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{SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT -1 40 "Partial Differential Equa
tions PowerTool" }}{PARA 19 "" 0 "" {TEXT -1 16 "by Dr. Jim Herod" }}}
{PARA 256 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT 260 42 "Sect
ion 4.3: Insulated Boundary Conditions" }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{SECT 0 {PARA 3 "" 0 "" {TEXT 276 30 "Maple Packages for Section 4
.3" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "with(PDEtools):" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 102 "In this worksheet, we begin an examination of \+
the heat equation with a variety of boundary conditions." }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 257 "" 0 "" {TEXT 271 17 "The First Probl
em" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 91 "For
 the purposes of this discussion, we write the equation to be consider
ed 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#/-%%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#-%%dif
fG6$%\"uG%\"xG" }{TEXT -1 16 " (t, 0) = 0 and " }{XPPEDIT 18 0 "diff(u
,x);" "6#-%%diffG6$%\"uG%\"xG" }{TEXT -1 11 " (t,1) = 0." }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 27 "The Initial Condit
ion:     " }{TEXT 279 1 "u" }{TEXT -1 4 "(0, " }{TEXT 280 1 "x" }
{TEXT -1 4 ") = " }{TEXT 281 1 "f" }{TEXT -1 1 "(" }{TEXT 282 1 "x" }
{TEXT -1 2 ")." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT 278 30 "Solution for the
 First Problem" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 53 "     We have discussed a physical interpretation for " }
{TEXT 256 7 "the PDE" }{TEXT 261 1 " " }{TEXT -1 8 "and for " }{TEXT 
257 21 "the initial condition" }{TEXT -1 40 ". Here is an idea to give
 intuition for " }{TEXT 258 24 "the boundary conditions " }{TEXT -1 
55 "in this problem. Having the derivative with respect to " }{TEXT 
262 1 "x" }{TEXT -1 93 " to be zero at these two ends suggests that th
ere is no movement of heat across the boundary " }{TEXT 263 1 "x" }
{TEXT -1 8 " = 0 or " }{TEXT 264 1 "x" }{TEXT -1 5 " = 1." }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 333 "     From this, w
hat do we expect for solutions of this problem? Since the problem acts
 as though there is no radial transfer of heat and no end point transf
er of heat, we expect the entire system to retain the total heat, and \+
to move to an equal distribution of heat throughout the rod. We will s
olve the equation with a particular " }{TEXT 265 1 "f" }{TEXT -1 115 "
 and sketch the graph of the solution. We expect to see that the solut
ion has limit a solution that is constant in " }{TEXT 266 1 "t" }
{TEXT -1 8 " and in " }{TEXT 267 1 "x" }{TEXT -1 86 ". Also, because t
he total heat in the system should remain 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$%\"tG%\"xG/F*;\"\"!\"\"\"" }{TEXT -1 
17 " , for all t > 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 
"" {TEXT -1 28 "Here goes making a solution." }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 93 "First, we find two ODE's, with \+
boundary conditions, which arise from separation of variables." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 17 "The PDE l
eads to " }}{PARA 0 "" 0 "" {TEXT -1 3 "   " }}{PARA 0 "" 0 "" {TEXT 
-1 26 "          X T ' = X '' T, " }}{PARA 0 "" 0 "" {TEXT -1 2 "or" }
}{PARA 0 "" 0 "" {TEXT -1 29 "          T ' / T = X '' / X." }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 38 "Since the left \+
side is independent of " }{TEXT 283 1 "x" }{TEXT -1 121 ", the right s
ide must be constant. In a similar manner, the left side is constant a
nd these constants are the same. Thus," }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 17 "          X '' = " }{XPPEDIT 18 0 "mu
;" "6#%#muG" }{TEXT -1 34 " X, with X'(0) = 0 and X '(1) = 0," }}
{PARA 0 "" 0 "" {TEXT -1 3 "and" }}{PARA 0 "" 0 "" {TEXT -1 16 "      \+
    T ' = " }{XPPEDIT 18 0 "mu;" "6#%#muG" }{TEXT -1 3 " T." }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 84 "Using methods w
e have seen earlier, this constant must be negative and have the form
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 10 "     \+
     " }{XPPEDIT 18 0 "mu = -n^2*Pi^2;" "6#/%#muG,$*&%\"nG\"\"#%#PiGF(
!\"\"" }{TEXT -1 8 "  with X" }{XPPEDIT 18 0 "` `[n];" "6#&%\"~G6#%\"n
G" }{TEXT -1 1 "(" }{TEXT 284 1 "x" }{TEXT -1 4 ") = " }{XPPEDIT 18 0 
"cos(n*Pi*x);" "6#-%$cosG6#*(%\"nG\"\"\"%#PiGF(%\"xGF(" }{TEXT -1 33 "
 , for each non-negative integer " }{TEXT 285 1 "n" }{TEXT -1 1 "." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 7 "Also, T" }
{XPPEDIT 18 0 "` `[n];" "6#&%\"~G6#%\"nG" }{TEXT -1 2 " (" }{TEXT 286 
1 "t" }{TEXT -1 9 ") = exp( " }{XPPEDIT 18 0 "-n^2*Pi^2*t;" "6#,$*(%\"
nG\"\"#%#PiGF&%\"tG\"\"\"!\"\"" }{TEXT -1 2 ")." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 45 "Thus, the general solutio
n 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)*co
s(n*Pi*x),n = 1 .. infinity);" "6#-%$sumG6$*(&%\"cG6#%\"nG\"\"\"-%$exp
G6#,$*(F*\"\"#%#PiGF1%\"tGF+!\"\"F+-%$cosG6#*(F*F+F2F+%\"xGF+F+/F*;F+%
)infinityG" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 50 "To find a particular solution, we need to speci
fy " }{TEXT 287 1 "f" }{TEXT -1 29 ". We do that now. Using this " }
{TEXT 288 1 "f" }{TEXT -1 74 ", we write a solution and check that the
 answer is right, and seems right!" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 16 "f:=x->x*(x-1)+1;" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 13 "Because this " }{TEXT 268 1 "f" }{TEXT 
-1 5 "  is " }{TEXT 289 1 "u" }{TEXT -1 4 "(0, " }{TEXT 290 1 "x" }
{TEXT -1 8 "), then " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 10 "          " }{TEXT 291 1 "f" }{TEXT -1 1 "(" }{TEXT 
292 1 "x" }{TEXT -1 5 ") =  " }{XPPEDIT 18 0 "c[0];" "6#&%\"cG6#\"\"!
" }{TEXT -1 3 " + " }{XPPEDIT 18 0 "sum(c[n]*cos(n*Pi*x),n = 1 .. infi
nity);" "6#-%$sumG6$*&&%\"cG6#%\"nG\"\"\"-%$cosG6#*(F*F+%#PiGF+%\"xGF+
F+/F*;F+%)infinityG" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}}{PARA 0 "" 0 "" {TEXT -1 23 "The computation of the " }{TEXT 293 1 "
c" }{TEXT -1 32 " 's is a job for Fourier Series." }}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 114 "for n from 0 to 10 do\n     c[n]:=int(f(x)*c
os(n*Pi*x),x=0..1)/\n             int(cos(n*Pi*x)^2,x=0..1);\nod;\nn:=
'n':" }}}{PARA 0 "" 0 "" {TEXT -1 36 "To see how good our fit is, we p
lot " }{TEXT 294 1 "f" }{TEXT -1 31 " and the Fourier Approximation." 
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "plot([f(x),c[0]+sum(c[n]*c
os(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 16 
"Now we make the " }{TEXT 295 1 "u" }{TEXT -1 1 "(" }{TEXT 296 1 "t" }
{TEXT -1 2 ", " }{TEXT 297 1 "x" }{TEXT -1 2 ")." }}{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);" }}}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT 269 18 "Check the solu
tion" }}{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 27 "D[2](u)(t,0
); D[2](u)(t,1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "plot([f
(x),u(0,x)],x=0..1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT 270 
18 "Graph the solution" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "pl
ot3d(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 p
redicted? What was the promise: that the total heat would stay about t
he 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 "This 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$%\"xG\"\"#" }{TEXT -1 
2 " ," }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 35 "
The Changed Boundary Conditions:   " }{XPPEDIT 18 0 "u;" "6#%\"uG" }
{TEXT -1 16 " (t, 0) = A and " }{XPPEDIT 18 0 "diff(u,x);" "6#-%%diffG
6$%\"uG%\"xG" }{TEXT -1 11 " (t,1) = 0." }}{PARA 0 "" 0 "" {TEXT -1 0 
"" }}{PARA 0 "" 0 "" {TEXT -1 27 "The Initial Condition:     " }{TEXT 
298 1 "u" }{TEXT -1 4 "(0, " }{TEXT 299 1 "x" }{TEXT -1 4 ") = " }
{TEXT 300 1 "f" }{TEXT -1 1 "(" }{TEXT 301 1 "x" }{TEXT -1 2 ")." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT 272 31 "
Solution for the Second Problem" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 165 "An interpretation for this problem is th
at there is no heat loss laterally, that the left end is held constant
 at temperature A, and that the right end is insulated." }}{PARA 0 "" 
0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 233 "We should recogniz
e that this problem does not have zero boundary conditions and so we s
hould break the problem into two: one problem gives a particular solut
ion and the other provides a general solution for the homogeneous equa
tion." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 5 "T
ake " }{TEXT 302 1 "v" }{TEXT -1 1 "(" }{TEXT 303 1 "x" }{TEXT -1 126 
") = A. This solutions satisfies the PDE, and the boundary conditions,
 but not the initial conditions, of course. The function " }{TEXT 304 
1 "v" }{TEXT -1 1 "(" }{TEXT 305 1 "x" }{TEXT -1 36 ") = A is the stea
dy state solution. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 25 "Now look for w to satisfy" }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 23 "Th
e same PDE:          " }{XPPEDIT 18 0 "diff(w,t) = diff(w,`$`(x,2));" 
"6#/-%%diffG6$%\"wG%\"tG-F%6$F'-%\"$G6$%\"xG\"\"#" }{TEXT -1 2 " ," }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 39 "The Homog
eneous Boundary Conditions:   " }{XPPEDIT 18 0 "w;" "6#%\"wG" }{TEXT 
-1 16 " (t, 0) = 0 and " }{XPPEDIT 18 0 "diff(w,x);" "6#-%%diffG6$%\"w
G%\"xG" }{TEXT -1 11 " (t,1) = 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 100 "Finding this w, we will add w and v and \+
then choose coefficients to satisfy the initial conditions. " }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 97 "By now, we see \+
that this homogeneous problem leads to two ODE 's, one having boundary
 conditions:" }}{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 -1 104 "Be reminded that in Section 3.3, w
e listed this differential equation and boundary conditions among the \+
" }{TEXT 273 25 "mixed boundary conditions" }{TEXT -1 37 ". We derive \+
the results listed there." }}{PARA 0 "" 0 "" {TEXT -1 38 "It is no sur
prise 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 "al
pha;" "6#%&alphaG" }{TEXT -1 5 " cos(" }{XPPEDIT 18 0 "lambda;" "6#%'l
ambdaG" }{TEXT -1 6 " x) + " }{XPPEDIT 18 0 "beta;" "6#%%betaG" }
{TEXT -1 5 " sin(" }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT -1 4 
" x)." }}{PARA 0 "" 0 "" {TEXT -1 43 "The requirement that X(0) = 0 im
plies that " }{XPPEDIT 18 0 "alpha;" "6#%&alphaG" }{TEXT -1 50 " = 0. \+
The requirement that X '(1) = 0 implies that" }}{PARA 0 "" 0 "" {TEXT 
-1 10 "          " }{XPPEDIT 18 0 "beta*lambda;" "6#*&%%betaG\"\"\"%'l
ambdaGF%" }{TEXT -1 5 " cos(" }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }
{TEXT -1 6 ") = 0." }}{PARA 0 "" 0 "" {TEXT -1 44 "To have a solution \+
that is not zero for all " }{TEXT 306 1 "x" }{TEXT -1 27 ", it must be
 true that cos(" }{XPPEDIT 18 0 "lambda;" "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#*&%#pi
G\"\"\"\"\"#!\"\"" }{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,!\"\"F,%#PiGF,F+F-" }{TEXT 
-1 21 " , n = 1, 2, 3, ... ." }}{PARA 0 "" 0 "" {TEXT -1 97 "Thus, the
se are the eigenvalues for this homogeneous problem corresponding to e
igenfunctions 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#*$&%'lambdaG6#%
\"nG\"\"#" }{TEXT -1 5 " t) ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 63 "We can now write the general solution for
 the original problem:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 9 "         " }{TEXT 307 1 "w" }{TEXT -1 9 "(t, x) + " }
{TEXT 308 1 "v" }{TEXT -1 6 "(x) = " }{XPPEDIT 18 0 "sum(c[n]*exp(-(2*
n-1)^2*Pi^2*t/4)*sin((2*n-1)*Pi*x/2),n);" "6#-%$sumG6$*(&%\"cG6#%\"nG
\"\"\"-%$expG6#,$**,&*&\"\"#F+F*F+F+F+!\"\"F3%#PiGF3%\"tGF+\"\"%F4F4F+
-%$sinG6#**,&*&F3F+F*F+F+F+F4F+F5F+%\"xGF+F3F4F+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 speci
fic, 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 0 142 "n:='n':\nfor n from 1
 to 5 do\n     c[n]:=int((f(x)-A)*sin((2*n-1)*Pi/2*x),x=0..1)/\n      \+
        int(sin((2*n-1)*Pi/2*x)^2,x=0..1);\nod;\nn:='n';" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "plot([f(x)-A,sum(c[n]*sin((2*n-1)*P
i/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 conve
rgence." }}{PARA 0 "" 0 "" {TEXT -1 19 "     We now define " }{TEXT 
309 1 "u" }{TEXT -1 1 "." }}{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);" }}}{SECT 0 {PARA 3 "" 0 "" {TEXT 274 18 "Check the solution" }}
{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 21 "u(t,0); D[2](u)(t,1
);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "plot([f(x),u(0,x)],x=
0..1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT 275 18 "Graph the s
olution" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 62 "plot3d(u(t,x),x=0
..1,t=0..1,axes=NORMAL,orientation=[-30,60]);" }}}{PARA 0 "" 0 "" 
{TEXT -1 187 "Does this graph have the properties we predicted? It sho
uld start off having a graph similar to the graph of the initial value
 and should move toward the steady state of being constant 1." }}
{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 "" }}{PARA 0 "" 0 
"" {TEXT -1 177 "In this Section, we have solved the simple heat equat
ion twice, once with Neumann boundary conditions, or insulated boundar
y conditions, and once with mixed boundary conditions." }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT 277 16 "Unassisted \+
Maple" }}{PARA 0 "" 0 "" {TEXT -1 215 "We check to see what Maple 8 ca
n do with this equation unassisted. As usual, we have called in the PD
E(tools) at the start of this Section. The partial differential equati
on has not changed from that of Section 4.1." }}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 7 "u:='u';" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
39 "PDE:=diff(u(t,x),t)-diff(u(t,x),x,x)=0;" }}}{PARA 0 "" 0 "" {TEXT 
-1 140 "By now, we are familiar with the suggestion that Maples makes \+
for the form for a solutions: products of two functions just as we use
d above." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "ans := pdsolve(P
DE);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 174 "It may be that the rea
der becomes impatient to see the other techniques that Maple uses. Tho
se techniques will use information about the boundary conditions. See \+
Section 8.2." }}}{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 258 "" 0 "" {TEXT -1 36 "Copyri
ght \251  2003  by James V. Herod" }}{PARA 258 "" 0 "" {TEXT -1 19 "Al
l rights reserved" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}}{MARK "0 0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }
{PAGENUMBERS 1 1 2 33 1 1 }