Consider the equation of the quadratic map. x n+1 = a ( x n ) 2. The graph for 0 < a < 2 looks like:

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1 The mathematics behind the quadratic map mirrors that of the system of semiconductor lasers. It is a basic example of a chaotic system that can be synchronized through coupling [4]. By definition, a map f is a generated sequence x n using a recursion x n+1 = f( x n ). f : R m R m. Consider the equation of the quadratic map x n+1 = a ( x n ) The graph for 0 < a < looks like: In the above figure we notice that there are areas on the graph where it splits. This is called bifurcation and it occurs at certain fixed points. Our fixed points are x n such that f( x n ) = x n. Analysis of fixed points helps describe the graph. In the neighborhood around one of our fixed points, if the map is iterated, the solution will act in 1 of ways. It will be attracted to the fixed point or repelled from it. (Attraction to the fixed point) (Repulsion from the fixed point)

2 In the case of the quadratic map, there is repulsion and attraction. If there is attraction to the fixed point, the fixed point is stable. If there is repulsion from the fixed point, the fixed point is unstable. In the case of the laser systems, the stable fixed points identify where laser activity doesn t change. In order to get a clear picture of what goes on in the quadratic map, the fixed points must be identified and their stability analyzed. Linearization is used. If x is a fixed point, so and x = a x 0 = x + x - a x ± = -1+ (1+4a) 1\ Are these fixed points stable? That is, is there attraction to the fixed points in the neighborhoods around them? Let x n = x ± + δ n, where δ is a small distance. Then x n+1 becomes x n+1 = a x n = a (x + δ n ) = a (x + δ n x + δ n ) = a x - δ n x - δ n Because x n+1 = x + δ n+1 and x = a - x, x + δ n+1 = x - δ n x - δ n Because δ n is a small length and δ n is even smaller, they are disregarded. Dropping higher ordered terms is called linearization. So, x + δ n+1 = x - δ n x - δ n δ n+1 = - δ n x

3 This is the equation that determines the stability of x. If δ n+1 < δ n, then there is attraction towards x and x is stable. If δ n+1 > δ n, then there is repulsion away from x and x is unstable. Since a, in our equation x = a - x, can be any constant, the value of a is crucial in determining the stability of x. Solving for a using our fixed points is necessary. x = -1± (1+4a) 1\ δ n+1 = δ n -1 - (1+4a) 1\ If the multiplier -1 - (1+4a) 1\ < 1 then x is stable. Solving for a, Since 0 < a <, x is stable for a < ¾. Since there is no a > 0 that satisfies -1 ± (1+4a) 1\ < 1 -¼ < a < ¾. x - =-1 - (1+4a) 1\ < 1 It is always unstable; and thus the only stable fixed point is x + = -1+ (1+4a) 1\ for a < ¾. The preceding figure shows that for a > ¾ the values of the map settle down to the period two behavior. Its values alternate. This means x n+ = f(f(x n )) = x n. Now we need to find fixed points for the twice iterated map. To do this, the relationship between x n+ and x n must be analyzed. Plug in (a x ) into the quadratic equation for x. This yields x = a (a - ax + x 4 ) 0 = a a +ax -x 4 -x We already know roots for this equation so we can divide

4 x + x a x 4 + 0x 3 x + ax + x + ( a 1) x a + a So to find the other two roots the quadratic formula is used on a - x + x - 1 = 0. This yields x ± = -1 + (4a - 3) 1\ Note, a > ¾ in order for x to be real. Which fixed points of the second iteration are stable? The same procedure, linearization, is used here. x n+ = a (a - ax n + x n 4 ) Since x = x + δ n, x n+ = x ± + δ n+ = a (a - ax n + x n 4 ) x ± + δ n+ = a a + a (x + δ n ) (x + δ n ) 4 =a-a -a(x +δ n x+δ n )+x 4 +4x 3 δn+6x δ n +4xδ n 3 +δ n 4 Everything multiplied by a δ n of degree higher than one can be neglected because of its small size. Thus, x + δ n+ = a-a -ax -4aδ n x+x 4 +4x 3 δ n Since x ± are fixed points f(f(x ± )) = x ±, we have x = a - a - ax + x 4 Yielding, δ n+ = -4aδ n x + 4x 3 δ n = (-4ax + 4x 3 )δ n The map is stable at x = x n+1 if -4ax + 4x 3 < 1 This is true iff ¾ < a < 1¼

5 The process of finding roots and determining stability can continue. However, to calculate more fixed points requires a computer. For each iteration n, the map has n fixed points. The system of the quadratic map is chaotic because it has the following characteristics. It is nonlinear. x n+1 = a ( x n ) can not be written in the form y = mx + b. It is deterministic. There is an equation that determines the behavior of the system. A very slight change value of x 0 can lead to significantly different behavior of the map The matlab code used to generate the map is as follows: % produce a plot of the set of values that the iteration settles % down to, showing 1, 4, 8 etc solutions as the value of a is varied % Invoke by typing gquad n0=5000; %Number of times to iterate to get past transient behavior n1=64; % number of values to plot for each a as=.01:.005:; % the range of values for a. A will vary from %.01 to in steps of size.005 plot(as,as,'linestyle','none'); %This is a dummy call to set % up a figure that points will be added to. xlim([0,]) % set the x limits of the plot

6 ylim([-.5,.5]); % set the y limits for a=as % loop over the values of a x=0; % initial value of x for j=1:n0 x=a-x^; % iterate the map n0 times, transient for j=1:n1 x=a-x^; line(a,x,'linestyle','none','marker','.','markersize',.01); % the line command here puts a dot at the a,x position in the plot % this is a bit more complicated than necessary, but runs faster % than the easier way (using hold on; plot(a,x,'.')) The quadratic map can be synchronized through coupling. We can couple two quadratic maps with a coupling coefficient δ. Each map uses as its input a weighted average of its own variable and the other variable. x n+1 = a - ((1 δ) x n + δ y n ) y n+1 = a - (δx n + (1 δ) y n ) We want to determine when x n y n 0. This is when the maps synchronize. If we let S n be the equation for the map and U n determine the amount of synchronization we have S n = x n + y n and U n = y n - x n

7 When δ = 0, if a is in the chaotic region then by definition a small U 0 will grow. On the other hand if δ =.5 and U n = 0, with n 1 we have synchronization. As we vary δ from 0 to.5,we must go from unsynchronization to synchronization.

8 Notice as δ goes from 0 to.5, U n = x-y goes to 0. Also as δ gets larger a has to be larger for the maps to synchronize. The matlab code used to generate the coupled map is function gcquad(d) % this is a function you can invoke by typing % gcquad(.) or gcquad(.4) etc. % the d value is the parameter that controls the amount of coupling of the % two quadratic maps % the program will produce a plot showing the behavior of y- x using the % value of d specified as the value of a varies from.01 to n0=5000; n1=64; as=.01:.005:; plot(as,as,'linestyle','none'); xlim([0,]) ylim([-3,3]); for a=as x=0; % initial value for x y=1; % initial value for y c=1-d; % since c+d=1, c=1-d for j=1:n0 % iterate both maps (with coupling) n0 times x=a-(c*x+d*y)^; y=a-(c*y+d*x)^; for j=1:n1 % iterate both maps n1 times, and plot the values of y-x x=a-(c*x+d*y)^;

9 y=a-(c*y+d*x)^; line(a,yx,'linestyle','none','marker','.','markersize',.01); We can use linearization to analyze the behavior of the coupled maps. Synchronization occurs when U n = 0. Because this is a chaotic system, if x 0 doesn t equal y 0 U n becomes large in comparison to 1. If x 0 = y 0 then U n will begin at and stay at 0. If δ = 0 and a is large there will never be coupling. Subbing in S n and U n into the coupled map, x n+1 = S n - U n y n+1 = S n + U n So the coupled map becomes S n+1 - U n+1 = a - (S n + (1 δ) U n ) S n+1 + U n+1 = a (S n -(1 δ) U n ) U n is assumed small but S n is not; we drop the terms containing U n,etc. that is, we linearize. S n+1 = a- S n Subtracting these equations and applying linearization yields U n+1 = ( δ-1)s n U n S n+1 = a- S n is the system of the coupled maps and U n+1 = ( δ-1)s n U n describes the difference between the maps. Applying this formula repeatedly we get: U n = ( δ-1)s n-1 U n-1 = ( δ-1)s n-1 ( δ-1)s n- U n- = ( δ-1)s n-1 ( δ-1)s n- ( δ-1)s n-3 U n-3 Eventually we obtain U n = (( δ-1)) n [S n-1 S n- S n-3 S 0 ]U 0

10 It is interesting to note that (( δ-1)) n is depent on δ and [S n-1 S n- S n-3 S 0 ] is depent on a. To see whether or not there is synchronization, the relationship between U n and U 0 is analyzed. If (( δ-1)) n [S n-1 S n- S n-3 S 0 ] < 0 the system will synchronize. If (( δ-1)) n [S n-1 S n- S n-3 S 0 ] >0 the system will not. can be rewritten as U n = (( δ-1)) n [S n-1 S n- S n-3 S 0 ]U 0 U n = [(( δ-1))[s n-1 S n- S n-3 S 0 ] 1/n ] n U 0 The multiplier [(( δ-1))[s n-1 S n- S n-3 S 0 ] 1/n ] n has two parts. ( δ-1) may shrink U = x-y; and [[S n-1 S n- S n-3 S 0 ] 1/n ] n causes chaos. This is because δ pairs the two maps and as δ grows to its limit the maps synchronize. Looking at the special case δ = 0 uncoupled maps we see that the factor is -[S n-1 S n- S n-3 S 0 ] 1/n and so if its absolute value is > 1, x-y grows, i.e. it is chaotic. This multiplier measures the rate that the small variation in initial conditions grows. It is the chaotic-ness when δ >0, that we can balance. It balances exactly when [(( δ-1))[s n-1 S n- S n-3 S 0 ] 1/n ] n = 1. This tells us what δ critical is. At this time it is not clear how [S n-1 S n- S n-3 S 0 ] 1/n ] n deps on a. Future Plans In further research, I would like to pursue the following: discovering how [S n-1 S n- S n- 3 S 0 ] 1/n ] n deps on a, applying the analysis used to study the quadratic map to the laser system, learning more about the physics behind lasers, and understanding the mathematics that model their behavior. The patience that Dr. Indik and Josh Soneson possess is unparalleled. I am very grateful for all that they have taught me.