nonlinear systems - dt - home page

CHAPTER II - Second order nonlinear systems

NONLINEAR SYSTEMS

JOSE C. GEROMEL

DSCE / School of Electrical and Computer EngineeringUNICAMP, CP 6101, 13081 - 970, Campinas, SP, Brazil,

[email protected]

Campinas, Brazil, August 2008

1 / 40


Contents

1 CHAPTER II - Second order nonlinear systemsNote to the readerPreliminariesLocal analysisGlobal analysisPeriodic solutions

Poincare’s indexExample

Problems

2 / 40


Note to the reader

Note to the reader

This text is based on the following main references :

J. C. Geromel e R. H. Korogui, “Controle Linear de SistemasDinamicos : Teoria, Ensaios Praticos e Exercıcios (inPortuguese), Edgard Blucher Ltda, 2011.

H. K. Khalil, “Nonlinear Systems”, Macmillan Publishing Co.,1992.

M. Vidyasagar, “Nonlinear Systems Analysis”, Prentice Hall,NJ, 1993.

3 / 40


Preliminaries

Preliminaries

Second order time-invariant systems behavior can be analyzedby the so called phase plane trajectory plots. Consider thesecond order system

z = −f (z , z)

where z(t) ∈ R× [0,T ]. Adopting the state variables x = zand y = z we have

x = y

y = −f (x , y)

The phase plane is defined by the (x , y) coordinatesparameterized with respect to t ∈ [0,T ]. Each solution (withdifferent initial conditions) defines a trajectory.

4 / 40


Preliminaries

Preliminaries

Since the nonlinear equation satisfies ydy + f (x , y)dx = 0, atrajectory Γ is characterized by

∮

Γydy + f (x , y)dx = 0

In some situations this equation can be integrated withoutdifficulty and provides Γ. Indeed, it is well known that if

∂y

∂x=

∂f (x , y)

∂y= 0

that is f (x , y) = g(x), the line integral does not depend onthe integration path.

5 / 40


Preliminaries

Preliminaries

In this case, from the initial point (x0, y0) ∈ Γ at t = 0 to ageneric one (x , y) ∈ Γ at any t ≥ 0, we have

∫ y

y0

ydy +

∫ x

x0

g(x)dx = 0

which corresponds to break the line integration from (x0, y0)to (x , y) in two line segments:

From (x0, y0) to (x0, y).From (x0, y) to (x , y).

It is important to stress that this is possible only because theline integral defining Γ does not depend on the particularintegration path used.

6 / 40


Preliminaries

Preliminaries

Example : For the pendulum discussed before

θ + ω20sin(θ) = 0

with x = θ, y = θ and g(x) = ω20sin(x) and

V (x , y) =

∫

ydy +

∫

ω20sin(x)dx

=y2

2− ω2

0cos(x)

any trajectory of the phase plane is given by

Γ ={(x , y) ∈ R

2 : V (x , y) = V0

}

where V0 = V (x0, y0) selects a particular trajectory passingthrough the point (x0, y0), for instance the initial condition.

7 / 40


Local analysis

Local analysis

In the general case, trajectories in the phase plane can bedetermined by numeric integration. However, in theneighborhood of each equilibrium point they can be drawnapproximatively. The equilibrium point (xe , ye) satisfies

f (xe , ye) = 0

ye = 0

which implies that they are located on the x axis. The linearapproximation, valid near (xe , ye), is given by

ξ = Aξ , A =

[0 1

− ∂f∂x

− ∂f∂y

]

(xe ,ye)

where ξ = [x − xe , y − ye ]′. Afterwards, cases of interest are

considered.8 / 40


Local analysis

Local analysis

Distinct real eigenvalues : Assume that λ 6= µ ∈ R areeigenvalues of matrix A ∈ R

2×2. Defining V = [vλ vµ] ∈ R2

the matrix of eigenvalues, we know that

V−1AV = Λ = diag{λ, µ}

which yields

ξ(t) =(eλtcλ

)vλ +

(eµtcµ

)vµ

where cλ and cµ are constants depending on the initialconditions. Consequently, near the equilibrium point we have

[x(t)− xey(t)− ye

]

=(eλtcλ

)vλ +

(eµtcµ

)vµ

9 / 40


Local analysis

Local analysis

−10 −8 −6 −4 −2 0 2 4 6 8 10−10

−8

−6

−4

−2

0

2

4

6

8

10

Stable node : Negative eigenvalues (λ < 0, µ < 0).

Unstable node : Positive eigenvalues (λ > 0, µ > 0).

10 / 40


Local analysis

Local analysis

−10 −8 −6 −4 −2 0 2 4 6 8 10−20

−15

−10

−5

0

5

10

15

20

Saddle : Negative and positive eigenvalues (λ < 0, µ > 0).

11 / 40


Local analysis

Local analysis

Complex eigenvalues : Assume that σ ± jω ∈ C areeigenvalues of matrix A ∈ R

2×2. Defining V = [vR vI ] ∈ R2

the matrix of eigenvalues, we know that

V−1AV = Λ =

[σ ω−ω σ

]

and changing to polar coordinates

ξ(t) = V

[r(t)cos φ(t)r(t)sin φ(t)

]

after simple derivation with respect to time of ξ = V η weobtain η = Λη and

η =

[cos φ −rsin φsin φ rcos φ

]

︸︷︷︸

U

[r

φ

]

12 / 40


Local analysis

Local analysis

Consequently

[r

φ

]

= U−1Λη

=

[σr−ω

]

yielding r(t) = eσtcσ and φ(t) = −ωt + cω where theconstants depend on the initial conditions. As in the firstcase, near the equilibrium point we have

[x(t)− xey(t)− ye

]

=(eσtcos(cω−ωt)cσ

)vR+

(eσtsin(cω−ωt)cσ)vI

13 / 40


Local analysis

Local analysis

−8 −6 −4 −2 0 2 4 6 8−8

−6

−4

−2

0

2

4

6

8

Stable focus : Complex eigenvalues (σ < 0).

Unstable focus : Complex eigenvalues (σ > 0).

Center : Pure imaginary eigenvalues (σ = 0).

14 / 40


Local analysis

Local analysis

The eigenvalues of matrix A ∈ R2×2 of the linearized system

at the equilibrium point x = xe are given by

s2 +Θs +∆ = 0

where Θ = −trace(A) and ∆ = det(A).

−4 −3 −2 −1 0 1 2 3 4−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

stable node

stable focus unstable focus

unstable node

saddle

∆ = Θ2/4

Θ

∆

15 / 40


Global analysis

Global analysis

The global analysis is done by drawing the whole phase planeas a composition of the parts obtained by the linearized modelvalid in the neighborhood of each equilibrium point. This isillustrated by means of the following nonlinear equation

x = −βx − y

y = x − g(y)

where g(y) is a piecewise linear function and two casescorresponding to β = 2 and β = 1/2 are considered. Theequilibrium points (xe , ye) are those satisfying

x = −y/β

x = g(y)

16 / 40


Global analysis

Global analysis

For β = 2 the equilibrium points are a, o and b.For β = 1/2 the equilibrium points are a′, o and b′. Theequilibrium points a′ and b′ are virtual.

y

g(y)10

−10

−20

x

a

a′

b

b′

o

17 / 40


Global analysis

Global analysis

The matrix A of the linearized system which depends on eachparticular equilibrium point has the general form

A =

[−β −11 −g ′(ye)

]

For β = 2 we have:

For points a and b, Θ = 3 and ∆ = 3 =⇒ Stable focusFor point o, Θ = 1 and ∆ = −1 =⇒ Saddle

For β = 1/2 we have:

For points a′ and b′, Θ = 3/2 and ∆ = 3/2 =⇒ Stable focusFor point o, Θ = −1/2 and ∆ = 1/2 =⇒ Unstable focus

18 / 40


Global analysis

Global analysis

−30 −20 −10 0 10 20 30−30

−20

−10

0

10

20

30

x

y

Globally, it is clear the behavior due to the saddle at the originand two real stable focus which are asymptotically reacheddepending on the initial conditions.

19 / 40


Global analysis

Global analysis

−30 −20 −10 0 10 20 30−30

−20

−10

0

10

20

30

x

y

The two virtual stable focus are never reached. The originbeing an unstable focus enables the appearance of a limitcycle (periodic solution).

20 / 40


Periodic solutions

Periodic solutions

Before proceed, we need some preliminary results.

Theorem (Green’s theorem)

Let D be a simply connected domain and C a simple closed curvesuch that R = intC ⊂ D.

∮

C

Pdx + Qdy =

∫ ∫

R

(∂Q

∂x−

∂P

∂y

)

dxdy

Moreover, it is well known that if

∂Q

∂x=

∂P

∂y,∀(x , y) ∈ R

then the line integral does not depend on the integration pathand the Green’s theorem implies that

∮

C

Pdx + Qdy = 021 / 40


Periodic solutions

Periodic solutions

We now discuss an important result used afterwards. Supposethat the equality

∂Q

∂x=

∂P

∂y,∀(x , y) ∈ R

holds excepted at some isolated point p1, · · · pN . Define thesimple closed curves Ci such that pi ∈ intCi for alli = 1, · · · ,N. It follows that

∮

C

Pdx + Qdy =N∑

i=1

∮

Ci

Pdx + Qdy

where each indicated line integral does not depend on theintegration path.

22 / 40


Periodic solutions

Periodic solutions

Example : Consider the line integral

J =

∮

C

−ydx + xdy

x2 + y2

Noticing that

∂Q

∂x=

∂P

∂y=

y2 − x2

x2 + y2, ∀x 6= 0, y 6= 0

and the differential of θ = tg−1(y/x) provides

dθ =xdy − ydx

x2 + y2

23 / 40


Periodic solutions

Periodic solutions

Two conclusions can be drawn:

If the origin is not in the interior of C then

J =

∫

dθ = 0

in accordance to the fact that the partial derivatives of P andQ are equal in all points inside C . From the Green’s theoremthe integral must be zero.If the origin is in the interior of C then

J =

∫

dθ = 2π

The Green’s theorem does not apply. The value of the integralis not null anymore but it still does not depend on theintegration path whenever the origin is in its interior.

24 / 40


Periodic solutions

Periodic solutions

Consider again the nonlinear second order system

x = y

y = −f (x , y)

The next two theorems are central on limit cycles analysis.

Theorem (Bendixson’s theorem)

If the function ∂f /∂y is not identically zero and has invariant signin all points (x , y) of a simply connect domain D then there is noperiodic solution (limit cycle) in the interior of D.

Indeed, it is known that any periodic solution satisfies∮

Γydy + f (x , y)dx = 0

for some closed curve Γ.25 / 40


Periodic solutions

Periodic solutions

Consider any closed curve Γ in the interior of D, from theGreen’s theorem we have

∮

Γydy + f (x , y)dx =

∫ ∫

intΓ

∂f

∂ydxdy

6= 0

because int Γ ⊂ D and ∂f /∂y is not identically zero and doesnot change sign in D. The only possible conclusion is thatthere are no closed trajectories inside D.

For the linear system characterized by f (x , y) = αx + βy , theBendixson’s theorem assures that if β 6= 0 then, no periodicsolutions exist.

26 / 40


Periodic solutions

Poincare’s index

Consider C a closed curve inside a simply connected domain.The Poincare’s index of C is

IC =1

2π

∮

C

dθ

where tg(θ) = −f (x , y)/y . The geometric interpretation is :

x

y

θ

ΓC

27 / 40


Periodic solutions

Poincare’s index

For the particular choice C = Γ, at any point of Γ the angle θis defined by the tangent line. Hence

IΓ = 1

For any other situation we need to calculate the line integral.With this purpose, we notice that

dθ =fdy − ydf

y2 + f 2

and

df =∂f

∂xdx +

∂f

∂ydy

28 / 40


Periodic solutions

Poincare’s index

As a result, it follows that

IC =1

2π

∮

C

Pdx + Qdy

where

P = −y

y2 + f 2∂f

∂x

and

Q =f

y2 + f 2−

y

y2 + f 2∂f

∂y

Very tedious algebraic manipulations put in evidence that

∂Q

∂x=

∂P

∂y

whenever y2 + f (x , y)2 6= 0.

29 / 40


Periodic solutions

Poincare’s index

The previous condition holds in all points excepted thosesatisfying

y = 0 , f (x , 0) = 0

that is, the equilibrium (isolated) points p1, p2, · · · of thenonlinear system under consideration.

Fact

For a given closed curve C, the following are true:

If there is no equilibrium point inside C then IC = 0.

If C and C ′ contain in their interior the same equilibrium point thenIC = IC ′ .

If a number of equilibrium points p1, p2 · · · are inside C thenIC =

∑

i ICiwhere Ci is any closed curve inside C containing only

the equilibrium point pi .

30 / 40


Periodic solutions

Poincare’s index

In conclusion the Poincare’s index is not a property of a givenclosed curve but the equilibrium point p in its interior. Hence,to calculate IC we can replace C by a closed curve C ′ in anarbitrarily small neighborhood of p and use the linearapproximation of the system, yielding

x = yy = −αx − βy

=⇒ A =

[0 1−α −β

]

where α and β are parameters given by

α =∂f

∂x

∣∣∣∣p

, β =∂f

∂y

∣∣∣∣p

31 / 40


Periodic solutions

Poincare’s index

Considering f = αx + βy we obtain

y2 + f 2 =

∥∥∥∥A

[xy

]∥∥∥∥

2

2

= r2

which if det(A) 6= 0 defines a closed curve C ′ containing theorigin for any r > 0 fixed. Moreover, simple calculations yield

IC =det(A)

2πr2

∮

C ′

xdy − ydx

=det(A)

πr2

∫ ∫

intC ′

dxdy

=det(A)

|det(A)|

32 / 40


Periodic solutions

Poincare’s index

Fact

Assume that C is a closed curve with a simply connected domaincontaining an equilibrium point:

If the equilibrium point is a center, a focus or a node thenIC = 1.

If the equilibrium point is a saddle then IC = −1.

Since we already know that IΓ = 1, a closed trajectory of thenonlinear system may exist only if :

At least one equilibrium point must be inside Γ.The sum of the Poincare’s indexes of the equilibrium pointsinside Γ must be one.

33 / 40


Periodic solutions

Example

Consider the nonlinear system

x = −x + xy

y = y − xy

which presents two equilibrium points, namely

p1 = (0, 0) =⇒ A =

[−1 00 1

]

, saddle, IC = −1

p2 = (1, 1) =⇒ A =

[0 1−1 0

]

, center, IC = 1

A closed trajectory (if any) which characterizes a periodicsolution or a limit cycle must contain p2 but not p1 in itsinterior!

34 / 40


Problems

Problems

1. Consider the line integral

J =

∮

C

x1dx2 − x2dx1

where C is the ellipsoid x ′Qx = r2 defined for r 6= 0 andQ ∈ R

2×2 positive definite.Determine J and interpret the result using the Green’stheorem.Determine J for Q = I , r = 2 and for Q = diag(1, 2), r = 1.

2. Let A ∈ Rn×n be given. Prove that if there exists P ∈ R

n×n

positive definite such that

(A+ αI )′P + P(A+ αI ) < 0

then maxi=1,··· ,n Re{λi (A)} < −α.

35 / 40


Problems

Problems

3. The following is known as the Duffing equation

z + z − z(1− z2/c2) = 0

where c 6= 0.

Determine the equilibrium points and types.Drawn the trajectories of the phase plane near the equilibriumpoints.

4. Consider the differential equation

z + ǫ(z)3 + z = 0

where |ǫ| < a for some a > 0 given.

Determine the type of the equilibrium point (0, 0).Linearizing, is it possible to conclude about the stability of thisequilibrium point?

36 / 40


Problems

Problems

5. Consider the block diagram where a symmetric nonlinearity isas indicated.

u e

3/2

1/2

1

y

f (·) 1−s/4

s2+s/2+1+−

For u = 0, determine the state space realization with x1 = eand x2 = e, the equilibrium points and draw the phase plane.Repeat the previous item for u = 3.

37 / 40


Problems

Problems

6. The following is the so called Van der Pol’s equation.

z − µ(1− z2)z + z = 0

where 0 < µ < 2.

Determine the state space realization for x1 = z and x2 = z .Determine the equilibrium points and their types.Draw the trajectories of the phase plane near the equilibriumpoints.Determine the Poincare’s index of each equilibrium point.Show that there is no closed trajectory in the region |z | < 1.

7. For the general nonlinear system x = g(x , y), y = −f (x , y)state and prove the Bendixson’s theorem.

38 / 40


Problems

Problems

8. A simplified model of a synchronous power machine is given by

δ + sin(δ) = 0.5

Determine:

the equilibrium points and their types.the trajectories in the phase plane and draw it for −π ≤ δ ≤ π.

9. Consider the nonlinear system

x = y

y = −2− x + 1/(1 − x)3

Determine and draw the trajectories in the phase plane.

39 / 40


Problems

Problems

10. Consider the nonlinear equation

z + 2ξωz + ω2z + ǫz3 = 0

with ω > 0. Determine and classify the equilibrium points forǫ < 0, ǫ = 0 and ǫ > 0.

11. Consider the second order system

x = −2x + y

y = x(x2 − 3x + 4)− y

Determine the equilibrium points and their types.Take a stable equilibrium point, calculate the linearapproximation and draw the phase plane near it.

40 / 40

nonlinear systems - dt - home page

Documents