Math 5110/6830Homework 4.1 Solutions

1. (a) A fixed point satisfies the equation

x∗ = r2x∗[1− (r + 1)x∗ + 2rx∗2 − rx∗3].

x∗ = 0 is immediately a solution. Now, notice that r−1r is a fixed point of the original logistic

equation, f(x). So, it satisfies x∗ = f(x∗). Since f2(x) = f(f(x)), and a fixed point off2(x) satisfies x∗ = f(f(x∗)), it follows that if x∗ is a fixed point of the original equation,f(f(x∗)) = f(x∗) = x∗, and so x∗ is a fixed point of the second-iterate equation. Thus,x∗ = r−1

r is a fixed point of f2(x) = r2x[1− (r + 1)x+ 2rx2 − rx3].As a fixed point, r−1

r must be root of the following equation:

−r2[1− (r + 1)x+ 2rx2 − rx3] + 1 = 0,

which implies that x− r−1r must be a factor of the left-hand side (LHS). We can therefore

divide the LHS by x− r−1r to determine the other two roots of the equation.

r3x2 − r2(r + 1)x+ r(r + 1)

x− r−1r

)r3x3 − 2r3x2 + r2(r + 1)x+ (1− r2)r3x3 − r2(r − 1)x2

− r2(r + 1)x2 + r2(r + 1)x

− r2(r + 1)x2 + r(r2 − 1)x

r(r + 1)x+ (1− r2)r(r + 1)x− (r2 − 1)

From this, it follows that r3x2 − r2(r + 1)x+ r(r + 1) = r2x2 − r(r + 1)x+ (r + 1) = 0,which has the roots

x1,2 =r(r + 1)±

√r2(r + 1)2 − 4r2(r + 1)

2r2

=r(r + 1)± r

√r2 + 2r + 1− 4r − 4

2r2

=r + 1±

√r2 − 2r − 3

2r

=r + 1±

√(r − 3)(r + 1)

2r,

and so the second-iterate map has the 4 fixed points x∗ = 0, r−1r ,

r+1±√

(r−3)(r+1)

2r .

(b) The fixed points u =r+1+√

(r−3)(r+1)

2r and v =r+1−√

(r−3)(r+1)

2r form the 2-cycle of theoriginal system. The 2-cycle exists when these two fixed points are distinct, real, defined, andpositive. They are distinct and real whenever (r−3)(r+1) > 0, which occurs when r < −1and r > 3. Since r > 0, the fixed points are defined, and r > 3 must be true. Further,

u is always positive, but v is only positive whenr+1−√

(r−3)(r+1)

2r > 0, which occurs when

r + 1 −√(r − 3)(r + 1) > 0. Multiplying both sides by the conjugate (which is positive)

1

gives the inequality

(r + 1−√

(r − 3)(r + 1))(r + 1 +√

(r − 3)(r + 1)) > 0

⇒ (r + 1)2 − (r + 1)(r − 3) > 0

⇒ 4(r + 1) > 0.

Since we already know that r > 3 must be true, this inequality will always hold. Thus, wehave a 2-cycle for r > 3. The other points, as mentioned earlier, correspond to the fixedpoints of the original logistic equation.

(c)d

dxf2(x) =

d

dx

[r2x(1− (r + 1)x+ 2rx2 − rx3)

]= r2(1− (r + 1)x+ 2rx2 − rx3) + r2x(−(r + 1) + 4rx− 3rx2)

= r2 − 2r2(r + 1)x+ 6r3x2 − 4r3x3

(d) To determine when the non-trivial 2-cycle is stable, we need to find out when the fixedpoints of the second-iterate map that corresponding to the 2-cycle are stable. Stability

of the 2-cycle occurs when∣∣∣ ddxf

2(x)∣∣x=u,v

∣∣∣ < 1. After some algebra, it turns out thatddxf

2(x)∣∣x=u

= ddxf

2(x)∣∣x=v

= −r2+2r+4. So, stability exists when | − r2+2r+4| < 1.We need to solve two inequalities:

i. −r2 + 2r + 4 < 1 ⇒ −r2 + 2r + 3 < 0 ⇒ r > 3

ii. −r2 + 2r + 4 > −1 ⇒ −r2 + 2r + 5 > 0 ⇒ 0 < r < 1 +√6

Therefore, the 2-cycle is stable for 3 < r < 1 +√6.

2. (a) For r = 3.2 and x1 = 0.513:

x1 x2 x3 x4 x5 x6 x7 x8 x9 x100.5130 0.7995 0.5130 0.7995 0.5130 0.7995 0.5130 0.7995 0.5130 0.7995

1 2 3 4 5 6 7 8 9 100

0.1

0.2

0.3

0.4

0.5

0.6

0.7

n

xn

0 0.2 0.4 0.6 0.8 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

xn+1

=f(xn)

xn+1

=xn

cobweb

The 2-cycle bounces between 0.5130 and 0.7995.

(b) For r = 3.55 and x1 = 0.8874:

x1 x2 x3 x4 x5 x6 x7 x8 x9 x100.8874 0.3547 0.8126 0.5407 0.8816 0.3705 0.8279 0.5057 0.8874 0.3548

2

1 2 3 4 5 6 7 8 9 100

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

n

xn

0 0.2 0.4 0.6 0.8 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

xn+1

=f(xn)

xn+1

=xn

cobweb

There may be a pattern that restarts at x9, depending on the decimal place to which thesolution is rounded.

(c) For r = 3.8 and x1 = 0.5:

x1 x2 x3 x4 x5 x6 x7 x8 x9 x100.5 0.95 0.1805 0.5621 0.9353 0.2298 0.6726 0.8369 0.5188 0.9487

1 2 3 4 5 6 7 8 9 100

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

n

xn

0 0.2 0.4 0.6 0.8 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

xn+1

=f(xn)

xn+1

=xn

cobweb

There doesn’t seem to be any observable pattern in the solution.

Matlab code

r = 3.2; % define parameter (change manually for parts b and c)

x = []; % defines x as an empty vector

x(1) = 0.513; % sets the first entry in x to be the given x_1 (change

% manually for parts b and c)

f = @(x) r*x.*(1-x); % define map as a function of x

N=10; % maximum n

% this for-loop evaluates the solution through n=10 and puts

% the solution at each successive time point in the vector x

for n=1:N-1

x(n+1) = f(x(n));

end

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% print the solution as a column vector in the command window

% (no semi-colon)

x'

% plot and label the solution in the specified range

plot(x,'-o')xlabel('n')ylabel('x_n')axis([1 N 0 max(x)])

Note: Any single quotations in the code will not copy correctly into Matlab (a LATEX issue), so youwill need to retype them after pasting the code into the Editor.

3. Feigenbaum diagram of the logistic map

0 0.5 1 1.5 2 2.5 3 3.5 40

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

r

x5

0

Stability of x*=0for 0<r<1

Stability of x*=(r−1)/rfor 1<r<3

Stable 2−cycle

Stable 4−cycle

chaos

4

Homework 4.2 Solutions

1. (a) Fixed points:p∗ = p∗2 − q∗

q∗ = p∗q∗ − 100p∗

has 3 solutions in the form P = (p∗, q∗): P1 = (0, 0), P2 = (11, 110) and P3 = (−9, 90) .

Stability: The Jacobian for this system is

J(p∗, q∗) =

[2p∗ −1

q∗ − 100 p∗

]

For P1 = (0, 0): J(0, 0) =

[0 −1−100 0

], which has eigenvalues λ1 = −10 and λ2 = 10.

Since |λ1,2| > 1, P1 is unstable.

For P2 = (11, 110): J(11, 110) =

[22 −110 11

], which has eigenvalues λ1 = 12 and λ2 = 21.

Since |λ1,2| > 1, P2 is unstable.

For P3 = (−9, 90): J(−9, 90) =

[−18 −1−10 −9

], which has eigenvalues λ1 = −19 and

λ2 = −8. Since |λ1,2| > 1, P3 is unstable.

(b) Fixed points:x∗ = rx∗ − x∗2 + x∗

has 2 solutions, x∗ = 0, r .Stability: Since this is a one-dimensional system, we can look at the derivative of the right-hand side evaluated at the fixed points to determine stability. If f(xn) = rxn − x2n + xn,then f ′(xn) = r − 2xn + 1.

For x∗ = 0: f ′(0) = r + 1, which implies stability provided |r + 1| < 1. But, since r > 0,this can never happen. Thus, the fixed point x∗ = 0 is always unstable.

For x∗ = r: f ′(r) = 1 − r, which implies stability provided |1 − r| < 1. So the fixed pointx∗ = r is stable when 0 < r < 2 and unstable for r > 2.

Bifurcation diagram(solid line: stable, dashed line: unstable)

5

0 1 2 3 4 50

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

r

x*

2. (a) When aR + pJ = 1 and aJ + pR = 1, Jn +Rn = Jo +Ro. We can see this by adding Rn+1

and Jn+1.Rn+1 + Jn+1 = (aR + pJ)Rn + (aJ + pR)Jn

Because the total amount of feeling (Jn + Rn) is preserved, this mode is called feelingpreserving.

(b) Fixed points:R∗ = aRR

∗ + pRJ∗

J∗ = pJR∗ + aJJ

∗

0 = (aR − 1)R∗ + pRJ∗

J∗ = pJR∗ + (aJ − 1)J∗[

aR − 1 pRpJ aJ − 1

] [R∗

J∗

]=

[00

]The determinant of this matrix is (aR− 1)(aJ − 1)− (pRpJ). Substituting 1− aR = pJ and1− aJ = pR, we see that the determinant is 0. Because the determinant is 0, we know thatthe two equations are redundant and we can write J∗ in terms of R∗. So J∗ = pJ

1−aJR∗.

(c) The Jacobian matrix is:

A =

[aR pRpJ aJ

]To find the eigenvalues we must find the determinant of A− λI.

A− λI =

[aR − λ pRpJ aJ − λ

]Then, the detA− λI = (aR − λ)(aJ − λ)− pRpJ ordetA− λI = λ2−λ(aR+aJ)+(−1+aJ +aR). To find the values of λ set detA− λI = 0.

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Therefore,

λ1,2 =1

2(aR + aJ ±

√(aR + aJ)2 − 4(−1 + aJ + aR))

λ1,2 =1

2(aR + aJ ±

√(2− aR − aJ)2)

λ1 = 1

λ2 = aR + aJ − 1

7