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:::::::eZ:Vy<>jx]:JBAj:J:DZJ^dcgg_WhZnc_whZNdigg[oGMK@;PBB:qAB:>l;ZC:UK:^:>X=j>>:_c<>y=F:;B:AocO;:sg:B:=J;Dlc`qsLqlp`h_:f?=J:B>N:F:nyyyyy]::yyyyyy::::::::::::::::::::::::::::::::::: ::::::::::::::::::::::::::::::::::::fyyyyya:nYf::G:jy;:::::::::::::::: :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::JcvGYMt >^:fBWMtNHm=;:::::::n:;JZC:bKB::>pQHd=IdJoq;V:>Z?B:<:=ja ^GE=;:::::::::N;?R:yyyyyyA:yayA:<::::::JDJ:j::F@[KaFFcmnnHEM:>:::::::o J;Zy=J:B::::::F:;JHYLkNG>::::::::N;N;;Jyk?J@j@>:W:YJ:>\\:B:]:_J:VQMCHRvHjAr<]R:?B \\_[DNZ@X_<>kpN\\=@iLVjsOZ@X?P=mu:Y:::::::FZ:nyyYZDjysyQj;J:> R<:TNC>:UTRcETcTX[US;SK]UW=EWMuUWm>j\\@A:::cYH_WV>Z::::::::jysyA:CB:F: ;jysy?:;:SG>nmr:SW:N:e:;jD:;B:C:?R:=jZ:^:n_;> =E:]c:=Z:f:V[f<>^In[:JMTjAN;>ryapyAvH^\\^;UTR:;J jf:J;D_mlVH[KRJ:<:::::::>=?R:>:?J:j=Z:J;vC JbNHVH>@>Z::::::::kJ:vYxI:;Z::::::Jyyyy[:::::::::::::vYxI:;Z:::::::::: ::::::::::yay=J:B::::::::jDB:qi:;fyB:>l;F:>ZcTTUUSaEBWTSiEB_tUUURWM :eXIOo>JSjPHj:jvNOJ SJKUj:>:k=OZ:N@f_DN:l;>:::JBAJ x;j>>:_;Ef:=J:F[Z:VY;RyB:>l;>::C:[q:F;;JSFq;n_ ;F:;B:?MyF;N@oOnIj:jmnI[Z:VY;><:[V:b:DZJVdscRYEUdKF:;B:uC:ap>JSj\\\\j:jOVTJSJbij:jOjDjw;<:[V:B:D:c\\_;rUJB_:f?;JJSjg`j:JU>r::_;oH==B:f@fl;>: X;j>>:_KtwfAF:HM:d:]e==:<:OB:n>N;yyyxy:J:><:sg:>Z:FZX=j;::::::::::::::::::::::::::::::::::::::::::: ::3:" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 48 "is a Lyapunov func tion for the nonlinear 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 first prove the equilibrium point is a stationary v alue of the Lyapunov function and then find the type of stationary poi nt." }}}{EXCHG {PARA 261 "" 0 "" {TEXT -1 36 "To prove (0,0) is a stat ionary 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:nyyy yy]::yyyyyy::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: ::::::::::::fyyyyya:nYf::wyyyqy;:::::::::::::::::::::::::::::::::::::: ::::::::::::::::::::::::::::::::::::::::::NDYmq^H;C:ELq^H_mvJ::::::::g jR<:T><::[dAatj]DNwW:A:;B;B:F:YLpfF>:::::::::J?NZ;vyyy yyY: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::::::::::>FAfO@J;ZrK ;TrXHJA:bO:`A?Z<>kjJb;B::::::::::FZ:nyyYZDjysyQj;J:>R<:TNC>:UTRcE TcTX[US;SK]UW=EWMuUWm>B:ETV:::^xsNpkK:<::::::::yayY:^Z:j:>:yayQ:>:neCJ @M:Kf:F:MZ=^b;B:AB:e:;j<>:Mb:>Z:^:B:;xyewy;DVMB:@[C:>Z::::::::kJ;@:;J;>Z:vYxY:>Z::::::jD_= a=[;;B:::::::JF>:yay=J:B::::::nYyA<::::::::::::jysy:>:<::::::::::::::: ::::vYxI:;Z::::::::JBBEZ:V\\<;B:qi:;fy>Z:JBA:DZaTXUeRYEUHLJSFk =fK=B:nP:_KjDjREj:>:ukcG;N@QM;Lj:JU;N@MNVLj:JU;N@;MhTj:>:u;e:qAB:>l;F: ;J:DJ:^dcSSaEBWTSiEB_tUUURWM@x;F:MJ:N@QmD@j:>:gLoG;N@Mn_@j: JvNOM:_ko>r>F:nD:_k>ox>F:FIOoDjw;<:[V:;Z:b::::sL>:_KjNkAF :>@;H:OZ:n>N;yyyxy:J:><X=j;: :::5:" }{TEXT -1 1 " " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 35 "and (0,0 ) is a stationary point of " }{TEXT 258 1 "V" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 259 42 "To determine t he type of stationary point." }}{PARA 0 "" 0 "" {TEXT -1 63 "First fin d the Hessian 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 " eig envals:=eigenvalues(H);" }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 74 "Since the eigenvalues are both positive t he stationary point is a minimum." }}{PARA 0 "" 0 "" {TEXT -1 60 "Thus the equilibrium point (0,0) is stable and an attractor." }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 111 "This can be confi rmed by finding the Jacobian matrix of the linearisation and using the Linearisation theorem. " }}{PARA 0 "" 0 "" {TEXT -1 25 "Try this meth od yourself." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 68 "A look at the phase portrait should also confirm the above anal ysis." }}{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,ste psize=0.1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK " 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 }