Discrete Dynamical Systems: Linearization and
cojugacy
Winfried Just
Department of Mathematics
Ohio University
March 20, 2015
Winfried Just at OU
Linearization and Conjugacy
Discrete-Time Dynamical Systems on Rn
The state space is Rn .
Time proceeds in discrete steps; t ∈ N or t ∈ Z.
The updating function is F (~x ) may or may not be linear.
The previous condition uniquely determines a state
~x (τ ) = F τ (~x (0)) for all nonnegative integers τ .
Let us assume that ~x ∗ is an equilibrium of this system. A suitable
change of variables allows us to assume wlog that ~x ∗ = ~0.
What can we say about the stability of ~x ∗ in this more
general setting?
Ohio University – Since 1804
Department of Mathematics
Winfried Just at OU
Linearization and Conjugacy
When is the equilibrium stable?
We will assume that F is differentiable at ~x ∗ = ~0 and let J denote
its Jacobian at ~0. Then
F (~x ) = J~x + R(~x ),
where
R(~x )
= 0.
xk
~x →~0 k~
lim
Assume there exists 0 ≤ µ < 1 such that kJ~x k ≤ µk~x k for all ~x .
x that are sufficiently close to but
Let α = 1−µ
2 . Then for all ~
kR(~x )k
~
different from 0 we have k~x k ≤ α, and it follows that for all
such ~x :
kF (x)k ≤ kJ~x k + kR(~x )k ≤ µk~x k + αk~x k = (µ + α)k~x k,
and we conclude that the equilibrium ~0 is asymptotically stable
whenever all eigenvalues of J satisfy |λ| < 1.
Ohio University – Since 1804
Department of Mathematics
Winfried Just at OU
Linearization and Conjugacy
When is the equilibrium stable?
We will assume that F is differentiable at ~x ∗ = ~0 and let J denote
its Jacobian at ~0. Then
F (~x ) = J~x + R(~x ),
where
R(~x )
= 0.
xk
~x →~0 k~
lim
Assume there exists an eigenvalue λ of J so that 1 < |λ|. For
simplicity assume λ is real, and let ~x be an eigenvector of J with
x that are sufficiently
this eigenvalue. Let α = |λ|−1
2 . Then for all ~
kR(~
x
)k
close to but different from ~0 we have k~x k ≤ α, and it follows
that for all such ~x :
kF (x)k ≥ kJ~x k − kR(~x )k ≥ |λ|k~x k − αk~x k = (|λ| − α)k~x k,
and we conclude that the equilibrium ~0 is unstable.
Ohio University – Since 1804
Department of Mathematics
Winfried Just at OU
Linearization and Conjugacy
When is the equilibrium stable?
Assume that F is differentiable at ~x ∗ = ~0 and let J denote its
Jacobian at ~0. Then
F (~x ) = J~x + R(~x ),
where
R(~x )
= 0.
xk
~x →~0 k~
lim
It follows that
If |λ| < 1 for all eigenvalues λ of J, then ~x ∗ = ~0 is locally
asymptotically stable, which means that limt→∞ ~x (t) = ~0 as
long as k~x (0)k is sufficiently small.
If |λ| > 1 for some eigenvalue λ of J, then ~x ∗ = ~0 is unstable.
If the maximum of |λ| for the eigenvalues λ of J is equal to 1,
then ~x ∗ = ~0 may be locally asymptotically stable, Lyapunov
stable, or unstable.
Ohio University – Since 1804
Winfried Just at OU
Department of Mathematics
Linearization and Conjugacy
Linearization at an equilibrium
Consider a system (Rn , F ) given by iterations of a function F ,
assume F (~x ∗ ) = ~x ∗ , and let J denote the Jacobian of F at ~x ∗ .
Then the iterations of
G (~x ) = J(~x − ~x ∗ ) + ~x ∗
define a linear system (Rn , F ) for which ~x ∗ is also an equilibrium.
The latter is called the linearization of (Rn , F ) at ~x ∗ .
The previous slide gives conditions under which stability of ~x ∗
behaves in the same way for (Rn , F ) and (Rn , G ).
Can we go one step further and assert that under some
conditions the systems (Rn , F ) and (Rn , G ) are the same in a
neighborhood of ~x ∗ ?
Ohio University – Since 1804
Department of Mathematics
Winfried Just at OU
Linearization and Conjugacy
Conjugacy
Consider two systems (X , F ) and (Y , G ), where X , Y are
topological spaces and F , G are continuous. These systems can be
considered the same if there exists a conjugacy, that is, a
homeomorphism h : Y → X such that h ◦ G = F ◦ h, that is, the
following diagram commutes:
G: Y
h: ↓
F : X
→ Y
↓
→ X
Being conjugate is an equivalence relation.
A conjugacy maps equilibria to equilibria and preserves their
stability.
More generally, a conjugacy maps trajectories to trajectories.
Ohio University – Since 1804
Department of Mathematics
Winfried Just at OU
Linearization and Conjugacy
An Example of a Conjugacy
Homework 1: (a) Consider two systems (Rn , F ) and (Rn , G ),
where F (~x ) = M~x + ~b and G (~x ) = M~x for some matrix M without
eigenvalue 1. Prove that there exists a conjugacy between these
two systems.
(b) Prove that if h is that being conjugate is an equivalence
relation.
(c) Prove that conjugacy preserves Lyapunov stability of equilibria.
Ohio University – Since 1804
Department of Mathematics
Winfried Just at OU
Linearization and Conjugacy
Conjugacies and preservation of eigenvalues
Consider two systems (R, F (x) = 2x) and (R, G (x) = 0.5x).
These systems are not conjugate.
On the other hand, the two systems (R, F (x) = 2x) and
(R, G (x) = 3x) are conjugate, as witnessed by
h(x) =
x log3 |x|
2
.
|x|
Notice that if we let h(0) = 0 then h is a homeomorphism and for
x >0
F ◦ h(x) = 2 · 2log3 x = 21+log3 x = 2log3 3+log3 x = 2log3 3x = h ◦ G (x).
In general, conjugacies of discrete dynamical systems don’t
preserve eigenvalues of the Jacobian at equilibria, but will usually
preserve the number of eigenvectors with eigenvalues |λ| > 1 and
eigenvalues |λ| < 1.
Ohio University – Since 1804
Department of Mathematics
Winfried Just at OU
Linearization and Conjugacy
Conjugacies and eigenvalues
Homework 2: Consider two systems (R2 , F (~x ) = M~x ) and
(R2 , G (~x ) = N~x ), where
0.5 0
0.5 1
M=
N=
0 0.5
1 0.5
Prove that the two systems are conjugate.
Homework 3: Consider two systems (Rn , F (~x ) = M~x ) and
(Rn , G (~x ) = N~x ), where M and N each have n pairwise distinct
real eigenvalues λ1 (M), . . . , λn (M), λ1 (N), . . . , λn (N) respectively,
so that for some k ∈ {0, . . . , n} we have 0 < λi (M), λi (N) < 1 for
all i ≤ k and 1 < λi (M), λi (N) for all i > k. Prove that the two
systems are conjugate.
Ohio University – Since 1804
Department of Mathematics
Winfried Just at OU
Linearization and Conjugacy
Are linearizations at an equilibrium locally conjugate to the
original system?
Consider a system (Rn , F ) with equilibrium at ~0, and let J denote
the Jacobian of F at ~0, and let (Rn , G ) be the corresponding
linearization.
When can we deduce that there exists a local conjugacy
between the two systems?
We are looking for a homeomorphism h between some
neighborhoods U1 , U2 , V1 , V2 of ~0 such that h ◦ G = F ◦ h, that is,
the following diagram commutes:
G : U1 → U2
h: ↓
↓
F : V1 → V2
Ohio University – Since 1804
Department of Mathematics
Winfried Just at OU
Linearization and Conjugacy
When will we be out of luck?
Consider a system (Rn , F ) with equilibrium at ~x ∗ , let J denote the
Jacobian of F at ~x ∗ , and let (Rn , G ) be the corresponding
linearization.
If J is not invertible, there usually will be no local conjugacy.
(Consider n = 1, F (x) = x 3 .)
If J has an eigenvalue λ = 1, there usually will be no local
conjugacy. (Consider n = 1, F (x) = x + x 3 . If h is a local
conjugacy, then G ◦ h = h, and we must have F (h(x)) = h(x)
for all x ∈ U. But 0 is the only fixed point of F , so the range
of h cannot be a neighborhood of 0.)
Similar problems may arise if J has any eigenvalues with
|λ| = 1. The equilibrium ~x ∗ is called hyperbolic if J has no
eigenvalues on the unit circle.
Ohio University – Since 1804
Department of Mathematics
Winfried Just at OU
Linearization and Conjugacy
The Hartman-Grobman Theorem for maps
Theorem
Consider a system (Rn , F ) with a hyperbolic equilibrium at ~x ∗ ,
assume that F is continuously differentiable, let J denote the
Jacobian of F at ~x ∗ , and let (Rn , G ) be the corresponding
linearization. Assume that J is invertible. Then there exists a
homeomorphism h between some neighborhoods U1 , U2 , V1 , V2
of ~x ∗ such that h ◦ G = F ◦ h, that is, the following diagram
commutes:
G : U1 → U2
h: ↓
↓
F : V1 → V2
Ohio University – Since 1804
Department of Mathematics
Winfried Just at OU
Linearization and Conjugacy
All assumptions of the Hartman-Grobman Theorem are
needed
We have already seen that the assumptions that ~x ∗ be hyperbolic
and J be invertible are needed in the Hartman-Grobman Theorem.
The conclusion of the Hartman-Grobman Theorem may also fail if
F is only assumed to be differentiable, but not continuously so.
For example, let G : R2 → R2 be defined by G (x, y ) = (2x, 0.5y ).
Construct a function H : R2 → R2 such that
F := G + H is differentiable (but not continuously).
G is the derivative of F at ~0.
F maps all points (x, y ) with |x| ≤ y 2 onto the y -axis.
Ohio University – Since 1804
Department of Mathematics
Winfried Just at OU
Linearization and Conjugacy
A consequence of the Hartman-Grobman Theorem
Suppose (Rn , F ), ~x ∗ , J, (Rn , G ) satisfy the assumptions and let h
be such that the following diagram commutes:
G : U1 → U2
h: ↓
↓
F : V1 → V2
Assume E s is the linear span of all (generalized) eigenvectors of J
with eigenvalues of J such that |λ| < 1 and E u is the linear span of
all (generalized) eigenvectors of J with eigenvalues of J such that
|λ| > 1. Then h maps E s ∩ U1 into the so-called stable manifold
W s of ~x ∗ and h maps E u ∩ U1 into the so-called unstable
manifold W u of ~x ∗ .
The stable manifold is forward invariant, and all trajectories that
start in it move towards ~x ∗ . The unstable manifold is backward
invariant, and all trajectories that start in it move away from ~x ∗ .
Ohio University – Since 1804
Department of Mathematics
Winfried Just at OU
Linearization and Conjugacy