{VERSION 6 0 "IBM INTEL NT" "6.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 2 1 2 0 0 0 1 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 2 2 2 0 0 0 1 }{CSTYLE "2D Comment" -1 18 "Times" 0 1 0 0 0 0 0 0 2 2 2 2 0 0 0 1 }{CSTYLE "_cstyle1" -1 205 "Times" 1 18 0 0 0 0 0 1 0 2 2 2 0 0 0 1 }{CSTYLE "_cstyle2" -1 206 "Times" 0 1 0 0 0 0 0 0 0 2 2 2 0 0 0 1 } {CSTYLE "_cstyle3" -1 207 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 0 0 0 1 } {CSTYLE "_cstyle4" -1 208 "Times" 1 18 0 0 0 0 0 1 0 2 2 2 0 0 0 1 } {CSTYLE "_cstyle5" -1 209 "Times" 0 1 0 0 0 0 0 0 1 2 2 2 0 0 0 1 } {PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 2 2 2 0 0 0 1 }0 0 0 -1 -1 -1 1 0 1 0 2 2 -1 1 }{PSTYLE "_pstyle1" -1 201 1 {CSTYLE "" -1 -1 "" 1 18 0 0 0 0 0 1 0 2 2 2 0 0 0 1 }3 0 0 -1 -1 -1 1 0 1 0 2 2 -1 1 }{PSTYLE "_pstyle2" -1 202 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 2 2 2 0 0 0 1 }0 0 0 -1 -1 -1 1 0 1 0 2 2 -1 1 }{PSTYLE "_pstyle3" -1 203 1 {CSTYLE "" -1 -1 "Courier" 0 1 255 0 0 1 0 1 0 2 1 2 0 0 0 1 }0 0 0 -1 -1 -1 1 0 1 0 2 2 -1 1 }{PSTYLE "_pstyle4" -1 204 1 {CSTYLE "" -1 -1 "" 1 18 0 0 0 0 0 1 0 2 2 2 0 0 0 1 }1 0 0 0 8 4 1 0 1 0 2 2 -1 1 }{PSTYLE "_pstyle5" -1 205 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 1 0 2 2 2 0 0 0 1 }0 0 0 -1 -1 -1 1 0 1 0 2 2 -1 1 } {PSTYLE "_pstyle6" -1 206 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 2 2 2 0 0 0 1 }0 0 0 -1 -1 -1 1 0 1 0 2 2 -1 1 }} {SECT 0 {PARA 201 "" 0 "" {TEXT 205 43 "Module 19: Convection Across t he Boundaries" }}{PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 "" {TEXT 206 103 " Suppose that we \+ take a well insulated rod -- sides and ends -- which has initial heat \+ distribution" }}{PARA 202 "" 0 "" {TEXT 206 52 " \+ 4 x (1 - x) + 2, 0 < x < 1." }}{PARA 202 "" 0 "" {TEXT 206 364 "We \+ know the 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 of the original distribution. After all, no heat \+ has been lost or gained in this well-insulated rod. Here is a graph of the initial distribution of heat, and the value of the total heat in \+ the system." }}{EXCHG {PARA 203 "> " 0 "" {MPLTEXT 1 0 37 "f:=x->4*x*( 1-x)+2;\nplot(f(x),x=0..1);" }}}{EXCHG {PARA 203 "> " 0 "" {MPLTEXT 1 0 17 "int(f(x),x=0..1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 " " }}}{EXCHG {PARA 203 "" 0 "" {TEXT -1 0 "" }}}{PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 "" {TEXT 206 240 "Maybe, we find this p roblem even boring now. Consider this alternative. Keep the rod well i nsulated, except at one end. There heat escapes according to Newton's \+ Law of Cooling, with an outside temperature of 2. More specifically, W e suppose" }}{PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 "" {TEXT 206 17 "The PDE: " } {TEXT 207 1 " " }{XPPEDIT 2 0 "diff(w,t) = diff(w,`$`(x,2))" "6#/-%%di ffG6$%\"wG%\"tG-F%6$F'-%\"$G6$%\"xG\"\"#" }{TEXT 206 2 " ," }}{PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 "" {TEXT 206 27 "The Bounda ry Conditions: " }{XPPEDIT 18 0 "diff(w,x);" "6#-%%diffG6$%\"wG%\"xG " }{TEXT 206 16 " (t, 0) = 0 and " }{XPPEDIT 18 0 "diff(w,x);" "6#-%%d iffG6$%\"wG%\"xG" }{TEXT 206 25 " (t,1) = - (w(t, 1) - 2)." }}{PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 "" {TEXT 206 51 "The Initia l Condition: w(0, x) = 4 x (1-x) + 2." }}{PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 "" {TEXT 206 681 "Now this problem is more in teresting. The heat begins to diffuse, causing temperatures to rise on the right and left of 1/2. As soon as the temperature goes above 2 on the right, the rod begins to cool by heat loss at that end. Heat will continue to leak out until the temperature over the entire rod is 2. \+ But what happens at the left side. Surely heat will start to spread th ere from the middle. With no leakage out the right side or anywhere el se, the temperature at the left would rise to 8/3. How high will it go now? Will it go above 2, and then settle back down? And what about th e hot spot? Will it move left as heat leaks out the right side, or wil l it drop straight down?" }}{PARA 202 "" 0 "" {TEXT 206 66 " Maybe you agree this problem will be interesting to consider." }}{PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 "" {TEXT 206 101 "The PDE is ho mogeneous, but the boundary conditions are not. Homogeneous boundary c onditions would be" }}{PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 "" {TEXT 206 27 "The Boundary Cond itions: " }{XPPEDIT 18 0 "diff(u,x);" "6#-%%diffG6$%\"uG%\"xG" } {TEXT 206 16 " (t, 0) = 0 and " }{XPPEDIT 18 0 "diff(u,x);" "6#-%%diff G6$%\"uG%\"xG" }{TEXT 206 19 " (t,1) = - u(t, 1)." }}{PARA 202 "" 0 " " {TEXT -1 0 "" }}{PARA 202 "" 0 "" {TEXT 206 137 "Thus, we need to fi nd a particular soluition, or, what's called the steady state solution . Is it clear that the steady state solution is " }}{PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 "" {TEXT 206 24 " v(x) = \+ 2?" }}{PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 "" {TEXT 206 153 "That solution satisfies the PDE and the non-homogeneous boundary \+ conditions. The w which satisfied the original problem is related to t he u by u = w - 2." }}{PARA 202 "" 0 "" {TEXT 206 95 " Now, we try to find the general solution for the PDE with homogeneous boundary co nditions." }}{PARA 202 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 204 "" 0 "" {TEXT 208 48 "The general solution for the homogeneous problem" }} {PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 "" {TEXT 206 100 "By \+ now, we know this partial differential equation changes into two ordin ary differential equations:" }}{PARA 202 "" 0 "" {TEXT -1 0 "" }} {PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 "" {TEXT 206 17 " \+ X '' = " }{XPPEDIT 18 0 "mu;" "6#%#muG" }{TEXT 206 40 " X, with \+ X'(0) = 0 and X '(1) +X(1) = 0," }}{PARA 202 "" 0 "" {TEXT 206 3 "and " }}{PARA 202 "" 0 "" {TEXT 206 16 " T ' = " }{XPPEDIT 18 0 " mu;" "6#%#muG" }{TEXT 206 3 " T." }}{PARA 202 "" 0 "" {TEXT -1 0 "" }} {PARA 202 "" 0 "" {TEXT 206 13 "The constant " }{XPPEDIT 18 0 "mu;" "6 #%#muG" }{TEXT 206 27 " must be negative, call it " }{XPPEDIT 18 0 "-l ambda^2;" "6#,$*$%'lambdaG\"\"#!\"\"" }{TEXT 206 42 " . Solutions for \+ the differential equation" }}{PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 "" {TEXT 206 28 " X '' = " }{XPPEDIT 18 0 "-lambda^2;" "6#,$*$%'lambdaG\"\"#!\"\"" }{TEXT 206 3 " X " }}{PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 "" {TEXT 206 19 "are \+ A sin( " }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT 206 13 " x) + B cos( " }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT 206 4 " x)." } }{PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 "" {TEXT 206 33 "We \+ invoke the boundary conditons." }}{EXCHG {PARA 203 "> " 0 "" {MPLTEXT 1 0 38 "X:=x->A*sin(lambda*x)+B*cos(lambda*x);" }}}{PARA 202 "" 0 "" {TEXT 206 42 "The left boundary condition is used first." }}{EXCHG {PARA 203 "> " 0 "" {MPLTEXT 1 0 10 "D(X)(0)=0;" }}}{PARA 202 "" 0 "" {TEXT 206 29 "This condition implies A = 0." }}{EXCHG {PARA 203 "> " 0 "" {MPLTEXT 1 0 5 "A:=0;" }}}{PARA 202 "" 0 "" {TEXT 206 44 "We now \+ use the remaining boundary condition." }}{EXCHG {PARA 203 "> " 0 "" {MPLTEXT 1 0 15 "D(X)(1)+X(1)=0;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 "" {TEXT 206 22 "This implies that tan(" }{XPPEDIT 18 0 "lambda;" "6#% 'lambdaG" }{TEXT 206 4 ") = " }{XPPEDIT 18 0 "1/lambda;" "6#*&\"\"\"F$ %'lambdaG!\"\"" }{TEXT 206 2 " ." }}{PARA 202 "" 0 "" {TEXT -1 0 "" }} {PARA 202 "" 0 "" {TEXT 206 13 "We must find " }{XPPEDIT 18 0 "lambda; " "6#%'lambdaG" }{TEXT 206 41 " 's that satisfy this requirement. Then , " }{XPPEDIT 18 0 "mu;" "6#%#muG" }{TEXT 206 3 " = " }{XPPEDIT 18 0 " -lambda^2;" "6#,$*$%'lambdaG\"\"#!\"\"" }{TEXT 206 59 " and the eigen function corresponding to this eigenvalue is" }}{PARA 202 "" 0 "" {TEXT 206 30 " cos( " }{XPPEDIT 18 0 "lambda; " "6#%'lambdaG" }{TEXT 206 4 " x)." }}{PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 "" {TEXT 206 28 "The general solution will be" }} {PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 "" {TEXT 206 20 " \+ u(t, x) = " }{XPPEDIT 18 0 "sum(c[n]*exp(-lambda[n]^2*t)*cos(lam bda[n]*x),n);" "6#-%$sumG6$*(&%\"cG6#%\"nG\"\"\"-%$expG6#,$*&&%'lambda G6#F*\"\"#%\"tGF+!\"\"F+-%$cosG6#*&&F26#F*F+%\"xGF+F+F*" }{TEXT 206 2 " ." }}{PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 "" {TEXT 206 25 "We have only to find the " }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT 206 22 " 's and then the c 's." }}{PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 "" {TEXT 206 88 " Is it now clear that there i s going to be no nice, closed form solution for these " }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT 206 160 " 's? We must find them numer ically. To help in doing this, we look to see about where they lie by \+ looking at where the graph of tan(x) crosses the graph of 1/x." }} {EXCHG {PARA 203 "> " 0 "" {MPLTEXT 1 0 67 "plot([tan(x),1/x],x=0..20, y=0..3/2,discont=true,color=[RED,BLACK]);" }}}{PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 "" {TEXT 206 57 "We see there are roots betwe en the places where tan(x) = " }{XPPEDIT 18 0 "infinity;" "6#%)infinit yG" }{TEXT -1 20 ", where cos(x) = 0. " }}{PARA 202 "" 0 "" {TEXT 206 31 " Maybe that will be enough." }}{EXCHG {PARA 203 "> " 0 "" {MPLTEXT 1 0 40 "lambda[1]:=fsolve(tan(x)=1/x,x,0..Pi/2);" }{MPLTEXT 1 0 46 "\nlambda[2]:=fsolve(tan(x)=1/x,x,Pi/2..3*Pi/2);" }{MPLTEXT 1 0 48 "\nlambda[3]:=fsolve(tan(x)=1/x,x,3*Pi/2..5*Pi/2);" }{MPLTEXT 1 0 48 "\nlambda[4]:=fsolve(tan(x)=1/x,x,5*Pi/2..7*Pi/2);" }{MPLTEXT 1 0 48 "\nlambda[5]:=fsolve(tan(x)=1/x,x,7*Pi/2..9*Pi/2);" }{MPLTEXT 1 0 49 "\nlambda[6]:=fsolve(tan(x)=1/x,x,9*Pi/2..11*Pi/2);" }{MPLTEXT 1 0 50 "\nlambda[7]:=fsolve(tan(x)=1/x,x,11*Pi/2..13*Pi/2);" }}}{PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 "" {TEXT 206 47 "We now hav e seven terms of the general solution" }}{PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 "" {TEXT 206 20 " u(t, x) = " }{XPPEDIT 18 0 "sum(c[n]*exp(-lambda[n]^2*t)*co s(lambda[n]*x),n);" "6#-%$sumG6$*(&%\"cG6#%\"nG\"\"\"-%$expG6#,$*&&%'l ambdaG6#F*\"\"#%\"tGF+!\"\"F+-%$cosG6#*&&F26#F*F+%\"xGF+F+F*" }{TEXT 206 2 " ." }}{PARA 202 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 203 "" 0 "" {TEXT -1 0 "" }}}} {PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 "" {TEXT -1 0 "" }} {PARA 202 "" 0 "" {TEXT 206 157 "We go back to solve the original prob lem. We need the general solution plus the particular solutions. These add together to give the soluion to this problem." }}{PARA 202 "" 0 " " {TEXT -1 0 "" }}{PARA 202 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 204 "" 0 "" {TEXT 208 37 "The solution for the original problem" }}{PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 "" {TEXT 206 75 "Now, we fi nd the c's by using the initial conditon: f(x) = u(0, x) + 2" }} {PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 "" {TEXT 206 22 "or \+ f(x) -2 = " }{XPPEDIT 18 0 "sum(c[n]*cos(lambda[n]*x),n);" "6# -%$sumG6$*&&%\"cG6#%\"nG\"\"\"-%$cosG6#*&&%'lambdaG6#F*F+%\"xGF+F+F*" }{TEXT 206 2 " ." }}{PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 " " {TEXT 206 24 "We compute coefficients." }}{EXCHG {PARA 203 "> " 0 " " {MPLTEXT 1 0 7 "n:='n':" }{MPLTEXT 1 0 21 "\nfor n from 1 to 7 do" } {MPLTEXT 1 0 49 "\n c[n]:=int((f(x)-2)*cos(lambda[n]*x), x=0..1)/" } {MPLTEXT 1 0 52 "\n int(cos(lambda[n]*x)^2,x=0..1); " }{MPLTEXT 1 0 4 "\nod;" }{MPLTEXT 1 0 8 "\nn:='n':" }}}{PARA 202 "" 0 "" {TEXT 206 43 "We check to see how close this is to our f." }} {EXCHG {PARA 203 "> " 0 "" {MPLTEXT 1 0 56 "plot([f(x),2+sum(c[n]*cos( lambda[n]*x),n=1..7)],x=0..1);" }}}{EXCHG {PARA 203 "" 0 "" {TEXT -1 0 "" }}}{PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 "" {TEXT 206 46 "We now define the solution and plot its graph." }}{PARA 202 "" 0 " " {TEXT -1 0 "" }}{EXCHG {PARA 203 "> " 0 "" {MPLTEXT 1 0 66 "w:=(t,x) ->2+sum(c[n]*exp(-lambda[n]^2*t)*cos(lambda[n]*x),n=1..7);" }}}{EXCHG {PARA 203 "> " 0 "" {MPLTEXT 1 0 65 "plot3d(w(t,x),x=0..1,t=0..1/10,ax es=NORMAL,orientation=[-20,60]);" }}}{EXCHG {PARA 203 "> " 0 "" {MPLTEXT 1 0 7 "n:='n':" }}}{PARA 202 "" 0 "" {TEXT 206 52 "This graph was the solution of the original problem." }}}{PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 "" {TEXT 206 97 "In order to watch the maximum move from one side to the \+ other, we animate u(t, x) as t increases." }}{EXCHG {PARA 203 "> " 0 " " {MPLTEXT 1 0 12 "with(plots):" }{MPLTEXT 1 0 33 "\nanimate(w(t,x),x= 0..1,t=0..1/8);" }}}{PARA 202 "" 0 "" {TEXT 206 71 "We ask: At what ti me would the left end point have maximum temperature?" }}{EXCHG {PARA 203 "> " 0 "" {MPLTEXT 1 0 27 "plot(w(t,0),t=0..1,y=0..3);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG }{PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 "" {TEXT 209 11 "Assignment:" }{TEXT 206 155 " Increase or decrease the rate of heat loss at the right side and see how the middle moves. Such a problem would change the boundar y conditions as follows:" }}{PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 "" {TEXT 206 27 "The Bounda ry Conditions: " }{XPPEDIT 18 0 "diff(u,x);" "6#-%%diffG6$%\"uG%\"xG " }{TEXT 206 16 " (t, 0) = 0 and " }{XPPEDIT 18 0 "diff(u,x);" "6#-%%d iffG6$%\"uG%\"xG" }{TEXT 206 11 " (t,1) = - " }{XPPEDIT 18 0 "alpha;" "6#%&alphaG" }{TEXT 206 15 " (u(t, 1) - 2) " }}{PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 "" {TEXT 206 6 "where " }{XPPEDIT 18 0 "alpha;" "6#%&alphaG" }{TEXT 206 36 " is more than one, or less than o ne." }}{PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 "" {TEXT 206 74 "Remark in Preparation: Here is syntax for drawing the graph of tan (s)-1/s." }}{EXCHG {PARA 203 "> " 0 "" {MPLTEXT 1 0 56 "plot(tan(x)-1/ x,x=0..20,y=-1..1,discont=true,color=RED);" }}}{EXCHG {PARA 203 "> " 0 "" {MPLTEXT 1 0 47 "plot([seq(w(p/15,x),p=0..7)],x=0..1,color=RED); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 203 " " 0 "" {TEXT -1 0 "" }}}{PARA 205 "" 0 "" {TEXT -1 0 "" }}{PARA 202 " " 0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 202 "" 0 "" {TEXT -1 0 "" }}{PARA 206 "" 0 "" {TEXT -1 0 "" }}}{MARK "0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 } {PAGENUMBERS 0 1 2 33 1 1 }