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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 256 37 "Section \+
4.1: The Simple Heat Equation" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{SECT 0 {PARA 3 "" 0 "" {TEXT 283 30 "Maple Packages for Section 4.1" 
}}{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 -1 227 "We consider the simple heat equation. This is a standard
 linear realization for diffusion in one dimension. Derivations for th
is model can be found in most texts in this subject. The equation typi
cally has the following form:  " }{TEXT 318 1 "u" }{TEXT -1 18 " is a \+
function of " }{TEXT 319 1 "t" }{TEXT -1 5 " and " }{TEXT 320 1 "x" }
{TEXT -1 19 " and is written as " }{TEXT 321 1 "u" }{TEXT -1 1 "(" }
{TEXT 322 1 "t" }{TEXT -1 2 ", " }{TEXT 323 1 "x" }{TEXT -1 19 "). We \+
suppose that " }{TEXT 324 1 "u" }{TEXT -1 36 " is differentiable as a \+
function of " }{TEXT 325 1 "t" }{TEXT -1 43 " and twice differentiable
 as a function of " }{TEXT 326 1 "x" }{TEXT -1 49 ". In this model, th
ese derivatives are related by" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 12 "            " }{XPPEDIT 18 0 "diff(u,t);
" "6#-%%diffG6$%\"uG%\"tG" }{TEXT -1 4 " =  " }{XPPEDIT 18 0 "diff(u,`
$`(x,2));" "6#-%%diffG6$%\"uG-%\"$G6$%\"xG\"\"#" }{TEXT -1 1 " " }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 149 "In the h
eat equation, there are boundary conditions and initial conditions. In
 this Section, we will take the boundary conditions to be specified at
 " }{TEXT 259 1 "x" }{TEXT -1 12 " = 0 and at " }{TEXT 260 1 "x" }
{TEXT -1 5 " = 1:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 7 "       " }{TEXT 261 1 "u" }{TEXT -1 15 "(t, 0) = 0 and " }
{TEXT 262 1 "u" }{TEXT -1 11 "(t, 1) = 0." }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 29 "We suppose that the value of " }
{TEXT 288 1 "u" }{TEXT -1 6 " when " }{TEXT 287 1 "t" }{TEXT -1 27 " =
 0 is specified as, say, " }{TEXT 257 1 "f" }{TEXT -1 1 "(" }{TEXT 
258 1 "x" }{TEXT -1 38 "). Thus, we have an initial condition:" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 6 "      " }
{TEXT 289 1 "u" }{TEXT -1 9 "(0, x) = " }{TEXT 263 1 "f" }{TEXT -1 1 "
(" }{TEXT 264 1 "x" }{TEXT -1 2 ")." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 348 "A physical realization of the above equ
ation is generally taken to be a thin, uniform bar over the interval [
0, 1]. In this simple model, one takes the lateral surface of the rod \+
to be insulated. Both ends of the rod are held at a fixed temperature.
 There are no internal or external heat sources. The rod has an initia
l heat distribution given by " }{TEXT 265 1 "f" }{TEXT -1 1 "(" }
{TEXT 266 1 "x" }{TEXT -1 457 "). The heat diffuses from the places wh
ere it is warm toward the places where it is cooler. The rate of diffu
sion is proportional to the curvature of the present heat distribution
. This statement about curvature takes the form of the second derivati
ve. Thus, if the current distribution is concave down, the second deri
vative is negative and the temperature at that point will be decreasin
g. If the distribution is concave up, the temperature is increasing." 
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 182 "We wil
l make the model more complicated later. For now, let's analyze this s
ituation with the experience and knowledge we have. To this point, the
 principle idea we have used is the " }{TEXT 290 12 "Fourier Idea" }
{TEXT -1 102 ". Here, we introduce the second idea: We suppose that th
e solution can be separate into a function of " }{TEXT 291 1 "t" }
{TEXT -1 14 " alone and of " }{TEXT 292 1 "x" }{TEXT -1 16 " alone, so
 that " }}{PARA 0 "" 0 "" {TEXT -1 10 "          " }{TEXT 293 1 "u" }
{TEXT -1 1 "(" }{TEXT 294 1 "t" }{TEXT -1 1 "," }{TEXT 295 1 "x" }
{TEXT -1 6 ") = X(" }{TEXT 296 1 "x" }{TEXT -1 4 ") T(" }{TEXT 297 1 "
t" }{TEXT -1 3 "). " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 54 "This notion for approaching the problem is called the \+
" }{TEXT 298 33 "Method of Separation of Variables" }{TEXT -1 1 "." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 73 "As we wil
l see, this assumption leads to ordinary differential equations." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "u:=(t,x)->X(x)*T(t);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "diff(u(t,x),t)=diff(u(t,x),x
,x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "
" {TEXT -1 33 "If we bring all the functions of " }{TEXT 271 1 "t" }
{TEXT -1 38 " to one side and all the functions of " }{TEXT 272 1 "x" 
}{TEXT -1 27 " to the other side, we have" }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 30 "          T ' / T  = X '' / X." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 38 "Since the
 left side is independent of " }{TEXT 299 1 "x" }{TEXT -1 10 ", then a
s " }{TEXT 300 1 "x" }{TEXT -1 172 " changes, X '' / X does not change
. This quotient is constant. In a similar manner, the quotient T ' / T
 is constant, and, from the equality, is the same constant. Call it " 
}{XPPEDIT 18 0 "mu;" "6#%#muG" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 36 "The boundary conditions g
ive that T(" }{TEXT 301 1 "t" }{TEXT -1 33 ") X(0) = 0. If there is a \+
single " }{TEXT 273 1 "t" }{TEXT -1 11 " so that T(" }{TEXT 302 1 "t" 
}{TEXT -1 114 ") is not zero, we can conclude that X(0) = 0. In a simi
lar manner, X(1) = 0. Thus, we have a differential equation" }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 17 "          X '' \+
= " }{XPPEDIT 18 0 "mu;" "6#%#muG" }{TEXT -1 25 " X, with X(0) = 0 = X
(1)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 104 "
We have already examined this differential equation. We found that the
re are an infinite number of such " }{XPPEDIT 18 0 "mu;" "6#%#muG" }
{TEXT -1 33 " 's and they are all of the form " }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 10 "          " }{XPPEDIT 18 
0 "mu = -n^2*Pi^2;" "6#/%#muG,$*&%\"nG\"\"#%#PiGF(!\"\"" }{TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 62 "and c
orresponding to each such constant, there is the solution" }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 11 "          X" }
{XPPEDIT 18 0 "` `[n];" "6#&%\"~G6#%\"nG" }{TEXT -1 2 " (" }{TEXT 303 
1 "x" }{TEXT -1 11 ") = sin( n " }{XPPEDIT 18 0 "pi;" "6#%#piG" }
{TEXT -1 1 " " }{TEXT 304 1 "x" }{TEXT -1 2 ")." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 86 "What about the possible T
 solutions? The quotient T '/ T is the same quotient, so that" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 16 "         \+
 T ' = " }{XPPEDIT 18 0 "-n^2*Pi^2;" "6#,$*&%\"nG\"\"#%#PiGF&!\"\"" }
{TEXT -1 3 " T." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 31 "Solutions for this equation are" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 11 "          T" }{XPPEDIT 18 0 "` \+
`[n];" "6#&%\"~G6#%\"nG" }{TEXT -1 2 " (" }{TEXT 305 1 "t" }{TEXT -1 
10 ") = exp(  " }{XPPEDIT 18 0 "-n^2*Pi^2;" "6#,$*&%\"nG\"\"#%#PiGF&!
\"\"" }{TEXT -1 1 " " }{TEXT 306 1 "t" }{TEXT -1 2 ")." }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 52 "Consequently, for ea
ch integer n, we have a solution" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 12 "          u(" }{TEXT 307 1 "t" }{TEXT -1 
2 ", " }{TEXT 308 1 "x" }{TEXT -1 10 ") = exp(  " }{XPPEDIT 18 0 "-n^2
*Pi^2;" "6#,$*&%\"nG\"\"#%#PiGF&!\"\"" }{TEXT -1 1 " " }{TEXT 309 1 "t
" }{TEXT -1 9 ") sin( n " }{XPPEDIT 18 0 "pi;" "6#%#piG" }{TEXT -1 1 "
 " }{TEXT 310 1 "x" }{TEXT -1 2 ")." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 198 "But, since this is a linear problem, co
nstant multiples of these solutions are also solutions, and (infinite)
 sums of solutions are solutions. Hence, a general solution can be wri
tten in the form of" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 12 "          u(" }{TEXT 274 1 "t" }{TEXT -1 2 ", " }{TEXT 
275 1 "x" }{TEXT -1 4 ") = " }{XPPEDIT 18 0 "sum(c[n]*exp(-n^2*Pi^2*t)
*sin(n*Pi*x),n);" "6#-%$sumG6$*(&%\"cG6#%\"nG\"\"\"-%$expG6#,$*(F*\"\"
#%#PiGF1%\"tGF+!\"\"F+-%$sinG6#*(F*F+F2F+%\"xGF+F+F*" }{TEXT -1 2 " .
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 136 "Ther
e is one more piece of information we have not used. We have not used \+
the initial value. This condition determines the coefficients:" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 10 "         \+
 " }{TEXT 276 1 "f" }{TEXT -1 1 "(" }{TEXT 277 1 "x" }{TEXT -1 9 ") = \+
u(0, " }{TEXT 278 1 "x" }{TEXT -1 4 ") = " }{XPPEDIT 18 0 "sum(c[n]*si
n(n*Pi*x),n);" "6#-%$sumG6$*&&%\"cG6#%\"nG\"\"\"-%$sinG6#*(F*F+%#PiGF+
%\"xGF+F+F*" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 55 "This looks like a job for Fourier Series. We kn
ow that " }{XPPEDIT 18 0 "c[n];" "6#&%\"cG6#%\"nG" }{TEXT -1 59 " can \+
be written as a quotient of integrals which resolve as" }}{PARA 0 "" 
0 "" {TEXT -1 8 "        " }{XPPEDIT 18 0 "c[n];" "6#&%\"cG6#%\"nG" }
{TEXT -1 5 "  =  " }{XPPEDIT 18 0 "int(f(x)*sin(n*Pi*x),x = 0 .. 1)/in
t(sin(n*Pi*x)^2,x = 0 .. 1);" "6#*&-%$intG6$*&-%\"fG6#%\"xG\"\"\"-%$si
nG6#*(%\"nGF,%#PiGF,F+F,F,/F+;\"\"!F,F,-F%6$*$-F.6#*(F1F,F2F,F+F,\"\"#
/F+;F5F,!\"\"" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 55 "It will be valuable at this point to take
 a particular " }{TEXT 311 1 "f" }{TEXT -1 65 ", solve the equation, a
nd draw some pictures. We take \n          " }{TEXT 267 1 "f" }{TEXT 
-1 1 "(" }{TEXT 268 1 "x" }{TEXT -1 4 ") = " }{XPPEDIT 18 0 "x*(1-x)^3
;" "6#*&%\"xG\"\"\"*$,&F%F%F$!\"\"\"\"$F%" }{TEXT -1 2 " ." }}{PARA 0 
"" 0 "" {TEXT -1 48 "To visualize the initial value, we draw a graph.
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "f:=x->x*(1-x)^3;" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "plot(f(x),x=0..1);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT 
-1 44 "We form the Fourier representation for this " }{TEXT 269 1 "f" 
}{TEXT -1 26 " and compare the graph of " }{TEXT 270 1 "f" }{TEXT -1 
172 " and of our series representation. First, note that the function \+
is continuous and its periodic extension is continuous, so that the Fo
urier series will converge uniformly." }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 101 "for n from 1 to 7 do\n   c[n]:=int(f(x)*sin(n*Pi*x),
x=0..1)/int(sin(n*Pi*x)^2,x=0..1):\nend do:\nn:='n';" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 40 "approx:=x->sum(c[n]*sin(n*Pi*x),n=1..7):
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "plot([f(x),approx(x)],x
=0..1,color=[BLACK,RED]);" }}}{EXCHG }{PARA 0 "" 0 "" {TEXT -1 91 "As \+
you can see, this is a very good fit. Now, we form the truncated serie
s expansion for u." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "u:=(t,
x)->sum(c[n]*exp(-n^2*Pi^2*t)*sin(n*Pi*x),n=1..7);" }}}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT 284 30 "Check that thi
s is a solution." }}{PARA 0 "" 0 "" {TEXT -1 108 "There are three thin
gs that should be checked. First, does this u satisfy the partial diff
erential equation?" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "diff(u
(t,x),t)-diff(u(t,x),x,x);" }}}{PARA 0 "" 0 "" {TEXT -1 48 "Second, do
es it satisfy the boundary conditions?" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 14 "u(t,0),u(t,1);" }}}{PARA 0 "" 0 "" {TEXT -1 62 "Final
ly, is it a good approximation for f as an initial value?" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "plot([f(x),u(0,x)],x=0..1,color=[re
d,green]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 
"" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT 285 25 "The grap
h of the solution" }}{PARA 0 "" 0 "" {TEXT -1 76 "Before drawing the g
raph of u, what is expected? We should expect that when " }{TEXT 313 
1 "t" }{TEXT -1 24 " = 0, the graph of u(0, " }{TEXT 314 1 "x" }{TEXT 
-1 26 ") looks like the graph of " }{TEXT 315 1 "f" }{TEXT -1 1 "(" }
{TEXT 316 1 "x" }{TEXT -1 15 "). Further, as " }{TEXT 317 1 "t" }
{TEXT -1 96 " increases, solution decreases to the stationary solution
 determined by the boundary conditions:" }}{PARA 0 "" 0 "" {TEXT -1 
15 "               " }{XPPEDIT 18 0 "u(infinity, x) = 0;" "6#/-%\"uG6$
%)infinityG%\"xG\"\"!" }{TEXT -1 2 " ." }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 65 "plot3d(u(t,x),x=0..1,t=0..1/10,axes=NORMAL,orientatio
n=[-20,55]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT 312 27 "A char
acter of the solution" }}{PARA 0 "" 0 "" {TEXT -1 251 "In designing th
is problem, I deliberately chose an initial distribution for which the
 highpoint of the graph is off center. It appears that this highpoint \+
moves in toward the center of the interval [0, 1]. We can look at this
 movement in an animation." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
43 "animate(u(t,x),x=0..1,t=0..1/3, frames=30);" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 261 "Observe that
 the point of highest temperature moves. We trace the movement of the \+
high point. First note that the high point of the initial distribution
 is at 1/4. Here's how to see that: It was clear that the first deriva
tive was zero in the interval [0, 1/2]." }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 32 "solve(\{diff(f(x),x)=0,x<1/2\},x);" }}}{EXCHG }{PARA 
0 "" 0 "" {TEXT -1 323 "Now we find where the temperature is highest a
s time evolves. Solving the equation for where the first derivative is
 zero in the approximation is harder to do. We make this determination
 numerically. That is, we get a floating point approximation for the d
erivatives. We expect to see the first derivatives converge to 1/2." }
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 73 "for p from 0 to 10 do\nfsol
ve(subs(t=p/30,diff(u(t,x),x))=0,x,0..1/2);\nod;" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 
"" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 151 "     In this wor
ksheet, we have solved a simple heat equation with zero boundary condi
tions and with an initial distribution. Here are the main steps: " }}
{PARA 0 "" 0 "" {TEXT 279 7 "Step 1:" }{TEXT -1 168 " By separation of
 variables, give two differential equations with boundary conditions w
hose solutions will determine the solution for the partial differentia
l equation." }}{PARA 0 "" 0 "" {TEXT 280 7 "Step 2:" }{TEXT -1 72 " So
lve the two ordinary differential equations with boundary conditions.
" }}{PARA 0 "" 0 "" {TEXT 281 7 "Step 3:" }{TEXT -1 94 " Construct the
 general solution for the partial differential equation with boundary \+
condition." }}{PARA 0 "" 0 "" {TEXT 282 7 "Step 4:" }{TEXT -1 67 " Fin
d the particular solution determined by the initial conditions." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT 286 16 "
Unassisted Maple" }}{PARA 0 "" 0 "" {TEXT -1 145 "Finally, it is of in
terest to see what progress Maple can make with this problem unassiste
d by humans. Note that we have read in the PDE package." }}{PARA 0 "" 
0 "" {TEXT -1 44 "We define the partial differential equation." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "PDE:=diff(v(t,x),t)-diff(v(t
,x),x,x)=0;" }}}{PARA 0 "" 0 "" {TEXT -1 49 "Look to see what form Map
le gets for a solutions." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "
ans := pdsolve(PDE);" }}}{PARA 0 "" 0 "" {TEXT -1 92 "Maple reminds us
 that we should take a product of solutions, just as we had suggested \+
above." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "build(ans);" }}}
{PARA 0 "" 0 "" {TEXT -1 257 "Maple gets generic solutions to these tw
o partial differential equations that it specified. Exercise care. Nei
ther the initial condition nor the boundary conditions have been used.
 This really is a solution, only it does not satisfy the boundary cond
itions." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "v:=(t,x)->A*exp(9
*t)*exp(3*x)+B*exp(9*t)*exp(-3*x);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 32 "diff(v(t,x),t)-diff(v(t,x),x,x);" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 15 "v(t,0);\nv(t,1);" }}}{PARA 0 "" 0 "" {TEXT 
-1 184 "We could take sums of these, but then we would have lost the o
rthogonality. It seems that sine functions makes a better choice. The \+
structure of the Fourier Series has many advantages." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 76 "Be aware that Maple can
 handle this problem very well with other techniques." }}{EXCHG }
{EXCHG }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 50 "
EMAIL: herod@math.gatech.edu   or   jherod@tds.net" }}{PARA 0 "" 0 "" 
{TEXT -1 38 "URL: http://www.math.gatech.edu/~herod" }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 36 "Copyright \251  2003 \+
 by James V. Herod" }}{PARA 256 "" 0 "" {TEXT -1 19 "All rights reserv
ed" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{MARK "0 0 0" 8 }{VIEWOPTS 1 1 0 1 1 
1803 1 1 1 1 }{PAGENUMBERS 1 1 2 33 1 1 }