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This model takes the object to be a point source. If \+ we assume that we have a rod that is not insulated, so that we have di ffusion and radiation cooling, the equation becomes slightly more comp licated. We have an equation with a diffusion term and a radiation coo ling term:" }}{PARA 0 "" 0 "" {TEXT -1 13 " " }{XPPEDIT 18 0 "diff(u(t,x),t) = diff(u(t,x),`$`(x,2))-k*(u(t,x)-beta);" "6#/-%% diffG6$-%\"uG6$%\"tG%\"xGF*,&-F%6$-F(6$F*F+-%\"$G6$F+\"\"#\"\"\"*&%\"k GF5,&-F(6$F*F+F5%%betaG!\"\"F5F<" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 128 "We suppose that the ends of the rod are at some constant temperature. For this example, take the end points to have temperatur e " }{XPPEDIT 18 0 "beta;" "6#%%betaG" }{TEXT -1 170 " . Also, the rod should have an initial temperature. So as not to be too specific, tak e the initial temperature to be distributed over the rod as expressed \+ by a function " }{TEXT 259 1 "f" }{TEXT -1 1 "(" }{TEXT 260 1 "x" } {TEXT -1 61 "). We write the boundary condition and initial conditions as " }}{PARA 0 "" 0 "" {TEXT -1 21 " u( t, 0) = " }{XPPEDIT 18 0 "beta;" "6#%%betaG" }{TEXT -1 17 " and u( t, L) = " }{XPPEDIT 18 0 "beta;" "6#%%betaG" }{TEXT -1 19 ", with u( 0, x) = " }{TEXT 261 1 "f" }{TEXT -1 1 "(" }{TEXT 262 1 "x" }{TEXT -1 2 ")." }}{PARA 0 "" 0 "" {TEXT -1 33 "Here, the length of the rod is L." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 51 "This is the problem \+ we address in this section. If " }{XPPEDIT 18 0 "beta;" "6#%%betaG" } {TEXT -1 28 " were zero, this would be a " }{TEXT 263 42 "homogeneous \+ partial differential equation." }{TEXT -1 4 " If " }{XPPEDIT 18 0 "bet a;" "6#%%betaG" }{TEXT -1 22 " is not zero, it is a " }{TEXT 264 45 "n on-homogeneous partial differential equation" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 329 "A review of Se ction 3.2, Problem 3 might be appropriate here. As indicated in the di scussion of that problem, we divide the equation into two parts: the s teady state equation and the transient equation. We solve both these e quations and find that their sum provides a solution to the original p roblem. Here are the two equations:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{PARA 0 "" 0 "" {TEXT 265 24 "Steady State Equation: " }}{PARA 0 "" 0 "" {TEXT -1 13 " " }{XPPEDIT 18 0 "0 = diff(v(x),`$`(x,2 ))-k*(v(x)-beta);" "6#/\"\"!,&-%%diffG6$-%\"vG6#%\"xG-%\"$G6$F,\"\"#\" \"\"*&%\"kGF1,&-F*6#F,F1%%betaG!\"\"F1F8" }{TEXT -1 16 " with v(t, 0) \+ = " }{XPPEDIT 18 0 "beta;" "6#%%betaG" }{TEXT -1 15 " and v(t, L) = " }{XPPEDIT 18 0 "beta;" "6#%%betaG" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 17 " " }}{PARA 0 "" 0 "" {TEXT 266 19 "Transi ent Equation:" }}{PARA 0 "" 0 "" {TEXT -1 13 " " } {XPPEDIT 18 0 "diff(w(t,x),t) = diff(w(t,x),`$`(x,2))-k*w(t,x);" "6#/- %%diffG6$-%\"wG6$%\"tG%\"xGF*,&-F%6$-F(6$F*F+-%\"$G6$F+\"\"#\"\"\"*&% \"kGF5-F(6$F*F+F5!\"\"" }{TEXT -1 34 " with w(t, 0) = 0 and w(t, L) = \+ 0." }}{PARA 0 "" 0 "" {TEXT -1 77 " The sum of these two solutions pro vides a solution for the original problem." }}{SECT 0 {PARA 3 "" 0 "" {TEXT 267 24 "Steady State + Transient" }}{PARA 0 "" 0 "" {TEXT -1 117 "We show here that the steady state solution plus the transient so lution provides a solution for the original problem." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "u:=(t,x)->v(x)+w(t,x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "diff(u(t,x),t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 93 "diff(v(x),x,x)+k*(v(x)-beta)+subs(\{diff(w(t,x), t)=diff(w(t,x),x,x)-k*w(t,x)\},diff(u(t,x),t));" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 13 "collect(%,k);" }}}{PARA 0 "" 0 "" {TEXT -1 66 "We see that, because u(t, x) = v(x) + w(t, x), this last becomes " } }{PARA 0 "" 0 "" {TEXT -1 13 " " }{XPPEDIT 18 0 "diff(u(t, x),t) = diff(u(t,x),`$`(x,2))-k*(u(t,x)-beta);" "6#/-%%diffG6$-%\"uG6$ %\"tG%\"xGF*,&-F%6$-F(6$F*F+-%\"$G6$F+\"\"#\"\"\"*&%\"kGF5,&-F(6$F*F+F 5%%betaG!\"\"F5F<" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 31 "Bou ndary conditions check, too." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 44 " We solve these two equations, one at a time." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT 269 38 "Solution for the Stead y State Equation" }}{PARA 0 "" 0 "" {TEXT -1 372 "The notion of a stea dy state solution is, first, physical. With a system such as that whic h modeled radiation cooling, we had a non-homogeneous partial differen tial equation. The long range forecast was that no matter what the ini tial condition, the solution for the equation will evolve in time, set tling down to a solution that is not changing in time. This solution i s " }{TEXT 268 6 "steady" }{TEXT -1 9 " in time." }}{PARA 0 "" 0 "" {TEXT -1 225 " Such a notion is easy to characterize mathematicall y. It must be that the steady solution is, in fact, a solution of the \+ partial differential equation, but also it is not changing in time. Th e derivative with respect to " }{TEXT 295 1 "t" }{TEXT -1 9 " is zero. " }}{PARA 0 "" 0 "" {TEXT -1 40 " We say this again with an equati on." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 9 " \+ 0 = " }{XPPEDIT 18 0 "diff(u(t,x),`$`(x,2))-k*u(t,x)+k*beta;" "6#,(- %%diffG6$-%\"uG6$%\"tG%\"xG-%\"$G6$F+\"\"#\"\"\"*&%\"kGF0-F(6$F*F+F0! \"\"*&F2F0%%betaGF0F0" }{TEXT -1 1 "," }}{PARA 0 "" 0 "" {TEXT -1 4 "w ith" }}{PARA 0 "" 0 "" {TEXT -1 8 " " }{XPPEDIT 18 0 "u(t,0) = \+ beta;" "6#/-%\"uG6$%\"tG\"\"!%%betaG" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "u(t,1) = beta;" "6#/-%\"uG6$%\"tG\"\"\"%%betaG" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 209 " S ince this is a steady state, and time is not a factor, we could consid er this as actually an ordinary differential equations in x. We write \+ it with that notation and call solution which is a function of " } {TEXT 296 1 "x" }{TEXT -1 4 " by " }{TEXT 297 1 "v" }{TEXT -1 1 "(" } {TEXT 298 1 "x" }{TEXT -1 2 ")." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 18 " 0 = v''(x) - " }{XPPEDIT 18 0 "k;" " 6#%\"kG" }{TEXT -1 8 " v(x) + " }{TEXT 285 2 "k " }{XPPEDIT 18 0 "beta ;" "6#%%betaG" }{TEXT -1 14 ", with s(0) = " }{XPPEDIT 18 0 "beta;" "6 #%%betaG" }{TEXT -1 12 " and s(1) = " }{XPPEDIT 18 0 "beta;" "6#%%beta G" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 38 "The steady state sol ution is computed." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 74 "SState:=dsolve(\{diff(v(x),x,x)-k*(v(x)-beta )=0,v(0)=beta,v(L)=beta\},v(x));" }}}{PARA 0 "" 0 "" {TEXT -1 73 "We s hould not be surprised that if the endpoints are held at temperature \+ " }{XPPEDIT 18 0 "beta;" "6#%%betaG" }{TEXT -1 44 " and the surroundin g medium has temperature " }{XPPEDIT 18 0 "beta;" "6#%%betaG" }{TEXT -1 62 " then the long range forecast is that the rod has temperature \+ " }{XPPEDIT 18 0 "beta;" "6#%%betaG" }{TEXT -1 1 "." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT 284 35 "Solution for the Transient Equation" } }{PARA 0 "" 0 "" {TEXT -1 64 "We solve the transient equation. Recall \+ the equation from above:" }}{PARA 0 "" 0 "" {TEXT -1 11 " " }{XPPEDIT 18 0 "diff(w(t,x),t) = diff(w(t,x),`$`(x,2))-k*w(t,x);" "6#/ -%%diffG6$-%\"wG6$%\"tG%\"xGF*,&-F%6$-F(6$F*F+-%\"$G6$F+\"\"#\"\"\"*&% \"kGF5-F(6$F*F+F5!\"\"" }{TEXT -1 34 " with w(t, 0) = 0 and w(t, L) = \+ 0." }}{PARA 0 "" 0 "" {TEXT -1 90 "As in Section 4.1, the method is se paration of variables. Suppose that w can be written as" }}{PARA 0 "" 0 "" {TEXT -1 19 " w(t,x) = T(" }{TEXT 270 1 "t" }{TEXT -1 3 ") X(" }{TEXT 271 1 "x" }{TEXT -1 2 ")." }}{PARA 0 "" 0 "" {TEXT -1 40 "I n this case, the above equation becomes" }}{PARA 0 "" 0 "" {TEXT -1 2 " " }{TEXT 277 4 " " }{TEXT -1 15 "T ' X = T X '' " }{TEXT 272 2 " - " }{TEXT 273 1 "k" }{TEXT 274 1 " " }{TEXT -1 3 "T X" }}{PARA 0 "" 0 "" {TEXT -1 2 "or" }}{PARA 0 "" 0 "" {TEXT -1 29 " T '/ T \+ = (X ''- " }{TEXT 275 1 "k" }{TEXT -1 8 " X) / X." }}{PARA 0 "" 0 " " {TEXT -1 137 "With the same arguments as used in Section 4.1, this l ast equation leads to two ordinary differential equations with boundar y conditions:" }}{PARA 0 "" 0 "" {TEXT -1 27 " X '' - " }{TEXT 276 1 "k" }{TEXT -1 5 " X = " }{XPPEDIT 18 0 "mu;" "6#%#mu G" }{TEXT -1 25 " X, with X(0) = X(L) = 0," }}{PARA 0 "" 0 "" {TEXT -1 24 "and T ' = " }{XPPEDIT 18 0 "mu;" "6#%#muG" } {TEXT -1 3 " T." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 73 "We solve these two differential equations. We can change \+ the first one to" }}{PARA 0 "" 0 "" {TEXT -1 27 " X \+ '' = (" }{TEXT 280 1 "k" }{TEXT -1 3 " + " }{XPPEDIT 18 0 "mu;" "6#%#m uG" }{TEXT -1 28 " ) X, with X(0) = X(L) = 0. " }}{PARA 0 "" 0 "" {TEXT -1 162 "From arguments similar to those in Section 3.3, we get t he eigenvalues and eigenfunctions for this problem. We find that there are an infinite of possible values:" }}{PARA 0 "" 0 "" {TEXT -1 16 " \+ (" }{TEXT 281 1 "k" }{TEXT -1 3 " + " }{XPPEDIT 18 0 "mu ;" "6#%#muG" }{TEXT -1 5 " ) = " }{XPPEDIT 18 0 "-n^2*Pi^2/(L^2);" "6# ,$*(%\"nG\"\"#%#PiGF&*$%\"LGF&!\"\"F*" }{TEXT -1 18 ", n = 1, 2, 3, .. ." }}{PARA 0 "" 0 "" {TEXT -1 22 "Corresponding to each " }{TEXT 278 1 "n" }{TEXT -1 10 ", we have " }}{PARA 0 "" 0 "" {TEXT -1 15 " \+ X" }{XPPEDIT 18 0 "` `[n];" "6#&%\"~G6#%\"nG" }{TEXT -1 2 " (" }{TEXT 299 1 "x" }{TEXT -1 11 ") = sin( n " }{XPPEDIT 18 0 "pi;" "6#%# piG" }{TEXT -1 1 " " }{TEXT 300 1 "x" }{TEXT -1 5 "/L )." }}{PARA 0 " " 0 "" {TEXT -1 26 "The equation for T becomes" }}{PARA 0 "" 0 "" {TEXT -1 25 " T ' = ( " }{XPPEDIT 18 0 "-k-n^2*Pi^2/(L ^2);" "6#,&%\"kG!\"\"*(%\"nG\"\"#%#PiGF(*$%\"LGF(F%F%" }{TEXT -1 5 " ) T." }}{PARA 0 "" 0 "" {TEXT -1 35 "Solutions for this equation will b e" }}{PARA 0 "" 0 "" {TEXT -1 18 " T" }{XPPEDIT 18 0 " ` `[n];" "6#&%\"~G6#%\"nG" }{TEXT -1 2 " (" }{TEXT 279 1 "t" }{TEXT -1 10 ") = exp( -" }{TEXT 282 1 "k" }{TEXT -1 1 " " }{TEXT 301 1 "t" } {TEXT -1 7 ") exp( " }{XPPEDIT 18 0 "-n^2*Pi^2/(L^2)*t;" "6#,$**%\"nG \"\"#%#PiGF&*$%\"LGF&!\"\"%\"tG\"\"\"F*" }{TEXT -1 3 " )." }}{SECT 0 {PARA 3 "" 0 "" {TEXT 283 38 "We check these solutions for the ODE's" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "X:=x->sin(n*Pi*x/L);\nT:=t ->exp(-k*t)*exp(-n^2*Pi^2*t/L^2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "simplify(diff(X(x),x,x)-k*X(x)+(k+n^2*Pi^2/L^2)*X(x)) ;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "X(0); X(L);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "simplify(diff(T(t),t)+(k+n^2 *Pi^2/L^2)*T(t));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 152 "Having t hese solutions for the ordinary differential equations, we can constru ct the general solution for the homogeneous partial differential equat ion." }}{PARA 0 "" 0 "" {TEXT -1 26 " u( t, x ) = " } {XPPEDIT 18 0 "exp(-k*t)*sum(A[n]*exp(-n^2*Pi^2/(L^2)*t)*sin(n*Pi*x/L) ,n = 1 .. infinity);" "6#*&-%$expG6#,$*&%\"kG\"\"\"%\"tGF*!\"\"F*-%$su mG6$*(&%\"AG6#%\"nGF*-F%6#,$**F4\"\"#%#PiGF9*$%\"LGF9F,F+F*F,F*-%$sinG 6#**F4F*F:F*%\"xGF*FF.F%/F6;F%%)infinityGF%F%" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 73 "In doing our calculations, we use only a finite approxima tion for this u." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 103 "u:=(t,x)->beta+exp(-k*t)*sum(A[n]*exp(-n^2*Pi ^2/L^2*t)*sin(n*Pi*x/L),\n n=1..10); " }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT 286 39 "Check the solution for the original PDE" }}{PARA 0 "" 0 "" {TEXT -1 96 "We check that this i s a solution for the partial differential equation with boundary condi tions." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "diff(u(t,x),t)-dif f(u(t,x),x,x)+k*(u(t,x)-beta);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "u(t,0); u(t,L);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 " " }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 67 "To \+ get the particular solution that satisfies the initial condition" }} {PARA 0 "" 0 "" {TEXT -1 33 " f(x) = u(0, x) = " } {XPPEDIT 18 0 "beta;" "6#%%betaG" }{TEXT -1 3 " + " }{XPPEDIT 18 0 "su m(A[n]*sin(n*Pi*x/L),n = 1 .. infinity);" "6#-%$sumG6$*&&%\"AG6#%\"nG \"\"\"-%$sinG6#**F*F+%#PiGF+%\"xGF+%\"LG!\"\"F+/F*;F+%)infinityG" } {TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 37 "We see that getting the \+ coefficients " }{XPPEDIT 18 0 "A[n];" "6#&%\"AG6#%\"nG" }{TEXT -1 82 " is simply a matter of computing the Fourier coefficients for the func tion f(x) - " }{XPPEDIT 18 0 "beta;" "6#%%betaG" }{TEXT -1 1 "." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 62 "At this p oint, we have not specified a particular L, or k, or " }{XPPEDIT 18 0 "beta;" "6#%%betaG" }{TEXT -1 110 ", or f. We do this now so that ca lculations may be made, solutions obtained, and graphs drawn. Take L \+ = 2, " }{XPPEDIT 18 0 "beta;" "6#%%betaG" }{TEXT -1 7 " = 3, " } {TEXT 287 4 "k = " }{TEXT -1 11 "0.005, and " }}{PARA 0 "" 0 "" {TEXT 288 47 " f" }{TEXT -1 1 " (" }{TEXT 289 1 "x" }{TEXT -1 4 ") = " }{XPPEDIT 18 0 "x^2*(2-x)+3;" " 6#,&*&%\"xG\"\"#,&F&\"\"\"F%!\"\"F(F(\"\"$F(" }{TEXT -1 1 "." }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "L:=2; beta:=3; k:=5/1000; f: =x->x^3*(2-x)+3;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "f(x)=u( 0,x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 122 "for n from 1 to 1 0 do\n A[n]:=int((f(x)-beta)*sin(n*Pi*x/L),x=0..L)/\n int (sin(n*Pi*x/L)^2,x=0..L);\nod;\nn:='n':" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 32 "Since we have checke d that this " }{TEXT 302 1 "u" }{TEXT -1 120 " satisfies the partial d ifferential equation and the boundary conditions, we only check that i t is an approximation for " }{TEXT 290 1 "f" }{TEXT -1 1 "(" }{TEXT 291 1 "x" }{TEXT -1 36 "). In fact, we off set the graph of " }{TEXT 303 1 "f" }{TEXT -1 47 " a little so that we can distinguish these two ." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "plot([u(0,x),f(x)+0.01] ,x=0..L,u=0..5);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {PARA 0 "" 0 "" {TEXT -1 28 "Finally, we draw a graph of " }{TEXT 304 1 "u" }{TEXT -1 88 ". What we should expect is that the solution moves from a shape similar to the graph of " }{TEXT 292 1 "f" }{TEXT -1 24 " to the steady state of " }{XPPEDIT 18 0 "beta;" "6#%%betaG" }{TEXT -1 1 "." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 63 "plot3d(u(t,x),x=0 ..2,t=0..1,axes=normal,orientation=[-120,60]);" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 453 "What was new in this section was how to handle a \+ non-homogeneous partial differential equation. The technique was to fo rmulate the steady state equation and the transient equation. It is a \+ good idea to verify that this has been done correctly by checking that the sum of these two is a solution for the original problem. The next step is to solve these two equations, add the solutions, and determin e the appropriate coefficients using the Fourier Idea." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT 293 16 "Unassisted \+ Maple" }}{PARA 0 "" 0 "" {TEXT -1 143 "We check to see what Maple 8 ca n do with this equation unassisted. Be aware that we have called in th e PDE(tools) at the start of this Section." }}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 7 "u:='u':" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "PDE:=diff(u(t,x),t)-diff(u(t,x),x,x)+k*(u(t,x)-beta);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "ans:=pdsolve(PDE);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "build(ans);" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 372 "We see that Maple successfully recognizes that the solut ion should be a sum of a solution for the steady state equation plus a product of functions which will be solutions for the transient equati on. Because the solution was obtained without any information about th e boundary conditions, this will not be the same solution that we huma ns got with the assistance of Maple." }}{PARA 0 "" 0 "" {TEXT -1 73 "M aple handles this problem deftly with other techniques. See Section 8. 2." }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 50 "EMAIL: herod@math.gatech.edu or jhe rod@tds.net" }}{PARA 0 "" 0 "" {TEXT -1 38 "URL: http://www.math.gatec h.edu/~herod" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 36 "Copyright \251 2003 by James V. Herod" }}{PARA 256 "" 0 "" {TEXT -1 19 "All rights reserved" }}{PARA 0 "" 0 "" {TEXT -1 1 " \+ " }}}{MARK "0 0 0" 10 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 1 1 2 33 1 1 }