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{SECT 0 {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT 256 14 "NBaddendum.mws" }{TEXT -1 129 "   Symbolic comput
ations with equilibrium points, Jacobians, eigenvalues.  The equations
 come from the Nicholson-Bailey model. . " }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "restart:  with(linal
g):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 57 "F:= K - alpha * q * \+
N^2;  G:= r * N - r* beta * N^2 / q ;" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 36 "J:= jacobian ( [ F , G ], [ q, N ]);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "equil:= solve( \{ F, G\}, \{ q, N \+
\} );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 157 "Maple unfortunately is \+
unhappy with the equation alpha * beta * Z^3 = K so we simply solve fo
r Z for ourselves, and evaluate the Jacobian at the equilibrium." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 85 "A:= subs(\{ N=(K/(alpha * be
ta))^(1/3 ), q=beta*(K/(alpha*beta))^(1/3 )\} , evalm(J) );" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "eigenvals(A); " }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 253 "Alternat
ively, we make up new variables a = alpha ^ (1/3),  b = beta^(1/3), k \+
= K^(1/3). Then we figure out how to rewrite A (you can actually force
 Maple to do all this, but it probably isn't worth the trouble since t
he hand work here is easy enough). " }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 57 "B:= matrix( [ [ -a*k^2/b^2, -2*a*b*k^2], [r/b^3, -r] \+
] );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "eigenvals(B);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 123 "H
ere's another approach.  Compute the eigenvalues of J in general; then
 substitute in the values of N and q at equilibrium." }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 21 "Jvals:= eigenvals(J);" }}}{EXCHG {PARA 0 
"" 0 "" {TEXT -1 91 "Get all the roots for the equilibrium point and s
elect the real one by cutting and pasting." }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 94 "real_
equil1:= \{N = (K*alpha^2*beta^2)^(1/3)/(alpha*beta), q = (K*alpha^2*b
eta^2)^(1/3)/alpha\};" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 55 "Another way to accomplish this is to use \+
the following." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 103 "real_equ
il2:= convert( \{N = RootOf(-K+alpha*beta*_Z^3), q = beta*RootOf(-K+al
pha*beta*_Z^3)\}, radical);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 109 "T
his seems to come out a tiny bit simpler than the other way.  Then eva
luate the Jacobian at the equilibrium." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 29 "subs( real_equil1, [Jvals] );" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 24 "simplify( % , radical );" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 172 "If the stuff inside the radical happens to be neg
ative, then this will tell you the eigenvalues are complex; in this ca
se you can simplify the real part pretty effectively." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 94 "(-1/2)*simplify( (K*alpha) + r*(K*a
lpha^2*beta^2)^(1/3)/(K*alpha^2*beta^2)^(1/3), symbolic);  " }}}
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