Chemistry Reference
In-Depth Information
It is straightforward to calculate that T
j
D
.3/
has the unique fixed point
L;
;
NL
1
ˇ
S
E
3
D
.3/
,
which is an equilibrium for the dynamical system (4.47) provided that E
3
2 D
that is if
z
.E
3
/>L. This condition is equivalent to
L<
A.1
ˇ
S
/.N
1
C
ˇ
S
/
cN
2
:
.2/
, so it is not
T
j
D
.2/
has the unique fixed point O
D
.0;0/ which is not inside
D
an equilibrium of the dynamic process.
The fixed points of T
j
D
.1/
are the solutions of the algebraic system
q
c
..N
1/x
C
ˇ
S
S/
..N
1/x
C
ˇ
S
S/;
.1
ˇ
S
/S
D
NŒ.1
a/x
C
ax:
x
D
From the second equation we get x
D
.1
ˇ
S
/S=N and after substituting this
expression into the first equation we get
A
Nc
..N
1/.1
ˇ
S
/S
C
Nˇ
S
S/
D
S
2
:
Using these relations it is easy to see that T
j
D
.1/
has the unique fixed point
E
1
D
x
1
; S
1
with
A.1
ˇ
S
/.N
1
C
ˇ
S
/
N
2
c
A.N
1
C
ˇ
S
/
Nc
S
1
D
x
1
D
,
.
.1/
,
in other words if 0<
z
.E
1
/<L. Using the fact that
z
.E
1
/
D
x
1
, this condition
becomes
2
D
This fixed point is also an equilibrium of the map T provided that E
1
L>
A.1
ˇ
S
/.N
1
C
ˇ
S
/
cN
2
:
Since
A.1
ˇ
S
/.N
1
C
ˇ
S
/
cN
2
<
A
4c
always holds, being equivalent to .N
2.1
ˇ
S
//
2
>0, we can summarize the
existence results for an equilibrium as follows:
If L>
A
.3/
is empty and the unique steady state is E
1
.
4c
, then the region
D
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