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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 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 262 46 "Section \+
5.1: The One Dimensional Wave Equation" }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{SECT 0 {PARA 3 "" 0 "" {TEXT 307 30 "Maple Packages for Section 5
.1" }}{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 52 "We now begin a study of the classical wave equation." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 48 "     The \+
classical, linearized wave equation is " }}{PARA 0 "" 0 "" {TEXT -1 
39 "                                       " }{XPPEDIT 18 0 "diff(u(t,
x),`$`(t,2)) = c^2*diff(u(t,x),`$`(x,2));" "6#/-%%diffG6$-%\"uG6$%\"tG
%\"xG-%\"$G6$F*\"\"#*&%\"cGF/-F%6$-F(6$F*F+-F-6$F+F/\"\"\"" }{TEXT -1 
1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 108 "
     We classify this PDE as a special case of the more general consta
nt coefficient, second order equation:" }}{PARA 0 "" 0 "" {TEXT -1 9 "
         " }{XPPEDIT 18 0 "a*diff(u(t,x),`$`(t,2))+b*diff(u(t,x),t,x)+
c*diff(u(t,x),`$`(x,2))+d*diff(u(t,x),t)+e*diff(u(t,x),x)+f*u(t,x);" "
6#,.*&%\"aG\"\"\"-%%diffG6$-%\"uG6$%\"tG%\"xG-%\"$G6$F-\"\"#F&F&*&%\"b
GF&-F(6%-F+6$F-F.F-F.F&F&*&%\"cGF&-F(6$-F+6$F-F.-F06$F.F2F&F&*&%\"dGF&
-F(6$-F+6$F-F.F-F&F&*&%\"eGF&-F(6$-F+6$F-F.F.F&F&*&%\"fGF&-F+6$F-F.F&F
&" }{TEXT -1 5 " = 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 
"" {TEXT 256 12 "Second order" }{TEXT -1 319 " refers to the lack of d
erivatives of more than second order.We will solve the wave equation b
y the method of separation of variables in this worksheet. We will fin
d there are alternate methods for this equation. We will also see that
 slightly more complicated situations lead to the more general second \+
order equation." }}{PARA 0 "" 0 "" {TEXT -1 183 "     We have seen tha
t the standard separation of variables methods work well for the heat \+
equation. This method is often good for linear equations. The method t
akes advantage of the " }{TEXT 257 22 "superposition property" }{TEXT 
-1 28 ". That is, if two functions " }{XPPEDIT 18 0 "u[1]" "6#&%\"uG6#
\"\"\"" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "u[2]" "6#&%\"uG6#\"\"#" }
{TEXT -1 66 " are solutions to a linear PDE, then so is any linear com
bination:" }}{PARA 0 "" 0 "" {TEXT -1 25 "                         " }
{XPPEDIT 18 0 "alpha*u[1]+beta*u[2]" "6#,&*&%&alphaG\"\"\"&%\"uG6#F&F&
F&*&%%betaGF&&F(6#\"\"#F&F&" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 178 "As with the heat equation, the
 wave equation commonly comes with boundary conditions and initial con
ditions. At the start, we take homogeneous boundary conditions, assumi
ng that " }{TEXT 308 1 "x" }{TEXT -1 22 " ranges between 0 and " }
{TEXT 309 1 "L" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 22 "Boundary Conditions:  " }{TEXT 263 1 "u" 
}{TEXT -1 1 "(" }{TEXT 264 1 "t" }{TEXT -1 14 ", 0) = 0, and " }{TEXT 
265 1 "u" }{TEXT -1 1 "(" }{TEXT 266 1 "t" }{TEXT -1 2 ", " }{TEXT 
267 1 "L" }{TEXT -1 6 ") = 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 40 "Because the equation is second order in \+
" }{TEXT 268 1 "t" }{TEXT -1 100 ", we expect two initial conditions. \+
 (Contrast this with the heat equation, which is first order in " }
{TEXT 310 1 "t" }{TEXT -1 43 " and so needed only one initial conditio
n.)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 21 "In
itial Conditions:  " }{TEXT 270 1 "u" }{TEXT -1 4 "(0, " }{TEXT 269 1 
"x" }{TEXT -1 4 ") = " }{TEXT 271 1 "f" }{TEXT -1 1 "(" }{TEXT 272 1 "
x" }{TEXT -1 6 ") and " }{XPPEDIT 18 0 "diff(u,t);" "6#-%%diffG6$%\"uG
%\"tG" }{TEXT -1 4 "(0, " }{TEXT 273 1 "x" }{TEXT -1 4 ") = " }{TEXT 
274 1 "g" }{TEXT -1 1 "(" }{TEXT 275 1 "x" }{TEXT -1 6 ") for " }
{TEXT 276 1 "x" }{TEXT -1 8 " in [0, " }{TEXT 277 1 "L" }{TEXT -1 2 "]
." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 119 "The
 usual physical realization made for this model is that of a string, h
eld fixed at two ends, displaced by an amount " }{TEXT 278 1 "f" }
{TEXT -1 1 "(" }{TEXT 279 1 "x" }{TEXT -1 43 ") initially, and given a
n initial velocity " }{TEXT 281 1 "g" }{TEXT -1 1 "(" }{TEXT 280 1 "x
" }{TEXT -1 135 "). This model will guide our intuition and, with smal
l displacements, gives an accurate impression for the vibrations of a \+
taut string." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 18 "     We often set " }{TEXT 282 
1 "c" }{TEXT -1 455 " = 1 for convenience. We do this in this workshee
t but remove the restriction later. It is of value to consider the dif
ference in the left and right side of the wave equation. That differen
ce reminds us that we are solving linear operator equations and lookin
g for the null space of these linear operators. This example is simply
 a linear equation which has an infinite number of solutions. We seek \+
one which satisfies the boundary and initial conditions." }}{EXCHG }
{PARA 0 "" 0 "" {TEXT -1 5 "     " }}{PARA 0 "" 0 "" {TEXT -1 245 "   \+
 We have had experience with the method of separation of variables and
 know that this method for solving partial differential equations is p
recisely what it says: One assumes that solutions can be written as pr
oducts of separate functions of " }{TEXT 283 1 "t" }{TEXT -1 5 " and \+
" }{TEXT 284 1 "x" }{TEXT -1 30 ". Thus, we make the assumption" }
{TEXT -1 6 " that " }{TEXT 285 1 "u" }{TEXT -1 1 "(" }{TEXT 286 1 "t" 
}{TEXT -1 2 ", " }{TEXT 287 1 "x" }{TEXT -1 25 ") is of the special fo
rm " }}{PARA 0 "" 0 "" {TEXT -1 17 "               T(" }{TEXT 289 1 "t
" }{TEXT -1 4 ") X(" }{TEXT 288 1 "x" }{TEXT -1 3 "), " }}{PARA 0 "" 
0 "" {TEXT -1 28 "known as a product solution." }}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 20 "u:=(t,x)->T(t)*X(x);" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 38 "eq:=diff(u(t,x),t,t)-diff(u(t,x),x,x);" }}}{PARA 0 
"" 0 "" {TEXT -1 47 "     This simplifies if we divide through by T(" 
}{TEXT 290 1 "t" }{TEXT -1 4 ") X(" }{TEXT 291 1 "x" }{TEXT -1 2 ")." 
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "eq/X(x)/T(t);\nexpand(%);
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "sep:=(%)+diff(X(x),x,x)
/X(x)=diff(X(x),x,x)/X(x);" }}}{PARA 0 "" 0 "" {TEXT -1 35 "     The l
eft side of the equation " }{TEXT 261 3 "sep" }{TEXT -1 17 " depends o
nly on " }{TEXT 259 1 "t" }{TEXT -1 36 " and the right side depends on
ly on " }{TEXT 260 1 "x" }{TEXT -1 142 ". Thus, each side must be cons
tant. We do not know the value of this constant, yet. As in the heat e
quation, it will be negative; we call it -" }{XPPEDIT 18 0 "mu^2" "6#*
$%#muG\"\"#" }{TEXT -1 1 "." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
39 "dsolve(diff(X(x),x,x)=-mu^2*X(x),X(x));" }}}{PARA 0 "" 0 "" {TEXT 
-1 46 "In order to satisfy the boundary condition at " }{TEXT 292 1 "x
" }{TEXT -1 18 " = 0, we must have" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 16 "X:=x->sin(mu*x);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 16 "solve(X(L)=0,L);" }}}{PARA 0 "" 0 "" {TEXT -1 78 "The
re are an infinite number of solutions for this equation, they change \+
with " }{TEXT 293 1 "L" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "mu;" "6#%#
muG" }{TEXT -1 45 ". Maple did not pick out any solution except " }
{TEXT 294 1 "L" }{TEXT -1 111 " = 0. We know, however where the sine f
unction is zero: the sine function is zero at all integral multiples o
f " }{XPPEDIT 18 0 "pi;" "6#%#piG" }{TEXT -1 14 " . Thus, take " }
{XPPEDIT 18 0 "mu*L=n*Pi" "6#/*&%#muG\"\"\"%\"LGF&*&%\"nGF&%#PiGF&" }
{TEXT -1 15 " and solve for " }{XPPEDIT 18 0 "mu" "6#%#muG" }{TEXT -1 
1 "." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "solve(mu*L=n*Pi,mu);
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "mu:=%;" }}}{PARA 0 "" 0 
"" {TEXT -1 32 "     We now have the function X(" }{TEXT 295 1 "x" }
{TEXT -1 2 ")." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "X(x);" }}}
{PARA 0 "" 0 "" {TEXT -1 19 "Check that X'/X is " }{XPPEDIT 18 0 "-mu^
2;" "6#,$*$%#muG\"\"#!\"\"" }{TEXT -1 1 "." }}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 20 "diff(X(x),x,x)/X(x);" }}}{PARA 0 "" 0 "" {TEXT -1 
71 "     We now look at the other part of the PDE which we \"separated
 off\"." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "dsolve(diff(T(t),
t,t)=-mu^2*T(t),T(t));" }}}{PARA 0 "" 0 "" {TEXT -1 229 "     Both the
 sine and cosine functions give possible solutions. We have no reason \+
to eliminate either of these. The general solution we find is a linear
 combination of the particular solutions we get by separating the equa
tion. " }}{PARA 0 "" 0 "" {TEXT -1 56 "     The most general linear co
mbination is of the form." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
103 "u:=(t,x)->sum((A[n]*cos(Pi*n*t/L)+B[n]*sin(Pi*n*t/L))*sin(n*Pi*x/
L),\n                   n=1..infinity);" }}}{PARA 0 "" 0 "" {TEXT -1 
154 "     The coefficients A and B have to be determined from the init
ial conditions in a manner that is familiar. We use the initial condit
ions. Suppose that " }{XPPEDIT 18 0 "u(0,x)=f(x)" "6#/-%\"uG6$\"\"!%\"
xG-%\"fG6#F(" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "u[t](0,x)=g(x)" "6#/
-&%\"uG6#%\"tG6$\"\"!%\"xG-%\"gG6#F+" }{TEXT -1 1 "." }}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 31 "u(0,x)=f(x);\nD[1](u)(0,x)=g(x);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT 
-1 97 "Doesn't this look like a job for Fourier Series? To compute the
 coefficients, recall that we have" }}{PARA 0 "" 0 "" {TEXT -1 25 "   \+
                      " }{XPPEDIT 18 0 "A[n];" "6#&%\"AG6#%\"nG" }
{TEXT -1 3 " = " }{XPPEDIT 18 0 "int(f(x)*sin(n*Pi*x/L),x = 0 .. L)/in
t(sin(n*Pi*x/L)^2,x = 0 .. L);" "6#*&-%$intG6$*&-%\"fG6#%\"xG\"\"\"-%$
sinG6#**%\"nGF,%#PiGF,F+F,%\"LG!\"\"F,/F+;\"\"!F3F,-F%6$*$-F.6#**F1F,F
2F,F+F,F3F4\"\"#/F+;F7F3F4" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
3 "and" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 24 
"                        " }{XPPEDIT 18 0 "B[n]*n*Pi/L;" "6#**&%\"BG6#
%\"nG\"\"\"F'F(%#PiGF(%\"LG!\"\"" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "in
t(g(x)*sin(n*Pi*x/L),x = 0 .. L)/int(sin(n*Pi*x/L)^2,x = 0 .. L);" "6#
*&-%$intG6$*&-%\"gG6#%\"xG\"\"\"-%$sinG6#**%\"nGF,%#PiGF,F+F,%\"LG!\"
\"F,/F+;\"\"!F3F,-F%6$*$-F.6#**F1F,F2F,F+F,F3F4\"\"#/F+;F7F3F4" }
{TEXT -1 4 " =  " }{XPPEDIT 18 0 "2/L;" "6#*&\"\"#\"\"\"%\"LG!\"\"" }
{TEXT -1 1 " " }{XPPEDIT 18 0 "int(g(x)*sin(n*Pi*x/L),x = 0 .. L);" "6
#-%$intG6$*&-%\"gG6#%\"xG\"\"\"-%$sinG6#**%\"nGF+%#PiGF+F*F+%\"LG!\"\"
F+/F*;\"\"!F2" }{TEXT -1 5 " , or" }}{PARA 0 "" 0 "" {TEXT -1 23 "    \+
                   " }{XPPEDIT 18 0 "B[n];" "6#&%\"BG6#%\"nG" }{TEXT 
-1 3 " = " }{XPPEDIT 18 0 "2/(n*Pi);" "6#*&\"\"#\"\"\"*&%\"nGF%%#PiGF%
!\"\"" }{TEXT -1 1 " " }{XPPEDIT 18 0 "int(g(x)*sin(n*Pi*x/L),x = 0 ..
 L);" "6#-%$intG6$*&-%\"gG6#%\"xG\"\"\"-%$sinG6#**%\"nGF+%#PiGF+F*F+%
\"LG!\"\"F+/F*;\"\"!F2" }{TEXT -1 3 "  ." }}{PARA 0 "" 0 "" {TEXT -1 
250 "Here is an example. We can animate the graph of the solution and \+
expect to see the vibrations of the string. The following two examples
 may give considerable insight for how the wave equation models the mo
tion of a string. In the first example, take " }{TEXT 296 1 "f" }
{TEXT -1 14 " as below and " }{TEXT 297 1 "g" }{TEXT -1 131 " to be ze
ro. This will mean that all the B coefficients will be zero and A coef
ficients will be determined by the Fourier quotient." }}{PARA 0 "" 0 "
" {TEXT -1 7 "       " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 96 "L:=
4;\nf:=x->(x-1)*(Heaviside(x-1)-Heaviside(x-2))+\n       (3-x)*(Heavis
ide(x-2)-Heaviside(x-3));" }}}{PARA 0 "" 0 "" {TEXT -1 37 "We sketch t
he graph of this function " }{TEXT 298 1 "f" }{TEXT -1 1 "." }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "plot(f(x),x=0..L);" }}}{PARA 0 "" 
0 "" {TEXT -1 44 "We compute the Fourier coefficients.  Since " }
{XPPEDIT 18 0 "g(x)=0" "6#/-%\"gG6#%\"xG\"\"!" }{TEXT -1 36 ", the B c
oefficients are also zero.." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
132 "for n from 1 to 5 do\n     A[n]:=int(f(x)*sin(n*Pi*x/L),x=0..L)/
\n              int(sin(n*Pi*x/L)^2,x=0..L);\n     B[n]:=0:\nod;\nn:='
n':" }}}{PARA 0 "" 0 "" {TEXT -1 23 "We define the solution " }{TEXT 
299 1 "u" }{TEXT -1 1 "." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 96 "
u:=(t,x)->sum((A[n]*cos(Pi*n*t/L)+B[n]*sin(Pi*n*t/L))*sin(n*Pi*x/L),\n
                   n=1..5);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 
0 {PARA 3 "" 0 "" {TEXT 304 29 "Check that this is a solution" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "diff(u(t,x),t,t)-diff(u(t,x)
,x,x);" }}}{PARA 0 "" 0 "" {TEXT -1 33 "Here are the boundary conditio
ns." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "u(t,0); u(t,L);" }}}
{PARA 0 "" 0 "" {TEXT -1 17 "We see how close " }{TEXT 300 1 "u" }
{TEXT -1 4 "(0, " }{TEXT 301 1 "x" }{TEXT -1 8 ") is to " }{TEXT 302 
1 "f" }{TEXT -1 1 "(" }{TEXT 303 1 "x" }{TEXT -1 2 ")." }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "plot([u(0,x),f(x)],x=0..L);" }}}
{PARA 0 "" 0 "" {TEXT -1 37 "We check the other initial condition." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "D[1](u)(0,x);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}}{PARA 0 "" 0 "" {TEXT -1 26 "Now, we draw the graph of " }{TEXT 305 
1 "u" }{TEXT -1 90 " and then we watch an animation. We should expect \+
to see the motion of a vibrating string." }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 64 "plot3d(u(t,x),x=0..L,t=0..2*L,axes=NORMAL,orientation
=[-45,70]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "animate(u(t,
x),x=0..L,t=0..2*L, frames=30);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 286 "In the second example, take \+
f to be zero and g to be as below. The formula for f and g suggests th
at the string is at rest and we give the middle an initial velocity.  \+
This will mean that all the B coefficients will be zero and A coeffici
ents will be determined by the Fourier quotient." }}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 45 "L:=4;\ng:=x->-(Heaviside(x-1)-Heaviside(x-3));
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "plot(g(x),x=0..L,discon
t=true,color=RED);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 106 "n:='
n':\nfor n from 1 to 5 do\n     A[n]:=0:\n     B[n]:=2/(n*Pi)*int(g(x)
*sin(n*Pi*x/L),x=0..L);\nod;\nn:='n':" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 96 "u:=(t,x)->sum((A[n]*cos(Pi*n*t/L)+B[n]*sin(Pi*n*t/L))
*sin(n*Pi*x/L),\n                   n=1..5);" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 65 "plot3d(u(t,x),x=0..L,t=0..2*L,axes=NORMAL,orient
ation=[-135,45]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "animat
e(u(t,x),x=0..L,t=0..2*L, frames=30);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 3 "
" 0 "" {TEXT 306 16 "Unassisted Maple" }}{PARA 0 "" 0 "" {TEXT -1 143 
"Again, it is of interest to see what progress Maple can make with thi
s problem unassisted by humans. Note that we have read in the PDE pack
age." }}{PARA 0 "" 0 "" {TEXT -1 44 "We define the partial differentia
l equation." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "PDE:=diff(v(t
,x),t,t)-diff(v(t,x),x,x)=0;" }}}{PARA 0 "" 0 "" {TEXT -1 49 "Look to \+
see what form Maple gets for a solutions." }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 20 "ans := pdsolve(PDE);" }}}{PARA 0 "" 0 "" {TEXT -1 
139 "Maple suggests that solutions have a particular form. There are n
o Fourier coefficients, no integrals to compute. The answer seems comp
act." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 155 "
No surprise. This solution for the one dimensional wave equation is cl
assical. It is called the d'Alembert solution. It is the subject of th
e next section." }}{EXCHG }{EXCHG }{EXCHG }{EXCHG }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 50 "EMAIL: herod@math.gatech.
edu   or   jherod@tds.net" }}{PARA 0 "" 0 "" {TEXT -1 38 "URL: http://
www.math.gatech.edu/~herod" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
256 "" 0 "" {TEXT -1 36 "Copyright \251  2003  by James V. Herod" }}
{PARA 256 "" 0 "" {TEXT -1 19 "All rights reserved" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{MARK "0 0 0" 0 }
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