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{SECT 0 {EXCHG {PARA 259 "" 0 "" {TEXT -1 31 "Local Stability Investig
ation 2" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 257 "" 1 "" {TEXT -1 
29 "Consider the nonlinear system" }}{PARA 258 "" 0 "" {OLE 1 4105 1 "
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{TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 181 "For this system \+
you are going to plot the phase portrait of the nonlinear system close
 to its equilibrium point (0,0) and also the phase portrait of the lin
earisation at this point." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 41 "First prove (0,0) is an equilibrium point
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "with(linalg):" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "with(DEtools):" }}{PARA 7 "
" 1 "" {TEXT -1 35 "Warning, new definition for adjoint" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 25 "At the equilibrium point " }{OLE 1 3589 
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" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "f:=2*x+6*
y-x^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fG,(%\"xG\"\"#%\"yG\"\"'
*$)F&F'\"\"\"!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "g:=-2
*y+y^2-x;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"gG,(%\"yG!\"#*$)F&\"
\"#\"\"\"\"\"\"%\"xG!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
23 "solve(\{f=0,g=0\},\{x,y\});" }}{PARA 12 "" 1 "" {XPPMATH 20 "6$<$/
%\"xG\"\"!/%\"yGF&<$/F%*&-%'RootOfG6#,*!\"#\"\"\"%#_ZG\"\"#*$)F2F3\"\"
\"!\"%*$)F2\"\"$F6F1F1,&F0F1F,F1F1/F(F," }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 42 "The point  (0,0) is an equilibrium point. " }}{PARA 0 "" 
0 "" {TEXT -1 102 "Note that their is another equilibrium point which \+
could be found by solving the given cubic equation." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 89 " Plot the phase
 plane of the nonlinear system in the region of the equilibrium point(
0,0)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "A:=diff(x(t),t)=2*x
+6*y-x^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"AG/-%%diffG6$-%\"xG6#
%\"tGF,,(F*\"\"#%\"yG\"\"'*$)F*F.\"\"\"!\"\"" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 27 "B:=diff(y(t),t)=-2*y+y^2-x;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"BG/-%%diffG6$-%\"yG6#%\"tGF,,(F*!\"#*$)F*\"\"#\"\"
\"\"\"\"%\"xG!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 78 "DEplo
t(\{A,B\},\{x(t),y(t)\},-10..50,[[0,.3,0]],x=-1.5..1,y=-0.5...1,stepsi
ze=.1);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 73 "The phase plot shows t
he equilibrium point to be unstable and a repeller." }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 52 "N
ow look at the phase portrait of the linearisation." }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 86 "First find the equations of the linearisation a
t (0,0)  by finding the Jacobian matrix" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 15 "f:=2*x+6*y+x^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%
\"fG,(%\"xG\"\"#%\"yG\"\"'*$)F&F'\"\"\"\"\"\"" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 14 "g:=-2*y+y^2-x;" }}{PARA 11 "" 1 "" {XPPMATH 20 "
6#>%\"gG,(%\"yG!\"#*$)F&\"\"#\"\"\"\"\"\"%\"xG!\"\"" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 9 "v:=[f,g];" }}{PARA 11 "" 1 "" {XPPMATH 20 
"6#>%\"vG7$,(%\"xG\"\"#%\"yG\"\"'*$)F'F(\"\"\"\"\"\",(F)!\"#*$)F)F(F-F
.F'!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "J:=jacobian(v,[
x,y]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"JG-%'matrixG6#7$7$,&\"\"
#\"\"\"%\"xGF+\"\"'7$!\"\",&!\"#F,%\"yGF+" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 49 "At the point (0,0) substituting for x and y gives" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 29 "J1:=subs(\{x=0,y=0\},evalm(J));" }}{PARA 11 "" 1 "" {XPPMATH 20 
"6#>%#J1G-%'matrixG6#7$7$\"\"#\"\"'7$!\"\"!\"#" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 20 "The linearisation is" }}}{EXCHG {PARA 256 "" 0 "" 
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}{TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 44 "To draw the phas
e plane of the linearisation" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "A1:=diff(x(t),t)=2*x+6*y;" }
}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#A1G/-%%diffG6$-%\"xG6#%\"tGF,,&F*
\"\"#%\"yG\"\"'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "B1:=diff
(y(t),t)=-x-2*y;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#B1G/-%%diffG6$-
%\"yG6#%\"tGF,,&%\"xG!\"\"F*!\"#" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 84 "DEplot(\{A1,B1\},\{x(t),y(t)\},-10..50,[[0,0,.8],[0,.
5,0]],x=-3..3,y=-2..2,stepsize=.1);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 106 "The phase po
rtrait shows that the equilibrium point of the linearisation is a cent
re and neutrally stable." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 153 " Thus in this example the li
nearisation theorem breaks down and the geometric behaviour of the non
linear system does not mimic that of its linearisation." }{MPLTEXT 1 
0 0 "" }}}}{MARK "10 0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 }