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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT 256 40 "Stability of equilibrium \+
investigation 4" }}{PARA 0 "" 0 "" {TEXT -1 19 "Consider the system" }
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:::3:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 257 29 
"To find the equilibrium point" }}{PARA 0 "" 0 "" {TEXT -1 13 "This is
 (0,0)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 259 
24 "To find the eigenvalues." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 8 "restart:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "with(linal
g):" }}{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 26 "A:=matrix(2,2,[1,1,-1,1]);" 
}}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 3 
"tr(" }{TEXT 265 1 "A" }{TEXT -1 3 ")=2" }}{PARA 0 "" 0 "" {TEXT -1 4 
"det(" }{TEXT 264 1 "A" }{TEXT -1 3 ")=2" }}{PARA 0 "" 0 "" {TEXT -1 
15 "discrim=4-8= -4" }}{PARA 0 "" 0 "" {TEXT -1 57 "Thus the eigenvalu
es are complex with positive real part." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 26 "eigenvalues:=eigenvals(A);" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 69 "The eigenvalues are complex so the eigenvectors will al
so be complex." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT 260 29 "To find the eigenvectors of A" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 37 "Eigensystem:=eigenvectors(A,radical);" }}
{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 63 "
Since the eigenvectors are complex there will be no directrix. " }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT 261 26 "To plot the phase portrait" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 14 "with(DEtools):" }}{PARA 7 "" 1 "" {TEXT -1 35 "Warnin
g, new definition for adjoint" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 20 "P:=diff(x(t),t)=x+y;" }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "Q:=diff(y(t),t)=-x+y;" }}
{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 48 "DEplot([P,Q],[x(t),y(t)],-2..2,x=-5..5,y=-5..5);" }}{PARA 13 "" 
1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 32 "Adding some \+
initial points gives" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 135 "DE
plot([P,Q],[x(t),y(t)],-2..2,[[x(0)=-1,y(0)=-1],[x(0)=-1,y(0)=1],[x(0)
=1,y(0)=-1],[x(0)=1,y(0)=1]],x=-5..5,y=-5..5,linecolour=black);" }}
{PARA 13 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 214 
"Notice that the trajectories spiral outwards from the equilibrium poi
nt. In general the trajectories will spiral outwards provided the real
 parts of the eigenvalues are positive. The equilibrium point is calle
d an " }{TEXT 262 25 "unstable spiral or focus." }}{PARA 0 "" 0 "" 
{TEXT -1 135 "If the eigenvalues have negative real part the trajector
ies will spiral towards the equilibrium point. The equilibrium point i
s then a " }{TEXT 263 24 "stable spiral or focus. " }{TEXT -1 111 "Inv
estigate this for yourself with the following example. Note that the e
quilibrium point is not at the origin." }}{PARA 0 "" 0 "" {TEXT -1 8 "
Consider" }}{PARA 257 "" 0 "" {OLE 1 4104 1 "[xm]Br=WfoRrB:::wk;nyyI;G
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