Modified and Extended Oligopolies - Nonlinear Oligopolies: Stability and Bifurcations - page 148

D ˛ 0 k .0/>0 for all firms, the different a k .1 C r k / values

Here we assume that a k

are

a 1 .1 C r 1 />a 2 .1 C r 2 /> >a s .1 C r s /;
and these values are repeated m 1 ;m 2 ;:::;m s times, respectively, among the N
firms. The value of j is the sum of all products r k a k such that a k .1 C r k / D a j .1 C
r j /.If j ¤ 0 and m j D 1,then1 a j .1 C r j / is not an eigenvalue of the Jacobian,
and if j D 0 or m j 2,then1 a j .1 C r j / is an eigenvalue. Since r j > 1,these
eigenvalues are inside the unit circle, if 1 a j .1 C r j /> 1,thatis,when

a j .1 C r j /<2:

Let g./ denote the bracketed factor in (4.12). It is easy to see that

lim
!˙1

g./ D 1;

(

˙1 ; if j <0;
1 ; if j >0:

lim g./ D

! 1 a j .1 C r j / ˙ 0

Since the derivative of g has no definite sign, no monotonocity property of g can
be established. Notice in addition, that all poles 1 a j .1 C r j / are less than 1. The
possible presence of complex conjugate roots makes stability analysis intractable
in the general case. In such cases computational methods can be used to find the
roots and check stability conditions. However, if there is at most one sign change in
sequence 1 ; 2 ;:::; s and it is from “ ”to“ C ”, then we always have only real
eigenvalues and we can derive simple stability conditions.
Case 1. All j >0. The graph of g./ is as shown in Fig. 4.3. Clearly all roots are
real, and all are between 1 and C 1 if all poles are larger than 1 and

g.1/>0:
Case 2. All j <0. Then the graph of g./ is as illustrated in Fig. 4.4. All roots are
real and they are between 1 and C 1 if all poles are larger than 1 and
g. 1/>0.
Case 3. There is a sign change in the sequence 1 ; 2 ;:::; s . The corresponding
graph of g is shown in Fig. 4.5. We have again s real roots and they are
between 1 and

C 1, if all poles are larger than 1 and both g. 1/ and

g.1/ are positive.
We note that conditions g. 1/>0 and g.1/>0 can be rewritten as (2.22) and

X

N

r k
1 C r k

<1;

kD1

respectively.
Example 4.2. Consider again the linear case examined in the previous example and
assume that h k .0/ D 0 for all k,thatis,q k D 0.ThenR 0 k

D 0 for all k,soR k