Page 353 - Advanced engineering mathematics

10.6 Phase Portraits 333

λt
μt
X(t) = c 1 E 1 e + c 2 E 2 e .
Examine trajectories along an arbitrary point P 0 other than the origin.
1. If P 0 is on L 1 , then c 2 = 0 and X(t) moves on part of L 1 away from the origin as t
increases, because e →∞ as t →∞.
λt
μt
2. If P 0 is on L 2 , then c 1 =0 and X(t) moves on part of L 2 toward the origin, because e →0
as t →∞. Thus, along these lines, some trajectories move toward the origin, others move
away.
3. Now suppose P 0 is on neither L 1 or L 2 . Then the trajectory through P 0 does not pass
arbitrarily close to the origin for any times but instead moves toward the origin asymptotic
to L 2 and then away from the origin asymptotic to L 1 as t increases. We may think
of L 1 and L 2 as separating the plane into four regions with each trajectory confined to
one region (because a trajectory starting in one of these regions cannot cross another
trajectory along one of L 1 or L 2 to pass into another region). The trajectories move along
L 1 away from the origin and along L 2 toward the origin or in one of the four regions these
lines determine, sweeping toward and then away from the origin asymptotic to these lines.
This is similar to Halley’s comet entering our solar system and moving toward the Sun,
then sweeping along a curve that takes it away from the Sun.
In this case, we call the origin a saddle point.
The behavior we have just described can be seen in the following example.

EXAMPLE 10.21
The system

−1 3

X = X
2 −2
has general solution

−1 −4t 3 t
X(t) = c 1 e + c 2 e .
1 2
The eigenvalues of A are −4 and 1, real and of opposite sign. Figure 10.18 shows a phase portrait,
with a saddle point at the origin.

Case 3: A Has Equal Eigenvalues
Suppose A has the eigenvalue λ of multiplicity 2. There are two possibilities.

Case 3(a): A Has Two Linearly Independent Eigenvectors E 1 and E 2
Now the general solution is
λt
X = (c 1 E 1 + c 2 E 2 )e .
If

a h
E 1 = and E 2 = ,
b k
then, in terms of components,
λt
kt
x(t) = (c 1 a + c 2 b)e and y(t) = (c 1 h + c 2 k)e .
Now
y(t)
= constant.
x(t)

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October 14, 2010 20:32 THM/NEIL Page-333 27410_10_ch10_p295-342