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Let ẋ = A(t)x and λk(t) be the eigenvalues of the matrix A(t). The main result of the Method of Freezing states that if supJ ‖A(t)‖ ≤ M , supJ max1≤k≤nRe λk(t) ≤ ρ and supJ (‖A(t) −A(s)‖/|t − s|) ≤ δ, then xmax ≤ ρ + 2Mλδ, for the highest exponent xmax of the system, where λδ = ( Cnδ 4M2 ) 1 n+1 . The previous best known value Cn = n(n+1) 2 and the substantially smaller values of Cn are reduced to the still smaller value. 1. Let us consider an n-dimensional system ẋ = A(t)x, t ∈ J = [t0,∞), (1) with (real or complex) bounded and continuous on J matrix function A(t) and let λk(t), k = 1, . . . , n, be the eigenvalues of A(t). The Lyapunov exponent χ[x(t)] of a solution x(t) of (1) and the highest Lyapunov exponent χmax of the system (1) are given by [3]: χ[x(t)] = lim t→∞ log ‖x(t)‖ t , χmax = sup x 6=0 χ[x(t)] = max x 6=0 χ[x(t)]. (2) (Throughout this paper ‖ · ‖ denotes the Euclidean norm for a vector or matrix argument:

An improved estimate for the highest Lyapunov exponent in the method of freezing