{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 "" 1 12 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 }{CSTYLE "" -1 259 "" 1 12 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 260 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 261 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 262 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 263 "" 1 12 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 264 "" 1 12 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 265 "" 1 12 0 0 0 0 0 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 266 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 267 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 268 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 269 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 270 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 271 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 272 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 273 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 274 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 275 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 276 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 277 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 278 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 279 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{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 "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 "Normal" -1 256 1 {CSTYLE "" -1 -1 "Times" 1 14 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 {EXCHG {PARA 18 "" 0 "" {TEXT -1 40 "Partial Differential Equa tions PowerTool" }}{PARA 19 "" 0 "" {TEXT -1 16 "by Dr. Jim Herod" }}} {PARA 256 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 41 "Secti on 4.4: Convection Across Boundaries" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT 257 30 "Maple Packages for Section 4.4 " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(plots):" }}}}{PARA 0 "" 0 "" {TEXT -1 5 " " }}{PARA 0 "" 0 "" {TEXT -1 136 "Suppose that we tak e a well insulated rod -- sides and ends insulated-- which has initial heat distribution as indicated by the function" }}{PARA 0 "" 0 "" {TEXT -1 27 " 4 " }{TEXT 266 1 "x" }{TEXT -1 6 " (1 - " }{TEXT 267 1 "x" }{TEXT -1 14 ") + 2, 0 < " }{TEXT 268 1 "x" }{TEXT -1 5 " < 1." }}{PARA 0 "" 0 "" {TEXT -1 385 "We can see t hat this rod will be hottest in the middle, and that as the heat diffu ses, a uniform distribution of heat will be achieved with total heat t he same as the total heat of the original distribution. After all, no \+ heat has been lost or gained in this well-insulated rod. Here is the d efinition of the initial distribution of heat, and the value of the to tal heat in the system." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "f :=x->4*x*(1-x)+2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "int(f( x),x=0..1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 294 "How to model the redistribution of heat was discussed in Section 4.3. Consider this al ternate problem. Keep the rod well insulated, except at one end. There heat escapes or gathers according as to whether it is more or less th an the outside temperature equal to 2. More specifically, We suppose" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 18 "The PDE: " }{XPPEDIT 18 0 "diff( u,t) = diff(u,`$`(x,2));" "6#/-%%diffG6$%\"uG%\"tG-F%6$F'-%\"$G6$%\"xG \"\"#" }{TEXT -1 2 " ," }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 27 "The Boundary Conditions: " }{XPPEDIT 18 0 "diff(u, x);" "6#-%%diffG6$%\"uG%\"xG" }{TEXT -1 16 " (t, 0) = 0 and " } {XPPEDIT 18 0 "diff(u,x);" "6#-%%diffG6$%\"uG%\"xG" }{TEXT -1 12 " (t, 1) = - (" }{TEXT 260 1 "u" }{TEXT -1 12 "(t, 1) - 2)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 27 "The Initial Condition: " }{TEXT 261 1 "u" }{TEXT -1 23 "(0, x) = 4 x (1-x) + 2." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 903 "Now this problem is more interesting. From the equation for the distribution f or the initial temperature, we can calculate that the temperature is 2 at the two ends and 3 in the exact middle. The heat begins to diffuse , causing temperatures to rise or fall along the rod. As soon as the t emperature goes above 2 on the right, the rod begins to cool by heat l oss at that end. Heat will continue to leak out until the temperature \+ over the entire rod is 2. But what happens at the left side. Surely he at will start to spread there from the middle. We calculated in the li nes above that with no leakage out the right side or anywhere else, th e temperature at the left would rise to 8/3. How high will it go now w ith this leakage out the right side? Will it go above 2, and then sett le back down? And what about the hot spot? Will it move left as heat l eaks out the right side, or will it drop straight down?" }}{PARA 0 "" 0 "" {TEXT -1 66 " Maybe you agree this problem will be interestin g to consider." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 101 "The PDE is homogeneous, but the boundary conditions are \+ not. Homogeneous boundary conditions would be" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 27 "The Boundary Conditions: " }{XPPEDIT 18 0 "diff(w,x);" "6#-%% diffG6$%\"wG%\"xG" }{TEXT -1 16 " (t, 0) = 0 and " }{XPPEDIT 18 0 "dif f(w,x);" "6#-%%diffG6$%\"wG%\"xG" }{TEXT -1 19 " (t,1) = - w(t, 1)." } }{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 160 "Thus, w e need to find a particular solution, or what we call the steady state solution, for the original problem. Is it clear that the steady state solution is " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 24 " v(x) = 2?" }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 159 "That solution satisfies the PDE and \+ the non-homogeneous boundary conditions. The u which satisfied the ori ginal problem is related to the transient solution w " }}{PARA 0 "" 0 "" {TEXT -1 13 "by u = w + 2." }}{PARA 0 "" 0 "" {TEXT -1 90 "Now, we \+ try to find the general solution for the PDE with homogeneous boundary conditions." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 " " {TEXT 264 48 "The general solution for the homogeneous problem" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 246 "By now, \+ we know to perform what was called Step 1 in Section 4.1: using separa tion of variables create two ordinary differential equations with boun dary conditions whose solutions will determine the solution for the pa rtial differential equation:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 17 " X '' = " }{XPPEDIT 18 0 "mu;" "6#%#mu G" }{TEXT -1 40 " X, with X'(0) = 0 and X '(1) +X(1) = 0," }}{PARA 0 " " 0 "" {TEXT -1 3 "and" }}{PARA 0 "" 0 "" {TEXT -1 16 " T ' = " }{XPPEDIT 18 0 "mu;" "6#%#muG" }{TEXT -1 3 " T." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 99 "Step 2 was to solve the t wo ordinary differential equations with boundary conditions. The const ant " }{XPPEDIT 18 0 "mu;" "6#%#muG" }{TEXT -1 27 " must be negative, \+ call it " }{XPPEDIT 18 0 "-lambda^2;" "6#,$*$%'lambdaG\"\"#!\"\"" } {TEXT -1 42 " . Solutions for the differential equation" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 28 " \+ X '' = " }{XPPEDIT 18 0 "-lambda^2;" "6#,$*$%'lambdaG\"\"#!\"\"" } {TEXT -1 3 " X " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 19 "are A sin( " }{XPPEDIT 18 0 "lambda;" "6#%'lambda G" }{TEXT -1 13 " x) + B cos( " }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG " }{TEXT -1 4 " x)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 34 "We invoke the boundary conditions." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "X:=x->A*sin(lambda*x)+B*cos(lambda*x);" }}} {PARA 0 "" 0 "" {TEXT -1 42 "The left boundary condition is used first ." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "D(X)(0)=0;" }}}{PARA 0 "" 0 "" {TEXT -1 29 "This condition implies A = 0." }}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 5 "A:=0;" }}}{PARA 0 "" 0 "" {TEXT -1 44 "We now use the remaining boundary condition." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "D(X)(1)+X(1)=0;" }}}{EXCHG }{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 22 "This implies that tan(" }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT -1 4 ") = " }{XPPEDIT 18 0 "1/lamb da;" "6#*&\"\"\"F$%'lambdaG!\"\"" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 13 "We must find " }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT -1 41 " 's that satisfy this requi rement. Then, " }{XPPEDIT 18 0 "mu;" "6#%#muG" }{TEXT -1 3 " = " } {XPPEDIT 18 0 "-lambda^2;" "6#,$*$%'lambdaG\"\"#!\"\"" }{TEXT -1 65 " \+ and the eigenfunction corresponding to this eigenvalue is cos( " } {XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT -1 30 " x). We have only \+ to find the " }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT -1 4 " 's. " }}{PARA 0 "" 0 "" {TEXT -1 88 " Is it now clear that there is go ing to be no nice, closed form solution for these " }{XPPEDIT 18 0 "l ambda;" "6#%'lambdaG" }{TEXT -1 121 " 's? We must find them numericall y. To help in doing this, we look at where the graph of tan(x) crosses the graph of 1/x." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "plot([ tan(x),1/x],x=0..20,y=0..3/2,discont=true,color=[RED,BLACK]);" }}} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 93 "We see th ere are roots between all the places where tan(x) is zero and where it is undefined." }}{PARA 0 "" 0 "" {TEXT -1 32 " Maybe seven will b e enough." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 83 "for n from 1 to 7 do\n lambda[n]:=fsolve(tan(x)=1/x,x,(n-1)*Pi..(2*n-1)*Pi/2);\nod; " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG }{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 163 "We now have approxi mations for seven of the eigenvalues. We are ready to take Step 3: con struct the general solution for the partial differential equation. We \+ have" }}{PARA 0 "" 0 "" {TEXT -1 20 " w(t, x) = " }{XPPEDIT 18 0 "sum(c[n]*exp(-lambda[n]^2*t)*cos(lambda[n]*x),n);" "6#-%$sumG6$* (&%\"cG6#%\"nG\"\"\"-%$expG6#,$*&&%'lambdaG6#F*\"\"#%\"tGF+!\"\"F+-%$c osG6#*&&F26#F*F+%\"xGF+F+F*" }{TEXT -1 2 " ." }}{EXCHG }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 158 "We go back to solve the original problem. We need the ge neral solution plus the particular solutions. These add together to gi ve the solution to this problem." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT 265 37 " The solution for the original problem" }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 76 "Now, we find the c's by using the ini tial condition: f(x) = w(0, x) + 2" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 12 "or " }{TEXT 270 1 "f" } {TEXT -1 1 "(" }{TEXT 271 1 "x" }{TEXT -1 7 ") -2 = " }{XPPEDIT 18 0 " sum(c[n]*cos(lambda[n]*x),n);" "6#-%$sumG6$*&&%\"cG6#%\"nG\"\"\"-%$cos G6#*&&%'lambdaG6#F*F+%\"xGF+F+F*" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 24 "We compute coefficients. " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 133 "for n from 1 to 7 do\n \+ c[n]:=int((f(x)-2)*cos(lambda[n]*x), x=0..1)/\n i nt(cos(lambda[n]*x)^2,x=0..1);\nod;\nn:='n':" }}}{PARA 0 "" 0 "" {TEXT -1 41 "We check to see how close this is to our " }{TEXT 269 1 " f" }{TEXT -1 1 "." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "plot([f (x),2+sum(c[n]*cos(lambda[n]*x),n=1..7)],x=0..1);" }}}{EXCHG }{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 90 "We now define the solution, check to see that it really is a solution, and plot its gra ph." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "u:=(t,x)->2+sum(c[n]*exp(-lambda[n]^2*t)*cos(lambda[n ]*x),n=1..7);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT 258 18 "Check the solution" }}{PARA 0 "" 0 "" {TEXT -1 59 " We check the differential equation and boundary conditions." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "diff(u(t,x),t)-diff(u(t,x),x,x);" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "D[2](u)(t,0); D[2](u)(t,1) +(u(t,1)-2);" }}}{PARA 0 "" 0 "" {TEXT -1 81 "The above answers might \+ be a surprise. They are not all zeros. Not to worry.\nThe " }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT -1 234 "'s we computed above were \+ only an approximation for the real ones. (Remember that we used fsolve ?) So, the numbers computed just above that we thought would be zero a re approximations for zero. Of course! Just look at their magnitude." }}{PARA 0 "" 0 "" {TEXT -1 65 "Finally, we observe how good a fit we h ave for the initial value." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "plot([f(x),u(0,x)],x=0..1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT 259 18 "Graph the solution" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 62 "plot3d(u(t,x),x=0..1,t=0..3,axes=NORMAL,orientation=[-20,60]); " }}}{PARA 0 "" 0 "" {TEXT -1 117 "Does this graph have the properties we predicted? Does it appear that the graph changes from being the sa me shape at " }{TEXT 272 1 "t" }{TEXT -1 21 " = 0 as the graph of " } {TEXT 273 1 "f" }{TEXT -1 1 "(" }{TEXT 274 1 "x" }{TEXT -1 56 ") and t hat the graph changes toward being constant 2 as " }{TEXT 275 1 "t" } {TEXT -1 9 " evolves." }}}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 89 "We are satisfied that we have an approximation for t he solution for the original problem." }}}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 74 "In or der to watch the maximum move from one side to the other, we animate \+ " }{TEXT 276 1 "u" }{TEXT -1 1 "(" }{TEXT 277 1 "t" }{TEXT -1 2 ", " } {TEXT 278 1 "x" }{TEXT -1 5 ") as " }{TEXT 279 1 "t" }{TEXT -1 89 " in creases. Using the cursor, touch the graph of the animation which come s next. Use the " }{TEXT 262 22 "Move To The Next Frame" }{TEXT -1 63 " button to watch how the maximum temperature moves to the left." }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 62 "animate(u(t,x),x=0..1,t=0..1 ,frames=100,view=[0..1,1.5..3.5]);" }}}{PARA 0 "" 0 "" {TEXT -1 218 "W e ask: At what time would the left end point and right end point have \+ maximum temperatures? This graph gives some answer. The red graph is t he left endpoint temperature and the green is the right endpoint tempe rature." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "plot([u(t,0),u(t, 1)],t=0..2,y=0..3,color=[red,green]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 353 "This problem was important for it led to an eigenvalue p roblem for which we could not find the eigenvalues in closed form. It \+ was necessary to use numerical methods to generate these eigenvalues. \+ From an applications perspective, the boundary conditions are more rea listic for it is unlikely that the end points can be held fixed or per fectly insulated." }}{PARA 0 "" 0 "" {TEXT 256 0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT 263 16 "Unassisted Maple" }}{PARA 0 "" 0 "" {TEXT -1 183 "We have seen how Maple suggests separation of variables in previo us sections for the diffusion equation and variations. In this section , we allow Maple freedom on a different problem." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 112 "Perhaps here the user fe lt uncomfortable with having to make approximations for where solution s for the equation" }}{PARA 0 "" 0 "" {TEXT -1 11 " tan(" } {XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT -1 6 ") = 1/" }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 51 "should lie. After all, if we wanted 20 or 200 such " }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT -1 164 "'s, we would not want to g o searching for the intervals in which to seek solutions. The question we address is how to let Maple find these without human assistance." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 110 "The an swer is to ask the right question. This question is suggested by the f ollowing graphs of 1/x and tan(x)." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "plot([tan(x),1/x],x=0..20,y=-2..2, discont=true);" }} }{PARA 0 "" 0 "" {TEXT -1 122 "This graph makes clear the obvious. The re is a solution for the equation tan(x) = 1/x between successive odd \+ multiples of " }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 115 "/2. Thus, we let Maple determine these values unassisted by our searching out s pecific intervals for each solution." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 85 "for n from 0 to 7 do\n mu[n]:=fsolve(tan(x)=1/x,x,ma x(0,(2*n-1)*Pi/2)..(2*n+1)*Pi/2);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 3 "od;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG }{EXCHG }{EXCHG }{EXCHG }}{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 "Copyri ght \251 2003 by James V. Herod" }}{PARA 257 "" 0 "" {TEXT -1 19 "Al l rights reserved" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{MARK "0 0 0" 6 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 } {PAGENUMBERS 1 1 2 33 1 1 }