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{SECT 0 {EXCHG {PARA 257 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 100 "First I decide how man
y Fourier modes I want to use.  I'll use 10.  If I want to use 20, the
n I need" }}{PARA 0 "" 0 "" {TEXT -1 114 "to change the 10 into a 20 i
n six places: where I define  a and b (below), where I fill them up (t
he two for loops" }}{PARA 0 "" 0 "" {TEXT -1 77 "right after) and wher
e I define the function (the two for loops after that.)." }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 109 "The a array is fo
r the sine transforms.  Since sin(0*x) = 0, we only have transforms fr
om integrating against" }}{PARA 0 "" 0 "" {TEXT -1 63 "sin(x), sin(2x)
,..., sin(10x).  So the a array is of length 10." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 113 "The b array is for the c
osine transforms.  Since cos(0*x) = 1, the first transform will pick u
p the mean value of" }}{PARA 0 "" 0 "" {TEXT -1 113 "the function.  We
 have transforms from integrating against 1, cos(x), cos(2x), ..., cos
(10x).  So the array is of" }}{PARA 0 "" 0 "" {TEXT -1 10 "length 11.
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 37 
"a := array(1..10): b := array(0..10):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}{PARA 0 "" 0 "" {TEXT -1 83 "I now choose initial \+
data.  In this case, it's pretty boring, just phi(x) = cos(x)." }
{MPLTEXT 1 0 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}{PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 17 "f := x -> cos(x);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"fG%$cosG" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 122 "Using the initial data, I define \+
the 10 coefficients for sin and the 11 coefficients for cos.   To do t
his I have to think" }}{PARA 0 "" 0 "" {TEXT -1 59 "first: sin(jx) and
 cos(jx) are supposed to have length 1.  " }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 110 "Since the integral from 0 to 2Pi \+
of sin^2(jx) equals Pi, I divide the integrals against sine and cosine
 by Pi." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
106 "On the other hand, cos(0x) should also have length 1.  The integr
al of cos(0x) from 0 to 2Pi is 2 Pi, so I" }}{PARA 0 "" 0 "" {TEXT -1 
43 "divide the integral against cos(0x) by 2Pi." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 41 "The arrays a and b are fi
lled as follows:" }}{PARA 0 "" 0 "" {TEXT -1 64 "a[1] = ahat_1, a[2] =
 ahat_2, a[3] = ahat_3, ... a[10] = ahat_10" }}{PARA 0 "" 0 "" {TEXT 
-1 64 "b[1] = bhat_0, b[2] = bhat_1, b[3] = bhat_2, ... b[11] = bhat_1
0" }{MPLTEXT 1 0 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}{PARA 0 
"> " 0 "" {MPLTEXT 1 0 83 "for j from 1 to 10 do a[j]:= evalf(Int(f(x)
*sin(j*x),x=0..2*Pi))/(Pi) od:\nprint(a);" }}{PARA 12 "" 1 "" 
{XPPMATH 20 "6#-%'VECTORG6#7,,$*$%#PiG!\"\"$!+ElR`Y!#D,$F($\"+#fD)Q=F-
,$F($\"+qzbKT!#E,$F($\"+V+8N?F4,$F($!+2KZC7F-,$F($!+%>@wA\"!#C,$F($!+P
IG&['F4\"\"!,$F($\"+]D\"><%!#AFB" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 127 "for j from 1 to 10 do b[j]:= evalf(Int(f(x)*cos(j*x)
,x=0..2*Pi))/(Pi) od:\nb[0] := evalf(Int(f(x),x=0...2*Pi))/(2*Pi): pri
nt(b);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#-%&ARRAYG6$7#;\"\"!\"#57-/6#
F(,$*$%#PiG!\"\"$\"+O])o1#!#A/6#\"\"\",$F.$\"+aEfTJ!\"*/6#\"\"#,$F.$\"
+(=.<8%F3/6#\"\"$,$F.$\"+dp$e8%F3/6#\"\"%,$F.$\"+7w.UTF3/6#\"\"&,$F.$
\"+9N![8%F3/6#\"\"',$F.$\"+NhgfTF3/6#\"\"(,$F.$\"+s+xLTF3/6#\"\"),$F.$
\"+(y'RYTF3/6#\"\"*,$F.$\"+#Qrh9%F3/6#F),$F.$\"+)egw9%F3" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}{PARA 0 "" 0 "" {TEXT -1 102 "No
w that I know the Fourier coefficients, I know exactly what the soluti
on does.  It takes the fourier" }}{PARA 0 "" 0 "" {TEXT -1 115 "coeffi
cient a(j) and multiplies it by  exp(-j^2 t).  So I do that multiplica
tion and then sum to find the solution " }}{PARA 0 "" 0 "" {TEXT -1 2 
"u." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 136 "u := b[0]:\nfor j from 1 to 10 do u := u + exp(-j^2*
t)*a[j]*sin(j*x) od:\nfor j from 1 to 10 do u := u + exp(-j^2*t)*b[j]*
cos(j*x) od:\nu;\n" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#,H*$%#PiG!\"\"$
\"+O])o1#!#A*(-%$expG6#,$%\"tGF&\"\"\"F%F&-%$sinG6#%\"xGF0$!+ElR`Y!#D*
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F26#,$F4\"\"'F0$!+%>@wA\"!#C*(-F,6#,$F/!#\\F0F%F&-F26#,$F4\"\"(F0$!+PI
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{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}{PARA 0 "" 0 "" {TEXT -1 80 "Fir
st let's see if we have the right initial data by plotting u(x,0) vers
us f(x)" }{MPLTEXT 1 0 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "plot(\{f(x),subs(t=0,u)\},x=0..2*Pi
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0 "" {TEXT -1 40 "Now plot the solution at t=0 versus t=.1" }{MPLTEXT 
1 0 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 43 "plot(\{subs(t=0,u),subs(t=.1,u)\},x=0..2*Pi);" }}
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0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 97 "f := x -> (x-Pi/2)*Heaviside(x-Pi/2) - 2*(x-Pi)*Heavi
side(x-Pi) + (x-3*Pi/2)*Heaviside(x-3*Pi/2);" }}{PARA 12 "" 1 "" 
{XPPMATH 20 "6#>%\"fG:6#%\"xG6\"6$%)operatorG%&arrowGF(,(*&,&9$\"\"\"%
#PiG#!\"\"\"\"#F0-%*HeavisideG6#F.F0F0*&,&F/F0F1F3F0-F66#F9F0!\"#*&,&F
/F0F1#!\"$F4F0-F66#F>F0F0F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 21 "plot(f(x),x=0..2*Pi);" }}{PARA 13 "" 1 "" {INLPLOT "6%-%'CURVESG
6$7Y7$\"\"!F(7$$\"15@;)eb&p8!#;F(7$$\"1D![Ce)>hDF,F(7$$\"1>FdfdK,RF,F(
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OURG6&%$RGBG$\"#5!\"\"F(F(-%+AXESLABELSG6$%\"xG%!G-%%VIEWG6$;F($\"+3`=
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45.000000 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 11376 0 0 0 0 
0 0 }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 20 "Homework Problem #1:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 84 "In the above, we used 10 fourier modes to appro
ximate the solution.  Choose the step" }}{PARA 0 "" 0 "" {TEXT -1 87 "
function above as initial data.  (The one that went from 0 to 1.)  App
roximate it with " }}{PARA 0 "" 0 "" {TEXT -1 79 "5, 10, 20, and 30 Fo
urier modes.  This determines four solutions u1, u2, u3, u4" }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 89 "a) Plot the rea
l initial data and the approximate initial data.  The real initial dat
a is" }}{PARA 0 "" 0 "" {TEXT -1 58 "nonnegative.  Is the approximate \+
initial data nonnegative?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 85 "b) Plot the four solutions at some time to compare
 how close they stay to each other." }}{PARA 0 "" 0 "" {TEXT -1 23 "\n
  Homework Problem #2:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 98 "Here we worked on solutions on the interval [0,2 Pi]
.  Now we consider solutions periodic on [0,L]" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 67 "a) What is the basis that
 the Fourier series expansion is based on?" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 93 "b) What is the new inner produc
t that makes the basis elements be orthogonal and unit length?" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 180 "c) Modif
y the above program to correctly find the Fourier coefficients for f's
 Fourier series approximation.  NOTE:  where you declare the arrays a \+
and b, you should declare L here" }}{PARA 0 "" 0 "" {TEXT -1 89 "as we
ll (L:=30, for example).  L will appear in your do loops and if you do
n't declare it" }}{PARA 0 "" 0 "" {TEXT -1 42 "at the very beginning, \+
maple will choke.  " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 88 "d)  Check that you did it right by testing it against f
(x) = 1 and f (x)= sin(x 2 Pi/L) " }}{PARA 0 "" 0 "" {TEXT -1 28 "and \+
f(x) = cos( x 2 Pi/L).  " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 96 "e) Modify the program to find the solution of the \+
heat equation.  Check that you did it right by" }}{PARA 0 "" 0 "" 
{TEXT -1 89 "testing it with initial data f(x) = 1 and f (x)= sin(x 2 \+
Pi/L) and f(x) = cos( x 2 Pi/L)." }}{PARA 0 "" 0 "" {TEXT -1 88 "I.e. \+
solve the equation by hand and make sure you get the right answer out \+
of maple, up " }}{PARA 0 "" 0 "" {TEXT -1 21 "to machine precision." }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 142 "f) Pick
 some amusing initial data and study it.  The more examples you do, th
e merrier.  Bonus points for initial data that looks like a cusp." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 19 "Homework \+
Problem #3" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 85 "In the above, we did the heat equation u_t = u_xx.  Now let's c
onsider u_t = c*u_xx. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 60 "a) Modify the above program so that you have incorpo
rated c." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
104 "b) Pick one set of initial data.  Now pick three different values
 of c.  This determines three solutions" }}{PARA 0 "" 0 "" {TEXT -1 
101 "u_1, u_2, and u_3.  Plot the three solutions at a fixed time to g
raphically demonstrate the effect of" }}{PARA 0 "" 0 "" {TEXT -1 76 "y
our choice of diffusion constant.  Explain why the behavior is as expe
cted." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}}{MARK "18 42 0" 0 }
{VIEWOPTS 1 1 0 1 1 1803 }