{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 } {CSTYLE "" -1 256 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 257 "" 1 14 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 258 "" 1 12 0 0 0 0 0 0 0 0 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 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 0 }1 0 0 0 8 4 0 0 0 0 0 0 -1 0 }{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 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Normal" -1 257 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 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 257 42 "Section 7.6: Vibrations of a C ircular Drum" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 " " {TEXT 258 34 "Maple Packages used in Section 7.6" }}{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 5 " " }}{PARA 0 "" 0 "" {TEXT -1 267 "A standard problem to be presented in a course on solving clas sical partial differential equations by the technique of separation of variables is that of a vibrating circular membrane. Here, we treat th e simple case in which the initial conditions are independent of " } {XPPEDIT 18 0 "theta;" "6#%&thetaG" }{TEXT -1 30 ". We recall the wav e equation" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 19 " " }{XPPEDIT 18 0 "diff(r*diff(u(t,r,theta),r ),r)/r+diff(u(t,r,theta),`$`(theta,2))/(r^2);" "6#,&*&-%%diffG6$*&%\"r G\"\"\"-F&6$-%\"uG6%%\"tGF)%&thetaGF)F*F)F*F)!\"\"F**&-F&6$-F.6%F0F)F1 -%\"$G6$F1\"\"#F**$F)F;F2F*" }{TEXT -1 4 " = " }{XPPEDIT 18 0 "1/(c^2 );" "6#*&\"\"\"F$*$%\"cG\"\"#!\"\"" }{TEXT -1 1 " " }{XPPEDIT 18 0 "di ff(u,`$`(t,2));" "6#-%%diffG6$%\"uG-%\"$G6$%\"tG\"\"#" }{TEXT -1 1 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 75 "Excep t, in this simplified case with the initial conditions independent of \+ " }{XPPEDIT 18 0 "theta;" "6#%&thetaG" }{TEXT -1 61 ", we can drop the second term of the left side. Thus, we have" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 17 " " }{XPPEDIT 18 0 "diff(r*diff(u(t,r),r),r)/r;" "6#*&-%%diffG6$*&%\"rG\"\"\"-F%6$-% \"uG6$%\"tGF(F(F)F(F)F(!\"\"" }{TEXT -1 5 " = " }{XPPEDIT 18 0 "1/(c ^2);" "6#*&\"\"\"F$*$%\"cG\"\"#!\"\"" }{TEXT -1 1 " " }{XPPEDIT 18 0 " diff(u,`$`(t,2));" "6#-%%diffG6$%\"uG-%\"$G6$%\"tG\"\"#" }{TEXT -1 1 " ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 27 "Boun dary conditions will be" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 69 " u(t, a) = 0, because the sides of the drum are nailed down," }}{PARA 0 "" 0 "" {TEXT -1 60 " u(0, r) = \+ f(r), which is the initial displacement," }}{PARA 0 "" 0 "" {TEXT -1 8 "and " }{XPPEDIT 18 0 "diff(u,t);" "6#-%%diffG6$%\"uG%\"tG" } {TEXT -1 45 "(0, r) = g(r), which is the initial velocity." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 123 " At this poi nt in our study of partial differential equations, we know to start by separation of variables. Assume that" }}{PARA 0 "" 0 "" {TEXT -1 35 " u(t, r) = T(t) R(r)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{PARA 0 "" 0 "" {TEXT -1 67 "Using this assumption in the partial dif ferential equation leads to" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 29 " 1/r (r R ') ' T = " }{XPPEDIT 18 0 " 1/(c^2);" "6#*&\"\"\"F$*$%\"cG\"\"#!\"\"" }{TEXT -1 10 " R T '', " }} {PARA 0 "" 0 "" {TEXT -1 64 "which, in turn, leads to the two ordinary differential equations" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 22 " (r R ') ' + " }{XPPEDIT 18 0 "lambda^2;" " 6#*$%'lambdaG\"\"#" }{TEXT -1 25 " r R = 0, with R(a) = 0 " }}{PARA 0 "" 0 "" {TEXT -1 3 "and" }}{PARA 0 "" 0 "" {TEXT -1 17 " T \+ '' + " }{XPPEDIT 18 0 "lambda^2;" "6#*$%'lambdaG\"\"#" }{TEXT -1 1 " \+ " }{XPPEDIT 18 0 "c^2;" "6#*$%\"cG\"\"#" }{TEXT -1 7 " T = 0." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 179 "We consi der the equation in R first. This equation leads to the Bessel functio ns and, in view of the physical requirement that solutions stay bounde d, we use only R(r) = BesselJ( " }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG " }{TEXT -1 135 " r ). By now, we should know how to check to see that these functions really are solutions for the ordinary differential eq uation in R." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "R:=r->Bessel J(0,lambda*r);" }}}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 52 "diff(r*diff(R(r),r),r)+lambda^2*r*R(r);\nsimpl ify(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 123 "To meet the boundary condition, recall that we use \+ the zero's of the Bessel function. Let's prepare about 10 of them. Tak e " }{TEXT 256 1 "a" }{TEXT -1 25 " to be 1 for specificity." }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 80 "c:=1;\nfor n from 1 to 10 do \n zero[n]:=evalf(BesselJZeros(0,n))/c;\nod;\nn:='n';" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 154 "N ow, we should find that the following functions satisfy the differenti al equation and the boundary conditions. We draw graphs of the first f ive of these." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "R:=(n,r)->B esselJ(0,zero[n]*r);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "plo t([seq(R(n,r),n=1..5)],r=0..c);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 80 " The equation in T is familiar. It leads to sines and cosines. Solutions for" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 26 " T '' + " }{XPPEDIT 18 0 "lambda^2;" "6 #*$%'lambdaG\"\"#" }{TEXT -1 1 " " }{XPPEDIT 18 0 "c^2;" "6#*$%\"cG\" \"#" }{TEXT -1 7 " T = 0," }}{PARA 0 "" 0 "" {TEXT -1 17 "or, in this \+ case," }}{PARA 0 "" 0 "" {TEXT -1 26 " T '' + " } {XPPEDIT 18 0 "zero[n]^2;" "6#*$&%%zeroG6#%\"nG\"\"#" }{TEXT -1 1 " " }{XPPEDIT 18 0 "c^2;" "6#*$%\"cG\"\"#" }{TEXT -1 6 " T = 0" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 8 "are " } {XPPEDIT 18 0 "a[n];" "6#&%\"aG6#%\"nG" }{TEXT -1 6 " cos( " } {XPPEDIT 18 0 "zero[n];" "6#&%%zeroG6#%\"nG" }{TEXT -1 10 " c t) + \+ " }{XPPEDIT 18 0 "b[n];" "6#&%\"bG6#%\"nG" }{TEXT -1 7 " sin( " } {XPPEDIT 18 0 "zero[n];" "6#&%%zeroG6#%\"nG" }{TEXT -1 6 " c t)." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 21 "Product s olutions are" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 63 "u:=(t,r)->R(n,r)*(a[n]*cos(zero[n]*c*t)+b[n]*sin(z ero[n]*c*t));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 63 "We verify that these satisfy the partial differ ential equation." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "1/r*diff (r*diff(u(t,r),r),r)-1/c^2*diff(u(t,r),t,t):\nsimplify(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{PARA 0 "" 0 "" {TEXT -1 36 "We now make sums of these solutions." }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 69 "u:=sum(R(n,r)*(a[n]*cos(zero [n]*c*t)+b[n]*sin(zero[n]*c*t)),n=1..10);" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 31 "The initial condition asks that" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 27 " f(r) = u(0, r) \+ = " }{XPPEDIT 18 0 "sum(R(n,r)*a[n],n = 1 .. 10);" "6#-%$sumG6$*&-%\"R G6$%\"nG%\"rG\"\"\"&%\"aG6#F*F,/F*;F,\"#5" }{TEXT -1 19 " and that g (r) = " }{XPPEDIT 18 0 "diff(v,t);" "6#-%%diffG6$%\"vG%\"tG" }{TEXT -1 10 "(0, r) = " }{XPPEDIT 18 0 "sum(R(n,r)*zero[n]*c*b[n],n = 1 .. \+ 10);" "6#-%$sumG6$**-%\"RG6$%\"nG%\"rG\"\"\"&%%zeroG6#F*F,%\"cGF,&%\"b G6#F*F,/F*;F,\"#5" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 41 "We compute these as fourier coefficients. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 10 " \+ " }{XPPEDIT 18 0 "a[n];" "6#&%\"aG6#%\"nG" }{TEXT -1 3 " = " } {XPPEDIT 18 0 "int(f(r)*R(n,r)*r,r = 0 .. 1)/int(R(n,r)^2*r,r = 0 .. 1 );" "6#*&-%$intG6$*(-%\"fG6#%\"rG\"\"\"-%\"RG6$%\"nGF+F,F+F,/F+;\"\"!F ,F,-F%6$*&-F.6$F0F+\"\"#F+F,/F+;F3F,!\"\"" }{TEXT -1 10 " and " } {XPPEDIT 18 0 "zero[n]*c*b[n];" "6#*(&%%zeroG6#%\"nG\"\"\"%\"cGF(&%\"b G6#F'F(" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "int(g(r)*R(n,r)*r,r = 0 .. \+ 1)/int(R(n,r)^2*r,r = 0 .. 1);" "6#*&-%$intG6$*(-%\"gG6#%\"rG\"\"\"-% \"RG6$%\"nGF+F,F+F,/F+;\"\"!F,F,-F%6$*&-F.6$F0F+\"\"#F+F,/F+;F3F,!\"\" " }{TEXT -1 3 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 74 "We do a specific example and watch the drum vibrate. We'll take f(r) = 1- " }{XPPEDIT 18 0 "r^2;" "6#*$%\"rG\"\"#" }{TEXT -1 96 " and g(r) = 0. For this example, take c = 1. In this case, all the b coef ficients will be zero." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "f:=r->r ^2-1;\nc:=1;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 108 "for n from 1 to 10 do\n a[n]:=int(f(r)*R(n,r)*r,r=0..1)/int(R(n,r)^2*r,r=0.. 1);\n b[n]:=0:\nod;\nn:='n';" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 32 "Check to see how well this fits." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 82 "u:=(t,r)->sum(R(n,r)*(a[n]*cos(zero[n]*c*t)+b[n] *sin(zero[n]*c*t)),\n n=1..10);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 57 "To separate the plot s, we offset the graph of f by 0.007." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "plot([u(0,r),f(r)+.007],r=-1..1,color=[black,red]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 57 "Of course, this should re ally be a graph in 3 dimensions." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 82 "plot3d([r,theta,u(0,r)],r=0..1,theta=-Pi..Pi,coords=cylindrica l,\n axes=normal);" }}}{PARA 0 "" 0 "" {TEXT -1 24 "Now, we do an a nimation." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 91 "animate3d([r,th eta,u(t,r)],r=0..1,theta=-Pi..Pi,t=0..4,\n coords=cylindrical,axes=n ormal);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 140 "In this final examp le, we take the initial distribution to be zero, but give the disk a w hack. That is, we take f(r) = 0 and g(r) not zero. " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "f:=r->0;\ng:=r->r^2-1;" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 183 "for n from 1 to 10 do\n a[n]:=int(f(r)* R(n,r)*r,r=0..1)/int(R(n,r)^2*r,r=0..1);\n b[n]:=int(g(r)*R(n,r)*r ,r=0..1)/int(R(n,r)^2*r,r=0..1)/\n (zero[n]*c);\nod;\nn: ='n';" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 102 "plot3d([r,theta,u (0,r)],r=0..1,theta=-Pi..Pi,coords=cylindrical,\n axes=normal,orien tation=[45,70]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 112 "animat e3d([r,theta,u(t,r)],r=0..1,theta=-Pi..Pi,t=0..Pi,\n coords=cylindri cal,orientation=[45,70],axes=normal);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 189 "There are further investigations that can be done at thi s point. For example, we might strike the drum at a point different fr om the center, so that the initial value is not independent of " } {XPPEDIT 18 0 "theta;" "6#%&thetaG" }{TEXT -1 88 ". We leave these sep aration of variable problems, however to consider numerical methods." }}{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 257 "" 0 "" {TEXT -1 36 "Copyright \251 2003 b y James V. Herod" }}{PARA 257 "" 0 "" {TEXT -1 19 "All rights reserved " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{MARK "0 0" 42 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 1 1 2 33 1 1 }