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{SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT -1 20 "Chapter 5, Problem 8" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 89 "restart : eqn := diff(diff(x
(t),t),t) + 9*x(t) - eps*(gamma*cos(t) - beta*x(t) + x(t)^3);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eqnG,(-%%diffG6$-%\"xG6#%\"tG-%\"$G
6$F,\"\"#\"\"\"F)\"\"**&%$epsGF1,(*&%&gammaGF1-%$cosGF+F1F1*&%%betaGF1
F)F1!\"\"*$)F)\"\"$\"\"\"F1F1F<" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 56 "eqn := subs(x(t) = x0(t) + eps*x1(t) + eps^2*x2(t),eqn):" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "expr := coeff(eqn,eps,0);" }
}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%exprG,&-%%diffG6$-%#x0G6#%\"tG-%\"
$G6$F,\"\"#\"\"\"F)\"\"*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 
"x0(t) := a0*cos(3*t) + b0*sin(3*t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "
6#>-%#x0G6#%\"tG,&*&%#a0G\"\"\"-%$cosG6#,$F'\"\"$F+F+*&%#b0GF+-%$sinGF
.F+F+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "expr := combine(co
eff(eqn,eps,1),trig);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%%exprG,<-%%
diffG6$-%#x1G6#%\"tG-%\"$G6$F,\"\"#\"\"\"F)\"\"**&%&gammaGF1-%$cosGF+F
1!\"\"*(%%betaGF1%#a0GF1-F66#,$F,\"\"$F1F1*(F9\"\"\"%#b0GF1-%$sinGF<F1
F1*&)F:F>F@-F66#,$F,F2F1#F7\"\"%*&FEF@F;F@#!\"$FJ*()F:F0F@FAF@-FCFGF1F
L*(FOF@FAF@FBF@FL*(F:F@F;F@)FAF0F@FL*(F:F@FSF@FFF@#F>FJ*&)FAF>F@FPF@#F
1FJ*&FWF@FBF@FL" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 52 "Now find the s
ecular terms from the O(eps) equation." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 29 "eqn1 := coeff(expr,cos(3*t));" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%%eqn1G,(*&%%betaG\"\"\"%#a0GF(F(*$)F)\"\"$\"\"\"#!\"$
\"\"%*&F)F-)%#b0G\"\"#F-F." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
29 "eqn2 := coeff(expr,sin(3*t));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>
%%eqn2G,(*&%%betaG\"\"\"%#b0GF(F(*&)%#a0G\"\"#\"\"\"F)F.#!\"$\"\"%*$)F
)\"\"$F.F/" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "solve(\{eqn1=
0,eqn2=0\},\{a0,b0\});" }}{PARA 12 "" 1 "" {XPPMATH 20 "6&<$/%#b0G\"\"
!/%#a0GF&<$/F(,$-%'RootOfG6#,&%%betaG!\"\"*$)%#_ZG\"\"#\"\"\"\"\"$F5F$
<$/F%F+F'<$/F%F%/F(-F-6#,(F2F7F0!\"%*$)F%F5F6F7" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 97 "So there are two constraints coming from the O(eps) \+
secular terms.  One is  (a0,b0) = (0,0).  The" }}{PARA 0 "" 0 "" 
{TEXT -1 100 "other is   a0^2 + b0^2 = 4/3 beta.  This means that the \+
amplitude of the O(1) solution is fixed, but" }}{PARA 0 "" 0 "" {TEXT 
-1 101 "the phase is free.  So we go back and take x0(t) = 2 sqrt(beta
/3) cos(3 t + phi_0) where phi_0 is the" }}{PARA 0 "" 0 "" {TEXT -1 
29 "additional degree of freedom." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 41 "x0(t) := 2*sqrt(beta/3)*cos(3*t + phi_0);" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>-%#x0G6#%\"tG,$*(-%%sqrtG6#\"\"$\"\"\"-F+6#
%%betaGF.-%$cosG6#,&F'F-%&phi_0G\"\"\"F7#\"\"#F-" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 39 "expr := combine(coeff(eqn,eps,1),trig);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%exprG,*-%%diffG6$-%#x1G6#%\"tG-%\"$
G6$F,\"\"#\"\"\"F)\"\"**&%&gammaGF1-%$cosGF+F1!\"\"*(-%%sqrtG6#\"\"$\"
\"\")%%betaG#F<F0F=-F66#,&F,F2%&phi_0GF<F1#!\"#F2" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 21 "dsolve(expr=0,x1(t));" }}{PARA 12 "" 1 "" 
{XPPMATH 20 "6#/-%#x1G6#%\"tG,**&,**&%&gammaG\"\"\"-%$cosG6#,$F'\"\"%F
-#F-\"#C*&F,\"\"\"-F/6#,$F'\"\"#F-#F-\"#7*(-%%sqrtG6#\"\"$F6)%%betaG#F
AF:F6-F/6#,&F'F<%&phi_0GFAF-#F-\"$C$*(F>F6FBF6-F/6#,&F'\"\"'FHFAF-#!\"
\"\"$i\"F--F/6#,$F'FAF-F-*&,**&F,F6-%$sinGF8F-F;*&F,F6-FZF0F-F3*(F>F6F
BF6-FZFMF-#F-FR*(F>F6FBF6-FZFFF-FIF--FZFTF-F-*&%$_C1GF-FSF6F-*&%$_C2GF
-F\\oF6F-" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "x1(t) := rhs(%
):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "x1(t) := subs(\{_C1=a
1,_C2=b1\},x1(t));" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>-%#x1G6#%\"tG,*
*&,**&%&gammaG\"\"\"-%$cosG6#,$F'\"\"%F-#F-\"#C*&F,\"\"\"-F/6#,$F'\"\"
#F-#F-\"#7*(-%%sqrtG6#\"\"$F6)%%betaG#FAF:F6-F/6#,&F'F<%&phi_0GFAF-#F-
\"$C$*(F>F6FBF6-F/6#,&F'\"\"'FHFAF-#!\"\"\"$i\"F--F/6#,$F'FAF-F-*&,**&
F,F6-%$sinGF8F-F;*&F,F6-FZF0F-F3*(F>F6FBF6-FZFMF-#F-FR*(F>F6FBF6-FZFFF
-FIF--FZFTF-F-*&%#a1GF-FSF6F-*&%#b1GF-F\\oF6F-" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 66 "We now look for the O(eps^2) secular terms to determ
ine a1 and b1." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "expr := c
ombine(coeff(eqn,eps,2),trig);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%%e
xprG,>-%%diffG6$-%#x2G6#%\"tG-%\"$G6$F,\"\"#\"\"\"F)\"\"**(%%betaGF1%&
gammaGF1-%$cosGF+F1#!\"\"\"\")*()F4#\"\"&F0\"\"\"-%%sqrtG6#\"\"$F?-F76
#,&F,F2%&phi_0GFCF1#F1\"$C$*(F4F?%#a1GF1-F76#,$F,FCF1F9*(F4F?%#b1GF1-%
$sinGFMF1F9*(F4F?F5F?-F76#,&F,F>FGF0F1F8*(F4F?F5F?-F76#,&F,\"\"(FGF0F1
F8*(F<F?F@F?-F76#,&F,FCFGF1F1FH*(F<F?F@F?-F76#,&F,\"#:FGF>F1FH*(F4F?FK
F?-F76#,&F,FCFGF0F1F9*(F4F?FKF?-F76#,&F,F2FGF0F1F9*(F4F?FPF?-FRFeoF1F9
*(F4F?FPF?-FRFaoF1F1" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "RHS
 := expr - diff(diff(x2(t),t),t) - 9*x2(t);" }}{PARA 12 "" 1 "" 
{XPPMATH 20 "6#>%$RHSG,:*(%%betaG\"\"\"%&gammaGF(-%$cosG6#%\"tGF(#!\"
\"\"\")*()F'#\"\"&\"\"#\"\"\"-%%sqrtG6#\"\"$F6-F+6#,&F-\"\"*%&phi_0GF:
F(#F(\"$C$*(F'F6%#a1GF(-F+6#,$F-F:F(F/*(F'F6%#b1GF(-%$sinGFEF(F/*(F'F6
F)F6-F+6#,&F-F4F?F5F(F.*(F'F6F)F6-F+6#,&F-\"\"(F?F5F(F.*(F2F6F7F6-F+6#
,&F-F:F?F(F(F@*(F2F6F7F6-F+6#,&F-\"#:F?F4F(F@*(F'F6FCF6-F+6#,&F-F:F?F5
F(F/*(F'F6FCF6-F+6#,&F-F>F?F5F(F/*(F'F6FHF6-FJF]oF(F/*(F'F6FHF6-FJFinF
(F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 120 "Now I have to be careful,
 if I ask for coeff(test,cos(3*t)), then it won't give me all the 3-mo
de stuff.  I want to make" }}{PARA 0 "" 0 "" {TEXT -1 117 "sure that R
HS of  \"d^2/dt^2 x2 + 9 x2 = RHS\" is orthogonal to cos(3t) and sin(3
t).  So I take the inner product below" }}{PARA 0 "" 0 "" {TEXT -1 51 
"and then demand that the two coefficients are zero." }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 0 "" }{MPLTEXT 1 0 53 "eqn1 := combine(int(RHS*cos(
3*t),t=0..2*Pi)/Pi,trig);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%eqn1G,
**(%%betaG\"\"\"%#a1GF(-%$cosG6#,$%&phi_0G\"\"#F(!\"\"*&F'\"\"\"F)F2F0
*()F'#\"\"&F/F2-%%sqrtG6#\"\"$F2-F+6#F.F(#F(\"$C$*(F'F2%#b1GF(-%$sinGF
,F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "eqn2 := combine(in
t(RHS*sin(3*t),t=0..2*Pi)/Pi,trig);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6
#>%%eqn2G,**(%%betaG\"\"\"%#a1GF(-%$sinG6#,$%&phi_0G\"\"#F(F(*&F'\"\"
\"%#b1GF(!\"\"*(F'F1F2F1-%$cosGF,F(F(*()F'#\"\"&F/F1-%%sqrtG6#\"\"$F1-
F+6#F.F(#F3\"$C$" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "sols :=
 solve(\{eqn1=0,eqn2=0\},\{a1,b1\});" }}{PARA 11 "" 1 "" {XPPMATH 20 "
6#>%%solsG<$/%#b1G,$*&,&*(%%betaG\"\"\"%#a1GF--%$cosG6#%&phi_0GF-\"$['
*&)F,#\"\"&\"\"#\"\"\"-%%sqrtG6#\"\"$F9!\"\"F9*&F,\"\"\"-%$sinGF1\"\"
\"!\"\"#F-F3/F.F." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 101 "Okay, so I'
m getting that a1 is free and b1 is determined by a1.  Which is the sa
me as we had before." }}{PARA 0 "" 0 "" {TEXT -1 97 "Although we don't
 have a1^2 + b1^2 = const, which we did have for a0 and b0.  (That's w
hy I could" }}{PARA 0 "" 0 "" {TEXT -1 46 "put the freedom in a0 and b
0 in a phase shift." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "b1 :
= rhs(sols[1]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#b1G,$*&,&*(%%bet
aG\"\"\"%#a1GF*-%$cosG6#%&phi_0GF*\"$['*&)F)#\"\"&\"\"#\"\"\"-%%sqrtG6
#\"\"$F6!\"\"F6*&F)\"\"\"-%$sinGF.\"\"\"!\"\"#F*F0" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 24 "combine(a1^2+b1^2,trig);" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#*&,(*&)%%betaG\"\"#\"\"\")%#a1GF(F)!'o*R\"**)F'#\"\"(
F(F)F+\"\"\"-%$cosG6#%&phi_0GF1-%%sqrtG6#\"\"$F)\"$K%*$)F'\"\"&F)!\"\"
F),&*$F&F)!&%)*p*&F&F)-F36#,$F5F(F1\"&%)*p!\"\"" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 29 "x1(t) := combine(x1(t),trig);" }}{PARA 12 "" 
1 "" {XPPMATH 20 "6#>-%#x1G6#%\"tG,$*&,.*(%%betaG\"\"\"%&gammaGF--%$si
nG6#,&F'F-%&phi_0GF-F-\"#\")*(F,\"\"\"F.F6-F06#,&F'F-F3!\"\"F-!#\")*()
F,#\"\"&\"\"#F6-%%sqrtG6#\"\"$F6-F06#,&F'\"\"*F3\"\"%F-!\"#*(F=F6FAF6-
F06#,&F'FHF3F@F-F@*(F,F6%#a1GF--F06#,&F'FDF3F-F-\"%'H\"*(-F06#,$F'FDF-
F=F6FAF6FJF6*&F,\"\"\"-F06#F3\"\"\"!\"\"#F-FT" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 26 "x(t) := x0(t) + eps*x1(t);" }}{PARA 12 "" 1 "" 
{XPPMATH 20 "6#>-%\"xG6#%\"tG,&*(-%%sqrtG6#\"\"$\"\"\"-F+6#%%betaGF.-%
$cosG6#,&F'F-%&phi_0G\"\"\"F7#\"\"#F-*&*&%$epsGF7,.*(F1F7%&gammaGF7-%$
sinG6#,&F'F7F6F7F7\"#\")*(F1F.F?F.-FA6#,&F'F7F6!\"\"F7!#\")*()F1#\"\"&
F9F.F*F.-FA6#,&F'\"\"*F6\"\"%F7!\"#*(FLF.F*F.-FA6#,&F'FRF6F9F7F9*(F1F.
%#a1GF7-FAF4F7\"%'H\"*(-FA6#,$F'F-F7FLF.F*F.FTF7F.*&F1\"\"\"-FA6#F6\"
\"\"!\"\"#F7Ffn" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 115 "Note that eve
n if you had assumed that at the O(1) level the solution had a0  = 2 s
qrt(beta/3) and b0 = 0, then you" }}{PARA 0 "" 0 "" {TEXT -1 84 "still
 get a degree of freedom at the O(eps) level from the O(eps^2) secular
 terms..." }{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" 
}}}}{MARK "28 1 0" 84 }{VIEWOPTS 1 1 0 1 1 1803 }