Dynamical Systems Analysis II:Evaluating Stability, Eigenvalues

By Peter Woolf (pwoolf@umich.edu)University of Michigan

Michigan Chemical Process Dynamics and Controls Open Textbook

version 1.0

Creative commons

Problem: Given a large and complex system of ODEs describing the dynamics and control of your process, you want to know:

(1)Where will it go?

(2)What will it do?

Is there anything fundamental you can say about it?

E.g. With my control architecture, this process will always ________.

Solution: Stability Analysis

Steady state from last lecture.

Topic for today!

Exponential increase Increase w/ oscillation

Stable oscillation

Periodic solution Non-periodic solution(chaotic)

Only possible for nonlinear systems

Decay w/ oscillationExponential decay

What will your system do?

How can we know where the system will go?

Possible approaches:

1. Simulate system and observe

Disadvantages:• Can’t provide guaranteed behavior, just samples of possible trajectories.• Requires simulations starting from many points• Assumes we have all variables defined, thus hard to use to design controllers.

Advantages:• Works for any system you can simulate• Intuitive--you see the results

How can we know where the system will go?

Possible approaches:

1. Simulate system and observe

2. Stability Analysis (this class)

Disadvantages:• Only works for linear models• Linear approximations of nonlinear models break down away from the point of linearization

Advantages:• Provides strong guarantees for linear systems• General

€

dA

dt= 3A − A2 − AB

dB

dt= 2B − AB − 2B2

Nonlinear modelLinear approximation at A=0, B=0

JacobianOr in a different format

€

dA

dt= 3A

dB

dt= 2B

From last class…

€

′ A

′ B

⎡

⎣ ⎢

⎤

⎦ ⎥=

3 0

0 2

⎡

⎣ ⎢

⎤

⎦ ⎥A

B

⎡

⎣ ⎢

⎤

⎦ ⎥+

0

0

⎡

⎣ ⎢

⎤

⎦ ⎥

Intuitively, what will the linear system do if A is perturbed slightly from 0?

€

dA

dt= 3(0 + Δ) Increase in A above 0

yields a positive derivative

Increase in A

Increase in slope of A

Exponential increase

But what if our model is more complex?

€

′ A

′ B

⎡

⎣ ⎢

⎤

⎦ ⎥=

3 −2

2 −2

⎡

⎣ ⎢

⎤

⎦ ⎥A

B

⎡

⎣ ⎢

⎤

⎦ ⎥+

3

4

⎡

⎣ ⎢

⎤

⎦ ⎥

E.g. (note: example below is made up)Or in a different format

€

dA

dt= 3A − 2B + 3

dB

dt= 2A − 2B + 4

What will happen if A or B are increased slightly from the steady state value of A=1, B=3?

Result: increase A, A and B increase!

Result: increase B, A and B decrease!

€

dA

dt= 3(1+ Δ) − 2(3) + 3 = +3Δ

dB

dt= 2(1+ Δ) − 2(3) + 4 = +2Δ

Increase A by :

€

dA

dt= 3(1) − 2(3+ Δ) + 3 = −2Δ

dB

dt= 2(1) − 2(3+ Δ) + 4 = −2Δ

Increase B by :

Observations:1. It is easy to predict where a linear system will go if the

variables are decoupled

2. Coupling between variables makes it harder to predict what will happen

3. Coupling is determined by the Jacobian

€

dA

dt= 3A

dB

dt= 2B

A only influences A, B only influences B.-> Variables are decoupled

€

dA

dt= 3A − 2B + 3

dB

dt= 2A − 2B + 4

Changes in A influence changes in A and B. Changes in B influence changes in A and B.--> Variables are coupled

Is it possible to change a coupled system to a decoupled one?

€

′ A

′ B

⎡

⎣ ⎢

⎤

⎦ ⎥=

k11 k12

k21 k22

⎡

⎣ ⎢

⎤

⎦ ⎥A

B

⎡

⎣ ⎢

⎤

⎦ ⎥+

k13

k23

⎡

⎣ ⎢

⎤

⎦ ⎥

€

k11 k12

k21 k22

⎡

⎣ ⎢

⎤

⎦ ⎥A

B

⎡

⎣ ⎢

⎤

⎦ ⎥

€

λ1 0

0 1

⎡

⎣ ⎢

⎤

⎦ ⎥A

B

⎡

⎣ ⎢

⎤

⎦ ⎥

??

Can we find a λ value that satisfies this relationship?

€

k11 k12

k21 k22

⎡

⎣ ⎢

⎤

⎦ ⎥− λ

1 0

0 1

⎡

⎣ ⎢

⎤

⎦ ⎥

⎛

⎝ ⎜

⎞

⎠ ⎟A

B

⎡

⎣ ⎢

⎤

⎦ ⎥= 0

€

k11 − λ k12

k21 k22 − λ

⎡

⎣ ⎢

⎤

⎦ ⎥

⎛

⎝ ⎜

⎞

⎠ ⎟A

B

⎡

⎣ ⎢

⎤

⎦ ⎥= 0€

k11 − λ( )A + k12B = 0

k21A + k22 − λ( )B = 0

Written differently..

This is an eigenvalue

€

k11 − λ( )A + k12B = 0

k21A + k22 − λ( )B = 0

€

k11A − λA + k12B = 0

k21A + Bk22 − λB = 0

expand

€

B =−k11A + λA

k12

k21A +−k11A + λA

k12

⎛

⎝ ⎜

⎞

⎠ ⎟k22 − λ

−k11A + λA

k12

⎛

⎝ ⎜

⎞

⎠ ⎟= 0

Solve for B

€

k21A −k11Ak22

k12

+λAk22

k12

+λk11A

k12

+λ2A

k12

= 0

€

A k21 −k11k22

k12

+λk22

k12

+λk11

k12

+λ2

k12

⎛

⎝ ⎜

⎞

⎠ ⎟= 0

€

A k21 −k11k22

k12

+λk22

k12

+λk11

k12

+λ2

k12

⎛

⎝ ⎜

⎞

⎠ ⎟= 0

Solve for λ

€

λ =1

2k11 + k22 ± k11

2 + 4k12k21 − 2k11k22 + k222

[ ]

Observations:1) Yes! There is always a way decouple a coupled linear

system2) Direct approach involves lots of algebra

There is an easier way..

A bit of linear algebra background

Goal: solve this system for λ

€

k11 − λ k12

k21 k22 − λ

⎡

⎣ ⎢

⎤

⎦ ⎥

⎛

⎝ ⎜

⎞

⎠ ⎟A

B

⎡

⎣ ⎢

⎤

⎦ ⎥= 0

Determinant: a property of any square matrix that describes the degree of coupling between the equations.

Determinant equals zero when the system is not linearly independent, meaning one of the equations can be cast as a linear combination of the others.

€

Det

a b c

d e f

g h i

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥= a* Det

e f

h i

⎡

⎣ ⎢

⎤

⎦ ⎥− b* Det

d f

g i

⎡

⎣ ⎢

⎤

⎦ ⎥+ c * Det

d e

g h

⎡

⎣ ⎢

⎤

⎦ ⎥

€

Deta b

c d

⎡

⎣ ⎢

⎤

⎦ ⎥= a * d − b *c

A bit of linear algebra background

Goal: solve this system for λ

€

k11 − λ k12

k21 k22 − λ

⎡

⎣ ⎢

⎤

⎦ ⎥

⎛

⎝ ⎜

⎞

⎠ ⎟A

B

⎡

⎣ ⎢

⎤

⎦ ⎥= 0

Determinant: a property of any square matrix that describes the degree of coupling between the equations.

Determinant equals zero when the system is not linearly independent, meaning one of the equations can be cast as a linear combination of the others.

€

Detk11 − λ k12

k21 k22 − λ

⎡

⎣ ⎢

⎤

⎦ ⎥= 0

Revised Goal: find λ that satisfies

€

k11 − λ( ) k22 − λ( ) − k12k21 = 0

€

λ =1

2k11 + k22 ± k11

2 + 4k12k21 − 2k11k22 + k222

[ ]

Similar Analysis can be done in Mathematica:

Det[{a,b},{c,d}] :Find the determinant of a matrix

Solve [{eqn1, eqn2,..},{var1, var2,..} ] : Solve algebraically

Eigenvalues[{a,b},{c,d}] : Automatically find the eigenvalues

What do eigenvalues tell us about stability?

Eigenvalues tell us the exponential part of the solution of the differential equation system

Three possible values for an eigenvalue1) Positive value: system will increase

exponentially2) Negative value: system will decay

exponentially3) Imaginary value: system will oscillate(note combinations of the above are possible)

What do eigenvalues tell us about stability?

Effect: If any eigenvalue has a positive real part, the system will tend to move away from the fixed point

Marble Analogy

Small perturbations left or right will cause the marble to decay back to the steady state position

Negative real eigenvalue

Small perturbations left or right will cause the marble to decay away from the steady state position (xss) Positive real eigenvalue

Small perturbations in y are stable, while perturbations in x are unstable (saddle point), thus overall point is unstable! Positive and negative real eigenvalues

xssx

Case I: stablexss

x

Case II: unstable

xss,,yss

x

y

Case III: Saddle point

€

′ A

′ B

⎡

⎣ ⎢

⎤

⎦ ⎥=

3 −2

2 −2

⎡

⎣ ⎢

⎤

⎦ ⎥A

B

⎡

⎣ ⎢

⎤

⎦ ⎥+

3

4

⎡

⎣ ⎢

⎤

⎦ ⎥

Revisit our example: What will happen here?

1) Calculate eigenvalues

Eigenvalues: λ1=2, λ2= -1

2) Classify stability:At least one eigenvalue is positive,

so the point is unstable and a saddle point.

Exponential increase

€

′ A

′ B

′ C

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥=

3 −2 1

2 2 −2

−1 2 0

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

A

B

C

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥+

8

−2

4

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

A more complex example: What will happen here?

1) Calculate eigenvalues

Force Mathematica to find a numerical value using N[ ]

Using the Eigenvalue[ ] function in Mathematica

Given these eigenvalues what will it do?

€

′ A

′ B

′ C

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥=

3 −2 1

2 2 −2

−1 2 0

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

A

B

C

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥+

8

−2

4

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

2) Classify stability:

• The real component of at least one eigenvalue is positive, so the system is unstable.

• There are imaginary eigenvalue components, so the response will oscillate.

Increase w/ oscillation

A more complex example: What will happen here?

Exponential increase Increase w/ oscillation

Stable oscillation

Decay w/ oscillationExponential decay

What will your system do?(according to eigenvalues)

All λs are real and negative

All λs are real and at least one positive

All λs have negative real parts, some imaginary parts

At least one λ has positive real parts, some imaginary parts

All λs have zero real parts and nonzero imaginary parts

Take Home Messages

• Stability of linear dynamical systems can be determined from eigenvalues

• Complicated sounding terms like eigenvalues and determinant can be derived from algebra alone--fear not!

• Stability of nonlinear dynamical systems can be locally evaluated using eigenvalues