DSpace at VNU: Degenerate cocycle with index-1 and Lyapunov exponents

DEGENERATE COCYCLE WITH INDEX-1 AND LYAPUNOVEXPONENTS NGUYEN HUU DU∗, TRINH KHANH DUY and VU TIEN VIET Faculty of Mathematics, Mechanics, and Informatics Hanoi National University, 334 N

DEGENERATE COCYCLE WITH INDEX-1 AND LYAPUNOV

EXPONENTS

NGUYEN HUU DU∗, TRINH KHANH DUY and VU TIEN VIET

Faculty of Mathematics, Mechanics, and Informatics Hanoi National University, 334 Nguyen Trai Thanh Xuan, Hanoi, Vietnam

∗ dunh@vnu.edu.vn

Received 7 February 2007 Revised 23 March 2007

This paper deals with the solvability of initial-value problem and with Lyapunov expo-nents for linear implicit random difference equations, i.e the difference equations where the leading term cannot be solved An index-1 concept for linear implicit random differ-ence equations is introduced and a formula of solutions is given Paper is also concerned with a version of the multiplicative theorem of Oseledets type.

Keywords: Random dynamical systems; linear implicit equation Lyapunov exponent;

index-1 tractable.

1 Introduction

Difference equations might define the simplest dynamical systems, but neverthe-less, they play an important role in the investigation of a dynamical system The difference equations arise naturally when we want to study the evolution of bio-logical population or economic models on a fixed period of time They can also be illustrated as discretization of continuous time systems in computing process

An important class of difference equations is the linear one which leads to prod-ucts of matrices The linear difference equations can be found when we handle linearizations of nonlinear system

f (X n+1 , X n , n) = 0, n ≥ 0 (1.1) along solution This leads to the difference equation

A n X n+1 = B n X n + q n , n ≥ 0, (1.2)

where A n= ∂f

∂xn+1 and B n= ∂f

∂xn

If in (1.2), the matrix A n is invertible for all n ∈ N, we can multiply both sides

of (1.2) by A −1 n to obtain

X n+1 = A −1 n B n X n + A −1 n q n , (1.3) which has been studied for a long time by many authors both in theory and practice

229

However, in case where the assumption for solvability of implicit functions (1.1) cannot be fulfilled (for example when ∂f

∂xn+1 is degenerate), the matrix A n may be

degenerate for some n and it is unable to solve explicitly the leading term X n+1

to obtain classical difference equations (1.3) This situation also occurs when we consider the backward form of (1.2), i.e

B n X n = A n X n+1 − q n , n = −1, −2, ,

and in general, there is no reason to assume that (B n) is invertible

The difference equation of the form (1.2) also appears in discretizing the linear differential algebraic equation (see [7])

A(t) ˙ X(t) = B(t)X(t) + q(t),

by explicit Euler method

In that cases, solving the system (1.1) and (1.2) becomes more complicated In fact, we are faced with an ill-posed problem where the solution of an equation may exist only on a submanifold or even, without any further assumption, the solution

of Cauchy problem does not exist

In this paper we are concerned with the solvability of initial value problem, dynamic property of the solution and with the Lyapunov spectrum for a real noise equation

A n X n+1 = B n X n ,

where (A n , B n) is a stationary process

When all matrices A n and B n are invertible, the solutions of (1.4) form a ran-dom dynamic system which is well investigated (see Refs [1 and 5] for example)

However, without the assumption of invertibility of A n and B n, most of works have

to consider as Eq (1.4) separately for n ≤ 0 or for n ≥ 0 and as far as we know,

there is no work dealing with the dynamics of (1.4) overZ

Therefore, the aim of this paper is to study dynamic property of (1.4) We develop techniques in [7] to difference equations First, we introduce a concept

of index-1 for Eq (1.4) By means of this concept, we give an expression of the solutions Hence, we can prove the dynamic property and consider their Lyapunov exponents

We are focused only on the case of index-1 tractable and we have to assume

that rank A0 is a nonrandom constant which implies that the sequence of

matri-ces (A n)n∈Z has constant rank with probability one So far there is no available

technique to solve this problem with rank A n varying in n Moreover, a higher

index concept of Eq (1.3) is not defined and it requires some further works In our opinion, this work can be considered as a pioneer one dealt with implicit random dynamical systems

The paper is organized as follows: In Sec 2 we deal with the index-1 tractable concept of the pencil of matrices {A n , B n } By this concept of index-1, we set

out the condition of solvability for Cauchy problem and give an explicit formula

of solutions In Sec 3 we prove the cocycle property of solutions over random dynamics Section 4 gives the proof of multiplicative ergodic theorem MET for implicit difference equation (1.4) Here, we show that the space Rm can be split into measurable subspaces Rm = ⊕ τ

i=0 W i where for any i, the space W0⊕ W i is

invariant and on the set W i , the solution (X n) of (1.4) has the same Lyapunov exponent

2 Solutions of Implicit Linear Difference Equation

Henceforth,Z denotes the set of all integers Assume that (Ω, F, P ) is a probability space satisfying the normal conditions (see [8]) Let θ : (Ω, F, P ) → (Ω, F, P ) be an invertible, P -preserving transformation Let A(·) and B(·) be two random variables valued in the space of m × m-matrices We consider the equation

A(θ n ω)X n+1 (ω) = B(θ n ω)X n (ω),

where θ n = θ ◦ θ n−1 Equation (2.1) is called real noise linear implicit difference equation

2.1. Some surveys on linear algebra

In this section, we survey some basic properties of linear algebra Let (A, A, B) be

a triple of matrices Assume that rank A = rank A = r and let T ∈ Gl(R m) such

that T | ker A is an isomorphism between ker A and ker A We can obtain such an operator T as follows: let Q (resp Q) be a projector onto ker A (resp onto ker A); find nonsingular matrices V and V such that Q = V Q(0)V −1 and Q = V Q(0)V −1

where Q(0) = diag(0, I m−r ) and finally we obtain T by putting T = V V −1

Denote S = {x : Bx ∈ im A} and let Q be a projector onto ker A.

Lemma 2.1 The following assertions are equivalent.

(a) S ∩ ker A = {0},

(b) the matrix G = A + BT Q is nonsingular,

(c) Rm = S ⊕ ker A.

Proof (a)⇒ (b): Let x ∈ R m such that (A + BT Q)x = 0 ⇐⇒ B(T Qx) = A(−x) This equation implies T Qx ∈ S Since S ∩ ker A = {0} and T Qx ∈ ker A, it follows that T Qx = 0 Hence, Qx = 0 which implies Ax = 0 This mean that x ∈ ker A Thus, x = Qx = 0, i.e the matrix G = A + BT Q is nonsingular.

(b)⇒ (c): It is obvious that x = (I − T QG −1 B)x + T QG −1 Bx We see that

T QG −1 Bx ∈ ker A and B(I − T QG −1 B)x = B − (A + BT Q)G −1 Bx + AG −1 Bx =

AG −1 Bx ∈ im A Thus, (I − T QG −1 B)x ∈ S and we have R m = S + ker A Let x ∈ S ∩ ker A, i.e x ∈ S and x ∈ ker A Since x ∈ S, there is a z ∈ R m such that Bx = Az = AP z and since x ∈ ker A, T −1 x ∈ ker A Therefore, T −1 x =

QT −1 x Hence, (A + BT Q)T −1 x = (A + BT Q)P z which follows that T −1 x = P z.

Thus, T −1 x = 0 and then x = 0 So, we have (c).

(c)⇒ (a) is obvious.

The lemma is proved

Lemma 2.2 Assume that the matrix G is nonsingular Then, the following

rela-tions hold:

(iii) Q := T QG −1 B is the projector onto ker A along S. (2.4)

(iv) If Q is a projector onto ker A, then

QG −1 B = QG −1 BP + T −1 Q, (2.6)

with P = I − Q.

Proof (i) Note that GP = (A + BT Q)P = AP = A we get (2.2).

(ii) From BT Q = (G − A), it follows that G −1 BT Q = (I − P ) = Q Thus, we

have (2.3)

(iii) By virtue of (2.3), Q2= T QG −1 BT QG −1 B by (2.3)= T QQG −1 B =

T QG −1 B =  Q and A  Q = AT QG −1 B = 0 This means that  Q is a projector

onto kerA From the proof of (c), Lemma 2.1, we see that  Q is the projector

onto kerA along S.

(iv) Since T −1 Qx ∈ ker A for any x,

P G −1 BQ = P G −1 BT T −1 Q = P G −1 (A + BT Q)QT −1 Q = 0.

Therefore, P G −1 B = P G −1 BP so we have (2.5) Finally,

QG −1 B = QG −1 BP + QG −1 BT T −1 Q = QG −1 BP + QG −1 (A + BT Q)QT −1 Q

= QG −1 BP + QT −1 Q = QG −1 BP + T −1 Q.

Lemma 2.2 is proved

2.2. Existence of solutions of (2.1) for n ≥ 0

Henceforth, we assume that rank A(ω) = k for P -a.s ω ∈ Ω where k (0 < k ≤ m)

is a nonrandom constant Let T be a random variable with values in Gl(R m) such

that T (ω)| ker A(ω) is an isomorphism between ker A(ω) and ker A(θ −1 ω) Let Q(ω)

be a measurable projector onto ker A(ω) Denote

S(ω) = {z : B(ω)z ∈ im A(ω)}.

Definition 2.3 The implicit difference equation (2.1) is said to be index-1

tractable if

S(ω) ∩ ker A(θ −1 ω) = {0} for a.s ω ∈ Ω. (2.7)

By virtue of Lemma 2.1 we see that Eq (2.1) is index-1 tractable if and only if the matrices

G(ω) := A(ω) + B(ω)T (ω)Q(ω)

are nonsingular with probability one

Remark 2.4 (1) In fact, what we are doing below is true if the space Ω can be

divided into θ-invariant subsets Ω i , i = 1, 2, , q such that rank A is constant on

every Ωi and (2.7) is satisfied for P -almost sure ω.

(2) In general, since A(ω) is degenerate, the dimension of the space of solutions varies in n If the dimension of the space of solutions at step n + 1 is greater than one at the step n, it may cause bifurcations or multi-valued functions Up

to now, there has been no systematic analysis of such systems As we can see

below, S(ω) is in fact the space of solutions Therefore, with the index-1 tractable assumption (condition (2.7)), dimS(ω) is constant That implies the regularity of

inherent ordinary difference equation

For the sake of simplicity we put

Q n (ω) = Q(θ n ω), P n = I − Q n , A n (ω) = A(θ n ω), B n (ω) = B(θ n ω),

G n (ω) = G(θ n ω), T n (ω) = T (θ n ω).

Throughout this paper, if there is no confusion, we will omit ω in formulas.

Assume that Eq (2.1) is an index-1 tractable By multiplying both sides of (2.1)

by P n G −1 n and Q n G −1 n respectively and applying (2.3)–(2.6) for A = A n , A = A n−1 and B = B n we come to the following equation:

P n X n+1 = P n G −1 n B n P n−1 X n ,

0 = Q n G −1 n B n P n−1 X n + T n −1 Q n−1 X n

Putting Y n = P n−1 X n ; Z n = Q n−1 X n, (2.1) is equivalent to







Y n+1 = P n G −1 n B n Y n ,

Z n=−T n Q n G −1 n B n Y n ,

X n = Y n + Z n , n = 0, 1, 2,

(2.8)

From (2.8) it follows that the solution with initial condition X0= x exists if

Q0G −10 B0x = 0

with probability 1 By (2.4) we see that the random space Jf(ω) = {ξ ∈ R m :

(Q0G −10 B0)(ω)ξ = 0} does not depend on the choice of the projector Q and the transformation T Moreover, by Lemma 2.2



is the projector onto ker A n−1 along S n ={ξ ∈ R m : B n ξ ∈ im A n } Hence, the

matrices



G n = A n + B n T n Qn are also nonsingular with probability one

Lemma 2.5 Q n−1 (ω), called canonical projector, and the matrix (  P n G −1 n B n )(ω)

are independent from the choice of Q(ω) and T (ω), where  P n = I −  Q n

Proof Since Q n−1 is the projector onto ker A n−1 along the space S n, it is

inde-pendent from the choice of Q and T

Let T  be another linear transformation fromRm ontoRm satisfying T  | ker A0

to be an isomorphism from ker A0 onto ker A −1 and Q  be another measurable

projector onto ker A Denote T n  (ω) = T  (θ n (ω)), Q  n (ω) = Q  (θ n (ω)) and G  n (ω) =

A n (ω) + B n (ω)T n  (ω)Q  n (ω) It is easy to see that



P n G −1 n B n= P n G −1 n G  n G −1 n B n= P n G −1 n (A n + B n T n  Q  n )G −1 n B n

= P n P n G −1 n B n+ P n G −1 n B n T n  Q  n G −1 n B n= P n G −1 n B n+ P n G −1 n B n Qn−1

= P n G −1 n B n+ P n P n G −1 n B n Qn−1see (2.5)

= Pn G −1

n B n

The lemma is proved

Using the canonical projectors we have Q n−1 (ω)X n (ω) = 0 which implies

X n (ω) =  P n−1 (ω)X n (ω) for a.s ω ∈ Ω Therefore, the forward equation of (2.1) for n ≥ 0 with the initial condition X0= x ∈ Jf is reduced to a classical difference equation

X n+1= P n G−1

n B n Pn−1 X n ,

X0= x, n = 0, 1, 2, , (2.10)

which follows that

X n (ω) =



i=n−1



P i G−1

i B i Pi−1

(ω)x, n ∈ N, X0(ω) = x. (2.11)

Summing up we see that the initial value problem of (2.1) for n ≥ 0 has a unique solution given by (2.11) provided x ∈ Jf.

Remark 2.6 Since ( P n G−1

n B n )(ω) = (  P n G −1 n B n )(ω), Eq (2.10) is rewritten

under the form

X n+1= P n G −1 n B n Pn−1 X n ,

X0= P −1 x, (x ∈ R m ), n = 0, 1, 2,

2.3. Solutions (2.1) for n < 0

For n < 0, Eq (2.1) turns into the implicit difference equation

B n X n = A n X n+1 ,

X0= x; n = −1, −2, (2.12)

We assume that rank B(ω) = r for P -a.s ω ∈ Ω where r (0 < r ≤ m) is a

nonrandom constant LetT be a random variable with values in Gl(R m) such that

T (ω)| ker B(ω) is an isomorphism between ker B(ω) and ker B(θω) Let Q(ω) be a measurable projector onto ker B(ω) Put

G(ω) := B(ω) + A(ω)T (ω)Q(ω).

Assume that G is nonsingular with probability 1 This assumption implies that

Eq (2.12) is index-1 tractable Let

Q n (ω) = Q(θ n ω); G n (ω) = G(θ n ω); P n = I − Q n; T n (ω) = T (θ n ω). (2.13)

For any n < 0, denote Y n = P n X n ; Z n = Q n X n, Eq (2.12) leads to







Y n = P n G −1 n A n Y n+1 ,

Z n+1=−T n Q n G −1 n A n Y n+1 ,

X n = Y n + Z n , n = −1, −2,

Q n := T n−1 Q n−1 G −1 n−1 A n−1which is

the projector onto ker B n along the space S n−1 = {ξ : A n−1 ξ ∈ im B n−1 } Let

G n (ω) = B n (ω) + A n (ω)T n Q n By using a similar argument as above we obtain

Q n P n G −1 n A n are

indepen-P n Q n

Q n Q n X n = 0 for any n < 0 Therefore,







X n= −1

i=n

P i G −1 i A i P i+1 x, n = −1, −2, ,

X0= x ∈ Jb,

(2.14)

where Jb ={ξ ∈ R m Q0ξ = 0} Similarly as in Sec 2.2 we see that the initial

value problem for n ≤ 0 has a unique solution given by (2.14) provided that x ∈ Jb

Remark 2.8 (1) Let us give some comments on the expressions of solutions (2.11)

and (2.14) To obtain (2.11) for n ≥ 0 we assume the initial condition X0= x ∈ J f (equivalently X0= P −1 x, x ∈ R m ) to be satisfied and obtain (2.14) for n ≤ 0 the equality X0 P0x is required Thus, there exists uniquely the solution of (2.1) for

n ∈ Z with the initial condition X0= x if and only if x ∈ Jf∩ Jb

(2) Unlike random ordinary difference equations, in general, the existence of solution

of random implicit difference equations implies that the initial condition must be a

measurable selection of the corresponding ω → Jf(ω) ∩ Jb(ω).

3 Dynamic Property

Q0 is the projector onto ker B0, Q −1 Q0 = T0Q0G −10 B0Q0 = 0 Similarly,

Q0Q−1= 0 Hence, the projectors P −1 P0commute, i.e.



with probability one

Put

Φ(n, ω) =















0

i=n−1

( P i G−1

i B i Pi−1 )(ω) = 0

i=n−1

( P i G −1 i B i )(ω) if n > 0,

−1

i=n P i G −1 i A i P i+1 )(ω) = −1

i=n P i G −1 i A i )(ω) if n < 0.

(3.2)

We are now in a position to give a fundamental expression in random dynamic theory, called cocycle property (cf L Arnold [1]):

Theorem 3.1 For any m, n ∈ Z the following relation holds:

Proof If m, n < 0 or m, n > 0 the relation (3.3) follows from properties of random

matrix products

Q0 is a projector onto ker B0,

Φ(1, ω)Φ(0, ω) =  P0G−1

0 B0P−1 P0= P0G−1

0 B0P0

= P0G−1

0 B0 Q0) = P0G−1

0 B0= Φ(1, ω).

Hence Φ(n, ω)Φ(0, ω) = Φ(n, ω) for all n > 0 The case n < 0 is similar because



Q −1 is the projector onto ker A −1

On the other hand, A0Φ(1, ω) = B0Φ(0, ω) Therefore, by multiplying both sides

of this relation withT0Q0G −10 we obtain

T0Q0G −10 A0Φ(1, ω) = T0Q0G −10 B0Φ(0, ω) = T0Q0P0Φ(0, ω) = 0.

Q1 = T0Q0G −10 A0 Q1Φ(1, ω) = 0, which leads to Φ(0, θω)· Φ(1, ω) = Φ(1, ω) This means that (3.3) is true for n = 0, m = 1 Hence

Φ(0, θ m ω)Φ(m, ω) = Φ(0, θ(θ m−1 ω))Φ(1, θ m−1 ω)Φ(m − 1, ω) = Φ(m, ω)

for n = 0 and m ≥ 1 The case n = 0, m < 0 is similar.

For n = −1, m = 1 we have

P0P0Φ(0, ω) = Φ(0, ω).

Suppose that n < 0 < m.

Φ(n, θ m ω)Φ(m, ω) = Φ(n + 1, θ m−1 ω)Φ(−1, θ m ω) · Φ(1, θ m−1 ω)Φ(m − 1, ω)

= Φ(n + 1, θ m−1 )Φ(0, θ m−1 ω)Φ(m − 1, ω)

= Φ(n + 1, θ m−1 ω)Φ(m − 1, ω).

By induction we obtain

Φ(n, θ m ω) · Φ(m, ω) =







Φ(0, θ n+m ω)Φ(n + m, ω) = Φ(n + m, ω) if n + m > 0,

Φ(n + m, ω)Φ(0, ω) = Φ(n + m, ω) if n + m < 0 The case n > 0 > m is proved similarly by noting that  P −1 P0 P0P−1, thus Theorem 3.1 is proved

Denote by X n (z(ω), ω) the solution of (2.1) satisfying X0(z(ω), ω) = z(ω) It is

clear that

X n (x(ω), ω) = Φ(n, ω)x with x(ω) = (  P −1 P0)(ω)x.

Example 3.2 Consider the initial value problem

X n+1 = B(θ n ω)X n ,

X0= x ∈ R m , n ∈ Z.

Assume that rank B = r is nonrandom constant Let Q(0)= diag(0, I m−r) Put



Q0 = 0, Q0 = V Q(0)V  and T = V (θ)V  where B = U

Σ 0

0 0 V  is a singular

value decomposition of B Suppose that G0 = B + T Q0 = B + V (θ)Q(0)V  is

Q0= V Q(0)V (θ −1) G −1 −1 P0 Q0

and then X n (ω, x) = Φ(n, ω)x with

Φ(n, ω) =











(B n−1 · B n−2 · · · B0)(ω) if n > 0,

−1

i=n P i G −1 i )(ω) if n < 0.

4 Lyapunov Exponents — Multiplicative Ergodic Theorem

Firstly, we recall some notions of generalized inverse of matrices Let M be an

m × m- matrix We say a matrix X is generalized inverse of M if the following

relations hold:

(i) M XM = M, (ii) XM X = X, (iii) (M X)  = M X, (iv) (XM )  = XM.

In [2] it is proved that such a matrix X, denoted by M+, exists uniquely If M

is nonsingular, then M+= M −1

For any number α we write

α+=

0 if α = 0,

α −1

By this notation we have



diag(α1, α2, , α m)+

= diag(α+1, α+2, , α+m ).

Moreover, if M is a diagonal-block matrix then



diag(M1, M2, , M m)+

= diag(M1+, M2+, , M τ+).

Using the Jordan form of matrices we see that

Lemma 4.1 λ is an eigenvalue with multiplier d of the matrix M iff λ+ is an eigenvalue with multiplier d of the generalized inverse M+.

Lemma 4.2 For any matrix M we have (M M )+= (M )+M+and ker(M M )+

coincides with ker M M 

We are now in position to study Lyapunov exponents of solutions (X n) of (2.1)

Suppose that θ is an ergodic transformation on (Ω, F, P ) and the following condition

is satisfied

Hypothesis 4.3.

ln  P0G−1

0 B0 ∈ L1 P0G −10 A0 ∈ L1(Ω, F, P ). (4.1)

We note that this assumption is independent of the choice of T and Q.

Since (Φn) is the product of ergodic stationary matrices P n G−1

n B n for n > 0

P n G −1 n A n for n < 0, by [6] we get

(i) Under the assumption (4.1), there exist the limits

lim

n→∞ (Φ(n, ω)  Φ(n, ω)) 1/2n =: ∆(ω) (4.2) and

lim

n→−∞ (Φ(n, ω)  Φ(n, ω)) 1/2|n| =: ∆(ω).

(ii) Let 0 < e λ1< e λ2< · · · < e λτ be the different nonzero eigenvalues of ∆ and

λ0=− ∞ We denote U0 = ker ∆(ω), and U i , i = 1, 2, , τ the eigenspace

with multipliers d i = dimU i corresponding to the eigenvalue e λi Then,

τ ; d i , λ i , i = 1, 2, , τ are nonrandom constants Let V k = U0⊕ U1⊕ · · · ⊕

U k , k = 0, 1, 2, , τ such that

{0} ⊂ V0⊂ V1⊂ · · · ⊂ V τ =Rm

4 Lyapunov Exponents — Multiplicative Ergodic Theorem

Firstly, we recall some notions of generalized inverse of matrices Let M be an

m × m- matrix We say a matrix... to study Lyapunov exponents of solutions (X n) of (2.1)

Suppose that θ is an ergodic transformation on (Ω, F, P ) and the following condition

is satisfied...

Using the Jordan form of matrices we see that

Lemma 4.1 λ is an eigenvalue with multiplier d of the matrix M iff λ+ is an eigenvalue with multiplier d of the