{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 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 257 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 258 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 259 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 260 "" 1 14 0 0 0 0 0 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 261 "" 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 1 }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 }1 0 0 0 8 4 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 0 }3 0 0 -1 12 12 0 0 0 0 0 0 19 0 }{PSTYLE "Author" 0 19 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 0 0 -1 8 8 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 256 1 {CSTYLE "" -1 -1 "" 1 18 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{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 }{PSTYLE "" 0 258 1 {CSTYLE "" -1 -1 "" 1 18 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {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 258 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT 260 36 "Sect ion 7.3: Heat Equation on a Disk" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {SECT 0 {PARA 3 "" 0 "" {TEXT 261 30 "Maple Packages for Section 7.3" }}{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 0 "" }}{PARA 0 "" 0 " " {TEXT -1 268 " We consider the diffusion equation on a disk. The physical situation is that of either a long circular beam, or a disk \+ with insulated top and bottom, but with a fixed temperature on the sid es, and an initial temperature throughout the disk. This can be writte n as" }}{PARA 0 "" 0 "" {TEXT -1 14 " 1/k " }{XPPEDIT 18 0 "d iff(u,t);" "6#-%%diffG6$%\"uG%\"tG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 " Delta;" "6#%&DeltaG" }{TEXT -1 3 " u," }}{PARA 0 "" 0 "" {TEXT -1 13 " with u(t, a, " }{XPPEDIT 18 0 "theta;" "6#%&thetaG" }{TEXT -1 8 " ) = \+ f( " }{XPPEDIT 18 0 "theta;" "6#%&thetaG" }{TEXT -1 15 " ), 0 < t, \+ " }{XPPEDIT 18 0 "-pi;" "6#,$%#piG!\"\"" }{TEXT -1 3 " < " }{XPPEDIT 18 0 "theta;" "6#%&thetaG" }{TEXT -1 3 " < " }{XPPEDIT 18 0 "Pi;" "6#% #PiG" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 15 "and u(0, r, " }{XPPEDIT 18 0 "theta;" "6#%&thetaG" }{TEXT -1 11 " ) = g( r, " } {XPPEDIT 18 0 "theta;" "6#%&thetaG" }{TEXT -1 6 " ), " }{XPPEDIT 18 0 "0 <= r;" "6#1\"\"!%\"rG" }{TEXT -1 9 " < a, " }{XPPEDIT 18 0 "-p i;" "6#,$%#piG!\"\"" }{TEXT -1 3 " < " }{XPPEDIT 18 0 "theta;" "6#%&th etaG" }{TEXT -1 3 " < " }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 2 " . " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 64 " \+ To do this problem, we first solve the steady state problem" }}{PARA 0 "" 0 "" {TEXT -1 14 " 0 = " }{XPPEDIT 18 0 "Delta;" "6#%&De ltaG" }{TEXT -1 10 " v, v(a, " }{XPPEDIT 18 0 "theta;" "6#%&thetaG" } {TEXT -1 8 " ) = f( " }{XPPEDIT 18 0 "theta;" "6#%&thetaG" }{TEXT -1 5 " ), " }{XPPEDIT 18 0 "-pi;" "6#,$%#piG!\"\"" }{TEXT -1 3 " < " } {XPPEDIT 18 0 "theta;" "6#%&thetaG" }{TEXT -1 3 " < " }{XPPEDIT 18 0 " Pi;" "6#%#PiG" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 32 "Then so lve the transient problem" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 14 " 1/k " }{XPPEDIT 18 0 "diff(u,t);" "6#-%% diffG6$%\"uG%\"tG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "Delta;" "6#%&Delt aG" }{TEXT -1 3 " u," }}{PARA 0 "" 0 "" {TEXT -1 13 "with u(t, a, " } {XPPEDIT 18 0 "theta;" "6#%&thetaG" }{TEXT -1 17 " ) = 0, 0 < t, " } {XPPEDIT 18 0 "-pi;" "6#,$%#piG!\"\"" }{TEXT -1 3 " < " }{XPPEDIT 18 0 "theta;" "6#%&thetaG" }{TEXT -1 3 " < " }{XPPEDIT 18 0 "Pi;" "6#%#Pi G" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 15 "and u(0, r, " } {XPPEDIT 18 0 "theta;" "6#%&thetaG" }{TEXT -1 11 " ) = g( r, " } {XPPEDIT 18 0 "theta;" "6#%&thetaG" }{TEXT -1 10 " ) - v(r, " } {XPPEDIT 18 0 "theta;" "6#%&thetaG" }{TEXT -1 5 ") , " }{XPPEDIT 18 0 "0 <= r;" "6#1\"\"!%\"rG" }{TEXT -1 7 " < a, " }{XPPEDIT 18 0 "-pi; " "6#,$%#piG!\"\"" }{TEXT -1 3 " < " }{XPPEDIT 18 0 "theta;" "6#%&thet aG" }{TEXT -1 3 " < " }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 76 " Fi nally, the solution for the original problem is the sum of these two. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 244 " \+ We have already studied how to solve the first problem in Sections 6. 4 and 6.5. Here is how to think of solving the second problem. We writ e out the partial differential equation in unabbreviated form to facil itate separation of variables." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 21 " " }{XPPEDIT 18 0 "di ff(r*diff(u(r,theta),r),r)/r+diff(u(r,theta),`$`(theta,2))/(r^2);" "6# ,&*&-%%diffG6$*&%\"rG\"\"\"-F&6$-%\"uG6$F)%&thetaGF)F*F)F*F)!\"\"F**&- F&6$-F.6$F)F0-%\"$G6$F0\"\"#F**$F)F:F1F*" }{TEXT -1 4 " = " } {XPPEDIT 18 0 "diff(u,t);" "6#-%%diffG6$%\"uG%\"tG" }{TEXT -1 1 "." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 167 "(We have dropped the constant of diffusion for purposes of illustration.) Sepa ration of variables leads to these three differential equations with b oundary conditions:" }}{PARA 0 "" 0 "" {TEXT -1 5 "(1) " }{XPPEDIT 18 0 "Theta;" "6#%&ThetaG" }{TEXT -1 6 " '' + " }{XPPEDIT 18 0 "mu^2*T heta;" "6#*&%#muG\"\"#%&ThetaG\"\"\"" }{TEXT -1 7 " = 0, " }{XPPEDIT 18 0 "Theta(Pi);" "6#-%&ThetaG6#%#PiG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "Theta(-Pi);" "6#-%&ThetaG6#,$%#PiG!\"\"" }{TEXT -1 3 ", " } {XPPEDIT 18 0 "Theta;" "6#%&ThetaG" }{TEXT -1 3 " '(" }{XPPEDIT 18 0 " Pi;" "6#%#PiG" }{TEXT -1 5 " ) = " }{XPPEDIT 18 0 "Theta;" "6#%&ThetaG " }{TEXT -1 3 " '(" }{XPPEDIT 18 0 "-Pi;" "6#,$%#PiG!\"\"" }{TEXT -1 2 ")," }}{PARA 0 "" 0 "" {TEXT -1 14 "(2) T ' = - " }{XPPEDIT 18 0 " lambda^2;" "6#*$%'lambdaG\"\"#" }{TEXT -1 9 " T, and " }}{PARA 0 "" 0 "" {TEXT -1 22 "(3) r ( r R ' ) ' - " }{XPPEDIT 18 0 "mu^2;" "6#*$ %#muG\"\"#" }{TEXT -1 7 " R = - " }{XPPEDIT 18 0 "lambda^2;" "6#*$%'la mbdaG\"\"#" }{TEXT -1 1 " " }{XPPEDIT 18 0 "r^2;" "6#*$%\"rG\"\"#" } {TEXT -1 39 " R, R(a) = 0, R is bounded on [0, a)." }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 84 "We know how to solve (1 ) and (2). Equation 3 leads to a study of Bessel's functions." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 74 " To b egin this study, we start with an example that is independent of " } {XPPEDIT 18 0 "Theta;" "6#%&ThetaG" }{TEXT -1 30 ". Here is the proble m we solve" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 15 " " }{XPPEDIT 18 0 "diff(u,t);" "6#-%%diffG6$%\"uG %\"tG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "Delta;" "6#%&DeltaG" }{TEXT -1 3 " u," }}{PARA 0 "" 0 "" {TEXT -1 15 "with u(t, a, " }{XPPEDIT 18 0 "theta;" "6#%&thetaG" }{TEXT -1 17 " ) = 0, 0 < t, " }{XPPEDIT 18 0 "-pi;" "6#,$%#piG!\"\"" }{TEXT -1 3 " < " }{XPPEDIT 18 0 "theta; " "6#%&thetaG" }{TEXT -1 3 " < " }{XPPEDIT 18 0 "Pi;" "6#%#PiG" } {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 15 "and u(0, r, " } {XPPEDIT 18 0 "theta;" "6#%&thetaG" }{TEXT -1 14 " ) = g( r), " } {XPPEDIT 18 0 "0 <= r;" "6#1\"\"!%\"rG" }{TEXT -1 7 " < a, " } {XPPEDIT 18 0 "-pi;" "6#,$%#piG!\"\"" }{TEXT -1 3 " < " }{XPPEDIT 18 0 "theta;" "6#%&thetaG" }{TEXT -1 3 " < " }{XPPEDIT 18 0 "Pi;" "6#%#Pi G" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 70 "Is it clear that the solution of this equation will be in dependent of " }{XPPEDIT 18 0 "theta;" "6#%&thetaG" }{TEXT -1 53 " ? T hus, the partial differential equation reduces to" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 16 " " } {XPPEDIT 18 0 "diff(r*diff(u(r,theta),r),r)/r;" "6#*&-%%diffG6$*&%\"rG \"\"\"-F%6$-%\"uG6$F(%&thetaGF(F)F(F)F(!\"\"" }{TEXT -1 4 " = " } {XPPEDIT 18 0 "diff(u,t);" "6#-%%diffG6$%\"uG%\"tG" }{TEXT -1 2 ". " } }{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 42 "Now we h ave the two differential equations" }}{PARA 0 "" 0 "" {TEXT -1 10 "(1) T ' + " }{XPPEDIT 18 0 "lambda^2;" "6#*$%'lambdaG\"\"#" }{TEXT -1 11 " T = 0, and" }}{PARA 0 "" 0 "" {TEXT -1 20 "(2) r ( r R ' ) ' + " } {XPPEDIT 18 0 "lambda^2;" "6#*$%'lambdaG\"\"#" }{TEXT -1 1 " " } {XPPEDIT 18 0 "r^2;" "6#*$%\"rG\"\"#" }{TEXT -1 47 " R = 0, with R(a) \+ = 0, and R bounded on [0, a)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 65 " We examine the second differential e quation. Expanded, it is" }}{PARA 0 "" 0 "" {TEXT -1 10 " " } {XPPEDIT 18 0 "r^2;" "6#*$%\"rG\"\"#" }{TEXT -1 16 " R '' + r R ' + " }{XPPEDIT 18 0 "lambda^2;" "6#*$%'lambdaG\"\"#" }{TEXT -1 1 " " } {XPPEDIT 18 0 "r^2;" "6#*$%\"rG\"\"#" }{TEXT -1 7 " R = 0." }}{PARA 0 "" 0 "" {TEXT -1 175 "We ask Maple to solve this equation. Commonly, t he solutions are introduced in elementary differential equations in a \+ discussion of series. We are, as you know, assuming that " }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT -1 89 " > 0. Maple would be assumi ng only that it is complex, unless we tell it to do otherwise." }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "assume(lambda>0);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 75 "dsolve(r^2*diff(R(r),r,r)+ r *diff(R(r),r)+lambda^2*r^2*R(r)=0,R(r),series);" }}}{PARA 0 "" 0 "" {TEXT -1 160 "Of course, Maple knows the solutions in the full details , without truncated series. We ask for solutions, again, and do not as k that the solutions be as series." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "dsolve(r^2*diff(R(r),r,r)+ r*diff(R(r),r)+lambda^2*r^ 2*R(r)=0,R(r));" }}}{PARA 0 "" 0 "" {TEXT -1 89 "It seems we have two \+ solutions. To understand the Bessel functions, we draw their graphs." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "plot([BesselY(0,r),BesselJ (0,r)],r=0..10,color=[BLACK,RED]);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{PARA 0 "" 0 "" {TEXT -1 39 "There are several observations to make: " }}{PARA 0 "" 0 "" {TEXT -1 217 "(1) BesselJ(0, r) is one at zero and BesselY(0, r) is unbounded at zero.\n(2) The two solutions oscillate, are not periodic, and have intertwining zeros.\n(3) These form eigenf unctions corresponding to the eigenvalues -" }{XPPEDIT 18 0 "lambda^2; " "6#*$%'lambdaG\"\"#" }{TEXT -1 30 " for the differential operator" } }{PARA 0 "" 0 "" {TEXT -1 35 " L(R) = 1/r ( r R ' ) '. " }} {PARA 0 "" 0 "" {TEXT -1 28 "Here's a check of this last." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 83 "R:=r->BesselY(0,lambda*r);\n1/r*(di ff(r*diff(R(r),r),r))+lambda^2*R(r);\nsimplify(%);" }}}{PARA 0 "" 0 " " {TEXT -1 1 " " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 83 "R:=r->Bes selJ(0,lambda*r);\n1/r*(diff(r*diff(R(r),r),r))+lambda^2*R(r);\nsimpli fy(%);" }}}{EXCHG }{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 135 " It is of value to compare these two functions with \+ the cosine and sine functions, which define the eigenfunctions for the problem " }{XPPEDIT 18 0 "Theta;" "6#%&ThetaG" }{TEXT -1 6 " '' + " } {XPPEDIT 18 0 "lambda^2;" "6#*$%'lambdaG\"\"#" }{TEXT -1 1 " " } {XPPEDIT 18 0 "Theta;" "6#%&ThetaG" }{TEXT -1 159 " = 0. The compariso n helps to crystallize the structure in your memory. Since we are int erested in only bounded solutions on the disk, we set aside BesselY. \+ " }}{PARA 0 "" 0 "" {TEXT -1 43 " Suppose a > 0. Take J(r) to be B esselJ" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 256 13 "Comparison 1:" }{TEXT -1 42 " There is an infinite increasing sequ ence " }{XPPEDIT 18 0 "lambda[1];" "6#&%'lambdaG6#\"\"\"" }{TEXT -1 4 " , " }{XPPEDIT 18 0 "lambda[2];" "6#&%'lambdaG6#\"\"#" }{TEXT -1 4 " , " }{XPPEDIT 18 0 "lambda[3];" "6#&%'lambdaG6#\"\"$" }{TEXT -1 22 " , ... such that J( " }{XPPEDIT 18 0 "lambda[p];" "6#&%'lambdaG6#%\" pG" }{TEXT -1 53 " a) = 0. For the cosine function, these numbers were " }{XPPEDIT 18 0 "n*Pi/a;" "6#*(%\"nG\"\"\"%#PiGF%%\"aG!\"\"" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 257 13 "Comparison 2:" }{TEXT -1 16 " The numbers - " }{XPPEDIT 18 0 "lam bda[j]^2;" "6#*$&%'lambdaG6#%\"jG\"\"#" }{TEXT -1 31 " are eigenvalues and R(r) = J( " }{XPPEDIT 18 0 "lambda[j];" "6#&%'lambdaG6#%\"jG" } {TEXT -1 94 " r) is an eigenfunction for the operator L(R) defined ab ove with R(a) = 0. The function cos( " }{XPPEDIT 18 0 "n*Pi/a;" "6#*(% \"nG\"\"\"%#PiGF%%\"aG!\"\"" }{TEXT -1 30 " x ) was an eigenfunction f or " }{XPPEDIT 18 0 "Theta;" "6#%&ThetaG" }{TEXT -1 9 " '' with " } {XPPEDIT 18 0 "Theta;" "6#%&ThetaG" }{TEXT -1 9 " (a) = 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 258 13 "Comparison 3:" } {TEXT -1 41 " Orthogonality persists in the sense that" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 10 " " } {XPPEDIT 18 0 "int(J(lambda[j]*x)*J(lambda[k]*x)*x,x = 0 .. a);" "6#-% $intG6$*(-%\"JG6#*&&%'lambdaG6#%\"jG\"\"\"%\"xGF/F/-F(6#*&&F,6#%\"kGF/ F0F/F/F0F//F0;\"\"!%\"aG" }{TEXT -1 8 " = 0 if " }{XPPEDIT 18 0 "j <> \+ k;" "6#0%\"jG%\"kG" }{TEXT -1 9 " and " }{XPPEDIT 18 0 "int(J(lamb da[j]*x)^2*x,x = 0 .. a);" "6#-%$intG6$*&-%\"JG6#*&&%'lambdaG6#%\"jG\" \"\"%\"xGF/\"\"#F0F//F0;\"\"!%\"aG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 " a^2/2;" "6#*&%\"aG\"\"#F%!\"\"" }{TEXT -1 4 " J '" }{XPPEDIT 18 0 "(la mbda[j]*a)^2" "6#*$*&&%'lambdaG6#%\"jG\"\"\"%\"aGF)\"\"#" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 259 13 "C omparison 4:" }{TEXT -1 45 " If f(r) is sectionally smooth on (0, a) t hen" }}{PARA 0 "" 0 "" {TEXT -1 4 " " }}{PARA 0 "" 0 "" {TEXT -1 17 " f(r) = " }{XPPEDIT 18 0 "sum(a[n]*J(lambda[n]*r),n);" "6 #-%$sumG6$*&&%\"aG6#%\"nG\"\"\"-%\"JG6#*&&%'lambdaG6#F*F+%\"rGF+F+F*" }{TEXT -1 7 " where " }{XPPEDIT 18 0 "a[n];" "6#&%\"aG6#%\"nG" }{TEXT -1 12 " = /" }{XPPEDIT 18 0 "abs(J(lambda[n]^2*r))^2;" "6#*$ -%$absG6#-%\"JG6#*&&%'lambdaG6#%\"nG\"\"#%\"rG\"\"\"F/" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 45 "In this case, the do tproduct is defined by " }{XPPEDIT 18 0 ";" "6#-%$<,>G6$%\"fG6# %\"gG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "int(f(x)*g(x)*x,x = 0 .. a); " "6#-%$intG6$*(-%\"fG6#%\"xG\"\"\"-%\"gG6#F*F+F*F+/F*;\"\"!%\"aG" } {TEXT -1 26 " with the associated norm." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 181 "It seems appropriate to do an examp le. We take the boundary conditions to be zero so that we will not hav e to do the steady state problem first. We take the initial condition \+ to be " }{XPPEDIT 18 0 "1-r^2;" "6#,&\"\"\"F$*$%\"rG\"\"#!\"\"" } {TEXT -1 49 " so that the initial condition is independent of " } {XPPEDIT 18 0 "theta;" "6#%&thetaG" }{TEXT -1 66 ". With a = 1, we wil l have a continuous function on the unit disk." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 52 " Here is a graph of \+ the initial condition. " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "f:=r->1-r^2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 76 "cylinde rplot([r,theta,f(r)],theta=-Pi..Pi,r=0..1,axes=NORMAL, shading=zhue); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 91 " Your expectation is \+ that the initial distribution will decay to the steady state u(r, " } {XPPEDIT 18 0 "theta;" "6#%&thetaG" }{TEXT -1 30 " ) = 0. Here are the details. " }}{PARA 0 "" 0 "" {TEXT -1 202 "The first thing to do is t o get the zeros of R(r) = BesselJ(r). We can create a program to compu te these. Go back to the graph and see how far the zeros are apart. Th en, solve for a zero in an interval." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 113 "zero[0]:=0;\nfor n from 1 to 4 do\n zero[n]:=fso lve(BesselJ(0,x)=0,x,zero[n-1]+1..zero[n-1]+4);\nend do;\nn:='n';" }}} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 85 "Pretty go od. No surprise. These zeros are so important that Maple already knows them." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 76 "for n from 1 to 10 do\n zero[n]:=evalf(BesselJZeros(0,n));\nend do;\nn:='n';" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 77 "To make further comparisons, we plot the first six of the eigen funtions cos( " }{XPPEDIT 18 0 "n*Pi/a;" "6#*(%\"nG\"\"\"%#PiGF%%\"aG! \"\"" }{TEXT -1 55 " x) and the first six of the eigenfunctions Bessel J(0, " }{XPPEDIT 18 0 "lambda[n]/a;" "6#*&&%'lambdaG6#%\"nG\"\"\"%\"aG !\"\"" }{TEXT -1 4 " x)." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 82 " plot([seq(cos(n*Pi*x),n=0..5)],x=0..1,color=[black,red,green,blue,yell ow, brown]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 91 "plot([seq(B esselJ(0,zero[n]*x),n=0..5)],x=0..1,color=[black,red,green,blue,yellow , brown]);" }}}{PARA 0 "" 0 "" {TEXT -1 263 "Let's check the orthogona lity. In order to meet the boundary conditions that R is bounded and R (a) is zero, we need the positive zeros of the BesselJ(0, x) function. What you should expect is that we don't get exactly zero, for each ze ro is only an approximation." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 146 "for n from 1 to 3 do\n for m from 1 to 3 do\n print(int(Bess elJ(0,zero[n]*x)*BesselJ(0,zero[m]*x)*x,x=0..1));\n end do; \nend do; \nn:='n':m:='m':" }}}{PARA 0 "" 0 "" {TEXT -1 40 "Now is time to compu te the coefficients." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 159 "for n from 1 to 10 do\n A[n]:=int(f(x)*BesselJ(0,zero[n]*x)*x,x=0..1)/ \n int(BesselJ(0,zero[n]*x)*BesselJ(0,zero[n]*x)*x,x=0.. 1);\nend do;\nn:='n':" }}}{PARA 0 "" 0 "" {TEXT -1 66 "The next two gr aphs are so close, I offset the first one by 0.007." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 91 "plot([f(x)+0.007,sum(A[n]*BesselJ(0,zero[n] *x),n=1..10)],x=0..1,\n color=[black,red]);" }}}{PARA 0 "" 0 "" {TEXT -1 27 "We now create the solution." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "u:=(t,r)->sum(A[n]*BesselJ(0,zero[n]*r)*exp(-zero[n]^ 2*t),n=1..10);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 78 "cylinderp lot([r,theta,u(1,r)],r=0..1,theta=-Pi..Pi,axes=NORMAL, shading=zhue); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 113 "animate3d([r,theta,u(t ,r)],r=0..1,theta=-Pi..Pi,t=0..1,\n coords=cylindrical, fra mes=30, shading=zhue);" }}}{PARA 0 "" 0 "" {TEXT -1 134 "We remark in \+ passing that the following is an alternate way to do a cylinder plot t hat does not require a call of the package \"plots\"." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 79 "plot3d([r,theta,u(0,r)],r=0..1,theta=-Pi. .Pi,coords=cylindrical, shading=zhue);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 3 " " 0 "" {TEXT -1 19 "Verify the Solution" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 86 "I emphasize that solutions should be checked. Here, I show that a solution of the form" }}{PARA 0 "" 0 "" {TEXT -1 15 " u(t, r) = " }{XPPEDIT 18 0 "sum(A[n]*BesselJ(0,zero[ n]*r)*exp(-zero[n]^2*t),n = 1 .. infinity);" "6#-%$sumG6$*(&%\"AG6#%\" nG\"\"\"-%(BesselJG6$\"\"!*&&%%zeroG6#F*F+%\"rGF+F+-%$expG6#,$*&&F26#F *\"\"#%\"tGF+!\"\"F+/F*;F+%)infinityG" }{TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 97 "is a solution for the equation for 0 < r < 1, and t > 0 . If the interval were 0 < r < a, we take" }}{PARA 0 "" 0 "" {TEXT -1 14 " u(t, r) = " }{XPPEDIT 18 0 "sum(A[n]*BesselJ(0, zero[n]*r/a )*exp(-zero[n]^2*t/(a^2)),n = 1 .. infinity);" "6#-%$sumG6$*(&%\"AG6#% \"nG\"\"\"-%(BesselJG6$\"\"!*(&%%zeroG6#F*F+%\"rGF+%\"aG!\"\"F+-%$expG 6#,$*(&F26#F*\"\"#%\"tGF+*$F5F>F6F6F+/F*;F+%)infinityG" }{TEXT -1 1 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 83 "Here \+ is a verification in case we sum only to 3. I incorporate the 1/a in t he term " }{XPPEDIT 18 0 "zero[n];" "6#&%%zeroG6#%\"nG" }{TEXT -1 2 " \+ ." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "restart; " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 71 "for n from 1 to 3 do\n zero[n]: =BesselJZeros(0,n)/a;\nend do;\nn:='n';" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 79 "u:=(t,r)->sum(A[n]*BesselJ(0,zero[n]*r)*exp(-zero[n]^ 2*t),\n n=1..3);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 62 "u(t,a);\nsimplify(diff(u(t,r),t)-1/r*diff(r*diff(u(t,r),r),r));" } }}{EXCHG }{EXCHG }{EXCHG }{EXCHG }}{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 257 "" 0 "" {TEXT -1 36 "Copyright \251 2003 by James V. Herod" }}{PARA 257 "" 0 "" {TEXT -1 19 "All rights reserved" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {MARK "0 0 0" 18 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 1 1 2 33 1 1 }