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{SECT 0 {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 256 45 "Section 4.2: Diff
usion with Radiation Cooling" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT
1 {PARA 3 "" 0 "" {TEXT 294 30 "Maple Packages for Section 4.2" }}
{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 75 "In Newton's Law of Cooling, it is supposed that if a bo
dy with temperature " }{XPPEDIT 18 0 "alpha;" "6#%&alphaG" }{TEXT -1
42 " is immersed in a medium with temperature " }{XPPEDIT 18 0 "beta;
" "6#%%betaG" }{TEXT -1 44 " then the body cools according to the mode
l," }}{PARA 0 "" 0 "" {TEXT -1 46 " \+
T ' = - " }{TEXT 257 1 "k" }{TEXT -1 7 " ( T - " }{XPPEDIT 18 0 "be
ta;" "6#%%betaG" }{TEXT -1 12 " ), T(0) = " }{XPPEDIT 18 0 "alpha;" "
6#%&alphaG" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 6 "Here, " }
{TEXT 258 1 "k" }{TEXT -1 308 " is the heat transfer coefficient. This
model takes the object to be a point source. If we assume that we hav
e a rod that is not insulated, so that we have diffusion and radiation
cooling, the equation becomes slightly more complicated. We have an e
quation with a diffusion term and a radiation cooling term:" }}{PARA
0 "" 0 "" {TEXT -1 13 " " }{XPPEDIT 18 0 "diff(u(t,x),t) =
diff(u(t,x),`$`(x,2))-k*(u(t,x)-beta);" "6#/-%%diffG6$-%\"uG6$%\"tG%
\"xGF*,&-F%6$-F(6$F*F+-%\"$G6$F+\"\"#\"\"\"*&%\"kGF5,&-F(6$F*F+F5%%bet
aG!\"\"F5F<" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 128 "We suppo
se that the ends of the rod are at some constant temperature. For this
example, take the end points to have temperature " }{XPPEDIT 18 0 "be
ta;" "6#%%betaG" }{TEXT -1 170 " . Also, the rod should have an initia
l temperature. So as not to be too specific, take the initial temperat
ure to be distributed over the rod as expressed by a function " }
{TEXT 259 1 "f" }{TEXT -1 1 "(" }{TEXT 260 1 "x" }{TEXT -1 61 "). We w
rite the boundary condition and initial conditions as " }}{PARA 0 ""
0 "" {TEXT -1 21 " u( t, 0) = " }{XPPEDIT 18 0 "beta;" "6#%%b
etaG" }{TEXT -1 17 " and u( t, L) = " }{XPPEDIT 18 0 "beta;" "6#%%bet
aG" }{TEXT -1 19 ", with u( 0, x) = " }{TEXT 261 1 "f" }{TEXT -1 1 "(
" }{TEXT 262 1 "x" }{TEXT -1 2 ")." }}{PARA 0 "" 0 "" {TEXT -1 33 "Her
e, the length of the rod is L." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 51 "This is the problem we address in this se
ction. If " }{XPPEDIT 18 0 "beta;" "6#%%betaG" }{TEXT -1 28 " were zer
o, this would be a " }{TEXT 263 42 "homogeneous partial differential e
quation." }{TEXT -1 4 " If " }{XPPEDIT 18 0 "beta;" "6#%%betaG" }
{TEXT -1 22 " is not zero, it is a " }{TEXT 264 45 "non-homogeneous pa
rtial differential equation" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 329 "A review of Section 3.2, Probl
em 3 might be appropriate here. As indicated in the discussion of that
problem, we divide the equation into two parts: the steady state equa
tion and the transient equation. We solve both these equations and fin
d that their sum provides a solution to the original problem. Here are
the two equations:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT 265 24 "Steady State Equation: " }}{PARA 0 "" 0 "" {TEXT -1
13 " " }{XPPEDIT 18 0 "0 = diff(v(x),`$`(x,2))-k*(v(x)-bet
a);" "6#/\"\"!,&-%%diffG6$-%\"vG6#%\"xG-%\"$G6$F,\"\"#\"\"\"*&%\"kGF1,
&-F*6#F,F1%%betaG!\"\"F1F8" }{TEXT -1 16 " with v(t, 0) = " }{XPPEDIT
18 0 "beta;" "6#%%betaG" }{TEXT -1 15 " and v(t, L) = " }{XPPEDIT 18
0 "beta;" "6#%%betaG" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 17 " \+
" }}{PARA 0 "" 0 "" {TEXT 266 19 "Transient Equation:
" }}{PARA 0 "" 0 "" {TEXT -1 13 " " }{XPPEDIT 18 0 "diff(w
(t,x),t) = diff(w(t,x),`$`(x,2))-k*w(t,x);" "6#/-%%diffG6$-%\"wG6$%\"t
G%\"xGF*,&-F%6$-F(6$F*F+-%\"$G6$F+\"\"#\"\"\"*&%\"kGF5-F(6$F*F+F5!\"\"
" }{TEXT -1 34 " with w(t, 0) = 0 and w(t, L) = 0." }}{PARA 0 "" 0 ""
{TEXT -1 77 " The sum of these two solutions provides a solution for t
he original problem." }}{SECT 1 {PARA 3 "" 0 "" {TEXT 267 24 "Steady S
tate + Transient" }}{PARA 0 "" 0 "" {TEXT -1 117 "We show here that th
e steady state solution plus the transient solution provides a solutio
n for the original problem." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0
22 "u:=(t,x)->v(x)+w(t,x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0
15 "diff(u(t,x),t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 93 "diff
(v(x),x,x)+k*(v(x)-beta)+subs(\{diff(w(t,x),t)=diff(w(t,x),x,x)-k*w(t,
x)\},diff(u(t,x),t));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "co
llect(%,k);" }}}{PARA 0 "" 0 "" {TEXT -1 66 "We see that, because u(t,
x) = v(x) + w(t, x), this last becomes " }}{PARA 0 "" 0 "" {TEXT -1
13 " " }{XPPEDIT 18 0 "diff(u(t,x),t) = diff(u(t,x),`$`(x,
2))-k*(u(t,x)-beta);" "6#/-%%diffG6$-%\"uG6$%\"tG%\"xGF*,&-F%6$-F(6$F*
F+-%\"$G6$F+\"\"#\"\"\"*&%\"kGF5,&-F(6$F*F+F5%%betaG!\"\"F5F<" }{TEXT
-1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 31 "Boundary conditions check, to
o." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 44 "We solve these two equati
ons, one at a time." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA
3 "" 0 "" {TEXT 269 38 "Solution for the Steady State Equation" }}
{PARA 0 "" 0 "" {TEXT -1 372 "The notion of a steady state solution is
, first, physical. With a system such as that which modeled radiation \+
cooling, we had a non-homogeneous partial differential equation. The l
ong range forecast was that no matter what the initial condition, the \+
solution for the equation will evolve in time, settling down to a solu
tion that is not changing in time. This solution is " }{TEXT 268 6 "st
eady" }{TEXT -1 9 " in time." }}{PARA 0 "" 0 "" {TEXT -1 225 " Suc
h a notion is easy to characterize mathematically. It must be that the
steady solution is, in fact, a solution of the partial differential e
quation, but also it is not changing in time. The derivative with resp
ect to " }{TEXT 295 1 "t" }{TEXT -1 9 " is zero." }}{PARA 0 "" 0 ""
{TEXT -1 40 " We say this again with an equation." }}{PARA 0 "" 0
"" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 9 " 0 = " }{XPPEDIT
18 0 "diff(u(t,x),`$`(x,2))-k*u(t,x)+k*beta;" "6#,(-%%diffG6$-%\"uG6$%
\"tG%\"xG-%\"$G6$F+\"\"#\"\"\"*&%\"kGF0-F(6$F*F+F0!\"\"*&F2F0%%betaGF0
F0" }{TEXT -1 1 "," }}{PARA 0 "" 0 "" {TEXT -1 4 "with" }}{PARA 0 ""
0 "" {TEXT -1 8 " " }{XPPEDIT 18 0 "u(t,0) = beta;" "6#/-%\"uG6
$%\"tG\"\"!%%betaG" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "u(t,1) = beta;
" "6#/-%\"uG6$%\"tG\"\"\"%%betaG" }{TEXT -1 1 "." }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 209 " Since this is a ste
ady state, and time is not a factor, we could consider this as actuall
y an ordinary differential equations in x. We write it with that notat
ion and call solution which is a function of " }{TEXT 296 1 "x" }
{TEXT -1 4 " by " }{TEXT 297 1 "v" }{TEXT -1 1 "(" }{TEXT 298 1 "x" }
{TEXT -1 2 ")." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 18 " 0 = v''(x) - " }{XPPEDIT 18 0 "k;" "6#%\"kG" }{TEXT
-1 8 " v(x) + " }{TEXT 285 2 "k " }{XPPEDIT 18 0 "beta;" "6#%%betaG" }
{TEXT -1 14 ", with s(0) = " }{XPPEDIT 18 0 "beta;" "6#%%betaG" }
{TEXT -1 12 " and s(1) = " }{XPPEDIT 18 0 "beta;" "6#%%betaG" }{TEXT
-1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 38 "The steady state solution is c
omputed." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 74 "SState:=dsolve(\{diff(v(x),x,x)-k*(v(x)-beta)=0,v(0)=
beta,v(L)=beta\},v(x));" }}}{PARA 0 "" 0 "" {TEXT -1 73 "We should not
be surprised that if the endpoints are held at temperature " }
{XPPEDIT 18 0 "beta;" "6#%%betaG" }{TEXT -1 44 " and the surrounding m
edium has temperature " }{XPPEDIT 18 0 "beta;" "6#%%betaG" }{TEXT -1
62 " then the long range forecast is that the rod has temperature " }
{XPPEDIT 18 0 "beta;" "6#%%betaG" }{TEXT -1 1 "." }}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1
{PARA 3 "" 0 "" {TEXT 284 35 "Solution for the Transient Equation" }}
{PARA 0 "" 0 "" {TEXT -1 76 "We solve the transient equation. Recall w
hat is the equation from the above:" }}{PARA 0 "" 0 "" {TEXT -1 11 " \+
" }{XPPEDIT 18 0 "diff(w(t,x),t) = diff(w(t,x),`$`(x,2))-k*w(
t,x);" "6#/-%%diffG6$-%\"wG6$%\"tG%\"xGF*,&-F%6$-F(6$F*F+-%\"$G6$F+\"
\"#\"\"\"*&%\"kGF5-F(6$F*F+F5!\"\"" }{TEXT -1 34 " with w(t, 0) = 0 an
d w(t, L) = 0." }}{PARA 0 "" 0 "" {TEXT -1 90 "As in Section 4.1, the \+
method is separation of variables. Suppose that w can be written as" }
}{PARA 0 "" 0 "" {TEXT -1 19 " w(t,x) = T(" }{TEXT 270 1 "t" }
{TEXT -1 4 ") X(" }{TEXT 271 1 "x" }{TEXT -1 2 ")." }}{PARA 0 "" 0 ""
{TEXT -1 40 "In this case, the above equation becomes" }}{PARA 0 "" 0
"" {TEXT -1 2 " " }{TEXT 277 4 " " }{TEXT -1 15 "T ' X = T X '' "
}{TEXT 272 2 "- " }{TEXT 273 1 "k" }{TEXT 274 1 " " }{TEXT -1 3 "T X"
}}{PARA 0 "" 0 "" {TEXT -1 2 "or" }}{PARA 0 "" 0 "" {TEXT -1 29 " \+
T '/ T = (X ''- " }{TEXT 275 1 "k" }{TEXT -1 8 " X) / X." }}
{PARA 0 "" 0 "" {TEXT -1 137 "With the same arguments as used in Secti
on 4.1, this last equation leads to two ordinary differential equation
s with boundary conditions:" }}{PARA 0 "" 0 "" {TEXT -1 27 " \+
X '' - " }{TEXT 276 1 "k" }{TEXT -1 5 " X = " }{XPPEDIT 18
0 "mu;" "6#%#muG" }{TEXT -1 25 " X, with X(0) = X(L) = 0," }}{PARA 0 "
" 0 "" {TEXT -1 24 "and T ' = " }{XPPEDIT 18 0 "mu;" "6#
%#muG" }{TEXT -1 3 " T." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 ""
0 "" {TEXT -1 73 "We solve these two differential equations. We can ch
ange the first one to" }}{PARA 0 "" 0 "" {TEXT -1 27 " \+
X '' = (" }{TEXT 280 1 "k" }{TEXT -1 3 " + " }{XPPEDIT 18 0 "mu;" "
6#%#muG" }{TEXT -1 28 " ) X, with X(0) = X(L) = 0. " }}{PARA 0 "" 0 "
" {TEXT -1 162 "From arguments similar to those in Section 3.3, we get
the eigenvalues and eigenfunctions for this problem. We find that the
re are an infinite of possible values:" }}{PARA 0 "" 0 "" {TEXT -1 16
" (" }{TEXT 281 1 "k" }{TEXT -1 3 " + " }{XPPEDIT 18 0 "
mu;" "6#%#muG" }{TEXT -1 5 " ) = " }{XPPEDIT 18 0 "-n^2*Pi^2/(L^2);" "
6#,$*(%\"nG\"\"#%#PiGF&*$%\"LGF&!\"\"F*" }{TEXT -1 18 ", n = 1, 2, 3, \+
..." }}{PARA 0 "" 0 "" {TEXT -1 22 "Corresponding to each " }{TEXT
278 1 "n" }{TEXT -1 10 ", we have " }}{PARA 0 "" 0 "" {TEXT -1 15 " \+
X" }{XPPEDIT 18 0 "` `[n];" "6#&%\"~G6#%\"nG" }{TEXT -1 2 "
(" }{TEXT 299 1 "x" }{TEXT -1 11 ") = sin( n " }{XPPEDIT 18 0 "pi;" "
6#%#piG" }{TEXT -1 1 " " }{TEXT 300 1 "x" }{TEXT -1 5 "/L )." }}{PARA
0 "" 0 "" {TEXT -1 26 "The equation for T becomes" }}{PARA 0 "" 0 ""
{TEXT -1 25 " T ' = ( " }{XPPEDIT 18 0 "-k-n^2*Pi^2/(L
^2);" "6#,&%\"kG!\"\"*(%\"nG\"\"#%#PiGF(*$%\"LGF(F%F%" }{TEXT -1 5 " )
T." }}{PARA 0 "" 0 "" {TEXT -1 35 "Solutions for this equation will b
e" }}{PARA 0 "" 0 "" {TEXT -1 18 " T" }{XPPEDIT 18 0 "
` `[n];" "6#&%\"~G6#%\"nG" }{TEXT -1 2 " (" }{TEXT 279 1 "t" }{TEXT
-1 10 ") = exp( -" }{TEXT 282 1 "k" }{TEXT -1 1 " " }{TEXT 301 1 "t" }
{TEXT -1 7 ") exp( " }{XPPEDIT 18 0 "-n^2*Pi^2/(L^2)*t;" "6#,$**%\"nG
\"\"#%#PiGF&*$%\"LGF&!\"\"%\"tG\"\"\"F*" }{TEXT -1 3 " )." }}{SECT 1
{PARA 3 "" 0 "" {TEXT 283 38 "We check these solutions for the ODE's"
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "X:=x->sin(n*Pi*x/L);\nT:=t
->exp(-k*t)*exp(-n^2*Pi^2*t/L^2);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 54 "simplify(diff(X(x),x,x)-k*X(x)+(k+n^2*Pi^2/L^2)*X(x))
;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "X(0); X(L);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "simplify(diff(T(t),t)+(k+n^2
*Pi^2/L^2)*T(t));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 152 "Having t
hese solutions for the ordinary differential equations, we can constru
ct the general solution for the homogeneous partial differential equat
ion." }}{PARA 0 "" 0 "" {TEXT -1 26 " u( t, x ) = " }
{XPPEDIT 18 0 "exp(-k*t)*sum(A[n]*exp(-n^2*Pi^2/(L^2)*t)*sin(n*Pi*x/L)
,n = 1 .. infinity);" "6#*&-%$expG6#,$*&%\"kG\"\"\"%\"tGF*!\"\"F*-%$su
mG6$*(&%\"AG6#%\"nGF*-F%6#,$**F4\"\"#%#PiGF9*$%\"LGF9F,F+F*F,F*-%$sinG
6#**F4F*F:F*%\"xGF*FF.F%/F6;F%%)infinityGF%F%" }{TEXT -1 0 "" }}{PARA 0 "" 0
"" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 73 "In doing our calculations, we use only a finite approxima
tion for this u." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 103 "u:=(t,x)->beta+exp(-k*t)*sum(A[n]*exp(-n^2*Pi
^2/L^2*t)*sin(n*Pi*x/L),\n n=1..10); " }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT 286 39 "Check the solution for
the original PDE" }}{PARA 0 "" 0 "" {TEXT -1 96 "We check that this i
s a solution for the partial differential equation with boundary condi
tions." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "diff(u(t,x),t)-dif
f(u(t,x),x,x)+k*(u(t,x)-beta);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT
1 0 15 "u(t,0); u(t,L);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "
" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 67 "To \+
get the particular solution that satisfies the initial condition" }}
{PARA 0 "" 0 "" {TEXT -1 33 " f(x) = u(0, x) = " }
{XPPEDIT 18 0 "beta;" "6#%%betaG" }{TEXT -1 3 " + " }{XPPEDIT 18 0 "su
m(A[n]*sin(n*Pi*x/L),n = 1 .. infinity);" "6#-%$sumG6$*&&%\"AG6#%\"nG
\"\"\"-%$sinG6#**F*F+%#PiGF+%\"xGF+%\"LG!\"\"F+/F*;F+%)infinityG" }
{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 37 "We see that getting the \+
coefficients " }{XPPEDIT 18 0 "A[n];" "6#&%\"AG6#%\"nG" }{TEXT -1 82 "
is simply a matter of computing the Fourier coefficients for the func
tion f(x) - " }{XPPEDIT 18 0 "beta;" "6#%%betaG" }{TEXT -1 1 "." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 62 "At this p
oint, we have not specified a particular L, or k, or " }{XPPEDIT 18 0
"beta;" "6#%%betaG" }{TEXT -1 110 ", or f. We do this now so that ca
lculations may be made, solutions obtained, and graphs drawn. Take L \+
= 2, " }{XPPEDIT 18 0 "beta;" "6#%%betaG" }{TEXT -1 7 " = 3, " }
{TEXT 287 4 "k = " }{TEXT -1 11 "0.005, and " }}{PARA 0 "" 0 "" {TEXT
288 47 " f" }{TEXT -1 1 "
(" }{TEXT 289 1 "x" }{TEXT -1 4 ") = " }{XPPEDIT 18 0 "x^2*(2-x)+3;" "
6#,&*&%\"xG\"\"#,&F&\"\"\"F%!\"\"F(F(\"\"$F(" }{TEXT -1 1 "." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "L:=2; beta:=3; k:=5/1000; f:
=x->x^3*(2-x)+3;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "f(x)=u(
0,x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 122 "for n from 1 to 1
0 do\n A[n]:=int((f(x)-beta)*sin(n*Pi*x/L),x=0..L)/\n int
(sin(n*Pi*x/L)^2,x=0..L);\nod;\nn:='n':" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 32 "Since we have checke
d that this " }{TEXT 302 1 "u" }{TEXT -1 120 " satisfies the partial d
ifferential equation and the boundary conditions, we only check that i
t is an approximation for " }{TEXT 290 1 "f" }{TEXT -1 1 "(" }{TEXT
291 1 "x" }{TEXT -1 36 "). In fact, we off set the graph of " }{TEXT
303 1 "f" }{TEXT -1 47 " a little so that we can distinguish these two
." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "plot([u(0,x),f(x)+0.01]
,x=0..L,u=0..5);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{PARA 0 "" 0 "" {TEXT -1 28 "Finally, we draw a graph of " }{TEXT 304
1 "u" }{TEXT -1 88 ". What we should expect is that the solution moves
from a shape similar to the graph of " }{TEXT 292 1 "f" }{TEXT -1 24
" to the steady state of " }{XPPEDIT 18 0 "beta;" "6#%%betaG" }{TEXT
-1 1 "." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 63 "plot3d(u(t,x),x=0
..2,t=0..1,axes=normal,orientation=[-120,60]);" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 453 "What was new in this section was how to handle a \+
non-homogeneous partial differential equation. The technique was to fo
rmulate the steady state equation and the transient equation. It is a \+
good idea to verify that this has been done correctly by checking that
the sum of these two is a solution for the original problem. The next
step is to solve these two equations, add the solutions, and determin
e the appropriate coefficients using the Fourier Idea." }}{PARA 0 ""
0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT 293 16 "Unassisted \+
Maple" }}{PARA 0 "" 0 "" {TEXT -1 143 "We check to see what Maple 8 ca
n do with this equation unassisted. Be aware that we have called in th
e PDE(tools) at the start of this Section." }}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 7 "u:='u':" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0
53 "PDE:=diff(u(t,x),t)-diff(u(t,x),x,x)+k*(u(t,x)-beta);" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "ans:=pdsolve(PDE);" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "build(ans);" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 372 "We see that \+
Maple successfully recognizes that the solution should be a sum of a s
olution for the steady state equation plus a product of functions whic
h will be solutions for the transient equation. Because the solution w
as obtained without any information about the boundary conditions, thi
s will not be the same solution that we humans got with the assistance
of Maple." }}{PARA 0 "" 0 "" {TEXT -1 73 "Maple handles this problem \+
deftly with other techniques. See Section 8.2." }}}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 ""
{TEXT -1 50 "EMAIL: herod@math.gatech.edu or jherod@tds.net" }}
{PARA 0 "" 0 "" {TEXT -1 38 "URL: http://www.math.gatech.edu/~herod" }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 36 "Copyri
ght \251 2003 by James V. Herod" }}{PARA 256 "" 0 "" {TEXT -1 19 "Al
l rights reserved" }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}}{MARK "0 1" 0 }
{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 1 1 2 33 1 1 }