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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 34 "Global Stability Worked Ex
ample 6." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
56 "Examine the long term behaviour of the nonlinear system " }}{PARA 
256 "" 0 "" {OLE 1 4126 1 "[xm]Br=WfoRrB:::wk;nyyI;G:;:j::>:B>N:F:nyyy
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::::::::::::::::::::::::::::::::3:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 58 "In any analysis the equilibrium points mu
st be found first" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT 256 30 "To find the equilibrium points" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "with(linalg):" }}
{PARA 7 "" 1 "" {TEXT -1 32 "Warning, new definition for norm" }}
{PARA 7 "" 1 "" {TEXT -1 33 "Warning, new definition for trace" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "f:=y;" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 11 "g:=x*(1-x);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 23 "solve(\{f=0,g=0\},\{x,y\});" }}{PARA 11 "" 1 "" 
{TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 43 "The equilibrium p
oints are (0,0) and (1,0)." }}{PARA 0 "" 0 "" {TEXT -1 272 "The next s
tep is to find the Jacobian matrices of the linearisations and classif
y the equilibrium points of the linearisations. Then, where possible, \+
use the Hartman Grobman linarisation theorem to determine the stabilit
y of the equilibrium points of the nonlinear system." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 257 "" 0 "" {TEXT -1 1 "T" }{TEXT 257 54 "o cl
assify the equilibrium points of the linearisation" }{TEXT -1 1 "s" }}
{PARA 0 "" 0 "" {TEXT -1 102 "To find the Jacobian matrix of the linea
risations  first represent the equations as vectors as follows" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 9 "v:=[f,g];" }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 31 "Then use the \"jacobian\" command" }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "J:=jacobi
an(v,[x,y]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 48 "At the point (0,0
) substitute in the coordinates" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 28 "J00:=subs(x=0,y=0,evalm(J));" }}{PARA 11 "" 1 "" {TEXT -1 0 "
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 96 "Place the arrow on the determ
inant and click the RHS button on the mouse to obtain commands for " }
{TEXT 261 5 "trace" }{TEXT -1 5 " and " }{TEXT 262 11 "determinant" }
{TEXT -1 3 " of" }{MPLTEXT 1 0 1 " " }{TEXT -1 0 "" }{MPLTEXT 1 0 1 " \+
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 3 "J00" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 6 "Since " }{OLE 1 4106 1 "[xm]Br=WfoR
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::::::::::::::::::::::::::::::::::::::::::::fyyyyya:nYf::G:jy;::::::::
::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
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::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::j:b:<Z::
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:::::::::::::::::::::::::::::::::::::::::::::5:" }{TEXT -1 220 "the eq
uilibrium point of the linearisation at the point (0,0) is unstable an
d a saddle. By the Hartman Grobman linearisation theorem the equilibri
um point (0,0) of the nonlinear system is unstable and a nonlinear sad
dle." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 48 "At the point (1,0) substitute in the coordinates" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "J10:=subs(x=1,y=0,evalm(J));
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }{TEXT -1 0 "" }}}
{EXCHG {PARA 258 "" 0 "" {TEXT -1 1 " " }{MPLTEXT 1 0 0 "" }{OLE 1 
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" 0 "" {TEXT -1 7 " Since " }{OLE 1 4106 1 "[xm]Br=WfoRrB:::wk;nyyI;G:
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:::::::::::::::::::::::::::::fyyyyya:nYf::G:jy;:::::::::::::::::::::::
::::::::::::::::::::::::::::::::::::::::::::::::::::::::JcvGYMt>^:fBWM
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JSFm;^b=F:<JknIM:_KGAJKTj:>:IMy:_kF_cEF:NF:_KGO_HF:vF><<jw?JB:>l;Z<b:<
kc`a\\wgf?ZI^:f?=J<Jr?:MJ:N`D>Jci;=Z:fAap>JSJxuj:jXjDjw;<:[V:B:D:c\\_;
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hngfgcUC:Uk:^Z:Jr?:AB:<J::::::::::::::::::::::::::::::::::::::::::::::
::::::::::::1:" }{TEXT -1 54 "the equilibrium point of the linearisati
on is centre. " }}{PARA 0 "" 0 "" {TEXT -1 76 "Thus the point (1,0) ne
eds further investigation to determine its stability." }}{PARA 0 "" 0 
"" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 82 " One method of doing t
his further investigation is to prove the system reversible." }}{PARA 
0 "" 0 "" {TEXT -1 71 "It then follows that the equilibrium point (1,0
) is a nonlinear centre." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT 258 33 "To prove the system is reversible" }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "A:=dif
f(x(t),t)=y;" }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 24 "B:=diff(y(t),t)=x*(1-x);" }}{PARA 11 "" 1 "" 
{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 81 "subs(\{t=-t
,y=-y,diff(x(t),t)=-diff(x(t),t),diff(y(t),t)=diff(y(t),t),x=x\},\{A,B
\});" }}{PARA 11 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 107 "It can be seen that these are identical with the origina
l equations and the system is therefore reversible." }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 64 "Thus the point \+
(1,0) is neutrally stable and a nonlinear centre." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 259 26 "To draw the phase portra
it" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "with(DEtools):" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 197 "DEplot(\{A,B\},\{x(t),y(t)
\},t=-10..10,[[x(0)=0,y(0)=1],[x(0)=2.5,y(0)=0],[x(0)=1,y(0)=0],[x(0)=
1.5,y(0)=0],[x(0)=.5,y(0)=0],[x(0)=-0.75,y(0)=0],[x(0)=-.1,y(0)=.1]],x
=-2..2.5,y=-2.5..2.5,stepsize=0.1);" }}{PARA 13 "" 1 "" {TEXT -1 0 "" 
}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 176 "Look at the phase portrait. Yo
u can see that the point (1,0) is a nonlinear centre and the point (0,
0) is a nonlinear saddle. Notice also that the x-axis is a line of sym
metry." }}{PARA 0 "" 0 "" {TEXT -1 236 "It is interesting that one tra
jectory which starts from the origin and returns to it divides the clo
sed periodic trajectories from the non-periodic ones. Such a trajector
y which starts and ends at a single equilibrium point is called a " }
{TEXT 260 18 "homoclinic orbit. " }{TEXT -1 99 "This trajectory is not
 periodic since it takes an infinite time to return to the equilibrium
 point." }}}}{MARK "0 0 0" 33 }{VIEWOPTS 1 1 0 1 1 1803 }