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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 34 " Global Stability Worked E
xample 3" }{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 76 "Investigate the long term \+
behaviour of  the pedator-prey system modelled by " }}{PARA 256 "" 0 "
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?=J<Jr?:MJ:N`DFDsT:=B:;B:uKfG;N@Nl<F:;jP<Jf;_;=b;=J:V?k@N@]Nn@j:jX:N@n
m<F:f?:_;mR<=:Q;e:qAB:>l;Z:b:^DP@::C:US:;J<JrG:MJ:N@]LoHj:>Z:>@;H:M:_;
WD;=:]C:;@N@v\\AF:>@^[:B:_;aB<AZ:>`jN@;PZk>JSFLgu:=B:;JR<JZg;N@Nh>^:>`
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c`qsLqlp`h_:f?=J<Jr?:AB:<J::::::::::::::3:" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 30 "To find the equilibrium points
" }}}{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 20 "f:=-0.1*x+0.002*x*y;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"fG,&%\"xG$!\"\"F(*&F&\"\"\"%\"yGF*$\"\"#!\"$" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "g:=-0.0025*x*y+0.2*y;" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"gG,&*&%\"xG\"\"\"%\"yGF($!#D!\"%F)
$\"\"#!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "solve(\{f=0,
g=0\},\{x,y\});" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$<$/%\"yG\"\"!/%\"xG
F&<$/F%$\"#]F&/F($\"#!)F&" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 
"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 45 "the equilibrium points are (
0,0) and (80,50)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 
258 "" 0 "" {TEXT -1 56 "To classify the equilibrium points of the lin
earisations" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 30 "First find the Jacobian matrix" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "v:=[f,g];" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#>%\"vG7$,&%\"xG$!\"\"F)*&F'\"\"\"%\"yGF+$\"\"#!\"$,&F
*$!#D!\"%F,$F.F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "J:=jaco
bian(v,[x,y]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"JG-%'matrixG6#7$
7$,&$!\"\"F,\"\"\"%\"yG$\"\"#!\"$,$%\"xGF/7$,$F.$!#D!\"%,&F3F6$F0F,F-
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 18 "At the point (0,0)" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 30 "J00:=subs(\{x=0,y=0\},evalm(J));" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%$J00G-%'matrixG6#7$7$$!\"\"F+\"\"!7$F,$\"\"#F+" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "R1 := linalg[trace](J00);" }
}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#R1G$\"\"\"!\"\"" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 23 "R0 := linalg[det](J00);" }}{PARA 11 "" 1 
"" {XPPMATH 20 "6#>%#R0G$!\"#F&" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
173 "The equilibrium point of the linearisation of the point (0,0) is \+
a saddle and by the linearisation theorem the point (0,0) shows simila
r behaviour and is a nonlinear saddle." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 20 "At the point \+
(80,50)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "J8050:=subs(\{x=
80,y=50\},evalm(J));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&J8050G-%'ma
trixG6#7$7$\"\"!$\"$g\"!\"$7$$!%]7!\"%F*" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 27 "R1 := linalg[trace](J8050);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#R1G\"\"!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
25 "R0 := linalg[det](J8050);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#R0
G$\"'++?!\"(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 231 "Therefore the equilibrium point of the l
inearisation at the point (80,50) is a centre and the linearisation th
eorem fails so further analysis is necessary to determine the stabilit
y of the equilibrium point of the nonlinear system." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 74 "This further analysis c
an be achieved by looking for a conserved quantity." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 258 30 "To find  a conserved q
uantity " }}{PARA 0 "" 0 "" {TEXT -1 53 "First find a differential equ
ation by dividing g by f" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 63 "deq:=diff(y(x),x)=((-.0025*x*y(x)+2
*y(x))/(-.1*x+.002*x*y(x)));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$deq
G/-%%diffG6$-%\"yG6#%\"xGF,*&,&*&F,\"\"\"F)F0$!#D!\"%F)\"\"#\"\"\",&F,
$!\"\"F8F/$F4!\"$!\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 45 "Use dso
lve to find a solution of the equation" }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "soln:=dsolve(deq,y(x),i
mplicit);" }}{PARA 12 "" 0 "" {TEXT -1 0 "" }}{PARA 12 "" 1 "" 
{XPPMATH 20 "6#>%%solnG/,,%\"xG$\"+++++D!#7-%#lnG6#F'$!\"#\"\"!-%\"yGF
-$\"+++++?F*-F,6#F1$!+++++5!#5%$_C1G\"\"\"F0" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 25 "soln1:=subs(y(x)=y,soln);" }}{PARA 12 "" 1 "" 
{XPPMATH 20 "6#>%&soln1G/,,%\"xG$\"+++++D!#7-%#lnG6#F'$!\"#\"\"!%\"yG$
\"+++++?F*-F,6#F1$!+++++5!#5%$_C1G\"\"\"F0" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 85 "Since the log function on
ly exists for positive values the LHS of the first integral " }{TEXT 
261 5 "soln1" }{TEXT -1 106 " is not a conserved quantity. It is neces
sary to convert it to an expression containing exponential terms." }}
{PARA 0 "" 0 "" {TEXT -1 102 "Since the RHS is a constant we can take \+
exponentials of both sides and the RHS will remain a constant." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 21 "LHS:=exp(lhs(soln1));" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%$LHSG
-%$expG6#,,%\"xG$\"+++++D!#7-%#lnG6#F)$!\"#\"\"!%\"yG$\"+++++?F,-F.6#F
3$!+++++5!#5%$_C1G\"\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 
"LHS1:=simplify(LHS);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%LHS1G*&-%$
expG6#,(%\"xG$\"+++++D!#7%\"yG$\"+++++?F-%$_C1G\"\"\"\"\"\"*&)F*\"\"#F
3)F.#\"\"\"\"#5F3!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "L
HS2:=1/LHS1;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%LHS2G*&*&)%\"xG\"\"
#\"\"\")%\"yG#\"\"\"\"#5F*F*-%$expG6#,(F($\"+++++D!#7F,$\"+++++?F6%$_C
1GF.!\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 18 "This simplifies to" 
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}{PARA 0 "" 0 "" {TEXT -1 17 "Thus the function" }}{PARA 257 "" 0 "" 
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E:;B:=J;Dlc`qsLqlp`h_:f?;J<JrG:AB:;B:;::::::::::::::::::::::::::::::::
:::::::::::::::::::::::::::::::::::::ZyyYB:<::3:" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 25 "exists for all values of \+
" }{TEXT 256 2 "x " }{TEXT -1 4 "and " }{TEXT 257 2 "y " }{TEXT -1 38 
"and is therefore a conserved quantity." }}{PARA 0 "" 0 "" {TEXT -1 0 
"" }}{PARA 0 "" 0 "" {TEXT -1 111 " Therefore the phase portrait of th
e nonlinear system shows the same geometric features as the level curv
es of " }{OLE 1 3632 1 "[xm]Br=WfoRrB:::wk;nyyI;G:;:j::>:B>N:F:nyyyyy]
::yyyyyy::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:::::::::fyyyyya:nYf::wyyyqy;:::::::::::::::::::::::::::::::::::::::::
:::::::::::::::::::::::::::::::::::::::NDYmq^H;C:ELq^H_mvJ::::::::gj<v
yyuyJ<j?V:;J:>R<:T><::;VHOER[iIwW:A:;v:B:F:YLpfF>:::::::::J?NZ;vyyyyyY
:vYxY:B:::::::c:;:=:jR>@Wlj^HMMufF;J:::::::N=?:xI:;Z::::::j:>:C=;:=j[v
GUMrvC?MoJ::::::::JCN:yyyxI:;Z::::::j;B:s<;:wA?Z::C:J=j=B:KJ:F;N;;j?>:
S:UJ:nYvY:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
::::::::::::::j:b:<Z::::::::::JjB:CR:Hj<dj:Dk;Qn;@Z:FZLVjrOZ<`_LVjsc:g
FD?b_M;Gj:VZJ?j`Fe_WHl^_bFtZAZ;>d:p^H[::::::::::j:<Jyyy;d:yayY;AZ:>:;`
:Z@O<;j`@Pt\\Pd`QrPPJPnrPqjLqnxPqF;fbk;::JtaMSAA;B::::::::vYxy;J<<:=J:
vYxY;J:JXP:qe:^w;j:F;HJuK;f<<J:f:;j>D:;B:C:?R:=j<j:Djyyyy=j<j>Z:^:n_;>
=E:]c:=Z:f:V[<Z:f:^[<>b<>r>n;;JMTjAN<yayAZ:>;;JZyYZyY:MZ<V;n>^;UTR:;Jj
<j<B:M:?B:YLpJbNHEms>@[;;B:::::::JFNZ;J:N:;B:yayA:;B::::::f:<JDD:<:?ja
:[LsjsJRJ:<:::::::>=;jysy:>:<::::::wqy[:::::::::::::vYxI:;Z:::::::::::
:::::::::yay=J:B::::::::jDB:qi:;fyB:>l;F:>Z<B>cTTUUSaEBWTSiEB_tUUURWMP
YS<>:US:F[:>Z:N`D>j;^C=J:<j\\<JrG;N@vw;F:>IKG:M:_;]t:=:mMNg<VYZ:JBAJ:B
:D:::jZ_:;jPF:C:[Y:F;;JSjNEj:>Z:^?;E:[Z:VY;RyB:>l;Z:b:<J]PPqHplHaM^:f?
;J<JrG:MJ:N@mV:_T:=J:<Jk>IM:_KGAJ@Lj:>:IMvN;<JMJ?vyyuy=:;JBB:qQBv:;J:^
q<J:<j:NZ\\VdsWhngfgcUC:Uk:^:>X;j;<Z:>::::::::::::::::::::::::::::::::
:::::5:" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 122 "If the equilibrium point (80,50) is a maximum or mi
nimum of E(x,y) then it will be surrounded  by closed curves of E(x,y)
." }}{PARA 0 "" 0 "" {TEXT -1 139 "Likewise the equilibrium point of t
he nonlinear system will be surrounded by closed curves and must be a \+
neutrally stable nonlinear centre." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT 259 58 "To prove the point (80,50) is a stationa
ry point of E(x,y)" }}{PARA 0 "" 0 "" {TEXT -1 52 ".It is necessary to
 prove that at the point (80,50) " }}{PARA 259 "" 0 "" {OLE 1 3620 1 "
[xm]Br=WfoRrB:::wk;nyyI;G:;:j::>:B>N:F:nyyyyy]::yyyyyy::::::::::::::::
:::::::::::::::::::::::::::::::::::::::::::::::::::::::fyyyyya:nYf::wy
yyqy;:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:::::::::::::::NDYmq^H;C:ELq^H_mvJ::::::::gj<vyyuyJ<j?V:;J:>R<:T><::;T
RaER[iIwW:A:;B;<Z:j:vCSmlJ::::::::::OJ;@jyyyyyy;jysy;Z:::::::^<>:F::]K
RnC=MtFGgml>:;::::::JGN:ry:>:<::::::=J:>H<:F:AlqfG[maNFO=;::::::::_J;v
yyuy:>:<::::::AZ:^E>:nYN:<:^::G:IZ:>;F;N;;j?>:S:UJ:n;v;;Jyky;:::::::::
::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::j:b:<Z
::::::::::Jp>:CR:Hj<dj:HJA>Z;NZ@ppLVj?S:?:\\A?b[us:W:Z?;RXN:DJ>==[DZGc
M>Z::::::j;JLxomJ:::FZ:nyyYZDjysyQj;J:>R<:TNC>:UTRcETcTX[US;SK]UW=EWMu
UWm>B:ETV:::^xsNpkK:<::::::::yayY:^Z:j:>:yayQ:>:VaCJ@M:Kf:F:MZ=^b;B:AB
:e:;j<>:Mb:>Z:^:<J;@j:f:FZ<vyyyy:f:F;B:C:WS:kj<jRDj:B:E:Qb:B:E:Sb:;C;;
E<W:Gc;YJCvYxY:<J>B:;xyewy;C<ex:Q:G;Sj`@@J:>VMB:<JwE:?B:YLpfF[LsfFaMR>
`:J:<:::::::>=?R:>:?J:<jysy;J:<::::::EZ:>p<Z:J;vCJbNHVH>@>Z::::::::kJ:
vYxI:;Z::::::JywYB:::::::::::::yay=J:B:::::::::::::::::::jysy:>:<:::::
:::[B:<:N>C:US:f:<JDDjxAJZf:<jCDjxAjSDj<JDjdIj<jCJ^Qj<j>D:<JB>Z:Vy<>jx
M:<:[V:b:DZJVdsgg\\wgfsBC:Uk:F;;JSdjo@jeF:nP;H:MB:N@Mv:=n:JU;N@;e:qf:=
J:fAkH:M:_ko^w=F:;JU;N@MNELj:JU;N@;MRTj:>:u;e:qAB:>l;F:;J:DJ:^dcSSaEBW
TSiEB_tUUURWmr^:f??J<JrAj:F;;JSVGiR:=J:fb:>XM:_k>Gb;F:>IKG:M:_ko^o>F:f
B:_k>ot>F:FIKoDjw;<:[V:;Z:b::::SK<jPF:C:[Q;<j>>:_KjneAF:;JR>r:N;;B:G;O
jyyiyI:>:[Z:VY[j=J:^q<Z:F:?bBaTXaEWEUUtP<jPN:C:[q:VZ:B:;:::::::::3:" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "E:=x^.2*y^.1*exp(-.002*y)*
exp(-.0025*x);" }}{PARA 11 "" 0 "" {TEXT -1 0 "" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"EG**)%\"xG$\"\"#!\"\"\"\"\")%\"yG$\"\"\"F*F+-%$expG
6#,$F-$!\"#!\"$F/-F16#,$F'$!#D!\"%F/" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 13 "P:=diff(E,x);" }}{PARA 12 "" 0 "" {TEXT -1 0 "" }}
{PARA 12 "" 1 "" {XPPMATH 20 "6#>%\"PG,&*&*()%\"yG$\"\"\"!\"\"\"\"\"-%
$expG6#,$F)$!\"#!\"$F+-F/6#,$%\"xG$!#D!\"%F+F-*$)F8$\"\")F,F-!\"\"$\"
\"#F,**)F8FAF-F(F-F.F-F5F-F9" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 13 "Q:=diff(E,y);" }}{PARA 12 "" 0 "" {TEXT -1 0 "" }}{PARA 12 "" 1 
"" {XPPMATH 20 "6#>%\"QG,&*&*()%\"xG$\"\"#!\"\"\"\"\"-%$expG6#,$%\"yG$
!\"#!\"$\"\"\"-F/6#,$F)$!#D!\"%F6F-*$)F2$\"\"*F,F-!\"\"$F6F,**F(F-)F2F
BF-F.F-F7F-F3" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 47 "To evaluate the \+
derivatives at the pont (80,50)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 27 "R:=subs(\{x=80,y=50\},\{P,Q\});" }}{PARA 12 "" 1 "" {XPPMATH 
20 "6#>%\"RG<$,$*&-%$expG6#$!$+\"!\"$\"\"\"-F)6#$!%+?!\"%F.$!\"#!#7,$F
'$F-F6" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "S:=simplify(R);" 
}}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"SG<$$!+Tkj\"[\"!#@$!+iYXAAF(" }}
}{EXCHG {PARA 12 "" 1 "" {TEXT -1 30 "therefore at the point (80,50)" 
}}{PARA 260 "" 0 "" {OLE 1 3620 1 "[xm]Br=WfoRrB:::wk;nyyI;G:;:j::>:B>
N:F:nyyyyy]::yyyyyy:::::::::::::::::::::::::::::::::::::::::::::::::::
::::::::::::::::::::fyyyyya:nYf::wyyyqy;::::::::::::::::::::::::::::::
::::::::::::::::::::::::::::::::::::::::::::::::::NDYmq^H;C:ELq^H_mvJ:
:::::::gj<vyyuyJ<j?V:;J:>R<:T><::;TRaER[iIwW:A:;B;<Z:j:vCSmlJ:::::::::
:OJ;@jyyyyyy;jysy;Z:::::::^<>:F::]KRnC=MtFGgml>:;::::::JGN:ry:>:<:::::
:=J:^F>:F:AlqfG[maNFO=;::::::::_J;vyyuy:>:<::::::AZ:^E>:nYN:<:^::G:IZ:
>;F;N;;j?>:S:UJ:n;v;;Jyky;::::::::::::::::::::::::::::::::::::::::::::
:::::::::::::::::::::::::::::j:b:<Z::::::::::JjB:CR:Hj<dj:HJA>Z;NZ@ppL
Vj?S:?:\\A?b[us:W:Z?;RXN:DJ>==[DZGcM>Z::::::::::j:<Jyyy;d:yayY;A:;JZC:
bK_J:fc[_hb_ds?h_?^?GhoGfnGgioGMZ:fbk;::JtaMSAA;B::::::::vYxy;J<<:=J:v
YxY;J:jW`:SF;>k<j:F;HJ\\@Z:VZ:f<>:EJ:F[<J:<J<B:?R:=j<j:Djyyyy=j<j>Z:^:
n_;>=E:]c:=Z:f:V[<Z:f:^[<>^>>fBn;n^@v;_jysy;B:KZ:>ryipyA^Bft=V;n>^;UTR
:;JjG[:B:oi:NZ:vCS=[LsfFaMR>`:J:<:::::::>=?R:>:?J:<jysy;J:<::::::EZ:>p
<Z:J;vCJbNHVH>@>Z::::::::kJ:vYxI:;Z::::::JywYB:::::::::::::yay=J:B::::
:::::::::::::::jysy:>:<::::::::[B:<:N>C:US:f:<JDDjxAJZf:<jCDjxAjSDj<JD
jdIj<jCJ^Qj<j>D:<JB>Z:Vy<>jxM:<:[V:b:DZJVdsgg\\wgfsBC:Uk:F;;JSdjo@jeF:
nP;H:MB:N@Mv:=n:JU;N@;e:qf:=J:fAkH:M:_ko^w=F:;JU;N@MNELj:JU;N@;MRTj:>:
u;e:qAB:>l;F:;J:DJ:^dcSSaEBWTSiEB_tUUURWmN^:f??J<JrAj:F;;JSVGiR:=J:fb:
>XM:_k>Gb;F:>IKG:M:_ko^o>F:fB:_k>ot>F:FIKoDjw;<:[V:;Z:b::::eN<jPF:C:[Q
;<j>>:_KjneAF:>@;H:OJ:<JMJ?vyyuy=:;JBB:qQBv:>:sg:B:=J;Dlc`qsLqlp`h_:f?
?J<JrG:AB:<J::::::::::1:" }}{PARA 0 "" 0 "" {TEXT -1 55 "Thus the poin
t (80,50) is a stationary point for E(x,y)" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 84 "The type of
 stationary point is determined by the eigenvalues of the Hessian matr
ix." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 54 "To find the hessian matrix
 use the \"hessian\" command ." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 20 "H:=hessian(E,[x,y]);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%\"HG
-%'matrixG6#7$7$,(*&*()%\"yG$\"\"\"!\"\"\"\"\"-%$expG6#,$F.$!\"#!\"$F0
-F46#,$%\"xG$!#D!\"%F0F2*$)F=$\"#=F1F2!\"\"$!#;F8*&*(F-F2F3F2F:F2F2*$)
F=$\"\")F1F2FE$!$+\"!\"&**)F=$\"\"#F1F2F-F2F3F2F:F2$\"$D'!\"),**&*&F3F
2F:F2F2*&)F=$\"\")F1F2)F.$\"\"*F1F2FE$FTF8*&*(FRF2F3F2F:F2F2*$)F.$\"\"
*F1F2FE$F?FPFH$F@F@FQ$\"#]!\"(7$FX,(*&*(FRF2F3F2F:F2F2*$)F.$\"#>F1F2FE
$!\"*F8F]oFdoFQ$\"\"%!\"'" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 20 "At t
he point (80,50)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "H1:=sub
s(\{x=80,y=50\},evalm(H));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#H1G-%
'matrixG6#7$7$,$*&-%$expG6#$!$+\"!\"$\"\"\"-F-6#$!%+?!\"%F2$!+cu556!#8
\"\"!7$F;,$F+$!+Xv$4U\"F:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 26 "Find
 the eigenvalues of H1" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "E
igenvals:=eigenvals(H1);" }}{PARA 8 "" 1 "" {TEXT -1 61 "Error, attemp
ting to assign to `Eigenvals` which is protected" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 1 " " }{TEXT 260 26 "To draw the phase portrait" }
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "with(DE
tools):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 95 "DEplot(\{A,B\},\{x(t),y(
t)\},-50..5,[[0,100,30],[0,80,100],[0,80,40],[0,10,50]],x=0..150,y=0..
100);" }}}}{MARK "35 11 0" 139 }{VIEWOPTS 1 1 0 1 1 1803 }