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{SECT 0 {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 264 74 "Section 9.1: An I
ntroduction to First Order Partial Differential Equations" }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT 265 31 "Maple Pac
kages for Section 9.1" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "res
tart:" }}}{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 introd
uce the beginnings of ideas about first order partial differential equ
ations." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1
58 "We begin with a first 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 "f
irst order" }{TEXT -1 108 " because the partial derivatives are only o
f order one. There are no second order partials in this equation." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 455 "Previous
ly, when considering a partial differential equation in these notes, w
e have discussed how a solution might be constructed, and then looked \+
to see what Maple could do with obtaining a solution and with little a
ssistant from the user. In this Section, we will break with the patter
n that we have used before, and look first at the solutions that Maple
obtains. We will try to point out the features to consider as we exam
ine these several examples." }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{SECT
1 {PARA 3 "" 0 "" {TEXT 267 10 "Example 1." }}{PARA 0 "" 0 "" {TEXT
-1 105 "Of course, the first thing is to declare what is the partial d
ifferential equation that we want to solve." }}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 49 "PDE := 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 thi
s partial differential equation is done with " }{TEXT 258 7 "pdsolve"
}{TEXT -1 1 "." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "sol1:=pdso
lve(PDE);" }}}{PARA 0 "" 0 "" {TEXT -1 256 "It is no surprise that, ju
st as with first order ordinary differential equations, there is one u
nknown in this general solution. It is to be found by using the initia
l condition. Toward using the initial condition to find _F1, we make t
he general solution." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "u1:=
unapply(rhs(sol1),(t,x));" }}}{PARA 0 "" 0 "" {TEXT -1 75 "Next, we us
e 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 special 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 check 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 visu
alize the initial value lying above the x axes when t = 0, and look to
see how the 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,o
rientation=[-70,70]);" }}}{PARA 0 "" 0 "" {TEXT -1 230 "Rotate that gr
aph around a little and what you should observe is that the initial va
lue moves toward zero at t increases and is pulled toward the \"northe
ast\". We try to develop a visualization of this by a sequence of grap
hs 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 " increase
s. Step through the animation and watch as the peak drops and moves in
to the first quadrant." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 123 "f
or n from 0 to 10 do\n J[n]:=spacecurve([x,n/10,u1s(n/10,x)],x=-3..3
,\n axes=normal,orientation=[-85,65]):\nod:" }}}{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 -1 246 "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
. And, 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 1 {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 82 "plot3d(u2s(t,x),x=-2*Pi..2*Pi,t=0..3,axes=NORMAL,\n
orientation=[65,65]);" }}}{PARA 0 "" 0 "" {TEXT -1 142 "Rot
ate the graph around. You should see that where x > 0 the graph moves \+
toward the Northeast, and where x < 0, it moves toward the Northwest.
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 159 "for n from 0 to 10 do\n
J[n]:=spacecurve([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]):\nod:
" }}}{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 som
e ideas about the structure of solutions for first order PDE's. And, t
he question begins to creep into the conscious: what happens if the gr
aph moves 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 1 {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 {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 154 "What is the status at this poi
nt. First, it should be clear that Maple can produce solutions and gra
phs for some partial differential 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 do with this type equation. We do
not choose that route. Rather, we introduce the notion of " }{TEXT
274 15 "characteristics" }{TEXT -1 132 " in the next section. This not
ion 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: herod@math.gatech.edu or jhero
d@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 "" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{MARK "0 1" 31 }{VIEWOPTS 1 1 0 1 1
1803 1 1 1 1 }{PAGENUMBERS 1 1 2 33 1 1 }