Math 211 Math 211 Lecture #30 Solutions of Systems and Stability - - PDF document

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Math 211 Math 211

Lecture #30 Solutions of Systems and Stability November 5, 2003

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Multiplicities Multiplicities

A an n × n matrix

• Distinct eigenvalues λ1, . . . , λk.
• The characteristic polynomial is

p(λ) = (λ − λ1)q1(λ − λ2)q2 · . . . · (λ − λk)qk.

• The algebraic multiplicity of λj is qj.
• The geometric multiplicity of λj is dj, the dimension of the

eigenspace of λj.

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Properties of Multiplicities Properties of Multiplicities

• q1 + q2 + · · · + qk = n.
• 1 ≤ dj ≤ qj.
• There are dj linearly independent exponential solutions

corresponding to λj.

• If dj = qj for all j we have n linearly independent

exponential solutions.

• If dj < qj we can use our Proposition.

1 John C. Polking

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Generalized Eigenvectors Generalized Eigenvectors

Definition: If λ is an eigenvalue of A and [A − λI]pv = 0 for some integer p ≥ 1, then v is called a generalized eigenvector associated with λ. Theorem: If λ is an eigenvalue of A with algebraic multiplicity q, then there is an integer p ≤ q such that null([A − λI]p) has dimension q.

• For each generalized eigenvector v we can compute etAv.
• We can find q linearly independent solutions associated with

the eigenvalue λ.

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Procedure for Solving x′ = Ax Procedure for Solving x′ = Ax

• Find the eigenvalues and their algebraic multiplicities.
• For each eigenvalue λ with algebraic multiplicity q find q

linearly independent solutions associated with λ:

Find the smallest integer p such that null([A − λI]p)

has dimension q.

Find a basis v1, v2, . . . , vq of null([A − λI]p). For j = 1, 2, . . . , q compute xj(t) = etAvj.

• This results in n linearly independent solutions.

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Procedure for a Complex Eigenvalue Procedure for a Complex Eigenvalue

• If λ is complex of algebraic multiplicity q. Then λ also has

multiplicity q.

Find the smallest integer p such that null([A − λI]p) has

dimension q.

Find a basis w1, w2, . . . , wq of null([A − λI]p). For j = 1, 2, . . . , q compute zj(t) = etAw. Compute xj(t) = Re(zj(t)) and yj(t) = Im(zj(t)).

• This results in 2q linearly independent real solutions

corresponding to the eigenvalues λ and λ.

2 John C. Polking

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Stability Stability

Autonomous system x′ = f(x) with an equilibrium point at x0. The basic question is: What happens to all solutions that start near x0 as t → ∞?

• x0 is stable if for every ǫ > 0 there is a δ > 0 such that a

solution x(t) with |x(0) − x0| < δ ⇒ |x(t) − x0| < ǫ for all t ≥ 0.

Every solution that starts close to x0 stays close to x0. In dimension 2 centers and sinks are stable.

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• x0 is asymptotically stable if it is stable and there is an

η > 0 such that if x(t) is a solution with |x(0) − x0| < η, then x(t) → x0 as t → ∞.

Every solution that starts close to x0 approaches x0. In d = 2 sinks are asymptotically stable, centers are not. x0 is called a sink.

• x0 is unstable if there is an ǫ > 0 such that for any δ > 0

there is a solution x(t) with |x(0) − x0| < δ with the property that there are values of t > 0 such that |x(t) − x0| > ǫ.

There are solutions starting arbitrarily close to x0 that

move away from x0.

In d =2 sources and saddles are unstable.

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Dimension 2 Dimension 2

• Sinks are asymptotically stable.

The eigenvalues have negative real part.

• Sources are unstable.

The eigenvalues have positive real part.

• Saddles are unstable.

One eigenvalue has positive real part.

• Centers are stable but not asymptotically stable.

The eigenvalues have real part = 0.

3 John C. Polking

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Stability Theorem Stability Theorem

Theorem: Let A be an n × n real matrix.

• Suppose the real part of every eigenvalue of A is negative.

Then 0 is an asymptotically stable equilibrium point for the system x′ = Ax.

• Suppose A has at least one eigenvalue with positive real
• part. Then 0 is an unstable equilibrium point for the

system x′ = Ax. Notice that if there are eigenvalues with real part equal to 0, no conclusion is made.

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Examples Examples

• Suppose the dimension is 2 and T 2 − 4D = 0.

T < 0 ⇒ sink. T > 0 ⇒ source.

• y′ = Ay,

A =

⎛ ⎜ ⎜ ⎝

−2 −18 −7 −14 1 6 2 5 2 2 −3 −2 −8 −1 −6

⎞ ⎟ ⎟ ⎠ .

A has eigenvalues −1, −2, & −1 ± i. 0 is asymptotically stable.

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Propostion Propostion

Proposition: Suppose that A is an n × n matrix, λ is a number, and v is a vector.

• 1. If [A − λI]v = 0, then etAv = eλtv.
• 2. If [A − λI]2v = 0, then etAv = eλt(v + t[A − λI]v).
• 3. If [A − λI]kv = 0, then

etAv = eλt

• v + t[A − λI]v + t2

2! [A − λI]2v + · · · + tk−1 (k − 1)![A − λI]k−1v

• .

4 John C. Polking