... Edgard Blucher Ltda, 2011. H. K. Khalil, “Nonlinear Systems”, Macmillan
Publishing Co.,. 1992. M. Vidyasagar, “Nonlinear Systems Analysis”, Prentice
Hall,.
CHAPTER II - Second order nonlinear systems
NONLINEAR SYSTEMS ´ C. GEROMEL JOSE DSCE / School of Electrical and Computer Engineering UNICAMP, CP 6101, 13081 - 970, Campinas, SP, Brazil,
[email protected]
Campinas, Brazil, August 2008
1 / 40
CHAPTER II - Second order nonlinear systems
Contents
1
CHAPTER II - Second order nonlinear systems Note to the reader Preliminaries Local analysis Global analysis Periodic solutions Poincare’s index Example
Problems
2 / 40
CHAPTER II - Second order nonlinear systems 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 Sistemas Dinˆamicos : Teoria, Ensaios Pr´aticos e Exerc´ıcios (in Portuguese), 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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CHAPTER II - Second order nonlinear systems Preliminaries
Preliminaries Second order time-invariant systems behavior can be analyzed by the so called phase plane trajectory plots. Consider the second order system z¨ = −f (z, z) ˙ where z(t) ∈ R × [0, T ]. Adopting the state variables x = z and y = z˙ we have x˙
= y
y˙
= −f (x, y )
The phase plane is defined by the (x, y ) coordinates parameterized with respect to t ∈ [0, T ]. Each solution (with different initial conditions) defines a trajectory. 4 / 40
CHAPTER II - Second order nonlinear systems Preliminaries
Preliminaries Since the nonlinear equation satisfies ydy + f (x, y )dx = 0, a trajectory Γ is characterized by I ydy + f (x, y )dx = 0 Γ
In some situations this equation can be integrated without difficulty and provides Γ. Indeed, it is well known that if ∂y ∂f (x, y ) = =0 ∂x ∂y that is f (x, y ) = g (x), the line integral does not depend on the integration path. 5 / 40
CHAPTER II - Second order nonlinear systems Preliminaries
Preliminaries In this case, from the initial point (x0 , y0 ) ∈ Γ at t = 0 to a generic one (x, y ) ∈ Γ at any t ≥ 0, we have Z y Z x ydy + g (x)dx = 0 y0
x0
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 the line integral defining Γ does not depend on the particular integration path used. 6 / 40
CHAPTER II - Second order nonlinear systems Preliminaries
Preliminaries Example : For the pendulum discussed before θ¨ + ω02 sin(θ) = 0 with x = θ, y = θ˙ and g (x) = ω02 sin(x) and Z Z V (x, y ) = ydy + ω02 sin(x)dx =
y2 − ω02 cos(x) 2
any trajectory of the phase plane is given by Γ = (x, y ) ∈ R2 : V (x, y ) = V0
where V0 = V (x0 , y0 ) selects a particular trajectory passing through the point (x0 , y0 ), for instance the initial condition. 7 / 40
CHAPTER II - Second order nonlinear systems Local analysis
Local analysis In the general case, trajectories in the phase plane can be determined by numeric integration. However, in the neighborhood of each equilibrium point they can be drawn approximatively. The equilibrium point (xe , ye ) satisfies f (xe , ye ) = 0 ye
= 0
which implies that they are located on the x axis. The linear approximation, valid near (xe , ye ), is given by 0 1 ξ˙ = Aξ , A = ∂f ∂f − ∂x − ∂y (xe ,ye )
where ξ = [x − xe , y − ye ]′ . Afterwards, cases of interest are considered. 8 / 40
CHAPTER II - Second order nonlinear systems Local analysis
Local analysis Distinct real eigenvalues : Assume that λ 6= µ ∈ R are eigenvalues of matrix A ∈ R2×2 . Defining V = [vλ vµ ] ∈ R2 the matrix of eigenvalues, we know that V −1 AV = Λ = diag{λ, µ} which yields ξ(t) = e λt cλ vλ + e µt cµ vµ
where cλ and cµ are constants depending on the initial conditions. Consequently, near the equilibrium point we have x(t) − xe = e λt cλ vλ + e µt cµ vµ y (t) − ye 9 / 40
CHAPTER II - Second order nonlinear systems 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
CHAPTER II - Second order nonlinear systems Local analysis
Local analysis
20
15
10
5
0
−5
−10
−15
−20 −10
−8
−6
−4
−2
0
2
4
6
8
10
Saddle : Negative and positive eigenvalues (λ < 0, µ > 0). 11 / 40
CHAPTER II - Second order nonlinear systems Local analysis
Local analysis Complex eigenvalues : Assume that σ ± jω ∈ C are eigenvalues of matrix A ∈ R2×2 . Defining V = [vR vI ] ∈ R2 the matrix of eigenvalues, we know that σ ω −1 V AV = Λ = −ω σ and changing to polar coordinates r (t)cos φ(t) ξ(t) = V r (t)sin φ(t) after simple derivation with respect to time of ξ = V η we obtain η˙ = Λη and r˙ cos φ −r sin φ η˙ = sin φ r cos φ φ˙ | {z } U
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CHAPTER II - Second order nonlinear systems Local analysis
Local analysis Consequently
r˙ φ˙
= U −1 Λη σr = −ω
yielding r (t) = e σt cσ and φ(t) = −ωt + cω where the constants depend on the initial conditions. As in the first case, near the equilibrium point we have x(t) − xe = e σt cos(cω −ωt)cσ vR + e σt sin(cω −ωt)cσ )vI y (t) − ye 13 / 40
CHAPTER II - Second order nonlinear systems 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
CHAPTER II - Second order nonlinear systems Local analysis
Local analysis The eigenvalues of matrix A ∈ R2×2 of the linearized system at the equilibrium point x = xe are given by s 2 + Θs + ∆ = 0 where Θ = −trace(A) and ∆ = det(A). 2
1.5
stable focus
unstable focus
1
0.5
∆
stable node
unstable node
0
−0.5
saddle
−1
∆ = Θ2 /4
−1.5
−2 −4
−3
−2
−1
0
1
2
3
4
Θ 15 / 40
CHAPTER II - Second order nonlinear systems Global analysis
Global analysis The global analysis is done by drawing the whole phase plane as a composition of the parts obtained by the linearized model valid in the neighborhood of each equilibrium point. This is illustrated by means of the following nonlinear equation x˙
= −βx − y
y˙
= x − g (y )
where g (y ) is a piecewise linear function and two cases corresponding to β = 2 and β = 1/2 are considered. The equilibrium points (xe , ye ) are those satisfying x
= −y /β
x
= g (y ) 16 / 40
CHAPTER II - Second order nonlinear systems 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 ′ . The equilibrium points a′ and b ′ are virtual. x a′ a
10
g (y )
o
−20
y
−10 b b′
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CHAPTER II - Second order nonlinear systems Global analysis
Global analysis The matrix A of the linearized system which depends on each particular equilibrium point has the general form −β −1 A= 1 −g ′ (ye ) For β = 2 we have: For points a and b, Θ = 3 and ∆ = 3 =⇒ Stable focus For 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 focus For point o, Θ = −1/2 and ∆ = 1/2 =⇒ Unstable focus
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CHAPTER II - Second order nonlinear systems Global analysis
Global analysis 30
20
10
x
0
−10
−20
−30 −30
−20
−10
0
10
20
30
y
Globally, it is clear the behavior due to the saddle at the origin and two real stable focus which are asymptotically reached depending on the initial conditions. 19 / 40
CHAPTER II - Second order nonlinear systems Global analysis
Global analysis 30
20
10
x
0
−10
−20
−30 −30
−20
−10
0
10
20
30
y
The two virtual stable focus are never reached. The origin being an unstable focus enables the appearance of a limit cycle (periodic solution). 20 / 40
CHAPTER II - Second order nonlinear systems 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 curve such that R = intC ⊂ D. I Z Z ∂Q ∂P − dxdy Pdx + Qdy = ∂x ∂y C R Moreover, it is well known that if ∂Q ∂P = , ∀(x, y ) ∈ R ∂x ∂y then the line integral does not depend on the integration path and the Green’s theorem implies that I Pdx + Qdy = 0 C
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CHAPTER II - Second order nonlinear systems Periodic solutions
Periodic solutions We now discuss an important result used afterwards. Suppose that the equality ∂Q ∂P = , ∀(x, y ) ∈ R ∂x ∂y holds excepted at some isolated point p1 , · · · pN . Define the simple closed curves Ci such that pi ∈ intCi for all i = 1, · · · , N. It follows that I
Pdx + Qdy = C
N I X i =1
Pdx + Qdy
Ci
where each indicated line integral does not depend on the integration path. 22 / 40
CHAPTER II - Second order nonlinear systems Periodic solutions
Periodic solutions Example : Consider the line integral I −ydx + xdy J= x2 + y2 C Noticing that ∂P y2 − x2 ∂Q = = 2 , ∀x 6= 0, y 6= 0 ∂x ∂y x + y2 and the differential of θ = tg−1 (y /x) provides dθ =
xdy − ydx x2 + y2 23 / 40
CHAPTER II - Second order nonlinear systems Periodic solutions
Periodic solutions Two conclusions can be drawn: If the origin is not in the interior of C then Z J = dθ = 0 in accordance to the fact that the partial derivatives of P and Q are equal in all points inside C . From the Green’s theorem the integral must be zero. If the origin is in the interior of C then Z J = dθ = 2π The Green’s theorem does not apply. The value of the integral is not null anymore but it still does not depend on the integration path whenever the origin is in its interior. 24 / 40
CHAPTER II - Second order nonlinear systems 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 sign in all points (x, y ) of a simply connect domain D then there is no periodic solution (limit cycle) in the interior of D. Indeed, it is known that any periodic solution satisfies I ydy + f (x, y )dx = 0 Γ
for some closed curve Γ. 25 / 40
CHAPTER II - Second order nonlinear systems Periodic solutions
Periodic solutions Consider any closed curve Γ in the interior of D, from the Green’s theorem we have I Z Z ∂f dxdy ydy + f (x, y )dx = Γ intΓ ∂y 6= 0 because int Γ ⊂ D and ∂f /∂y is not identically zero and does not change sign in D. The only possible conclusion is that there are no closed trajectories inside D. For the linear system characterized by f (x, y ) = αx + βy , the Bendixson’s theorem assures that if β 6= 0 then, no periodic solutions exist.
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CHAPTER II - Second order nonlinear systems Periodic solutions
Poincare’s index Consider C a closed curve inside a simply connected domain. The Poincare’s index of C is I 1 IC = dθ 2π C where tg(θ) = −f (x, y )/y . The geometric interpretation is : y Γ
C
θ x
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CHAPTER II - Second order nonlinear systems Periodic solutions
Poincare’s index For the particular choice C = Γ, at any point of Γ the angle θ is 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 = and df =
fdy − ydf y2 + f 2
∂f ∂f dx + dy ∂x ∂y 28 / 40
CHAPTER II - Second order nonlinear systems Periodic solutions
Poincare’s index As a result, it follows that 1 IC = 2π
I
where P=−
Pdx + Qdy
C
y ∂f y 2 + f 2 ∂x
and
f y ∂f − y 2 + f 2 y 2 + f 2 ∂y Very tedious algebraic manipulations put in evidence that Q=
∂Q ∂P = ∂x ∂y whenever y 2 + f (x, y )2 6= 0. 29 / 40
CHAPTER II - Second order nonlinear systems Periodic solutions
Poincare’s index The previous condition holds in all points excepted those satisfying y = 0 , f (x, 0) = 0 that is, the equilibrium (isolated) points p1 , p2 , · · · of the nonlinear 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 then IC = IC ′ . If a number of equilibrium points p1 , p2 · · · are inside C then P IC = i ICi where Ci is any closed curve inside C containing only the equilibrium point pi . 30 / 40
CHAPTER II - Second order nonlinear systems Periodic solutions
Poincare’s index In conclusion the Poincare’s index is not a property of a given closed curve but the equilibrium point p in its interior. Hence, to calculate IC we can replace C by a closed curve C ′ in an arbitrarily small neighborhood of p and use the linear approximation of the system, yielding x˙ = y 0 1 =⇒ A = y˙ = −αx − βy −α −β where α and β are parameters given by ∂f ∂f α= , β= ∂x p ∂y p 31 / 40
CHAPTER II - Second order nonlinear systems Periodic solutions
Poincare’s index Considering f = αx + βy we obtain
2
x 2
y +f =
A y = r 2 2
2
which if det(A) 6= 0 defines a closed curve C ′ containing the origin for any r > 0 fixed. Moreover, simple calculations yield I det(A) IC = xdy − ydx 2πr 2 C ′ Z Z det(A) = dxdy πr 2 intC ′ det(A) = |det(A)| 32 / 40
CHAPTER II - Second order nonlinear systems Periodic solutions
Poincare’s index Fact Assume that C is a closed curve with a simply connected domain containing an equilibrium point: If the equilibrium point is a center, a focus or a node then IC = 1. If the equilibrium point is a saddle then IC = −1. Since we already know that IΓ = 1, a closed trajectory of the nonlinear system may exist only if : At least one equilibrium point must be inside Γ. The sum of the Poincare’s indexes of the equilibrium points inside Γ must be one.
33 / 40
CHAPTER II - Second order nonlinear systems Periodic solutions
Example Consider the nonlinear system x˙
= −x + xy
y˙
= y − xy
which presents two equilibrium points, namely −1 0 p1 = (0, 0) =⇒ A = , saddle, IC = −1 0 1 0 1 p2 = (1, 1) =⇒ A = , center, IC = 1 −1 0 A closed trajectory (if any) which characterizes a periodic solution or a limit cycle must contain p2 but not p1 in its interior! 34 / 40
CHAPTER II - Second order nonlinear systems Problems
Problems 1. Consider the line integral I J= x1 dx2 − x2 dx1 C
where C is the ellipsoid x ′ Qx = r 2 defined for r 6= 0 and Q ∈ R2×2 positive definite. Determine J and interpret the result using the Green’s theorem. 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 ∈ Rn×n positive definite such that (A + αI )′ P + P(A + αI ) < 0 then maxi =1,··· ,n Re{λi (A)} < −α. 35 / 40
CHAPTER II - Second order nonlinear systems Problems
Problems 3. The following is known as the Duffing equation z¨ + z˙ − z(1 − z 2 /c 2 ) = 0 where c 6= 0. Determine the equilibrium points and types. Drawn the trajectories of the phase plane near the equilibrium points.
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 this equilibrium point? 36 / 40
CHAPTER II - Second order nonlinear systems Problems
Problems 5. Consider the block diagram where a symmetric nonlinearity is as indicated. u
y
e 1−s/4 s 2 +s/2+1
f (·)
+ −
1 3/2 1/2
For u = 0, determine the state space realization with x1 = e and x2 = e, ˙ the equilibrium points and draw the phase plane. Repeat the previous item for u = 3. 37 / 40
CHAPTER II - Second order nonlinear systems Problems
Problems 6. The following is the so called Van der Pol’s equation. z¨ − µ(1 − z 2 )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 equilibrium points. 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
CHAPTER II - Second order nonlinear systems 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
CHAPTER II - Second order nonlinear systems Problems
Problems 10. Consider the nonlinear equation z¨ + 2ξω z˙ + ω 2 z + ǫz 3 = 0 with ω > 0. Determine and classify the equilibrium points for ǫ < 0, ǫ = 0 and ǫ > 0. 11. Consider the second order system x˙
= −2x + y
y˙
= x(x 2 − 3x + 4) − y
Determine the equilibrium points and their types. Take a stable equilibrium point, calculate the linear approximation and draw the phase plane near it. 40 / 40
