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{SECT 0 {PARA 256 "" 0 "" {TEXT -1 62 "Module 14 First Order, Non-homo
geneous, Initial Value Problems" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 290 " In \+
this outline, we solve our first partial differential equation. It is \+
the intent that by first solving two ordinary differential equations w
hich have the same flavor, the partial differential equation will be s
een as a natural extension of the ordinary differential equations idea
s." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 260 10 "Pr
oblem 1." }{TEXT -1 76 " Graph the solution for the differential equat
ion y ' = - 2 y + 3, y(0) = 5." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT 256 7 "Step 1:" }{TEXT -1 49 " Find a particular
(independent of time) solution" }}{PARA 0 "" 0 "" {TEXT -1 44 " \+
0 = -2 y + 3, so that y = 1/2 * 3." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT 257 7 "Step 2:" }{TEXT -1 55 " Find the gene
ral solution for the homogeneous equation" }}{PARA 0 "" 0 "" {TEXT -1
50 " y ' = -2 y, so that y(t) = exp(-2t) C." }}{PARA 0 ""
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 258 7 "Step 3:" }{TEXT -1
83 " Add these two to get the general solution to the non-homogeneous \+
equation, so that" }}{PARA 0 "" 0 "" {TEXT -1 35 " y(t) = exp
(-2 t) C + 3/2." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT 259 7 "Step 4:" }{TEXT -1 98 " Find the solution for the non-hom
ogeneous equation which satisfies the initial condition, so that" }}
{PARA 0 "" 0 "" {TEXT -1 41 " 5 = y(0) = C + 3/2 and C = 7/2.
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 60 "Here \+
is a check of the solution we have found and the graph." }}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "y1:=t->exp(-2*t)*7/2 + 3/2;" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "y1(0); diff(y1(t),t)+2*y1(t)
-3;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "plot(y1(t),t=0..1,y1
=0..6);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 ""
0 "" {TEXT 272 7 "Remark:" }{TEXT -1 167 " Of course, Maple would find
this solution without help from the user. We go through the details a
bove so that the concepts will be revealed. Here is Maple's solution.
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "dsolve(\{diff(y(t),t)=-2
*y(t)+3,y(0)=5\},y(t));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "
" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 265 10 "Pro
blem 2." }{TEXT -1 124 " Suppose that A is a matrix which has an inve
rse and that v is a vector. Graph the solution for the differential e
quation " }}{PARA 0 "" 0 "" {TEXT -1 55 " Z \+
' = A Z + v, Z(0) = [2,3]." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0
"" 0 "" {TEXT 261 7 "Step 1:" }{TEXT -1 49 " Find a particular (indepe
ndent of time) solution" }}{PARA 0 "" 0 "" {TEXT -1 35 " 0 = \+
A Z + v, so that Z = " }{XPPEDIT 18 0 "-A^(-1);" "6#,$)%\"AG,$\"\"\"!
\"\"F(" }{TEXT -1 2 "v." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 ""
0 "" {TEXT 262 7 "Step 2:" }{TEXT -1 55 " Find the general solution fo
r the homogeneous equation" }}{PARA 0 "" 0 "" {TEXT -1 49 " Z
' = A y, so that Z(t) = exp(A t) C." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT 263 7 "Step 3:" }{TEXT -1 83 " Add these two
to get the general solution to the non-homogeneous equation, so that
" }}{PARA 0 "" 0 "" {TEXT -1 28 " Z(t) = exp(A t) C " }
{XPPEDIT 18 0 "-A^(-1);" "6#,$)%\"AG,$\"\"\"!\"\"F(" }{TEXT -1 4 " v .
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 264 7 "Step \+
4:" }{TEXT -1 98 " Find the solution for the non-homogeneous equation \+
which satisfies the initial condition, so that" }}{PARA 0 "" 0 ""
{TEXT -1 27 " [2, 3] = Z(0) = C" }{XPPEDIT 18 0 "-A^(-1);" "6
#,$)%\"AG,$\"\"\"!\"\"F(" }{TEXT -1 21 "v and C = [2, 3] + " }
{XPPEDIT 18 0 "A^(-1);" "6#)%\"AG,$\"\"\"!\"\"" }{TEXT -1 2 "v." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 78 "We provid
e a particular A and a particular v in order to work out the details.
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "restart: with(linalg):"
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "A:=matrix([[-3,1],[1,-3]]
);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "v:=[5,7];" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "C:=evalm([2,3]+inverse(A)&*v);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "Z:=evalm(exponential(A,t)&*C
-inverse(A)&*v);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "x2:=una
pply(Z[1],t);\ny2:=unapply(Z[2],t);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 12 "x2(0),y2(0);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1
0 42 "plot([x2(t),y2(t),t=0..20],x=0..4,y=0..4);" }}}{EXCHG {PARA 0 ">
" 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 69 "Again, Maple
will solve this equation without any help from the user." }}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 142 "dsolve(\{diff(x(t),t)=A[1,1]*x(t)+
A[1,2]*y(t)+v[1],\n diff(y(t),t)=A[2,1]*x(t)+A[2,2]*y(t)+v[2], \+
x(0)=2, y(0)=3\},\n \{x(t),y(t)\});" }}}{EXCHG {PARA 0 "> "
0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1
0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT 271 11 "Problem 3: " }{TEXT -1 57 " Graph th
e solution for the partial differential equation" }}{PARA 0 "" 0 ""
{TEXT -1 10 " " }{XPPEDIT 18 0 "diff(u,t) = diff(u(x),`$`(x,2
));" "6#/-%%diffG6$%\"uG%\"tG-F%6$-F'6#%\"xG-%\"$G6$F-\"\"#" }{TEXT
-1 48 " u(t, 0) = 3, u(t, 2) = 5, with u(0, x) = f(x)." }}{PARA 0 ""
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 17 "This problem has " }
{TEXT 266 35 "non-homogeneous boundary conditions" }{TEXT -1 101 ". Ho
mogeneous boundary conditions would be that the solution is zero at x \+
= 0 and at x = 2 for all t." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 267 7 "Step 1:" }
{TEXT -1 49 " Find a particular (independent of time) solution" }}
{PARA 0 "" 0 "" {TEXT -1 9 " 0 = " }{XPPEDIT 18 0 "diff(u(x),`$`(x
,2));" "6#-%%diffG6$-%\"uG6#%\"xG-%\"$G6$F)\"\"#" }{TEXT -1 28 " with \+
u(0) = 3 and u(2) = 5." }}{PARA 0 "" 0 "" {TEXT -1 116 "This is just a
n ordinary differential equation 0 = u '', with boundary conditions. T
he solution for this equation is" }}{PARA 0 "" 0 "" {TEXT -1 19 " \+
u(x) = x + 3." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 268 7 "Step 2:" }{TEXT -1 55 " F
ind the general solution for the homogeneous equation" }}{PARA 0 "" 0
"" {TEXT -1 6 " " }{XPPEDIT 18 0 "diff(u,t);" "6#-%%diffG6$%\"uG%
\"tG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "diff(u(x),`$`(x,2))" "6#-%%dif
fG6$-%\"uG6#%\"xG-%\"$G6$F)\"\"#" }{TEXT -1 34 " with u(t, 0) = 0 and \+
u(t, 2) = 0." }}{PARA 0 "" 0 "" {TEXT -1 65 "We will see later that th
e general solution for this equation is " }}{PARA 0 "" 0 "" {TEXT -1
15 " u(t, x) = " }{XPPEDIT 18 0 "sum(C[n]*exp(-n^2*Pi^2*t/4)*sin(n
*Pi*x/2));" "6#-%$sumG6#*(&%\"CG6#%\"nG\"\"\"-%$expG6#,$**F*\"\"#%#PiG
F1%\"tGF+\"\"%!\"\"F5F+-%$sinG6#**F*F+F2F+%\"xGF+F1F5F+" }{TEXT -1 2 "
." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 88 "Rem
ark: to make this last statement believable, we will make the followin
g illustration." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 61 "u:=(t,x)-
>sum(C[n]*exp(-n^2*Pi^2*t/4)*sin(n*Pi*x/2),n=1..20);" }}}{EXCHG {PARA
0 "> " 0 "" {MPLTEXT 1 0 15 "u(t,0); u(t,2);" }}}{EXCHG {PARA 0 "> "
0 "" {MPLTEXT 1 0 32 "diff(u(t,x),t)-diff(u(t,x),x,x);" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 269 7 "Step 3:" }{TEXT -1 83 " A
dd these two to get the general solution to the non-homogeneous equati
on, so that" }}{PARA 0 "" 0 "" {TEXT -1 24 " u(t, x) = x + 3 + "
}{XPPEDIT 18 0 "sum(C[n]*exp(-n^2*Pi^2*t/4)*sin(n*Pi*x/2));" "6#-%$sum
G6#*(&%\"CG6#%\"nG\"\"\"-%$expG6#,$**F*\"\"#%#PiGF1%\"tGF+\"\"%!\"\"F5
F+-%$sinG6#**F*F+F2F+%\"xGF+F1F5F+" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 270 7 "Step 4:" }{TEXT -1 98 " F
ind the solution for the non-homogeneous equation which satisfies the \+
initial condition, so that" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0
"" 0 "" {TEXT -1 20 " f(x) = x + 3 + " }{XPPEDIT 18 0 "sum(C[n]*si
n(n*Pi*x/2));" "6#-%$sumG6#*&&%\"CG6#%\"nG\"\"\"-%$sinG6#**F*F+%#PiGF+
%\"xGF+\"\"#!\"\"F+" }{TEXT -1 16 " for 0 < x < 2." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 71 "This looks like a job f
or Fourier Series. Perhaps you would agree that " }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 5 " " }{XPPEDIT 18 0 "C[n
];" "6#&%\"CG6#%\"nG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "int((f(x)-x-3)
*sin(n*Pi*x/2),x = 0 .. 2)/int(sin(n*Pi*x/2)^2,x = 0 .. 2);" "6#*&-%$i
ntG6$*&,(-%\"fG6#%\"xG\"\"\"F,!\"\"\"\"$F.F--%$sinG6#**%\"nGF-%#PiGF-F
,F-\"\"#F.F-/F,;\"\"!F6F--F%6$*$-F16#**F4F-F5F-F,F-F6F.F6/F,;F9F6F." }
{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 96 "In order to draw a graph, we need to be specific about f.
For simplicity, choose f to be f(x) = " }{XPPEDIT 18 0 "x^2-x+3;" "6#
,(*$%\"xG\"\"#\"\"\"F%!\"\"\"\"$F'" }{TEXT -1 111 " . We know how the \+
graph of this looks on [0, 2]. We will compute five terms of the serie
s and draw that graph." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "f:
=x->x^2-x+3;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 125 "for n from
1 to 10 do\n C[n]:=int((f(x)-x-3)*sin(n*Pi*x/2),x=0..2)/\n \+
int(sin(n*Pi*x/2)^2,x=0..2);\nod;\nn:='n';" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 65 "u:=(t,x)->x+3+sum(C[n]*exp(-n^2*Pi^2*t/4)*si
n(n*Pi*x/2),n=1..10);" }}}{PARA 0 "" 0 "" {TEXT -1 204 "Before drawing
the graph, think about what you expect: when t = 0, you should expect
a quadratic. There after, the graph stays fixed on the ends and has l
imit the time independent solution which is x + 3." }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 63 "plot3d(u(t,x
),x=0..2,t=0..1,axes=NORMAL,orientation=[-135,45]);" }}}{EXCHG {PARA
0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT -1 125 "At t = 0, do you see the \+
quadratic? Sure. At t = 1, can you see the staight line that represent
s x=3. Look closely. Oh, yeah!" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 98 "Assignment: Sketch the graph for the solu
tion of the non-homogeneous partial differential equation" }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 9 " " }
{XPPEDIT 18 0 "diff(u,t) = diff(u(x),`$`(x,2));" "6#/-%%diffG6$%\"uG%
\"tG-F%6$-F'6#%\"xG-%\"$G6$F-\"\"#" }{TEXT -1 43 " u(t, 0) = 3, u(t, \+
2) = 7, with u(0, x) = " }{XPPEDIT 18 0 "x^2+3;" "6#,&*$%\"xG\"\"#\"\"
\"\"\"$F'" }{TEXT -1 2 " ." }}}{MARK "0 0" 62 }{VIEWOPTS 1 1 0 1 1
1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }