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{SECT 0 {PARA 0 "" 0 "" {TEXT 256 71 "Section 7.2: Two Dimensional Dif
fusion with Neumann Boundary Conditions" }}{PARA 0 "" 0 "" {TEXT -1 0
"" }}{SECT 1 {PARA 3 "" 0 "" {TEXT 274 30 "Maple Packages for Section \+
7.2" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart:" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(plots):" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 65 "In Lecture 32, we considered boundary conditions t
hat are called " }{TEXT 258 29 "Dirichlet Boundary Conditions" }{TEXT
-1 36 ". Here, we will work a problem with " }{TEXT 259 18 "Neumann Co
nditions" }{TEXT -1 64 " corresponding to an insulated surface in the \+
diffusion of heat." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 54 "We illustrate this type problem with a simple example." }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 257 36 "Conside
r the heat conduction problem" }{TEXT -1 100 " in a rectangle where th
ere is insulation on all boundaries and the initial condition is as fo
llows:" }}{PARA 0 "" 0 "" {TEXT -1 48 " \+
" }{XPPEDIT 18 0 "du/dt = Delta(u);" "6#/*&%#duG\"\"
\"%#dtG!\"\"-%&DeltaG6#%\"uG" }{TEXT -1 1 "," }}{PARA 0 "" 0 "" {TEXT
-1 26 "Insulated boundaries: " }{XPPEDIT 18 0 "du/dy;" "6#*&%#duG
\"\"\"%#dyG!\"\"" }{TEXT -1 10 "(t,x,0) = " }{XPPEDIT 18 0 "du/dy;" "6
#*&%#duG\"\"\"%#dyG!\"\"" }{TEXT -1 15 "(t,x,1)= 0, " }{XPPEDIT 18
0 "du/dx;" "6#*&%#duG\"\"\"%#dxG!\"\"" }{TEXT -1 10 "(t,0,y) = " }
{XPPEDIT 18 0 "du/dx;" "6#*&%#duG\"\"\"%#dxG!\"\"" }{TEXT -1 11 "(t,1,
y)= 0." }}{PARA 0 "" 0 "" {TEXT -1 65 "Initial condition: \+
u(0, x, y) = 100 x (1-x) y (1-y)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 38 "Here are the tasks we will accomplish:" }
}{PARA 0 "" 0 "" {TEXT -1 109 "(A) Give three ODE's, together with bou
ndary conditions, whose solutions lead to a solution for this problem.
" }}{PARA 0 "" 0 "" {TEXT -1 42 "(B) Give the solution for the three O
DE's." }}{PARA 0 "" 0 "" {TEXT -1 42 "(C) Give the general solution fo
r the PDE." }}{PARA 0 "" 0 "" {TEXT -1 78 "(D) Approximate the particu
lar solution that goes with this initial condition." }}{PARA 0 "" 0 "
" {TEXT -1 36 "(E) Animate a graph of the solution." }}{PARA 0 "" 0 "
" {TEXT -1 55 "(F) Compute the total heat for the system for all time.
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 50 "(A.) Here are three ODE's and boundary co
nditions:" }}{PARA 0 "" 0 "" {TEXT -1 17 " X '' + " }
{XPPEDIT 18 0 "lambda^2;" "6#*$%'lambdaG\"\"#" }{TEXT -1 29 " X = 0, \+
X '(0) = X '(1) = 0." }}{PARA 0 "" 0 "" {TEXT -1 16 " Y '' + \+
" }{XPPEDIT 18 0 "mu^2;" "6#*$%#muG\"\"#" }{TEXT -1 29 " Y = 0, Y '(0
) = Y '(1) = 0." }}{PARA 0 "" 0 "" {TEXT -1 18 " T ' + ( " }
{XPPEDIT 18 0 "lambda^2+mu^2;" "6#,&*$%'lambdaG\"\"#\"\"\"*$%#muGF&F'
" }{TEXT -1 9 " ) T = 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 38 "(B.) Solutions for these equations are" }}{PARA 0
"" 0 "" {TEXT -1 22 " X(x) = cos( n " }{XPPEDIT 18 0 "Pi;" "6#%
#PiG" }{TEXT -1 19 " x), Y(y) = cos(m " }{XPPEDIT 18 0 "Pi;" "6#%#PiG
" }{TEXT -1 21 " y), T(t) = exp( - (" }{XPPEDIT 18 0 "n^2+m^2;" "6#,&
*$%\"nG\"\"#\"\"\"*$%\"mGF&F'" }{TEXT -1 3 " ) " }{XPPEDIT 18 0 "Pi^2;
" "6#*$%#PiG\"\"#" }{TEXT -1 4 " t )" }}{PARA 0 "" 0 "" {TEXT -1 0 ""
}}{PARA 0 "" 0 "" {TEXT -1 40 "(C.) The general solution for the pde i
s" }}{PARA 0 "" 0 "" {TEXT -1 17 " u(t,x,y) = " }{XPPEDIT 18 0 "s
um(sum(A[mn]*cos(n*Pi*x)*cos(m*Pi*y)*exp(-(n^2+m^2)*Pi^2*t),n),m);" "6
#-%$sumG6$-F$6$**&%\"AG6#%#mnG\"\"\"-%$cosG6#*(%\"nGF-%#PiGF-%\"xGF-F-
-F/6#*(%\"mGF-F3F-%\"yGF-F--%$expG6#,$*(,&*$F2\"\"#F-*$F8FAF-F-*$F3FAF
-%\"tGF-!\"\"F-F2F8" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 62 "T
his is so important I will check an approximation of the sum." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 15 "Here defi
nes u." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 100 "u:=(t,x,y)->sum(s
um(A[m,n]*cos(n*Pi*x)*cos(m*Pi*y)*\n exp(-(n^2+m^2)*Pi^2*t),n=0.
.10),m=0..10);" }}}{PARA 0 "" 0 "" {TEXT -1 36 "Here checks the bounda
ry conditions." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 63 "D[3](u)(t,
x,0);\nD[3](u)(t,x,1);\nD[2](u)(t,0,y);\nD[2](u)(t,1,y);" }}}{PARA 0 "
" 0 "" {TEXT -1 20 "Here checks the PDE." }}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 49 "D[1](u)(t,x,y)-D[2,2](u)(t,x,y)-D[3,3](u)(t,x,y);" }}
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 57 "(D.) We \+
compute several terms of the particular solution." }}{EXCHG {PARA 0 ">
" 0 "" {MPLTEXT 1 0 29 "100*x*(1-x)*y*(1-y)=u(0,x,y):" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 197 "for n from 0 to 10 do\nfor m from \+
0 to 10 do\nA[n,m]:=100*int(x*(1-x)*cos(n*Pi*x),x=0..1)/int(cos(n*Pi*x
)^2,x=0..1)*int(y*(1-y)*cos(m*Pi*y),y=0..1)/int(cos(m*Pi*y)^2,y=0..1);
\nod: \nod:\nn:='n'; m:='m';" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1
0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 98 "In order to check my solution, \+
I compare the graph of the initial value with the initial function." }
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "plot3d(u(0,x,y),x=0..1,y=0.
.1,axes=normal);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "plot3d(
100*x*(1-x)*y*(1-y),x=0..1,y=0..1,axes=normal);" }}}{PARA 0 "" 0 ""
{TEXT -1 160 "Note that there is a small error along the boundaries. T
his error could be made smaller by computing more than the 121 coeffic
ients that we have computed above." }}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 31 "plot([u(0,x,0)],x=0..1,y=0..1);" }}}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 42 "(E.) Here is an animation
of the solution." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "animate
3d(u(t,x,y),x=0..1,y=0..1,t=0..1/8,axes=normal);" }}}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 88 "(F.) We compute the total
heat and verify that it is the same as the initial total heat." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 87 "int(int(u(t,x,y),x=0..1),y=0
..1);\nTotHeat:=int(int(100*x*(1-x)*y*(1-y),x=0..1),y=0..1);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 153 "Finally, we recall how to make
changes in case the boundary conditions were not homogeneous. A quick
review for working such a problem seems appropriate." }}{PARA 0 "" 0
"" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 49 "Non homogeneou
s Neumann Type Boundary Conditions." }}{PARA 0 "" 0 "" {TEXT -1 79 "Su
ppose the partial differential equation and boundary conditions have t
he form" }}{PARA 0 "" 0 "" {TEXT -1 48 " \+
" }{XPPEDIT 18 0 "du/dt = Delta(u);" "6#/*&%#duG\"\"
\"%#dtG!\"\"-%&DeltaG6#%\"uG" }{TEXT -1 1 "," }}{PARA 0 "" 0 "" {TEXT
-1 25 "Boundary Conditions: " }{XPPEDIT 18 0 "du/dy;" "6#*&%#duG\"
\"\"%#dyG!\"\"" }{TEXT -1 14 "(t,x,0) = 0, " }{XPPEDIT 18 0 "du/dy;"
"6#*&%#duG\"\"\"%#dyG!\"\"" }{TEXT -1 16 "(t,x,1)= -1, " }{XPPEDIT
18 0 "du/dx;" "6#*&%#duG\"\"\"%#dxG!\"\"" }{TEXT -1 14 "(t,0,y) = 0, \+
" }{XPPEDIT 18 0 "du/dx;" "6#*&%#duG\"\"\"%#dxG!\"\"" }{TEXT -1 11 "(t
,1,y)= 1." }}{PARA 0 "" 0 "" {TEXT -1 64 "Initial condition: \+
u(0, x, y) = x (1-x) y (1-y) + (" }{XPPEDIT 18 0 "x^2-y^2;" "6#,&*
$%\"xG\"\"#\"\"\"*$%\"yGF&!\"\"" }{TEXT -1 6 ") / 2." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 260 12 "Note changes" }{TEXT
-1 1 " " }{TEXT 261 26 "from the previous problem:" }}{PARA 0 "" 0 ""
{TEXT -1 10 " The " }{TEXT 262 19 "boundary conditions" }{TEXT
-1 88 " have been changed to model heat coming in at the top and going
out the right side. The " }{TEXT 263 18 "initial conditions" }{TEXT
-1 100 " have been changed so that the results computed at the beginni
ng of this worksheet can be used here." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 148 "The pattern for working problems of \+
this type have been set. First, we solve the steady state problem. The
steady state problem will have this form:" }}{PARA 0 "" 0 "" {TEXT
-1 59 "PDE: 0 = " }
{XPPEDIT 18 0 "Delta;" "6#%&DeltaG" }{TEXT -1 4 " v, " }}{PARA 0 "" 0
"" {TEXT -1 25 "Boundary Conditions: " }{XPPEDIT 18 0 "dv/dy;" "6#
*&%#dvG\"\"\"%#dyG!\"\"" }{TEXT -1 14 "(t,x,0) = 0, " }{XPPEDIT 18 0
"dv/dy;" "6#*&%#dvG\"\"\"%#dyG!\"\"" }{TEXT -1 16 "(t,x,1)= -1, " }
{XPPEDIT 18 0 "dv/dx;" "6#*&%#dvG\"\"\"%#dxG!\"\"" }{TEXT -1 14 "(t,0,
y) = 0, " }{XPPEDIT 18 0 "dv/dx;" "6#*&%#dvG\"\"\"%#dxG!\"\"" }{TEXT
-1 11 "(t,1,y)= 1." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 106 "We have addressed this problem previously. We found the \+
following solution which we define and check here." }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "v:=(x,y)->(x
^2-y^2)/2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "D[2](v)(x,0);
\nD[2](v)(x,1);\nD[1](v)(0,y);\nD[1](v)(1,y);" }}}{EXCHG {PARA 0 "> "
0 "" {MPLTEXT 1 0 34 "D[1,1](v)(t,x,y)+D[2,2](v)(t,x,y);" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 50 "Ne
xt we find the transient solution which we call " }{TEXT 264 1 "w" }
{TEXT -1 151 ". To do this we use homogeneous boundary conditions and \+
an initial condition to reflect that the solution u for the original p
roblem will be defined by" }}{PARA 0 "" 0 "" {TEXT -1 22 " \+
" }{TEXT 265 1 "u" }{TEXT -1 12 "(t, x, y) = " }{TEXT 266 1
"w" }{TEXT -1 12 "(t, x, y) + " }{TEXT 267 1 "v" }{TEXT -1 7 "(x, y).
" }}{PARA 0 "" 0 "" {TEXT -1 145 "The initial conditions of this non-h
omogeneous problem have been created so that the transient solution is
exactly the problem we worked above: " }{TEXT 273 1 "w" }{TEXT -1
12 "(t, x, y) = " }{TEXT 271 1 "u" }{TEXT -1 12 "(t, x, y) - " }{TEXT
272 6 "v(x,y)" }{TEXT -1 28 ". We use these results here." }}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 98 "w:=(t,x,y)->sum(sum(A[m,n]*cos(n*Pi
*x)*cos(m*Pi*y)*\n exp(-(n^2+m^2)*Pi^2*t),n=0..4),m=0..4);" }}}
{PARA 0 "" 0 "" {TEXT -1 70 "The solution for the problem in this sect
ion should now be the sum of " }{TEXT 268 1 "w" }{TEXT -1 5 " and " }
{TEXT 269 1 "v" }{TEXT -1 16 ". We check this." }}{EXCHG {PARA 0 "> "
0 "" {MPLTEXT 1 0 36 "u:=unapply(w(t,x,y)+v(x,y),(t,x,y)):" }}}{PARA
0 "" 0 "" {TEXT -1 33 "We check the boundary conditions." }}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 63 "D[3](u)(t,x,0);\nD[3](u)(t,x,1);\nD
[2](u)(t,0,y);\nD[2](u)(t,1,y);" }}}{PARA 0 "" 0 "" {TEXT -1 17 "We ch
eck the PDE." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "D[1](u)(t,x,
y)-D[2,2](u)(t,x,y)-D[3,3](u)(t,x,y);" }}}{PARA 0 "" 0 "" {TEXT -1 59
"We check the initial conditions by comparing the graphs of " }{TEXT
270 1 "u" }{TEXT -1 14 "(0, x, y) and " }}{PARA 0 "" 0 "" {TEXT -1 33
" 100* x (1-x) y (1-y) - (" }{XPPEDIT 18 0 "x^2-y^2;" "6#,&*$%
\"xG\"\"#\"\"\"*$%\"yGF&!\"\"" }{TEXT -1 95 ") / 2.\nIn order to see t
hat there are two graphs here, we displace the second by an amount 0.1
." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 91 "plot3d(\{u(0,x,y),100*x
*(1-x)*y*(1-y)+(x^2-y^2)/2+0.1\},\n x=0..1,y=0..1,axes=normal)
;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 79 "plot3d(\{u(1,x,y),TotH
eat+(x^2-y^2)/2+0.1\},\n x=0..1,y=0..1,axes=normal);" }}}
{PARA 0 "" 0 "" {TEXT -1 132 "Finally, to give us intuition, we draw t
he animation as the solution moves from the initial condition to the s
teady state condition." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 57 "an
imate3d(u(t,x,y),x=0..1,y=0..1,t=0..1/20, axes=normal);" }}}}{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 256 "" 0 "" {TEXT -1 36 "Copyright \251 2003 by James V. Herod
" }}{PARA 256 "" 0 "" {TEXT -1 19 "All rights reserved" }}}{MARK "0 0
" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 1 1 2 33 1 1 }