Lyapunov exponents of the predator prey model Ljapunovovy ...

21The following example uses the alternative definition to compute the Lyapunov exponentof the Tent map.Example 3.10 (Tent map)Let T be a tent map, defined byT (x) =2x x ∈ 0, 1 2,−2x + 2 x ∈ 12 , 1The Lyapunov exponent of T can be computed asn−11 L (T ) = lim log T ′ T k (x).n→∞ nk=0for some initial x.Since |T ′ (x)| = 2 for any x ∈ (0, 1/2) ∪ (1/2, 1) it is log |T ′ T i (x) | = log 2. Hence theLyapunov exponent of T is equal to the mean value of log 2.L (T ) = log 2.This is in agreement with Theorem 3.9 because the Tent map is topologically conjugateto the Logistic map by conjugacy h (x) = sin 2 π2 x which is piecewise differentiable on[0, 1].The geometrical representation of Lyapunov exponent is the divergence speed of twoneighboring points. The following definition gives us the way to numerically approximatethe value of the Lyapunov exponent by computing the distances of orbits of twoclose points.Definition 3.11 [6, page 276] Let x ∈ a, b − 10 −k , a, b ∈ R and ˜x = x + 10 −k . DefineH n,k (x) = 1 n log |f n (x) − f n (˜x)| + k n .Theorem 3.12 Suppose f is an ergodic and piecewise continuously differentiable map on X =[a, b]. For large enough k, n is H n,k (x) close to the Lyapunov exponent of f.Proof. Let x ∈ a, b − 10 −k , a, b ∈ R and ˜x = x + 10 −k as in Definition 3.11The following inequality is equivalent to the definition of derivation of f(∀ε > 0) (∃k 0 > 0) (∀k ≥ k 0 ) :|f (x) − f (˜x)||x − ˜x|It is easy to see that the following statement holds.(∀n ∈ N) (∀ε > 0) (∃k n > 0) (∀k ≥ k n ) :|f n (x) − f n (˜x)||x − ˜x|−− f ′ (x) ≤ ε 2 .n−1i=0f ′ f i x ≤ ε 2 .