One-Dimensional Nonlinear Cobweb Model

Abstract

This chapter provides an introductory exposition of nonlinear discrete dynamics, which takes readers into the vast ocean outside the equilibrium paradigm, by presenting a one-dimensional nonlinear cobweb model. The model is a simple extension of the standard cobweb model. The new ingredients are an iso-elastic inverse demand function (i.e., an inverse demand function with constant price elasticity) and boundedly rational producers who gradually adjust their production toward the target levels based upon naive price expectations. Thus, the key parameters of the model are the price elasticity of demand and the production adjustment speed of producers. With the aid of mathematical analysis and numerical simulations, it is shown that for a large set of parameter values, the cobweb market exhibits observable chaos (a strange attractor) as well as topological chaos (a horseshoe) associated with homoclinic points.

Nature creates curved lines while humans create straight lines.

Hideki Yukawa Kyokubi no Sekai [The Ultramicroscopic World] (1950)

Nature abhors a straight line. William Kent

The orginal version of this chapter was revised: Incorect figure labelling has been corrected. The eratum to this chapter is available at https://doi.org/10.1007/978-4-431-54971-0_8

Appendix

A.1 Proof of Proposition 2.1

Before proving Proposition 2.1, we prove the following lemma.

Lemma 2.2

$f^{\, 2}(x)<x$for any $x>1$.

Proof

Transforming (2.16) as

$$\begin{aligned} \phi =\frac{\theta ^{1+\eta }}{\theta ^{1+\eta }+\eta } \end{aligned}$$

and substituting it into (2.15) and (2.17) yields

$$\begin{aligned} f(x)=\left( \frac{\eta }{\theta ^{1+\eta }+\eta }\right) x+\left( \frac{\theta ^{1+\eta }}{\theta ^{1+\eta }+\eta }\right) \frac{1}{x^{\eta }} \end{aligned}$$

(2.20)

and

$$\begin{aligned} \frac{\eta (1-\theta ^{1+\eta })}{\theta ^{1+\eta }+\eta } > -1. \end{aligned}$$

(2.21)

By using (2.20), the second iteration of f can be calculated as follows:

$$\begin{aligned} f^{\, 2}(x)= & {} f(f(x)) \\= & {} \frac{\eta }{\theta ^{1+\eta }+\eta }f(x)+\frac{\theta ^{1+\eta }}{\theta ^{1+\eta }+\eta }\frac{1}{f(x)^{\eta }} \\= & {} \frac{\eta }{\theta ^{1+\eta }+\eta }\left( \frac{\eta }{\theta ^{1+\eta }+\eta }x+\frac{\theta ^{1+\eta }}{\theta ^{1+\eta }+\eta }\frac{1}{x^{\eta }}\right) \\&+\frac{\theta ^{1+\eta }}{\theta ^{1+\eta }+\eta }\frac{1}{\left( \frac{\eta }{\theta ^{1+\eta }+\eta }x+\frac{\theta ^{1+\eta }}{\theta ^{1+\eta }+\eta }\frac{1}{x^{\eta }} \right) ^{\eta }} \\= & {} \frac{\eta x}{\theta ^{1+\eta }+\eta }\left( \frac{\eta }{\theta ^{1+\eta }+\eta }+\frac{1}{\theta ^{1+\eta }+\eta }\left( \frac{\theta }{x}\right) ^{1+\eta }\right) \\&+\frac{x\left( \frac{\theta }{x}\right) ^{1+\eta }}{\theta ^{1+\eta }+\eta }\frac{1}{\left( \frac{\eta }{\theta ^{1+\eta }+\eta }+\frac{1}{\theta ^{1+\eta }+\eta }\left( \frac{\theta }{x} \right) ^{1+\eta } \right) ^{\eta }}. \end{aligned}$$

Now let

$$\begin{aligned} t=\frac{\eta }{\theta ^{1+\eta }+\eta }+\frac{1}{\theta ^{1+\eta }+\eta }\left( \frac{\theta }{x}\right) ^{1+\eta }, \end{aligned}$$

so that

$$\begin{aligned} f^{\, 2}(x)=xF(t), \end{aligned}$$

where

$$\begin{aligned} F(t)=\frac{\eta }{\theta ^{1+\eta }+\eta }t+\frac{1}{t^{\,-1+\eta }}-\frac{\eta }{(\theta ^{1+\eta }+\eta )\, t^{\,\eta }}. \end{aligned}$$

Since $x>1$,

$$\begin{aligned} \frac{\eta }{\theta ^{1+\eta }+\eta }< t < 1. \end{aligned}$$

Therefore, it suffices to show $F(t)<1$ for any $\frac{\eta }{\theta ^{1+\eta }+\eta }< t < 1$.

Now, we calculate the first and second derivatives of F : 

$$\begin{aligned} F^{\prime }(t)= & {} \frac{\eta }{\theta ^{1+\eta }+\eta }-\frac{(-1+\eta )}{t^{\,\eta }}+\frac{\eta ^{2}}{(\theta ^{1+\eta }+\eta )t^{1+\eta }} \\ F^{\prime \prime }(t)= & {} \frac{(-1+\eta )\,\eta }{t^{1+\eta }}-\frac{\eta ^{2}(1+\eta )}{(\theta ^{1+\eta }+\eta )\,t^{\, 2+\eta }} \\= & {} \frac{\eta \, (-1+\eta )}{t^{\, 2+\eta }}\left\{ t-\frac{\eta \, (1+\eta )}{(-1+\eta )(\theta ^{1+\eta }+\eta )}\right\} . \end{aligned}$$

Case I: $\eta \in (0,1]$

In this case, since $F^{\prime }(t)>0$ for any $t>0$, F is strictly increasing on $[0, \infty )$. Hence

$$\begin{aligned} F(t)<F(1)=1\quad \text {for any}\ \ t<1. \end{aligned}$$

Case II: $\eta \in (1,\infty )$

From simple calculations, (2.21) is equivalent to the condition

$$\begin{aligned} \frac{\eta (1+\eta )}{(-1+\eta )(\theta ^{1+\eta }+\eta )}\ge 1. \end{aligned}$$

Therefore, $F''(x)>0$ for any $\frac{\eta }{\theta ^{1+\eta }+\eta }<t<1$, so that $F'$ is strictly decreasing on $\left[ \frac{\eta }{\theta ^{1+\eta }+\eta }, 1\right]$. Since $F^{\prime }(1)=(-1+\eta )\left\{ \frac{\eta (1+\eta )}{(\theta ^{1+\eta }+\eta )(-1+\eta )}-1\right\} \ge 0$,

$$\begin{aligned} F^{\prime }(t)>0\quad \text {for any}\ \ \frac{\eta }{\theta ^{1+\eta }+\eta }<t<1. \end{aligned}$$

Hence F is strictly increasing on $\left[ \frac{\eta }{\theta ^{1+\eta }+\eta }, 1\right]$, so that

$$\begin{aligned} F(t)<F(1)=1\quad \text {for any}\ \ \frac{\eta }{\theta ^{1+\eta }+\eta }<t<1, \end{aligned}$$

which completes the proof. $\square$

Proof of Proposition 2.1.    The function f has the following properties:

• P1. $f(x)\ge f(\theta )$ for any $x>0$,

• P2. f is strictly decreasing on $(0,\theta ]$, and

• P3. f is strictly increasing on $[\theta ,\infty )$ with

  $$\begin{aligned} 0 \le f^{\prime }(x)=\frac{\eta }{\theta ^{1+\eta }+\eta }\left( 1-\frac{\theta ^{1+\eta }}{x^{1+\eta }} \right)<\frac{\eta }{\theta ^{1+\eta }+\eta } < 1. \end{aligned}$$

By P1, $f({\mathbb R_{++})}=[f(\theta ), \infty )$. Thus, it suffices to show that

$$\begin{aligned} \lim _{n\rightarrow \infty }f^{n}(x) = 1\quad \text {for any}\ \ x \in [f(\theta ), \infty ). \end{aligned}$$

Let $x \in [f(\theta ) , \infty )$; then, $f^n(x) \in [f(\theta ),\infty )$ for any positive integer n by P1.

Case I: $0<\theta \le 1$

In this case, $\frac{(1+\eta )}{\theta ^{1+\eta }+\eta }>1$, so that $\theta \le f(\theta )=\frac{(1+\eta )}{\theta ^{1+\eta }+\eta }\theta \le 1$. Hence, $f^n(x) \in [f(\theta ), \infty )\subset [\theta , \infty )$ for any nonnegative integer n. By the mean value theorem and P3,

$$\begin{aligned} \left| 1-f^n(x) \right|= & {} \left| f^n(1)-f^n(x) \right| = \left| f'(c) \right| \left| f^{n-1}(1)-f^{n-1}(x) \right| \\< & {} \frac{\eta }{\theta ^{1+\eta }+\eta } \left| f^{n-1}(1)-f^{n-1}(x) \right| , \end{aligned}$$

where $c \in \left( 1, f^{n-1}(x) \right)$ or $c \in \left( f^{n-1}(x), 1 \right)$. Following the same argument,

$$\begin{aligned} |1-f^n(x)|< & {} \frac{\eta }{\theta ^{1+\eta }+\eta } \left| f^{n-1}(1)-f^{n-1}(x) \right| \\< & {} \cdots < \left( \frac{\eta }{\theta ^{1+\eta }+\eta } \right) ^n|1-x\,| \end{aligned}$$

This implies

$$\begin{aligned} \lim _{n\rightarrow \infty } \left| 1-f^n(x) \right| = 0 \ \ \ \Leftrightarrow \ \ \ \lim _{n\rightarrow \infty }f^n(x)=1. \end{aligned}$$

Case II: $\theta \ge 1$

We have the following two possibilities:

• (i) $1 \le f^{2n}(x)$ for any nonnegative integer n, or

• (ii) There is some nonnegative integers k such that $f(\theta ) \le f^{2k}(x)<1$.

In the first case (i), by Lemma 2.2,

$$\begin{aligned} 1<\cdots<f^{2n}(x)<f^{2n-2}(x)<\cdots<f^{2}(x)<x, \end{aligned}$$

so that a sequence $\{f^{2n}(x)\}$ converges to some number $a \ge 1$. It is clear that the number a is 1, otherwise

$$\begin{aligned} a >f^{2}(a)= & {} f^{2}\left( \lim _{n \rightarrow \infty }f^{2n}(x)\right) \\= & {} \lim _{n \rightarrow \infty }f^{2n+2}(x) = a, \end{aligned}$$

which is a contradiction.

We show that the sequence $\{f^n(x)\}$ converges to 1. Let $\varepsilon > 0$. By the continuity of f at 1, there is a $\delta ' > 0$ such that

$$\begin{aligned} | 1 - y\, |< \delta ' \ \ \ \Rightarrow \ \ \ | f(1) - f(y) | < \varepsilon . \end{aligned}$$

Let $\delta = \min \{\varepsilon , \delta '\}$. Since $\lim _{n \rightarrow \infty }f^{2n}(x)=1$, there is a natural number N such that for any $2n \ge N$

$$\begin{aligned} \left| f(1)-f^{2n}(x) \right| < \delta . \end{aligned}$$

This implies that

$$\begin{aligned} \left| 1 - f^{2n+1}(x) \right| = \left| f(1) - f^{2n+1}(x) \right| = \left| f(1) - f(f^{2n}(x)) \right| < \varepsilon . \end{aligned}$$

Therefore, for any $m \ge N$

$$\begin{aligned} \left| 1 - f^m(x) \right| < \varepsilon , \end{aligned}$$

that is, $\lim _{n \rightarrow \infty }f^{m}(x)=1$.

In the second case (ii), that is, $f^{2k}(x) \in [f(\theta ), 1)$, let $y=f^{2k+1}(x)$. Then, by P2, we obtain

$$\begin{aligned} f([f(\theta , 1))= & {} (1, f^2(\theta )]\\ f^2((1, \theta ])= & {} f([f(\theta ), 1))=(1, f^2(\theta )]. \end{aligned}$$

Hence $1< y \le f^2(\theta )$ and $1< f^2(y) \le f^2(\theta )$. By Lemma 2.2, $1< y \le \theta$ and $1<f^2(y) \le y$.

Inductively we have

$$\begin{aligned} 1< \cdots< f^{2n}(y)< f^{2n-2}(y)< \cdots< f^{2}(y) < y. \end{aligned}$$

As in the first case, a sequence $\{f^n(y)\}$, that is, $\{f^n(x)\}$ converges to 1 as $n \rightarrow \infty$. $\square$

A.2 Proof of Proposition 2.4

By Theorem 2.12 in Sect. 2.4.3, it suffices to show that the map f has a nondegenerate homoclinic orbit. To show this, we must make some preparations.

Lemma 2.3

The following statements hold:

• 1. $0<f(\theta )<1<\theta$ , and

• 2. There is a unique point $q=q(\eta )>\theta$ such that $f(q)=1$.

Proof

(i) From (2.18), we get the last inequality. Since f has its global minimum at $\theta$, we get the second inequality (see Fig. 2.4). (ii) Since $f(\theta )<1$ and $f(x)>1$ for large x, there is $q>\theta$ with $f(q)=1$. Since f(x) is strictly increasing on the interval $(\theta ,\infty )$, uniqueness follows (see Fig. 2.4). $\square$

Let us define a piecewise linear map $L_{\eta }:\mathbb {R}\rightarrow \mathbb {R}$ by

$$\begin{aligned} L_{\eta }(x) := {\left\{ \begin{array}{ll} \ l_{1}(x)=f^{\prime }(1)(x-1)+1, \quad &{} x \le 1, \\ \ l_{2}(x)=-(x-1)+1, &{} x > 1. \end{array}\right. } \end{aligned}$$

Clearly, $L(1)=1$ and $L^{-n}(x) \rightarrow 1$ ($n \rightarrow \infty$) for each $x \in \mathbb {R}$ (see Fig. 2.20).

Fig. 2.20

Graph of the map (2.15) and the piecewise linear map $L_{\eta }(x)$

Lemma 2.4

There is a number $\eta _{1}>(2-\phi )/\phi$ such that for every $\eta \ge \eta _{1}$ and for every $x \in I := [f(\theta ),\theta ]$ there is a unique sequence $\{ x_{-i}\} _{i=0}^{\infty }\subset I$ such that $x_{0} = x$, $f(x_{-i-1}) = x_{-i}$ for $i \ge 0$, and $x_{-i} \rightarrow 1$ as $i \rightarrow \infty$.

Proof

Since f is one-to-one on the interval $[f(\theta ),\theta ]$, the conclusion holds if given sufficiently large $\eta$, $f(x)>l_{1}(x)$ for $x\in [f(\theta ), 1)$ and $f(x)<l_{2}(x)$ for $x \in (1,\theta ]$. By the strict convexity of f and the construction of L, it is sufficient to show that for a sufficiently large $\eta$, the inequality $f(\theta )<l_{2}(\theta )$, that is,

$$\begin{aligned} f(\theta (\eta ))+\theta (\eta )<2 \end{aligned}$$

holds. This is verified by the fact

$$\begin{aligned} \lim _{\eta \rightarrow \infty }\theta (\eta )&= \lim _{\eta \rightarrow \infty }\left( \frac{\phi \eta }{1-\phi } \right) ^{\frac{1}{\eta +1}}=1, \\ \lim _{\eta \rightarrow \infty }f(\theta (\eta ))&= \lim _{\eta \rightarrow \infty }\left( (1-\phi )\theta (\eta )+\phi \theta (\eta )^{-\eta } \right) \\&= 1-\phi \end{aligned}$$

which completes the proof. $\square$

Lemma 2.5

There is a number $\eta _{2}>(2-\phi )/\phi$ such that for every $\eta \ge \eta _{2}$, the following inequality holds:

$$\begin{aligned} 0< f(\theta (\eta ))< 1<\theta (\eta )< q(\eta ) < f^{\, 2}(\theta (\eta )). \end{aligned}$$

Proof

By Lemma 2.3, it suffices to show that for any arbitrarily large $\eta$, the following inequality holds:

$$\begin{aligned} q<f^{\, 2}(\theta ). \end{aligned}$$

(2.22)

Let us define a function l by

$$\begin{aligned} l(x) := f^{\prime }(1)(x-1)+1=(1-\phi -\phi \eta )(x-1)+1. \end{aligned}$$

Again, by the strict convexity of f, we obtain

$$\begin{aligned} l\circ f(\theta (\eta ))<f^{\, 2}(\theta (\eta )). \end{aligned}$$

Note that $f(x)>(1-\phi )x$ holds for every $x \in \mathbb {R}_{++}$ (see Fig. 2.4), so we have

$$\begin{aligned} q(\eta )<\frac{1}{1-\phi }. \end{aligned}$$

To obtain the inequality (2.22), it is therefore sufficient to show

$$\begin{aligned} \frac{1}{1-\phi }\le l\circ f(\theta (\eta )) \end{aligned}$$

for any sufficiently large $\eta$. Noting that

$$\begin{aligned} l\circ f(\theta (\eta ))=(1-\phi -\phi \eta )\big ( (1-\phi )\theta (\eta )+\phi \theta (\eta )^{-\eta }-1\big ) +1, \end{aligned}$$

we get $\lim _{\eta \rightarrow \infty } \{ l\circ f(\theta (\eta )) \}=\infty$. Hence, the lemma follows. $\square$

Proof of Proposition 2.4. Let $\overline{\eta }=\max (\eta _{1},\eta _{2})$ and pick $\eta \ge \overline{\eta }$ arbitrarily. By Lemma 2.5, there is a point $q^{\prime }\in (f(\theta ), 1)$ such that $f(q^{\prime })=q$. By Lemma 2.4, there is a sequence $\{q_{-i}^{\prime }\}_{i=0}^{\infty }\subset I$ such that $q_{0}^{\prime }=q^{\prime },\, f(q_{-i-1}^{\prime })=q_{-i}^{\prime }\,\,(i\ge 0)$, and $q_{-i}^{\prime }\rightarrow 1$ $(i\rightarrow \infty )$. Hence, together with $f^{\, 2}(q^{\prime })=1$, $q^{\prime }$ is a homoclinic point to the repellor 1. Clearly, for any homoclinic point $z\in \varvec{\mathcal {HO}}_{f}(q^{\prime }, 1)=\{q^{\prime },f(q^{\prime }),f^{\, 2}(q^{\prime })=1\}\cup \{q_{-i}^{\prime }\}_{i=0}^{\infty }$, we have $x\ne \theta$ and so $f^{\prime }(x)\ne 0$, which implies that the homoclinic orbit of $q^{\prime }$ to 1, $\varvec{\mathcal {HO}}_{f}(q^{\prime }, 1)$, is nondegenerate. By Theorem 2.12, the statement is thus proved. $\square$

A.3 Proof of Proposition 2.5

To find the abundance of strange attractors for the family of maps $\{f\}_{\eta }$, we exploit Theorem 2.15.

Proof of Proposition 2.5. We show that the nondegenerate critical point $\theta (\eta )$ is contained in the homoclinic orbit of the repelling fixed point $x^{*}=1$ for some sequence of $\eta$ values. Note first that the critical point $\theta$ is always nondegenerate, that is, $f^{\prime \prime }(\theta )\ne 0$, since $f^{\prime \prime }(x)>0$ for all $x\in \mathbb {R}_{++}$.

We can observe that there is a sequence of eventually fixed points depending smoothly on $\eta$,

$$\begin{aligned} Q(\eta )=\{q_{i}(\eta )\mid q_{i}(\eta )=f(q_{i+1}(\eta ))\,\, \text{ for }\,\,i\in \mathbb {N},\,\, q=q_{1}<q_{2}<...<q_{n}<...\}, \end{aligned}$$

where $q_{i}(\eta )\rightarrow (1-\phi )^{-i}<\infty$ as $\eta \rightarrow \infty$ for every $i\in \mathbb {N}$.

Let us fix $\phi \in (0,1)$ arbitrarily and take $\eta =\eta _{1}$ as in Lemma 2.4. Then, from the observation above, there is $q_{i}(\eta _{1})\in Q(\eta _{1})$ such that $f^{\, 2}(\theta (\eta _{1}))<q_{i}(\eta _{1})$. Since $f(\theta (\eta ))\rightarrow \infty$ as $\eta \rightarrow \infty$ by the proof of Lemma 2.5, and $q_{i}(\eta )\rightarrow (1-\phi )^{-i}$ as $\eta \rightarrow \infty$, there is $\eta ^{*}>\eta _{1}$ such that $f^{\, 2}(\theta (\eta ^{*}))=q_{i}(\eta ^{*})$, which implies that the backward orbit of $\theta (\eta ^{*})$ converges to the repelling fixed point $x^{*}=1$ and there is an integer n such that $f^{\, n}(\theta (\eta ^{*}))=1$ and $f^{m}(\theta (\eta ^{*}))\ne 1$ for $m<n$. Hence, the nondegenerate critical point $\theta (\eta ^{*})$ lies in a homoclinic orbit to the fixed point $x^{*}=1$ (homoclinic tangency). Since $\frac{\partial }{\partial \eta }f(x,\eta )=0$ if and only if $x=1$, we may generically assume that $\frac{d}{d\eta }f^{\, n}(\theta (\eta ))\ne 0$ at $\eta =\eta ^{*}$. By Theorem 2.13, the statement of Proposition 2.5 thus follows. $\square$