{VERSION 4 0 "IBM INTEL NT" "4.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 1 }{CSTYLE "" -1 256 "" 1 14 0 0 0 0 0 1 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 257 "" 1 14 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 258 "" 1 14 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 259 "" 1 14 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 260 "" 1 14 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }0 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 }1 0 0 0 8 4 0 0 0 0 0 0 -1 0 }{PSTYLE "Heading 2" 3 4 1 {CSTYLE "" -1 -1 "" 1 14 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }0 0 0 -1 8 2 0 0 0 0 0 0 -1 0 }} {SECT 0 {PARA 0 "" 0 "" {TEXT -1 5 " " }}{PARA 0 "" 0 "" {TEXT -1 31 "This is all the Maple syntax in" }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{PARA 4 "" 0 "" {TEXT -1 13 "Chapter 10.4." }}{PARA 4 "" 0 "" {TEXT -1 75 "An Introduction to the Mathematics of Biology, with Computer Al gebra Models" }}{PARA 4 "" 0 "" {TEXT -1 34 "by Yeargers, Shonkwiler, \+ & Herod. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 34 "The Syntax is written for Maple 6." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT 258 24 "Syntax for page 342-344." }{TEXT -1 107 "\nWith the data of Table 10.4.1, we calculate the stationary valu e of To using equation (10.4.3) as follows." }}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 28 "f:=T->s+r*T*(1-T/Tmax)-mu*T;" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 35 "s:=10; r:=.03; mu:=.02; Tmax:=1500;" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "fzero:=solve(f(T)=0,T);\nT0: =max(fzero[1],fzero[2]);" }}}{PARA 0 "" 0 "" {TEXT -1 109 " Next we calculate and display trajectories from various starting values. This will produce Figure 10.4.1." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "deq:=\{diff(T(t),t)=f(T(t))\};" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "with(DEtools):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "inits:=\{[0,0],[0,T0/4],[0,T0/2],[0,(T0+Tmax)/2],[0,Tmax]\};" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 61 "phaseportrait(deq,T(t),t= 0..25,inits,stepsize=1,arrows=NONE);" }}}{PARA 0 "" 0 "" {TEXT -1 53 " The syntax on page 343 can be revised as follows" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "s:=0; r:=.06; mu:=.02; Tmax:=1500;" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "fzero:=solve(f(T)=0,T);\nT0: =max(fzero[1],fzero[2]);" }}}{PARA 0 "" 0 "" {TEXT -1 44 " What fo llows will produce Figure10.4.2." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 149 "deq:=\{diff(T(t),t)=f(T(t))\};\ninits:=\{[0,0],[0,T0/4],[0,T0 /2],[0,(T0+Tmax)/2],[0,Tmax]\};\nphaseportrait(deq,T(t),t=0..25,inits, stepsize=1,arrows=NONE);" }}}{PARA 0 "" 0 "" {TEXT -1 3 "\n\n\n" } {TEXT 257 20 "Exercise 1, page 348" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 104 "s:=10; r:=.03; mu:=.02; Tmax:=1500;\nfzero:=solve(s+ r*T*(1-T/Tmax)-mu*T=0,T);\nT0:=max(fzero[1],fzero[2]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 103 "s:=0; r:=.06; mu:=.02; Tmax:=1500; \nfzero:=solve(s+r*T*(1-T/Tmax)-mu*T=0,T);\nT0:=max(fzero[1],fzero[2]) ;" }}}{PARA 0 "" 0 "" {TEXT -1 707 " The solution for the equation \+ f(T) = s + r T (1 - T/Tmax) - mu T = 0 is important for the differenti al equation (10.4.1). The zeros of the equation are the steady states \+ for the equation (10.4.1).\n In the context of the stability, als o examined in Section 4.4, we establish that this positive solution fo r \n T'(t) \+ = f(T(t))\nis a stable stationary point -- in the sense that solutions with values nearby have trajectories that converge to this zero for f . The tool for verifying this is to see that the derivative of f at th e stationary point is negative. Again, reference is made to the ideas \+ in Section 4.4. We establish that f'(T0) < 0." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "diff(f(T),T);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "subs(T=T0,%);" }}}{PARA 0 "" 0 "" {TEXT -1 161 " T he figures 10.4.1 and 10.4.2 suggest what is the case: with the values above, T0 is an attracting stationary point for all positive initial \+ values for T(t)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 1 "\n" }{TEXT 256 21 "Exercise 2, page 348." }{TEXT -1 68 "\n We increase r by 10% and determine how large the increase in T0 is." } }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 112 "s:=10; r:=(1+.10)*.03; mu: =.02; Tmax:=1500;\nfzero:=solve(s+r*T*(1-T/Tmax)-mu*T=0,T);\nT0:=max(f zero[1],fzero[2]);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 259 21 "Exercise 3, page 348. " }{TEXT -1 53 "\nWe set parameters and determine pertinant constants. " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 88 "s:=10; r:=0.03; mu:=0.02 ; Tmax:=1500;\nb:=.24; a:=2.4; k1:=0.000024; k2:=0.003; N:=1400;\n" }} }{PARA 0 "" 0 "" {TEXT -1 89 "We give the value of Tmas and of the uni nfected steady state with these parameter values." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "Tmax;\nsolve (s+r*T*(1-T/Tmax)-mu*T=0,T);\nT0:=max(%);" }}}{PARA 0 "" 0 "" {TEXT -1 34 "We give the infected steady state." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 143 "eq:=s-mu*T+r*T*(1-(T+TL+TA)/Tmax)-k1*V*T=0,\n \+ k1*V*T-mu*TL-k2*TL=0, k2*TL-b*TA=0,\n N*b*TA-k1*V*T-a*V=0;\nsolv e(\{eq,V>0\},\{T,TL,TA,V\});" }}}{PARA 0 "" 0 "" {TEXT -1 27 "We find \+ the value of Ncrit." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "Ncrit:=(k2+mu)*(a+k1*T0)/(k2*k1*T0);\nis(N< Ncrit);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 137 "\n\n Because N > Ncrit, we do not expect the unin fected steady state to be stable. That is, we expect to have V > 0 in \+ the steady state. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 260 21 "Exercise 4, page 348." }{TEXT -1 1 "\n" }}{PARA 0 "" 0 "" {TEXT -1 69 "What follows is Maple V, release 5, syntax for Exercise \+ 4, page 348." }}{PARA 0 "" 0 "" {TEXT -1 274 "These computations will \+ take a long time. The number of digits kept by Maple is increased to 1 7. This will change the default relative error in the Fehlberg fourth- fifth order Runge-Kutta method. To obtain an explaination of the contr ols for the method, type ?dsolve[rkf45]." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 171 "deq:=diff(T(t),t)= s-mu*T+r*T*(1-(T+TL+TA)/Tmax)-k1* V*T, \n diff(TL(t),t)= k1*V*T-mu*TL-k2*TL,\n diff(TA(t),t) = k2*TL-b*TA,\n diff(V(t),t)= N*b*TA-k1*V*T-a*V;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 96 "s: =10; r:=0.03; Tmax:=1500; mu:=0.02; N:=1400;\nb:=.24; a:=2.4; k1:=0.00 0024; k2:=0.003; N:=1400;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "init:=T(0)=1000,TL(0)=0, TA(0)=0,V(0)=0.001;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 95 "Digits:=16;\nsol:=dsolve(\{deq,init\},\{T(t), TL(t),TA(t),V(t)\},numeric,\n output=listprocedure);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 91 "Tsol:=subs(sol,T(t));\nTAsol:=subs( sol,TA(t));\nTLsol:=subs(sol,TL(t));\nVsol:=subs(sol,V(t));" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "plot('Tsol(t)','t'=0..900); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "plot(\{'TLsol(t)'\},'t' =0..600);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "plot(\{'TAsol( t)'\},'t'=0..365);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "plot( \{'Vsol(t)'\},'t'=0..365);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{MARK "3 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }