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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 64 "To examine the phase diagr
am close to the equilibrium point(0,0)" }}{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" }}{PARA 7 "" 
1 "" {TEXT -1 33 "Warning, new definition for trace" }}}{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; " }{TEXT 
-1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "B:=diff(y(t),t)=.
1*y-.001*y^2-.003*x*y;" }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 112 "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
=-0.1..1,y=-0.1..1,stepsize=.1);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 68 "To examine th
e phase diagram of the linearisation at the point (0,0)" }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 39 "First find t
he 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 28 "J00:=subs(x=0,y=0,evalm(J));
" }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
30 "The linearisation is given by " }}{PARA 256 "" 0 "" {OLE 1 4104 1 
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::::::::::::::::::::::::::::::::::::::::3:" }{TEXT -1 0 "" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 46 "To plot the phase diagram of the linearis
ation" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 23 "A1:=diff(x(t),t)=0.2*x;" }}{PARA 11 "" 1 "" {TEXT -1 
0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "B1:=diff(y(t),t)=0.1
*y;" }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 114 "DEplot(\{A1,B1\},\{x(t),y(t)\},-50..50,[[0,1,1],[0,-
1,1],[0,0.1,-0.1],[0,-0.1,-0.1]],x=-0.1..1,y=-0.1..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 n
onlinear system both show the same geometric behaviour close to the eq
uilibrium point?" }}}}{MARK "4 0 0" 39 }{VIEWOPTS 1 1 0 1 1 1803 }