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{SECT 0 {PARA 207 "" 0 "" {TEXT 206 57 "Module 12: Review of Elementa
ry Differential Equations I" }}{PARA 208 "" 0 "" {TEXT -1 0 "" }}
{PARA 208 "" 0 "" {TEXT 207 184 " We will use information about el
ementary, ordinary differential equations in the process of solving pa
rtial differential equations. It is well to remember some of the basic
ideas." }}{PARA 208 "" 0 "" {TEXT -1 0 "" }}{PARA 208 "" 0 "" {TEXT
207 119 " Simple, constant coefficient, initial value problems are
classical problems in elementary differential equations. " }}{PARA
208 "" 0 "" {TEXT -1 0 "" }}{PARA 208 "" 0 "" {TEXT -1 0 "" }}{PARA
209 "" 0 "" {TEXT 208 73 "General Solutions for Linear Constant Coeffi
cient Differential Equations." }}{PARA 208 "" 0 "" {TEXT -1 0 "" }}
{PARA 208 "" 0 "" {TEXT 207 108 "The methods for solving these are a p
art of the standard tools. Here is an example to recall the techniques
." }}{PARA 208 "" 0 "" {TEXT 207 40 " Consider the differential eq
uation," }}{PARA 208 "" 0 "" {TEXT 207 33 " y '' + 3 y ' + 2 \+
y = 0." }}{PARA 208 "" 0 "" {TEXT 207 192 "Using the ideas that have c
ome before, the consideration of such a differential equation could be
viewed as the consideration of the nullspace of the linear operator L
(y) = y '' + 3 y ' + 2 y." }}{PARA 208 "" 0 "" {TEXT 207 193 " Rec
all that there are two linearly independent solutions for this second \+
order equation. Said in our context, the nullspace of this linear oper
ation had a basis consisting of two elements." }}{EXCHG {PARA 210 "> \+
" 0 "" {MPLTEXT 1 0 42 "deq:=diff(y(x),x,x)+3*diff(y(x),x)+2*y(x);" }}
}{EXCHG {PARA 210 "> " 0 "" {MPLTEXT 1 0 30 "dsolve(deq,y(x),output=ba
sis);" }}}{EXCHG {PARA 210 "" 0 "" {TEXT -1 0 "" }}}{PARA 208 "" 0 ""
{TEXT -1 0 "" }}{PARA 208 "" 0 "" {TEXT 207 153 "The nullspace is two \+
dimensional and any solution of the equation -- any element of the nul
lspace -- can be written as a linear combination of these two." }}
{EXCHG {PARA 210 "> " 0 "" {MPLTEXT 1 0 17 "dsolve(deq,y(x));" }}}
{EXCHG {PARA 210 "" 0 "" {TEXT -1 0 "" }}}{PARA 208 "" 0 "" {TEXT -1
0 "" }}{PARA 211 "" 0 "" {TEXT 209 52 "Linear, Constant Coefficient, I
nitial Value Problems" }}{PARA 208 "" 0 "" {TEXT -1 0 "" }}{PARA 208 "
" 0 "" {TEXT 207 157 "If we want to know a particular solution, say th
e solution of an initial value problem, we must choose the constants s
o that the initial values are attained." }}{EXCHG {PARA 210 "> " 0 ""
{MPLTEXT 1 0 24 "inits:=y(0)=1,D(y)(0)=5;" }}}{EXCHG {PARA 210 "> " 0
"" {MPLTEXT 1 0 31 "Part:=dsolve(\{deq,inits\},y(x));" }}}{EXCHG
{PARA 210 "" 0 "" {TEXT -1 0 "" }}}{PARA 208 "" 0 "" {TEXT -1 0 "" }}
{PARA 208 "" 0 "" {TEXT 207 54 "We can now draw the graph of this part
icular solution." }}{EXCHG {PARA 210 "> " 0 "" {MPLTEXT 1 0 25 "y1:=un
apply(rhs(Part),x);" }}}{EXCHG {PARA 210 "> " 0 "" {MPLTEXT 1 0 26 "pl
ot(y1(x),x=0..1,y=0..3);" }}}{EXCHG {PARA 210 "" 0 "" {TEXT -1 0 "" }}
}{PARA 208 "" 0 "" {TEXT 207 187 "You might choose to draw a family of
solutions. For example, we draw a family of solutions for the differe
ntial equation that begin at zero with value 1 and have different init
ial slopes." }}{EXCHG {PARA 210 "> " 0 "" {MPLTEXT 1 0 24 "inits:=y(0)
=1,D(y)(0)=b;" }}}{EXCHG {PARA 210 "> " 0 "" {MPLTEXT 1 0 31 "Part:=ds
olve(\{deq,inits\},y(x));" }}}{EXCHG {PARA 210 "> " 0 "" {MPLTEXT 1 0
48 "plot([seq(subs(b=n,rhs(Part)),n=-2..2)],x=0..4);" }}}{EXCHG {PARA
210 "" 0 "" {TEXT -1 0 "" }}}{PARA 208 "" 0 "" {TEXT -1 0 "" }}{PARA
212 "" 0 "" {TEXT 210 54 "Linear, Constant Coefficient, Boundary Value
Problems." }}{PARA 208 "" 0 "" {TEXT -1 0 "" }}{PARA 208 "" 0 ""
{TEXT 207 243 " Boundary value problems are a little more complic
ated in concept. There are second order, constant coefficient differen
tial equations for which there is exactly one solution, or there is no
solution, or there is an infinity of solutions." }}{PARA 208 "" 0 ""
{TEXT 207 85 " Suppose you want solutions for the above differenti
al equations to start off at " }{TEXT 211 1 "a" }{TEXT 207 15 " and en
d up at " }{TEXT 211 1 "b" }{TEXT 207 14 ". Not so hard." }}{EXCHG
{PARA 210 "> " 0 "" {MPLTEXT 1 0 7 "Part2:=" }{MPLTEXT 1 0 71 "\n dso
lve(\{diff(y(x),x,x)+3*diff(y(x),x)+2*y(x)=0,y(0)=a,y(1)=b\},y(x));" }
}}{PARA 208 "" 0 "" {TEXT 207 55 "For example, we draw the graph in ca
se a = 1 and b = 1." }}{EXCHG {PARA 210 "> " 0 "" {MPLTEXT 1 0 47 "plo
t(subs(\{a=1,b=1\},rhs(Part2)),x=0..1,y=0..2);" }}}{EXCHG {PARA 210 "
" 0 "" {TEXT -1 0 "" }}}{PARA 208 "" 0 "" {TEXT 207 95 "As an alternat
e, you might choose the slope at either end, instead of where the solu
tions lies." }}{EXCHG {PARA 210 "> " 0 "" {MPLTEXT 1 0 74 "dsolve(\{di
ff(y(x),x,x)+3*diff(y(x),x)+2*y(x)=0,D(y)(0)=a,D(y)(1)=b\},y(x));" }}}
{PARA 208 "" 0 "" {TEXT 207 130 "It is of value to note that we can ma
ke second order, constant coefficient, differential equations for whic
h there is no solution." }}{EXCHG {PARA 210 "> " 0 "" {MPLTEXT 1 0 74
"dsolve(\{diff(y(x),x,x)+3*diff(y(x),x)+0*y(x)=0,D(y)(0)=a,D(y)(1)=b\}
,y(x));" }}}{PARA 208 "" 0 "" {TEXT 207 230 "There was no solution for
this system. That is why Maple is silent. Maybe it would have been be
tter programming if Maple had responded with something such as \"I can
not find a solution for that equation. Are you sure there is one?\"" }
}{PARA 208 "" 0 "" {TEXT -1 0 "" }}{PARA 208 "" 0 "" {TEXT 207 118 "An
d, there are second order, constant coefficient, differential equation
s for which there are an infinity of solution." }}{EXCHG {PARA 210 "> \+
" 0 "" {MPLTEXT 1 0 74 "dsolve(\{diff(y(x),x,x)+0*diff(y(x),x)+0*y(x)=
0,D(y)(0)=0,D(y)(1)=0\},y(x));" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }
{TEXT -1 94 "For this system, you pick an number. The function y(x) = \+
that number is a solution. Just look." }}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 210 "" 0 "" {TEXT -1 0 "" }}}{PARA
208 "" 0 "" {TEXT -1 0 "" }}{PARA 213 "" 0 "" {TEXT 212 41 "Linear Sys
tems of Differential Equations." }}{PARA 208 "" 0 "" {TEXT -1 0 "" }}
{PARA 208 "" 0 "" {TEXT 207 241 " Often, one has a coupled system \+
of equations instead of a single differential equation. Review almost \+
any text on elementary differential equations, and you will find syste
ms of equations, with initial conditions, such as the following:" }}
{PARA 208 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 210 "> " 0 "" {MPLTEXT
1 0 35 "sys:=diff(x(t),t)=-2*x(t)-11*y(t), " }{MPLTEXT 1 0 34 "\n \+
diff(y(t),t)=11*x(t)-2*y(t);" }}}{EXCHG {PARA 210 "> " 0 "" {MPLTEXT
1 0 21 "inits:=x(0)=1,y(0)=2;" }}}{EXCHG {PARA 210 "> " 0 "" {MPLTEXT
1 0 32 "dsolve(\{sys,inits\},\{x(t),y(t)\});" }}}{EXCHG {PARA 210 "> \+
" 0 "" {MPLTEXT 1 0 10 "assign(%);" }}}{EXCHG {PARA 210 "> " 0 ""
{MPLTEXT 1 0 25 "plot([x(t),y(t)],t=0..1);" }}}{EXCHG {PARA 210 "> "
0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 210 "" 0 "" {TEXT -1 0
"" }}}{PARA 208 "" 0 "" {TEXT -1 0 "" }}{PARA 214 "" 0 "" {TEXT 213
24 "Non-homogeneous Systems." }}{PARA 208 "" 0 "" {TEXT -1 0 "" }}
{PARA 208 "" 0 "" {TEXT 207 303 "We might have a differential equation
for which the right hand side is not zero. For example, think of the \+
simplified Newton's Law of Cooling. It says that a body cools at a rat
e proportioned to the difference in its temperature and the temperatur
e of the surrounding medium. We could write this idea as" }}{PARA 208
"" 0 "" {TEXT 207 33 " T ' = K (A-T), T(0) = C" }}{PARA 208 "
" 0 "" {TEXT 207 267 "where A is the ambient temperature and C is the \+
initial temperature of the object. The number K is the constant of pro
portionality. We see that if K is positive and the object is hotter th
an the ambient temperature, it will cool. This equations could be rewr
itten as " }}{PARA 208 "" 0 "" {TEXT 207 41 " T ' + K T = K A
, with T(0) = C." }}{PARA 208 "" 0 "" {TEXT 207 158 "This introduces a
differential equation with right hand side not zero. Such systems are
common. Techniques for obtaining their solutions need to be recalled.
" }}{EXCHG {PARA 210 "> " 0 "" {MPLTEXT 1 0 48 "dsolve(\{diff(T(t),t)
+2*T(t)=2*3/2,T(0)=5\},T(t));" }}}{EXCHG {PARA 210 "> " 0 "" {MPLTEXT
1 0 31 "plot(3/2+7/2*exp(-2*t),t=0..4);" }}}{PARA 208 "" 0 "" {TEXT
207 47 "The right side may, itself, be a function of t." }}{EXCHG
{PARA 210 "> " 0 "" {MPLTEXT 1 0 49 "dsolve(\{diff(T(t),t)+2*T(t)=sin(
t),T(0)=5\},T(t));" }}}{EXCHG {PARA 210 "" 0 "" {TEXT -1 0 "" }}}
{PARA 208 "" 0 "" {TEXT -1 0 "" }}{PARA 208 "" 0 "" {TEXT 207 54 "Inde
ed, the right hand side may not be differentiable." }}{EXCHG {PARA
210 "> " 0 "" {MPLTEXT 1 0 56 "dsolve(\{diff(T(t),t)+2*T(t)=signum(1/2
-t),T(0)=5\},T(t));" }}}{EXCHG {PARA 210 "" 0 "" {TEXT -1 0 "" }}}
{PARA 208 "" 0 "" {TEXT -1 0 "" }}{PARA 215 "" 0 "" {TEXT 214 46 "Nume
rical solutions for Differential Equations" }}{PARA 208 "" 0 "" {TEXT
-1 0 "" }}{PARA 208 "" 0 "" {TEXT 207 130 "One might choose to obtain \+
a numerical solution. We do that for the last equation. Look for a cha
nge in the derivative at t = 1/2." }}{EXCHG {PARA 210 "> " 0 ""
{MPLTEXT 1 0 96 "Cool:=dsolve(\{diff(T(t),t)+2*T(t)=signum(1/2-t),T(0)
=5\},T(t),type=numeric,output=listprocedure);" }}}{PARA 208 "" 0 ""
{TEXT -1 0 "" }}{EXCHG {PARA 210 "> " 0 "" {MPLTEXT 1 0 20 "y2:=subs(C
ool,T(t));" }}}{EXCHG {PARA 210 "> " 0 "" {MPLTEXT 1 0 21 "plot('y2(t)
',t=0..1);" }}}{PARA 208 "" 0 "" {TEXT -1 0 "" }}{PARA 208 "" 0 ""
{TEXT 207 62 "Numerical solutions can be used for systems of equations
, too." }}{EXCHG {PARA 210 "> " 0 "" {MPLTEXT 1 0 49 "Drug1:=dsolve(\{
diff(x(t),t)=sin(t)-2*x(t)-3*y(t)," }{MPLTEXT 1 0 43 "\n \+
diff(y(t),t)=3*x(t)-2*y(t)," }{MPLTEXT 1 0 67 "\n x(0)=1,y(0)=0\},
\{x(t),y(t)\},type=numeric,output=listprocedure);" }}}{EXCHG {PARA
210 "> " 0 "" {MPLTEXT 1 0 21 "x3:=subs(Drug1,x(t));" }{MPLTEXT 1 0
22 "\ny3:=subs(Drug1,y(t));" }}}{EXCHG {PARA 210 "> " 0 "" {MPLTEXT 1
0 33 "plot(['x3(t)','y3(t)'],'t'=0..5);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 210 "" 0 "" {TEXT -1 0 "" }}}{PARA
208 "" 0 "" {TEXT -1 0 "" }}{PARA 208 "" 0 "" {TEXT 207 50 "Assignment
: Draw one graph with five solutions for" }}{PARA 208 "" 0 "" {TEXT
-1 0 "" }}{PARA 208 "" 0 "" {TEXT 207 65 " x '' + 2 x ' + 2 x
= 20 cos(t), x(0) = a, x '(0) = 0," }}{PARA 208 "" 0 "" {TEXT -1 0
"" }}{PARA 208 "" 0 "" {TEXT 207 259 "using five different values of a
. Looking at the graph, it seems that all solutions to dampen to the s
ame curve. Can you find that steady periodic solution for the equation
? What is its initial value? (This problem is found on page 194 of the
attractive text " }{TEXT 211 50 "Differential Equations and Boundary \+
Value Problems" }{TEXT 207 55 ", by Edwards and Penny and published by
Prentice Hall.)" }}{PARA 208 "" 0 "" {TEXT -1 0 "" }}{PARA 216 "" 0 "
" {TEXT -1 0 "" }}}{MARK "0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }
{PAGENUMBERS 0 1 2 33 1 1 }