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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
oints" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart:" }}}
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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\});" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 45 "the equilibrium points are (0,0) and (80,50)." }}{PARA 
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{MPLTEXT 1 0 30 "J00:=subs(\{x=0,y=0\},evalm(J));" }}}{EXCHG {PARA 
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{TEXT -1 173 "The equilibrium point of the linearisation of the point \+
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"" {TEXT -1 231 "Therefore the equilibrium point of the linearisation \+
at the point (80,50) is a centre and the linearisation theorem fails s
o further analysis is necessary to determine the stability of the equi
librium point of the nonlinear system." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 74 "This further analysis can be achieved
 by looking for a conserved quantity." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 264 "" 0 "" {TEXT -1 29 "To find  a conserved quantity" }
{TEXT 263 1 " " }}{PARA 0 "" 0 "" {TEXT -1 53 "First find a differenti
al equation by dividing g by f" }}{PARA 0 "" 0 "" {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 to find a solution
 of the equation" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{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 this is  the value o
f the first integral. LambertW is a special function which will disapp
ear 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 -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 85 "Since the log function only exists for positive valu
es the LHS of the first integral " }{TEXT 256 5 "soln1" }{TEXT -1 106 
" is not a conserved quantity. It is necessary to convert it to an exp
ression containing exponential terms." }}{PARA 0 "" 0 "" {TEXT -1 6 "S
ince " }{TEXT 260 3 "cq1" }{TEXT -1 45 "  is a constant we can take ex
ponentials and " }{TEXT 261 3 "exp" }{TEXT -1 1 "(" }{TEXT 262 3 "cq1
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"" }}}{EXCHG {PARA 261 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 17 "Thus the function" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 257 
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:::vYxI:;Z:::::::::eZ:Vy<>jx]:JBAj:J:DZJ^dcgg_WhZnc_whZNdigg[oGEO<jP@:
MB:;B:_c<;W:Sl:njgN: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:fFf<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:vF
wMBB:qQ:[:JBA:DZ<B>aTXDpql`N^:f?=J<Jr?:MJ:N`D>NwH;=Z:fAaPBB:qQJZ:JBA:D
Z<B>jq:f?;J<JrG:MJ:N`D^XIh<=Z:f_:VIM:_;qf==B:f?f<VYZ:JBA:<Z<JdRSZZ:fNj
P@j:^:>X;j>>:_;Mi;=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::::::::::::::::::
::::::::::::::::::::::::::::::::::::::::::::::::::::6:" }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 25 "exists for all value
s 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 por
trait of the nonlinear system shows the same geometric features as the
 level curves of " }{OLE 1 3655 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><::;ugUFL_DNwW:A:;v:B:F:YLpfF>:::::::::J
?NZ;vyyyyyY:vYxY:B:::::::c:;:=:jR>@Wlj^HMMufF;J:::::::N=?:xI:;Z::::::j
:>:C=;:=j[vGUMr<jaNFO=;::::::::_J;vyyuy:>:<::::::AZ:^E>:nYN:<:^::G:IZ:
>;;j>J?>:QJ:^;f;;Jyky;::::::::::::::::::::::::::::::::::::::::::::::::
:::::::::::::::::::::::::::=Z<B:<::::::::::;E:<J<@Z=fZDFZLVj?WZ;B:=b>A
FX?bZSc>AVXDJE]L;TsNJ=F:ABN;fCmTSa]FSCD]Hr;R:[DZGcMB::::::::::FZ:nyyYZ
DjysyQj;B:;JZC:bK_J:fc[_hb_ds?h_?^?GhoGfnGgioGM:ETV:::^xsNpkK:<:::::::
:yayY:^Z:j:>:yayQ:>:^a?jwD:SY:F:MZ=nXN:eZ:>:EJ:F[<J:<J<J;@j:f:FZ<vyyyy
:f:F;B:C:WS:kj<jRDj:B:E:Qb:B:E:Sb:;d:;H;WJ:n^@v;_jysy;B:KJ:>ryAryAj>b:
Q:G;Sj`@@J:>f:f:<j>J;<ja^G>D_mlVH[KRJ:<:::::::>]:NZ;J:N:;B:yayA:;B::::
::f:<JDD:<:?ja:[LsjsJRJ:<:::::::>=;jysy:>:<::::::wqy[:::::::::::::vYxI
:;Z::::::::::::::::::::yay=J:B::::::::jDB:qi:;fyB:>l;F:>Z<B>cTTUUSaEBW
TSiEB_tUUURWMPYS<>:US:F[:>Z:N`D>j;^C=J:<j\\<JrG;N@vw;F:>IKG:M:_;]t:=:m
MNg<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:>::::::::::::::::::
:::::::::::::::::::ZG^KB:<::6:" }{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 c
losed 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 stationary 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 3643 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><::;ugUFL_DNwW:A:;B;B:F:YLpfF>:::::::::J?N
Z;vyyyyyY:vYxY:B:::::::c:;:=:jR>@Wlj^HMMufF;J:::::::N=?:xI:;Z::::::j:>
:[]::=j[vGUMr<jaNFO=;::::::::_J;vyyuy:>:<::::::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>==[DZGcM>Z::::::j;JLxomJ:::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;S
j`@@J:>VMB:<JwE:?B:YLpfF[LsfFaMR>`:J:<:::::::>]:NZ;J:N:;B:yayA:;B:::::
:f:<JRE:<:?ja:[LsjsJRJ:<:::::::>=;jysy:>:<::::::wqy[:::::::::::::vYxI:
;Z::::::::::::::::::::yay=J:B::::::::>\\:B:JK^:f_;j<B:cb:uY:;l<B:ab:uY
:ac:E:c:ex:E:a:KX;E:Mb:B:[J:<jwEJ:uI;B:>l;Z<b:<kc`qpDpql`]^:f?=j>>:_c<
QU:il:JU?r:F[:JSFk=FJ=:g?JS>f<Vm<F:;jX>u:F;N@QMpIj:>:g?JSFKgB;=:g?JS>F
[c;=J:fAf<VYZ:JBAj:>:;b:;Jd``_WhZnc_whZNdigg[oG]M<jPN:C:[Y:=j>>:_kov\\
;F:;j\\<JrG;N@MnZ@j:Jv>o:F;N@QMPMj:j\\:N@MNeMj:jv>Oe:qAB:>l;>:<Z<:::^?
C:Uk:^:>X?B:MJ:N@;MiXj:>:[KZ=J?>Z:n>N;yyyxy:J:><<jw?<I:;JXE:<j:NZ\\Vds
WhngfgcUC:UK;^:>X=j;<Z:>:::::::::::::4:" }}}{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:=subs(\{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 "therefore at the point (80,50)" }}{PARA 260 "" 0 "
" {OLE 1 3643 1 "[xm]Br=WfoRrB:::wk;nyyI;G:;:j::>:B>N:F:nyyyyy]::yyyyy
y:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
::fyyyyya:nYf::wyyyqy;::::::::::::::::::::::::::::::::::::::::::::::::
::::::::::::::::::::::::::::::::NDYmq^H;C:ELq^H_mvJ::::::::gj<vyyuyJ<j
?V:;J:>R<:T><::tmA]Nc<OYAj;J:L:<:=ja^GE=;:::::::::N;?R:yyyyyyA:yayA:<:
:::::JDJ:j::F@[KaFFcmnnHEM:>:::::::oJ;Zy=J:B::::::F:;JlJ:j:VBYmp>h:vC?
MoJ::::::::JCN:yyyxI:;Z::::::j;B:s<;:wA?Z::C:J=j=B:K:M:OJ:V;;J@j@>:W:Y
J: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:YLpJbN
HEms>@[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>aTXUeRYEUHL<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>J
SVGSG;=:E<JSFKgH;=:mMNg<VYZ:JBAJ:B:D:::jD_:f?=J<JrOZ:F;;JS>Fwt;=:[KZ=J
?>Z:n>N;yyyxy:J:><<jw?<I:;JXE:<j:NZ\\VdsWhngfgcUC:UK;^:>X=j;<Z:>::::::
:::::::4:" }}{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 poi
nt is determined by the eigenvalues of the Hessian matrix." }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 54 "To find the hessian matrix use the \"hess
ian\" command ." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "H:=hessi
an(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 eigenv
alues 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 poi
nt is a maximum and the point (80,50) is a nonlinear centre and neutra
lly 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+.00
2*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 0 127 "DEplot(\{A
,B\},\{x(t),y(t)\},-50..5,[[x(0)=100,y(0)=30],[x(0)=80,y(0)=100],[x(0)
=80,y(0)=40],[x(0)=10,y(0)=50]],x=0..150,y=0..100);" }}{PARA 13 "" 1 "
" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}
{MARK "0 0 0" 34 }{VIEWOPTS 1 1 0 1 1 1803 }