# periodic point of a quadratic map

I am analyzing a discrete dynamic quadratic map. The bifurcation diagram shows the existence of an asymptotic 3- cycle periodic point for a given parameter. But the graphic of the lyapunov exponent gives a positive coefficient.
My map quad[{x,y}] is {x + K x (2. - 0.76 x - 0.4 y), y + K (2. - 0.6 x - 0.5 y) y}. I am working on 3-periodic cycle. The bifurcation diagram shoes that there an asymptotic convergent 3-cycle period at the value of the parameter K=K = 1.417. When I take the map quad3 and analyze fixed point by writing quad3[{x_, y_}] = Simplify[quad[quad[quad[{x, y}]]]]; and then : FindRoot[quad3[{x, y}] == {x, y}, {{x, 1.85}, {y, 2.96}}]. I obtain {x -> 1.85234, y -> 2.96374} hence x=1.8523360794543522 and y=2.9637377271382412. When writing x = 1.8523360794543522; y = 2.9637377271382412; and quad3[{x, y}] I obtain x=1.8523360794681456 and y=2.9637377271328504. there is small difference. How to be sure that the fixed point of quad3 exits or equivalently a 3-period cycle exists ?

• What have you tried? What is your code? And can you try to give a specific question, too. – aardvark2012 Nov 26 '17 at 10:22
• My map f[{x,y}] is {x + K x (2. - 0.76 x - 0.4 y), y + K (2. - 0.6 x - 0.5 y) y}. I am working on 3-periodic cycle. The bifurcation diagram shoes that there an asymptotic convergent 3-cycle period at the value of the parameter K=K = 1.417. When I take the map f3 and analyze fixed point by – khaled Nov 26 '17 at 12:24
• You could try increasing the WorkingPrecision of FindRoot. See also AccuracyGoal and PrecisionGoal. – aardvark2012 Nov 26 '17 at 12:50
• How'd you get the bifurcation diagram and Lyapunov exponents? – Chris K Nov 26 '17 at 15:25

I don't think the 3-cycle you found at K=1.417 is stable. Here are two lines of evidence:

First, simulate the map (warmup for 100 steps, then plot the next 20) and see:

quad[{x_, y_}] := {x + K x (2. - 0.76 x - 0.4 y), y + K (2. - 0.6 x - 0.5 y) y};
K=1.417; Doesn't seem to be settling down to a cycle.

Second, find the eigenvalues of the Jacobian matrix of the 3-step map quad3:

quad3[{x_, y_}] = Simplify[quad[quad[quad[{x, y}]]]];
j3 := D[quad3[{x, y}], {{x, y}, 1}]
eq3 = FindRoot[quad3[{x, y}] == {x, y}, {{x, 1.85}, {y, 2.96}}]
(* {x -> 1.85234, y -> 2.96374} *)
Eigenvalues[j3 /. eq3]
(* {-1.42176, -0.280642} *)


The eigenvalue of -1.42176 indicates instability of the 3-cycle eq3.

One last idea: simulate starting near the unstable 3-cycle to watch the divergence.

ListLinePlot[Transpose[NestList[quad, {x + 10^-4, y} /. eq3, 100]]] 