recall: finding eigvals and eigvecs
DESCRIPTION
Recall: Finding eigvals and eigvecs. Recall: Newton’s 2 nd Law for Small Oscillations. Equilibrium: F=0. ~0. Systems of 1st-order, linear, homogeneous equations. How we solve it (the basic idea). Why it matters. How we solve it (details, examples). Solution: the basic idea. - PowerPoint PPT PresentationTRANSCRIPT
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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
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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
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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).
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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
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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
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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
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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?
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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
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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.
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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
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2-eqns: ecosystem modeling( ) = foxes( ) = hares
F t
dH rH aHFdt
dF sF bHFdt
H t
reproduction
starvation eating
getting eaten
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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
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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
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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
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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
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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
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Solution0
0
eigvec of eigval growth rate (generally complex)
tdv Av v v edt
v A
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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
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Interpreting σ
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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
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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
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b. repellor (unstable)a. attractor (stable) c. saddle (unstable)
d. limit cycle (neutral) e. unstable spiral f. stable spiral
Interpreting two σ’s
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Strange Attractor
Need N>3
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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
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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)
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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
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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
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Limit cycle
(1) (2)
(1) (2)0r r
i i ab
R
J
i ab
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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
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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
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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
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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
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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
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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
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R
J R JR J
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R
J
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R
J
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Why a saddle is unstable
R
J
No matter where you start, things eventually blow up.