4A - Lyapunov Stability

c©Professor Victor Solo

School of Electrical Engineering

University of New South Wales

Sydney, AUSTRALIA

ELEC 9732 Nonlinear Control Term III

– p. 1/2

Topics

1 Define Stability and Asymptotic Stability

2 Lyapunov Indirect Method (Linearization)of Stability Analysis

3 Lyapunov Direct Method (Energy Method)of Stability Analysis

4 LaSalle’s Methods of Stability Analysis

ReferencesSlotineLi ch3 and Khalil 4.1,4.2,4.3AcronymsLAS - local asymptotic stabilityGAS - global asymptotic stabilityES - exponential stabilityLPF - Lyapunov Function

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1 Stability

Consider the autonomous ODE (AODE)x = f(x), x(0) = xo

Supposexe is an EP so thatf(xe) = 0.If ‖ x(t) ‖≤ c for all t we sayx(t) is bounded.

We sayxe is a stableEP if a trajectory starting closeto xe remains close for allt.

Formallyxe is a stable EP if for allǫ > 0 there is aδ = δ(ǫ) > 0 such that‖ xo − xe ‖≤ δ(ǫ)⇒‖ x(t)− xe ‖≤ ǫ for all t.

If xe is not a stable EP we say it is an unstable EP.

0 10 20 30 40 50 60 70 80 90 100−4

−2

0

2

4

time

x(t)

2-1

2 Stability II

For linear systems: unstable≡ blows upBut for NL systems this is not so.

ExampleVan der Pol System.x + ǫx(x2 − 1) + x = 0EP obeysxe = 0 = xe ⇒ xe = 0

Now recall the VdP phase plane plot.All trajectories starting nearxe = 0 do not remain

close; rather they end up coalescing onto the LC - so noblow up.

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0

2

4

time

x(t)

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3 Local Asymptotic Stability (LAS)

Consider the AODEx = f(x), x(0) = xo

and supposexe is an EP, sof(xe) = oWe sayxe is a LASEP if it is a stable EP and there

is aδ > 0 so that‖ x0 − xe ‖≤ δ ⇒ x(t)→ xe ast→∞

The regionx :‖ x − xe ‖≤ δ is called a regionofattraction(ROA) for the EP. The largest region of attrac-tron is called theROA of the EPxe.

NB. LAS includes a stability requirement since wecan have cases wherex(t) → xe but does not stay closeto xe all the time.

If x(t) → xe as t → ∞ for all x0 we sayxe is aglobally asymptotic stable (GAS) EP.

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4 Local Asymptotic Stability II

Example:x + x2 = 0EP obeysxe = 0⇒ xe = 0.PP plot is easly assembled

0 10 20 30 40 50 60 70 80 90 100−4

−2

0

2

4

time

x(t)

Solution:dx + x2dt = 0⇒ −dx/x2 = dtIntegrate⇒ 1

x ]xx0= t = 1

x − 1x0

⇒ x(t) = x0

1+tx0

if x0 < 0 thenx(t)→∞ ast→ 1|x0|

i.e. finite time blow up!

if x0 > 0 thenx(t)→ xe = 0 as t→∞⇒ xe unstable!

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5 Exponential Stability

Consider the AODEx = f(x), x(0) = xo.

We sayxe is an exponentiallystable(ES) EP if thereexista > 0, λ > 0 with‖ x(t)− xe ‖≤ a ‖ xo − xe ‖ e−λt for all t.

ES is critical for reliable operation of control systems.

Example. x = −(1 + x2)x.

IntroduceH(t) = t +∫ t

0x2(s)ds⇒ H = 1 + x2.

⇒ x + Hx = 0⇒ eH x + HeHx = 0⇒ d

dt(eHx) = 0⇒ eH(t)x(t)− x(0) = 0

⇒ x(t) = e−H(t)x(0)

But H(t) ≥ t⇒ −t ≤ −H(t)⇒ e−t ≥ e−H(t)

⇒ |x(t)| ≤ |x(0)|e−t ⇒ ES.

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6 Stability Analysis

How to analyse stability?Lyapunov 1890

′

s delivered two methods.

(i) Indirect Method or

Linearization method

- relates behaviour near an EP to behaviour of lin-earized system.

(ii) Direct Method or Energy Method

Find a system energy and show it goes→ 0 ast→∞.

Usually this will⇒ system is LAS.

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7 Lyapunov Linearization Theorem

Consider the AODEx = f(x), x(0) = xe

Supposexe is an EP sof(xe) = 0

We have already introduced, via Taylor series, the as-sociated linear system

˙x = Ax, x = x− xe

A = ∂f∂xT |x=xe

So rewrite the AODE as˙x = x = A(x− xe) + g(x)g(x) = f(x)−A(x− xe)

Lyapunov Linearization Theorem

Suppose‖g(x)‖‖x−xe‖ → 0 as‖ x− xe ‖→ 0.

Then if the linearized systemsz = Azhas a focus, node or saddle point,

so doesx = f(x) atxe

If the linear system has a center the nonlinear systemhas a center or a focus.

Proof : Later.

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8 Lyapunov Stability - Example

Recall mass spring with linear viscosity and hardeningspring (used by SlotineLi fig 2.2)

x + ǫx + 3x(1 + |x|3 ) = 0

or x + ǫx + 3x + x2 = 0

State space system is(

x1

x2

)

=(

x2

−ǫx2−3x1−x2

1

)

EPs obeyx2e = 03xe,1 + x2

e,1 = 0⇒ xe,1 = 0or − 3.So two EPS

(

00

)

,(−3

0

)

Find Linearized systemA(x) = ∂f

∂xT =(

∂f1/∂x1

∂f2/∂x1

∂f1/∂x2

∂f2/∂x2

)

=(

0−3−2x1

1−ǫ

)

⇒ A0 = A(

0

0

) =(

0−3

1−ǫ

)

A−3 = A(

−3

0

) =(

03

1−ǫ

)

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9 Lyapunov Stability - Example con-tinued

Behaviour nearxe =(

00

)

We need to analyse stability of linearized system byfinding eigenvalues of A0⇒ |λI −Ao| = 0⇒

∣

∣

(

λ3

−1λ+ǫ

)∣

∣ = 0⇒ λ(λ + ǫ) + 3 = 0⇒ λ2 + ǫλ + 3 = 0⇒ λ = −ǫ±

√ǫ2−122

≈ − ǫ2 ± j

√3

⇒stable focus

Now check conditions of the theorem.go(x) = f(x)− Ao(x− xe)

=(

x2

−ǫx2−3x1−x2

1

)

−(

0−3

1−ǫ

)(

x1

x2

)

=(

x2

−ǫx2−3x1−x2

1

)

−(

x2

−3x1−ǫx2

)

=(

0−x2

1

)

⇒‖ go(x) ‖=√

02 + (−x21)

2 = x21

⇒ ‖g0(x)‖‖x−xe‖ =

x2

1√x2

1+x2

2

≤ |x1| → 0 as‖ x ‖→ 0

So NL system has stable focus at(

00

)

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10 Lyapunov Stability Analysis - III

Behaviour nearxe =(−3

0

)

.Eigenvalues obey|λI −A3| = 0⇒

∣

∣

(

λ−3

−1λ+ǫ

)∣

∣ = 0⇒ λ(λ + ǫ)− 3 = 0⇒ λ2 + ǫλ− 3 = 0⇒ λ = −ǫ±

√ǫ2+122

≈ −ǫ2 ±

√3

⇒one stable root, one unstable⇒ saddlepointNow check conditionsg3(x) = f(x)−A3(x− xe)=

(

x2

−ǫx2−3x1−x2

1

)

−(

03

1−ǫ

)(

x1+3x2

)

=(

x2

−ǫx2−3x1−x2

1

)

−(

x2

3(x1+3)−ǫx2

)

=(

0−x2

1−6x1−9

)

=(

0(x1+3)2

)

⇒ ‖g3(x)‖‖x−xe‖ = (x1+3)2√

(x1+3)2+x2

2

≤ |x1 + 3| → 0

as‖ x− xe ‖→ 0So NL system has a saddlepoint at

(−30

)

.We already saw this from the PP plot.

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11 Lyapunov’s Direct Method

Based on the idea that stability follows if system energyis continually dissipated.

Example. Nonlinear mass spring system.mx + bx|x|+ k0x + k1x

3 = 0Pull the mass to the right & let go : What happens?

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x(t)

System mechanical energy =potential energy + kinetic energy

=∫ x

0(k0x + k1x

3)dx + 12mx2

⇒ V (x, x) = 12mx2 + 1

2k0x2 + 1

4k1x4

(i) zero energy= V (x, x) = 0 ⇒ x = 0, x = 0 i.e.EP.

(ii) LAS⇒ V (t) = V (x(t), x(t))→ 0 ast→∞.(iii) Instability is intuitively associated with a growth

of energy.

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12 Mass Spring II

The rate of energy dissipation is

dVdt = mxx + k0xx + k1x

3x= x[mx + k0x + k1x

3]= −b|x|3≤ 0

So energy V is indeed dissipated untilx = 0 i.e. massceases motion.

This does not prove energy dissipation⇒ LAS;but intuitively links the the two.

We now look at a more formal result.

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13 Lyapunov Functions

A scalar continuous functionV (x) is said to be locallypositivedefinite(LPD) if

(i) V (xe) = 0(ii) V (x) > 0, x 6= xe, x ∈ Ω ( a region of interest)

If (ii) holds for all x 6= xe we sayV (x) is a globallypositive definite (GPD) function.

Example: Mass spring in SS form.(

x1

x2

)

=(

x2

−bx2|x2|−k0x1−k1x3

1

)

V (x) = V (x1,x2) = 12mx2

2 + k0x21 + 1

4k1x41

EPs obeyxe,2 = 0k0xe,1 + k1x

3e,1 = 0⇒ xe,1 = 0

or x2e,1k1 + k0 = 0 which is not possible

ThusV (xe) = 0;

and clearlyV (x) = 0⇒ x = xe

soV (x) > 0, x 6= xe

SoV (x) is a GPD function.

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14 Lyapunov Functions - II

Example 2: PendulumForce = mass× acceleration, sol2mθ = −mglsin(θ)− kθ (← hinge friction)

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4

time

x(t)

The Pendulum system energy isV = kinetic + potential (from height above equilib-

rium)= V (θ, θ) = 1

2ml2θ2 + mgl(1− cos θ)

EP obeysθe = θe = 0⇒ sin θe = 0⇒ cos θe = 1⇒ V (θe, θe) = 0 + 0 = 0There are of course multiple EPsθe = 0, θe = ±π

Also clearly for,θ 6= θe, θ 6= θe, V (θ, θ) > 0

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15 Lyapunov Direct Method

Consider the AODEx = f(x), x(0) = xo

Supposexe is an EP.

SupposeV (x) is LPD ie.(i) V (xe) = 0(ii) V (x) > 0, x ∈ Ω (a region of interest), x 6= xe

(iii) 1 V (t) ≤ 0, x(t) ∈ Ω

Thenxe is a stable EP andV (x) is called a Lyapunovfunction (LPF)

If also(iv) V (x(t)) < 0, x(t) ∈ Ω, x 6= xe

thenxe is a LAS EP.

1We should writeV (x(t)) but often as a shorthand we writeV (t)

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16 Direct Method II

Note that (iii) entailsV = d

dtV (x(t))

= ∂V∂xT

dxdt

= ∂V∂xT x = ∂V

∂xT f(x) ≤ 0

So angle between normal∂V∂x and fieldf(x) is nega-

tive.Energy Conservation i.e.V = 0⇒ V (x(t)) =constantV < 0 Energy dissipation.

Boundedness. Integrate⇒V (x(t))− V (x(0)) =

∫ t

0V (x(s))ds ≤ 0

⇒ V (x(t)) ≤ V (x(0))So whole trajectory lies in the setz : V (z) ≤ V (x0)So need level sets or contours to be bounded.

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17 Direct Method - Example

Pendulum: The energy LPF we constructed above is12 lmθ2 + mgl(1− cos θ)

We already checkedV (θ, θ) is LPD i.e (i), (ii) holdfor θ ∈ Ω = θ : −π < θ < π

Now calculateV = dV

dt = mlθ + mglθsin(θ)

= θ(mlθ + mgl sin θ)= θ(−kθ)= −kθ2

⇒ V ≤ 0

Frictionless Case: k = 0 ⇒ V = 0 and we deduceθeθe = 0, 0 is locally stable.

We cannot deduce LAS since (iv) fails. In any casewe know independently LAS does not hold since the mo-tion is purely oscillatory.

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18 Pendulum Example - II

With friction we haveV ≤ 0So again we deduceθe, θe = 0, 0 is locally stable.

Can we deduce LAS?It clearly holds physicallydue to the friction.

But V does not obey (iv) sinceV (θe, θe) = −kθ2 = 0 for θ = 0whatever the value ofθ i.e. not just forθ = θe.So we cannot conclude LAS with this LPF.

This makes it clear that the direct method is sufficientfor stability results but not necessary. We can try to finda better LPF (not easy) or try other methods.

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