Applying Snapback Repellers in Ecology

Transcription

1 Applying Snapback Repellers in Ecology Shu-Ming Chang 張書銘 Department of Applied Mathematics National Chiao Tung University December 11, 2010

2 Outline

3 Model Generalized Resource Budget Model A Satake and Y Iwasa, Pollen Coupling of Forest Trees: Forming Synchronized and Periodic Reproduction out of Chaos, J theor Biol, Vol 203 (2000), Resource Budget Model Y Isagi, K Sugimura, A Sumida and H Ito, How does masting happen and synchronize?, J theor Biol, Vol 187 (1997),

4 Model Isagi s Resource Budget Model S (t) : Amount of energy at the beginning of year t P s : Photosynthate (constant from year to year) L T : Resource threshold C f : Flowering energy C a : Fruiting energy R c: The ratio of C a/c f

5 Model Satake s Generalized Resource Budget Model Main problem Y (t+1) = { Y (t) + 1, if Y (t) 0, ky (t) + 1, if Y (t) > 0, where k: depletion coefficient, t = 0, 1, 2,

6 Model Non-dimensionalization S (t) : energy reserved at the t-th year Isagi s Resource Budget Model { S (t+1) S (t) + P = s, if S (t) + P s L T, S (t) + P s C f C a, if S (t) + P s > L T Let a > 0, C f a(s (t) + P s L T ) Satake s Generalized Resource Budget Model { S (t+1) S (t) + P = s, S (t) + P s a(r c + 1)(S (t) + P s L T )

7 Model Non-dimensionalization S (t+1) = = = { S (t) + P s, S (t) + P s a(r c + 1)(S (t) + P s L T ) { S (t) + P s, S (t) + P s L T a(r c + 1)(S (t) + P s L T )+L T { S (t) + P s, (1 a(r c + 1)) (S (t) + P s L T ) + L T

8 Model Non-dimensionalization S (t+1) +P s L T = (S (t) + P s L T )+P s, (1 a(r c + 1)) (S (t) + P s L T ) +L T L T +P s, { ) (S (S (t) + P (t+1) s L T )/P s + 1, + P s L T /P s = (1 a(r c + 1)) (S (t) + P s L T )/P s + 1 Let Y (t) = (S (t) + P s L T )/P s, k a(r c + 1) 1

9 Model Coupled Systems where Y (t+1) i = { Y (t) i + 1 if Y (t) i 0, κp (t) i Y (t) i + 1 if Y (t) i > 0, P (t) 1 i = N 1 N j i [Y (t) j ] + β

10 Model Satake and Iwasa proved by computing the positive Lyapunov exponent that if the depletion coefficient k is greater than one, then the generalized budget resource model is chaotic However, a positive Lyapunov exponent means only sensitivity in Devaney s chaos When the depletion coefficient k is a positive integer, Satake and Iwasa proved that the generalized budget resource model is periodic 1 Snapback repeller method Devaney s chaos 2 Investigate the difference between odd depletion coefficients and even depletion coefficients

11 Technique Snapback Repeller F R Marotto, Snap-Back Repellers Imply Chaos in R n, Math Anal Appl, Vol 63 (1978), expanding fixed point Let p R n, suppose f : R n R n be diff in B r (p ), if f(p ) = p and σ(df(x)) > 1 x B r (p ) p : snapback repeller Suppose p is an expanding fixed point of f in B r (p ) for some r > 0, if there exists a point x 0 B r (p ) with x 0 p and some positive integer m such that f m (x 0 ) = p and det(df m (x 0 )) 0

12 Technique

13 Technique Let snapback repeller p, f, m, and x 0 be the same as above If f is C 1 in some neighborhood of x j, det(df(x j )) 0, 0 j m 1, and f has a snapback repeller p, then f is chaotic in the sense of Devaney Y Shi and G Chen Chaos of discrete dynamical systems in complete metric spaces Chaos, Solitons and Fractals, 22: , 2004 Y Shi and G Chen Discrete chaos in Banach spaces Science in China Ser A Mathematics, 48(2): , 2005 Y Shi, P Yu, and G Chen Chaotification of discrete dynamical systems in Banach spaces Internat J Bifur Chaos, 16: , 2006 Chen-Chang Peng Numerical Computation of Orbits and Rigorous Verification of Existence of Snapback Repellers Chaos, 17:013107, 2007 M C Li, M J Lyu and P Zgliczyński Topological entropy for multidimensional perturbations of snap-back repellers and one-dimensional maps Nonlinearity, 21: , 2008 Y Shi and P Yu Chaos induced by regular snap-back repellers J Math Anal Appl, 337: , 2008 M C Li and M J Lyu A simple proof for persistence of snap-back repellers Journal of Mathematical Analysis and Applications, 352: , 2009 Z Li, Y Shi, and W Liang Discrete chaos induced by heteroclinic cycles connecting repellers in Banach spaces Nonlinear Analysis, 72(2): , 2010

14 Chaos Devaney s Chaos Let X be a metric space, f : X X conti sensitivity δ > 0 st x X, any nbd(x), y nbd(x) and n N such that f n (x) f n (y) > δ; transitivity for any pair of nonempty open sets U, V X, k > 0 st f k (U) V ; density of periodic points

15 Chaos Li & Yorke s Chaos Let I be an interval, f : I I conti, if f has an uncountable scrambled set S I which satisfies the following conditions: (i) p, q S with p q, lim sup f n (p) f n (q) > 0, n lim inf f n (p) f n (q) = 0; n (ii) p S and periodic point q I, lim sup f n (p) f n (q) > 0 n

16 Entropy Topological Entropy f : X X conti with metric d S X: (n, ϵ)-separated for f with n Z + and ϵ > 0 provided that for every pair of distinct points x, y S, x y, there is at least one k with 0 k < n st d(f k (x), f k (y)) > ϵ The number of different orbits of length n (as measured by ϵ) is defined by r(n, ϵ, f) = {#(S) : S X is a (n, ϵ)-separated set for f }, where #(S) is the cardinality of elements in S Let h top (ϵ, f) = lim sup n log(r(n, ϵ, f)), n and define the topological entropy of f as h top (f) = lim h top(ϵ, f) ϵ 0,ϵ>0

17 Entropy Theorem Assume f : X X is uniformly continuous or X is compact, and k is an integer with k 1 Then h top (f k ) = k h top (f) C Robinson Dynamical Systems: Stability, Symbolic Dynamics, and Chaos, 2nd Ed, CRC, Boca Raton, Florida, 1998

18 Lyapunov Exponent Let f : R R be a C 1 function For each point x 0, define the Lyapunov exponent of x 0, λ(x 0 ), as follows: where x j = f j (x 0 ) λ(x 0 ) = lim sup n = lim sup n 1 n log( (f n ) (x 0 ) ) 1 n 1 n k=0 log( f (x k ) ),

19 Important relations

20 Y (t+1) = { Y (t) + 1, if Y (t) 0, ky (t) + 1, if Y (t) > 0 k 1 Y (t) tends to infinity 1 < k < 1 Y (t) converges to the stable equilibrium 1 k + 1 k = 1 Y (t) converges to a period 2

21 Main results ( ) 1/3 ( ) 1/3 1 (I) 23 k > k the system is chaotic in Devaney s sense (II) 1 < k k the system is chaotic in Devaney s sense, too

22 Y n (x) = (Y Y)(x) p N {0} L 2 p(x), x Y 2p (x) = R 2 p(x), x L 1 (x) = x + 1, R 1 (x) = kx + 1, [ ( ) ( )] 1 1 C p 3, C p 2, k k [ ( ) ] 1 C p 2, 1, k R 2 p(x) = (L 2 p 1 R 2 p 1)(x), kr 2 p(x) + k + 1, p odd, L 2 p(x) = R 2 p(x) + k + 1, k p even

23 j N, C j 1 A A C j 1, C j = C j 1 B C j 1, j odd, j even with C 0 (x) = B(x), C 1 (x) = x, C 2 = 0, C 3 = k + 1, where A(x) = 1 (1 x), B(x) k = 1 (2 x) k

24 Main Result (I) p = 0, 1 k 0 = x + 1, x [ k + 1, 0], Y(x) = kx + 1, x [0, 1] ; k [ k 2 x k + 1, x 0, 1 ], k Y 2 (x) = [ ] 1 kx + 2, x k, 1

25 Main Result (I) Proof: Part I: k > ( ) 1/3 ( ) 1/ Let p = k, g = Y 1 y=1 ( g^2(p^*), g^2(p^*) ) ( g(p^*), g(p^*) ) (p^*, p^*) ( g^2(p^*), g(p^*) ) x = -k+1

26 Main Result (I) Since g (p ) < 1, there exists r > 0 with U = (p r, p + r), U (0, 1) such that lim m gm (x) = p if x U Choose g(p ) = k 1 + k < 0 and g2 (p ) = 2k + 1 k 2 + k > 0 Solve g(p ) > k + 1 and g 2 (p ) < 1, choosing k > allows j to 2 be found such that g j (p ) > 0 for all j 3 by the definition of Y Computing g j (p ) p, yield g j (p ) p = 1 0 as j That is, for this r, there exists a j 1 k natural number J > 0 such that g j (p ) U as j J Fix J and let x 0 = g J (p ), then x 0 U and Y J (x 0 ) = p Since Y (p) = k > 1 for all p U, and (Y J ) (x 0 ) 0, p is a snapback repeller of Y

27 Main Result (I) Let p = k, h = (Y2 ) 1 y=1 ( h^2(p^**), h^2(p^**) ) ( p^**, p^** ) ( h^2(p^**), h(p^**) ) ( h(p^**), h(p^**) ) x = -k+1

28 Main Result (I) Since h (p ) < 1, there exists r > 0 with V = (p r, p + r), V ( 1, 1) such that k lim m hm (x) = p if x V Choose h(p ) < 1 k and h 2 (p ) > 1 k Solve h(p ) > k + 2 and h 2 (p ) < 1, choosing ( ) 1/3 ( ) 1/ k > allows j to be found such that 108 h j (p ) > 1 k for all j 3 by the definition of Y 2

29 Main Result (I) Computing g j (p ) p, yield h j (p ) p = k 1 0 as j j+1 k That is, for this r, there exists a natural number J > 0 such that h j (p ) V as j J Fix this J, let y 0 = h J (p ), then y 0 V and (Y 2 ) J (y 0 ) = p Since [ (Y 2 ) (p) = k > 1 for all p V, and (Y 2 ) J ] (y0 ) 0, p is a snapback repeller of Y 2 Y 2 is chaotic in the Devaney sense h top (Y 2 ) > 0 h top (Y 2 ) = 2 h top (Y) > 0 h top (Y) > 0 h top (Y) > 0 Y is chaotic in Devaney s sense as ( ) 1/3 ( ) 1/ k >

30 Main Result (II) p = 2 Y 22 (x) = k 2 x 2k + 2, k 3 x + 2k 2 k + 1, [ ( )] 1 1 x k, B, k [ ( ) ] 1 x B, 1 k

31 Main Result (II) p = 3 Y 23 (x) = k 6 x 2k 5 + k 4 k 3 + 2k 2 k + 1, ( ) 1 x B k, BAAB }{{} C 1 k 5 x + 2k 4 k 3 + k 2 2k + 2, [ ( ) ] 1 x C 1, 1 k ( ) 1, k

32 Numerical Computation MatLab p k p

33 Numerical Computation Maple p k p

34 Y(t+1) = ky(t) + 1, Y(t) > Y (t) Depletion coefficient k

35 Y (t+1) = ky (t) + 1, Y (t) > 0 k =2 1 1 k = Y (t) 05 Y (t) t t k =4 1 k = Y (t) 15 Y (t) t t

36 What happen in Y (t)? For any initial value x Q k N Y (t) (x) is periodic eventually

37 What happen in Y (t)? Binary representation with finite digits k N \ {1} k even Y (t) always converges to the periodic cycle S { k + 1, k + 2,, 0, 1} with period k + 1 for any initial value k odd Y (t) can not converge to the periodic cycle S { k + 1, k + 2,, 0, 1} as the initial value x / S

38 Thank you for your attention!