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C21 Nonlinear Systems

Mark Cannon

4 lectureswebpage: www.eng.ox.ac.uk/conmrc/nlc

Michaelmas Term 2013

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Lecture 2

Linearization and Lyapunovs direct method

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Linearization and Lyapunovs direct method

Review of stability definitions

Linearization method

Direct method for stability

Direct method for asymptotic stability

Linearization method revisited

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Review of stability definitions

System: x= f(x) unforced system (i.e. closed-loop) consider stability of individual equilibrium points

0

r

R

x(0)

x(t)

0

r

x(0)

x(t)

0 is astableequilibrium if:

x(0) r = x(t) Rfor any R >0

0 isasymptoticallystable if:

x(0) r = x(t) 0

as t Stability local propertyAsymptotic stability global ifr= allowed

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Historical development of Stability Theory

Potential energy in conservative mechanics(Lagrange 1788):

An equilibrium point of a conservative system is stable if it

corresponds to a minimum of the potential energy stored in the

system

Energy storage analogy for general ODEs(Lyapunov 1892)

Invariant sets(Lefschetz, La Salle 1960s)

J-L. Lagrange 1736-1813 A. M. Lyapunov 1857-1918 S. Lefschetz 1884-1972

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Lyapunovs linearization method

Determine stability of equilibrium at x= 0 by analyzing the stability ofthe linearized system at x= 0.

Jacobian linearization:

x= f(x) original nonlinear dynamics

=f(0) +f

x

x=0

x + R1 Taylors series expansion,R1 =O(x2)

Ax since f(0) = 0

where

A= f

x x=0 =

f1x1

f1

xn..

.

. . ...

.fnx1

fn

xn

, fx assumed continuous at x= 0

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Conditions on A for stability of original nonlinear system at x= 0:

stability of linearization stability of nonlinear system atx= 0

Re

(A)

< 0 asymptotically stable(locally)

max Re

(A)

= 0 stableorunstable

max Re

(A)

> 0 unstable

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Some examples

(stable) x= x3 linearize

x= 0 (indeterminate)

(unstable) x= x3 linearize

x= 0 (indeterminate)

higher order terms determine stability

Why does linear control work?

1. Linearize the model:

x= f(x, u)

Ax + Bu, A= f

x(0, 0), B =

f

u(0, 0)

2. Design a linear feedback controller using the linearized model:

u= Kx, max Re

(A BK)

< 0

closed-loop linear model strictly stable

nonlinear system x= f(x, Kx) is locallyasymptotically stable at x= 0

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Lyapunovs direct method: mass-spring-damper example

m

k(y)

c(y) y

y

y

c(y)

k(y)

0

0

Equation of motion: my+ c(y) + k(y) = 0

Stored energy: V = K.E.+P.E.

K.E. = 12

my2

P.E. =

y0

k(y) dy

Rate of energy dissipation V = 12

my d

dyy2 + y

d

dy

y0

k(y) dy

= myy+ yk(y)

butmy+ k(y) =c(y), so V = c(y)y

0 since sign

c(y)

= sign(y)

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Mass-spring-damper example contd.

System state: e.g. x= [y y]T

V(x) 0 implies that x= 0 is stable

V(x(t)) must decrease over timebut

V(x) increases with increasing x

0 y

y

contoursofV(x)

Formal argument:

for any given R >0:

x< R whenever V(x)< V for some V

and V(x)< V whenever x< r for somer

x(0)< r = Vx(0)< V= V

x(t)

< V for allt >0

= x(t)< R for allt >0

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Positive definite functions

What ifV(x) is not monotonically increasing in x?

Same arguments apply ifV(x) is continuous and positive definite, i.e.

(i). V(0) = 0

(ii). V(x)> 0 for allx= 0

r r x

V

V

for any given V >0,can always find r so that

V(x)< V whenever x< r

V / V / x

V

V

V(x) xfor some constant , so

x< V / whenever V(x)< V

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Lyapunov stability theorem

If there exists a continuous functionV(x) such that

V(x) is positive definite

V(x) 0

thenx= 0 isstable.

To show that this implies x(t)< R for all t >0 whenever x(0)< rfor any R and some r:

1. choose V as the minimum ofV(x) forx= R

2. findr so that V(x)< V wheneverx< r

3. then V(x) 0 ensures that

V

x(t)

< V t >0 ifx(0)< r

x(t)< R t >0

0

r

R

V(x) =V

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Lyapunovs direct method also applies ifV(x) is locally positive definite,i.e. if

(i). V(0) = 0(ii). V(x)> 0 forx= 0 andx< R0

thenx= 0 is stable ifV(x) 0 whenever x< R0.

Apply the theorem without determining R, r

only need to find p.d. V(x) satisfying V(x) 0.

Examples

(i). x= a(t)x, a(t)> 0

V = 12

x2 = V =xx

=a(t)x2 0

(ii). x= a(x), sign

a(x)

= sign(x)

V = 12

x2 = V =xx

=a(x)x 0

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More examples

(iii). x= a(x),

x0

a(x) dx >0

V = x0

a(x) dx = V =a(x)x

=a2(x) 0

(iv). + sin = 0

V = 12

2 +

0

sin d = V =+ sin

= 0

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Asymptotic stability theorem



V(x) is negative definite

thenx= 0 is locally asymptotically stable.(V negative definite V positive definite)

Asymptotic convergence x(t) 0 as t can be shown by contradiction:

ifx(t)> R for all t 0, then

V(x)< W

V(x) V

for allt 0

contradiction

0R

r=x(0)

V(x) =V

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Linearization method and asymptotic stability

Asymptotic stability result also applies ifV(x) is onlylocallynegativedefinite.

Why does the linearization method work?

consider 1st order system: x = f(x)linearize about x= 0: = ax + R R= O(x2)

assume a >0 and try Lyapunov functionV:

V(x) = 12

x2

V(x) = xx= ax2 + Rx= x2(a R/x) x2(a |R/x|)

but R= O(x2) implies |R| x2 for some constant , so

V x2(a |x|) x2 if|x| (a )/

= V negative definite for |x| small enough= x= 0 locally asymptotically stable

Generalization to nth order systems is straightforward

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Global asymptotic stability theorem



V(x) is negative definite for all x

V(x) as x

thenx= 0 is globally asymptotically stable

IfV(x) as x , then V(x) is radially unbounded

Test whether V(x) is radially unbounded by checking ifV(x) aseach individual element ofx tends to infinity (necessary).

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Global asymptotic stability theorem

Global asymptotic stability requires:

x(t) finite

for allt >0

for all x(0)

not guaranteed by V negative definite

in addition to asymptotic stability ofx= 0

Hence add extra condition: V(x) as x

equiv. to

level sets {x : V(x) V} are finite

equiv. to

x is finite whenever V(x) is finite

prevents x(t) drifting away from 0 despite V

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Asymptotic stability example

System: x1=(x2 1)x3

1

x2= x41

(1 + x21

)2

x21 + x2

2

Trial Lyapunov function V(x) =x21+ x2

2:

V(x) = 2x1 x1+ 2x2 x2

=2x41+ 2x2x4

1 2 x2x

4

1

(1 + x21

)2 2

x221 + x2

2

0

change V to makethese terms cancel

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Asymptotic stability example

New trial Lyapunov function V(x) = x211 + x2

1

+ x22:

V(x) = 2 x1

1 + x21

x31

(1 + x21

)2

x1+ 2x2 x2

=2 x41

(1 + x21

)2 2

x221 + x2

2

0

V(x) positive definite, V(x) negative definite = x= 0 a.s.ButV(x) not radially unbounded, so cannot conclude global a.s.

State trajectories:

10 5 0 5 102

1.5

1

0.5

0

0.5

1

1.5

2

x1

x2

V0 = 0.5

V0 = 3

V0 = 1

V0 = 1

V0 = 3

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Summary

Positive definite functions

Derivative ofV(x) along trajectories ofx= f(x)

Lyapunovs direct method for: stabilityasymptotic stabilityglobal stability


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