Chapter 2: Second Order Systems - Lehigh Universityeus204/teaching/ME450_NSC/lectures/lecture01.pdfBehavior of Second Order Systems Consider the following linear system (1) x˙ =Ax

Lecture 1Chapter 2: Second Order Systems

Eugenio Schuster

[email protected]

Mechanical Engineering and Mechanics

Lehigh University

Lecture 1 – p. 1/26

Behavior of Second Order Systems

Consider the following linear system

x = Ax(1)

The solution of (1) for an initial condition x0 is given by

x(t) = M exp(Jrt)M−1x0

where Jr is the real Jordan form of A and M is a realnonsingular matrix such that

Jr = M−1AM



There are three possible Jordan forms for A:

Different real eigenvalues

Equal real eigenvalues

Complex conjugate eigenvalues

[

λ1 0

0 λ2

] [

λ k

0 λ

] [

α −β

β α

]

In addition, we need to consider the case where at leastone of the eigenvalues is zero.



Different real eigenvalues: λ1 6= λ2 (non-zero)In this case,

M = [ v1 v2 ]

where v1 and v2 are the real eigenvectors of A associatedwith λ1 and λ2, respectively.The change of coordinate z = M−1x transforms the systeminto two decoupled first-order differential equations, i.e.,

z1 = λ1z1, z2 = λ2z2

with solution

z1(t) = z10eλ1t, z2(t) = z20e

λ2t ⇒ z2 =z20

(z10)λ2/λ1

zλ2/λ1

1



Stable Node: λ1, λ2 < 0

Figure 1: Modal (left) and original (right) coordinates.



Unstable Node: λ1, λ2 > 0

Figure 2: Original coordinates.



Saddle Point: λ1 > 0, λ2 < 0




Complex conjugate eigenvalues: λ1,2 = α± jβ

In this case,M = [ v1 v2 ]

where v1 and v2 are the real eigenvectors of A associatedwith λ1 and λ2, respectively.The change of coordinate z = M−1x transforms the systeminto the form

z1 = αz1 − βz2, z2 = βz1 + αz2

Defining the change of coordinates

r =√

z21 + z22 , θ = tan−1

(

z2

z1

)



we can write the dynamic equations in polar coordinates as

r = αr, θ = β

with solution

r(t) = r0eαt, θ(t) = θ0 + βt



Stable Focus: α < 0




Unstable Focus: α > 0




Center: α = 0




Equal real eigenvalues: λ1 = λ2 = λ (non-zero)In this case,

M = [ v1 v2 ]


z1 = λz1 + kz2, z2 = λz2

with solution

z1(t) = (z10+kz20t)eλt, z2(t) = z20e

λt ⇒ z1 = z2

[

z10

z20+

k

λln

(

z2

z20

)]



Case 1: k = 0

Figure 7: (a) λ < 0, (b) λ > 0.



Case 2: k = 1

Figure 8: (a) λ < 0, (b) λ > 0.



One or both zero eigenvalue: λ1 = 0, λ2 6= 0 or λ1 = λ2 = 0In this case A has a non-trivial null space.When λ1 = 0 and λ2 6= 0,

M = [ v1 v2 ]

where v1 and v2 are the real eigenvectors of A associatedwith λ1 and λ2, respectively. Note that v1 spans the nullspace of A.The change of coordinate z = M−1x transforms the systeminto the form

z1 = 0, z2 = λ2z2

with solutionz1(t) = z10, z2(t) = z20e

λ2t



When λ1 = λ2 = 0,

M = [ v1 v2 ]


z1 = z2, z2 = 0

with solutionz1(t) = z10 + z20t, z2(t) = z20



Case 1: λ1 = 0 and λ2 6= 0

Figure 9: (a) λ1 = 0, λ2 < 0, (b) λ1 = 0, λ2 > 0.



Case 2: When λ1 = λ2 = 0

Figure 10: λ1 = λ2 = 0.


Qualitative Behavior Near Equilibria

Given the nonlinear system

x1 = f1(x1, x2)

x2 = f2(x1, x2)(2)

let us assume p = (p1, p2) is an equilibrium point of (2), i.e.,

f1(p1, p2) = f2(p1, p2) = 0

Let us know expand the right-hand side of (2) around theequilibrium point p, i.e.,

x = f(p) +∂f(x)

∂x

∣

∣

∣

∣

x=p

(x− p) +H.O.T.



where

x =

[

x1

x2

]

, f(x) =

[

f1(x)

f2(x)

]

and∂f(x)

∂x

∣

∣

∣

∣

x=p

=

[

∂f1(x)x1

∂f1(x)x2

∂f2(x)x1

∂f2(x)x2

]

x=p

is the Jacobian evaluated at the equilibrium point p. Sincewe are interested in the behavior near p, we define

A ≡∂f(x)

∂x

∣

∣

∣

∣

x=p

, y = x− p

and we obtain



y ≈ Ay

which represents the Jacobi linearization.

Q: Is the system’s linearization a good approximation of its localbehavior?A: Yes, but provided the linearization has no eigenvalue on the imaginaryaxis, i.e., provided the equilibrium is hyperbolic.

Therefore, as long as f1(x) and f2(x) have continuous firstpartial derivatives, we can conclude that

stable/unstable node stable/unstable nodestable/unstable focus remains stable/unstable focussaddle point saddle point



Example: hyperbolic case - Pendulum with friction

θ = −b

ml2θ −

g

lsin(θ)



−6 −4 −2 0 2 4 6−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

x1

x2

Figure 11: Pendulum: Saddle + Stable Focus.



Example: non-hyperbolic case

x1 = −x2 − µx1(x21 + x22)

x2 = x1 − µx2(x21 + x22)



−2 −1.5 −1 −0.5 0 0.5 1 1.5 2−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

x1

x2

Figure 12: Stable/Unstable Focus.


Chapter 2: Second Order Systems - Lehigh Universityeus204/teaching/ME450_NSC/lectures/lecture01.pdfBehavior of Second Order Systems Consider the following linear system (1) x˙ =Ax

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