Applied Nonlinear Dynamics (SS 2020) - Exercise 1: Basic

Applied Nonlinear Dynamics (SS 2020) - Exercise 1: Basic concepts

Dr. Alexander Schaum

Chair of Automatic Control , Christian-Albrechts-University Kiel

1. Get yourselve familiarized with a simulation software for numerically solving nonlinear ordinarydifferential equations. You can use python2 with scipy together with matplotlib and study thetemplate/example odesolver.py provided on OLAT, that gives you a solution for the system

x1 = a11x31 + a12x2u, x1(0) = x10

x2 = a21x21 + a22x

32, x2(0) = x20

with initial conditions x10 = 1.0, x20 = −2.0 and u(t) = sin(t).

2. Do the following systems have a unique solution for any initial condition? Justify your answer.

(a) x = x1/4

(b) x = ax, a ∈ R.

(c) x = x1+x2

Hint: Recall the theorem of Picard and Lindelof.

3. Is it possible that for a two-dimensional time-invariant system with unique solutions the trajectoriescross at some point x∗, i.e. a situation shown in Figure 1 happens?

Figure 1: Crossing trajectories in the plane?

1

Applied Nonlinear Dynamics (SS 2020) - Solution for exercise 1



2. Do the following systems have a unique solution for any initial condition? Justify your answer.

(a) x = x1/4

(b) x = ax, a ∈ R.

(c) x = x1+x2

Hint: Recall the theorem of Picard and Lindelof.

Solution:

(a) The slope of the right-hand side function f(x) = x1/4 is given by f ′(x) = 14x−3/4 and goes to

infinity for x → 0. Thus, the function f(x) is not Lipschitz continuous at x = 0. In accordance,the solution with initial condition x = 0 is not unique. Actually, besides the trivial solutionx(t) = 0 one can derive a second solution by the variation of constants formula:∫ x(t)

0

dx

x1/4=

∫ t

0

dt

⇔ 4

3x3/4(t) = t

⇔ x(t) =

(3t

4

)4/3

.

(b) For the linear system it holds that f(x) = ax and thus f ′(x) = a implying the Lipschitzcontinuity of f(x) and thus, by the Theorem of Picard-Lindelof the uniqueness of the solution.The unique solution is given by x(t) = x0e

at.

(c) For the case that f(x) = x1+x2 it results that f ′(x) = 1−x2

(1+x2)2 showing that f ′(x) attains finite

values for all x ∈ R. Thus, f(x) is globally continuously differentiable, implying its Lipschitzcontinuity. The uniqueness of the solutions follows from the Theorem of Picard and Lindelof.

3. Is it possible that for a two-dimensional time-invariant system with unique solutions the trajectoriescross at some point x∗, i.e. a situation shown in Figure 1 happens?

Figure 1: Crossing trajectories in the plane?

Solution:Consider that two trajectories in the plane would cross and take the intersection point as initialcondition. Then, one trajectory starting at this point would go into one direction and the other inanother direction, implying the non-uniqueness in contradiction to the given assumption. Thus, theonly possibility for an appearent intersection is an equilibrium point where trajectories convergeto (or are repulsed away from), but an equilibrium point implies that trajectories starting in theequilibrium stay there for all future time, and thus this does not correspond to an intersection oftrajectories. In conclusion, it is not possible that trajectories for a system with uniquesolutions do intersect.

1

Applied Nonlinear Dynamics (SS 2020) - Exercise 2: Stability and qualitative solution



1. Which of the following assertions is true:

(a) A stable equilibrium point is attractive.

(b) An attractive equilibrium point is stable.

(c) An attractive equilibrium point can be unstable.

2. Analyze graphically the stability of the origin for the dynamics

a) x = x− x3

b) x = −x− x3

by plotting x versus x, and interpreting the vector field. Verify your results by numerical simulation.

3. Consider a one-dimensional system x = f(x) with f(0) = 0. What can you say about the relationbetween the slope f ′(0) = and the stability of the equilibrium point x = 0?

4. Sketch a possible graph of the function f(x) for a scalar dynamical system x = f(x) with exactly3 attractor equilibrium points.

1




1. Which of the following assertions is true:

(a) A stable equilibrium point is attractive.

(b) An attractive equilibrium point is stable.

(c) An attractive equilibrium point can be unstable.

Solution:

(a) This is false. A counterexample is a center, with persistent oscillations around the equilibriumpoint.

(b) This is false. A counterexample is given by the system introduced by Vinograd.

(c) This is true. The system introduced by Vinograd provides an example.

2. Analyze graphically the stability of the origin for the dynamics

a) x = x− x3

b) x = −x− x3

by plotting x versus x, and interpreting the vector field. Verify your results by numerical simulation.

Solution:

− 2.0 − 1.5 − 1.0 − 0.5 0.0 0.5 1.0 1.5 2.0

x

− 6

− 4

− 2

0

2

4

6

◌x

− 2.0 − 1.5 − 1.0 − 0.5 0.0 0.5 1.0 1.5 2.0

x

− 10.0

− ◌.5

− 5.0

− 2.5

0.0

2.5

5.0

◌.5

10.0

◌x

(a) x = x(1− x2) (b) x = −x(1 + x2)

(a) The origin is a repulsor. Thus it is unstable. In particular, for small ε < 1 it holds that for allδ > 0 trajectories starting in the δ-neigborhood always leave the ε-neighborhood after sometime.

(b) The origin is a (gloabl) attractor. Furthermore it is stable in the sense of Lyapunov: ∀ε > 0there exists δ > 0 so that any trajectory starting in the δ neighborhood remains within theε-neighborhood for all times. In particular this holds for all δ ≤ ε. In consequence, anyε-neighborhood of the origin is a positively invariant set.

3. Consider a one-dimensional system x = f(x) with f(0) = 0. What can you say about the relationbetween the slope f ′(0) = and the stability of the equilibrium point x = 0?

Solution:

1




• For f ′(x∗) > 0 and f(x∗) = 0 it follows that for any δ > 0 the value f(x∗δ) is positive andf(x∗ − δ) is negative. This implies, that trajectories starting to the right of x∗ move furtherto the right and those starting to the left move furhter to the left. Thus, for small ε it followsthat for any δ > 0 the trajectories starting in the δ-neighborhood will leave eventually theε-neighboord and thus the equilibrium point is unstable.

• For f ′(x∗) < 0 and f(x∗) = 0 it follows that for any δ > 0 the value f(x∗δ) is negativeand f(x∗ − δ) is positive. This implies, that trajectories starting to the right of x∗ move tothe left and converge to x∗, while trajectories starting on the left move to the right and alsoconverge to x∗. This implies that x∗ is attractive. Furthermore, for any ε > 0 one can find aδ > 0 small enough, so that trajectories starting within the δ-neighborhood stay within theε-neighborhood of x∗ and thus x∗ is stable. In consequence it is asymptotically stable.1

• For f ′(x∗) = 0 one cannot conclude anything about the stability because f can eithter havea minimum, maximum or inflection point at x∗.

4. Sketch a possible graph of the function f(x) for a scalar dynamical system x = f(x) with exactly3 attractor equilibrium points.

− 2 − 1 0 1 2 3 4

x

− 200

− 100

0

100

200

◌x

x = (x− 4)(x− 3)(x− 2)(x− 1)x(x+ 1)(x+ 2)

1This argumentation, which derives the stability from the attractivity does only work for time-invariant systems ina one-dimensional state space, i.e. with state dimension n = 1. For n = 2 the system proposed by Vinograd gives acounterexample (see Problem 1 of this exercise).

2

Applied Nonlinear Dynamics (SS 2020) - Exercise 3: Linear Dynamics


Chair of Automatic Control, Christian-Albrechts-University Kiel

1. Consider a dynamical system in (x, y)-coordinates. Show that in polar coordinates, the correspond-ing dynamics can be determined using the relations (cp. e.g. [1])

rr = xx+ yy, θ =xy − yxr2

.

2. Show that for any 2× 2-matrix A = {aij}i,j=1,2 the eigenvalues are given by the relation

λ12 =1

2

(tr(A)±

√tr(A)2 − 4det(A)

)where tr(A) = a11 + a22 is the trace and det(A) = a11a22 − a21a12 the determinant of A.

3. For the following example systems, determine the transformation into diagonal, Jordan or real Jor-dan normal form, solve the system in normal form and apply the back transformation to obtain thesolution in the original (x1, x2)-coordinates. Compare the solutions with numerical approximationsusing some standard ode-solver.

a) x =

[−4 2−3 1

]x

b) x =

[−1 1−1 −1

]x

c) x =

[−2 1−1 −4

]x

d) x =

[2 05 −1

]x

What can you say about stability in each case?

4. Consider the linear control system

x =

[1 20 1

]︸︷︷︸

A

x +

[01

]︸︷︷︸b

u, x(0) = x0.

Determine the eigenvalues and eigenvector(s) of the dynamic matrix A. Design a state-feedbackcontrol u = −kTx with k ∈ R2 such that the eigenvalues of the feedback matrix A−bkT are givenby λ1 = −1 and λ2 = −2. What are the eigenvectors of the modified dynamics matrix A − bkT ?What happens to the eigenvector structure when k is chosen such that both eigenvalues are thesame, say λ1 = λ2 = −3?

References

[1] S. H. Strogatz. Nonlinear Dynamics and Chaos, with applications to physics, biology, chemistry, andengineering. Perseus, Massachusetts,, 1994.

1

Applied Nonlinear Dynamics (SS 2020) - Solutions for exercise 3



1. Consider a dynamical system in (x, y)-coordinates. Show that in polar coordinates, the correspond-ing dynamics can be determined using the relations (cp. e.g. (?))

rr = xx+ yy, θ =xy − yxr2

.

Solution:Use r2 = x2+y2 and θ = arctan

(yx

), derive with respect to time (chain rule) and do some algebraic

manipulations. Hint: It holds that d arctan(x)dx = 1

1+x2 .

2. Show that for any 2× 2-matrix A = {aij}i,j=1,2 the eigenvalues are given by the relation

λ12 =1

2

(tr(A)±

√tr(A)2 − 4det(A)

)where tr(A) = a11 + a22 is the trace and det(A) = a11a22 − a21a12 the determinant of A.

Solution:The derivation can be found almost completely in the lecture notes.

3. For the following example systems, determine the transformation into diagonal, Jordan or real Jor-dan normal form, solve the system in normal form and apply the back transformation to obtain thesolution in the original (x1, x2)-coordinates. Compare the solutions with numerical approximationsusing some standard ode-solver.

a) x =

[−4 2−3 1

]x

b) x =

[−1 1−1 −1

]x

c) x =

[−2 1−1 −4

]x

d) x =

[2 05 −1

]x

What can you say about stability in each case?

Solution:

a) The eigenvalues are given by λ1 = −1 and λ2 = −2. Correspondingly, the origin is asymptot-ically stable, or more precisely, a stable node. The eigenvectors are given by

v1 =

[23

], v2 =

[11

]leading to the transformation matrix T =

[v1 v2

]. A simple calculation shows that

T−1AT =1

(−1)

[1 −1−3 2

] [−4 2−3 1

] [2 13 1

]=

[−1 00 −2

]

1




Setting ξ = T−1x it follows that

ξ =

[−1 00 −2

]ξ, ξ(0) = ξ0 = T−1x0, ⇒ ξ(t) =

[e−t 00 e−2t

]ξ0

and thus

x(t) = S(t)x0 = Tξ(t) = T

[e−t 00 e−2t

]T−1x0︸︷︷︸

=ξ0

=

[2 13 1

] [e−t 00 e−2t

]1

(−1)

[1 −1−3 2

]︸︷︷︸

=S(t)

x0

=

[2e−t e−2t

3e−t e−2t

] [−1 13 −2

]x0 =

[−2e−t + 3e−2t 2e−t − 2e−2t

−3e−t + 3e−2t 3e−t − 2e−2t

]︸︷︷︸

=S(t)

x0

The solutions are illustrated in Figure 1. It can be seen, that a behavior similar to a Jordanblock is achieved, given that the eigenvectors are quite close to another.

Figure 1: Solution curves for problem a) showing a stable node.

b) The dynamics matrix is given in the real Jordan normal form with eigenvalues λ12 = −1± i.Correspondinlgy the origin is a stable spiral. Using the transformation into polar coordinates(cp. E 1.2) we obtain

r = −r, r(0) = r0 =√x201 + x202

θ = −1, θ(0) = θ0 = arctan

(x02x01

)with solutions

r(t) = r0e−t, θ(t) = θ0 − t

or equivalently in cartesian coordinates

x1(t) = r(t) cos(θ(t)) = r0e−t cos(θ0 − t)

= r0e−t (cos(θ0) cos(t) + sin(θ0) sin(t))

= e−t (x01 cos(t) + x02 sin(t))

x2(t) = r(t) sin(θ(t)) = r0e−t sin(θ0 − t)

= r0e−t (sin(θ0) cos(t)− cos(θ0) sin(t))

= e−t (−x01 sin(t) + x02 cos(t))

2




or summarizing in vector notation

x(t) = e−t

[cos(t) sin(t)− sin(t) cos(t)

]︸︷︷︸

=S(t)

x0.

The solution are illustrated in Figure 2 showing the predicted behavior of a stable spiral.

Figure 2: Solution curves for problem b) showing a stable spiral.

c) The unique eigenvalue is given by λ1 = −3 with algebraic multiplicity α1 = 2. Consideringthe matrix (λ1I −A) yields

(λ1I −A) =

[−1 −11 1

], rank (λ1I −A) = 1

implying that the geometric multiplicity is γ1 = 1 (The kernel of (λ1I − A) is of dimension1). Thus, generalized eigenvectors need to be constructed using the associated Jordan chain

(λ1I −A)v1 = 0

(λ1I −A)h1,2 = v1

yielding

v1 =

[1−1

], h1,2 =

[10

].

The associated transformation matrix

T =[v1 h1,2

]=

[1 1−1 0

], T−1 =

[0 −11 1

]yields to the Jordan block

J1 = T−1AT =

[0 −11 1

] [−2 1−1 4

] [1 1−1 0

]=

[−3 10 −3

].

The transformed dynamics

ξ = T−1x, ξ = T−1ATξ = J1ξ, ξ(0) = ξ0 = T−1x0

3




or equivalently

ξ1 = −3ξ1 + ξ2, ξ1(0) = ξ10

ξ2 = −3ξ2, ξ2(0) = ξ20

has solutions (using the variations of constants formula for the solution of ξ1)

ξ(t) = e−3t

[ξ10 + tξ20

ξ20

]=

[e−3t te−3t

0 e−3t

]ξ0.

The solution in original coordinates is thus given by

x(t) = S(t)x0 = T

[e−3t te−3t

0 e−3t

]T−1x0

=

[1 1−1 0

] [e−3t te−3t

0 e−3t

] [0 −11 1

]x0

=

[1 1−1 0

] [te−3t −(1− t)e−3t

e−3t e−3t

]x0

=

[(1 + t)e−3t te−3t

−te−3t (1− t)e−3t

]︸︷︷︸

=S(t)

x0

Figure 3 shows the behavior of the solutions illustrating that the origin is a degenerated node(the transition between a stable node and a spiral).

Figure 3: Solution curves for problem c) showing a degenerated stable node.

d) Given the lower triangular structure of the matrix A the eigenvalues are given by the diagonalelements, i.e. λ1 = 2 and λ2 = −1. Correspondingly, the origin is an unstable saddle. Theassociated eigenvectors are given by

v1 =

[153

], v2 =

[01

]leading to the transformation matrix T =

[v1 v2

]. A simple calculation shows that

T−1AT =

[1 0− 5

3 1

] [2 05 −1

] [1 053 1

]=

[2 00 −1

]

4




Setting ξ = T−1x it follows that

ξ =

[2 00 −1

]ξ, ξ(0) = ξ0 = T−1x0, ⇒ ξ(t) =

[e2t 00 e−t

]ξ0

and thus

x(t) = S(t)x0 = Tξ(t) = T

[e2t 00 e−t

]T−1x0︸︷︷︸

=ξ0

=

[1 053 1

] [e2t 00 e−t

] [1 0− 5

3 1

]︸︷︷︸

=S(t)

x0

=

[e2t 053e

2t e−t

] [1 0− 5

3 1

]x0 =

[e2t 0

53

(e2t − e−t

)e−t

]︸︷︷︸

=S(t)

x0

The solutions are illustrated in Figure 1. It can be seen, that a behavior similar to a Jordanblock is achieved, given that the eigenvectors are quite close to another.

-1 -0.5 0 0.5-2

-1

0

1

2

x2

x1

Figure 4: Solution curves for problem d) showing an unstable saddle.

Follows similar to Problem a). The eigenvalues are given by λ1 = 2 and λ2 = −1 implying an(unstable) saddle.

4. Consider the linear control system

x =

[1 20 1

]︸︷︷︸

A

x+

[01

]︸︷︷︸b

u, x(0) = x0.

Determine the eigenvalues and eigenvector(s) of the dynamic matrix A. Design a state-feedbackcontrol u = −kTx with k ∈ R2 such that the eigenvalues of the feedback matrix A−bkT are givenby λ1 = −1 and λ2 = −2. What are the eigenvectors of the modified dynamics matrix A − bkT ?What happens to the eigenvector structure when k is chosen such that both eigenvalues are thesame, say λ1 = λ2 = −3?

Solution:The matrix has triangular structure and thus the eigenvalues are given by λ1 = λ2 = 1, implyingan algebraic multiplicity of α = 2. Given that

rank(A− λI) = rank

([0 20 0

])= 1

5




it follows for the geometric multiplicity that γ = 1. Accordingly there is only one eigenvector andthus only one positively invariant subspace, going along with typical overshoot phenomena.

Considering the closed-loop dynamics matrix

A− bkT =

[1 2−k1 1− k2

]with characteristic polynomial

(λ− 1)(λ− 1 + k2) + 2k1 = λ2 + (k2 − 2)λ+ 2k1 + 1− k2 = 0.

The characteristic polynomial associated to the desired eigenvalues λ1 = −1, λ2 = −2 is given by

(λ+ 1)(λ+ 2) = λ2 + 3λ+ 2 = 0.

A comparison of the coefficients yields

k1 = 3, k2 = 5 ↔ A− bkT =

[1 2−3 −4

].

The associated eigenvectors are given by

v1 =

[11

], v2 =

[1− 3

2

].

For the case that λ1 = λ2 = −3, implying an algebraic multiplicity of α = 2 one obtains thecharacteristic polynomial

(λ+ 3)2 = λ2 + 6λ+ 9

which corresponds to

k1 = 8, k2 = 8 ↔ A− bkT =

[1 2−8 −7

].

Given that rank(A− bkT

)= rank

([4 2−8 −4

])= 1 it turns out that the geometric multiplicity

γ = 1, implying that only one eigenvalue exists. This is given by

w =

[12

].

which does not correspond to either of the preceding eigenvalues. This shows that by changing theeigenvalues for the state-feedback based closed-loop control also the eigenvector structure and thusthe geometry of the solutions changes.

6

Applied Nonlinear Dynamics (SS 2020) - Exercise 4



1. Let v1,v2 be eigenvectors of the dynamics matrix A of a linear system x = Ax, i.e. Avi = λivi

with the eigenvalue λi ∈ R. Show that the set

Σ = span(v1,v2) = {x ∈ Rn |x = a1v1 + a2v2, a1, a2 ∈ R},

i.e. the span of the eigenvectors, is a positively invariant set.

2. Show that for λ = 0 the dynamics

x =

[λ 10 λ

]x, x(0) = x0

is unstable in the sense of Lyapunov, i.e. ∃ ε > 0 ∀ δ > 0 : ∃ ‖x0‖ ≤ δ ⇒ ‖x(t;x0)‖ > ε for somet(x0) > 0.

3. Show that for the following system

x =

[α β−β α

]x, x(0) = x0

the origin is stable in the sense of Lyapunov for any β ∈ R and α ≤ 0.

1




1. Let v1,v2 be eigenvectors of the dynamics matrix A of a linear system x = Ax, i.e. Avi = λiviwith the eigenvalue λi ∈ R. Show that the set

Σ = span(v1,v2) = {x ∈ Rn |x = a1v1 + a2v2, a1, a2 ∈ R},

i.e. the span of the eigenvectors, is a positively invariant set.

Solution:Let at some t ≥ 0 the solution x(t) ∈ Σ = span(v1,v2), meaning that there exist a1, a2 ∈ R sothat x(t) = a1v1 + a2v2. It follows that

x(t) = Ax(t) = a1Av1 + a2Av2 = a1λ1v1 + a2λ2v2 ∈ Σ.

Given that this reasoning is independent of x ∈ Σ it follows that the solution will stay in Σ for allt∗ ≥ t, implying the positive invariance of Σ. �

2. Show that for λ = 0 the dynamics

x =

[λ 10 λ

]x, x(0) = x0

is unstable in the sense of Lyapunov, i.e. ∃ ε > 0 ∀ δ > 0 : ∃ ‖x0‖ ≤ δ ⇒ ‖x(t;x0)‖ > ε for somet(x0) > 0.

Solution:Let ε > 0 be arbitrary and let δ < ε. Let ‖x0‖ =

√x210 + x220 ≤ δ. The solution for λ = 0 reads

x2(t) = x20 and x1(t) = x10 +∫ t0x20dτ = x10 + tx20 (by the variation of constants formula). Thus

‖x(t)‖ =√x210 + 2tx20 + t2x220 + x220 ≥

√2tx20 + t2x220 > ε

if it holds that

ε2 < 2tx20 + t2x220.

Considering x20 6= 0 this holds true for all t > t∗(x0) where

(t∗)2 +2

x20t− ε2

x220= 0

with positive solution t∗ = 1+√1+ε2

x20. �

3. Show that for the following system

x =

[α β−β α

]x, x(0) = x0

the origin is stable in the sense of Lyapunov for any β ∈ R and α ≤ 0.

Solution:(Note: There are several ways for showing this fact.)

In polar coordinates the dynamics read

r = αr, r(0) = ‖x0‖

φ = β, φ(0) = arctan

(x20x10

),

1




with solutions r(t) = eαtr(0) and φ(t) = φ(0) + βt. Accordingly, the norm of x(t) is equivalentlygiven by r(t) which for α ≤ 0 is non-decreasing.

Let ε > 0 be given. For all δ ≤ ε it holds that ‖x(t)‖ = r(t) = eαtr(0) = eαt‖x0‖ ≤ eαtδ ≤ δ ≤ ε,implying the stability in the sense of Lyapunov. �

2




1. Determine the analytical solution of the following dynamics and compare them with numericalsimulations:

(a) x(t) = −tx(t), x(t0) = x0

(b) x(t) = −(1 + t)x(t), x(t0) = x0

(c) x(t) = − sin(t)x(t), x(t0) = x0

2. Determine the monodromy matrix and the Floquet multiplier µ and Lyapunov exponent σ for case(c) of the preceding problem.

3. Consider the linear time-varying system

x = A(t)x(t), x(t0) = x0, A(t) =

[a 00 a− b cos(t)

](1)

• Determine the fundamental solution for t0 = 0 and P (t0) = I.

• Calculate the analytical solution.

• Determine the monodromy matrix M and the Floquet mulipliers. What is the condition forasymptotic stability?

• Determine the Floquet and Lyapunov exponents.

• Solve the dynamics (1) numerically for a = −0.05, b = 0.5 and compare the discrete-timesolution obtained using the monodromy matrix M .

1




1. Determine the analytical solution of the following dynamics and compare them with numericalsimulations:

(a) x(t) = −tx(t), x(t0) = x0

(b) x(t) = −(1 + t)x(t), x(t0) = x0

(c) x(t) = − sin(t)x(t), x(t0) = x0

Solution:

(a) Using the variation of constants formula it follows that∫ x(t)

x(t0)

dx

x= −

∫ t

t0

τdτ

or equivalently

x(t) = e

(− t2−t20

2

)x0.

(b) In the same way as in the preceding example it follows that

x(t) = e

(−(t−t0)−

t2−t202

)x0.

(c) Following the same reasoning it follows that

x(t) = e−∫ tt0

sin(τ)dτx0 = e(cos(t)−cos(t0))x0

The solutions are compared with numerical solutions obtained using Python in Figure 1.

2. Determine the monodromy matrix and the Floquet multiplier µ and Lyapunov exponent σ for case(c) of the preceding problem.

Solution:The coefficient − sin(t) is periodic with period T = 2π. The differential equation for the funda-mental matrix is given by

p(t) = − sin(t)p(t), p(t0) = p0.

Setting t0 = 0 and p0 = 1 it follows from the above considerations that

p(t) = e(cos(t)−1).

The monodromy matrix is thus given by

m = p(t0 + T ) = p(2π) = e(cos(2π)−1) = 1. (1)

Given that m is scalar it corresponds to the Floquet multiplier µ = m. The associated Lyapunovexponent is given by σ = cos(2π) − 1 = 0.

3. Consider the linear time-varying system

x = A(t)x(t), x(t0) = x0, A(t) =

[a 00 a− b cos(t)

](2)

1




0.0 2.5 5.0 7.5 10.0 12.5 15.0 17.5 20.0t

0.0

0.5

1.0

x 1

0.0 2.5 5.0 7.5 10.0 12.5 15.0 17.5 20.0t

0.0

0.5

1.0

x 2

0.0 2.5 5.0 7.5 10.0 12.5 15.0 17.5 20.0t

0.5

1.0

x 2

Figure 1: Comparison of numerical solutions (blue lines) and analytic solutions (orange, discontinuouslines) for the problems a,b,c in the top, center and bottom plot, respectively with x(t0) = 1 and t0 = 0.

• Determine the fundamental solution for t0 = 0 and P (t0) = I.

• Calculate the analytical solution.

• Determine the monodromy matrix M and the Floquet mulipliers. What is the condition forasymptotic stability?

• Determine the Floquet and Lyapunov exponents.

• Solve the dynamics (2) numerically for a = −0.05, b = 0.5 and compare the discrete-timesolution obtained using the monodromy matrix M .

Solution:

• The differential equations for the coefficients of the fundamental matrix with P (t0) = I andt0 = 0 are given by

p11(t) = ap11(t), p11(0) = 1

p12(t) = ap12(t), p12(0) = 0

p21(t) = (a− b cos(t))p21(t), p21(0) = 0

p22(t) = (a− b cos(t))p22(t), p22(0) = 1.

Accordingly it holds that

p11(t) = exp (at)

p12(t) = 0

p21(t) = 0

p22(t) = exp (at− b sin(t))

Summarizing, it holds that

P (t) =

[eat 00 e(at−b sin(t))

]• With the fundamental solution P (t) and t0 = 0, P (t0) = I the analytic solution reads

x(t) = P (t)x0

2




• Given the period T = 2π it follows that

M =

[e2aπ 0

0 e2aπ

]with the Floquet multipliers (the eigenvalues of M) µ1 = e2aπ, µ2 = e2aπ. Accordingly, theasymptotic stability of the origin is ensured for a < 0.

• Given the structure of the monodromy matrix its logarithm exists, so that one can computethe matrix

B =1

Tlog(M) =

[a 00 a

].

Accordingly, the Floquet exponents are given by

σ1,2 = a.

As the Floquet exponents are real they correspond to the Lyapunov exponents.

• With the choice P (t0) = I above it follows that

x(t0 + kT ) = Mx(t0 + (k − 1)T ). (3)

A comparison of numerical solutions of the continuous-time system (2) and (3) are shown inFigure 2. The simulation verifies the exact discrete-time approximation by the monodromymatrix.

0 20 40 60 80 100t

0.0

0.2

0.4

0.6

0.8

1.0

x 1

0 20 40 60 80 100t

−1.25−1.00−0.75−0.50−0.250.00

x 2

Figure 2: Comparison of the numerical solutions of the continuous-time model (2) and the discretizedversion (3) based on the monodromy matrix.

3




1. Consider the nonlinear dynamics

x1 = − sin(x1) + x2x2 = −x31 + sin(x2).

Determine the possible equilibrium points (at least approximately) and analyze their local stabilityproperties. Verify your results numerically.

2. Consider the dynamics of the undriven pendulum with friction

(J +ml2s)d2θ

dτ2+ c

dθ

dτ−mgls sin(θ) = 0

with the moment of inertia J , the mass m, the dis-tance from the rotation point to the center of massls, the friction constant c and g the gravitational ac-celeration.Show that with appropriate nondimensionalization the pendulum dynamics can be written in thecompact form

θ + δθ − sin(θ) = 0, δ > 0.

where x = dx/dt with t = κτ being an appropriate dimensionless time with scaling factor κ andδ a constant. Write the dynamics in state space form, and determine the equilibrium points andtheir local stability properties. Verify your results numerically.

3. Consider the oscillator dynamics with nonlinear damping

x− (1− x2)x+ x = 0.

Write the dynamics in state space form. What can you say about the local stability of the origin?

Simulate the dynamics close to the origin (i.e., with initial values x(0) = x0 satisfying ||x0|| � 1,say ||x0|| ≤ 0.1) to verify your analytic result.

What happens when you consider initial values with ||x0|| ≥ 1 (say 0 � ||x0|| ≤ 3)? Explain theconcept of locality using your observations.

1

Applied Nonlinear Dynamics (SS 2020) - Solution to exercise 6



1. Consider the dynamics

x1 = − sin(x1) + x2

x2 = −x31 + sin(x2)(1)

Determine the possible equilibrium points (at least approximately) and analyze their local stabilityproperties. Verify your results numerically. Solution:The equilibrium points of the system are obtained by solving

x1 ≡ 0 = − sin(x1) + x2

x2 ≡ 0 = −x31 + sin(x2).(2)

The first equilibrium point is given by x?1 = [0 0]T . The other two equilibrium points are obtainedwhen x31 = sin(sin(x1)) holds true (due to x2 = sin(x1)). The functions x3 and sin(sin(x)) havetheir intersections approximately at 0, 0.9 and −0.9 which is illustrated in Figure 1.

x

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

x3

sin(sin(x))

X: 0.8879

Y: 0.7002

X: -0.8879

Y: -0.7002

Figure 1: Intersections of the functions x3 and sin(sin(x)).

The equilibrium points x? of system (2) are therefore given by

x?1 = [0 0], x?2 = [0.9 0.8], x?3 = [−0.9 − 0.8].

The linearization of system (2) is given by

Jn =∂f(x)

∂x

∣∣∣∣x=x?

n

=

[− cos(x1) 1−3x21 cos(x2)

] ∣∣∣∣x=x?

n

and reads evaluated at the equilibirum points

J1 =

[−1 10 1

], J2 =

[−0.6216 1−2.43 0.6967

], J3 =

[−0.6216 1−2.43 0.6967

].

1

with the eigenvalues

s1,1 = 1, s1,2 = −1

s2,1 = 0.038 + 1.413i, s2,2 = 0.038− 1.413i

s3,1 = 0.038 + 1.413i, s3,2 = 0.038− 1.413i

as the solution of the characteristic polynomial det(sI−Jn) = 0. The first equilibrium point in theorigin is a settle, because of one eigenvalue with negative and one eigenvalue with positive real part.The other two equilibrium points are unstable spirals due to the complex conjugated eigenvalueswith positive real parts. Figure 2 shows the vector field of system (2) and the trajectories of initialconditions starting close to each equilibrium point.

-2 -1 0 1

-2

-1

0

1

2

x2

x1

Figure 2: Vector field of system (2).

2. Consider the dynamics of the undriven pendulum with friction

(J +ml2s)d2θ

dτ2+ c

dθ

dτ−mgls sin(θ) = 0 (3)

with the moment of inertia J , the mass m, the dis-tance from the rotation point to the center of massls, the friction constant c and g the gravitational ac-celeration.Show that with appropriate nondimensionalization the pendulum dynamics can be written in thecompact form

θ + δθ − sin(θ) = 0, δ > 0. (4)

where x = dx/dt with t = κτ being an appropriate dimensionless time with scaling factor κ andδ a constant. Write the dynamics in state space form, and determine the equilibrium points andtheir local stability properties. Verify your results numerically.

Remark: When the stability property of a physical system is analyzed it can often be helpfull tonondimensional normalize the system equations with respect to certain parameter or states, suchthat a simpler form of the system is achieved. This is possible because the stability property holdsfor t→∞ and therefore we can scale the time integration without having an impact on the stabilityproperty.

Solution:

(J +ml2s)d2θ

d t2+ c

dθ

d t−mgls sin(θ) = 0

∣∣ : mgls

(J +ml2s)

mgls

d2θ

d k2τ2+

c

mgls

dθ

d kτ− sin(θ) = 0

∣∣∣∣dxdt =dx

dkτ

choosing k =

√(J +ml2s)

mgls

leads to θ +c

mgls

√mgls

(J +ml2s)︸︷︷︸δ

θ − sin(θ) = 0

and δ =c

mgls

√mgls

(J +ml2s).

To transform the dynamics in state space form we rewrite the second order differential equationwith respect to second order time derivative

θ = −δθ + sin(θ)

and introduce the state vector

x =

[x1x2

]=

[θ

θ

].

In state space x = f(x) the system then reads

x =

[x1x2

]=

[x2

−δx2 + sin(x1)

]. (5)

The system (5) has infinite equilibrium points which are given by x? = [kπ 0]T with k ∈ R.The Jacobian is given by

J =∂f(x)

∂x

∣∣∣∣x=x?

=

[0 1

cos(kx1) −δ

].

Since cos(kπ) has alternating sign the Jacobians for and even k read

Jodd =

[0 1−1 −δ

], Jeven =

[0 11 −δ

].

The eigenvalues are given by the solution of

det(sI − Jodd) = s2 + δs+ 1 = 0

det(sI − Jeven) = s2 + δs− 1 = 0

and read

sodd = −δ2±√δ2

4− 1

seven = −δ2±√δ2

4+ 1.

The equilibrium points for even k are attractors because they always have negative real parts since

δ

2>

√δ2

4− 1

holds true for δ > 0. The equilibrium points for odd k are saddles because they always have anegative and a postive real part, i.e for δ > 0

δ

2<

√δ2

4+ 1 (6)

holds true. The solutions together with the vector field are shown in Figure 3. The infinite numberof possible equilibrium points and the local validity of the stability assessment on the basis of theHartman-Grobman theorem can be appreciated.

Figure 3: State space with solution trajectories and vector field for 4.

3. Consider the oscillator dynamics with nonlinear damping

x− (1− x2)x+ x = 0. (7)

Write the dynamics in state space form. What can you say about the local stability of the origin?

Simulate the dynamics close to the origin (i.e., with initial values x(0) = x0 satisfying ||x0|| � 1,say ||x0|| ≤ 0.1) to verify your analytic result.

What happens when you consider initial values with ||x0|| ≥ 1 (say 0 � ||x0|| ≤ 3)? Explain theconcept of locality using your observations.

Solution:To transform the dynamics in state space form we rewrite the second order differential equationwith respect to second order time derivative

x = (1− x2)x− x

and introduce the state vector

x =

[x1x2

]=

[xx

].

In state space x = f(x) the system then reads

x =

[x1x2

]=

[x2

(1− x21)x2 − x1

]. (8)

and the Jacobian at x? = [0 0]T is given by

J =∂f(x)

∂x

∣∣∣∣x=x?

=

[0 1

2x1x2 − 1 1− x21

] ∣∣∣∣x=x?

=

[0 1−1 1

].

The eigenvalues read

s2 − s+ 1 = 0 ⇒ s =1

2±√

1

4− 1 =

1

2± j√

3

4.

and reveal that the origin is an unstable spiral because of the conjugated complex eigenvalues withpositive real parts. The solutions together with the vector field are shown in Figure 4. It can beeclearly seen that the preceding analysis is only valid in a small neighborhood of the origin.

-5 -2.5 0 2.5 5

-5

-2.5

0

2.5

5

x2

x1

Figure 4: State space with solution trajectories and vector field for 7.




1. Analyze the following systems near the origin using center manifold theory [1, 2, 3].

(a)x1 = −x32x2 = −x2 + x21

(b)x1 = −x2 + x1x3x2 = x1 + x2x3x3 = −x3 − (x21 + x22) + x23

Approximate the center manifold and establish if the origin is a local attractor. Compare theapproximate center manifold close to the origin with the one resulting from numerical integrationof the complete nonlinear dynamics using some numerical integration package.

2. Analyze the following systems near the origin using center manifold theory [1, 2, 3].

x1 = ax21 + bx1x2 + cx22x2 = −x2 + dx21 + ex1x2 + fx22.

For which parameter values is the origin a local attractor? Verify your results with numericalsimulations.

3. For the following control system

x1 = −x31 + αx22x2 = x31 + u

with proportional (P) control u = −kx2, determine the stability conditions in terms of the pair(α, k) using Center Manifold Theory. Determine the domain of attraction of the origin correspond-ing to the approximate dynamics on the center manifold for different α. Verify your result usingnumerical integration. Does the domain of attraction coincide with its prediction?

References

[1] S. Sastry. Nonlinear Systems: Analysis, Stability, and Control. Springer, New-York, 1999.

[2] L. Perko. Differential Equations and Dynamical Systems. Springer, New York, 3rd ed. edition, 2001.

[3] A. Schaum. Applied Nonlinear Dynamics. Lecture Notes, 2020.

1




1. Analyze the following system fa(x) near the origin using center manifold theory

x1 = −x32x2 = −x2 + x21.

(1)

Approximate the center manifold and establish if the origin is a local attractor. Compare theapproximate center manifold close to the origin with the one resulting from numerical integrationof the complete nonlinear dynamics using some numerical integration package.

Solution: Frist linearize the System fa(x) around the origin

∂fa(x)

∂x

∣∣∣∣x=[0 0]T

=

[−3x22 0

2x1 −1

] ∣∣∣∣x=[0 0]T

=

[0 00 −1

]. (2)

The eigenvalues of the linearized system are in general given by the solution of the characteristicequation

det(sI −A) = 0 (3)

where A is the linearized system dynamic matrix and I the unity matrix.

det(sI −A) = s(s+ 1) = 0 (4)

leads to s1 = 0 and s2 = −1. Therefore the system fa(x) can be split into one stable component,i.e. x2 with eigenvalue −1 and one center component (zentrum in german) x1 with eigenvalue zero

xs = −xs + x2z

xz = −x3s.(5)

According to the Center-manifold theorem the x2 component locally converges onto a manifold onwhich it holds that

xs = h(xz). (6)

We are now looking for an appropriate polynomial approximation of the center manifold whichsatisfies

h(0) = 0,∂h(0)

∂xz= 0 (7)

xs∣∣xs=h(xz)

=∂h(xz)

∂xzxz. (8)

Since the dimension of the center component for the given system is one, we already know thata polynomial of at least one variable is required. To approximate the center manifold we locallyassume

h(xz) = c0 + c1xz + c2x2z, (9)

with c0, c1, c2 being constant values. Taking condition (7) into account it gets directly clear that

c0 = c1 = 0 because otherwise h(0) = 0 and ∂h(0)∂xz

= 0 would not hold. Condition (8) for theSystem (5) and the polynomial defined in (9) reads

−h(xz) + x2z!≈ 2c2xz(−h(xz)

3) (10)

−c2x2z + x2z!≈ −2c22x

7z. (11)

1

-1 -0.5 0 0.5 1

xz

-0.5

-0.4

-0.3

-0.2

-0.1

0

0.1

xz

Figure 1: xz vs. xz plot for the dynamics in equation (12).

Figure 2: Vector field and numerical solutions for (1).

Neglecting the higher order term by −2c22x7z ≈ 0 an approximation of the center manifold for c2 = 1

up to the order of 7 is achieved. The dynamics of the center component are finally given by

xz = −x3s = −h(x)3 = −x6z. (12)

and reveal that the origin is not locally stable. The associated vector field with some particularsolutions is shown in Figure 2.

2. Analyze the following system fb(x) near the origin using center manifold theory

x1 = −x2 + x1x3x2 = x1 + x2x3x3 = −x3 − (x21 + x22) + x23

(13)

Approximate the center manifold and establish if (and for which parameters) the origin is a localattractor. Compare the approximate center manifold close to the origin with the one resultingfrom numerical integration of the complete nonlinear dynamics using some numerical integrationpackage.

Solution: The Jacobian evaluated at the origin is given by

J [0] =

0 −1 01 0 00 0 −1

(14)

with a real Jordan block for the (x1, x2)-component indicating a center and the asymptotically sta-

ble component x3. Thus, setting xs = x3 and xz =[x1 x2

]Tone yields the system representation

xs = −xs + x2s − (x21 + x22)x1 = −x2 + x1xsx2 = x1 + x2xs.

(15)

According to the Center-manifold theorem the x3 component locally converges onto a manifold onwhich it holds that

xs = h(xz). (16)

Combining (16) with (15) it turns out that locally it must hold that

xs = −h(xz) + h(xz)2 − (x21 + x22) =

∂h

∂xz.(xz)

[−x2 + x1h(xz)x1 + x2h(xz)

](17)

To approximate the center manifold locally set

h(xz) ≈ ax21 + bx1x2 + cx22. (18)

With this approximation (17) evaluates to

− (ax21 + bx1x2 + cx22)− (ax21 + bx1x2 + cx22)2 − (x21 + x22)

!≈[2ax1 + bx2 bx1 + 2cx2

] [−x2 + x1(ax21 + bx1x2 + cx22)x1 + x2(ax21 + bx1x2 + cx22)

].

This can be written as

x21(−a− b− 1) + x1x2(2a− b− 2c) + x22(b− c− 1) = O4(xz)

and thus a fourth order approximation (i.e. locally so that xz is such that O4(xz) ≈ 0) is obtainedif a, b, c are chosen such that the coefficients in the polynomial on the left hand side vanish. Thisleads to a system of linear equations for (a, b, c) which can be written in matrix form as−1 −1 0

2 −1 −20 1 −1

abc

=

101

⇔

abc

=

−10−1

implying that

h(xz) ≈ −(x21 + x22) (19)

in a neighborhood of the origin small enough to ensure that O4(xz) ≈ 0. The dynamics on thisapproximated center manifold are obtained from substituting (19) into the (x1, x2) component of(15), i.e.

x1 = −x2 − x1(x21 + x22)x2 = x1 − x2(x21 + x22).

(20)

Motivated by the real Jordan block in the Jacobian matrix (14) and the fact that h(xz) ≈ −r2where r =

√x21 + x22 is the distance of a vector in the (x1, x2) plan to the origin, which corresponds

to the radius of a cyclic motion, write the system in polar coordinates (r, θ). For this purpose notethat

rr = x1x1 + x2x2 = −(x21 + x22)(x21 + x22) = r4

r2θ = x1x2 − x1x2 = x21 + x22 = r2

implying that

r = −r3, r(0) =√x210 + x220

θ = 1, θ(0) = arctan

(x20x10

)(21)

This shows that actually the origin of the dynamics on h(xz) is a (locally asymptotically) stablespiral. In comparison to the linearization by the Jacobian (14) it must be pointed out that the centerpredicted by the linearization does not correspond to the actual behavior of a stable spiral. This isin accordance with Hartman-Grobmann’s theorem (the real part of the corresponding eigenvaluesis zero). On the other hand, the convergence velocity is asymptotically slowing down with orderO3(r) and thus a rather slow convergence will be observed. Figure 3 shows two trajectories for

the dynamics (13) with x10 =

[.5 .5 1

]Tand x2

0 =[−.1 .1 −1

]Tin the three-dimensional

state space as well as the approximated center manifold h(xz). It can be seen that the trajectoriesquickly converge onto the center manifold and then spiral towards the origin quite slowly. Figure

−1.5

−1

−0.5

0

0.5

1

1.5

−1.5

−1

−0.5

0

0.5

1

1.5

−2

−1.5

−1

−0.5

0

0.5

1

x1 x2

x3

Figure 3: Solutions of (13) in state space for two different initial conditions and local approximation ofthe center manifold.

4 shows the associated time responses, illustrating again the quick convergence onto the centermanifold in the x3 component and the slowly converging spiralling behavior on the manifold.

0 5 10 15 20 25 30 35 40−1

−0.5

0

0.5

0 5 10 15 20 25 30 35 40−1

0

1

2

0 5 10 15 20 25 30 35 40−1

−0.5

0

0.5

1

x1

x2

x3

Time

Figure 4: Solutions of (13) in state space and local approximation of the center manifold for two differentinitial conditions.

3. Analyze the following system near the origin using center manifold theory

x1 = ax21 + bx1x2 + cx22

x2 = −x2 + dx21 + ex1x2 + fx22(22)

Solution: We proceed as before by calculating the Jacobian

J∣∣x=[0 0]T

=∂f(x)

∂x

∣∣∣∣x=[0 0]T

=

[2ax1 + bx2 bx1 + 2cx22dx1 + ex2 −1 + ex1 + 2fx2

] ∣∣∣∣x=[0 0]T

=

[0 00 −1

]. (23)

The Eigenvalues are independently of the parameter set given by s1 = 0 and s2 = −1, such thatin stable and center components xs, xz the system reads

xz = ax2z + bxzxs + cx2s

xs = −xs + dx2z + exzxs + fx2s.(24)

Again we are looking for an appropriate approximation of the center manifold by

xs = h(xz) (25)

which satisfies the conditions

h(0) = 0,∂h(0)

∂xz= 0 (26)

xs∣∣xs=h(xz)

=∂h(xz)

∂xzxz. (27)

Note that the dimension of the center component xz is given by one and the highest order of thecenter component in the derivative of the stable component is given by two. Therefore we definelocally

h(xz) = c2x2z (28)

and evaluate the conditions (26) and (27). Obviously condition (26) holds true. Condition (27)reads

−h(xz) + dx2z + exzh(xz) + fh(xz)2 !≈ 2c2xz(ax

2z + bxzh(xz) + ch(xz)

2) (29)

−c2x2z + dx2z + ec2x3z + fc22x

4z

!≈ 2ac2x

3z + 2bc22x

4z + 2c32cx

5z. (30)

Rearranging the equation and neglecting terms with an order higher than two leads to

−c2x2z + dx2z!≈ 0 (31)

and for c2 = d an appropriate approximation of the center manifold up to the order of two is found.The dynamics of the center component is then given by

xz = ax2z + bdx3z + cd2x4z. (32)

The lowest odd power of equation (32) is given by bdx3z. Therefore in order to satisfy stability ofthe origin a = 0 and bd < 0 has to hold. Since the Term cd2x4z is dominated by the third orderterm bdx3z (note that x is close to zero) the sign of cd does not change the stability properties.Furthermore the parameters f and e do not show up and can be choosen arbitrarily.

The associated vector field with some particular solutions is shown in Figure 5.

Figure 5: Vector field and numerical solutions for (22) with a = 1, b = 1, c = −1, d = 1, e = 1, f = 1.

4. For the following control system

x1 = −x31 + ax22

x2 = x31 + u(33)

with proportional (P) control u = −kx2, determine the stability conditions in terms of the pair (a, k)using the Center Manifold Theory. Determine the domain of attraction of the origin correspondingto the approximated dynamics on the center manifold for a different a. Verify your results usingnumerical integration. Does the domain of attraction coincide with its prediction?

Solution: For u = −kx2 the system dynamics read

x1 = −x31 + ax22

x2 = x31 − kx2,(34)

with the Jacobian evaluated at the origin given by

J∣∣x=[0 0]T

=

[−3x21 3ax2x31 −k

] ∣∣∣∣x=[0 0]T

=

[0 00 −k

]. (35)

It can be concluded that due to the negative linear feed back at least one component can bestabilized, namely x2. We split the system in the center and stable components

xz = −x3z + ax2s

xs = x3z − kxs.(36)

Note that the highest order of the center component in the derivative of the stable component isgiven by three. We locally assume

h(xz) = c3x3z (37)

and evaluate the conditions

h(0) = 0,∂h(0)

∂xz= 0 (38)

xs∣∣xs=h(xz)

=∂h(xz)

∂xzxz. (39)

Obviously (38) holds true for the polynomial defined in (37). Condition (39) reads

x3z − kc3x3z!≈ 3c3x

2z(−x3z + ac23x

6z) (40)

x3z − kc3x3z!≈ −3c3x

5z + 3ac33x

8z. (41)

Neglecting the terms with a higher order than three, an appropriate approximation of the centermanifold is found for c3 = 1

k . The dynamics of the center component are then given by

xz = −x3z + a(x3zk

)2 = −x3z +a

k2x6z = −x3z(1−

a

k2x3z︸︷︷︸

=β

). (42)

Stability of the origin is thus achieved as long as β remains positive. Therefore

a

k2x3z < 1 (43)

has to hold and the domain of attraction, i.e. the initial conditions xz,0 for which the origin is anattractor, is given by

xz <3

√k2

a. (44)

Figure 6 shows the associated vector field and some numerical solutions for k = 3.

Figure 6: Vector field and numerical solutions for (33) with u = −3x2.




1. Show that a saddle equilibrium x∗ ∈ R2 has index I[x∗] = −1.

2. Consider the approximated glycolisis cycle [1]

x1 = −x1 + ax2 + x21x2

x2 = b− ax2 − x21x2

with a, b > 0. Determine the equilibrium points and characterize their local stability in dependenceof a and b. Analyze the possible existence of a limit cycle using index theory. Use some numericalintegration tool to analyze the existence of a stable limit cycle.

3. (Brusselator) Consider the autocatalytic chemical reaction scheme (see e.g. [2])

A→ XB +X → Y + C2X + Y → 3XX → D

with mass-action kinetics-based dynamics in dimensionless form

x1 = a− (b+ 1)x1 + x21x2, x1(0) = x10

x2 = bx1 − x21x2, x2(0) = x20

with a, b > 0. Find the equilibrium point x∗ of these dynamics and determine its local stability.What is the index of the equilibrium point? Characterize and scetch in the parameter space (a, b)the local behavior (spiral or node) close to x∗. Verify your results numerically. What can youconclude about the possible existence of limit cycles using index theory?

References

[1] E. E. Selkov. Self-oscillations in glycolisis. European J. Biochem., 4:79–86, 1968.

[2] R. J. Field. Das Experiment: Eine oszillierende Reaktion. Chemie in unserer Zeit, 7 (6):171–176,1973.

1




1. Show that a saddle equilibrium x∗ ∈ R2 has index I[x∗] = −1.

Solution:This can be shown using the vector mapping technique [1] illustrated in the lecture notes [2].

2. Consider the approximated glycolisis cycle [3]

x1 = −x1 + ax2 + x21x2

x2 = b− ax2 − x21x2

with a, b > 0. Determine the equilibrium points and characterize their local stability in dependenceof a and b. Analyze the possible existence of a limit cycle using index theory. Use some numericalintegration tool to analyze the existence of a stable limit cycle.

Solution:The associated null-clines ni(x1), i = 1, 2 (i.e., the curves in state space along which xi = 0 holdstrue) are given by

x2 = n1(x1) =x1

a+ x21

x2 = n2(x1) =b

a+ x21

.

The intersection corresponds to the equilibrium point and is given by

x∗ =

[

x∗

1

x∗

2

]

, x∗

1 = b, x∗

2 = n1(x1) = n2(x1) =b

a+ b2.

The Jacobian at this point is given by

J(x∗) =

[

∂f1∂x1

∂f1∂x2

∂f2∂x1

∂f2∂x2

]

x=x∗

=

[

−1 + 2x1x2 a+ x21

−2x1x2 −(a+ x21)

]

x=x∗

=

[

−1 + 2 b2

a+b2a+ b2

−2 b2

a+b2−(a+ b2)

]

with characteristic equation

0 =

(

λ+ 1− 2b2

a+ b2

)

(λ+ (a+ b2)) + 2b2

= λ2 + λ

(

1− 2b2

a+ b2+ a+ b2

)

+ a+ b2

= λ2 + λ1

a+ b2

(

a+ b2 − 2b2 + (a+ b2)2)

+ a+ b2

= λ2 + λ1

a+ b2

(

a− b2 + (a+ b2)2)

+ a+ b2

with solutions

λ1,2 = −a− b2 + (a+ b2)2

2(a+ b2)±

√

(a− b2 + (a+ b2)2)2

4(a+ b2)2− (a+ b2).

Accordingly, as a, b > 0 the eigenvalues will always have the same sign 1 and thus correspond eitherto a locally asymptotically stable or an unstable equilibrium point. The index is

I(x∗) = 1

1This holds true because

√

(a−b2+(a+b2)2)2

4(a+b2)2− (a + b2) <

√

(a−b2+(a+b2)2)2

4(a+b2)2=

a−b2+(a+b

2)2

2(a+b2).

1




in both cases, so that according to index theory a limit cycle surrounding x∗ is possible. For

(a− b2 + (a+ b2)2)2

4(a+ b2)2− (a+ b2) > 0

both eigenvalues are real and if

a− b2 + (a+ b2)2 > 0

the equilibrium point x∗ is locally asymptotically stable and in case that this sign is inverted it isunstable. On the other hand, if

(a− b2 + (a+ b2)2)2

4(a+ b2)2− (a+ b2) < 0

x∗ corresponds to a (stable or unstable) spiral. Numerical solutions using plotdf in Wxmaxima

are shown in Figure 1

0 0.5 1 1.5

0

0.25

0.5

0.75

1

x2

x1

0 0.5 1 1.5

0

0.4

0.8

1.2

1.6

2

2.4

2.8

x2

x1

Figure 1: Numerical solution for the glycolisis cycle with a = 1, b = 1 yielding a stable spiral (left), andfor a = 0.08, b = 0.6 yielding an asymptotically stable limit cycle (right)

3. (Brusselator) Consider the autocatalytic chemical reaction scheme (see e.g. [4])

A → X

B +X → Y + C

2X + Y → 3XX → D

with mass-action kinetics-based dynamics in dimensionless form

x1 = a− (b+ 1)x1 + x21x2, x1(0) = x10

x2 = bx1 − x21x2, x2(0) = x20

with a, b > 0. Find the equilibrium point x∗ of these dynamics and determine its local stability.

What is the index of the equilibrium point? Characterize and scetch in the parameter space (a, b)the local behavior (spiral or node) close to x

∗. Verify your results numerically. What can youconclude about the possible existence of limit cycles using index theory?

Solution:The associated null-clines ni(x1), i = 1, 2 are given by

x2 = n1(x1) =(b+ 1)x1 − a

x21

x2 = n2(x1) =b

x1

.

2




The equilibrium point is thus given by

x∗ =

[

x∗

1

x∗

2

]

, x∗

1 = a, x∗

2 =b

a.

The Jacobian at this point is given by

J(x∗) =

[

−(b+ 1) + 2x1x2 x21

b − 2x1x2 −x21

]

x=x∗

=

[

b− 1 a2

−b −a2

]

with trace and determinant given by

tr(J(x∗)) = b− 1− a2, det(J(x∗)) = a2 > 0.

Given that the determinant is positive, x∗ is either an asymptotically stable or an unstable equi-librium but will nether correspond to a saddles. Thus the index is

I(x∗) = 1

and according to index theory a limit cylce is possible. Numerical simulation results are illustratedin Figure 2 showing that for specific parameters a limit cycle exists.

0 1 2 3 4 5

0

2.5

5

7.5

x2

x1

0 1 2 3 4 5

0

2.5

5

7.5

x2

x1

Figure 2: Numerical solution for the Brusselator with a = 2, b = 4 yieding a stable spiral (left), and fora = 0.6, b = 2 yielding an asymptotically stable limit cycle (right)

References

[1] S. H. Strogatz. Nonlinear Dynamics and Chaos, with applications to physics, biology, chemistry, and

engineering. Perseus, Massachusetts,, 1994.


[3] E. E. Selkov. Self-oscillations in glycolisis. European J. Biochem., 4:79–86, 1968.

[4] R. J. Field. Das Experiment: Eine oszillierende Reaktion. Chemie in unserer Zeit, 7 (6):171–176,1973.

3





x1 = x1 − x2 − x1(x21 + x22), x1(0) = x10

x2 = x1 + x2 − x2(x21 + x22), x2(0) = x20

and show the following:

• The set C1+δ = {x ∈ R2 | ||x|| ≤ 1 + δ} is positively invariant for any δ > 0.

• The unique SS solution in C1+δ is the origin.

• There exists an attractive limit cycle surrounding the origin.

Hint: Use polar coordinates and employ the Poincare-Bendixson theorem.

2. [1] Show that the following dynamics have no closed orbits in the first quadrant

x1 = x1(2− x1 − x2)x2 = x2(4x1 − x21 − 3)

and verify your result using numerical integration.

3. Using the Bendixson-Dulac criterion, show that the dynamics

x1 = −x2 + µx1(x21 + x22)

x2 = x1 + µx2(x21 + x22)

can not have periodic (closed) trajectories for µ 6= 0.

4. (Active (anti-)damping control) Consider the controlled oscillator

x+ µux+ x = 0

x(0) = x0, x(0) = v0. Design an active anti-damping control of the form u = ϕ(x) so that theclosed-loop system has a unique attractive limit cycle. Can you design the controller so that theclosed-loop system behaves like a van der Pol oscillator?

References


1





x1 = x1 − x2 − x1(x21 + x22), x1(0) = x10

x2 = x1 + x2 − x2(x21 + x22), x2(0) = x20

and show the following:

• The set C1+δ = {x ∈ R2 | ||x|| ≤ 1 + δ} is positively invariant for any δ > 0.

• The unique SS solution in C1+δ is the origin.

• There exists an attractive limit cycle surrounding the origin.

Hint: Use polar coordinates and employ the Poincare-Bendixson theorem.

Solution:Transforming the system into polar coordinates (r, φ) yields

rr = x1x1 + x2x2

= x1[x1 − x2 − x1(x21 + x22)

]+ x2

[x1 + x2 − x2(x21 + x22)

]= x21 − x1x2 − x21r2 + x1x2 + x22 − x22r2

= r2(1− r2)

r 6=0⇐⇒ r = r(1− r2)

r2φ = x1x2 − x2x1= x1

[x1 + x2 − x2(x21 + x22)

]− x2

[x1 − x2 − x1(x21 + x22)

]= x21 + x1x2 − x1x2r2 − (x1x2 − x22 − x1x2r2)

= x21 + x22

= r2

r 6=0⇐⇒ φ = 1.

Note that the same dynamics is obtained considering the limit for r → 0 using the rule of Hopital-Bernoulli. Accordingly for all r > 1 it holds that for all r < 0, so that with any δ > 0 theset

C1+δ = {x ∈ R2 | ||x|| ≤ 1 + δ}

is positively invariant.

From the dynamics for (r, φ) it follows that the unique SS solution in C1+δ is the origin (r = 0 andφ ∈ R).

According to the dynamics of r the origin is unstable, so that by the Poincare-Bendixson theoremit follows that there exists an attractive limit cycle surrounding the origin1.

2. [1] Show that the following dynamics have no closed orbits in the first quadrant

x1 = x1(2− x1 − x2)x2 = x2(4x1 − x21 − 3)

1Note that the same result directly can be obtained from the dynamics in polar coordinates, given that r = 1 is theunique attractive solution for the radius r in the set C1+δ and φ = 1 holds for any r ∈ R+.

1

and verify your result using numerical integration.

Solution:Using the scalar differentiable function

ϕ(x) =1

x1x2

and calculating the divergence of ϕ(x)f(x) yields

∇ · ϕ(x)f(x) =∂

∂x1

(2− x1 − x2

x2

)+

∂

∂x2

(4x1 − x21 − 3

x1

)= − 1

x2< 0 ∀x2 ∈ [0,∞).

It follows that ∇·ϕ(x)f(x) does not change its sign in the first quadrant and thus, according to thecriterion of Bendixson-Dulac there are no closed trajectories lying entirely in the first quadrant.

3. Using the Bendixson-Dulac criterion, show that the dynamics

x1 = −x2 + µx1(x21 + x22)

x2 = x1 + µx2(x21 + x22)

can not have periodic (closed) trajectories for µ 6= 0.

Solution:The divergence of the vector field f(x) is given by

∇ · f(x) =∂

∂x1f1(x) +

∂

∂x2fx(x)

= 4µ(x21 + x22) = 4µr2 6= 0, ∀ r 6= 0,

with the radius r = x21 + x22 (in polar coordinates). Accordingly, for any µ 6= 0 the divergence∇ · f(x) does not change sign for any x ∈ R2 (i.e., any radius r ∈ R+) and thus the Bendixson-Dulac criterion implies that no closed trajectories exist.

4. (Active (anti-)damping control) Consider the controlled oscillator

x+ µux+ x = 0

x(0) = x0, x(0) = v0. Design an active anti-damping control of the form u = ϕ(x) so that theclosed-loop system has a unique attractive limit cycle. Can you design the controller so that theclosed-loop system behaves like a van der Pol oscillator?

Solution:With u = ϕ(x) the dynamics correspond to the class of Lienard systems, so that for ϕ(x) chosensuch that the conditions of Theorem 4.2.2 are satisfied the existence of a unique limit cycle directlyfollows. For the identification with the van der Pol oscillator a direct comparison shows that forϕ(x) = x2 − 1 an identical solution behavior is obtained.

References





1. Consider the simplified model of a laser (cp. [1, 2])

n = GnN − kn

where n is the number of photons in the lase field, N is the number of excited atoms, G is apositive gain coefficient, and k is the loss of photons due to escape from the laser field. Consideringthe number of excited atoms is given by N(t) = N0 − αn where N0 is the number of atoms thepump keeps excited in absence of laser action, and α the rate with which atoms drop back to theirground states, emitting a photon. Find an autonomous differential equation for n and analyze thebifurcation behavior of the laser model.

2. Consider the fishing model (cp. [2])

N = rN

(1 − N

K

)−H

N

A+N

where N is the population number, r the logistic growth coefficient, K the carrier capacity, H thefishing rate, and A a saturation constant. Classify the equilibrium points and bifurcations.

3. Consider the dimensionless model of the chemostat with continuous substrate supply and conversion

b = −db+ µ(s)b, b(0) = b0s = d(1 − s) − γµ(s)b, s(0) = s0

where b, s, d, µ and γ are the biomass and substrate concentrations, the dilution rate, the biomassgrowth rate and yield, respectively. Consider the Haldane growth rate

µ(s) =k0s

s2/ki + s+ ks

with maximum rate constant k0, half saturation constant ks, and inhibition constant ki. Analyzethe bifurcation behavior of the bioreactor model in dependence of the dilution rate d.

References

[1] H. Haken. Synergetics. Springer, Berlin, 3rd. edition, 1983.


1

Applied Nonlinear Dynamics (SS 2020) - Solution to selected problems from exercise 10



• Consider the simplified model of a laser (cp. [1, 2])

n = GnN − kn

where n is the number of photons in the lase field, N is the number of excited atoms, G is apositive gain coefficient, and k is the loss of photons due to escape from the laser field. Consideringthe number of excited atoms is given by N(t) = N0 − αn where N0 is the number of atoms thepump keeps excited in absence of laser action, and α the rate with which atoms drop back to theirground states, emitting a photon. Find an autonomous differential equation for n and analyze thebifurcation behavior of the laser model.

Solution:Gathering the given information and rewriting the system in a compact manner yields

n = Gn(N0 − αn)− kn = (G− k)n

= (GN0 − k)n− αGn2

= (GN0 − k − αGn)n.

Introducing the scaled time variable

τ = αGt ⇔ dt

dτ=

1

αG

yields that in the τ time scale the dynamics are given by

dn

dτ=dn

dt

dt

dτ=

1

αG(GN0 − k − αGn)n =

(GN0 − kαG

− n)n.

With the parameter

p =GN0 − kαG

these dynamics are written as

dn

dτ= n(p− n) = pn− n2

which corresponds to the normal form of the transcritical bifurcation. Accordingly, a transcriticalbifurcation takes place at p = 0, i.e. for G∗ = k

N0and for G > G∗ the positive asymptotically

stable solution n = p = GN0−kαG exists.

• Consider the fishing model (cp. [2])

N = rN

(1− N

K

)−H N

A+N

where N is the population number, r the logistic growth coefficient, K the carrier capacity, H thefishing rate, and A a saturation constant. Classify the equilibrium points and bifurcations.

Solution:Considering the dynamics

N = rN

(1− N

K

)−H N

A+N

1




the equilibrium points are given by the intersections of the two curves

η1(N) = rN

(1− N

K

), η2(N) = H

N

A+N.

Analyzing the two curves shows that over the real numbers N ∈ R there are either one or threeintersections, with only two being physically meaningful (N ≥ 0). The condition for the switchbetween one and three intersections is that the tangents of both curves are equal at N = 0, i.e.

η′1(0) = r = η′2(0) =H

A.

Accordingly, a bifurcation takes place at the value H∗ for the fishing rate given by

H = H∗ = rA.

Given that one can write the original dynamics in terms of the curves η1(N) and η2(N) as

N = η1(N)− η2(N)

shows that for all N so that η1(N) > η2(N) the number of fish N increase and for η1(N) < η2(N)it decreases. Thus the origin is an attractor for H > H∗ and a repulsor for H < H∗, while theother steady state is an attractor for H < H∗ and does not exist for H > H∗. This bifurcationbehavior corresponds to a supercritical pitchfork bifurcation with respect to −H, or with respectto r.

• Consider the dimensionless model of the chemostat with continuous substrate supply and conversion

b = −db+ µ(s)b, b(0) = b0s = d(1− s)− γµ(s)b, s(0) = s0

where b, s, d, µ and γ are the biomass and substrate concentrations, the dilution rate, the biomassgrowth rate and yield, respectively. Consider the Haldane growth rate

µ(s) =k0s

s2/ki + s+ ks

with maximum rate constant k0, half saturation constant ks, and inhibition constant ki. Analyzethe bifurcation behavior of the bioreactor model in dependence of the dilution rate d.

Solution:Considering the dimensionless model of the chemostat

b = −db+ µ(s)b, b(0) = b0s = d(1− s)− γµ(s)b, s(0) = s0

with Haldane growth rate

µ(s) =k0s

s2/ki + s+ ks

with maximum µ∗ at s =√kski the equilibria satisfy

b = 0, s = 1 or d = µ(s), b =1− sγ

.

Taking into account the non-monotonic behavior of µ(s) shows that there are up to three equilib-rium points and two bifurcation values d∗ = µ(1) and d∗ = µ∗ with

1. One asymptotically stable equilibrium at b = bh (high biomass) and sh = 1−γbh for d < d∗ =µ(1)

2. Transcritical bifurcation at d = d∗ with two equilibria at b = bh and b = 0 (washout)

3. Three equilibria (two locally asymptotically stable solutions and one respulsive intermediatesolution) for d∗ < d < d∗ with high-biomass b = bh (attractive), intermediate biomass b = bi(respulsive) and no biomass (attractive washout) b = 0 and s given by s = 1−γb, respectively

4. Saddle-node bifurcation at d = d∗ = µ∗ with coincidence of bh and bi

5. One asymptotically stable at b = 0 (washout) for d > d∗.

2




References

[1] H. Haken. Synergetics. Springer, Berlin, 3rd. edition, 1983.


3




Poincare maps are defined by the successive intersections of the trajectories with a lower dimensionalsurface as illustrated in Figure 1. In two dimensional systems this can be e.g. the intersection withthe positive x1-axis. This naturally leads to a description of the continuous-time dynamics in form of adiscrete-time system.

xx

>

>

<

xx x x1

x2

Figure 1: Illustration of the idea of a Poincare map.

For periodic solutions like limit cycles considering these intersections after each period leads to adescription that is locally equivalent to the one using the monodromy matrix introduced in the chapterabout Floquet theory.

In this exercise the approach based on Poincare maps is practiced and put into perspective of theFloquet theory.

1. (Cp. Example 8.7.1 in [1]) Consider the following dynamics given in polar coordinates

r =r(1 − r2), r(0) = r0

θ =1, θ(0) = θ0.

Determine the discrete-time system provided by the Poincare map defined by the successive inter-sections of the trajectories with the positive x-axis. Show, by analyzing the discrete-time dynamicsthe existence of a unique limit cycle. Determine the Floquet multipliers and check the stability ofthe limit cycle.

2. (Cp. Example 8.7.2 in [1]) Consider the RC-circuit depicted in Figure 2

Figure 2: RC circuit with input u(t).

a) Show by scaling that the dynamics can be written in the formdv

dτ+ v(τ) = u(τ).

1




b) Determine the Poincare map and show that a unique limit cycle exists for u(τ) = A sin(ωτ).

c) Analyze the local stability by calculating the Floquet multipliers.

Hint for part c: Perform a change of variables so that in the new coordinates the origin is at thefixed point of the Poincare map.

References


2




1. Having as starting point the dynamics in polar coordinates

r =r(1− r2), r(0) = r0

θ =1, θ(0) = θ0

the Poincare map with the positive x1 (half-) axis can be obtained directly by noting that thesolution for the phase is given by θ(t) = θ0 + t so that a period is equal to T = 2π. Starting at thepositive x1-axis corresponds to choosing θ0 = 0. The change in r after one period can be calculatedby evaluating the following integral equation

T =

∫ T

0

dt =

∫ r(T )

r0

dr

r(1− r2)=

∫ r(T )

r0

dr

r+

∫ r(T )

r0

rdr

1− r2.

The first integral evaluates to ∫ r(T )

r0

dr

r= ln

(r(T )

r0

).

The second integral can be evaluated as follows:∫ r(T )

r0

rdr

1− r2ρ=r2

=1

2

∫ ρ(T )

ρ0

dρ

1− ρz=1−ρ

= −1

2

∫ z(T )

z0

dz

z

= −1

2ln (z(T )) +

1

2ln (z0) = ln (

√z0)− ln

(√z(T )

)= ln

( √z0√z(T )

)= ln

( √1− r20√

1− r2(T )

)Summarizing

T = ln

(r(T )

r0

)+ ln

( √1− r20√

1− r2(T )

).

Resolving for r(T ) yields

r(T ) =

√e2T r20

1− r20 (1− e2T )= P (r0).

Using the notation

r(tn) = r(nT )

and recalling that T = 2π it follows that the Poincare map is given by

r(tn+1) = P (r(tn)) =

√e4πr(tn)2

1− r(tn)2 (1− e4π).

The associated fixed-points are given by r = 0 and r = 1 and the respective slope of P (r) at r = 1is given by

0 < P ′(1) = e−4π � 1

showing that r → 1 exponentially fast with the associated Floquet multiplier µ being given by

µ = P ′(1).

1




2. (a) Using Ohm’s and Kirchhoff’s laws yields u = v + Ric what together with ic = C dvdt = Cv

directly implies that the voltage v at the capacitance obeys the differential equation

RCv + v = u, v(0) = v0.

Introducing the new time-scale

τ =t

RC,

d

dτ= RC

d

dt

one finds that RCv = dvdτ and thus

dv

dτ+ v = u, v(0) = v0.

(b) The input in the new time scale is given by u(τ) = A sin (ωRCt) = A sin(ω′τ) with ω′ = ωRC.The associated solution for v can be found by using the variation of constants formula (seee.g. [1])

v(t) = e−τv0 +

∫ τ

0

e−(τ−s)A sin(ω′s)ds = e−τ(v0 +A

∫ τ

0

es sin(ω′s)ds

)It holds that ∫ τ

0

es sin(ω′s)ds =

[es

sin(ω′s)− ω′ cos(ω′s)

1 + ω′2

]τ0

=1

1 + ω′2(eτ (sin(ω′τ)− ω′ cos(ω′τ)) + ω′) .

Accordingly, the general solution is given by

v(τ) = e−τ[v0 +

A

1 + ω′2(eτ (sin(ω′τ)− ω′ cos(ω′τ)) + ω′)

]The period (in τ -time) is given by T ′ = 2π

ω′ and it holds that sin(ω′T ′) = 0, cos(ω′T ′) = 1 sothat

v(T ′) = e−2πω′ v0 −

Aω′

1 + ω′2

(1− e− 2π

ω′)

Accordingly the Poincare map is given by

v(tn+1) = P (v(tn)) = e−2πω′ v(tn)−K, 0 < K =

Aω′

1 + ω′2

(1− e− 2π

ω′)

and the associated fixed point

v∗ =Aω′

1 + ω′2

is an attractor given that the slope P ′ = e−2πω′ < 1. The Floquet multiplier µ corresponds to

the slope of the Poincare map, i.e. µ = P ′.

References


2

Documents

Applied Nonlinear Dynamics (SS 2020) - Exercise 1: Basic