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{SECT 0 {PARA 256 "" 0 "" {TEXT -1 41 "Section 4.4: Convection Across \+
Boundaries" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 ""
{TEXT 257 30 "Maple Packages for Section 4.4" }}{EXCHG {PARA 0 "> " 0
"" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0
12 "with(plots):" }}}}{PARA 0 "" 0 "" {TEXT -1 5 " " }}{PARA 0 ""
0 "" {TEXT -1 136 "Suppose that we take a well insulated rod -- sides \+
and ends insulated-- which has initial heat distribution as indicated \+
by the function" }}{PARA 0 "" 0 "" {TEXT -1 27 " \+
4 " }{TEXT 266 1 "x" }{TEXT -1 6 " (1 - " }{TEXT 267 1 "x" }{TEXT
-1 14 ") + 2, 0 < " }{TEXT 268 1 "x" }{TEXT -1 5 " < 1." }}{PARA 0
"" 0 "" {TEXT -1 385 "We can see that this 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 heat of the ori
ginal distribution. After all, no heat has been lost or gained in this
well-insulated rod. Here is the definition of the initial distributio
n of heat, and 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);" }}}{EXCHG {PARA
0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 294 "How to model the redistribution of heat \+
was discussed in Section 4.3. Consider this alternate problem. Keep th
e rod well insulated, except at one end. There heat escapes or gathers
according as to whether it is more or less than the outside temperatu
re equal to 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$%\"xG\"\"#" }{TEXT -1 2 " ," }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 27 "The Bound
ary Conditions: " }{XPPEDIT 18 0 "diff(u,x);" "6#-%%diffG6$%\"uG%\"x
G" }{TEXT -1 16 " (t, 0) = 0 and " }{XPPEDIT 18 0 "diff(u,x);" "6#-%%d
iffG6$%\"uG%\"xG" }{TEXT -1 12 " (t,1) = - (" }{TEXT 260 1 "u" }{TEXT
-1 12 "(t, 1) - 2)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 27 "The Initial Condition: " }{TEXT 261 1 "u" }{TEXT
-1 23 "(0, x) = 4 x (1-x) + 2." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 903 "Now this problem is more interesting. Fr
om the equation for the distribution for the initial temperature, we c
an calculate that the temperature is 2 at the two ends and 3 in the ex
act middle. The heat begins to diffuse, causing temperatures to rise o
r fall along the rod. As soon as the temperature goes above 2 on the r
ight, the rod begins to cool by heat loss at that end. Heat will conti
nue to leak out until the temperature over the entire rod is 2. But wh
at happens at the left side. Surely heat will start to spread there fr
om the middle. We calculated in the lines above that with no leakage o
ut the right side or anywhere else, the temperature at the left would \+
rise to 8/3. How high will it go now with this leakage out the right s
ide? Will it go above 2, and then settle back down? And what about the
hot spot? Will it move left as heat leaks out the right side, or will
it drop straight down?" }}{PARA 0 "" 0 "" {TEXT -1 66 " Maybe you
agree this problem will be interesting to consider." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 101 "The PDE is homogeneous
, but the boundary conditions are not. Homogeneous boundary conditions
would be" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1
0 "" }}{PARA 0 "" 0 "" {TEXT -1 27 "The Boundary Conditions: " }
{XPPEDIT 18 0 "diff(w,x);" "6#-%%diffG6$%\"wG%\"xG" }{TEXT -1 16 " (t,
0) = 0 and " }{XPPEDIT 18 0 "diff(w,x);" "6#-%%diffG6$%\"wG%\"xG" }
{TEXT -1 19 " (t,1) = - w(t, 1)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 160 "Thus, we need to find a particular solut
ion, or what we call the steady state solution, for the original probl
em. Is it clear that the steady state solution is " }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 24 " v(x) = 2?
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 172 "That
solution satisfies the PDE and the non-homogeneous boundary condition
s. The u which satisfied the original problem is related to the transi
ent solution w by u = w + 2." }}{PARA 0 "" 0 "" {TEXT -1 95 " Now,
we try to find the general solution for the PDE with homogeneous boun
dary conditions." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "
" 0 "" {TEXT 264 48 "The general solution for the homogeneous problem
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 246 "By n
ow, we know to perform what was called Step 1 in Section 4.1: using se
paration of variables create two ordinary differential equations with \+
boundary conditions whose solutions will determine the solution for th
e partial differential equation:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 17 " X '' = " }{XPPEDIT 18 0 "mu;" "
6#%#muG" }{TEXT -1 40 " X, with X'(0) = 0 and X '(1) +X(1) = 0," }}
{PARA 0 "" 0 "" {TEXT -1 3 "and" }}{PARA 0 "" 0 "" {TEXT -1 16 " \+
T ' = " }{XPPEDIT 18 0 "mu;" "6#%#muG" }{TEXT -1 3 " T." }}{PARA
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 99 "Step 2 was to s
olve the two ordinary differential equations with boundary conditions.
The constant " }{XPPEDIT 18 0 "mu;" "6#%#muG" }{TEXT -1 27 " must be \+
negative, call it " }{XPPEDIT 18 0 "-lambda^2;" "6#,$*$%'lambdaG\"\"#!
\"\"" }{TEXT -1 42 " . Solutions for the differential equation" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 28 " \+
X '' = " }{XPPEDIT 18 0 "-lambda^2;" "6#,$*$%'lambdaG\"\"#
!\"\"" }{TEXT -1 3 " X " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 ""
0 "" {TEXT -1 19 "are A sin( " }{XPPEDIT 18 0 "lambda;" "6#%'l
ambdaG" }{TEXT -1 13 " x) + B cos( " }{XPPEDIT 18 0 "lambda;" "6#%'lam
bdaG" }{TEXT -1 4 " x)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 ""
0 "" {TEXT -1 34 "We invoke the boundary conditions." }}{EXCHG {PARA
0 "> " 0 "" {MPLTEXT 1 0 38 "X:=x->A*sin(lambda*x)+B*cos(lambda*x);" }
}}{PARA 0 "" 0 "" {TEXT -1 42 "The left boundary condition is used fir
st." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "D(X)(0)=0;" }}}{PARA
0 "" 0 "" {TEXT -1 29 "This condition implies A = 0." }}{EXCHG {PARA
0 "> " 0 "" {MPLTEXT 1 0 5 "A:=0;" }}}{PARA 0 "" 0 "" {TEXT -1 44 "We \+
now use the remaining boundary condition." }}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 15 "D(X)(1)+X(1)=0;" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 22 "This implies that tan(" }{XPPEDIT 18 0 "lambda;" "6#%'lam
bdaG" }{TEXT -1 4 ") = " }{XPPEDIT 18 0 "1/lambda;" "6#*&\"\"\"F$%'lam
bdaG!\"\"" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0
"" 0 "" {TEXT -1 13 "We must find " }{XPPEDIT 18 0 "lambda;" "6#%'lamb
daG" }{TEXT -1 41 " 's that satisfy this requirement. Then, " }
{XPPEDIT 18 0 "mu;" "6#%#muG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "-lambd
a^2;" "6#,$*$%'lambdaG\"\"#!\"\"" }{TEXT -1 65 " and the eigenfunctio
n corresponding to this eigenvalue is cos( " }{XPPEDIT 18 0 "lambda;"
"6#%'lambdaG" }{TEXT -1 30 " x). We have only to find the " }{XPPEDIT
18 0 "lambda;" "6#%'lambdaG" }{TEXT -1 4 " 's." }}{PARA 0 "" 0 ""
{TEXT -1 88 " Is it now clear that there is going to be no nice, c
losed form solution for 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 about where they lie by looking at where the grap
h of tan(x) crosses the 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 93 "We see there are roots between all the places where tan(x
) is zero and where it is undefined." }}{PARA 0 "" 0 "" {TEXT -1 32 " \+
Maybe seven will be enough." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT
1 0 83 "for n from 1 to 7 do\n lambda[n]:=fsolve(tan(x)=1/x,x,(n-1)*
Pi..(2*n-1)*Pi/2);\nod;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "
" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 163 "We \+
now have approximations for seven of the eigenvalues. We are ready to \+
take Step 3: construct the general solution for the partial differenti
al equation. We have" }}{PARA 0 "" 0 "" {TEXT -1 20 " w(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+-%$cosG6#*&&F26#F*F+%\"xGF+F+F*" }{TEXT -1 2 " ." }}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT -1 0 ""
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 158 "We go \+
back to solve the original problem. We need the general solution plus \+
the particular solutions. These add together to give the solution to t
his problem." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT 265 37 "The solution for the o
riginal problem" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 76 "Now, we find the c's by using the initial condition: \+
f(x) = w(0, x) + 2" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 12 "or " }{TEXT 270 1 "f" }{TEXT -1 1 "(" }{TEXT
271 1 "x" }{TEXT -1 7 ") -2 = " }{XPPEDIT 18 0 "sum(c[n]*cos(lambda[n]
*x),n);" "6#-%$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 compute coefficients." }}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 133 "for n from 1 to 7 do\n c[n]:=int((f(x)-2)*c
os(lambda[n]*x), x=0..1)/\n int(cos(lambda[n]*x)^2,
x=0..1);\nod;\nn:='n':" }}}{PARA 0 "" 0 "" {TEXT -1 41 "We check to se
e how close this is to our " }{TEXT 269 1 "f" }{TEXT -1 1 "." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "plot([f(x),2+sum(c[n]*cos(la
mbda[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 90
"We now define the solution, check to see that it really is a 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);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1
{PARA 3 "" 0 "" {TEXT 258 18 "Check the solution" }}{PARA 0 "" 0 ""
{TEXT -1 59 "We check the differential equation and boundary condition
s." }}{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 38 "D[2](u)(t,0);
D[2](u)(t,1)+(u(t,1)-2);" }}}{PARA 0 "" 0 "" {TEXT -1 81 "The above a
nswers might be a surprise. They are not all zeros. Not to worry.\nThe
" }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT -1 234 "'s we compute
d above were only an approximation for the real ones. (Remember that w
e used fsolve?) So, the numbers computed just above that we thought wo
uld be zero are approximations for zero. Of course! Just look at their
magnitude." }}{PARA 0 "" 0 "" {TEXT -1 65 "Finally, we observe how go
od a fit we have for the initial value." }}{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 1
{PARA 3 "" 0 "" {TEXT 259 18 "Graph the solution" }}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 62 "plot3d(u(t,x),x=0..1,t=0..3,axes=NORMAL,orient
ation=[-20,60]);" }}}{PARA 0 "" 0 "" {TEXT -1 117 "Does this graph hav
e the properties we predicted? Does it appear that the graph changes f
rom being the same shape at " }{TEXT 272 1 "t" }{TEXT -1 21 " = 0 as t
he graph of " }{TEXT 273 1 "f" }{TEXT -1 1 "(" }{TEXT 274 1 "x" }
{TEXT -1 56 ") and that the graph changes toward being constant 2 as \+
" }{TEXT 275 1 "t" }{TEXT -1 9 " evolves." }}}{PARA 0 "" 0 "" {TEXT
-1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 89 "We are satisfied that we have \+
an approximation for the solution for the original problem." }}}{PARA
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 ""
0 "" {TEXT -1 74 "In order to watch the maximum move from one side to \+
the other, we animate " }{TEXT 276 1 "u" }{TEXT -1 1 "(" }{TEXT 277 1
"t" }{TEXT -1 2 ", " }{TEXT 278 1 "x" }{TEXT -1 5 ") as " }{TEXT 279
1 "t" }{TEXT -1 89 " increases. Using the cursor, touch the graph of t
he animation which comes next. Use the " }{TEXT 262 22 "Move To The Ne
xt Frame" }{TEXT -1 63 " button to watch how the maximum temperature m
oves to the left." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "animate
(u(t,x),x=0..1,t=0..1,frames=100,view=[0..1,0..3.5]);" }}}{PARA 0 ""
0 "" {TEXT -1 218 "We ask: At what time would the left end point and r
ight end point have maximum temperatures? This graph gives some answer
. The red graph is the left endpoint temperature and the green is the \+
right endpoint temperature." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0
54 "plot([u(t,0),u(t,1)],t=0..2,y=0..3,color=[red,green]);" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 353 "This problem was important for it led t
o an eigenvalue problem for which we could not find the eigenvalues in
closed form. It was necessary to use numerical methods to generate th
ese eigenvalues. From an applications perspective, the boundary condit
ions are more realistic for it is unlikely that the end points can be \+
held fixed or perfectly insulated." }}{PARA 0 "" 0 "" {TEXT 256 0 "" }
}{SECT 1 {PARA 3 "" 0 "" {TEXT 263 16 "Unassisted Maple" }}{PARA 0 ""
0 "" {TEXT -1 183 "We have seen how Maple suggests separation of varia
bles in previous sections for the diffusion equation and variations. I
n this section, we allow Maple freedom on a different problem." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 112 "Perhaps \+
here the user felt uncomfortable with having to make approximations fo
r where solutions for the equation" }}{PARA 0 "" 0 "" {TEXT -1 11 " \+
tan(" }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT -1 6 ") = 1/"
}{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT -1 1 " " }}{PARA 0 "" 0
"" {TEXT -1 51 "should lie. After all, if we wanted 20 or 200 such " }
{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT -1 164 "'s, we would not \+
want to go searching for the intervals in which to seek solutions. The
question we address is how to let Maple find these without human assi
stance." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1
110 "The answer is to ask the right question. This question is suggest
ed by the following graphs of 1/x and tan(x)." }}{EXCHG {PARA 0 "> "
0 "" {MPLTEXT 1 0 35 "plot([tan(x),1/x],x=0..20,y=-2..2);" }}}{PARA 0
"" 0 "" {TEXT -1 122 "This graph makes clear the obvious. There is a s
olution for the equation tan(x) = 1/x between successive odd multiples
of " }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 115 "/2. Thus, we let M
aple determine these values unassisted by our searching out specific i
ntervals for each solution." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0
85 "for n from 0 to 7 do\n mu[n]:=fsolve(tan(x)=1/x,x,max(0,(2*n-1)*P
i/2)..(2*n+1)*Pi/2);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 3 "od;" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 0 "" }}}}{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 257 "" 0 "" {TEXT -1 36 "Copyri
ght \251 2003 by James V. Herod" }}{PARA 257 "" 0 "" {TEXT -1 19 "Al
l rights reserved" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}}{MARK "0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }
{PAGENUMBERS 1 1 2 33 1 1 }