Recall: Pendulum

s

mg

22

22

sin

sin

Newton:ma F

d sm mgdt

dm mgdt

2 3 52 sin ...3! 5!

provided is small

g gddt

g

Oscillation with angular frequency g

F

m

Unstable Pendulum

mg

22 sin

/2

Newton:

Expand about

dm mgdt

2 3 52 sin ...3! 5!

g gddt

g

g gt tAe Be

Exponential growth dominates.Equilibrium is unstable.

Recall:Finding eigvals and eigvecs

( ) ( ) ( ) ( ) ( )

( )

th

( ) 0

homogeneous | | 0

order polynomial equation for , solutions.

n n n n n

n

Av v A v

A

NN

Nonlinear systems: the qualitative theoryDay 8: Mon Sep 20

Two unknown function ( ), ( )s:

OR

f t g t

f f g

g f g

f fg

d

g

a bdtd c ddt

a bddt c d

1. How we solve it (the basic idea).2. Why it matters.3. How we solve it (details, examples).

Systems of 1st-order, linear, homogeneous equations

where time-dependent vector;

In gene

(

ral:

) constant, mat

rix.

d Adt

Av t

N

v

N

v

Solution: the basic idea

0

0

0 0

0 0

Try: ( )

OR .

t

t

t t

dv Avdt

v t v e

dv v edt

v e Av e

Av v

0 eigvec of eigval

v A

General solution

.

0 0

0

, where

At 0,

What if the initial state is NOT an eigve

( ) e

c

igvec.

?

tv t v e v

v vt

General solution

.

0 0

0

, where

At 0,

What if the initial state is NOT an eigve

( ) e

c

igvec.

?

tv t v e v

v vt

1 2(1) (2) ( )1 0 2 0 0

Linear, homogeneous superposition

Works for any initial condition (usually)

( )

.

Nt t tNNv t a v e a v e a v e

Systems of 1st-order, linear, homogeneous equations

1. Higher order equations can be converted to 1st order equations.2. A nonlinear equation can be linearized.3. Method extends to inhomogenous equations.

Why important?

1. 2. 3.

Conversion to 1st order

Define such that .

2nd order 2 1st order

0 11 0

tt

t

tt t

t

t

f f

g f gf g

f gg f

d f fg gdt

Another example

3 2

Define: ,

3 2

3rd order 3 1st order

xxx x

x xx x

x

x

x

f ff

f g f g h

f gg hh fg

Any higher order equation can be converted to a set of 1st order equations.

dx Py Pxdtdy rx y xzdtdz xy bzdt

Nonlinear systems: qualitative solution

x

y

e.g. Lorentz: 3 eqnschaos

Stability of equilibria is alinear problem°qualitative description of solutions

phase planediagram

2-eqns: ecosystem modeling( ) = foxes( ) = hares

F t

dH rH aHFdt

dF sF bHFdt

H t

reproduction

starvation eating

getting eaten

Ecosystem modeling( ) = foxes( ) = hares

F t

dH rH aHFdt

dF sF bHFdt

H t

reproduction

starvation eating

getting eaten

( )

( )

dH r aF Hdt

dF s bH Fdt

OR: Reproduction rate reduced

Starvation rate reduced

Equilibria( ) = foxes( ) = hares

F t

dH rH aHFdt

dF sF bHFdt

H t

0 0

0 0

0 0

0 0

equilibria , :

0 ( )

0 ( )

1) 0

F H

H r aF

F s bH

F H

H

F

Equilibria( ) = foxes( ) = hares

F t

dH rH aHFdt

dF sF bHFdt

H t

0 0

0 0

0 0

0 0

0 0

equilibria , :

0 ( )

0 ( )

1) 0

2) / ; /

F H

H r aF

F s bH

F H

F r a H s b

H

F

/r a

/s b

Linearizing about an equilibrium

dH rH aHFdt

dF sF bHFdt

2nd-order (quadratic) nonlinearity

Linearizing about an equilibrium

dH rH aHFdt

dF sF bHFdt

2nd-order (quadratic) nonlinearity

0 0

0 0

0 0 0 0 0 0

Expand about equilibrium state , : ; ( )' ' prime = small perturbation

' ' ') ')( ' '(

H FH H F F

HF H

H F

H F FF HH F HH FF

0 0 0 0 ' ' H F H F HF

small small reallysmall

The linearized system

0 0

0 0

( )

(' ' '

'

)

' 'H H F

F

d r aF aHdt

d s bH bFdt F H

dH rH aHFdt

dF sF bHFdt

0 0

0

0

0 0 0 0

0 0( ) (

( ) (

' '

' ')

' )

'

dH r H adt

d

H F H F

H F H FF s F b

H F H

F Fd Ht

cancel

The linearized system

0 0

0 0

( )

(' ' '

'

)

' 'H H F

F

d r aF aHdt

d s bH bFdt F H

0 0e.g. 0

' ' '~

' ' '~

rt

st

H H H

H F

d rdt

d sF F edt

e

F

'H

'F

Phase plane diagram

The “other” equilibrium

H

F

/r a

/s b

'H

'F

'F

'H

Section 6Problem 4

0 0/ ; /F r a H s b

?

Linear, homogeneous systems

0 0

0 0

0 0

0 0

For example:

' ( ) ' '

' ( ) ' '

' ' .' '

dv Avdt

dH r aF H aH Fdt

dF s bH F bF Hdt

r aF aHH Hddt F FbF s bH

Solution0

0

eigvec of eigval growth rate (generally complex)

tdv Av v v edt

v A

0

( )

In general

so (cos sin )r i r ri

t

i t i tt tti

r i

i

v v e

i

e e e e e t i t

Interpreting σ

growth rate oscillation frequencyir

Interpreting σ

General solution

1 2

1 2

(1) (2) ( )1 0 2 0 0

(1) (2)1 0 2 0

( )

( )2:

Nt t tNN

t t

v t a v e a v e a v e

v t a v vN e a e

N=2 case1 2(1) (2)

1 0 2 0( ) t tv t a v e a v e

( )

th

1

21,2

1

2

,2

| | 0

order polynomial equation for , solutions.

2 quadratic for 2 solutions:

42

Two cases:

, both real

= complex conjugate pair

i

n

r

A

NN

N

b b aca

Recall

b. repellor (unstable)a. attractor (stable) c. saddle (unstable)

d. limit cycle (neutral) e. unstable spiral f. stable spiral

Interpreting two σ’s

Strange Attractor

Need N>3

b. repellor

(1) (2)0, 0r r (1) (2)0, 0r r

(1) (2)0, 0r r

a. attractor c. saddle

Interpreting two σ’sboth real

(1) (2) 0i i

(1) (2) 0r r (1) (2) 0r r

(1) (2) 0r r

d. limit cycle e. unstable spiral f. stable spiral

Interpreting two σ’s:complex conjugate pair

(1) (2) 0i i

b. repellor

(1) (2)

(1) (2)0, 00i i

r r

(1) (2

(1) (2

)

) 00

i i

r r

(1) (2

(1) (2

)

) 00

i i

r r

(1) (2

(1) (2

)

) 00

i i

r r

(1) (2)

(1) (2)0, 00i i

r r

(1) (2)

(1) (2)0, 00i i

r r

a. attractor c. saddle

d. limit cycle e. unstable spiral f. stable spiral

Interpreting two σ’s

The mathematics of love affairs

dR aR bJdt

R(t)=Romeo’s affection for JulietJ(t) = Juliet’s affection for Romeo

Response toown feelings(><0)

Response toother person(><0)

Strogatz, S., 1988, Math. Magazine 61, 35.

The mathematics of love affairs(S. Strogatz)

dR aR bJdt

R(t)=Romeo’s affection for JulietJ(t) = Juliet’s affection for Romeo

Response toown feelings(><0)

Response toother person(><0)

Likewise: dJ cJ dRdt R a b Rddt J Jc d

Example: Out of touch with feelings

2

0; , or,0

0, 0.

0

0 1 0 00 10

adR dJaJ bR Adt dt b

a b

A

aa abbb

i ab

Limit cycle(1) (2)

(1) (2)0r r

i i ab

R

J

i ab

Sense of circulation:

; ,

So, e.g.: 0, 0 0

clockwise

dR dJaJ bRdt dt

dRJ R dt

Example: Birds of a feather

2 2

2 2

2 2

(1) (2)

; , or,

0, 0.

0

1 0 ( ) 00 1

( ) 0

( )

,

a

a

bdR dJaR bJ bR aJ Adt dt b a

a b

A

b a b a bb a b a

a b

a b

a b

a b a b

Example: Birds of a feather(1) (2), a b a b

negativepositive if b>anegative if b<a

b<a: both negative (romance fizzles)b>a: one positive, one negative (saddle …?)

both real

(1) (2)

(1) (2)0, 0

0r r

i i

c. saddle

growth eigvec

decay eigvec

Example: Birds of a feather

(1) (2)

(1)

(2)

( ) 0

0

,

(

(

1

)1

)

a

J a

b RJ

a a b a

a a b a b

b a

a b a b

a R bJ

R

J

b

R

J

b

R

R

J(1)

(2)

Decaying case: a>b

R

J R JR J

Saddle: a<b

R

J

R

J

Homework

Sec. 6, p. 89

#4: Sketch the full phase diagram:

H

F

/r a

/s b

'H

'F

'F

'H

?

?

#6: Optional

Why a saddle is unstable

R

J

No matter where you start, things eventually blow up.