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 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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