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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 34 " Global Stability Worked E
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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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{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 30 "To find the equilibrium p
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1 "" {TEXT -1 32 "Warning, new definition for norm" }}{PARA 7 "" 1 "" 
{TEXT -1 33 "Warning, new definition for trace" }}}{EXCHG {PARA 0 "> \+
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{MPLTEXT 1 0 21 "g:=-0.0025*x*y+0.2*y;" }}}{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 45 "the equilibrium p
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{EXCHG {PARA 258 "" 0 "" {TEXT -1 56 "To classify the equilibrium poin
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" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 30 "First find the Jacobian matri
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" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "J:=j
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"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 173 "The equilibrium point of th
e linearisation of the point (0,0) is a saddle and by the linearisatio
n theorem the point (0,0) shows similar behaviour and is a nonlinear s
addle." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 20 "At the point (80,50)" }}}{EXCHG {PARA 0 "> " 0 
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{EXCHG {PARA 0 "" 0 "" {TEXT -1 231 "Therefore the equilibrium point o
f the linearisation at the point (80,50) is a centre and the linearisa
tion theorem fails so further analysis is necessary to determine the s
tability of the equilibrium point of the nonlinear system." }}{PARA 0 
"" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 74 "This further anal
ysis can be achieved by looking for a conserved quantity." }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 264 "" 0 "" {TEXT -1 29 "To find  a conse
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{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "deq:=diff(y
(x),x)=((-.0025*x*y(x)+0.2*y(x))/(-.1*x+.002*x*y(x)));" }}{PARA 11 "" 
1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 45 "Use dsolve t
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{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "soln:=dsolve(deq,y(x));" }}
{PARA 12 "" 0 "" {TEXT -1 0 "" }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 144 "You now need to solve for _C1 as \+
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on which will disappear when you do this." }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 20 "cq:=solve(soln,_C1);" }}{PARA 11 "" 1 "" {TEXT -1 
0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "cq1:=subs(y(x)=y,cq)
;" }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
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soln1" }{TEXT -1 106 " is not a conserved quantity. It is necessary to
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"" 0 "" {TEXT -1 6 "Since " }{TEXT 260 3 "cq1" }{TEXT -1 45 "  is a co
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{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
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1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 11 "" 1 "" {TEXT -1 0 "" }}}{EXCHG 
{PARA 261 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 17 "Thus th
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:::eZ:Vy<>jx]:JBAj:J:DZJ^dcgg_WhZnc_whZNdigg[oGEO<jP@:MB:;B:_c<;W:Sl:n
jgN:NJJ>==_c:KVG_rZkE:>ZcogZVbnf_QgbmZ::::::cWy=T:cWy=L:>Z:::::::::j\\
<JoG;N@vw;F:>IOo>JSjbHj:jvNOJSJLTj:>:k=JSJW\\j:jv:N@F[DF:fFO?N@fuGF:fF
f<VY;syB:>l;:;b::::]M<jPF:C:[Y:<j>>:_KtWfGF:<jvVFM:_;_V>=J:>IAmDjw?JB:
>l;>:<Z<:::^@C:UK:^:>X=j>>:_KJAjNEj:>Z:^?;mDjw?>B:>l;Z:b:^D::FHjPF:C:[
Q:F;;JS^Xqt;=J:<JQ<JVF;N@FhCF:n?:_;cw<=:W;JSJ>yj:JQJBB:qQ:pI<:[V:B:DZ:
nb_og]gfmC?C:UK:^:>X=j>>:_kVAJc@j:>Z:NFwm>JSNQSB;=J:vFwMBB:qQ:[:JBA:DZ
<B>aTXDpql`N^:f?=J<Jr?:MJ:N`D>NwH;=Z:fAaPBB:qQJZ:JBA:DZ<B>jq:f?;J<JrG:
MJ:N`D^XIh<=Z:f_:VIM:_;qf==B:f?f<VYZ:JBA:<Z<JdRSZZ:fNjP@j:^:>X;j>>:_;M
i;=Z:>@[m>JSjDYj:Z:>HZDjy]j:JR:N@fmCF:F`:>HJSJ[hj:JRJ@<J:<JSJqij;B:[CF
_;[=M:_;gr==:[C:[m@JSJ[yJ<JR<]o>h:>H[=OB:<JMJ?vyyuy=:;JBB:qQBv:<J:^q<J
:<j:NZ\\VdsWhngfgcUC:UK:^:>X=j;<J:<J::::::::::::::::::::::::::::::::::
::::::::::::::::::::::::::::::::::::5:" }}{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 t
herefore a conserved quantity." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 111 " Therefore the phase portrait of the non
linear system shows the same geometric features as the level curves of
 " }{OLE 1 3654 1 "[xm]Br=WfoRrB:::wk;nyyI;G:;:j::>:B>N:F:nyyyyy]::yyy
yyy:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
::::fyyyyya:nYf::wyyyqy;::::::::::::::::::::::::::::::::::::::::::::::
::::::::::::::::::::::::::::::::::NDYmq^H;C:ELq^H_mvJ::::::::gj<vyyuyJ
<j?V:;J:>R<:T><::[rvQd[]DNwW:A:;v:B:F:YLpfF>:::::::::J?NZ;vyyyyyY:vYxY
:B:::::::c:;:=:jR>@Wlj^HMMufF;J:::::::N=?:xI:;Z::::::j:>:C=;:=j[vGUMrv
C?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:gFD?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:[Lsj
sJRJ:<:::::::>=;jysy:>:<::::::wqy[:::::::::::::vYxI:;Z::::::::::::::::
::::yay=J:B::::::::jDB:qi:;fyB:>l;F:>Z<B>cTTUUSaEBWTSiEB_tUUURWMPYS<>:
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<Jr
G:MJ:N@mV:_T:=J:<Jk>IM:_KGAJ@Lj:>:IMvN;<JMJ?vyyuy=:;JBB:qQBv:;J:^q<J:<
j:NZ\\VdsWhngfgcUC:Uk:^:>X;j;<Z:>:::::::::::::::::::::::::::::::::::::
ZG^KB:<::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 minimum 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 the 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 258 58 "To prove the point (80,50) is a sta
tionary point of E(x,y)" }}{PARA 0 "" 0 "" {TEXT -1 52 ".It is necessa
ry to prove that at the point (80,50) " }}{PARA 259 "" 0 "" {OLE 1 
3642 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><::[rvQd[]DNwW:A:;B;B:F:YLpfF>:::::::::J?NZ;vyyyyyY:vYxY:B:::::::c:
;:=:jR>@Wlj^HMMufF;J:::::::N=?:xI:;Z::::::j:>:[]::=j[vGUMrvC?MoJ::::::
::JCN:yyyxI:;Z::::::j;B:s<;:wA?Z::C:J=j=B:K:M:OJ:V;;J@j@>:W:YJ:nYvY:::
::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:F:DZ:B::::::::::^G;J<@Z=fZDFZ=n;;R:?b[gg>AVK@J;ZrK;TrXHJA:RK:`A?Z<>kj
Jb<RMt>;B::::::V:^^YwF>:::=B:wyyAb<vYxy?V:>:;`:Z@O<;j`@Pt\\Pd`QrPPJPnr
PqjLqnxPqF;<j\\@A:::cYH_WV>Z::::::::jysyA:CB:F:;jysy?:;:qS<^k>J>E:=j>r
:CT:<j;<jDJ:f:;j>D:;B:CZ:NZ;F:E:=b:yyyyI:E:M:<J<JQ@JFf:F`<F:<j<j?D:<j<
J@DJJLJj\\JAJMTjAN<yayAZ:>;<JZyyTyYJ\\jdIj?JMJ@fc[;>:;q><Z:Ny<J;<ja^GE
MbNHEms>@[C:>Z::::::::kJ;@:;J;>Z:vYxY:>Z::::::j<B:[g:B:N:Y<>D_=a=[;;B:
::::::JF>:yay=J:B::::::nYyA<::::::::::::jysy:>:<:::::::::::::::::::vYx
I:;Z::::::::JB<Z::?K<jP@:EZ:^\\<fy;>BEZ:V\\<fy;V`<f:^<ft=f:V<>s?f:F[<Z
:><;B:qi:;fy>Z:JBA:DZ<B>aTXUeRYEUHL<jPF:MJ:N`DVg;vD=:gOZ=j><:_k>Ij:G:n
P:_KjDjGEj:>:uKf=j>JSVGSy:=J:nP:_k>o\\>F:nP:_Kj>`@F:;jXjDjw;<:[V:=J:>Z
<>:cTTPps<LaPpu<Lcxpp@PqFHC:UK;^:>x;F:MJ:N@QmE@j:>:ED:[q>JSFK=T:=:kMN=
j>JSVGSG;=:E<JSFKgH;=:mMNg<VYZ:JBAJ:B:D:::JP^:f?=J<JrOZ:F;;JS>Fwt;=J:>
@;H:OJ:<JMJ?vyyuy=:;JBB:qQBv:>:sg:B:=J;Dlc`qsLqlp`h_:f??J<JrG:AB:<J:::
::::::::::5:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "E:=x^0.2*y^
.1*exp(-.002*y)*exp(-.0025*x);" }}{PARA 11 "" 0 "" {TEXT -1 0 "" }}
{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 13 "P:=diff(E,x);" }}{PARA 12 "" 0 "" {TEXT -1 0 "" }}{PARA 11 "" 1 
"" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "Q:=diff(
E,y);" }}{PARA 12 "" 0 "" {TEXT -1 0 "" }}{PARA 11 "" 1 "" {TEXT -1 0 
"" }}}{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:=s
ubs(\{x=80,y=50\},\{P,Q\});" }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "S:=simplify(R);" }}{PARA 11 
"" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 12 "" 1 "" {TEXT -1 30 "therefor
e at the point (80,50)" }}{PARA 260 "" 0 "" {OLE 1 3642 1 "[xm]Br=WfoR
rB:::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><::[WXUd[]DNwW:A
:;B;B:F:YLpfF>:::::::::J?NZ;vyyyyyY:vYxY:B:::::::c:;:=:jR>@Wlj^HMMufF;
J:::::::N=?:xI:;Z::::::j:>:C=;:=j[vGUMrvC?MoJ::::::::JCN:yyyxI:;Z:::::
:j;B:s<;:wA?Z::C:J=j=B:K:M:OJ:V;;J@j@>:W:YJ:nYvY::::::::::::::::::::::
::::::::::::::::::::::::::::::::::::::::::::::::::::F:DZ:B::::::::::>F
<J<@Z=fZDFZ=n;;R:?b[gg>AVK@J;ZrK;TrXHJA:RK:`A?Z<>kjJb<RMt>;B::::::::::
FZ:nyyYZDjysyQj;J:>R<:TNC>:UTRcETcTX[US;SK]UW=EWMuUWm>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:YLpJbNHEms>@[C:>Z::::::::kJ;@:;J;>
Z:vYxY:>Z::::::j<B:[g:B:N:Y<>D_=a=[;;B:::::::JF>:yay=J:B::::::nYyA<:::
:::::::::jysy:>:<:::::::::::::::::::vYxI:;Z::::::::JB<Z::?K<jP@:EZ:^\\
<fy;>BEZ:V\\<fy;V`<f:^<ft=f:V<>s?f:F[<Z:><;B:qi:;fy>Z:JBA:DZ<B>aTXUeRY
EUHL<jPF:MJ:N`DVg;vD=:gOZ=j><:_k>Ij:G:nP:_KjDjGEj:>:uKf=j>JSVGSy:=J:nP
:_k>o\\>F:nP:_Kj>`@F:;jXjDjw;<:[V:=J:>Z<>:cTTPps<LaPpu<Lcxpp@PqF?C:UK;
^:>x;F:MJ:N@QmE@j:>:ED:[q>JSFK=T:=:kMN=j>JSVGSG;=:E<JSFKgH;=:mMNg<VYZ:
JBAJ:B:D:::jD_:f?=J<JrOZ:F;;JS>Fwt;=:[KZ=J?>Z:n>N;yyyxy:J:><<jw?<I:;JX
E:<j:NZ\\VdsWhngfgcUC:UK;^:>X=j;<Z:>:::::::::::::3:" }}{PARA 0 "" 0 "
" {TEXT -1 55 "Thus the point (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 ei
genvalues of the Hessian matrix." }}}{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 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 20 "
At the point (80,50)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "H1:
=subs(\{x=80,y=50\},evalm(H));" }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 26 "Find the eigenvalues of H1" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "Eigenvalues:=eigenvals(H1);
" }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
130 "Since these are both negative the equilibrium point is a maximum \+
and the point (80,50) is a nonlinear centre and neutrally stsble." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 1 " \+
" }{TEXT 259 26 "To draw the phase portrait" }{MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "with(DEtools):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "A:=diff(x(t),t)=-.1*x+.002*x*y;" }}
{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 32 "B:=diff(y(t),t)=-.0025*x*y+.2*y;" }}{PARA 11 "" 1 "" {TEXT -1 0 
"" }}}{EXCHG {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);" }}{PARA 13 "" 1 "" {TEXT -1 0 "" }}}}{MARK "54 1 0" 0 }
{VIEWOPTS 1 1 0 1 1 1803 }