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{SECT 0 {PARA 201 "" 0 "" {TEXT 207 29 "Worksheet 16: The Simple Heat
" }{TEXT 208 1 " " }{TEXT 207 8 "Equation" }}{PARA 202 "" 0 "" {TEXT
-1 0 "" }}{PARA 202 "" 0 "" {TEXT 209 433 " We consider the simple
heat equation. This is a standard linear realization for diffusion in
one dimension. Derivations for this model can be found in most texts \+
in this subject. 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 \+
differentiable as a function of t and twice differentiable as a functi
on of x. In this model, these derivatives are related by" }}{PARA 202
"" 0 "" {TEXT -1 0 "" }}{PARA 203 "" 0 "" {TEXT 200 10 " " }
{XPPEDIT 2 0 "diff(u,t)" "6#-%%diffG6$%\"uG%\"tG" }{TEXT 200 2 "= " }
{XPPEDIT 2 0 "diff(u,`$`(x,2))" "6#-%%diffG6$%\"uG-%\"$G6$%\"xG\"\"#"
}{TEXT 200 2 " ." }}{PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 "
" {TEXT 209 170 "In the heat equation, there are boundary conditions a
nd initial conditions. In this worksheet, we will take the boundary co
nditions to be specified at x = 0 and at x = 1:" }}{PARA 202 "" 0 ""
{TEXT -1 0 "" }}{PARA 202 "" 0 "" {TEXT 209 35 " u(t, 0) = 0 and
u(t, 1) = 0." }}{PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 ""
{TEXT 209 69 "We suppose that the value of u when t = 0 is specified a
s, say, f(x):" }}{PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 ""
{TEXT 209 21 " u(0, x) = f(x)." }}{PARA 202 "" 0 "" {TEXT -1 0 "
" }}{PARA 202 "" 0 "" {TEXT 209 805 " An intuitive way to think of
this equation is that there is a thin, uniform bar over the interval \+
[0, 1]. In this simple problem, one takes the lateral surface of the r
od to be insulated. Both ends of the rod are held at a fixed temperatu
re. There are no internal or external heat sources. The rod has an ini
tial heat distribution given by f(x). The heat diffuses from the place
s where it is warm toward the places where it is cooler. The rate of d
iffusion is proportioned to the curvature of the present heat distribu
tion. This statement about curvature takes the form of the second deri
vative. Thus, if the current distribution is concave down, the second \+
derivative is negative and the temperature at that point will be decre
asing. if the distribution is concave up, the temperature is increasin
g." }}{PARA 202 "" 0 "" {TEXT 209 345 " We will make the model mor
e complicated later. For now, let's make an analysis of this situation
with the experience and knowledge we have. To this point, the princip
le idea we have used is the Fourier Idea. Here, we introduce the secon
d idea: We suppose that the solution can be separate into a function o
f t alone and of x alone, so that " }}{PARA 202 "" 0 "" {TEXT 209 30 "
u(t,x) = X(x) T(t). " }}{PARA 202 "" 0 "" {TEXT -1 0 "" }}
{PARA 202 "" 0 "" {TEXT 209 57 "This assumption leads to ordinary diff
erential equations." }}{EXCHG {PARA 204 "> " 0 "" {MPLTEXT 1 0 20 "u:=
(t,x)->X(x)*T(t);" }}}{EXCHG {PARA 204 "> " 0 "" {MPLTEXT 1 0 32 "diff
(u(t,x),t)=diff(u(t,x),x,x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1
0 0 "" }}}{EXCHG {PARA 204 "" 0 "" {TEXT -1 0 "" }}}{PARA 202 "" 0 ""
{TEXT 209 100 "If we bring all the functions of t to one side and all \+
the functions of x to the other side, we have" }}{PARA 202 "" 0 ""
{TEXT -1 0 "" }}{PARA 202 "" 0 "" {TEXT 209 30 " T ' / T = X
'' / X." }}{PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 "" {TEXT
209 215 "Since the right side is independent of x, then as x changes, \+
X '' / X does not change. This quotient is constant. In a similar mann
er, the quotient T ' / T is constant, and, from the equality, is the s
ame constant. " }}{PARA 202 "" 0 "" {TEXT 209 202 " The boundary c
onditions 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 202 "" 0 ""
{TEXT -1 0 "" }}{PARA 202 "" 0 "" {TEXT 209 17 " X '' = " }
{XPPEDIT 18 0 "mu;" "6#%#muG" }{TEXT 209 25 " X, with X(0) = 0 = X(1).
" }}{PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 "" {TEXT 209 104
"We have already examined this differential equation. We found that th
ere are an infinite number of such " }{XPPEDIT 18 0 "mu;" "6#%#muG" }
{TEXT 209 33 " 's and they are all of the form " }}{PARA 202 "" 0 ""
{TEXT -1 0 "" }}{PARA 202 "" 0 "" {TEXT 209 10 " " }{XPPEDIT
18 0 "mu = -n^2*Pi^2;" "6#/%#muG,$*&%\"nG\"\"#%#PiGF(!\"\"" }}{PARA
202 "" 0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 "" {TEXT 209 62 "and corres
ponding to each such constant, there is the solution" }}{PARA 202 ""
0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 "" {TEXT 209 24 " X(x) = \+
sin( n " }{XPPEDIT 18 0 "pi;" "6#%#piG" }{TEXT 209 4 " x)." }}{PARA
202 "" 0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 "" {TEXT 209 91 " What \+
about the possible T solutions? The quotient T '/ T is the same quotie
nt, so that" }}{PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 ""
{TEXT 209 16 " T ' = " }{XPPEDIT 18 0 "-n^2*Pi^2;" "6#,$*&%\"
nG\"\"#%#PiGF&!\"\"" }{TEXT 209 3 " T." }}{PARA 202 "" 0 "" {TEXT -1
0 "" }}{PARA 202 "" 0 "" {TEXT 209 31 "Solutions for this equation are
" }}{PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 "" {TEXT 209 23 "
T(t) = exp( " }{XPPEDIT 18 0 "-n^2*Pi^2;" "6#,$*&%\"nG\"\"#
%#PiGF&!\"\"" }{TEXT 209 4 " t)." }}{PARA 202 "" 0 "" {TEXT -1 0 "" }}
{PARA 202 "" 0 "" {TEXT 209 52 "Consequently, for each integer n, we h
ave a solution" }}{PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 ""
{TEXT 209 26 " u(t, x) = exp( " }{XPPEDIT 18 0 "-n^2*Pi^2;"
"6#,$*&%\"nG\"\"#%#PiGF&!\"\"" }{TEXT 209 11 " t) sin( n " }{XPPEDIT
18 0 "pi;" "6#%#piG" }{TEXT 209 4 " x)." }}{PARA 202 "" 0 "" {TEXT -1
0 "" }}{PARA 202 "" 0 "" {TEXT 209 192 "But, since this is a linear pr
oblem, multiples of these solutions by a constant are also solutions, \+
and sums of solutions are solutions. Hence, a general solution can be \+
written in the form of" }}{PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 202
"" 0 "" {TEXT 209 20 " u(t, x) = " }{XPPEDIT 18 0 "sum(c[n]*e
xp(-n^2*Pi^2*t)*sin(n*Pi*x),n);" "6#-%$sumG6$*(&%\"cG6#%\"nG\"\"\"-%$e
xpG6#,$*(F*\"\"#%#PiGF1%\"tGF+!\"\"F+-%$sinG6#*(F*F+F2F+%\"xGF+F+F*" }
{TEXT 209 2 " ." }}{PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 "
" {TEXT 209 136 "There is one more piece of information we have not us
ed. We have not used the initial value. This condition determines the \+
coefficients:" }}{PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 ""
{TEXT 209 27 " f(x) = u(0, x) = " }{XPPEDIT 18 0 "sum(c[n]*si
n(n*Pi*x),n);" "6#-%$sumG6$*&&%\"cG6#%\"nG\"\"\"-%$sinG6#*(F*F+%#PiGF+
%\"xGF+F+F*" }{TEXT 209 1 "." }}{PARA 202 "" 0 "" {TEXT -1 0 "" }}
{PARA 202 "" 0 "" {TEXT 209 55 "This looks like a job for Fourier Seri
es. We know that " }{XPPEDIT 18 0 "c[n];" "6#&%\"cG6#%\"nG" }{TEXT
209 59 " can be written as a quotient of integrals which resolve as" }
}{PARA 202 "" 0 "" {TEXT 209 3 " " }}{PARA 202 "" 0 "" {TEXT 209 8 "
" }{XPPEDIT 18 0 "c[n];" "6#&%\"cG6#%\"nG" }{TEXT 209 4 " = \+
" }{XPPEDIT 18 0 "int(f(x)*sin(n*Pi*x),x = 0 .. 1);" "6#-%$intG6$*&-%
\"fG6#%\"xG\"\"\"-%$sinG6#*(%\"nGF+%#PiGF+F*F+F+/F*;\"\"!F+" }{TEXT
209 1 " " }{TEXT 209 2 "/ " }{XPPEDIT 18 0 "int(sin(n*Pi*x)^2,x = 0 ..
1);" "6#-%$intG6$*$-%$sinG6#*(%\"nG\"\"\"%#PiGF,%\"xGF,\"\"#/F.;\"\"!
F," }{TEXT 209 2 " ." }}{PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 202 "
" 0 "" {TEXT 209 110 "It will be valuable at this point to take a part
icular f, solve the equation, and draw some pictures. We take " }
{TEXT 209 18 "\n f(x) = " }{XPPEDIT 18 0 "x*(1-x)^3;" "6#*&%
\"xG\"\"\"*$,&F%F%F$!\"\"\"\"$F%" }{TEXT 209 2 " ." }}{PARA 202 "" 0 "
" {TEXT 209 48 "To visualize the initial value, we draw a graph." }}
{EXCHG {PARA 204 "> " 0 "" {MPLTEXT 1 0 8 "restart:" }{MPLTEXT 1 0 17
"\nf:=x->x*(1-x)^3;" }}}{EXCHG {PARA 204 "> " 0 "" {MPLTEXT 1 0 18 "pl
ot(f(x),x=0..1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 204 "" 0 "" {TEXT -1 0 "" }}}{PARA 202 "" 0 "" {TEXT 209
249 " We form the Fourier representation for this f and compare th
e graph of f and of our series representation. First, note that the fu
nction is continuous and its periodic extension is continuous, so that
the Fourier series will converge uniformly." }}{EXCHG {PARA 204 "> "
0 "" {MPLTEXT 1 0 20 "for n from 1 to 7 do" }{MPLTEXT 1 0 65 "\n c[n
]:=int(f(x)*sin(n*Pi*x),x=0..1)/int(sin(n*Pi*x)^2,x=0..1):" }{MPLTEXT
1 0 4 "\nod:" }{MPLTEXT 1 0 8 "\nn:='n';" }}}{EXCHG {PARA 204 "> " 0 "
" {MPLTEXT 1 0 37 "Fou:=x->sum(c[n]*sin(n*Pi*x),n=1..7):" }}}{EXCHG
{PARA 204 "> " 0 "" {MPLTEXT 1 0 45 "plot([f(x),Fou(x)],x=0..1,color=[
BLACK,RED]);" }}}{EXCHG {PARA 204 "" 0 "" {TEXT -1 0 "" }}}{PARA 202 "
" 0 "" {TEXT 209 96 " As you can see, this is a very good fit. Now
, we form the truncated series expansion for u." }}{EXCHG {PARA 204 ">
" 0 "" {MPLTEXT 1 0 56 "u:=(t,x)->sum(c[n]*exp(-n^2*Pi^2*t)*sin(n*Pi*
x),n=1..7);" }}}{PARA 202 "" 0 "" {TEXT 209 243 " Before drawing t
he graph, what is expected? We should expect that when t = 0, the grap
h of u(0, x) looks like the graph of f(x). Further, as t increases, so
lution decreases to the stationary solution determined by the boundary
conditions:" }}{PARA 202 "" 0 "" {TEXT 209 15 " " }
{XPPEDIT 18 0 "u(infinity,x) = 0;" "6#/-%\"uG6$%)infinityG%\"xG\"\"!"
}{TEXT 209 2 " ." }}{EXCHG {PARA 204 "> " 0 "" {MPLTEXT 1 0 65 "plot3d
(u(t,x),x=0..1,t=0..1/10,axes=NORMAL,orientation=[-20,55]);" }}}
{EXCHG {PARA 204 "" 0 "" {TEXT -1 0 "" }}}{PARA 202 "" 0 "" {TEXT 209
256 " In designing this problem, I deliberately chose an initial d
istribution for which the highpoint of the graph is off center. It app
ears that this highpoint moves in toward the center of the interval [0
, 1]. We can look at this movement in an animation." }}{EXCHG {PARA
204 "> " 0 "" {MPLTEXT 1 0 12 "with(plots):" }}}{EXCHG {PARA 204 "> "
0 "" {MPLTEXT 1 0 32 "animate(u(t,x),x=0..1,t=0..1/3);" }}}{EXCHG
{PARA 204 "" 0 "" {TEXT -1 0 "" }}}{PARA 202 "" 0 "" {TEXT 209 189 " \+
We trace the movement of the high point. First note that the high p
oint of the initial distribution is at 1/4. It was clear that the firs
t derivative was zero in the interval [0, 1/2]." }}{EXCHG {PARA 204 ">
" 0 "" {MPLTEXT 1 0 32 "solve(\{diff(f(x),x)=0,x<1/2\},x);" }}}
{EXCHG {PARA 204 "" 0 "" {TEXT -1 0 "" }}}{PARA 202 "" 0 "" {TEXT 209
263 "Solving the equation for where the first derivative is zero in th
e approximation is harder to find. We make this determination numerica
lly. That is, we get a floating point approximation for the derivative
s. We expect to see the first derivatives converge to 1/2." }}{EXCHG
{PARA 204 "> " 0 "" {MPLTEXT 1 0 21 "for p from 0 to 10 do" }{MPLTEXT
1 0 48 "\nfsolve(subs(t=p/30,diff(u(t,x),x))=0,x,0..1/2);" }{MPLTEXT
1 0 4 "\nod;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG
{PARA 204 "" 0 "" {TEXT -1 0 "" }}}{PARA 202 "" 0 "" {TEXT -1 0 "" }}
{PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 "" {TEXT 209 125 " \+
In this worksheet, we have solved a simple heat equation with zero b
oundary conditions and with an initial distribution." }}{PARA 202 ""
0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 "" {TEXT 209 89 "Assignment: Solve
the same distribution with this different initial distribution: f(x) \+
= " }}{EXCHG {PARA 204 "> " 0 "" {MPLTEXT 1 0 16 "f:=x->x^3*(1-x);" }}
}{EXCHG {PARA 204 "> " 0 "" {MPLTEXT 1 0 18 "plot(f(x),x=0..1);" }}}
{EXCHG {PARA 204 "" 0 "" {TEXT -1 0 "" }}}{PARA 202 "" 0 "" {TEXT -1
0 "" }}{PARA 202 "" 0 "" {TEXT 209 107 "Note in preparation: Here is h
ow I drew the sequence of snapshots to observe the movement of the hig
hpoint." }}{EXCHG {PARA 204 "> " 0 "" {MPLTEXT 1 0 7 "n:='n':" }
{MPLTEXT 1 0 21 "\nfor i from 1 to 5 do" }{MPLTEXT 1 0 37 "\n P[i]:=
plot(u((i-1)/15,x),x=0..1):" }{MPLTEXT 1 0 4 "\nod:" }}}{EXCHG {PARA
204 "> " 0 "" {MPLTEXT 1 0 7 "i:='i':" }{MPLTEXT 1 0 13 "\nwith(plots)
:" }{MPLTEXT 1 0 29 "\ndisplay([seq(P[i],i=1..5)]);" }}}{EXCHG {PARA
0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 202 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT -1 537 "This was too simple. Vari
ous questions should come to mind. Suppose heat is loss laterally. Or,
suppose the end points are not zero. the first time I made these lect
ures, I handled these questions in the next Lecture 17 by reducing the
problem to the simple one. Students hated that lecture. So, I made a \+
simpler version. Look at the original Lecture 17, but don't complain l
oudly. Then, look at the revision which you get to see. Then, tell me \+
you are glad that I experimented to find the best way to do this befor
e you took the course." }}{PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 205
"" 0 "" {TEXT -1 0 "" }}}{MARK "0 2" 8 }{VIEWOPTS 1 1 0 1 1 1803 1 1
1 1 }{PAGENUMBERS 0 1 2 33 1 1 }