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{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 -1 0 "" }}{PARA 257 "" 0 "" {TEXT -1 39 "Secti
on 5.2: The Solution of d'Alembert" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{SECT 0 {PARA 3 "" 0 "" {TEXT 260 30 "Maple Packages for Section 5.2" 
}}{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(PDEtools):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 396 " The solution we obtaine
d for the wave equation in the previous module is a natural extension \+
of the development for solutions of the heat diffusion model. The math
ematics is accurate, and the method is satisfactory in that the soluti
ons obtained for the model meet with our expectations. There was no ne
w mathematical insight, except that we can do the process again for th
is different equation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 32 "Here is a different perspective." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 146 "In the 18th Century, Jea
n Le Rond d'Alembert had formulated a different way to come to a solut
ion of the wave equation. Morris Kline in his book, " }{TEXT 259 49 "M
athematical Thought from Ancient to Modern Times" }{TEXT -1 531 ", rem
inds us that several mathematical concepts had to evolve with time. Kl
ine says that d'Alembert accepted only analytic curves and solutions. \+
He wanted functions that could be given with equations of relations, f
or example, between the elementary algebraic and trigonometric functio
ns. A part of the undeveloped understanding was that an initial functi
on need not be expressible by a single analytic expression. There was \+
also the struggle for the concept of an infinite sum -- a series. It i
s easy to imagine the difficulty in " }}{PARA 0 "" 0 "" {TEXT -1 52 "(
1) having the idea of an infinite sum of functions," }}{PARA 0 "" 0 "
" {TEXT -1 91 "(2) having a series of cosine functions to converge to,
 for example, the sine function, and" }}{PARA 0 "" 0 "" {TEXT -1 122 "
(3) having the series to converge to an analytic function on a finite \+
interval, but not to the function off that interval." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 325 "     When a creative p
erson is stuck on an idea, interesting things can happen. We present d
'Alembert's solution here. If he could see what follows, d'Alembert mi
ght not admit that the presentation is the way he had thought of the s
olution. But, we should take advantage of the fact that we stand on th
e shoulders of giants! " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 
{PARA 3 "" 0 "" {TEXT 261 23 "Transforming Equations." }}{PARA 0 "" 0 
"" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 57 "Suppose that u is a fu
nction which satisfies the equation" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 13 "(*)          " }{XPPEDIT 18 0 "diff(u,`$
`(t,2))-diff(u,`$`(x,2)) = 0;" "6#/,&-%%diffG6$%\"uG-%\"$G6$%\"tG\"\"#
\"\"\"-F&6$F(-F*6$%\"xGF-!\"\"\"\"!" }{TEXT -1 1 "." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 8 " Define " }{TEXT 264 1 "
v" }{TEXT -1 1 "(" }{TEXT 265 1 "w" }{TEXT -1 2 ", " }{TEXT 266 1 "z" 
}{TEXT -1 12 ") as follows" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
29 "v:=(w,z)->u((w-z)/2,(w+z)/2);" }{TEXT -1 0 "" }}}{PARA 0 "" 0 "" 
{TEXT -1 9 "That is, " }{TEXT 267 1 "t" }{TEXT -1 4 " = (" }{TEXT 268 
1 "w" }{TEXT -1 3 " - " }{TEXT 269 1 "z" }{TEXT -1 10 ")/2 , and " }
{TEXT 270 1 "x" }{TEXT -1 4 " = (" }{TEXT 271 1 "w" }{TEXT -1 3 " + " 
}{TEXT 272 1 "z" }{TEXT -1 27 ")/2, or, what is the same, " }{TEXT 
273 1 "w" }{TEXT -1 3 " = " }{TEXT 274 1 "t" }{TEXT -1 3 " + " }{TEXT 
275 1 "x" }{TEXT -1 5 " and " }{TEXT 276 1 "z" }{TEXT -1 3 " = " }
{TEXT 277 1 "x" }{TEXT -1 3 " - " }{TEXT 278 1 "t" }{TEXT -1 1 "." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "solve(\{t=(w-z)/2,x=(z+w)/2
\},\{w,z\});" }}}{PARA 0 "" 0 "" {TEXT -1 16 "As we see, this " }
{TEXT 279 1 "v" }{TEXT -1 35 " will satisfy a different equation." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "diff(v(w,z),w,z);" }}}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 39 "We see that, be
cause u satisfies (*),  " }{TEXT 280 1 "v" }{TEXT -1 27 " defined as a
bove satisfies" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 27 "(**)                       " }{XPPEDIT 18 0 "diff(v,w,z)=
0" "6#/-%%diffG6%%\"vG%\"wG%\"zG\"\"!" }{TEXT -1 1 "." }}{PARA 0 "" 0 
"" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 25 "Conversely, suppose th
at " }{TEXT 281 1 "v" }{TEXT -1 40 " satisfies this last equation (**)
 and  " }{TEXT 282 1 "u" }{TEXT -1 15 " is defined by " }}{PARA 0 "" 
0 "" {TEXT -1 24 "                        " }{TEXT 283 1 "u" }{TEXT 
-1 2 "( " }{TEXT 284 1 "t" }{TEXT -1 3 " , " }{TEXT 285 1 "x" }{TEXT 
-1 7 " )  =  " }{TEXT 286 1 "v" }{TEXT -1 3 "( (" }{TEXT 287 1 "x" }
{TEXT -1 1 "+" }{TEXT 288 1 "t" }{TEXT -1 6 ")/2, (" }{TEXT 289 1 "x" 
}{TEXT -1 1 "-" }{TEXT 290 1 "t" }{TEXT -1 6 ")/2 )." }}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 7 "v:='v';" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 25 "u:=(t,x)->v((x+t),(x-t));" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 34 "diff(u(t,x),x,x)-diff(u(t,x),t,t);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 181 "T
hus, u solves equation (*) because v satisfies (**). This observation \+
enables us to provide a solution for the wave equation that can be mad
e quite independent of the Fourier Idea." }}{PARA 0 "" 0 "" {TEXT -1 
69 "     You see, Equation (**) says that the derivative with respect \+
to " }{TEXT 291 1 "z" }{TEXT -1 6 " of   " }{XPPEDIT 18 0 "diff(v,w);
" "6#-%%diffG6$%\"vG%\"wG" }{TEXT -1 119 "  is zero. We know all funct
ions whose derivative is zero. They are constant. If taking the deriva
tive with respect to " }{TEXT 292 1 "z" }{TEXT -1 20 " gives zero, the
n   " }{XPPEDIT 18 0 "diff(v,w);" "6#-%%diffG6$%\"vG%\"wG" }{TEXT -1 
17 "  is constant in " }{TEXT 293 1 "z" }{TEXT -1 8 ". We say" }}
{PARA 0 "" 0 "" {TEXT -1 12 "            " }{XPPEDIT 18 0 "diff(v,w);
" "6#-%%diffG6$%\"vG%\"wG" }{TEXT -1 9 "  = C(w)." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 35 "Now integrate this with r
espect to " }{TEXT 294 1 "w" }{TEXT -1 30 " and you get some function \+
of " }{TEXT 295 1 "w" }{TEXT -1 42 ", plus a constant -- a constant as
 far as " }{TEXT 296 1 "w" }{TEXT -1 23 " is concerned. We write" }}
{PARA 0 "" 0 "" {TEXT -1 14 "              " }{TEXT 297 1 "v" }{TEXT 
-1 1 "(" }{TEXT 298 1 "w" }{TEXT -1 2 ", " }{TEXT 299 1 "z" }{TEXT -1 
4 ") = " }{XPPEDIT 18 0 "psi;" "6#%$psiG" }{TEXT -1 2 "( " }{TEXT 300 
1 "w" }{TEXT -1 5 " ) + " }{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 
2 "( " }{TEXT 301 1 "z" }{TEXT -1 3 " )." }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 29 "To get this back in terms of " }
{TEXT 302 1 "t" }{TEXT -1 5 " and " }{TEXT 303 1 "x" }{TEXT -1 20 ", r
ecall above that " }{TEXT 304 1 "w" }{TEXT -1 3 " =(" }{TEXT 305 1 "x
" }{TEXT -1 3 " + " }{TEXT 306 1 "t" }{TEXT -1 6 ") and " }{TEXT 307 
1 "z" }{TEXT -1 5 " = ( " }{TEXT 308 1 "x" }{TEXT -1 3 " - " }{TEXT 
309 1 "t" }{TEXT -1 9 "). Thus, " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 10 "          " }{TEXT 316 1 "u" }{TEXT -1 1 
"(" }{TEXT 315 1 "t" }{TEXT -1 2 ", " }{TEXT 314 1 "x" }{TEXT -1 5 ") \+
=  " }{XPPEDIT 18 0 "psi;" "6#%$psiG" }{TEXT -1 2 "( " }{TEXT 313 1 "x
" }{TEXT -1 3 " + " }{TEXT 312 1 "t" }{TEXT -1 5 " ) + " }{XPPEDIT 18 
0 "phi;" "6#%$phiG" }{TEXT -1 2 "( " }{TEXT 311 1 "x" }{TEXT -1 3 " - \+
" }{TEXT 310 1 "t" }{TEXT -1 3 " )." }}{PARA 0 "" 0 "" {TEXT -1 64 "Th
ese are the solutions d'Alembert found. We pursue these ideas." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT 262 32 "Solutions for The Wave
 Equation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 190 "     Take note that nothing has been said about initial condit
ions or boundary conditions for this wave equation. One might conceive
 these function as solutions of the wave equation for all " }{TEXT 
317 1 "t" }{TEXT -1 13 " > 0 and all " }{TEXT 318 1 "x" }{TEXT -1 67 "
. We found that functions satisfying the wave equation had the form" }
}{PARA 0 "" 0 "" {TEXT -1 11 "           " }{TEXT 319 1 "u" }{TEXT -1 
1 "(" }{TEXT 320 1 "t" }{TEXT -1 2 ", " }{TEXT 321 1 "x" }{TEXT -1 5 "
) =  " }{XPPEDIT 18 0 "psi;" "6#%$psiG" }{TEXT -1 2 "( " }{TEXT 322 1 
"x" }{TEXT -1 3 " + " }{TEXT 323 1 "t" }{TEXT -1 5 " ) + " }{XPPEDIT 
18 0 "phi;" "6#%$phiG" }{TEXT -1 2 "( " }{TEXT 324 1 "x" }{TEXT -1 3 "
 - " }{TEXT 325 1 "t" }{TEXT -1 3 " )." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 69 "Furthermore, all functions defined in
 that form satisfy the equation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 58 "We conceive of finding all solutions for \+
the wave equation" }}{PARA 0 "" 0 "" {TEXT -1 28 "                    \+
        " }{XPPEDIT 18 0 "diff(u(t,x),`$`(t,2)) = diff(u(t,x),`$`(x,2)
);" "6#/-%%diffG6$-%\"uG6$%\"tG%\"xG-%\"$G6$F*\"\"#-F%6$-F(6$F*F+-F-6$
F+F/" }}{PARA 0 "" 0 "" {TEXT -1 46 "as the problem for finding the nu
llspace of a " }{TEXT 256 13 "wave operator" }{TEXT -1 1 ":" }}{PARA 
0 "" 0 "" {TEXT -1 27 "                           " }{XPPEDIT 18 0 "di
ff(u(t,x),`$`(t,2))-diff(u(t,x),`$`(x,2)) = 0;" "6#/,&-%%diffG6$-%\"uG
6$%\"tG%\"xG-%\"$G6$F+\"\"#\"\"\"-F&6$-F)6$F+F,-F.6$F,F0!\"\"\"\"!" }
{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 156 "We define a wave operat
or in Maple. We test that functions of the special form satisfy the wa
ve equation -- they are in the null space of the wave operator." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 29 "u:=(t,x)->sin(exp(cos(x-t)));" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 34 "diff(u(t,x),t,t)-diff(u(t,x),x,x);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT 
-1 26 "     We have seen that if " }{TEXT 257 1 "f" }{TEXT -1 46 " is \+
any function depending on the combination " }{TEXT 326 1 "x" }{TEXT 
-1 3 " + " }{TEXT 327 1 "t" }{TEXT -1 4 " or " }{TEXT 328 1 "x" }
{TEXT -1 3 " - " }{TEXT 329 1 "t" }{TEXT -1 6 ", then" }{TEXT 258 2 " \+
f" }{TEXT -1 48 " will lie in the nullspace of the wave operator." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 109 "f:=(t,x)->psi(x+t);\ndiff(f
(t,x),t,t)-diff(f(t,x),x,x);\nf:=(t,x)->phi(x-t);\ndiff(f(t,x),t,t)-di
ff(f(t,x),x,x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{PARA 0 "" 0 "" {TEXT -1 145 "     Since the wave equation is linear, \+
it is not a surprise that given any two twice differentiable functions
 of a single variable, the function" }}{PARA 0 "" 0 "" {TEXT -1 23 "  \+
                     " }{XPPEDIT 18 0 "psi(x+t)+phi(x-t);" "6#,&-%$psi
G6#,&%\"xG\"\"\"%\"tGF)F)-%$phiG6#,&F(F)F*!\"\"F)" }}{PARA 0 "" 0 "" 
{TEXT -1 167 "is also a solution. D'Alembert had the insight that not \+
only is this combination a solution, it is the general solution. To un
derstand this we need to think about how " }{XPPEDIT 18 0 "phi" "6#%$p
hiG" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "psi" "6#%$psiG" }{TEXT -1 40 
" could be determine for a specific case." }}{PARA 0 "" 0 "" {TEXT -1 
45 "     This is the next subject for discussion." }}}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT 263 41 "Initial conditio
ns for the wave equation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 95 "     Suppose that we have an infinitely long strin
g, with initial conditions of the usual form:" }}{PARA 0 "" 0 "" 
{TEXT -1 29 "                             " }{XPPEDIT 18 0 "u(0,x) = f
(x)" "6#/-%\"uG6$\"\"!%\"xG-%\"fG6#F(" }{TEXT -1 4 ",   " }{XPPEDIT 
18 0 "u[t](0,x)=g(x)" "6#/-&%\"uG6#%\"tG6$\"\"!%\"xG-%\"gG6#F+" }
{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 8 "Here, - " }{XPPEDIT 18 0 
"infinity" "6#%)infinityG" }{TEXT -1 3 " < " }{TEXT 330 1 "x" }{TEXT 
-1 3 " < " }{XPPEDIT 18 0 "infinity" "6#%)infinityG" }{TEXT -1 106 ", \+
so that we have no boundary conditions to impose. The question is: How
 do we find a solution of the form" }}{PARA 0 "" 0 "" {TEXT -1 29 "   \+
                          " }{TEXT 331 1 "u" }{TEXT -1 1 "(" }{TEXT 
332 1 "t" }{TEXT -1 1 "," }{TEXT 333 1 "x" }{TEXT -1 3 ")= " }
{XPPEDIT 18 0 "psi(x+c*t)+phi(x-c*t);" "6#,&-%$psiG6#,&%\"xG\"\"\"*&%
\"cGF)%\"tGF)F)F)-%$phiG6#,&F(F)*&F+F)F,F)!\"\"F)" }}{PARA 0 "" 0 "" 
{TEXT -1 59 "which solves this problem?  The initial conditions are th
at" }}{PARA 0 "" 0 "" {TEXT -1 10 "          " }{TEXT 334 1 "f" }
{TEXT -1 1 "(" }{TEXT 335 1 "x" }{TEXT -1 4 ") = " }{TEXT 336 1 "u" }
{TEXT -1 4 "(0, " }{TEXT 337 1 "x" }{TEXT -1 4 ") = " }{XPPEDIT 18 0 "
psi;" "6#%$psiG" }{TEXT -1 1 "(" }{TEXT 338 1 "x" }{TEXT -1 4 ") + " }
{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 1 "(" }{TEXT 339 1 "x" }
{TEXT -1 2 ") " }}{PARA 0 "" 0 "" {TEXT -1 3 "and" }}{PARA 0 "" 0 "" 
{TEXT -1 10 "          " }{TEXT 340 1 "g" }{TEXT -1 1 "(" }{TEXT 341 
1 "x" }{TEXT -1 4 ") = " }{XPPEDIT 18 0 "u[t];" "6#&%\"uG6#%\"tG" }
{TEXT -1 4 "(0, " }{TEXT 342 1 "x" }{TEXT -1 4 ") = " }{TEXT 343 1 "c
" }{TEXT -1 1 " " }{XPPEDIT 18 0 "psi;" "6#%$psiG" }{TEXT -1 3 " '(" }
{TEXT 344 1 "x" }{TEXT -1 4 ") - " }{TEXT 345 1 "c" }{TEXT -1 1 " " }
{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 3 " '(" }{TEXT 346 1 "x" }
{TEXT -1 2 ")." }}{PARA 0 "" 0 "" {TEXT -1 46 "In the second equation,
 get an antiderivative:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 12 "          G(" }{TEXT 349 1 "x" }{TEXT -1 10 ") + C  \+
=  " }{XPPEDIT 18 0 "psi;" "6#%$psiG" }{TEXT -1 1 "(" }{TEXT 348 1 "x
" }{TEXT -1 4 ") - " }{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 1 "(" 
}{TEXT 347 1 "x" }{TEXT -1 47 "),  where C is the constant of integrat
ion and " }}{PARA 0 "" 0 "" {TEXT -1 11 "         G(" }{TEXT 350 1 "x
" }{TEXT -1 4 ") = " }{XPPEDIT 18 0 "1/C*int(g(y),y = 0 .. x);" "6#*(
\"\"\"F$%\"CG!\"\"-%$intG6$-%\"gG6#%\"yG/F-;\"\"!%\"xGF$" }{TEXT -1 2 
" ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 23 "Pu
tting this value for " }{XPPEDIT 18 0 "psi;" "6#%$psiG" }{TEXT -1 1 "(
" }{TEXT 351 1 "x" }{TEXT -1 4 ") - " }{XPPEDIT 18 0 "phi;" "6#%$phiG
" }{TEXT -1 1 "(" }{TEXT 352 1 "x" }{TEXT -1 21 ") with the value for \+
" }{XPPEDIT 18 0 "psi;" "6#%$psiG" }{TEXT -1 1 "(" }{TEXT 353 1 "x" }
{TEXT -1 4 ") + " }{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 1 "(" }
{TEXT 354 1 "x" }{TEXT -1 11 ") , we have" }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 5 "     " }{XPPEDIT 18 0 "psi;" "6#%$p
siG" }{TEXT -1 1 "(" }{TEXT 355 1 "x" }{TEXT -1 4 ") = " }{XPPEDIT 18 
0 "(f(x)+G(x)+C)/2;" "6#*&,(-%\"fG6#%\"xG\"\"\"-%\"GG6#F(F)%\"CGF)F)\"
\"#!\"\"" }{TEXT -1 7 "   and " }{XPPEDIT 18 0 "phi;" "6#%$phiG" }
{TEXT -1 1 "(" }{TEXT 356 1 "x" }{TEXT -1 5 ") =  " }{XPPEDIT 18 0 "(f
(x)-G(x)-C)/2;" "6#*&,(-%\"fG6#%\"xG\"\"\"-%\"GG6#F(!\"\"%\"CGF-F)\"\"
#F-" }{TEXT -1 3 "  ." }}{PARA 0 "" 0 "" {TEXT -1 5 "     " }}{PARA 0 
"" 0 "" {TEXT -1 16 "For example, if " }{TEXT 357 1 "f" }{TEXT -1 34 "
 is a function with one bump near " }{TEXT 358 1 "x" }{TEXT -1 247 " =
 0 and there is no initial velocity, we can anticipate how this initia
l condition will evolve with increasing time: we should see the bump b
reak into the superposition of two bumps of half the height, one movin
g to the right and one to the left." }}{PARA 0 "" 0 "" {TEXT -1 24 "  \+
   Take, for example, " }{XPPEDIT 18 0 "f(x)=exp(-x^2)" "6#/-%\"fG6#%
\"xG-%$expG6#,$*$F'\"\"#!\"\"" }{TEXT -1 125 " as the initial function
. We construct a solution to the wave equation with this initial value
 and watch it evolve with time." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 36 "f:=x->exp(-x^2);\nplot(f(x),x=-4..4);" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 43 "animate((f(x+t)+f(x-t))/2,x=-4..4,t=0..10);" }}}
{PARA 0 "" 0 "" {TEXT -1 41 "     Sometimes it is instructive to view \+
" }{TEXT 377 1 "u" }{TEXT -1 15 " as a graph in " }{TEXT 378 1 "t" }
{TEXT -1 5 " and " }{TEXT 379 1 "x" }{TEXT -1 26 ", but not as an anim
ation." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 87 "plot3d((f(x+t)+f(x
-t))/2,x=-10..10,t=0..6,axes=NORMAL,\n         orientation=[-120,65]);
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 49 "     We now investigate a
 second example -- that " }{TEXT 380 1 "f" }{TEXT -1 1 "(" }{TEXT 381 
1 "x" }{TEXT -1 10 ") = 0 and " }{XPPEDIT 18 0 "g(x) <> 0;" "6#0-%\"gG
6#%\"xG\"\"!" }{TEXT -1 4 "..  " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 85 "We have seen that the solution for the wa
ve equation with these initial conditions is" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 39 "                               \+
        " }{XPPEDIT 18 0 "u(t,x)= int(g(s),s=x-c*t..x+c*t)/(2*c)" "6#/
-%\"uG6$%\"tG%\"xG*&-%$intG6$-%\"gG6#%\"sG/F0;,&F(\"\"\"*&%\"cGF4F'F4!
\"\",&F(F4*&F6F4F'F4F4F4*&\"\"#F4F6F4F7" }{TEXT -1 1 "." }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 33 "     Here is an exam
ple. We take " }{XPPEDIT 18 0 "f(x) = 0 " "6#/-%\"fG6#%\"xG\"\"!" }
{TEXT -1 5 " and " }{XPPEDIT 18 0 "g(x) = x/(1+x+x^2)" "6#/-%\"gG6#%\"
xG*&F'\"\"\",(F)F)F'F)*$F'\"\"#F)!\"\"" }{TEXT -1 12 ". We make G." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
18 "g:=x->x/(1+x+x^2);" }}}{PARA 0 "" 0 "" {TEXT -1 19 "Now, we can de
fine " }{TEXT 382 1 "u" }{TEXT -1 1 "." }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 31 "u:=(t,x)->int(g(s),s=x-t..x+t);" }}}{PARA 0 "" 0 "" 
{TEXT -1 45 "Here is Maple's integration of this function." }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "u(t,x);" }}}{PARA 0 "" 0 "" {TEXT 
-1 91 "The following animation will illustrate how the wave propagates
 with this initial velocity." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
33 "animate(u(t,x),x=-10..10,t=0..6);" }}}{PARA 0 "" 0 "" {TEXT -1 41 
"     Sometimes it is instructive to view " }{TEXT 361 1 "u" }{TEXT 
-1 15 " as a graph in " }{TEXT 360 1 "t" }{TEXT -1 5 " and " }{TEXT 
359 1 "x" }{TEXT -1 26 ", but not as an animation." }}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 126 "plot3d(u(t,x),x=-10..10,t=0..6,axes=NORMAL,
\n         orientation=[-120,65], grid=[50,50],lightmodel=light1,style
=patchnogrid);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT 367 22 "
Propagation of corners" }}{PARA 0 "" 0 "" {TEXT -1 247 "One of the fea
tures of the heat equation, you will recall, is that the solution was \+
infinitely differentiable for t > 0. This is not true for solutions of
 the wave equation. Consider the following. We take the initial veloci
ty to be zero, so that " }{TEXT 363 1 "g" }{TEXT -1 42 " = 0. We take \+
the initial distribution of " }{TEXT 364 1 "u" }{TEXT -1 138 " to have
 three sharp corners: at -1, 0, and 1. Watch how these corners distrib
ute. First, the initial distribution is defined and graphed." }{TEXT 
362 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "f:=x->(signum(x-1
)-signum(x+1))/2*(abs(x)-1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 19 "plot(f(x),x=-3..3);" }}}{PARA 0 "" 0 "" {TEXT -1 17 "Now, we mak
e the " }{TEXT 365 1 "u" }{TEXT -1 67 ", which by d'Alembert's method \+
is a solution for the wave equation." }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 24 "u:=(t,x)->f(t+x)+f(t-x);" }}}{PARA 0 "" 0 "" {TEXT 
-1 48 "We first see an animation and then the graph of " }{TEXT 366 1 
"u" }{TEXT -1 1 "." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "animat
e((f(x+t)+f(x-t))/2,x=-4..4,t=0..10);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 108 "plot3d(u(t,x),x=-5..5,t=0..2,axes=NORMAL,numpoints=1
00^2,                orientation=[-20,35],color=yellow);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 164 "In this \+
Section, we have set up what is the form for d'Alembert's solution for
 the one dimensional wave equation. The solution that we considered is
 for all number " }{TEXT 368 1 "x" }{TEXT -1 9 " and for " }{TEXT 369 
1 "t" }{TEXT -1 304 " > 0. What needs to be considered next is how is \+
this form modified if the domain of the initial functions is not the r
eal line, but a half infinite interval, or a finite interval. In the n
ext Section, we examine solutions of the wave equation on a half infin
ite line and on an interval of finite length." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 "
" {TEXT 376 16 "Unassisted Maple" }}{PARA 0 "" 0 "" {TEXT -1 174 "Ther
e are two items here. First, Maple could have done that change of vari
able to get the form for d'Alembert's solution with out any assistance
. Watch. First define the PDE." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 41 "PDE:=diff(u(t,x),t,t)-diff(u(t,x),x,x)=0;" }}}{PARA 0 "" 0 "" 
{TEXT -1 37 "The list what transformation to make." }}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 27 "tr:=\{t=(w-z)/2, x=(w+z)/2\};" }}}{PARA 0 "
" 0 "" {TEXT -1 46 "Now, let Maple make the transformation change." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "dchange(tr,PDE);" }}}{PARA 
0 "" 0 "" {TEXT -1 152 "This is precisely the change we got above. Now
, we would be in a position to solve the PDE using only information ab
out ordinary differential equations." }}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}}{PARA 0 "" 0 "" {TEXT -1 89 "Here is the second item to note here. A
bove, we illustrated that an initial displacement " }{TEXT 373 2 "f " 
}{TEXT -1 69 "might occur over an interval of finite length. So might \+
the function " }{TEXT 374 1 "g" }{TEXT -1 130 " have a non-zero value \+
on an interval of finite length. One inexperienced with Maple might th
ink that the integral used to define " }{TEXT 375 1 "u" }{TEXT -1 134 
" might need to be defined piecewise. This is not the case. Maple can \+
handle such integrals unassisted. Consider the following example." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "g:=x-
>signum(x-1)-signum(x+1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
19 "plot(g(x),x=-3..3);" }}}{PARA 0 "" 0 "" {TEXT -1 17 "Now, we defin
e u." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "u:=(t,x)->int(g(s),s
=x-t..x+t);" }}}{PARA 0 "" 0 "" {TEXT -1 140 "Here is Maple's integrat
ion of this function. Is it clear how to read this answer? We illustra
te with a few values before drawing the graph." }}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 7 "u(t,x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 23 "u(1,2); u(2,2); u(2,3);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 91 "The following animation will illustrate h
ow the wave propagates with this initial velocity." }}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 34 "animate(u(t,x),x=-20..20,t=0..15);" }}}
{PARA 0 "" 0 "" {TEXT -1 41 "     Sometimes it is instructive to view \+
" }{TEXT 372 1 "u" }{TEXT -1 15 " as a graph in " }{TEXT 371 1 "t" }
{TEXT -1 5 " and " }{TEXT 370 1 "x" }{TEXT -1 26 ", but not as an anim
ation." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 76 "plot3d(u(t,x),x=-1
0..10,t=0..6,axes=NORMAL,\n         orientation=[-120,65]);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 50 "EMAIL: he
rod@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 Jame
s V. Herod" }}{PARA 256 "" 0 "" {TEXT -1 19 "All rights reserved" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{MARK "0 0 0" 18 }{VIEWOPTS 1 1 0 1 
1 1803 1 1 1 1 }{PAGENUMBERS 1 1 2 33 1 1 }