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{SECT 0 {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT 256 15 "volterra.mws   " }{TEXT -1 22 " Predator-prey mod
els\n" }{TEXT 261 1 "\n" }}{PARA 0 "" 0 "" {TEXT -1 90 "We use Maple's
 DEtools to study solutions of the Lotka-Volterra system and its refin
ements" }{TEXT -1 61 ".  As you play with the models, keep these quest
ions in mind:" }}{PARA 0 "" 0 "" {TEXT -1 731 "\n1.  What is the long \+
term behavior of the system?\n2.  In the case of oscillations, what is
 the period (time interval from peak to peak or trough to trough), and
 what is the amplitude?\n3.  How does changing the initial conditions \+
affect your answers to qustions 1 and 2?  \n4.  Does the system have a
ny steady states (equilibria)?  Do these appear to be stable or unstab
le?\n5.  If there are steady states, are they in any way related to th
e long term behavior?\n\nHere h represents the hare population and u r
epresents 60 times the lynx population (since the lynx population is n
umerically much smaller than the hare population, we scale it up to fi
t on the same graph).  We are going to work with three different initi
al conditions. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "restart:
 with(plots):  with(DEtools):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 141 "rate_eqn1:= diff(h(t),t)=(0.1)*h(t)-(0.005)*h(t)*(1/60)*u(t); r
ate_eqn2:=diff(u(t),t)=(0.00004)*h(t)*u(t)-(0.04)*u(t);\nvars:= [h(t),
 u(t)];  " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 110 "init1:=[h(0)=
2000,u(0)=600]; init2:=[h(0)=2000,u(0)=1200]; init3:=[h(0)=2000, u(0)=
3000]; \ndomain := 0 .. 320;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 368 "
\n We plot the hare and lynx populations jointly against time using th
e first of the given initial conditions.   You should repeat this with
 the other initial conditions.  Get a feeling for the accuracy of the \+
computations by changing the step size, and for the long term behavior
 by changing the time interval.  Keep a record of the results with que
stions 1-5 in mind!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 207 "Lpl
ot:= DEplot(\{rate_eqn1, rate_eqn2\}, vars, domain,\{init1 \}, stepsiz
e=0.5, scene=[t, u], arrows=NONE):\nHplot:= DEplot(\{rate_eqn1, rate_e
qn2\}, vars, domain,\{init1 \}, stepsize=0.5, scene=[t, h], arrows=NON
E):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 69 "display( \{Lplot, Hp
lot\} , title = `Hares and 60 * Lynxes vs. time` );" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 109 "
\nWhich graph is which?  You may want to inset options such as linecol
or= or thickness= to distinguish them.  " }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 341 "Next we plot the hare an
d lynx populations against one another in what is called a PHASE PORTR
AIT.  We do this for three different initial conditions.  ***Can you i
dentify which curve goes with which initial condition?  How is the ind
ependent variable t showing up in these pictures?  (Hint: try it again
 with time interval t = 0 .. 20.) ***" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 161 "DEplot(\{rate_eqn1, rate_eqn2\}, vars, t= 0 .. 160, \+
\{init1, init2, init3\}, stepsize=0.5, scene=[h,u], title=`Hares vs. 6
0 * Lynxes for t = 0 .. 160`, arrows=NONE);" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 115 "What is the significan
ce of the next calculation?  (Hint: try using these values of h and u \+
as initial conditions.)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 85 "
equil:= solve( \{(0.1)*H-(0.005)*H*(1/60)*U = 0, (0.00004)*H*U-(0.04)*
U = 0\}, \{H, U\});" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 72 "CQfu
nction:= (H, U) -> 0.00004*H + (0.005/60)*U - 0.04*ln(H)- 0.1*ln(U);" 
}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 97 "Fill in (H, U) coordinates of s
everal points from one of the \"orbits\" above. What do you observe?" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 109 "CQvalue:= CQfunction(  %
? , %? ); # use the TAB key to get to the %? and replace them with act
ual coordinates" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 137 "implici
tplot( 0.00004*H + (0.005/60)*U - 0.04*ln(H)- 0.1*ln(U) = CQvalue, H =
 0 .. 4200, U = 0 .. 3500, thickness = 2, numpoints = 1000);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 178 "Does this last plot help convince you that the \"orbits
\" really do close up?  (Hint: along each orbit there appears to be a \+
conserved value, CQ. What is the significance of this?)" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 92 "\n***Discuss the answers to questions 1-5 above in light of you
r examination of the model.***" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 420 " Next we study solutions of the Lotk
a-Volterra system where the prey is assumed to grow logistically in th
e absence of any predators.  Can you see how the rate equations have b
een changed from the original L-V model to incorporate this assumption
?  This time h represents the hare (rabbit) population and u represent
s 100 times the lynx (fox) population.   We are going to work with thr
ee different initial conditions.  " }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 37 "restart: with(plots):  with(DEtools):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 163 "rate_eq1:= diff(h(t),t) = (0.1)*h(
t)-.00001*(h(t))^2-(0.005)*h(t)*(1/100)*u(t) ;\nrate_eq2:= diff(u(t),t
) = (0.00004)*h(t)*u(t)-(0.04)*u(t) ;\nvars:= [h(t), u(t)];  " }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 108 "init1:=[h(0)=2000,u(0)=500]
; init2:=[h(0)=2000,u(0)=1000]; init3:=[h(0)=2000,u(0)=4000]; \ndomain
:= 0 .. 380;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 237 "\n First we plot
 the hare and lynx populations jointly against time using the first of
 the given initial conditions.   As above you should repeat this with \+
the other initial conditions, different time intervals, different step
 sizes, etc." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 204 "Lplot:= DEplot([ra
te_eq1, rate_eq2], vars, domain,\{init1 \},stepsize=0.5, scene=[t, u],
 arrows=NONE):\nHplot:= DEplot([ rate_eq1, rate_eq2], vars, domain, \{
init1 \}, stepsize=0.5, scene=[t, h], arrows=NONE):" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 69 "display( \{Lplot, Hplot\}, title = `Hares
 and 100 * Lynxes vs. time` );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 139 "Repeat these plots wi
th the other initial conditions. What do you observe, especially as re
gards the frequency and amplitude of the cycles?" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 366 "\nNext we plot the hare and lynx populations agai
nst one another in what is called a PHASE PORTRAIT.  We do this for tw
o different initial conditions.  Can you identify which curve goes wit
h which initial condition?  How is the independent variable t showing \+
up in these pictures?  (Hint: try it again with time interval t = 0 ..
 20, or use the option arrows = SLIM.)" }}{PARA 0 "> " 0 "" {MPLTEXT 
1 0 202 "DEplot([ rate_eq1, rate_eq2 ], vars, 0 .. 480, [init1, init2,
 init3], linecolor=[blue, black, green], arrows = SLIM, stepsize= 0.5,
 title =\"Hares vs. 100 * Lynxes for t = 0 .. 480\", dirgrid = [15, 15
]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 89 "equil:= solve( \{(0.
1)*H-.00001*H^2-(0.005)*H*(1/100)*U, (0.00004)*H*U-(0.04)*U\}, \{H, U
\});" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 95 "\nWhat is the significanc
e of this last calculation?  ***Answer questions 1-5 for this model.**
*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
494 "Finally we study solutions of the May system (problem 5 on page 1
89 of CIC; see also E-K page 265 problem 32 where the predators are wh
ales and the prey are krill).   Here x is measured in units of \"hecto
hares\" (i.e., the number of hares in units of 100) and y is the numbe
r of lynxes.  Choose a variety of initial conditions, time intervals, \+
stepsizes, and so forth.  ***Are there any QUALITATIVE similarities an
d/or differences that you notice between the May model and the two L-V
 models?***" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 82 "eq1:=diff( x
(t), t) = 0.6 * x(t) *(1 -(x(t) / 10)) - 0.5 * x(t) * y(t) /(x(t)+ 1);
" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 61 "eq2:=diff( y(t), t) = 0.1 * y(t
) * ( 1 - y(t) / (2 * x(t)) );" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "v
ars:= [x(t), y(t)];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 124 "ini
t1:=[x(0)=8, y(0)=10]; init2:=[x(0)=8, y(0)=15]; init3:=[x(0)=8, y(0)=
5]; init4:=[x(0)=1.4, y(0)=2.5]; domain:= 0 .. 150;" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 94 "Xplot:= DEplot([eq1, eq2], vars, domain, \+
\{init1 \}, stepsize=0.5, scene=[t,x], linecolor=blue):" }}{PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 95 "Yplot:= DEplot([eq1, eq2], vars, domain, \{ini
t1 \}, stepsize=0.5, scene=[t,y], linecolor=green):" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 79 "display( \{Xplot, Yplot\}, title = `May m
odel: Rabbits/100 and Foxes vs. time` );" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 139 "Repeat these
 plots with the other initial conditions. What do you observe, especia
lly as regards the frequency and amplitude of the cycles?" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 188 "DEplot([eq1, eq2], vars, 0 .. 280,
 [init1, init2, init3, init4], linecolor=[blue, black, aquamarine, plu
m],  stepsize=0.25, scene=[x,y], title = `May model phase portrait`, a
rrows = SLIM);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 102 "equil:= solve( \{0.6 * X *(
1 -(X / 10)) - 0.5 * X * Y /(X + 1), 0.1 * Y * ( 1 - Y / (2 * X))\}, \+
\{X, Y\});" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 382 "\nWhat is the sign
ificance of the last calculation?  *** You may have heard the term \"l
imit cycle\".  Can you explain where this term comes from?  Predict wh
at will happen if you use initial conditions close to, but not at, the
 equilibrium values.  Test your prediction!***  Suggest two very diffe
rent examples of global trajectories with the same local behavior near
 the equilibrium." }}{PARA 0 "" 0 "" {TEXT -1 816 "\n***At last it's t
ime to judge the models.  What sort of field measurements would you wa
nt to have in order to choose one over another?  Are there any purely \+
mathematical features of the predictions of the models that might help
?  (One can critique the models on the basis of unreasonable assumptio
ns that might go into their construction--this is a separate matter.) \+
 For example, you have surely observed that each of the three models (
L-V with unbounded prey growth, L-V with bounded prey growth, and May)
 exhibits a different possibility for long term behavior.  Why is this
 considered to be such an important aspect of the model?  What about s
ensitivity to initial conditions--how does long term behaviour depend \+
on initial conditions, and what does this mean in terms of actual obse
rvation of populations?***  " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 0 "" }}}{SECT 0 {PARA 3 "" 0 "" {TEXT 258 57 "Lotka-Volterra type mo
dels with local eigenvalue analysis" }}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 209 "\nHere h represents the hare population and u represents 60 ti
mes the lynx population (since the lynx population is numerically much
 smaller than the hare population, we scale it up to fit on the same g
raph).  " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 38 "restart: with(linalg):  with(plots):  " }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 101 "rate_eqn1:= diff(h(t),t)=(0.1)*h-(
0.005)*h*(1/60)*u; rate_eqn2:= diff(u(t),t)=(0.00004)*h*u-(0.04)*u;" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "equil:= solve( \{rhs(rate_
eqn1), rhs(rate_eqn2)\}, \{h , u \});" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 40 "F:= rhs(rate_eqn1);  G:= rhs(rate_eqn2);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "J:= jacobian([F, G], [h, u] );" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 88 "sel_equil:=  ; # select the \+
desired equilibrium point (copy and paste from output above)" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "A:= evalf(subs(sel_equil, ev
alm(J)));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "eigenvals( A )
;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 98 "What is this calculation telling you about the natur
e of the equilibrium point in phase space for " }{TEXT 259 24 "the lin
earized system?  " }{TEXT -1 73 "What conclusion can you draw, if any,
 for the original non-linear system?" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 357 " Next we study solutions of the Lotka-Volterra sy
stem where the prey is assumed to grow logistically in the absence of \+
any predators.  Can you see how the rate equations have been changed f
rom the original L-V model to incorporate this assumption?  This time \+
h represents the hare (rabbit) population and u represents 100 times t
he lynx (fox) population.   " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 118 "rate_eq1:= diff(h(t),t) = (
0.1)*h-.00001*h^2-(0.005)*h*(1/100)*u ;\nrate_eq2:= diff(u(t),t) = (0.
00004)*h*u-(0.04)*u  ;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "e
quil:= solve( \{rhs(rate_eq1), rhs(rate_eq2)\}, \{h , u\});" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "F:= rhs(rate_eq1);  G:= rhs(
rate_eq2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "J:= jacobian(
[F, G], [h, u] );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 120 "sel_e
quil:=  ; # select the desired equilibrium point (copy and paste from \+
output above, complete with the curly braces)" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 37 "A:= evalf(subs(sel_equil, evalm(J)));" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "eigenvals( A );" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 53 "\nWhat is the significance of this last c
alculation?  " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 375 "F
inally we study solutions of the May system.   Here x is measured in u
nits of \"hectohares\" (i.e., the number of hares in units of 100) and
 y is the number of lynxes.  Choose a variety of initial conditions, t
ime intervals, stepsizes, and so forth.  ***Are there any QUALITATIVE \+
similarities and/or differences that you notice between the May model \+
and the two L-V models?***" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "eq1:=diff( x(t), t) = 0.6 *
 x *(1 -(x / 10)) - 0.5 * x * y /(x + 1);" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 52 "eq2:=diff( y(t), t) = 0.1 * y * ( 1 - y / (2 * x) );
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "equil:= solve( \{rhs(eq
1), rhs(eq2)\}, \{x, y\});" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
28 "F:= rhs(eq1);  G:= rhs(eq2);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 30 "J:= jacobian([F, G], [x, y] );" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 89 "sel_equil:=   ; # select the desired equilibri
um point (copy and paste from output above)" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 37 "A:= evalf(subs(sel_equil, evalm(J)));" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "eigenvals( A );" }}}{EXCHG {PARA 0 
"" 0 "" {TEXT -1 314 "\nWhat is the significance of the last calculati
on?  ***You may have heard the term \"limit cycle\".  Can you explain \+
where this term comes from?  Predict what will happen if you use initi
al conditions close to, but not at, the equilibrium values.  Test your
 prediction, using the first section of this worksheet. ***" }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" 
}}}}}{MARK "0 0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 
0 1 2 33 1 1 }