{VERSION 3 0 "APPLE_PPC_MAC" "3.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 }{CSTYLE "2D Comment" 2 18 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 258 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 }{CSTYLE "" -1 259 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 }{CSTYLE "" -1 260 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 }{CSTYLE "" -1 261 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 } {CSTYLE "" -1 262 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 }{CSTYLE "" -1 263 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 }{CSTYLE "" -1 264 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 }{CSTYLE "" -1 266 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 267 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 268 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Heading 1" 0 3 1 {CSTYLE "" -1 -1 "" 1 18 0 0 0 0 0 1 0 0 0 0 0 0 0 }1 0 0 0 8 4 0 0 0 0 0 0 -1 0 }{PSTYLE "Heading 2" 3 4 1 {CSTYLE "" -1 -1 "" 1 14 0 0 0 0 0 0 0 0 0 0 0 0 0 }0 0 0 -1 8 2 0 0 0 0 0 0 -1 0 }{PSTYLE "Title" 0 18 1 {CSTYLE "" -1 -1 "" 1 18 0 0 0 0 0 1 1 0 0 0 0 0 0 }3 0 0 -1 12 12 0 0 0 0 0 0 19 0 }{PSTYLE "" 4 259 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 260 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {SECT 0 {PARA 18 "" 0 "" {TEXT -1 37 "Linear Methods of Applied Mathem atics" }}{PARA 259 "" 0 "" {TEXT -1 12 "Warm Spheres" }}{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 21 "Spherical coordinates" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 92 "Another important coordinate sys tem is that of a sphere. In that coordinate system we have " }}{PARA 0 "" 0 "" {TEXT -1 5 " " }{XPPEDIT 18 0 "rho;" "6#%$rhoG" }{TEXT -1 34 " is the distance from the origin," }}{PARA 0 "" 0 "" {TEXT -1 5 " " }{XPPEDIT 18 0 "theta;" "6#%&thetaG" }{TEXT -1 66 " is the \+ angle in the x - y plane; that is, it measures longitude." }}{PARA 0 " " 0 "" {TEXT -1 5 " " }{XPPEDIT 18 0 "phi;" "6#%$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#%$rhoG" } {TEXT -1 5 " sin(" }{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 8 " ) si n( " }{XPPEDIT 18 0 "theta;" "6#%&thetaG" }{TEXT -1 2 " )" }}{PARA 0 " " 0 "" {TEXT -1 9 " z = " }{XPPEDIT 18 0 "rho;" "6#%$rhoG" }{TEXT -1 6 " cos( " }{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 2 " )" }} {PARA 0 "" 0 "" {TEXT -1 1 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 84 "plot3d(1,t heta=0..2*Pi,phi=0..Pi, coords=spherical, style=wireframe,axes=NORM AL);" }}}{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 77 " is a half ring runn ing from the north pole to the south pole of the sphere." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 95 "plot3d([r,Pi/4,phi],r = 1..1.2,phi= 0..Pi, coords=spherical,axes=NORMAL,orientation=[15,75]);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{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 "" {MPLTEXT 1 0 0 "" }}}{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#*&\"\"\"\"\"\"*$%$rh oG\"\"#!\"\"" }{TEXT -1 2 " " }{TEXT 267 1 "\{" }{TEXT -1 2 " " } {XPPEDIT 18 0 "diff(rho^2*diff(u,rho),rho);" "6#-%%diffG6$*&%$rhoG\"\" #-F$6$%\"uGF'\"\"\"F'" }{TEXT -1 5 " + " }{XPPEDIT 18 0 "1/sin(phi); " "6#*&\"\"\"\"\"\"-%$sinG6#%$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#*&\"\"\"\"\"\"*$-%$sinG6#%$phiG\"\"#!\"\"" }{TEXT -1 1 " " } {XPPEDIT 18 0 "diff(u,`$`(theta,2));" "6#-%%diffG6$%\"uG-%\"$G6$%&thet aG\"\"#" }{TEXT -1 1 " " }{TEXT 266 1 "\}" }{TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 15 "We solve 0 = " }{XPPEDIT 18 0 "Delta;" "6#%&DeltaG" } {TEXT -1 20 " u, u(1, phi) = f( " }{XPPEDIT 18 0 "phi;" "6#%$phiG" } {TEXT -1 30 " ), where u is independent of " }{XPPEDIT 18 0 "theta;" " 6#%&thetaG" }{TEXT -1 28 " . From the assumption 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#%$phiG" }{TEXT -1 3 " , " }}{PARA 0 "" 0 "" {TEXT -1 54 "we are led to this separation of variables situation: " } }{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 12 " \+ ( " }{XPPEDIT 18 0 "rho^2;" "6#*$%$rhoG\"\"#" }{TEXT -1 6 " R '( " } {XPPEDIT 18 0 "rho;" "6#%$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#%$ph iG" }{TEXT -1 2 ") " }{XPPEDIT 18 0 "Phi;" "6#%$PhiG" }{TEXT -1 6 " = 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 49 "Th is suggests the ordinary differential equations" }}{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 "r ho;" "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 -1 117 "Neither equation ha s a boundary condition. We have conditions of boundness. In the second equation, we take x = cos( " }{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 36 " ) changing the equation as follows:" }}{PARA 0 "" 0 "" {TEXT -1 5 " " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "x:=phi->cos(phi);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "Phi:=phi->y(x(phi));" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "diff(sin(phi)*diff(Phi(phi), phi),phi);" }}}{PARA 0 "" 0 "" {TEXT -1 15 "Hence, ( sin( " } {XPPEDIT 18 0 "phi;" "6#%$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#%$phiG" }{TEXT -1 10 " ) y '(x)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 46 "The second differential equation a bove becomes" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 5 " " }{XPPEDIT 18 0 "sin(phi)^2;" "6#*$-%$sinG6#%$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 exami ne solutions of this differential equation." }}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 8 "restart;" }}}{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 Polynomials. We digress to recall these func tions." }}}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{SECT 1 {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 diffe rential 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 54 "with(orthopoly);\nplot([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 satisfies 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*di ff(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 Schmid t Process to 1, x, " }{XPPEDIT 18 0 "x^2;" "6#*$%\"xG\"\"#" }{TEXT -1 7 ", ... ." }}{PARA 0 "" 0 "" {TEXT -1 98 "We examined these ideas ear lier. Here, we illustrate that the Legendre polynomials are orthogonal ." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "int(P(3,x)*P(5,x),x=-1. .1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 260 9 "Method 3." }{TEXT -1 64 " We could generate these by taking the appropriate derivatives." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 16 " We s how that " }{XPPEDIT 18 0 "1/(2^n*n!);" "6#*&\"\"\"\"\"\"*&)\"\"#%\"nG F%-%*factorialG6#F)F%!\"\"" }{TEXT -1 2 " " }{XPPEDIT 18 0 "d^n*(x^2- 1)^n/(d*x^n);" "6#*()%\"dG%\"nG\"\"\"),&*$%\"xG\"\"#F'\"\"\"!\"\"F&F'* &F%F')F+F&F'F." }{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$n);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "for n from 1 to 3 do\n expand(Q(n,x)); P(n,x);\nod; " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 261 9 "Method 3:" }{TEXT -1 35 " Use the recursion formulas (n+1) " }{XPPEDIT 18 0 "P[n+1](x);" "6#-& %\"PG6#,&%\"nG\"\"\"\"\"\"F)6#%\"xG" }{TEXT -1 8 " + n " }{XPPEDIT 18 0 "P[n-1](x);" "6#-&%\"PG6#,&%\"nG\"\"\"\"\"\"!\"\"6#%\"xG" }{TEXT -1 15 " = (2 n + 1) " }{XPPEDIT 18 0 "P[n](x);" "6#-&%\"PG6#%\"nG6#% \"xG" }{TEXT -1 4 " x." }}{PARA 0 "" 0 "" {TEXT -1 51 "With this meth od, we assume you know the first two." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 110 "P3[0]:=1;\nP3[1]:=x;\nf or n from 1 to 3 do\n P3[n+1]:=expand(1/(n+1)*((2*n+1)*P3[n]*x-n*P3 [n-1]));\nod;\nn:='n';" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "f or n from 0 to 4 do\n P3[n]/P(n,x);\nod;\nn:='n';" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT 262 14 "Observation 1:" }{TEXT -1 22 " if 0 < m < n, then " }{XPPEDIT 18 0 "int(x^m*P(n,x),x = -1 .. 1);" "6#-%$intG 6$*&)%\"xG%\"mG\"\"\"-%\"PG6$%\"nGF(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$%\"nG %\"xG\"\"#/F+;,$\"\"\"!\"\"\"\"\"" }{TEXT -1 16 " = 2/(2 n + 1)." }} {PARA 0 "" 0 "" {TEXT -1 21 "We check three cases." }}{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 approximations f or functions on [-1, 1]. We approximate cos( " }{XPPEDIT 18 0 "Pi/2;" "6#*&%#PiG\"\"\"\"\"#!\"\"" }{TEXT -1 47 " x ) with the first three Le gendre 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;\n od; \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 disco urse on Legendre Polynomials. We return to the partial differential eq uation." }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 33 "Separation of spherical variables" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 99 "Recall where we were. \+ The partial differential equation led to two ordinary differential equ ations." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 40 "The second 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$F%\"\"\"F%F%" }{TEXT -1 23 " y(x) = 0, -1 < x < 1" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 25 "and Legendre Polynomials." }}{PARA 0 "" 0 "" {TEXT -1 70 "Here is the first of the differential equations from abov e, with this " }{XPPEDIT 18 0 "mu;" "6#%#muG" }{TEXT -1 2 " :" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 12 " \+ ( " }{XPPEDIT 18 0 "rho^2;" "6#*$%$rhoG\"\"#" }{TEXT -1 6 " R '( " } {XPPEDIT 18 0 "rho;" "6#%$rhoG" }{TEXT -1 9 ") ) ' - " }{XPPEDIT 18 0 "n*(n+1);" "6#*&%\"nG\"\"\",&F$F%\"\"\"F%F%" }{TEXT -1 4 " R( " } {XPPEDIT 18 0 "rho;" "6#%$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$F%\"\"\"F%F%" }{TEXT -1 4 " R( " }{XPPEDIT 18 0 "rho;" "6#%$ rhoG" }{TEXT -1 12 ") = 0, 0 < " }{XPPEDIT 18 0 "rho;" "6#%$rhoG" } {TEXT -1 5 " < 1." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 14 "We solve this." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "dsolve(r^2*diff(R(r),r,r)+2*r*diff(R(r),r)-n*(n+1)*R(r)=0,R(r));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 43 "We see that the only bounded solutions are " }{XPPEDIT 18 0 "r^n;" "6#)%\"rG%\"nG" }{TEXT -1 94 " . Hence, we are ready to ma ke up the general solution for the partial differential eqeuation." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 260 "" 0 "" {TEXT -1 17 "General Solution:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 58 "First, we verify that products of solutions are solutions." }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "with(orthopoly);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "n:=4;\nu:=(r,phi)->r^n*P(n,cos(phi) );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 95 "diff(r^2*diff(u(r,phi ),r),r)+\n 1/sin(phi)*diff(sin(phi)*diff(u(r,phi),phi),phi):\nsimpli fy(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 73 "I checked this with n = 4. You check it at other place s. What about sums?" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "n:='n ';\nu:=(r,phi)->sum(a[n]*r^n*P(n,cos(phi)),n=0..4);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 95 "diff(r^2*diff(u(r,phi),r),r)+\n 1/sin(p hi)*diff(sin(phi)*diff(u(r,phi),phi),phi):\nsimplify(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{PARA 0 "" 0 "" {TEXT -1 79 "We are now ready to compute the coeffici ents for a boundary condition. We solve" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 14 " 0 = " }{XPPEDIT 18 0 "Delt a;" "6#%&DeltaG" }{TEXT -1 12 "u with 0 < " }{XPPEDIT 18 0 "rho;" "6# %$rhoG" }{TEXT -1 11 " < 1, 0 < " }{XPPEDIT 18 0 "phi;" "6#%$phiG" } {TEXT -1 3 " < " }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 6 ", 0 < " } {XPPEDIT 18 0 "theta;" "6#%&thetaG" }{TEXT -1 5 " < 2 " }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 29 " with boundary conditon u(1, " } {XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 7 " ) = f(" }{XPPEDIT 18 0 " phi;" "6#%$phiG" }{TEXT -1 3 " )." }}{PARA 0 "" 0 "" {TEXT -1 24 "The \+ coefficients will be" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 10 " " }{XPPEDIT 18 0 "a[n];" "6#&%\"aG6#%\"nG" } {TEXT -1 3 " = " }{XPPEDIT 18 0 "(2*n+1)/2;" "6#*&,&*&\"\"#\"\"\"%\"nG F'F'\"\"\"F'F'\"\"#!\"\"" }{TEXT -1 2 " " }{XPPEDIT 18 0 "int(f(phi)* P(n,cos(phi))*sin(phi),phi = 0 .. Pi);" "6#-%$intG6$*(-%\"fG6#%$phiG\" \"\"-%\"PG6$%\"nG-%$cosG6#F*F+-%$sinG6#F*F+/F*;\"\"!%#PiG" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 36 "L et's take the special case that f( " }{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 10 " ) = cos( " }{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 3 " )." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 93 "for n from 0 to 4 \+ do\n a[n]:=(2*n+1)/2*int(cos(phi)*P(n,cos(phi))*sin(phi),phi=0..Pi );\nod;" }}}{PARA 0 "" 0 "" {TEXT -1 49 "Thus, the solution for this p roblem is\n u(r, " }{XPPEDIT 18 0 "phi;" "6#%$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 illustr ate what we have?" }}{PARA 0 "" 0 "" {TEXT -1 69 "(1) Each cross secti onal plane parallel to the x-y plane has value z." }}{PARA 0 "" 0 "" {TEXT -1 31 "(2) If we hold r fixed and let " }{XPPEDIT 18 0 "phi;" "6 #%$phiG" }{TEXT -1 14 " go from 0 to " }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 30 ", we get an interesting graph." }}{PARA 0 "" 0 "" {TEXT -1 30 "We give a plot for values of u" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(plots):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "u:=(r,phi)->r*P(1,cos(phi));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 73 "plot([u(1,phi),u(1/2,phi),u(1/4,phi)],phi=0..Pi,color =[black,red,green]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }} }{PARA 0 "" 0 "" {TEXT -1 53 "Maybe this gives an understanding for th e solution u." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 8 "Exercise" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 25 "Repeat \+ the analysis if f(" }{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 5 " ) = " }{XPPEDIT 18 0 "cos(phi)^2;" "6#*$-%$cosG6#%$phiG\"\"#" }{TEXT -1 2 " ." }}}}{PARA 0 "" 0 "" {TEXT 268 0 "" }}}}{MARK "9" 0 }{VIEWOPTS 1 1 0 1 1 1803 }