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

June 26, 2007 4:2 WSPC/168-SD 00201 Stochastics and Dynamics, Vol 7, No (2007) 229–245 c World Scientific Publishing Company 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 February 2007 Revised 23 March 2007 This paper deals with the solvability of initial-value problem and with Lyapunov exponents 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 difference 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 Introduction Difference equations might define the simplest dynamical systems, but nevertheless, 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 biological 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 products of matrices The linear difference equations can be found when we handle linearizations of nonlinear system f (Xn+1 , Xn , n) = 0, n≥0 (1.1) along solution This leads to the difference equation An Xn+1 = Bn Xn + qn , n ≥ 0, (1.2) ∂f where An = ∂x∂f and Bn = ∂x n+1 n If in (1.2), the matrix An is invertible for all n ∈ N, we can multiply both sides of (1.2) by A−1 n to obtain −1 Xn+1 = A−1 n Bn Xn + An qn , (1.3) which has been studied for a long time by many authors both in theory and practice 229 June 26, 2007 4:2 WSPC/168-SD 230 00201 N H Du, T K Duy & V T Viet However, in case where the assumption for solvability of implicit functions (1.1) is degenerate), the matrix An may be cannot be fulfilled (for example when ∂x∂f n+1 degenerate for some n and it is unable to solve explicitly the leading term Xn+1 to obtain classical difference equations (1.3) This situation also occurs when we consider the backward form of (1.2), i.e Bn Xn = An Xn+1 − qn , n = −1, −2, , and in general, there is no reason to assume that (Bn ) 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 An Xn+1 = Bn Xn , X = x ∈ Rm , n = 0, ±1, ±2, , (1.4) where (An , Bn ) is a stationary process When all matrices An and Bn are invertible, the solutions of (1.4) form a random dynamic system which is well investigated (see Refs [1 and 5] for example) However, without the assumption of invertibility of An and Bn , most of works have to consider as Eq (1.4) separately for n ≤ or for n ≥ and as far as we know, there is no work dealing with the dynamics of (1.4) over Z 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 matrices (An )n∈Z has constant rank with probability one So far there is no available technique to solve this problem with rank An 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 we deal with the index-1 tractable concept of the pencil of matrices {An , Bn } By this concept of index-1, we set June 26, 2007 4:2 WSPC/168-SD 00201 Degenerate Cocycle with Index-1 and Lyapunov Exponents 231 out the condition of solvability for Cauchy problem and give an explicit formula of solutions In Sec we prove the cocycle property of solutions over random dynamics Section 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 = ⊕τi=0 Wi where for any i, the space W0 ⊕ Wi is invariant and on the set Wi , the solution (Xn ) of (1.4) has the same Lyapunov exponent 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(θn ω)Xn+1 (ω) = B(θn ω)Xn (ω), X0 = x ∈ Rm a.s, n ∈ Z, (2.1) where θn = θ ◦ θn−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(Rm ) 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); −1 find nonsingular matrices V and V such that Q = V Q(0) V −1 and Q = V Q(0) V where Q(0) = diag(0, Im−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 ∈ Rm such that (A + BT Q)x = ⇐⇒ 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 = Hence, Qx = which implies Ax = 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−1Bx + AG−1 Bx = AG−1 Bx ∈ im A Thus, (I − T QG−1 B)x ∈ S and we have Rm = S + ker A Let x ∈ S ∩ ker A, i.e x ∈ S and x ∈ ker A Since x ∈ S, there is a z ∈ Rm such that Bx = Az = AP z and since x ∈ ker A, T −1 x ∈ ker A Therefore, T −1 x = June 26, 2007 4:2 WSPC/168-SD 232 00201 N H Du, T K Duy & V T Viet 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 = and then x = 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 relations hold: (i) (ii) (iii) (iv) P = G−1 A where P = I − Q G−1 BT Q = Q Q := T QG−1 B is the projector onto ker A along S If Q is a projector onto ker A, then (2.2) (2.3) (2.4) P G−1 B = P G−1 BP , −1 QG −1 B = QG BP + T (2.5) −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) by (2.3) = T QQG−1B = (iii) By virtue of (2.3), Q2 = T QG−1 BT QG−1 B −1 −1 T QG B = Q and AQ = AT QG B = This means that Q is a projector onto ker A From the proof of (c), Lemma 2.1, we see that Q is the projector onto ker A along S (iv) Since T −1 Qx ∈ ker A for any x, P G−1 BQ = P G−1 BT T −1Q = P G−1 (A + BT Q)QT −1Q = 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 −1Q = 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 ≥ 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(Rm ) 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)} June 26, 2007 4:2 WSPC/168-SD 00201 Degenerate Cocycle with Index-1 and Lyapunov Exponents 233 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 Ωi , 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 + 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 Qn (ω) = Q(θn ω), Pn = I − Qn , Gn (ω) = G(θn ω), Tn (ω) = T (θn ω) An (ω) = A(θn ω), Bn (ω) = B(θn 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) −1 by Pn G−1 n and Qn Gn respectively and applying (2.3)–(2.6) for A = An , A = An−1 and B = Bn we come to the following equation: Pn Xn+1 = Pn G−1 n Bn Pn−1 Xn , −1 = Qn G−1 n Bn Pn−1 Xn + Tn Qn−1 Xn Putting Yn = Pn−1 Xn ; Zn = Qn−1 Xn , (2.1) is equivalent to  −1   Yn+1 = Pn Gn Bn Yn , Zn = −Tn Qn G−1 n Bn Yn ,   X = Y + Z , n = 0, 1, 2, n n (2.8) n From (2.8) it follows that the solution with initial condition X0 = x exists if Q0 G−1 B0 x = with probability By (2.4) we see that the random space Jf (ω) = {ξ ∈ Rm : (Q0 G−1 B0 )(ω)ξ = 0} does not depend on the choice of the projector Q and the transformation T Moreover, by Lemma 2.2 Qn−1 := Tn Qn G−1 n Bn , (2.9) June 26, 2007 4:2 WSPC/168-SD 234 00201 N H Du, T K Duy & V T Viet is the projector onto ker An−1 along Sn = {ξ ∈ Rm : Bn ξ ∈ im An } Hence, the matrices Gn = An + Bn Tn Qn are also nonsingular with probability one Lemma 2.5 Qn−1 (ω), called canonical projector, and the matrix (Pn G−1 n Bn )(ω) are independent from the choice of Q(ω) and T (ω), where Pn = I − Qn Proof Since Qn−1 is the projector onto ker An−1 along the space Sn , it is independent from the choice of Q and T Let T be another linear transformation from Rm onto Rm 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 Tn (ω) = T (θn (ω)), Qn (ω) = Q (θn (ω)) and Gn (ω) = An (ω) + Bn (ω)Tn (ω)Qn (ω) It is easy to see that −1 Pn G−1 n Bn = Pn Gn Gn G = Pn Pn G = Pn G −1 n Bn −1 n Bn −1 n Bn −1 n Bn −1 Pn G n Bn = Pn G−1 n (An + Bn Tn Qn )G + Pn G−1 n Bn Tn Qn G + Pn Pn G−1 n Bn Qn−1 −1 n Bn see (2.5) = = Pn G + Pn G−1 n Bn Qn−1 −1 n Bn The lemma is proved Using the canonical projectors we have Qn−1 (ω)Xn (ω) = which implies Xn (ω) = Pn−1 (ω)Xn (ω) for a.s ω ∈ Ω Therefore, the forward equation of (2.1) for n ≥ with the initial condition X0 = x ∈ Jf is reduced to a classical difference equation Xn+1 = Pn G−1 n Bn Pn−1 Xn , X0 = x, (2.10) n = 0, 1, 2, , which follows that Xn (ω) = Pi G−1 i Bi Pi−1 n ∈ N, (ω)x, X0 (ω) = x (2.11) i=n−1 Summing up we see that the initial value problem of (2.1) for n ≥ has a unique solution given by (2.11) provided x ∈ Jf −1 Remark 2.6 Since (Pn G−1 n Bn )(ω) = (Pn Gn Bn )(ω), Eq (2.10) is rewritten under the form Xn+1 = Pn G−1 n Bn Pn−1 Xn , X0 = P−1 x, (x ∈ Rm ), n = 0, 1, 2, June 26, 2007 4:2 WSPC/168-SD 00201 Degenerate Cocycle with Index-1 and Lyapunov Exponents 235 2.3 Solutions (2.1) for n < For n < 0, Eq (2.1) turns into the implicit difference equation Bn Xn = An Xn+1 , X0 = x; (2.12) n = −1, −2, We assume that rank B(ω) = r for P -a.s ω ∈ Ω where r (0 < r ≤ m) is a nonrandom constant Let T be a random variable with values in Gl(Rm ) 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 This assumption implies that Eq (2.12) is index-1 tractable Let Qn (ω) = Q(θn ω); Gn (ω) = G(θn ω); P n = I − Qn ; T n (ω) = T (θn ω) (2.13) For any n < 0, denote Yn = P n Xn ; Zn = Qn Xn , Eq (2.12) leads to  −1   Yn = P n Gn An Yn+1 , −1 Zn+1 = −T n Qn Gn An Yn+1 ,   X = Y + Z , n = −1, −2, n n n −1 We find the canonical projector Denote Qn := T n−1 Qn−1 Gn−1 An−1 which is the projector onto ker Bn along the space S n−1 = {ξ : An−1 ξ ∈ im Bn−1 } Let Gn (ω) = Bn (ω) + An (ω)T n Qn By using a similar argument as above we obtain −1 Lemma 2.7 The canonical projector Qn and the matrix Pn Gn An are independent of the choice of Q and T , where Pn = I − Qn With the canonical projector Qn we have Qn Xn = for any n < Therefore,  −1  X = Pi G−1 n = −1, −2, , n i Ai Pi+1 x, (2.14) i=n  X0 = x ∈ Jb , where Jb = {ξ ∈ Rm : Q0 ξ = 0} Similarly as in Sec 2.2 we see that the initial value problem for n ≤ 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 ≥ we assume the initial condition X0 = x ∈ Jf (equivalently X0 = P−1 x, x ∈ Rm ) to be satisfied and obtain (2.14) for n ≤ the equality X0 = P0 x 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) June 26, 2007 4:2 WSPC/168-SD 236 00201 N H Du, T K Duy & V T Viet Dynamic Property Since Q0 is the projector onto ker B0 , Q−1 Q0 = T0 Q0 G−1 B0 Q0 = Similarly, Q0 Q−1 = Hence, the projectors P−1 and P0 commute, i.e P−1 P0 = P0 P−1 with probability one Put  0   (Pi G−1 Bi Pi−1 )(ω) = (Pi G−1  i i Bi )(ω)   i=n−1 i=n−1    Φ(n, ω) = (P−1 P0 )(ω)     −1 −1  −1  −1  (Pi Gi Ai )(ω)  (Pi Gi Ai Pi+1 )(ω) = i=n (3.1) if n > 0, if n = 0, (3.2) if n < i=n 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: Φ(n + m, ω) = Φ(n, θm ω) · Φ(m, ω) (3.3) Proof If m, n < or m, n > the relation (3.3) follows from properties of random matrix products We consider the case m = Since Q0 is a projector onto ker B0 , −1 Φ(1, ω)Φ(0, ω) = P0 G−1 B0 P−1 P0 = P0 G0 B0 P0 −1 = P0 G−1 B0 (I − Q0 ) = P0 G0 B0 = Φ(1, ω) Hence Φ(n, ω)Φ(0, ω) = Φ(n, ω) for all n > The case n < is similar because Q−1 is the projector onto ker A−1 On the other hand, A0 Φ(1, ω) = B0 Φ(0, ω) Therefore, by multiplying both sides of this relation with T Q0 G−1 we obtain −1 T Q0 G−1 A0 Φ(1, ω) = T Q0 G0 B0 Φ(0, ω) = T Q0 P0 Φ(0, ω) = Noting that Q1 = T Q0 G−1 A0 we have Q1 Φ(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 = Hence Φ(0, θm ω)Φ(m, ω) = Φ(0, θ(θm−1 ω))Φ(1, θm−1 ω)Φ(m − 1, ω) = Φ(m, ω) for n = and m ≥ The case n = 0, m < is similar For n = −1, m = we have −1 Φ(−1, θω)Φ(1, ω) = P0 G−1 A0 Φ(1, ω) = P0 G0 B0 Φ(0, ω) = P0 P0 Φ(0, ω) = Φ(0, ω) June 26, 2007 4:2 WSPC/168-SD 00201 Degenerate Cocycle with Index-1 and Lyapunov Exponents 237 Suppose that n < < m Φ(n, θm ω)Φ(m, ω) = Φ(n + 1, θm−1 ω)Φ(−1, θm ω) · Φ(1, θm−1 ω)Φ(m − 1, ω) = Φ(n + 1, θm−1 )Φ(0, θm−1 ω)Φ(m − 1, ω) = Φ(n + 1, θm−1 ω)Φ(m − 1, ω) By induction we obtain   Φ(0, θn+m ω)Φ(n + m, ω) = Φ(n + m, ω)  Φ(n, θm ω) · Φ(m, ω) = Φ(0, ω)Φ(0, ω) = Φ(0, ω)   Φ(n + m, ω)Φ(0, ω) = Φ(n + m, ω) if n + m > 0, if n + m = 0, if n + m < The case n > > m is proved similarly by noting that P−1 P0 = P0 P−1 , thus Theorem 3.1 is proved Denote by Xn (z(ω), ω) the solution of (2.1) satisfying X0 (z(ω), ω) = z(ω) It is clear that Xn (x(ω), ω) = Φ(n, ω)x with x(ω) = (P−1 P0 )(ω)x Example 3.2 Consider the initial value problem Xn+1 = B(θn ω)Xn , X = x ∈ Rm , n ∈ Z Assume that rank B = r is nonrandom constant Let Q(0) = diag(0, Im−r ) Put Q0 = 0, Q0 = V Q(0) V and T = V (θ)V where B = U Σ 0 V is a singular value decomposition of B Suppose that G0 = B + T Q0 = B + V (θ)Q(0) V is −1 nonsingular with probability We see that Q0 = V Q(0) V (θ−1 ) G−1 ; P0 = I − Q0 and then Xn (ω, x) = Φ(n, ω)x with   (Bn−1 · Bn−2 · · · B0 )(ω) if n > 0,      if n = 0, Φ(n, ω) = P0 (ω)    −1  −1   (Pi Gi )(ω) if n < i=n 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 June 26, 2007 4:2 WSPC/168-SD 238 00201 N H Du, T K Duy & V T Viet 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+ = if α = 0, −1 α if α = By this notation we have diag(α1 , α2 , , αm ) + + + = diag(α+ , α2 , , αm ) Moreover, if M is a diagonal-block matrix then diag(M1 , M2 , , Mm ) + = diag(M1+ , M2+ , , Mτ+ ) 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 (Xn ) of (2.1) Suppose that θ is an ergodic transformation on (Ω, F , P ) and the following condition is satisfied Hypothesis 4.3 −1 ln P0 G−1 B0 ∈ L1 (Ω, F , P ) and ln P0 G0 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 Pn G−1 n Bn for n > −1 and Pn Gn An for n < 0, by [6] we get (i) Under the assumption (4.1), there exist the limits lim (Φ(n, ω) Φ(n, ω))1/2n =: ∆(ω) n→∞ (4.2) and lim (Φ(n, ω) Φ(n, ω))1/2|n| =: ∆(ω) n→−∞ (ii) Let < eλ1 < eλ2 < · · · < eλτ be the different nonzero eigenvalues of ∆ and λ0 = − ∞ We denote U0 = ker ∆(ω), and Ui , i = 1, 2, , τ the eigenspace with multipliers di = dim Ui corresponding to the eigenvalue eλi Then, τ ; di , λi , i = 1, 2, , τ are nonrandom constants Let Vk = U0 ⊕ U1 ⊕ · · · ⊕ Uk , k = 0, 1, 2, , τ such that {0} ⊂ V0 ⊂ V1 ⊂ · · · ⊂ Vτ = Rm June 26, 2007 4:2 WSPC/168-SD 00201 Degenerate Cocycle with Index-1 and Lyapunov Exponents 239 defines a filtration of Rm For each x ∈ Rm the Lyapunov exponent λ(ω, x) = lim n→∞ ln Φ(n, ω)x n (4.3) exists and λ(ω, x) = λk ⇐⇒ x ∈ (Vk \Vk−1 )(ω) for k = 1, 2, , τ Furthermore, λ(ω, x) = −∞ ⇔ x ∈ V0 and the spaces Vi are invariant in the sense Φ(n, ω)Vi (ω) ⊂ Vi (θn ω) (4.4) for any n ≥ 0; i = 0, 1, 2, , τ (iii) A similar result can be formulated for the limits with n → −∞ That is, let < eλτ < eλτ −1 < · · · < eλ1 be the different nonzero eigenvalues of ∆, λτ +1 = −∞ Denote U τ +1 = ker ∆(ω) and U τ , , U the corresponding eigenspaces with multipliers di = dim U i , i = 1, 2, , τ Then τ ; di , λi , i = 1, 2, , τ are nonrandom constants Let V k = U τ +1 ⊕ U τ ⊕ · · · ⊕ U k , k = τ + 1, τ , , 1, such that {0} ⊂ V τ +1 ⊂ V τ ⊂ · · · ⊂ V = Rm (4.5) defines another filtration of Rm For each x ∈ Rm the Lyapunov exponent λ(ω, x) = lim ln Φ(n, ω)x n→−∞ |n| exists and λ(ω, x) = λk ⇔ x ∈ (V k \ V k+1 )(ω) for k = 1, 2, , τ , and λ(ω, x) = −∞ ⇔ x ∈ V τ +1 Moreover, the spaces V i is invariant in the sense Φ(n, ω)V i (ω) ⊂ V i (θn ω), ∀n 0) of (Φ(n, ω) Φ(n, ω))1/2n converges to the ith one of ∆(ω)’s Moreover, for any n, from (3.3) we see that (Φ(n, ω) Φ(n, ω))x = if and only if π0 x = Hence, ker π0 = ker ∆ = ker ∆ which implies that V0 = V τ +1 Furthermore, πn Φ(n, ω) = Φ(n, ω)π0 (ω) Indeed, from (3.3) we have π0 (ω) = Φ(−n, θn ω)Φ(n, ω) which follows that Φ(n, ω)π0 (ω) = Φ(n, ω)Φ(−n, θn ω)Φ(n, ω) = Φ(n, θ−n θn ω)Φ(−n, θn ω)Φ(n, ω) = πn (ω)Φ(n, ω) June 26, 2007 4:2 WSPC/168-SD 240 00201 N H Du, T K Duy & V T Viet Therefore, by (4.4) we conclude that Φ(n, ω)π0 (ω)Vi (ω) = πn (ω)Φ(n, ω)Vi (ω) ⊂ πn (ω)Vi (θn ω) for P -almost ω ∈ Ω Further, dim Φ(n, ω)Vi (ω) = dim πn (ω)Vi (θn ω) which implies Φ(n, ω)π0 (ω)Vi (ω) = πn (ω)Vi (θn ω), ∀ n ≥ Hence, for i = 1, 2, , τ , Φ(n, ω)π0 (ω)(Vi \Vi−1 )(ω) = πn (ω)(Vi (θn ω)\Vi−1 (θn ω)) ∀ n ≥ 0, (4.7) λ(x, ω) = λi ⇔ x ∈ Vi (ω)\Vi−1 (ω), a.s ω For n < 0, from (4.7) we have Φ(−n, θn ω)π0 (θn (ω))(Vi (θn ω)\Vi−1 (θn ω)) = π0 (ω)(Vi (ω)\Vi−1 (ω)) Therefore, Φ(n, ω)π0 (ω)(Vi (ω)\Vi−1 (ω)) = Φ(n, ω)Φ(−n, θn ω)πn (ω)(Vi (θn ω)\Vi−1 (θn ω)) = πn (ω)(Vi (θn ω)\Vi−1 (θn ω)) Thus, (4.7) is true for n ∈ Z Similarly, Φ(n, ω)π0 (V i \V i+1 )(ω) = πn (V i (θn ω)\V i+1 (θn ω)) ∀ n ∈ Z, (4.8) λ(x, ω) = λi ⇔ x ∈ V i (ω)\V i+1 (ω) In the case where π0 is symmetric, by (3.3) and the relations Φ(−n, θn ω)Φ(n, ω)Φ(−n, θn ω) = Φ(−n, θn ω)πn = Φ(−n, θn ω), Φ(n, ω)Φ(−n, θn ω)Φ(n, ω) = Φ(n, ω)π0 = Φ(n, ω), it follows that Φ(−n, θn ω) = (Φ(n, ω))+ Therefore, by virtue of Lemma 4.2 we have Φ(−n, θn ω)Φ (−n, θn ω) = Φ (n, ω)Φ(n, ω) + Hence, by Lemma 4.1 we can use the same argument as in [1] to obtain τ = τ, λi = −λi , di = di , i = 1, 2, , τ (4.9) In fact, this property is valid without the symmetry of π0 Since π0 is a projector, we can find a random variable V (ω) with values in Gl(Rm ) such that V (ω)π0 V −1 (ω) = diag(I, 0) Put Ψ(n, ω) = V (θn ω)Φ(n, ω)V −1 (ω), n ∈ Z, ω ∈ Ω It is easy to see that Ψ(n + m, ω) = Ψ(n, θm (ω))Ψ(m, ω), Ψ(0, ω) = diag(I, 0) June 26, 2007 4:2 WSPC/168-SD 00201 Degenerate Cocycle with Index-1 and Lyapunov Exponents 241 This means that Ψ(n, ω) is a cocycle with symmetric projector diag(I, 0) Moreover, Ψ(−n, ω) = Ψ(n, θ−n ω)+ (4.10) However, the integrable conditions of this cocycle are not satisfied Therefore, we are unable to immediately deduce the conclusion (4.9) It notes that rank Φ(n, ω) =: d is a nonrandom constant Indeed, from the relations rank (M N ) ≤ min{rank M, rank N } and Φ(n, ω) = Φ(n − m, θm ω)Φ(m, ω), it follows that rank Φ(n, ω) ≤ rank Φ(m, ω) for all m, n ∈ Z Therefore rank Φ(n, ω) = rank Φ(0, ω) for all n ∈ Z On the other hand, Φ(0, ω) = Φ(1, θ−1 ω)Φ(−1, ω) which implies that rank Φ(0, ω) ≤ rank Φ(1, θ−1 ω) = rank Φ(0, θ−1 ω) Similarly, rank Φ(0, ω) ≤ rank Φ(0, θω) Hence, rank Φ(0, ω) = rank Φ(0, θω) This means that rank Φ(0, ω) is an invariant function By ergodicity of θ it follows that rank Φ(0, ω) is constant Remark 4.4 It is easy to see that dim Jf = dim Sn = rank An for all n > and dim Jb = dim S n = rank Bn for all n < However, there is no relation between rank Φ(0, ω) and rank An , rank Bn We denote by Λk the measurable function successively defined by Λ1 (ω) + · · · + Λk (ω) = lim n→∞ ln ∧k Φ(n, ω) , n k = 1, , d From [1] we get Λk (ω) = lim n→∞ ln δk (Φ(n, ω)), n k = 1, , d, where δk (Φ(n, ω)) are the nonzero singular values of Φ(n, ω), and Λ1 (ω) ≥ Λ2 (ω) ≥ · · · ≥ Λd (ω) We see that λ1 < · · · < λτ are the different numbers in the sequence Λ1 ≥ Λ2 ≥ · · · ≥ Λd (Λk = const in the ergodic case) and di is the frequence of appearance of λi in this sequence Similarly, λτ < · · · < λ1 are the different numbers in the − − sequence Λ− ≥ Λ2 ≥ · · · ≥ Λd , where Λ− k (ω) = lim n→∞ ln δk (Φ(−n, ω)), n k = 1, , d We have lim n→∞ ln Φ(n, ω) = Λ1 (ω) P -a.s n Therefore, due to P -preserving property of the transformation θ we get 1 ln V (θn ω) + ln Φ(n, ω) + ln V −1 (ω) −→ Λ1 (ω) in probability, n n n 1 − ln V −1 (θn ω) + ln Φ(n, ω) − ln V (ω) −→ Λ1 (ω) in probability n n n June 26, 2007 4:2 WSPC/168-SD 242 00201 N H Du, T K Duy & V T Viet Hence, by using the inequalities − ln V −1 (θn ω) + ln Φ(n, ω) − ln V (ω) ≤ ln Ψ(n, ω) ≤ ln V (θn ω) + ln Φ(n, ω) + ln V −1 (ω) , it follows that ln Ψ(n, ω) → Λ1 (ω) in probability as n → +∞ n Similarly, we can also prove that for k = 1, , d ln ∧k Ψ(n, ω) → Λ1 (ω) + · · · + Λk (ω) n Hence, ln δi (Ψ(n, ω)) −→ Λi (ω) n This relation implies that in probability as n → +∞ (4.11) (4.12) in probability as n → +∞, i = 1, , d ln δi (Ψ(n, θ−n ω)) −→ Λi (ω) in probability as n → +∞, i = 1, , d (4.13) n By the same argument as above we obtain ln δi (Ψ(−n, ω)) −→ Λ− i (ω) n From (4.10) we get in probability as n → +∞, i = 1, , d (4.14) δi (Ψ(−n, ω)) = 1/δd+1−i (Ψ(n, θ−n ω)) Using (4.13) and (4.14) we obtain Λ− i = −Λd+1−i In particular λi = −λi , di = di , τ = τ We now show that P {ω : V i (ω) ∩ Vi−1 (ω) = V0 (ω)} = for any i = 1, 2, , τ Suppose on the contrary, K1 := {ω : V i ∩ (Vi−1 (ω) \ V0 ) = ∅} then P {K1 } = β > (4.15) By definition of Vi−1 and V i we see that for n > the sets K2 (n) := {ω : Φ(n, ω)x ≤ exp((λi−1 + δ)n) x , ∀ x ∈ Vi−1 }, K3 (n) := {ω : Φ(−n, ω)y ≤ exp((−λi + δ)n) y , ∀ y ∈ V i }, (4.16) have the following properties P {K2 (n)} ≥ − β , if δ = 4−1 |λi − λi−1 | for n > N (β) P {K3 (n)} ≥ − β (4.17) June 26, 2007 4:2 WSPC/168-SD 00201 Degenerate Cocycle with Index-1 and Lyapunov Exponents 243 On the other hand, θ is P -preserving transformation then P {θ−n K3 (n)} = P {ω : Φ(−n, θn ω)y ≤ exp((−λi + δ)n) y β ∀y ∈ V i (θn ω)} = P {K3 (n)} ≥ − It is clear that for all n > N (β) β We now take ω ∈ K1 (n) ∩ K2 (n) ∩ θ−n K3 (n) (it is evident P {K1 (n) ∩ K2 (n) ∩ θ−n K3 (n)} > β2 ) If y ∈ (Vi−1 ∩ V i ) \ V0 ), then π0 y = By the definition of K2 (n) and θ−n K3 (n) we have P {K2 (n) ∩ θ−n K3 (n)} ≥ − π0 y = Φ(−n, θn ω)Φ(n, ω)y ≤ exp((λi−1 − λi + 2δ)n) π0 y , that is a contradiction because this relation is impossible Summing up we obtain Theorem 4.5 (MET for implicit linear difference equation) Suppose that ln P0 G−1 B0 , ln P0 G−1 A0 ∈ L1 Then, (i) λ1 , λ2 , , λτ and d1 , d2 , , dτ are nonrandom numbers (ii) For any i = 1, 2, , τ, the set Vi ∩ V i is invariant (iii) There exist subspaces W0 = V0 , W1 , , Wτ such that Rm = ⊕τi=0 Wi , Wi ⊕ V0 = Vi ∩ V i , dim Wi = di , i = 1, 2, , τ, (4.18) and for any x ∈ Wi \{0} 1 ln Φ(n, ω)x(ω) = λi = lim ln Xn (x(ω), ω) , lim n→±∞ n n→±∞ n where x = P−1 P0 x Proof Only the relation (4.18) is to be proved Assume that Wi ⊕V0 = Vi ∩V i By definition, dim Vi−1 + dim V i = m + dim V0 Therefore, dim Wi = dim Vi + dim V i − dim V0 − m = di Thus we have (4.18) Example 4.6 We give a simple example with A(·) and B(·) depending on a stationary process ξ(n) = ξ(θn ) Let A(ω) = A(ξ(ω)) and B(ω) = B(ξ(ω)) (there is an abuse of only notations used here), where,     0 −1 −1 A(+) := 0 1 , A(−) =  0 , 0 0 0     0 0 1 B(+) :=  −1 , B(−) := 0 −1 −1 0 June 26, 2007 4:2 WSPC/168-SD 244 00201 N H Du, T K Duy & V T Viet By calculations we obtain   0 Q0 (++) = −1 −1 , 1   0 −1  Q0 (−+) = 0 0 , 0  Q0 (+−) = −1  0  Q0 (−−) = 0 0  0 −1 ,  −1 0 Therefore, Jf (ω) = Similarly, expand{(0, 1, 0); (1, 0, 1)} if ξ(ω) = +, expand{(0, 1, 0); (1, 0, 0)} if ξ(ω) = −    0 Q0 (++) = −1 0 , −1 0   0 Q0 (+−) = 0 0 , 0 −1  Q0 (−+) = 1  1 Q0 (−−) = 0 0  −1 −1 1 , 1  0 Hence, Jb (ω) = expand{(0, 1, 0); (0, 0, 1)} if ξ(ω) = +, expand{(−1, 0, 1); (−1, 1, 0)} if ξ(ω) = − Thus, the solution Xn of Eq (2.1) exists on the whole Z if X0 ∈ expand{(0, 1, 0)} with ξ(ω) = + and X0 ∈ expand{(−1, 1, 0)} with ξ(ω) = − It is easy to see that the matrix Pn G−1 n Bn := Hθ n+1 ω,θ n ω,θ n−1 ω , where     0 0 0 H+++ = −2 1 , H++− = −1/2 −1/2 , 0 0 0  H−+−  0 = −1/2 −1/2 , 0 H+−+  H+−−  −1/2 = 0 0 , −1/2  H−++  H−−−  −1 = 0  , 0   −1/2 −1/2 = 0 1 , −1/2 −1/2 H−−+  0 = −2 1 , 0   −1 −1 = 0 1  0 June 26, 2007 4:2 WSPC/168-SD 00201 Degenerate Cocycle with Index-1 and Lyapunov Exponents 245 (n) Since rank (H) = 1, it is clear that for any n, the largest eigenvalue λ1 of Φ(n) Φ(n) equals to trace(Φ(n) Φ(n)) but it is difficult to compute trace(Φ(n) Φ(n)) Suppose that (ξ(n)) is an i.i.d sequence with P (ξ(n) = +) = 0.4, P (ξ(n) = −) = 0.6 A simulation program on the Maple software allows us to estimate the unique finite Lyapunov exponent of this system λ[π0 x] ≈ 0.098 for any x ∈ R3 such that π0 x = Discussion So far we are able to solve the initial value problem for implicit difference equations only in case where rankA(ω) and rankB(ω) are nonrandom Moreover, an index-1 tractable of the pencil matrices (A, B) is required It is interesting if we can free these assumptions or how can we define a higher index of (2.1) Acknowledgments The authors would like to extend their appreciations to the anonymous referee(s) for his very helpful suggestions which greatly improve this paper References L Arnold, Random Dynamical Systems (Springer-Verlag, 1998) L S Campbell and D Meyer, Generalized Inverses of Linear Transformations (Dover, 1991) P K Anh, N H Du and L C Loi, On linear implicit non-autonomous systems of difference equations, J Diff Eqns Appl (2002) 1085–1105 H Furstenberg and Y Kiffer, Random matrix products and measures on projective spaces, Isr J Math 46 (1983) 12–32 I Ya Goldsheid and G A Margulis, Lyapunov exponents of random matrices product, Usp Mat Nauk 44 (1989) 13–60 V M Gundlach and O Steinkamp, Product of random rectangular matrices, Math Nachr 212 (2000) 54–76 R Mă arz, On linear dierential algebraic equations and linearization, Appl Numer Math 18 (1995) A N Shiryaev, Probability, 2nd edn (Springer-Verlag, 1995)  ... such that {0} ⊂ V0 ⊂ V1 ⊂ · · · ⊂ Vτ = Rm June 26, 2007 4:2 WSPC/168-SD 00201 Degenerate Cocycle with Index-1 and Lyapunov Exponents 239 defines a filtration of Rm For each x ∈ Rm the Lyapunov. .. see that Ψ(n + m, ω) = Ψ(n, θm (ω))Ψ(m, ω), Ψ(0, ω) = diag(I, 0) June 26, 2007 4:2 WSPC/168-SD 00201 Degenerate Cocycle with Index-1 and Lyapunov Exponents 241 This means that Ψ(n, ω) is a cocycle. .. WSPC/168-SD 00201 Degenerate Cocycle with Index-1 and Lyapunov Exponents 231 out the condition of solvability for Cauchy problem and give an explicit formula of solutions In Sec we prove the cocycle property