nonlinear systems - unicampgeromel/non_introd.pdf · h. k. khalil, “nonlinear systems”,...

CHAPTER I - Introduction

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

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Contents

1 CHAPTER I - IntroductionNote to the readerIntroduction

PreliminariesLinear differential equationsTransfer function and frequency response

Nonlinear differential equationsPreliminariesExistence and uniquenessPeriodic solutionsEquilibrium pointsLinearization

Problems

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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.

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Introduction

Preliminaries

Nonlinear systems analysis is based on the followingobservations:

Classes of nonlinear systems are identified allowing theadoption of specific results to analyze them. For example, theLur’e and Persidiskii nonlinear systems.Some approximations are adopted. For instance, in timedomain (power series) and in frequency domain (Fourie series).Global and local analysis. Putting in evidence the validity ofproperties in the overall state space or only in some regions orneighborhoods.

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Introduction

Preliminaries

The next figure shows a pendulum of length ℓ and mass m

mg

F

x

y θ

From Newton´s Law we obtain the model

md2

dt2(ℓsinθ) + F sinθ = 0

md2

dt2(ℓcosθ) + F cosθ = mg

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Introduction

Preliminaries

Eliminating the radial force F we obtain the swinging equation

θ + (g/ℓ)sinθ = 0

which is nonlinear. Trajectories with (θ(t), θ(t)) small enoughfor all t ≥ 0, can be obtained from the linear approximation

θ + (g/ℓ)θ = 0

which is solved with no difficulty, that is

θ(t) = A sin

(√g

ℓt + B

)

where the constants A and B are determined from the initialconditions (θ(0), θ(0)).

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Introduction

Preliminaries

The nonlinear equation does not admit a solution θ(t) inclosed form. However, defining the angular velocity ν = θ,noticing that

dν

dt=

dν

dθν

and integrating νdν + (g/ℓ)sinθdθ = 0 we obtain thetrajectories in the plane (θ, ν) given by

ν2 − 2(g/ℓ)cosθ = const

Moreover, adopting the same reasoning to the linearapproximation we have

ν2 + (g/ℓ)θ2 = const

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Introduction

Preliminaries

The next figure shows the trajectories in the plane (ν, θ) forthe nonlinear system (on the left side) and for the linearapproximation (on the right side):

−4 −2 0 2 4−4

−3

−2

−1

0

1

2

3

4

−4 −2 0 2 4−4

−3

−2

−1

0

1

2

3

4

θθ

ν=

θ

Notice the difference for (ν, θ) far from the origin.

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Introduction

Linear differential equations

Our interest are linear differential equations with state spacerealization

x = A(t)x + B(t)u, x(0) = x0

y = C (t)x + D(t)u

where x(t) ∈ Rn, u(t) ∈ R

m and y(t) ∈ Rr for all t ∈ [0, T ]

for some T > 0.

Generally, the final time is unbounded, that is T = +∞.Matrices (A(t),B(t),C (t),D(t)) are time varying withcompatible dimensions.Matrices (A(t),B(t),C (t),D(t)) are continuous functions ofthe independent variable t ≥ 0.

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Introduction


Fact (Existence and unicity)

Under the previous assumptions, for each given initial condition

x0 ∈ Rn, the liner differential equation admits an unique solution in

the time interval [0,T ].

The solution follows from the so called state transition matrixΦ(t, τ) ∈ Rn×n for all (t, τ) ∈ [0,T ] × [0,T ] which satisfiesthe partial differential equation

∂Φ

∂t(t, τ) = A(t)Φ(t, τ) ,Φ(τ, τ) = I

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Introduction


Theorem (General solution)

The general solution of the linear differential equation is

x(t) = Φ(t, 0)x0 +

∫ t

0Φ(t, τ)B(τ)u(τ)dτ

To prove this theorem we need the following result related tothe time derivative of the function

g(t) =

∫ t

0f (t, τ)dτ

which is given by

dg

dt(t) = f (t, t) +

∫ t

0

∂f

∂t(t, τ)dτ

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Introduction


The proof follows from the calculation

x(t) = A(t)Φ(t, 0) + B(t)u(t) +

∫ t

0A(t)Φ(t, τ)B(τ)u(τ)dτ

= A(t)

{

Φ(t, 0) +

∫ t

0Φ(t, τ)B(τ)u(τ)dτ

}

+ B(t)u(t)

= A(t)x(t) + B(t)u(t)

and from the fact that the proposed solution satisfies theinitial condition, that is

x(0) = Φ(0, 0)x0 = x0

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Introduction


For the time invariant case, characterized by constantmatrices A(t) = A, B(t) = B , C (t) = C , D(t) = D we have

Φ(t, τ) = eA(t−τ)

where the matrix exponential for any X ∈ Rn×n is

eX =

∞∑

i=0

X i

i !

yielding the following well known solution

x(t) = eAtx0 +

∫ t

0eA(t−τ)Bu(τ)dτ

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Introduction

Transfer function and frequency response

For time invariant systems it is useful to write the output y(t)as follows

y(t) = CeAtx0 +

∫ t

0CeA(t−τ)Bu(τ)dτ + Du(t)

= CeAtx0︸︷︷︸

H0(t)

+

∫ t

0

{

CeA(t−τ)B + Dδ(t − τ)}

︸︷︷︸

H(t−τ)

u(τ)dτ

where H(t) = CeAtB + Dδ(t) is the impulse response. Noticethat

y(t) = H(t) ∗ u(t)under zero initial condition.

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Introduction


The transfer function H(s) is obtained from the Laplacetransform of the impulse response H(t), yielding

H(s) = C (sI − A)−1B + D

For SISO systems we can write H(s) = N(s)/D(s) whereN(s) and D(s) are polynomials.

The poles of H(s) are the roots of D(s) = det(sI − A) = 0.The zeros of H(s) are the roots of

N(s) = det(sI − A)H(s)

= det

[sI − A −B

C D

]

= 0

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Introduction


The frequency response of the system with transfer functionH(s) is simply

H(jω) , ∀ω ∈ R

which clearly imposes that jω ∈ D(H). The imaginary axismust belong to the domain of H(s) and consequently all polesof H(s) must be located in the region Re(s) < 0.

In this case, with the input u(t) = e jωt ,∀t ≥ 0 we obtain

y(s) =H(s)

(s − jω)

= R(s) +H(jω)

(s − jω)

where R(s) and H(s) have the same poles.

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Introduction


The fact r(t) = L−1(R(s)) vanishes as t → ∞ implies that

y(t) ≈ yperm(t) = H(jω)e jωt

for t ≥ 0 big enough. The quantity yperm(t) is the steadystate response of the system. Based on this we conclude thatthe steady state response corresponding to the inputu(t) = sin(ωt), ∀t ≥ 0 is

yperm(t) = |H(jω)|sin(ωt + ∠H(jω))

The output oscillates with the same frequency as the inputwith amplitude and phase depending on H(jω) exclusively.

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Introduction


The Laplace transform domain D(f ) of a function f (t) withtime domain t ≥ 0 assures that the equality

f (s) =

∫∞

0f (t)e−stdt

holds for all s ∈ D(f ). If jω ∈ D(f ) then the equality

f (jω) =

∫∞

0f (t)e−jωtdt

holds and provides the Fourier transform of f (t). On thecontrary, if jω /∈ D(f ) the same equality is not verified.However, the quantity f (jω) can be calculated whenever f (s)does not have poles on the imaginary axis.

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Introduction


In minimal state space realization, the poles of H(s) are suchthat det(sI − A) = 0. With A ∈ R

n×n the pairs of eigenvaluesand eigenvectors satisfy

Avi = λivi , i = 1, · · · , n , vi 6= 0

where λi ∈ C and vi ∈ Cn.

Lemma

If the eigenvalues of A ∈ Rn×n are distinct then the eigenvectors

v1, · · · , vn are linearly independent.

The proof is by absurd, suppose there exists p ≤ n − 1 suchthat

vk =

p∑

i=1

αivi

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Introduction


Multiplying this equality to the left by A and by λk we obtain

p∑

i=1

αi (λi − λk)vi = 0

showing that the vectors v1, · · · , vp are linear dependent aswell. Applying this argument successively, until p = 1, weconclude that vp = 0 for some p = 1, · · · , n which is anabsurd since vp is an eigenvector.

⇓

Matrix V = [vi · · · vn] ∈ Cn×n is nonsingular!

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Introduction


Matriz V defines the so called Similarity Transformation withthe following properties :

AV = VΛ where Λ ∈ Cn×n is the diagonal matrix

Λ = diag(λ1, · · · , λn)

The state space representation (A,B,C ,D) expressed in x

coordinates is converted to (V−1AV ,V−1B,CV ,D) expressedin x = V−1x coordinates with invariant transfer function, thatis

H(s) = CV (sI − V−1AV )−1V−1B + D

= C (sI − A)−1B + D

= H(s)

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Introduction


It is important to notice that if λ ∈ C is an eigenvalue ofA ∈ R

n×n then the same occurs to the complex conjugate λ∗.Hence, by definition Av = λv implies Av∗ = λ∗v∗ anddenoting λ = σ + jω and v = vR + jvI we have

A[vR vI ] = [vR vI ]

[σ ω−ω σ

]

If a matrix A ∈ Rn×n has n − 2 real eigenvalues and λn−1, λn

complex eigenvalues, then the similarity transformation withV = [v1, · · · , vn−2, vR , vI ] ∈ R

n×n provides

Λ = diag

(

λ1, · · · , λn−2,

[σ ω−ω σ

])

∈ Rn×n

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Nonlinear differential equations

Preliminaries

The nonlinear differential equation under analysis is as follows

x(t) = f (t, x(t)) , x(0) = x0

where

x(t) ∈ Rn × [0,T ]f (t, x) : [0,T ]× Rn → Rn

The final time T may be unbounded

The existence of a solution follows from the Picard’s method.By simple integration we obtain the equivalent version

x(t) = x0 +

∫ t

0f (τ, x(τ))dτ , t ∈ [0,T ]

depending only on the trajectory x(t) for all t ∈ [0,T ].

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Preliminaries

Picard’s method is an important device to solve nonlineardifferential equations. It is based on the sequence oftrajectories xk(t),∀t ∈ [0,T ] for k ∈ N defined as

xk+1 = Pxk

where P is the nonlinear operator

Px = x(0) +

∫ t

0f (τ, x(τ))dτ , t ∈ [0,T ]

Clearly, a solution of the previous nonlinear equation is anequilibrium trajectory of the operator P , that is x∗ = Px∗.The method starts with any trajectory x0(t) satisfying theinitial condition x0(0) = x0.

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Preliminaries

For illustration let us solve the linear equation x = Ax withinitial condition x(0) = x0 by Picard’s method. As it can beeasily verified, it provides the sequence

xk(t) =

k∑

i=0

(At)i

i !x0

which, as expected, converges (globally) to x∗(t) = eAtx0 forall t ∈ [0,∞]. Each iteration adds a new term in the Taylorseries of eAt developed at t = 0.

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Preliminaries

The operator P is not unique. For the important class ofnonlinear differential equation

x(t) = Ax(t) + Bg(t, x(t)) , x(0) = x0

the Picard’s method can be applied with:the operator obtained from direct integration

Px = x(0) +

∫ t

0

(Ax(τ) + Bg(τ, x(τ))dτ

the operator obtained from the solution of the linear part

Px = eAtx(0) +

∫ t

0

eA(t−τ )Bg(τ, x(τ))dτ

This is particulary useful for time-delay systems whereg(t, x(t)) = x(t − d) for some d > 0.

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Preliminaries

Before proceed we need to introduce some well knownconcepts. A linear vector space is a set of elements togetherwith the addition and multiplication by scalar operations. Asfor instance, the set Rn and the operations

x + y =

x1 + y1...

xn + yn

, αx =

αx1...

αxn

, α ∈ R

or the set of functions F n defined on the interval [0,T ] andthe operations

(x+y)(t) =

x1(t) + y1(t)...

xn(t) + yn(t)

, (αx)(t) =

αx1(t)...

αxn(t)

, α ∈ R

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Preliminaries

A normed linear space is a linear vector space X equippedwith a real valued function called norm ‖ · ‖ : X → R whichsatisfies the following axioms:

‖x‖ ≥ 0, ∀x ∈ X and ‖x‖ = 0 iff x = 0 ∈ X

‖αx‖ = |α|‖x‖ for any scalar α‖x + y‖ ≤ ‖x‖+ ‖y‖, ∀x , y ∈ X

Consider the vector linear vector space Rn and

‖x‖p =

(n∑

i=1

|xi |p)1/p

valid for p = 1, 2, · · · . The more important are

‖x‖1 =n∑

i=1

|xi |, ‖x‖2 =

√√√√

n∑

i=1

|xi |2, ‖x‖∞ = maxi=1,··· ,n

|xi |

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Preliminaries

Consider a sequence {xk}∞k=0 in a normed linear space(X , ‖ · ‖). It converges to x∗ ∈ X if for every ǫ > 0 there existan integer N(ǫ) such that

‖xk − x∗‖ < ǫ, ∀k ≥ N(ǫ)

Notice that, to apply this test, x∗ must be known.

A sequence {xk}∞k=0 in a normed linear space (X , ‖ · ‖) is aCauchy sequence if for every ǫ > 0 there exist an integer N(ǫ)such that

‖xk − x r‖ < ǫ, ∀k , r ≥ N(ǫ)

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Preliminaries

Any convergent sequence is a Cauchy sequence. The contraryin not true since there exist Cauchy sequences which do notconverge to some element x∗ ∈ X .

A normed linear space (X , ‖ · ‖) is a complete normed linearspace or a Banach space if each Cauchy sequence convergesto some element of X .

Important example : X = C[0,T ] is the set of continuousfunction f (t) : [0,T ] → R and ‖f ‖ = maxt∈[0,T ] |f (t)| Noticethat the first two axioms hold and

‖f1 + f2‖ = maxt∈[0,T ]

|f1(t) + f2(t)|

≤ maxt∈[0,T ]

|f1(t)|+ |f2(t)|

≤ ‖f1‖+ ‖f2‖30 / 58



Preliminaries

On the other hand, with ǫ > 0 arbitrarily small and anyconvergent sequence f k ∈ X , the constraint

‖f k − f ∗‖ = maxt∈[0,T ]

|f k(t)− f ∗(t)| < ǫ

implies that f ∗ is continuous and consequently (X , ‖ · ‖) is aBanach space.

The same set of continuous functions X = C[0,T ], togetherwith the Euclidean norm

‖f ‖ =

√∫ T

0f (τ)2dτ

is not complete. Hence, it is not a Banach space.

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Preliminaries

Indeed, this follows from the fact that it is possible to generatea sequence of continuous function f k ∈ X that converges to adiscontinuous function f ∗ /∈ X . The next figure shows in solidline the discontinuous function in the interval X = C[−1,1]

f ∗(t) =

{1 0 < t ≤ 1−1 −1 ≤ t < 0

−1 0 1

−1

0

1

f ∗f k

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Preliminaries

The same figure shows in dashed lines the sequence ofcontinuous functions

f k(t) =

{1− e−t/τk 0 < t ≤ 1

−1 + et/τk −1 ≤ t < 0

where τk > 0 goes to zero as k goes to infinity. With theEuclidean norm we obtain

‖f k − f ∗‖2 = τk(1− e−2/τk )

< τk

which shows that the sequence of continuous functions f k

converges to the discontinuous function f ∗.

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Preliminaries

Consider (X , ‖ · ‖) and (Y , ‖ · ‖) normed linear spaces. Thefunction f : X → Y is uniformly continuous at z ∈ X if forevery ǫ > 0 there exist δ(ǫ) such that

‖f (x)− f (z)‖ < ǫ whenever ‖x − z‖ < δ(ǫ)

If f (x) is continuous at x0 ∈ X then a convergent sequencexk → x0 provides a convergent sequence f (xk) → f (x0).

A particularly important concept is the inner product ofelements x , y ∈ X . For instance:

For X = Rn then x · y =∑n

i=1 xiyi = x ′y .

For X = C n[0,T ] then x · y =

∫ T

0 x(t)′y(t)dt.

The quantity√x · x is a norm of x ∈ X .

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Preliminaries

A complete normed linear space (X , ‖ · ‖) with ‖x‖ =√x · x

is denominated Hilbert space.

Notice that:

The normed linear space (Rn, ‖ · ‖) with ‖x‖ =√x ′x is a

Hilbert space.The normed linear space (C n

[0,T ], ‖ · ‖) with

‖x‖ =

√∫ T

0 x(t)′x(t)dt is not a Hilbert space.

The normed linear space (C n[0,T ], ‖ · ‖) with

‖x‖ = maxt∈[0,T ]

√

x(t)′x(t) is not a Hilbert space but it is aBanach space.

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Existence and uniqueness

The analysis of existence and uniqueness of a solution of agiven nonlinear differential equation depends on twoimportant results to be discussed in the sequel.

Theorem (Contraction mapping theorem)

Let (X , ‖ · ‖) be a Banach space and Q : X → X be an operator.

If there exists 0 ≤ ρ < 1 such that

‖Qx − Qy || ≤ ρ‖x − y‖, ∀x , y ∈ X

then the following hold:

There exists an unique x∗ ∈ X satisfying x∗ = Qx∗.

The sequence xk+1 = Qxk , x0 = x0 converges to x∗ for all x0 ∈ X.

For all k ≥ 1 we have ‖xk − x∗‖ ≤ (ρk/(1− ρ))‖Qx0 − x0‖.

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Proof : Each item is proven as follows:If here exist x , y ∈ X such that Qx = x and Qy = y then

‖Qx − Qy‖ = ‖x − y‖ ≤ ρ‖x − y‖

which together with ρ < 1 yields x = y .For all k , r ≥ 1, we have ‖xk+1 − xk‖ ≤ ρk‖Qx0 − x0‖ and

‖xk+r − xk‖ ≤r−1∑

i=0

‖xk+i+1 − xk+i‖

≤ ρk‖Qx0 − x0‖r−1∑

i=0

ρi

≤ (ρk/(1− ρ))‖Qx0 − x0‖

The sequence is a Cauchy sequence and converges to somex∗ ∈ X due to the fact that (X , ‖ · ‖) is a Banach space.

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The third item is important because it provides an estimate ofthe rate of convergence to the fixed point x∗ ∈ X .

Form the previous calculations we have

‖x∗ − xk‖ = limr→∞

‖xk+r − xk‖

≤ ρk‖Qx0 − x0‖

1− ρ

and the proof is concluded.

Example : Consider X = R and f (x) : R → R continuouslydifferentiable. Since f (x) = f (y) + f ′(ξ)(x − y) for someξ ∈ [x , y ] ⊂ R, then |f ′(ξ)| < 1, ∀ξ ∈ R implies that thesequence xk+1 = f (xk) always converges to the uniquesolution of the equation x = f (x).

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Theorem (Gronwall’s inequality)

Let a(t) be a continuous function and c ≥ 0, b scalars. The

inequality

a(t) ≤ b + c

∫ t

0a(τ)dτ,∀t ≥ 0

implies that a(t) ≤ bect for all t ≥ 0.

Proof : Defining q(t) = b + c∫ t

0 a(τ)dτ , it follows thata(t) ≤ q(t) and q(t) = ca(t) ≤ cq(t). With x(t) = cx(t) andx(0) = q(0) = b it is seen that x(t) ≥ q(t), ∀t ≥ 0 becausethe existence of t > 0 such that x(t) = q(t) impliesx(t) = cx(t) = cq(t) ≥ q(t). Hence

a(t) ≤ q(t) ≤ x(t) = bect , ∀t ≥ 0

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Let us now consider the nonlinear differential equation

x(t) = f (t, x(t)) , x(0) = x0

where x(t) ∈ Rn and t ∈ [0,T ] with T > 0. Defining the

nonlinear operator

Px = x(0) +

∫ t

0f (τ, x(τ))dτ

it is clear that a solution of the above differential equationsatisfies x = Px which can be calculated iteratively by thePicard’s method

xk+1 = Pxk

from an initial trajectory x0.

40 / 58




The next theorem is important for further developments.

Theorem

Assume there exist positive constants κ and µ such that

‖f (t, x)− f (t, y)‖ ≤ κ‖x − y‖, ∀x , y ∈ Rn, t ∈ [0,T ]

‖f (t, x0)‖ ≤ µ, ∀t ∈ [0,T ].

The differential equation x = f (t, x), x(0) = x0 admits an unique

solution in the time interval [0,T ].

Proof : For X = Cn[0,T ] and ‖x‖C = maxt∈[0,T ‖x(t)‖ the

normed linear space is a Banach space. On the other hand

Px(t)− Py(t) =

∫ t

0

(f (τ, x(τ)) − f (τ, y(τ))

)dτ

since x(0) = y(0) = x0.

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Yielding

‖Px(t)− Py(t)‖ ≤ κ

∫ t

0‖x(τ)− y(τ)‖dτ

≤ (κt)‖x − y‖Cand, with the same reasoning,

‖P2x(t)− P2y(t)‖ ≤ κ

∫ t

0‖Px(τ)− Py(τ)‖dτ

≤ (κ2t2/2)‖x − y‖Cwhich implies, by induction, that

‖Pmx − Pmy‖C ≤ (κmTm/m!)‖x − y‖Cfor all m ≥ 1.

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Choosing an integer m ≥ 1 such that (it always exists)

ρ = κmTm/m! < 1

and considering Q = Pm the proof follows from thecontraction mapping theorem provided that ‖Qx0 − x0‖ is abounded quantity which is assured by the second assumption.

This theorem puts in evidence two possible procedures (to bediscussed afterwards) to solve numerically a given differentialequation:

Time decompositionSpace decomposition

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Time decomposition : Divide the time interval [0,T ] intosub-intervals of length ∆T such that ρ = κ∆T < 1 and applythe Picard’s method in each time interval, successively.

Spacial decomposition : Determine m ≥ 1 such thatρ = κmTm/m! < 1 and apply the Picard’s method once tothe time interval [0,T ].

In both cases, global convergence is assured by the fact that

‖Qx − Qy‖C ≤ ρ‖x − y‖C

holds for ρ < 1 and Q = P for the first case and Q = Pm forthe second one.

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Remarks :

The conditions of the previous theorem are only sufficient. Forinstance, the differential equation x = −x2, x(0) = 1 admitsthe solution x(t) = 1/(1 + t) ∈ C[0,∞) but does not satisfy thefirst condition.The first condition of the previous theorem is also calledLipschitz condition.In the linear case f (x) = Ax , we have

‖f (x)− f (y)‖ = ‖A(x − y)‖≤ ‖A‖‖x − y‖

providing µ = κ = ‖A‖.

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Periodic solutions

Linear and nonlinear differential equations may admit periodicsolutions. However, they are intrinsically different as we willshow in the sequel.

Consider the linear differential equation

x1 = x2

x2 = −x1

which can be expressed in polar coordinates (r , φ) wherer2 = x21 + x22 and tg(φ) = x2/x1. Indeed, by simple derivationwith respect to time we obtain

r =x1x1 + x2x2

r

φ =x1x2 − x2x1

r2

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Periodic solutions

By substitution we have two decoupled equations r = 0 andφ = −1 which provide the solution

r(t) = r(0) , φ(t) = φ(0)− t, ∀t ≥ 0

This is a periodic solution represented in the (x1, x2) plane bya circle. Notice however that the radius and consequently deamplitude of the oscillation depends on the initial conditionsthrough

r(0)2 = x1(0)2 + x2(0)

2

As it will be seen, the behavior of the periodic solutions ofnonlinear systems is quite different.

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Periodic solutions

Consider now the nonlinear differential equation

x1 = x2 + x1(1− x21 − x22 )

x2 = −x1 + x2(1− x21 − x22 )

Adopting again the polar coordinates (r , φ), the timederivatives provide

r = r(1− r2)

φ = −1

which as it can be verified, admits the solution

r(t) =1√

1 + ce−2t, φ(t) = φ(0)− t, ∀t ≥ 0

where c = −1 + 1/r(0)2.

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Periodic solutions

On the left part of the next figure, the linear equation providesperiodic solutions. Notice that different periodic solutions areobtained corresponding to each initial condition. On the rightpart, for the nonlinear equation, only one periodic solution isobtained starting from any initial condition.

0 2 4 6 8 10−3

−2

−1

0

1

2

3

0 2 4 6 8 10−3

−2.5

−2

−1.5

−1

−0.5

0

0.5

1

1.5

tt

x1(t)

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Equilibrium points

A point xe ∈ Rn is said to be an equilibrium point of the

autonomous differential equation

x = f (x), x(0) = x0

where f (·) : Rn → Rn, if the initial condition x(0) = xe

implies that the corresponding solution is x(t) = xe ∀t ≥ 0.From this definition, all equilibrium points are characterizedby the algebraic equation

f (xe) = 0

which, clearly, may have multiple solutions.

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Linearization

Consider again the differential equation x = f (x). In aneighborhood of some equilibrium point xe ∈ R

n, eachcomponent fi(x), i = 1, · · · , n, can be approximated by theTaylor’s series

fi (x) ≈ fi(xe) +n∑

j=1

∂fi∂xj

(xe)(xj − xej), i = 1, · · · , n

yieldingf (x) ≈ f (xe) + A(x − xe)

where An×n is a square matrix with elements

aij =∂fi∂xj

(xe) , i , j = 1, · · · , n

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Linearization

Defining the new variable z = x − xe , taking into account thatf (xe) = 0, an approximate linear differential equation isobtained

z = Az , z(0) = x0 − xe

which provides the approximate solution

x(t) = xe + eAt(x0 − xe),∀t ≥ 0

Remarks :

Of course this is only an approximation. Hence, it is valid onlynear the equilibrium point.The matrix A ∈ R

n×n depends on each equilibrium pointxe ∈ Rn.

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Problems

Problems

1. Show that the state transmission matrix Φ(t, τ) can bewritten as

Φ(t, τ) = Θ(t)Θ(τ)−1

and determine Θ(t) for all t ≥ 0.

2. Consider X = Rn. Prove that

‖x‖∞ = maxi=1,··· ,n

|xi |

3. Determine and plot the phase plane (ν, θ) where ν = θ for thedifferential equation

θ + 2θ3 = 0

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Problems

Problems

4. Consider A ∈ Rn×n, X = Rn and define the induced norm

‖A‖ = maxx∈X

{‖Ax‖p : ‖x‖p = 1}

for some nonnegative integer p.

Prove that ‖A‖p satisfies the norm axioms.Prove that for all A,B ∈ Rn×n we have ‖AB‖ ≤ ‖A‖p‖B‖pDetermine ‖A‖p for p = 1, 2,∞.

5. Consider the normed linear space (X , ‖ · ‖). Prove that thefunction f (·) = ‖x‖ : X → R is convex.

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Problems

Problems

6. Consider R ∈ Rn×n a symmetric matrix. Prove that :

All eigenvalues and eigenvectors are real.Determine V ∈ Rn×n and Λ ∈ Rn×n diagonal such that

V−1RV = Λ , V−1 = V ′

For all x 6= 0 ∈ Rn it is true that

λmin ≤ x ′Rx

x ′x≤ λmax

where λmin and λmax are the minimum and the maximumeigenvalue of R , respectively.

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Problems

Problems

7. Consider the nonlinear differential equation

x − e−x = 0 , x(0) = 1

and the time interval t ∈ [0, 10]. Determine:

its solution, analytically.its solution by the Picard’s method.

8. Consider the nonlinear equation θ + θ + sin(θ) = 0 with initial

conditions θ(0) = π/4 and θ(0) = 0.

Solve it numerically.Determine κ and Tmax such that Q = P2 is a contractionmapping.For T = Tmax/2, solve it by the Picard’s method.

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Problems

Problems

9. For the dynamic system given in the next figure, determine :

Its mathematical model in terms of the state variables(x , x , θ, θ).The linearized model valid around the equilibrium point(0, 0, 0, 0).

m

M

ℓ

f

θ

x

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Problems

Problems

10. For the same system car - pendulum considered in theprevious problem, determine:

The transfer function from the force f and the angulardisplacement of the pendulum θ.The impulse response.The solution of the nonlinear model corresponding to f (t) = 1for all t ≥ 0 and initial conditions

x(0) = x(0) = 0 , θ(0) = π/4, θ(0) = 0

The solution of the linear model and compare.

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nonlinear systems - unicampgeromel/non_introd.pdf · h. k. khalil, “nonlinear systems”,...

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