BASINS OF ATTRACTION IN A COURNOT DUOPOLY MODEL OF ...

BASINS OF ATTRACTION IN A COURNOT DUOPOLY MODEL OF KOPEL DOUGLAS R. ANDERSON, NICHOLAS G. MYRAN, AND DUSTIN L. WHITE

Abstract. For a nonlinear Cournot duopoly map of Kopel, we show that a circle, lines, and rectangles play a key role in determining the basins of attraction in the case of three nontrivial Nash equilibria.

1. A Nonlinear Cournot Duopoly Map of Kopel In [1, p. 191] we found the discrete map xn+1 = (1 − A)xn + AByn (1 − yn ) yn+1 = (1 − A)yn + ABxn (1 − xn )

(1)

that models a nonlinear Cournot duopoly game [2] with normalized outputs x, y ∈ [0, 1] for the duopolists with adjustment coeﬃcient A ∈ [0, 1] and parameter B > 0, a measure of the positive external influence one player exerts on the other player. The right-hand sides represent the reaction functions of the players given initial beliefs (x0 , y0 ), and determine the subsequent trajectory of beliefs (xn , yn ). Game (1) was introduced by Kopel [3], and initially analyzed in terms of fixed points (often called Nash equilibria or equilibrium beliefs) and chaos control by Agiza [4]. Bischi and Kopel [5] then provided an extensive economic rationale for the model together with an analysis of parameter space in which three nontrivial consistent beliefs, corresponding to Nash equilibria in the product quantity space, coexist, so that long-term outcomes are greatly determined by the players’ choice of initial beliefs (the seed of the map in state space). In this note we explore further the basins of attraction for these three coexistent belief equilibria by in particular proving that 2000 Mathematics Subject Classification. 39A10. Key words and phrases. discrete dynamical systems, asymptotic behavior, fixed point, attractor, repeller. Supported by the Concordia College 2004-2005 Carl L. Bailey Centennial Faculty Research Scholarship.

2

ANDERSON, M YRAN, AND WHITE

key basin boundaries are lines and circles in the xy plane. When A = 1, our results are a special case of more general maps discussed by Bischi, Mammana, and Gardini [6]. 2. Stability Analysis of Belief Equilibria For A ∈ [0, 1) The following results on fixed points and the associated eigensystem from the linearized Jacobian matrix for (1) can also be found in [4]; they are repeated here for convenience. The null state equilibrium for all parameter values is fixed point one at the origin: 

     1 −1 fp1 = (0, 0), with eigenpairs 1 − A + AB,   , 1 − A − AB,   . 1 1

The second belief equilibrium is       µ ¶ 1 −1 B−1 B−1 fp2 = , : 1 + A − AB,   , 1 − 3A + AB,   ; B B 1 1 fp3 =

Ã

1+B−

! p p (B − 3)(B + 1) 1 + B + (B − 3)(B + 1) , 2B 2B

fp4 =

Ã

1+B+

! p p (B − 3)(B + 1) 1 + B − (B − 3)(B + 1) , 2B 2B

and

are symmetric across the y = x line in the first quadrant, both with the same eigenvalues p λ = 1 − A ± A 4 + 2B − B 2 .

In the case of fp3, the player associated with y dominates, whereas for fp4 player x dominates. A further convenience will be to sometimes write (1) as the two-variable vector function duop(x, y) = ((1 − A)x + ABy(1 − y), (1 − A)y + ABx(1 − x)). Note that if y0 = x0 , then yn+1 = xn+1 = (1 − A)xn + ABxn (1 − xn ),

n = 0, 1, 2, · · · .

Define g(x) := (1 − A)x + ABx(1 − x),

f(x) := rx(1 − x).

(2)

COURNOT DUOPOLY

3

If φ(x) :=

rx , AB

r = 1 − A + AB,

(3)

then φ(f (x)) = g(φ(x)), in other words φ is a topological conjugacy between the discrete logistic equation f and (1) restricted to the line y = x represented by g. The fixed points 0 and (r−1)/r for f correspond to φ(0) = 0 and φ((r−1)/r) = (B−1)/B for g, or fp1 and fp2 for (1). Knowing the rich dynamics [7, p. 97] associated with the discrete logistic map f as the parameter r ranges within [0, 4] gives us an idea of the complexity of the dynamics to be found for (1). In parameter space, let (A, B) ∈ [0, 1] × [3, min{1 + 2/A, 1 +

p 4 + 2/A }];

(4)

then according to the eigenvalues for the linearization stated earlier fp3 and fp4 p √ exist as local sinks (spiral sinks for B ∈ [1 + 5, 1 + 4 + 2/A ]), with fp2 a saddle

or a saddle with reflection, and fp1 either a saddle or source with reflection.

Lemma 1. Assume (4) holds. Then for any initial beliefs (x0 , y0 ) satisfying the linear equation y = −x +

AB + A − 1 AB

(5)

we have x1 = y1 ; in other words, (5) is in the preimage under (1) of the line y = x. Proof. The proof is a straightforward exercise in algebra.

¤

The following theorem is similar to part of Proposition 3 in [5]. Proposition 1. If the parameters satisfy (A, B) ∈ (0, 1/6) × (3, 1 +

p 4 + 2/A)

or

(A, B) ∈ [1/6, 1/4) × (3, −1 + 1/A),

then the basin of attraction for fp3 is the connected region in the first quadrant above the line y = x, the basin of attraction for fp4 is the connected region in the first quadrant below the line y = x, and fp2 attracts the line y = x. Proof. Under these assumptions (4) holds. The eigenvalues of fp1 satisfy 0 < |1 − A − AB| < 1,

1 < 1 − A + AB,

4


so that the origin fp1 is a saddle point with y = x in its unstable set. Similarly fp2 has eigenvalues satisfying 0 < |1 + A − AB| < 1,

1 < 1 − 3A + AB,

whence fp2 is a saddle point with y = x in its stable set; see the eigenvectors listed previously. Since y = x is an invariant set for (1), it is clearly a boundary for any possible basins of attraction for fp3 and fp4. For these values of A, B we have B < −1 + 1/A, so that the preimage line (5) of y = x has negative y intercept and slope and thus avoids the first quadrant; preimages of the line y = x are either on y = x or on line (5). Therefore there are no other basin boundaries except y = x, ¤

and the result holds. Proposition 2. Let (4) hold, with AB + A − 1 > 0.

(6)

Then the line (5) cuts through the square S = [0, 1] × [0, 1] in the phase portrait, and the circle µ ¶2 µ ¶2 AB − A + 1 3 − 4A + A2 (B − 1)2 AB − A + 1 + y− = x− 2AB 2AB 2A2 B 2 is in the preimage of (5) and also crosses through S. Moreover, the points Ã ! µ p p ¶ γ ± γ(AB−3A+3) γ ∓ γ(AB −3A+3) AB −A+1 AB −A+1 , , , 2AB 2AB AB AB

(7)

(8)

where γ = AB + A − 1, are real preimages of fp1 = (0, 0). Proof. The assumption B > −1 + 1/A implies 0 < 1 + (A − 1)/(AB) < 1, which is the y intercept of line (5), ensuring it passes through S. Condition (4) and inequality (6) likewise guarantee that the center coordinates satisfy 0
2AB 2A2 B 2 that is to say the distance from the center of the circle to the origin is greater than the radius of the circle, so that at least its lower left arc crosses through

COURNOT DUOPOLY

5

[0, 1] × [0, 1]. Let (x, y) = (x0 , y0 ) satisfy (7); solve (7) for y0 in terms of x0 and substitute the result into (1) to obtain (x1 , y1 ). After algebraic simplification it is seen that y1 + x1 = 1 + (A − 1)/(AB), putting (x1 , y1 ) on the line (5). By (6), the three points (x, y) given in (8) are all real, and we have duop(x, y) = (0, 0) using ¤

(2).

Proposition 3. Assume (4) and (6) hold. Then there exists an infinite sequence of nodes {(an , an )}∞ n=0 in S on the y = x line such that a0 =

AB+A−1 2AB ,

duop(an , an ) = (an−1 , an−1 ) using (2), {an } is decreasing with lim an = 0,

n→∞

and

lim

n→∞

µ

an an+1

¶

= 1 − A + AB,

the unstable multiplier for the origin fp1. Proof. The intersection of y = x and line (5) is the initial node coordinate a0 := AB+A−1 2AB .

Using (2), solve duop(x, x) = (y, y) in terms of x; the results as single-

variable functions are

p (1 − A + AB)2 − 4ABx , inv− (x) := 2AB p 1 − A + AB + (1 − A + AB)2 − 4ABx inv+ (x) := . 2AB 1 − A + AB −

(9) (10)

Set an+1 = inv− (an ),

n = 0, 1, 2, · · · .

>From (4) and (6) the parameters satisfy A ∈ (0, 1] and B ≥ 3, so that a0 ∈ (0, 1/2] and inv− (x) < x. Since inv− (x) = 0 if and only if x = 0 and inv− has no fixed points between 0 and (B − 1)/B, {an }∞ n=0 is a strictly decreasing infinite sequence with limit 0. Using l’Hôpital’s rule, we have lim

n→∞

an a = lim an+1 a→0 inv− (a)

2ABa p 1 − A + AB − (1 − A + AB)2 − 4ABa p 2AB (1 − A + AB)2 − 4ABa = lim a→0 2AB

= lim

a→0

= 1 − A + AB.

6


Note that 1 − A + AB > 1 is the multiplier for the linearization of (1) about fp1,

¤

and the inverse of the multiplier for inv− at 0. Proposition 4. Assume (4) and (6) hold. Let b0 , c0 , d0 = there exists an infinite sequence of nodes lim bn =

n→∞

{(bn , bn )}∞ n=0

AB+A−1 2AB

= a0 . Then

on the y = x line with

AB − A + 1 , AB

∞ and infinite sequences of nodes {(cn , dn )}∞ n=0 , {(dn , cn )}n=0 on line (5) with Ã ! p p γ − γ(AB −3A+3) γ + γ(AB −3A+3) , lim (cn , dn ) = , n→∞ 2AB 2AB

γ = AB + A − 1; i.e., (bn , bn ), (cn , dn ), (dn , cn ) go to preimages (8) of (0, 0) in the limit. Moreover, duop(bn , bn ) = duop(cn , dn ) = duop(dn , cn ) = duop(an , an ) = (an−1 , an−1 ) using (2). Proof. Let b0 = a0 , and set bn+1 = inv+ (an ),

n = 0, 1, 2, 3, · · ·

for an given in Proposition 3 and the function inv+ given in (10). Since {an } is a strictly decreasing sequence and inv+ is a decreasing function, {bn } is a strictly increasing sequence, with lim bn = lim inv+ (x) =

n→∞

x→0

AB − A + 1 . AB

Again let c0 = d0 = a0 , and consider the functions p AB + A − 1 − A2 (B 2 − 2B − 3) − 3 + A(6 + 2B − 4By) pre− (y) := , 2AB p AB + A − 1 + A2 (B 2 − 2B − 3) − 3 + A(6 + 2B − 4By) pre+ (y) := . 2AB

(11) (12)

Let cn+1 = pre− (an ),

dn+1 = pre+ (an ),

>From (11) and (12) it is clear that cn + dn =

n = 0, 1, 2, 3, · · · .

AB+A−1 AB

for n = 0, 1, 2, · · · , so that

(cn , dn ) and (dn , cn ) are indeed points on line (5). Because pre+ is a decreasing

COURNOT DUOPOLY

7

function and {an } is a strictly decreasing sequence, {dn } is a strictly increasing sequence, with lim dn = lim pre+ (y) =

n→∞

y→0

AB + A − 1 +

p (AB + A − 1)(AB − 3A + 3) ; 2AB

similarly, {cn } is a strictly decreasing sequence, with p AB + A − 1 − (AB + A − 1)(AB − 3A + 3) . lim cn = lim pre− (y) = n→∞ y→0 2AB ¤ Proposition 5. If the parameters satisfy p (A, B) ∈ [1/6, 1/4) × [−1 + 1/A, 1 + 4 + 2/A] or p √ [1/4, (−1 + 17)/4] × (3, 1 + 4 + 2/A] or √ [(−1 + 17)/4, 1) × (3, 1 + 2/A], then line (5), circle (7), the sequence of nodes ∞ ∞ ∞ {(an , an )}∞ n=0 , {(bn , bn )}n=0 , {(cn , dn )}n=0 , {(dn , cn )}n=0 ,

(13)

and their respective preimages form the boundaries and intersection points of boundaries for the disconnected basins of attraction for fp3 and fp4. The boundaries themselves are the basin of attraction for fp2. Proof. It is straightforward to check that (4) and (6) hold for these parameter regions of A and B, ensuring that fp3 and pf4 are attractors (asymptotically stable), and fp2 is a saddle or saddle with reflection. Therefore the hypotheses of Propositions 2,3, and 4 hold, putting line (5), at least an arc of circle (7), and the nodes {(an , an )} in S. All nodes (13) serve as intersection points of basin boundaries with the boundary lines y = x and (5). Clearly the line y = x is invariant for (1), so that that line and all its preimages, including line (5), the circle (7), and the nodes ¤

(13), make up the basin of attraction for fp2. 3. The Case of A = 1, B ∈ (3, 1 +

√ 6)

When A = 1 the map (1) becomes xn+1 = Byn (1 − yn ),

yn+1 = Bxn (1 − xn ).

8


Proposition 6. Assume A = 1 and and B ∈ (3, 1 +

√ 6). Then fp3 and fp4 are

attractors with basins of attraction consisting of unions of rectangles with boundaries parallel to the x and y axes, fp2 is a repeller, and there are six pairs of two-cycles given by

¶ µ ¶ µ B−1 B−1 ,0 , 0, & B B ¶ µ ¶ µ 1+B±σ 1+B∓σ ,0 , 0, & 2B 2B µ µ

where σ =

B−1 1+B±σ , B 2B

¶

1+B −σ 1+B −σ , 2B 2B

¶

&

µ

1+B∓σ B−1 , 2B B

¶

&

µ

1+B+σ 1+B+σ , 2B 2B

, ¶

,

p (B − 3)(B + 1), that attract along the rectangular boundaries.

Proof. For these values of B, by (3) and the well-known discrete logistic map we know that fp2 is a repeller and there exists an attracting two-cycle on the y = x line. It is straightforward to check that the various combinations of coordinates from the fixed points given in the statement of the theorem generate other period-two points. By Propositions 3 and 4 from [6], the image and preimage of a horizontal line is a vertical line and vice versa; since A = 1, the preimages of (0, 0) given in (8) are now (1, 0), (0, 1), (1, 1), leading to this straight-line alignment. Moreover, we have that for any period-n point P = (x0 , y 0 ), the horizontal and vertical lines y = y 0 and x = x0 are trapping sets for the nth composite of the map. These lines are found as follows: Note that for A = 1, from the pairs (9), (11) and (10), (12) we have that inv± ≡ pre± . Let p0 = p2m = inv+ (p2m−1 ),

B−1 B ,

p1 = inv− (p0 ), and

p2m+1 = inv− (p2m−1 ),

m = 1, 2, · · · .

The infinitely-many horizontal and vertical lines in S given by y = 0,

y = pn ,

x = 0,

x = pn ,

n = 0, 1, 2, · · ·

(14)

form the trapping sets for the two-cycles predicted by [6] and serve as the boundaries of the rectangular basins of attraction for fp3 and fp4.

¤

COURNOT DUOPOLY

9

1

7

3

8 10

5

8

2

9

6

10

9

4

1

7

5

6

1

4. Examples Example 1. Let A = 1.0 and B = 3.3. As shown in Figure 1, there are six pairs of two cycles as predicted in Proposition 6 (labeled 5 through 10). Several of the horizontal and vertical lines from (14) that serve as basin boundaries are also depicted. Example 2. Let A = 0.7 and B = 3.4. Then (4) and (6) are satisfied, and Propositions 3,4, and 5 can be applied. As shown in Figure 2, the white regions are the basin of attraction for fp3, the shaded regions for fp4, while the circle and lines with various nodes marked are attracted to fp2; note that we have depicted the entire object containing the various basins of attraction, not just that contained in S. The following key summarizes the labeled points in the figure. 1 fp1= (0, 0) ¡ ¢ B−1 2 fp2= µB−1 B , √ B ¶ √ 1+B− (B−3)(B+1) 1+B+ (B−3)(B+1) , 3 fp3= 2B 2B

10


1 21 15 0.8

12 11 10

18

17

9

3 13

2

20

0.6

4

5 0.4

19

6 0.2

18

14

7 0.2

4 fp4=

µ

0.4

√

1+B+

0.6

√

(B−3)(B+1) 1+B− , 2B

5 (a0 , a0 ) = (b0 , b0 ) =

0.8

¡1 2

+

A−1 1 2AB , 2

(B−3)(B+1) 2B

+

A−1 2AB

6 (a1 , a1 ) = (inv− (a0 ), inv− (a0 )); see (9)

¢

16 1 22

¶

7 (a2 , a2 ) = (inv2− (a0 ), inv2− (a0 ))

8 (a3 , a3 ) = (inv3− (a0 ), inv3− (a0 )) 9 (b1 , b1 ) = (inv+ (a0 ), inv+ (a0 )); see (10) 10 (b2 , b2 ) = (inv+ (a1 ), inv+ (a1 )) 11 (b3 , b3 ) = (inv+ (a2 ), inv+ (a2 )) 12 ( AB−A+1 , AB−A+1 ); see (8) AB AB 13 (c1 , d1 ) = (pre− (a0 ), pre+ (a0 )); see (11) 14 (d1 , c1 ) = (pre+ (a0 ), pre− (a0 )); see (12) 15 (c2 , d2 ) = (pre− (a1 ), pre+ (a1 )) 16 (d2 , c2 ) = (pre+ (a1 ), pre− (a1 )) 17 preimage of (c1 , d1 ), intersection of circle and its preimage 18 another preimage of (c1 , d1 ), intersection of circle and its preimage 19 preimage of (d1 , c1 ), intersection of circle and its preimage

COURNOT DUOPOLY

11

20 µ another preimage of (d1 , c1 ), intersection of circle and its preimage¶ √ √ AB+A−1+ (AB+A−1)(AB−3A+3) AB+A−1− (AB+A−1)(AB−3A+3) 21 , 2AB 2AB µ ¶ √ √ AB+A−1− (AB+A−1)(AB−3A+3) AB+A−1+ (AB+A−1)(AB−3A+3) 22 , 2AB 2AB References [1] M. R. S. Kulenovic and O. Merino, Discrete Dynamical Systems and Diﬀerence Equations with Mathematica, Chapman & Hall /CRC, Boca Raton, 2002. [2] A. Cournot, Recherche sur la Principes Matematiques de la Theorie de las Richesse, Hachette, Paris, 1838. [3] M. Kopel, Simple and Complex Adjustment Dynamics in Cournot Duopoly Models, Chaos, Solitons & Fractals 7:12 (1996) 2031—2048. [4] H. N. Agiza, On the Analysis of Stability, Bifurcation, Chaos and Chaos Control of Kopel Map, Chaos, Solitons & Fractals 10:11 (1999) 1909—1916. [5] G. I. Bischi and M. Kopel, Equilibrium selection in a nonlinear duopoly game with adaptive expectations, Journal of Economic Behavior & Organization 46 (2001) 73—100. [6] G. I. Bischi, C. Mammana, and L. Gardini, Multistability and cyclic attractors in duopoly games, Chaos, Solitons & Fractals 11 (2000) 543—564. [7] R. A. Holmgren, A First Course in Discrete Dynamical Systems, 2nd edition, Springer, New York, 1996.

Department of Mathematics and Computer Science, Concordia College, Moorhead, MN 56562 USA E-mail address : [email protected], [email protected], [email protected]