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{SECT 0 {PARA 18 "" 0 "" {TEXT -1 37 "Linear Methods of Applied Mathem
atics" }}{PARA 256 "" 0 "" {TEXT -1 56 "Potential equations and the he
at equation on a rectangle" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 
{PARA 3 "" 0 "" {TEXT -1 56 "Copyright 2000 by Evans M. Harrell II and
 James V. Herod" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "
" 0 "" {TEXT -1 61 "Laplace's Equation (or the Potential Equation) on \+
a rectangle" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 122 "     From what has come before, we can expect that a stu
dy of the heat equation on a rectangle to have a predictable form:" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 36 "         \+
                           " }{XPPEDIT 18 0 "diff(u,t) = diff(u,`$`(x,
2))+diff(u,`$`(y,2));" "6#/-%%diffG6$%\"uG%\"tG,&-F%6$F'-%\"$G6$%\"xG
\"\"#\"\"\"-F%6$F'-F-6$%\"yG\"\"#F1" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 74 "Boundary conditions:                u(t,
 x, 0) = f(x),   u(t, x, L) = g(x)" }}{PARA 0 "" 0 "" {TEXT -1 88 "   \+
                                               u(t, 0, y) = h(y),   u(
t, M, y) = J(y)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 64 "Initial condition:                         u(0, x, y) = K
(x, y)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
83 "It is not a surprise, either, that one should solve the steady-sta
te problem first." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 34 "                                  " }{XPPEDIT 18 0 "0 = d
iff(u,`$`(x,2))+diff(u,`$`(y,2));" "6#/\"\"!,&-%%diffG6$%\"uG-%\"$G6$%
\"xG\"\"#\"\"\"-F'6$F)-F+6$%\"yG\"\"#F/" }}{PARA 0 "" 0 "" {TEXT -1 0 
"" }}{PARA 0 "" 0 "" {TEXT -1 68 "Boundary conditions:                \+
u(x, 0) = f(x),   u(x, L) = g(x)" }}{PARA 0 "" 0 "" {TEXT -1 82 "     \+
                                             u(0, y) = h(y),   u(M, y)
 = J(y)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
31 "and then the transition problem" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 34 "                                  " }
{XPPEDIT 18 0 "diff(u,t)=diff(u,x,x)+diff(u,y,y)" "6#/-%%diffG6$%\"uG%
\"tG,&-F%6%F'%\"xGF,\"\"\"-F%6%F'%\"yGF0F-" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 68 "Boundary conditions:           \+
     u(t, x, 0) = 0,   u(t, x, L) = 0" }}{PARA 0 "" 0 "" {TEXT -1 82 "
                                                  u(t, 0, y) = 0,   u(
t, M, y) = 0" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 64 "Initial condition:                         u(0, x, y) = k(x, y)
." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 111 "The
 wave equation leads to this same steady state problem. We consider th
is problem in this part of the notes. " }}{PARA 0 "" 0 "" {TEXT -1 98 
"     We begin consideration of this steady-state problem. It is what \+
is called Laplace's equation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 83 "This steady state problem is commonly bro
ken into two problems. Here is one of them" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 47 
"                                               " }{XPPEDIT 18 0 "0 = \+
diff(u,`$`(x,2))+diff(u,`$`(y,2));" "6#/\"\"!,&-%%diffG6$%\"uG-%\"$G6$
%\"xG\"\"#\"\"\"-F'6$F)-F+6$%\"yG\"\"#F/" }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 68 "Boundary conditions:              \+
  u(x, 0) = f(x),   u(x, L) = g(x)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 76 "                                         \+
         u(0, y) = 0,   u(M, y) = 0" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 54 "Here are the steps: (1) Make a separatio
n of variables" }}{PARA 0 "" 0 "" {TEXT -1 111 "                      \+
        (2) Identify the resulting ordinary differential equation and \+
boundary conditions" }}{PARA 0 "" 0 "" {TEXT -1 55 "                  \+
            (3) Solve these equations" }}{PARA 0 "" 0 "" {TEXT -1 80 "
                              (4) Construct the general solution for t
he problem" }}{PARA 0 "" 0 "" {TEXT -1 111 "                          \+
    (5) Use the boundary conditions to get the solution for this parti
cular equation." }}{PARA 0 "" 0 "" {TEXT -1 27 "Step 1: Separate varia
bles." }}{PARA 0 "" 0 "" {TEXT -1 96 "  The partial differential equat
ion leads to X '' Y + X Y'' = 0, with X(0) Y(y) = 0 = X(M) Y(y)." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 44 "Step 2: I
dentify the differential equations." }}{PARA 0 "" 0 "" {TEXT -1 39 " T
he differential equations are X '' + " }{XPPEDIT 18 0 "lambda^2;" "6#*
$%'lambdaG\"\"#" }{TEXT -1 36 " X = 0, X(0) = 0 = X(M), and Y '' = " }
{XPPEDIT 18 0 "lambda^2;" "6#*$%'lambdaG\"\"#" }{TEXT -1 3 " Y." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 30 "Step 3: S
olve these equations." }}{PARA 0 "" 0 "" {TEXT -1 5 "     " }{XPPEDIT 
18 0 "lambda;" "6#%'lambdaG" }{TEXT -1 5 " = n " }{XPPEDIT 18 0 "Pi;" 
"6#%#PiG" }{TEXT -1 18 "/M ,  X(x) = sin(n" }{XPPEDIT 18 0 "pi;" "6#%#
piG" }{TEXT -1 23 "x/M), and Y(y) = sinh(n" }{XPPEDIT 18 0 "pi;" "6#%#
piG" }{TEXT -1 15 "y/M) and sinh(n" }{XPPEDIT 18 0 "pi;" "6#%#piG" }
{TEXT -1 9 "(L-y)/M)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 
"" {TEXT -1 58 "Step 4: Construct the general solution. If L &  M were
 1, " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 15 " \+
    u(x, y) = " }{XPPEDIT 18 0 "sum(a[n]*sinh(n*Pi*y)+b[n]*sinh(n*Pi*(
1-y))*sin(n*Pi*x),n);" "6#-%$sumG6$,&*&&%\"aG6#%\"nG\"\"\"-%%sinhG6#*(
F+F,%#PiGF,%\"yGF,F,F,*(&%\"bG6#F+F,-F.6#*(F+F,F1F,,&\"\"\"F,F2!\"\"F,
F,-%$sinG6#*(F+F,F1F,%\"xGF,F,F,F+" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 53 "Step 5: Use the boundary \+
conditions to determine the " }{TEXT 256 1 "a" }{TEXT -1 8 " 's and " 
}{TEXT 257 1 "b" }{TEXT -1 4 " 's." }}{PARA 0 "" 0 "" {TEXT -1 2 "  " 
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 
24 "Details for this problem" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 155 "We are now ready to apply this to a situation \+
with and both f and g specified.  We consider the problem with f = g a
nd as given below. We choose L = M = 1." }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
102 "u:=(x,y)->sum((a[n]*sinh(n*Pi*y)+b[n]*sinh(n*Pi*(1-y)))*sin(n*Pi*
x),\n                  n=1..infinity);" }}}{PARA 0 "" 0 "" {TEXT -1 
64 "We wish to determine the coefficients. How will they be defined?" 
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "f(x)=u(x,0);\ng(x)=u(x,1);
" }}}{PARA 0 "" 0 "" {TEXT -1 83 "This reminds us to use the Fourier c
oefficients. ( Don't forget to divide by sinh(n" }{XPPEDIT 18 0 "pi;" 
"6#%#piG" }{TEXT -1 4 "). )" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
35 "f:=x->piecewise(x<1/2,2*x,2*(1-x));" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 18 "plot(f(x),x=0..1);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 93 "for n from 1 to 3 do\n   b[n]:=2*int('f(x)'*sin(n*Pi*
x),x=0..1)/sinh(n*Pi);\n   a[n]:=b[n]:\nod;" }}}{PARA 0 "" 0 "" {TEXT 
-1 35 "We incorporate these values into u." }}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 118 "u:=(x,y)->8/Pi^2*sum(sin(n*Pi/2)/n^2*\n         (s
inh(n*Pi*y)+sinh(n*Pi*(1-y)))/sinh(n*Pi)*sin(n*Pi*x),n=1..3);\nn:='n':
" }}}{PARA 0 "" 0 "" {TEXT -1 59 "As a safety check, we plot the ends \+
against the graph of f." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "p
lot([u(x,0),'f(x)'],x=0..1);" }}}{PARA 0 "" 0 "" {TEXT -1 56 "We are n
ow ready to see a plot of u. What do you expect?" }}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 41 "plot3d(u(x,y),x=0..1,y=0..1,axes=NORMAL);" }}}
{PARA 0 "" 0 "" {TEXT -1 46 "It is of value sometimes to see coutour p
lots." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(plots):" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "contourplot(u(x,y),x=0..1,y=
0..1);" }}}{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 85 "We now do a problem with two sides insula
ted. Here are the equations for the problem." }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 47 "                               \+
                " }{XPPEDIT 18 0 "0 = diff(u,`$`(x,2))+diff(u,`$`(y,2)
);" "6#/\"\"!,&-%%diffG6$%\"uG-%\"$G6$%\"xG\"\"#\"\"\"-F'6$F)-F+6$%\"y
G\"\"#F/" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
36 "Boundary conditions:                " }{XPPEDIT 18 0 "u;" "6#%\"uG
" }{TEXT -1 17 "(x, 0) = f(x),   " }{XPPEDIT 18 0 "u;" "6#%\"uG" }
{TEXT -1 13 "(x, L) = g(x)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 
"" 0 "" {TEXT -1 50 "                                                 \+
 " }{XPPEDIT 18 0 "u[x];" "6#&%\"uG6#%\"xG" }{TEXT -1 14 "(0, y) = 0, \+
  " }{XPPEDIT 18 0 "u[x];" "6#&%\"uG6#%\"xG" }{TEXT -1 11 "(M, y) = 0.
" }}{PARA 0 "" 0 "" {TEXT -1 168 "This says there is no passage across
 the left, or right, boundary of the rectangle, but the top and bottom
 of the distribution are held as g(x) and f(x), respectively. " }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 111 "spacecurve(\{[x,0,f(x)],[x,
1,f(x)],[1,x,0],[x,1,0]\},x=0..1,color=BLACK,\n     axes=NORMAL,orient
ation=[-25.,50]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 31 "We perfor
m the same five steps." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 27 "Step 1: Separate variables." }}{PARA 0 "" 0 "" 
{TEXT -1 100 "  The partial differential equation leads to X '' Y + X \+
Y'' = 0, with X '(0) Y(y) = 0 = X '(M) Y(y)." }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 44 "Step 2: Identify the differenti
al equations." }}{PARA 0 "" 0 "" {TEXT -1 39 " The differential equati
ons are X '' + " }{XPPEDIT 18 0 "lambda^2;" "6#*$%'lambdaG\"\"#" }
{TEXT -1 40 " X = 0, X '(0) = 0 = X '(M), and Y '' = " }{XPPEDIT 18 0 
"lambda^2;" "6#*$%'lambdaG\"\"#" }{TEXT -1 3 " Y." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 30 "Step 3: Solve these equat
ions." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 7 " \+
   if " }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT -1 38 " = 0, the
n X(x) = 1 and Y(y) = 1 or y." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 6 "   if " }{XPPEDIT 18 0 "lambda <> 0;" "6#0
%'lambdaG\"\"!" }{TEXT -1 8 ", then  " }{XPPEDIT 18 0 "lambda;" "6#%'l
ambdaG" }{TEXT -1 5 " = n " }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 
18 "/M ,  X(x) = cos(n" }{XPPEDIT 18 0 "pi;" "6#%#piG" }{TEXT -1 23 "x
/M), and Y(y) = sinh(n" }{XPPEDIT 18 0 "pi;" "6#%#piG" }{TEXT -1 15 "y
/M) and sinh(n" }{XPPEDIT 18 0 "pi;" "6#%#piG" }{TEXT -1 9 "(L-y)/M).
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 39 "Step \+
4: Construct the general solution." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 15 "     u(x, y) = " }{XPPEDIT 18 0 "sum((a[n
]*sinh(n*Pi*y/M)+b[n]*sinh(n*Pi*(L-y)/M))*cos(n*Pi*x/L),n);" "6#-%$sum
G6$*&,&*&&%\"aG6#%\"nG\"\"\"-%%sinhG6#**F,F-%#PiGF-%\"yGF-%\"MG!\"\"F-
F-*&&%\"bG6#F,F--F/6#**F,F-F2F-,&%\"LGF-F3F5F-F4F5F-F-F--%$cosG6#**F,F
-F2F-%\"xGF-F>F5F-F," }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 53 "Step 5: Use the boundary conditions to d
etermine the " }{TEXT 258 1 "a" }{TEXT -1 8 " 's and " }{TEXT 259 1 "b
" }{TEXT -1 4 " 's." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 
3 "" 0 "" {TEXT -1 24 "Details for this problem" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 155 "We are now ready to appl
y this to a situation with and both f and g specified.  We consider th
e problem with f = g and as given below. We choose L = M = 1." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 120 "u:=(x,y)->a[0]+(b[0]-a[0])*y +\n   sum((a[n
]*sinh(n*Pi*y)+b[n]*sinh(n*Pi*(1-y)))*cos(n*Pi*x),\n            n=1..i
nfinity);" }}}{PARA 0 "" 0 "" {TEXT -1 64 "We wish to determine the co
efficients. How will they be defined?" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 25 "f(x)=u(x,0);\ng(x)=u(x,1);" }}}{PARA 0 "" 0 "" {TEXT 
-1 83 "This reminds us to use the Fourier coefficients. ( Don't forget
 to divide by sinh(n" }{XPPEDIT 18 0 "pi;" "6#%#piG" }{TEXT -1 4 "). )
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "f:=x->piecewise(x<1/2,2*
x,2*(1-x));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "plot(f(x),x=
0..1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 129 "b[0]:=int(f(x),x
=0..1);\na[0]:=b[0];\nfor n from 1 to 3 do\n   b[n]:=2*int('f(x)'*cos(
n*Pi*x),x=0..1)/sinh(n*Pi);\n   a[n]:=b[n]:\nod;" }}}{PARA 0 "" 0 "" 
{TEXT -1 35 "We incorporate these values into u." }}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 123 "u:=(x,y)->a[0]+(b[0]-a[0])*y +sum((a[n]*sinh(
n*Pi*y)+b[n]*sinh(n*Pi*(1-y)))*cos(n*Pi*x),\n                  n=1..3)
;\nn:='n':" }}}{PARA 0 "" 0 "" {TEXT -1 59 "As a safety check, we plot
 the ends against the graph of f." }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 29 "plot([u(x,0),'f(x)'],x=0..1);" }}}{PARA 0 "" 0 "" 
{TEXT -1 56 "We are now ready to see a plot of u. What do you expect?
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "plot3d(u(x,y),x=0..1,y=0
..1,axes=NORMAL);" }}}{PARA 0 "" 0 "" {TEXT -1 46 "It is of value some
times to see coutour plots." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
12 "with(plots):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "contour
plot(u(x,y),x=0..1,y=0..1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 8 "Exercise" }}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 32 "Graph a solution for the problem" 
}}{PARA 0 "" 0 "" {TEXT -1 32 "                                " }
{XPPEDIT 18 0 "0 = diff(u,`$`(x,2))+diff(u,`$`(y,2));" "6#/\"\"!,&-%%d
iffG6$%\"uG-%\"$G6$%\"xG\"\"#\"\"\"-F'6$F)-F+6$%\"yG\"\"#F/" }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 50 "Boundary condit
ions:                u(x, 0) = sin(" }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }
{TEXT -1 18 " x),   u(x, 1) = 0" }}{PARA 0 "" 0 "" {TEXT -1 64 "      \+
                                            u(0, y) = sin(" }{XPPEDIT 
18 0 "Pi;" "6#%#PiG" }{TEXT -1 18 " y),   u(1, y) = 0" }}}}}{SECT 1 
{PARA 3 "" 0 "" {TEXT -1 42 "Laplace's equation with insulated boundar
y" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
29 "We work the following problem" }}{PARA 0 "" 0 "" {TEXT -1 42 "    \+
                                      " }{XPPEDIT 18 0 "0 = diff(u,`$`
(x,2))+diff(u,`$`(y,2));" "6#/\"\"!,&-%%diffG6$%\"uG-%\"$G6$%\"xG\"\"#
\"\"\"-F'6$F)-F+6$%\"yG\"\"#F/" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 31 "Boundary conditions:           " }
{XPPEDIT 18 0 "u(0,y) = sin(Pi*y);" "6#/-%\"uG6$\"\"!%\"yG-%$sinG6#*&%
#PiG\"\"\"F(F." }{TEXT -1 5 " and " }{XPPEDIT 18 0 "u[x](1,y)=0" "6#/-
&%\"uG6#%\"xG6$\"\"\"%\"yG\"\"!" }}{PARA 0 "" 0 "" {TEXT -1 45 "      \+
                                       " }{XPPEDIT 18 0 "u[y](x,0)=0" 
"6#/-&%\"uG6#%\"yG6$%\"xG\"\"!F+" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "
u[y](x,1)=0" "6#/-&%\"uG6#%\"yG6$%\"xG\"\"\"\"\"!" }{TEXT -1 1 "." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(plots):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 113 "spacecurve(\{[x,0,0],[x,1,0],[1,x,
0],[0,x,sin(Pi*x)]\},x=0..1,color=BLACK,\n     axes=NORMAL,orientation
=[-25.,50]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 
0 "" 0 "" {TEXT -1 212 "We will not need to break this into a set of t
wo PDE's with boundary conditions for one of them will have only the z
ero solution.  The other one will break into two differential equation
s with boundary conditons:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 
"" 0 "" {TEXT -1 23 "                X '' - " }{XPPEDIT 18 0 "lambda^2
;" "6#*$%'lambdaG\"\"#" }{TEXT -1 20 " X = 0,  and Y '' + " }{XPPEDIT 
18 0 "lambda^2;" "6#*$%'lambdaG\"\"#" }{TEXT -1 28 " Y = 0, Y '(0) = 0
 = Y '(1)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 29 "These two will have solutions" }}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}}{PARA 0 "" 0 "" {TEXT -1 20 "                    " }{XPPEDIT 18 0 "x
[0](x)" "6#-&%\"xG6#\"\"!6#F%" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "A[0]*
x+B[0]" "6#,&*&&%\"AG6#\"\"!\"\"\"%\"xGF)F)&%\"BG6#F(F)" }{TEXT -1 4 "
,   " }{XPPEDIT 18 0 "X[n](x)" "6#-&%\"XG6#%\"nG6#%\"xG" }{TEXT -1 10 
" = sinh(n " }{XPPEDIT 18 0 "pi;" "6#%#piG" }{TEXT -1 16 " x)  or  sin
h(n " }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 13 " (1-x))      " }
{XPPEDIT 18 0 "Y[n](y)=cos(n*Pi*y)" "6#/-&%\"YG6#%\"nG6#%\"yG-%$cosG6#
*(F(\"\"\"%#PiGF/F*F/" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 
"" {TEXT -1 5 "Thus," }}{PARA 0 "" 0 "" {TEXT -1 32 "                 \+
      u(x,y) = " }{XPPEDIT 18 0 "A[0]*x+B[0]+sum((A[n]*sinh(n*Pi*x)+B[
n]*sinh(n*Pi*(1-x)))*cos(n*Pi*y),n=1..infinity)" "6#,(*&&%\"AG6#\"\"!
\"\"\"%\"xGF)F)&%\"BG6#F(F)-%$sumG6$*&,&*&&F&6#%\"nGF)-%%sinhG6#*(F6F)
%#PiGF)F*F)F)F)*&&F,6#F6F)-F86#*(F6F)F;F),&\"\"\"F)F*!\"\"F)F)F)F)-%$c
osG6#*(F6F)F;F)%\"yGF)F)/F6;\"\"\"%)infinityGF)" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 66 "We use the remaining boundary conditions to compute coefficient
s: " }}{PARA 0 "" 0 "" {TEXT -1 23 "Because           sin( " }
{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 6 " y) = " }{XPPEDIT 18 0 "B[0
]+sum(B[n]*sinh(n*Pi)*cos(n*Pi*y),n=1..infinity" "6#,&&%\"BG6#\"\"!\"
\"\"-%$sumG6$*(&F%6#%\"nGF(-%%sinhG6#*&F/F(%#PiGF(F(-%$cosG6#*(F/F(F4F
(%\"yGF(F(/F/;\"\"\"%)infinityGF(" }}{PARA 0 "" 0 "" {TEXT -1 25 "We c
an compute the B's:  " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 
"" {TEXT -1 18 "                  " }{XPPEDIT 18 0 "B[0];" "6#&%\"BG6#
\"\"!" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "int(sin(Pi*y),y = 0 .. 1);" "
6#-%$intG6$-%$sinG6#*&%#PiG\"\"\"%\"yGF+/F,;\"\"!\"\"\"" }{TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 18 "                  " }{XPPEDIT 18 0 "B
[n]*sinh(n*Pi) = 2*int(sin(Pi*y)*cos(n*Pi*y),y = 0 .. 1);" "6#/*&&%\"B
G6#%\"nG\"\"\"-%%sinhG6#*&F(F)%#PiGF)F)*&\"\"#F)-%$intG6$*&-%$sinG6#*&
F.F)%\"yGF)F)-%$cosG6#*(F(F)F.F)F9F)F)/F9;\"\"!\"\"\"F)" }{TEXT -1 14 
"    for n > 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 22 "because           0 = " }{XPPEDIT 18 0 "u[x](1,y)" "6#-&%
\"uG6#%\"xG6$\"\"\"%\"yG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "A[0]" "6#&
%\"AG6#\"\"!" }{TEXT -1 3 " + " }{XPPEDIT 18 0 "sum(n*Pi*(cosh(n*Pi)*A
[n]-B[n])*cos(n*Pi*y),n=1..infinity);" "6#-%$sumG6$**%\"nG\"\"\"%#PiGF
(,&*&-%%coshG6#*&F'F(F)F(F(&%\"AG6#F'F(F(&%\"BG6#F'!\"\"F(-%$cosG6#*(F
'F(F)F(%\"yGF(F(/F';\"\"\"%)infinityG" }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 20 "then                " }{XPPEDIT 18 0 
"A[0]=0" "6#/&%\"AG6#\"\"!F'" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "A[n]
=B[n]/cosh(n*Pi)" "6#/&%\"AG6#%\"nG*&&%\"BG6#F'\"\"\"-%%coshG6#*&F'F,%
#PiGF,!\"\"" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 21 "Here are the details." }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 7 "Details" }}{PARA 0 "" 0 
"" {TEXT -1 50 "We proceed to compute coefficients:  first the B's" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "B[0]:=int(sin(Pi*y)*1,y=0..1
);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 81 "for i from 1 to 6 do
\n   B[i]:=2*int(sin(Pi*y)*cos(i*Pi*y),y=0..1)/sinh(i*Pi);\nod;" }}}
{PARA 0 "" 0 "" {TEXT -1 11 "Now the A'x" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 51 "for i from 1 to 6 do\n    A[i]:=B[i]/cosh(i*Pi);\nod;
" }}}{PARA 0 "" 0 "" {TEXT -1 35 "Here is a candidate for a solution.
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 96 "u:=(x,y)->B[0]+sum((A[n]
*sinh(n*Pi*x)+B[n]*sinh(n*Pi*(1-x)))*cos(n*Pi*y),\n              n=1..
6);" }}}{PARA 0 "" 0 "" {TEXT -1 56 "When x = 0, this u should approxi
mate the sine function." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "p
lot([sin(Pi*y),u(0,y)],y=0..1);" }}}{PARA 0 "" 0 "" {TEXT -1 68 "Here'
s checking the boundary condition, and THAT U SATISFIES THE PDE" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 82 "D[1](u)(1,y);\nD[2](u)(x,0);
\nD[2](u)(x,1);\nsimplify(D[1,1](u)(x,y)+D[2,2](u)(x,y));" }}}{PARA 0 
"" 0 "" {TEXT -1 75 "Finally, draw the graph. Guess what it should loo
k like before you draw it." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
63 "plot3d(u(x,y),x=0..1,y=0..1,axes=NORMAL,orientation=[-135,45]);" }
}}{PARA 0 "" 0 "" {TEXT -1 47 "Can you guess what the contour lines lo
ok like?" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "plots[contourplo
t](u(x,y),x=0..1,y=0..1);" }}}{PARA 0 "" 0 "" {TEXT -1 74 "Would you g
uess that for each x, the total \"heat\" is the same as at x = 0?" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "int(u(x,y),y=0..1);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 28 "We work this second pro
blem:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 41 "
                                         " }{XPPEDIT 18 0 "0 = diff(u,
`$`(x,2))+diff(u,`$`(y,2));" "6#/\"\"!,&-%%diffG6$%\"uG-%\"$G6$%\"xG\"
\"#\"\"\"-F'6$F)-F+6$%\"yG\"\"#F/" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 31 "Boundary conditions:           " }
{XPPEDIT 18 0 "u[x](0,y) = 0;" "6#/-&%\"uG6#%\"xG6$\"\"!%\"yGF*" }
{TEXT -1 5 " and " }{XPPEDIT 18 0 "u[x](1,y)=0" "6#/-&%\"uG6#%\"xG6$\"
\"\"%\"yG\"\"!" }}{PARA 0 "" 0 "" {TEXT -1 45 "                       \+
                      " }{XPPEDIT 18 0 "u[y](x,0) = 1-cos(2*Pi*x);" "6
#/-&%\"uG6#%\"yG6$%\"xG\"\"!,&\"\"\"\"\"\"-%$cosG6#*(\"\"#F.%#PiGF.F*F
.!\"\"" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "u[y](x,1) = 1;" "6#/-&%\"u
G6#%\"yG6$%\"xG\"\"\"\"\"\"" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 456 "Here is a way to think of this
 problem: first, it is not changing in time. And, second, material is \+
flowing into the bottom of the rectangle and flowing out the top. Sinc
e nothing crosses the left or right side, there had better be as much \+
going in the bottom as is going out the top. The total going out the t
op is 1. We compute how much is going in the bottom, confident I have \+
made the problem well. Also, we draw the graph of the input from the b
ottom." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "int(1-cos(2*Pi*y),
y=0..1);\nplot(1-cos(2*Pi*y),y=0..1);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 69 "We know how to separate variables. The differential equat
ions will be" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 17 "          X '' + " }{XPPEDIT 18 0 "lambda^2;" "6#*$%'lambdaG\"
\"#" }{TEXT -1 44 " X = 0, X '(0) = 0 = X '(1),   and   Y '' = " }
{XPPEDIT 18 0 "lambda^2;" "6#*$%'lambdaG\"\"#" }{TEXT -1 3 " Y." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 51 "Solutions
 for this will be       X(x) = 1 or cos(n " }{XPPEDIT 18 0 "pi;" "6#%#
piG" }{TEXT -1 13 " x) if n > 0 " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 4 "and " }}{PARA 0 "" 0 "" {TEXT -1 69 "     \+
                                      Y(y) = 1 and y  or  cosh(n" }
{XPPEDIT 18 0 "pi;" "6#%#piG" }{TEXT -1 14 " y) and cosh(n" }{XPPEDIT 
18 0 "pi;" "6#%#piG" }{TEXT -1 17 " (1-y)) if n > 0." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 36 "We can construct a gene
ral solution." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 15 "     U(x, y) = " }{XPPEDIT 18 0 "a[0];" "6#&%\"aG6#\"\"!
" }{TEXT -1 3 " + " }{XPPEDIT 18 0 "b[0];" "6#&%\"bG6#\"\"!" }{TEXT 
-1 5 " y + " }{XPPEDIT 18 0 "sum(a[n]*cosh(n*Pi*y)+b[n]*cosh(n*Pi*(1-y
))*cos(n*Pi*x),n);" "6#-%$sumG6$,&*&&%\"aG6#%\"nG\"\"\"-%%coshG6#*(F+F
,%#PiGF,%\"yGF,F,F,*(&%\"bG6#F+F,-F.6#*(F+F,F1F,,&\"\"\"F,F2!\"\"F,F,-
%$cosG6#*(F+F,F1F,%\"xGF,F,F,F+" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 25 "We have only to find the \+
" }{TEXT 260 1 "a" }{TEXT -1 8 " 's and " }{TEXT 261 2 "b " }{TEXT -1 
4 " 's." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
31 "  Coming in from the bottom is " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 9 "  f(x) = " }{XPPEDIT 18 0 "1-cos(2*Pi*x)
" "6#,&\"\"\"\"\"\"-%$cosG6#*(\"\"#F%%#PiGF%%\"xGF%!\"\"" }{TEXT -1 3 
" = " }{XPPEDIT 18 0 "diff(u,y);" "6#-%%diffG6$%\"uG%\"yG" }{TEXT -1 
9 "(x, 0) = " }{XPPEDIT 18 0 "b[0];" "6#&%\"bG6#\"\"!" }{TEXT -1 4 "  \+
- " }{XPPEDIT 18 0 "sum(b[n]*n*Pi*sinh(n*Pi)*cos(n*Pi*x),n);" "6#-%$su
mG6$*,&%\"bG6#%\"nG\"\"\"F*F+%#PiGF+-%%sinhG6#*&F*F+F,F+F+-%$cosG6#*(F
*F+F,F+%\"xGF+F+F*" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 20 "Going out the top is" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 9 "     1 = " }{XPPEDIT 18 0 
"diff(u,y);" "6#-%%diffG6$%\"uG%\"yG" }{TEXT -1 10 "(x, 1) =  " }
{XPPEDIT 18 0 "b[0];" "6#&%\"bG6#\"\"!" }{TEXT -1 3 " + " }{XPPEDIT 
18 0 "sum(a[n]*sinh(n*Pi)*cos(n*Pi*x),n);" "6#-%$sumG6$*(&%\"aG6#%\"nG
\"\"\"-%%sinhG6#*&F*F+%#PiGF+F+-%$cosG6#*(F*F+F0F+%\"xGF+F+F*" }{TEXT 
-1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
172 "Finding the coefficients is now a job for Fourier Series. For thi
s example, we can see what the coefficients are by inspection. Is it n
ot clear from the last equation that " }{XPPEDIT 18 0 "b[0];" "6#&%\"b
G6#\"\"!" }{TEXT -1 17 " = 1 and all the " }{TEXT 262 1 "a" }{TEXT -1 
77 " 's are zero? From the next to last equation, we can see by inspec
tion that  " }{XPPEDIT 18 0 "-b[2]*2*Pi*sinh(n*Pi) = -1;" "6#/,$**&%\"
bG6#\"\"#\"\"\"\"\"#F*%#PiGF*-%%sinhG6#*&%\"nGF*F,F*F*!\"\",$\"\"\"F2
" }{TEXT -1 85 " and all the other b 's are zero. We can now make the \+
u that satisfies this equation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{SECT 1 {PARA 3 "" 0 "" {TEXT -1 7 "Details" }}{PARA 0 "" 0 "" {TEXT 
-1 73 "Here we define u, check that it satisfies the equation, and dra
w a graph." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "u:=(x,y)->y+co
sh(2*Pi*(1-y))*cos(2*Pi*x)/(2*Pi*sinh(2*Pi));" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 34 "diff(u(x,y),x,x)+diff(u(x,y),y,y);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "D[1](u)(0,y);\nD[1](u)(1,y);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "D[2](u)(x,0);\nD[2](u)(x,1);
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 37 "     We draw a graph of t
he solution." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "plot3d(u(x,y
),x=0..1,y=0..1,axes=NORMAL,orientation=[-100,100]);" }{TEXT -1 0 "" }
}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 8 
"Exercise" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 17 "Solve the problem" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 42 "         \+
                                 " }{XPPEDIT 18 0 "0 = diff(u,`$`(x,2)
)+diff(u,`$`(y,2));" "6#/\"\"!,&-%%diffG6$%\"uG-%\"$G6$%\"xG\"\"#\"\"
\"-F'6$F)-F+6$%\"yG\"\"#F/" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 
"" 0 "" {TEXT -1 31 "Boundary conditions:           " }{XPPEDIT 18 0 "
u(0,y)=sin(y)" "6#/-%\"uG6$\"\"!%\"yG-%$sinG6#F(" }{TEXT -1 5 " and " 
}{XPPEDIT 18 0 "u[x](1,y)=0" "6#/-&%\"uG6#%\"xG6$\"\"\"%\"yG\"\"!" }}
{PARA 0 "" 0 "" {TEXT -1 45 "                                         \+
    " }{XPPEDIT 18 0 "u(x,0) = sin(x);" "6#/-%\"uG6$%\"xG\"\"!-%$sinG6
#F'" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "u[y](x,1)=0" "6#/-&%\"uG6#%\"
yG6$%\"xG\"\"\"\"\"!" }{TEXT -1 1 "." }}}}}{SECT 1 {PARA 3 "" 0 "" 
{TEXT -1 26 "The structure of solutions" }}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 127 "The Maximum Principle fo
r solutions of Laplace's Equation is a generalization of the following
 familiar idea from the calculus." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT 263 9 "Theorem 1" }{TEXT -1 15 ": Suppose that \+
" }{XPPEDIT 18 0 "0 <= `u ''`;" "6#1\"\"!%%u~''G" }{TEXT -1 57 "  on [
0,1]. Then the maximum value of u occurs at 0 or 1." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 140 "The next theorem is how the idea expressed in previous T
heorem changes for a function u defined on a bounded region D. As is c
ommon, we use " }{XPPEDIT 18 0 "Delta;" "6#%&DeltaG" }{TEXT -1 17 "(u)
 to represent " }{XPPEDIT 18 0 "diff(u,`$`(x,2))+diff(u,`$`(y,2));" "6
#,&-%%diffG6$%\"uG-%\"$G6$%\"xG\"\"#\"\"\"-F%6$F'-F)6$%\"yG\"\"#F-" }
{TEXT -1 62 "  and bdry(D) to represent the boundary of u. Most texts \+
call " }{XPPEDIT 18 0 "Delta;" "6#%&DeltaG" }{TEXT -1 6 "u the " }
{TEXT 270 15 "Laplacian of u." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT 264 9 "Theorem 2" }{TEXT -1 96 ": Suppose that f is co
ntinuous on a closed and bounded rectangle D. Let u be continuous on D
 and" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 16 " \+
               " }{XPPEDIT 18 0 "Delta;" "6#%&DeltaG" }{TEXT -1 27 "u \+
= f on the interior of D." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 5 "If   " }{XPPEDIT 18 0 "0 <= f(x,y);" "6#1\"\"!-%\"f
G6$%\"xG%\"yG" }{TEXT -1 46 "  then u assumes its maximum value on bdr
y(D)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 77 "
Proof: Suppose 0 < f. Any function continuous on a closed and bounded \+
set in " }{XPPEDIT 18 0 "R^n" "6#)%\"RG%\"nG" }{TEXT -1 122 " has a ma
ximum on that set. Since u is continuous on the closed and bounded set
 D, it has a maximum on this set. Suppose \{" }{XPPEDIT 18 0 "alpha,be
ta" "6$%&alphaG%%betaG" }{TEXT -1 12 "\} is in the " }{TEXT 269 8 "int
erior" }{TEXT -1 86 " of D and the maximum of u occurs at this place. \+
We hope to establish a contradiction." }}{PARA 0 "" 0 "" {TEXT -1 33 "
We are supposing that max(u) = u(" }{XPPEDIT 18 0 "alpha,beta" "6$%&al
phaG%%betaG" }{TEXT -1 47 "). It follows that the first partials of u \+
at \{" }{XPPEDIT 18 0 "alpha" "6#%&alphaG" }{TEXT -1 2 ", " }{XPPEDIT 
18 0 "beta" "6#%%betaG" }{TEXT -1 80 "\} are zero and the second parti
als are not positive. This is a contradiction to " }{XPPEDIT 18 0 "Del
ta;" "6#%&DeltaG" }{TEXT -1 158 "u = f > 0. It must be that the point \+
at which the maximum occurs is not in the interior of D. The only choi
ce is that the maximum occurs on the boundary of D." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 17 "Suppose now that " }
{XPPEDIT 18 0 "0 <= f;" "6#1\"\"!%\"fG" }{TEXT -1 83 ". We will show t
hat it remains true that u must have its maximum value on bdry(D). " }
}{PARA 0 "" 0 "" {TEXT -1 4 "Let " }{XPPEDIT 18 0 "V[epsilon]" "6#&%\"
VG6#%(epsilonG" }{TEXT -1 17 "(x,y) = u(x,y) + " }{XPPEDIT 18 0 "epsil
on" "6#%(epsilonG" }{TEXT -1 2 " (" }{XPPEDIT 18 0 "x^2+y^2" "6#,&*$%
\"xG\"\"#\"\"\"*$%\"yG\"\"#F'" }{TEXT -1 8 "). Then " }{XPPEDIT 18 0 "
Delta;" "6#%&DeltaG" }{TEXT -1 1 " " }{XPPEDIT 18 0 "v[epsilon]" "6#&%
\"vG6#%(epsilonG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "Delta;" "6#%&Delta
G" }{TEXT -1 5 "u + 4" }{XPPEDIT 18 0 "epsilon" "6#%(epsilonG" }{TEXT 
-1 8 " = f + 4" }{XPPEDIT 18 0 "epsilon" "6#%(epsilonG" }{TEXT -1 52 "
. > 0 on the interior of D. By the previous result, " }{XPPEDIT 18 0 "
v[epsilon]" "6#&%\"vG6#%(epsilonG" }{TEXT -1 50 " has its maximum on b
dry(D). Let this maximum be \{" }{XPPEDIT 18 0 "c[epsilon],d[epsilon]
" "6$&%\"cG6#%(epsilonG&%\"dG6#F&" }{TEXT -1 12 "\} and let u(" }
{XPPEDIT 18 0 "alpha,beta" "6$%&alphaG%%betaG" }{TEXT -1 15 ") = max u
. Then" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 2 "
u(" }{XPPEDIT 18 0 "alpha,beta" "6$%&alphaG%%betaG" }{TEXT -1 6 ") < u
(" }{XPPEDIT 18 0 "alpha,beta" "6$%&alphaG%%betaG" }{TEXT -1 4 ") + " 
}{XPPEDIT 18 0 "epsilon" "6#%(epsilonG" }{TEXT -1 2 " (" }{XPPEDIT 18 
0 "alpha^2+beta^2" "6#,&*$%&alphaG\"\"#\"\"\"*$%%betaG\"\"#F'" }{TEXT 
-1 4 ") = " }{XPPEDIT 18 0 "V[epsilon](alpha,beta) <= v[epsilon](c[eps
ilon],d[epsilon]);" "6#1-&%\"VG6#%(epsilonG6$%&alphaG%%betaG-&%\"vG6#F
(6$&%\"cG6#F(&%\"dG6#F(" }{TEXT -1 7 "  =  u(" }{XPPEDIT 18 0 "c[epsil
on],d[epsilon]" "6$&%\"cG6#%(epsilonG&%\"dG6#F&" }{TEXT -1 4 ") + " }
{XPPEDIT 18 0 "epsilon" "6#%(epsilonG" }{TEXT -1 2 " (" }{XPPEDIT 18 
0 "c[epsilon]^2+d[epsilon]^2" "6#,&*$&%\"cG6#%(epsilonG\"\"#\"\"\"*$&%
\"dG6#F(\"\"#F*" }{TEXT -1 2 ")." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 28 "Also, because D is bounded, " }{XPPEDIT 
18 0 "lim[epsilon->0]" "6#&%$limG6#R6#%(epsilonG7\"6$%)operatorG%&arro
wG6\"\"\"!F-F-F-" }{TEXT -1 1 " " }{XPPEDIT 18 0 "epsilon*(c[epsilon]^
2+d[epsilon]^2)" "6#*&%(epsilonG\"\"\",&*$&%\"cG6#F$\"\"#F%*$&%\"dG6#F
$\"\"#F%F%" }{TEXT -1 11 " = 0. Thus," }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 15 "             u(" }{XPPEDIT 18 0 "alph
a;" "6#%&alphaG" }{TEXT -1 2 ", " }{XPPEDIT 18 0 "beta;" "6#%%betaG" }
{TEXT -1 4 ") = " }{XPPEDIT 18 0 "max(u) <= u;" "6#1-%$maxG6#%\"uGF'" 
}{TEXT -1 1 " " }{XPPEDIT 18 0 "lim*(c[epsilon], d[epsilon]) <= max;" 
"6#1*&%$limG\"\"\"6$&%\"cG6#%(epsilonG&%\"dG6#F+F&%$maxG" }{TEXT -1 4 
"(u)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 30 "
The conclusion is that u(lim[ " }{XPPEDIT 18 0 "c[epsilon],d[epsilon]
" "6$&%\"cG6#%(epsilonG&%\"dG6#F&" }{TEXT -1 49 "]) = max(u) and, beca
use bdry(D) is closed, lim[ " }{XPPEDIT 18 0 "c[epsilon],d[epsilon]" "
6$&%\"cG6#%(epsilonG&%\"dG6#F&" }{TEXT -1 26 "] is on the boundary of \+
D." }}{PARA 0 "" 0 "" {TEXT -1 13 "END OF PROOF." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 259 "Actually, this maximum principle holds for any reasonably nice
, bounded region with boundary. The ideas can be made precise for more
 general regions. It is intellectually fun to do that, but we resist h
ere. Note this example, however where D is the unit disk." }}{PARA 0 "
" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT 265 10 "Example 1." }{TEXT -1 16 "  Let u satisfy " }{XPPEDIT 
18 0 "Delta;" "6#%&DeltaG" }{TEXT -1 32 "u = 0 on the unit disk with u
(1," }{XPPEDIT 18 0 "theta" "6#%&thetaG" }{TEXT -1 8 ") = sin(" }
{XPPEDIT 18 0 "theta" "6#%&thetaG" }{TEXT -1 11 ").  Then u(" }
{XPPEDIT 18 0 "r,theta" "6$%\"rG%&thetaG" }{TEXT -1 4 ") = " }
{XPPEDIT 18 0 "r*sin(theta)" "6#*&%\"rG\"\"\"-%$sinG6#%&thetaGF%" }
{TEXT -1 160 ". It is clear that this u takes on  its maximum value on
 the boundary of the unit disk -- on the unit circle. Wait until the n
ext module to see how to evaluate " }{XPPEDIT 18 0 "Delta;" "6#%&Delta
G" }{TEXT -1 23 "u in polar coordinates." }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 76 "Th
e notion of boundedness is important in the result. Consider this exam
ple." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}}{PARA 0 "" 0 "" {TEXT 266 10 "Example 2." }{TEXT -1 48 " Let D be th
e half infinite strip with x in [0, " }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }
{TEXT -1 52 " ], y > 0, and g(x,y) = sin(x) on the boundary. Then" }}
{PARA 0 "" 0 "" {TEXT -1 14 "     u(x,y) = " }{XPPEDIT 18 0 "sin(x)*ex
p(y)" "6#*&-%$sinG6#%\"xG\"\"\"-%$expG6#%\"yGF(" }{TEXT -1 30 ".  Note
 that u goes unbounded." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 84 "The maximum principle implies the following. This re
sult is important for engineers." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 267 9 "Theorem 3
" }{TEXT -1 111 ": (Continuous Dependence on the data) Let D and u be \+
as in Theorem 2. Let g be u on the boundary of D and f be " }{XPPEDIT 
18 0 "Delta;" "6#%&DeltaG" }{TEXT -1 53 "u on the interior of D. There
 is a number k such that" }}{PARA 0 "" 0 "" {TEXT -1 22 "             \+
  | u |  " }{XPPEDIT 18 0 "` ` <= ` `" "6#1%\"~GF$" }{TEXT -1 25 "  ma
x| g | + k max | f |." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 
"" {TEXT -1 9 "Proof:  0" }{XPPEDIT 18 0 "` ` <= ` `;" "6#1%\"~GF$" }
{TEXT -1 14 "f + max |f| = " }{XPPEDIT 18 0 "Delta;" "6#%&DeltaG" }
{TEXT -1 14 "u + max |f| = " }{XPPEDIT 18 0 "Delta;" "6#%&DeltaG" }
{TEXT -1 7 "( u + (" }{XPPEDIT 18 0 "x^2+y^2" "6#,&*$%\"xG\"\"#\"\"\"*
$%\"yG\"\"#F'" }{TEXT -1 13 ") max |f|/4 )" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 4 "And," }}{PARA 0 "" 0 "" {TEXT 
-1 11 "          0" }{XPPEDIT 18 0 "` ` <= ` `;" "6#1%\"~GF$" }{TEXT 
-1 18 "- f + max |f| = - " }{XPPEDIT 18 0 "Delta;" "6#%&DeltaG" }
{TEXT -1 14 "u + max |f| = " }{XPPEDIT 18 0 "Delta;" "6#%&DeltaG" }
{TEXT -1 9 "( - u + (" }{XPPEDIT 18 0 "x^2+y^2" "6#,&*$%\"xG\"\"#\"\"
\"*$%\"yG\"\"#F'" }{TEXT -1 13 ") max |f|/4 )" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 24 "By Theorem 2, both u + (
" }{XPPEDIT 18 0 "x^2+y^2" "6#,&*$%\"xG\"\"#\"\"\"*$%\"yG\"\"#F'" }
{TEXT -1 24 ") max |f|/4  and  -u + (" }{XPPEDIT 18 0 "x^2+y^2" "6#,&*
$%\"xG\"\"#\"\"\"*$%\"yG\"\"#F'" }{TEXT -1 80 ") max |f|/4 assume thei
r maximum value on ¶D. Since D is bounded, let   4 k ³  (" }{XPPEDIT 
18 0 "x^2+y^2" "6#,&*$%\"xG\"\"#\"\"\"*$%\"yG\"\"#F'" }{TEXT -1 19 ") \+
on bdry(D).  Then" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 25 "                |u(x,y)| " }{XPPEDIT 18 0 "` ` <= ` `;" "
6#1%\"~GF$" }{TEXT -1 14 " max( |u| +  (" }{XPPEDIT 18 0 "x^2+y^2" "6#
,&*$%\"xG\"\"#\"\"\"*$%\"yG\"\"#F'" }{TEXT -1 30 ") max |f|/4 ) ² g + \+
k max |f|." }}{PARA 0 "" 0 "" {TEXT -1 13 "END OF PROOF." }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 78 "We will discuss in class why this previous result is im
portant for engineers. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT 268 8 "Theorem " }{TEXT -1 106 "4: Suppose D is as in Theor
em 2 and both u and v satisfy the same PDE and boundary conditions. Th
en u = v." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
39 "Indication of proof: Use the Theorem 3." }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 277 0 
"" }{TEXT 271 10 "Example 3." }{TEXT 278 1 " " }{TEXT -1 23 "We consid
er the problem" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 10 "          " }{XPPEDIT 18 0 "u[xx]+u[yy];" "6#,&&%\"uG6#%#
xxG\"\"\"&F%6#%#yyGF(" }{TEXT -1 5 " = 0." }}{PARA 0 "" 0 "" {TEXT -1 
23 "Verify these solutions." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
13 "with(linalg):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "laplac
ian(u(x,y),[x,y]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "lapla
cian(sin(2*x)*sinh(2*y),[x,y]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 40 "laplacian(sin(3*x)*sinh(3*(7-y)),[x,y]);" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT 272 11 " Example 4:" }{TEXT -1 81 " Here are three proble
ms for the Laplace Equation. We choose three solutions for " }
{XPPEDIT 18 0 "Delta;" "6#%&DeltaG" }{TEXT -1 28 "u = 0  on the rectan
gle [0, " }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 5 "]x[0," }
{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 68 "] by the \"Behold Method!
\" It is good to see the logic of the method." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 273 21 "Boundary Conditions 1" }
{TEXT -1 35 "          u(x, 0) = 0   and   u(x, " }{XPPEDIT 18 0 "Pi;
" "6#%#PiG" }{TEXT -1 14 ") = 2 sin(3 x)" }}{PARA 0 "" 0 "" {TEXT -1 
73 "                                                   u(0, y) = 0   a
nd   u(" }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 9 ", y) = 0." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "u1:=(x,y)->2*sin(3*x)*sinh(3
*y)/sinh(3*Pi);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "laplacia
n(u1,[x,y]);\nu1(x,0);\nu1(x,Pi);\nu1(0,y);\nu1(Pi,y);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "u2:=(x,y)" }}}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT 274 21 "Boundary Conditions 2" }{TEXT 
-1 35 "          u(x, 0) = 0   and   u(x, " }{XPPEDIT 18 0 "Pi;" "6#%#
PiG" }{TEXT -1 14 ") = 5 sin(7 x)" }}{PARA 0 "" 0 "" {TEXT -1 73 "    \+
                                               u(0, y) = 0   and   u(
" }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 9 ", y) = 0." }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "u2:=(x,y)->5*sin(7*x)*sinh(7*y)/sin
h(7*Pi);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "laplacian(u2,[x
,y]);\nu2(x,0);\nu2(x,Pi);\nu2(0,y);\nu2(Pi,y);" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT 275 21 "Boundary Conditions 3" }{TEXT -1 35 "          u(
x, 0) = 0   and   u(x, " }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 5 ")
 = 0" }}{PARA 0 "" 0 "" {TEXT -1 83 "                                 \+
                  u(0, y) = 11 sin(13y)   and   u(" }{XPPEDIT 18 0 "Pi
;" "6#%#PiG" }{TEXT -1 9 ", y) = 0." }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 44 "u3:=(x,y)->11*sin(13*y)*sinh(Pi-x)/sinh(Pi);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "laplacian(u3,[x,y]);\nu3(x,0
);\nu3(x,Pi);\nu3(0,y);\nu3(Pi,y);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }{TEXT 276 21 "Boundary Conditions 4" }{TEXT -1 35 "    \+
      u(x, 0) = 0   and   u(x, " }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }
{TEXT -1 25 ") = 2 sin(3 x)+5 sin(7 x)" }}{PARA 0 "" 0 "" {TEXT -1 83 
"                                                   u(0, y) = 11 sin(1
3y)   and   u(" }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 9 ", y) = 0.
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
{TEXT 279 8 "Theorem " }{TEXT -1 47 "4: Suppose that u is a solution f
or the problem" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 13 "             " }{XPPEDIT 18 0 "Delta;" "6#%&DeltaG" }
{TEXT -1 8 " u = 0, " }}{PARA 0 "" 0 "" {TEXT -1 8 "with    " }
{XPPEDIT 18 0 "u[y];" "6#&%\"uG6#%\"yG" }{TEXT -1 20 " (x, 0) = f(x), \+
and " }{XPPEDIT 18 0 "u[y];" "6#&%\"uG6#%\"yG" }{TEXT -1 15 " (x, M) =
 g(x)," }}{PARA 0 "" 0 "" {TEXT -1 11 "           " }{XPPEDIT 18 0 "u[
x];" "6#&%\"uG6#%\"xG" }{TEXT -1 19 " (0, y) = h(y) and " }{XPPEDIT 
18 0 "u[x];" "6#&%\"uG6#%\"xG" }{TEXT -1 15 " (L, y) = j(y)." }}{PARA 
0 "" 0 "" {TEXT -1 6 "Then, " }}{PARA 0 "" 0 "" {TEXT -1 10 "         \+
 " }{XPPEDIT 18 0 "int(g(x),x = 0 .. L);" "6#-%$intG6$-%\"gG6#%\"xG/F)
;\"\"!%\"LG" }{TEXT -1 3 " - " }{XPPEDIT 18 0 "int(f(x),x = 0 .. L);" 
"6#-%$intG6$-%\"fG6#%\"xG/F);\"\"!%\"LG" }{TEXT -1 3 " + " }{XPPEDIT 
18 0 "int(j(y),y = 0 .. M);" "6#-%$intG6$-%\"jG6#%\"yG/F);\"\"!%\"MG" 
}{TEXT -1 3 " - " }{XPPEDIT 18 0 "int(h(y),y = 0 .. M);" "6#-%$intG6$-
%\"hG6#%\"yG/F);\"\"!%\"MG" }{TEXT -1 5 " = 0." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 7 "Proof: " }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT 280 7 "Remark:" }{TEXT -1 122 " It is clear, is it not, that if \+
u is a solution for a Neumann problem, and C is a constant, the u + C \+
is also a solution?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 
4 "" 0 "" {TEXT -1 8 "Exercise" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 19 "
 Solve the problem " }{XPPEDIT 18 0 "Delta;" "6#%&DeltaG" }{TEXT -1 
38 " u = 0 with Neumann conditions on [0, " }{XPPEDIT 18 0 "2*Pi;" "6#
*&\"\"#\"\"\"%#PiGF%" }{TEXT -1 11 " ] x  [0,  " }{XPPEDIT 18 0 "2*Pi;
" "6#*&\"\"#\"\"\"%#PiGF%" }{TEXT -1 68 " ] and with. f(x) = sin(x) an
d g(x) = cos(x), h(y) = 0 and j(y) = 0." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(pl
ots):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 93 "spacecurve(\{[x,0,
sin(x)],[0,x,0],[Pi,x,0]\},x=0..Pi,axes=NORMAL,\n       orientation=[-
50,50]);" }}}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 32 "The heat equation \+
on a rectangle" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 486 "We consider the
 diffusion of heat into a long beam with cross section a rectangle. Th
e supposition that the beam is \"long\" is to produce the mathematical
 idea that heat diffusing to a point comes essentially only from the s
ides and that the ends are so far away that heat coming from above or \+
below can be ignored. This is the same problem as considering heat dif
fusion in a thin rectangular plate that is insulated at the top and bo
ttom. These two problems lead to an equation such as " }}{PARA 0 "" 0 
"" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 23 "                      \+
 " }{XPPEDIT 18 0 "u[t](t,x,y) = diff(u,`$`(x,2))+diff(u,`$`(y,2));" "
6#/-&%\"uG6#%\"tG6%F(%\"xG%\"yG,&-%%diffG6$F&-%\"$G6$F*\"\"#\"\"\"-F.6
$F&-F16$F+\"\"#F4" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 24 "with boundary conditions" }}{PARA 0 "" 0 
"" {TEXT -1 34 "                    u( t, x, 0) = " }{XPPEDIT 18 0 "f[
0]" "6#&%\"fG6#\"\"!" }{TEXT -1 18 ",   u( t, x, b) = " }{XPPEDIT 18 
0 "f[1]" "6#&%\"fG6#\"\"\"" }}{PARA 0 "" 0 "" {TEXT -1 34 "           \+
         u( t, 0, y) = " }{XPPEDIT 18 0 "g[0]" "6#&%\"gG6#\"\"!" }
{TEXT -1 18 ",   u( t, a, 0) = " }{XPPEDIT 18 0 "g[1]" "6#&%\"gG6#\"\"
\"" }}{PARA 0 "" 0 "" {TEXT -1 22 "and initial conditions" }}{PARA 0 "
" 0 "" {TEXT -1 44 "                     u( 0, x, y) = h( x, y)." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 93 "The probl
em is broken into two parts: the steady state v(x,y) and the transient
, w( t, x, y)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 27 "The steady-state problem is" }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 29 "                         0 = " }
{XPPEDIT 18 0 "diff(v,`$`(x,2))+diff(v,`$`(y,2));" "6#,&-%%diffG6$%\"v
G-%\"$G6$%\"xG\"\"#\"\"\"-F%6$F'-F)6$%\"yG\"\"#F-" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 26 "with boundary conditions \+
 " }}{PARA 0 "" 0 "" {TEXT -1 33 "                   v( t, x, 0) = " }
{XPPEDIT 18 0 "f[0]" "6#&%\"fG6#\"\"!" }{TEXT -1 18 ",   v( t, x, b) =
 " }{XPPEDIT 18 0 "f[1]" "6#&%\"fG6#\"\"\"" }}{PARA 0 "" 0 "" {TEXT 
-1 34 "                    v( t, 0, y) = " }{XPPEDIT 18 0 "g[0]" "6#&%
\"gG6#\"\"!" }{TEXT -1 18 ",   v( t, a, 0) = " }{XPPEDIT 18 0 "g[1]" "
6#&%\"gG6#\"\"\"" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 84 "We have discussed how to break this problem into two part
s previously. They would be" }}{PARA 0 "" 0 "" {TEXT -1 8 "    0 = " }
{XPPEDIT 18 0 "diff(v1,`$`(x,2))+diff(v1,`$`(y,2));" "6#,&-%%diffG6$%#
v1G-%\"$G6$%\"xG\"\"#\"\"\"-F%6$F'-F)6$%\"yG\"\"#F-" }{TEXT -1 2 ", " 
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 24 "with bo
undary conditions" }}{PARA 0 "" 0 "" {TEXT -1 28 "                v1( \+
x, 0) = " }{XPPEDIT 18 0 "f[0]" "6#&%\"fG6#\"\"!" }{TEXT -1 16 ",   v1
( x, b) = " }{XPPEDIT 18 0 "f[1]" "6#&%\"fG6#\"\"\"" }}{PARA 0 "" 0 "
" {TEXT -1 28 "                v1( 0, y) = " }{XPPEDIT 18 0 "0;" "6#\"
\"!" }{TEXT -1 16 ",   v1( a, 0) = " }{XPPEDIT 18 0 "0;" "6#\"\"!" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 3 "and" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 8 "    0 = " }{XPPEDIT 18 0 "diff(v2,`$`(x,2))+diff
(v2,`$`(y,2));" "6#,&-%%diffG6$%#v2G-%\"$G6$%\"xG\"\"#\"\"\"-F%6$F'-F)
6$%\"yG\"\"#F-" }{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 24 "with boundary conditions" }}{PARA 0 "" 0 
"" {TEXT -1 46 "                v2( x, 0) = 0,  v2( x, b) = 0," }}
{PARA 0 "" 0 "" {TEXT -1 28 "                v2( 0, y) = " }{XPPEDIT 
18 0 "g[0];" "6#&%\"gG6#\"\"!" }{TEXT -1 16 ",   v2( a, 0) = " }
{XPPEDIT 18 0 "g[1];" "6#&%\"gG6#\"\"\"" }{TEXT -1 1 "." }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 78 "The solution for the
se two problems are added to get the steady state solution" }}{PARA 0 
"" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 45 "               v(
x, y) = v1(x, y) + v2(x, y)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 170 "We will not detail how to make this computatio
n for the transient solution, since we have covered how to compute sol
utions for Laplace's Equation on a rectangle earlier. " }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 24 "The transient proble
m is" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 25 " \+
                        " }{XPPEDIT 18 0 "w[t]" "6#&%\"wG6#%\"tG" }
{TEXT -1 3 " = " }{XPPEDIT 18 0 "diff(w,`$`(x,2))+diff(w,`$`(y,2));" "
6#,&-%%diffG6$%\"wG-%\"$G6$%\"xG\"\"#\"\"\"-F%6$F'-F)6$%\"yG\"\"#F-" }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 26 "with bou
ndary conditions  " }}{PARA 0 "" 0 "" {TEXT -1 53 "                   \+
w( t, x, 0) = 0,   w( t, x, b) = 0" }}{PARA 0 "" 0 "" {TEXT -1 54 "   \+
                 w( t, 0, y) = 0,   w( t, a, 0) = 0" }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 22 "with initial condition
" }}{PARA 0 "" 0 "" {TEXT -1 55 "                     w( 0, x, y) = h(
 x, y) - v( x, y)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 91 "Think about why u(t, x, y) = w(t, x, y) + v(x, y) is the \+
solution for the original problem." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 80 "We break this last PDE into three ODE's, \+
two of which have boundary conditions. " }}{PARA 0 "" 0 "" {TEXT -1 0 
"" }}{PARA 0 "" 0 "" {TEXT -1 8 "X '' = -" }{XPPEDIT 18 0 "lambda^2;" 
"6#*$%'lambdaG\"\"#" }{TEXT -1 24 " X,             Y '' = -" }
{XPPEDIT 18 0 "mu^2;" "6#*$%#muG\"\"#" }{TEXT -1 27 " Y               \+
T ' = - ( " }{XPPEDIT 18 0 "lambda^2+mu^2;" "6#,&*$%'lambdaG\"\"#\"\"
\"*$%#muG\"\"#F'" }{TEXT -1 4 " ) T" }}{PARA 0 "" 0 "" {TEXT -1 39 "X(
0) = X(a) = 0        Y(0) = Y(b) = 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 70 "The r
esult is a general solution of the transient equation in the form" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 21 "       w(
 t, x, y) = " }{XPPEDIT 18 0 "sum(sum(A[n,m]*sin(n*Pi*x/a)*sin(m*Pi*y/
b)*exp(-(n^2+m^2)*Pi^2*t),n),m" "6#-%$sumG6$-F$6$**&%\"AG6$%\"nG%\"mG
\"\"\"-%$sinG6#**F,F.%#PiGF.%\"xGF.%\"aG!\"\"F.-F06#**F-F.F3F.%\"yGF.%
\"bGF6F.-%$expG6#,$*(,&*$F,\"\"#F.*$F-\"\"#F.F.*$F3\"\"#F.%\"tGF.F6F.F
,F-" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 32 "Th
e coefficients are obtained as" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 16 "                " }{XPPEDIT 18 0 "A[n,m]=
4*int(int(w(0,x,y)*sin(n*Pi*x/a)*sin(m*Pi*y/b),y=0..b),x=0..a)/(a*b)" 
"6#/&%\"AG6$%\"nG%\"mG*(\"\"%\"\"\"-%$intG6$-F-6$*(-%\"wG6%\"\"!%\"xG%
\"yGF+-%$sinG6#**F'F+%#PiGF+F6F+%\"aG!\"\"F+-F96#**F(F+F<F+F7F+%\"bGF>
F+/F7;F5FB/F6;F5F=F+*&F=F+FBF+F>" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}
{SECT 1 {PARA 3 "" 0 "" {TEXT -1 7 "Example" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 62 "     We illustrate how an anima
tion of such a result behaves. " }}{PARA 0 "" 0 "" {TEXT -1 37 "\nTake
 boundary conditions as follows:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 58 "                   u( t, x, 0) = sin(x), \+
  u( t, x, 1) = 0" }}{PARA 0 "" 0 "" {TEXT -1 60 "                   u
( t, 0, y) = 0,          u( t, 1, 0) = 0" }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 30 "Take the initial condition as " }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 40 "         \+
          u( 0, x, y) = sin(x)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 137 "By now, maybe we can do the steady state
 without the computation of any integrals. The formula for the steady \+
state, and a check follows." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 46 "v:=(x,y)->(sinh(Pi*(1-y)))*sin(Pi*x)/sinh(Pi);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "diff(v(x,y),y,y)+diff(v(x,y)
,x,x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "v(0,y);v(1,y);v(x
,0);v(x,1);" }}}{PARA 0 "" 0 "" {TEXT -1 93 "The function w will be th
e transient solution. The initial condition for w is k, given below." 
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "k:=(x,y)->sin(Pi*x)-v(x,y)
;\nk(0,y);k(1,y);k(x,0);k(x,1);" }}}{PARA 0 "" 0 "" {TEXT -1 312 "We n
ow compute the  coefficients as directed above. I set this up to do 10
0 double integrals. You could do more, or less depending on the speed \+
of your computer. The point is that k(x,y) is not continuous on the cl
osed rectangle making the domain of u. Thus, we will not have a good a
pproximation for u(0, x, y)." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
127 "for n from 1 to 10 do\nfor m from 1 to 10 do 4*int(int(k(x,y)*sin
(n*Pi*x)*sin(m*Pi*y),x=0..1),y=0..1);\nA[n,m]:=evalf(%);\nod;\nod;" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "n:='n'; m:='m';" }}}{PARA 
0 "" 0 "" {TEXT -1 33 "We make the transient solution w." }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 87 "sum(sum(A[n,m]*sin(n*Pi*x)*sin(m*Pi
*y)*exp(-(n^2+m^2)*Pi^2*t),\n      n=1..10),m=1..10):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "w:=unapply(%,(t,x,y)):" }}}{PARA 0 
"" 0 "" {TEXT -1 60 "From w and v, the solution for the original probl
em is made." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "u:=(t,x,y)->w
(t,x,y)+v(x,y);" }}}{PARA 0 "" 0 "" {TEXT -1 84 "As a check to see how
 accurate all this is, u( 0, x, y) should be 1 in this problem." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "plot3d(u(1/50,x,y),x=0..1,y=
0..1,axes=NORMAL);" }}}{PARA 0 "" 0 "" {TEXT -1 78 "Finally, here is a
n animation of the temperature decaying to the steady state." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "plots[animate3d](u(t,x,y),x=
0..1,y=0..1,t=0..1/10);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "
" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT 
-1 28 "Example: Speed of Diffusion." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 89 "If we wa
nt to incorporate speed of diffusion, we should change the model by in
cluding c.:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 23 "                       " }
{XPPEDIT 18 0 "u[t](t,x,y) = c*(diff(u,`$`(x,2))+diff(u,`$`(y,2)));" "
6#/-&%\"uG6#%\"tG6%F(%\"xG%\"yG*&%\"cG\"\"\",&-%%diffG6$F&-%\"$G6$F*\"
\"#F.-F16$F&-F46$F+\"\"#F.F." }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" 
{TEXT -1 31 "This time, we choose conditions" }}{PARA 0 "" 0 "" {TEXT 
-1 59 "                     u( t, x, 0) = 100,   u( t, x, 1) = 100" }}
{PARA 0 "" 0 "" {TEXT -1 59 "                     u( t, 0, y) = 100,  \+
 u( t, 1, 0) = 100" }}{PARA 0 "" 0 "" {TEXT -1 22 "and initial conditi
ons" }}{PARA 0 "" 0 "" {TEXT -1 37 "                     u( 0, x, y) =
 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 86 "We
 ask, for various values of c, what is the value of t such that u(t, 1
/2, 1/2) = 50." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 81 "Is it clear that the steady state solutions is 100? And, \+
that general solution is" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 20 "      u( t, x, y) = " }{XPPEDIT 18 0 "sum(sum(A[n,
m]*sin(n*Pi*x/a)*sin(m*Pi*y/b)*exp(-c*(n^2+m^2)*Pi^2*t),n),m);" "6#-%$
sumG6$-F$6$**&%\"AG6$%\"nG%\"mG\"\"\"-%$sinG6#**F,F.%#PiGF.%\"xGF.%\"a
G!\"\"F.-F06#**F-F.F3F.%\"yGF.%\"bGF6F.-%$expG6#,$**%\"cGF.,&*$F,\"\"#
F.*$F-\"\"#F.F.F3\"\"#%\"tGF.F6F.F,F-" }{TEXT -1 7 " + 100," }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 6 "where " }}{PARA 
0 "" 0 "" {TEXT -1 12 "            " }{XPPEDIT 18 0 "A[n,m] = -400*int
(int(sin(n*Pi*x)*sin(m*Pi*y),y = 0 .. 1),x = 0 .. 1);" "6#/&%\"AG6$%\"
nG%\"mG,$*&\"$+%\"\"\"-%$intG6$-F.6$*&-%$sinG6#*(F'F,%#PiGF,%\"xGF,F,-
F46#*(F(F,F7F,%\"yGF,F,/F<;\"\"!\"\"\"/F8;F?\"\"\"F,!\"\"" }{TEXT -1 
8 " = -400 " }{XPPEDIT 18 0 "(cos(n*Pi)-1)*(cos(m*Pi)-1)/(n*Pi*m*Pi);
" "6#*(,&-%$cosG6#*&%\"nG\"\"\"%#PiGF*F*\"\"\"!\"\"F*,&-F&6#*&%\"mGF*F
+F*F*\"\"\"F-F***F)F*F+F*F2F*F+F*F-" }{TEXT -1 1 "." }}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 126 "for n from 1 to 20 do\nfor m from 1 to 20 \+
do\n   A[n,m]:=-400*(cos(n*Pi)-1)/(n*Pi)*(cos(m*Pi)-1)/(m*Pi):\nod: od
:\nn:='n'; m:='m';" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 89 "sum(s
um(A[n,m]*sin(n*Pi*x)*sin(m*Pi*y)*exp(-c*(n^2+m^2)*Pi^2*t),\n      n=1
..20),m=1..20):" }}}{PARA 0 "" 0 "" {TEXT -1 53 "We write this u as a \+
function of t, x, y, and also c." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 28 "u:=unapply(%,(c,t,x,y))+100:" }}}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 108 "Here, we plot the initial value for \+
c = 1. The purpose is to see how well we have fit the initial conditio
n." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "plot3d(u(1,0,x,y),x=0.
.1,y=0..1,axes=NORMAL);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 101 
"simplify(diff(u(c,t,x,y),t)-\n         c*diff(u(c,t,x,y),x,x)-\n     \+
          c*diff(u(c,t,x,y),y,y));" }}}{PARA 0 "" 0 "" {TEXT -1 39 "He
re, we check the boundary conditions." }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 44 "u(1,t,x,0);u(1,t,x,1);u(1,t,0,y);u(1,t,1,y);" }}}
{PARA 0 "" 0 "" {TEXT -1 85 "This is an animation so that you can see \+
how the solution has limit the steady state." }}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 65 "plots[animate3d](u(1,t,x,y),x=0..1,y=0..1,t=0..1/1
0,axes=NORMAL);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{PARA 0 "" 0 "" {TEXT -1 207 "The goal now is to choose a sequence of \+
c 's and, for each c, to find t so that the point [1/2, 1/2] is at tem
perature half way between 0 and 100. We restrict solutions for t to be
ing in the interval [0, 5]." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
80 "for p from 1 to 10 do\n   s[p]:=fsolve(u(p/10,t,1/2,1/2)=50.,t,0..
5);\nod;\np:='p';" }}}{PARA 0 "" 0 "" {TEXT -1 136 "Here, we plot the \+
values of t found above. The purpose is to see the kind of relationshi
p there is between the speed of diffusion and c." }}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 42 "plots[pointplot]([seq([p,s[p]],p=1..10)]);" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 
4 "" 0 "" {TEXT -1 8 "Exercise" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 205 
"Repeat the previous problem with values on the boundary to being 50, \+
instead of 100. Compute the values of the variable t to achieve a temp
erature of 25 in the middle. Compare the results with those above." }}
}}}}{MARK "1 0" 56 }{VIEWOPTS 1 1 0 1 1 1803 }