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{SECT 0 {EXCHG {PARA 256 "" 0 "" {TEXT 256 34 "Trajectoires de champs \+
de vecteurs" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 257 "" 0 "" 
{TEXT 260 26 "Felice Ronga - 18 mai 1999" }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}}{SECT 0 {PARA 3 "" 0 "" {TEXT 257 12 "Introduction" }}{PARA 
258 "" 0 "" {TEXT 258 145 "On vous propose d'etudier experimentalement
 l'allure des trajectoires de champs de vecteurs et des phenomenes de \+
bifurcation, a l'aide de Maple. " }}{PARA 259 "" 0 "" {TEXT -1 0 "" }}
{PARA 260 "" 0 "" {TEXT 259 27 "D'abord un peu de theorie. " }}{PARA 
260 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 4 "" 0 "" {TEXT 261 56 "Syst
emes lineaires a coefficients constants dans le plan" }}{PARA 261 "" 
0 "" {TEXT 265 75 "Considerons un systeme de 2  equations lineaires a \+
coefficients constants :" }}{PARA 262 "" 0 "" {TEXT -1 0 "" }}{PARA 
263 "" 0 "" {TEXT 262 16 "                " }{TEXT 264 22 " x'(t)=a x(
t) + b y(t)" }}{PARA 0 "" 0 "" {TEXT -1 33 "            y'(t)=c x(t) +
 d y(t)" }}{PARA 264 "" 0 "" {TEXT -1 0 "" }}{PARA 265 "" 0 "" {TEXT 
263 97 "L'allure des trajectoires est determinee par les valeurs propr
es de la matrice des coefficients: " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "A:=array(1..2,1..2,[[a,b],[
c,d]]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"AG-%'MATRIXG6#7$7$%\"aG
%\"bG7$%\"cG%\"dG" }}}{PARA 265 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 33 "Voici la liste des cas possible :" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 38 "1. A est diagonalisable s
ur les reels." }}{PARA 0 "" 0 "" {TEXT -1 68 "   1a. les deux valeurs \+
propres sont positives ou les deux negatives" }}{PARA 0 "" 0 "" {TEXT 
-1 73 "   1b. une valeur propre est positive (resp. negative), l'autre
 est nulle" }}{PARA 0 "" 0 "" {TEXT -1 59 "   1c. une valeur propre es
t positive, l'autre est negative" }}{PARA 0 "" 0 "" {TEXT -1 392 "2. A
 possede une valeur propre lambda non reelle, de la forme a+Ib; alors \+
le conjugue a-Ib de lambda est distinct de lambda (i.e. b est differen
t de 0), et est aussi valeur propre. Donc A est diagonalisable sur les
 complexes. Soit v un vecteur propre complexe de A correspondant a la \+
valeur propre a-Ib; en prenant pour base de R^2 la partie relle et ima
ginaire de v, A se met sous la forme :" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 35 "A:=array(1..2,1..2,[[a,-b],[b,a]]);" }}{PARA 11 "" 1 
"" {XPPMATH 20 "6#>%\"AG-%'MATRIXG6#7$7$%\"aG,$%\"bG!\"\"7$F,F*" }}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 22 "Il y a de
ux sous-cas :" }}{PARA 0 "" 0 "" {TEXT -1 44 "    2a. lambda est purem
ent imaginaire (a=0)" }}{PARA 0 "" 0 "" {TEXT -1 90 "    2b. la partie
 reelle a de lambda est strictement positive (resp. strictement negati
ve)" }}{PARA 0 "" 0 "" {TEXT -1 147 "3. A n'est pas diagonalisable. Al
ors elle possede necessairement une seule valeur propre, qui doit etre
 reelle. Elle peut se mettre sous la forme :" }}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 34 "A:=array(1..2,1..2,[[1,a],[0,1]]);" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#>%\"AG-%'MATRIXG6#7$7$\"\"\"%\"aG7$\"\"!F*" }}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 16 "av
ec a non nul. " }}{PARA 0 "" 0 "" {TEXT -1 186 "Et voici l'allure des \+
trajectoires des equations correspondantes. Maple met a disposition la
 procedure \"dsolve\" pour calculer des solutions d'equations differen
tielles. Voici son usage :" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 
"" 0 "" {TEXT -1 24 "on donne les equations :" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 50 "e1:=diff(x(t),t)=-2*x(t);e2:=diff(y(t),t)=-3*y(t
);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#e1G/-%%diffG6$-%\"xG6#%\"tGF,
,$F)!\"#" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#e2G/-%%diffG6$-%\"yG6#%
\"tGF,,$F)!\"$" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 35 "on donne les co
nditions initiales :" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "ini
1:=x(0)=1,y(0)=1;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%ini1G6$/-%\"xG
6#\"\"!\"\"\"/-%\"yGF)F+" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 114 "on d
emande de resoudre; les solutions seront sous forme d'une liste, qu'on
 assignera a la nouvelle variable \"s1\" :" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 37 "s1:=dsolve([e1,e2,ini1],[x(t),y(t)]);" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#>%#s1G<$/-%\"yG6#%\"tG-%$expG6#,$F*!\"$/-%\"xGF)
-F,6#,$F*!\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "ini2:=x(0)
=1,y(0)=-1;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%ini2G6$/-%\"xG6#\"\"
!\"\"\"/-%\"yGF)!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "s2
:=dsolve([e1,e2,ini2],[x(t),y(t)]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6
#>%#s2G<$/-%\"xG6#%\"tG-%$expG6#,$F*!\"#/-%\"yGF),$-F,6#,$F*!\"$!\"\"
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 107 "Maple a donc pu nous calcule
r les solutions explicitement. On va calculer une solution dans chaque
 cadran :" }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 117 "ini3:=x(0)=-1,y(0)=1;ini4:=x(0)=-1,y(0)=-1;s3:=dso
lve([e1,e2,ini3],[x(t),y(t)]);s4:=dsolve([e1,e2,ini4],[x(t),y(t)]);" }
}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%ini3G6$/-%\"xG6#\"\"!!\"\"/-%\"yGF
)\"\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%ini4G6$/-%\"xG6#\"\"!!\"
\"/-%\"yGF)F+" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#s3G<$/-%\"xG6#%\"t
G,$-%$expG6#,$F*!\"#!\"\"/-%\"yGF)-F-6#,$F*!\"$" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#s4G<$/-%\"xG6#%\"tG,$-%$expG6#,$F*!\"#!\"\"/-%\"yGF)
,$-F-6#,$F*!\"$F1" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 278 289 "ATTENTION \+
: Maple renverse parfois  l'ordre de x et y dans le resultat de l'appe
l a \"dsolve\", ce qui a pour consequence de fausser les graphiques (v
oir s1 ci-dessus). Ce phenomene est aleatoire et n'est pas reproductib
le sur demande. Il fau en tenir compter lorsqu'on trace les graphiques
." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 193 "On \+
va calculer les graphiques des 4 trajectoires pour t=-1 a t=0. Pour ce
la, on applique la commande \"rhs\" (=\"right hand side\") a s1,..,s4 \+
 pour extraire les parametrisations des trajectoires :" }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "rhs(s1
[1]);rhs(s1[2]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$expG6#,$%\"tG!
\"$" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$expG6#,$%\"tG!\"#" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 294 "with(plots):p1:=plot([rhs(s
1[2]),rhs(s1[1]),t=0..1],color=blue):p2:=plot([rhs(s2[1]),rhs(s2[2]),t
=0..1],color=red):p3:=plot([rhs(s3[1]),rhs(s3[2]),t=0..1],color=green)
:p4:=plot([rhs(s4[1]),rhs(s4[2]),t=0..1],color=yellow):ini:=pointplot(
[[1,1],[-1,1],[-1,-1],[1,-1]],symbol=circle,color=black):" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 308 "Il est preferable de terminer ces comman
des par \":\" plutot que \";\", car la reponse est un texte interminab
le decrivant le graphique (essayez). Maintenant on peut afficher les g
raphiques calcules; les conditions initiales sont representes par de p
etits cercles et permettent d'identifier le sens de parcours :" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 97 "display(p1,p2,p3,p4,ini,titl
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}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 251 "Si t augmente, on s'approche d
e l'origine. On dit qu'un tel point singulier est un attracteur; on pa
rle de repulseur lorsque, au contraire, toutes les trajectoires s'eloi
gnent de l'origine; c'est le cas si les 2 valeurs propres sont reelles
 positives." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 84 "Le cas 1b est laisse au lecteur. Passons a 1c :
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{MPLTEXT 1 0 22 "e2:=diff(y(t),t)=y(t);" }}{PARA 11 "" 1 "" {XPPMATH 
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{MPLTEXT 1 0 108 "ini1:=x(0)=1/10,y(0)=1/10;ini2:=x(0)=-1/10,y(0)=1/10
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{MPLTEXT 1 0 148 "s1:=dsolve([e1,e2,ini1],[x(t),y(t)]);s2:=dsolve([e1,
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{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#s2G<$/-%\"xG6#%\"tG,$-%$expG6#,$F*!
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s(s3[1]),rhs(s3[2]),t=0..1],color=green):p4:=plot([rhs(s4[1]),rhs(s4[2
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{PARA 0 "" 0 "" {TEXT -1 18 "Passons au cas 2 :" }}{PARA 0 "" 0 "" 
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Q%eE?R#Fdal7$Fi\\p$!1m=0ak*))H#Fdal-F\\]p6&F^]pF_]pF(F(-%'POINTSG6&F'F
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res~complexes~conjuguees~de~partie~reelle~negative~(spirale)G-%%FONTG6
%%*HELVETICAG%%BOLDG\"\")" 2 324 321 321 2 0 1 0 2 9 0 4 2 1.000000 
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0 0 0 0 0 }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 129 "et on obtient des sp
irales, qui tendent vers l'origine lorsque t augmente. Dans ce dernier
 cas, la matrice des coefficients est :" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 43 "A:=array(1..2,1..2,[[-1/10,-1],[1,-1/10]]);" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>%\"AG-%'MATRIXG6#7$7$#!\"\"\"#5F+7$\"\"\"F*
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 39 "Verifions que l'on est dans l
e cas 2 b:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "with(linalg);
" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#7^r%.BlockDiagonalG%,GramSchmidtG%
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G%(adjointG%&angleG%(augmentG%(backsubG%%bandG%&basisG%'bezoutG%,block
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ausselimG%*gaussjordG%(geneqnsG%*genmatrixG%%gradG%)hadamardG%(hermite
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0 "" {MPLTEXT 1 0 15 "eigenvalues(A);" }}{PARA 11 "" 1 "" {XPPMATH 20 
"6$,&#!\"\"\"#5\"\"\"%\"IGF',&F$F'F(F%" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 91 "Si les deux valeurs propres sont purement imaginaires, le
s trajectoires sont des ellipses :" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "e1:=diff(x(t),t)=x(t)-5*y(t
);e2:=diff(y(t),t)=2*x(t)-y(t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#
e1G/-%%diffG6$-%\"xG6#%\"tGF,,&F)\"\"\"-%\"yGF+!\"&" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#>%#e2G/-%%diffG6$-%\"yG6#%\"tGF,,&-%\"xGF+\"\"#F)!\"
\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 94 "s1:=dsolve([e1,e2,x(0
)=2/3,y(0)=0],[x(t),y(t)]);s2:=dsolve([e1,e2,x(0)=1,y(0)=0],[x(t),y(t)
]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#s1G<$/-%\"xG6#%\"tG,&-%$cosG
6#,$F*\"\"$#\"\"#F0-%$sinGF.#F2\"\"*/-%\"yGF),$F3#\"\"%F6" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#>%#s2G<$/-%\"xG6#%\"tG,&-%$cosG6#,$F*\"\"$\"\"
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{MPLTEXT 1 0 206 "p1:=plot([rhs(s1[1]),rhs(s1[2]),t=0..2*Pi/3],color=b
lue):p2:=plot([rhs(s2[1]),rhs(s2[2]),t=0..2*Pi/3],color=red):display(p
1,p2,title=`2 valeurs propres imaginaires conjuguees (centre)`,scaling
=constrained);" }}{PARA 13 "" 1 "" {INLPLOT "6&-%'CURVESG6$7S7$$\"1mmm
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$\"1))z^9Z=<@F*$!1t3fP<BsdF*7$$\"1xY/WYTqMF*$!1\"f=W_ryF&F*7$$\"1A/^uj
s^ZF*$!10\"[i.[^p%F*7$$\"1'4*)4bll&eF*$!1(GRkpDs3%F*7$$\"1G^cK&Ge,(F*$
!1'H;k(z,<LF*7$$\"1twZnZMQzF*$!1^$4(yBJtDF*7$$\"1()pdhUH'y)F*$!1pNG?#*
yO<F*7$$\"1y=\"=x*H`%*F*$!1B>q4`6t!*F17$$\"10og++++5Fi[l$\"1(zny>4O@\"
!#C-Fjz6&F\\[lF][lF+F+-%&TITLEG6#%R2~valeurs~propres~imaginaires~conju
guees~(centre)G-%(SCALINGG6#%,CONSTRAINEDG" 2 356 356 356 2 0 1 0 2 9 
0 4 1 1.000000 45.000000 45.000000 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 
1 1 0 0 0 0 0 0 0 0 0 0 0 }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 48 "Ici o
n dit que le point singulier est un centre." }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{SECT 0 {PARA 5 "" 0 "" {TEXT -1 10 "Exercice 1" }}
{PARA 0 "" 0 "" {TEXT -1 267 "Tracer les trajectoires des systemes sui
vants; determiner la nature du point critique (0,0). Controlez vos rep
onses en calculant les valeurs propres de la matrice du systeme. De pr
eference, ecrivez une procedure pour dessiner des trajectoires a parti
r des  equations." }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" 
{TEXT -1 13 "a)   x'=-x-2y" }}{PARA 0 "" 0 "" {TEXT -1 14 "      y'=4x
.5y" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 13 "b)
  x'=-3x+4y" }}{PARA 0 "" 0 "" {TEXT -1 14 "     y'=-2x+3y" }}{PARA 0 
"" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 12 "c)  x'=5x+2y" }}
{PARA 0 "" 0 "" {TEXT -1 14 "    y'=-17x-5y" }}{SECT 0 {PARA 20 "" 0 "
" {TEXT -1 9 "Solutions" }}{PARA 0 "" 0 "" {TEXT -1 93 "La procedure s
uivante, quoique un peu rustique, permet de visualiser l'allure des so
lutions :" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "calin := proc(e
q1,eq2,a,b)" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "local e1,e2,s1,s2,s3
,s4,p1,p2,p3,p4;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "e1:=diff(x(t),t
)=eq1;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "e2:=diff(y(t),t)=eq2;" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "s1:=dsolve([e1,e2,x(0)=1,y(0)=1],[x
(t),y(t)]);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "s2:=dsolve([e1,e2,x(
0)=-1,y(0)=1],[x(t),y(t)]);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "s3:=
dsolve([e1,e2,x(0)=-1,y(0)=-1],[x(t),y(t)]);" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 47 "s4:=dsolve([e1,e2,x(0)=1,y(0)=-1],[x(t),y(t)]);" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 289 "p1:=plot([rhs(s1[1]),rhs(s1[2]),t=
a..b],color=red,scaling=constrained):p2:=plot([rhs(s2[1]),rhs(s2[2]),t
=a..b],color=blue,scaling=constrained):p3:=plot([rhs(s3[1]),rhs(s3[2])
,t=a..b],scaling=constrained,color=black):p4:=plot([rhs(s4[1]),rhs(s4[
2]),t=a..b],scaling=constrained,color=green);" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 21 "display(p1,p2,p3,p4);" }}{PARA 0 "> " 0 "" {MPLTEXT 
1 0 4 "end;" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%&calinG:6&%$eq1G%$eq2
G%\"aG%\"bG6,%#e1G%#e2G%#s1G%#s2G%#s3G%#s4G%#p1G%#p2G%#p3G%#p4G6\"F6C-
>8$/-%%diffG6$-%\"xG6#%\"tGFA9$>8%/-F<6$-%\"yGF@FA9%>8&-%'dsolveG6$7&F
9FD/-F?6#\"\"!\"\"\"/-FIFSFU7$F>FH>8'-FN6$7&F9FD/FR!\"\"FVFX>8(-FN6$7&
F9FDFhn/FWFinFX>8)-FN6$7&F9FDFQF_oFX>8*-%%plotG6%7%-%$rhsG6#&FL6#FU-F
\\p6#&FL6#\"\"#/FA;9&9'/%&colorG%$redG/%(scalingG%,constrainedG>8+-Fho
6%7%-F\\p6#&FZF_p-F\\p6#&FZFcpFep/Fjp%%blueGF\\q>8,-Fho6%7%-F\\p6#&F[o
F_p-F\\p6#&F[oFcpFepF\\q/Fjp%&blackG>8--Fho6%7%-F\\p6#&FaoF_p-F\\p6#&F
aoFcpFepF\\q/Fjp%&greenG-%(displayG6&FfoF`qF]rFjrF6F6" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "calin(-x(t)-2*y(t),4*x(t)-5*y(t),0,
1);" }}{PARA 13 "" 1 "" {INLPLOT "6'-%'CURVESG6$7S7$$\"\"\"\"\"!F(7$$
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\\F.7$$\"1n3]tX.oVF.$\"1)\\\"z@bMnYF.7$$\"1(e\\PeM0I%F.$\"1w!30S,QT%F.
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1.)y(*4G@+%F.$\"1`yQ>%Gaj$F.7$$\"1u+SM'yv)QF.$\"1'==w7]TS$F.7$$\"1$*)e
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1V72BRu`BF.7$$\"1-TQ$3aF3$F.$\"1B^c1%>8=#F.7$$\"1d4UE7=UHF.$\"1.T>6z[5
?F.7$$\"1KL,u%)[0GF.$\"1`l(>EaD&=F.7$$\"1?'[nD^Qm#F.$\"1$pfnI*e'p\"F.7
$$\"1))4`TmUHDF.$\"1k'G$y_=b:F.7$$\"1vAqN.[%R#F.$\"1B)p`xN#>9F.7$$\"1e
:dWmfjAF.$\"1O!4dijFH\"F.7$$\"1_s^9RAY@F.$\"1N\"*zb3i$=\"F.7$$\"1M#oN=
@a,#F.$\"1otb-<am5F.7$$\"1<4%>L??!>F.$\"1%e.Xwfyo*Ffs7$$\"1!y./.R]y\"F
.$\"1\\EY><)[r)Ffs7$$\"1inwH%4qn\"F.$\"1rL%\\x#pZyFfs7$$\"1=_q/\\Kl:F.
$\"13V*4`xB)pFfs-Fiz6&F[[lF*F\\[lF*-%(SCALINGG6#%,CONSTRAINEDG" 2 324 
324 324 2 0 1 0 2 9 0 4 1 1.000000 45.000000 45.000000 10030 0 10056 
10074 0 0 0 20030 0 12010 0 0 0 0 0 0 0 1 1 0 0 0 56 288 0 0 0 0 0 0 }
}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 37 "Hein ? C'est choquant ! (pourquo
i ?)." }}}{SECT 0 {PARA 20 "" 0 "" {TEXT -1 10 "Pourquoi ?" }}{PARA 0 
"" 0 "" {TEXT -1 107 "Parce-que il y a un point par lequel passent 2 t
rajectoires. Ne me demandez pas pourquoi Maple a fait cela." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "restart;with(plots):with(lin
alg):" }}{PARA 7 "" 1 "" {TEXT -1 34 "Warning, `new definition for nor
m`" }}{PARA 7 "" 1 "" {TEXT -1 35 "Warning, `new definition for trace`
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 73 "On remonte de quelques lignes
 pour recharger la procedure \"calin\". Puis :" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 38 "calin(-x(t)-2*y(t),4*x(t)-5*y(t),0,1);" }}
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$$!1Udj*yY0d\"F.$\"1Sr#y(*yh+\"Fj[l7$$!1(HvwwR*)Q'Fhs$\"1/_Y'*>#>!**F.
7$$\"1n!\\\\qko9#Fhs$\"1mv!odLDq*F.7$$\"1vx@Ff<G'*Fhs$\"1tR#yMuWZ*F.7$
$\"1H<lvi%Rj\"F.$\"11h<!GTV@*F.7$$\"1j8Map**[@F.$\"1,'e#\\yWn*)F.7$$\"
17$*fn%p0l#F.$\"14LTs.Pt')F.7$$\"1T+`)fHy2$F.$\"1e/9rx_k$)F.7$$\"1nSR+
OfBMF.$\"1jV<(\\Us0)F.7$$\"1QwO*f'R'o$F.$\"1h!H#3&HBx(F.7$$\"1aP#Q-x7%
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\"1%z!\\/149UF.$\"1()))HOu`[TF.7$$\"1OPXj:>7TF.$\"1]lHlQ?%)QF.7$$\"1.)
y(*4G@+%F.$\"1`yQ>%Gaj$F.7$$\"1u+SM'yv)QF.$\"1'==w7]TS$F.7$$\"1%*)e#>A
#Hv$F.$\"1zJ.ytLeJF.7$$\"1K!)pi,oEOF.$\"1_)z'pQOZHF.7$$\"1NAK=f!y[$F.$
\"1_n(Q8KCt#F.7$$\"1R%G5n\"RfLF.$\"15M0GA#oa#F.7$$\"1#yK4iis@$F.$\"1V7
2BRu`BF.7$$\"1-TQ$3aF3$F.$\"1B^c1%>8=#F.7$$\"1e4UE7=UHF.$\"1.T>6z[5?F.
7$$\"1LL,u%)[0GF.$\"1`l(>EaD&=F.7$$\"1?'[nD^Qm#F.$\"1$pfnI*e'p\"F.7$$
\"1))4`TmUHDF.$\"1k'G$y_=b:F.7$$\"1vAqN.[%R#F.$\"1B)p`xN#>9F.7$$\"1e:d
WmfjAF.$\"1O!4dijFH\"F.7$$\"1_s^9RAY@F.$\"1N\"*zb3i$=\"F.7$$\"1M#oN=@a
,#F.$\"1otb-<am5F.7$$\"1<4%>L??!>F.$\"1%e.Xwfyo*Fhs7$$\"1!y./.R]y\"F.$
\"1\\EY><)[r)Fhs7$$\"1jnwH%4qn\"F.$\"1rL%\\x#pZyFhs7$$\"1=_q/\\Kl:F.$
\"13V*4`xB)pFhs-Fiz6&F[[lF*F\\[lF*-%(SCALINGG6#%,CONSTRAINEDG" 2 324 
324 324 2 0 1 0 2 9 0 4 1 1.000000 45.000000 45.000000 10030 0 10056 
10074 0 0 0 20030 0 12010 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 }}
{PARA 0 "" 0 "" {TEXT -1 41 "C'est mieux ! Voyons les valurs propres :
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "eigenvalues([[-1,-2],[4
,-5]]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$,&!\"$\"\"\"%\"IG\"\"#,&F$F
%F&!\"#" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 122 "et on constate avec s
atisfaction qu'elles ont toutes deux des parties reelles negatives. b)
 et c) sont laisses au lecteur." }}}}}}{SECT 0 {PARA 4 "" 0 "" {TEXT 
266 21 "Sytemes non lineaires" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 202 "
Lorsqu'on a une equation non lineaire, Maple n'est pas capable de trou
ver explicitement des solution. Il faut lui demander  de calculer les \+
solutions numeriquement par l'option \"numeric\" dans \"dsolve\" :" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 110 "nonlin:=dsolve(\{diff(x(t),t)=-x(t)+y(t)^2,diff(y(t),t)=-2*y(t)
+x(t)^2,x(0)=0.1,y(0)=0.1\},\{x(t),y(t)\},numeric);" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#>%'nonlinG:6#%(rkf45_xG6'%\"iG%(rkf45_sG%)outpointG%#
r1G%#r2G6#%aoCopyright~(c)~1993~by~the~University~of~Waterloo.~All~rig
hts~reserved.G6\"C&>8&-%&evalfG6#9$@$52-%$absG6#,$F3!\"\"-F<6#,&&%,loc
_controlG6#\"\"#\"\"\"F3F?4-%'memberG6$&FD6#\"\"'<*F?FG!\"#FF$FG\"\"!$
F?FR$FPFR$FFFRC%>FD-%%copyG6#=F06#;FG\"#LE\\[lB\"#I\"\"!\"#:F[o\"#JF[o
\"#;F[o\"\"\"\"\"#\"#KF[o\"#<F[oF`oF[o\"#LF[o\"#=F[o\"\"$F[o\"#>F[o\"
\"%$F_o!\")\"#?$F_o!\"\"\"\"&Fho\"#@F[p\"\"'F_o\"#AF[o\"\"($F_o!\"*\"#
BF[o\"\")\"&++$\"#CF[o\"\"*\"%+5\"#DF[o\"#5F[o\"#EF[o\"#6F[o\"#FF[o\"#
7F[o\"#GF[o\"#8F[o\"#HF[o\"#9F[o>%'loc_y0G-FY6#=F06#;FGFFE\\[l#F_oF[pF
`oF[p>%'loc_y1G-FY6#=F0FiqE\\[l!@$0F;FRC$>&FD6#\"\"$F3@%1%'DigitsG-%'e
valhfG6#F[sC$>8%-%*traperrorG6#-F]s6#-%=dsolve/numeric_solnall_rkf45G6
,%&loc_FG-%$varG6#FD-F\\t6#Feq-F\\t6#F]r-F\\t6#%'loc_F1G-F\\t6#%'loc_F
2G-F\\t6#%'loc_F3G-F\\t6#%'loc_F4G-F\\t6#%'loc_F5G-F\\t6#%)loc_workG@$
/Fas%*lasterrorGC%>8'-%+searchtextG6$.F]s-%(convertG6$-%#opG6$FG7#Fas%
%nameG>8(-F[v6$.%)hardwareGF^v@%50FiuFR0FgvFR-Fhs6,FjsFDFeqF]rFdtFgtFj
tF]uF`uFcu-%&ERRORG6#FasF`w7$/%\"tGF7-%$seqG6$/&%$ordG6#,&8$FGFGFG&Feq
6#F`x/F`xFjqF06%FDFeqF]r" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 159 "La r
eponse est une procedure capable de calculer des valeurs numeriques de
 la solution ayant les conditions initiales donnees, pour des valeurs \+
de t choisies :" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "nonlin(0
);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7%/%\"tG\"\"!/-%\"xG6#F%$\"\"\"!
\"\"/-%\"yGF*F+" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 68 "Pour tracer la
 solution, on utiliser la procedure ad hoc \"odeplot\" :" }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "odep
lot(nonlin,[x(t),y(t)],0..1,color=blue);" }}{PARA 13 "" 1 "" {INLPLOT 
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$\"1shrKB3P'*F.$\"1Er0q#pQD*F.7$$\"1')z**yQ/f%*F.$\"1O\"y![H*>!*)F.7$$
\"1o`3C/O$G*F.$\"1^E\"op?Nc)F.7$$\"1w)pM4x+6*F.$\"1hp[\"oPzB)F.7$$\"1h
M8l6BR*)F.$\"1MwM5$\\Z#zF.7$$\"1;f!zu_3x)F.$\"1$G*y!R![BwF.7$$\"1/pC4a
'\\g)F.$\"1%)\\YHVnLtF.7$$\"1W2t_neT%)F.$\"1bo(pO#*[0(F.7$$\"1/%=.&*G2
G)F.$\"15NjZGr'y'F.7$$\"1)*)[FB*RA\")F.$\"1k=6A0tGlF.7$$\"1\\'z7M+m'zF
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`K!p_Y@\"eF.7$$\"1CKG.sN9vF.$\"1=*=c!f?\"f&F.7$$\"1v#*eFJjotF.$\"1gJ2:
+my`F.7$$\"1)=BQq*RDsF.$\"143OQ!*=u^F.7$$\"1e())fSSY3(F.$\"1\"ofEZ&[x
\\F.7$$\"1Q*R<_Oj%pF.$\"1(*)f6k`#)y%F.7$$\"1ba\"[Qn/\"oF.$\"1?=mD#4ig%
F.7$$\"1')Hm)e5qn'F.$\"13@&)=)y5V%F.7$$\"1P&)p#=Ufa'F.$\"1RPK&\\*fiUF.
7$$\"1hg`_oB<kF.$\"1*[e/S=05%F.7$$\"1E-U&3o3H'F.$\"1R%RjN#fWRF.7$$\"1(
4,?K3o;'F.$\"10q!R[(e%z$F.7$$\"1$f-b4H]/'F.$\"1=[hS)y-l$F.7$$\"1*Qc#f6
]DfF.$\"12>)G2]9^$F.7$$\"1X*)\\.Y>3eF.$\"1k^n#3$*yP$F.7$$\"1h4Ei*yIp&F
.$\"1c6f2xS\\KF.7$$\"1Ab66L7!e&F.$\"1)GRQT,e7$F.7$$\"1$*G\\cjHpaF.$\"1
P\"f:+*)o+$F.7$$\"1dh@<lcg`F.$\"1PP$>K#\\#*GF.7$$\"1k@'))*>!RD&F.$\"1H
xe0+W#y#F.7$$\"1Wj]h3F\\^F.$\"13[y,scwEF.7$$\"1kvO\"3Tm/&F.$\"1\")G'zK
:Zd#F.7$$\"1`*Hog!)f%\\F.$\"1/z;H=twCF.7$$\"1![#G4uDZ[F.$\"1+'op%*pCQ#
F.7$$\"1?N=H&R/v%F.$\"1YUX(\\)y\"H#F.7$$\"1*zEx6&\\bYF.$\"1+m$zl^X?#F.
7$$\"1\\kXuCRiXF.$\"1%y%*>wG17#F.7$$\"1HD:\"3+6Z%F.$\"1t(=D5%*)R?F.7$$
\"12RGKme\"Q%F.$\"1cHUUnAi>F.7$$\"1YGIi5#QH%F.$\"1pirK.^()=F.7$$\"1JZ.
qDx2UF.$\"1Z&Hw$Hj:=F.7$$\"11rRS1TBTF.$\"1*>9l'o[Y<F.7$$\"1<Vmj]qSSF.$
\"1-!RA^o*z;F.7$$\"1uvZ_fifRF.$\"1&)Q#e>yfh\"F.7$$\"1()\\ycP9!)QF.$\"1
&*41<+Ua:F.7$$\"1w+Pv#HA!QF.$\"1feu4<?&\\\"F.-%'COLOURG6&%$RGBG\"\"!F_
[l$\"*++++\"!\")" 2 356 356 356 2 0 1 0 2 6 0 4 2 1.000000 45.000000 
45.000000 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 
}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 130 "En fait, l'equation qu'on a pr
is est une perturbation de l'equation lineaire x'(t)=-x(t),'(t)=-2y(t)
 par des termes quadratiques. " }}{PARA 0 "" 0 "" {TEXT -1 385 "On app
elle point critique ou singulier d'un champs de vecteurs (f(x,y),g(x,y
)) un points (a,b) ou' il est nul : f(a,b)=g(a,b)=0; dans l'exemple pr
ecedent, (0,0) est un point critique du champs (-x+y^2,-2y+x^2). La pa
rtie lineaire du champs (f(x,y),g(x,y)) en un point critique est le te
rme d'ordre 1 du developpement de Taylor, c'est-a-dire l'application l
ineaire dont la matrice est :" }}{PARA 0 "" 0 "" {TEXT -1 13 "        \+
     " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 63 "array(1..2,1..2,[[Diff(f,x),Diff(f,y)],[Diff(g,x),Dif
f(g,y)]]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'MATRIXG6#7$7$-%%DiffG
6$%\"fG%\"xG-F)6$F+%\"yG7$-F)6$%\"gGF,-F)6$F3F/" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 42 "les derivees etant prises au point (a,b). " }}{PARA 
0 "" 0 "" {TEXT -1 3 "Le " }{TEXT 267 21 "theoreme de Liapunov " }
{TEXT -1 320 "affirme que lorsque les parties reelles des valeurs prop
res de la partie lineaire d'un champ de vecteur en un point critique s
ont non nulles, les trajectoires du champs de vecteur ressemblent a ce
lles de sa partie lineaire. C'est ce que montre l'exemple precedent, q
ue l'on reprend maintenant avec plus de trajectoires :" }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 162 "e1:=d
iff(x(t),t)=-x(t)+y(t)^2;e2:=diff(y(t),t)=-2*y(t)+x(t)^2;ini1:=x(0)=0.
1,y(0)=0.1;ini2:=x(0)=0.1,y(0)=-0.1;ini3:=x(0)=-0.1,y(0)=-0.1;ini4:=x(
0)=-0.1,y(0)=0.1;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#e1G/-%%diffG6$
-%\"xG6#%\"tGF,,&F)!\"\"*$-%\"yGF+\"\"#\"\"\"" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#e2G/-%%diffG6$-%\"yG6#%\"tGF,,&F)!\"#*$-%\"xGF+\"\"#
\"\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%ini1G6$/-%\"xG6#\"\"!$\"
\"\"!\"\"/-%\"yGF)F+" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%ini2G6$/-%
\"xG6#\"\"!$\"\"\"!\"\"/-%\"yGF)$F-F-" }}{PARA 11 "" 1 "" {XPPMATH 20 
"6#>%%ini3G6$/-%\"xG6#\"\"!$!\"\"F,/-%\"yGF)F+" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%%ini4G6$/-%\"xG6#\"\"!$!\"\"F,/-%\"yGF)$\"\"\"F," }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 409 "p1:=odeplot(dsolve([e1,e2,i
ni1],[x(t),y(t)],numeric),[x(t),y(t)],0..10,color=red):p2:=odeplot(dso
lve([e1,e2,ini2],[x(t),y(t)],numeric),[x(t),y(t)],0..10,color=blue):p3
:=odeplot(dsolve([e1,e2,ini3],[x(t),y(t)],numeric),[x(t),y(t)],0..10,c
olor=green):p4:=odeplot(dsolve([e1,e2,ini4],[x(t),y(t)],numeric),[x(t)
,y(t)],0..10,color=brown):ini:=pointplot([[0.1,0.1],[-0.1,0.1],[-0.1,-
0.1],[0.1,-0.1]],symbol=circle):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 45 "display(p1,p2,p3,p4,ini,scaling=constrained);" }}
{PARA 13 "" 1 "" {INLPLOT "6(-%'CURVESG6$7T7$$\"\"\"!\"\"F(7$$\"1)HD2&
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e&F.$\"1\"*>nv7!e7$F.7$$\"1U!=-Z#RiXF.$\"1k6$QhG17#F.7$$\"15P\"pm`es$F
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\\0*\\VKIFJ7$$\"1X)HMRxaM\"F.$\"118f[z6`?FJ7$$\"1$G/<$*\\r4\"F.$\"1Ug
\\-?i*Q\"FJ7$$\"1DEKa'ej%*)FJ$\"1Jv#[X?DS*!#>7$$\"1^3$30N\\H(FJ$\"1]=)
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\\2HFco7$$\"1%*yuZX)[&RFJ$\"1j^PB+-l>Fco7$$\"1TX(z*3![A$FJ$\"1**=/!))*
pF8Fco7$$\"1J/)z9\"\\HEFJ$\"1T\\D'RV&o*)!#?7$$\"1*Q-A!p2W@FJ$\"1_'fR70
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p*)o]Fbq$\"1:l!)f%4(3ZF_w7$$!1W<$e)p:LTFbq$\"1A(H6mFb;$F_w7$$!1U,N4'e,
P$Fbq$\"1D%3BlLy7#F_w7$$!1RPj\")H,[FFbq$\"1`<]Ss8I9F_w7$$!1*=xb3=2C#Fb
q$\"1zxMGF(4h*F^y7$$!1a#Qr)>2F=Fbq$\"1oM#4fU\"ekF^y7$$!1,pc>ny*[\"Fbq$
\"1i*eOY*4RVF^y7$$!1v*y$[dw97Fbq$\"1IRan$Q]\"HF^y7$$!1[;s$>[^!**Fas$\"
19e(f\"\\8e>F^y7$$!1qp1mchw!)Fas$\"14nxv8@:8F^y7$$!1S4)z-Qce'Fas$\"1dc
WgW\"H$))Fhz7$$!1*o6(47!*p`Fas$\"1\"fvPlf:$fFhz7$$!1V3dmXfyVFas$\"1u_y
sL#G)RFhz-Fd[l6&Ff[l$\")#)eqkFi[l$\"))eqk\"Fi[lFa[o-%'POINTSG6'F'Fg[nF
\\\\mF^\\l-%'SYMBOLG6#%'CIRCLEG-%(SCALINGG6#%,CONSTRAINEDG" 2 399 399 
399 2 0 1 0 2 9 0 4 1 1.000000 45.000000 45.000000 0 0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 }}}{SECT 0 {PARA 5 "" 0 "" 
{TEXT -1 11 "Exercice 2." }}{PARA 0 "" 0 "" {TEXT -1 209 "Calculer les
 points critiques des systemes suivants et determiner leur nature, en \+
tracant des trajectoires et en calculant les valeurs propres des parti
es lineaires des champs de vecteurs aux points critiques." }}{PARA 0 "
" 0 "" {TEXT -1 13 "a) x'=-x+y^2;" }}{PARA 0 "" 0 "" {TEXT -1 14 "    \+
y'=-x+y^2;" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 16 "b) x'=x^2-y^2-x;" }}{PARA 0 "" 0 "" {TEXT -1 13 "    y'=2xy-y;
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 20 "" 0 "" {TEXT -1 
8 "Solution" }}{PARA 0 "" 0 "" {TEXT -1 50 "D'abord une procedure util
e; les parametres sont :" }}{PARA 0 "" 0 "" {TEXT -1 78 "eq1,eq2 : par
tie de droite de la premiere, respectivement la deuxieme equation" }}
{PARA 0 "" 0 "" {TEXT -1 63 "x0,y0 : coordonnee x, respectivement y de
 la condition initiale" }}{PARA 0 "" 0 "" {TEXT -1 49 "a,b : temps de \+
depart et d'arrivee respectivement" }}{PARA 0 "" 0 "" {TEXT -1 14 "cou
l : couleur" }}{PARA 0 "" 0 "" {TEXT -1 185 "npoints : nombre de point
s de la trajectoire qui seront calcules. Plus (b-a) est grand, plus il
 faudra calculer beaucoup de points pour avoir une bonne approximation
 de la trajectoire. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 43 "palin:=proc(eq1,eq2,x0,y0,a,b,coul,npoint
s)" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "local e1,e2,ini;" }}{PARA 0 "
> " 0 "" {MPLTEXT 1 0 21 "e1:=diff(x(t),t)=eq1;" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 21 "e2:=diff(y(t),t)=eq2;" }}{PARA 0 "> " 0 "" {MPLTEXT 
1 0 21 "ini:=x(0)=x0,y(0)=y0;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 115 "o
deplot(dsolve([e1,e2,ini],[x(t),y(t)],numeric),[x(t),y(t)],a..b,scalin
g=constrained,color=coul,numpoints=npoints);" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 4 "end;" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%&palinG:6*%$
eq1G%$eq2G%#x0G%#y0G%\"aG%\"bG%%coulG%(npointsG6%%#e1G%#e2G%$iniG6\"F3
C&>8$/-%%diffG6$-%\"xG6#%\"tGF>9$>8%/-F96$-%\"yGF=F>9%>8&6$/-F<6#\"\"!
9&/-FFFM9'-%(odeplotG6(-%'dsolveG6%7%F6FAFI7$F;FE%(numericGFZ;9(9)/%(s
calingG%,constrainedG/%&colorG9*/%*numpointsG9+F3F3" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 38 "Cherchons les points critiques de a) :" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "solve(\{-x+y^2,-y+x^2\},\{x,
y\});" }}{PARA 12 "" 1 "" {XPPMATH 20 "6%<$/%\"yG\"\"!/%\"xGF&<$/F%\"
\"\"/F(F+<$/F%-%'RootOfG6#,(*$%#_ZG\"\"#F+F4F+F+F+/F(,&!\"\"F+F/F8" }}
}{EXCHG {PARA 0 "" 0 "" {TEXT -1 195 "Les racines de z^2+z+1 ne sont p
as reelles; il y a donc 2 points critiques (0,0) et (1,1). En (0,0) la
 partie lineaire est (-x,-y), donc pres de (0,0) les trajectoires revi
nennent vers ce point. " }}{PARA 0 "" 0 "" {TEXT -1 127 "On trace quel
ques trajectoires pres de (1,1); apres tatonnement, on arrive a choisi
r des valeurs de x0,y0, a et b convenables :" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 329 "p1:=palin(-x(t)+y(t)^2,-y(t)+x(t)^2,0.9,1.2,0,1
,red,10):p2:=palin(-x(t)+y(t)^2,-y(t)+x(t)^2,0.7,1.2,0,1,blue,10):p3:=
palin(-x(t)+y(t)^2,-y(t)+x(t)^2,0.9,0.8,0,1,green,10):p4:=palin(-x(t)+
y(t)^2,-y(t)+x(t)^2,1.2,0.9,0,1,brown,10):display(p1,p2,p3,p4,pointplo
t([[0.9,1.2],[0.7,1.2],[0.9,0.8],[1.2,0.9]],color=black,symbol=circle)
);" }}{PARA 13 "" 1 "" {INLPLOT "6(-%'CURVESG6$7-7$$\"\"*!\"\"$\"#7F*7
$$\"1Zr'\\>tRZ*!#;$\"1h'>hGps;\"!#:7$$\"1fR**)[&\\U)*F0$\"1oV'**of^9\"
F37$$\"1&)RqH6s85F3$\"1lEh>HIJ6F37$$\"1@cBcr8Q5F3$\"1M3FUy\"R7\"F37$$
\"1o'39;I#f5F3$\"1%e<@h)p@6F37$$\"1/#G/e:$y5F3$\"1B#[Fc;P7\"F37$$\"1nm
%Q3=k4\"F3$\"1Cbtf4MH6F37$$\"1Q2srUO96F3$\"1,$yRUy\"Q6F37$$\"1\"3`;c_G
8\"F3$\"1w@WO!H+:\"F37$$\"1Lo9;!3D:\"F3$\"1+T:**o&[;\"F3-%'COLOURG6&%$
RGBG$\"*++++\"!\")\"\"!Fbo-F$6$7-7$$\"\"(F*F+7$$\"18F\"H,j&HwF0$\"1r9+
>02P6F37$$\"1)Q\"*3t3%y!)F0$\"12Tw(3Hy3\"F37$$\"1\"\\;EKDRR)F0$\"1hXg
\"yf!\\5F37$$\"1*G!G]UF5')F0$\"1&)R)eYl\"=5F37$$\"1E.<Rf8_()F0$\"1l#3U
2=5$**F07$$\"1+di;NQP))F0$\"1mOS0:$Gs*F07$$\"1'\\U7s\")*y))F0$\"1CmW9o
*[a*F07$$\"1%zn<oQj)))F0$\"1@o1Gs'yQ*F07$$\"1c!H<vti'))F0$\"1^/c(Q(eW#
*F07$$\"1&\\S\\W:P#))F0$\"1v9!4;s&4\"*F0-F\\o6&F^oFboFboF_o-F$6$7-7$F(
$\"\")F*7$$\"1:F?^j<_()F0$\"1;;*\\];y)zF07$$\"1l\">F#[,B&)F0$\"1W.<:(*
=PzF07$$\"1B%*G^?a0$)F0$\"1vr)pP[_&yF07$$\"1S#e0'R\\%4)F0$\"1eD;kWIZxF
07$$\"1-_%3YOf)yF0$\"12-Cv<O<wF07$$\"1BA*Q6vpn(F0$\"17*GEn*[ouF07$$\"1
-^P5]`luF0$\"1N#pR<*4.tF07$$\"1%Q^mk%>]sF0$\"1/:r@v9BrF07$$\"1gxvM/1Iq
F0$\"1:l,$o*GIpF07$$\"1qzr(>rY!oF0$\"1=,!f#)*)fs'F0-F\\o6&F^oFboF_oFbo
-F$6$7-7$F+F(7$F1F.7$F7F57$F<F:7$FAF?7$FFFD7$FKFI7$FPFN7$FUFS7$FZFX7$F
inFgn-F\\o6&F^o$\")#)eqkFao$\"))eqk\"FaoFiw-%'POINTSG6(F'FfoF`sFjv-F\\
o6&F^oFboFboFbo-%'SYMBOLG6#%'CIRCLEG-%(SCALINGG6#%,CONSTRAINEDG" 2 
352 352 352 2 0 1 0 2 9 0 4 1 1.000000 45.000000 45.000000 0 0 0 0 0 
0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 218 "ce qui suggere que (1,1) est un point critique de
 type \"selle\" : 2 valeurs propres reelles, de signe oppose. Voici en
core, pour aider a comprendre, des trajectoires partant de divers poin
ts de l'axe OX et de l'axe OY :" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 383 "p1:=palin(-x(t)+y(t)^2,-y(t)+x(t)^2,0,1.2,0,10,red,50):p2:=pa
lin(-x(t)+y(t)^2,-y(t)+x(t)^2,0,1.7,0,3,blue,20):p3:=palin(-x(t)+y(t)^
2,-y(t)+x(t)^2,0,1.8,0,3,green,20):p4:=palin(-x(t)+y(t)^2,-y(t)+x(t)^2
,1.7,0,0,2,brown,20):p5:=palin(-x(t)+y(t)^2,-y(t)+x(t)^2,1.8,0,0,1.5,b
lack,20):display(p1,p2,p3,p4,p5,pointplot([[0,1.2],[0,1.7],[0,1.8],[1.
7,0],[1.8,0]],color=black,symbol=circle));" }}{PARA 13 "" 1 "" 
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-1 75 "Calculons tout de meme les valeurs propres de la partie lineair
e en (1,1) :" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "A:=jacobian
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{TEXT -1 113 "On evalue en (1,1); notez l'utilisation de \"op\" qui pe
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257 "p1:=palin(champs,1,-0.1,0,10,red,50):p2:=palin(champs,0.9,-0.1,0,
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0.7,0.3,0,10,blue,50):display(p1,p2,p3,p4,pointplot([[1,-0.1],[0.9,-0.
1],[1.1,0.1],[0.7,0.3]],color=black,symbol=circle));" }}{PARA 13 "" 1 
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CLEG-%(SCALINGG6#%,CONSTRAINEDG" 2 409 409 409 2 0 1 0 2 9 0 4 1 
1.000000 45.000000 45.000000 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 
0 0 0 0 0 0 0 0 0 0 }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 272 "Ce sont en
 fait des portions de cercles passont par (0,0) et (1,0). Les trajecto
ires repoussent les points proches de (1,0) vers le point (0,0). On pe
ut donc parier qu'en (0,0) et (1,0) les valeurs propres sont reelles, \+
les 2 positives en (1,0), les 2 negatives en (0,0). " }{TEXT 279 12 "V
erifiez-le." }}}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}}{SECT 0 {PARA 3 "
" 0 "" {TEXT 268 19 "Bifurcation de Hopf" }}{PARA 3 "" 0 "" {TEXT 277 
68 "Nous allons etudier l'allure des trajectoires du champs de vecteur
s " }}{PARA 0 "" 0 "" {TEXT -1 23 "    x'=-y+x(e-x^2-y^2);" }}{PARA 0 
"" 0 "" {TEXT -1 22 "    y'=x+y(e-x^2-y^2);" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 57 "lorsque e varie de -0.1 a 0.1 e
n passant par la valeur 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 
"" 0 "" {TEXT 269 10 "Exercice. " }{TEXT -1 95 "Montrer que les valeur
s propres de la partie lineaire de ce champs en (0,0)  valent e+I et e
-I." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 41 "champs:=-y+x*(e-x^2-y^2),x+y*(e-x^2-y^2);" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>%'champsG6$,&%\"yG!\"\"*&%\"xG\"\"\",(%\"eG
F+*$F*\"\"#F(*$F'F/F(F+F+,&F*F+*&F'F+F,F+F+" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 48 "f1:=subs(e=0.1,champs[1]),subs(e=0.1,champs[2]);" 
}}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#f1G6$,&%\"yG!\"\"*&%\"xG\"\"\",($
F+F(F+*$F*\"\"#F(*$F'F/F(F+F+,&F*F+*&F'F+F,F+F+" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 123 "p1:=palin(f1,0.2,0,0,10,blue,50):p2:=palin(f1
,sqrt(0.1),0,0,10,red,50):p3:=palin(f1,0.4,0,0,10,green,50):display(p1
,p2,p3);" }}{PARA 13 "" 1 "" {INLPLOT "6&-%'CURVESG6$7U7$$\"\"#!\"\"\"
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{TEXT -1 193 "On voit que la trajectoire par (sqrt(e),0) est un cercle
 attracteur, car les trajectoires proches du cercle, a l'interieur ou \+
a l'exterieur, sont des spirales qui s'enroulent autour du cercle. " }
}{PARA 0 "" 0 "" {TEXT -1 86 "En realite, les trajectoire exterieures \+
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{TEXT -1 106 " ses trajectoires sont des cercles centres a l'origine. \+
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$$!1p?f&R8f&RF/$!15N$4e#*Ge$F/7$$!1@1Gc#QB3$F/$!1k5G@&)z?UF/7$$!1?=Z\"
R+\"=@F/$!1s!pQSP#fYF/7$$!1UpWyAY16F/$!1+k#*yjT))[F/7$$!1+T\"*fk`&3*!#
>$!1xy!)>i^2\\F/7$$\"1EpOqE/p))FS$!1X.3*4eVs%F/7$$\"10k@<:E)y\"F/$!16F
eK1taVF/7$$\"1J^pdvOzDF/$!1NvEi%=9#QF/7$$\"1zDwNQFKKF/$!1ro.tT)H:$F/7$
$\"1iuvkh$es$F/$!1*)yTELO#Q#F/7$$\"1J&p=n.j/%F/$!1TW$)>wLX:F/7$$\"1X)R
-F\"e(=%F/$!1Ud`6a`*y'FS7$$\"1V'*)Q7o6:%F/$\"1J?R!)fY+=FS7$$\"1,#e7s(y
XRF/$\"1oE8pc<n**FS7$$\"1?S)\\:Pne$F/$\"19Hq[+FR<F/7$$\"1%=8)yN*\\4$F/
$\"1txb7_C!Q#F/7$$\"1`)4t@Ch\\#F/$\"1&4x5Rku*GF/7$$\"1UkHvC0>=F/$\"1I!
fn1CZF$F/7$$\"1R:Xokt%4\"F/$\"1#*\\:\"*z@-NF/7$$\"1f')Rbh<[NFS$\"1!RK*
e,owNF/7$$!1fo%efTqp$FS$\"1*)[ZPSD,NF/7$$!1CwzIar\\5F/$\"18l/UW5&G$F/7
$$!1,d!op1\"f;F/$\"1\"*\\sC$)zUHF/7$$!1;&Gif@d<#F/$\"1ckN_\"*[$\\#F/7$
$!1&G^gz0@e#F/$\"1+GK*>Y*f>F/7$$!1Fd5MH.mGF/$\"1@U$GxuuO\"F/7$$!1!3-u.
&z?IF/$\"1A(**)y+wFuFS7$$!1P4@#*\\DXIF/$\"12j9:R&z7\"FS7$$!1&R4%)[oO%H
F/$!14$)\\iL#R'\\FS7$$!188t3?HDFF/$!1)\\61m<11\"F/7$$!1WX:H:!QS#F/$!1P
PV%*4`e:F/-F^[l6&F`[lFa[lF+F+-%(SCALINGG6#%,CONSTRAINEDG" 2 335 335 
335 2 0 1 0 2 9 0 4 1 1.000000 45.000000 45.000000 0 0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 115 "Les trajectoires sont des spirales qui tendent rapidemen
t vers l'origine, comme prevu par le theoreme de Liapounov." }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 182 "On voit donc que lorsque e varie \+
de -0.1 a +0.1, le point critique attracteur (l'origine) donne naissan
ce, lorsque e=0, a un petit cercle attracteur. Ce type de phenomene es
t appele " }{TEXT 271 19 "bifurcation de Hopf" }{TEXT -1 58 " (Eberhar
dt), du nom de celui qui les a etudie en premier." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 329 "Premiere variation sur c
e theme : l'equation suivante correspond a un champs de vecteurs qui e
st aussi une deformation de (-y,x), mais c'est une deformation differe
nte. Neanmoins, qualitativement cet exemple ressemble au precedent,  c
omme vous etes invites a vous en rendre compte en etudiant les traject
oires pres de l'origine :" }}{PARA 0 "" 0 "" {TEXT -1 17 "  x'=-y+x(e-
x^2);" }}{PARA 0 "" 0 "" {TEXT -1 16 "  y'=x+y(e-y^2);" }}{PARA 0 "" 
0 "" {TEXT -1 107 "En particulier, pour e>0 on devrait avoir un cycle \+
attracteur, qui ressemble a une patate de rayon sqrt(e)." }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}}{SECT 0 {PARA 4 "" 0 "" {TEXT 270 8 "Reponse \+
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 33 "champs:=-y+x*(e-x^2),x+y*(e-y^2);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%'champsG6$,&%\"yG!\"\"*&%\"xG\"\"\",&%\"eGF+*$F*\"\"#
F(F+F+,&F*F+*&F'F+,&F-F+*$F'F/F(F+F+" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 48 "f1:=subs(e=0.4,champs[1]),subs(e=0.1,champs[2]);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#f1G6$,&%\"yG!\"\"*&%\"xG\"\"\",&$\"
\"%F(F+*$F*\"\"#F(F+F+,&F*F+*&F'F+,&$F+F(F+*$F'F0F(F+F+" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 115 "Par tatonnement a partir du point (sqrt(
0.4),0), on voit que le point (sqrt(0.4)+0.09,0) semble etre sur un cy
cle :" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "palin(f1,sqrt(0.4)
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{PARA 0 "" 0 "" {TEXT -1 104 "On voit donc que cette famille de champs
 de vecteurs presente un comportement semblable a la precedente." }}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 23 "D
euxieme variation : l'" }{TEXT 272 24 "equation de Van der Pol." }}
{PARA 0 "" 0 "" {TEXT -1 91 "Il s'agit de l'equation suivante, qui pro
vient de l'etude d'un certain circuit electrique :" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 20 "  x''+e(x^2-1)x'+x=0" }}
{PARA 0 "" 0 "" {TEXT -1 37 "Elle se ramene au systeme d'ordre 1 :" }}
{PARA 0 "" 0 "" {TEXT -1 7 "  x'=y;" }}{PARA 0 "" 0 "" {TEXT -1 17 " y
'=-x-e(x^2-1)y;" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 47 "Si on prend e=0, on trouve le systeme lineaire " }}{PARA 
0 "" 0 "" {TEXT -1 10 "   x'=-y; " }}{PARA 0 "" 0 "" {TEXT -1 8 "   y'
=x;" }}{PARA 0 "" 0 "" {TEXT -1 159 " dont les trajectoires sont des c
ercles centres a l'origine. De nouveau, si on donne a e des valeurs po
sitives un cycle limite attracteur apparait. Montrez-le." }}}{SECT 0 
{PARA 4 "" 0 "" {TEXT 273 7 "Reponse" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 25 "champs:=y,-x-e*(x^2-1)*y;" }}{PARA 11 "" 1 "" 
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{PARA 0 "> " 0 "" {MPLTEXT 1 0 82 "p1:=palin(f1,0.1,0,0,20,blue,200):p
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0 {PARA 5 "" 0 "" {TEXT -1 8 "Exercice" }}{PARA 0 "" 0 "" {TEXT -1 
289 "Calculer les valeurs propres de la partie lineaire du champs de v
ecteurs precedent. Montrer que leur partie reelle passe de valeurs neg
atives a des valeurs positives lorsque e traverse la valeur 0. Montrer
 (experimentalement) que pour e=-0.1 le champs possede un cycle repuls
eur par (2,0)." }}{SECT 0 {PARA 20 "" 0 "" {TEXT -1 7 "Reponse" }}
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\\&px\\?AF\\\\m$!1@C46cgWbF.7$$!1,[(*R4tlAF\\\\m$!1e$=>`S@[$F.7$$!1g'=
xb^(*G#F\\\\m$!1@#o(y`=.8F.7$$!1#z/S9R9H#F\\\\m$\"1R2W'zx9#)*F17$$!1j,
q+Q\")pAF\\\\m$\"1jo7')pgbLF.7$$!1/x@:T7CAF\\\\m$\"1V9g:Vg*y&F.7$$!1N>
=nK%R:#F\\\\m$\"1&*yx$pJoC)F.7$$!1ZSu*[j#f?F\\\\m$\"1sM?2B5o5F\\\\m7$$
!19$>JAw0%>F\\\\m$\"1'GtR=2RI\"F\\\\m-F][m6&F_[mF*F`[mF*-%(SCALINGG6#%
,CONSTRAINEDG" 2 351 351 351 2 0 1 0 2 9 0 4 1 1.000000 45.000000 
45.000000 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 
}}}}}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT 
274 18 "Equation de Lorenz" }}{PARA 0 "" 0 "" {TEXT -1 141 "C'est un s
ysteme lineaire de l'espace R^3 dependant d'un parametre r, qui pour c
ertains valeurs de celui-ci presente un caractere chaotique :" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 12 " x'=10(y-
x);" }}{PARA 0 "" 0 "" {TEXT -1 13 " y'=r*x-y-zx;" }}{PARA 0 "" 0 "" 
{TEXT -1 16 " z'= -(8/3)z+xy;" }}{PARA 4 "" 0 "" {TEXT 275 8 "Exercice
" }}{PARA 0 "" 0 "" {TEXT -1 39 "Montrer que les points critiques sont
 ." }}{PARA 0 "" 0 "" {TEXT -1 79 "    x=y=z=0,     x=y=sqrt(8/3(r-1))
, z=r-1  et     x=y= - sqrt(8/3(r-1)), z=r-1" }}{PARA 0 "" 0 "" {TEXT 
-1 10 "==========" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 112 "Il se trouve que les 2 points singuliers differents de l
'origine sont des attracteur seulement si 1 < r < 24.74." }}{PARA 0 "
" 0 "" {TEXT -1 131 "Tracez des trajectoires pour r=10 et r=20 pour le
 verifier. Observez que pour r=40 les trajectoires ont un caractere im
predictible." }}{PARA 0 "" 0 "" {TEXT -1 27 "Voici une procedure utile
 :" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "lorenz:=proc(r,x0,y0,z
0,a,b,coul,npoints)" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "local e1,e2,
e3,ini;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 32 "e1:=diff(x(t),t)=10*(y(t)-x(t));" }}{PARA 0 "> " 0 "
" {MPLTEXT 1 0 39 "e2:=diff(y(t),t)=r*x(t)-y(t)-z(t)*x(t);" }}{PARA 0 
"> " 0 "" {MPLTEXT 1 0 39 "e3:=diff(z(t),t)=-(8/3)*z(t)+x(t)*y(t);" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "ini:=x(0)=x0,y(0)=y0,z(0)=z0;" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 140 "odeplot(dsolve([e1,e2,e3,ini],[x(t
),y(t),z(t)],numeric),[x(t),y(t),z(t)],a..b,numpoints=npoints,color=co
ul,scaling=constrained,axes=normal);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 
0 4 "end;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}{PARA 12 "" 1 "" 
{XPPMATH 20 "6#>%'lorenzG:6*%\"rG%#x0G%#y0G%#z0G%\"aG%\"bG%%coulG%(npo
intsG6&%#e1G%#e2G%#e3G%$iniG6\"F4C'>8$/-%%diffG6$-%\"xG6#%\"tGF?,&-%\"
yGF>\"#5F<!#5>8%/-F:6$FAF?,(*&9$\"\"\"F<FMFMFA!\"\"*&-%\"zGF>FMF<FMFN>
8&/-F:6$FPF?,&FP#!\")\"\"$*&F<FMFAFMFM>8'6%/-F=6#\"\"!9%/-FBF[o9&/-FQF
[o9'-%(odeplotG6)-%'dsolveG6%7&F7FFFSFgn7%F<FAFP%(numericGF[p;9(9)/%*n
umpointsG9+/%&colorG9*/%(scalingG%,constrainedG/%%axesG%'normalGF4F4" 
}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 67 "Verifions que  x=y=sqrt(8/3(r-1
)), z=r-1 est attracteur pour r=10 :" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 31 "lorenz(10,1,0,8,0,10,blue,300);" }}{PARA 13 "" 1 "" 
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305 305 305 1 0 1 0 2 6 0 4 1 1.000000 80.000000 103.000000 14040 0 
14086 14104 14112 15010 15070 20030 0 17050 0 0 0 0 0 0 0 1 1 216 
5383 0 0 0 0 0 0 0 0 0 }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 195 "Le poin
t de vue a ete change a theta=103 et phi=80. On devine que l'orbite he
site entre 2 attracteurs. Il semble impossible de prevoir, meme approx
imativement, ou on se trouvera a un temps donne." }}}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}}}}{MARK "4 8 0" 37 }{VIEWOPTS 1 1 0 1 1 1803 }