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{SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT -1 40 "Partial Differential Equa
tions PowerTool" }}{PARA 19 "" 0 "" {TEXT -1 16 "by Dr. Jim Herod" }}}
{PARA 256 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT 265 25 "Sect
ion 7.5: Warm Spheres" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 
{PARA 3 "" 0 "" {TEXT 266 36 "Maple Packages needed in Section 7.5" }}
{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 17 "with(orthopoly);\n" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 92 "Another important coordinate system is that of a sphere. \+
 In that coordinate system we have " }}{PARA 0 "" 0 "" {TEXT -1 5 "   \+
  " }{XPPEDIT 18 0 "rho;" "6#%$rhoG" }{TEXT -1 34 "  is the distance f
rom the origin," }}{PARA 0 "" 0 "" {TEXT -1 5 "     " }{XPPEDIT 18 0 "
theta;" "6#%&thetaG" }{TEXT -1 66 "  is the angle in the x - y plane; \+
that is, it measures longitude." }}{PARA 0 "" 0 "" {TEXT -1 5 "     " 
}{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 59 "  is the angle from the
 top; that is, it measures latitude." }}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}}{PARA 0 "" 0 "" {TEXT -1 6 "Thus, " }{XPPEDIT 18 0 "rho;" "6#%$rhoG
" }{TEXT -1 12 " = 1,   0 < " }{XPPEDIT 18 0 "theta;" "6#%&thetaG" }
{TEXT -1 5 " < 2 " }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 9 ",    0 \+
< " }{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 3 " < " }{XPPEDIT 18 0 
"Pi;" "6#%#PiG" }{TEXT -1 30 "    is a sphere with radius 1." }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 46 "The connection \+
with rectangular coordinates is" }}{PARA 0 "" 0 "" {TEXT -1 9 "     x \+
= " }{XPPEDIT 18 0 "rho;" "6#%$rhoG" }{TEXT -1 5 " sin(" }{XPPEDIT 18 
0 "phi;" "6#%$phiG" }{TEXT -1 7 ") cos( " }{XPPEDIT 18 0 "theta;" "6#%
&thetaG" }{TEXT -1 2 " )" }}{PARA 0 "" 0 "" {TEXT -1 9 "     y = " }
{XPPEDIT 18 0 "rho;" "6#%$rhoG" }{TEXT -1 5 " sin(" }{XPPEDIT 18 0 "ph
i;" "6#%$phiG" }{TEXT -1 8 " ) sin( " }{XPPEDIT 18 0 "theta;" "6#%&the
taG" }{TEXT -1 2 " )" }}{PARA 0 "" 0 "" {TEXT -1 9 "     z = " }
{XPPEDIT 18 0 "rho;" "6#%$rhoG" }{TEXT -1 6 " cos( " }{XPPEDIT 18 0 "p
hi;" "6#%$phiG" }{TEXT -1 2 " )" }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 104 "plot3d(1,theta=0..2*Pi,phi=
0..Pi, coords=spherical,    style=wireframe,axes=NORMAL,orientation=[3
5,70]);" }}}{PARA 0 "" 0 "" {TEXT -1 6 "Also, " }{XPPEDIT 18 0 "rho;" 
"6#%$rhoG" }{TEXT -1 8 " = 1,   " }{XPPEDIT 18 0 "theta;" "6#%&thetaG
" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "Pi/4;" "6#*&%#PiG\"\"\"\"\"%!\"\"
" }{TEXT -1 9 " ,   0 < " }{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 
3 " < " }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 89 "  is a half ring \+
running from the north pole to the south pole of the sphere. I draw th
e " }{XPPEDIT 18 0 "rho;" "6#%$rhoG" }{TEXT -1 86 " fat so you can see
 it. You might experiment by making r just 1 to see what I avoided." }
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 130 "plot3d([r,Pi/4,phi],r = 1.
.1.2,phi=0..Pi, coords=spherical,axes=NORMAL,orientation=[15,75],style
=patchnogrid,scaling=constrained);" }}}{EXCHG }{PARA 0 "" 0 "" {TEXT 
-1 9 "Finally, " }{XPPEDIT 18 0 "rho;" "6#%$rhoG" }{TEXT -1 6 " = 1, \+
" }{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 95 " = 49 degrees, define
s a part of the boundary between Western Canada and Western United Sta
tes." }}{EXCHG }{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 52 "In this coordinate system, the Laplacian Operator is" }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 6 "      " }
{XPPEDIT 18 0 "Delta;" "6#%&DeltaG" }{TEXT -1 5 " u = " }{XPPEDIT 18 
0 "1/(rho^2);" "6#*&\"\"\"F$*$%$rhoG\"\"#!\"\"" }{TEXT -1 2 "  " }
{TEXT 257 1 "\{" }{TEXT -1 2 "  " }{XPPEDIT 18 0 "diff(rho^2*diff(u,rh
o),rho);" "6#-%%diffG6$*&%$rhoG\"\"#-F$6$%\"uGF'\"\"\"F'" }{TEXT -1 5 
"  +  " }{XPPEDIT 18 0 "1/sin(phi);" "6#*&\"\"\"F$-%$sinG6#%$phiG!\"\"
" }{TEXT -1 1 " " }{XPPEDIT 18 0 "diff(sin(phi)*diff(u,phi),phi);" "6#
-%%diffG6$*&-%$sinG6#%$phiG\"\"\"-F$6$%\"uGF*F+F*" }{TEXT -1 6 "   +  \+
" }{XPPEDIT 18 0 "1/(sin(phi)^2);" "6#*&\"\"\"F$*$-%$sinG6#%$phiG\"\"#
!\"\"" }{TEXT -1 1 " " }{XPPEDIT 18 0 "diff(u,`$`(theta,2));" "6#-%%di
ffG6$%\"uG-%\"$G6$%&thetaG\"\"#" }{TEXT -1 1 " " }{TEXT 256 1 "\}" }
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 15 "We solve   0 = " }{XPPEDIT 18 
0 "Delta;" "6#%&DeltaG" }{TEXT -1 20 " u,  u(1, phi) = f( " }{XPPEDIT 
18 0 "phi;" "6#%$phiG" }{TEXT -1 30 " ), where u is independent of " }
{XPPEDIT 18 0 "theta;" "6#%&thetaG" }{TEXT -1 28 " . From the assumpti
on that " }}{PARA 0 "" 0 "" {TEXT -1 13 "          u( " }{XPPEDIT 18 
0 "rho;" "6#%$rhoG" }{TEXT -1 3 " , " }{XPPEDIT 18 0 "phi;" "6#%$phiG
" }{TEXT -1 8 " ) = R( " }{XPPEDIT 18 0 "rho;" "6#%$rhoG" }{TEXT -1 3 
" ) " }{XPPEDIT 18 0 "Phi(phi);" "6#-%$PhiG6#%$phiG" }{TEXT -1 3 " , \+
" }}{PARA 0 "" 0 "" {TEXT -1 54 "we are led to this separation of vari
ables situation: " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 12 "          ( " }{XPPEDIT 18 0 "rho^2;" "6#*$%$rhoG\"\"#" }
{TEXT -1 6 " R '( " }{XPPEDIT 18 0 "rho;" "6#%$rhoG" }{TEXT -1 21 ") )
 ' / R  +  ( sin( " }{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 3 " ) \+
" }{XPPEDIT 18 0 "Phi;" "6#%$PhiG" }{TEXT -1 12 " ' ) '/ sin(" }
{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 2 ") " }{XPPEDIT 18 0 "Phi;
" "6#%$PhiG" }{TEXT -1 6 "  = 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 49 "This suggests the ordinary differential e
quations" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
12 "          ( " }{XPPEDIT 18 0 "rho^2;" "6#*$%$rhoG\"\"#" }{TEXT -1 
6 " R '( " }{XPPEDIT 18 0 "rho;" "6#%$rhoG" }{TEXT -1 9 ") ) '  - " }
{XPPEDIT 18 0 "mu^2;" "6#*$%#muG\"\"#" }{TEXT -1 4 " R( " }{XPPEDIT 
18 0 "rho;" "6#%$rhoG" }{TEXT -1 12 ") = 0,  0 < " }{XPPEDIT 18 0 "rho
;" "6#%$rhoG" }{TEXT -1 5 " < 1," }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 17 "          ( sin( " }{XPPEDIT 18 0 "phi;" 
"6#%$phiG" }{TEXT -1 3 " ) " }{XPPEDIT 18 0 "Phi;" "6#%$PhiG" }{TEXT 
-1 11 " ' ) '  +  " }{XPPEDIT 18 0 "mu^2;" "6#*$%#muG\"\"#" }{TEXT -1 
6 " sin( " }{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 3 " ) " }
{XPPEDIT 18 0 "Phi;" "6#%$PhiG" }{TEXT -1 12 "  = 0,  0 < " }{XPPEDIT 
18 0 "phi;" "6#%$phiG" }{TEXT -1 3 " < " }{XPPEDIT 18 0 "Pi;" "6#%#PiG
" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 119 "Neither equation has a boundary condition. We have condi
tions of boundedness. In the second equation, we take x = cos( " }
{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 36 " ) changing the equation
 as follows:" }}{PARA 0 "" 0 "" {TEXT -1 5 "     " }}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 17 "x:=phi->cos(phi);" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 20 "Phi:=phi->y(x(phi));" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 38 "diff(sin(phi)*diff(Phi(phi),phi),phi);" }}}{PARA 0 "
" 0 "" {TEXT -1 15 "Hence,  ( sin( " }{XPPEDIT 18 0 "phi;" "6#%$phiG" 
}{TEXT -1 3 " ) " }{XPPEDIT 18 0 "Phi;" "6#%$PhiG" }{TEXT -1 11 " ' ) \+
'  =  " }{XPPEDIT 18 0 "sin(phi)^3;" "6#*$-%$sinG6#%$phiG\"\"$" }
{TEXT -1 19 " y ''(x)  -  2 sin(" }{XPPEDIT 18 0 "phi;" "6#%$phiG" }
{TEXT -1 7 ") cos( " }{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 10 " )
 y '(x)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
46 "The second differential equation above becomes" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 5 "     " }{XPPEDIT 18 0 "sin
(phi)^2;" "6#*$-%$sinG6#%$phiG\"\"#" }{TEXT -1 18 " y ''(x) - 2 cos( \+
" }{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 12 " ) y '(x) + " }
{XPPEDIT 18 0 "mu^2;" "6#*$%#muG\"\"#" }{TEXT -1 10 " y(x) = 0." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 21 "In terms \+
of x alone, " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 10 "     (1 - " }{XPPEDIT 18 0 "x^2;" "6#*$%\"xG\"\"#" }{TEXT -1 
26 ") y ''(x) - 2 x y '(x) +  " }{XPPEDIT 18 0 "mu^2;" "6#*$%#muG\"\"#
" }{TEXT -1 26 " y(x) = 0,   -1 < x < 1.  " }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 51 "We examine solutions of this di
fferential equation." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "x:='x
':" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "dsolve((1-x^2)*diff(y
(x),x,x)- 2*x*diff(y(x),x)+mu^2*y(x)=0,y(x));" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 100 "There is the suggestion that we consider Legendre P
olynomials. We digress to recall these functions." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 "
" {TEXT -1 38 "A recollection of Legendre Polynomials" }}{PARA 0 "" 0 
"" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 89 "Here are three ways to
 conceive of the Legendre Polynomials -- four, if we include Maple." }
}{PARA 0 "" 0 "" {TEXT 258 9 "Method 1:" }{TEXT -1 33 " solve the diff
erential equation." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 69 "dsolve
((1-x^2)*diff(y(x),x,x)- 2*x*diff(y(x),x)+n*(n+1)*y(x)=0,y(x));" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "plot([LegendreP(1,x),Legendr
eQ(1,x)],x=-1..1,color=[red,black]);" }}}{PARA 0 "" 0 "" {TEXT -1 116 
"We got only the first graph. The following shows that the second grap
hs are not defined on the interval of interest." }}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 69 "plot([LegendreQ(1,x),LegendreQ(2,x),LegendreQ(
3,x)],x=0..5,y=0..1/2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "
" }}}{PARA 0 "" 0 "" {TEXT -1 69 "Here are the first three Legendre Po
lynomials with an easier calling." }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 37 "plot([P(0,x),P(1,x),P(2,x)],x=-1..1);" }}}{PARA 0 "" 
0 "" {TEXT -1 83 "Here is a check that, for example, the 5th one satis
fies the differential equation." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 81 "y:=x->P(5,x);\n(1-x^2)*diff(y(x),x,x)- 2*x*diff(y(x),x)+5*(5+1
)*y(x);\nsimplify(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 259 9 "Method \+
2:" }{TEXT -1 51 " We could apply the Gramm Schmidt Process to 1, x, \+
" }{XPPEDIT 18 0 "x^2;" "6#*$%\"xG\"\"#" }{TEXT -1 7 ", ... ." }}
{PARA 0 "" 0 "" {TEXT -1 98 "We examined these ideas earlier. Here, we
 illustrate that the Legendre polynomials are orthogonal." }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "int(P(3,x)*P(5,x),x=-1..1);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT 260 9 "Method 3." }{TEXT -1 64 "  We c
ould generate these by taking the appropriate derivatives." }}{PARA 0 
"" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 16 "   We show that \+
" }{XPPEDIT 18 0 "1/(2^n*n!);" "6#*&\"\"\"F$*&)\"\"#%\"nGF$-%*factoria
lG6#F(F$!\"\"" }{TEXT -1 2 "  " }{XPPEDIT 18 0 "d^n*(x^2-1)^n/(d*x^n);
" "6#*()%\"dG%\"nG\"\"\"),&*$%\"xG\"\"#F'F'!\"\"F&F'*&F%F')F+F&F'F-" }
{TEXT -1 9 "  is the " }{XPPEDIT 18 0 "n^th;" "6#)%\"nG%#thG" }{TEXT 
-1 12 " polynomial." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 59 "f:=(n,x)->(x^2-1)^n;\nQ:=(n,x)->1/(2^n*n!)*
diff(f(n,x),x$n);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 57 "for n \+
from 1 to 3 do\n   expand(Q(n,x)),`and`, P(n,x);\nod;" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT 261 9 "Method 4:" }{TEXT -1 35 " Use the recurs
ion formulas  (n+1) " }{XPPEDIT 18 0 "P[n+1](x);" "6#-&%\"PG6#,&%\"nG
\"\"\"F)F)6#%\"xG" }{TEXT -1 8 "   +  n " }{XPPEDIT 18 0 "P[n-1](x);" 
"6#-&%\"PG6#,&%\"nG\"\"\"F)!\"\"6#%\"xG" }{TEXT -1 15 "  =  (2 n + 1) \+
" }{XPPEDIT 18 0 "P[n](x);" "6#-&%\"PG6#%\"nG6#%\"xG" }{TEXT -1 4 "  x
." }}{PARA 0 "" 0 "" {TEXT -1 51 "With this method, we assume you know
 the first two." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 110 "P3[0]:=1;\nP3[1]:=x;\nfor n from 1 to 3 do\n \+
   P3[n+1]:=expand(1/(n+1)*((2*n+1)*P3[n]*x-n*P3[n-1]));\nod;\nn:='n';
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "for n from 0 to 4 do\n \+
  P3[n]/P(n,x);\nod;\nn:='n';" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 262 
14 "Observation 1:" }{TEXT -1 22 "  if 0 < m < n, then  " }{XPPEDIT 
18 0 "int(x^m*P(n,x),x = -1 .. 1);" "6#-%$intG6$*&)%\"xG%\"mG\"\"\"-%
\"PG6$%\"nGF(F*/F(;,$F*!\"\"F*" }{TEXT -1 40 "  = 0. We check this for
 a special case." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "int(x^3*
P(5,x),x=-1..1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{PARA 0 "" 0 "" {TEXT 263 14 "Observation 2:" }{TEXT -1 45 "  We have \+
a formula for the norm of P(n,x):  " }{XPPEDIT 18 0 "int(P(n,x)^2,x = \+
-1 .. 1);" "6#-%$intG6$*$-%\"PG6$%\"nG%\"xG\"\"#/F+;,$\"\"\"!\"\"F0" }
{TEXT -1 16 "  = 2/(2 n + 1)." }}{PARA 0 "" 0 "" {TEXT -1 64 "We check
 three cases by comparing the integral with 2/(2 n + 1)." }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "for n from 0 to 5 do\n   int(P(n,x)
^2,x=-1..1)-2/(2*n+1);\nod;\nn:='n';" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT 264 14 "Observation 3:" }{TEXT -1 85 " We can make polynomial ap
proximations for functions on [-1, 1]. We approximate cos( " }
{XPPEDIT 18 0 "Pi/2;" "6#*&%#PiG\"\"\"\"\"#!\"\"" }{TEXT -1 47 " x ) w
ith the first three Legendre polynomials." }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 85 "for n from 0 to 2 do\n   a[n]:=int(cos(Pi/2*x)*P(n,x)
,x=-1..1)*(2*n+1)/2;\nod; \nn:='n';" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 81 "plot([a[0]+a[1]*P(1,x)+a[2]*P(2,x),cos(Pi/2*x)],x=-1.
.1,\n     color=[black,red]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
101 "This completes our discourse on Legendre Polynomials. We return t
o the partial differential equation." }}}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 99 "Recall where we were. The partial dif
ferential equation led to two ordinary differential equations." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 40 "The secon
d differential equation  became" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 5 "     " }{XPPEDIT 18 0 "sin(phi)^2;" "6#*$-
%$sinG6#%$phiG\"\"#" }{TEXT -1 18 " y ''(x) - 2 cos( " }{XPPEDIT 18 0 
"phi;" "6#%$phiG" }{TEXT -1 12 " ) y '(x) + " }{XPPEDIT 18 0 "mu^2;" "
6#*$%#muG\"\"#" }{TEXT -1 10 " y(x) = 0." }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 21 "In terms of x alone, " }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 10 "     (1 - " }
{XPPEDIT 18 0 "x^2;" "6#*$%\"xG\"\"#" }{TEXT -1 26 ") y ''(x) - 2 x y \+
'(x) +  " }{XPPEDIT 18 0 "mu^2;" "6#*$%#muG\"\"#" }{TEXT -1 26 " y(x) \+
= 0,   -1 < x < 1.  " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 12 "This led to " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 15 "          (1 - " }{XPPEDIT 18 0 "x^2;" "6#*$%\"
xG\"\"#" }{TEXT -1 26 ") y ''(x) - 2 x y '(x) +  " }{XPPEDIT 18 0 "n*(
n+1);" "6#*&%\"nG\"\"\",&F$F%F%F%F%" }{TEXT -1 23 " y(x) = 0,   -1 < x
 < 1" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 25 "a
nd Legendre Polynomials." }}{PARA 0 "" 0 "" {TEXT -1 70 "Here is the f
irst of the differential equations from above, with this " }{XPPEDIT 
18 0 "mu;" "6#%#muG" }{TEXT -1 2 " :" }}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}}{PARA 0 "" 0 "" {TEXT -1 12 "          ( " }{XPPEDIT 18 0 "rho^2;" "
6#*$%$rhoG\"\"#" }{TEXT -1 6 " R '( " }{XPPEDIT 18 0 "rho;" "6#%$rhoG
" }{TEXT -1 9 ") ) '  - " }{XPPEDIT 18 0 "n*(n+1);" "6#*&%\"nG\"\"\",&
F$F%F%F%F%" }{TEXT -1 4 " R( " }{XPPEDIT 18 0 "rho;" "6#%$rhoG" }
{TEXT -1 12 ") = 0,  0 < " }{XPPEDIT 18 0 "rho;" "6#%$rhoG" }{TEXT -1 
5 " < 1," }}{PARA 0 "" 0 "" {TEXT -1 2 "or" }}{PARA 0 "" 0 "" {TEXT 
-1 10 "          " }{XPPEDIT 18 0 "rho^2;" "6#*$%$rhoG\"\"#" }{TEXT 
-1 10 " R '' + 2 " }{XPPEDIT 18 0 "rho;" "6#%$rhoG" }{TEXT -1 7 " R ' \+
- " }{XPPEDIT 18 0 "n*(n+1);" "6#*&%\"nG\"\"\",&F$F%F%F%F%" }{TEXT -1 
4 " R( " }{XPPEDIT 18 0 "rho;" "6#%$rhoG" }{TEXT -1 12 ") = 0,  0 < " 
}{XPPEDIT 18 0 "rho;" "6#%$rhoG" }{TEXT -1 5 " < 1." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 14 "We solve this." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "dsolve(r^2*diff(R(r),r,r)+2*
r*diff(R(r),r)-n*(n+1)*R(r)=0,R(r));" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 43 "We see that the only
 bounded solutions are " }{XPPEDIT 18 0 "r^n;" "6#)%\"rG%\"nG" }{TEXT 
-1 93 " . Hence, we are ready to make up the general solution for the \+
partial differential equation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 257 "" 0 "" {TEXT -1 17 "General Solution:" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 58 "First, we verify that pro
ducts of solutions are solutions." }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 36 "n:=4;\nu:=(r,phi)->r^n*P(n,cos(phi));" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 95 "diff(r^2*diff(u(r,phi),r),r)+\n   1
/sin(phi)*diff(sin(phi)*diff(u(r,phi),phi),phi):\nsimplify(%);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT 
-1 73 "I checked this with n = 4. You check it at other places. What a
bout sums?" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "n:='n';\nu:=(r
,phi)->sum(a[n]*r^n*P(n,cos(phi)),n=0..4);" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 95 "diff(r^2*diff(u(r,phi),r),r)+\n   1/sin(phi)*diff(s
in(phi)*diff(u(r,phi),phi),phi):\nsimplify(%);" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 79 "We are now ready to compute the coefficients for a
 boundary condition. We solve" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 14 "          0 = " }{XPPEDIT 18 0 "Delta;" "
6#%&DeltaG" }{TEXT -1 12 "u  with 0 < " }{XPPEDIT 18 0 "rho;" "6#%$rho
G" }{TEXT -1 11 " < 1,  0 < " }{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT 
-1 3 " < " }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 6 ", 0 < " }
{XPPEDIT 18 0 "theta;" "6#%&thetaG" }{TEXT -1 5 " < 2 " }{XPPEDIT 18 
0 "Pi;" "6#%#PiG" }{TEXT -1 30 " with boundary condition u(1, " }
{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 7 " ) = f(" }{XPPEDIT 18 0 "
phi;" "6#%$phiG" }{TEXT -1 3 " )." }}{PARA 0 "" 0 "" {TEXT -1 24 "The \+
coefficients will be" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 10 "          " }{XPPEDIT 18 0 "a[n];" "6#&%\"aG6#%\"nG" }
{TEXT -1 3 " = " }{XPPEDIT 18 0 "(2*n+1)/2;" "6#*&,&*&\"\"#\"\"\"%\"nG
F'F'F'F'F'F&!\"\"" }{TEXT -1 2 "  " }{XPPEDIT 18 0 "int(f(phi)*P(n,cos
(phi))*sin(phi),phi = 0 .. Pi);" "6#-%$intG6$*(-%\"fG6#%$phiG\"\"\"-%
\"PG6$%\"nG-%$cosG6#F*F+-%$sinG6#F*F+/F*;\"\"!%#PiG" }{TEXT -1 1 "." }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 36 "Let's ta
ke the special case that f( " }{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT 
-1 10 " ) = cos( " }{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 3 " )." 
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 93 "for n from 0 to 4 do\n    \+
 a[n]:=(2*n+1)/2*int(cos(phi)*P(n,cos(phi))*sin(phi),phi=0..Pi);\nod;
" }}}{PARA 0 "" 0 "" {TEXT -1 49 "Thus, the solution for this problem \+
is\n     u(r, " }{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 16 " ) = r*
 P(1,cos(" }{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 11 ")) = r cos(
" }{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 2 ")." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 35 "How can we illustrate wha
t we have?" }}{PARA 0 "" 0 "" {TEXT -1 94 "(1) On each cross sectional
 plane parallel to the x-y plane, u has the same value, namely u(r," }
{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 86 ") = z. We draw a portion
 of the sphere where all the points have the same temperature." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 295 "addcoords(z_cylindrical,[z,
r,theta],[r*cos(theta),r*sin(theta),z]);\ndis:=plot3d(1/2,r=0..0.86,th
eta=0..2*Pi,coords=z_cylindrical, orientation=[35,70],axes=normal):\ns
ph:=plot3d(1,theta=0..2*Pi,phi=0..Pi, coords=spherical,    style=wiref
rame,axes=NORMAL,orientation=[35,70]):\ndisplay3d(\{dis,sph\});" }}}
{PARA 0 "" 0 "" {TEXT -1 54 "(2) If we hold r fixed between 0 and 1, a
nd if we let " }{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 14 " go from
 0 to " }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 128 ", we get imbedde
d spheres. Here are three of these where only half of the middle one i
s drawn so that the innermost can be seen." }}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 324 "S1:=plot3d(1,theta=0..2*Pi,phi=0..Pi, coords=spher
ical,    style=wireframe,axes=NORMAL,orientation=[-35,70]):\nS2:=plot3
d(3/4,theta=0..Pi,phi=0..Pi, coords=spherical,axes=NORMAL,orientation=
[-35,70]):\nS3:=plot3d(1/2,theta=0..2*Pi,phi=0..Pi, coords=spherical,a
xes=NORMAL,orientation=[-35,70],color=black):\ndisplay(\{S1,S2,S3\});
" }}}{PARA 0 "" 0 "" {TEXT -1 162 "The temperature on each such sphere
 varies from the hottest point at the top and the coldest at the botto
m. Here are graphs of u on the three embedded spheres as " }{XPPEDIT 
18 0 "phi;" "6#%$phiG" }{TEXT -1 19 " changes from 0 to " }{XPPEDIT 
18 0 "Pi;" "6#%#PiG" }{TEXT -1 43 " ... from the north pole to the sou
th pole." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "u:=(r,phi)->r*P(
1,cos(phi));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 73 "plot([u(1,p
hi),u(3/4,phi),u(1/2,phi)],phi=0..Pi,color=[black,red,green]);" }}}
{EXCHG }{EXCHG }{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 242 "In this Section, we have made analytic what we know intuitivel
y: if we know the temperature distribution on the surface of a sphere \+
which has no sources or sinks, then we can determine the temperature d
istribution on the inside of the sphere." }}{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 258 "" 0 "
" {TEXT -1 36 "Copyright \251  2003  by James V. Herod" }}{PARA 258 "
" 0 "" {TEXT -1 19 "All rights reserved" }}}{MARK "0 0 0" 0 }
{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 1 1 2 33 1 1 }