Destabilization and chaos induced by harvesting: insights from one-dimensional discrete-time models

Abstract

One-dimensional discrete-time population models are often used to investigate the potential effects of increasing harvesting on population dynamics, and it is well known that suitable harvesting rates can stabilize fluctuations of population abundance. However, destabilization is also a possible outcome of increasing harvesting even in simple models. We provide a rigorous approach to study when harvesting is stabilizing or destabilizing, considering proportional harvesting and constant quota harvesting, that are usual strategies for the management of exploited populations. We apply our results to some of the most popular discrete-time population models (quadratic, Ricker and Bellows maps). While the usual case is that increasing harvesting is stabilizing, we prove, somehow surprisingly, that increasing values of constant harvesting can destabilize a globally stable positive equilibrium in some cases; moreover, we give a general result which ensures that global stability can be shifted to observable chaotic dynamics by increasing one model parameter, and apply this result to some of the considered harvesting models.

Appendices

A Appendix: results on stability switches

In this appendix, we prove the results related to the destabilizing effects of harvesting. Our results on stabilization/destabilization are based on the following characterization of flip bifurcations:

Theorem A.1

(Sharkovsky et al. 1997, Theorem 8.2) Let $f_{\gamma }:I\rightarrow I$ be a family of $\mathcal {C}^3$ maps with smooth dependence on the parameter $\gamma$, where I is a real interval. Assume that $f_{\gamma _0}$ has a fixed point $x_{\gamma _0}$ and the following conditions hold:

• (a) $f_{\gamma _0}'(x_{\gamma _0})=-1$,

• (b) $(Sf_{\gamma _0})(x_{\gamma _0})<0$, where Sf is the Schwarzian derivative of f, and

• (c) $\displaystyle \frac{\partial }{\partial \gamma } \left( f_{\gamma }'(x)\right) <0$ at $\gamma =\gamma _0$ and $x=x_{\gamma _0}$.

Then, there are $\varepsilon >0$ and $\delta >0$ such that

• (i) for $\gamma \in (\gamma _0-\delta ,\gamma _0)$, $f_{\gamma }$ has exactly one fixed point $x_{\gamma }\in (x_{\gamma _0}-\varepsilon ,x_{\gamma _0}+\varepsilon )$ and $x_{\gamma }$ is asymptotically stable;

• (ii) for $\gamma \in (\gamma _0,\gamma _0+\delta )$, there are three fixed points of $f_{\gamma }^2$ in $(x_{\gamma _0}-\varepsilon ,x_{\gamma _0}+\varepsilon )$. Moreover, the middle point is an unstable fixed point of $f_{\gamma }$, and the other two points form an asymptotically stable cycle of $f_{\gamma }$ of period two.

If the inequality in (c) has the opposite sign, then the conclusions remain valid, but the 2-cycle appears as $\gamma$ decreases.

Proof of Proposition 3.1

The proof follows easily from the application of Theorem A.1 to the family $f_{\gamma }(x)=(1-\gamma )f(x)$. Denote by $G(x,\gamma )=f_{\gamma }(x)-x$, so that the equilibria of (2.2) are defined by $G(x,\gamma )=0$.

Using implicit differentiation, we have:

$$\begin{aligned} \frac{\partial }{\partial \gamma } \left( f_{\gamma }'(x_{\gamma })\right)&=\frac{\partial }{\partial \gamma } \left( (1-\gamma )f'(x_{\gamma })\right) \\&=-f'(x_{\gamma }) -(1-\gamma )f''(x_{\gamma })\left( \frac{\partial G/\partial \gamma }{\partial G/\partial x}\right) (x_{\gamma })\\&=-f'(x_{\gamma })+(1-\gamma )f''(x_{\gamma })\frac{f(x_{\gamma })}{(1-\gamma )f'(x_{\gamma })-1}. \end{aligned}$$

If $f_{\gamma }'(x_{\gamma })=-1$, we get $(1-\gamma )f'(x_{\gamma })=-1$, and therefore

$$\begin{aligned} \frac{\partial }{\partial \gamma } \left( f_{\gamma }'(x_{\gamma })\right) =\frac{1}{1-\gamma }-\frac{f''(x_{\gamma }) x_{\gamma }}{2}. \end{aligned}$$

Thus,

$$\begin{aligned} \frac{\partial }{\partial \gamma } \left( f_{\gamma }'(x)\right) <0\Longleftrightarrow (1-\gamma )x_{\gamma }f''(x_{\gamma })> 2, \end{aligned}$$

from which the result follows. $\square$

Proof of Theorem 4.2

Denote $f(x)=ax/(1+x^m)$, so that

$$\begin{aligned} f'(x)=\frac{-a\left( (m-1)x^m-1\right) }{\left( 1+x^m\right) ^2}\quad ;\quad f''(x)=\frac{a m x^{m-1}\left( m(x^m-1)-1-x^m\right) }{\left( 1+x^m\right) ^3}. \end{aligned}$$

If p is the positive fixed point of f and $x_{\gamma }$ is the largest positive equilibrium of (4.2), then it is clear that $x_{\gamma }<p$ if $\gamma >0$, and $x_{\gamma }>p$ if $\gamma <0$.

First we notice that condition $a\le m/(m-2)$ is necessary to ensure that the fixed point p of f(x) is asymptotically stable. Next, condition $a>4m/(m-1)^2$ is necessary to ensure that there is a value of $\gamma$ for which $f'(x_{\gamma })=-1$ and $x_{\gamma }=f(x_{\gamma })-\gamma$. Indeed, the first equality leads to equation

$$\begin{aligned} a\left( (m-1)x_{\gamma }^m-1\right) =\left( 1+x_{\gamma }^m\right) ^2. \end{aligned}$$

Denoting $z=1+x_{\gamma }^m$, we get that z must be a zero of the polynomial $q(z)=z^2-a(m-1)z+a m$. Since the discriminant of q(z) is $\Delta =a^2(m-1)^2-4 a m$, we have that $\Delta>0\Longleftrightarrow a>4m/(m-1)^2$. In case $\Delta =0$, we can verify that $f'(x)> -1$ for all $x\ne x_\gamma$, and therefore destabilization by CH is not possible.

Hence, under the conditions of Theorem 4.2 (a) and (b), equation $q(z)=0$ has two positive solutions $z_1>z_2$, that provide two values $\gamma _1, \gamma _2$ for which $f'(x_{\gamma _i})=-1$ and $x_{\gamma _i}=f(x_{\gamma _i})-\gamma _i$, $i=1,2$. Moreover, condition $a\le m/(m-2)$ implies that $z_1>z_2\ge a$ if $m>3$, and $z_2<z_1\le a$ if $m<3$, the equality corresponding to the case $a=m/(m-2)$.

If $m<3$, then $z_2<z_1\le a$, which is equivalent to $x_{\gamma _2}^m<x_{\gamma _1}^m\le p$ and then $f(x_{\gamma _2}^m)>x_{\gamma _1}^m\ge p$. Therefore, $0\le \gamma _1<\gamma _2$. We prove that $f''(x_{\gamma _1})>0$ and $f''(x_{\gamma _2})<0$, which implies by Proposition 4.1 that increasing harvesting is first destabilizing and then stabilizing again.

Using the expression of $f''(x)$, we have:

$$\begin{aligned} f''(x_{\gamma _1})>0>f''(x_{\gamma _2})&\Longleftrightarrow m(x_{\gamma _1}^m-1)-1-x_{\gamma _1}^m>0> m(x_{\gamma _2}^m-1)-1-x_{\gamma _2}^m\\&\Longleftrightarrow 1+x_{\gamma _2}^m=z_2<\frac{2m}{m-1}<1+x_{\gamma _1}^m=z_1. \end{aligned}$$

Thus, we only need to prove that $q(2m/(m-1))<0$. Indeed,

$$\begin{aligned} q(2m/(m-1))=m\left( \frac{4m}{(m-1)^2}-a\right) <0, \end{aligned}$$

because $a>4m/((m-1)^2)$.

If $m>3$, then $a\le z_2<z_1$ and $\gamma _1<\gamma _2\le 0$. Repeating the above argument, it follows that $f''(x_{\gamma _2})<0$ and $f''(x_{\gamma _1})>0$, which implies in this case that increasing immigration is first destabilizing and then stabilizing again. $\square$

Proof of Proposition 4.4

Denote by $f(x)=\alpha x+(1-\alpha ) x e^{r(1-x)}$.

The system of equations

$$\begin{aligned} f'(x_{\gamma })=-1\quad ,\quad x_{\gamma }=f(x_{\gamma })-\gamma \end{aligned}$$

(A.10)

leads to

$$\begin{aligned} (r x_{\gamma }-1)\left( (1-\alpha )x_{\gamma }+\gamma \right) =(1+\alpha )x_{\gamma }. \end{aligned}$$

(A.20)

Hence, system (A.10) has a positive solution, which is given by the positive root of the quadratic equation

$$\begin{aligned} Q(x_{\gamma })=r(1-\alpha )x_{\gamma }^2+(\gamma r-2)x_{\gamma }-\gamma =0. \end{aligned}$$

Since $f''(x_{\gamma })>0$ if and only if $r x_{\gamma }>2$, we have, using (A.20), that

$$\begin{aligned} f''(x_{\gamma })>0&\Longleftrightarrow r x_{\gamma }>2 \Longleftrightarrow (1-\alpha )x_{\gamma }+\gamma<(1+\alpha )x_{\gamma }\\&\Longleftrightarrow x_{\gamma }>\frac{\gamma }{2\alpha }\Longleftrightarrow Q(\gamma /(2 \alpha ))<0 \Longleftrightarrow \gamma r<4\alpha . \end{aligned}$$

Therefore, Proposition 4.1 ensures that the positive equilibrium $p=1$ of the Ricker–Clark equation can be destabilized if $\gamma r<4\alpha$ and can be stabilized if $\gamma r>4\alpha$.

We next show that for a given $r>2$, destabilization occurs if and only if $\alpha \in (\alpha _1(r),\alpha _2(r))$, with $\alpha _1(r)=(r-2)/r$ and $\alpha _2(r)=(-1+e^{r-2})/(1+e^{r-2}).$

Notice first that $p=1$ must be asymptotically stable for the Ricker–Clark equation, that is, $f'(1)\ge -1$. This condition is equivalent to $r>2$ and $\alpha \ge (r-2)/r$.

The value $\alpha _2(r)$ is determined finding the intersection point between the flip bifurcation curve defined by (A.10) and the line $\gamma r=4\alpha$ (see Fig. 6).

Using the expression of the positive root of the quadratic polynomial Q(x) and substituting $\gamma$ by $4\alpha /r$, we easily get $x_{\gamma }=2/r$.

Now, using the formulas $r x_{\gamma }=2$ and $\gamma =4\alpha /r$ in the equilibrium equation $x_{\gamma }=f(x_{\gamma })-\gamma$, it follows that $\alpha =(-1+e^{r-2})/(1+e^{r-2}).$

Let us emphasize, to conclude, that (as in Sect. 4.1) 2/r is the only zero of $f''$, allowing two $\gamma$’s with corresponding $x_\gamma$ satisfying $f_\gamma '(x_\gamma )=-1$ for every $\alpha \in (\alpha _1(r),\alpha _2(r))$. $\square$

B Appendix: results on chaos

In this appendix, we state and prove our main results related to the existence of observable chaos in discrete-time population models with harvesting.

We follow the notations and definitions stated at the beginning of Sect. 2. The celebrated Allwright-Singer theorem (Allwright 1978; Singer 1978) establishes that if f has a stable fixed point p, $f(x)>x$ (respectively $f(x)<x$) whenever $x<p$ (respectively, $x>p$), and $Sf(x)<0$ except for at most one critical point (an extremum) of f, then all orbits of f converge to p.

Let $f\in C(I)$ with I compact. The (topological) entropy of f, $h(f)\in [0,\infty ]$, is defined by

$$\begin{aligned} h(f)=\lim _{\epsilon \rightarrow 0}\limsup _{n\rightarrow \infty }\frac{1}{n}\log s_n(f,\epsilon ), \end{aligned}$$

where $s_n(f,\epsilon )$ is the maximal cardinality of the sets E having the property that for all distinct points $x,y\in E$ there is $0\le i<n$ such that $|f^i(x)-f^i(y)|>\epsilon$. If f consists of finitely many pieces of monotonicity, and $c_n$ is the number of laps of $f^n$, that is, the minimal cardinality of an interval partition of I such that $f^n$ is monotone on each of the intervals of the partition, then $h(f)=\lim _{n\rightarrow \infty } (1/n)\log c_n$ (Misiurewicz and Szlenk 1980). Positive entropy admits several useful characterizations. Namely, if $f\in C(I)$, then the following statements are equivalent (Li et al. 1982; Ruette 2017, Theorem 4.58, p. 84):

• $h(f)>0$;

• f has a periodic point of period not a nonnegative power of 2;

• there are a point x, a positive integer m and an odd number k such that either $f^{km}(x)\le x<f^m(x)$ or $f^{km}(x)\ge x>f^m(x)$;

• there are subintervals J, K of I with disjoint interiors and positive integers m, l such that $f^m(J)\cap f^l(K)\supset J\cup K$.

On the other hand, if the orbits of all points of I are attracted by periodic orbits (that is, for any x there is a periodic point p such that $\lim _{n\rightarrow \infty }|f^n(x)-f^n(p)|=0$), then $h(f)=0$ (Ruette 2017, Lemma 5.5, p. 110, and Theorem 5.17, p. 116). A sufficient condition for this to happen (Sharkovsky et al. 1997, pp. 73–74) is that the set of period of periodic points of f is bounded —equivalently, see below, f has type less than $2^\infty$ in the Sharkovsky order—. Therefore, monotone maps have zero entropy (they can only have periodic points of periods 1 and 2) and so they have the restrictions to invariant compact intervals of maps satisfying the hypotheses of the Allwright-Singer theorem.

Thus, if a map has positive entropy, then it is, in a sense, dynamically “bad-behaved”. It is important to stress that this complicated behaviour needs not be “observable” in practice. For instance, for the quadratic map $f(x)=r x(1-x)$, $r\approx 3.83187\ldots$, $I=[0,1]$, the point 1/2 is 3-periodic, hence $h(f)>0$, but the orbits of almost all points (in the sense of Lebesgue measure) are attracted by the 3-point orbit of 1/2 (Guckenheimer 1979). The following definition takes care of this problem:

Let $f\in C(I)$. We say that $(x,y)\in I^2$ is a Li-Yorke pair (for f) if $\limsup _{n\rightarrow \infty }|f^n(x)-f^n(y)|>0$ and $\liminf _{n\rightarrow \infty }|f^n(x)-f^n(y)|=0$. We say that f has observable chaos if the set of Li-Yorke pairs has (two-dimensional) positive Lebesgue measure.

A sufficient condition (under mild additional assumptions) to guarantee a very strong form of observable chaos is the existence of an absolutely continuous invariant measure (acip) for f. By this we mean a Borel probability measure $\mu$ such that $\mu (f^{-1}(B))=\mu (B)$ for any Borel set B and, moreover, satisfying $\mu (B)=0$ whenever $\lambda (B)=0$, $\lambda$ denoting the Lebesgue measure. For instance, if I is compact, $f\in C^3(I)$ and $f''(c)\ne 0$ for any critical point c of f, then the existence of an acip for f implies (besides $h(f)>0$) the existence of a subinterval J of I, invariant for some iterate $f^r$ of f, such that:

• $\lambda ^2$-almost every $(x,y)\in (f^i(J))^2$, $0\le i<r$, is a Li-Yorke pair;

• the orbit of $\lambda$-almost every $x\in K=\bigcup _{i=0}^{r-1} J$ is dense in K;

• $\lim _{n\rightarrow \infty } \log |(f^{n})'(x)|/n>0$ for $\lambda$-almost every $x\in K$;

• $\lim _{n\rightarrow \infty } \lambda (f^{rn+i}(A))=\lambda (f^i(J))$ for any measurable set $A\subset J$ of positive Lebesgue measure.

For more details on this, including many relevant references, see Bruin and Jiménez López (2010); Barrio Blaya and Jiménez López (2012).

Under the previous smoothness and compactness assumptions, a sufficient hypothesis implying the existence of an acip is the so-called Misiurewicz condition, that is, the orbits of critical points do not accumulate on the set of critical points and all periodic points are unstable (van Strien 1990). If f has just one critical point c and negative Schwarzian derivate outside c, this condition can be somewhat relaxed (see Lemma B.5); in particular, if there are an integer k and an unstable periodic point p such that $f^k(c)=p$ (when we say that f satisfies the strong Misiurewicz condition), then f has an acip.

The following theorem is a kind of mixture of (Misiurewicz 1981a, Theorem 7.9) and (Thieullen et al. 1994, Theorem I.3), and then not substantially new, but some conditions there are not really necessary, so this concrete formulation may be more useful in applications.

Theorem B.1

Let the family $f_\gamma \in C^3([a(\gamma ),b(\gamma )])$, $\gamma \in [\gamma _0,\gamma _1]$, satisfy the following conditions:

• (i) the maps $a,b:[\gamma _0,\gamma _1]\rightarrow \mathbb {R}$ are continuous;

• (ii) if $M=\{(x,\gamma ): \gamma \in [\gamma _0,\gamma _1], x\in [a(\gamma ),b(\gamma )]\}$, then both $F(x,\gamma )=f_\gamma (x)$ and $F'(x,\gamma )=f_\gamma '(x)$ are continuous in M;

• (iii) every map $f_\gamma$ has at most one critical point $c(\gamma )$ and $f_\gamma ''(c(\gamma ))\ne 0$ when such a point exists;

• (iv) $Sf_\gamma (x)<0$ for any $x\ne c(\gamma )$.

Let $\Lambda \supset \Lambda _S$ be, respectively, the sets of parameters $\gamma$ such that $f_\gamma$ has an acip (and then observable chaos) and $f_\gamma$ satisfies the strong Misiurewicz condition. If one of the maps $f_{\gamma _0}, f_{\gamma _1}$ has zero entropy, and the other one positive entropy, then $\Lambda$ is uncountable and $\Lambda _S$ is infinite.

Let $\rho \in \Lambda _S$, assume additionally that F is $C^3$ near $[a(\rho ),b(\rho )]\times \{\rho \}$, let $k,r\ge 1$ be such that $q=f_{\rho }^k(c(\rho ))$ is an unstable r-periodic point of $f_\rho$ and write $F_n(x,\gamma )=f_\gamma ^n(x)$, $d=c(\rho )$. If

$$\begin{aligned} \left( 1-(f_{\rho }^r)'(q)\right) \frac{\partial F_k}{\partial \gamma }(d,\rho ) \ne \frac{\partial F_r}{\partial \gamma }(q,\rho ), \end{aligned}$$

(B.10)

then $\rho$ is a Lebesgue density point of $\Lambda$ (hence the set of parameters $\gamma$ such that $f_\gamma$ has observable chaos has positive measure).

Remark B.2

By “F is $C^3$ near $[a(\rho ),b(\rho )]\times \{\rho \}$” we mean that if $\epsilon >0$ is small enough, then the restriction of F to $M_\epsilon =\{(x,\gamma ): \gamma \in [\rho -\epsilon ,\rho +\epsilon ], x\in [a(\gamma ),b(\gamma )]\}$ can be extended to a $C^3$-map defined on an open set containing $M_\epsilon$. We say that $\rho$ is a Lebesgue density point of $\Lambda$ if there is a Borel set $\Sigma \subset \Lambda$ such that $\lim _{\epsilon \rightarrow 0^+}\lambda (\Sigma \cap [\rho -\epsilon ,\rho +\epsilon ])/(2\epsilon )=1$ (or $\lim _{\epsilon \rightarrow 0^+}\lambda (\Sigma \cap [\rho -\epsilon ,\rho ])/\epsilon =1$ if $\rho =\gamma _1$, and analogously for $\gamma _0$).

Before proving Theorem B.1, a number of additional notions and results will be needed. Some of them involve finite and infinite sequences $\alpha =\alpha _1\alpha _2\cdots$ of symbols L, C, R, with $|\alpha |$ denoting the (finite or infinite) length of $\alpha$.

The shift operator ${\mathcal {S}}$ is given by ${\mathcal {S}}(\alpha )=\alpha _2\alpha _3\cdots$, hence it is well defined except if $|\alpha |=1$. The renormalization operator ${\mathcal {R}}$ is given by ${\mathcal {R}}(\alpha )=R\overline{\alpha _1}R\overline{\alpha _2}\cdots$, where we mean $\overline{L}=R$, $\overline{C}=C$ and $\overline{R}=L$. As usual, we denote by ${\mathcal {S}}^n$ (whenever it makes sense) and ${\mathcal {R}}^n$ the n-iterates of these operators. If $\alpha$ (respectively, $\beta$) is finite (respectively, finite or infinite), then $\alpha \beta$ is the concatenation of $\alpha$ and $\beta$, with $\alpha ^n=\alpha \cdots \alpha$ (n times) and $\alpha ^\infty =\alpha \alpha \cdots$ (infinitely many times).

We say that $\alpha$ is admissible if it is either an infinite sequence of L’s and R’s, or a (maybe empty) finite sequence of L’s and R’s, followed (and finished) by C. Observe that if $\alpha$ is admissible, then ${\mathcal {S}}(\alpha )$ and ${\mathcal {R}}(\alpha )$ are admissible as well.

We introduce a total order < in the set of admissible sequences as follows. Firstly, $L<C<R$. Now, if $\alpha \ne \beta$ and k is the first index such that $\alpha _k\ne \beta _k$ (such an index does exist because $\alpha$ and $\beta$ are admissible), then $\alpha <\beta$ if either there is an even number of R’s in $\alpha _1\cdots \alpha _{k-1}=\beta _1\cdots \beta _{k-1}$ and $\alpha _k<\beta _k$, or there is an odd number of R’s in $\alpha _1\cdots \alpha _{k-1}=\beta _1\cdots \beta _{k-1}$ and $\alpha _k>\beta _k$. An admissible sequence $\alpha$ is said to be maximal if ${\mathcal {S}}^n(\alpha )\le \alpha$ for any $n\le |\alpha |-1$ (for all n if $|\alpha |=\infty$).

Let $f\in C([a,b])$. We say that f is unimodal if there is $a<c<b$ such that both $f|_{[a,c]}$ and $f|_{[c,b]}$ are strictly monotone and c is a turning point of f, when f is called upper-unimodal (respectively, lower-unimodal) if c is a maximum (respectively, a minimum). If additionally $\{f^2(c),f(c)\}=\{a,b\}$, then we say that f is strictly unimodal.

The kneading invariant of a unimodal map f, K(f), is an admissible sequence defined as follows. If $f^n(c)\ne c$ for any positive integer n, then $|K(f)|=\infty$ and $K(f)_n$ equals L or R according to whether f is increasing or decreasing on $f^n(c)$ (that is, whether $f^n(c)<c$ or $f^n(c)>c$ when f is upper-unimodal, and whether $f^n(c)>c$ or $f^n(c)<c$ when f is lower-unimodal). If, otherwise, k is the first index such that $f^k(c)=c$, then $|K(f)|=k$ and $K(f)_n$ ($1\le n<k$) equals L or R according to whether f is increasing or decreasing on $f^n(c)$, with $K(f)_k=C$. It can be proved, although this will be of no consequence here, that K(f) is always maximal. Observe that if $K(f)=R\cdots$, then there is exactly one fixed point on which f is decreasing. We call it the essential fixed point of f. If, moreover, $K(f)=RL\cdots$, and $f^3(c)$ belongs to the interval with endpoints p and f(c), then we say that f is renormalizable. If f is renormalizable, then the interval with endpoints $f^2(c)$ and p is invariant for $f^2$ and the restriction g of $f^2$ to this interval is unimodal: we call g the renormalization of f and write $g=N(f)$. Observe that if f is upper-unimodal then g is lower-unimodal, and conversely. Of course, the renormalization of a map f needs not be renormalizable itself: if the n-iteration of the operator N, $N^n$, is well defined on f for all n, then we call f infinitely renormalizable.

Lemma B.3

Let $f\in C([a,b])$ be a unimodal map and assume that $K(f)={\mathcal {R}}(\alpha )$ for some admissible sequence $\alpha =RL\cdots \ne RL^\infty$. Then f is renormalizable and $K(N(f))=\alpha$.

Proof

Assume, for instance, that f is upper-unimodal. We have $\beta =K(f)=RLRR\cdots$, so to prove that f is renormalizable we just need to show that $f^3(c)> p$. But this is clear, because otherwise the first index $k\ge 5$ such that $\beta _k=L$ (such an index does exist because $\alpha \ne RL^\infty$) would be odd, contradicting that $\beta ={\mathcal {R}}(\alpha )$. The second statement of the lemma easily follows from the first one, because if f is renormalizable, then $f^n(x)\ge p$ for any odd integer n and any $x\in [f^2(c),p]$.

The Sharkovsky order is a total order $\lhd$ in the set $\mathbb {Z}^+\cup \{2^\infty \}$ defined as follows:

$$\begin{aligned}&1\lhd 2\lhd 4\lhd \cdots \lhd 2^n\lhd \cdots \lhd 2^\infty \lhd \cdots \lhd \cdots \lhd \\&\quad (2k+1)2^n\lhd \cdots \lhd 7\cdot 2^n \lhd 5\cdot 2^n \lhd 3\cdot 2^n \lhd \cdots \lhd \\&\quad (2k+1)4\lhd \cdots \lhd 7\cdot 4 \lhd 5\cdot 4 \lhd 3\cdot 4 \lhd \cdots \lhd \\&\quad (2k+1)2\lhd \cdots \lhd 7\cdot 2 \lhd 5\cdot 2 \lhd 3\cdot 2 \lhd \cdots \lhd \\&\quad 2k+1\lhd \cdots \lhd 7 \lhd 5 \lhd 3. \end{aligned}$$

We say that $f\in C(I)$ is of type $t\in \mathbb {Z}^+\cup \{2^\infty \}$ (in the Sharkovsky order), and then we write $T(f)=t$, if the set of periods of all periodic points of f is exactly $\{r\in \mathbb {Z}^+:r\unlhd t\}$. According to the famous Sharkovsky theorem (see, e.g., Ruette 2017, Theorem 3.13, p. 40), every $f\in C(I)$ has a type. Therefore, the type of f is larger than, equal to, or smaller than $2^\infty$ according to, respectively, f has some periodic point of period not a power of 2 —equivalently, $h(f)>0$—, f has periodic points of periods all powers of 2, and no other periods, or the set of periods of periodic points of f is bounded.

Let $\omega =\omega _1\omega _2\dots$ denote the infinite sequence given by $\omega _n=R$ or $\omega _n=L$ according to (after writing $n=r 2^m$ for some odd number r and some nonnegative integer m) whether m is even or odd. Clearly,

$$\begin{aligned} \omega =RLRRRLRLRLRRRLRRRLRRRLRLRLRRRLRL\cdots \end{aligned}$$

is characterized by the property ${\mathcal {R}}(\omega )=\omega$.

Lemma B.4

Let $f\in C([a,b])$ be a unimodal map. Then the following statements are equivalent:

• (i) $T(f)=2^\infty$;

• (ii) f is infinitely renormalizable;

• (iii) $K(f)=\omega$.

Proof

We assume, without loss of generality, that f is upper-unimodal.

(i) $\Rightarrow$ (ii): if $f(c)\le c$ then $T(f)=1$ and if $f^2(c)\ge c$ then $T(f)\unlhd 2$, contradicting the hypothesis. Finally, if $f^3(c)<p$, with p the essential fixed point of f, then $f^2([f^2(c),c])\cap f^2([c,p])\supset [f^2(c),p]$. This, as we mentioned before, means that $T(f)\rhd 2^\infty$, again a contradiction. Therefore, f is renormalizable. But if $T(f)=2^\infty$, then it is clear that $T(N(f))=2^\infty$. Repeating the previous argument, we conclude that f is infinitely renormalizable.

(ii) $\Rightarrow$ (i): If f is renormalizable, then it cannot have periodic points of odd period (except 1), that is, $T(f)\unlhd 6$. Analogously, $T(N(f))\unlhd 6$ because N(f) is renormalizable, hence $T(f)\unlhd 12$ and, in general, $T(f)\unlhd 3\cdot 2^n$ for any n, that is, $T(f)\unlhd 2^\infty$. On the other hand, if $T(f)= 2^k$, then $T(N(f))=2^{k-1}$ and, by induction, $T(N^k(f))=1$. But if $g=N^k(f)$, then, because g is renormalizable, the signs of the numbers $g^2(c)-c$ and $g^2(g^2(c))-g^2(c)$ differ, hence there is q between $g^2(c)$ and c such that $g^2(q)=q$ (and $g(q)\ne q$), contradicting $T(g)=1$. Thus $T(f)=2^\infty$, as we desired to prove.

(ii) $\Rightarrow$ (iii): If $\beta =K(f)$ and $\alpha =K(N(f))$, then $\beta _{2m-1}=R$ and $\alpha _m=\overline{\beta _{2m}}$ for all $m\ge 1$. But N(f) is renormalizable as well, so $\alpha _{2m-1}=R$, that is, $\beta _{2(2m-1)}=L$ for all $m\ge 1$. In fact N(f), as f, is infinitely renormalizable, and we can analogously get $\alpha _{2(2m-1)}=L$ and then $\beta _{4(2m-1)}=R$ for all m. Proceeding in this way we conclude $\beta =\omega$.

(iii) $\Rightarrow$ (ii): This follows immediately from Lemma B.3.

Lemma B.5

Let $f\in C^3([a,b])$ be unimodal with turning point c and assume $f''(c)\ne 0$ and $Sf(x)<0$ for any $x\ne c$. If the orbit of c does not accumulate at c and is not attracted by a stable periodic orbit, then f has an acip.

Proof

Assume as usual that f is upper-unimodal. Since the orbit of c is not attracted by any stable periodic orbit, it is easy to check that $f^2(c)<c<f(c)$, and $f^3(c)\ge f^2(c)$. Therefore, $g=f|_{[f^2(c),f(c)]}$ is a well defined strictly unimodal map, and negative Schwarzian derivative, together with Theorem 6.1 in (de Melo and van Strien 1993, p. 145), guarantee that all periodic points of g are unstable, which after adding the hypothesis that the orbit of c does not accumulate at c means the g satisfies the Misiurewicz condition. Therefore, by Misiurewicz (1981a) or van Strien (1990) (here $g''(c)\ne 0$ is needed), g has an acip. This easily implies (just giving zero measure to $[a,b]{\setminus } [f^2(c),f(c)]$) that f has an acip as well.

Proof of Theorem B.1

To prove the first part of the theorem we assume, without loss of generality, that $h(f_{\gamma _0})=0<h(f_{\gamma _1})$. We claim that there is $\gamma _0\le \gamma _0'<\gamma _1$ such that $h(f_{\gamma _0'})=0$ (with $T(f_{\gamma _0'})=2^\infty$) and $h(f_{\gamma })>0$ for any $\gamma \in (\gamma _0',\gamma _1]$. To prove the claim, let $\psi _{a,b}:[a,b]\rightarrow [-1,1]$ be the affine diffeomorphism given by $\psi _{a,b}(x)=-1+2(x-a)/(b-a)$, $a<b$, and write $g_\gamma =\psi _{a(\gamma ),b(\gamma )}\circ f_\gamma \circ \psi _{a(\gamma ),b(\gamma )}^{-1}$. Then (i) and (ii) just mean that $\gamma \mapsto g_\gamma$ maps continually the interval $[\gamma _0,\gamma _1]$ into the subspace of $C^1([-1,1])$ (endowed with the $C^1$-topology) of $C^1$-selfmaps on $[-1,1]$ with at most two pieces of monotonicity. But topological entropy is continuous in this space (Misiurewicz 1995), hence there is a parameter $\gamma _0\le \gamma _0'<\gamma _1$ such that $h(g_{\gamma _0'})=0$ and $h(g_{\gamma })>0$ for any $\gamma \in (\gamma _0',\gamma _1]$. Moreover, the set of $C^1$-maps of type less than $2^\infty$ is open in the $C^1$-topology (Misiurewicz 1981b), so $T(f_{\gamma _0'})=2^\infty$. Since $h(f_\gamma )=h(g_\gamma )$ and $T(f_\gamma )=T(g_\gamma )$ for all $\gamma$, the claim follows.

As a consequence, all maps $f_{\gamma }$, $\gamma \in [\gamma _0',\gamma _1]$, are unimodal, that is, $c(\gamma )$ exists and is not an endpoint of $[a(\gamma ),b(\gamma )]$; moreover, $c(\gamma )$ lies between $f_\gamma ^2(c(\gamma ))$ and $f_\gamma (c(\gamma ))$, that is, $K(f_\gamma )=RL\cdots$. Additionally, (i) and (ii) imply that the sets $\Gamma _{\text {up}}$ and $\Gamma _{\text {low}}$ of parameters $\gamma$ with $f_\gamma$ being, respectively, upper- or lower-unimodal, are both open in $[\gamma _0',\gamma _1]$. Therefore, by connectedness, either $\Gamma _{\text {up}}=[\gamma _0',\gamma _1]$ or $\Gamma _{\text {low}}=[\gamma _0',\gamma _1]$. We assume, without loss of generality, that all these maps are upper-unimodal. The continuity conditions (i) and (ii) also imply that the map $c:[\gamma _0',\gamma _1]\rightarrow \mathbb {R}$ is continuous, and so are the maps $\tilde{a}(\gamma )=f_\gamma ^2(c(\gamma ))$, $\tilde{b}(\gamma )=f_\gamma (c(\gamma ))$. Let $\varphi _c:[-1,1]\rightarrow [-1,1]$ be the increasing diffeomorphism defined by $\varphi _c(x)=(x-c)/(1-cx)$, $-1<c<1$, and redefine, for any $\gamma \in [\gamma _0',\gamma _1]$, $g_\gamma \in C^3([-1,1])$ by

$$\begin{aligned} g_\gamma =\phi _\gamma \circ f_\gamma |_{[\tilde{a}(\gamma ),\tilde{b}(\gamma )]}\circ \phi _\gamma ^{-1}, \end{aligned}$$

where we mean $\phi _\gamma =\varphi _{\tilde{c}(\gamma )}\circ \psi _{\tilde{a}(\gamma ),\tilde{b}(\gamma )}$, with $\tilde{c}(\gamma )=\psi _{\tilde{a}(\gamma ),\tilde{b}(\gamma )}(c(\gamma ))$. Once again, $\gamma \mapsto g_\gamma$ maps continually $[\gamma _0',\gamma _1]$ into $C^1([-1,1])$, the maps $g_\gamma$ are strictly unimodal and have 0 as its critical point (with $(g_\gamma )''(0)\ne 0$ by (iii)). Notice that the maps $\psi _{a,b}$, $\varphi _c$ have zero Schwarzian derivative: since the composition of maps with nonpositive Schwarzian derivative has negative Schwarzian derivative provided that one of then has (de Melo and van Strien 1993, p. 144), the maps $g_\gamma$ have negative Schwarzian derivative by (iv).

Let $\chi =\chi _1\chi _2\cdots$ be an arbitrary infinite sequence of 0’s and 1’s, and let $\alpha =\alpha (\chi )=RLRRL\alpha _1\alpha _2\cdots$, where $\alpha _n=RLRR$ or $\alpha _n=RR$ according to, respectively, $\chi _n=0$ or $\chi _n=1$. Also, let $\beta =(RLR)^\infty$. It is easy to check that both $\alpha$ and $\beta$ are maximal and ${\mathcal {R}}(\alpha )<\beta$. Fix $m\ge 1$ such that $3\cdot 2^m\lhd T(g_{\gamma _1})=T(f_{\gamma _1})$. As shown in Collet and Eckmann (1980) (proof of Theorem II.3.10, p. 92), ${\mathcal {R}}^m(\beta )<K(g_{\gamma _1})$. Now Proposition II.2.2 in (Collet and Eckmann 1980, p. 74) implies ${\mathcal {R}}^{m+1}(\alpha )<{\mathcal {R}}^m(\beta )$, hence ${\mathcal {R}}^{m+1}(\alpha )<K(g_{\gamma _1})$. Similarly, we have $\omega <\alpha$ and then $\omega ={\mathcal {R}}^{m+1}(\omega )<{\mathcal {R}}^{m+1}(\alpha )$. Since $K(g_{\gamma _0'})=K(f_{\gamma _0'})=\omega$ by Lemma B.4, and ${\mathcal {R}}^{m+1}(\alpha )$ is maximal (Collet and Eckmann 1980, Corollary II.2.4, p. 75), we can apply Theorem III.1.1 in (Collet and Eckmann 1980, p. 173) (here the continuity of $\gamma \mapsto g_\gamma$ in the $C^1$-topology is essential) to find $\gamma _0'<\gamma <\gamma _1$ such that $K(g_{\gamma })={\mathcal {R}}^{m+1}(\alpha )$.

Let $g=g_{\gamma }$ and $\tilde{g}=N^{m+1}(g)$, which is well defined by Lemma B.3 (with $K(\tilde{g})=\alpha$) and assume, for instance, that m is odd, that is, $\tilde{g}$ is upper-unimodal. Observe that 0 is still the critical point of $\tilde{g}$, let p be the essential fixed point of $\tilde{g}$ and let $v=\tilde{g}(0)$, $u=\tilde{g}^2(0)$, when $K(\tilde{g})=\alpha$ implies $u<0<v$. Observe that if U is a sufficiently small neighbourhood of 0 and $y\in U$, then $\tilde{g}(y), \tilde{g}^3(y),\tilde{g}^4(y)>0$ and $\tilde{g}^2(y),\tilde{g}^5(y)<0$, so $\tilde{g}^n(0)\notin U$ for any $n\ge 1$ and the $\tilde{g}-$orbit of 0 does not accumulate at 0. In fact, because of the way the renormalization operator is defined, the g-orbit of 0 cannot accumulate at 0 either.

At this point recall that $\alpha =\alpha (\chi )$ and assume that one of the following possibilities holds: (a) $\chi$ is not eventually periodic; (b) $\chi$ is eventually 1. In we are in case (a), then the $\tilde{g}$-orbit of 0 cannot be attracted by any periodic orbit, and the same thing can be said for $g=g_\gamma$. By Lemma B.5, $g_\gamma$ has an acip $\nu$, which becomes an acip $\mu$ for $f_\gamma$ by writing $\mu (B)=\nu (\phi _\gamma (B))$ if B is a Borel subset of $[\tilde{a}(\gamma ),\tilde{b}(\gamma )]$ and giving $\mu$-zero measure to $[a(\gamma ),b(\gamma )]{\setminus } [\tilde{a}(\gamma ),\tilde{b}(\gamma )]$. In other words, $\gamma \in \Lambda$. But there are uncountably many sequences $\chi$ satisfying (a), so $\Lambda$ is uncountable.

In case (b) we can say more: $\tilde{g}$ satisfies the strong Misiurewicz condition. Indeed, let $q\in (p,v)$ be such that $\tilde{g}(q)=0$. Then $\tilde{g}^2$ has positive derivative in (0, q) and $\tilde{g}^2(0)=u<0$, $\tilde{g}^2(q)=v>q$. Therefore, according to the minimum principle (de Melo and van Strien 1993, Lemma 6.1, p. 144) (which establishes that if a map has negative Schwarzian derivative in a compact interval, then the absolute value of its derivative attains its minimum value at some endpoint of the interval), $(\tilde{g}^2)'(p)>1$, and $\tilde{g}^2(y)<y$ (respectively, $\tilde{g}^2(y)>y$) for any $y\in (0,p)$ (respectively, for any $y\in (p,q)$. Now, realize that the kneading sequence of $\tilde{g}$ guarantees that $\tilde{g}^n(0)$ stays at (0, q) for every n large enough, which is impossible unless 0 is eventually mapped by $\tilde{g}$ to p, that is, the strong Misiurewicz condition is satisfied (when the same happens to $g_\gamma$ and $f_\gamma$). Since there are infinitely many sequences $\chi$ satisfying (b), we conclude that $\Lambda _S$ is infinite, which finishes the proof of the first part of Theorem B.1.

Now we prove the second part of Theorem B.1. Recall, first of all, that $f_\rho$ has an acip and then $h(f_\rho )>0$, so (by the continuity of entropy) $h(f_\gamma )>0$ if $\gamma$ is close to $\rho$, when the strictly unimodal maps $g_\gamma \in C^3([-1,1]$ with $g_\gamma (0)=1$, $g_\gamma (1)=-1$, can be defined as above. Further, the additional hypothesis on F and (iii) imply that $c(\gamma )$ is $C^2$ near $\rho$, so each $G_n(y,\gamma )=g_\gamma ^n(y)$ is $C^2$ in an open set containing a small rectangle $[-1,1]\times [\rho -\epsilon ,\rho +\epsilon ]$. On the other hand, the conditions $q=p(\rho )$ and $f_\gamma ^r(p(\gamma ))=p(\gamma )$, together with $(f_\rho ^r)'(q)\ne 1$, define uniquely a continuous (in fact, $C^3$) map $p(\gamma )$ near $\rho$. (This works even if $q=\tilde{a}(\rho )$ or $q=\tilde{b}(\rho )$, because the maps $f_\gamma ^r$ can be seen as defined on slightly larger open intervals than $[\tilde{a}(\gamma ),\tilde{b}(\gamma )]$.)

We claim that if $\gamma$ is close enough to $\rho$, then $g_\gamma (x)>x$ for any $x\in (-1,0]$. If $g_\rho (-1)>-1$, then the claim follows for $\gamma =\rho$ from the strong Misiurewicz condition, and for all $\gamma$ near $\rho$ from the continuity of the family. Now assume $g_\rho (-1)=-1$, when $g_\rho '(-1)>1$ by the strong Misiurewicz condition and then $g_\gamma '(x)>1$ whenever x is close enough to $-1$ and $\gamma$ is close enough to $\rho$. Then the claim follows from the minimum principle.

Let $z(\gamma )=F_k(c(\gamma ),\gamma )$ and realize that (B.10) amounts to say (because $(\partial F_k/\partial x)(d,\rho )=0$) that $z'(\rho )\ne p'(\rho )$. Write $\phi (x,\gamma )=\phi _\gamma (x)$, $Z(\gamma )=\phi (z(\gamma ),\gamma )=G_k(0,\gamma )$, $P(\gamma )=\phi (p(\gamma ),\gamma )=G_r(P(\gamma ),\gamma )$. Then

$$\begin{aligned} Z'(\rho )-P'(\rho )= \frac{\partial \phi }{\partial x}(q,\rho )(z'(\rho )-p'(\rho ))\ne 0 \end{aligned}$$

and all conditions in (Thieullen et al. 1994, Theorem I.3) are satisfied, which ensures that $\rho$ is a Lebesgue density point of a Borel set $\Sigma \subset (\gamma _0',\gamma _1]$ such that $g_\gamma$ (and then $f_\gamma$) has an acip whenever $\gamma \in \Sigma$. This finishes the proof. $\square$