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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 66 "To examine the phase diagr
am close to the equilibrium point(20,40)" }}{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" }}}
{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 "" {MPLTEXT 1 0 40 "A:=diff(x(t),t)=.2*x-.004*x^2-.003*
x*y; " }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 39 "B:=diff(y(t),t)=.1*y-.001*y^2-.003*x*y;" }}{PARA 11 "
" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 138 "DEp
lot(\{A,B\},\{x(t),y(t)\},-50..50,[[x(0)=18,y(0)=36],[x(0)=15,y(0)=35]
,[x(0)=21,y(0)=40],[x(0)=21,y(0)=43]],x=15..25,y=35..45,stepsize=.1);
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 70 "To examine the phase diagram of the linearisation at
 the point (20,40)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 
"" 0 "" {TEXT -1 39 "First find the linearisation as follows" }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "f
:=.2*x-.004*x^2-.003*x*y; " }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "g:=.1*y-.001*y^2-.003*x*y;" 
}}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 9 "v:=[f,g];" }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 21 "J:=jacobian(v,[x,y]);" }}{PARA 11 "" 1 "" 
{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "JC:=subs(x=
20,y=40,evalm(J));" }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 30 "The linearisation is given by " }}{PARA 256 "" 
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`n>r:>RJMT:OjyyiyIZ:B:;JBB:qQBv:>:sg:B:=J;Dlc`qsLqlp`h_:f?=J<Jr?:AB:;B
:;::::::::::::::::::::5:" }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 46 "To plot the phase diagram of the linearisation" }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "A
1:=diff(x(t),t)=-.08*x-.06*y;" }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "B1:=diff(y(t),t)=-.12*x-.04*
y;" }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 138 "DEplot(\{A1,B1\},\{x(t),y(t)\},-50..100,[[x(0)=1,y(0
)=1],[x(0)=0,y(0)=0.6],[x(0)=0,y(0)=-0.6],[x(0)=-1,y(0)=-1]],x=-1..1,y
=-1..1,stepsize=.1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}
}{EXCHG {PARA 0 "" 0 "" {TEXT -1 142 "Compare the two diagrams. Do the
 linearisation and the nonlinear system both show the same geometric b
ehaviour close to the equilibrium point?" }}}}{MARK "17 1 0" 0 }
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