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

Recall: Newton’s 2nd Law for Small Oscillations

(3) ( )22

32 1 1 1''(0) (0) (0)2! 3! ! = (0) '(0) n nF x F x F xnd xd

Fm F xt

Equilibrium:F=0 ~0

22 '(0) 0 = '(0) oscillation

What if '(0) 0 ??

d xm F x Fdt

F

x

0x

mF

Systems of 1st-order, linear, homogeneous equations

where ( ) time-dependent vector; consta

OR

In general:

nt, matrix.

dv Avd

df af bgdtdg cf dgdt

a bd f fg gdt c d

t

v tA N N

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

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?

Conversion to 1st order

Define such that .

2nd order 2 1st order

0 11 0

xx

x

xx x

x

x

f f

g f gf g

f gg f

d f fg gdx

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

Ecosystem modeling( ) = foxes( ) = hares

F t

dH rH aHFdt

dF sF bHFdt

H t

equilibria:

0 ( )

0 ( )

1) 0

2) / ; /

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

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

0 0e.g. 0

' ' '~

' ' '~

rt

st

H H H

H F

d rdt

d sF F edt

e

F

'H

'F

Phase plane diagram

0 0Homework: / ; /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

a rF 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 case

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

( )

th

1 2

1,2

| | 0

order polynomial equation for , solutions.

2 2 solutions:

Either , both realor

n

r i

A

NN

N

yesterday

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)

(1) (2)0, 0

0r r

i i

(1) (2)

(1) (2)0

0r r

i i

(1) (2)

(1) (2)0

0r r

i i

(1) (2)

(1) (2)0

0r r

i i

(1) (2)

(1) (2)0, 0

0r r

i i

(1) (2)

(1) (2)0, 0

0r r

i i

a. attractor c. saddle

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

Interpreting two σ’s

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)

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

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), 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

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

Example: Birds of a feather

(1) (2)

(1

2

)

2

) (1

growth eigvec:

0

0

,

:

0

.

( )

b R RJ Jb a

b RJb a

a b a b

a a a b

b b R Det b bJb b

a a

b

a

b

a

b

R

b

0 or J R J

Example: Birds of a feather

(

(

2) (2)

2 2

1) (2)

decay eigvec:: ( )

0 0.

0

,

0

a a a b a a b b

b b R Det b bJb b

bR

b R RJ Jb a

b RJ

a

b a

a b a

bJ

b

a

or R J

R

J R JR J

R

J

R

J

Why a saddle is unstable

R

J

No matter where you start, things eventually blow up.