{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 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 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 264 "" 1 14 0 0 0 0 0 1 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 "" 1 12 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 268 "" 1 12 0 0 0 0 0 0 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 "" 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 }{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 264 74 "Section 9.1: An Introduction to First Or der Partial Differential Equations" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {SECT 0 {PARA 3 "" 0 "" {TEXT 265 31 "Maple Packages for Section 9.1 " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(plots):" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 107 "In this Section 9.1, we introduce the beginnings \+ of ideas about first order partial differential equations." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 58 "We begin with a f irst order partial differential equation:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 8 " " }{XPPEDIT 18 0 "1*diff(u, t);" "6#*&\"\"\"F$-%%diffG6$%\"uG%\"tGF$" }{TEXT -1 5 " + 2 " } {XPPEDIT 18 0 "diff(u,x);" "6#-%%diffG6$%\"uG%\"xG" }{TEXT -1 5 " = - \+ " }{TEXT 256 1 "u" }{TEXT -1 13 "(t, x) with " }{TEXT 257 1 "u" } {TEXT -1 14 "(0, x) = exp(-" }{XPPEDIT 18 0 "x^2;" "6#*$%\"xG\"\"#" } {TEXT -1 2 ")." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 23 "The equation is called " }{TEXT 266 11 "first order" } {TEXT -1 108 " because the partial derivatives are only of order one. \+ There are no second order partials in this equation." }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 451 "Previously, when consi dering a partial differential equation in these notes, we have discuss ed how a solution might be constructed, and then looked to see what Ma ple could do with obtaining a solution with little assistant from the \+ user. In this Section, we will break with the pattern that we have use d before, and look first at the solutions that Maple obtains. We will \+ try to point out the features to consider as we examine these several \+ examples." }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{SECT 0 {PARA 3 "" 0 "" {TEXT 267 10 "Example 1." }}{PARA 0 "" 0 "" {TEXT -1 105 "Of course, t he first thing is to declare what is the partial differential equation that we want to solve." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "P DE := diff(u(t,x),t)+2*diff(u(t,x),x)=-3*u(t,x);" }}}{PARA 0 "" 0 "" {TEXT -1 83 "The construction of a solution for this partial different ial equation is done with " }{TEXT 258 7 "pdsolve" }{TEXT -1 1 "." }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "sol1:=pdsolve(PDE);" }}} {PARA 0 "" 0 "" {TEXT -1 256 "It is no surprise that, just as with fir st order ordinary differential equations, there is one unknown in this general solution. It is to be found by using the initial condition. T oward using the initial condition to find _F1, we make the general sol ution." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "u1:=unapply(rhs(so l1),(t,x));" }}}{PARA 0 "" 0 "" {TEXT -1 75 "Next, we use the initial \+ condition to determine this unknown function, _F1." }}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 18 "u1(0,x)=exp(-x^2);" }}}{PARA 0 "" 0 "" {TEXT -1 110 "Thus, we substitute this value for _F1, and make the spe cial solution to the pde which has this initial value." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 61 "u1s:=unapply(subs(_F1(x-2*t)=exp(-( x-2*t)^2),u1(t,x)),(t,x));" }}}{PARA 0 "" 0 "" {TEXT -1 42 "Why not ch eck to see that all is in order?" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 57 "simplify(diff(u1s(t,x),t)+3*u1s(t,x)+2*diff(u1s(t,x),x));" }}} {PARA 0 "" 0 "" {TEXT -1 149 "We plot the solution. We visualize the i nitial value lying above the x axes when t = 0, and look to see how th e initial value evolves as t increases." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "plot3d(u1s(t,x),x=-3..3,t=0..2,axes=NORMAL,orientatio n=[-70,70]);" }}}{PARA 0 "" 0 "" {TEXT -1 230 "Rotate that graph aroun d a little and what you should observe is that the initial value moves toward zero at t increases and is pulled toward the \"northeast\". We try to develop a visualization of this by a sequence of graphs of u( " }{XPPEDIT 18 0 "t[n];" "6#&%\"tG6#%\"nG" }{TEXT -1 9 " , x) as " } {XPPEDIT 18 0 "t[n];" "6#&%\"tG6#%\"nG" }{TEXT -1 101 " increases. Ste p through the animation and watch as the peak drops and moves into the first quadrant." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 127 "for n f rom 0 to 10 do\n J[n]:=spacecurve([x,n/10,u1s(n/10,x)],x=-3..3,\n \+ axes=normal,orientation=[-85,65]):\nend do:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "display3d([seq(J[n],n=0..10)],inseq uence=true);" }}}{PARA 0 "" 0 "" {TEXT -1 220 "We ask: what changes in the pde could make the peak of the solution move toward the \"northwe st\", into the second quadrant. Here is an example which shows one pos sible change to accomplish that modification of the graph.." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "PDE := diff(u(t,x),t)-2*diff(u(t,x) ,x)=-3*u(t,x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "sol1:=pds olve(PDE);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "u1:=unapply(r hs(sol1),(t,x));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "u1(0,x) =exp(-x^2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 61 "u1s:=unapply (subs(_F1(x+2*t)=exp(-(x+2*t)^2),u1(t,x)),(t,x));" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 57 "simplify(diff(u1s(t,x),t)-2*diff(u1s(t,x),x) +3*u1s(t,x));" }}}{PARA 0 "" 0 "" {TEXT -1 15 "We changed one " } {TEXT 261 4 "plus" }{TEXT -1 6 " to a " }{TEXT 262 5 "minus" }{TEXT -1 115 " and expect the peak to change directions. Here is the graph t hat will show that this change accomplishes the task." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "plot3d(u1s(t,x),x=-3..3,t=0..2,axes=NORMA L,orientation=[-70,70]);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 98 "What changes in the pde could make the peak increa se instead of decreasing. Here is a possibility." }}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 49 "PDE := diff(u(t,x),t)+2*diff(u(t,x),x)= 3*u(t, x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "sol1:=pdsolve(PDE); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "u1:=unapply(rhs(sol1),( t,x));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "u1(0,x)=exp(-x^2) ;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 61 "u1s:=unapply(subs(_F1( x-2*t)=exp(-(x-2*t)^2),u1(t,x)),(t,x));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 57 "simplify(diff(u1s(t,x),t)+2*diff(u1s(t,x),x)-3*u1s(t, x));" }}}{PARA 0 "" 0 "" {TEXT -1 23 "We changed a different " }{TEXT 259 4 "plus" }{TEXT -1 6 " to a " }{TEXT 260 5 "minus" }{TEXT -1 48 " \+ and expect the peak to grow. Here is the graph." }}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 65 "plot3d(u1s(t,x),x=-3..3,t=0..1,axes=NORMAL,ori entation=[-70,70]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 207 "As a che ck to see that you understand this idea, you might go back and change \+ the PDE to get an increasing maximum moving into the second quadrant, \+ instead of into the first quadrant as in the above picture." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 241 "Just as in a first study of ordinary differential e quations, so here we started this examination with constant coefficien ts. What about variable coefficients? Some such equations can be tough , some are standard. Consider this second example:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 12 " " }{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 6 " = - \+ t" }{TEXT 270 2 " u" }{TEXT -1 12 "(t, x) with " }{TEXT 263 1 "u" } {TEXT -1 16 "(0, x) = cos(x)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 275 "The term on the right is negative. Shoul d we expect the peaks to decay to zero? The two partials are added if \+ x is positive and subtracted if x is negative. Should we expect the pe aks to move to the northeast in the right quadrant and to the northwes t in the second quadrant?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 29 "We look at this example next." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT 268 9 "Example 2" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 57 "We follow the pattern we have had before without comment." }}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 49 "PDE := diff(u(t,x),t)+x*diff(u(t,x),x)=-t*u(t, x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "sol2:=pdsolve(PDE); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "u2:=unapply(rhs(sol2),( t,x));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "u2(0,x)=cos(x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "u2s:=unapply(subs(_F1=cos ,u2(t,x)),(t,x));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "diff(u 2s(t,x),t)+x*diff(u2s(t,x),x)+t*u2s(t,x);" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 136 "plot3d(u2s(t,x),x=-2*Pi..2*Pi,t=0..3,axes=NORMAL, \n orientation=[65,65], lightmodel=light1, style=patchnogrid , numpoints=1000);" }}}{PARA 0 "" 0 "" {TEXT -1 142 "Rotate the graph \+ around. You should see that where x > 0 the graph moves toward the Nor theast, and where x < 0, it moves toward the Northwest." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 163 "for n from 0 to 10 do\n J[n]:=sp acecurve([x,n/10,u2s(n/10,x)],x=-2*Pi..2*Pi,\n axes=normal ,orientation=[-85,65],view=[-2*Pi..2*Pi,0..2,-1..1]):\nend do:" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "display3d([seq(J[n],n=0..10) ],insequence=true);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {PARA 0 "" 0 "" {TEXT -1 245 "These two example begin to give some ide as about the structure of solutions for first order PDE's. And, the qu estion begins to creep into the conscious: what happens if the graph m oves to the Northeast when x < 0 and to the Northwest when x > 0?" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 278 "We have illustrated a constant coefficie nt, first order pde and a variable coefficient, first order pde. We no w illustrate that some nonhomogeneous first order pde's can be solved. Recall that with ordinary differential equations, one had to exercise care if the coefficient of " }{TEXT 271 5 "du/dt" }{TEXT -1 139 " was zero at t = 0. That happens in this next example. Just as with those \+ examples in ordinary differential equations, so here we give the " } {TEXT 272 13 "initial value" }{TEXT -1 10 " at t = 1." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT 269 9 "Example 3" }} {PARA 0 "" 0 "" {TEXT -1 80 "Since the steps are the same as in the ab ove, we proceed without comment, again." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "PDE := t*diff(u(t,x),t)+x*diff(u(t,x),x)=t*x*(t^2+1); " }}}{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 12 "u3(1,x)=x^2;" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "u3s:=unapply(subs(_F1(x/t)=( x/t)^2-3*x/t/4,u3(t,x)),(t,x));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "simplify(t*diff(u3s(t,x),t)+x*diff(u3s(t,x),x)-t*x*(t^2+1));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "plot3d(u3s(t,x),x=-Pi..Pi ,t=1..2,axes=NORMAL,orientation=[65,65]);" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 157 "for n from 10 to 20 do\n J[n]:=spacecurve([x,n/1 0,u3s(n/10,x)],x=-3..3,\n axes=normal,orientation=[35,65], view=[-3..3,0..2,-5..5],color=red):\nod:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "display3d([seq(J[n],n=10..20)],insequence=true);" }}} {EXCHG }{EXCHG }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 154 "What is the status at this point. First, it should be cl ear that Maple can produce solutions and graphs for some partial diffe rential equation in the form" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 12 " p(t, x) " }{XPPEDIT 18 0 "u[t];" "6#&%\"uG6 #%\"tG" }{TEXT -1 13 " + q(t, x) " }{XPPEDIT 18 0 "u[x];" "6#&%\"uG6 #%\"xG" }{TEXT -1 11 " + r(t, x) " }{TEXT 273 1 "u" }{TEXT -1 16 "(t,x ) = s(t, x)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 153 "It would be interesting to find the limits of what Maple can d o with this type equation. We do not choose that route. Rather, we int roduce the notion of " }{TEXT 274 15 "characteristics" }{TEXT -1 132 " in the next section. This notion sheds some light on the geometry of \+ solutions for such first order partial differential equations." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 50 "EMAIL: he rod@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 256 "" 0 "" {TEXT -1 36 "Copyright \251 2003 by Jame s V. Herod" }}{PARA 256 "" 0 "" {TEXT -1 19 "All rights reserved" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{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 }