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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 "" 0 328 1 {CSTYLE "" -1 -1 "" 1 8 0 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 }} {SECT 0 {EXCHG {PARA 259 "" 0 "" {TEXT -1 43 "MATH 2280 - 1 \+ SECON" }{TEXT 274 0 "" }{TEXT -1 18 "D MAPLE ASSIGNMENT " }}{PARA 260 "" 0 "" {TEXT -1 67 "Treibergs \+ Due Wednesday, Sept. 24, 2003" }}{PARA 261 "" 0 "" {TEXT -1 0 "" }} {PARA 262 "" 0 "" {TEXT -1 471 "In this lab we explore some of the way s computers approximate the solution of first order ODE's. In the bas ic Euler method, the computer calculates the slope at a point and draw s a short straight line with that slope from that point to get the nex t point. The modified Euler Method and the Runge-Kutta methods are imp roved versions, that take into account the derivative at more than jus t the left endpoint. This material is discussed in sections 2.4-2.6 of the text by " }{TEXT 256 18 "Edwards and Penney" }{TEXT -1 105 ". \n \nThe initial value problem we wish to address is the general first or der equation with initial point.\n" }}{PARA 263 "" 0 "" {TEXT -1 0 "" }{XPPEDIT 18 0 "diff(y(x),x) = f(x,y(x)),y(x0) = y0;" "6$/-%%diffG6$-% \"yG6#%\"xGF*-%\"fG6$F*-F(6#F*/-F(6#%#x0G%#y0G" }}{PARA 264 "" 0 "" {TEXT -1 0 "" }}{PARA 265 "" 0 "" {TEXT 286 15 "Euler's Method." } {TEXT -1 5 "\nLet " }{TEXT 312 8 "(x0,y0) " }{TEXT -1 49 "denote the i nitial point of the solution curve. W" }}{PARA 328 "" 0 "" {TEXT -1 60 "e try to find an approximation the solution on the interval " } {TEXT 265 7 "[x0,x1]" }{TEXT -1 72 ". Suppose that we wish to take n \+ equally spaced points in the interval " }{TEXT 263 22 "[x0,x1], x[i]:= x0+i*h," }{TEXT -1 22 " where i=0,...,n and " }{TEXT 264 12 "h:=(x1-x 0)/n" }{TEXT -1 25 ". Suppose that the point " }{TEXT 266 11 "[x[i],y[ i]]" }{TEXT -1 48 " has been determined. Then we compute the slope " } {TEXT 267 15 "k:=f(x[i],y[i])" }{TEXT -1 87 " and approximate the solu tion of the differential equation by a line segment of slope " } {TEXT 268 1 "k" }{TEXT -1 20 " to the new point " }{TEXT 269 16 "y[i +1]:=y[i]+k*h" }{TEXT -1 1 "." }}{PARA 266 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 267 "> " 0 "" {MPLTEXT 1 0 55 "restart:with(DEtools):with (plots): #clear the memory" }}{PARA 7 "" 1 "" {TEXT -1 50 "Warning, the name changecoords has been redefined\n" }}}{EXCHG {PARA 268 "> " 0 "" {MPLTEXT 1 0 59 "f:= (x,y)-> -3*x*(x-1.25)+y/(3+x); #enter slo pe function" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fGf*6$%\"xG%\"yG6\" 6$%)operatorG%&arrowGF),&*(\"\"$\"\"\"9$F0,&F1F0$\"$D\"!\"#!\"\"F0F6*& 9%F0,&F/F0F1F0F6F0F)F)F)" }}}{EXCHG {PARA 269 "> " 0 "" {MPLTEXT 1 0 57 "x0:=0.0; y0:= 1.0; x1:= 2.0; n:=16; #IC,endpt,num.steps" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%#x0G$\"\"!F&" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#y0G$\"#5!\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%# x1G$\"#?!\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"nG\"#;" }}} {EXCHG {PARA 270 "> " 0 "" {MPLTEXT 1 0 57 " h:= (x1-x0)/n; y[0]:=y0; \+ #step size, copy IC" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>% \"hG$\"++++]7!#5" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"yG6#\"\"!$\"# 5!\"\"" }}}{EXCHG {PARA 271 "> " 0 "" {MPLTEXT 1 0 59 "eqn:= diff(w(t) ,t) = f(t,w(t)); #enter DE for solver " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eqnG/-%%diffG6$-%\"wG6#%\"tGF,,&*(\"\"$\"\"\"F,F0,&F ,F0$\"$D\"!\"#!\"\"F0F5*&F)F0,&F/F0F,F0F5F0" }}}{EXCHG {PARA 272 "> " 0 "" {MPLTEXT 1 0 55 "soln:=dsolve(\{eqn,w( x0 )=y0\}); #symboli c soln DE" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%solnG/-%\"wG6#%\"tG*&, ,*(\"\"$\"\"\"\"\"#!\"\"F)F/F0*(\"#^F.\"\"%F0F)F.F.*&#\"$`\"F3F.-%#lnG 6#,&F-F.F)F.F.F0*&#F6F3F.-F86#F-F.F.#F.F-F.F.F:F." }}}{EXCHG {PARA 273 "> " 0 "" {MPLTEXT 1 0 122 "printf(\"EULER METHOD\");\nprintf(\"x[ i] y[i] Actual y error\"); \nfor i from 0 to n do ; # loop. " }}{PARA 274 "> " 0 "" {MPLTEXT 1 0 53 " x[i]:= x0+h*i: # GUTS of the program" }}{PARA 275 "> " 0 "" {MPLTEXT 1 0 62 " k:= f(x[i],y[i]): # Slope of line at (x [i],y[i])" }}{PARA 276 "> " 0 "" {MPLTEXT 1 0 61 " y[i+1]:=y[i]+k*h; # straight line approximation" }}{PARA 277 "> " 0 "" {MPLTEXT 1 0 72 " valu:=subs(t=x[i],rhs(soln)):# error between appro ximation and actual" }}{PARA 278 "> " 0 "" {MPLTEXT 1 0 19 " erro:=y [i]-valu:" }}{PARA 279 "> " 0 "" {MPLTEXT 1 0 40 " printf(\"%6.3f%15 .7f %15.7f %15.7f\\n\"," }}{PARA 280 "> " 0 "" {MPLTEXT 1 0 44 " x[i ],y[i],valu, erro);#right side of soln" }}{PARA 281 "> " 0 "" {MPLTEXT 1 0 3 "od:" }{TEXT -1 0 "" }}{PARA 6 "" 1 "" {TEXT -1 12 "EUL ER METHOD" }}{PARA 6 "" 1 "" {TEXT -1 44 "x[i] y[i] Act ual y error" }}{PARA 6 "" 1 "" {TEXT -1 53 " 0.000 1.0000000 1.0000000 0.0000000" }}{PARA 6 "" 1 "" {TEXT -1 53 " .12 5 1.0416667 1.0693893 -.0277226" }}{PARA 6 "" 1 "" {TEXT -1 53 " .250 1.1360677 1.1877118 -.0516441" }} {PARA 6 "" 1 "" {TEXT -1 53 " .375 1.2735126 1.3447510 \+ -.0712384" }}{PARA 6 "" 1 "" {TEXT -1 53 " .500 1.4437266 \+ 1.5297446 -.0860180" }}{PARA 6 "" 1 "" {TEXT -1 53 " .625 \+ 1.6359133 1.7314427 -.0955294" }}{PARA 6 "" 1 "" {TEXT -1 53 " .750 1.8388085 1.9381594 -.0993509" }}{PARA 6 "" 1 "" {TEXT -1 53 " .875 2.0407271 2.1378153 -.0 970882" }}{PARA 6 "" 1 "" {TEXT -1 53 " 1.000 2.2296039 2.3 179763 -.0883724" }}{PARA 6 "" 1 "" {TEXT -1 53 " 1.125 2.3 930290 2.4658862 -.0728572" }}{PARA 6 "" 1 "" {TEXT -1 53 " 1.250 2.5182794 2.5684972 -.0502178" }}{PARA 6 "" 1 "" {TEXT -1 53 " 1.375 2.5923464 2.6124940 -.020147 6" }}{PARA 6 "" 1 "" {TEXT -1 53 " 1.500 2.6019604 2.584318 2 .0176422" }}{PARA 6 "" 1 "" {TEXT -1 53 " 1.625 2.533612 0 2.4701880 .0634240" }}{PARA 6 "" 1 "" {TEXT -1 53 " 1.7 50 2.3735724 2.2561158 .1174566" }}{PARA 6 "" 1 "" {TEXT -1 53 " 1.875 2.1079098 1.9279256 .1799842" }} {PARA 6 "" 1 "" {TEXT -1 53 " 2.000 1.7225057 1.4712663 \+ .2512394" }}}{EXCHG {PARA 282 "" 0 "" {TEXT -1 1 "\n" }{TEXT 287 44 "Plotting Euler Polygons and Actual Solution." }{TEXT -1 317 "\nWe \+ superimpose the actual solution of the IVP that MAPLE found by symboli c manipulation on top of the numerical solution. To do this, we store \+ the plot structures from the plot of the curve, the plot of the polygo n, and the plot of the points and display them together. The x and y s cales are the same if we include " }{TEXT 257 19 "scaling=constrained " }{TEXT -1 61 ". The equation giving the solution is stored in the va riable " }{TEXT 258 4 "soln" }{TEXT -1 3 ". " }{TEXT 259 10 "rhs(soln ) " }{TEXT -1 116 "picks off the expression of the function. Note that it is not a MAPLE function. Thus to evaluate it at the variable " } {TEXT 260 4 "x[i]" }{TEXT -1 22 " we had to substitute " }{TEXT 261 6 "t=x[i]" }{TEXT -1 11 " using the " }{TEXT 262 4 "subs" }{TEXT -1 75 " command. Using colons prevents the display of the tedious plot struct ures!" }}}{EXCHG {PARA 283 "> " 0 "" {MPLTEXT 1 0 147 "P1:=plot([seq([ x[i],y[i]],i=0..n)],color=magenta):\nP2:=plot([seq([x[i],y[i]],i=0..n) ],style=point,\n symbol=circle,symbolsize=5,color=red):" }} {PARA 284 "> " 0 "" {MPLTEXT 1 0 40 "P3:=plot(rhs(soln),t=x0..x1,color =blue):" }}{PARA 285 "> " 0 "" {MPLTEXT 1 0 69 "display(\{P1,P2,P3\},s caling=constrained,title=\"Euler Polygon Method\");" }}{PARA 13 "" 1 " " {GLPLOT2D 444 350 350 {PLOTDATA 2 "6)-%'CURVESG6$737$$\"\"!F)$\"\"\" F)7$$\"3+++++++]7!#=$\"35+++nmmT5!#<7$$\"3++++++++DF/$\"3)*******3x1O6 F27$$\"3+++++++]PF/$\"31+++@E^t7F27$$\"3++++++++]F/$\"3++++ImsV9F27$$ \"3+++++++]iF/$\"3%******\\H8fj\"F27$$\"3++++++++vF/$\"30+++t%3)Q=F27$ $\"3+++++++]()F/$\"3A+++*3F2/#F27$F*$\"3/+++qQgHAF27$$\"3+++++++D6F2$ \"3)******4**GIR#F27$$F.F2$\"31+++'Rz#=DF27$$\"3*************\\P\"F2$ \"3y*****pVYBf#F27$$\"3++++++++:F2$\"3&******HNg>g#F27$$\"3+++++++D;F2 $\"3$*******G?hLDF27$$\"3+++++++]o0Ds\"F2-%'COLOURG 6&%$RGBG$\"*++++\"!\")F(Fip-F$6&F&-Ffp6&FhpFipF(F(-%&STYLEG6#%&POINTG- %'SYMBOLG6$%'CIRCLEG\"\"&-F$6$7UF'7$$\"39LLLL3VfV!#>$\"3M?$oPzG!=5F27$ $\"3&pmm;H[D:)F_r$\"3PIo(>2.#R5F27$$\"3LLLLe0$=C\"F/$\"3J?)\\H.m(o5F27 $$\"3HLLL3RBr;F/$\"3!>R>-^#H/6F27$$\"3Ymm;zjf)4#F/$\"3'>g%H#[$*\\9\"F2 7$$\"3=LL$e4;[\\#F/$\"3q:-lIQ8(=\"F27$$\"3p****\\i'y]!HF/$\"3oASpA<([B \"F27$$\"3,LL$ezs$HLF/$\"3ih[Sg/C)G\"F27$$\"3^****\\7iI_PF/$\"3Y'R![$) )p]M\"F27$$\"3#pmmm@Xt=%F/$\"3$\\x-VRioS\"F27$$\"3QLLL3y_qXF/$\"3i*=J \")HBPY\"F27$$\"3i******\\1!>+&F/$\"3@30%*=8/I:F27$$\"3')******\\Z/NaF /$\"3!*4vMCWj)f\"F27$$\"3&*******\\$fC&eF/$\"3[\"*)yax'=m;F27$$\"3ELL$ ez6:B'F/$\"3'*zUcONRGF27$$\"3slmmm (y8!zF/$\"38(\\si5HO+#F27$$\"3V++]i.tK$)F/$\"3!H3hR18F2#F27$$\"39++](3 zMu)F/$\"3KIvK%Q8o8#F27$$\"3\"pmm;H_?<*F/$\"3&yS)eGZX,AF27$$\"3dmm;zih l&*F/$\"3->$pBWp$eAF27$$\"39LLL3#G,***F/$\"3)3t`azimJ#F27$$\"3;LLezw5V 5F2$\"3:\\[y'*GGtBF27$$\"3*)***\\PQ#\\\"3\"F2$\"33@$R>-t)=CF27$$\"3ALL $e\"*[H7\"F2$\"3&zMnQz)yjCF27$$\"3#*******pvxl6F2$\"3%*3H6r,)\\]#F27$$ \"3z****\\_qn27F2$\"3wsq$\\ZZ'RDF27$$\"3%)***\\i&p@[7F2$\"3h'4c&QOTnDF 27$$\"3#)****\\2'HKH\"F2$\"3!*[bH\"GE5f#F27$$\"3_mmmwanL8F2$\"3[\"fgJk k_g#F27$$\"3&******\\2goP\"F2$\"3\\[&=*ewj7EF27$$\"3CLLeR<*fT\"F2$\"3k [(*3,`(=h#F27$$\"3'******\\)Hxe9F2$\"37N+2Q6V-EF27$$\"3Ymm\"H!o-*\\\"F 2$\"3aI;\"*4;&[e#F27$$\"3))***\\7k.6a\"F2$\"3\"[dzmk`pb#F27$$\"3emmmT9 C#e\"F2$\"3fR!f\"Hkw>DF27$$\"3!****\\i!*3`i\"F2$\"3b)=D7)yypCF27$$\"3P LLL$*zym;F2$\"3Xirwb8W5CF27$$\"3GLL$3N1#4F27$$\"3 /++DOl5;>F2$\"3#*=!*Ga_j#z\"F27$$\"3/+++q`KO>F2$\"3kE-,Qt$3s\"F27$$\"3 /++v.Uac>F2$\"36Rh\"R\"G\\X;F27$$\"3-+](=5s#y>F2$\"3xT8^f5]g:F27$Fap$ \"3zw4U@hEr9F2-Ffp6&FhpF(F(Fip-%&TITLEG6#Q5Euler~Polygon~Method6\"-%(S CALINGG6#%,CONSTRAINEDG-%+AXESLABELSG6%Q!F[blFcbl-%%FONTG6#%(DEFAULTG- %%VIEWG6$;F($\"#?!\"\"Fgbl" 1 2 0 1 10 0 2 9 1 4 1 1.000000 69.000000 69.000000 0 0 "Curve 1" "Curve 2" "Curve 3" }}}}{EXCHG {PARA 286 "" 0 "" {TEXT -1 408 "The Improved Euler Method is a predictor - corrector \+ method. The problem is that the solution depends on the values of f(x, y) along the whole interval, but the Euler polygon just uses the left \+ endpoint to construct the approximation. In numerical analysis classes , you will derive the fact that the Improved Euler Method has higher o rder of accuracy. The basic idea is that one uses an Euler step with s lope " }{TEXT 270 2 "k1" }{TEXT -1 151 " to find a first guess (the pr edictor) and then use the first guess to evaluate the function near th e actual right hand endpoint to find another slope " }{TEXT 271 2 "k2 " }{TEXT -1 75 " (the corrector.) The better approximation of the slop e along the interval " }{TEXT 272 13 "[x[i],x[i+1]]" }{TEXT -1 34 " is the average of the two slopes " }{TEXT 273 12 "k:=(k1+k2)/2" }{TEXT -1 234 ". The resulting improvement is similar to the improvement from the left-hand rule to the trapezoid rule in numerical integration. To implement this improvement merely requires that the guts of the prece ding code be slightly uprgraded." }}}{EXCHG {PARA 287 "> " 0 "" {MPLTEXT 1 0 100 "z[0]:=y0:\nprintf(\"IMPROVED EULER METHOD\");\nprint f(\"x[i] Impr. Euler y Actual y error\"); " }}{PARA 288 "> " 0 "" {MPLTEXT 1 0 33 "for i from 0 to n do; " }}{PARA 289 "> " 0 "" {MPLTEXT 1 0 48 " k1:= f(x[i],z[i]): # GUTS \+ " }}{PARA 290 "> " 0 "" {MPLTEXT 1 0 42 " k2:= f(x[i+1],z[i ]+k1*h): " }}{PARA 291 "> " 0 "" {MPLTEXT 1 0 28 " z[i+ 1]:=z[i]+(k1+k2)*h/2;" }}{PARA 292 "> " 0 "" {MPLTEXT 1 0 33 " valu: =subs(t=x[i],rhs(soln)): " }}{PARA 293 "> " 0 "" {MPLTEXT 1 0 19 " e rro:=z[i]-valu:" }}{PARA 294 "> " 0 "" {MPLTEXT 1 0 66 " printf(\"%6 .2f%15.7f %15.7f %15.7f\\n\",\n x[i],z[i],valu, erro);" }}{PARA 295 "> " 0 "" {MPLTEXT 1 0 3 "od:" }}{PARA 6 "" 1 "" {TEXT -1 21 "IMPROVED EULER METHOD" }}{PARA 6 "" 1 "" {TEXT -1 44 "x[i] Impr. Euler y \+ Actual y error" }}{PARA 6 "" 1 "" {TEXT -1 53 " 0.00 1.000 0000 1.0000000 0.0000000" }}{PARA 6 "" 1 "" {TEXT -1 53 " \+ .13 1.0680339 1.0693893 -.0013554" }}{PARA 6 "" 1 " " {TEXT -1 53 " .25 1.1850115 1.1877118 -.0027003" }}{PARA 6 "" 1 "" {TEXT -1 53 " .38 1.3407234 1.3447510 \+ -.0040276" }}{PARA 6 "" 1 "" {TEXT -1 53 " .50 1.5244131 \+ 1.5297446 -.0053315" }}{PARA 6 "" 1 "" {TEXT -1 53 " .63 \+ 1.7248356 1.7314427 -.0066071" }}{PARA 6 "" 1 "" {TEXT -1 53 " .75 1.9303088 1.9381594 -.0078506" }} {PARA 6 "" 1 "" {TEXT -1 53 " .88 2.1287565 2.1378153 \+ -.0090588" }}{PARA 6 "" 1 "" {TEXT -1 53 " 1.00 2.3077471 \+ 2.3179763 -.0102292" }}{PARA 6 "" 1 "" {TEXT -1 53 " 1.13 \+ 2.4545269 2.4658862 -.0113593" }}{PARA 6 "" 1 "" {TEXT -1 53 " 1.25 2.5560492 2.5684972 -.0124480" }}{PARA 6 "" 1 "" {TEXT -1 53 " 1.38 2.5990005 2.6124940 -.0 134935" }}{PARA 6 "" 1 "" {TEXT -1 53 " 1.50 2.5698235 2.5 843182 -.0144948" }}{PARA 6 "" 1 "" {TEXT -1 53 " 1.63 2.4 547368 2.4701880 -.0154512" }}{PARA 6 "" 1 "" {TEXT -1 53 " 1.75 2.2397539 2.2561158 -.0163619" }}{PARA 6 "" 1 "" {TEXT -1 53 " 1.88 1.9106990 1.9279256 -.017226 6" }}{PARA 6 "" 1 "" {TEXT -1 53 " 2.00 1.4532216 1.471266 3 -.0180447" }}}{EXCHG {PARA 296 "> " 0 "" {MPLTEXT 1 0 145 "P1: =plot([seq([x[i],z[i]],i=0..n)],color=coral):\nP2:=plot([seq([x[i],z[i ]],i=0..n)],style=point,\n symbol=circle,symbolsize=5,color=r ed):" }}{PARA 297 "> " 0 "" {MPLTEXT 1 0 40 "P3:=plot(rhs(soln),t=x0.. x1,color=blue):" }}{PARA 298 "> " 0 "" {MPLTEXT 1 0 70 "display(\{P1,P 2,P3\},scaling=constrained,title=\"Improved Euler Method\");" }}{PARA 13 "" 1 "" {GLPLOT2D 483 397 397 {PLOTDATA 2 "6)-%'CURVESG6$7U7$$\"\"! F)$\"\"\"F)7$$\"39LLLL3VfV!#>$\"3M?$oPzG!=5!#<7$$\"3&pmm;H[D:)F/$\"3PI o(>2.#R5F27$$\"3LLLLe0$=C\"!#=$\"3J?)\\H.m(o5F27$$\"3HLLL3RBr;F;$\"3!> R>-^#H/6F27$$\"3Ymm;zjf)4#F;$\"3'>g%H#[$*\\9\"F27$$\"3=LL$e4;[\\#F;$\" 3q:-lIQ8(=\"F27$$\"3p****\\i'y]!HF;$\"3oASpA<([B\"F27$$\"3,LL$ezs$HLF; $\"3ih[Sg/C)G\"F27$$\"3^****\\7iI_PF;$\"3Y'R![$))p]M\"F27$$\"3#pmmm@Xt =%F;$\"3$\\x-VRioS\"F27$$\"3QLLL3y_qXF;$\"3i*=J\")HBPY\"F27$$\"3i***** *\\1!>+&F;$\"3@30%*=8/I:F27$$\"3')******\\Z/NaF;$\"3!*4vMCWj)f\"F27$$ \"3&*******\\$fC&eF;$\"3[\"*)yax'=m;F27$$\"3ELL$ez6:B'F;$\"3'*zUcONRG< F27$$\"3Rmmm;=C#o'F;$\"3+7*p\\MhH!=F27$$\"3,mmmm#pS1(F;$\"3@**RY1`Cm=F 27$$\"3]****\\i`A3vF;$\"3k!fF0o4&R>F27$$\"3slmmm(y8!zF;$\"38(\\si5HO+# F27$$\"3V++]i.tK$)F;$\"3!H3hR18F2#F27$$\"39++](3zMu)F;$\"3KIvK%Q8o8#F2 7$$\"3\"pmm;H_?<*F;$\"3&yS)eGZX,AF27$$\"3dmm;zihl&*F;$\"3->$pBWp$eAF27 $$\"39LLL3#G,***F;$\"3)3t`azimJ#F27$$\"3;LLezw5V5F2$\"3:\\[y'*GGtBF27$ $\"3*)***\\PQ#\\\"3\"F2$\"33@$R>-t)=CF27$$\"3ALL$e\"*[H7\"F2$\"3&zMnQz )yjCF27$$\"3#*******pvxl6F2$\"3%*3H6r,)\\]#F27$$\"3z****\\_qn27F2$\"3w sq$\\ZZ'RDF27$$\"3%)***\\i&p@[7F2$\"3h'4c&QOTnDF27$$\"3#)****\\2'HKH\" F2$\"3!*[bH\"GE5f#F27$$\"3_mmmwanL8F2$\"3[\"fgJkk_g#F27$$\"3&******\\2 goP\"F2$\"3\\[&=*ewj7EF27$$\"3CLLeR<*fT\"F2$\"3k[(*3,`(=h#F27$$\"3'*** ***\\)Hxe9F2$\"37N+2Q6V-EF27$$\"3Ymm\"H!o-*\\\"F2$\"3aI;\"*4;&[e#F27$$ \"3))***\\7k.6a\"F2$\"3\"[dzmk`pb#F27$$\"3emmmT9C#e\"F2$\"3fR!f\"Hkw>D F27$$\"3!****\\i!*3`i\"F2$\"3b)=D7)yypCF27$$\"3PLLL$*zym;F2$\"3Xirwb8W 5CF27$$\"3GLL$3N1#4F27$$\"3/++DOl5;>F2$\"3#*=!*Ga _j#z\"F27$$\"3/+++q`KO>F2$\"3kE-,Qt$3s\"F27$$\"3/++v.Uac>F2$\"36Rh\"R \"G\\X;F27$$\"3-+](=5s#y>F2$\"3xT8^f5]g:F27$$\"\"#F)$\"3zw4U@hEr9F2-%' COLOURG6&%$RGBGF(F($\"*++++\"!\")-F$6$73F'7$$\"3+++++++]7F;$\"31+++aQ. o5F27$$\"3++++++++DF;$\"3.+++=:,&=\"F27$$\"3+++++++]PF;$\"3'******HVB2 M\"F27$$\"3++++++++]F;$\"35+++gITC:F27$$\"3+++++++]iF;$\"3++++Sc$[s\"F 27$$\"3++++++++vF;$\"3++++D)3.$>F27$$\"3+++++++]()F;$\"3&******\\`c(G@ F27$F*$\"33+++Yru2BF27$$\"3+++++++D6F2$\"3,+++')o_aCF27$$F_\\lF2$\"3>+ ++#=\\gb#F27$$\"3*************\\P\"F2$\"3A+++P0+*f#F27$$\"3++++++++:F2 $\"3*)******\\M#)pDF27$$\"3+++++++D;F2$\"31+++%yOZX#F27$$\"3+++++++]F27$F_[ l$\"3!*******z:A`9F2-Fd[l6&Ff[lFg[l$\")AR!)\\Fi[lF(-F$6&F\\\\l-Fd[l6&F f[lFg[lF(F(-%&STYLEG6#%&POINTG-%'SYMBOLG6$%'CIRCLEG\"\"&-%&TITLEG6#Q6I mproved~Euler~Method6\"-%(SCALINGG6#%,CONSTRAINEDG-%+AXESLABELSG6%Q\"t F]blQ!F]bl-%%FONTG6#%(DEFAULTG-%%VIEWG6$;F($\"#?!\"\"Fjbl" 1 2 0 1 10 0 2 9 1 4 1 1.000000 47.000000 46.000000 0 0 "Curve 1" "Curve 2" "Curv e 3" }}}}{EXCHG {PARA 299 "" 0 "" {TEXT -1 143 "The Runge-Kutta Method is a fourth order method. This time, there are four evaluations of th e function along the interval. The left hand slope " }{TEXT 278 2 "k1 " }{TEXT -1 104 " is used to find an prediction of the midpoint. The s econd evaluation finds the slope near the midpoint " }{TEXT 275 2 "k2 " }{TEXT -1 44 " and is used to find a better approximation " }{TEXT 277 2 "k3" }{TEXT -1 96 " near the midpoint. This is used to predict t he point near the right end where the fourth slope " }{TEXT 276 2 "k4 " }{TEXT -1 136 " is computed. Finally, the new point is computed by t aking an average of aloop four slopes. This resembles Simpson's rule i n quadrature." }}}{EXCHG {PARA 300 "> " 0 "" {MPLTEXT 1 0 108 "\nprint f(\"RUNGE - KUTTA METHOD\");\nprintf(\"x[i] Runge-Kutta y Actua l y error\");\nrk[0]:=y0:h2:=h/2:" }}{PARA 301 "> " 0 "" {MPLTEXT 1 0 54 "for i from 0 to n do; #RUNGE - KUTTA METHOD " }}{PARA 302 "> " 0 "" {MPLTEXT 1 0 33 " k1:= f(x[i],rk[i]): \+ " }}{PARA 303 "> " 0 "" {MPLTEXT 1 0 106 " k2:= f(x[i]+h2,rk[i] +k1*h2):\n k3:= f(x[i]+h2,rk[i]+k2*h2):\n k4:= f(x[i+1],rk[i]+k3*h ): " }}{PARA 304 "> " 0 "" {MPLTEXT 1 0 40 " rk[i+1]:=rk [i]+(k1+2*k2+2*k3+k4)*h/6;" }}{PARA 305 "> " 0 "" {MPLTEXT 1 0 33 " \+ valu:=subs(t=x[i],rhs(soln)): " }}{PARA 306 "> " 0 "" {MPLTEXT 1 0 20 " erro:=rk[i]-valu:" }}{PARA 307 "> " 0 "" {MPLTEXT 1 0 86 " print f(\"%6.2f%15.7f %15.7f %15.7f\\n\",\n x[i],rk[i],valu, erro);#right \+ side of soln" }}{PARA 308 "> " 0 "" {MPLTEXT 1 0 3 "od:" }}{PARA 6 "" 1 "" {TEXT -1 21 "RUNGE - KUTTA METHOD" }}{PARA 6 "" 1 "" {TEXT -1 44 "x[i] Runge-Kutta y Actual y error" }}{PARA 6 "" 1 "" {TEXT -1 53 " 0.00 1.0000000 1.0000000 0.0000000" }} {PARA 6 "" 1 "" {TEXT -1 53 " .13 1.0693891 1.0693893 \+ -.0000002" }}{PARA 6 "" 1 "" {TEXT -1 53 " .25 1.1877115 \+ 1.1877118 -.0000003" }}{PARA 6 "" 1 "" {TEXT -1 53 " .38 \+ 1.3447506 1.3447510 -.0000004" }}{PARA 6 "" 1 "" {TEXT -1 53 " .50 1.5297440 1.5297446 -.0000006" }}{PARA 6 "" 1 "" {TEXT -1 53 " .63 1.7314421 1.7314427 -.0 000006" }}{PARA 6 "" 1 "" {TEXT -1 53 " .75 1.9381588 1.9 381594 -.0000006" }}{PARA 6 "" 1 "" {TEXT -1 53 " .88 2.1 378147 2.1378153 -.0000006" }}{PARA 6 "" 1 "" {TEXT -1 53 " 1.00 2.3179755 2.3179763 -.0000008" }}{PARA 6 "" 1 "" {TEXT -1 53 " 1.13 2.4658855 2.4658862 -.000000 7" }}{PARA 6 "" 1 "" {TEXT -1 53 " 1.25 2.5684963 2.568497 2 -.0000009" }}{PARA 6 "" 1 "" {TEXT -1 53 " 1.38 2.612493 1 2.6124940 -.0000009" }}{PARA 6 "" 1 "" {TEXT -1 53 " 1. 50 2.5843173 2.5843182 -.0000009" }}{PARA 6 "" 1 "" {TEXT -1 53 " 1.63 2.4701869 2.4701880 -.0000011" }} {PARA 6 "" 1 "" {TEXT -1 53 " 1.75 2.2561147 2.2561158 \+ -.0000011" }}{PARA 6 "" 1 "" {TEXT -1 53 " 1.88 1.9279244 \+ 1.9279256 -.0000012" }}{PARA 6 "" 1 "" {TEXT -1 53 " 2.00 \+ 1.4712650 1.4712663 -.0000013" }}}{EXCHG {PARA 309 "> \+ " 0 "" {MPLTEXT 1 0 148 "P1:=plot([seq([x[i],rk[i]],i=0..n)],color=mar oon):\nP2:=plot([seq([x[i],rk[i]],i=0..n)],style=point,\n sym bol=circle,symbolsize=8,color=red):" }}{PARA 310 "> " 0 "" {MPLTEXT 1 0 40 "P3:=plot(rhs(soln),t=x0..x1,color=blue):" }}{PARA 311 "> " 0 "" {MPLTEXT 1 0 47 "display(\{P1,P2,P3\},title=\"Runge-Kutta Method\");" }}{PARA 13 "" 1 "" {GLPLOT2D 512 392 392 {PLOTDATA 2 "6(-%'CURVESG6$73 7$$\"\"!F)$\"\"\"F)7$$\"3+++++++]7!#=$\"3)******R3*Qp5!#<7$$\"3+++++++ +DF/$\"34+++<:r(=\"F27$$\"3+++++++]PF/$\"3/+++61vW8F27$$\"3++++++++]F/ $\"35+++'*RuH:F27$$\"3+++++++]iF/$\"3'*******R@WJF27$$\"3+++++++]()F/$\"3<+++hY\"y8#F27$F*$\"31+++Ob(zJ#F 27$$\"3+++++++D6F2$\"3-+++Db)eY#F27$$F.F2$\"3/+++[j\\oDF27$$\"3******* ******\\P\"F2$\"35+++WJ\\7EF27$$\"3++++++++:F2$\"32+++MtJ%e#F27$$\"3++ +++++D;F2$\"3y*****>%p=qCF27$$\"3+++++++]Z6cAF27$$\"3***** ********\\(=F2$\"3&******RNCz#>F27$$\"\"#F)$\"3)******f-l7Z\"F2-%'COLO URG6&%$RGBG$\")viob!\")$\")!\\DP\"F[q$\")%yg>%F[q-F$6$7UF'7$$\"39LLLL3 VfV!#>$\"3M?$oPzG!=5F27$$\"3&pmm;H[D:)Ffq$\"3PIo(>2.#R5F27$$\"3LLLLe0$ =C\"F/$\"3J?)\\H.m(o5F27$$\"3HLLL3RBr;F/$\"3!>R>-^#H/6F27$$\"3Ymm;zjf) 4#F/$\"3'>g%H#[$*\\9\"F27$$\"3=LL$e4;[\\#F/$\"3q:-lIQ8(=\"F27$$\"3p*** *\\i'y]!HF/$\"3oASpA<([B\"F27$$\"3,LL$ezs$HLF/$\"3ih[Sg/C)G\"F27$$\"3^ ****\\7iI_PF/$\"3Y'R![$))p]M\"F27$$\"3#pmmm@Xt=%F/$\"3$\\x-VRioS\"F27$ $\"3QLLL3y_qXF/$\"3i*=J\")HBPY\"F27$$\"3i******\\1!>+&F/$\"3@30%*=8/I: F27$$\"3')******\\Z/NaF/$\"3!*4vMCWj)f\"F27$$\"3&*******\\$fC&eF/$\"3[ \"*)yax'=m;F27$$\"3ELL$ez6:B'F/$\"3'*zUcONRGF27$$\"3slmmm(y8!zF/$\"38(\\si5HO+#F27$$\"3V++]i.tK$)F/$\"3! H3hR18F2#F27$$\"39++](3zMu)F/$\"3KIvK%Q8o8#F27$$\"3\"pmm;H_?<*F/$\"3&y S)eGZX,AF27$$\"3dmm;zihl&*F/$\"3->$pBWp$eAF27$$\"39LLL3#G,***F/$\"3)3t `azimJ#F27$$\"3;LLezw5V5F2$\"3:\\[y'*GGtBF27$$\"3*)***\\PQ#\\\"3\"F2$ \"33@$R>-t)=CF27$$\"3ALL$e\"*[H7\"F2$\"3&zMnQz)yjCF27$$\"3#*******pvxl 6F2$\"3%*3H6r,)\\]#F27$$\"3z****\\_qn27F2$\"3wsq$\\ZZ'RDF27$$\"3%)*** \\i&p@[7F2$\"3h'4c&QOTnDF27$$\"3#)****\\2'HKH\"F2$\"3!*[bH\"GE5f#F27$$ \"3_mmmwanL8F2$\"3[\"fgJkk_g#F27$$\"3&******\\2goP\"F2$\"3\\[&=*ewj7EF 27$$\"3CLLeR<*fT\"F2$\"3k[(*3,`(=h#F27$$\"3'******\\)Hxe9F2$\"37N+2Q6V -EF27$$\"3Ymm\"H!o-*\\\"F2$\"3aI;\"*4;&[e#F27$$\"3))***\\7k.6a\"F2$\"3 \"[dzmk`pb#F27$$\"3emmmT9C#e\"F2$\"3fR!f\"Hkw>DF27$$\"3!****\\i!*3`i\" F2$\"3b)=D7)yypCF27$$\"3PLLL$*zym;F2$\"3Xirwb8W5CF27$$\"3GLL$3N1#4F27$$\"3/++DOl5;>F2$\"3#*=!*Ga_j#z\"F27$$\"3/+++q`KO> F2$\"3kE-,Qt$3s\"F27$$\"3/++v.Uac>F2$\"36Rh\"R\"G\\X;F27$$\"3-+](=5s#y >F2$\"3xT8^f5]g:F27$Fap$\"3zw4U@hEr9F2-Ffp6&FhpF(F($\"*++++\"F[q-F$6&F &-Ffp6&FhpF^alF(F(-%&STYLEG6#%&POINTG-%'SYMBOLG6$%'CIRCLEG\"\")-%&TITL EG6#Q3Runge-Kutta~Method6\"-%+AXESLABELSG6%Q!FablFebl-%%FONTG6#%(DEFAU LTG-%%VIEWG6$;F($\"#?!\"\"Fibl" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" "Curve 2" "Curve 3" }}}}{EXCHG {PARA 312 "" 0 "" {TEXT -1 87 "\nWe tabulate the three methods at some of the x[i]'s to compare the numerical methods. " }}}{EXCHG {PARA 313 "> " 0 "" {MPLTEXT 1 0 328 "\nprintf(\"x[i] Actual y Euler \+ E-error Impr.Euler IE-error Runge-Kutta RK-error\" );\nfor j from 2 to n by 2 do;\nvalu:=subs(t=x[j],rhs(soln)): \ner1:=v alu-y[j]:\ner2:=valu-z[j]:\ner3:=valu-rk[j]:\nprintf(\"%6.3f %12.8f %1 2.8f %12.8f %12.8f %12.8f %12.8f %12.8f\\n\", \n x[j],valu,y[j],er1,z [j],er2,rk[j],er3);\nod:" }}{PARA 6 "" 1 "" {TEXT -1 95 "x[i] Actua l y Euler E-error Impr.Euler IE-error Runge-Kutt a RK-error" }}{PARA 6 "" 1 "" {TEXT -1 97 " .250 1.18771180 1.1 3606771 .05164410 1.18501152 .00270030 1.18771152 .000000 30" }}{PARA 6 "" 1 "" {TEXT -1 97 " .500 1.52974460 1.44372663 \+ .08601800 1.52441306 .00533150 1.52974400 .00000060" }} {PARA 6 "" 1 "" {TEXT -1 97 " .750 1.93815940 1.83880847 .0993 5090 1.93030883 .00785060 1.93815879 .00000060" }}{PARA 6 " " 1 "" {TEXT -1 97 " 1.000 2.31797630 2.22960387 .08837240 2. 30774715 .01022920 2.31797554 .00000080" }}{PARA 6 "" 1 "" {TEXT -1 97 " 1.250 2.56849720 2.51827940 .05021780 2.5560491 8 .01244800 2.56849635 .00000090" }}{PARA 6 "" 1 "" {TEXT -1 97 " 1.500 2.58431820 2.60196035 -.01764220 2.56982345 .014 49480 2.58431733 .00000090" }}{PARA 6 "" 1 "" {TEXT -1 97 " 1.750 2.25611580 2.37357241 -.11745660 2.23975392 .01636190 2. 25611472 .00000110" }}{PARA 6 "" 1 "" {TEXT -1 97 " 2.000 1.47126 630 1.72250568 -.25123940 1.45322158 .01804470 1.47126503 \+ .00000130" }}}{EXCHG {PARA 314 "" 0 "" {TEXT -1 132 "\nFinally, in o rder to study the effect of decreasing h, we solve a new equation by t he Euler method several times for different h's." }}}{EXCHG {PARA 315 "> " 0 "" {MPLTEXT 1 0 55 "restart:with(DEtools):with(plots): #clea r the memory" }}{PARA 7 "" 1 "" {TEXT -1 50 "Warning, the name changec oords has been redefined\n" }}}{EXCHG {PARA 316 "> " 0 "" {MPLTEXT 1 0 60 "f:= (x,y)-> cos(3*x)-2*y; #enter slope function" }} {PARA 317 "> " 0 "" {MPLTEXT 1 0 30 "x0:=0.0; y0:= -2/13; x1:= 2.0;" } }{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fGf*6$%\"xG%\"yG6\"6$%)operatorG% &arrowGF),&-%$cosG6#,$*&\"\"$\"\"\"9$F4F4F4*&\"\"#F49%F4!\"\"F)F)F)" } }{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#x0G$\"\"!F&" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#y0G#!\"#\"#8" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#x 1G$\"#?!\"\"" }}}{EXCHG {PARA 318 "> " 0 "" {MPLTEXT 1 0 59 "eqn:= dif f(w(t),t) = f(t,w(t)); #enter DE for solver " }}{PARA 319 "> " 0 "" {MPLTEXT 1 0 47 "soln:=dsolve(\{eqn, w(x0)=y0\});#symbolic soln D E" }}{PARA 320 "" 0 "" {TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6 #>%$eqnG/-%%diffG6$-%\"wG6#%\"tGF,,&-%$cosG6#,$*&\"\"$\"\"\"F,F4F4F4*& \"\"#F4F)F4!\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%solnG/-%\"wG6#% \"tG,(*&#\"\"#\"#8\"\"\"-%$cosG6#,$*&\"\"$F/F)F/F/F/F/*&#F5F.F/-%$sinG F2F/F/*&#\"\"%F.F/-%$expG6#,$*&F-F/F)F/!\"\"F/FB" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 321 "" 0 "" {TEXT -1 50 "The problem is rerun using the Euler Method with " }{TEXT 289 30 "n = 2^ loop, loop=3,4,5,...,10." }{TEXT -1 11 " For each " }{TEXT 290 1 "n" }{TEXT -1 77 ", the value of the solution is sampled at eight equally spaced points along " }{TEXT 291 7 "[x0,x1]" }{TEXT -1 2 " (" }{TEXT 292 6 "=[0,2]" }{TEXT -1 44 " in my example.) Below each value we prin t " }{TEXT 293 9 "error/h^1" }{TEXT -1 8 ", where " }{TEXT 294 5 "h=2 /n" }{TEXT -1 23 " is the step size and " }{TEXT 295 28 "error = appr oximation-actual" }{TEXT -1 10 " at those " }{TEXT 296 1 "x" }{TEXT -1 45 "'s. Since this is approximately constant as " }{TEXT 297 1 "n " }{TEXT -1 66 " increases, it shows that the method is first-order a ccurate. If " }{TEXT 298 9 "error/h^2" }{TEXT -1 22 " remained constan t as " }{TEXT 299 1 "n" }{TEXT -1 137 " increases, then the method is \+ second order accurate, and so on. A remark about the program. As the n umber of steps increases, there are " }{TEXT 300 6 "jjskip" }{TEXT -1 45 " intermediate points computed between each " }{TEXT 301 1 "x" } {TEXT -1 25 " that is printed. Thus " }{TEXT 302 31 "x0+jj*(x1-x0)/8 .0=x[jj*jjskip] " }{TEXT -1 44 " (.250,.500,.750,... in the example.). The " }{TEXT 315 11 " evalf(...)" }{TEXT -1 118 " is used to force \+ evaluation as floating point number, rather than as a symbolic express ion which takes a lot longer." }}}{EXCHG {PARA 322 "> " 0 "" {MPLTEXT 1 0 198 "printf(\"\\n Comparison of different step sizes in the EULER METHOD\");\nprintf(\" ---------------------------------------------- -------------------------------------------------\\n\"); \+ " }{TEXT -1 0 "" }{MPLTEXT 1 0 239 "printf(\" n h | x =%5.3f %7.3f %7.3f %7.3f %7.3f %7.3f %7.3f %7.3f\\n\",\n seq(x0+jj*(x1-x0)/8.0,jj=1..8)); printf(\" -------------------------- --------------------------------------------------------------------- \\n\");" }}{PARA 323 "> " 0 "" {MPLTEXT 1 0 603 "for loop from 3 to 10 do:\nn:=2^loop:\nh:=(x1-x0)/n:\ny[0]:=y0:\n for i from 0 to n do: \n x[i]:=x0+i*h:\n k:=evalf( f(x[i],y[i]) ):\n y[i+1]:=y[i]+k *h;\n od:\nlis[loop]:=[seq([x[j],y[j]],j=0..n)]:\njjskip:=2^(loop-3 ):\nprintf(\"%5d %10.6f | %8.6f%10.6f%10.6f%10.6f%10.6f%10.6f%10.6f%10 .6f\\n\",n,h,seq(y[jj*2^(loop-3) ],jj=1..8));\nprintf(\" D/h = \+ |%8.4f%10.4f%10.4f%10.4f%10.4f%10.4f%10.4f%10.4f\\n\",\nseq( ( y[j j*jjskip] - evalf(subs(t=x[jj*jjskip],rhs(soln))) )/h ,jj=1..8));\nod : printf(\" --------------------------------------------------------- --------------------------------------\\n\");" }}{PARA 6 "" 1 "" {TEXT -1 0 "" }}{PARA 6 "" 1 "" {TEXT -1 56 " Comparison of different step sizes in the EULER METHOD" }}{PARA 6 "" 1 "" {TEXT -1 97 " ---- ---------------------------------------------------------------------- ---------------------" }}{PARA 6 "" 1 "" {TEXT -1 97 " n h \+ | x= .250 .500 .750 1.000 1.250 1.500 1.75 0 2.000" }}{PARA 6 "" 1 "" {TEXT -1 97 " ------------------------ ---------------------------------------------------------------------- -" }}{PARA 6 "" 1 "" {TEXT -1 97 " 8 .250000 | .173077 .26946 1 .152415 -.080836 -.287916 -.349098 -.227248 .014397" }} {PARA 6 "" 1 "" {TEXT -1 96 " D/h = | .3593 .5663 \+ .5526 .3222 -.0181 -.3031 -.3941 -.2528" }}{PARA 6 "" 1 "" {TEXT -1 97 " 16 .125000 | .123525 .191976 .077176 -.1 24168 -.284267 -.306101 -.171707 .050311" }}{PARA 6 "" 1 "" {TEXT -1 96 " D/h = | .3222 .5128 .5033 .2977 \+ -.0070 -.2621 -.3439 -.2183" }}{PARA 6 "" 1 "" {TEXT -1 97 " 32 .062500 | .102459 .158559 .044445 -.143418 -.28357 1 -.288699 -.148956 .064829" }}{PARA 6 "" 1 "" {TEXT -1 96 " \+ D/h = | .3074 .4909 .4830 .2874 -.0028 -.2459 -.3237 -.2044" }}{PARA 6 "" 1 "" {TEXT -1 97 " 64 .031250 \+ | .092642 .142908 .029061 -.152548 -.283426 -.280787 -.138556 .071413" }}{PARA 6 "" 1 "" {TEXT -1 96 " D/h = | .3007 \+ .4809 .4737 .2827 -.0010 -.2385 -.3146 -.1981 " }}{PARA 6 "" 1 "" {TEXT -1 97 " 128 .015625 | .087893 .135319 .021590 -.157001 -.283398 -.277006 -.133573 .074555" }}{PARA 6 "" 1 "" {TEXT -1 96 " D/h = | .2975 .4761 .4693 \+ .2804 -.0001 -.2351 -.3103 -.1950" }}{PARA 6 "" 1 "" {TEXT -1 97 " 256 .007813 | .085556 .131582 .017907 -.159200 -.283393 -.275156 -.131132 .076090" }}{PARA 6 "" 1 "" {TEXT -1 96 " D/h = | .2960 .4738 .4671 .2793 .0003 -.2334 -.3082 -.1936" }}{PARA 6 "" 1 "" {TEXT -1 97 " 512 \+ .003906 | .084397 .129726 .016078 -.160293 -.283394 -.274241 -.129924 .076849" }}{PARA 6 "" 1 "" {TEXT -1 96 " D/h = \+ | .2952 .4726 .4660 .2787 .0005 -.2325 -.3071 \+ -.1929" }}{PARA 6 "" 1 "" {TEXT -1 97 " 1024 .001953 | .083820 \+ .128802 .015167 -.160838 -.283394 -.273786 -.129324 .077226 " }}{PARA 6 "" 1 "" {TEXT -1 96 " D/h = | .2948 .4721 \+ .4655 .2785 .0006 -.2321 -.3066 -.1925" }}{PARA 6 "" 1 "" {TEXT -1 97 " --------------------------------------------- --------------------------------------------------" }}}{EXCHG {PARA 324 "> " 0 "" {MPLTEXT 1 0 227 "P1:=plot([seq(lis[jj],jj=3..10)]):\nP3 :=plot(rhs(soln),t=x0..x1,color=black):\nP2:=plot([seq(lis[jj],jj=3..1 0)],style=point,\n symbol=circle,symbolsize=8):\ndisplay(\{P1 ,P2,P3\},title=\"Decreasing step size in Euler method\");" }}{PARA 13 "" 1 "" {GLPLOT2D 496 380 380 {PLOTDATA 2 "66-%'CURVESG6&7+7$$\"\"!F)$ 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4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" "Curve 2" "Curve 3" "Curve \+ 4" "Curve 5" "Curve 6" "Curve 7" "Curve 8" "Curve 9" "Curve 10" "Curve 11" "Curve 12" "Curve 13" "Curve 14" "Curve 15" "Curve 16" "Curve 17 " }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 " " }}}{EXCHG {PARA 325 "" 0 "" {TEXT -1 33 "SECOND MAPLE ASSIGNMENT QUE STIONS" }}{PARA 327 "" 0 "" {TEXT -1 0 "" }}{PARA 326 "" 0 "" {TEXT -1 21 "1. Choose a function " }{TEXT 303 7 "f(x,y) " }{TEXT -1 19 "and initial values " }{TEXT 304 7 "(x0,y0)" }{TEXT -1 174 " (not from the book or this assignment!) such that the solution of the IVP has an in teresting but not too complicated a trajectory. Find the actual soluti on by hand or using " }{TEXT 279 6 "dsolve" }{TEXT -1 23 ". Fix a medi um smaloop " }{TEXT 305 1 "h" }{TEXT -1 84 ". Solve your IVP using Eul er's Method. Tabulate your solution for several values of " }{TEXT 280 4 "x[i]" }{TEXT -1 91 ". Graph the numerical solution and the actu al solution on the same graph.\n\n2. For the same " }{TEXT 309 1 "f" } {TEXT -1 9 ", IC and " }{TEXT 310 1 "h" }{TEXT -1 110 " as in Prob. 1, solve the IVP using the improved Euler's Method. Tabulate your soluti on for several values of " }{TEXT 281 5 "x[i]." }{TEXT -1 90 " Graph t he numerical solution and the actual solution on the same graph.\n\n3. For the same " }{TEXT 308 1 "f" }{TEXT -1 9 ", IC and " }{TEXT 311 1 "h" }{TEXT -1 105 " as in Prob. 1, solve the IVP using the Runge-Kutta Method. Tabulate your solution for several values of " }{TEXT 282 5 " x[i]." }{TEXT -1 405 " Graph the numerical solution and the actual sol ution on the same graph.\n\n4. Compare the three methods using your ta bles and graphs. Compare the amount of computation versus the accuracy of the answer.\n\n5. Choose another initial value problem (not from t he book or this assignment!) For one of the two methods, Improved Eule r or Runge-Kutta, compute the solution of the IVP with at least eight \+ different " }{TEXT 306 1 "h" }{TEXT -1 57 "'s. Tabulate the approxima te solution at eight different " }{TEXT 307 4 "x[i]" }{TEXT -1 37 "'s. compute the errors and the ratio " }{TEXT 283 9 "error/h^2" }{TEXT -1 23 " for improved Euler or " }{TEXT 284 9 "error/h^4" }{TEXT -1 54 " for Runge-Kutta. \n (You have to change the power of " }{TEXT 288 1 "h" }{TEXT -1 115 ".)\n\n6. Now consider an unstable problem with a \+ spout-type instability (see page 138.) For example, you may analyze " }{TEXT 285 55 "f:= (x,y)-> cos(3*x)+2*y; x0:=0.0; y0:= -2/13; x1:= 2.0 " }{TEXT -1 163 ". Draw the graphs and tabulate the numerical approxim ations. Show that the numerical methods don't do a very good job of a pproximating the solution in this case. " }}}}{MARK "26 0 0" 0 } {VIEWOPTS 1 1 0 3 2 1804 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }