{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 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 257 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 258 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 259 "" 0 1 0 0 0 0 1 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 0 1 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 14 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 264 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 265 "" 0 1 0 0 0 0 1 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 "" 1 12 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 268 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 269 "" 1 12 0 0 0 0 0 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 "" 1 12 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 272 "" 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 "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 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 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 263 75 "Section \+ 9.2: Characteristics for First Order Partial Differential Equations" } }{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT 267 30 "Maple Packages for Section 9.2" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(plo ts):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "with(PDEtools):" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 164 "In this Section 9.2, we try to understand some of the underlying geometric structure for solutions o f a first order partial differential equation which has the form" }} {PARA 0 "" 0 "" {TEXT -1 18 " p(t, x) " }{XPPEDIT 18 0 "u[t]; " "6#&%\"uG6#%\"tG" }{TEXT -1 16 "(t,x) + q(t, x) " }{XPPEDIT 18 0 "u[ x];" "6#&%\"uG6#%\"xG" }{TEXT -1 28 "(t,x) + r(t, x) u(t, x) = 0." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 217 "To try t o get an understanding of this geometric structure, we suggest three o rdinary differential equations. We use functions T, X and Z as the unk nown functions in the three ordinary differential equations. We use [ " }{TEXT 265 1 "t" }{TEXT -1 2 ", " }{XPPEDIT 18 0 "x;" "6#%\"xG" } {TEXT -1 59 "] to designate an arbitrary point in the upper half plane : " }{TEXT 264 1 "t" }{TEXT -1 61 " will tell how far the point is loc ated up the time axes and " }{XPPEDIT 18 0 "x;" "6#%\"xG" }{TEXT -1 192 " will give the x - coordinate of the point. Having those letters \+ reserved compels us to choose a different variable for the independent variable in the definition of T, X, and Z. We choose s." }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 49 "Here, then, are th e three differential equations:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 21 " " }{XPPEDIT 18 0 "di ff(T,s);" "6#-%%diffG6$%\"TG%\"sG" }{TEXT -1 8 "(s) = p(" }{TEXT 256 1 "T" }{TEXT -1 26 "(s), X(s)), with T(0) = 0," }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 22 " " } {XPPEDIT 18 0 "diff(X,s);" "6#-%%diffG6$%\"XG%\"sG" }{TEXT -1 8 "(s) = q(" }{TEXT 257 1 "T" }{TEXT -1 24 "(s), X(s)), with X(0) = " } {XPPEDIT 18 0 "eta;" "6#%$etaG" }{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 3 "and" }}{PARA 0 "" 0 "" {TEXT -1 22 " \+ " }{XPPEDIT 18 0 "diff(Z,s);" "6#-%%diffG6$%\"ZG%\"sG" }{TEXT -1 15 "( s) = v(s) Z(s)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 77 "where v(s) will be defined below. We ask the following fo r functions p and q:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT 268 25 "Requirements for p and q:" }}{PARA 0 "" 0 "" {TEXT -1 9 "(a) p(t, " }{XPPEDIT 18 0 "x;" "6#%\"xG" }{TEXT -1 28 ") > 0 for al l t > 0 and all " }{XPPEDIT 18 0 "x;" "6#%\"xG" }{TEXT -1 2 ", " }} {PARA 0 "" 0 "" {TEXT -1 88 "(b) solutions for the first two ordinary \+ differential equations exist for all s > 0, and" }}{PARA 0 "" 0 "" {TEXT -1 74 "(c) if T(s) and X(s) are solutions for the first two equa tions and if [t, " }{XPPEDIT 18 0 "x;" "6#%\"xG" }{TEXT -1 56 "] is a \+ point in the plane with t > 0, then the equations" }}{PARA 0 "" 0 "" {TEXT -1 46 " T(s) = t and X(s) = " } {XPPEDIT 18 0 "x;" "6#%\"xG" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 17 "can be solved for" }{TEXT 266 1 " " }{TEXT -1 6 "s and " }{XPPEDIT 18 0 "eta;" "6#%$etaG" }{TEXT -1 120 ". In some sense, this last condition tells how long it takes and \+ where to start on the x axes in order to arrive at [t, " }{XPPEDIT 18 0 "x;" "6#%\"xG" }{TEXT -1 42 "] along the parametric curve [T(s), X(s )]." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 123 "T he effect of requirement (a) is that T(s) will be increasing and the e ffect of requirement (c) is that if you pick a point" }}{PARA 0 "" 0 " " {TEXT -1 15 " [t, " }{XPPEDIT 18 0 "x;" "6#%\"xG" }{TEXT -1 2 "] " }}{PARA 0 "" 0 "" {TEXT -1 59 "in the upper half plane, then some trajectory determined by" }}{PARA 0 "" 0 "" {TEXT -1 39 " \+ [T(s), X(s)]" }}{PARA 0 "" 0 "" {TEXT -1 77 "will \+ intersect that point. Furthermore, it can be determined from what poin t " }{XPPEDIT 18 0 "eta;" "6#%$etaG" }{TEXT -1 50 " on the x axes that trajectory began and for what " }{TEXT 258 1 " " }{TEXT -1 26 "time s it arrived at [t, " }{XPPEDIT 18 0 "x;" "6#%\"xG" }{TEXT -1 2 "]." } }{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 36 "Geometr ically, this locates a point " }{XPPEDIT 18 0 "eta;" "6#%$etaG" } {TEXT -1 131 " on the x axes so that a point following along the traje ctory given parametrically by the functions T(s) and X(s) and beginnin g at " }{XPPEDIT 18 0 "eta;" "6#%$etaG" }{TEXT -1 16 " arrives at [t, \+ " }{XPPEDIT 18 0 "x;" "6#%\"xG" }{TEXT -1 59 "] in time s. This parame tric trajectory, [T(s), X(s)] is a " }{TEXT 259 20 "characteristic cur ve" }{TEXT -1 40 " for the partial differential equation. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 69 "Let's draw some c haracteristic curves that arise with the simple pde." }}{PARA 0 "" 0 " " {TEXT -1 15 " 1 " }{XPPEDIT 18 0 "diff(u,t);" "6#-%%diff G6$%\"uG%\"tG" }{TEXT -1 5 " + 2 " }{XPPEDIT 18 0 "diff(u,x);" "6#-%%d iffG6$%\"uG%\"xG" }{TEXT -1 5 " + 3 " }{TEXT 260 1 "u" }{TEXT -1 50 "( t, x) = 0, with initial condition u(0, x) = exp( " }{XPPEDIT 18 0 "-x^ 2;" "6#,$*$%\"xG\"\"#!\"\"" }{TEXT -1 3 " )." }}{PARA 0 "" 0 "" {TEXT -1 88 "The ordinary differential equations associated with the charact eristics for this pde are" }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 38 " dT/ds = 1, T(0) = 0 and " }}{PARA 0 "" 0 "" {TEXT -1 32 " dX/ds = 2, X(0) = " }{XPPEDIT 18 0 "eta;" "6#%$etaG" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 53 "These have easy solutions: T(s) = s and X (s) = 2 s + " }{XPPEDIT 18 0 "eta;" "6#%$etaG" }{TEXT -1 1 "." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 22 "We draw s ome of these." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 137 "for n from -6 to 3 do\n J[n]:=spacecurve([2*s+n,s,0],s=0..3,axes=normal,\n \+ orientation=[-70,70],view=[-7..7,0..3,0..1]):\nend do:" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "display3d([seq(J[n],n=-6..3) ]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 174 "Making connections back \+ with the previous Section 9.1, we would have predicted that solutions \+ \"moved\" to the northeast in this pde, and the characteristics confir m that idea." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 251 "We derive the solution surface by thinking of the smooth surfa ce u(T(s), X(s)) above the characteristics. Here is a way to describe \+ the evolution of the surface. Choose r(t, x) from the partial differen tial equation and solve the differential equation" }}{PARA 0 "" 0 "" {TEXT -1 29 " " }{XPPEDIT 18 0 "diff(Z,s); " "6#-%%diffG6$%\"ZG%\"sG" }{TEXT -1 15 "(s) = v(s) Z(s)" }}{PARA 0 " " 0 "" {TEXT -1 6 "where " }}{PARA 0 "" 0 "" {TEXT -1 44 " \+ v(s) = r( T(s), X(s) )." }}{PARA 0 "" 0 "" {TEXT -1 25 "Then, define u implicitly" }}{PARA 0 "" 0 "" {TEXT -1 74 " \+ u(T(s), X(s) ) = Z(s), or one might say u(T, X) = Z(s)." }}{PARA 0 " " 0 "" {TEXT -1 15 "It follows that" }}{PARA 0 "" 0 "" {TEXT -1 26 " \+ p(T, X) " }{XPPEDIT 18 0 "u[t];" "6#&%\"uG6#%\"tG" } {TEXT -1 11 " + q(T, X) " }{XPPEDIT 18 0 "u[x];" "6#&%\"uG6#%\"xG" } {TEXT -1 24 " + r(T, X) u = T ' (t) " }{XPPEDIT 18 0 "u[t];" "6#&%\"u G6#%\"tG" }{TEXT -1 11 " + X ' (t) " }{XPPEDIT 18 0 "u[x];" "6#&%\"uG6 #%\"xG" }{TEXT -1 14 " + r(T, X) u =" }}{PARA 0 "" 0 "" {TEXT -1 26 " \+ " }{XPPEDIT 18 0 "diff(` `,t);" "6#-%%diffG6$ %\"~G%\"tG" }{TEXT -1 45 "u(T(t),X(t)) + r(T(t), X(t)) u(T(t), X(t)) = " }{XPPEDIT 18 0 "diff(Z,t);" "6#-%%diffG6$%\"ZG%\"tG" }{TEXT -1 17 " + v(t) Z(t) = 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 38 "Because Z(0) = u( T(0), X(0) ) = u(0, " }{XPPEDIT 18 0 "e ta;" "6#%$etaG" }{TEXT -1 58 "), we know how the initial condition for Z should be made:" }}{PARA 0 "" 0 "" {TEXT -1 31 " \+ Z(0) = u(0, " }{XPPEDIT 18 0 "eta;" "6#%$etaG" }{TEXT -1 2 ")." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 261 7 "SUMMARY" }}{PARA 0 "" 0 "" {TEXT -1 80 "We have a recipe for providing a soluti on for the partial differential equation." }}{PARA 0 "" 0 "" {TEXT -1 72 "(1) Solve for the characteristics T(s) and X(s) with initial condi tions." }}{PARA 0 "" 0 "" {TEXT -1 44 "(2) Define v(s) by v(s) = r( T(s), X(s) )." }}{PARA 0 "" 0 "" {TEXT -1 59 "(3) Solve the differenti al equation for Z with Z(0) = u(0, " }{XPPEDIT 18 0 "eta;" "6#%$etaG" }{TEXT -1 2 ")." }}{PARA 0 "" 0 "" {TEXT -1 26 "(4) Solve the equation s T(" }{XPPEDIT 18 0 "s;" "6#%\"sG" }{TEXT -1 9 ") = t, X(" }{XPPEDIT 18 0 "s;" "6#%\"sG" }{TEXT -1 4 ") = " }{XPPEDIT 18 0 "x;" "6#%\"xG" } {TEXT -1 11 " for s and " }{XPPEDIT 18 0 "eta;" "6#%$etaG" }{TEXT -1 10 " creating " }{XPPEDIT 18 0 "s;" "6#%\"sG" }{TEXT -1 10 "(t,x) and \+ " }{XPPEDIT 18 0 "eta;" "6#%$etaG" }{TEXT -1 6 "(t,x)." }}{PARA 0 "" 0 "" {TEXT -1 44 "(5) Compose u(T, X) = U(T(s), X(s)) = Z(s)." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 50 "To help m ake this clear, we work several examples." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT 269 9 "Example 1" }}{PARA 0 "" 0 "" {TEXT -1 26 "The characteristics for 1 " }{XPPEDIT 18 0 "diff(u,t); " "6#-%%diffG6$%\"uG%\"tG" }{TEXT -1 5 " + 2 " }{XPPEDIT 18 0 "diff(u, x);" "6#-%%diffG6$%\"uG%\"xG" }{TEXT -1 5 " + 3 " }{TEXT 262 1 "u" } {TEXT -1 11 "(t, x) = 0," }}{PARA 0 "" 0 "" {TEXT -1 29 "are T(s) = s, X(s) = 2 s + " }{XPPEDIT 18 0 "eta;" "6#%$etaG" }{TEXT -1 1 "." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 68 "The diffe rential equation determining u along the characteristics is" }}{PARA 0 "" 0 "" {TEXT -1 14 " " }{XPPEDIT 18 0 "diff(Z,s);" "6# -%%diffG6$%\"ZG%\"sG" }{TEXT -1 30 " + 3 Z = 0, with Z(0) = exp(- " } {XPPEDIT 18 0 "eta^2;" "6#*$%$etaG\"\"#" }{TEXT -1 2 ")," }}{PARA 0 " " 0 "" {TEXT -1 36 "which has solution exp(-3 s) exp(- " }{XPPEDIT 18 0 "eta^2;" "6#*$%$etaG\"\"#" }{TEXT -1 3 "). " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 44 "We invert the relationshi p s = t and 2 s + " }{XPPEDIT 18 0 "eta;" "6#%$etaG" }{TEXT -1 3 " = \+ " }{XPPEDIT 18 0 "x;" "6#%\"xG" }{TEXT -1 23 " to get that t = s and \+ " }{XPPEDIT 18 0 "eta;" "6#%$etaG" }{TEXT -1 2 "= " }{XPPEDIT 18 0 "x; " "6#%\"xG" }{TEXT -1 7 " - 2 t." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 51 "Using the solution for the z equation, we have that" }}{PARA 0 "" 0 "" {TEXT -1 60 " u(t, x) = u(T(s), X (s) ) = Z(s) = exp(- 3 s) exp( - " }{XPPEDIT 18 0 "eta^2;" "6#*$%$etaG \"\"#" }{TEXT -1 21 ") = exp(-3 t) exp( - " }{XPPEDIT 18 0 "(x-2*t)^2; " "6#*$,&%\"xG\"\"\"*&\"\"#F&%\"tGF&!\"\"F(" }{TEXT -1 1 ")" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 23 "We check this s olution." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "u:=(t,x)->exp(-3 *t)*exp(-(x-2*t)^2);" }}}{PARA 0 "" 0 "" {TEXT -1 81 "First, we see th at it really is a solution for the partial differential equation." }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "simplify(diff(u(t,x),t)+2*di ff(u(t,x),x)+3*u(t,x));" }}}{PARA 0 "" 0 "" {TEXT -1 54 "Then, we draw a graph using the format we always have." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 63 "plot3d(u(t,x),x=-3..3,t=0..2,axes=normal,orientation= [-45,65]);" }}}{PARA 0 "" 0 "" {TEXT -1 72 "It is of interest to see t hat Maple can do all of graphing. Start again." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "u:='u';" }}}{PARA 0 "" 0 "" {TEXT -1 15 "Define t he PDE." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "pde:=diff(u(t,x), t)+2*diff(u(t,x),x)+3*u(t,x)=0;" }}}{PARA 0 "" 0 "" {TEXT -1 28 "Give \+ the initial conditions." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "i nit:=[0,x,exp(-x^2)];" }}}{PARA 0 "" 0 "" {TEXT -1 39 "Make the graph \+ of the characteristics.." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 101 "PDEplot(pde,u(t,x),init,x=-3..3,axes=normal,scene=[x,t,u],t=0..2,orie ntation=[-65,65],basechar=only);" }}}{PARA 0 "" 0 "" {TEXT -1 29 "Make a graph of the solution." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 91 "PDEplot(pde,u(t,x),init,x=-3..3,axes=normal,scene=[x,t,u],\n t=0..2 ,orientation=[-65,65]);" }}}{PARA 0 "" 0 "" {TEXT -1 60 "Show both the graph of the solution and the characteristics." }}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 121 "PDEplot(pde,u(t,x),init,x=-3..3,axes=normal,s cene=[x,t,u],\n t=0..2,orientation=[-65,65],basechar=true,style=wire frame);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 199 "Finally, in the previous Section 9.1, we used Maple to provide an analytic solution to this partial differential equation with little assistance required from the user. Those ideas could be r eviewed." }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT 271 9 "Example 2" }}{PARA 0 "" 0 "" {TEXT -1 255 "We explain how humans might get the analytic solution to Example 2 from the previous Section 9.1. Already there, we saw that Maple could make the solution . To emphasize that these problems have a pattern, we almost repeat th e entire scheme from Example 1." }}{PARA 0 "" 0 "" {TEXT -1 26 "The ch aracteristics for 1 " }{XPPEDIT 18 0 "diff(u,t);" "6#-%%diffG6$%\"uG% \"tG" }{TEXT -1 5 " + x " }{XPPEDIT 18 0 "diff(u,x);" "6#-%%diffG6$%\" uG%\"xG" }{TEXT -1 5 " + t " }{TEXT 270 1 "u" }{TEXT -1 11 "(t, x) = 0 ," }}{PARA 0 "" 0 "" {TEXT -1 24 "are T(s) = s, X(s) = " }{XPPEDIT 18 0 "eta;" "6#%$etaG" }{TEXT -1 8 " exp(s)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 68 "The differential equation deter mining u along the characteristics is" }}{PARA 0 "" 0 "" {TEXT -1 14 " " }{XPPEDIT 18 0 "diff(Z,s);" "6#-%%diffG6$%\"ZG%\"sG" } {TEXT -1 29 " + s Z = 0, with Z(0) = cos( " }{XPPEDIT 18 0 "eta;" "6#% $etaG" }{TEXT -1 2 ")." }}{PARA 0 "" 0 "" {TEXT -1 24 "This has soluti on exp( " }{XPPEDIT 18 0 "-s^2/2;" "6#,$*&%\"sG\"\"#F&!\"\"F'" } {TEXT -1 8 " ) cos( " }{XPPEDIT 18 0 "eta;" "6#%$etaG" }{TEXT -1 3 "). " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 45 "We i nvert the relationship s = t and exp(s) " }{XPPEDIT 18 0 "eta;" "6#%$ etaG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "x;" "6#%\"xG" }{TEXT -1 23 " t o get that t = s and " }{XPPEDIT 18 0 "eta;" "6#%$etaG" }{TEXT -1 2 "= " }{XPPEDIT 18 0 "x*exp(-t);" "6#*&%\"xG\"\"\"-%$expG6#,$%\"tG!\"\"F% " }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 51 "Using the solution for the z equation, we have that" }} {PARA 0 "" 0 "" {TEXT -1 47 " u(t, x) = u(T(s), X(s) ) = Z(s) = exp( " }{XPPEDIT 18 0 "-s^2/2;" "6#,$*&%\"sG\"\"#F&!\"\"F'" }{TEXT -1 7 ") cos( " }{XPPEDIT 18 0 "eta;" "6#%$etaG" }{TEXT -1 8 ") = exp( " }{XPPEDIT 18 0 "-t^2/2;" "6#,$*&%\"tG\"\"#F&!\"\"F'" }{TEXT -1 7 ") \+ cos( " }{XPPEDIT 18 0 "x*exp(-t);" "6#*&%\"xG\"\"\"-%$expG6#,$%\"tG!\" \"F%" }{TEXT -1 3 " )." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 23 "We check this solution." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "u:=(t,x)->exp(-t^2/2)*cos(x*exp(-t));" }}}{PARA 0 "" 0 "" {TEXT -1 81 "First, we see that it really is a solution for the p artial differential equation." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "simplify(diff(u(t,x),t)+x*diff(u(t,x),x)+t*u(t,x));" }}}{PARA 0 "" 0 "" {TEXT -1 54 "Then, we draw a graph using the format we always \+ have." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 63 "plot3d(u(t,x),x=-3. .3,t=0..2,axes=normal,orientation=[-45,65]);" }}}{PARA 0 "" 0 "" {TEXT -1 72 "It is of interest to see that Maple can do all of graphin g. Start again." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "u:='u';" } }}{PARA 0 "" 0 "" {TEXT -1 15 "Define the PDE." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "pde:=diff(u(t,x),t)+x*diff(u(t,x),x)+t*u(t,x)=0; " }}}{PARA 0 "" 0 "" {TEXT -1 28 "Give the initial conditions." }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "init:=[0,x,cos(x)];" }}} {PARA 0 "" 0 "" {TEXT -1 39 "Make the graph of the characteristics.." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 101 "PDEplot(pde,u(t,x),init,x =-3..3,axes=normal,scene=[x,t,u],t=0..2,orientation=[-65,65],basechar= only);" }}}{PARA 0 "" 0 "" {TEXT -1 29 "Make a graph of the solution. " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 91 "PDEplot(pde,u(t,x),init, x=-3..3,axes=normal,scene=[x,t,u],\n t=0..2,orientation=[-65,65]);" }}}{PARA 0 "" 0 "" {TEXT -1 167 "Show both the graph of the solution a nd the characteristics. Rotate the graph around a little to distinguis h between the characteristics and the graph of the solution." }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 121 "PDEplot(pde,u(t,x),init,x=- 3..3,axes=normal,scene=[x,t,u],\n t=0..2,orientation=[-65,65],basech ar=true,style=wireframe);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 159 "Finally, in the previous Section 9 .1, we used Maple to provide an analytic to this partial differential \+ equation with little assistance required from the user." }}}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT 272 9 "Example 3 " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 258 "We w ill not step through the way humans solve the first order PDE in this \+ example. We let Maple do all the work. The importance of this example \+ is to get a glimpse of what happens when characteristics coalesce. Her e is the example. We start with a generic u." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "u:='u':" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "pde:=diff(u(t,x),t)-x*diff(u(t,x),x)+0*u(t,x)=0;" }}{PARA 0 "" 0 " " {TEXT -1 31 "We give the initial conditions." }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 40 "init:=[0,x,exp(-(x-3)^2)+exp(-(x+3)^2)];" }}} {PARA 0 "" 0 "" {TEXT -1 50 "Maple will make the graph of the characte ristics.." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 101 "PDEplot(pde,u( t,x),init,x=-5..5,axes=normal,scene=[x,t,u],t=0..2,orientation=[-65,65 ],basechar=only);" }}}{PARA 0 "" 0 "" {TEXT -1 32 "Here is a graph of \+ the solution." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 91 "PDEplot(pde ,u(t,x),init,x=-5..5,axes=normal,scene=[x,t,u],\n t=0..2,orientation =[-65,65]);" }}}{PARA 0 "" 0 "" {TEXT -1 175 "Next we show both the gr aph of the solution and the characteristics. Rotate the graph around a little to distinguish between the characteristics and the graph of th e solution." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 121 "PDEplot(pde, u(t,x),init,x=-5..5,axes=normal,scene=[x,t,u],\n t=0..2,orientation= [-65,65],basechar=true,style=wireframe);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 141 "Now, we recall that , with just a little help, Maple can solve this pde analytically.Recal l Section 9.1 from which the following syntax comes." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "sol3:=pdsolve(pde);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "u3:=unapply(rhs(sol3),(t,x));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "u3(0,x)=exp(-(x-3)^2)+exp(-(x+3)^2) ;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 105 "u3s:=unapply(subs(_F1 (x*exp(t))=\n exp(-(x*exp(t)-3)^2)+exp(-(x*exp(t)+3)^2),\n \+ u3(t,x)),(t,x));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "simplif y(diff(u3s(t,x),t)-x*diff(u3s(t,x),x)+0*u3s(t,x));\nu3s(0,x);" }}} {PARA 0 "" 0 "" {TEXT -1 184 "Finally, we plot our analytic solution. \+ It will make you happy to have the above PDEplot, even though the numb er of points on the plot of the analytic solution is stepped up to 160 0 = " }{XPPEDIT 18 0 "40^2;" "6#*$\"#S\"\"#" }{TEXT -1 1 "." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 89 "plot3d(u3s(t,x),x=-5..5,t=0..2,axes =NORMAL,\n orientation=[-55,65],numpoints=1600);" }}}{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 -1 178 "We have just begun to introduce the notions from first o rder partial differential equations. With this introduction, the inter ested reader will be able to make other experiments." }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 50 "EMAIL: herod@math.gatec h.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 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 0 "" }}}{MARK "0 0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 } {PAGENUMBERS 1 1 2 33 1 1 }