Solutions to Assignment 2

y
0.4
y=sin(4x)/4
0.3
0.2
0.1
0
-0.1
-0.2
-0.3
-0.4
-1
-0.5
0
x
0.5
1
y
0.2
0.15
0.1
0.05
0
y=sin(4x)/4
-0.05
-0.3
-0.2
-0.1
0
x
0.1
0.2
0.3
0.4
Figure 1: 1. Cobwebbing of f (x) = sin(4x)/4 : showing 200 iterations for orbits starting close to
x = 0 with initial conditions x0 = 0.3 and x0 = −0.05. The bottom graph is a zoom-in of the graph
on the top. The orbits seem to be approaching 0, but very slowly, and even after 200 iterations one
cannot be sure that they are actually converging to 0, without a proof! See the proof that in fact
they are converging to 0 on the next page.
y
2
2
f(x)=-1.05x+x
1.5
3
3
1
1
0.5
0.5
0
0
-0.5
-0.5
-1
-1
-1.5
-1.5
-2
-2
-1.5
-1
f(x)=-0.95x+x
1.5
-0.5
0
x
0.5
1
1.5
y
0.6
2
-2
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
y
0.6
3
f(x)=0.9x+x
3
f(x)=1.1x+x
0.4
0.4
0.2
0.2
0
0
-0.2
-0.2
-0.4
-0.4
-0.6
-0.6
-0.6
-0.4
-0.2
x
0
0.2
0.4
-0.4
-0.2
0
x
0.2
0.4
Figure 2: 3(a). Graphs supporting stability conclusions. f (x) = µx + x3 . (TOP-LEFT) For µ = −1.05
(2)
(2)
the cycle 2, x+ , x− is asymptotically stable. The fixed points: x̄ = 0, x− and x+ are repelling.
(TOP-RIGHT) For µ = −.95 the fixed point x̄ = 0 is asymptotically stable. The fixed points x− and x+
are repelling. (BOTTOM-LEFT) For µ = 0.9 the fixed point x̄ = 0 is asymptotically stable. The fixed
points x− and x+ are repelling. (BOTTOM-RIGHT) For µ = 1.1 the fixed point x̄ = 0 is repelling.
y
2
First and second iterates of Gμ(ξ) = μξ + ξ
1.5
3
1
0.5
0
-0.5
-1
There are 6 points of prime period 2 for μ < −2 !
-1.5
-2
-2
-1.5
-1
-0.5
0
x
0.5
1
1.5
2
y
2
3
G-2.2(x) = -2.2 x + x
1.5
1
0.5
0
-0.5
-1
-1.5
-2
There are 2 more 2 cycles for μ < −2 !
-2
-1.5
-1
-0.5
0
x
0.5
1
1.5
2
Figure 3: 3(b). Graphs demonstrating that the original 2-cycle created at the period doubling
bifurcation at µ = −1 undergoes a secondary bifurcation at µ = −2, a pitchfork bifurcation. (TOP )
At µ = −2.2 there are 4 additional periodic points of prime period 2. (BOTTOM) The two attracting
2-cycles created as µ decreases through -2. These are two distinct 2-cycles, NOT a 4-cycle. To
which of these 2-cycles an orbit converges, depends on the initial condition x0 . This is an example
of bi-stability: two co-existing attractors.