{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 "" -1 261 "" 1 14 0 0 0 0 0 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 262 "" 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 "" 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 }} {SECT 0 {PARA 256 "" 0 "" {TEXT 261 36 "Section 7.3: Heat Equation on \+ a Disk" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT 262 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 t hat of either a long circular beam, or a disk with insulated top and b ottom, but with a fixed temperature on the sides, and an initial tempe rature throughout the disk. This can be written as" }}{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#%&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 -1 253 " \+ 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 unabbreviated form to facilita te undertaking separation of variables." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 21 " " }{XPPEDIT 18 0 "diff(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 -1 180 "(We have dropped the constant of diffusion for purposes of illustration of tec hnique.) Separation of variables leads to these three differential equ ations with boundary conditions:" }}{PARA 0 "" 0 "" {TEXT -1 5 "(1) \+ " }{XPPEDIT 18 0 "Theta;" "6#%&ThetaG" }{TEXT -1 6 " '' + " }{XPPEDIT 18 0 "mu^2*Theta;" "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 "lambd a^2;" "6#*$%'lambdaG\"\"#" }{TEXT -1 1 " " }{XPPEDIT 18 0 "r^2;" "6#*$ %\"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 h ow to solve (1) and (2). Equation 3 leads to a study of Bessel's funct ions." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 74 " To begin this study, we start with an example that is independent of " }{XPPEDIT 18 0 "Theta;" "6#%&ThetaG" }{TEXT -1 30 ". Here is the problem 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#%#Pi G" }{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#*$%\"rG\"\"#" }{TEXT -1 7 " R = 0." }}{PARA 0 "" 0 "" {TEXT -1 175 "We ask Maple to solve this equation. Commonly, t he solutions are introduced in elementary differential equations in a \+ discussion of series. We are, as you know, assuming that " }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT -1 89 " > 0. Maple would be assumi ng only that it is complex, unless we tell it to do otherwise." }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "assume(lambda>0);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 75 "dsolve(r^2*diff(R(r),r,r)+ r *diff(R(r),r)+lambda^2*r^2*R(r)=0,R(r),series);" }}}{PARA 0 "" 0 "" {TEXT -1 160 "Of course, Maple knows the solutions in the full details , without truncated series. We ask for solutions, again, and do not as k that the solutions be as series." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "dsolve(r^2*diff(R(r),r,r)+ r*diff(R(r),r)+lambda^2*r^ 2*R(r)=0,R(r));" }}}{PARA 0 "" 0 "" {TEXT -1 103 "It seems we have two solutions. As a start to understanding the Bessel functions, we draw \+ their graphs." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "plot([Besse lY(0,r),BesselJ(0,r)],r=0..10,color=[BLACK,RED]);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 39 "There are several observa tions to make:" }}{PARA 0 "" 0 "" {TEXT -1 217 "(1) BesselJ(0, r) is o ne at zero and BesselY(0, r) is unbounded at zero.\n(2) The two soluti ons oscillate, are not periodic, and have intertwining zeros.\n(3) The se form eigenfunctions corresponding to the eigenvalues -" }{XPPEDIT 18 0 "lambda^2;" "6#*$%'lambdaG\"\"#" }{TEXT -1 30 " for the different ial 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*(diff(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->BesselJ(0,lambda*r);\n1/r*(diff(r*diff(R(r),r),r))+lambda^ 2*R(r);\nsimplify(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" } }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 158 " I t is of value to think of the comparison here of these two functions w ith 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 62 "cylinde rplot([r,theta,f(r)],theta=-Pi..Pi,r=0..1,axes=NORMAL);" }}}{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 initia l distribution will decay to the steady state u(r, " }{XPPEDIT 18 0 "t heta;" "6#%&thetaG" }{TEXT -1 30 " ) = 0. Here are the details. " }} {PARA 0 "" 0 "" {TEXT -1 202 "The first thing to do is to get the zero s of R(r) = BesselJ(r). We can create a program to compute these. Go b ack to the graph and see how far the zeros are apart. Then, solve for \+ a zero in an interval." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 109 "z ero[0]:=0;\nfor n from 1 to 4 do\n zero[n]:=fsolve(BesselJ(0,x)=0, x,zero[n-1]+1..zero[n-1]+4);\nod;\nn:='n';" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 85 "Pretty good. No surprise. These zeros are so important that Maple already knows them." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 72 "for n from 1 to 10 do\n zero[n]: =evalf(BesselJZeros(0,n));\nod;\nn:='n';" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 77 "To make further comp arisons, we plot the first six of the eigenfuntions cos( " }{XPPEDIT 18 0 "n*Pi/a;" "6#*(%\"nG\"\"\"%#PiGF%%\"aG!\"\"" }{TEXT -1 55 " x) an d the first six of the eigenfunctions BesselJ(0, " }{XPPEDIT 18 0 "lam bda[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,yellow, brown]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 91 "plot([seq(BesselJ(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 orthogonality. In order to meet t he 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 zero is only an approximat ion." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 131 "for n from 1 to 3 d o\nfor m from 1 to 3 do\n print(int(BesselJ(0,zero[n]*x)*BesselJ(0,z ero[m]*x)*x,x=0..1));\nod; od;\nn:='n';m:='m';" }}}{PARA 0 "" 0 "" {TEXT -1 40 "Now is time to compute the coefficients." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 155 "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);\nod;\nn:='n':" }}}{PARA 0 "" 0 "" {TEXT -1 66 "The next two graphs are so close, I offset the first one \+ by 0.007." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 91 "plot([f(x)+0.00 7,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 71 "plots[cylinderplot]([r,theta,u(1,r)],r=0..1,theta=- Pi..Pi,axes=NORMAL);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 88 "ani mate3d([r,theta,u(t,r)],r=0..1,theta=-Pi..Pi,t=0..1,\n coor ds=cylindrical);" }}}{PARA 0 "" 0 "" {TEXT -1 134 "We remark in passin g that the following is an alternate way to do a cylinder plot that do es not require a call of the package \"plots\"." }}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 65 "plot3d([r,theta,u(0,r)],r=0..1,theta=-Pi..Pi,c oords=cylindrical);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 19 "V erify the Solution" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 86 "I emphasize that solutions should be checked. Here, I sho w 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 s olution for the equation for 0 < r < 1, and t > 0. If the interval we re 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+-%$expG6#,$*(&F26#F*\"\"#%\"tG F+*$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 the 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 67 "for n from 1 to 3 do\n zero[n]:=BesselJZeros(0,n) /a;\nod;\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(d iff(u(t,r),t)-1/r*diff(r*diff(u(t,r),r),r));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" } }}}{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" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 1 1 2 33 1 1 }