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{TEXT -1 40 "Partial Differential Equa tions PowerTool" }}{PARA 19 "" 0 "" {TEXT -1 16 "by Dr. Jim Herod" }}} {PARA 256 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT 257 47 "Sect ion 5.3: D'Alembert's Solution on Intervals" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT 258 30 "Maple Packages for Sec tion 5.3" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart:" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(plots):" }{TEXT -1 0 " " }}}{PARA 0 "" 0 "" {TEXT -1 138 "The remaining part of this portion \+ of Section 5.3 recreates the methods to make extensions. Details for t hese techniques are discussed in " }{TEXT 380 23 "Section 2.2: Extensi ons" }{TEXT -1 27 ". Recall that the function " }{TEXT 379 2 "HS" } {TEXT -1 104 " defined below is very closely related to what is called the Heaviside function. It has the value 1 for " }{TEXT 390 1 "x" } {TEXT -1 22 " > 0, the value 0 for " }{TEXT 389 1 "x" }{XPPEDIT 18 0 " ` ` < ` `;" "6#2%\"~GF$" }{TEXT -1 118 "0. The difference from the the Heaviside is structural and conceptual. We discussed this difference \+ in more detail in " }{TEXT 381 23 "Section 2.2: Extensions" }{TEXT -1 23 " in the portion titled " }{TEXT 382 16 "Unassisted Maple" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "HS:=x->(1+signum(x))/2;" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "HS(1); HS(0); HS(-1);" }}} {PARA 0 "" 0 "" {TEXT -1 33 "The next two lines have as input " } {TEXT 391 1 "x" }{TEXT -1 84 " and a function defined for x > 0. They \+ produce an even function or an odd function." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "evnf:=(f,x)->HS(x)*f(x)+HS(-x)*f(-x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "oddf:=(f,x)->HS(x)*f(x)-HS(-x)*f(-x );" }}}{PARA 0 "" 0 "" {TEXT -1 67 "We will also need to extend functi ons from intervals of the form [-" }{TEXT 424 1 "L" }{TEXT -1 2 ", " } {TEXT 425 1 "L" }{TEXT -1 94 "] to the entire real line. Recall that w e have also referred to a procedure for doing this in " }{TEXT 377 23 "Section 2.2: Extensions" }{TEXT -1 1 "." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 170 "PeriodicExtender:=proc(f,d::range)\nsubs( \{'F' = f, 'L'=lhs(d),\n 'D'=rhs(d)-lhs(d)\},\nproc(x::algebraic) local y ;\n y:=floor((x-L)/D);\n F(x-y*D);\nend)\nend proc:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 448 "In Section 5.2, we derived d'Alembert' s solution for the wave equation on the entire real line. We will now \+ get the solution for the wave equation on finite intervals. One way to think about the process is this: suppose that the initial functions a re given on a finite interval. Extend these functions appropriately to functions on the entire real line and then use the techniques of the \+ previous section.. We make these extensions in this Section." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 81 "Hereafter we ac knowledge from Section 5.2 that d'Alembert's solution has the form" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 22 " \+ u(t, x) = " }{XPPEDIT 18 0 "psi(x+c*t);" "6#-%$psiG6#,&%\"xG\"\"\"* &%\"cGF(%\"tGF(F(" }{TEXT -1 3 " + " }{XPPEDIT 18 0 "phi(x-c*t);" "6#- %$phiG6#,&%\"xG\"\"\"*&%\"cGF(%\"tGF(!\"\"" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 189 "We have not used any information about boundary conditions to arrive at this result. W e only used information about the initial conditions. We review the si tuation with the assumption that " }{TEXT 259 1 "u" }{TEXT -1 3 "(0," }{TEXT 260 1 "x" }{TEXT -1 4 ") = " }{TEXT 261 1 "f" }{TEXT -1 1 "(" } {TEXT 262 1 "x" }{TEXT -1 6 ") and " }{XPPEDIT 18 0 "u[t];" "6#&%\"uG6 #%\"tG" }{TEXT -1 10 " (0, x) = " }{TEXT 263 1 "g" }{TEXT -1 1 "(" } {TEXT 264 1 "x" }{TEXT -1 2 ")." }}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 1 " " }{TEXT 316 40 "A general solution for the wave equation" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 171 " We use th e ideas of the previous Section 5.2 to construct a general solution fo r the wave equation. If necessary, review that section. The definition of u depends on " }{TEXT 269 1 "f" }{TEXT -1 2 ", " }{TEXT 268 1 "g" }{TEXT -1 2 ", " }{TEXT 267 1 "t" }{TEXT -1 2 ", " }{TEXT 266 1 "x" } {TEXT -1 6 ", and " }{TEXT 265 1 "c" }{TEXT -1 43 ". These are all the items used to make the " }{TEXT 270 1 "u" }{TEXT -1 1 "." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 87 "u:=(f,g,t,x,c)->(f(x+c*t)+f(x-c*t)) /2\n +int(g(s),s=x-c*t..x+c*t)/(2*c);" }}}{PARA 0 "" 0 "" {TEXT -1 91 " To illustrate this general solution, we take a \+ particular example. We will take c = 1." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "f:=x->4*sin(Pi*x);\ng:=x->cos(x);" }}}{PARA 0 "" 0 " " {TEXT -1 66 " Here is an evaluation and animation of this genera l solution." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "u(f,g,t,x,1); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "animate(u(f,g,t,x,1),x= -2..2,t=0..4, frames=30);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT 317 21 "Half Infinite Strings" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 43 "The solution just given is for all number s " }{TEXT 271 1 "x" }{TEXT -1 105 ". It is as though we had a string \+ that was infinite in both directions. That is, this solution holds for " }{XPPEDIT 18 0 "-infinity < x;" "6#2,$%)infinityG!\"\"%\"xG" } {TEXT -1 5 " and " }{XPPEDIT 18 0 "x < infinity;" "6#2%\"xG%)infinityG " }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 88 " Our next idea is to suppose that we want to solve the equation on the interval [0, " } {XPPEDIT 18 0 "infinity;" "6#%)infinityG" }{TEXT -1 123 " ). This woul d model a long string tied down at one end. We begin by imposing only \+ one boundary condition. We suppose that " }{TEXT 272 1 "x" }{TEXT -1 73 " > 0. With this assumption, we have a boundary condition at the le ft end." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 20 "Boundary Condition: " }{TEXT 274 1 "u" }{TEXT -1 1 "(" }{TEXT 273 1 "t" }{TEXT -1 9 ", 0) = 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 20 "Initial Conditions: " }{TEXT 275 1 "u" }{TEXT -1 4 "(0, " }{TEXT 276 1 "x" }{TEXT -1 4 ") = " }{TEXT 277 1 "f" } {TEXT -1 1 "(" }{TEXT 278 1 "x" }{TEXT -1 6 ") and " }{XPPEDIT 18 0 "u [t];" "6#&%\"uG6#%\"tG" }{TEXT -1 4 "(0, " }{TEXT 279 1 "x" }{TEXT -1 4 ") = " }{TEXT 280 1 "g" }{TEXT -1 1 "(" }{TEXT 281 1 "x" }{TEXT -1 6 ") for " }{TEXT 282 1 "x" }{TEXT -1 5 " > 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 74 "With no initial condition s or boundary conditions, we have concluded that " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 22 " " } {TEXT 283 1 "u" }{TEXT -1 1 "(" }{TEXT 284 1 "t" }{TEXT -1 2 ", " } {TEXT 285 1 "x" }{TEXT -1 4 ") = " }{XPPEDIT 18 0 "psi(x+c*t);" "6#-%$ psiG6#,&%\"xG\"\"\"*&%\"cGF(%\"tGF(F(" }{TEXT -1 3 " + " }{XPPEDIT 18 0 "phi(x-c*t);" "6#-%$phiG6#,&%\"xG\"\"\"*&%\"cGF(%\"tGF(!\"\"" } {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 3 "as " }{TEXT 286 1 "t" }{TEXT -1 81 " increases. With initi al conditions, as in Section 5.2 and the above, we got that" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 21 " \+ " }{TEXT 287 1 "u" }{TEXT -1 1 "(" }{TEXT 288 1 "t" }{TEXT -1 2 ", " }{TEXT 289 1 "x" }{TEXT -1 4 ") = " }{XPPEDIT 18 0 "(f(x+c*t)+f(x -c*t))/2;" "6#*&,&-%\"fG6#,&%\"xG\"\"\"*&%\"cGF*%\"tGF*F*F*-F&6#,&F)F* *&F,F*F-F*!\"\"F*F*\"\"#F2" }{TEXT -1 3 " + " }{XPPEDIT 18 0 "(G(x+c*t )-G(x-c*t))/2;" "6#*&,&-%\"GG6#,&%\"xG\"\"\"*&%\"cGF*%\"tGF*F*F*-F&6#, &F)F**&F,F*F-F*!\"\"F2F*\"\"#F2" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 32 "where G is an antiderivat ive of " }{TEXT 290 1 "g" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 44 " As we have said, we are suppo sing that " }{TEXT 291 1 "x" }{TEXT -1 47 " > 0. The solution is not d efined for negative " }{TEXT 392 1 "x" }{TEXT -1 38 ". However, we mus t do an extension of " }{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 13 " because, as " }{TEXT 292 1 "t" }{TEXT -1 23 " increases, the number \+ " }{TEXT 293 3 "c t" }{TEXT -1 13 " will exceed " }{TEXT 294 1 "x" } {TEXT -1 36 " and we will need the definition of " }{XPPEDIT 18 0 "phi ;" "6#%$phiG" }{TEXT -1 25 " for the negative number " }{TEXT 318 2 " \+ x" }{TEXT -1 3 " - " }{TEXT 319 3 "c t" }{TEXT -1 8 ". While " }{TEXT 383 1 "x" }{TEXT -1 14 " is positive, " }{TEXT 384 1 "x" }{TEXT -1 3 " - " }{TEXT 385 3 "c t" }{TEXT -1 138 " may be negative. Happily, the re is information available through the boundary conditions which comp els how the extension should be made." }}{PARA 0 "" 0 "" {TEXT -1 36 " The first boundary condition is" }}{PARA 0 "" 0 "" {TEXT -1 16 " \+ 0 = u(" }{TEXT 295 1 "t" }{TEXT -1 7 ", 0) = " }{XPPEDIT 18 0 "psi;" "6#%$psiG" }{TEXT -1 3 " ( " }{TEXT 296 3 "c t" }{TEXT -1 5 " ) + " }{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 4 " (- " }{TEXT 297 3 "c t" }{TEXT -1 5 " ) = " }{XPPEDIT 18 0 "(f(c*t)+G(c*t))/2;" "6#*&, &-%\"fG6#*&%\"cG\"\"\"%\"tGF*F*-%\"GG6#*&F)F*F+F*F*F*\"\"#!\"\"" } {TEXT -1 6 " + " }{XPPEDIT 18 0 "(f(-c*t)-G(-c*t))/2;" "6#*&,&-%\"f G6#,$*&%\"cG\"\"\"%\"tGF+!\"\"F+-%\"GG6#,$*&F*F+F,F+F-F-F+\"\"#F-" } {TEXT -1 3 " ." }}{PARA 0 "" 0 "" {TEXT -1 6 " Thus," }}{PARA 0 "" 0 "" {TEXT -1 19 " 0 = " }{TEXT 298 1 "f" }{TEXT -1 2 "( \+ " }{TEXT 299 3 "c t" }{TEXT -1 5 " ) + " }{TEXT 300 1 "f" }{TEXT -1 4 "( - " }{TEXT 301 3 "c t" }{TEXT -1 8 " ) + G( " }{TEXT 302 3 "c t" } {TEXT -1 10 " ) - G( - " }{TEXT 303 3 "c t" }{TEXT -1 3 " )." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 25 "Since this hold s for all " }{TEXT 304 1 "f" }{TEXT -1 13 " and for all " }{TEXT 305 1 "g" }{TEXT -1 22 ", then it must be that" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 10 " " }{TEXT 306 1 "f" } {TEXT -1 2 "( " }{TEXT 307 3 "c t" }{TEXT -1 7 " ) = - " }{TEXT 308 1 "f" }{TEXT -1 4 "( - " }{TEXT 309 3 "c t" }{TEXT -1 13 " ) and G( \+ " }{TEXT 310 3 "c t" }{TEXT -1 10 " ) = G( - " }{TEXT 311 3 "c t" } {TEXT -1 11 " ) for all " }{TEXT 312 1 "t" }{TEXT -1 5 " > 0." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 6 "Thus, " } {TEXT 313 1 "f" }{TEXT -1 66 " should have an odd extension and G shou ld have an even extension." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT 320 30 "Making even and odd extensions" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 181 "It's clear tha t in order to illustrate these ideas, we are going to need to be able \+ to make even and odd extensions for functions. There is a procedure fo r doing this introduced in " }{TEXT 393 22 "Section 2.2 Extensions" } {TEXT -1 93 " and recalled at the beginning of this Section. We illust rate this extension here. We define " }{TEXT 314 1 "f" }{TEXT -1 27 " \+ and g on the interval [0, " }{XPPEDIT 18 0 "infinity;" "6#%)infinityG " }{TEXT -1 50 "). Then we construct G. Finally we draw graphs of " } {TEXT 315 1 "f" }{TEXT -1 81 " and G, and we draw graphs of the odd ex tension of f and the even extension of G." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "f:=x-> (signum(x-2)-signum(x-3))*(x-3)*(x-2);\nplot(f (x),x=-6..6);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 87 "g:=x->(sig num(x-2)-signum(x-3));\nG:=x->int(g(s),s=0..x);\nplot([x,G(x),x=0..6], x=-6..6);" }}}{PARA 0 "" 0 "" {TEXT -1 43 "Having executed the top lin es contained in " }{TEXT 386 30 "Maple Packages for Section 5.3" } {TEXT -1 24 " we draw the extensions." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "plot(oddf(f,x),x=-6..6);" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 24 "plot(evnf(G,x),x=-6..6);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 58 "Having reviewed t his part we proceed with the development." }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT 321 64 "Comparing Infinite Int erval and Half-Infinite Interval Problems." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 141 "It is wise to contrast this \" half\" infinite string with the infinite string of the previous module . Let's do one example twice: first we let " }{TEXT 394 1 "x" }{TEXT -1 87 " range over all numbers and use just the original d'Alembert so lution, and then we let " }{TEXT 395 1 "x" }{TEXT -1 47 " only be posi tive and use the odd extension of " }{TEXT 396 1 "f" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 387 65 "Exampl e for Comparison: No Boundary Conditions, Infinite Interval" }}{PARA 0 "" 0 "" {TEXT -1 5 " " }}{PARA 0 "" 0 "" {TEXT -1 15 "Here's f a nd g." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "f:=x->-(HS(2*Pi-x)- HS(Pi-x))*sin(x);\ng:=x->0;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "plot(f(x),x=0..10);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 115 "Now, we solve the PDE with initial conditions and no boundary conditions. We use the d'Alembert solution, and take " } {TEXT 397 1 "c" }{TEXT -1 35 " = 1. The function G will be zero." }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "u:=(t,x)->(f(x+t)+f(x-t))/2; " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 80 "plot3d(u(t,x),x=-10..10 ,t=0..5,orientation=[-105,50],axes=normal, grid=[50,50]);" }}}{PARA 0 "" 0 "" {TEXT -1 85 "What we expect to see is that the wave splits apa rt and moves off in both directions." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "animate(u(t,x),x=-10..20,t=0..15, frames=40, numpoint s=200);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 388 70 "Example for Compari son: One Boundary Condition, Half Infinite Interval" }}{PARA 0 "" 0 " " {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 72 "Next, we solve the PDE with initial conditions and boundary condition u(" }{TEXT 398 1 "t" } {TEXT -1 15 ", 0) = 0, same " }{TEXT 399 1 "f" }{TEXT -1 7 " and G." } }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "u:=(t,x)->(oddf(f,x+t)+oddf (f,x-t))/2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 95 "plot3d(u(t,x ),x=0..4*Pi,t=0..10,orientation=[-35,45],\n axes=normal, \+ grid=[50,50]);" }}}{PARA 0 "" 0 "" {TEXT -1 116 "In this example, wher e the boundary is held at zero at the left end, we expect the wave to \+ bounce off the left side." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "animate(u(t,x),x=0..4*Pi,t=0..10,numpoints=75,color=RED, frames=40); " }}}{PARA 0 "" 0 "" {TEXT -1 12 "Pretty nice!" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 110 "The problem gets more co mplicated if both ends of the string are fixed. This is the next stage of development." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT 322 14 "Finite Strings" }} {PARA 0 "" 0 "" {TEXT -1 57 " If we have a finite string, we have \+ an end point at " }{TEXT 323 1 "x" }{TEXT -1 12 " = 0 and at " }{TEXT 324 1 "x" }{TEXT -1 3 " = " }{TEXT 325 1 "L" }{TEXT -1 30 ". We have u sed information at " }{TEXT 326 1 "x" }{TEXT -1 100 " = 0. We have nev er used the other end point. At that end point, we can get in a simila r manner that" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 12 " 0 = " }{TEXT 327 1 "u" }{TEXT -1 3 "(t," }{TEXT 328 1 "L" }{TEXT -1 4 ") = " }{TEXT 329 1 "f" }{TEXT -1 2 "( " }{TEXT 330 1 "L" }{TEXT -1 3 " + " }{TEXT 331 3 "c t" }{TEXT -1 5 " ) + " } {TEXT 332 1 "f" }{TEXT -1 2 "( " }{TEXT 333 1 "L" }{TEXT -1 4 " - " } {TEXT 334 3 "c t" }{TEXT -1 8 " ) + G( " }{TEXT 335 1 "L" }{TEXT -1 3 " + " }{TEXT 336 3 "c t" }{TEXT -1 8 " ) - G( " }{TEXT 337 1 "L" } {TEXT -1 3 " - " }{TEXT 338 3 "c t" }{TEXT -1 3 " )," }}{PARA 0 "" 0 " " {TEXT -1 3 "and" }}{PARA 0 "" 0 "" {TEXT -1 10 " " }{TEXT 339 1 "f" }{TEXT -1 2 "( " }{TEXT 340 1 "L" }{TEXT -1 3 " + " }{TEXT 341 3 "c t" }{TEXT -1 7 " ) = - " }{TEXT 342 1 "f" }{TEXT -1 2 "( " } {TEXT 343 1 "L" }{TEXT -1 3 " - " }{TEXT 344 3 "c t" }{TEXT -1 13 " ) \+ and G( " }{TEXT 345 1 "L" }{TEXT -1 3 " + " }{TEXT 346 3 "c t" } {TEXT -1 8 " ) = G( " }{TEXT 347 1 "L" }{TEXT -1 3 " - " }{TEXT 348 3 "c t" }{TEXT -1 11 " ) for all " }{TEXT 349 1 "t" }{TEXT -1 5 " > 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 15 "The odd ness of " }{TEXT 350 1 "f" }{TEXT -1 35 " and evenness of G changes th ese to" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 10 " " }{TEXT 351 1 "f" }{TEXT -1 2 "( " }{TEXT 352 1 "L" } {TEXT -1 3 " + " }{TEXT 353 3 "c t" }{TEXT -1 5 " ) = " }{TEXT 354 1 " f" }{TEXT -1 4 "(- " }{TEXT 355 1 "L" }{TEXT -1 3 " + " }{TEXT 356 3 "c t" }{TEXT -1 13 " ) and G( " }{TEXT 357 1 "L" }{TEXT -1 3 " + " }{TEXT 358 3 "c t" }{TEXT -1 10 " ) = G( - " }{TEXT 359 1 "L" }{TEXT -1 3 " + " }{TEXT 360 3 "c t" }{TEXT -1 11 " ) for all " }{TEXT 361 1 "t" }{TEXT -1 5 " > 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 9 "That is, " }{TEXT 362 1 "f" }{TEXT -1 21 " and G have period 2 " }{TEXT 363 1 "L" }{TEXT -1 44 ". Now we know how to make t he extensions of " }{TEXT 364 1 "f" }{TEXT -1 27 " and of G so we can \+ define " }{TEXT 365 1 "u" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 87 " We work a first easy example. Then, we will do a series of mo re interesting ones. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 5 "Take " }{TEXT 366 1 "c" }{TEXT -1 35 " = 1 here. For th e first example, " }{XPPEDIT 18 0 "f(x) <> 0;" "6#0-%\"fG6#%\"xG\"\"! " }{TEXT -1 5 " and " }{TEXT 367 1 "g" }{TEXT -1 1 "(" }{TEXT 368 1 "x " }{TEXT -1 21 ") = 0. Let us choose " }{TEXT 369 1 "f" }{TEXT -1 1 "( " }{TEXT 370 1 "x" }{TEXT -1 9 ") = sin( " }{TEXT 371 1 "x" }{TEXT -1 24 " ) on the interval [ 0, " }{XPPEDIT 18 0 "pi;" "6#%#piG" }{TEXT -1 50 " ]. This function is already odd and has period 2 " }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 100 " We have found how to construct a \+ general solution for the wave equation on the interval [0, L]." }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "f:=x->sin(x); g:=x->0; L:=Pi ; c:=1;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "u:=(t,x)->(f(x+c *t)+f(x-c*t))/2;" }}}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 53 "Check the boundary conditions and initial conditions." }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "u(t,0);\nu(t,L);\nu(0,x);\ns implify(subs(t=0,diff(u(t,x),t)));" }}}{PARA 0 "" 0 "" {TEXT -1 102 "W e expect that this string should be fixed as zero at both ends and vib rate back and forth in between." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "animate(u(t,x),x=0..2*Pi,t=0..2*Pi,color=RED);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "plot3d(u(t,x),x=0..L,t=0..2*Pi,axes =NORMAL,orientation=[-155,65]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT 418 52 "Even periodic extensions and odd periodic extensions" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 226 "As we sh owed just above, in order to be able to graph solutions for the finite string, we need to be able to make a different extensions of function s other than simply even extensions or odd extensions. We need to make odd, 2 " }{TEXT 400 1 "L" }{TEXT -1 43 " periodic extensions and to \+ make even, 2 " }{TEXT 372 1 "L" }{TEXT -1 63 " periodic extensions o f functions defined on the interval [0, " }{TEXT 373 1 "L" }{TEXT -1 114 "]. Here are examples where we do this. To illustrate, we define a function f on the interval [0,1]. In this case, " }{TEXT 401 1 "L" } {TEXT -1 5 " = 1." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "L:=1;\n f:=x->x*(1-x);" }}}{PARA 0 "" 0 "" {TEXT -1 24 "Next, we make an odd, \+ 2 " }{TEXT 374 1 "L" }{TEXT -1 74 " extension and draw the graph to v erify that the extension looks correct." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "odf:=x->oddf(f,x);\nplot(odf(x),x=-L..L);" }}}{PARA 0 "" 0 "" {TEXT -1 125 "Here is where we make extension of the functio n to all numbers and draw the graph to verify that the extension looks correct." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "opf:=PeriodicEx tender(odf,-L..L):\nplot(opf(x),x=-2*L..2*L);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 19 "This was the od d, 2" }{TEXT 402 1 "L" }{TEXT -1 104 " periodic extension of the speci fied function f. We now do a similar process, only we create the even, 2" }{TEXT 403 1 "L" }{TEXT -1 19 " periodic function." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 70 "First we make the ex tension of a function defined on the interval [0, " }{TEXT 375 1 "L" } {TEXT -1 34 "] to a function on the interval [-" }{TEXT 404 1 "L" } {TEXT -1 2 ", " }{TEXT 376 1 "L" }{TEXT -1 96 "] aware that the goal i s to get an even extension. We also draw a graph to assist our intuiti on." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "evf:=x->evnf(f,x);\np lot(evf(x),x=-L..L);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 125 "Here is where we make extension of the function to al l numbers and draw the graph to verify that the extension looks correc t." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "epf:=PeriodicExtender( evf,-L..L):\nplot(epf(x),x=-2*L..2*L);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 117 "It seems we are ready to make solutions for the wave equ ation on a finite interval and with zero boundary conditions." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT 417 13 "Illustrations" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 256 9 "Examples:" }{TEXT -1 65 " Take c = 1 and L, f, and g are specif ied:\n1. L = 2, f(x) = sin(" }{XPPEDIT 18 0 "pi;" "6#%#piG" }{TEXT -1 47 "x), g(x) = 0.\n2. L = 2, f(x) = 0, g(x) = sin(" }{XPPEDIT 18 0 "pi" "6#%#piG" }{TEXT -1 11 "x).\n3. L =" }{XPPEDIT 18 0 "pi" "6#%# piG" }{TEXT -1 9 ", f(x) = " }{XPPEDIT 18 0 "pi" "6#%#piG" }{TEXT -1 8 "/2 - |x-" }{XPPEDIT 18 0 "pi" "6#%#piG" }{TEXT -1 14 "/2|, g(x) = 0 ." }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 18 "Her e is Example 1." }}{PARA 0 "" 0 "" {TEXT -1 18 "We define L and f." }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "L:=2;\nf:=x->sin(Pi*x);" }}} {PARA 0 "" 0 "" {TEXT -1 44 "We make the odd extension on the interval [-" }{TEXT 405 1 "L" }{TEXT -1 2 ", " }{TEXT 406 1 "L" }{TEXT -1 2 "] ." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "odf:=x->oddf(f,x);" }}} {PARA 0 "" 0 "" {TEXT -1 44 "We extend this odd function so that it is 2 " }{TEXT 407 1 "L" }{TEXT -1 10 " periodic." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "opf:=PeriodicExtender(odf,-L..L):" }}}{PARA 0 " " 0 "" {TEXT -1 65 "We draw the graph of this extension to check that \+ it is odd and 2" }{TEXT 408 1 "L" }{TEXT -1 10 " periodic." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "plot(opf(x),x=-2*L..2*L);" }}} {PARA 0 "" 0 "" {TEXT -1 23 "We define the solution " }{TEXT 409 1 "u " }{TEXT -1 1 "." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "u1:=(t,x )->(opf(x+t)+opf(x-t))/2;" }}}{PARA 0 "" 0 "" {TEXT -1 19 "We draw a g raph of " }{TEXT 410 1 "u" }{TEXT -1 76 " and make an animation to see that the solution meets with our expectations." }}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 66 "plot3d(u1(t,x),x=0..L,t=0..2*L,axes=NORMAL,ori entation=[-135,45]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "ani mate(u1(t,x),x=0..L,t=0..L);" }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 18 "Here is example 2." }} {PARA 0 "" 0 "" {TEXT -1 18 "We define L and g." }}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 22 "L:=2;\ng:=x->sin(Pi*x);" }}}{PARA 0 "" 0 "" {TEXT -1 30 "We make G from the function g." }}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 29 "int(g(x),x);\nG:=unapply(%,x);" }}}{PARA 0 "" 0 "" {TEXT -1 32 "We make the even extension of G." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "evG:=x->evnf(G,x);" }}}{PARA 0 "" 0 "" {TEXT -1 13 "We make the 2" }{TEXT 411 1 "L" }{TEXT -1 61 " periodic extension \+ of this even function and draw the graph." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "epG:=PeriodicExtender(evG,-L..L):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "plot(epG(x),x=-2*L..2*L);" }}}{PARA 0 "" 0 "" {TEXT -1 23 "We define the solution " }{TEXT 412 1 "u" }{TEXT -1 1 "." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "u2:=(t,x)->(epG(x+t)-e pG(x-t))/2;" }}}{PARA 0 "" 0 "" {TEXT -1 19 "We draw a graph of " } {TEXT 413 1 "u" }{TEXT -1 76 " and make an animation to see that the s olution meets with our expectations." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "plot3d(u2(t,x),x=0..L,t=0..2*L,axes=NORMAL,orientatio n=[-135,45]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "animate(u2 (t,x),x=0..L,t=0..L);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" } }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 18 "Here is example 3." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "L:=Pi;\nf:=x- >Pi/2-abs(x-Pi/2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "plot( oddf(f,x),x=-L..L);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "odf: =x->oddf(f,x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "opf:=Peri odicExtender(odf,-L..L):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "plot(opf(x),x=-2*L..2*L);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "u3:=(t,x)->(opf(x+t)+opf(x-t))/2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "plot3d(u3(t,x),x=0..L,t=0..2*L,axes=NORMAL,orientatio n=[-135,45]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "animate(u3 (t,x),x=0..L,t=0..2*L);" }{TEXT -1 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 18 "Here is example 4." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "L:=4;\nf:=x->(x-1)*(HS(x-1)-HS(x-2)) + \n \011\011\011(3-x)*(HS(x-2)-HS(x-3));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "odf:=x->oddf(f,x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "plot(odf(x),x=-L..L);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "opf:=PeriodicExtender(odf,-L..L):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "plot(opf(x),x=-2*L..2*L);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "u4:=(t,x)->(opf(t+x)+opf(x-t))/2;" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 79 "plot3d(u4(t,x),x=0..L,t=0..2 *L,axes=NORMAL,orientation=[-35,60], grid=[50,50]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "animate(u4(t,x),x=0..L,t=0..4*L, frames = 80);" }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }} }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT 416 22 "Structure of Solutions" }}{PARA 0 "" 0 "" {TEXT -1 137 "Before we \+ leave these examples too far behind, they call our attention to at lea st four properties of the wave equation that are central." }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 163 "(1) Bumps in the \+ initial distribution will split into two parts each having half the he ight of the original. One bump moves to the right and one moves to the left." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 299 "(2) We are willing to talk about functions being solutions to the wave equation that are not even differentiable once, much less twice. This luxury is seen as a result of having d'Alembert's formulation of solutions. Such an extension of the idea of solution can be made prec ise with the concept of " }{TEXT 414 14 "weak solutions" }{TEXT -1 44 ". We will not pursue this idea further here." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 18 "(3) Choose points " } {XPPEDIT 18 0 "[t[0], x[0]];" "6#7$&%\"tG6#\"\"!&%\"xG6#F'" }{TEXT -1 54 " out in the plane and ask what points of the extended " }{TEXT 426 1 "f" }{TEXT -1 5 " and " }{TEXT 427 1 "g" }{TEXT -1 76 " influenc e the behavior of the solution at this chosen point. We see that u(" } {XPPEDIT 18 0 "t[0],x[0];" "6$&%\"tG6#\"\"!&%\"xG6#F&" }{TEXT -1 25 ") involves the values of " }{TEXT 428 1 "f" }{TEXT -1 6 " at " } {XPPEDIT 18 0 "x[0]+c*t[0];" "6#,&&%\"xG6#\"\"!\"\"\"*&%\"cGF(&%\"tG6# F'F(F(" }{TEXT -1 8 " and " }{XPPEDIT 18 0 "x[0]-c*t[0];" "6#,&&%\" xG6#\"\"!\"\"\"*&%\"cGF(&%\"tG6#F'F(!\"\"" }{TEXT -1 29 " and involve s the values of " }{TEXT 429 1 "g" }{TEXT -1 20 " over the interval [ " }{XPPEDIT 18 0 "x[0]-c*t[0];" "6#,&&%\"xG6#\"\"!\"\"\"*&%\"cGF(&%\"t G6#F'F(!\"\"" }{TEXT -1 3 " , " }{XPPEDIT 18 0 "x[0]+c*t[0]" "6#,&&%\" xG6#\"\"!\"\"\"*&%\"cGF(&%\"tG6#F'F(F(" }{TEXT -1 32 " ]. This interva l, then, is the " }{TEXT 415 20 "domain of dependence" }{TEXT -1 5 " f or " }{XPPEDIT 18 0 "[t[0], x[0]];" "6#7$&%\"tG6#\"\"!&%\"xG6#F'" } {TEXT -1 134 " . If we change the initial conditions outside this inte rval and leave it the same inside, the change will not influence the v alue of " }{TEXT 430 1 "u" }{TEXT -1 15 " at this point." }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 7 "(4) If " }{TEXT 431 1 "f" }{TEXT -1 5 " and " }{TEXT 432 1 "g" }{TEXT -1 34 " are zero outside some interval [ " }{TEXT 433 1 "a" }{TEXT -1 2 ", " }{TEXT 434 1 "b" }{TEXT -1 27 "], then u will be zero for " }{XPPEDIT 18 0 "b +c*t <= x;" "6#1,&%\"bG\"\"\"*&%\"cGF&%\"tGF&F&%\"xG" }{TEXT -1 9 " an d for " }{XPPEDIT 18 0 "x <= a-c*t;" "6#1%\"xG,&%\"aG\"\"\"*&%\"cGF'% \"tGF'!\"\"" }{TEXT -1 50 " . Thus, information travels no faster than speed " }{TEXT 435 1 "c" }{TEXT -1 29 " to the left or to the right. " }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 265 "In \+ this Section, we have defined and illustrated solutions for the simple wave equation on half infinite intervals, and on finite intervals. In the next Section, we make the wave equation more complicated by suppo sing that the vibration happens in a viscous medium." }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT 423 16 "Unassisted Map le" }}{PARA 0 "" 0 "" {TEXT -1 37 "We did not check a single one of th e " }{TEXT 419 1 "u" }{TEXT -1 148 "'s generated in the Illustrations. Think what a mess it would be for a human to compute derivatives of t hese. Here is the first example. Stand back!" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "u1(t,x);" }}}{PARA 0 "" 0 "" {TEXT -1 36 "Who wants to take one derivative of " }{TEXT 420 1 "u" }{TEXT -1 26 "? Or, two \+ with respect to " }{TEXT 421 1 "t" }{TEXT -1 30 " and then two with re spect to " }{TEXT 422 1 "x" }{TEXT -1 185 "? Maple will do this comple tely unemotionally and unassisted. It may be hard to see what is obtai ned is zero, so graph the difference in the derivatives and see how cl ose it it to zero." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 104 "simpl ify(diff(u1(t,x),t,t)-diff(u1(t,x),x,x)):\nplot3d(%,x=0..2,t=0..2,axes =normal,orientation=[-30,75]);" }}}{PARA 0 "" 0 "" {TEXT -1 46 "We rep eat this with the other three solutions." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 104 "simplify(diff(u2(t,x),t,t)-diff(u2(t,x),x,x));\nplot 3d(%,x=0..2,t=0..2,axes=normal,orientation=[-30,75]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 104 "simplify(diff(u3(t,x),t,t)-diff(u3 (t,x),x,x)):\nplot3d(%,x=0..2,t=0..2,axes=normal,orientation=[-30,75]) ;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 104 "simplify(diff(u4(t,x) ,t,t)-diff(u4(t,x),x,x)):\nplot3d(%,x=0..2,t=0..2,axes=normal,orientat ion=[-30,75]);" }}}{SECT 0 {PARA 3 "" 0 "" {TEXT 441 38 "Maple's Solut ion for the Wave Equation" }}{PARA 0 "" 0 "" {TEXT -1 229 "The d'Alemb ert solution for the one dimensional wave equation is so simple in str ucture, it would be a surprise if the solution was not built into Mapl e. We illustrate how to use this built in solution with the following \+ problem." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 12 "PDE: " }{XPPEDIT 18 0 "diff(u,`$`(t,2));" "6#-%%diffG6$%\"u G-%\"$G6$%\"tG\"\"#" }{TEXT -1 5 " = " }{XPPEDIT 18 0 "diff(u,`$`(x, 2));" "6#-%%diffG6$%\"uG-%\"$G6$%\"xG\"\"#" }{TEXT -1 2 " ," }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 42 "Boundary condit ions: u(t,0) = 0 = u(t,1)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 46 "Initial conditions: u(0,x) = 4 x (1 - x), \+ " }{XPPEDIT 18 0 "diff(u,t);" "6#-%%diffG6$%\"uG%\"tG" }{TEXT -1 11 "( 0,x) = 0. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 129 "The solution is obtained by using the partial differential equ ation tools package. We need to read in this package at this stage." } }{PARA 0 "" 0 "" {TEXT -1 1 " " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "with(PDEtools):\nu:='u':" }}}{PARA 0 "" 0 "" {TEXT -1 41 "We rea d in the pde that Maple will solve." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "pdeM:=diff(u(t,x),t,t)-c^2*diff(u(t,x),x,x);" }}} {PARA 0 "" 0 "" {TEXT -1 75 "We now require that Maple should solve th is equation and call the solution " }{TEXT 436 3 "wav" }{TEXT -1 1 ". " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "wav:=pdsolve(pdeM,u(t,x) );" }}}{PARA 0 "" 0 "" {TEXT -1 229 "Recall from the discussion in thi s section that, because this is a finite \"string\", we must make an o dd extension of u(0,x) with period 2. The other initial condition is z ero, so that we need no other extensions for this problem." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 1 " " }{TEXT 437 7 "W ARNING" }{TEXT -1 16 ": the procedure " }{TEXT 440 17 "Periodic Extend er" }{TEXT -1 77 " is needed in what follows and can be found at the b eginning of this Section." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "f:=x->4*x*(1-x);" }}}{PARA 0 "" 0 " " {TEXT -1 79 "We now make the odd, periodic extension of f. First, he re is the odd extension." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 " F:=unapply(oddf(f,x),x);" }}}{PARA 0 "" 0 "" {TEXT -1 105 "As a check, the first two of these should be equal and the next two should be the negative of each other." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 " f(1/2),F(1/2);\nf(1/2),F(-1/2);" }}}{PARA 0 "" 0 "" {TEXT -1 16 "Next, we extend " }{TEXT 439 1 "F" }{TEXT -1 15 " with period 2." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "Fext:=PeriodicExtender(F,-1..1):" } }}{PARA 0 "" 0 "" {TEXT -1 21 "As a check, we graph " }{TEXT 438 4 "Fe xt" }{TEXT -1 71 " on the interval [-2, 2]. We should find an odd func tion with period 2." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "plot( Fext(x),x=-2..2);" }}}{PARA 0 "" 0 "" {TEXT -1 96 "We put this odd, pe riod 2 function into the solution for the wave equation Maple provided above." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "eval(wav,\{c=1,_F 1=Fext/2,_F2=Fext/2\}):" }}}{PARA 0 "" 0 "" {TEXT -1 30 "This will def ine the solution." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "u:=unap ply(rhs(%),(t,x)):" }}}{PARA 0 "" 0 "" {TEXT -1 18 "We draw the graph. " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "plot3d(u(t,x),x=0..1,t=0 ..1,axes=normal);" }}}{PARA 0 "" 0 "" {TEXT -1 45 "And we provide an a nimation for the solution." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "animate(u(t,x),x=0..1,t=0..4, frames=30);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" } }}{PARA 0 "" 0 "" {TEXT -1 175 "It is hoped that the explanation in th is Section 5.3 for how to construct a solution for this simple wave eq uation makes this built in solution understandable and appreciated." } }{EXCHG }}{EXCHG }{EXCHG }{EXCHG }}{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 257 "" 0 "" {TEXT -1 36 "Copyright \251 2003 b y James V. Herod" }}{PARA 257 "" 0 "" {TEXT -1 19 "All rights reserved " }}{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 }