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**Unformatted text preview: **t→∞ for every initial vector c. In this case u = Au is said to be a stable
system, and A is called a stable matrix.
If Re (λi ) > 0 for some i, then components of u(t) can become
unbounded as t → ∞, in which case the system u = Au as well
as the underlying matrix A are said to be unstable.
If Re (λi ) ≤ 0 for each i, then the components of u(t) remain
ﬁnite for all t, but some can oscillate indeﬁnitely. This is called a
semistable situation. Example 7.4.2
Predator–Prey Application. Consider two species of which one is the predator and the other is the prey, and assume there are initially 100 in each population. Let u1 (t) and u2 (t) denote the respective population of the predator and Copyright c 2000 SIAM Buy online from SIAM
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7.4 Systems of Diﬀerential Equations
http://www.amazon.com/exec/obidos/ASIN/0898714540 545 prey species at time t, and suppose their growth rates are given by
u1 = u1 + u2 ,
u2 = −u1 + u2 .
Problem: Determine the size of each population at all future times, and decide
if (and when) either population will eventually become extinct. It is illegal to print, duplicate, or distribute this material
Please report violations to meyer@ncsu.edu Solution: Write the system as u = Au, u(0) = c, where
A= 1
−1 1
1 , u= u1
u2 , and 100
100 c= . D
E The characteristic equation for A is p(λ) = λ2 − 2λ + 2 = 0, so the eigenvalues
for A are λ1 = 1 + i and λ2 = 1 − i. We know from (7.4.7) that
u(t) = eλ1 t v1 + eλ2 t v2 (where vi = Gi c ) (7.4.9) T
H is the solution to u = Au, u(0) = c. The spectral theorem on p. 517 implies
A − λ2 I = (λ1 − λ2 )G1 and I = G1 + G2 , so (A − λ2 I)c = (λ1 − λ2 )v1 and
c = v1 + v2 , and consequently
(A − λ2 I)c
v1 =
= 50
(λ1 − λ2 ) λ2
λ1 and λ1
λ2 v2 = c − v1 = 50 IG
R . With the aid of (7.4.8) we obtain the solution components from (7.4.9) as
u1 (t) = 50 λ2 eλ1 t + λ1 eλ2 t = 100et (cos t + sin t)
and u2 (t) = 50 λ1 eλ1 t + λ2 eλ2 t = 100et...

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