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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 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 58 "In any analysis the equilibrium points must be fou
nd 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 points are (0,0) and (1,0
)." }}{PARA 0 "" 0 "" {TEXT -1 272 "The next step is to find the Jacob
ian matrices of the linearisations and classify the equilibrium points
 of the linearisations. Then, where possible, use the Hartman Grobman \+
linarisation theorem to determine the stability of the equilibrium poi
nts of the nonlinear system." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
257 "" 0 "" {TEXT -1 1 "T" }{TEXT 257 54 "o classify the equilibrium p
oints of the linearisation" }{TEXT -1 1 "s" }}{PARA 0 "" 0 "" {TEXT 
-1 102 "To find the Jacobian matrix of the linearisations  first repre
sent 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 u
se the \"jacobian\" command" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "J:=jacobian(v,[x,y]);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 48 "At the point (0,0) substitute in t
he 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 6 "Since " }{TEXT -1 220 "the equilibrium point of \+
the linearisation at the point (0,0) is unstable and a saddle. By the \+
Hartman Grobman linearisation theorem the equilibrium point (0,0) of t
he nonlinear system is unstable and a nonlinear saddle." }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 48 "A
t 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 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 7 " Since " }{TEXT -1 54 "the equilibrium point of the linearisatio
n is centre. " }}{PARA 0 "" 0 "" {TEXT -1 76 "Thus the point (1,0) nee
ds further investigation to determine its stability." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 82 " One method of doing th
is 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 "11 0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 }