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0 0 -1 0 }{PSTYLE "" 0 284 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 285 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 286 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 287 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 288 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 289 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 290 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 291 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 292 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 293 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 294 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 66 "Unit 5: Boundary Value Pro blems for Partial Differential Equations" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 25 "Chapter 24: Wave Equation" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 33 "Section 2 4.1: the plucked string" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 9 "Copyright" }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 48 "Copyright * 2001 by Addison Wesley Lo ngman, Inc." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 302 "All rights reserved. No part of this publication may be repro duced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherw ise, without the prior written permission of the publisher. Printed in the United States of America." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {SECT 1 {PARA 3 "" 0 "" {TEXT -1 15 "Initializations" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "interface(showassumed=0); \nassume(n,integer);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "wit h(plots):\nwith(plottools):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 37 "The following Maple code d efines the " }{TEXT 265 2 "PX" }{TEXT -1 118 " command which creates a periodic extension of a function. See lesson 2-4-1 for details of in stallation and behavior." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 499 "PX := proc(h, g)\nlocal f, var, d; \n if type(h, procedure) then RETURN(subs(\n \{'D' = rhs(g) \+ - lhs(g), 'F' = h, 'L' = lhs(g)\}, proc(\n x::algebraic)\n \+ local y;\n y := floor((x - L)/D); F(x - y*D)\n en d))\n fi;\n if type(g, equation) then var := lhs(g); d := rhs(g) fi;\n f := unapply(h, var);\n subs(\{'L' = lhs(d), 'D' = rhs(d) - lhs(d), 'F' = f\}, proc(\n x::algebraic)\n local y;\n y := floor((x - L)/D); F(x - y*D)\n end)\nend:" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 3 "" 0 " " {TEXT -1 34 "Wave Equation on the Finite String" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 82 "An idealized strin g, uniform and without internal friction, is stretched along an " } {XPPEDIT 18 0 "x" "6#%\"xG" }{TEXT -1 14 "-axis from to " }{XPPEDIT 18 0 "x=L" "6#/%\"xG%\"LG" }{TEXT -1 163 ". The ends are fixed, the s tring is taut, and it is constrained to move in a vertical plane. The equilibrium position for the string coincides with the interval " } {XPPEDIT 18 0 "[0,L]" "6#7$\"\"!%\"LG" }{TEXT -1 8 " on the " } {XPPEDIT 18 0 "x" "6#%\"xG" }{TEXT -1 66 "-axis, and displacements in \+ the string are measured vertically by " }{XPPEDIT 18 0 "u" "6#%\"uG" } {TEXT -1 11 ". At time " }{XPPEDIT 18 0 "t" "6#%\"tG" }{TEXT -1 18 ", and at location " }{XPPEDIT 18 0 "x" "6#%\"xG" }{TEXT -1 4 " in " } {XPPEDIT 18 0 "[0,L]" "6#7$\"\"!%\"LG" }{TEXT -1 53 ", the displacemen t of the string from equilibrium is " }{XPPEDIT 18 0 "u(x,t)" "6#-%\"u G6$%\"xG%\"tG" }{TEXT -1 13 ". For fixed " }{XPPEDIT 18 0 "t=t^`^`" " 6#/%\"tG)F$%\"^G" }{TEXT -1 38 ", the shape of the string is given by \+ " }{XPPEDIT 18 0 "u(x,t^`^`)" "6#-%\"uG6$%\"xG)%\"tG%\"^G" }{TEXT -1 3 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 74 "Figure 24.1, below, shows the shape of the string at a particular ins tant " }{XPPEDIT 18 0 "t^`^`" "6#)%\"tG%\"^G" }{TEXT -1 1 " " } {XPPEDIT 18 0 "`>`*0" "6#*&%\">G\"\"\"\"\"!F%" }{TEXT -1 1 "." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 445 "p1 := plot(x*(1-x),x=0..1,color=black):\np2 := textplot([1,-.05 ,`L`], font=[TIMES,ITALIC,12]):\np3 := arrow([.5,.09],[.5,0],.01,.03,. 2,color=black):\np4 := arrow([.5,.16],[.5,.25],.01,.03,.2,color=black) :\np5 := textplot([.5,.125,`u(x,t)`]):\n### WARNING: note that `I` is \+ no longer of type `^`\np6 := textplot([.525,.136,`^`]):\ndisplay([p||( 1..6)],scaling=constrained, labels=[x,u], xtickmarks=[-1,2], ytickmark s=[-1,1], labelfont=[TIMES,ITALIC,12]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 44 "Wave Equation (1) " }{XPPEDIT 18 0 "u[tt]=c^2*u[xx]" "6#/ &%\"uG6#%#ttG*&%\"cG\"\"#&F%6#%#xxG\"\"\"" }{TEXT -1 8 " for " } {XPPEDIT 18 0 "t*`>`*0" "6#*(%\"tG\"\"\"%\">GF%\"\"!F%" }{TEXT -1 1 " \+ " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 47 "fixed left end (2) " }{XPPEDIT 18 0 " u(0,t)=0 " "6#/-%\"uG6$\"\"!%\"tGF'" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 46 "fixed right end (3) " }{XPPEDIT 18 0 "u(L,t) =0" "6#/-%\"uG6$%\"LG%\"tG\"\"!" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 49 "initial shape \+ (4) " }{XPPEDIT 18 0 "u(x,0)=f(x)" "6#/-%\"uG 6$%\"xG\"\"!-%\"fG6#F'" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 50 "initial velocity (5) " }{XPPEDIT 18 0 "u [t](x,0)=g(x)" "6#/-&%\"uG6#%\"tG6$%\"xG\"\"!-%\"gG6#F*" }{TEXT -1 2 " " }}{PARA 0 "" 0 "" {TEXT -1 50 "___________________________________ _______________" }}{PARA 0 "" 0 "" {TEXT -1 48 " \+ Table 24.1 " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 79 "As we will derive in Section 24.4, displacements \+ in the string are governed by " }{XPPEDIT 18 0 "u[tt]=c^2*u[xx]" "6#/& %\"uG6#%#ttG*&%\"cG\"\"#&F%6#%#xxG\"\"\"" }{TEXT -1 6 ", the " }{TEXT 268 4 "wave" }{TEXT -1 1 " " }{TEXT 269 8 "equation" }{TEXT -1 31 " (1 ) in Table 24.1. This is a " }{TEXT 270 7 "partial" }{TEXT -1 1 " " } {TEXT 271 12 "differential" }{TEXT -1 1 " " }{TEXT 272 8 "equation" } {TEXT -1 40 " (PDE) in the two independent variables " }{XPPEDIT 18 0 "x" "6#%\"xG" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "t" "6#%\"tG" }{TEXT -1 17 ". The parameter " }{XPPEDIT 18 0 "c" "6#%\"cG" }{TEXT -1 25 " \+ will turn out to be the " }{TEXT 275 4 "wave" }{TEXT -1 1 " " }{TEXT 276 5 "speed" }{TEXT -1 219 ", the speed at which a disturbance can pr opagate horizontally along the string. The boundary conditions (2) an d (3) in Table 24.1 assure that the endpoints of the string remain fiz ed. Because they proscribe values of " }{XPPEDIT 18 0 "u" "6#%\"uG" } {TEXT -1 35 " at the endpoints, they are called " }{TEXT 273 9 "Dirich let" }{TEXT -1 73 " conditions. Since the prescribed values are zero, these conditions are " }{TEXT 274 11 "homogeneous" }{TEXT -1 148 " Di richlet conditions. The initial conditions (4) and (5) in Table 24.1 \+ give the initial shape and velocity of the string. At first, we will \+ take " }{XPPEDIT 18 0 "g(x)=0" "6#/-%\"gG6#%\"xG\"\"!" }{TEXT -1 61 ", so that the string is given a (small) initial displacement " } {XPPEDIT 18 0 "f(x)" "6#-%\"fG6#%\"xG" }{TEXT -1 108 ", and released. \+ This corresponds to a gentle plucking of the string, much like the ac tion in a harpsichord." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 202 "Together, the five equations in Table 24.1 constitu te an initial-boundary value problem (or BVP) for the finite string. \+ Our goal is to obtain a solution to this BVP, and to understand how th e function " }{XPPEDIT 18 0 "u(x,t)" "6#-%\"uG6$%\"xG%\"tG" }{TEXT -1 59 " so determined describes the physical motion of the string." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 12 "Example 24.1" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 18 "We will show that " }{XPPEDIT 18 0 "u(x,t )=sin(x)*cos(3*t)" "6#/-%\"uG6$%\"xG%\"tG*&-%$sinG6#F'\"\"\"-%$cosG6#* &\"\"$F-F(F-F-" }{TEXT -1 45 " is a solution to the BVP in Table 24.1 \+ when " }{XPPEDIT 18 0 "c=3,L=Pi,g(x)=0" "6%/%\"cG\"\"$/%\"LG%#PiG/-%\" gG6#%\"xG\"\"!" }{TEXT -1 6 ", and " }{XPPEDIT 18 0 "f(x)=sin(x)" "6#/ -%\"fG6#%\"xG-%$sinG6#F'" }{TEXT -1 22 ". A calculation shows" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 270 "" 0 "" {TEXT -1 1 " " } {XPPEDIT 18 0 "u[tt]=-9*sin(x)*cos(3*t)" "6#/&%\"uG6#%#ttG,$*(\"\"*\" \"\"-%$sinG6#%\"xGF+-%$cosG6#*&\"\"$F+%\"tGF+F+!\"\"" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 3 "and" }}{PARA 271 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "c^2*u[xx]=3^2*(-sin(x))*cos(3*t)" "6#/*&%\"cG\"\"#&%\" uG6#%#xxG\"\"\"*(\"\"$F&,$-%$sinG6#%\"xG!\"\"F+-%$cosG6#*&F-F+%\"tGF+F +" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 44 "To implement this calculation in Maple, set " }{XPPEDIT 18 0 "c=3" "6#/%\"cG\"\"$" }{TEXT -1 4 " so " }{XPPEDIT 18 0 "c^2=9" " 6#/*$%\"cG\"\"#\"\"*" }{TEXT -1 16 " in the equation" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "q := diff (u(x,t),t,t) = 9*diff(u(x,t),x,x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 6 "Write " } {XPPEDIT 18 0 "u(x,t)" "6#-%\"uG6$%\"xG%\"tG" }{TEXT -1 3 " as" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "U := sin(x)*cos(3*t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 25 "and make the substitution " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "simplify(subs(u(x,t)=U,q));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 10 "Moreover , " }{XPPEDIT 18 0 "u(0,t)=u(Pi,t)" "6#/-%\"uG6$\"\"!%\"tG-F%6$%#PiGF( " }{TEXT -1 18 " = 0, as we see by" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "eval(subs(x=0,U));" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 3 "and" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 19 "eval(subs(x=Pi,U));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 9 "Finally, " }{XPPEDIT 18 0 "u(x,0)=sin(x)" "6#/-%\"uG6$%\"xG\"\"!-%$sinG6#F'" } {TEXT -1 3 " = " }{XPPEDIT 18 0 "f(x)" "6#-%\"fG6#%\"xG" }{TEXT -1 14 ", as we see by" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 18 "eval(subs(t=0,U));" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 105 "Figure 24. 2 consisting of Figures 24.2a and 24.2 b, shows the shape of the strin g at two different times." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 15 "Figure 24.2a is" }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 271 "U := sin(x)*cos(3*t):\np1 := plot(subs(t=.3,U),x=0..Pi,color=black, scaling=constrained, xtickm arks=[0], ytickmarks=[-1,1], labels=[x,u], labelfont=[TIMES,ITALIC,12] ):\np2 := textplot([Pi,-.1,`p`],font=[SYMBOL,10]):\np3 := textplot([2. 7,.56,`t = .3`]):\ndisplay([p||(1..3)]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 19 "and Figure 24 .2b is" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 252 "p1 := plot(subs(t=1.5,U),x=0..Pi,color=black, scalin g=constrained, xtickmarks=[0], ytickmarks=[-1,1], labels=[x,u], labelf ont=[TIMES,ITALIC,12]):\np2 := textplot([Pi,-.1,`p`],font=[SYMBOL,10]) :\np3 := textplot([2.7,-.25,`t = 1.5`]):\ndisplay([p||(1..3)]);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 47 "Figure 24.3 shows the solution surface whereby " } {XPPEDIT 18 0 "u(x,t)" "6#-%\"uG6$%\"xG%\"tG" }{TEXT -1 32 " is graphe d as a surface in the " }{XPPEDIT 18 0 "xt" "6#%#xtG" }{TEXT -1 7 "-pl ane." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 175 "p1 := plot3d(U,x=0..Pi,t=0..4*Pi/3, axes=frame, labe ls=[` x`,` t`,`u `], labelfont=[TIMES,ITALIC,12], color=red, style =hidden, tickmarks=[3,4,3], orientation=[45,60]):\np1;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 87 "Figure 24.4 shows several snapshots of the string superimposed \+ on the solution surface." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 197 "F := z -> spacecurve([x,z,sin(x)*c os(3*z),x=0..Pi], color=black, thickness=3):\np2 := display3d([seq(F(2 *Pi/30*k),k=0..20)]):\ndisplay3d([p1,p2],orientation=[25,80], scaling= constrained, axes=frame);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 297 "The solution surface is tra ced out in time by a succession of moving images of the dynamic string , or equivalently, the plane sections of the solution surface are snap shots in time of the physical motion of the string. The obvious anima tion of the string tracing out the solution surface follows." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 135 " p2 := display3d([seq(F(2*Pi/30*k),k=0..20)], insequence=true):\ndispla y3d([p1,p2],orientation=[25,80], scaling=constrained, axes=frame);" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 164 "Figures 24.3 and 24.4 show that the motion of the string is an example of a standing wave because the points of zero displacem ent do not translate along the string." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 69 "In fact, a complete animation of the \+ motion of the string is given by" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 126 "animate(U,x=0..Pi,t=0..2*Pi /3, frames=20, color=black, labels=[x,u], labelfont=[TIMES,ITALIC,12], xtickmarks=3, ytickmarks=3);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 12 "Example 24.2" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 154 "A wave \+ free to move along the string is called a traveling wave. Suppose the initial disturbance in the string has the shape shown in figure 24.5, below." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 197 "g := x -> piecewise(x<3*Pi/8,0, x " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 282 "Under the action of the wave equation, that in itial disturbance would travel along the string, reflecting at the end points, and bouncing back and forth between between them. Figure 24.6 (consisting of Figures 24.6a and 24.6b) shows the traveling wave at t wo successive times after " }{XPPEDIT 18 0 "t=0" "6#/%\"tG\"\"!" } {TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 15 "Figure 24.6a is" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 311 "g1 := piecewise(x<0,-g(-x), x " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 21 "while Figure 24.6b is" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 228 "pp3 := plo t(subs(t=1.47,U1),x=0..Pi, color=red, xtickmarks=3, ytickmarks=[-.4,0, .4], scaling=constrained, view=[0..Pi,-.4.. .4]):\npp4:=textplot([2.9, .3,`t = 1.47`]):\ndisplay([pp3,pp4], labels=[x,u], labelfont=[TIMES,IT ALIC,12]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 133 "The complete animation showing how the i nitial energy in the string is distributed under the action of the wav e equation is given by" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 242 "g1 := piecewise(x<0,-g(-x), x " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 271 "The initial energy splits into two equal parts, wit h each new wave traveling in opposite directions. When the wave reach es the boundary, it is absorbed, and then re-transmitted, only now, wi th a reflection. Figure 24.7 shows the solutoin surface for the trave ling wave." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 217 "plot3d(U1,x=0..Pi,t=0..2*Pi,axes=boxed, labels=[x, t,u], labelfont=[TIMES,ITALIC,12], color=black, style=hidden, shading= zgrayscale, tickmarks=[[0,1,2,3],5,[0]], grid=[25,50], scaling=constra ined, orientation=[50,60]);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 168 "The initial disturbance splits into two waves of half th e initial height, one traveling left, and one traveling right. Withou t changing shape, these waves move in the " }{XPPEDIT 18 0 "xt" "6#%#x tG" }{TEXT -1 13 "-plane along " }{TEXT 277 15 "characteristics" } {TEXT -1 12 ", the lines " }{XPPEDIT 18 0 "x" "6#%\"xG" }{TEXT -1 1 " \+ " }{TEXT 278 1 "+" }{TEXT -1 1 " " }{XPPEDIT 18 0 "c*t" "6#*&%\"cG\"\" \"%\"tGF%" }{TEXT -1 73 " = constant, which are therefore said to \"ca rry the initial information.\"" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 96 "A more dynamic view of the propagation of the initial shape is given by the following animation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 254 "ff := z -> spacecurve([x,z,subs(t=z,U1),x=0..Pi], color=black, thickness=3, tickmarks=[2,2,[0]]):\ndisplay3d([seq(ff(2*Pi/50*k),k=0..50)], insequ ence=true, axes=boxed, labels=[x,t,u], labelfont=[TIMES,ITALIC,14], sc aling=constrained, orientation=[55,65]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 35 "Solution by Separation of Variables" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 64 "The classical method of solution for such BVPs is the method of " }{TEXT 258 23 "separation of variables" } {TEXT -1 28 ". Assume a solution of the " }{TEXT 279 9 "separated" } {TEXT -1 5 " form" }}{PARA 272 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 " u(x,t)=X(x)*T(t)" "6#/-%\"uG6$%\"xG%\"tG*&-%\"XG6#F'\"\"\"-%\"TG6#F(F- " }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 42 "a product of one function just containing " }{TEXT 259 1 "x" }{TEXT -1 34 " and one function just containing " }{TEXT 260 1 "t " }{TEXT -1 29 ". Recall the initial example" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 273 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "u(x,t )=sin(x)*cos(3*t)" "6#/-%\"uG6$%\"xG%\"tG*&-%$sinG6#F'\"\"\"-%$cosG6#* &\"\"$F-F(F-F-" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 11 "from ab ove." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 36 "Now, generalize this to the product " }{XPPEDIT 18 0 "u(x,t)=X( x)*T(t)" "6#/-%\"uG6$%\"xG%\"tG*&-%\"XG6#F'\"\"\"-%\"TG6#F(F-" }{TEXT -1 13 ", that is, to" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "U := X(x)*T(t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 63 "to which we next apply the fixed endpoint conditions, obtaining" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 274 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 " X(0)*T(t)=0" "6#/*&-%\"XG6#\"\"!\"\"\"-%\"TG6#%\"tGF)F(" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 3 "and" }}{PARA 275 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "X(L)*T(t)=0" "6#/*&-%\"XG6#%\"LG\"\"\"-%\"TG6#%\"tG F)\"\"!" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 8 "that is," }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "subs(x=0,U) = 0;\nsubs(x=L,U) = 0;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 22 "Ruling out th e choice " }{XPPEDIT 18 0 "T(t)=0" "6#/-%\"TG6#%\"tG\"\"!" }{TEXT -1 22 " because that implies " }{XPPEDIT 18 0 "u(x,t)" "6#-%\"uG6$%\"xG% \"tG" }{TEXT -1 32 " = 0 (identically), we conclude " }{XPPEDIT 18 0 " X(0)=X(L)" "6#/-%\"XG6#\"\"!-F%6#%\"LG" }{TEXT -1 5 " = 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 75 "Under the assumpt ion of a separated solution, the initial conditions become" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 258 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "u(x,0)=X(x)*T(0)" "6#/-%\"uG6$%\"xG\"\"!*&-%\"XG6#F'\"\"\"-%\"TG6#F (F-" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "f(x)" "6#-%\"fG6#%\"xG" }} {PARA 0 "" 0 "" {TEXT -1 3 "ans" }}{PARA 259 "" 0 "" {TEXT -1 1 " " } {XPPEDIT 18 0 "u[t](x,0)=X(x)" "6#/-&%\"uG6#%\"tG6$%\"xG\"\"!-%\"XG6#F *" }{TEXT -1 9 " T'(0) = " }{XPPEDIT 18 0 "g(x)" "6#-%\"gG6#%\"xG" }} {PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 14 "Unless e ither " }{XPPEDIT 18 0 "f(x)" "6#-%\"fG6#%\"xG" }{TEXT -1 4 " or " } {XPPEDIT 18 0 "g(x)" "6#-%\"gG6#%\"xG" }{TEXT -1 131 " is zero, no con clusion can be drawn from these two equations. Assuming for simplicit y that the initial velocity is zero, so that " }{XPPEDIT 18 0 "g(x)=0 " "6#/-%\"gG6#%\"xG\"\"!" }{TEXT -1 13 ", we conclude" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 262 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "u[ t](x,0)" "6#-&%\"uG6#%\"tG6$%\"xG\"\"!" }{TEXT -1 32 " = X(x) T'(0) = \+ 0 => T'(0) = 0" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 22 "The initial condition " }{XPPEDIT 18 0 "u(x,0)=f(x)" "6#/ -%\"uG6$%\"xG\"\"!-%\"fG6#F'" }{TEXT -1 142 " can be satisfied with th e aid of a Fourier series. Before we can see that, we must apply the \+ separation assumption to the PDE itself, namely" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "q := diff(u (x,t),t,t) = c^2*diff(u(x,t),x,x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 6 "giving" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 276 "" 0 "" {TEXT -1 1 " " } {XPPEDIT 18 0 "X(x)*`T''`(t)=c^2*`X''`(x)*T(t)" "6#/*&-%\"XG6#%\"xG\" \"\"-%$T''G6#%\"tGF)*(%\"cG\"\"#-%$X''G6#F(F)-%\"TG6#F-F)" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 8 "that is," }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "q1 := eval(subs(u (x,t)=U,q));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 105 "We have just deliberately committed a no tational no-no. The derivatives on the left are with respect to " } {XPPEDIT 18 0 "t" "6#%\"tG" }{TEXT -1 57 ", while the derivatives on t he right are with respect to " }{XPPEDIT 18 0 "x" "6#%\"xG" }{TEXT -1 25 ". Yet, we have used the " }{TEXT 280 4 "same" }{TEXT -1 107 " sym bol for each differentiation, a mathematical imprecision and ambiguity that is too convenient to avoid." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 16 "Next, divide by " }{XPPEDIT 18 0 "c^2*u(x ,t)" "6#*&%\"cG\"\"#-%\"uG6$%\"xG%\"tG\"\"\"" }{TEXT -1 7 ", with " } {XPPEDIT 18 0 "u(x,t)" "6#-%\"uG6$%\"xG%\"tG" }{TEXT -1 13 " in the fo rm " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 277 "" 0 "" {XPPEDIT 18 0 "u(x,t)=X(x)*T(t)" "6#/-%\"uG6$%\"xG%\"tG*&-%\"XG6#F'\"\"\"-%\"TG6#F (F-" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 79 "This general strategy, which seems to work with linear \+ second-order PDEs, gives" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 278 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "`T''`(t)/c^2/T(t)=`X''`(x)/X(x) " "6#/*(-%$T''G6#%\"tG\"\"\"*$%\"cG\"\"#!\"\"-%\"TG6#F(F-*&-%$X''G6#% \"xGF)-%\"XG6#F5F-" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 8 "that is," }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "q2 := q1/(c^2*U);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 4 "the " }{TEXT 256 14 "separated form" }{TEXT -1 139 " of the PDE wherein each side o f the equation is a function of only one of the variables. Since on t he left there is a function purely of " }{TEXT 261 1 "t" }{TEXT -1 41 ", and on the right, a function purely of " }{TEXT 262 1 "x" }{TEXT -1 44 ", and these two functions must be equal for " }{TEXT 257 3 "all " }{TEXT -1 11 " values of " }{TEXT 263 1 "x" }{TEXT -1 5 " and " } {TEXT 264 1 "t" }{TEXT -1 42 ", each must be equal to the same constan t." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 136 "Th is argument is generally not transparent to the novice. Consider the \+ following. Student A is asked to state a \"favorite function of " } {XPPEDIT 18 0 "t" "6#%\"tG" }{TEXT -1 56 ".\" Student B is asked to s tate a \"favorite function of " }{XPPEDIT 18 0 "x" "6#%\"xG" }{TEXT -1 73 ".\" Typically, the functions now written on the board are some thing like " }{XPPEDIT 18 0 "t^2" "6#*$%\"tG\"\"#" }{TEXT -1 5 " and \+ " }{XPPEDIT 18 0 "sin(x)" "6#-%$sinG6#%\"xG" }{TEXT -1 53 ". Student \+ C is asked to select a \"favorite value of " }{XPPEDIT 18 0 "t" "6#%\" tG" }{TEXT -1 51 ",\" and student D is asked for a \"favorite value of " }{XPPEDIT 18 0 "x" "6#%\"xG" }{TEXT -1 18 ".\" The functions " } {XPPEDIT 18 0 "t^2" "6#*$%\"tG\"\"#" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "sin(x)" "6#-%$sinG6#%\"xG" }{TEXT -1 36 " are now computed at value s such as " }{XPPEDIT 18 0 "t=1" "6#/%\"tG\"\"\"" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "x=Pi" "6#/%\"xG%#PiG" }{TEXT -1 42 ". The students th en realize that because " }{XPPEDIT 18 0 "1^2<>sin(Pi)" "6#0*$\"\"\"\" \"#-%$sinG6#%#PiG" }{TEXT -1 37 " = 0, the only way for a function of \+ " }{XPPEDIT 18 0 "t" "6#%\"tG" }{TEXT -1 39 " to equal, consistently, \+ a function of " }{XPPEDIT 18 0 "x" "6#%\"xG" }{TEXT -1 41 " is for the two functions to be the same " }{TEXT 281 8 "constant" }{TEXT -1 10 " function." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 36 "This constant is typically taken as " }{XPPEDIT 18 0 "lambda" " 6#%'lambdaG" }{TEXT -1 19 " and is called the " }{TEXT 282 9 "Bernoull i" }{TEXT -1 1 " " }{TEXT 283 10 "separation" }{TEXT -1 1 " " }{TEXT 284 8 "constant" }{TEXT -1 57 ". Hence, we have the two ordinary diff erential equations" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 261 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "`X''` (x)/X(x)=lambda" "6#/*&-%$X''G6#%\"xG\"\"\"-%\"XG6#F(!\"\"%'lambdaG" } }{PARA 0 "" 0 "" {TEXT -1 3 "and" }}{PARA 260 "" 0 "" {TEXT -1 1 " " } {XPPEDIT 18 0 "`T''`(t)/c^2/T(t)=lambda" "6#/*(-%$T''G6#%\"tG\"\"\"*$% \"cG\"\"#!\"\"-%\"TG6#F(F-%'lambdaG" }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{PARA 0 "" 0 "" {TEXT -1 61 "which, together with the relevant data, \+ form the two problems" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 279 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "`X''`(x)-lambda*X(x)=0" "6#/,&-%$X ''G6#%\"xG\"\"\"*&%'lambdaGF)-%\"XG6#F(F)!\"\"\"\"!" }{TEXT -1 1 " " } }{PARA 280 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "X(0)=X(L)" "6#/-%\"X G6#\"\"!-F%6#%\"LG" }{TEXT -1 5 " = 0 " }}{PARA 0 "" 0 "" {TEXT -1 3 " and" }}{PARA 281 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "`T''`(t)-c^2*l ambda*T(t)=0" "6#/,&-%$T''G6#%\"tG\"\"\"*(%\"cG\"\"#%'lambdaGF)-%\"TG6 #F(F)!\"\"\"\"!" }{TEXT -1 1 " " }}{PARA 282 "" 0 "" {TEXT -1 1 " " } {XPPEDIT 18 0 "`T'`(0)=0" "6#/-%#T'G6#\"\"!F'" }{TEXT -1 1 " " }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 112 "The firs t problem is a Sturm-Liouville eigenvalue problem, while the second is a modified initial value problem." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 118 "Obtaining the two separated ordinary dif ferential equations in Maple is not a simple task. For example, we be gin with" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "q3 := rhs(q2) = lambda;\nq4 := lhs(q2) = lambda;" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 26 "Then, re-arranging, we get" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 85 "q5 := lhs(X(x)*q3) - r hs(X(x)*q3) = 0;\nq6 := lhs(c^2*T(t)*q4) - rhs(c^2*T(t)*q4) = 0;" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 167 "In effect, we did all the work. Maple really only playe d the role of bookkeeper, because it does not have any built-in code f or performing manipulations of this type." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 86 "The solution of this Sturm-Liouvil le eigenvalue problem, obtained in Section 16.1, is " }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 283 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "X[n ](x)=B[n]*sin(n*Pi*x/L)" "6#/-&%\"XG6#%\"nG6#%\"xG*&&%\"BG6#F(\"\"\"-% $sinG6#**F(F/%#PiGF/F*F/%\"LG!\"\"F/" }{TEXT -1 1 " " }}{PARA 0 "" 0 " " {TEXT -1 16 "with eigenvalues" }}{PARA 284 "" 0 "" {TEXT -1 1 " " } {XPPEDIT 18 0 "lambda[n]=-n^2*Pi^2/L^2,n=1,2,`...`" "6&/&%'lambdaG6#% \"nG,$*(F'\"\"#%#PiGF**$%\"LGF*!\"\"F./F'\"\"\"F*%$...G" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 107 "Aga in, Maple presently does not have the tools with which to construct th ese solutions, so we enter them as" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "Xn := B[n]*sin(n*Pi*x/L);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 25 "with eigenvalues given by" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "lambda[n] := -(n*Pi/ L)^2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 4 "for " }{XPPEDIT 18 0 "n=1,2" "6$/%\"nG\"\"\"\"\" #" }{TEXT -1 6 ", ...." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 18 "For each value of " }{XPPEDIT 18 0 "n" "6#%\"nG" } {TEXT -1 17 ", the function T(" }{XPPEDIT 18 0 "t" "6#%\"tG" }{TEXT -1 21 ") is the solution of " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 285 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "`T''`(t)+c^2" "6#,&-%$T''G6 #%\"tG\"\"\"*$%\"cG\"\"#F(" }{TEXT -1 1 " " }{XPPEDIT 18 0 "n^2*Pi^2/L ^2" "6#*(%\"nG\"\"#%#PiGF%*$%\"LGF%!\"\"" }{TEXT -1 1 " " }{XPPEDIT 18 0 "T(t)=0" "6#/-%\"TG6#%\"tG\"\"!" }{TEXT -1 1 " " }}{PARA 286 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "`T'`(0)=0" "6#/-%#T'G6#\"\"!F'" } {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 11 "so we write" }}{PARA 287 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "T[n](t)=C[n]*cos(c*n*Pi*t/L )" "6#/-&%\"TG6#%\"nG6#%\"tG*&&%\"CG6#F(\"\"\"-%$cosG6#*,%\"cGF/F(F/%# PiGF/F*F/%\"LG!\"\"F/" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 69 "a solution we obtain in Maple by writ ing the differential equation as" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "q7 := subs(lambda=lambda[n], q6);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 17 "then solving with" }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "q8 := dsolve(\{q7,D(T)( 0)=0\},T(t));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 30 "Emphasizing the dependence on " } {XPPEDIT 18 0 "n" "6#%\"nG" }{TEXT -1 10 ", we write" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "Tn := sub s(_C2=C[n],rhs(q8));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }} }{EXCHG {PARA 0 "" 0 "" {TEXT -1 1 "(" }{TEXT 267 7 "Caution" }{TEXT -1 110 ": If Maple uses _C1 instead of _C2 for the constant of integr ation, please adjust the worksheet accordingly.)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 24 "A single eigenfunction, \+ " }{XPPEDIT 18 0 "u[n](x,t)" "6#-&%\"uG6#%\"nG6$%\"xG%\"tG" }{TEXT -1 26 " is therefore the product " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 288 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "u[n](x,t)=B[n]*sin(n* Pi*x/L)*C[n]*cos(c*n*Pi*t/L)" "6#/-&%\"uG6#%\"nG6$%\"xG%\"tG**&%\"BG6# F(\"\"\"-%$sinG6#**F(F0%#PiGF0F*F0%\"LG!\"\"F0&%\"CG6#F(F0-%$cosG6#*,% \"cGF0F(F0F5F0F+F0F6F7F0" }{TEXT -1 1 " " }}}{EXCHG {PARA 294 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 8 "that is," }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "Un := Xn *Tn;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 83 "which contains two constants where only one is ne eded. Hence, replace the product " }{XPPEDIT 18 0 "B[n]*C[n]" "6#*&&% \"BG6#%\"nG\"\"\"&%\"CG6#F'F(" }{TEXT -1 26 " with the single constant " }{XPPEDIT 18 0 "b[n]" "6#&%\"bG6#%\"nG" }{TEXT -1 4 " via" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "u n := algsubs(B[n]*C[n]=b[n],Un);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 26 "The set of ei gensolutions " }{XPPEDIT 18 0 "\{u[n]\}" "6#<#&%\"uG6#%\"nG" }{TEXT -1 336 " can be thought of as the analog of a fundamental set for a li near ODE. The general solution of a linear ODE is a linear combinatio n of all the members of the fundamental set. Similarly, the most gene ral solution of the linear PDE is a linear combination of all the memb ers of its \"fundamental set,\" namely, the set of eigensolutions " } {XPPEDIT 18 0 "\{u[n](x,t)\}" "6#<#-&%\"uG6#%\"nG6$%\"xG%\"tG" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 94 "Since there are an infinite number of eigensolutions, our general \+ solution is the infinite sum" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 289 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "u(x,t)=Sum(u[n](x,t),n=1..i nfinity)" "6#/-%\"uG6$%\"xG%\"tG-%$SumG6$-&F%6#%\"nG6$F'F(/F/;\"\"\"%) infinityG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "Sum(b[n]*sin(n*Pi*x/L)*co s(c*n*Pi*t/L),n=1..infinity)" "6#-%$SumG6$*(&%\"bG6#%\"nG\"\"\"-%$sinG 6#**F*F+%#PiGF+%\"xGF+%\"LG!\"\"F+-%$cosG6#*,%\"cGF+F*F+F0F+%\"tGF+F2F 3F+/F*;F+%)infinityG" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 8 "th at is," }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "U := Sum(un,n=1..infinity);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 80 "The fina l condition to be applied in this BVP is the initial shape requirement , " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 290 "" 0 "" {TEXT -1 1 " \+ " }{XPPEDIT 18 0 "u(x,0)=Sum(b[n]*sin(n*Pi*x/L),n=1..infinity)" "6#/-% \"uG6$%\"xG\"\"!-%$SumG6$*&&%\"bG6#%\"nG\"\"\"-%$sinG6#**F0F1%#PiGF1F' F1%\"LG!\"\"F1/F0;F1%)infinityG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "f(x )" "6#-%\"fG6#%\"xG" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 8 "tha t is," }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "simplify(subs(t=0,U)) = f(x);" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 17 "The co efficients " }{XPPEDIT 18 0 "b[n]" "6#&%\"bG6#%\"nG" }{TEXT -1 73 " mu st therefore be the Fourier sine series coefficients for the function \+ " }{XPPEDIT 18 0 "f(x)" "6#-%\"fG6#%\"xG" }{TEXT -1 96 ". There is no other choice for these constants if this final condition is to be sat isfied. If " }{XPPEDIT 18 0 "f(x)" "6#-%\"fG6#%\"xG" }{TEXT -1 136 " \+ is to be equal to a sum of sine terms, then the coefficients in that s um must be the Fourier sine series coefficients. Therefore, the " } {XPPEDIT 18 0 "b[n]" "6#&%\"bG6#%\"nG" }{TEXT -1 53 "'s are determined by the Fourier sine series formulas" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 263 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "b[n]=2/L" "6#/&%\"b G6#%\"nG*&\"\"#\"\"\"%\"LG!\"\"" }{TEXT -1 1 " " }{XPPEDIT 18 0 "Int(f (x)*sin(n*Pi*x/L),x=0..L)" "6#-%$IntG6$*&-%\"fG6#%\"xG\"\"\"-%$sinG6#* *%\"nGF+%#PiGF+F*F+%\"LG!\"\"F+/F*;\"\"!F2" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 17 "and the solution " }{XPPEDIT 18 0 "u(x,t)" "6#-%\"uG6$%\"xG%\"tG" }{TEXT -1 32 " is given by the in finite series" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 264 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "u(x,t)=Sum(b[n]*sin(n*Pi*x/L)*cos(c*n*P i*t/L),n=1..infinity)" "6#/-%\"uG6$%\"xG%\"tG-%$SumG6$*(&%\"bG6#%\"nG \"\"\"-%$sinG6#**F0F1%#PiGF1F'F1%\"LG!\"\"F1-%$cosG6#*,%\"cGF1F0F1F6F1 F(F1F7F8F1/F0;F1%)infinityG" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 61 "This infinite sum, along with the defining inte grals for the " }{XPPEDIT 18 0 "b[n]" "6#&%\"bG6#%\"nG" }{TEXT -1 83 " , constitutes the formal solution to the given BVP for the finite pluc ked string. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 12 "Exampl e 24.3" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 19 "A string of length " }{XPPEDIT 18 0 "L=Pi" "6#/%\"LG%#PiG " }{TEXT -1 23 " has its ends fixed at " }{XPPEDIT 18 0 "x=0" "6#/%\"x G\"\"!" }{TEXT -1 6 ", and " }{XPPEDIT 18 0 "x=Pi" "6#/%\"xG%#PiG" } {TEXT -1 7 " on an " }{TEXT 266 1 "x" }{TEXT -1 80 "-axis. The string is under tension, so that when it is given the initial shape " } {XPPEDIT 18 0 "f(x) = x*(Pi-x)/10" "6#/-%\"fG6#%\"xG*(F'\"\"\",&%#PiGF )F'!\"\"F)\"#5F," }{TEXT -1 115 ", as shown in Figure 24.8 (below), an d released, it oscillates. We solve for the motion of the string at a ny time " }{XPPEDIT 18 0 "t*`>`*0" "6#*(%\"tG\"\"\"%\">GF%\"\"!F%" } {TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 59 "The associated BVP that models the motion of this string \+ is" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 265 "" 0 "" {TEXT -1 1 " \+ " }{XPPEDIT 18 0 "u[tt](x,t) = c^2*u[xx](x,t)" "6#/-&%\"uG6#%#ttG6$%\" xG%\"tG*&%\"cG\"\"#-&F&6#%#xxG6$F*F+\"\"\"" }{TEXT -1 7 ", t > 0" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 266 "" 0 "" {TEXT -1 1 " " } {XPPEDIT 18 0 "u(0,t) = 0" "6#/-%\"uG6$\"\"!%\"tGF'" }}{PARA 267 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "u(pi,t)=0" "6#/-%\"uG6$%#piG%\"tG\" \"!" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 268 "" 0 "" {TEXT -1 28 " " }{XPPEDIT 18 0 "u(x,0)=f(x)" "6#/-%\"uG6 $%\"xG\"\"!-%\"fG6#F'" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "x*(Pi-x)/10" "6#*(%\"xG\"\"\",&%#PiGF%F$!\"\"F%\"#5F(" }{TEXT -1 1 " " }}{PARA 269 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "u[t](x,0)=0" "6#/-&%\"uG6#%\"tG 6$%\"xG\"\"!F+" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 23 "The solution, given by " }}{PARA 291 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "u(x,t)=Sum(b[n]*sin(n*Pi*x/L)*cos(c*n*Pi*t/L),n=1.. infinity)" "6#/-%\"uG6$%\"xG%\"tG-%$SumG6$*(&%\"bG6#%\"nG\"\"\"-%$sinG 6#**F0F1%#PiGF1F'F1%\"LG!\"\"F1-%$cosG6#*,%\"cGF1F0F1F6F1F(F1F7F8F1/F0 ;F1%)infinityG" }}{PARA 0 "" 0 "" {TEXT -1 3 "and" }}{PARA 292 "" 0 " " {TEXT -1 1 " " }{XPPEDIT 18 0 "b[n]=2/L" "6#/&%\"bG6#%\"nG*&\"\"#\" \"\"%\"LG!\"\"" }{TEXT -1 1 " " }{XPPEDIT 18 0 "Int(f(x)*sin(n*Pi*x/L) ,x=0..L)" "6#-%$IntG6$*&-%\"fG6#%\"xG\"\"\"-%$sinG6#**%\"nGF+%#PiGF+F* F+%\"LG!\"\"F+/F*;\"\"!F2" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 66 "is implemented in Maple as follows. The initial s hape function is" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "f := x*(Pi-x)/10;" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 42 "and Figure \+ 24.8 (which shows its graph) is" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 120 "plot(f,x=0..Pi,color=black, scaling=constrained, xtickmarks=3, ytickmarks=2, labels=[x,u], labelf ont=[TIMES,ITALIC,12]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 " " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 50 "The Fourier coefficients are \+ given by the integral" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "q := (2/Pi)*Int(f*sin(n*x),x=0..Pi) ;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 14 "whose value is" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "b := simplify(value(q));" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 37 "A peek at the first ten coefficients " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "seq(subs(n= k,b),k=1..10);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 180 "reveals that the odd-indexed term s \"survive,\" the even-indexed terms are zero. Moreover, the coeffic ients rapidly converge to zero for this function. In fact, they go to zero as " }{XPPEDIT 18 0 "1/n^3" "6#*&\"\"\"F$*$%\"nG\"\"$!\"\"" } {TEXT -1 80 ", a rapid decrease, as the following conversion to floati ng point numbers shows." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "seq(evalf(subs(n=k,b)),k=1..10);" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 70 "Hence, a very few terms will be needed to get a good ap proximation to " }{XPPEDIT 18 0 "u(x,t)" "6#-%\"uG6$%\"xG%\"tG" } {TEXT -1 86 ". We can test this by examining how many terms it takes \+ to get the Fourier series of " }{XPPEDIT 18 0 "f(x)" "6#-%\"fG6#%\"xG " }{TEXT -1 16 " to approximate " }{XPPEDIT 18 0 "f(x)" "6#-%\"fG6#%\" xG" }{TEXT -1 70 " well. The first two distinct partial sums in the F ourier series for " }{XPPEDIT 18 0 "f(x)" "6#-%\"fG6#%\"xG" }{TEXT -1 4 " are" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "p1 := sum(b*sin(n*x),n=1..1);\np3 := sum(b*sin(n*x),n =1..3);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 5 "with " }{XPPEDIT 18 0 "f(x)" "6#-%\"fG6#% \"xG" }{TEXT -1 18 " (thin black) and " }{XPPEDIT 18 0 "p[1]" "6#&%\"p G6#\"\"\"" }{TEXT -1 41 " (thick red) shown in Figure 24.9, below." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 149 "plot([f,p1],x=0..Pi,color=[black,red], thickness=[1,3], scaling =constrained, xtickmarks=3, ytickmarks=2, labels=[x,u], labelfont=[TI MES,ITALIC,12]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 58 "This experiment suggests that two \+ terms of the series for " }{XPPEDIT 18 0 "u(x,t)" "6#-%\"uG6$%\"xG%\"t G" }{TEXT -1 54 " might yield a reasonably good approximation, so writ e" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 293 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "U=sum(b[n]*sin(n*x)*cos(c*n*t),n=1..3)" "6#/%\"UG-%$su mG6$*(&%\"bG6#%\"nG\"\"\"-%$sinG6#*&F,F-%\"xGF-F--%$cosG6#*(%\"cGF-F,F -%\"tGF-F-/F,;F-\"\"$" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "4/135/Pi" "6# *(\"\"%\"\"\"\"$N\"!\"\"%#PiGF'" }{TEXT -1 1 " " }{XPPEDIT 18 0 "``(27 *sin(x)*cos(c*t)+sin(3*x)*cos(3*c*t))" "6#-%!G6#,&*(\"#F\"\"\"-%$sinG6 #%\"xGF)-%$cosG6#*&%\"cGF)%\"tGF)F)F)*&-F+6#*&\"\"$F)F-F)F)-F/6#*(F8F) F2F)F3F)F)F)" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 8 "that is," }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "U := sum(b*sin(n*x)*cos(c*n*t),n=1..3);" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 42 "as an \+ approximation to the solution. Set " }{XPPEDIT 18 0 "c=1" "6#/%\"cG\" \"\"" }{TEXT -1 10 ", that is," }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "U1 := subs(c=1,U);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 72 "to obtain the graph of the solution surface seen in Figur e 24.10, below." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 162 "plot3d(U1, x=0..Pi, t=0..4*Pi, axes=frame, la bels=[x,`t `,`u `], labelfont=[TIMES,ITALIC,12], style=hidden, color =red, tickmarks=[3,6,3], orientation=[-45,60]);" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 19 "The pl ane sections " }{XPPEDIT 18 0 "t=t[k]" "6#/%\"tG&F$6#%\"kG" }{TEXT -1 85 " are snapshots in time, freezing the physical motion of the string for each value of " }{XPPEDIT 18 0 "t[k]" "6#&%\"tG6#%\"kG" }{TEXT -1 167 ". Exhibiting these shapshots in succession forms a movie, or \+ animation, of the physical motion of the string. This animation is gi ven by the following Maple command." }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 151 "animate(U1,x=0..Pi,t=0..2 *Pi, frames=60, color=black, scaling=constrained, xtickmarks=3, ytickm arks=2, labels=[x,`u `], labelfont=[TIMES,ITALIC,12]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 105 "As the string moves in space-time, its history generates a sur face, a small part of which we now animate." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 193 "F := z -> plot3d(U1, x=0..Pi, t=0..z):\n display3d([seq(F(Pi/10*k),k=1..30)],insequence=true, axes=boxed, label s=[x,t,u], labelfont=[TIMES,ITALIC,12], style=hidden, color=red, tickm arks=[3,5,3]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}} {MARK "1" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }