{VERSION 5 0 "IBM INTEL NT" "5.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 1 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 1 }{CSTYLE "2D Comment" 2 18 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 } {PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Heading 1" -1 3 1 {CSTYLE "" -1 -1 "Times" 1 18 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 8 4 1 0 1 0 2 2 0 1 }{PSTYLE "Normal" -1 256 1 {CSTYLE "" -1 -1 "Times" 1 18 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }3 1 0 0 0 0 1 0 1 0 2 2 0 1 }} {SECT 0 {PARA 256 "" 0 "" {TEXT -1 54 "Module 31: Laplace's Equation o n a Ring or a Half Disk" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 40 " Suppose that u is the solution for " }{XPPEDIT 18 0 "Delta;" "6#%&DeltaG" }{TEXT -1 27 " u = 0 on any region. Take " } {XPPEDIT 18 0 "p[0];" "6#&%\"pG6#\"\"!" }{TEXT -1 77 " to be any poin t on the interior of the region. Take C to be a circle about " } {XPPEDIT 18 0 "p[0];" "6#&%\"pG6#\"\"!" }{TEXT -1 116 " which is smal l enough that the circle lies in the region. We draw the picture as th ough the region were a square, " }{XPPEDIT 18 0 "p[0];" "6#&%\"pG6#\" \"!" }{TEXT -1 57 " were in the 3rd quadrant and C is a small circle with " }{XPPEDIT 18 0 "p[0];" "6#&%\"pG6#\"\"!" }{TEXT -1 39 " as its center and lying in the square." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 152 "plot([[t,-1,t=-1..1],[t,1,t=-1..1],[-1,t,t=-1..1],[1,t,t=-1.. 1],\n [-1/2+cos(t)/4,1/2+sin(t)/4,t=-Pi..Pi]],\n color=blac k,scaling=constrained);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 " " }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 52 " \+ Agree that the integral of u about that circle " }}{PARA 0 "" 0 "" {TEXT -1 8 " " }{XPPEDIT 18 0 "1/(2*Pi);" "6#*&\"\"\"F$*&\"\"#F $%#PiGF$!\"\"" }{TEXT -1 2 " " }{XPPEDIT 18 0 "int(u(x(t),y(t)),t = - Pi .. Pi);" "6#-%$intG6$-%\"uG6$-%\"xG6#%\"tG-%\"yG6#F,/F,;,$%#PiG!\" \"F3" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 88 "would give the ave rage value of u on that circle. We show below that the value of u at \+ " }{XPPEDIT 18 0 "p[0];" "6#&%\"pG6#\"\"!" }{TEXT -1 19 " is this inte gral. " }}{PARA 0 "" 0 "" {TEXT -1 92 " You ask what is the signif icance of this result? It would imply that the value of u at " } {XPPEDIT 18 0 "p[0];" "6#&%\"pG6#\"\"!" }{TEXT -1 55 " is the average \+ of the values of u on any circle about " }{XPPEDIT 18 0 "p[0];" "6#&% \"pG6#\"\"!" }{TEXT -1 133 " which lies in the region on which u is de fined. This gives an understanding of the Maximum Principle. How could u have a maximum at " }{XPPEDIT 18 0 "p[0];" "6#&%\"pG6#\"\"!" } {TEXT -1 46 " if it is the average of values all around it?" }}{PARA 0 "" 0 "" {TEXT -1 46 " The idea generalizes to three dimensions. " }}{PARA 0 "" 0 "" {TEXT -1 456 " Think of a cube with specified, unchanging temperature, perhaps different on each face. The temperatu re at each point on the interior will be exactly the average of the su rrounding temperatures. This seems to be an interesting way to conceiv e the construction of the heat distribution. One might think it would \+ not be possible to make such a distributions -- each point has the ave rage value property --knowing only the temperatures on the boundaries. " }}{PARA 0 "" 0 "" {TEXT -1 105 " That it is possible is surely a tribute to our predecessors: Fourier, Cauchy, Laplace, Poisson, etc. " }}{PARA 0 "" 0 "" {TEXT -1 53 " Here is verification of this pro perty: Think of " }{XPPEDIT 18 0 "p[0];" "6#&%\"pG6#\"\"!" }{TEXT -1 65 " as being the center of a circle with radius r. Then u satisfies \+ " }{XPPEDIT 18 0 "Delta;" "6#%&DeltaG" }{TEXT -1 74 "u = 0 in this cir cle and defines boundary conditions on the circle. Then, " }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 5 " " }{XPPEDIT 18 0 "1/(2*Pi);" "6#*&\"\"\"F$*&\"\"#F$%#PiGF$!\"\"" }{TEXT -1 1 " " } {XPPEDIT 18 0 "int(u(r,theta),theta = -Pi .. Pi);" "6#-%$intG6$-%\"uG6 $%\"rG%&thetaG/F*;,$%#PiG!\"\"F." }{TEXT -1 3 " = " }{XPPEDIT 18 0 "1/ (2*Pi);" "6#*&\"\"\"F$*&\"\"#F$%#PiGF$!\"\"" }{TEXT -1 1 " " } {XPPEDIT 18 0 "int(a[0],theta = -Pi .. Pi);" "6#-%$intG6$&%\"aG6#\"\"! /%&thetaG;,$%#PiG!\"\"F." }{TEXT -1 3 " + " }{XPPEDIT 18 0 "sum(a[n]*r ^n*int(cos(n*theta),theta = -Pi .. Pi),n);" "6#-%$sumG6$*(&%\"aG6#%\"n G\"\"\")%\"rGF*F+-%$intG6$-%$cosG6#*&F*F+%&thetaGF+/F5;,$%#PiG!\"\"F9F +F*" }{TEXT -1 3 " + " }{XPPEDIT 18 0 "sum(b[n]*r^n*int(sin(n*theta),t heta = -Pi .. Pi),n);" "6#-%$sumG6$*(&%\"bG6#%\"nG\"\"\")%\"rGF*F+-%$i ntG6$-%$sinG6#*&F*F+%&thetaGF+/F5;,$%#PiG!\"\"F9F+F*" }{TEXT -1 2 " . " }}{PARA 0 "" 0 "" {TEXT -1 39 " \+ = " }{XPPEDIT 18 0 "a[0];" "6#&%\"aG6#\"\"!" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 28 "Laplace 's Equation on a Ring" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 37 "We find a function u which satisfies " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 30 " \+ " }{XPPEDIT 18 0 "Delta;" "6#%&DeltaG" }{TEXT -1 14 " u = 0, \+ for 1" }{XPPEDIT 18 0 "` ` <= ` `;" "6#1%\"~GF$" }{TEXT -1 1 "r" } {XPPEDIT 18 0 "` ` <= ` `;" "6#1%\"~GF$" }{TEXT -1 6 "2, - " } {XPPEDIT 18 0 "Pi;" "6#%#PiG" }{XPPEDIT 18 0 "` ` <= ` `;" "6#1%\"~GF$ " }{XPPEDIT 18 0 "theta;" "6#%&thetaG" }{XPPEDIT 18 0 "` ` <= ` `;" "6 #1%\"~GF$" }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 1 "." }}{PARA 0 " " 0 "" {TEXT -1 4 "with" }}{PARA 0 "" 0 "" {TEXT -1 37 " \+ u(1," }{XPPEDIT 18 0 "theta" "6#%&thetaG" }{TEXT -1 8 ") = u(2," }{XPPEDIT 18 0 "theta" "6#%&thetaG" }{TEXT -1 8 ") = s in(" }{XPPEDIT 18 0 "theta" "6#%&thetaG" }{TEXT -1 9 ") for - " } {XPPEDIT 18 0 "Pi;" "6#%#PiG" }{XPPEDIT 18 0 "` ` <= ` `;" "6#1%\"~GF$ " }{XPPEDIT 18 0 "theta;" "6#%&thetaG" }{XPPEDIT 18 0 "` ` <= ` `;" "6 #1%\"~GF$" }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 1 "." }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 80 "We have seen that \+ the two differential equations associated with this system are" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 5 " " } {XPPEDIT 18 0 "theta;" "6#%&thetaG" }{TEXT -1 6 " '' + " }{XPPEDIT 18 0 "lambda^2*theta;" "6#*&%'lambdaG\"\"#%&thetaG\"\"\"" }{TEXT -1 6 " = 0, " }{XPPEDIT 18 0 "theta;" "6#%&thetaG" }{TEXT -1 4 " (- " } {XPPEDIT 18 0 "pi;" "6#%#piG" }{TEXT -1 4 ") = " }{XPPEDIT 18 0 "theta ;" "6#%&thetaG" }{TEXT -1 3 " ( " }{XPPEDIT 18 0 "pi;" "6#%#piG" } {TEXT -1 7 "), and " }{XPPEDIT 18 0 "theta;" "6#%&thetaG" }{TEXT -1 5 " '(- " }{XPPEDIT 18 0 "pi;" "6#%#piG" }{TEXT -1 4 ") = " }{XPPEDIT 18 0 "theta;" "6#%&thetaG" }{TEXT -1 4 " '( " }{XPPEDIT 18 0 "pi;" "6# %#piG" }{TEXT -1 2 ")." }}{PARA 0 "" 0 "" {TEXT -1 3 "and" }}{PARA 0 " " 0 "" {TEXT -1 18 " r( rR ') ' - " }{XPPEDIT 18 0 "lambda^2;" "6# *$%'lambdaG\"\"#" }{TEXT -1 7 " R = 0." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 77 "The solutions for the first equation, with eigenvalues and eigenfunctions are" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 5 " " }{XPPEDIT 18 0 "lambda;" "6# %'lambdaG" }{TEXT -1 9 " = 0, " }{XPPEDIT 18 0 "theta;" "6#%&thetaG " }{TEXT -1 5 " = 1." }}{PARA 0 "" 0 "" {TEXT -1 5 " " }{XPPEDIT 18 0 "-lambda^2;" "6#,$*$%'lambdaG\"\"#!\"\"" }{TEXT -1 3 " = " } {XPPEDIT 18 0 "-n^2;" "6#,$*$%\"nG\"\"#!\"\"" }{TEXT -1 5 ", " } {XPPEDIT 18 0 "Theta;" "6#%&ThetaG" }{TEXT -1 1 "(" }{XPPEDIT 18 0 "th eta;" "6#%&thetaG" }{TEXT -1 10 ") = cos(n " }{XPPEDIT 18 0 "theta;" " 6#%&thetaG" }{TEXT -1 13 ") or sin(n " }{XPPEDIT 18 0 "theta;" "6#%& thetaG" }{TEXT -1 2 ")." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 28 "The second equation is now " }{XPPEDIT 18 0 "r^2;" "6#*$%\"rG\"\"#" }{TEXT -1 16 " R '' + r R ' - " }{XPPEDIT 18 0 "n^2; " "6#*$%\"nG\"\"#" }{TEXT -1 6 "R = 0." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 81 "This has led to two familiar ordinary differential equations which have solutions" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 39 " R(r) = 1 and ln( r), " }{XPPEDIT 18 0 "Theta(theta);" "6#-%&ThetaG6#%&thetaG" } {TEXT -1 20 " = 1, in case n = 0," }}{PARA 0 "" 0 "" {TEXT -1 3 "and" }}{PARA 0 "" 0 "" {TEXT -1 21 " R(r) = " }{XPPEDIT 18 0 " r^n" "6#)%\"rG%\"nG" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "r^(-n)" "6#)% \"rG,$%\"nG!\"\"" }{TEXT -1 2 ", " }{XPPEDIT 18 0 "Theta(theta)" "6#-% &ThetaG6#%&thetaG" }{TEXT -1 9 " = sin(n " }{XPPEDIT 18 0 "theta" "6#% &thetaG" }{TEXT -1 11 ") and cos(n" }{XPPEDIT 18 0 "theta" "6#%&thetaG " }{TEXT -1 10 ") in case " }{XPPEDIT 18 0 "n <> 0;" "6#0%\"nG\"\"!" } {TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 20 "Combine these to get" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 10 " u(r ," }{XPPEDIT 18 0 "theta" "6#%&thetaG" }{TEXT -1 4 ") = " }{XPPEDIT 18 0 "a[0];" "6#&%\"aG6#\"\"!" }{TEXT -1 3 " + " }{XPPEDIT 18 0 "b[0]; " "6#&%\"bG6#\"\"!" }{TEXT -1 1 " " }{XPPEDIT 18 0 "ln(r);" "6#-%#lnG6 #%\"rG" }{TEXT -1 3 " + " }{XPPEDIT 18 0 "sum((a[p]*r^p+a[-p]*r^(-p))* cos(p*theta)+(b[p]*r^p+b[-p]*r^(-p))*sin(p*theta),p)" "6#-%$sumG6$,&*& ,&*&&%\"aG6#%\"pG\"\"\")%\"rGF-F.F.*&&F+6#,$F-!\"\"F.)F0,$F-F5F.F.F.-% $cosG6#*&F-F.%&thetaGF.F.F.*&,&*&&%\"bG6#F-F.)F0F-F.F.*&&FA6#,$F-F5F.) F0,$F-F5F.F.F.-%$sinG6#*&F-F.F " 0 "" {MPLTEXT 1 0 54 "solve(\{b[1]+b[ -1]=1,2*b[1]+1/2*b[-1]=1\},\{b[1],b[-1]\});" }}}{PARA 0 "" 0 "" {TEXT -1 18 "We can now make u." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "u:=(r,theta)->(r+2/r)/3*sin(theta);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "u(1,theta); u(2,theta);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "diff(r*diff(u(r,theta),r),r)/r+diff(u(r,theta),theta, theta)/r^2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "simplify(%); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(plots):" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 99 "cylinderplot([r,theta,u(r,th eta)],r=1..2,theta=0..2*Pi,\n orientation=[-5,70],axes=NORMAL );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 33 "Laplace's Equatio n on a Half Disk" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 28 "We suppose that u satisfies " }{XPPEDIT 18 0 "Delta;" "6# %&DeltaG" }{TEXT -1 48 " u = 0 for 0 < r < 1, with u( r, 0) = 0 = u(r , " }{XPPEDIT 18 0 "pi;" "6#%#piG" }{TEXT -1 10 "), u( 1, " } {XPPEDIT 18 0 "theta;" "6#%&thetaG" }{TEXT -1 9 " ) = 1. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 26 "We know what is t he PDE: " }}{PARA 0 "" 0 "" {TEXT -1 19 " " } {XPPEDIT 18 0 "diff(r*diff(v(r,theta),r),r)/r+diff(v(r,theta),`$`(thet a,2))/(r^2);" "6#,&*&-%%diffG6$*&%\"rG\"\"\"-F&6$-%\"vG6$F)%&thetaGF)F *F)F*F)!\"\"F**&-F&6$-F.6$F)F0-%\"$G6$F0\"\"#F**$F)F:F1F*" }{TEXT -1 8 " = 0, " }}{PARA 0 "" 0 "" {TEXT -1 83 "and that this leads to the ordinary differential equations with boundary conditions" }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 5 " " }{XPPEDIT 18 0 "Theta;" "6#%&ThetaG" }{TEXT -1 6 " '' + " }{XPPEDIT 18 0 "lambda ^2*Theta;" "6#*&%'lambdaG\"\"#%&ThetaG\"\"\"" }{TEXT -1 7 " = 0, " } {XPPEDIT 18 0 "Theta;" "6#%&ThetaG" }{TEXT -1 11 " (0) = 0, " } {XPPEDIT 18 0 "Theta;" "6#%&ThetaG" }{TEXT -1 1 "(" }{XPPEDIT 18 0 "Pi ;" "6#%#PiG" }{TEXT -1 5 ") = 0" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 17 " r( rR ') ' - " }{XPPEDIT 18 0 "lambda ^2;" "6#*$%'lambdaG\"\"#" }{TEXT -1 7 " R = 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 18 "It follows that " } {XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT -1 7 " = n, " }{XPPEDIT 18 0 "Theta;" "6#%&ThetaG" }{TEXT -1 1 "(" }{XPPEDIT 18 0 "theta;" "6# %&thetaG" }{TEXT -1 10 ") = sin(n " }{XPPEDIT 18 0 "theta;" "6#%&theta G" }{TEXT -1 14 "), and R(r) = " }{XPPEDIT 18 0 "r^n;" "6#)%\"rG%\"nG " }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 24 " General solutio n is" }}{PARA 0 "" 0 "" {TEXT -1 20 " u(r, " }{XPPEDIT 18 0 "theta;" "6#%&thetaG" }{TEXT -1 5 " ) = " }{XPPEDIT 18 0 "sum(a[n ]*r^n*sin(n*theta),n);" "6#-%$sumG6$*(&%\"aG6#%\"nG\"\"\")%\"rGF*F+-%$ sinG6#*&F*F+%&thetaGF+F+F*" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 31 "The requirement that 1 = u( 1, " }{XPPEDIT 18 0 "theta;" "6#%&t hetaG" }{TEXT -1 5 " ) = " }{XPPEDIT 18 0 "sum(a[n]*sin(n*theta),n);" "6#-%$sumG6$*&&%\"aG6#%\"nG\"\"\"-%$sinG6#*&F*F+%&thetaGF+F+F*" } {TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 21 "We computer the a 's." } }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "restart:\nwith(plots):" }}} {PARA 0 "" 0 "" {TEXT -1 100 "We know how to compute the a 's in the a bove to get a Fourier sine series for the function 1 on (0, " } {XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 59 " ). Since each term of the sum will be zero at zero and at " }{XPPEDIT 18 0 "Pi;" "6#%#PiG" } {TEXT -1 101 ", we expect the convergence to be NOT uniform. In fact, \+ it is quite slow. Look at these calculations." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 91 "assume(n,integer):\nint(1*sin(n*theta),theta=0.. Pi)/int(sin(n*theta)^2,theta=0..Pi);\nn:='n';" }}}{PARA 0 "" 0 "" {TEXT -1 92 "Thus, we compute 40 coefficients -- more than usual. Ther e are so many, we don't print them." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 126 "for n from 1 to 40 do\n a[n]:=int(1*sin(n*theta) ,theta=0..Pi)/\n int(sin(n*theta)^2,theta=0..Pi):\nod: \nn:='n':" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "plot(sum(a[n]*sin(n*t),n=1..40),t=0 ..Pi);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "u:=(r,theta)->sum (a[n]*r^n*sin(n*theta),n=1..40);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 98 "cylinderplot([r,theta,u(r,theta)],r=0..1,theta=0..Pi, \n orientation=[-30,65],axes=NORMAL);" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 0 "" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 " " }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{MARK "0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }