{VERSION 3 0 "IBM INTEL LINUX" "3.0" }
{USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 
1 0 0 0 0 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 
0 }{CSTYLE "2D Comment" 2 18 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }
{CSTYLE "2D Output" 2 20 "" 0 1 0 0 255 1 0 0 0 0 0 0 0 0 0 }{PSTYLE "
Normal" -1 0 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }0 0 
0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Text Output" -1 2 1 {CSTYLE "" 
-1 -1 "Courier" 1 10 0 0 255 1 0 0 0 0 0 1 3 0 3 }1 0 0 -1 -1 -1 0 0 
0 0 0 0 -1 0 }{PSTYLE "Heading 1" 0 3 1 {CSTYLE "" -1 -1 "" 1 18 0 0 
0 0 0 1 0 0 0 0 0 0 0 }1 0 0 0 8 4 0 0 0 0 0 0 -1 0 }{PSTYLE "Maple Ou
tput" 0 11 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 3 0 
-1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Maple Plot" 0 13 1 {CSTYLE "" -1 
-1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }
{PSTYLE "" 3 256 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }
3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 11 257 1 {CSTYLE "" -1 -1 
"" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }1 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }
{PSTYLE "" 11 258 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 
}1 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }}
{SECT 0 {EXCHG {PARA 256 "" 0 "" {TEXT -1 83 "Verhulst logistic equati
on - stable steady states, periodic solutions, bifurcations" }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 85 "Verhulst logistic equation is the followi
ng first order nonlinear difference equation" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#/-%\"xG6#,&%\"tG\"\"\"F)F)*(%\"rGF)-F%6#F(F),&F)F)F,!\"
\"F)" }}{PARA 0 "" 0 "" {TEXT -1 35 "where r is a nonnegative paramete
r." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 53 "First, we define the function on the right hand side:" }}{PARA 
0 "> " 0 "" {MPLTEXT 1 0 20 "f:=(r,x)->r*x*(1-x);" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 70 "A st
eady state (equilibrium) is a solution of the (algebraic) equation" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"fG6$%\"rG%\"xGF(" }}{PARA 0 "" 0 
"" {TEXT -1 39 "Its stability depends on the eigenvalue" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#/%'lambdaG-%%diffG6$-%\"fG6$%\"rG%\"xGF," }}
{PARA 0 "" 0 "" {TEXT -1 3 "If " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#2-%
$absG6#%'lambdaG\"\"\"" }}{PARA 0 "" 0 "" {TEXT -1 46 "then the steady
 state is asymptoticaly stable." }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 52 "Plot the implicit functions defined by the equations" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"fG6$%\"rG%\"xGF(" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#/-%%diffG6$-%\"fG6$%\"rG%\"xGF+\"\"\"" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#/-%%diffG6$-%\"fG6$%\"rG%\"xGF+!\"\"" }}{PARA 
0 "" 0 "" {TEXT -1 72 "that is, the parameter dependent steady states \+
and critical eigenvalues." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
12 "with(plots):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 79 "p1:=implicitplo
t(f(r,x)=x,r=0..6,x=0..1,numpoints=10000,color=red,thickness=2):" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 77 "p2:=implicitplot(diff(f(r,x),x)=1,r
=0..6,x=0..1,numpoints=10000,color=green):" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 77 "p3:=implicitplot(diff(f(r,x),x)=-1,r=0..6,x=0..1,nump
oints=10000,color=blue):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "display
(p1,p2,p3);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 2 "" 1 "
" {TEXT -1 39 "Red lines represent the steady states; " }}{PARA 2 "" 
0 "" {TEXT -1 32 "a steady state is asymptoticaly " }{TEXT 18 0 "" }
{TEXT -1 47 "stable if it lies between blue and green lines." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 125 "Calculate the value of the parameter r and the correspon
ding steady state where the positive equilibrium loses its stability.
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "eqns:=\{f(r,x)=x,diff(f
(r,x),x)=-1,r>0,x>0\}:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "solve(eqn
s,\{r,x\});" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 112 "See the dependence of the function f (co
nsidered as a function of single variable x) on the growing parameter \+
r." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "animate(\{f(r,x),x\},
x=0..1,r=0..4,color=red,frames=100);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 142 "Plot the fun
ction f for the two critical values of parameter r - appearence of the
 positive equilibrium (r=1) and loss of its stability (r=3)." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "p1:=plot(f(1,x),x=0..1,color
=magenta):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "t1:=textplot([0.5,0.2
8,`r=1`],align=RIGHT):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "p2:=plot(
f(3,x),x=0..1,color=brown):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "t2:=
textplot([0.5,0.8,`r=3`],align=RIGHT):" }}{PARA 0 "> " 0 "" {MPLTEXT 
1 0 19 "p3:=plot(x,x=0..1):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "disp
lay(p1,p2,p3,t1,t2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}
}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 64 "Now, we
 will investigate the second iteration, i.e. the equation" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#/-%\"xG6#,&%\"tG\"\"\"F)F)-%\"fG6$%\"rG-F+6$F--
F%6#F(" }}{PARA 0 "" 0 "" {TEXT -1 23 "or, in a shortened form" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"xG6#,&%\"tG\"\"\"F)F)-%#f2G6$%\"r
G-F%6#F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 73 "We define the function f2 and simplify th
e left hand side of the equation" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/,
&-%#f2G6$%\"rG%\"xG\"\"\"F)!\"\"\"\"!" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 23 "f2:=(r,x)->f(r,f(r,x));" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 8 "f2(r,x);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "factor(
x-f2(r,x));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 34 "We find equilibria
 of the equation" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"xG6#,&%\"tG\"
\"\"F)F)-%#f2G6$%\"rG-F%6#F(" }}{PARA 0 "" 0 "" {TEXT -1 104 "and esta
blish their stability. They represents 2-periodic solutions of the ori
ginal difference equation." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 66 "Plot the implicit function
 for equilibria and critical eigenvalues" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 80 "p1:=implicitplot(f2(r,x)=x,r=0..6,x=0..1,numpoints=10
000,color=red,thickness=2):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 78 "p2:=
implicitplot(diff(f2(r,x),x)=1,r=0..6,x=0..1,numpoints=10000,color=gre
en):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 78 "p3:=implicitplot(diff(f2(r,
x),x)=-1,r=0..6,x=0..1,numpoints=10000,color=blue):" }}{PARA 0 "> " 0 
"" {MPLTEXT 1 0 18 "display(p1,p2,p3);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 2 "" 1 "" {TEXT -1 39 "Red lines represent the s
teady states; " }}{PARA 2 "" 0 "" {TEXT -1 32 "a steady state is asymp
toticaly " }{TEXT 18 0 "" }{TEXT -1 47 "stable if it lies between blue
 and green lines." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 139 "Calculate the values of the param
eter r and the corresponding steady states where the equilibria (2-per
iodic solutions) lose the stability." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 37 "eqns:=\{f2(r,x)=x,diff(f2(r,x),x)=-1\}:" }}{PARA 0 ">
 " 0 "" {MPLTEXT 1 0 39 "fsolve(eqns,\{r,x\},\{r=3..4,x=0.4..0.6\});" 
}}{PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "fsolve(eqns,\{r,x\},\{r=3..4,x=0.
8..1.0\});" }}}{EXCHG {PARA 2 "" 0 "" {TEXT -1 89 "We see that the bot
h equilibria lose the stability for a unique value of the parameter r.
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 97 "See the dependence of the function f2 (considered as
 a function of x) on the growing parameter r." }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 56 "animate(\{f2(r,x),x\},x=0..1,r=1..4,color=red,fr
ames=100);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 132 "Plot the function f2 for the two critica
l values of parameter r - appearence of the 2-periodic solutions andd \+
loss of its stability." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "p
1:=plot(f2(3,x),x=0..1,color=magenta):" }}{PARA 0 "> " 0 "" {MPLTEXT 
1 0 42 "t1:=textplot([0.1,0.4,`r=3`],align=RIGHT):" }}{PARA 0 "> " 0 "
" {MPLTEXT 1 0 47 "p2:=plot(f2(3.449489743,x),x=0..1,color=brown):" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "t2:=textplot([0.18,0.89,`r=3.449489
743`],align=RIGHT):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "p3:=plot(x,x
=0..1):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "display(p1,p2,p3,t1,t2);
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 64 "Now, we will investigate the fourth iteration, i.e. the equatio
n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"xG6#,&%\"tG\"\"\"F)F)-%#f2G6
$%\"rG-F+6$F--F%6#F(" }}{PARA 0 "" 0 "" {TEXT -1 23 "or, in a shortene
d form" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"xG6#,&%\"tG\"\"\"F)F)-%
#f4G6$%\"rG-F%6#F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 26 "We define the function f4:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "f4:=(r,x)->f2(r,f2(r,x));" }
}{PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "f4(r,x);" }{TEXT -1 0 "" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 45 "We find equilibria of the difference equation" }}{PARA 11 "" 1 
"" {XPPMATH 20 "6#/-%\"xG6#,&%\"tG\"\"\"F)F)-%#f4G6$%\"rG-F%6#F(" }}
{PARA 257 "" 0 "" {TEXT -1 104 "and establish their stability. They re
presents 4-periodic solutions of the original difference equation." }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 66 "Plot the implicit function for equilibria and critical ei
genvalues" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 80 "p1:=implicitpl
ot(f4(r,x)=x,r=0..6,x=0..1,numpoints=10000,color=red,thickness=2):" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 78 "p2:=implicitplot(diff(f4(r,x),x)=1,
r=0..6,x=0..1,numpoints=10000,color=green):" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 78 "p3:=implicitplot(diff(f4(r,x),x)=-1,r=0..6,x=0..1,num
points=10000,color=blue):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "displa
y(p1,p2,p3);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 2 "" 1 "
" {TEXT -1 39 "Red lines represent the steady states; " }}{PARA 2 "" 
0 "" {TEXT -1 32 "a steady state is asymptoticaly " }{TEXT 18 0 "" }
{TEXT -1 47 "stable if it lies between blue and green lines." }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 2 "" 0 "" {TEXT -1 92 "We can see an a
ppearence of 4-periodic solutions either as a doubling of the 2-period
ic ones" }}{PARA 2 "" 0 "" {TEXT -1 31 "or as \"independent emergence
\". " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 
"" 0 "" {TEXT -1 91 "See a detail of the previous plot - appearence of
 a steady state and loss of its stability." }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 45 "rmin:=3.44:rmax:=3.56:xmin:=0.81:xmax:=0.855:" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 92 "p1:=implicitplot(f4(r,x)=x,r=rmin..
rmax,x=xmin..xmax,numpoints=10000,color=red,thickness=2):" }}{PARA 0 "
> " 0 "" {MPLTEXT 1 0 90 "p2:=implicitplot(diff(f4(r,x),x)=1,r=rmin..r
max,x=xmin..xmax,numpoints=10000,color=green):" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 90 "p3:=implicitplot(diff(f4(r,x),x)=-1,r=rmin..rmax,x=xm
in..xmax,numpoints=10000,color=blue):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 
0 18 "display(p1,p2,p3);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 268 "This plot enables us to esti
mate the value of parameter r - the bifurcation value, where 4- period
ic solution lose its stability, i.e. a 8-periodic solution appears. Th
is rough estimation we use as an input parameter for numerical computa
tion of the bifurcation value." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 37 "eqns:=\{f4(r,x)=x,diff(f4(r,x),x)=-1\}:" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 47 "fsolve(eqns,\{r,x\},\{r=3.54..3.55,x=0.81..0.83\});" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 108 "See another detail of the entire plot - \"independent \+
emergence\" of a steady state and loss of its stability." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "rmin:=3.96:rmax:=3.961:xmin:=0.12:x
max:=0.15:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 92 "p1:=implicitplot(f4(r
,x)=x,r=rmin..rmax,x=xmin..xmax,numpoints=10000,color=red,thickness=2)
:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 90 "p2:=implicitplot(diff(f4(r,x),
x)=1,r=rmin..rmax,x=xmin..xmax,numpoints=10000,color=green):" }}{PARA 
0 "> " 0 "" {MPLTEXT 1 0 90 "p3:=implicitplot(diff(f4(r,x),x)=-1,r=rmi
n..rmax,x=xmin..xmax,numpoints=10000,color=blue):" }}{PARA 0 "> " 0 "
" {MPLTEXT 1 0 18 "display(p1,p2,p3);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 221 "This plot en
ables us to estimate the value of parameter r - the bifurcation value,
 where 4- periodic solution emerges. This rough estimation we use as a
n input parameter for numerical computation of the bifurcation value.
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "eqns:=\{f4(r,x)=x,diff(
f4(r,x),x)=1\}:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "fsolve(eqns,\{r,
x\},\{r=3.96..3.9602,x=0.149..0.1495\});" }{TEXT -1 0 "" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 97 "See the dependence of the function f4 (considered as a function
 of x) on the growing parameter r." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 70 "animate(\{f4(r,x),x\},x=0..1,r=1..4,color=red,frames=
100,numpoints=500);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 187 "Plot the function f4 for the two \+
critical values of parameter r - appearence of the 4-periodic solution
s by \"doubling bifurcation\" and \"independent emergence\" of the 4-p
eriodic solutions." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "p1:=p
lot(f4(3.449489743,x),x=0..1,color=magenta):" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 54 "t1:=textplot([0.75,0.02,`r=3.449489743`],align=RIGHT)
:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "p2:=plot(f4(3.960101883,x),x=0
..1,color=plum):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "t2:=textplot([0
.42,1.03,`r=3.960101883`],align=RIGHT):" }}{PARA 0 "> " 0 "" {MPLTEXT 
1 0 19 "p3:=plot(x,x=0..1):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "disp
lay(p1,p2,p3,t1,t2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}
}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 63 "Now, we will investigate the third iteration, i
.e. the equation" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"xG6#,&%\"tG\"
\"\"F)F)-%\"fG6$%\"rG-F+6$F--F+6$F--F%6#F(" }}{PARA 0 "" 0 "" {TEXT 
-1 23 "or, in a shortened form" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%
\"xG6#,&%\"tG\"\"\"F)F)-%#f3G6$%\"rG-F%6#F(" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 26 "We define \+
the function f3:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "f3:=(r,
x)->f2(r,f(r,x));" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "f3(r,x);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 45 "We find equilibria of the difference equation" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#/-%\"xG6#,&%\"tG\"\"\"F)F)-%#f3G6$%\"rG-F%6#
F(" }}{PARA 258 "" 1 "" {TEXT -1 104 "and establish their stability. T
hey represents 3-periodic solutions of the original difference equatio
n." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 66 "Plot the implicit function for equilibria and crit
ical eigenvalues" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 80 "p1:=imp
licitplot(f3(r,x)=x,r=0..6,x=0..1,numpoints=10000,color=red,thickness=
2):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 78 "p2:=implicitplot(diff(f3(r,x
),x)=1,r=0..6,x=0..1,numpoints=10000,color=green):" }}{PARA 0 "> " 0 "
" {MPLTEXT 1 0 78 "p3:=implicitplot(diff(f3(r,x),x)=-1,r=0..6,x=0..1,n
umpoints=10000,color=blue):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "disp
lay(p1,p2,p3);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 2 "" 
1 "" {TEXT -1 39 "Red lines represent the steady states; " }}{PARA 2 "
" 0 "" {TEXT -1 32 "a steady state is asymptoticaly " }{TEXT 18 0 "" }
{TEXT -1 47 "stable if it lies between blue and green lines." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 91 "See a detail of the previous plot - appearence of a stead
y state and loss of its stability." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 48 "rmin:=3.828:rmax:=3.842:xmin:=0.148:xmax:=0.162:" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 92 "p1:=implicitplot(f3(r,x)=x,r=rmin..
rmax,x=xmin..xmax,numpoints=10000,color=red,thickness=2):" }}{PARA 0 "
> " 0 "" {MPLTEXT 1 0 90 "p2:=implicitplot(diff(f3(r,x),x)=1,r=rmin..r
max,x=xmin..xmax,numpoints=10000,color=green):" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 90 "p3:=implicitplot(diff(f3(r,x),x)=-1,r=rmin..rmax,x=xm
in..xmax,numpoints=10000,color=blue):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 
0 18 "display(p1,p2,p3);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 255 "This plot enables us to esti
mate the values of parameter r - the bifurcation values, where 3- peri
odic solution emerges and where it lose its stability. This rough esti
mation we use as an input parameter for numerical computation of the b
ifurcation value." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "eqns:=
\{f3(r,x)=x,diff(f3(r,x),x)=1\}:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 50 
"fsolve(eqns,\{r,x\},\{r=3.828..3.83,x=0.158..0.162\});" }}{PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 37 "eqns:=\{f3(r,x)=x,diff(f3(r,x),x)=-1\}:" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "fsolve(eqns,\{r,x\},\{r=3.841..3.84
2,x=0.148..0.15\});" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 97 "See the dependence of the function
 f3 (considered as a function of x) on the growing parameter r." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 70 "animate(\{f3(r,x),x\},x=0..1
,r=1..4,color=red,frames=100,numpoints=500);" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 99 "Plot the
 function f3 for the critical value of parameter r - appearence of the
 3-periodic solution." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "p1
:=plot(f3(3.828427125,x),x=0..1,color=magenta):" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 54 "t1:=textplot([0.32,1.02,`r=3.828427125`],align=RIGHT)
:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "p2:=plot(x,x=0..1):" }}{PARA 
0 "> " 0 "" {MPLTEXT 1 0 18 "display(p1,p2,t1);" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 172 "In a similar way you can investigate appearenc
e of periodic solutions with the others periods. The success of such a
 bussiness depends on speed and memory of your computer." }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 137 "At the end, we can simulate the s
olutions of the Verhulst logistic equation for various values of param
eter r and various initial values." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 23 "r:=3.84;  # parameter r" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 25 "x0:=0.1;  # initial value" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 40 "tmax:=60: # maximal number of iterations" }}{PARA 0 "
> " 0 "" {MPLTEXT 1 0 27 "x:=array(0..tmax):x[0]:=x0:" }}{PARA 0 "> " 
0 "" {MPLTEXT 1 0 26 "   for i from 1 to tmax do" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 21 "   x[i]:=f(r,x[i-1]):" }}{PARA 0 "> " 0 "" {MPLTEXT 
1 0 6 "   od:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 72 "p1:=pointplot([[t,
x[t]]$t=0..tmax],style=point,symbol=circle,color=red):" }}{PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 65 "p2:=pointplot([[t,x[t]]$t=0..tmax],style=line,
labels=[`t`,`x `]):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "display(p1,p
2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"rG$\"$%Q!\"#" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#>%#x0G$\"\"\"!\"\"" }}{PARA 13 "" 1 "" {INLPLOT 
"6%-%'POINTSG6\\o7$\"\"!$\"\"\"!\"\"7$$F)F'$\"%cM!\"%7$$\"\"#F'$\"+w&o
Xo)!#57$$\"\"$F'$\"+$\\)z'Q%F57$$\"\"%F'$\"+?)4cX*F57$$\"\"&F'$\"+&zbm
(>F57$$\"\"'F'$\"+!p0+4'F57$$\"\"(F'$\"+P[wV\"*F57$$\"\")F'$\"+#y<k+$F
57$$\"\"*F'$\"+&*=%Q2)F57$$\"#5F'$\"+mWxrfF57$$\"#6F'$\"+&GrtB*F57$$\"
#7F'$\"+$))e^q#F57$$\"#8F'$\"+JCuxvF57$$\"#9F'$\"+#o8%[qF57$$\"#:F'$\"
+Llt))zF57$$\"#;F'$\"+7B!*phF57$$\"#<F'$\"+w,Vu!*F57$$\"#=F'$\"+(>B_A$
F57$$\"#>F'$\"+XUY!R)F57$$\"#?F'$\"+Y[#e=&F57$$\"#@F'$\"+X,u'e*F57$$\"
#AF'$\"++pL@:F57$$\"#BF'$\"+m(yJ&\\F57$$\"#CF'$\"+%=e\"*f*F57$$\"#DF'$
\"+'pLvZ\"F57$$\"#EF'$\"+dnTN[F57$$\"#FF'$\"+Y$)f*e*F57$$\"#GF'$\"+<aE
6:F57$$\"#HF'$\"+2FBE\\F57$$\"#IF'$\"+=/\"zf*F57$$\"#JF'$\"+*RS>[\"F57
$$\"#KF'$\"+^0LZ[F57$$\"#LF'$\"+X(\\5f*F57$$\"#MF'$\"++([h]\"F57$$\"#N
F'$\"+)=8D\"\\F57$$\"#OF'$\"+&)31(f*F57$$\"#PF'$\"+))*R\\[\"F57$$\"#QF
'$\"+&fJa&[F57$$\"#RF'$\"+)Qu>f*F57$$\"#SF'$\"+[\"))G]\"F57$$\"#TF'$\"
+Q,w.\\F57$$\"#UF'$\"+\\LW'f*F57$$\"#VF'$\"+z+7([\"F57$$\"#WF'$\"+p]Jh
[F57$$\"#XF'$\"+IVh#f*F57$$\"#YF'$\"+\\6j+:F57$$\"#ZF'$\"+vjp(*[F57$$
\"#[F'$\"+W5)ff*F57$$\"#\\F'$\"+l>v)[\"F57$$\"#]F'$\"+*p;d'[F57$$\"#^F
'$\"+3d2$f*F57$$\"#_F'$\"+GP+*\\\"F57$$\"#`F'$\"+S;K$*[F57$$\"#aF'$\"+
x*Hcf*F57$$\"#bF'$\"+%3\"***[\"F57$$\"#cF'$\"+mv0p[F57$$\"#dF'$\"+lfT$
f*F57$$\"#eF'$\"+LM!y\\\"F57$$\"#fF'$\"+vP4!*[F57$$\"#gF'$\"+@:O&f*F5-
%'COLOURG6&%$RGBG$\"*++++\"!\")F'F'-%'SYMBOLG6#%'CIRCLEG-%&STYLEG6#%&P
OINTG-%'CURVESG6$7inF&F+F0F6F;F@FEFJFOFTFYFhnF]oFboFgoF\\pFapFfpF[qF`q
FeqFjqF_rFdrFirF^sFcsFhsF]tFbtFgtF\\uFauFfuF[vF`vFevFjvF_wFdwFiwF^xFcx
FhxF]yFbyFgyF\\zFazFfzF[[lF`[lFe[lFj[lF_\\lFd\\lFi\\lF^]lFc]lFh]lF]^l-
F^_l6#%%LINEG-%+AXESLABELSG6$%\"tG%#x~G" 2 736 297 297 2 0 1 0 2 9 0 
4 2 1.000000 45.000000 45.000000 10030 10061 10056 10074 0 0 0 20530 
0 12020 0 0 0 0 0 0 0 1 1 0 0 0 16413 400 0 0 0 0 0 0 }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "0 0 0" 0 }{VIEWOPTS 1 1 0 1 
1 1803 }