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{SECT 0 {PARA 0 "" 0 "" {TEXT 262 46 "Section 6.1: Laplace's Equation \+
on a Rectangle" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 ""
0 "" {TEXT 261 30 "Maple Packages for Section 6.1" }}{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 1 " " }}{PARA 0 "" 0 "" {TEXT -1
0 "" }}{PARA 0 "" 0 "" {TEXT -1 121 " From what has come before, we
can expect that a study 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 "dif
f(u,t) = diff(u,`$`(x,2))+diff(u,`$`(y,2));" "6#/-%%diffG6$%\"uG%\"tG,
&-F%6$F'-%\"$G6$%\"xG\"\"#\"\"\"-F%6$F'-F-6$%\"yGF0F1" }}{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 86 " 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-state problem first." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 34 " " }
{XPPEDIT 18 0 "0 = diff(u,`$`(x,2))+diff(u,`$`(y,2));" "6#/\"\"!,&-%%d
iffG6$%\"uG-%\"$G6$%\"xG\"\"#\"\"\"-F'6$F)-F+6$%\"yGF.F/" }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 68 "Boundary condition
s: u(x, 0) = f(x), u(x, L) = g(x)" }}{PARA 0 "" 0 ""
{TEXT -1 80 " u(0, y) =
h(y), u(M, y) = J(y)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 ""
0 "" {TEXT -1 31 "and then the transition problem" }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 34 " \+
" }{XPPEDIT 18 0 "diff(u,t) = diff(u,`$`(x,2))+diff(u,`$`(y,2
));" "6#/-%%diffG6$%\"uG%\"tG,&-F%6$F'-%\"$G6$%\"xG\"\"#\"\"\"-F%6$F'-
F-6$%\"yGF0F1" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 68 "Boundary conditions: u(t, x, 0) = 0, u(t
, x, L) = 0" }}{PARA 0 "" 0 "" {TEXT -1 81 " \+
u(t, 0, y) = 0, u(t, M, y) = 0" }}{PARA 0 ""
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 64 "Initial condition: \+
u(0, x, y) = k(x, y)." }}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 186 "The wave equation leads to thi
s same steady state problem. We consider the steady state problem in t
his part of the notes. It is what is called Laplace's Equation, or Poi
sson's Equation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 137 "This steady state problem is commonly broken into two pr
oblems. Here is one of them. Note that half the boundary conditions ar
e zero now." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 47 " \+
" }{XPPEDIT 18 0 "0 = diff(u,`$`(x,2))+diff(u,`$`(y,2)
);" "6#/\"\"!,&-%%diffG6$%\"uG-%\"$G6$%\"xG\"\"#\"\"\"-F'6$F)-F+6$%\"y
GF.F/" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 69 "
Boundary conditions: u(x, 0) = f(x), u(x, L) = g(x),
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 78 " \+
u(0, y) = 0, u(M, y) =
0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 103 "T
he other problem looks exactly the same except the zero boundary condi
tions are on the other two ends:" }}{PARA 0 "" 0 "" {TEXT -1 77 " \+
u(x, 0) = 0, u(x, L) =
0" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 79 " \+
u(0, y) = h(y), u(M, y) \+
=J(y)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 26
"We work the first problem." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA
0 "" 0 "" {TEXT -1 54 "Here are the steps: (1) Make a separation of va
riables" }}{PARA 0 "" 0 "" {TEXT -1 111 " \+
(2) Identify the resulting ordinary differential equation and boundar
y conditions" }}{PARA 0 "" 0 "" {TEXT -1 55 " \+
(3) Solve these equations" }}{PARA 0 "" 0 "" {TEXT -1 80 " \+
(4) Construct the general solution for the prob
lem" }}{PARA 0 "" 0 "" {TEXT -1 111 " (5)
Use the boundary conditions to get the solution for this particular e
quation." }}{PARA 0 "" 0 "" {TEXT -1 27 "Step 1: Separate variables."
}}{PARA 0 "" 0 "" {TEXT -1 96 " The partial differential equation lea
ds 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 t
he differential equations." }}{PARA 0 "" 0 "" {TEXT -1 39 " The differ
ential equations are X '' + " }{XPPEDIT 18 0 "lambda^2;" "6#*$%'lambda
G\"\"#" }{TEXT -1 36 " X = 0, X(0) = 0 = X(M), and Y '' = " }{XPPEDIT
18 0 "lambda^2;" "6#*$%'lambdaG\"\"#" }{TEXT -1 3 " Y." }}{PARA 0 ""
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 30 "Step 3: Solve these \+
equations." }}{PARA 0 "" 0 "" {TEXT -1 5 " " }{XPPEDIT 18 0 "lambd
a;" "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 81 "If the solution for th
e Y equation is puzzling, review Question 5 of Section 3.1." }}{PARA
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 58 "Step 4: Constru
ct 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$,&*&&%\"aG6#%\"nG\"\"\"-%%sinhG6#*(F+F,%#PiGF,%\"yGF,F,F,*(&%\"b
G6#F+F,-F.6#*(F+F,F1F,,&F,F,F2!\"\"F,F,-%$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 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT
260 28 "Check the resulting solution" }}{PARA 0 "" 0 "" {TEXT -1 0 ""
}}{PARA 0 "" 0 "" {TEXT -1 248 "Here, I suppose that n is an integer a
nd check that the terms u(x,y) = X(x) Y(y) form a solution to Laplace'
s equation with zero boundary conditions at x = 0 and at x = M. I also
check that at y = 0 and at y = L, we are set to use a Fourier Series.
" }}{PARA 0 "" 0 "" {TEXT -1 2 " " }}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 18 "assume(n,integer);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 77 "u:=(x,y)->(a*sinh(n*Pi*y/M)+b*sinh(n*Pi*(L-y)/M))*\n \+
sin(n*Pi*x/M);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0
44 "simplify(diff(u(x,y),x,x)+diff(u(x,y),y,y));" }}}{EXCHG {PARA 0 ">
" 0 "" {MPLTEXT 1 0 15 "u(0,y); u(M,y);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 15 "u(x,0); u(x,L);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 7 "n:='n';" }}}{PARA 0 "" 0 "" {TEXT -1 42 "The point is \+
to check and check and check." }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT 263 24 "
Details for this problem" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 151 "We are now ready to apply this to a situation wit
h both f and g specified. We consider the problem with f = g and as g
iven below. We choose L = M = 1." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT
1 0 101 "u:=(x,y)->sum((a[n]*sinh(n*Pi*y)+b[n]*sinh(n*Pi*(1-y)))*\n \+
sin(n*Pi*x),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 84 "This reminds us to use the Fourie
r coefficients. ( Don't forget to divide by sinh(n " }{XPPEDIT 18 0 "p
i;" "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 0 135 "for n from 1 to 3 do\n b[n]:=int('f(x)'*sin(n*Pi*x
),x=0..1)/\n int(sin(n*Pi*x)^2,x=0..1)/sinh(n*Pi);\n a[n]:=b
[n]:\nod;\nn:='n':" }}}{PARA 0 "" 0 "" {TEXT -1 35 "We incorporate the
se values into u." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 111 "u1:=(x
,y)->8/Pi^2*sum(sin(n*Pi/2)/n^2*\n (sinh(n*Pi*y)+sinh(n*Pi*(1-
y)))/sinh(n*Pi)*sin(n*Pi*x),n=1..3);" }}}{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 30 "plot([u1(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 42 "plot3d(u1
(x,y),x=0..1,y=0..1,axes=NORMAL);" }}}{PARA 0 "" 0 "" {TEXT -1 46 "It \+
is of value sometimes to see contour plots." }}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 35 "contourplot(u1(x,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 85 "We now do a problem with two sides insu
lated. 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$%\"yGF.F/" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT
-1 36 "Boundary conditions: " }{XPPEDIT 18 0 "u;" "6#%
\"uG" }{TEXT -1 17 "(x, 0) = f(x), " }{XPPEDIT 18 0 "u;" "6#%\"uG" }
{TEXT -1 13 "(x, L) = g(x)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0
"" 0 "" {TEXT -1 50 " \+
" }{XPPEDIT 18 0 "u[x];" "6#&%\"uG6#%\"xG" }{TEXT -1 14 "(0, y) = 0, \+
" }{XPPEDIT 18 0 "u[x];" "6#&%\"uG6#%\"xG" }{TEXT -1 11 "(M, y) = 0.
" }}{PARA 0 "" 0 "" {TEXT -1 168 "This says there is no passage across
the left, or right, boundary of the rectangle, but the top and bottom
of the distribution are held as g(x) and f(x), respectively. " }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 117 "spacecurve(\{[x,0,f(x)],[x,
1,f(x)],[1,x,0],[x,1,0]\},x=0..1,\n color=BLACK,axes=NORMAL,
orientation=[-25.,50]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "
" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 31 "We p
erform the same five steps." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA
0 "" 0 "" {TEXT -1 27 "Step 1: Separate variables." }}{PARA 0 "" 0 ""
{TEXT -1 100 " The partial differential equation leads to X '' Y + X \+
Y'' = 0, with X '(0) Y(y) = 0 = X '(M) Y(y)." }}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 44 "Step 2: Identify the differenti
al equations." }}{PARA 0 "" 0 "" {TEXT -1 39 " The differential equati
ons are X '' + " }{XPPEDIT 18 0 "lambda^2;" "6#*$%'lambdaG\"\"#" }
{TEXT -1 40 " X = 0, X '(0) = 0 = X '(M), and Y '' = " }{XPPEDIT 18 0
"lambda^2;" "6#*$%'lambdaG\"\"#" }{TEXT -1 3 " Y." }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 30 "Step 3: Solve these equat
ions." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 7 " \+
if " }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT -1 38 " = 0, the
n X(x) = 1 and Y(y) = 1 or y." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 6 " if " }{XPPEDIT 18 0 "lambda <> 0;" "6#0
%'lambdaG\"\"!" }{TEXT -1 8 ", then " }{XPPEDIT 18 0 "lambda;" "6#%'l
ambdaG" }{TEXT -1 5 " = n " }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1
18 "/M , X(x) = cos(n" }{XPPEDIT 18 0 "pi;" "6#%#piG" }{TEXT -1 23 "x
/M), and Y(y) = sinh(n" }{XPPEDIT 18 0 "pi;" "6#%#piG" }{TEXT -1 15 "y
/M) and sinh(n" }{XPPEDIT 18 0 "pi;" "6#%#piG" }{TEXT -1 9 "(L-y)/M).
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 39 "Step \+
4: Construct the general solution." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 14 " u(x, y) =" }{XPPEDIT 18 0 "a[0]+(b[0
]-a[0])*y/L;" "6#,&&%\"aG6#\"\"!\"\"\"*(,&&%\"bG6#F'F(&F%6#F'!\"\"F(%
\"yGF(%\"LGF0F(" }{TEXT -1 3 " + " }{XPPEDIT 18 0 "sum((a[n]*sinh(n*Pi
*y/M)+b[n]*sinh(n*Pi*(L-y)/M))*cos(n*Pi*x/M),n);" "6#-%$sumG6$*&,&*&&%
\"aG6#%\"nG\"\"\"-%%sinhG6#**F,F-%#PiGF-%\"yGF-%\"MG!\"\"F-F-*&&%\"bG6
#F,F--F/6#**F,F-F2F-,&%\"LGF-F3F5F-F4F5F-F-F--%$cosG6#**F,F-F2F-%\"xGF
-F4F5F-F," }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 53 "Step 5: Use the boundary conditions to determine t
he " }{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 apply this to a \+
situation with and both f and g specified. We consider the problem wi
th f = g and as given below. We choose L = M = 1." }}{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..in
finity);" }}}{PARA 0 "" 0 "" {TEXT -1 64 "We wish to determine the coe
fficients. 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 0 180 "b[0]:=int(f(x),x
=0..1)/int(1^2,x=0..1);\na[0]:=b[0];\nfor n from 1 to 3 do\n b[n]:=i
nt('f(x)'*cos(n*Pi*x),x=0..1)/\n int(cos(n*Pi*x)^2,x=0..1)/si
nh(n*Pi);\n a[n]:=b[n]:\nod;" }}}{PARA 0 "" 0 "" {TEXT -1 35 "We inc
orporate these values into u." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1
0 124 "u2:=(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 again
st the graph of f." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "plot([
u2(x,0),'f(x)'],x=0..1);" }}}{PARA 0 "" 0 "" {TEXT -1 56 "We are now r
eady to see a plot of u. What do you expect?" }}{EXCHG {PARA 0 "> " 0
"" {MPLTEXT 1 0 42 "plot3d(u2(x,y),x=0..1,y=0..1,axes=NORMAL);" }}}
{PARA 0 "" 0 "" {TEXT -1 46 "It is of value sometimes to see contour p
lots." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "contourplot(u2(x,y)
,x=0..1,y=0..1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}
{PARA 0 "" 0 "" {TEXT -1 2 " " }}{PARA 0 "" 0 "" {TEXT -1 332 "In thi
s section, we have examined Laplace's equation on a rectangle. We work
ed two problems. One had opposite ends zero and the other had the oppo
site ends insulate. You can imagine that there are many variations on \+
how the problems can be configured. In the next section, we revisit th
is problem with different boundary conditions." }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT 264 16 "Unassisted Maple
" }}{PARA 0 "" 0 "" {TEXT -1 118 "To verify that the functions u1 and \+
u2 developed above satisfied Laplace's Equation, we formed the partial
derivatives" }}{PARA 0 "" 0 "" {TEXT -1 22 " " }
{XPPEDIT 18 0 "diff(u,`$`(x,2));" "6#-%%diffG6$%\"uG-%\"$G6$%\"xG\"\"#
" }{TEXT -1 3 " + " }{XPPEDIT 18 0 "diff(u,`$`(y,2));" "6#-%%diffG6$%
\"uG-%\"$G6$%\"yG\"\"#" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 5 "
This " }{TEXT 265 18 "Laplacian Operator" }{TEXT -1 103 " is used so o
ften, it is no surprise that Maple has in already programed. We find t
he Laplacian in the " }{TEXT 266 15 "Vector Calculus" }{TEXT -1 9 " pa
ckage." }}{PARA 0 "" 0 "" {TEXT -1 49 " We read in the Vector Calc
ulus package here." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "with(V
ectorCalculus):" }}}{PARA 0 "" 0 "" {TEXT -1 111 "Now, compute the Lap
lacian of the function u1 obtained above. It is a complicated mess tha
t Maple can simplify." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "Lap
lacian(u1(x,y),[x,y]);\nsimplify(%);" }}}{PARA 0 "" 0 "" {TEXT -1 23 "
We repeat this with u2." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "L
aplacian(u2(x,y),[x,y]);\nsimplify(%);" }}}{PARA 0 "" 0 "" {TEXT -1
328 "The reader might wonder why we did not bring in the Vector Calcul
us package at the beginning of this Section as is our habit. The reaso
n is that some notational changes occur. It did not seem appropriate t
o make these notational changes at this point in our developement. To \+
see the character of the changes, contrast these two." }}{EXCHG {PARA
0 "> " 0 "" {MPLTEXT 1 0 10 "f:=x->2*x;" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 6 "2 . x;" }}}{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 50 "EMAIL: herod@math.gatech.edu or jhero
d@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
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