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SKj><<jw?Zw]:JBAJ:B:DZ:nb_og]gfmc@C:UK:^:>X=j>>:_kVAJQ@j:>Z:NFwm>JSNQK
y:=:IMy><<jw?JB:>l;Z<b:<kc`a\\wgfc@C:Uk:^:>X;j>>:_c<;OWMj:>Z:fa:VTM:_;
ED<=:Qkc;_;[x<=:Q;e:qAB:>l;Z:b:^DP@Zi;[N<jP>:C:[q:F;;JSJhQj:B:eC:;H:M:
_;qT<=:gKZ=:N@NuFFZ:jTjDjw?^y]:JBAJ:B:D:js::;>f?=J<Jr?:MJ:N@cqmXj:B:<J
rF;b<FhHF:B:N[:JMJ?vyyuy=:;JBB:qQBv:<J:^q<J:<j:DJ;Dlc`qsLqlp`h_:f?;J<J
rG:AB:<J:::::::::::::::::::::::::::::::::::::::::::::5:" }{TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 48 "is a Lyapunov function for the nonlin
ear system." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 259 "" 0 
"" {TEXT -1 30 "To find the equilibrium point." }}}{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 30 "The equilibrium
 point is (0,0)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 179 "To classify the equilibrium point of the linearisation f
irst prove the equilibrium point is a stationary value of the Lyapunov
 function and then find the type of stationary point." }}}{EXCHG 
{PARA 261 "" 0 "" {TEXT -1 36 "To prove (0,0) is a stationary point" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "subs(\{x=0,y=0\},Vx);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "subs(\{x=0,y=0\},Vy);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 4 "Thus" }}}{EXCHG {PARA 262 "" 0 "" 
{OLE 1 3590 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><::;fQGhWAfPwW:A:;B;B:F:YLpfF>:::::::::J?NZ;vyyyyyY:vYxY:B::
:::::c:;:=:jR>@Wlj^HMMufF;J:::::::N=?:xI:;Z::::::j:>:C=;:=j[vGUMrvC?Mo
J::::::::JCN:yyyxI:;Z::::::j;B:s<;:wA?Z::C:J=j=B:KJ:F;N;;j?>:S:UJ:n;v;
;Jyky;::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:::::::::j:b:<Z::::::::::JjB:CR:Hj<dj:HJA>Z;NZ@ppLVjPS:?:\\A?b[us:W:ZP
;RXN:DJ>==[DZGcM>Z::::::::::j:<Jyyy;d:yayY;A:;JZC:bK_J:fc[_hb_ds?h_?^?
GhoGfnGgioGMZ:fbk;::JtaMSAA;B::::::::vYxy;J<<:=J:vYxY;J:Ji`: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:^[<>^>>jBn;n^@v;_jysy;B:KZ:>ryipyAbBft=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:<jCDjxAj^Dj<JDjoIj<jCJtQj<j>D:<JB>
Z:Vy<>jxM:<:[V:b:DZJVdsgg\\wgfsBC:Uk:F;;JSdjo@jeF:nPwo>JSFk=fK=B:nP:_K
jDjREj:>:ukcG;N@QM;Lj:JU;N@MNVLj:JU;N@;MhTj:>:u;e:qAB:>l;F:;J:DJ:^dcSS
aEBWTSiEB_tUUURWM@<U<jPNZ:^:>x;F:MJ:N@QmD@j:>:gLoG;N@Mn_@j:JvNOM:_ko>r
>F:nD:_k>ox>F:FIOoDjw;<:[V:;Z:b::::sL<jPF:C:[Q;<j>>:_KjNkAF:>@;H:OZ:n>
N;yyyxy:J:><<jw?<I:;JXE:<j:NZ\\VdsWhngfgcUC:UK;^:>X=j;<Z:>::::5:" }
{TEXT -1 1 " " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 35 "and (0,0) is a s
tationary point of " }{TEXT 258 1 "V" }{TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 259 42 "To determine the type \+
of stationary point." }}{PARA 0 "" 0 "" {TEXT -1 63 "First find the He
ssian matrix and then look at its eigenvalues." }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 20 "H:=hessian(V,[x,y]);" }}{PARA 11 "" 1 "" 
{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 " eigenvals:
=eigenvalues(H);" }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 74 "Since the eigenvalues are both positive the statio
nary point is a minimum." }}{PARA 0 "" 0 "" {TEXT -1 60 "Thus the equi
librium point (0,0) is stable and an attractor." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 111 "This can be confirmed by
 finding the Jacobian matrix of the linearisation and using the Linear
isation theorem. " }}{PARA 0 "" 0 "" {TEXT -1 25 "Try this method your
self." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 68 "
A look at the phase portrait should also confirm the above analysis." 
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 26 "A:=diff(x(t),t)=-10*x-6*y;" }}{PARA 11 "" 1 "" {TEXT -1 0 "" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "B:=diff(y(t),t)=-6*x-10*y;
" }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 14 "with(DEtools):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 131 "DEplot(\{A,B\},[x(t),y(t)],-50..50,[[x(0)=0,y(0)=1],[x(0)=0,y
(0)=-1],[x(0)=-1,y(0)=0],[x(0)=1,y(0)=0]],x=-1..1,y=-1..1,stepsize=0.1
);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "13 0 0" 
0 }{VIEWOPTS 1 1 0 1 1 1803 }