{VERSION 2 3 "DEC ALPHA UNIX" "2.3" } {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 "Maple Output" 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 }} {SECT 0 {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 123 "As an example of Heun's method for syste ms of first-order equations,\nconsider the following nonlinear predato r-prey model.\n" }{XPPEDIT 18 0 "dF/dt = (-0.03 + 0.001*R)*F" "/*&%#dF G\"\"\"%#dtG!\"\"*&,&$\"\"$!\"#F'*&$\"\"\"!\"$F%%\"RGF%F%F%%\"FGF%" } {TEXT -1 14 " " }{XPPEDIT 18 0 "F(0) = 20" "/-%\"FG6#\"\" !\"#?" }{MPLTEXT 1 0 1 "\n" }{XPPEDIT 18 0 "dR/dt = (0.05 - 0.001*F)*R " "/*&%#dRG\"\"\"%#dtG!\"\"*&,&$\"\"&!\"#F%*&$\"\"\"!\"$F%%\"FGF%F'F%% \"RGF%" }{TEXT -1 90 " R(0) = 50\n\nFirst, define the \+ functions that give the value of the slope:\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "f := (F,R,t) -> (-0.03 + 0.001*R)*F;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fG:6%%\"FG%\"RG%\"tG6\"6$%)operatorG%&ar rowGF**&,&$!\"$!\"#\"\"\"9%$F3F1F39$F3F*F*" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 36 "g := (F,R,t) -> (0.05 - 0.001*F)*R;\n" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#>%\"gG:6%%\"FG%\"RG%\"tG6\"6$%)operatorG%&arrowG F**&,&$\"\"&!\"#\"\"\"9$$!\"\"!\"$F39%F3F*F*" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 78 "Then, define the step size and the initial values for e ach variable and for t." }{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 7 "h := 1;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"hG\" \"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "t[0] := 0; F[0] := 20; R[0] := 50;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"tG6#\"\"!F' " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"FG6#\"\"!\"#?" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"RG6#\"\"!\"#]" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 201 "Then for each variable, the new value is ``old value + s tep size times slope.''\nThe slope is the average of the derivative at two different points.\nAfter one step, the values are calculated as f ollows:" }{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 320 "t[1] := t[0] + h:\n\nslope1F := f(F[0],R[0],t[0]):\nslope1R := g( F[0],R[0],t[0]):\n\nslope2F := f(F[0]+h*slope1F,R[0]+h*slope1R,t[1]): \nslope2R := g(F[0]+h*slope1F,R[0]+h*slope1R,t[1]):\n\nslopeF := ( slo pe1F + slope2F )/2:\nslopeR := ( slope1R + slope2R )/2:\n\nF[1] := F[0 ] + h*slopeF:\nR[1] := R[0] + h*slopeR:\n\n[F[1],R[1],t[1]];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7%$\"+++$>/#!\")$\"+++A^^F&\"\"\"" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 391 "Recall that Euler's method gave F [1] := 20.4 and R[1] := 51.5.\nThe true values are approximately F[1] \+ := 20.4194 and R[1] := 51.5121. You can\nsee that Heun's method, whi le requiring about twice as many calculations, returns a\nmuch more ac curate answer.\n\nAs in the Euler example, let's create three sequence s of points: one in the t-F plane,\none in the t-R plane, and one in \+ the F-R plane." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 56 "We need to defi ne the number of steps and the step size:" }{MPLTEXT 1 0 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "N := 200; h := 1;" }}{PARA 0 "" 0 "" {TEXT -1 51 "and the first point in each of the three sequences:" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "heun_ptsF := [t[0],F[0]];" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "heun_ptsR := [t[0],R[0]];" }}{PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 26 "heun_ptsFR := [F[0],R[0]];" }}{PARA 0 "" 0 "" {TEXT -1 37 "Now we can proceed with the for loop." }{MPLTEXT 1 0 0 " " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "for n from 0 to N-1 do" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 23 " t[n+1] := t[n] + h;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 32 " s lope1F := f(F[n],R[n],t[n]);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 32 " \+ slope1R := g(F[n],R[n],t[n]);" }}{PARA 0 "" 0 "" {TEXT -1 65 " \+ First slope is the slope from Euler's method." }{MPLTEXT 1 0 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 54 " slope2F := f(F[n]+h*s lope1F,R[n]+h*slope1R,t[n+1]);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 54 " \+ slope2R := g(F[n]+h*slope1F,R[n]+h*slope1R,t[n+1]);" }}{PARA 0 "" 0 "" {TEXT -1 91 " Second slope is calculated at the \+ points resulting from Euler's method." }{MPLTEXT 1 0 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 37 " slopeF := ( slope1F + slope2F )/2;" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 " slopeR := ( slope1R + slope2R )/ 2;" }}{PARA 0 "" 0 "" {TEXT -1 74 " Final value for \+ slope is the average of the two slopes." }{MPLTEXT 1 0 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 29 " F[n+1] := F[n] + h*slopeF;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 29 " R[n+1] := R[n] + h*slopeR;" }}{PARA 0 "" 0 "" {TEXT -1 60 " new value = old value + stepsize * slope" }{MPLTEXT 1 0 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 33 " n ew_ptF := [ t[n+1], F[n+1] ];" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 33 " \+ new_ptR := [ t[n+1], R[n+1] ];" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 34 " new_ptFR := [ F[n+1], R[n+1] ];" }}{PARA 0 "" 0 "" {TEXT -1 78 " \+ define the newest points in the three sequences of po ints." }{MPLTEXT 1 0 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 35 " heun _ptsF := heun_ptsF, new_ptF;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 35 " \+ heun_ptsR := heun_ptsR, new_ptR;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 38 " heun_ptsFR := heun_ptsFR, new_ptFR;" }}{PARA 0 "" 0 "" {TEXT -1 90 " modify each of the three sequences by adding t he new point at the end." }{MPLTEXT 1 0 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 3 "od:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"NG\"$+#" }}{PARA 11 "" 1 "" {XPPMATH 20 "6# >%\"hG\"\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%*heun_ptsFG7$\"\"!\" #?" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%*heun_ptsRG7$\"\"!\"#]" }} {PARA 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