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{SECT 0 {PARA 18 "" 0 "" {TEXT -1 37 "Linear Methods of Applied Mathem
atics" }}{PARA 256 "" 0 "" {TEXT -1 27 "The Mathematics of Hot Rods" }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 62 "
Copyright 1998, 2000 by Evans M. Harrell II and James V. Herod" }}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 24 "T
he simple heat equation" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 433 " We consider the simple heat equatio
n. This is a standard linear realization for diffusion in one dimensio
n. Derivations for this model can be found in most texts in this subje
ct. The equation typically has the following form: u is a function of
t and x and is written as u(t, x). We suppose that u is differentiabl
e as a function of t and twice differentiable as a function of x. In t
his model, these derivatives are related by" }}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 10 " " }{XPPEDIT 18 0 "dif
f(u,t);" "6#-%%diffG6$%\"uG%\"tG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "di
ff(u,`$`(x,2));" "6#-%%diffG6$%\"uG-%\"$G6$%\"xG\"\"#" }{TEXT -1 2 " .
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 170 "In t
he heat equation, there are boundary conditions and initial conditions
. In this worksheet, we will take the boundary conditions to be specif
ied at x = 0 and at x = 1:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0
"" 0 "" {TEXT -1 35 " u(t, 0) = 0 and u(t, 1) = 0." }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 69 "We suppose that th
e value of u when t = 0 is specified as, say, f(x):" }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 21 " u(0, x) = f(x)."
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 804 " A
n intuitive way to think of this equation is that there is a thin, uni
form bar over the interval [0, 1]. In this simple problem, one takes t
he lateral surface of the rod to be insulated. Both ends of the rod ar
e held at a fixed temperature. There are no internal or external heat \+
sources. The rod has an intial heat distribution given by f(x). The he
at diffuses from the places where it is warm toward the places where i
t is cooler. The rate of diffusion is proportioned to the curvature of
the present heat distribution. This statement about curvature takes t
he form of the second derivative. Thus, if the current distribution is
concave down, the second derivative is negative and the temperature a
t that point will be decreasing. if the distribution is concave up, th
e temperature is increasing." }}{PARA 0 "" 0 "" {TEXT -1 345 " We \+
will make the model more complicated later. For now, let's make an ana
lysis of this situation with the experience and knowledge we have. To \+
this point, the principle idea we have used is the Fourier Idea. Here,
we introduce the second idea: We suppose that the solution can be sep
arate into a function of t alone and of x alone, so that " }}{PARA 0 "
" 0 "" {TEXT -1 30 " u(t,x) = X(x) T(t). " }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 57 "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 -1 100 "If we brin
g all the functions of t to one side and all the functions of x 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 -1 215 "Since the right side is \+
independent of x, then as x changes, X '' / X does not change. This qu
otient is constant. In a similar manner, the quotient T ' / T is const
ant, and, from the equality, is the same constant. " }}{PARA 0 "" 0 "
" {TEXT -1 202 " The boundary conditions give that T(t) X(0) = 0. \+
If there is a single t so that T(t) is not zero, 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 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\"\"#%#PiG\"\"#!
\"\"" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 62 "and corresponding to each such constant, there is the s
olution" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1
24 " X(x) = sin( n " }{XPPEDIT 18 0 "pi;" "6#%#piG" }{TEXT
-1 4 " x)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT
-1 91 " What about the possible T solutions? The quotient T '/ T i
s 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#,$*&%\"nG\"\"#%#PiG\"\"#!\"\"" }{TEXT -1 3 " T." }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 31 "Solutions for this equati
on are" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 23
" T(t) = exp( " }{XPPEDIT 18 0 "-n^2*Pi^2;" "6#,$*&%\"nG\"\"
#%#PiG\"\"#!\"\"" }{TEXT -1 4 " t)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 52 "Consequently, for each integer n, we hav
e a solution" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT
-1 26 " u(t, x) = exp( " }{XPPEDIT 18 0 "-n^2*Pi^2;" "6#,$*&
%\"nG\"\"#%#PiG\"\"#!\"\"" }{TEXT -1 11 " t) sin( n " }{XPPEDIT 18 0 "
pi;" "6#%#piG" }{TEXT -1 4 " x)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 192 "But, since this is a linear problem, mul
tiples of these solutions by a constant are also solutions, and sums o
f solutions are solutions. Hence, a general solution can be written in
the form of" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT
-1 20 " u(t, x) = " }{XPPEDIT 18 0 "sum(c[n]*exp(-n^2*Pi^2*t)
*sin(n*Pi*x),n);" "6#-%$sumG6$*(&%\"cG6#%\"nG\"\"\"-%$expG6#,$*(F*\"\"
#%#PiG\"\"#%\"tGF+!\"\"F+-%$sinG6#*(F*F+F2F+%\"xGF+F+F*" }{TEXT -1 2 "
." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 136 "Th
ere is one more piece of information we have not used. We have not use
d the initial value. This condition determines the coefficients:" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 27 " \+
f(x) = u(0, x) = " }{XPPEDIT 18 0 "sum(c[n]*sin(n*Pi*x),n);" "6#-%$su
mG6$*&&%\"cG6#%\"nG\"\"\"-%$sinG6#*(F*F+%#PiGF+%\"xGF+F+F*" }{TEXT -1
1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 55 "T
his looks like a job for Fourier Series. We know that " }{XPPEDIT 18
0 "c[n];" "6#&%\"cG6#%\"nG" }{TEXT -1 59 " can be written as a quotien
t of integrals which resolve as" }}{PARA 0 "" 0 "" {TEXT -1 3 " " }}
{PARA 0 "" 0 "" {TEXT -1 8 " " }{XPPEDIT 18 0 "c[n];" "6#&%\"cG
6#%\"nG" }{TEXT -1 5 " = 2 " }{XPPEDIT 18 0 "int(f(x)*sin(n*Pi*x),x = \+
0 .. 1);" "6#-%$intG6$*&-%\"fG6#%\"xG\"\"\"-%$sinG6#*(%\"nGF+%#PiGF+F*
F+F+/F*;\"\"!\"\"\"" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 0 ""
}}{PARA 0 "" 0 "" {TEXT -1 128 "It will be valuable at this point to t
ake a particular f, solve the equation, and draw some pictures. We tak
e \n f(x) = " }{XPPEDIT 18 0 "x*(1-x)^3;" "6#*&%\"xG\"\"\"*$,
&\"\"\"F%F$!\"\"\"\"$F%" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1
48 "To visualize the initial value, we draw a graph." }}{EXCHG {PARA
0 "> " 0 "" {MPLTEXT 1 0 25 "restart:\nf:=x->x*(1-x)^3;" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "plot(f(x),x=0..1);" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 249 " \+
We form the Fourier representation for this f and compare the grap
h of f and of our series representation. First, note that the function
is continuous and its periodic extension is continuous, so that the F
ourier series will converge uniformly." }}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 73 "for n from 1 to 7 do\n c[n]:=2*int(f(x)*sin(n*Pi*x)
,x=0..1):\nod:\nn:='n';" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "
Fou:=x->sum(c[n]*sin(n*Pi*x),n=1..7):" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 45 "plot([f(x),Fou(x)],x=0..1,color=[BLACK,RED]);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }{TEXT -1 0 "" }}}{PARA 0 "
" 0 "" {TEXT -1 96 " As you can see, this is a very good fit. Now,
we form the truncated series expansion for u." }}{EXCHG {PARA 0 "> "
0 "" {MPLTEXT 1 0 56 "u:=(t,x)->sum(c[n]*exp(-n^2*Pi^2*t)*sin(n*Pi*x),
n=1..7);" }}}{PARA 0 "" 0 "" {TEXT -1 242 " Before drawing the gra
ph, what is expected? We should expect that when t = 0, the graph of u
(0, x) looks like the graph of f(x). Further, as t increases, solution
decreases to the stationary solution determined by the boundary condt
ions:" }}{PARA 0 "" 0 "" {TEXT -1 15 " " }{XPPEDIT 18 0
"u(t,infinity) = 0;" "6#/-%\"uG6$%\"tG%)infinityG\"\"!" }{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 -1 254 " In designing
this problem, I delibertly chose an initial distribution for which th
e 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 thi
s movement in an animation." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0
12 "with(plots):" }}}{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 -1 189 " We trace the movement of the \+
high point. First note that the high point of the initial distribution
is at 1/4. It was clear that the first derivative was zero in the int
erval [0, 1/2]." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "solve(\{d
iff(f(x),x)=0,x<1/2\},x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0
"" }}}{PARA 0 "" 0 "" {TEXT -1 263 "Solving the equation for where the
first derivative is zero in the approximation is harder to find. We m
ake this determination numerically. That is, we get a floating point a
pproximation for the derivatives. We expect to see the first derivativ
es converge to 1/2." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 73 "for p
from 0 to 10 do\nfsolve(subs(t=p/30,diff(u(t,x),x))=0,x,0..1/2);\nod;
" }}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 8 "Exercise" }}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 122 " In this sect
ion, we have solved a simple heat equation with zero boundary conditon
s and with an initial distribution." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 89 "Assignment: Solve the same distribution \+
with this different initial distribution: f(x) = " }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 16 "f:=x->x^3*(1-x);" }}}{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 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 107 "Note in preparation: Here is how I drew the sequence of \+
snapshots to observe the movement of the highpoint." }}{EXCHG {PARA 0
"> " 0 "" {MPLTEXT 1 0 69 "n:='n':\nfor i from 1 to 5 do\n P[i]:=plo
t(u((i-1)/15,x),x=0..1):\nod:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1
0 49 "i:='i':\nwith(plots):\ndisplay([seq(P[i],i=1..5)]);" }}}}}{SECT
1 {PARA 3 "" 0 "" {TEXT -1 31 "Variations on the heat equation" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}
{PARA 0 "" 0 "" {TEXT -1 169 " We illustrate the generality of the
simple heat equation with this problem which has the appearance of be
ing more general. Here is the equation we will investigate:" }}{PARA
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 44 "In this workshe
et we call this equation (*4)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 13 " " }{XPPEDIT 18 0 "diff(Z,t);
" "6#-%%diffG6$%\"ZG%\"tG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "diff(Z,`$
`(x,2));" "6#-%%diffG6$%\"ZG-%\"$G6$%\"xG\"\"#" }{TEXT -1 14 " - 4 (Z
- 32)" }}{PARA 0 "" 0 "" {TEXT -1 25 " " }}
{PARA 0 "" 0 "" {TEXT -1 31 " Z(t,0) = 50 = Z(t,1)," }}{PARA
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 23 " Z(0, \+
x) = 50." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1
458 " To give us intuition for the behavior of solutions, we think
of this equation as being a model for the diffusion of heat along the
rod and for the radiation of heat to the environment where the temper
ature is 32. ( For radiation into the environment, we assume Newton's \+
Law of Cooling, which says that the rate of cooling is proportioned to
the difference in the temperature and the temperature of the surrondi
ng mediam. ) The endpoints are held at 50." }}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 54 "There are two techniques for ho
w to proceed from here." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 ""
0 "" {TEXT -1 101 "Method 1. modify the solution for each of the follo
wing equations to produce a solution for the next." }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 18 "Equation (*1) " }
{XPPEDIT 18 0 "diff(u,t);" "6#-%%diffG6$%\"uG%\"tG" }{TEXT -1 3 " = "
}{XPPEDIT 18 0 "diff(u,`$`(x,2));" "6#-%%diffG6$%\"uG-%\"$G6$%\"xG\"\"
#" }{TEXT -1 49 " , with u(t, 0) = 0, u(t, 1) = 0, u(0, x) = " }
{XPPEDIT 18 0 "f[1];" "6#&%\"fG6#\"\"\"" }{TEXT -1 4 "(x)." }}{PARA 0
"" 0 "" {TEXT -1 19 "Equation (*2) " }{XPPEDIT 18 0 "diff(v,t);"
"6#-%%diffG6$%\"vG%\"tG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "diff(v,`$`(
x,2));" "6#-%%diffG6$%\"vG-%\"$G6$%\"xG\"\"#" }{TEXT -1 51 " - 4 v, w
ith v(t, 0) = 0, v(t, 1) = 0, v(0, x) = " }{XPPEDIT 18 0 "f[2];" "6#&
%\"fG6#\"\"#" }{TEXT -1 23 "(x).\nEquation (*3) " }{XPPEDIT 18 0 "
diff(w,t);" "6#-%%diffG6$%\"wG%\"tG" }{TEXT -1 3 " = " }{XPPEDIT 18 0
"diff(w,`$`(x,2));" "6#-%%diffG6$%\"wG-%\"$G6$%\"xG\"\"#" }{TEXT -1
54 " - 4 w , with w(t, 0) = 18, w(t, 1) = 18, w(0, x) = " }{XPPEDIT
18 0 "f[3];" "6#&%\"fG6#\"\"$" }{TEXT -1 4 "(x)." }}{PARA 0 "" 0 ""
{TEXT -1 18 "Equation (*4) " }{XPPEDIT 18 0 "diff(Z,t);" "6#-%%dif
fG6$%\"ZG%\"tG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "diff(Z,`$`(x,2));" "
6#-%%diffG6$%\"ZG-%\"$G6$%\"xG\"\"#" }{TEXT -1 62 " - 4 (Z - 32), wit
h Z(t, 0) = 50, Z(t, 1) = 50, Z(0, x) = 50." }}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 401 "Method 2. This method is not s
o different from the previous, Only, the construction begins with equa
tion (*4). Suppose you know a solution for (*3) and modify it to get a
solution for (*4). Then suppose you know a solution for (*2) and modi
fy it to get a solution for (*3). Finally, you know a solution for (*1
), so modify it to make a solution for (*2). Now, you know solution fo
r all the equations. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0
"" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 233 "We derive the solutio
n directly, or we present how we came to that derivation. Thus, at thi
s point, we make a choice for how to proceed: Move directly to the der
ivation of a solution. Later, we explain how the solution was conceive
d." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT
-1 113 "How to Construction the Solution for The Heat Equation with Ra
diation Cooling from the simple heat equation (*1)." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 156 "We begin with the simple heat equation and gradually mak
e more complicated equations. At each step, we will know how to solve \+
the more complicated equation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 76 "First, there is the simple heat equation \+
(*1) with zero boundary conditions." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 10 " \+
" }{XPPEDIT 18 0 "diff(u,t);" "6#-%%diffG6$%\"uG%\"tG" }{TEXT -1 3 "
= " }{XPPEDIT 18 0 "diff(u,`$`(x,2));" "6#-%%diffG6$%\"uG-%\"$G6$%\"x
G\"\"#" }{TEXT -1 40 " , with u(t, 0) = 0 and u(t, 1) = 0." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 25 "This has \+
general solution" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 18 " u(t, x) = " }{XPPEDIT 18 0 "sum(c[n]*exp(-n^2*Pi^
2*t)*sin(n*Pi*x),n);" "6#-%$sumG6$*(&%\"cG6#%\"nG\"\"\"-%$expG6#,$*(F*
\"\"#%#PiG\"\"#%\"tGF+!\"\"F+-%$sinG6#*(F*F+F2F+%\"xGF+F+F*" }{TEXT
-1 2 " ," }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1
54 "as we derived in the last worksheet. The coefficients " }{XPPEDIT
18 0 "c[n];" "6#&%\"cG6#%\"nG" }{TEXT -1 226 " are determined by know
ing the initial condition. At this time, we take the initial condition
to be u(0, x) = f(x) and decide later what f should be inorder to att
ain the desired solution. This u satisfies the equation (*1)." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 30 "We creat
e v(t, x) as follows:" }}{PARA 0 "" 0 "" {TEXT -1 5 " " }}{PARA 0
"" 0 "" {TEXT -1 44 " v(t, x) = exp(-4 t) u(t, x). " }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 82 "We ask: w
hat partial differential equation and boundary conditions will v satis
fy?" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "v:=(t,x)->exp(-4*t)*u
(t,x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "diff(v(t,x),t);"
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 55 "Because of the equation (
*1), we have that this last is" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 41 " -4 exp(4 t) u(t, x) + exp(4 t) \+
" }{XPPEDIT 18 0 "diff(u,`$`(x,2));" "6#-%%diffG6$%\"uG-%\"$G6$%\"xG\"
\"#" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "diff(v,`$`(x,2));" "6#-%%diffG6
$%\"vG-%\"$G6$%\"xG\"\"#" }{TEXT -1 7 " - 4 v." }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 40 "That is, v(t, x) satisfie
s equation(*2):" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 14 " " }{XPPEDIT 18 0 "diff(v,t);" "6#-%%diffG6$
%\"vG%\"tG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "diff(v,`$`(x,2));" "6#-%
%diffG6$%\"vG-%\"$G6$%\"xG\"\"#" }{TEXT -1 55 " - 4 v, with boundary \+
conditions v(t, 0) = 0 = v(t, 1)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 53 "and initial condition v(0,x) = exp(0) u(0
, x) = f(x)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT
-1 34 "We now make function w as follows:" }}{PARA 0 "" 0 "" {TEXT -1
0 "" }}{PARA 0 "" 0 "" {TEXT -1 30 " w(t, x) = v(t, x) + " }
{XPPEDIT 18 0 "18*(sinh(2*x)-sinh(2*(x-1)))/sinh(2);" "6#*(\"#=\"\"\",
&-%%sinhG6#*&\"\"#F%%\"xGF%F%-F(6#*&\"\"#F%,&F,F%\"\"\"!\"\"F%F3F%-F(6
#\"\"#F3" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 120 "Now we repeat the question for this function: wha
t partial differential equation and boundary conditions will w satisfy
?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 300 " \+
The best way to think of this is to recall that v(t, x) satisfies eq
uation (*2), So, to take the derivative of w with respect to t, is onl
y to take the derivative to v with respect to t, as the second term in
independent of t. The derivative of the second term with repect to t \+
is zero. That is," }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 6 " " }{XPPEDIT 18 0 "diff(w,t);" "6#-%%diffG6$%\"wG%\"t
G" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "diff(v,`$`(x,2));" "6#-%%diffG6$%
\"vG-%\"$G6$%\"xG\"\"#" }{TEXT -1 11 " - 4 v + 0." }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 63 "What about the derivative
of the second term with respect to x." }}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 48 "wterm2:=x->18*(sinh(2*x)-sinh(2*(x-1)))/sinh(2);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "simplify(diff(wterm2(x),x,x)
-4*wterm2(x));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 72 "Thus, the
zero on the right side of the last equation can be replaced by" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 12 " 0
= " }{XPPEDIT 18 0 "diff(wterm2(x),`$`(x,2));" "6#-%%diffG6$-%'wterm2
G6#%\"xG-%\"$G6$F)\"\"#" }{TEXT -1 15 " - 4 wterm2(x)," }}{PARA 0 ""
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 33 "so that w satisfies \+
equation (*3)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 11 " " }{XPPEDIT 18 0 "diff(w,t);" "6#-%%diffG6$%\"
wG%\"tG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "diff(w,`$`(x,2));" "6#-%%di
ffG6$%\"wG-%\"$G6$%\"xG\"\"#" }{TEXT -1 8 " - 4 w ," }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 57 "with changed boundary c
onditons: w(t, 0) = 18 = w(t, 1)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 15 "We verify this." }}{EXCHG {PARA 0 "> " 0
"" {MPLTEXT 1 0 55 "w:=(t,x)->v(t,x)+ 18*(sinh(2*x)-sinh(2*(x-1)))/sin
h(2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "simplify(diff(w(t,
x),t)-diff(w(t,x),x,x)+4*w(t,x));" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 15 "w(t,0);\nw(t,1);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 7 "w(0,x);" }}}{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 30 " One step remains. Define " }}{PARA
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 29 " Z(t, x) = \+
w(t, x) + 32. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> "
0 "" {MPLTEXT 1 0 20 "Z:=(t,x)->w(t,x)+32;" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 56 "simplify(diff(Z(t,x),t)-diff(Z(t,x),x,x)+4*(Z(t,x)-
32));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "Z(t,0);\nZ(t,1);"
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "Z(0,x);" }}}{EXCHG {PARA
0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 173 "Behold! with the knowledge the u is the \+
solution to the simple heat equation with zero boundary condtions, we \+
have that Z satisfies (*1), provided we choose f(x) as follows:" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 22 " \+
50 = f(x) + " }{XPPEDIT 18 0 "18*(sinh(2*x)-sinh(2*(x-1)))/sinh(2);"
"6#*(\"#=\"\"\",&-%%sinhG6#*&\"\"#F%%\"xGF%F%-F(6#*&\"\"#F%,&F,F%\"\"
\"!\"\"F%F3F%-F(6#\"\"#F3" }{TEXT -1 6 " + 32." }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 9 "That is, " }}{PARA 0 "" 0
"" {TEXT -1 25 " f(x) = 18 - " }{XPPEDIT 18 0 "18*(sinh(2*
x)-sinh(2*(x-1)))/sinh(2);" "6#*(\"#=\"\"\",&-%%sinhG6#*&\"\"#F%%\"xGF
%F%-F(6#*&\"\"#F%,&F,F%\"\"\"!\"\"F%F3F%-F(6#\"\"#F3" }{TEXT -1 2 " .
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 26 "We dr
aw a graph of this f." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "f:=
x->18 - 18*(sinh(2*x)-sinh(2*(x-1)))/sinh(2);" }}}{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 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 92 "To make this specific, we
solve the simple heat equation (*2) with zero boundary conditions." }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 10 " " }{XPPEDIT 18 0 "diff(u,t);" "
6#-%%diffG6$%\"uG%\"tG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "diff(u,`$`(x
,2));" "6#-%%diffG6$%\"uG-%\"$G6$%\"xG\"\"#" }{TEXT -1 40 " , with
u(t, 0) = 0 and u(t, 1) = 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 26 "which has general solution" }}{PARA 0 ""
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 18 " u(t, x) = "
}{XPPEDIT 18 0 "sum(c[n]*exp(-n^2*Pi^2*t)*sin(n*Pi*x),n);" "6#-%$sumG6
$*(&%\"cG6#%\"nG\"\"\"-%$expG6#,$*(F*\"\"#%#PiG\"\"#%\"tGF+!\"\"F+-%$s
inG6#*(F*F+F2F+%\"xGF+F+F*" }{TEXT -1 2 " ," }}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 52 "The coefficients are determined
by f(x) given above." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA
0 "> " 0 "" {MPLTEXT 1 0 86 "N:=5;\nfor n from 1 to N do\n c[n]:=2*e
valf(Int(f(x)*sin(n*Pi*x),x=0..1));\nod;\nn:='n';" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 62 "Maybe we sh
ould check to see how well this fits the initial f." }}{EXCHG {PARA 0
"> " 0 "" {MPLTEXT 1 0 49 "plot([f(x),sum(c[n]*sin(n*Pi*x),n=1..N)],x=
0..1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0
"" {TEXT -1 96 "A good approximation for the graph! We now are ready \+
to draw the graph of Z. First we define 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
..N);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 268 "How do you expect the graph to look. At t = zero, the gr
aph should be the straight line Z(0, x) = 50. As time moves forward, t
he ends of the graph should remain at 50 but the middle should cool, s
hould decrease, toward 32. Who knows how cold it will get in the middl
e?" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "plot3d(Z(t,x),x=0..1,t
=0..1/2,axes=normal);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }
}}{PARA 0 "" 0 "" {TEXT -1 142 "Let's try to answer that question. How
cold will the middle of the bar get? Agree that the cold spot is at x
= 1/2. Then, we compute Z(1,1/2)." }}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 16 "evalf(Z(1,1/2));" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 27 "limit(Z(t,1/2),t=infinity);" }}}{EXCHG {PARA 0 "> "
0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 ""
0 "" {TEXT -1 119 "Of course, we only used 5 terms of the series. Usin
g more terms will change this last result. Hopefully, only a little."
}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 128 "In this \+
Method 1, we have provided a solution for the equation (*4) in terms o
f the solution for (*1). The solution for (*4) is" }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 42 " Z(t, x) = exp(-
4 t) u(t, x) + " }{XPPEDIT 18 0 "18*(sinh(2*x)-sinh(2*(x-1)))/sinh(2
);" "6#*(\"#=\"\"\",&-%%sinhG6#*&\"\"#F%%\"xGF%F%-F(6#*&\"\"#F%,&F,F%
\"\"\"!\"\"F%F3F%-F(6#\"\"#F3" }{TEXT -1 8 " + 32." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 49 "How to Conceive t
he Construction of the solution." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{PARA 0 "" 0 ""
{TEXT -1 36 "To explain how the solution for (*4)" }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT
-1 13 " " }{XPPEDIT 18 0 "diff(Z,t);" "6#-%%diffG6$%\"ZG%
\"tG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "diff(Z,`$`(x,2));" "6#-%%diffG
6$%\"ZG-%\"$G6$%\"xG\"\"#" }{TEXT -1 14 " - 4 (Z - 32)" }}{PARA 0 ""
0 "" {TEXT -1 25 " " }}{PARA 0 "" 0 "" {TEXT
-1 31 " Z(t,0) = 50 = Z(t,1)," }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 23 " Z(0, x) = 50," }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 196 "can be conceived,
I thought of the problem this way: I want to change this problem to t
he simple heat equation. The first idea was to get rid of the \" - 32
\" term on the right side. Thus, I make w:" }}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 27 " w(t, x) = Z(t, x) - 32" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 140 "and note
that if Z satisfies the differential equation for which we seek a sol
ution, then there is a differential equation that w satisfies." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0
8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "Z:=(t,x)->W(
t,x)+32;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "diff(Z(t,x),t)=
diff(Z(t,x),x,x)-4*(Z(t,x)-32);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT
1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 111 "We see that if Z satisfies (
*4) then w satisfies the following partial differential equation which
we call (*3)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 14 " " }{XPPEDIT 18 0 "diff(w,t);" "6#-%%diffG6$
%\"wG%\"tG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "diff(w,`$`(x,2));" "6#-%
%diffG6$%\"wG-%\"$G6$%\"xG\"\"#" }{TEXT -1 7 " - 4 w" }}{PARA 0 "" 0
"" {TEXT -1 50 "Of course, w(t, 0) = Z(t, 0) - 32 = 50 - 32 = 18, " }}
{PARA 0 "" 0 "" {TEXT -1 59 " w(t, 1) = Z(t, 1) - 32 =
50 - 32 = 18, and" }}{PARA 0 "" 0 "" {TEXT -1 37 " w(
0, x) = f(x) - 32." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 17 "The Creation of v" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 253 "Our second step is to find the steady st
ate solution for equation (*3), in order to take care of the boundary \+
conditions. To do this, we make make a function which does not depend \+
on time, but which satisfies the boundary conditions. There should be \+
no " }{TEXT 261 7 "initial" }{TEXT -1 19 " condition, for an " }{TEXT
262 7 "initial" }{TEXT -1 53 " distribution would suggest the involvem
ent of time. " }}{PARA 0 "" 0 "" {TEXT -1 15 " 0 = " }
{XPPEDIT 18 0 "diff(s,`$`(x,2));" "6#-%%diffG6$%\"sG-%\"$G6$%\"xG\"\"#
" }{TEXT -1 51 " - 4 s, s(0) = 18 = s(1). We solve this equation." }
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 69 "OneSol:=dsolve(\{diff(s(x),
x,x)-4*s(x)= 0,s(0) = 18, s(1) = 18\},s(x));" }}}{EXCHG {PARA 0 "> "
0 "" {MPLTEXT 1 0 25 "plot(rhs(OneSol),x=0..1);" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "
" }}}{PARA 0 "" 0 "" {TEXT -1 178 " The solution that Maple produc
ed is certainly a perfectly nice solution. However, as a human, perhap
s you will find the following solution has a symmetry that is attracti
ve:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 54 " \+
s(x) = 18 (sinh(2 x)-sinh(2 (x-1))/ sinh(2)." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 66 "Maybe we should first v
erify that this is a steady state solution." }}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 43 "s:=x->18*(sinh(2*x)-sinh(2*(x-1)))/sinh(2);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "s(0);\ns(1);" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "simplify(diff(s(x),x,x)-4*s(x));" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT
-1 4 " " }}{PARA 0 "" 0 "" {TEXT -1 25 " Thus, we define v by"
}}{PARA 0 "" 0 "" {TEXT -1 35 " w(t, x) = v(t, x) + s(x)." }}
{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 25 "To get this in terms of Z" }}{PARA 0 ""
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 59 " v(t, x) = \+
w(t, x) - s(x) = Z(t, x) - 32 - s(x)." }}{PARA 0 "" 0 "" {TEXT -1 0
"" }}{PARA 0 "" 0 "" {TEXT -1 36 "Check the boundary conditions first.
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "Z:=(t,x)->v(t,x)+32+s(x)
;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "Z(t,0);\nZ(t,1);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "Z(0,x);" }}}{EXCHG {PARA 0 ">
" 0 "" {MPLTEXT 1 0 57 "simplify(diff(Z(t,x),t)-diff(Z(t,x),x,x) +4*(
Z(t,x)-32));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 30 "Thus, we have t
hat v satisfies" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 15 " " }{XPPEDIT 18 0 "diff(v(t,x),t) = diff(v(
t,x),`$`(x,2))-4*v(t,x);" "6#/-%%diffG6$-%\"vG6$%\"tG%\"xGF*,&-F%6$-F(
6$F*F+-%\"$G6$F+\"\"#\"\"\"*&\"\"%F5-F(6$F*F+F5!\"\"" }{TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 24 "with bou
ndary conditions" }}{PARA 0 "" 0 "" {TEXT -1 38 " v(t,0) = Z(t, 0)
- 32 - S(0) = 0 " }}{PARA 0 "" 0 "" {TEXT -1 4 "and " }}{PARA 0 "" 0
"" {TEXT -1 37 " v(t,1) = Z(t, 1) - 32 - s(1) = 0." }}{PARA 0 "" 0
"" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 39 "Define u by u(t, x) = \+
exp(4 t) v(t, x) " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 2 "or" }}{PARA 0 "" 0 "" {TEXT -1 45 " Z(t, x) = exp(-4 t)
u(t, x) + 32 + s(x). " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "Z:=(t,x)->exp(-4*t)*u(t,x)+32+s(x);
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 57 "simplify(diff(Z(t,x),t)
-diff(Z(t,x),x,x) +4*(Z(t,x)-32));" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 12 "simplify(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1
0 15 "Z(t,0);\nZ(t,1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "Z(
0,x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "
" {TEXT -1 80 "We find that we have Z in terms of u. But we know u. So
we are finished (again)." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "
" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 4 "" 0 "" {TEXT
-1 8 "Exercise" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 5 "Solve" }}{PARA 0
"" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 13 " " }{XPPEDIT 18 0 "diff(Z,t);" "6#-%%diffG6
$%\"ZG%\"tG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "diff(Z,`$`(x,2));" "6#-
%%diffG6$%\"ZG-%\"$G6$%\"xG\"\"#" }{TEXT -1 15 " - 4 (Z - 100)" }}
{PARA 0 "" 0 "" {TEXT -1 25 " " }}{PARA 0 ""
0 "" {TEXT -1 31 " Z(t,0) = 50 = Z(t,1)," }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 23 " Z(0, x) = 50."
}}}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 29 "Insulated boundary condition
s" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1
103 " In this section, we begin an examination of the heat equation \+
with a variety of boundary conditions." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 257 "" 0 "" {TEXT -1 17 "The First Problem" }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 91 "For the purposes of thi
s discussion, we write the equation to be considered in three parts:"
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 18 "The PDE
: " }{XPPEDIT 18 0 "diff(u,t) = diff(u,`$`(x,2));" "6#/-%%dif
fG6$%\"uG%\"tG-F%6$F'-%\"$G6$%\"xG\"\"#" }{TEXT -1 2 " ," }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 27 "The Boundary Condi
tions: " }{XPPEDIT 18 0 "diff(u,x);" "6#-%%diffG6$%\"uG%\"xG" }
{TEXT -1 16 " (t, 0) = 0 and " }{XPPEDIT 18 0 "diff(u,x);" "6#-%%diffG
6$%\"uG%\"xG" }{TEXT -1 11 " (t,1) = 0." }}{PARA 0 "" 0 "" {TEXT -1 0
"" }}{PARA 0 "" 0 "" {TEXT -1 42 "The Initial Condition: u(0, x) =
f(x)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 30 "Solution for the First Probl
em" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 47 " \+
We have discussed the physical meaning of " }{TEXT 263 7 "The PDE" }
{TEXT -1 5 " and " }{TEXT 264 21 "The Initial Condition" }{TEXT -1 40
". Here is an idea to give intuition for " }{TEXT 265 24 "The Boundary
Conditions " }{TEXT -1 160 "in this problem. Having the derivative wi
th respect to x to be zero at these two ends mean that there is no mov
ement of heat across the boundary x = 0 or x = 1." }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 545 " From this, what do \+
we expect for solutions of this problem? Since the problem acts as tho
ugh there is no radial transfer of heat and no end point transfer of h
eat, we expect the entire system to retain the total heat, but to move
to an equal distribution of heat throughout the rod. We will solve th
e equation with a particular f and sketch the graph of the solution. W
e expect to see that the solution has limit a solution that is constan
t in t and in x. Also, because the total heat in the system should rem
ain constant, it should be that" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 15 " " }{XPPEDIT 18 0 "int(f(x)
,x = 0 .. 1);" "6#-%$intG6$-%\"fG6#%\"xG/F);\"\"!\"\"\"" }{TEXT -1 3 "
= " }{XPPEDIT 18 0 "int(u(t,x),x = 0 .. 1);" "6#-%$intG6$-%\"uG6$%\"t
G%\"xG/F*;\"\"!\"\"\"" }{TEXT -1 17 " , for all t > 0." }}{PARA 0 ""
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 28 "Here goes making a s
olution." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1
92 "First, we find two ODE's, with boundary conditons, which arise fro
m separation of variables." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0
"" 0 "" {TEXT -1 17 "The PDE leads to " }}{PARA 0 "" 0 "" {TEXT -1 3 "
" }}{PARA 0 "" 0 "" {TEXT -1 26 " X T ' = X '' T, " }}
{PARA 0 "" 0 "" {TEXT -1 2 "or" }}{PARA 0 "" 0 "" {TEXT -1 29 " \+
T ' / T = X '' / X." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 ""
0 "" {TEXT -1 160 "Since the left side is independent of x, the right \+
side must be constant. In a similar manner, the left side is constant \+
and these constants are the same. Thus," }}{PARA 0 "" 0 "" {TEXT -1 0
"" }}{PARA 0 "" 0 "" {TEXT -1 17 " X '' = " }{XPPEDIT 18 0 "m
u;" "6#%#muG" }{TEXT -1 34 " X, with X'(0) = 0 and X '(1) = 0," }}
{PARA 0 "" 0 "" {TEXT -1 3 "and" }}{PARA 0 "" 0 "" {TEXT -1 16 " \+
T ' = " }{XPPEDIT 18 0 "mu;" "6#%#muG" }{TEXT -1 3 " T." }}{PARA
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 83 "From methods we
have seen earlier, this constant must be negative and have the form"
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 10 " \+
" }{XPPEDIT 18 0 "mu = -n^2*Pi^2;" "6#/%#muG,$*&%\"nG\"\"#%#PiG\"\"
#!\"\"" }{TEXT -1 14 " with X(x) = " }{XPPEDIT 18 0 "cos(n*Pi*x);" "6
#-%$cosG6#*(%\"nG\"\"\"%#PiGF(%\"xGF(" }{TEXT -1 35 " , for each non-n
egative integer n." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 18 "Also, T(t) = exp( " }{XPPEDIT 18 0 "-n^2*Pi^2*t;" "6#,$*(
%\"nG\"\"#%#PiG\"\"#%\"tG\"\"\"!\"\"" }{TEXT -1 2 ")." }}{PARA 0 "" 0
"" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 45 "Thus, the general solu
tion for the problem is" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 ""
0 "" {TEXT -1 20 " u(t ,x) = " }{XPPEDIT 18 0 "c[0];" "6#&%\"
cG6#\"\"!" }{TEXT -1 3 " + " }{XPPEDIT 18 0 "sum(c[n]*exp(-n^2*Pi^2*t)
*cos(n*Pi*x),n = 1 .. infinity);" "6#-%$sumG6$*(&%\"cG6#%\"nG\"\"\"-%$
expG6#,$*(F*\"\"#%#PiG\"\"#%\"tGF+!\"\"F+-%$cosG6#*(F*F+F2F+%\"xGF+F+/
F*;\"\"\"%)infinityG" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 155 "To find a particular solution, we ne
ed to specify f. We do that now. Using this f, we write a solution and
check that the answer is right, and seems right!" }}{EXCHG {PARA 0 ">
" 0 "" {MPLTEXT 1 0 16 "f:=x->x*(x-1)+1;" }}}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 33
"Because this f is u(0, x), then " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 18 " f(x) = " }{XPPEDIT 18 0 "c[0];
" "6#&%\"cG6#\"\"!" }{TEXT -1 3 " + " }{XPPEDIT 18 0 "sum(c[n]*cos(n*P
i*x),n = 1 .. infinity);" "6#-%$sumG6$*&&%\"cG6#%\"nG\"\"\"-%$cosG6#*(
F*F+%#PiGF+%\"xGF+F+/F*;\"\"\"%)infinityG" }{TEXT -1 2 " ." }}{PARA 0
"" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 56 "The computation o
f the c 's is a job for Fourier Series." }}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 53 "assume(n,integer):\nint(cos(n*Pi*x)^2,x=0..1);\nn:='n
':" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 100 "c[0]:=int(f(x),x=0..
1);\nfor n from 1 to 10 do\n c[n]:=2*int(f(x)*cos(n*Pi*x),x=0..1);
\nod;\nn:='n':" }}}{PARA 0 "" 0 "" {TEXT -1 68 "To see how good our fi
t is, we plot f and the Fourier Approximation." }}{EXCHG {PARA 0 "> "
0 "" {MPLTEXT 1 0 55 "plot([f(x),c[0]+sum(c[n]*cos(n*Pi*x),n=1..10)],x
=0..1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 ""
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 24 "Now we make the u(t,
x)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 63 "u:=(t,x)-> c[0]+sum(c[n]*exp(-n^2*Pi^2*t)*cos(n*Pi*x)
,n=1..10);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "plot3d(u(t,x)
,x=0..1,t=0..1/20,axes=NORMAL,orientation=[45,55]);" }}}{PARA 0 "" 0 "
" {TEXT -1 136 "Does this graph have the properties we predicted? What
was the promise: that the total heat would stay about the same? Let's
check that." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "int(f(x),x=0
..1); \nint(u(t,x),x=0..1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0
0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 48 "T
his completes a solution of this first problem." }}}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT
-1 23 "We do a second problem." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 23 "The same \+
PDE: " }{XPPEDIT 18 0 "diff(u,t) = diff(u,`$`(x,2));" "6#/-%%
diffG6$%\"uG%\"tG-F%6$F'-%\"$G6$%\"xG\"\"#" }{TEXT -1 2 " ," }}{PARA
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 35 "The Changed Bou
ndary Conditions: " }{XPPEDIT 18 0 "u;" "6#%\"uG" }{TEXT -1 16 " (t,
0) = A and " }{XPPEDIT 18 0 "diff(u,x);" "6#-%%diffG6$%\"uG%\"xG" }
{TEXT -1 11 " (t,1) = 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 42 "The Initial Condition: u(0, x) = f(x)." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 31 "S
olution for the Second Problem" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 171 " An interpretation for this problem \+
is that there is no heat loss laterally, that the left end is held con
stant at temperature A, and that the right end is insulated. " }}
{PARA 0 "" 0 "" {TEXT -1 227 " We should recognize that this probl
em does not have zero boundary conditons and so we should break the pr
oblem into two: one has gives a particular solution and one provides a
general solution for the homogeneous equation." }}{PARA 0 "" 0 ""
{TEXT -1 151 " Take v(x) = A. This solutions satisfies the PDE, an
d the boundary conditions, but not the initial conditions, of course. \+
Now look for u to satisfy" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 23 "The same PDE: \+
" }{XPPEDIT 18 0 "diff(u,t) = diff(u,`$`(x,2));" "6#/-%%diffG6$%
\"uG%\"tG-F%6$F'-%\"$G6$%\"xG\"\"#" }{TEXT -1 2 " ," }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 39 "The Homogeneous Boundar
y Conditions: " }{XPPEDIT 18 0 "u;" "6#%\"uG" }{TEXT -1 16 " (t, 0) \+
= 0 and " }{XPPEDIT 18 0 "diff(u,x);" "6#-%%diffG6$%\"uG%\"xG" }{TEXT
-1 11 " (t,1) = 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 196 "Finding this u, we will add u and v and then choose coef
ficients to satisfy the initial conditons. By now, we see that this ho
mogeneous problem leads to two ODE 's, one having boundary conditions:
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 17 " X '' = " }{XPPEDIT 18 0 "mu;" "
6#%#muG" }{TEXT -1 33 " X, with X(0) = 0 and X '(1) = 0," }}{PARA 0 "
" 0 "" {TEXT -1 3 "and" }}{PARA 0 "" 0 "" {TEXT -1 16 " T ' =
" }{XPPEDIT 18 0 "mu;" "6#%#muG" }{TEXT -1 3 " T." }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 38 "It is no surprise that we
should take " }{XPPEDIT 18 0 "mu = -lambda^2;" "6#/%#muG,$*$%'lambdaG
\"\"#!\"\"" }{TEXT -1 31 ". The solution for the X '' = " }{XPPEDIT
18 0 "-lambda^2;" "6#,$*$%'lambdaG\"\"#!\"\"" }{TEXT -1 6 " X is " }}
{PARA 0 "" 0 "" {TEXT -1 10 " " }{XPPEDIT 18 0 "alpha;" "6#%&
alphaG" }{TEXT -1 5 " cos(" }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }
{TEXT -1 6 " x) + " }{XPPEDIT 18 0 "beta;" "6#%%betaG" }{TEXT -1 5 " s
in(" }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT -1 4 " x)." }}
{PARA 0 "" 0 "" {TEXT -1 43 "The requirement that X(0) = 0 implies tha
t " }{XPPEDIT 18 0 "alpha;" "6#%&alphaG" }{TEXT -1 50 " = 0. The requi
rement that X '(1) = 0 implies that" }}{PARA 0 "" 0 "" {TEXT -1 10 " \+
" }{XPPEDIT 18 0 "beta*lambda;" "6#*&%%betaG\"\"\"%'lambdaGF%
" }{TEXT -1 5 " cos(" }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT
-1 6 ") = 0." }}{PARA 0 "" 0 "" {TEXT -1 72 "To have a solution that i
s not zero for all x, it must be true that cos(" }{XPPEDIT 18 0 "lambd
a;" "6#%'lambdaG" }{TEXT -1 23 ") = 0. This happens if " }{XPPEDIT 18
0 "lambda;" "6#%'lambdaG" }{TEXT -1 23 " is an odd multiple of " }
{XPPEDIT 18 0 "pi/2;" "6#*&%#piG\"\"\"\"\"#!\"\"" }{TEXT -1 10 ". That
is," }}{PARA 0 "" 0 "" {TEXT -1 15 " " }{XPPEDIT 18 0 "
lambda[n] = (2*n-1)*Pi/2;" "6#/&%'lambdaG6#%\"nG*(,&*&\"\"#\"\"\"F'F,F
,\"\"\"!\"\"F,%#PiGF,\"\"#F." }{TEXT -1 21 " , n = 1, 2, 3, ... ." }}
{PARA 0 "" 0 "" {TEXT -1 97 "Thus, these are the eigenvalues for this \+
homogeneous problem corresponding to eigenfunctions sin(" }{XPPEDIT
18 0 "lambda[n];" "6#&%'lambdaG6#%\"nG" }{TEXT -1 76 " x). It is clear
that the corresponding solution for the T equation is exp(-" }
{XPPEDIT 18 0 "lambda[n]^2;" "6#*$&%'lambdaG6#%\"nG\"\"#" }{TEXT -1 5
" t) ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 63
"We can now write the general solution for the original problem:" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 26 " \+
u(t, x) + v(x) = " }{XPPEDIT 18 0 "sum(c[n]*exp(-(2*n-1)^2*Pi^2*t/4)*s
in((2*n-1)*Pi*x/2),n);" "6#-%$sumG6$*(&%\"cG6#%\"nG\"\"\"-%$expG6#,$**
,&*&\"\"#F+F*F+F+\"\"\"!\"\"\"\"#%#PiG\"\"#%\"tGF+\"\"%F5F5F+-%$sinG6#
**,&*&\"\"#F+F*F+F+\"\"\"F5F+F7F+%\"xGF+\"\"#F5F+F*" }{TEXT -1 8 " + \+
A.\n" }}{PARA 0 "" 0 "" {TEXT -1 59 "To determine the c 's, let t = 0
, and use the Fourier idea." }}{PARA 0 "" 0 "" {TEXT -1 47 "To be spec
ific, we choose a particular f and A." }}{PARA 0 "" 0 "" {TEXT -1 1 " \+
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "A:=1;\nf:=x->cos(Pi/2*x)
;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 95 "n:='n':\nfor n from 1 \+
to 5 do\n c[n]:=2*int((f(x)-A)*sin((2*n-1)*Pi/2*x),x=0..1);\nod;\n
n:='n';" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "plot([f(x)-A,sum
(c[n]*sin((2*n-1)*Pi/2*x),n=1..5)],x=0..1);" }}}{EXCHG {PARA 0 "> " 0
"" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 43 "In passing, note \+
the nature of convergence." }}{PARA 0 "" 0 "" {TEXT -1 33 " We now
u and draw the graph." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 74 "u:
=(t,x)->A+sum(c[n]*exp(-(2*n-1)^2*Pi^2*t/4)*sin((2*n-1)*Pi/2*x),n=1..5
);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 63 "plot3d(u(t,x),x=0..1,
t=0..1,axes=NORMAL,orientation=[-100,65]);" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 51 "This completes the solution for the second problem." }}
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 8 "E
xercise" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 44 "Solve and graph the sol
ution for the problem" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0
"" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 18 "The PDE: " }
{XPPEDIT 18 0 "diff(u,t) = diff(u,`$`(x,2));" "6#/-%%diffG6$%\"uG%\"tG
-F%6$F'-%\"$G6$%\"xG\"\"#" }{TEXT -1 2 " ," }}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 27 "The Boundary Conditions: " }
{XPPEDIT 18 0 "diff(u,x);" "6#-%%diffG6$%\"uG%\"xG" }{TEXT -1 16 " (t,
0) = 0 and " }{XPPEDIT 18 0 "u;" "6#%\"uG" }{TEXT -1 11 " (t,1) = 1.
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 42 "The I
nitial Condition: u(0, x) = cos( " }{XPPEDIT 18 0 "2*Pi*x;" "6#*(
\"\"#\"\"\"%#PiGF%%\"xGF%" }{TEXT -1 3 " )." }}}}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 32 "Convection acros
s the boundaries" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 ""
0 "" {TEXT -1 103 " Suppose that we take a well insulated rod -- s
ides and ends -- which has initial heat distribution" }}{PARA 0 "" 0 "
" {TEXT -1 52 " 4 x (1 - x) + 2, 0 < x < 1." }
}{PARA 0 "" 0 "" {TEXT -1 364 "We know the rod will be hottest in the \+
middle, and that as the heat diffuses, a uniform distribution of heat \+
will be achieved with total heat the same as the total of the original
distribution. After all, no heat has been lost or gained in this well
-insulated rod. Here is a graph of the initial distribution of heat, a
nd the value of the total heat in the system." }}}{EXCHG {PARA 0 "> "
0 "" {MPLTEXT 1 0 18 "f:=x->4*x*(1-x)+2;" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 17 "int(f(x),x=0..1);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 240 "Maybe, we find this problem even boring
now. Consider this alternative. Keep the rod well insulated, except a
t one end. There heat escapes according to Newton's Law of Cooling, wi
th an outside temperature of 2. More specifically, We suppose" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA
0 "" 0 "" {TEXT -1 18 "The PDE: " }{XPPEDIT 18 0 "diff(u,t) =
diff(u,`$`(x,2));" "6#/-%%diffG6$%\"uG%\"tG-F%6$F'-%\"$G6$%\"xG\"\"#
" }{TEXT -1 2 " ," }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 27 "The Boundary Conditions: " }{XPPEDIT 18 0 "diff(u,x);"
"6#-%%diffG6$%\"uG%\"xG" }{TEXT -1 16 " (t, 0) = 0 and " }{XPPEDIT 18
0 "diff(u,x);" "6#-%%diffG6$%\"uG%\"xG" }{TEXT -1 25 " (t,1) = - (u(t,
1) - 2)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1
51 "The Initial Condition: u(0, x) = 4 x (1-x) + 2." }}{PARA 0 ""
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 681 "Now this problem is
more interesting. The heat begins to diffuse, causing temperatures to
rise on the right and left of 1/2. As soon as the temperature goes ab
ove 2 on the right, the rod begins to cool by heat loss at that end. H
eat will continue to leak out until the temperature over the entire ro
d is 2. But what happens at the left side. Surely heat will start to s
pread there from the middle. With no leakage out the right side or any
where else, the temperature at the left would rise to 8/3. How high wi
ll it go now? Will it go above 2, and then settle back down? And what \+
about the hot spot? Will it move left as heat leaks out the right side
, or will it drop straight down?" }}{PARA 0 "" 0 "" {TEXT -1 66 " \+
Maybe you agree this problem will be interesting to consider." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 101 "The PDE \+
is homogeneous, but the boundary conditions are not. Homogeneous bound
ary conditions would be" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 ""
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 27 "The Boundary Conditi
ons: " }{XPPEDIT 18 0 "diff(u,x);" "6#-%%diffG6$%\"uG%\"xG" }{TEXT
-1 16 " (t, 0) = 0 and " }{XPPEDIT 18 0 "diff(u,x);" "6#-%%diffG6$%\"u
G%\"xG" }{TEXT -1 19 " (t,1) = - u(t, 1)." }}{PARA 0 "" 0 "" {TEXT -1
0 "" }}{PARA 0 "" 0 "" {TEXT -1 137 "Thus, we need to find a particula
r soluition, or, what's called the steady state solution. Is it clear \+
that the steady state solution is " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 24 " v(x) = 2?" }}{PARA 0 "" 0
"" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 76 "That solution satisfie
s the PDE and the non-homogeneous boundary conditions." }}{PARA 0 ""
0 "" {TEXT -1 95 " Now, we try to find the general solution for th
e PDE with homogeneous boundary conditions." }}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 48 "The general solution fo
r the homogeneous problem" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 100 "By now, we know this partial differential equatio
n changes into two ordinary differential equations:" }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 17 " X '' = " }{XPPEDIT 18 0 "mu;" "6#%#muG" }{TEXT
-1 40 " X, with X'(0) = 0 and X '(1) +X(1) = 0," }}{PARA 0 "" 0 ""
{TEXT -1 3 "and" }}{PARA 0 "" 0 "" {TEXT -1 16 " T ' = " }
{XPPEDIT 18 0 "mu;" "6#%#muG" }{TEXT -1 3 " T." }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 13 "The constant " }{XPPEDIT
18 0 "mu;" "6#%#muG" }{TEXT -1 27 " must be negative, call it " }
{XPPEDIT 18 0 "-lambda^2;" "6#,$*$%'lambdaG\"\"#!\"\"" }{TEXT -1 42 " \+
. Solutions for the differential equation" }}{PARA 0 "" 0 "" {TEXT -1
0 "" }}{PARA 0 "" 0 "" {TEXT -1 28 " X '' = " }
{XPPEDIT 18 0 "-lambda^2;" "6#,$*$%'lambdaG\"\"#!\"\"" }{TEXT -1 3 " X
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 19 "are \+
A sin( " }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT -1 13 "
x) + B cos( " }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT -1 4 " x)
." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 33 "We i
nvoke the boundary conditons." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1
0 38 "X:=x->A*sin(lambda*x)+B*cos(lambda*x);" }}}{PARA 0 "" 0 ""
{TEXT -1 42 "The left boundary condition is used first." }}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "D(X)(0)=0;" }}}{PARA 0 "" 0 ""
{TEXT -1 29 "This condition implies A = 0." }}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 5 "A:=0;" }}}{PARA 0 "" 0 "" {TEXT -1 44 "We now use th
e remaining boundary condition." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT
1 0 15 "D(X)(1)+X(1)=0;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "
" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 22 "This
implies that tan(" }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT -1
4 ") = " }{XPPEDIT 18 0 "1/lambda;" "6#*&\"\"\"\"\"\"%'lambdaG!\"\"" }
{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 13 "We must find " }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }
{TEXT -1 41 " 's that satisfy this requirement. Then, " }{XPPEDIT 18
0 "mu;" "6#%#muG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "-lambda^2;" "6#,$*
$%'lambdaG\"\"#!\"\"" }{TEXT -1 59 " and the eigenfunction correspond
ing to this eigenvalue is" }}{PARA 0 "" 0 "" {TEXT -1 30 " \+
cos( " }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT -1
4 " x)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1
28 "The general solution will be" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 20 " u(t, x) = " }{XPPEDIT 18 0 "sum
(c[n]*exp(-lambda[n]^2*t)*cos(lambda[n]*x),n);" "6#-%$sumG6$*(&%\"cG6#
%\"nG\"\"\"-%$expG6#,$*&&%'lambdaG6#F*\"\"#%\"tGF+!\"\"F+-%$cosG6#*&&F
26#F*F+%\"xGF+F+F*" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 25 "We have only to find the " }{XPPEDIT 18
0 "lambda;" "6#%'lambdaG" }{TEXT -1 22 " 's and then the c 's." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 88 " Is i
t now clear that there is going to be no nice, closed form solution fo
r these " }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT -1 160 " 's? \+
We must find them numerically. To help in doing this, we look to see a
bout where they lie by looking at where the graph of tan(x) crosses th
e graph of 1/x." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "plot([tan
(x),1/x],x=0..20,y=0..3/2,discont=true,color=[RED,BLACK]);" }}}{PARA
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 153 "We see there a
re roots between 0 and 2, between 3 and 4, between 6 and 7, between 9 \+
and 10, between 12 and 13, between 15 and 16, between18 and 20, ... .
" }}{PARA 0 "" 0 "" {TEXT -1 31 " Maybe that will be enough." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 272 "lambda[1]:=fsolve(tan(x)=1/
x,x,0..2);\nlambda[2]:=fsolve(tan(x)=1/x,x,3..4);\nlambda[3]:=fsolve(t
an(x)=1/x,x,6..7);\nlambda[4]:=fsolve(tan(x)=1/x,x,9..10);\nlambda[5]:
=fsolve(tan(x)=1/x,x,12..13);\nlambda[6]:=fsolve(tan(x)=1/x,x,15..16);
\nlambda[7]:=fsolve(tan(x)=1/x,x,18..20);" }}}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 47 "We now have seven terms of the \+
general solution" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 20 " u(t, x) = " }
{XPPEDIT 18 0 "sum(c[n]*exp(-lambda[n]^2*t)*cos(lambda[n]*x),n);" "6#-
%$sumG6$*(&%\"cG6#%\"nG\"\"\"-%$expG6#,$*&&%'lambdaG6#F*\"\"#%\"tGF+!
\"\"F+-%$cosG6#*&&F26#F*F+%\"xGF+F+F*" }{TEXT -1 2 " ." }}{PARA 0 ""
0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA
0 "" 0 "" {TEXT -1 157 "We go back to solve the original problem. We n
eed the general solution plus the particular solutions. These add toge
ther to give the soluion to this problem." }}{PARA 0 "" 0 "" {TEXT -1
0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT
-1 37 "The solution for the original problem" }}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 75 "Now, we find the c's by using t
he initial conditon: f(x) = u(0, x) + 2" }}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 22 "or f(x) -2 = " }
{XPPEDIT 18 0 "sum(c[n]*cos(lambda[n]*x),n);" "6#-%$sumG6$*&&%\"cG6#%
\"nG\"\"\"-%$cosG6#*&&%'lambdaG6#F*F+%\"xGF+F+F*" }{TEXT -1 2 " ." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 24 "We comput
e coefficients." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 141 "n:='n':
\nfor n from 1 to 7 do\n c[n]:=int((f(x)-2)*cos(lambda[n]*x), x=0..1
)/\n int(cos(lambda[n]*x)^2,x=0..1);\nod;\nn:='n':
" }}}{PARA 0 "" 0 "" {TEXT -1 43 "We check to see how close this is to
our f." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "plot([f(x),2+sum(
c[n]*cos(lambda[n]*x),n=1..7)],x=0..1);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 46 "We now define the solution and plot its graph." }}{PARA
0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "u:
=(t,x)->2+sum(c[n]*exp(-lambda[n]^2*t)*cos(lambda[n]*x),n=1..7);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "plot3d(u(t,x),x=0..1,t=0..1/
10,axes=NORMAL,orientation=[-20,60]);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 7 "n:='n':" }}}{PARA 0 "" 0 "" {TEXT -1 52 "This graph wa
s the solution of the original problem." }}}{PARA 0 "" 0 "" {TEXT -1
0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 97 "In
order to watch the maximum move from one side to the other, we animat
e u(t, x) as t increases." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45
"with(plots):\nanimate(u(t,x),x=0..1,t=0..1/8);" }}}{PARA 0 "" 0 ""
{TEXT -1 71 "We ask: At what time would the left end point have maximu
m temperature?" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "plot(u(t,0
),t=0..1,y=0..3);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 266 11 "Assignme
nt:" }{TEXT -1 155 " Increase or decrease the rate of heat loss at the
right side and see how the middle moves. Such a problem would change \+
the boundary conditions as follows:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 27 "The Boun
dary Conditions: " }{XPPEDIT 18 0 "diff(u,x);" "6#-%%diffG6$%\"uG%\"
xG" }{TEXT -1 16 " (t, 0) = 0 and " }{XPPEDIT 18 0 "diff(u,x);" "6#-%%
diffG6$%\"uG%\"xG" }{TEXT -1 11 " (t,1) = - " }{XPPEDIT 18 0 "alpha;"
"6#%&alphaG" }{TEXT -1 15 " (u(t, 1) - 2) " }}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 6 "where " }{XPPEDIT 18 0 "alpha;"
"6#%&alphaG" }{TEXT -1 36 " is more than one, or less than one." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 74 "Remark in
Preparation: Here is syntax for drawing the graph of tan(s)-1/s." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "plot(tan(x)-1/x,x=0..20,y=-1
..1,discont=true,color=RED);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1
0 47 "plot([seq(u(p/15,x),p=0..7)],x=0..1,color=RED);" }}}}{SECT 1
{PARA 3 "" 0 "" {TEXT -1 26 "The structure of solutions" }}{EXCHG
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 60 " We r
eview structure for solutions of the heat equation:" }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 44 "The Partial Differentia
l Equation: " }{XPPEDIT 18 0 "diff(u,t);" "6#-%%diffG6$%\"uG%
\"tG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "diff(u,`$`(x,2));" "6#-%%diffG
6$%\"uG-%\"$G6$%\"xG\"\"#" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 41 "The Boundary Conditions: \+
u(t, 0) = " }{XPPEDIT 18 0 "T[0];" "6#&%\"TG6#\"\"!" }{TEXT -1 16 " a
nd u(t, 1) = " }{XPPEDIT 18 0 "T[1];" "6#&%\"TG6#\"\"\"" }{TEXT -1 1 "
." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 86 "The \+
Initial Conditions: u(0, x) = f(x), where f(x) is continuous on [0, a]
and f(0) = " }{XPPEDIT 18 0 "T[0];" "6#&%\"TG6#\"\"!" }{TEXT -1 10 " \+
, f(a) = " }{XPPEDIT 18 0 "T[1];" "6#&%\"TG6#\"\"\"" }{TEXT -1 1 "." }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 267 23 "Existen
ce of solutions:" }{TEXT -1 86 " There is one and only one solution f
or the above problem and it exists for t in [0, " }{XPPEDIT 18 0 "infi
nity;" "6#%)infinityG" }{TEXT -1 92 " ) and for x in [0, a]. Moreover,
the solution u is infinitely differentiable for t in [0, " }
{XPPEDIT 18 0 "infinity;" "6#%)infinityG" }{TEXT -1 23 " ) and for x i
n [0, a]." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1
388 "Note that this implies that the solutions have an intolerance for
sharp corners. The initial conditions may have some points where the \+
derivative fails to exist, but the solutions become infinitely differe
ntiable instantaneously. We illustrate with an initial value having a \+
sharp corner and examine the solution near t = 0. We expect to see tha
t the sharp corner is immediately smoothed." }}}{EXCHG {PARA 0 "> " 0
"" {MPLTEXT 1 0 8 "restart:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0
21 "f:=x->1/2-abs(x-1/2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0
18 "plot(f(x),x=0..1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 74 "f
or n from 1 to 20 do\n c[n]:=2*int(f(x)*sin(n*Pi*x),x=0..1):\nod:\nn
:='n';" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 57 "u:=(t,x)->sum(c[n
]*exp(-n^2*Pi^2*t)*sin(n*Pi*x),n=1..20);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 66 "plot3d(u(t,x),x=0..1,t=0..1/10,axes=NORMAL,orientatio
n=[-125,65]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 268 36 "Asymptotic Pro
perties for Solutions:" }{TEXT -1 138 " As t increases, u(t, x) approa
ches a solution which is independent of t. For the above equation, thi
s time independent solution would be " }}{PARA 0 "" 0 "" {TEXT -1 17 "
( " }{XPPEDIT 18 0 "T[1]-T[0];" "6#,&&%\"TG6#\"\"\"\"\"
\"&F%6#\"\"!!\"\"" }{TEXT -1 7 " ) x + " }{XPPEDIT 18 0 "T[0];" "6#&%
\"TG6#\"\"!" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA
0 "" 0 "" {TEXT -1 194 "That is, no matter what initial value, f(x), w
e begin with, all solutions for the above equation would have limit th
is steady state solution. We compute the solution for the above equati
on with " }}{PARA 0 "" 0 "" {TEXT -1 37 " f(x) =
(1+sin(" }{XPPEDIT 18 0 "pi;" "6#%#piG" }{TEXT -1 15 " x / 2)) (x+1) \+
" }}{PARA 0 "" 0 "" {TEXT -1 10 "and with " }{XPPEDIT 18 0 "T[0];" "6
#&%\"TG6#\"\"!" }{TEXT -1 6 " = 1, " }{XPPEDIT 18 0 "T[2];" "6#&%\"TG6
#\"\"#" }{TEXT -1 53 " = 2. We compute the solution and check the res
ults." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 8 "restart:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42
"f:=x->(sin(Pi*x)+1)*(x+1);\ng:=x->f(x)-x-1;" }}}{EXCHG {PARA 0 "> "
0 "" {MPLTEXT 1 0 74 "for n from 1 to 20 do\n c[n]:=2*int(g(x)*sin(n
*Pi*x),x=0..1):\nod:\nn:='n';" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1
0 61 "u:=(t,x)->1+x+sum(c[n]*exp(-n^2*Pi^2*t)*sin(n*Pi*x),n=1..20);" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "plot3d(u(t,x),x=0..1,t=0..
1/2,axes=NORMAL,orientation=[-125,65]);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 48 "u(t,0);\nu(t,1);\ndiff(u(t,x),t)-diff(u(t,x),x,x);" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "plot([u(1/1000,x),f(x)],x=
0..1,color=[red,black]);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 287 " One might tr
y the following experiment. Take an initial distribution of heat with \+
two or more peaks and see if, as the heat diffuses, the heat from one \+
peak might combine with the heat from another peak to produce a temper
ature greater than that for either of the two initial peaks." }}{PARA
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 29 "Too bad. That w
ill not work. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT 269 40 "Maximal Principle for the Heat Equation:" }{TEXT -1 201
" The solution for the above heat equation for t in [0, T] and x in [0
, a] attains it maximum and minimum on one of four sides of the bounda
ry of the region in the plane described as all points given by" }}
{PARA 0 "" 0 "" {TEXT -1 30 " t = 0 and x in [0, a], or" }}{PARA
0 "" 0 "" {TEXT -1 30 " t in [0, T] and x = 0, or" }}{PARA 0 "" 0
"" {TEXT -1 30 " t in [0, T] and x = a, or" }}{PARA 0 "" 0 ""
{TEXT -1 27 " t = T and x in [0, a]." }}{PARA 0 "" 0 "" {TEXT -1
0 "" }}{PARA 0 "" 0 "" {TEXT -1 398 "That is, the temperature in the i
nterval [0, a] can never exceed the larger of the maximum of the initi
al temperature or the maximum temperature at the ends of the interval \+
[0, a] as time increases. In a similar manner, the temperature in the \+
interval [0, a] is not smaller than the minimum of the initial tempera
ture or the minimum temperature at the ends of the interval [0, a] as \+
time increases." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 85 "Here's an example to try out for t in the interval [0, 10
] and x in the interval [0, " }{XPPEDIT 18 0 "pi;" "6#%#piG" }{TEXT
-1 73 "/2]. We ask what is the maximum value of u(t, x) and where does
it occur?" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "u:=(t,x)->exp(
-t)*sin(x)+x+1;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "diff(u(t
,x),t)-diff(u(t,x),x,x);\nu(t,0);\nu(t,Pi/2);\nu(0,x);" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 158 "maximize(u(0,x),\{x\},\{x=0..evalf
(Pi/2)\});\nmaximize(u(t,0),\{t\},\{t=0..10\});\nmaximize(u(t,Pi/2),\{
t\},\{t=0..10\});\nmaximize(u(t,x),\{t,x\},\{t=0..10,x=0..evalf(Pi/2)
\});" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "plot3d(u(t,x),x=0..
Pi/2,t=0..10,axes=NORMAL);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0
0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 90 "S
uppose that u satisfies the differential equation above and that v sat
isfies the equation" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 44 "The Partial Differentia
l Equation: " }{XPPEDIT 18 0 "diff(v,t);" "6#-%%diffG6$%\"vG%
\"tG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "diff(v,`$`(x,2));" "6#-%%diffG
6$%\"vG-%\"$G6$%\"xG\"\"#" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 40 "The Boundary Condtions: v
(t, 0) = " }{XPPEDIT 18 0 "T[0];" "6#&%\"TG6#\"\"!" }{TEXT -1 16 " an
d v(t, 1) = " }{XPPEDIT 18 0 "T[1];" "6#&%\"TG6#\"\"\"" }{TEXT -1 1 ".
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 87 "The I
nitial Conditions: v(0, x) = g(x), where f(x) is continuous on [0, a] \+
and g(0) = ." }{XPPEDIT 18 0 "T[0];" "6#&%\"TG6#\"\"!" }{TEXT -1 10 " \+
, g(a) = " }{XPPEDIT 18 0 "T[1];" "6#&%\"TG6#\"\"\"" }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 18 "Suppose also that " }
{XPPEDIT 18 0 "f <= g;" "6#1%\"fG%\"gG" }{TEXT -1 62 ". We argue that \+
for all t and x in the appropriate intervals, " }{XPPEDIT 18 0 "u(t,x)
<= v(t,x);" "6#1-%\"uG6$%\"tG%\"xG-%\"vG6$F'F(" }{TEXT -1 175 " . The
suggestion for doing this problem is to consider the function w(t, x)
= u(t, x) - v(t, x) and ask what equation w satisfies. You will see t
hat w satisfies the equation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 44 "The Parti
al Differential Equation: " }{XPPEDIT 18 0 "diff(w,t);" "6#-%
%diffG6$%\"wG%\"tG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "diff(w,`$`(x,2))
;" "6#-%%diffG6$%\"wG-%\"$G6$%\"xG\"\"#" }{TEXT -1 2 " ." }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 40 "The Boundary Condt
ions: w(t, 0) = " }{XPPEDIT 18 0 "0;" "6#\"\"!" }{TEXT -1 16 " \+
and v(t, 1) = " }{XPPEDIT 18 0 "0;" "6#\"\"!" }{TEXT -1 1 "." }}{PARA
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 46 "The Initial Con
ditions: w(0, x) = f(x) - g(x)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 14 "Conclude that " }{XPPEDIT 18 0 "w(t,x) <=
0;" "6#1-%\"wG6$%\"tG%\"xG\"\"!" }{TEXT -1 1 "." }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 8 "Exercise" }}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 149 "Show that each of these satisfies
the heat equation and find where the maximum occurs for t in the inte
rval [0, 10] and x in the interval [0, 1].\n1. " }{XPPEDIT 18 0 "t+x^2
/2+5;" "6#,(%\"tG\"\"\"*&%\"xG\"\"#\"\"#!\"\"F%\"\"&F%" }{TEXT -1 1 ".
" }}{PARA 0 "" 0 "" {TEXT -1 24 "2. exp(- 25 t) sin(5 x)," }}{PARA 0 "
" 0 "" {TEXT -1 14 "3. exp(t - x)." }}}}}}{MARK "9" 0 }{VIEWOPTS 1 1
0 1 1 1803 }