MAS 20001 CD

Differential Equations and Applications

§11.2

Gyo Taek Jin

Spring 2025

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 Chapter 11
Systems of Nonlinear Differential
Equations
§11.1 Autonomous Systems
§11.2 Stability of Linear Systems
§11.3 Linearization and Local Stability
§11.4 Autonomous Systems as Mathematical
Models
§11.5 Periodic Solutions, Limit Cycles, and
Global Stability

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Some Fundamental Questions
Suppose that X = X(t) is a solution of a plane
autonomous system X0 = (P(x , y ), Q(x , y )) with
initial condition X(0) = X0 . We have the follow-
ing questions when X0 is near a critical point X1 :
(i ) Will the particle return to the critical point?
More precisely, if X = X(t) is the solution that
satisfies X(0) = X0 , is limt→∞ X(t) = X1 ?
(ii ) If the particle does not return to the critical
point, does it remain close to the critical point
or move away from the critical point?
If (a) or (b) in the figure always occurs in a neigh-
borhood of a critical point, we call the critical
point locally stable. If an initial point X0 as in
(c) can be found in any neighberhood of a critical
point, we call the critical point unstable.

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Stability Analysis
We consider a linear first-order autonomous system

x 0 = ax + by
(1)
y 0 = cx + dy

The stability of such a system depends on the property of the

coefficient matrix !
a b
A= .
c d
To ensure that X0 = (0, 0) is the only critical point, we assume
det A = ∆ = ad − bc 6= 0. If trace(A) = τ = a + d, then the
characteristic equation can be written as

det(A − λI) = λ2 − τ λ + ∆ = 0
√
τ ± τ 2 − 4∆
The eigenvalues of A are λ = .
2
There are three distinct cases depending on the sign of τ 2 − 4∆.
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Example (1) (Eigenvalues and the Shape of Solutions)
Find the eigenvalues of the linear system
x 0 = −x + y
y 0 = cx − y
in terms of c, and use a numerical solver to discover the shapes of
solutions corresponding to the cases c = 41 , 4, 0, and −9.
Solution !
−1 1
A= , ∆ = 1 − c, τ = −2
c −1
√
√
p
τ ± τ 2 − 4∆ −2 ± 4 − 4(1 − c)
λ= = = −1 ± c
2 2
c = 14 : λ = − 21 and λ = − 23 .
c = 4: λ = 1 and λ = −3.
c = 0: λ = −1.
c = −9: λ = −1 + 3i and λ = −1 − 3i. (Continued)
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Solution (Continued)
The figures below are phase portraits of the above system for the
four values of the parameter c.
1
c= 4 c=4 c=0 c = −9

λ = − 12 , − 23 λ = 1, −3 λ = −1 λ = −1 ± 3i

Each solution curve X(t) in the figures starts from a point

X(0) = X0 lying on the boundary of the square [−1, 1] × [−1, 1].
Each green arrow indicates the direction in which the variable t
increases.
Notice that all except the case of c = 4 are locally stable and have
real part of the eigenvalues negative.
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Case I: Real Distinct Eigenvalues λ1 , λ2 (τ 2 − 4∆ > 0)
The general solution of X0 = AX is given by
X(t) = c1 K1 e λ1 t + c2 K2 e λ2 t (2)
h i
= e λ1 t c1 K1 + c2 K2 e (λ2 −λ1 )t (3)
where K1 , K2 are the corresponding eigenvectors and λ1 > λ2 .
Depending on the signs of λ1 , λ2 , there are three distinct cases:
(a) λ2 < λ1 < 0 (b) 0 < λ2 < λ1 (c) λ2 < 0 < λ1
τ 2 − 4∆ > 0 τ 2 − 4∆ > 0 τ 2 − 4∆ > 0
τ < 0, ∆ > 0 τ > 0, ∆ > 0 ∆<0

Stable Node Unstable Node Saddle point

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Example (2a) (Real Distinct Eigenvalues)
Classify the critical point (0, 0) of the linear system as either a
stable node, an unstable node, or a saddle point:
!
0 2 3
X = X
2 1
Solution
2−λ 3
= λ2 − 3λ − 4 = (λ − 4)(λ + 1) = 0,
2 1−λ
λ1 = 4, λ2 = −1. The critical point (0, 0) is a saddle point.
! ! ! !
2−4 3 k1 0 3
= , K1 =
2 1−4 k2 0 2
! ! ! !
2 − (−1) 3 k1 0 1
= , K2 =
2 1 − (−1) k2 0 −1

(Continued)
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Solution (Continued) ! !
3 4t 1
The general solution is X(t) = c1 e + c2 e −t .
2 −1

When c1 = 0, c2 6= 0, then the initial point X(0) is on the line

y = −x and X(t) approaches 0. For any other initial point X(0),
X(t) become unbounded as t increases.

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Example (2b) (Real Distinct Eigenvalues)
Classify the critical point (0, 0) of the linear system as either a
stable node, an unstable node, or a saddle point:
!
0 −10 6
X = X
15 −19
Solution
−10 − λ 6
= λ2 + 29λ + 100 = (λ + 4)(λ + 25) = 0,
15 −19 − λ
λ1 = −4, λ2 = −25. The critical point (0, 0) is a stable node.
! ! ! !
−10 − (−4) 6 k1 0 1
= , K1 =
15 −19 − (−4) k2 0 1
! ! ! !
−10 − (−25) 6 k1 0 2
= , K2 =
15 −19 − (−25) k2 0 −5

(Continued)
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Solution (Continued) ! !
1 −4t 2
The general solution is X(t) = c1 e + c2 e −25t .
1 −5

All solutions approach (0, 0) as t increases.

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Case II: A Repeated Real Eigenvalue λ1 (τ 2 − 4∆ = 0)

There are two different forms depending on the number of linearly

independent eigenvectors corresponding to λ1 .
(a) Two linearly independent eigenvectors K1 and K2
The general solution is given by

X(t) = c1 K1 e λ1 t + c2 K2 e λ1 t = (c1 K1 + c2 K2 )e λ1 t

If λ1 < 0, then X(t) approaches 0 along the lines

determined by the vector c1 K1 + c2 K2 as in the
figure. The critical point is called a degenerate
stable node.
If λ1 > 0, the arrows in the figure are reversed
and the critical point is called a degenerate un-
stable node. (Continued)

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Case II: (Continued)
(b) A single linearly independent eigenvector K1
The general solution is given by

X(t) = c1 K1 e λ1 t + c2 (K1 te λ1 t + Pe λ1 t )

where (A − λ1 I)P = K1 and the solution may be rewritten as

c1 c2
 
λ1 t
X(t) = te c2 K1 + K1 + P
t t
If λ1 < 0, then limt→∞ te λ 1 t = 0 and it follows
that X(t) approaches 0 in one of the directions
determined by the vector K1 as in the figure. The
critical point is also called a degenerate stable
node.
If λ1 > 0, the arrows in the figure are reversed
and the critical point is called a degenerate un-
stable node.
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Example (3a) (A Repeated Eigenvalue)
Classify the critical point (0, 0) of the linear system
!
0 3 −18
X = X.
2 −9

Discuss the nature of the solution that satisfies X(0) = (1, 0).
Determine a parametric equation of the solution.
Solution
τ = 3 − 9 = −6, ∆ = −27 + 36 = 9, λ2 + 6λ + 9 = (λ + 3)2 = 0
λ1 = λ2 = −3 < 0, (0, 0) is a degenerate stable node.
! ! ! !
3+3 −18 k1 0 3
(A − λI)K = 0, = , K=
2 −9 + 3 k2 0 1
! ! ! !
1
6 −18 p1 3 2
(A − λI)P = K, = , P=
2 −6 p2 1 0

(Continued) 14 / 22
Solution (Continued)
The general solution is
! " ! ! #
1
3 −3t 3
X(t) = c1 e + c2 te −3t + 2 e −3t
1 1 0
! !
3c1 + c2 /2 1
X(0) = = , c1 = 0, c2 = 2
c1 0

x (t) = (6t + 1)e −3t , y (t) = 2te −3t

As in the figure, X(t) approaches (0, 0) in the direction specified
by the line y = x /3.

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Case III: Complex Eigenvalue λ1 , λ2 = λ̄1 (τ 2 − 4∆ < 0)
If λ1 = α + iβ, λ̄1 = α − iβ are the complex eigenvalues and
K1 = B1 + iB2 is a complex eigenvector corresponding to λ1 , the
general solution can be written as X(t) = c1 X1 (t) + c2 X2 (t) where

X1 (t) = (B1 cos βt − B2 sin βt)e αt

X2 (t) = (B2 cos βt + B1 sin βt)e αt .

Therefore the solution is in the form

x (t) = e αt (c11 cos βt + c12 sin βt)
y (t) = e αt (c21 cos βt + c22 sin βt)

(a) Pure imaginary roots (τ 2 − 4∆ < 0, τ = 0)

This is the case when α = 0. The solution curve is periodic with
period 2π/β and is an ellipse if c11 c22 − c12 c21 6= 0. The critical
point 0 is called a center. (Continued)

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Case III: (Continued)
(b) Nonzero real part (τ 2 − 4∆ < 0, τ 6= 0)
The critical point 0 is called a stable spiral point if α < 0 and an
unstable spiral point if α > 0.

In the phase portraits below, the direction of rotation is determined

by the signs of c11 c22 − c12 c21 and β.

(a) α = 0 (b1) α < 0 (b2) α > 0

Center Stable spiral point Unstable spiral point

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Example (3b) (Complex Eigenvalues)
Classify the critical point (0, 0) of the linear system
!
0 −1 2
X = X.
−1 1

Discuss the nature of the solution that satisfies X(0) = (1, 0).
Determine a parametric equation of the solution.

Solution
τ = −1 + 1 = 0, ∆ = −1 + 2 = 1, λ2 + 1 = 0
λ1 = i, λ2 = λ̄1 = −i, (0, 0) is a center.
! ! ! ! ! !
−1 − i 2 k1 0 1−i 1 −1
= , K1 = = +i
−1 1−i k2 0 1 1 0
! !
1 −1
B1 = , B2 =
1 0

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Solution (Continued)
" ! ! # " ! ! #
1 −1 −1 1
X(t) = c1 cos t − sin t + c2 cos t + sin t
1 0 0 1
! !
cos t + sin t − cos t + sin t
= c1 + c2
cos t sin t
! ! !
1 −1 1
X(0) = c1 + c2 = , c1 = 0, c2 = −1
1 0 0

x (t) = cos t − sin t, y (t) = − sin t

As in the figure, X(t) turns around the ellipse clockwise with
period 2π.

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Classifying Critical Points

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Example (4) (Classifying Critical Points)
Classify the critical point (0, 0) of each of the following linear
system X0 = AX
! !
1.01 3.10 −ax̂ −abx̂
(a) A = (b) A=
−1.10 −1.02 −cd ŷ −d ŷ

for positive constants a, b, c, d, x̂ , ŷ .

Solution
(a) τ = −0.01, ∆ = 2.3798, τ 2 − 4∆ < 0. The critical point is a
stable spiral point.
(b) τ = −(ax̂ + d ŷ ) < 0, ∆ = ad(1 − bc)x̂ ŷ .
If bc > 1, then ∆ < 0 and the critical point is a saddle point.
If bc < 1, then ∆ > 0 and the critical point is either a stable node,
a degenerate stable node, or a stable spiral point. In all three of
these cases limt→∞ X(t) = 0.

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Stability Criteria

Theorem (11.2.1) (Stability Criteria for Linear Systems)

For a linear plane autonomous system X0 = AX with det A 6= 0,
let X = X(t) denote the solution that satisfies the initial
condition X(0) = X0 , where X0 6= 0.
(a) limt→∞ X(t) = 0 if and only if the eigenvalues of A have
negative real parts. This will occur when ∆ > 0 and τ < 0.
(b) X(t) is periodic if and only if the eigenvalues of A are pure
imaginery. This will occur when ∆ > 0 and τ = 0.
(c) In all other cases, given any neighborhood of the origin,
there is at least one X0 in the neighborhood for which
X(t) becomes unbounded as t increases.

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