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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 4109 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 4105 1 "[xm]Br=WfoRrB
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:::::::::::::::4:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 160 "The equilib
rium point of the linearisation is nonsimple and nothing can be deduce
d about the equiliibrium point of the nonlinear system from its linear
isation. " }}{PARA 0 "" 0 "" {TEXT -1 175 " Look at the phase portrait
 below to see if you can determine anything about the stsbility of thi
s equilibrium point.the equilibrium point of the nonlinear system is u
nstable." }}}{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 4617 1 "[xm]Br=WfoRrB:
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::::::::::::::::::::::::::::::::::::::::::::::::2:" }{MPLTEXT 1 0 0 "
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 166 "The equilibrium point of the
 linearisation is a stable node and by the linearisation theorem the e
quilibrium point of the nonlinear system is stable and an attractor." 
}}}{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 4105 1 "[xm]Br=WfoRrB:::wk;nyyI;G:;:
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FAj\\@j:>Z:NFwm>JSNm;vj=F:vFw=N@mnR]j:Jk:N@onsej:<:I=[Z:VY;><:[V:b:DJ:
<kc`a\\wgfCAC:Uk:^:>X;j>>:_c<;nfLj:>Z:fAap>JSj\\ij:jX:N@VyFF:f_:VTe:qA
B:>l;Z:b:^DP@:J@<U<jP@J:^:>X=j>>:_;=d;=J:<JP>f:N;n>N;yyyxy:J:><<jw?<I:
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::::::::::::::::::::::::::2:" }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 164 "The equilibrium point of the linearisation is a saddle
 and by the linearisation theorem the point (0,-1) of the nonlinear sy
stem is unstable and a nonlinear saddle." }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 136 "Plot the phase diagram of the sys
tem 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=.1
);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "30 0 1" 
0 }{VIEWOPTS 1 1 0 1 1 1803 }