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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 48 "Unit 3: Ordinary Different
ial Equations - Part 2" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 28 "Chapter 14: Series Solutions" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 58 "Section 14.5:  the nonlin
ear spring and Lindstedt's method" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}
{SECT 1 {PARA 3 "" 0 "" {TEXT -1 9 "Copyright" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 48 "Copyright * 2001 by Addis
on Wesley Longman, Inc." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 302 "All rights reserved.  No part of this publication m
ay be reproduced, stored in a retrieval system, or transmitted, in any
 form or by any means, electronic, mechanical, photocopying, recording
, or otherwise, without the prior written permission of the publisher.
 Printed in the United States of America." }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 15 "Initializations" }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "res
tart;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "with(plots):\nwith
(DEtools):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "interface(sho
wassumed=0);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 
1 {PARA 3 "" 0 "" {TEXT -1 22 "A Nonlinear Oscillator" }}{PARA 0 "" 0 
"" {TEXT -1 0 "" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 25 "The differentia
l equation" }}{PARA 258 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "`x''`(t
)+x(t)+a*x(t)^3=0" "6#/,(-%$x''G6#%\"tG\"\"\"-%\"xG6#F(F)*&%\"aGF)*$-F
+6#F(\"\"$F)F)\"\"!" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 8 "tha
t is," }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 42 "q := diff(x(t),t,t) + x(t) + a*x(t)^3 = 0;" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 79 "governs an undamped oscillator whose spring obeys a nonli
near restoration law, " }{XPPEDIT 18 0 "F=-(x+a*x^3)" "6#/%\"FG,$,&%\"
xG\"\"\"*&%\"aGF(*$F'\"\"$F(F(!\"\"" }{TEXT -1 7 ".  For " }{XPPEDIT 
18 0 "a*`>`*0" "6#*(%\"aG\"\"\"%\">GF%\"\"!F%" }{TEXT -1 41 ", the spr
ing is called \"hard,\" while for " }{XPPEDIT 18 0 "a<0" "6#2%\"aG\"\"
!" }{TEXT -1 158 ", it is called \"soft.\"  The magnitude of the resto
rative force for the hard spring is greater than that of the linear sp
ring (whose restorative force is just " }{XPPEDIT 18 0 "-k*x" "6#,$*&%
\"kG\"\"\"%\"xGF&!\"\"" }{TEXT -1 143 ").  Likewise, the magnitude of \+
the restorative force for the soft spring is less than that of the lin
ear spring for small enough displacements." }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 5 "When " }{XPPEDIT 18 0 "a=0" "6#/
%\"aG\"\"!" }{TEXT -1 118 " the system is just a simple harmonic oscil
lator which clearly supports periodic solutions.  We wish to show that
 for " }{XPPEDIT 18 0 "a<>0" "6#0%\"aG\"\"!" }{TEXT -1 278 " the syste
m also exhibits periodic solutions.  We show this by obtaining a first
 integral (a function constant on the trajectories in the phase plane)
 and showing that its level curves, the trajectories, are closed curve
s.  To this end, multiply the differential equation by x'(" }{TEXT 
256 1 "t" }{TEXT -1 33 "), and integrate with respect to " }{XPPEDIT 
18 0 "t" "6#%\"tG" }{TEXT -1 11 ", obtaining" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "q2 := map(Int,exp
and(q*diff(x(t),t)),t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 59 "The antiderivative on the rig
ht is a constant we will call " }{TEXT 270 1 "E" }{TEXT -1 71 ", and e
ach term on the left is an exact differential.  Hence, we obtain" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 25 "q3 := lhs(value(q2)) = E;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 18 "For each value of " }
{XPPEDIT 18 0 "a" "6#%\"aG" }{TEXT -1 22 ", the level curves of " }
{XPPEDIT 18 0 "E=E(x,`x'`)" "6#/%\"EG-F$6$%\"xG%#x'G" }{TEXT -1 73 " d
efine solutions of the differential equation.  Indeed, differentiating
 " }{TEXT 271 1 "E" }{TEXT -1 17 " with respect to " }{XPPEDIT 18 0 "t
" "6#%\"tG" }{TEXT -1 7 " yields" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "q4 := collect(diff(q3,t),dif
f);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 58 "which is the original differential equation multip
lied by " }{XPPEDIT 18 0 "`x'`(t)" "6#-%#x'G6#%\"tG" }{TEXT -1 1 "." }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 67 "Hence, a
 solution of the differential equation is a level curve of " }
{XPPEDIT 18 0 "E(x,`x'`)" "6#-%\"EG6$%\"xG%#x'G" }{TEXT -1 22 " and a \+
level curve of " }{XPPEDIT 18 0 "E=E(x,`x'`)" "6#/%\"EG-F$6$%\"xG%#x'G
" }{TEXT -1 59 " defines a solution of the nonlinear differential equa
tion." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 41 "
It is easier to plot the level curves of " }{XPPEDIT 18 0 "E(x,`x'`)" 
"6#-%\"EG6$%\"xG%#x'G" }{TEXT -1 36 " if we change variables according
 to" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 258 "" 0 "" {TEXT -1 1 " \+
" }{TEXT 262 1 "u" }{TEXT -1 5 " = x(" }{TEXT 257 1 "t" }{TEXT -1 1 ")
" }}{PARA 258 "" 0 "" {TEXT -1 1 " " }{TEXT 263 1 "v" }{TEXT -1 6 " = \+
x'(" }{TEXT 258 1 "t" }{TEXT -1 1 ")" }}{PARA 0 "" 0 "" {TEXT -1 8 "so
 that " }}{PARA 258 "" 0 "" {TEXT 259 1 "u" }{TEXT -1 9 "' = x' = " }
{TEXT 260 15 "v              " }}{PARA 258 "" 0 "" {TEXT 261 1 "v" }
{TEXT -1 10 "' = x'' = " }{XPPEDIT 18 0 "-u-a*u^3" "6#,&%\"uG!\"\"*&%
\"aG\"\"\"*$F$\"\"$F(F%" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 17 "which then gives " }{XPPEDIT 18 0 "E(u,v)" "6#-%\"EG
6$%\"uG%\"vG" }{TEXT -1 12 " in the form" }}{PARA 258 "" 0 "" {TEXT 
-1 1 " " }{XPPEDIT 18 0 "E=1/2" "6#/%\"EG*&\"\"\"F&\"\"#!\"\"" }{TEXT 
-1 1 " " }{XPPEDIT 18 0 "``(u^2+v^2)" "6#-%!G6#,&*$%\"uG\"\"#\"\"\"*$%
\"vGF)F*" }{TEXT -1 3 " + " }{XPPEDIT 18 0 "a/4" "6#*&%\"aG\"\"\"\"\"%
!\"\"" }{TEXT -1 1 " " }{XPPEDIT 18 0 "u^4" "6#*$%\"uG\"\"%" }{TEXT 
-1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 8 "that is," }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "E := subs(d
iff(x(t),t)=v,x(t)=u,lhs(q3));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 18 "For each value of " }
{XPPEDIT 18 0 "a" "6#%\"aG" }{TEXT -1 24 ", the level curves of E(" }
{TEXT 264 1 "u" }{TEXT -1 1 "," }{TEXT 265 1 "v" }{TEXT -1 144 ") = co
nstant define solutions of the differential equation.  Some notion of \+
how the level curves (which are phase plane trajectories) vary with " 
}{XPPEDIT 18 0 "a" "6#%\"aG" }{TEXT -1 87 " is obtained from the follo
wing animation in which three contours of E are shown while " }
{XPPEDIT 18 0 "a" "6#%\"aG" }{TEXT -1 22 " varies from 0.2 to 2." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 154 "f := z -> contourplot(subs(a=z,E),u=-1..1,v=-1..1, contours=[.1
,.2,.3], color=black):\ndisplay([seq(f(k/5),k=1..10)],insequence=true,
 scaling=constrained);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" 
}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 144 "We rely on the graphs in the a
nimation to support the claim that the trajectories for this different
ial equation are closed curves.  Hence, for " }{XPPEDIT 18 0 "a<>0" "6
#0%\"aG\"\"!" }{TEXT -1 79 " the undamped oscillator with the nonlinea
r spring supports periodic solutions." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 3 "
" 0 "" {TEXT -1 47 "Dependence of Frequency on Initial Displacement" }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 232 "
In the nonlinear case, the (angular) frequency depends on the initial \+
displacement.  If we let that intial displacement be first 1, then 2, \+
(along with an initial velocity x'(0) = 0), and obtain graphs of the n
umeric solutions when " }{XPPEDIT 18 0 "a=1" "6#/%\"aG\"\"\"" }{TEXT 
-1 90 ", we obtain Figure 14.10, the following graph, which shows two \+
solutions corresponding to " }{XPPEDIT 18 0 "a=1" "6#/%\"aG\"\"\"" }
{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 256 "p1:=DEplot(subs(a=1,q),x(t),t=0..10,[[x(0)=1,D(
x)(0)=0], [x(0)=2,D(x)(0)=0]], stepsize=.1,linecolor=black):\np2:=subs
(THICKNESS(3)=THICKNESS(1),p1):\ndisplay(p2, scaling=constrained, xtic
kmarks=5, ytickmarks=5, labels=[`t   `,x],labelfont=[TIMES,ITALIC,12])
;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 143 "The graph clearly indicates different periods, and \+
hence frequencies, for the two solutions which start with different in
itial displacements.  " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 75 "This is in distinction to the behavior of the linear
 oscillator where, for " }{XPPEDIT 18 0 "a=0" "6#/%\"aG\"\"!" }{TEXT 
-1 26 ", the angular frequency is" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 258 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "omega=sqrt(k/m)" "6#/
%&omegaG-%%sqrtG6#*&%\"kG\"\"\"%\"mG!\"\"" }{TEXT -1 4 " = 1" }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 49 "a constant depe
nding on just the spring constant " }{XPPEDIT 18 0 "k" "6#%\"kG" }
{TEXT -1 14 " and the mass " }{XPPEDIT 18 0 "m" "6#%\"mG" }{TEXT -1 
15 ".  Indeed, for " }{XPPEDIT 18 0 "a=0" "6#/%\"aG\"\"!" }{TEXT -1 8 
" we have" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 258 "" 0 "" {TEXT 
-1 1 " " }{XPPEDIT 18 0 "x(t)=c[1]*cos(t)+c[2]*sin(t)" "6#/-%\"xG6#%\"
tG,&*&&%\"cG6#\"\"\"F--%$cosG6#F'F-F-*&&F+6#\"\"#F--%$sinG6#F'F-F-" }
{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 8 "that is," }}{PARA 0 "" 0 
"" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "dsolve(s
ubs(a=0,q),x(t));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 26 "and the angular frequency " }
{XPPEDIT 18 0 "omega=1" "6#/%&omegaG\"\"\"" }{TEXT -1 22 " is the mult
iplier of " }{XPPEDIT 18 0 "t" "6#%\"tG" }{TEXT -1 39 " in the trig fu
nctions in the solution." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 131 "In the nonlinear case, the dependence of the angu
lar frequency on the initial displacement can be computed from the fir
st integral " }{XPPEDIT 18 0 "E(u,v)" "6#-%\"EG6$%\"uG%\"vG" }{TEXT 
-1 32 ".  First, evaluate the constant " }{TEXT 272 1 "E" }{TEXT -1 
29 " from the initial conditions " }{XPPEDIT 18 0 "``(u,v)=``(x,`x'`)
" "6#/-%!G6$%\"uG%\"vG-F%6$%\"xG%#x'G" }{TEXT -1 3 " = " }{XPPEDIT 18 
0 "``(A,0)" "6#-%!G6$%\"AG\"\"!" }{TEXT -1 11 ", obtaining" }}{PARA 0 
"" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "E0 \+
:= subs(v=0,u=A,E);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 25 "Next, solve the equation " }
{XPPEDIT 18 0 "E(u,v)=E[0]" "6#/-%\"EG6$%\"uG%\"vG&F%6#\"\"!" }{TEXT 
-1 9 " for v = " }{XPPEDIT 18 0 "dx/dt" "6#*&%#dxG\"\"\"%#dtG!\"\"" }
{TEXT -1 29 ", obtaining the two solutions" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 258 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "v(u)" "6#-%
\"vG6#%\"uG" }{TEXT -1 3 " = " }{TEXT 273 1 "+" }{TEXT -1 1 " " }
{XPPEDIT 18 0 "1/2" "6#*&\"\"\"F$\"\"#!\"\"" }{TEXT -1 1 " " }
{XPPEDIT 18 0 "sqrt(2*A^4*a-4*u^2-2*u^4*a+4*A^2)" "6#-%%sqrtG6#,**(\"
\"#\"\"\"*$%\"AG\"\"%F)%\"aGF)F)*&F,F)*$%\"uGF(F)!\"\"*(F(F)*$F0F,F)F-
F)F1*&F,F)*$F+F(F)F)" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 8 "th
at is," }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 19 "V := solve(E=E0,v);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 10 "If we set " }
{XPPEDIT 18 0 "v(u)=dx/dt" "6#/-%\"vG6#%\"uG*&%#dxG\"\"\"%#dtG!\"\"" }
{TEXT -1 4 "  = " }{XPPEDIT 18 0 "du/dt" "6#*&%#duG\"\"\"%#dtG!\"\"" }
{TEXT -1 31 " and separate variables, we get" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 258 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "t=Int(du/v(
u),u=A..u[final])" "6#/%\"tG-%$IntG6$*&%#duG\"\"\"-%\"vG6#%\"uG!\"\"/F
.;%\"AG&F.6#%&finalG" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 68 "On a closed trajectory in the phase plane, motion is cl
ockwise.  At " }{XPPEDIT 18 0 "``(u,v)=``(A,0)" "6#/-%!G6$%\"uG%\"vG-F
%6$%\"AG\"\"!" }{TEXT -1 28 ", the differential equations" }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 258 "" 0 "" {TEXT 266 1 "u" }{TEXT -1 9 "
' = x' = " }{TEXT 267 15 "v              " }}{PARA 258 "" 0 "" {TEXT 
268 1 "v" }{TEXT -1 10 "' = x'' = " }{XPPEDIT 18 0 "-u-a*u^3" "6#,&%\"
uG!\"\"*&%\"aG\"\"\"*$F$\"\"$F(F%" }}{PARA 0 "" 0 "" {TEXT -1 4 "give
" }}{PARA 258 "" 0 "" {TEXT -1 1 " " }{TEXT 269 1 "v" }{TEXT -1 4 "' =
 " }{XPPEDIT 18 0 "-A-a*A^3<0" "6#2,&%\"AG!\"\"*&%\"aG\"\"\"*$F%\"\"$F
)F&\"\"!" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 125 "so motion at this point is downward, and hence, c
lockwise around the trajectory.  Thus, we pick the negative square roo
t for " }{XPPEDIT 18 0 "v(u)" "6#-%\"vG6#%\"uG" }{TEXT -1 19 ", and in
tegrate to " }{XPPEDIT 18 0 "u[final]=0" "6#/&%\"uG6#%&finalG\"\"!" }
{TEXT -1 86 ", one-quarter of the way around the trajectory.  Thus, th
e total time for one orbit is" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "T := Int(4/V[2],u=A..0);" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 38 "To third-order, a series expansion of " }{XPPEDIT 18 0 "o
mega=2*Pi/T" "6#/%&omegaG*(\"\"#\"\"\"%#PiGF'%\"TG!\"\"" }{TEXT -1 14 
" in powers of " }{XPPEDIT 18 0 "a" "6#%\"aG" }{TEXT -1 8 " is then" }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 258 "" 0 "" {TEXT -1 1 " " }
{XPPEDIT 18 0 "omega=1+3/8" "6#/%&omegaG,&\"\"\"F&*&\"\"$F&\"\")!\"\"F
&" }{TEXT -1 1 " " }{XPPEDIT 18 0 "A^2*a-21/256" "6#,&*&%\"AG\"\"#%\"a
G\"\"\"F(*&\"#@F(\"$c#!\"\"F," }{TEXT -1 1 " " }{XPPEDIT 18 0 "A^4*a^2
+81/2048" "6#,&*&%\"AG\"\"%%\"aG\"\"#\"\"\"*&\"#\")F)\"%[?!\"\"F)" }
{TEXT -1 1 " " }{XPPEDIT 18 0 "A^6*a^3" "6#*&%\"AG\"\"'%\"aG\"\"$" }
{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 21 "obtained in Maple via" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "assume(A>0):\nomega = serie
s(2*Pi/T,a,4);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 240 "Thus, for the nonlinear oscillato
r, the angular frequency depends on the initial displacement, and we h
ave obtained a series expansion of this dependence.  Shortly, we will \+
compute this same expression by Lindstedt's perturbation technique." }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 0 "" }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 20 "Regular Perturbation" 
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 342 
"As we saw in Section 14.4, a regular perturbation will lead to an exp
ansion containing secular terms which are not periodic.  Hence, the re
gular perturbation series of Poincare cannot approximate the periodic \+
solutions of the nonlinear oscillator.  We therefore fail to learn abo
ut the depencence of the frequency on the initial displacement." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 236 "However,
 for the sake of completeness, we will develop the regular perturbatio
n.  The purpose is to justify the claim that it contains secular terms
 which prevent it from approximating the periodic solutions of the non
linear oscillator." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 48 "Assume a regular perturbation series of the form" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 29 "X := add(x[k](t)*a^k,k=0..2);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 62 "and substitut
e this into the differential equation, obtaining " }{XPPEDIT 18 0 "sum
(A[k]*a^k,k=0..2)=0" "6#/-%$sumG6$*&&%\"AG6#%\"kG\"\"\")%\"aGF+F,/F+;
\"\"!\"\"#F1" }{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 8 "that is,
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 42 "q5 := collect(simplify(subs(x(t)=X,q)),a);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 23 "where the coefficients " }{XPPEDIT 18 0 "A[k]" "6#&%\"AG6
#%\"kG" }{TEXT -1 13 " are given by" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "for k from 0 to 2 do\nA[k]
 = map(coeff,q5,a,k);\nod;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 31 "The identical vanishing of
 the " }{XPPEDIT 18 0 "A[k]" "6#&%\"AG6#%\"kG" }{TEXT -1 38 " gives di
fferential equations for the " }{XPPEDIT 18 0 "x[k]" "6#&%\"xG6#%\"kG
" }{TEXT -1 21 ".  The first equation" }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 258 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "`x''`[0](t)+x[0](
t)=0" "6#/,&-&%$x''G6#\"\"!6#%\"tG\"\"\"-&%\"xG6#F)6#F+F,F)" }{TEXT 
-1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
16 "depends only on " }{XPPEDIT 18 0 "x[0](t)" "6#-&%\"xG6#\"\"!6#%\"t
G" }{TEXT -1 80 ", so can be solved immediately.  For simplicity, we c
hoose the initial point as " }{XPPEDIT 18 0 "``(1,0)" "6#-%!G6$\"\"\"
\"\"!" }{TEXT -1 19 ", thereby obtaining" }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 258 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "x[0](t)=cos(t)
" "6#/-&%\"xG6#\"\"!6#%\"tG-%$cosG6#F*" }{TEXT -1 1 " " }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 172 "Each successive dif
ferential equation contains terms depending on the solution of the pre
vious equations.  We solve this sequence of equations, taking the init
ial point as " }{XPPEDIT 18 0 "``(0,0" "6#-%!G6$\"\"!F&" }{TEXT -1 35 
" for each successive approximation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}}{PARA 0 "" 0 "" {TEXT -1 91 "Clearly, this is a job for Maple.  To f
acilitate our work in Maple, we replace the symbols " }{XPPEDIT 18 0 "
A[k]" "6#&%\"AG6#%\"kG" }{TEXT -1 6 " with " }{XPPEDIT 18 0 "Q[k]" "6#
&%\"QG6#%\"kG" }{TEXT -1 13 " via the loop" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 51 "for k from 0 to 2 do\nQ||k := map(coeff,q5,a,k);\nod;
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 15 "We then obtain " }{XPPEDIT 18 0 "x[0](t)" "6#-&%\"xG
6#\"\"!6#%\"tG" }{TEXT -1 59 ", the solution of the first equation and
 initial condition " }{XPPEDIT 18 0 "``(1,0)" "6#-%!G6$\"\"\"\"\"!" }
{TEXT -1 4 ", as" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 50 "s0 := dsolve(\{Q0,x[0](0)=1,D(x[0])(0)=0\},x[
0](t));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 228 "Each successive differential equation co
ntains terms depending on the solutions of previous equations.  We sol
ve this sequence of equations, taking the initial point as (0,0) for e
ach successive approximation.  We thereby obtain" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 119 "for k from
 1 to 2 do\ns||k := combine(dsolve(\{subs(seq(s||n,n=0..k-1),Q||k),x[k
](0)=0,D(x[k])(0)=0\}, x[k](t)),trig);\nod;" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 23 "The soluti
ons for both " }{XPPEDIT 18 0 "x[1](t)" "6#-&%\"xG6#\"\"\"6#%\"tG" }
{TEXT -1 5 " and " }{XPPEDIT 18 0 "x[2](t)" "6#-&%\"xG6#\"\"#6#%\"tG" 
}{TEXT -1 251 " contain secular terms, as predicted.  The regular pert
urbation series is not useful for studying the dependence of the frequ
ency on the initial displacement.  We therefore consider Lindstedt's m
ethod for generating a more useful asymptotic expansion." }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 18 "Lindstedt's Method" }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 66 "In Lindsted
t's method, the angular frequency is made to depend on " }{XPPEDIT 18 
0 "a" "6#%\"aG" }{TEXT -1 12 " by assuming" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 258 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "omega=omega
(a)" "6#/%&omegaG-F$6#%\"aG" }{TEXT -1 7 " = 1 + " }{XPPEDIT 18 0 "sum
(omega[k]*a^k,k=1..infinity)" "6#-%$sumG6$*&&%&omegaG6#%\"kG\"\"\")%\"
aGF*F+/F*;F+%)infinityG" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 
"" }}{PARA 0 "" 0 "" {TEXT -1 32 "for which a truncated version is" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 34 "W := 1 + add(omega[k]*a^k,k=1..3);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 81 "We then carry
 out a perturbation solution of the differential equation by writing" 
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 258 "" 0 "" {TEXT -1 1 " " }
{XPPEDIT 18 0 "x(t)=Sum(x[k](t)*a^k,k=0..infinity)" "6#/-%\"xG6#%\"tG-
%$SumG6$*&-&F%6#%\"kG6#F'\"\"\")%\"aGF/F1/F/;\"\"!%)infinityG" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 28 "Since the
 angular frequence " }{XPPEDIT 18 0 "omega" "6#%&omegaG" }{TEXT -1 26 
" is related to the period " }{XPPEDIT 18 0 "T" "6#%\"TG" }{TEXT -1 3 
" by" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 258 "" 0 "" {TEXT -1 1 "
 " }{XPPEDIT 18 0 "omega*T=2*Pi" "6#/*&%&omegaG\"\"\"%\"TGF&*&\"\"#F&%
#PiGF&" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 81 "the two expansions in the Lindstedt method are merge
d by the change of variables " }{XPPEDIT 18 0 "tau=omega*t" "6#/%$tauG
*&%&omegaG\"\"\"%\"tGF'" }{TEXT -1 15 ".  In terms of " }{XPPEDIT 18 
0 "tau" "6#%$tauG" }{TEXT -1 31 ", the solution assumes the form" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 258 "" 0 "" {TEXT -1 1 " " }
{XPPEDIT 18 0 "x(t)=x(tau/omega" "6#/-%\"xG6#%\"tG-F%6#*&%$tauG\"\"\"%
&omegaG!\"\"" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "U(tau)=Sum(u[k](tau)*a
^k,k=0..infinity)" "6#/-%\"UG6#%$tauG-%$SumG6$*&-&%\"uG6#%\"kG6#F'\"\"
\")%\"aGF0F2/F0;\"\"!%)infinityG" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 37 "and the differential equation becomes" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 258 "" 0 "" {TEXT -1 1 " " }
{XPPEDIT 18 0 "omega^2*`U''`(tau)+U(tau)+a*U(tau)^3=0" "6#/,(*&%&omega
G\"\"#-%$U''G6#%$tauG\"\"\"F,-%\"UG6#F+F,*&%\"aGF,*$-F.6#F+\"\"$F,F,\"
\"!" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 57 "where now, primes denote differentiation with respect t
o " }{XPPEDIT 18 0 "tau" "6#%$tauG" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 59 "The change of variables r
ests on the chain rule which gives" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 258 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "dx/dt=dU/d/tau" "6#/*
&%#dxG\"\"\"%#dtG!\"\"*(%#dUGF&%\"dGF(%$tauGF(" }{TEXT -1 1 " " }
{XPPEDIT 18 0 "d*tau/dt=dU/d/tau" "6#/*(%\"dG\"\"\"%$tauGF&%#dtG!\"\"*
(%#dUGF&F%F)F'F)" }{TEXT -1 1 " " }{XPPEDIT 18 0 "omega" "6#%&omegaG" 
}{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 54 "A second application of the chain rule then gives the " }
{XPPEDIT 18 0 "omega^2" "6#*$%&omegaG\"\"#" }{TEXT -1 17 " multiplying
 U''(" }{XPPEDIT 18 0 "tau" "6#%$tauG" }{TEXT -1 2 ")." }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 48 "In fact, with the di
fferential equation given by" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 2 "q;" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 47 "the change
 of variables is effected in Maple by" }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "q6 := Dchangevar(\{t=ta
u/omega,x(t)=U(tau)\},q,t,tau);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 57 "We next write the firs
t few terms in the expansion for U(" }{XPPEDIT 18 0 "tau" "6#%$tauG" }
{TEXT -1 2 ")." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 32 "UU := add(u[k](tau)*a^k,k=0..3);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 47 "Then, substitute into the differential equation" }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 3 "q6;" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 24 "the expansions for both " }{XPPEDIT 18 0 "U(tau)" "6#-%\"
UG6#%$tauG" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "omega(tau)" "6#-%&omeg
aG6#%$tauG" }{TEXT -1 41 ", collect coefficients of like powers of " }
{XPPEDIT 18 0 "a" "6#%\"aG" }{TEXT -1 17 ", and thus obtain" }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "q
7 := collect(simplify(subs(omega=W,U(tau)=UU,q6)),a);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 27 "Putting this into the form " }{XPPEDIT 18 0 "sum(sigma[k]*a^k,k
=0..3)=0" "6#/-%$sumG6$*&&%&sigmaG6#%\"kG\"\"\")%\"aGF+F,/F+;\"\"!\"\"
$F1" }{TEXT -1 13 ", we get for " }{XPPEDIT 18 0 "sigma[k]" "6#&%&sigm
aG6#%\"kG" }{TEXT -1 15 " the expresions" }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "for k from 0 to 3 do
\nsigma[k] = map(coeff,q7,a,k);\nod;" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 31 "The identical
 vanishing of the " }{XPPEDIT 18 0 "sigma[k]" "6#&%&sigmaG6#%\"kG" }
{TEXT -1 37 " give differential equations for the " }{XPPEDIT 18 0 "u[
k]" "6#&%\"uG6#%\"kG" }{TEXT -1 69 ".  To facilitate this calculation \+
in Maple, we rewrite the equations " }{XPPEDIT 18 0 "sigma[k]=0" "6#/&
%&sigmaG6#%\"kG\"\"!" }{TEXT -1 4 " as " }{XPPEDIT 18 0 "Q[k]=0" "6#/&
%\"QG6#%\"kG\"\"!" }{TEXT -1 24 " via the following loop." }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "for \+
k from 0 to 3 do\nQ||k := map(coeff,q7,a,k);\nod;" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 33 "The \+
first equation contains only " }{XPPEDIT 18 0 "u[0](tau)" "6#-&%\"uG6#
\"\"!6#%$tauG" }{TEXT -1 47 ", so this equation, and the initial condi
tions " }{XPPEDIT 18 0 "u[0](0)=A" "6#/-&%\"uG6#\"\"!6#F(%\"AG" }
{TEXT -1 2 ", " }{XPPEDIT 18 0 "u[0]" "6#&%\"uG6#\"\"!" }{TEXT -1 30 "
'(0) = 0, lead to the solution" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "s0 := dsolve(\{Q0,u[0](0)=A,
D(u[0])(0)=0\},u[0](tau));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 17 "The equation for " }
{XPPEDIT 18 0 "u[1](tau)" "6#-&%\"uG6#\"\"\"6#%$tauG" }{TEXT -1 19 " c
ontains terms in " }{XPPEDIT 18 0 "u[0](tau)" "6#-&%\"uG6#\"\"!6#%$tau
G" }{TEXT -1 58 ".  Incorporating this solution and the initial condit
ions " }{XPPEDIT 18 0 "u[1](0)=0" "6#/-&%\"uG6#\"\"\"6#\"\"!F*" }
{TEXT -1 2 ", " }{XPPEDIT 18 0 "u[0]" "6#&%\"uG6#\"\"!" }{TEXT -1 19 "
'(0) = 0, we obtain" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 86 "s1 := combine(dsolve(\{simplify(subs(s0,Q
1)),u[1](0)=0,D(u[1])(0)=0\}, u[1](tau)),trig);" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 93 "As wit
h the regular perturbation, we have secular terms appearing.  However,
 the coefficient " }{XPPEDIT 18 0 "omega[1]" "6#&%&omegaG6#\"\"\"" }
{TEXT -1 146 " is as yet undetermined, so we choose a value for it whi
ch will guarantee the vanishing of the secular term.  Thus, collect al
l secular terms with" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 28 "q8 := collect(s1,[tau,sin]);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 58 "and set the coefficient of the secular term equal to zero." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 29 "q9 := map(coeff,q8,sin(tau));" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 25 "Solving for t
he value of " }{XPPEDIT 18 0 "omega[1]" "6#&%&omegaG6#\"\"\"" }{TEXT 
-1 40 " which eliminates the secular term from " }{XPPEDIT 18 0 "u[1](
tau)" "6#-&%\"uG6#\"\"\"6#%$tauG" }{TEXT -1 9 ", we find" }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "w[1]
 := solve(q9,omega[1]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 8 "so that " }{XPPEDIT 18 0 "u[1]
(tau)" "6#-&%\"uG6#\"\"\"6#%$tauG" }{TEXT -1 8 " becomes" }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "subs
(omega[1]=w[1],s1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 17 "The equation for " }{XPPEDIT 18 0 
"u[2](tau)" "6#-&%\"uG6#\"\"#6#%$tauG" }{TEXT -1 26 " contains terms i
nvolving " }{XPPEDIT 18 0 "u[0](tau),u[1](tau)" "6$-&%\"uG6#\"\"!6#%$t
auG-&F%6#\"\"\"6#F)" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "omega[1]" "6#
&%&omegaG6#\"\"\"" }{TEXT -1 73 ".  Thus, making the appropriate subst
itutions, we solve for the function " }{XPPEDIT 18 0 "u[2](tau)" "6#-&
%\"uG6#\"\"#6#%$tauG" }{TEXT -1 35 " satisfying the initial conditions
 " }{XPPEDIT 18 0 "u[2](0)=0,u[2]" "6$/-&%\"uG6#\"\"#6#\"\"!F*&F&6#F(
" }{TEXT -1 19 "'(0) = 0, obtaining" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 105 "s2 := combine(dsolve(\{si
mplify(subs(s0,s1,omega[1]=w[1],Q2)), u[2](0)=0, D(u[2])(0)=0\}, u[2](
tau)),trig);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 51 "Again, there are secular terms which we c
ollect via" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 29 "q10 := collect(s2,[tau,sin]);" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 18 "We th
en determine " }{XPPEDIT 18 0 "omega[2]" "6#&%&omegaG6#\"\"#" }{TEXT 
-1 54 " by the requirement that there be no secular terms in " }
{XPPEDIT 18 0 "u[2](tau)" "6#-&%\"uG6#\"\"#6#%$tauG" }{TEXT -1 27 ".  \+
Hence, form the equation" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "q11 := map(coeff,q10,sin(tau));" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 23 "from which we determine" }}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "w[2] := solve(q11,omega[2
]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 13 "The function " }{XPPEDIT 18 0 "u[2](tau)" "6#-&%\"
uG6#\"\"#6#%$tauG" }{TEXT -1 13 " thus becomes" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "subs(omega[
2]=w[2],s2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 17 "The equation for " }{XPPEDIT 18 0 "u[3](t
au)" "6#-&%\"uG6#\"\"$6#%$tauG" }{TEXT -1 19 " contains terms in " }
{XPPEDIT 18 0 "u[0](tau),u[1](tau),u[2](tau),omega[1]" "6&-&%\"uG6#\"
\"!6#%$tauG-&F%6#\"\"\"6#F)-&F%6#\"\"#6#F)&%&omegaG6#F-" }{TEXT -1 5 "
 and " }{XPPEDIT 18 0 "omega[2]" "6#&%&omegaG6#\"\"#" }{TEXT -1 61 ". \+
 Making the appropriate substitutions, and solving for the " }
{XPPEDIT 18 0 "u[3](tau)" "6#-&%\"uG6#\"\"$6#%$tauG" }{TEXT -1 40 " wh
ich satisfies the initial conditions " }{XPPEDIT 18 0 "u[3](0)=0,u[3]
" "6$/-&%\"uG6#\"\"$6#\"\"!F*&F&6#F(" }{TEXT -1 19 "'(0) = 0, we obtai
n" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 123 "s3 := combine(dsolve(\{simplify(subs(s0,s1,s2,omega[
1]=w[1], omega[2]=w[2], Q3)), u[3](0)=0,D(u[3])(0)=0\}, u[3](tau)),tri
g);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 65 "Once again, there are secular terms which we gathe
r together with" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 63 "q12 := collect(subs(omega[1]=w[1],omega[2]=w[2
],s3),[tau,sin]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 14 "To choose the " }{XPPEDIT 18 0 "om
ega[3]" "6#&%&omegaG6#\"\"$" }{TEXT -1 40 " which eliminates the secul
ar term from " }{XPPEDIT 18 0 "u[3](tau)" "6#-&%\"uG6#\"\"$6#%$tauG" }
{TEXT -1 19 ", form the equation" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "q13 := map(coeff,q12,sin(tau
));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 20 "and solve, obtaining" }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "w[3] := solve(q13,om
ega[3]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 31 "Consequently, the solution for " }
{XPPEDIT 18 0 "u[3](tau)" "6#-&%\"uG6#\"\"$6#%$tauG" }{TEXT -1 3 " is
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 23 "subs(omega[3]=w[3],s3);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 46 "and, the firs
t few terms of the expansion for " }{XPPEDIT 18 0 "omega" "6#%&omegaG
" }{TEXT -1 4 " are" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 33 "omega = 1 + add(w[k]*a^k,k=1..3);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 41 "This compares favorably to the expansion " }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "series
(2*Pi/T,a,4);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 28 "found in an earlier section." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 40 "The Linds
tedt solution is then given by " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "U(tau) = subs(s||(0..3),seq(
omega[k]=w[k],k=1..3),UU);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 6 "where " }{XPPEDIT 18 0 "U(t
au)=U(t*omega)" "6#/-%\"UG6#%$tauG-F%6#*&%\"tG\"\"\"%&omegaGF," }
{TEXT -1 3 " = " }{XPPEDIT 18 0 "x(t)" "6#-%\"xG6#%\"tG" }{TEXT -1 6 "
, and " }{XPPEDIT 18 0 "omega" "6#%&omegaG" }{TEXT -1 107 " is as abov
e.  To judge the viability of the Lindstedt expansion, write the appro
ximation as a function of " }{XPPEDIT 18 0 "t" "6#%\"tG" }{TEXT -1 17 
", obtaining, for " }{XPPEDIT 18 0 "x(t)" "6#-%\"xG6#%\"tG" }}{PARA 0 
"" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 85 "q14
 := subs(s||(0..3),seq(omega[k]=w[k],k=1..3), \ntau=t*(1+add(w[k]*a^k,
k=1..3)),UU);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 82 "Since our test of the utility of t
he expansion will be a graph, assign the values " }{XPPEDIT 18 0 "A=2
" "6#/%\"AG\"\"#" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "a=1/10" "6#/%\"a
G*&\"\"\"F&\"#5!\"\"" }{TEXT -1 101 " to the initial displacement and \+
spring parameter, respectively.  Thus, the approximate solution for " 
}{XPPEDIT 18 0 "x(t) " "6#-%\"xG6#%\"tG" }{TEXT -1 3 " is" }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "q15 \+
:= subs(A=2,a=1/10,q14);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 13 "To establish " }{XPPEDIT 18 
0 "xx(t)" "6#-%#xxG6#%\"tG" }{TEXT -1 84 " as a numeric reference solu
tion against which to test Lindstedt's solution, execute" }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 88 "F :=
 dsolve(\{subs(a=1/10,q),x(0)=2,D(x)(0)=0\},x(t),numeric):\nxx := z ->
 subs(F(z),x(t)):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 140 "and then draw Figure 14.11, the g
raph of the Lindstedt solution (the dotted red curve), and a numeric s
olution (the solid black curve), for " }{XPPEDIT 18 0 "100<=t" "6#1\"$
+\"%\"tG" }{XPPEDIT 18 0 "``<=110" "6#1%!G\"$5\"" }{TEXT -1 1 "." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 142 "plot(['xx'(t),q15],t=100..110,color=[black,red], linestyle=[1,2
], xtickmarks=6, ytickmarks=5, labels=[t,`x   `], labelfont=[TIMES,ITA
LIC,12]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 23 "In the neighborhood of " }{XPPEDIT 18 0 "
t=105" "6#/%\"tG\"$0\"" }{TEXT -1 104 ", the two solutions are still i
n agreement.  Figure 14.12, the next plot, shows these two solutions f
or " }{XPPEDIT 18 0 "150<=t" "6#1\"$]\"%\"tG" }{XPPEDIT 18 0 "``<=160
" "6#1%!G\"$g\"" }{TEXT -1 26 ".  In the neighborhood of " }{XPPEDIT 
18 0 "t=155" "6#/%\"tG\"$b\"" }{TEXT -1 39 ", the two solutions start \+
separating.  " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 142 "plot(['xx'(t),q15],t=150..160,color=[black,red]
, linestyle=[1,2], xtickmarks=6, ytickmarks=5, labels=[t,`x   `], labe
lfont=[TIMES,ITALIC,12]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 
"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 14 "However, with " }{XPPEDIT 
18 0 "a=1/10" "6#/%\"aG*&\"\"\"F&\"#5!\"\"" }{TEXT -1 5 " and " }
{XPPEDIT 18 0 "t=100" "6#/%\"tG\"$+\"" }{TEXT -1 52 ", the Lindstedt e
xpansion gives remarkable accuracy." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}}{MARK "1" 0 }
{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }