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{SECT 0 {PARA 0 "" 0 "" {TEXT 256 37 "Section 4.1: The Simple Heat Equ
ation" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {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(PD
Etools):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 227 "We consider the s
imple heat equation. This is a standard linear realization for diffusi
on in one dimension. Derivations for this model can be found in most t
exts in this subject. The equation typically 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, these 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 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 153 "In the heat equation, there are boundary
conditions and initial conditions. In this Section 4.1, 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 equation is generally taken to be
a thin, uniform bar over the interval [0, 1]. In this simple model, o
ne 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 exte
rnal heat sources. The rod has an initial heat distribution given by \+
" }{TEXT 265 1 "f" }{TEXT -1 1 "(" }{TEXT 266 1 "x" }{TEXT -1 519 "). \+
The heat diffuses from the places where it is warm toward the places w
here it is cooler and from places where it is cold to places where it \+
is warmer. The rate of diffusion is proportioned to the curvature of t
he present heat distribution. This statement about curvature takes the
form of the second derivative. Thus, if the current distribution is c
oncave down, the second derivative is negative and the temperature at \+
that point will be decreasing. If the distribution is concave up, the \+
temperature is increasing." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0
"" 0 "" {TEXT -1 194 "We will make the model more complicated later. F
or now, let's make an analysis of this situation with the experience a
nd knowledge we have. To this point, the principle idea we have used i
s the " }{TEXT 290 12 "Fourier Idea" }{TEXT -1 102 ". Here, we introdu
ce the second idea: We suppose that the 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 approac
hing the problem is called the " }{TEXT 298 33 "Method of Separation o
f Variables" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA
0 "" 0 "" {TEXT -1 73 "As we will see, this assumption leads to ordina
ry 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 as " }{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 equ
ality, 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 give 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 zer
o, we can conclude that X(0) = 0. In a similar 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 there are an infinite numbe
r 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,
$*&%\"nG\"\"#%#PiGF(!\"\"" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1
0 "" }}{PARA 0 "" 0 "" {TEXT -1 62 "and corresponding to each such con
stant, there is the solution" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA
0 "" 0 "" {TEXT -1 11 " X" }{XPPEDIT 18 0 "` `[n];" "6#&%\"~G
6#%\"nG" }{TEXT -1 1 " " }{TEXT -1 1 "(" }{TEXT 303 1 "x" }{TEXT -1
11 ") = sin( n " }{XPPEDIT 18 0 "pi;" "6#%#piG" }{TEXT -1 1 " " }
{TEXT 304 1 "x" }{TEXT -1 2 ")." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 86 "What about the possible T solutions? The \+
quotient T '/ T is the same quotient, so that" }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 16 " T ' = " }
{XPPEDIT 18 0 "-n^2*Pi^2;" "6#,$*&%\"nG\"\"#%#PiGF&!\"\"" }{TEXT -1 3
" T." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 31 "S
olutions 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 1 " " }{TEXT -1 1 "(" }{TEXT 305 1 "t" }
{TEXT -1 10 ") = exp( " }{XPPEDIT 18 0 "-n^2*Pi^2;" "6#,$*&%\"nG\"\"#
%#PiGF&!\"\"" }{TEXT -1 1 " " }{TEXT 306 1 "t" }{TEXT -1 2 ")." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 52 "Consequen
tly, for each integer n, we have a solution" }}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 12 " u(" }{TEXT 307 1 "t"
}{TEXT -1 2 ", " }{TEXT 308 1 "x" }{TEXT -1 10 ") = exp( " }{XPPEDIT
18 0 "-n^2*Pi^2;" "6#,$*&%\"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 -1 187 "But, since this is a lin
ear problem, constant multiples of these solutions are also solutions,
and sums of solutions are solutions. Hence, a general solution can be
written 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*P
i^2*t)*sin(n*Pi*x),n);" "6#-%$sumG6$*(&%\"cG6#%\"nG\"\"\"-%$expG6#,$*(
F*\"\"#%#PiGF1%\"tGF+!\"\"F+-%$sinG6#*(F*F+F2F+%\"xGF+F+F*" }{TEXT -1
2 " ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 136
"There is one more piece of information we have not used. We have not \+
used the initial value. This condition determines the coefficients:" }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 10 " \+
" }{TEXT 276 1 "f" }{TEXT -1 1 "(" }{TEXT 277 1 "x" }{TEXT -1 9 ") =
u(0, " }{TEXT 278 1 "x" }{TEXT -1 4 ") = " }{XPPEDIT 18 0 "sum(c[n]*s
in(n*Pi*x),n);" "6#-%$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 97 "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):\nod:\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 "" {MPLTEXT 1 0 0 "" }
{TEXT -1 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 91 "As you can see, this is \+
a very good fit. Now, we form the truncated series 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 1 {PARA 3 "" 0 "" {TEXT 284 30 "Check that this is a solution."
}}{PARA 0 "" 0 "" {TEXT -1 108 "There are three things that should be \+
checked. First, does this u satisfy the partial differential 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, does it satisfy the \+
boundary conditions?" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "u(t,
0),`and`, u(t,1);" }}}{PARA 0 "" 0 "" {TEXT -1 62 "Finally, is it a go
od 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=[red,green]);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT 285 25 "The graph of the solut
ion" }}{PARA 0 "" 0 "" {TEXT -1 76 "Before drawing the graph of u, wha
t 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 ") look
s 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 " incre
ases, 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,orientation=[-20,55]);" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT -1 0 ""
}}{SECT 1 {PARA 3 "" 0 "" {TEXT 312 27 "A character of the solution" }
}{PARA 0 "" 0 "" {TEXT -1 251 "In designing this problem, I deliberate
ly chose an initial distribution for which the highpoint of the graph \+
is off center. It appears that this highpoint moves in toward the cent
er of the interval [0, 1]. We can look at this movement in an animatio
n." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "animate(u(t,x),x=0..1,
t=0..1/3);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "
" 0 "" {TEXT -1 261 "Observe that the point of highest temperature mov
es. 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: I
t was clear that the first derivative 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 "" {MPLTEXT 1 0 0 "" }}}{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 1 {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 -1 119 "Be aware that Maple ca
n handle this problem very well with other techniques. See Section 8.2
for an alternate approach." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0
0 "" }}}}{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 }{VIEWOPTS 1 1 0 1 1
1803 1 1 1 1 }{PAGENUMBERS 1 1 2 33 1 1 }