{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 14 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 257 "" 1 14 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 258 "" 0 14 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 259 "" 1 12 0 0 0 0 0 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 "Headi ng 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 18 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 "" 256 257 1 {CSTYLE "" -1 -1 "" 1 14 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Nor mal" -1 258 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 "" 256 259 1 {CSTYLE "" -1 -1 "" 1 14 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }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 259 "" 0 "" {TEXT -1 0 "" }}{PARA 257 "" 0 "" {TEXT 256 38 "Sect ion 7.1: The Heat Equation on a" }{TEXT 257 1 " " }{TEXT 258 9 "Rec tangle" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT 259 30 "Maple Packages for Section 7.1" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 486 "We consider the diffusion of heat into a long beam with cross section a rectangle. The supposition that the beam is \"long\" is to produce \+ the mathematical idea that heat diffusing to a point comes essentially only from the sides and that the ends are so far away that heat comin g from above or below can be ignored. This is the same problem as cons idering heat diffusion in a thin rectangular plate that is insulated a t the top and bottom. These two problems lead to an equation such as \+ " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 23 " \+ " }{XPPEDIT 18 0 "diff(u,t) = diff(u,`$`(x,2))+diff( u,`$`(y,2));" "6#/-%%diffG6$%\"uG%\"tG,&-%%diffG6$%\"uG-%\"$G6$%\"xG\" \"#\"\"\"-F+6$F--F/6$%\"yGF2F3" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 24 "with boundary conditions " }}{PARA 0 "" 0 "" {TEXT -1 34 " u( t, x, 0) = " } {XPPEDIT 18 0 "f[0]" "6#&%\"fG6#\"\"!" }{TEXT -1 18 ", u( t, x, b) = " }{XPPEDIT 18 0 "f[1]" "6#&%\"fG6#\"\"\"" }}{PARA 0 "" 0 "" {TEXT -1 34 " u( t, 0, y) = " }{XPPEDIT 18 0 "g[0]" "6#&% \"gG6#\"\"!" }{TEXT -1 18 ", u( t, a, 0) = " }{XPPEDIT 18 0 "g[1]" " 6#&%\"gG6#\"\"\"" }}{PARA 0 "" 0 "" {TEXT -1 22 "and initial condition s" }}{PARA 0 "" 0 "" {TEXT -1 44 " u( 0, x, y) = h ( x, y)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 93 "The problem is broken into two parts: the steady state v(x,y) and \+ the transient, w( t, x, y)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 27 "The steady-state problem is" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 29 " \+ 0 = " }{XPPEDIT 18 0 "diff(v,`$`(x,2))+diff(v,`$`(y,2));" "6#,&-%%diff G6$%\"vG-%\"$G6$%\"xG\"\"#\"\"\"-F%6$F'-F)6$%\"yGF,F-" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 26 "with boundary conditio ns " }}{PARA 0 "" 0 "" {TEXT -1 33 " v( t, x, 0) = \+ " }{XPPEDIT 18 0 "f[0]" "6#&%\"fG6#\"\"!" }{TEXT -1 18 ", v( t, x, b ) = " }{XPPEDIT 18 0 "f[1]" "6#&%\"fG6#\"\"\"" }}{PARA 0 "" 0 "" {TEXT -1 34 " v( t, 0, y) = " }{XPPEDIT 18 0 "g[0] " "6#&%\"gG6#\"\"!" }{TEXT -1 18 ", v( t, a, 0) = " }{XPPEDIT 18 0 " g[1]" "6#&%\"gG6#\"\"\"" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 84 "We have discussed how to break this \+ problem into two parts previously. They would be" }}{PARA 0 "" 0 "" {TEXT -1 8 " 0 = " }{XPPEDIT 18 0 "diff(v1,`$`(x,2))+diff(v1,`$`(y, 2));" "6#,&-%%diffG6$%#v1G-%\"$G6$%\"xG\"\"#\"\"\"-F%6$F'-F)6$%\"yGF,F -" }{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 24 "with boundary conditions" }}{PARA 0 "" 0 "" {TEXT -1 28 " v1( x, 0) = " }{XPPEDIT 18 0 "f[0]" "6#&%\"fG6#\"\"!" }{TEXT -1 16 ", v1( x, b) = " }{XPPEDIT 18 0 "f[1]" "6#&%\"fG6#\"\" \"" }}{PARA 0 "" 0 "" {TEXT -1 28 " v1( 0, y) = " } {XPPEDIT 18 0 "0;" "6#\"\"!" }{TEXT -1 16 ", v1( a, 0) = " } {XPPEDIT 18 0 "0;" "6#\"\"!" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 3 "and" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 8 " 0 = " } {XPPEDIT 18 0 "diff(v2,`$`(x,2))+diff(v2,`$`(y,2));" "6#,&-%%diffG6$%# v2G-%\"$G6$%\"xG\"\"#\"\"\"-F%6$F'-F)6$%\"yGF,F-" }{TEXT -1 2 ", " }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 24 "with boun dary conditions" }}{PARA 0 "" 0 "" {TEXT -1 46 " v2( x, 0) = 0, v2( x, b) = 0," }}{PARA 0 "" 0 "" {TEXT -1 28 " \+ v2( 0, y) = " }{XPPEDIT 18 0 "g[0];" "6#&%\"gG6#\"\"!" }{TEXT -1 16 ", v2( a, 0) = " }{XPPEDIT 18 0 "g[1];" "6#&%\"gG6#\"\"\"" } {TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 78 "The solution for these two problems are added to get the \+ steady state solution" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 45 " v(x, y) = v1(x, y) + v2(x, y)." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 173 "We will \+ not detail how to make this computation for the steady state solution, since we have covered how to compute solutions for Laplace's Equation on a rectangle earlier. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 24 "The transient problem is" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 25 " " } {XPPEDIT 18 0 "diff(w,t);" "6#-%%diffG6$%\"wG%\"tG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "diff(w,`$`(x,2))+diff(w,`$`(y,2));" "6#,&-%%diffG6$%\" wG-%\"$G6$%\"xG\"\"#\"\"\"-F%6$F'-F)6$%\"yGF,F-" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 26 "with boundary conditions \+ " }}{PARA 0 "" 0 "" {TEXT -1 53 " w( t, x, 0) = 0, \+ w( t, x, b) = 0" }}{PARA 0 "" 0 "" {TEXT -1 54 " \+ w( t, 0, y) = 0, w( t, a, 0) = 0" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 22 "with initial condition" }}{PARA 0 "" 0 " " {TEXT -1 55 " w( 0, x, y) = h( x, y) - v( x, y). " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 91 "Think about why u(t, x, y) = w(t, x, y) + v(x, y) is the solution for the o riginal problem." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 139 "We break this last PDE into three ODE's (separation of t hree variables, assuming w = X(x)Y(y)T(t)), two of which have boundary conditions. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 8 "X '' = -" }{XPPEDIT 18 0 "lambda^2;" "6#*$%'lambdaG\"\"#" }{TEXT -1 24 " X, Y '' = -" }{XPPEDIT 18 0 "mu^2;" "6#*$%# muG\"\"#" }{TEXT -1 27 " Y T ' = - ( " }{XPPEDIT 18 0 "l ambda^2+mu^2;" "6#,&*$%'lambdaG\"\"#\"\"\"*$%#muGF&F'" }{TEXT -1 4 " ) T" }}{PARA 0 "" 0 "" {TEXT -1 39 "X(0) = X(a) = 0 Y(0) = Y(b) \+ = 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 70 "The result is a general solution of the transient equation in the form" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 21 " w( t, x, y) = " }{XPPEDIT 18 0 "su m(sum(A[n,m]*sin(n*Pi*x/a)*sin(m*Pi*y/b)*exp(-(n^2/(a^2)+m^2/(b^2))*Pi ^2*t),n),m);" "6#-%$sumG6$-F$6$**&%\"AG6$%\"nG%\"mG\"\"\"-%$sinG6#**F, F.%#PiGF.%\"xGF.%\"aG!\"\"F.-F06#**F-F.F3F.%\"yGF.%\"bGF6F.-%$expG6#,$ *(,&*&F,\"\"#*$F5FCF6F.*&F-FC*$F;FCF6F.F.*$F3FCF.%\"tGF.F6F.F,F-" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 50 "The coeff icients are obtained by Fourier series as" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 16 " " }{XPPEDIT 18 0 " A[n,m] = int(int(w(0,x,y)*sin(n*Pi*x/a)*sin(m*Pi*y/b),y = 0 .. b),x = \+ 0 .. a)/(int(sin(n*Pi*x)^2,x = 0 .. a)*int(sin(m*Pi*y)^2,y = 0 .. b)); " "6#/&%\"AG6$%\"nG%\"mG*&-%$intG6$-F+6$*(-%\"wG6%\"\"!%\"xG%\"yG\"\" \"-%$sinG6#**F'F6%#PiGF6F4F6%\"aG!\"\"F6-F86#**F(F6F;F6F5F6%\"bGF=F6/F 5;F3FA/F4;F3F " 0 "" {MPLTEXT 1 0 111 "u:=(t,x,y)->sum(sum(A[n,m]* sin(n*Pi*x/a)*sin(m*Pi*y/b)*\n exp(-(n^2/a^2+m^2/b^2)*Pi^2*t),n =1..3),m=1..3);" }}}{PARA 0 "" 0 "" {TEXT -1 33 "Here are the boundary conditions." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "u(t,0,y); u( t,a,y);\nu(t,x,0); u(t,x,b);" }}}{PARA 0 "" 0 "" {TEXT -1 36 "This wou ld be the initial condition." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "u(0,x,y);" }}}{PARA 0 "" 0 "" {TEXT -1 21 "And, here is the PDE." } }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "simplify(diff(u(t,x,y),t)-d iff(u(t,x,y),x,x)-diff(u(t,x,y),y,y));" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 7 "Example" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 57 "We illustrate how an anim ation of such a result behaves. " }}{PARA 0 "" 0 "" {TEXT -1 37 "\nTak e boundary conditions as follows:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 37 " u( t, x, 0) = sin(" } {XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 22 " x), u( t, x, 1) = 0" }} {PARA 0 "" 0 "" {TEXT -1 60 " u( t, 0, y) = 0, \+ u( t, 1, y) = 0" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 30 "Take the initial condition as " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 37 " u( 0, \+ x, y) = sin(" }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 3 "x)." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 137 "By now, \+ maybe we can do the steady state without the computation of any integr als. The formula for the steady state, and a check follows." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "res tart;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "v:=(x,y)->(sinh(Pi *(1-y)))*sin(Pi*x)/sinh(Pi);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "diff(v(x,y),y,y)+diff(v(x,y),x,x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "v(0,y);v(1,y);v(x,0);v(x,1);" }}}{PARA 0 "" 0 "" {TEXT -1 93 "The function w will be the transient solution. The initia l condition for w is k, given below." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "k:=(x,y)->sin(Pi*x)-v(x,y);\nk(0,y);k(1,y);k(x,0);k(x ,1);" }}}{PARA 0 "" 0 "" {TEXT -1 312 "We now compute the coefficient s as directed above. I set this up to do 100 double integrals. You cou ld do more, or less depending on the speed of your computer. The point is that k(x,y) is not continuous on the closed rectangle making the d omain of u. Thus, we will not have a good approximation for u(0, x, y) ." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 202 "for n from 1 to 10 do \n for m from 1 to 10 do int(int(k(x,y)*sin(n*Pi*x)*sin(m*Pi*y),x=0.. 1),y=0..1)/\n (int(sin(n*Pi*x)^2,x=0..1)*int(sin(m*Pi*y)^2,y=0..1) ):\n A[n,m]:=evalf(%):\n end do:\nend do:" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 15 "n:='n': m:='m':" }}}{PARA 0 "" 0 "" {TEXT -1 33 "We make the transient solution w." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 87 "sum(sum(A[n,m]*sin(n*Pi*x)*sin(m*Pi*y)*exp(-(n^2+m^2) *Pi^2*t),\n n=1..10),m=1..10):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "w:=unapply(%,(t,x,y)):" }}}{PARA 0 "" 0 "" {TEXT -1 60 "From w and v, the solution for the original problem is made." }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "u:=(t,x,y)->w(t,x,y)+v(x,y); " }}}{PARA 0 "" 0 "" {TEXT -1 70 "As a check to see how accurate all t his is, u( 0, x, y) should be sin(" }{XPPEDIT 18 0 "Pi;" "6#%#PiG" } {TEXT -1 20 " x) in this problem." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "plot3d(u(1/100,x,y),x=0..1,y=0..1,axes=NORMAL);" }}} {PARA 0 "" 0 "" {TEXT -1 78 "Finally, here is an animation of the temp erature decaying to the steady state." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 62 "plots[animate3d](u(t,x,y),x=0..1,y=0..1,t=0..1/10, fr ames=50);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 28 "Example: S peed of Diffusion." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 89 "If we want to incorporate speed of diffusion, we should change the model by including c.:" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 23 " " }{XPPEDIT 18 0 "u[t](t ,x,y) = c*(diff(u,`$`(x,2))+diff(u,`$`(y,2)));" "6#/-&%\"uG6#%\"tG6%F( %\"xG%\"yG*&%\"cG\"\"\",&-%%diffG6$F&-%\"$G6$F*\"\"#F.-F16$F&-F46$F+F6 F.F." }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 31 "This time, we ch oose conditions" }}{PARA 0 "" 0 "" {TEXT -1 59 " u ( t, x, 0) = 100, u( t, x, 1) = 100" }}{PARA 0 "" 0 "" {TEXT -1 59 " u( t, 0, y) = 100, u( t, 1, y) = 100" }}{PARA 0 "" 0 "" {TEXT -1 22 "and initial conditions" }}{PARA 0 "" 0 "" {TEXT -1 37 " u( 0, x, y) = 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 86 "We ask, for various value s of c, what is the value of t such that u(t, 1/2, 1/2) = 50." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 81 "Is it cle ar that the steady state solutions is 100? And, that general solution \+ is" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 20 " \+ u( t, x, y) = " }{XPPEDIT 18 0 "sum(sum(A[n,m]*sin(n*Pi*x/a)*sin(m* Pi*y/b)*exp(-c*(n^2+m^2)*Pi^2*t),n),m);" "6#-%$sumG6$-F$6$**&%\"AG6$% \"nG%\"mG\"\"\"-%$sinG6#**F,F.%#PiGF.%\"xGF.%\"aG!\"\"F.-F06#**F-F.F3F .%\"yGF.%\"bGF6F.-%$expG6#,$**%\"cGF.,&*$F,\"\"#F.*$F-FDF.F.F3FD%\"tGF .F6F.F,F-" }{TEXT -1 7 " + 100," }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 6 "where " }}{PARA 0 "" 0 "" {TEXT -1 12 " \+ " }{XPPEDIT 18 0 "A[n,m] = -400*int(int(sin(n*Pi*x)*sin(m*Pi* y),y = 0 .. 1),x = 0 .. 1);" "6#/&%\"AG6$%\"nG%\"mG,$*&\"$+%\"\"\"-%$i ntG6$-F.6$*&-%$sinG6#*(F'F,%#PiGF,%\"xGF,F,-F46#*(F(F,F7F,%\"yGF,F,/F< ;\"\"!F,/F8;F?F,F,!\"\"" }{TEXT -1 8 " = -400 " }{XPPEDIT 18 0 "(cos(n *Pi)-1)*(cos(m*Pi)-1)/(n*Pi*m*Pi);" "6#*(,&-%$cosG6#*&%\"nG\"\"\"%#PiG F*F*F*!\"\"F*,&-F&6#*&%\"mGF*F+F*F*F*F,F***F)F*F+F*F1F*F+F*F," }{TEXT -1 1 "." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 141 "for n from 1 to \+ 20 do\n for m from 1 to 20 do\n A[n,m]:=-400*(cos(n*Pi)-1)/(n*Pi) *(cos(m*Pi)-1)/(m*Pi):\n end do: \nend do:\nn:='n': m:='m':" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 89 "sum(sum(A[n,m]*sin(n*Pi*x)*s in(m*Pi*y)*exp(-c*(n^2+m^2)*Pi^2*t),\n n=1..20),m=1..20):" }}} {PARA 0 "" 0 "" {TEXT -1 53 "We write this u as a function of t, x, y, and also c." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "u:=unapply(% ,(c,t,x,y))+100:" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 108 "Here, we plot the initial value for c = 1. The purpose i s to see how well we have fit the initial condition." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 61 "plot3d(u(1,0,x,y),x=0..1,y=0..1,axes=NORM AL, numpoints=1000);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 101 "si mplify(diff(u(c,t,x,y),t)-\n c*diff(u(c,t,x,y),x,x)-\n \+ c*diff(u(c,t,x,y),y,y));" }}}{PARA 0 "" 0 "" {TEXT -1 39 "Here, we check the boundary conditions." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "u(1,t,x,0);u(1,t,x,1);u(1,t,0,y);u(1,t,1,y);" }}} {PARA 0 "" 0 "" {TEXT -1 85 "This is an animation so that you can see \+ how the solution has limit the steady state." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 76 "plots[animate3d](u(1,t,x,y),x=0..1,y=0..1,t=0..1/1 0,axes=NORMAL, frames=50);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 207 "The goal now is to choose a sequ ence of c 's and, for each c, to find t so that the point [1/2, 1/2] i s at temperature half way between 0 and 100. We restrict solutions for t to being in the interval [0, 5]." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 80 "for p from 1 to 10 do\n s[p]:=fsolve(u(p/10,t,1/2,1 /2)=50.,t,0..5);\nod;\np:='p';" }}}{PARA 0 "" 0 "" {TEXT -1 136 "Here, we plot the values of t found above. The purpose is to see the kind o f relationship there is between the speed of diffusion and c." }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "plots[pointplot]([seq([p,s[p ]],p=1..10)]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG }{PARA 0 "" 0 "" {TEXT -1 236 "What this graph shows is that as c increases the time for the middle point to reach the temperate 50 d ecreases. The decrease seems like an exponential decay. Can you justif y that in the mathematics, or in an exponential fit to the data?" }} {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 258 "" 0 "" {TEXT -1 36 "Copyri ght \251 2003 by James V. Herod" }}{PARA 258 "" 0 "" {TEXT -1 19 "Al l rights reserved" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{MARK "0 0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 1 1 2 33 1 1 }