Walrasian versus Cournot behavior in an oligopoly of boundedly rational firms

Abstract

An evolutionary oligopoly game, where firms can select between the best-reply rule and the Walrasian rule, is considered. The industry is characterized by a finite number of ex-ante homogeneous firms that, characterized by naïve expectations, decide next-period output by employing one of the two behavioral rules. The inverse demand function is linear and all firms have the same quadratic and convex cost function (decreasing return to scale). Based upon realized profits, the distribution of behavioral rules is updated according to a replicator dynamics. The model is characterized by two equilibria: the Cournot-Nash equilibrium, where all firms adopt the best-reply rule and produce the Cournot-Nash quantity, and the Walrasian equilibrium, where all firms adopt the Walrasian rule and produce the Walrasian quantity. The analysis reveals that the Walrasian equilibrium is globally stable as long as the rate of change of marginal cost exceeds the sum of residual market price sensitivities to output. If not, the Walrasian equilibrium loses stability and an attractor, representing complicated dynamics with evolutionary stable heterogeneity, arises through a bifurcation. As the propensity of firms to select the more profitable behavioral rule increases, the attractor disappears through a global bifurcation and the Cournot-Nash equilibrium can become a global Milnor attractor. To sum up, the best-reply rule can be evolutionary dominant over the Walrasian rule and this can lead an oligopoly to select the Cournot-Nash equilibrium.

Appendix

Proof of Lemma 1

From (13), it is straightforward to see that the steady states of the evolutionary model are obtained for r = 0, r = 1or any r ^∗∈ (0, 1)for which Δπ = 0.With r = 0(all Walrasian firms), map (13) becomes:

$$ T^{0}:=\left\{\begin{array}{l} x\left( t+1\right)=\max\left[\frac{a-b\left( N-1\right)y\left( t\right)}{2\left( b+c\right)},0\right]\\ \\ y\left( t+1\right)=\max\left[\frac{a-bNy\left( t\right)}{2c},0\right] \end{array}\right. $$

(22)

where the second component is uncoupled from the first one, i.e. it is a one-dimensionaldifference equation (master equation) and it is piecewise linear, whereas the first componentdepends only on the second variable (slave equation). Thus, by straightforward algebra, E ₀is the only equilibriumwhen r = 0and bysimple calculations $p_{W}^{*}$and π(E ₀)are obtained.

With r = 1(all best-reply firms), map (13) becomes:

$$ T^{1}:=\left\{\begin{array}{l} x\left( t+1\right)=\max\left[\frac{a-b\left( N-1\right)x\left( t\right)}{2\left( b+c\right)},0\right]\\ \\ y\left( t+1\right)=\max\left[\frac{a-bNx\left( t\right)}{2c},0\right] \end{array}\right. $$

(23)

where the first component is uncoupled from the second one, i.e. it is a one-dimensional differenceequation (master equation) and it is piecewise linear, whereas the second componentdepends only on the first variable (slave equation). Thus, by straightforward algebra, E ₁is the onlyequilibrium when r = 1and defining

$$ K = \frac{bN^{2}+2c\left( N-1\right)}{\left( 2c+b\left( N+1\right)\right)^{2}}b^{2}>0 $$

(24)

by simple calculation $p^{*}_{CN}$and π(E ₁)are obtained.

Let us now investigate the existence of other equilibria (x ^∗, y ^∗, r ^∗)with r ^∗∈ (0, 1). If (x ^∗, y ^∗, r ^∗)∈Ω₁,then

$$ \left\{\begin{array}{l} x^{*}=R\left( x^{*},y^{*},r^{*}\right)\\ y^{*}=W\left( x^{*},y^{*},r^{*}\right)\\ {\Delta}{\Pi}\left( x^{*},y^{*},r^{*}\right)=0 \end{array} \right. $$

(25)

From the third equation of the system we obtain either x ^∗ = y ^∗or$r^{*}=\frac {c\left (x^{*}+y^{*}\right )+bNy^{*}-a}{bN\left (y^{*}-x^{*}\right )}$. Solving the first andthe second equation for x ^∗ = y ^∗,we obtain x ^∗≠y ^∗,which is a contradiction. Analogously, substituting$r^{*}=\frac {c\left (x^{*}+y^{*}\right )+bNy^{*}-a}{bN\left (y^{*}-x^{*}\right )}$in the first and secondequation we obtain $x^{*}=-y^{*}-\frac {a}{\left (N-1\right )c}$and x ^∗ = −y ^∗, respectively. Thus, afurther equilibrium in Ω₁cannot exist. An equilibrium (x ^∗, y ^∗, r ^∗)in Ω₂implies y ^∗ = 0, x ^∗ = R(x ^∗, 0, r ^∗)and Δπ(x ^∗, y ^∗, r ^∗) = 0. Solving,we obtain $x^{*}=\frac {a}{2Nb+cN+c}$and $r^{*}=\frac {2b+c}{b}$. Since r ^∗ > 1, an equilibrium in Ω₂cannot exist. Moreover, theonly possible equilibrium in Ω₃is (0, 0, r ^∗)which does not belongto such region. Thus, E ₀and E ₁are the only possibleequilibria of the model (13). □

Proof of Proposition 1

A straightforward computation allows us to obtain the Jacobian matrix of model (13),when r = 0and (x, y, r) ∈Ω₁,given by

$$ J\left( x,y,0\right)=\left[ \begin{array}{ccc} 0 & \frac{-b\left( N-1\right)}{2\left( b+c\right)} & \frac{b\left( N-1\right)\left( y-x\right)}{2\left( b+c\right)} \\ 0 & \frac{-bN}{2c} & \frac{bN\left( y-x\right)}{2c} \\ 0 & 0 & J_{33}\left( x,y,0\right) \end{array} \right]\quad \text{ where }\quad J_{33}\left( x,y,0\right)=e^{-\beta \left( a-bNy-c\left( x+y\right)\right)\left( y-x\right)} $$

(26)

which is singular and upper triangular. Thus, the associated eigenvalues are the diagonal entriesand the eigenvalues for the Jacobian matrix computed at the Walrasian equilibrium, i.e. J(E ₀),are:

$$ {\lambda^{W}_{1}}=0\text{, }\quad {\lambda^{W}_{2}}=\frac{-bN}{2c}=\frac{\frac{\partial P\left( E_{0}\right)}{\partial y}}{\frac{\partial^{2} C}{\partial y^{2}}}\text{, }{\quad\lambda^{W}_{3}}=J_{33}\left( E_{0}\right)=e^{-\beta c\left( \frac{ab}{\left( 2c+bN\right)2\left( b+c\right)}\right)^{2}} $$

(27)

Imposing the well-known conditions for the local asymptotic stability of a fixed point, i.e. alleigenvalues of J(E ₀)inside the unit circle, we obtain the first part of the proposition.

Similar computation allows us to obtain the Jacobian matrix of model (13), when r = 1and (x, y, r) ∈Ω₁, givenby

$$ J\left( x,y,1\right)=\left[ \begin{array}{ccc} \frac{\partial R\left( x,y,1\right)}{\partial x} & 0 & \frac{b\left( N-1\right)\left( y-x\right)}{2\left( b+c\right)} \\ \frac{-bN}{2c} & 0 & \frac{bN\left( y-x\right)}{2c} \\ 0 & 0 & J_{33}\left( x,y,1\right) \end{array} \right]\quad \text{ where }\quad J_{33}\left( x,y,1\right)=e^{-\beta \left( a-bNx-c\left( x+y\right)\right)\left( y-x\right)} $$

(28)

Thus, the eigenvalues for the Jacobian matrix computed at the Cournot-Nash equilibrium, i.e. J(E ₁),are:

$$ \lambda_{1}^{CN}=\frac{\partial R\left( E_{1}\right)}{\partial x}=\frac{-b\left( N-1\right)}{2\left( b+c\right)} , \quad \lambda^{CN}_{2}=0, \quad \lambda^{CN}_{3}=e^{\beta \frac{a^{2}b^{2}}{4c\left( 2c+b\left( N+1\right)\right)^{2}}} $$

(29)

Noting that $\lambda ^{CN}_{3}>1$,the transverse instability of the Cournot-Nash equilibrium follows. □

Proof of Proposition 2

The restriction of map (13) to the invariant plane r = 0reduces to the two-dimensional map$T^{0}:\mathbb {R}^{2}_{+}\rightarrow \mathbb {R}^{2}_{+}$defined in Eq. 22. Let us divide the plane$\mathbb {R}^{2}_{+}$into the following three regions:

$$ \begin{array}{l} {{\Omega}^{0}_{1}}=\left\{\left( x,y\right)| 0\leq y<\frac{a}{bN}\right\}\\ {{\Omega}^{0}_{2}}=\left\{\left( x,y\right)| \frac{a}{bN}\leq y<\frac{a}{b\left( N-1\right)}\right\}\\ {{\Omega}^{0}_{3}}=\left\{\left( x,y\right)| \frac{a}{b\left( N-1\right)}\leq y\right\} \end{array} $$

(30)

The lines

$$ \tilde{BC}_{1} := y=\frac{a}{bN}\quad \text{ and }\quad \tilde{BC}_{2} := y=\frac{a}{b\left( N-1\right)} $$

(31)

mark the borders of non differentiability between regions${{\Omega }^{0}_{1}}$and${{\Omega }^{0}_{2}}$andbetween ${{\Omega }^{0}_{2}}$and ${{\Omega }^{0}_{3}}$,respectively.

By straightforward considerations, see Lemma 1, it follows that$\tilde {E}_{0}=\left (q^{*}_{W}\frac {b+2c}{2\left (b+c\right )},q^{*}_{W}\right ){\in {\Omega }_{1}^{0}}$, with$q^{*}_{W}=\frac {a}{2c+bN}$, is the uniquefixed point of T ⁰.

For bN < 2c, we have thatall points of region ${{\Omega }^{0}_{2}}$and ${{\Omega }^{0}_{3}}$are mapped inone iteration in ${{\Omega }^{0}_{1}}$. Infact, all points of region ${{\Omega }^{0}_{2}}$are mapped in $\left \{\left (y,x\right )|y=0\text {,} \ x\geq 0\right \}{\in {\Omega }^{0}_{1}}$and allpoints of ${{\Omega }^{0}_{3}}$are mappedin $\left (0,0\right ){\in {\Omega }^{0}_{1}}$. Moreover, themap T ⁰is linear in${{\Omega }^{0}_{1}}$with a unique innerfixed point $\tilde {E}_{0}$. Theeigenvalues associated to $\tilde {E}_{0}$are $\tilde {\lambda }_{1}^{W}=0$and $\tilde {\lambda }_{2}^{W}=\frac {-bN}{2c}$with $-1<\tilde {\lambda }_{2}^{W}<0$.Thus, $\tilde {E}_{0}$is locally asymptotically stable and the image of region${{\Omega }^{0}_{1}}$is the manifold spannedby the eigenvector $v=\left [1,\frac {N\left (b+c\right )}{c\left (N-1\right )}\right ]^{\prime }$associated to $\tilde {\lambda }_{2}^{W}$:

$$ x=\frac{c\left( N-1\right)}{N\left( b+c\right)}y+\frac{a}{2N\left( b+c\right)} $$

(32)

Noting that after one iteration $y\in \left [0,\frac {a}{2c}\right ]$,it follows that the following region:

$$ \tilde{A}=\left\{\left( x,y\right)| y\in\left[0,\frac{a}{2c}\right]\text{,} \ x=\frac{c\left( N-1\right)}{N\left( b+c\right)}y+\frac{a}{2N\left( b+c\right)}\right\} $$

(33)

is invariant and attracts all points of $\mathbb {R}^{2}_{+}$in at most two iterations. In addition, $\tilde {E}_{0}\in \tilde {A}$.

For bN = 2c, the eigenvaluesassociated to $\tilde {E}_{0}\in \tilde {A}$are $\tilde {\lambda }^{W}_{1}=0$and$\tilde {\lambda }^{W}_{2}=-1$. Then, thelinearity of T ⁰in $\tilde {A}$implies a degenerate flip bifurcation through which an infinity of 2-cycles that fill region$\tilde {A}$are originated.It follows that $\tilde {\mathcal {C}}_{1}^{W}=\left \{\mathcal {\bar {C}}_{0}^{W},\mathcal {\underline {C}}_{0}^{W}\right \}$,where

$$ \mathcal{\bar{C}}_{0}^{W}=\left( \frac{a}{2\left( b+c\right)},\frac{a}{2c}\right) \quad \text{and} \quad \mathcal{\underline{C}}_{0}^{W}=\left( a\frac{2c-b\left( N-1\right)}{4c\left( b+c\right)},0\right) $$

(34)

are the borders of region $\tilde {A}$, isa 2-cycle. Since $\mathcal {\bar {C}}_{0}^{W}$lies on theborder of non differentiability $\tilde {BC}_{1}$,we have that $\tilde {\mathcal {C}}_{1}^{W}$undergoes a persistence border collision.

For b(N − 1) < 2c < bN, we havethat $T^{0}\left (\mathcal {\underline {C}}_{0}^{W}\right )=\mathcal {\bar {C}}_{0}^{W}{\in {\Omega }^{0}_{2}}$and all pointsof subregion ${{\Omega }^{0}_{2}}$aremapped in $\mathcal {\underline {C}}_{0}^{W}$in oneiteration. Thus, $\tilde {\mathcal {C}}_{1}^{W}$is a2-cycle for T ⁰. Moreover,we have that $\tilde {\lambda }^{W}_{1}=0$and $\tilde {\lambda }^{W}_{2}<-1$. Thus,$\tilde {E}_{0}$is a repellor and by the linearityof map T ⁰is possible to excludethe existence of 2-cycles in ${{\Omega }^{0}_{1}}$.This implies that, except for $\tilde {\mathcal {C}}_{1}^{W}$, all2-cycles that filled region $\tilde {A}$havedisappeared. Since all points of ${{\Omega }^{0}_{2}}$are mapped in $\mathcal {\underline {C}}_{0}^{W}$,all points of ${{\Omega }^{0}_{3}}$aremapped in (0, 0),$T^{0}\left (0,0\right )=\mathcal {\bar {C}}_{0}^{W}$and in${{\Omega }^{0}_{3}}$the map T ⁰is linear with insidea unique repellor $\tilde {E}_{0}$, itfollows that $\tilde {\mathcal {C}}_{1}^{W}$attractsall the point of $\mathbb {R}^{2}_{+}$except for $\tilde {E}_{0}$.

For b(N − 1) = 2c,$\mathcal {\bar {C}}_{0}^{W}$collides with the borderof non-differentiability $\tilde {BC}_{2}$.Thus $\tilde {\mathcal {C}}_{1}^{W}$undergoes a border collision bifurcation and disappears. Moreover, let us define$\mathcal {\hat {C}}_{0}^{W}=\left (0,0\right )$. Since$T^{0}\left (\mathcal {\hat {C}}_{0}^{W}\right )=\mathcal {\bar {C}}_{0}^{W}$,$T^{0}\left (\mathcal {\bar {C}}_{0}^{W}\right )=\mathcal {\hat {C}}_{0}^{W}$and$\mathcal {\bar {C}}_{0}^{W}\in \tilde {BC}_{2}$, it followsthat the 2-cycle $\tilde {\mathcal {C}}_{2}^{W}=\left \{\mathcal {\bar {C}}_{0}^{W},\mathcal {\hat {C}}_{0}^{W}\right \}$appears through a border collision bifurcation.

For 2c < b(N − 1),as $T^{0}\left (\mathcal {\hat {C}}_{0}^{W}\right )=\mathcal {\bar {C}}_{0}^{W}$and$T^{0}\left (\mathcal {\bar {C}}_{0}^{W}\right )=\mathcal {\hat {C}}_{0}^{W}$, it follows that$\tilde {\mathcal {C}}_{2}^{W}$is a 2-cycle for T ⁰. In addition,since all points of ${{\Omega }^{0}_{3}}$are plotted in $\mathcal {\hat {C}}_{0}^{W}$,all points of ${{\Omega }^{0}_{2}}$aremapped in ${{\Omega }^{0}_{1}}$,and in this latter region the map is linear with a unique inner repellor$\tilde {E}_{0}$, it followsthat $\tilde {\mathcal {C}}_{2}^{W}$attractsall the points of $\mathbb {R}^{2}_{+}$except for $\tilde {E}_{0}$.

Since $\frac {\partial ^{2} C}{\partial y^{2}}=2c$,$\frac {\partial P\left (E_{0}\right )}{\partial y}=-bN$and T ⁰is the restriction of map(13) on the invariant plane r = 0, the results on E ₀,$\mathcal {C}_{1}^{W}$,$\mathcal {C}_{2}^{W}$and region A, defined as in the statement of the Proposition, follow by the results on$\tilde {E}_{0}$,$\tilde {\mathcal {C}}_{1}^{W}$,$\tilde {\mathcal {C}}_{2}^{W}$and$\tilde {A}$.

By the smoothness of map (13) in a neighborhood of$\mathcal {\bar {C}}^{W}$and$\mathcal {\underline {C}}^{W}$, and from the Jacobianmatrix of this map when r = 0(see, e.g., (26)) follows that the transverse eigenvalue of$\mathcal {C}_{1}^{W}$is given by$J_{33}\left (\mathcal {\bar {C}}^{W}\right )J_{33}\left (\mathcal {\underline {C}}^{W}\right )$. Since$J_{33}\left (\mathcal {\bar {C}}^{W}\right )>1$and$J_{33}\left (\mathcal {\underline {C}}^{W}\right )\geq 1$the transverseinstability of $\mathcal {C}_{1}^{W}$follows. By similar considerations and calculations we have that$J_{33}\left (\mathcal {\bar {C}}^{W}\right )J_{33}\left (\mathcal {\hat {C}}^{W}\right )>1$, from which the transverseinstability of the 2-cycle $\mathcal {C}_{2}^{W}$follows. □

Proof of Proposition 2

The restriction of map (13) to the invariant plane r = 1reduces to the two-dimensional map$T^{1}:\mathbb {R}^{2}_{+}\rightarrow \mathbb {R}^{2}_{+}$defined in Eq. 23. Let us divide the plane$\mathbb {R}^{2}_{+}$into the following three regions:

$$ \begin{array}{l} {{\Omega}^{1}_{1}}=\left\{\left( x,y\right)| 0\leq x<\frac{a}{bN}\right\}\\ {{\Omega}^{1}_{2}}=\left\{\left( x,y\right)| \frac{a}{bN}\leq x<\frac{a}{b\left( N-1\right)}\right\}\\ {{\Omega}^{1}_{3}}=\left\{\left( x,y\right)| \frac{a}{b\left( N-1\right)}\leq x\right\} \end{array} $$

(35)

The lines

$$ \bar{BC}_{1} := x=\frac{a}{bN}\quad \text{ and }\quad \bar{BC}_{2} := x=\frac{a}{b\left( N-1\right)} $$

(36)

mark the borders of non differentiability between regions${{\Omega }^{1}_{1}}$and${{\Omega }^{1}_{2}}$andbetween ${{\Omega }^{1}_{2}}$and ${{\Omega }^{1}_{3}}$,respectively.

By straightforward considerations, see Lemma 1, it follows that$\tilde {E}_{1}=\left (q^{*}_{CN},q^{*}_{CN}\frac {b+2c}{2c}\right ){\in {\Omega }_{1}^{1}}$, with$q^{*}_{CN}=\frac {a}{2c+b\left (N+1\right )}$, is the uniquefixed point of T ¹.

For b(N − 3) < 2c, we have thatall points of region ${{\Omega }^{1}_{2}}$and ${{\Omega }^{1}_{3}}$are mapped inone iteration in ${{\Omega }^{1}_{1}}$. Infact, all points of region ${{\Omega }^{1}_{2}}$are mapped in $\left \{\left (y,x\right )|y=0\text {,} \ x\geq 0\right \}{\in {\Omega }^{1}_{1}}$and allpoints of ${{\Omega }^{1}_{3}}$are mappedin $\left (0,0\right ){\in {\Omega }^{1}_{1}}$. Moreover, themap T ¹is linear in${{\Omega }^{1}_{1}}$with a unique innerfixed point $\tilde {E}_{1}$. Theeigenvalues associated to $\tilde {E}_{1}$are $\tilde {\lambda }_{1}^{CN}=\frac {-b\left (N-1\right )}{2\left (b+c\right )}$and $\tilde {\lambda }_{2}^{CN}=0$with $-1<\tilde {\lambda }_{1}^{CN}<0$.Thus, $\tilde {E}_{1}$is locally asymptotically stable and the image of region${{\Omega }^{1}_{1}}$is the manifold spannedby the eigenvector $v=\left [1,\frac {N\left (b+c\right )}{c\left (N-1\right )}\right ]^{\prime }$associated to $\tilde {\lambda }_{1}^{CN}$:

$$ y=\frac{N\left( b+c\right)}{c\left( N-1\right)}x-\frac{a}{2c\left( N-1\right)} $$

(37)

Noting that, after one iteration, $x\in \left [0,\frac {a}{2\left (b+c\right )}\right ]$and y cannot be negative, it follows that the region:

$$ \tilde{B}=\left\{\left( x,y\right)| x\in\left[0,\frac{a}{2\left( b+c\right)}\right]\text{,} \ y=\max\left[\frac{N\left( b+c\right)}{c\left( N-1\right)}x-\frac{a}{2c\left( N-1\right)},0\right]\right\} $$

(38)

is invariant and attracts all points of $\mathbb {R}^{2}_{+}$in at most two iterations. In addition, $\tilde {E}_{1}\in \tilde {B}$.

For b(N − 3) = 2c, the eigenvaluesassociated to $\tilde {E}_{1}\in \tilde {B}$are $\tilde {\lambda }^{CN}_{1}=-1$and$\tilde {\lambda }^{CN}_{2}=0$. Then, the linearity(with respect to x) of T ¹in $\tilde {B}$implies a degenerate flip bifurcation through which an infinity of 2-cycles that fill region$\tilde {B}$are originated.It follows that $\tilde {\mathcal {C}}^{CN}=\left \{\mathcal {\bar {C}}_{1}^{CN},\mathcal {\underline {C}}_{1}^{CN}\right \}$,where

$$ \mathcal{\bar{C}}_{1}^{CN}=\left( \frac{a}{2\left( b+c\right)},\frac{a}{2c}\right) \quad \text{and} \quad \mathcal{\underline{C}}_{1}^{CN}=\left( 0,0\right) $$

(39)

are the borders of region $\tilde {B}$, isa 2-cycle. Since $\mathcal {\bar {C}}_{1}^{CN}$lies on theborder of non differentiability $\bar {BC}_{2}$,we have that $\tilde {\mathcal {C}}^{CN}$undergoes a persistence border collision.

For 2c < b(N − 3), we havethat $T^{1}\left (\mathcal {\underline {C}}_{1}^{CN}\right )=\mathcal {\bar {C}}_{1}^{CN}{\in {\Omega }^{1}_{3}}$and all pointsof subregion ${{\Omega }^{1}_{3}}$aremapped in $\mathcal {\underline {C}}_{1}^{CN}$in oneiteration. Thus, $\tilde {\mathcal {C}}^{CN}$is a2-cycle for T ¹. Moreover,we have that $\tilde {\lambda }^{CN}_{1}<-1$and $\tilde {\lambda }^{CN}_{2}=0$. Thus,$\tilde {E}_{1}$is a repellor and bythe linearity of map T ¹(with respect to x) it is possible to exclude the existence of 2-cycles in${{\Omega }^{1}_{1}}$and${{\Omega }^{1}_{2}}$. This implies that, exceptfor $\tilde {\mathcal {C}}_{1}^{CN}$, all 2-cycles that filledregion $\tilde {B}$have disappeared.Moreover, since all points of ${{\Omega }^{1}_{2}}$are mapped in ${{\Omega }_{1}^{1}}$,all points of ${{\Omega }^{1}_{3}}$aremapped in $\mathcal {\underline {C}}_{1}^{CN}$,$T^{1}\left (\mathcal {\underline {C}}_{1}^{CN}\right )=\mathcal {\bar {C}}_{1}^{CN}$and in${{\Omega }^{1}_{1}}$the map T ¹is linear with insidea unique repellor $\tilde {E}_{1}$, itfollows that $\tilde {\mathcal {C}}^{CN}$attractsall the point of $\mathbb {R}^{2}_{+}$except for $\tilde {E}_{1}$.

Since $\frac {\partial R\left (E_{1}\right )}{\partial x}=\frac {-b\left (N-1\right )}{2\left (b+c\right )}$and T ¹is the restriction of map(13) on the invariant plane r = 1,the results on E ₁,$\mathcal {C}^{CN}$and region B, defined as in the statement of the Proposition, follow by the results on$\tilde {E}_{1}$,$\tilde {\mathcal {C}}^{CN}$and$\tilde {B}$.

By the smoothness of map (13) in a neighborhood$\mathcal {\bar {C}}^{CN}$and$\mathcal {\underline {C}}^{CN}$, and from the Jacobianmatrix of this map when r = 1,see, e.g., (28), follows that the transverse eigenvalue of$\mathcal {C}^{CN}$is given by$J_{33}\left (\mathcal {\bar {C}}^{CN}\right )J_{33}\left (\mathcal {\underline {C}}^{CN}\right )$. Since$J_{33}\left (\mathcal {\bar {C}}^{CN}\right )<1$and$J_{33}\left (\mathcal {\underline {C}}^{CN}\right )\leq 1$, the transverseattractiveness of $\mathcal {C}_{1}^{CN}$follows. □