MA282 – Spring 2013
11.2 Stability of Linear Systems
Introduction
In section 11.1 we saw that the plane autonomous system
𝑑𝑥
= 𝑃(𝑥, 𝑦)
𝑑𝑡
𝑑𝑦
= 𝑄(𝑥, 𝑦)
𝑑𝑡
gives rise to a vector field
𝑽(𝑥, 𝑦) = (𝑃(𝑥, 𝑦), 𝑄(𝑥, 𝑦)). In this section we examine the
behavior of solution when 𝑿(0) = 𝑿0 is close to a critical point of the system.
Definitions We interpret the solution as a parametric curve, that is, a path of a moving particle.
If the particle returns to the critical point, i.e., lim⁡𝑿(𝑡) = 𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙⁡𝑝𝑜𝑖𝑛𝑡 then we call the
critical point locally stable. However if the particle gets away from the critical point, we call the
critical point unstable.
We investigate the stability for linear plane autonomous systems:
𝑥′
𝑎
( )=(
𝑦′
𝑐
𝑏 𝑥
) ( ) ⁡⁡𝑿′ = ⁡𝑨𝑿
𝑑 𝑦
Note that any linear plane autonomous system has the only equilibrium solution at the origin 0.
I.
Distinct Real Eigen-values
Suppose that the matrix 𝑨 has two distinct eigenvalues, 𝜆1 ⁡and⁡⁡𝜆2 ⁡with associated
eigenvectors 𝐾1 ⁡and⁡⁡𝐾2 , respectively. The general solution is given by
𝑿 = 𝑐1 𝑲1 𝑒 𝜆1 𝑡 + 𝑐2 𝑲2 𝑒 𝜆2 𝑡
Case 1. 𝜆1 < 𝜆2 < 0
Example
Once again we consider
𝑑𝒀
−3
=⁡(
−1
𝑑𝑡
1
)𝒀
0
In this example, the eigenvalues are
𝜆=
Both are negative.
−3 ± √5⁡⁡⁡
2
The slope of the eigenline that⁡corresponds⁡to⁡the⁡“fast”⁡eigenvalue⁡ 𝜆1 = (−3 − √5)/2⁡ is
approximately 0.4; the slope of the eigenline that⁡corresponds⁡to⁡the⁡“slow”⁡eigenvalue⁡ 𝜆2 =
(−3 + √5)/2⁡ is approximately 2.6.
All solutions tend to the origin. A critical point is called a stable node.
Once we understand the phase portrait we can sketch the component graphs by hand.
Let’s sketch the 𝑥(𝑡) − and 𝑦(𝑡) −graphs that correspond to the initial conditions (−3, 2)
And (3, 2) for
𝑑𝒀
−3
=⁡(
−1
𝑑𝑡
1
)𝒀
0
Case 2. 𝜆1 < 0 < 𝜆2
Example
Consider
𝑑𝒀
4 −5
=⁡(
)𝒀
−2 1
𝑑𝑡
The eigenvalues are
𝜆1 = ⁡ −1,⁡⁡⁡𝜆2 = 6
The slope of the eigenline that⁡corresponds⁡to⁡the⁡“slow”⁡eigenvalue⁡ 𝜆1 = −1⁡is 1; the slope of
the eigenline that⁡corresponds⁡to⁡the⁡“fast”⁡eigenvalue⁡ 𝜆2 = 6⁡ is −2/5.
All solutions come from ∞ and approach 0 along the line determined by the eigenvector 𝑲1
and tend back to ∞ along the line determined by the eigenvector 𝑲2 . This unstable critical
point is called a saddle point.
Case 3. 0 < 𝜆1 < 𝜆2
Example
Consider
𝑑𝒀
3
=⁡(
1
𝑑𝑡
−1
)𝒀
0
The eigenvalues are
𝜆=
3 ± √5⁡⁡⁡
2
The slope of the eigenline that⁡corresponds⁡to⁡the⁡“slow”⁡eigenvalue⁡ 𝜆1 = (3 − √5)/2⁡ is
approximately 2.62; the slope of the eigenline that⁡corresponds⁡to⁡the⁡“fast”⁡eigenvalue⁡ 𝜆2 =
(3 + √5)/2⁡ is approximately 0.38.
All solutions tend away from the origin. This type of critical point, corresponding to the case
when both eigenvalues are positive, is called an unstable node.
Case 4. 𝜆1 ≠ 0, 𝜆2 = 0
Example
Consider
𝑑𝒀
3
−2 ⁡⁡1
=⁡(
) 𝒀,⁡⁡⁡𝒀(0) = ( )
0
⁡⁡⁡2 −1
𝑑𝑡
The eigenvalues are
𝜆 = −3, 0
The slope of the eigenline that corresponds to 𝜆1 = −3⁡ is −1; the slope of the eigenline
that corresponds to 𝜆2 = 0⁡ is 2.
Phase Portrait:
x- and y- graphs:
II.
Repeated Eigen-value
Suppose that the matrix 𝑨 has a repeated eigenvalue λ1 .
(a) Two linearly independent eigenvectors
If 𝐾1 ⁡and⁡⁡𝐾2 are two linearly independent eigenvectors, the general solution is given by
𝑿 = 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 line determined by the vector 𝑐1 𝑲1 + 𝑐2 𝑲2
and the critical point is called a degenerate stable node.
When λ1 > 0. the arrows are and we have a degenerate unstable node.
(b) A single linearly independent eigenvector
When only a single linearly independent eigenvector 𝑲1 exists, the general solution
is given by
𝑿 = 𝑐1 𝑲1 𝑒 𝜆1 𝑡 + 𝑐2 (𝑲1 𝑡𝑒 𝜆1 𝑡 + 𝑷𝑒 𝜆1 𝑡 )
where (𝐀 − λ1 𝐈)𝐏 = 𝐊1 .
If λ1 < 0 then lim 𝑡𝑒 𝜆1 𝑡 = 0⁡and it follows that 𝑿(𝑡) approaches 0 in one of the directions
t→∞
determined by the vector 𝑲1 .
When λ1 > 0, the solutions look like those in the figure above with the arrows reversed. The
critical point is again called a degenerate unstable node.
III.
Complex Eigenvalues
If 𝜆1 = 𝛼 + 𝛽𝑖⁡is a complex eigenvalue and 𝑲𝟏 ⁡is a complex eigenvector corresponding to 𝜆1 ,
the general solution can be written as
𝑿 = 𝑐1 Re𝑿1 + 𝑐2 Im𝑿1
where 𝐗1 = 𝐊1 eλ1t = 𝐊1 𝑒 𝛼 (cos⁡𝛽𝑡 + 𝑖⁡sin⁡𝛽𝑡).
(a) If 𝛼 > 0, 𝑒 𝛼𝑡⁡ increases as 𝑡 → ∞, and the solutions spiral away from the origin. The
critical
point is called a unstable spiral point.
(b) If 𝛼 < 0, 𝑒 𝛼𝑡⁡ decreases as 𝑡 → ∞, and the solutions spiral toward the origin. The critical
point is called a stable spiral point.
(c) If 𝛼 = 0, the solutions are periodic. The critical point (0, 0) is called a center.
𝐄𝐱𝐚𝐦𝐩𝐥𝐞⁡⁡Classify the critical point (0, 0) of each of the following linear systems 𝑿′ = 𝑨𝑿.
(a) 𝑨 = (
3 −18
)
2 −9
(b) 𝑨 = (
−1 2
)
−1 1
In each case discuss the nature of the solution that satisfies 𝑿(0) = (1, 0). Determine
parametric equations for each solution.
⁡