By Mustafa R.S. Kulenovic, G. Ladas
This self-contained monograph presents systematic, instructive research of second-order rational distinction equations. each one bankruptcy features a component of open difficulties and conjectures, lots of them perfect for examine tasks or Ph.D. theses. The suggestions and leads to this monograph are tremendous valuable in interpreting the equations within the mathematical types of assorted organic structures and different functions. transparent, uncomplicated, and direct exposition mixed with considerate uniformity within the presentation make the publication beneficial as a textual content as a reference for researchers, and as a complement to each textbook on distinction equations.
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Additional resources for Dynamics of Second Order Rational Difference Equations: With Open Problems and Conjectures
42 Case 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Chapter 2. 9. 9 43 Open Problems and Conjectures What is it that makes all solutions of a nonlinear difference equation periodic with the same period? For a linear equation, every solution is periodic with period p ≥ 2, if and only if every root of the characteristic equation is a pth root of unity. 1 Assume that f ∈ C 1 [(0, ∞) × (0, ∞), (0, ∞)] is such that every positive solution of the equation xn+1 = f (xn , xn−1 ), n = 0, 1, . . 37) is periodic with period p ≥ 2.
30) holds, the two cycle is . . , φ, γ−A γ−A − φ, φ, − φ, . . C C with 0 ≤ φ < γ−A . 31) holds, the values φ and ψ of the two cycle are given by the two roots of the quadratic equation t2 − γ−β−A αC + β(γ − β − A) t+ =0 C C(B − C) and we should also require that the discriminant of this quadratic equation be positive. That is, (γ − β − A)[B(γ − β − A) − C(γ + 3β − A)] α< . 6 Local Asymptotic Stability of a Two Cycle Let . . , φ, ψ, φ, ψ, . . 1). Set un = xn−1 and vn = xn for n = 0, 1, . . 6.
We will assume that xN −1 < x ≤ xN . All other cases are similar and will be omitted. Then by using the monotonic character of f (x, y) we have xN +1 = f (xN , xN −1 ) < f (x, x) = x and xN +2 = f (xN +1 , xN ) > f (x, x) = x. Thus xN +1 < x < xN +2 26 Chapter 1. Preliminary Results and the proof follows by induction. ✷ The next result applies when the function f is decreasing in both arguments. 2 Assume that f ∈ C[(0, ∞) × (0, ∞), (0, ∞)] and that f (x, y) is decreasing in both arguments. 33).
Dynamics of Second Order Rational Difference Equations: With Open Problems and Conjectures by Mustafa R.S. Kulenovic, G. Ladas