Nonlinear Dynamics and Chaos
Nonlinear Dynamics and Chaos
2nd Edition
ISBN: 9780813349107
Author: Steven H. Strogatz
Publisher: PERSEUS D
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Chapter 6.3, Problem 11E
Interpretation Introduction

Interpretation:

To calculate r(t) and θ(t) explicitly for the given initial conditions (r00). The system is in polar coordinates given by r˙ = - r, θ˙ = 1lnr. Prove that r(t)0 and |θ(t)| as t and origin is a stable spiral. Show that the linearized system at the origin is x˙=-x,y˙=-y and origin is a stable star for the linearized system.

Concept Introduction:

The linearized system for x˙ = f(x, y),y˙ = g(x, y) can be expressed as (x˙y˙)=(x˙xx˙yy˙xy˙y)(xy).

The matrix (x˙xx˙yy˙xy˙y) is a Jacobian matrix at any fixed point (x*,y*).

Expert Solution & Answer
Check Mark

Answer to Problem 11E

Solution:

r(t) = r0e-t, θ(t) = ln (ln(r0)|ln(r0) - t|)+ θ0.

r(t)0 and |θ(t)| as t is proved and origin is a stable spiral is proved.

The system in Cartesian coordinate is x˙ = -x -2yln(x2+y2), y˙ = -y +2xln(x2+y2).

The linearized system at the origin is x˙=-x,y˙=-y is proved and origin is a stable star.

Explanation of Solution

Using the first equation

r˙  = - rdrdt = -rdrr = -dtr0rdrr= - 0tdtlnr - lnr0= -tlnrr0=-tr(t) = r0e-t

θ˙ = 1lnrr(t) = r0e-tθ˙ = 1ln(r0e-t)θ˙ =1ln(r0) + lne-tθ˙ =1ln(r0) - t

Integrating both sides,

θ(t) = -ln|ln(r0) - t| + ln(ln(r0)) + θ0θ(t) = ln (ln(r0)|ln(r0) - t|)+ θ0

r(t) = r0e-t

When t, r(t)0 and θ(t)0.

Solve the stability of the origin by solving limtr(t).

limtr(t)=limtr0e- t0

Thus the origin is a stable point.

Solve the nature of the origin by solving limtθ(t).

limt|θ(t)|limtln |ln(r0)|ln(r0) - t|+ θ0|

Thus the origin is spiral.

The system in x, y can be written as,

r = x2+y2θ = arctan(yx)r˙ =ddtx2+y2r˙ = 2xx˙+2yy˙2x2+y2r˙ = xx˙+yy˙x2+y2θ˙ =ddtarctan(yx)θ˙ = xy˙-yx˙x2+y2

Comparing the above polar coordinates with the given coordinates,

r˙ = -r r˙ =xx˙+yy˙x2+y2 r = -(xx˙+yy˙x2+y2)r =x2+y2x2+y2 = - xx˙ - yy˙x2+y2x2+y2= -(xx˙ + yy˙).......................1)θ˙ =1lnr θ˙ = xy˙ - yx˙x2+ y2xy˙-yx˙x2+y2=1ln rr =x2+y2xy˙-yx˙x2+y2 = 1ln(x2+y2)xy˙-yx˙ = 2(x2+y2)ln(x2+y2)......................2)

Multiply equation 1) by x and equation 2) by y and subtract,

x(xx˙ + yy˙)-y(xy˙-yx˙)= - x (x2+y2) - y2(x2+y2)ln(x2+y2)(x2+y2)x˙= -x(x2+y2) - y2(x2+y2)ln(x2+y2)x˙ = -x -2yln(x2+y2)

Multiply equation 1) by y and equation 2) by x and add,

y(xx˙ + yy˙) + x(xy˙-yx˙) = - y (x2+y2) + x2(x2+y2)ln(x2+y2)(x2+y2)y˙ = -y(x2+y2) + x2(x2+y2)ln(x2+y2)y˙ = -y +2xln(x2+y2)

The Jacobian matrix at any fixed point (x*,y*) given as,

A=(x˙xx˙yy˙xy˙y)

Substituting values of x˙ , y˙  of part c) in the above Jacobian matrix,

A=((-x -2yln(x2+y2))x(-x -2yln(x2+y2))y(-y +2xln(x2+y2))x(-y +2xln(x2+y2))y)A=(-1 +4xy(x2+y2)ln2(x2+y2) 4y2(x2+y2)ln2(x2+y2)2ln(x2+y2)2ln(x2+y2)4x2(x2+y2)ln2(x2+y2)-1 - 4xy(x2+y2)ln2(x2+y2))A(0,0)=(-1  00-1)A(xy)=(x˙y˙)(-1  00-1)(xy)=(x˙y˙)(-x-y)=(x˙y˙)

It is proved and origin is a stable star for the linearized system.

Conclusion

Integrating given polar equations we can find r(t) and θ(t) explicitly for the given initial conditions (r00). Also, r(t)0 and |θ(t)| as t is proved. In this case, the origin is a stable spiral. The linearized system at the origin is x˙=-x,y˙=-y is proved and origin is a stable star.

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Chapter 6 Solutions

Nonlinear Dynamics and Chaos

Ch. 6.1 - Prob. 11ECh. 6.1 - Prob. 12ECh. 6.1 - Prob. 13ECh. 6.1 - Prob. 14ECh. 6.2 - Prob. 1ECh. 6.2 - Prob. 2ECh. 6.3 - Prob. 1ECh. 6.3 - Prob. 2ECh. 6.3 - Prob. 3ECh. 6.3 - Prob. 4ECh. 6.3 - Prob. 5ECh. 6.3 - Prob. 6ECh. 6.3 - Prob. 7ECh. 6.3 - Prob. 8ECh. 6.3 - Prob. 9ECh. 6.3 - Prob. 10ECh. 6.3 - Prob. 11ECh. 6.3 - Prob. 12ECh. 6.3 - Prob. 13ECh. 6.3 - Prob. 14ECh. 6.3 - Prob. 15ECh. 6.3 - Prob. 16ECh. 6.3 - Prob. 17ECh. 6.4 - Prob. 1ECh. 6.4 - Prob. 2ECh. 6.4 - Prob. 3ECh. 6.4 - Prob. 4ECh. 6.4 - Prob. 5ECh. 6.4 - Prob. 6ECh. 6.4 - Prob. 7ECh. 6.4 - Prob. 8ECh. 6.4 - Prob. 9ECh. 6.4 - Prob. 10ECh. 6.4 - Prob. 11ECh. 6.5 - Prob. 1ECh. 6.5 - Prob. 2ECh. 6.5 - Prob. 3ECh. 6.5 - Prob. 4ECh. 6.5 - Prob. 5ECh. 6.5 - Prob. 6ECh. 6.5 - Prob. 7ECh. 6.5 - Prob. 8ECh. 6.5 - Prob. 9ECh. 6.5 - Prob. 10ECh. 6.5 - Prob. 11ECh. 6.5 - Prob. 12ECh. 6.5 - Prob. 13ECh. 6.5 - Prob. 14ECh. 6.5 - Prob. 15ECh. 6.5 - Prob. 16ECh. 6.5 - Prob. 17ECh. 6.5 - Prob. 18ECh. 6.5 - Prob. 19ECh. 6.5 - Prob. 20ECh. 6.6 - Prob. 1ECh. 6.6 - Prob. 2ECh. 6.6 - Prob. 3ECh. 6.6 - Prob. 4ECh. 6.6 - Prob. 5ECh. 6.6 - Prob. 6ECh. 6.6 - Prob. 7ECh. 6.6 - Prob. 8ECh. 6.6 - Prob. 9ECh. 6.6 - Prob. 10ECh. 6.6 - Prob. 11ECh. 6.7 - Prob. 1ECh. 6.7 - Prob. 2ECh. 6.7 - Prob. 3ECh. 6.7 - Prob. 4ECh. 6.7 - Prob. 5ECh. 6.8 - Prob. 1ECh. 6.8 - Prob. 2ECh. 6.8 - Prob. 3ECh. 6.8 - Prob. 4ECh. 6.8 - Prob. 5ECh. 6.8 - Prob. 6ECh. 6.8 - Prob. 7ECh. 6.8 - Prob. 8ECh. 6.8 - Prob. 9ECh. 6.8 - Prob. 10ECh. 6.8 - Prob. 11ECh. 6.8 - Prob. 12ECh. 6.8 - Prob. 13ECh. 6.8 - Prob. 14E
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