Nonlinear Dynamics and Chaos
Nonlinear Dynamics and Chaos
2nd Edition
ISBN: 9780429972195
Author: Steven H. Strogatz
Publisher: Taylor & Francis
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Chapter 7.6, Problem 1E
Interpretation Introduction

Interpretation:

To show that x(t,ε) = (1- ε2)-1/2e-εtsin[(1-ε2)1/2t] is expanded as a power series in ε, we recover x(t, ε) = (1- ε2)-1/2e-εtsin t - εt sin t+Ο(ε2).

Concept Introduction:

Usepower series expansion.

Expert Solution & Answer
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Answer to Problem 1E

Solution:

It is shown that for the given equation x(t,ε) = (1- ε2)-1/2e-εtsin[(1-ε2)1/2t], power series expansion is x(t,τ) = sin t - εt sin t+Ο(ε2).

Explanation of Solution

The power series expansion of the function f(x)= an(x- c)n is given as,

0an(x - c)n=a0+ a1(x- c)+ a2(x- c)2+.....

Here, c is the constant and an the coefficient of the nth term.

From the equation 7.6.7

x(t,ε) = (1- ε2)-1/2e-εtsin[(1-ε2)1/2t]

By differentiating the above equation with respect to ε,

dx(t,ε) = d((1- ε2)-1/2e-εtsin[(1-ε2)1/2t])

dx(t,ε) = e-εtsin[(1-ε2)1/2t]d((1- ε2)-1/2)+(1- ε2)-1/2d(e-εtsin[(1-ε2)1/2t])

dx(t,ε) = e-εtsin[(1-ε2)1/2t](12(1- ε2)-3/2(2ε))+(1- ε2)-1/2d(e-εtsin[(1-ε2)1/2t])

dx(t,ε) = e-εtsin[(1-ε2)1/2t](ε)(1- ε2)-3/2+(1- ε2)-1/2d(e-εtsin[(1-ε2)1/2t])

dx(t,ε) = e-εtsin[(1-ε2)1/2t](ε)(1- ε2)-3/2+(1- ε2)-1/2e-εtd(sin[(1-ε2)1/2t])+(1- ε2)-1/2sin[(1-ε2)1/2t]d(e-εt)

dx(t,ε) = e-εtsin[(1-ε2)1/2t](ε)(1- ε2)-3/2+(1- ε2)-1/2e-εtcos[(1- ε2)1/2t]t12(1-ε2)-1/2(2ε)               + (1- ε2)-1/2sin[(1-ε2)1/2t]e-εt(-t)

dx(t,ε) = e-εtsin[(1-ε2)1/2t](ε)(1- ε2)-3/2+(1- ε2)-1/2e-εtcos[(1- ε2)1/2t]t(1-ε2)-1/2(ε)               + (1- ε2)-1/2sin[(1-ε2)1/2t]e-εt(-t)

dx(t,ε) = e-εtsin[(1-ε2)1/2t](ε)(1- ε2)-3/2+(1- ε2)-1e-εtcos[(1- ε2)1/2t]t(1-ε2)-1/2(ε)               + (1- ε2)-1/2sin[(1-ε2)1/2t]e-εt(-t)

dx(t,ε) = ε(1- ε2)-3/2e-εtsin[(1-ε2)1/2t]εt(1- ε2)-1e-εtcos[(1- ε2)1/2t] - t(1- ε2)-1/2e-εtsin[(1-ε2)1/2t]

The value of x(t,ε) at ε0 is calculated as,

Substitute 0 for ε in the above expression x(t,ε)

x(t,0) = sin t

Value of dx(t,ε), as ε0

d(x(t, 0))=-t sin t

The power series expansion of x(t,ε) is given as:

x(t,τ)= x(t,ε)|ε=0+εdx(t,ε)dε|ε=0+Ο(ε2)

x(t,τ) = x(t,0)+εdx(t, 0)dε+Ο(ε2)

x(t,τ) = sin t+ε(-t sin t)+Ο(ε2)

x(t,τ) = sin t - εt sin t+Ο(ε2)

Hence, the power series expansion is x(t,τ) = sin t - εt sin t+Ο(ε2) for the given equation x(t,ε) = (1- ε2)-1/2e-εtsin[(1-ε2)1/2t].

Conclusion

It is shown that for the givenequation x(t,ε) = (1- ε2)-1/2e-εtsin[(1-ε2)1/2t], the power series expansion is x(t,τ) = sin t - εt sin t+Ο(ε2)

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