{VERSION 5 0 "IBM INTEL NT" "5.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 1 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 1 }{CSTYLE "2D Comment" 2 18 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 256 "" 1 14 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 257 "" 1 12 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 258 "" 1 12 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "T imes" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 } {PSTYLE "Heading 1" -1 3 1 {CSTYLE "" -1 -1 "Times" 1 18 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 8 4 1 0 1 0 2 2 0 1 }{PSTYLE "Title" -1 18 1 {CSTYLE "" -1 -1 "Times" 1 18 0 0 0 1 2 1 1 2 2 2 1 1 1 1 }3 1 0 0 12 12 1 0 1 0 2 2 19 1 }{PSTYLE "Author" -1 19 1 {CSTYLE "" -1 -1 "Times " 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }3 1 0 0 8 8 1 0 1 0 2 2 0 1 } {PSTYLE "Normal" -1 256 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }3 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Normal" -1 257 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }} {SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT -1 40 "Partial Differential Equa tions PowerTool" }}{PARA 19 "" 0 "" {TEXT -1 16 "by Dr. Jim Herod" }}} {PARA 257 "" 0 "" {TEXT 256 41 "Section 5.4: A String in a Viscous Med ium" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT 257 30 "Maple Packages for Section 5.4" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(plots):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "with(PDEto ols):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 80 "The variation on the w ave equation we consider in this section is the following:" }}{PARA 0 "" 0 "" {TEXT -1 29 " " }{XPPEDIT 18 0 "di ff(u,`$`(x,2))-k*diff(u,t)-g = diff(u,`$`(t,2));" "6#/,(-%%diffG6$%\"u G-%\"$G6$%\"xG\"\"#\"\"\"*&%\"kGF.-F&6$F(%\"tGF.!\"\"%\"gGF4-F&6$F(-F* 6$F3F-" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 65 " u(t, 0) = 0 = u(t, L)" }} {PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 59 " \+ u( 0, x) = F(x)" }}{PARA 0 "" 0 " " {TEXT -1 44 " " } {XPPEDIT 18 0 "u[t];" "6#&%\"uG6#%\"tG" }{TEXT -1 17 " ( 0 , x) = G(x) ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 333 "Thi s problem has a physical realization. It models a string for which the re is a downward pull, perhaps gravity. Also, there is a force pulling in the opposite direction from the velocity. Thus, if the vibration o f the string occurs in a viscous medium, there is a resistance proport ional to the velocity of the motion of the string." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 286 "If you were in a Mathema tics office or an Engineering office and stepped out into the hall fro m your workplace, stopped the first person you saw, and asked them how to solve this PDE, they would respond something such as, \"Separate v ariables. Isn't that how you do all those problems?\" " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 208 "I say to you: \"Sur ely. Separation of variables is one of the main ideas of these notes. \" So, let's solve the equation as the person we might stop in the hal l would. Let's use the techniques of Fourier Series." }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 73 "To be specific, we will get solutions that we can graph for the equation:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 26 " \+ " }{XPPEDIT 18 0 "diff(w,`$`(x,2))-diff(w,t)/5-32 = diff(w,`$`(t,2)); " "6#/,(-%%diffG6$%\"wG-%\"$G6$%\"xG\"\"#\"\"\"*&-F&6$F(%\"tGF.\"\"&! \"\"F4\"#KF4-F&6$F(-F*6$F2F-" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 64 " w(t , 0) = 0 = w(t, " }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 1 ")" }} {PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 61 " \+ w( 0, x) = sin(x)" }}{PARA 0 "" 0 "" {TEXT -1 44 " " } {XPPEDIT 18 0 "w[t]" "6#&%\"wG6#%\"tG" }{TEXT -1 14 " ( 0 , x) = 0." } }{PARA 0 "" 0 "" {TEXT -1 151 "Recognize that this is a non-homogeneou s problem. The -32 term makes it so. We must separate out the steady s tate solution and the transient solution. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 100 "The steady state solution v(x) is independent of time. Thus, that solution will satisfy the equation " }}{PARA 0 "" 0 "" {TEXT -1 17 " " }{XPPEDIT 18 0 "di ff(v(x),`$`(x,2));" "6#-%%diffG6$-%\"vG6#%\"xG-%\"$G6$F)\"\"#" }{TEXT -1 30 " - 32 = 0, with v(0) = 0 = v(" }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 2 ")." }}{PARA 0 "" 0 "" {TEXT -1 58 "We solve this second o rder ordinary differential equation." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "dsolve(\{diff(v(x),x,x)=32,v(0)=0,v(Pi)=0\},v(x));" } }}{PARA 0 "" 0 "" {TEXT -1 42 "We find that the steady state solution \+ is " }{XPPEDIT 18 0 "v(x) = 16*x^2-16*Pi*x;" "6#/-%\"vG6#%\"xG,&*&\"#; \"\"\"*$F'\"\"#F+F+*(F*F+%#PiGF+F'F+!\"\"" }{TEXT -1 1 "." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "v:=x->16*x^2-16*Pi*x;" }}}{EXCHG } {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 63 "Next we s olve the transient equation. The transient equation is" }}{PARA 0 "" 0 "" {TEXT -1 24 " " }{XPPEDIT 18 0 "diff(w,`$` (x,2))-diff(w,t)/5 = diff(w,`$`(t,2));" "6#/,&-%%diffG6$%\"wG-%\"$G6$% \"xG\"\"#\"\"\"*&-F&6$F(%\"tGF.\"\"&!\"\"F4-F&6$F(-F*6$F2F-" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 63 " \+ w(t, 0) = 0 = w(t, " }{XPPEDIT 18 0 "Pi; " "6#%#PiG" }{TEXT -1 1 ")" }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 70 " w( \+ 0, x) = sin(x/2) - v(x)" }}{PARA 0 "" 0 "" {TEXT -1 44 " \+ " }{XPPEDIT 18 0 "w[t]" "6#&%\"wG6#%\"tG " }{TEXT -1 14 " ( 0 , x) = 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 119 "How do we solve this transient equation? Just like the hall-walker said: we separate variables. This leads to \+ equations" }}{PARA 0 "" 0 "" {TEXT -1 44 " X''/X = \+ (T'' +T '/5 )/T." }}{PARA 0 "" 0 "" {TEXT -1 128 "In the usual manner, we arrive at two ordinary differential equations, the solutions of wh ich lead to the solutions for the PDE:" }}{PARA 0 "" 0 "" {TEXT -1 19 " X '' = " }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT -1 20 " X, with X(0) = X( " }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 6 " ) = 0" }}{PARA 0 "" 0 "" {TEXT -1 3 "and" }}{PARA 0 "" 0 "" {TEXT -1 26 " T '' + T'/5 = " }{XPPEDIT 18 0 "lambda;" "6#%'lambd aG" }{TEXT -1 3 " T." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 56 "We've seen the X equation enough to know all solutions: " }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "-n^2;" "6#,$*$%\"nG\"\"#!\"\"" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "X[n];" "6#&%\"XG6#%\"nG" }{TEXT -1 10 "(x) = sin(" }{XPPEDIT 18 0 " n*Pi*x/L;" "6#**%\"nG\"\"\"%#PiGF%%\"xGF%%\"LG!\"\"" }{TEXT -1 10 ") = sin(n " }{XPPEDIT 18 0 "x;" "6#%\"xG" }{TEXT -1 24 "). Check this by \+ asking:" }}{PARA 0 "" 0 "" {TEXT -1 7 "(1) Is " }{XPPEDIT 18 0 "X[n]; " "6#&%\"XG6#%\"nG" }{TEXT -1 5 "'' = " }{XPPEDIT 18 0 "lambda;" "6#%' lambdaG" }{TEXT -1 2 " " }{XPPEDIT 18 0 "X[n];" "6#&%\"XG6#%\"nG" } {TEXT -1 2 " ?" }}{PARA 0 "" 0 "" {TEXT -1 7 "(2) Is " }{XPPEDIT 18 0 "X[n];" "6#&%\"XG6#%\"nG" }{TEXT -1 15 "(0) = 0 ? Is " }{XPPEDIT 18 0 "X[n];" "6#&%\"XG6#%\"nG" }{TEXT -1 1 "(" }{XPPEDIT 18 0 "Pi;" "6#%# PiG" }{TEXT -1 7 ") = 0 ?" }}{PARA 0 "" 0 "" {TEXT -1 41 "The answer t o the three questions is yes." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 76 "For each n, we solve the T equation. Be s ure we know what the T equation is:" }}{PARA 0 "" 0 "" {TEXT -1 38 " \+ T '' + T'/5 = " }{XPPEDIT 18 0 "-n^2;" "6#,$*$% \"nG\"\"#!\"\"" }{TEXT -1 3 " T." }}{PARA 0 "" 0 "" {TEXT -1 141 "Sinc e this is a second order equation in T, we expect two constants. These will be constants that we will determine by some Fourier process. " } }{PARA 0 "" 0 "" {TEXT -1 94 " How many terms shall we do? Let's c all the number of terms we choose to do by the name N." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "N:=5;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 93 "for n from 1 to N do\n dsolve(diff(T(t),t,t)+diff(T (t),t)/5=-n^2*T(t),T(t));\nend do;\nn:='n':" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 109 "It would be nice to have a good formula for those trig terms. I need a formula that re presents this sequence:" }}{PARA 0 "" 0 "" {TEXT -1 17 " \+ " }{XPPEDIT 18 0 "sqrt(99),sqrt(399),sqrt(899),sqrt(1599),sqrt(2499 );" "6'-%%sqrtG6#\"#**-F$6#\"$*R-F$6#\"$**)-F$6#\"%*f\"-F$6#\"%*\\#" } {TEXT -1 7 ", ... ." }}{PARA 0 "" 0 "" {TEXT -1 3 " " }}{PARA 0 "" 0 "" {TEXT -1 237 "The question is: how can I write that multiple of t inside those sine or cosine function as a function of n. (Such questi ons are often on IQ tests.) Here's a way to find the answer. I'll solv e the equation again without saying what n is." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "dsolve(diff(T(t),t,t)+diff(T(t),t)/5=-n^2*T(t),T (t));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 " " {TEXT -1 133 "It is clear what the terms are now, isn't it? Using th e Euler identity relating exponential and trig functions, each T is of the form" }}{PARA 0 "" 0 "" {TEXT -1 13 " " }{XPPEDIT 18 0 "T[n](t) = A[n]*exp(-t/10)*cos(t*sqrt(100*n^2-1)/10)+B[n]*exp(-t/10) *sin(t*sqrt(100*n^2-1)/10);" "6#/-&%\"TG6#%\"nG6#%\"tG,&*(&%\"AG6#F(\" \"\"-%$expG6#,$*&F*F0\"#5!\"\"F7F0-%$cosG6#*(F*F0-%%sqrtG6#,&*&\"$+\"F 0*$F(\"\"#F0F0F0F7F0F6F7F0F0*(&%\"BG6#F(F0-F26#,$*&F*F0F6F7F7F0-%$sinG 6#*(F*F0-F=6#,&*&FAF0*$F(FCF0F0F0F7F0F6F7F0F0" }{TEXT -1 1 "." }} {PARA 0 "" 0 "" {TEXT -1 68 "I can now make the general solution for t he PDE. I hope you can too." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 140 "w:=(t,x)->\n sum(A[n]*exp(-t/10)*c os(t*sqrt(100*n^2-1)/10)*sin(n*x),n=1..N)+\n sum(B[n]*exp(-t/10)*sin(t *sqrt(100*n^2-1)/10)*sin(n*x),n=1..N);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 162 "We check that this \+ is a general solution. Ask: does it satisfy the PDE and the boundary c onditions? (The initial conditions will determine the values of A and \+ B.)" }}{PARA 0 "" 0 "" {TEXT -1 8 "The PDE:" }}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 61 "simplify(diff(w(t,x),x,x)-diff(w(t,x),t)/5-diff(w(t ,x),t,t));" }}}{PARA 0 "" 0 "" {TEXT -1 20 "Boundary conditions:" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "w(t,0); w(t,Pi);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 57 "It remains to determine the A's and B's. Observe what is " }{XPPEDIT 18 0 "u(0,x);" "6#-%\"uG6$\"\"!%\"xG" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "u[t](0,x);" "6#-&%\"uG6#%\"tG6$\"\"!%\"xG" }{TEXT -1 1 "." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "w(0,x);" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 13 "D[1](w)(0,x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 101 "DerivTerm:=collect(collect(collect(collect(collect(% ,sin(x)),sin(2*x)),sin(3*x)),sin(4*x)),sin(5*x));" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 247 "It seems c lear that we can get the A's through a simple Fourier Series. Having t he A's we can get the B's. We do that here. This is the first time in \+ this problem to use the initial conditions.\n\nRecall that the solutio n for the original problem is" }}{PARA 0 "" 0 "" {TEXT -1 33 " \+ u(t, x) = v(x) + w(t, x)." }}{PARA 0 "" 0 "" {TEXT -1 42 "Thus, u(0, x) = v(x) + w(0, x) and " }{XPPEDIT 18 0 "diff(u,t);" "6#-%%diff G6$%\"uG%\"tG" }{TEXT -1 8 "(0,x) = " }{XPPEDIT 18 0 "diff(w,t);" "6#- %%diffG6$%\"wG%\"tG" }{TEXT -1 6 "(0,x)." }}{PARA 0 "" 0 "" {TEXT -1 17 "This means that " }}{PARA 0 "" 0 "" {TEXT -1 3 " " }{XPPEDIT 18 0 "A[n] = int((f(x)-v(x))*sin(n*x),x)/int(sin(n*x)^2,x);" "6#/&%\"A G6#%\"nG*&-%$intG6$*&,&-%\"fG6#%\"xG\"\"\"-%\"vG6#F1!\"\"F2-%$sinG6#*& F'F2F1F2F2F1F2-F*6$*$-F86#*&F'F2F1F2\"\"#F1F6" }{TEXT -1 1 " " }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 3 "and" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 3 " " } {XPPEDIT 18 0 "(sqrt(100*n^2-1)*B[n]-A[n])/10;" "6#*&,&*&-%%sqrtG6#,&* &\"$+\"\"\"\"*$%\"nG\"\"#F,F,F,!\"\"F,&%\"BG6#F.F,F,&%\"AG6#F.F0F,\"#5 F0" }{TEXT -1 5 " = " }{XPPEDIT 18 0 "int(g(x)*sin(n*x),x)/int(sin(n *x)^2,x());" "6#*&-%$intG6$*&-%\"gG6#%\"xG\"\"\"-%$sinG6#*&%\"nGF,F+F, F,F+F,-F%6$*$-F.6#*&F1F,F+F,\"\"#-F+6\"!\"\"" }{TEXT -1 2 " ." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 10 "We comput e" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "fmv:=x->sin(x)-v(x);" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 87 "for n from 1 to N do\n A [n]:=int(fmv(x)*sin(n*x),x=0..Pi)/int(sin(n*x)^2,x=0..Pi);\nod;" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "g:=x->0;" }}}{EXCHG }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 140 "for n from 1 to N do\n B[n]:=(10 *int(g(x)*sin(n*x),x=0..Pi)/int(sin(n*x)^2,x=0..Pi)+A[n])\n \+ /sqrt(100*n^2-1);\nend do;\nn:='n';" }}}{EXCHG }{PARA 0 "" 0 "" {TEXT -1 105 "We now have the transient solution completely determined . We create the solution to the original problem:" }}{PARA 0 "" 0 "" {TEXT -1 53 " u(t,x) = w(t,x) + v(x)." }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "u:=(t,x)->w(t,x)+v(x);" }}} {EXCHG }{PARA 0 "" 0 "" {TEXT -1 97 "It seems appropriate to check tha t this is really the solution. We check the boundary conditions." }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "u(t,0); u(t,Pi);" }}}{PARA 0 "" 0 "" {TEXT -1 17 "We check the PDE." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "simplify(diff(u(t,x),x,x)-diff(u(t,x),t)/5-32-diff(u( t,x),t,t));" }}}{PARA 0 "" 0 "" {TEXT -1 107 "We check the initial con ditions. For each one, we draw graphs to see how close we are to the i nitial value." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "plot([sin(x ),u(0,x)],x=0..Pi,color=[black,red]);" }}}{PARA 0 "" 0 "" {TEXT -1 107 "We draw a graph to see how close the initial velocity is to zero. Off-set the graph so that it can be seen." }}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 59 "plot(eval(subs(t=0,diff(u(t,x),t)))+0.001,x=0..Pi,y =-1..1);" }}}{PARA 0 "" 0 "" {TEXT -1 261 "We now show an animation. C an you predict the movement? It should be a vibrating string; the firs t derivative term is a retarding force so the oscillations should decr ease in size; and the downward force should pull the string down to th at hanging, steady state" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 " with(plots):\nanimate(u(t,x),x=0..Pi,t=0..20, frames=40);" }{TEXT -1 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 371 " In this Section, we have made the physical model more interesting. We \+ assumed that the string lies in a viscous medium and is subject to a c onstant force pulling it downward. This made the partial differential \+ equation more interesting. There were more terms to consider than the \+ simple wave equation. To solve the problem, we used the methods of sep aration of variables." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT 258 16 "Unassisted Maple" }}{PARA 0 "" 0 "" {TEXT -1 267 "Because the problem of this section has been a wave equa tion problem and because we did not use d'Alembert's techniques, we wo nder what Maple will suggest when unassisted. Will it suggest a variat ion on d'Alembert's Method, or will it suggest separation of variables ?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 102 "Not e that at the beginning, we called up the package PDEtools. We make a \+ generic u and define the PDE." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "u:='u';" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "PDE:=diff(u (t,x),x,x)-diff(u(t,x),t,t)-diff(u(t,x),t)=0;" }}}{PARA 0 "" 0 "" {TEXT -1 49 "Look to see what form Maple gets for a solutions." }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "ans := pdsolve(PDE);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "build(ans);" }}}{PARA 0 "" 0 "" {TEXT -1 297 "This process provides solutions as exponentials. Mo st important, it suggests that one can obtain a solution to this probl em by performing a d'Alembert type transformation to get a different s econd order partial differential equation. Going in these directions m ight be pursued at a different time." }}{EXCHG }{EXCHG }{EXCHG } {EXCHG }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 50 " EMAIL: herod@math.gatech.edu or jherod@tds.net" }}{PARA 0 "" 0 "" {TEXT -1 38 "URL: http://www.math.gatech.edu/~herod" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 36 "Copyright \251 2003 \+ by James V. Herod" }}{PARA 256 "" 0 "" {TEXT -1 19 "All rights reserv ed" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{MARK "0 0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 1 1 2 33 1 1 }