Lyapunov exponents of the predator prey model Ljapunovovy ...

19We use substitution t = 2x − 1 and get the result2 log 2π 102dx = 2 log 2 [arcsin (2x − 1)] 11 − (2x − 1) 2 0π= 2 log 2Since both log |1 − 2x| and x (1 − x) are symmetric with respect to x = 1/2, we canwriteL (F 4 (x)) = 2 1/2log |1 − 2x| dx + 2 log 2.π x (1 − x)Now the task to solve is the following integral 1/200log |1 − 2x|x (1 − x)dx.The expression under the square root can be completed to the square, 1/20 1/2log |1 − 2x| dx =x (1 − x) 0log |1 − 2x|− 1 4 (1 − 2x)2 + 1 4Put 1 − 2x = sin θ, 0 ≤ θ ≤ π 2 . Then dx = − 1 2cos θ dθ and 1/2012log |1 − 2x|dx =1 − (1 − 2x) 2 π/2Hence the Lyapunov exponent of the Logistic map isL (F 4 (x)) = log 2.0dx.log sin θ dθ = − π log 2.2The second equality is proved in [6, page 273, Lemma 9.10.]. The lemma with proof isgiven for completeness.Lemma 3.8 π/2log sin θ dθ = − π log 2.20Proof. Consider an analytic function f (z) = log (1 − z) , |z| ≤ r < 1. Its real partu (z) = log |1 − z| is harmonic, and hence the Mean Value Theorem for harmonic functionsimplies that0 = u (0) = 12π 2π0u r e i θ dθ = 1 2π2π 0log 1 − r e i θ dθ.Letting r → 1, we have 0 = log 1 − r e i θ dθ. Since 1 − e i θ = 2 sin θ 2, 0 ≤ θ ≤ 2π, wehave−2π log 2 = 2π0log sin θ 2 dθ.