Chemistry Reference
In-Depth Information
If this condition is violated with strict inequality, then the equilibrium is unstable.
Proof.
The Jacobian of the system (4.82) is given by
0
@
1
A
1
a
1
.1
r
1x
/
a
1
r
1Q
:::
a
1
r
1Q
a
2
r
2Q
1
a
2
.1
r
2x
/:::
a
2
r
2Q
;
:
:
:
:
:
:
:
:
:
:::1
a
N
.1
r
Nx
/
a
N
r
NQ
a
N
r
NQ
where the previous notation of this section is used. The eigenvalue equation can be
easily determined by using relation (E.5) and turns out to be
.1
a
k
.1
r
kx
C
r
kQ
/
/
"
1
C
#
Y
N
X
N
a
k
r
kQ
1
a
k
.1
r
kx
C
r
kQ
/
D
0:
kD1
kD1
(4.83)
As before, assume that a
k
>0 for all k,andlet1
a
j
.1
r
jx
C
r
jQ
/.j
D
1;2;:::;s/ denote the different 1
a
k
.1
r
kx
C
r
kQ
/ values and assume that
they are repeated m
1
;m
2
;:::;m
s
times among the N firms. By adding the terms
with identical denominators in the bracketed expression and denoting by
j
the sum
of the corresponding numerators a
k
r
kQ
, one obtains
2
3
s
s
Y
X
j
1
a
j
.1
r
jx
C
r
jQ
/
4
1
C
5
D
0;
.1
a
j
.1
r
jx
C
r
jQ
/
/
m
j
j
D
1
j
D
1
(4.84)
where
j
0 for all j.If
j
D
0 or m
j
2,then1
a
j
.1
r
jx
C
r
jQ
/ is an
eigenvalue of the Jacobian, and this eigenvalue is between
1 and
C
1,if
2
1
r
jx
C
r
jQ
:
0<a
j
<
(4.85)
All other eigenvalues are the roots of the equation
s
X
j
1
a
j
.1
r
jx
C
r
jQ
/
D
0;
1
C
j D1
when it is assumed that all
j
values are nonzero. Under assumption (4.85), the
graph of the left hand side is the same as the one shown in Fig. 2.1, so all roots are
real, all poles are between
1 and
C
1, furthermore all roots are between
1 and
C
1,if
X
N
a
k
r
kQ
2
a
k
.1
r
kx
C
r
kQ
/
>0:
1
C
(4.86)
kD1
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