Biology Reference
In-Depth Information
I
may never be observed in nature because they are unstable. We repeat
an intuitive example by revisiting Figure 1-12 in Section VII of Chapter 1.
Observe that points 1, 2, 3, and 4 are equilibrium points, of which
1 and 4 are unstable, and 2 and 3 are stable.
50
In a two-dimensional system, we can give a similar heuristic definition
for stability, and the phase plane can sometimes be used to classify
equilibrium states. For stable equilibrium points, all trajectories that
initiate sufficiently close to those point ''remain close'' for any t
>
0. This
0
200
S
A
is not the case for unstable equilibrium points.
I
B. Finding the Equilibrium Points of a Two-Component System
Recall that for a one component system, we had:
dx
dt
¼
50
ð
Þ;
f
x
and found that the steady-states are values of x
0
, where f (x
0
)
¼
0.
In an analogous two-component system, if the components are x and y,
the equations are
0
200
S
B
dx
dt
¼
I
f
ð
x
;
y
Þ;
dy
dt
¼
g
ð
x
;
y
Þ;
50
and the equilibrium states are points (x
0
,y
0
), where f (x
0
,y
0
)
¼
0 and
g(x
0
,y
0
)
¼
0.
Example 2-1
.......................
0
200
S
C
If x
0 and y
0, find the equilibrium states for the system:
FIGURE 2-11.
Possible behaviors of the phase trajectories
corresponding to the directional diagrams in
Figure 2-10. Panel A: A trajectory that spirals
away from the equilibrium point (200,50); panel
B: A trajectory that spirals into the equilibrium
point; panel C: A trajectory that orbits around
the equilibrium point.
dx
dt
y
2
¼
f
ð
x
;
y
Þ¼
y
þ
x
1
dy
dt
¼
g
ð
x
;
y
Þ¼
y
x
:
S
OLUTION
:
We need to find the point (x
0
,y
0
) where f (x
0
,y
0
)
¼
0 and
g(x
0
,y
0
)
¼
0 Thus, we need to solve the system of equations:
y
2
y
þ
x
1
¼
0
y
x
¼
0
:
¼
The second equations yields y
x. After substituting in the first
equation, we obtain x
2
-x
0, (i.e. x
2
-1
þ
¼
¼
x -1
0, ). This equation has
x
¼
1 and x
¼
1 as solutions. Since we have the restriction that
