Biology Reference
In-Depth Information
We consider the following special case:
dS
dt
¼
0
:
00001IS
þ
0
:
2
dI
dt
¼
0
:
00001IS
0
:
02I
:
We want to determine the equilibrium points and whether they are
asymptotically stable. It is easy to find that this system of equations has
one equilibrium point, S
¼
2000 and I
¼
10. Converting to the earlier
notations, we can write:
f
ð
S
;
I
Þ¼
0
:
00001IS
þ
0
:
2
g
ð
S
;
I
Þ¼
0
:
00001IS
0
:
02I
;
and thus:
@
f
@
f
dI
¼
S
¼
0
:
00001I
0
:
00001S
@
@
g
@
g
S
¼
0
:
00001I
I
¼
0
:
00001S
0
:
02
:
@
@
The Jacobian associated with the equilibrium state (S,I)
¼
(2000,10) is:
0
1
@
f
@
f
S
ð
2000
;
10
Þ
I
ð
2000
;
10
Þ
@
A
¼
@
@
0
:
0001
0
:
02
J
¼
:
@
g
@
g
0
:
0001
0
S
ð
2000
;
10
Þ
I
ð
2000
;
10
Þ
@
@
Computing the trace and the determinant of J gives trace ( J)
¼
0. According to the criteria
given in the previous theorem, the equilibrium point is asymptotically
stable.
0.0001
<
0 and det ( J)
¼
0.000002
>
To summarize, we have now made a certain amount of progress in
analyzing phenomena that are governed by the system of differential
equations given by Eq. (2-8), where f (x,y) and g(x,y) have continuous
partial derivatives. Namely, we know:
1. Given an initial point (x
0
, y
0
) there is exactly one solution to the
system with initial point (x
0
, y
0
).
2. We can find the equilibrium points of the system.
3. We can distinguish between types of equilibrium points according
to their stability.
When we study predator-prey models, we shall return to the idea of
stability for two-component systems and give additional descriptions of
the equilibrium states.
