r-2)/(r-1)$.

(iii) :

For a given value of $T\in (0,1)$, increasing q is stabilizing.

(iv) :

For a fixed value of $q\in (0,1)$, we have three cases:

(a):

If $q\ge 1-\mathrm {e}^{2-r}$ ($r\le 2-\ln (1-q)$), then $K_1$ is asymptotically stable, independently of T.

(b):

If $(r-2)/(r-1)<q< 1-\mathrm {e}^{2-r}$, then increasing T is stabilizing.

(c):

If $q\le (r-2)/(r-1)$ ($r\ge (2-q)/(1-q)$), then $K_1$ is unstable, independently of T.

Proof

(i) :

For $T=0$, we have $g(r,q,T)=g(r,q,0)=2/r$, and therefore (A.2) reads $1-q\le \mathrm {e}^{2-r}$.

(ii) :

For $T=1$, the solution of (A.4) is $x=1$, and therefore the limit form of the asymptotic stability condition for $K_1$ as $T\rightarrow 1^-$ is given by $-1<(1-q)f'(1)=(1-q)(1-r),$ which is equivalent to $q> (r-2)/(r-1)$.

(iii) :

and (iv) are a straightforward consequence of Proposition A.5 and the previous items (i) and (ii).

$\square$

1.6 A.6 Stability switches for (2.3) with a quadratic map

In this subsection we prove some analytic results for (2.3) with the quadratic map $f(x)=rx-(r-1)x^2$. By Proposition A.2, $r>3$ is a necessary condition for the existence of a stability switch at the positive equilibrium $K_1$.

Proposition A.7

Assume that $r>3$, and $q\in (0,1)$. As $T\rightarrow 1^-$, the limit form of the asymptotic stability condition for the equilibrium $K_1$ is given by the inequality $q> (r-3)/(r-2)$.

Proof

It is clear that the unique equilibrium of $F(x)=(1-q) f(x)+q$ is $K_1=1$. Thus, the limit form of the asymptotic stability condition is $-1< (1-q)f'(1)=(1-q)(2-r)$, which is equivalent to $q>(r-3)/(r-2)$. $\square$

A consequence of Propositions A.2 and A.7 is that, for a given value of $r>3$, a necessary condition for the existence of a stability switch is $q< (r-3)/(r-2)$.

Proposition A.8

Assume that $r>3$, $0\le T<1$, and $q< (r-3)/(r-2)$.

(I) :

The boundary of the stability region of the positive equilibrium $K_1$ of (2.3) with $f(x)=rx-(r-1)x^2$ is implicitly defined by the equation

$$\begin{aligned} ((1-q)r-1)^2+4q(1-q)(r-1)T=4. \end{aligned}$$

(A.8)

Moreover, for each $r>3$, (A.8) defines an increasing curve $q=q(T)$ in the plane (T, q), $0<T<1$.

(II) :

In the limit case $T=0$, the stability switch is defined by $q=(r-3)/r$.

(III) :

A stability switch as T is decreased occurs for all q such that

$$\begin{aligned} \frac{r-3}{r}<q<\frac{r-3}{r-2}. \end{aligned}$$

Proof

(I) :

The positive equilibrium $K_1$ is the unique positive root of the quadratic equation $x=(1-q)f(x)+qT$, whose expression is

$$\begin{aligned} K_1=\frac{(1-q)r-1+\sqrt{((1-q)r-1)^2+4q(1-q)(r-1)T}}{2(1-q)(r-1)}. \end{aligned}$$

Notice that $q< (r-3)/(r-2)$ implies that $q<(r-1)/r$, and therefore $(1-q)r-1>0$. A stability switch occurs when $(1-q)f'(K_1)=-1$, which leads to (A.8). To show that $q=q(T)$ is increasing, we use implicit differentiation with the map

$$\begin{aligned} H(q,T)=((1-q)r-1)^2+4q(1-q)(r-1)T-4. \end{aligned}$$

It is clear that $\partial H/\partial T>0$ for all $r>1$. On the other hand,

$$\begin{aligned} \partial H/\partial q=-2r((1-q)r-1)+4T(r-1)(1-2q)<0 \end{aligned}$$

if $q\ge 1/2$. If $q<1/2$, since $T<1$, we have

$$\begin{aligned} \partial H/\partial q< & {} -2r((1-q)r-1)+4(r-1)(1-2q)\\= & {} -2((1-q)r^2-(3-4q)r+2(1-2q)), \end{aligned}$$

and we can easily check that the last expression is negative for all $r>3$. Hence, $\partial q/\partial T=-(\partial H/\partial T)/(\partial H/\partial q)>0$.

(II) :

For $T=0$, (A.8) gives $(1-q)r-1=2$, which is equivalent to $q=(r-3)/r$.

(III) :

is a consequence of (II) and Proposition A.7.$\square$

1.7 A.7 Condition for the MSY in (2.3) with a Ricker map

Proposition A.9

System (4.1) with $f(x)=x \,\mathrm {e}^{r(1-x)}$ has a solution (x, T) with $x>0$, $T\in (0,K)$, if and only if $r<q-\ln (1-q)$.

Proof

For $T=0$ and a fixed $q\in (0,1)$, we can find the unique solution $(r^*,x^*)$ of system (4.1). Indeed, system (4.1) with $T=0$ becomes

$$\begin{aligned} f'(x)=1;\quad \quad (1-q)f(x)=x. \end{aligned}$$

(A.9)

Using the expression of f in (A.9), we get

$$\begin{aligned} (1-rx)\mathrm {e}^{r(1-x)}=1=(1-q)\mathrm {e}^{r(1-x)}, \end{aligned}$$

which leads to $rx=q$. Replacing $x=q/r$ into $(1-q)\mathrm {e}^{r(1-x)}=1$ gives $(1-q)\mathrm {e}^{r-q}=1$, which is equivalent to $r=q-\ln (1-q)$. Thus, the solution is $r^*=q-\ln (1-q)$, $x^*=q/r^*=q/(q-\ln (1-q))$.

Now, if $T>0$, then system (4.1) leads to

$$\begin{aligned} (1-rx)\mathrm {e}^{r(1-x)}=1;\quad \quad (1-q)x\mathrm {e}^{r(1-x)}+qT=x, \end{aligned}$$

which, by elementary calculus, gives

$$\begin{aligned} rx=q\frac{x-T}{x-qT}:=q_1<q, \end{aligned}$$

and therefore $r=q_1-\ln (1-q_1)<r^*$, since the map $v(q)=q-\ln (1-q)$ is increasing for $q\in (0,1)$. $\square$