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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 32 "Global Stability Investiga
tion 1" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 68 
"You are going to look at the phase portrait of the nonlinear system \+
" }}{PARA 256 "" 0 "" {OLE 1 4113 1 "[xm]Br=WfoRrB:::wk;nyyI;G:;:j::>:
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-1 77 "and compare it with the level curves of a conserved quantity fo
r that system." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 257 "What can you deduce about the phase portrait of the nonl
inear system before you plot it? You should know how to find the equil
ibrium points and how to classify the equilibrium points of their line
arisations by using the Hartman Grobman linearsation theorem." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 29 "To find he equilibrium po
ints" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "with(linalg):" }}}
{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\});" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 43 " The equilibrium points are (0,0) an (1,0)." }}{PARA 0 
"> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 179 "Thus the phase portrait has an eq
uilibrium point at (0,0) which is a nonlinear saddle at the point (0,0
) and an equilibrium point at (1,0) which as yet you are unable to cla
ssify." }}{PARA 0 "" 0 "" {TEXT -1 93 "Plot the phase portrait and exa
mine it carefully. Does it show the features you would expect?" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 15 "with(DEtools): " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "A:
=diff(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 0 197 "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);" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 129 "To plot \+
the level curves of a conserved quantity for the system first you need
 to find an expression for the  conserved quantity." }}{PARA 0 "" 0 "
" {TEXT -1 61 "To do this you need to find a first integral for the sy
stem. " }}{PARA 0 "" 0 "" {TEXT -1 78 "Remember to do this you first f
ind a differential equation by dividing g by f." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "deq:=diff(y
(x),x)=(-x*(1-x))/y(x);" }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 28 " Then use the dsolve command" }}{PARA 0 "
" 0 "" {TEXT -1 1 "." }{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 26 "dsolve(deq,y(x),implicit);" }}{PARA 11 "" 1 "" {TEXT 
-1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 37 "Plot the level curves o
f the function" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 12 "with(plots):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 21 "Y:=y^2/2-x^2/2+x^3/3;" }}{PARA 11 "" 1 "" {TEXT -1 0 
"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 95 "implicitplot(\{Y=0.5,Y
=12.5,Y=-1/6,Y=9/4,Y=-1/12,Y=27/64,Y=1/3000,Y=0\},x=-2.5..2.5,y=-2.5..
2.5);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 69 
"Compare this diagram with the phase portrait of the nonlinear system.
" }}{PARA 0 "" 0 "" {TEXT -1 41 "Do they have the same equilibrium poi
nts?" }}{PARA 0 "" 0 "" {TEXT -1 41 "Do they have the same geometric f
eatures?" }}}{MARK "17 0 0" 25 }{VIEWOPTS 1 1 0 1 1 1803 }