Homework 3

Problem 2.1

a)

æ ç è 4 30 1 3 ö ÷ ø

0 = |A-lI| = (4-l)(3-l)-30 = l2-7l-18 = (l-9)(l+2)

So, l = -2,9, and the origin is a source.

b)

æ ç ç ç ç ç ç ç ç è 1 1 2 1 4 3 4 ö ÷ ÷ ÷ ÷ ÷ ÷ ÷ ÷ ø

0 = |A-lI| = (1-l)( 3 4 -l)- 1 8 = l2- 7 4 l+ 5 8 = (l- 5 4 )(l- 2 4 )

So, l = ¹/₂,⁵/₄, and the origin is a saddle.

c)

æ ç è -0.4 2.4 -0.4 1.6 ö ÷ ø

0 = |A-lI| = (-0.4-l)(1.6-l)+2.4(0.4) = l2-1.2l+0.32 = (l-0.4)(l-0.8)

So, l = 0.4,0.8, and the origin is a sink.

Problem 2.2

Find the limit

So, we have a repeated eigenvalue of l = ¹/₂. This means that matrix A is similar to a matrix of the form

Problem 2.7

In this problem we investigate the Henon Map for b = 0.3, given by f(x,y) = (a-x²+by,x).

a)

First, determine the fixed points: F(x,y) = (x,y). This means, y = x and x²+(1-b)x-a = 0. Solving this equation yields:

x± = -0.35±0.5   _______ Ö0.49+4a   .

So the fixed points for the Henon map are (x₊,x₊) and (x_-,x_-). Note that these fixed points only exist for 0.49+4a ³ 0, or a ³ -[0.49/ 4] = -0.1225.

We are seeking a range of values for a such that one of the fixed points is a saddle and the other is a sink. The stability of the fixed points is found by studying the Jacobian matrix for the Henon map:

DF(x,y) = æ ç è -2x b 1 0 ö ÷ ø .

The eigenvalues are found as:

0 = |DF-lI| = (-2x-l)(-l)-b = l2+2xl-b.

So,

l± = -x ±   ______ Öx2+0.3   ,

where x has to be evaluated at x_± to determine the stability at each fixed point.

Before proceeding, we note that l₊ > 0 and l_- < 0 for all values of x. This is easily seen by noting Ö[(x²+.3)] > Ö[(x²)] = |x|. Thus, to determine the behavior of l for various values of x, we only need to consider l₊ > 1 and l_- < -1.

i) l₊ > 1

-x+   _____ Öx2+.3   > 1   _____ Öx2+.3   > 1+x x2+.3 > (1+x)2 = x2+2x+1 .3 > 1+ 2x x < -0.35

So, we know that 0 < l₊ < 1 for x > -0.35.

ii) l_- < -1.

-x-   _____ Öx2+.3   < -1   _____ Öx2+.3   > 1-x x2+.3 > (1-x)2 = x2-2x+1 .3 > 1- 2x x > .35

So, we know that -1 < l_- < 0 for x < 0.35.

Thus, for a saddle, |x| > .35. Note that x = x_± = -0.35±0.5Ö[(0.49+4a)], which implies that x_- < -.35 as long as .49+4a > 0 Þ a > -.1225 Þ l₊ > 1 and 0 > l_- > -1. So, we have a saddle at x_- for a > -.1225.

For a sink, we then need |x₊| < .35. This implies that

-.35 < -.35+.5   ______ Ö.49+4a   < .35.

The lower limit automatically gives , a > -.1225. The upper limit gives

.5   ______ Ö.49+4a   < .7 Þ 4a < 4(.7)2-(.7)2Þ a < 3 4 (.7)2 = .3675

b)

We now look for a period 2 sink. A period two orbit can be found from the equation F(F(x,y)) = (x,y). Looking at the orbit, we have

(x1,y1) = (a-x02+by0,x0) (x0,y0) = (a-x12+by1,x1) = (a-(a-x02+by0)2+bx0,a-x02+by0).

Eliminating y₀ leads to and equation for x₀, which can be factored as

(x2-(1-b)*x-a+(1-b)2)(x2+(1-b)x-a) = 0.

The second term is zero for fixed points, as seen previously. So, we need only set the first factor equal to zero to find the period 2 orbits. For b = 0.3, we have to solve

x2-.7x-a+.49 = 0

to obtain

x± = 1 2 [.7 ±   ________ Ö4a-3(.49)   ].

[Technically, we need both x₀ and y₀ in order to specify the period two orbit. However, y₀ will not be needed for the stability analysis.]

We need to determine when this period two orbit is a sink. This is accomplished from the3 Jacobian matrix for the second iterate of the map: F(F(x,y)). By the chain rule, we have D(F(F(x,y))) = DF(F(x))DF(x), So, we need only to study the matrix

A = DF(x+)DF(x-) = æ ç è -2x+ b 1 0 ö ÷ ø æ ç è -2x- b 1 0 ö ÷ ø = æ ç è 4x+x-+b -2x+b -2x- b ö ÷ ø .

The eigenvalues of this matrix need to have magnitude less than one for a sink. The eigenvalue equation is given by

(4x+x-+b-l)(b-l)-4x+x-b = 0.

This might look ugly, but it can be simplified using the substitutions a = 4x₊x_- and m = b-l. Then, we have

(a+m)m-ab = 0,

m2+am-ab = 0,

Þ m = 1 2 [-a± Ö a2+ab   ].

Solving for l, we find

l = b+ 1 2 [a± Ö a2+ab   ].

Note that for our b value

a = 4x+x- = .49-(4a-3(.49) = 1.96-4a.

Simplifying the expression for l,

l = 2[.64-a ±Ö(a2-1.28a+.3871)].

We are interested in when the above expression is real and has magnitude less than one. Noting that a²-1.28a+.3871 = (a-.49)(a-.79) It is easy to see that a < .49 or a > .79 will make l real. We further need to satisfy the inequalities:

-1 < l < 1 -1 < 2[.64-a±   ______________ Öa2-1.28a+.3871   ] < 1 -.5 < .64-a±   ______________ Öa2-1.28a+.3871   < .5 a-1.14 < ±   ______________ Öa2-1.28a+.3871   < a-.14.

One finds that there are only solutions for a < 1.14 and a > .14, so we need to consider the two inequalities separately:

a-1.14 < -   ______________ Öa2-1.28a+.3871   1.14-a >   ______________ Öa2-1.28a+.3871   (1.14-a)2 > a2-1.28a+.3871 a2-2.28a+1.2996 > a2-1.28a+.3871 .9125 > a.

______________ Öa2-1.28a+.3871   < a-.14   ______________ Öa2-1.28a+.3871   < (a-.14)2 a2-1.28a+.3871 < a2-.28a+.0196 .3675 < a.

Therefore, we find the existence of a sink for a Î (.3675,.49]È[.79,.9125).

Problem 2.8

a)

A = æ ç è 2 0.5 2 -0.5 ö ÷ ø

AAT = æ ç è 2 0.5 2 -0.5 ö ÷ ø æ ç è 2 2 0.5 -0.5 ö ÷ ø = æ ç ç ç ç ç ç ç ç è 17 4 15 4 15 4 17 4 ö ÷ ÷ ÷ ÷ ÷ ÷ ÷ ÷ ø .

Looking for the eigenvalues, we obtain

0 = |AAT-lI| = ( 17 4 -l)2- æ ç è 15 4 ö ÷ ø 2   17 4 -l = ± 15 4 l = 8, 1 2 .

The lengths of the principal axes of the ellipse are then given by the square roots of the eigenvalues: 2Ö2,[1/( Ö2)].

The area of the ellipse can be found one of two ways. The simplest is to note that the image of an area under the linear map A is scaled by |det(A)|. Since a unit circle has area p, the area of the ellipse is just |det(A)|p. In this case, we obtain and area of 2p.

The area of an ellipse is also given in terms of the semimajor and semiminor axes of an ellipse: pab. Here these lengths are just the lengths determined in the first part of the problem. So, the area is pab = 2Ö2[1/( Ö2)]p = 2p.

b)

A = æ ç è 2 1 -2 2 ö ÷ ø

This problem is done like the one above. The area of the ellipse is given as |det(A)|p = 6p.

Looking for the eigenvalues, we obtain