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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 33 "Local Stability worked exa
mple 2." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 43 "Find the equilibria of the nonlinear syst
em" }}{PARA 256 "" 0 "" {OLE 1 4108 1 "[xm]Br=WfoRrB:::wk;nyyI;G:;:j::
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{TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 30 "and determine the
ir stability." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 34 "First find the equilibrium points." }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 8 "restart:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 13 "with(linalg):" }}{PARA 7 "" 1 "" {TEXT -1 32 "Warning, new def
inition for norm" }}{PARA 7 "" 1 "" {TEXT -1 33 "Warning, new definiti
on for trace" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "with(DEtool
s):" }}{PARA 7 "" 1 "" {TEXT -1 35 "Warning, new definition for adjoin
t" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "f:=-2*x*y;" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "g:=y-x+x*y-y^3;" }{TEXT -1 0 "" }}}
{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 
48 "The equilibrium points are (0,0), (0,1), (0,-1)." }}{PARA 0 "" 0 "
" {TEXT -1 78 "Note that the final bracket only gives the point (0,0) \+
which is already given." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 27 "To fin
d the Jacobian matrix" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "f:
=-2*x*y;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "g:=y-x+x*y-y^3;
" }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "v:=[f,g];
" }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "J:=jacob
ian(v,[x,y]);" }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 18 "At the point (0,0)" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 29 "J0:=subs(\{x=0,y=0\},evalm(J));" }}{PARA 11 "" 1 "" 
{TEXT -1 0 "" }}}{EXCHG {PARA 257 "" 0 "" {OLE 1 4104 1 "[xm]Br=WfoRrB
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:::::::::::1:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 160 "The equilibrium
 point of the linearisation is nonsimple and nothing can be deduced ab
out the equiliibrium point of the nonlinear system from its linearisat
ion. " }}{PARA 0 "" 0 "" {TEXT -1 175 " Look at the phase portrait bel
ow to see if you can determine anything about the stsbility of this eq
uilibrium point.the equilibrium point of the nonlinear system is unsta
ble." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "A:=diff(x(t),t)=-2*
x*y;" }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 28 "B:=diff(y(t),t)=y-x+x*y-y^3;" }}{PARA 11 "" 1 "" 
{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}{PARA 0 
"> " 0 "" {MPLTEXT 1 0 118 "DEplot(\{A,B\},\{x(t),y(t)\},-50..50,[[0,1
,1],[0,-.1,.1],[0,0.1,-0.1],[0,-0.1,-0.1]],x=-0.5..0.5,y=-0.5..0.5,ste
psize=.1);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 86 "The phase portrait \+
suggests the equilibrium point of the nonlinear system is unstable." }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 18 "
At the point (0,1)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "J1:=s
ubs(\{x=0,y=1\},evalm(J));" }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}
{EXCHG {PARA 258 "" 0 "" {TEXT -1 1 " " }{OLE 1 4616 1 "[xm]Br=WfoRrB:
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0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 166 "The equilibrium point o
f the linearisation is a stable node and by the linearisation theorem \+
the equilibrium point of the nonlinear system is stable and an attract
or." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 19 "At the point (0.-1)" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "J2:=subs(\{x=0,y=-1\},evalm(
J));" }}}{EXCHG {PARA 259 "" 0 "" {OLE 1 4104 1 "[xm]Br=WfoRrB:::wk;ny
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Z:N`D>j;^@?J:>:::::::::::^hlWGen>>Z:N@nw;F:<J_NOSJ:<JSj\\Uj;JlPPt>r:NO
Qm>JSJg`j:J_jDjw;<:[V:;B:<Z<B>::jf^:f?=J<JrA:MB:;JSJZEj:<J:<Z:>RM:dZ;v
j?F:>@;@N@^_DF:B:;@b<@Jaqj:JT>R[Z:VY;RyB:>l;Z:b:<J]PPqHplHaV^:;jP>:C:[
q:F;;JSFm;fb;F:;B:?MyF;N@oV:Iv:=:IMy:_kFGpBF:NF:_KGWxDFZ:jmJBB:qQ:[:JB
A:DZ<>ZJVdscRYEUlK<jPF:C:[Q:F;;JSdJ:Ge>F:;B:ukcG;N@frEF:fA:_;qI==:UC:a
pDjw;<:[V:B:D:c\\_;:^[j`:f_;>:C:[q:F;;JSjZTj:>Z:^?;E:O:G;OjyyiyI:>:[Z:
VY[j=J:^q<J:<j:NZ\\VdsWhngfgcUC:Uk:^:>X;j;<Z:>::::::::::::::::::::::::
::::::::::::::::::::::::::::::::5:" }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 164 "The equilibrium point of the linearisation is a s
addle and by the linearisation theorem the point (0,-1) of the nonline
ar system is unstable and a nonlinear saddle." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 136 "Plot the phase diagram o
f the system and check that the equilibrium points show the geometric \+
features determined by analytical methods." }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 190 "DEplot(\{A,B\},\{x(t),y(t)\},-10..50,[[0,.1,0.5],[
0,-.1,1.9],[0,0.1,1.9],[0,0.1,.5],[0,-.01,-.5],[0,.01,-.5],[0,-.1,-1.1
],[0,0.1,-1.1],[0,.02,0.1],[0,.02,-0.1]],x=-0.1..0.1,y=-2..2,stepsize=
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