{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 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1
257 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 258 "" 0 1 0 0
0 0 0 0 1 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 "" 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 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 268 "" 0 1 0 0
0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 269 "" 0 1 0 0 0 0 1 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 }{CSTYLE "" -1 275 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }
{CSTYLE "" -1 276 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1
277 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 278 "" 0 1 0 0
0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 279 "" 0 1 0 0 0 0 1 0 0 0 0 0
0 0 0 0 }{CSTYLE "" -1 280 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }
{CSTYLE "" -1 281 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1
282 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 283 "" 0 1 0 0
0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 284 "" 0 1 0 0 0 0 0 1 0 0 0 0
0 0 0 0 }{CSTYLE "" -1 285 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }
{CSTYLE "" -1 286 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1
287 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 288 "" 0 1 0 0
0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 289 "" 0 1 0 0 0 0 1 0 0 0 0 0
0 0 0 0 }{CSTYLE "" -1 290 "" 0 1 0 0 0 0 0 1 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 "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 }
{PSTYLE "Heading 1" -1 257 1 {CSTYLE "" -1 -1 "Times" 1 14 0 0 0 1 2
1 2 2 2 2 1 1 1 1 }3 1 0 0 8 4 1 0 1 0 2 2 0 1 }{PSTYLE "Normal" -1
258 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1
0 0 0 0 1 0 1 0 2 2 0 1 }}
{SECT 0 {PARA 257 "" 0 "" {TEXT -1 90 "Periodic Solutions for a First \+
Order Differential Equation \nwith Periodic Forcing Function" }}{PARA
256 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 9 "Jim Herod" }
}{PARA 256 "" 0 "" {TEXT -1 21 "Georgia Tech, Retired" }}{PARA 256 ""
0 "" {TEXT -1 14 "jherod@tds.net" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT 290 8 "Summary:" }{TEXT -1 127 " This Maple \+
Worksheet provides a method for finding solutions for non-homogeneous \+
partial differential equation of the form" }}{PARA 0 "" 0 "" {TEXT -1
22 " " }{XPPEDIT 18 0 "diff(u,t);" "6#-%%diffG6$%
\"uG%\"tG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "diff(u,`$`(x,2));" "6#-%%
diffG6$%\"uG-%\"$G6$%\"xG\"\"#" }{TEXT -1 5 " + " }{XPPEDIT 18 0 "F(
t,x);" "6#-%\"FG6$%\"tG%\"xG" }{TEXT -1 25 ", u(t, 0) = 0 = u(t, 1).
" }}{PARA 0 "" 0 "" {TEXT -1 336 "More important, if the function F(t,
x) is periodic as a function of t, it provides a method for finding a \+
periodic solution for the partial differential equation. Even more. Te
chniques are used to emphasize a general structure suitable for findin
g a periodic solution for ordinary differential equations having the s
ame general form of " }}{PARA 0 "" 0 "" {TEXT -1 49 " \+
Y' = AY + F" }}{PARA 0 "" 0 "" {TEXT -1 20 "where
F is periodic." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 ""
0 "" {TEXT -1 22 "Section 1:Introduction" }}{SECT 1 {PARA 3 "" 0 ""
{TEXT -1 24 "Statement of the problem" }}{PARA 0 "" 0 "" {TEXT -1 73 "
In this discussion, we consider solutions Y for the differential equat
ion" }}{PARA 0 "" 0 "" {TEXT -1 32 "(1.1) Y' = AY + F."
}}{PARA 0 "" 0 "" {TEXT -1 396 "In case A is a number and F is a real \+
valued function, students in an introductory calculus course often can
make a solution for this equation. If A is a matrix, students in an \+
introductory ordinary differential equations course likely can formula
te a solution. The principal interest here is the beginning of an unde
rstanding for how to handle this problem in case A is a differential o
perator." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1
219 "We want more than to compute solutions. Our interest is in period
ic solutions. We choose a function A in each of the above contexts and
suppose that F is a periodic function from R into the space on which \+
A is defined. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 56 "We seek periodic solutions for the equation Y' = AY + F.
" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 35 "Different Contexts for the P
roblem " }}{PARA 0 "" 0 "" {TEXT -1 77 "We illustrate the ideas of the
problem we have presented with three examples." }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 256 9 "Example 1" }{TEXT -1 69 "
: Find a periodic solution for the equation y'(t) = -2 y(t) + sin(t).
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 257 9 "Examp
le 2" }{TEXT -1 54 ": Find a periodic solution for the system of equat
ions" }}{PARA 0 "" 0 "" {TEXT -1 40 " x' = -2 x + y + \+
cos(t)," }}{PARA 0 "" 0 "" {TEXT -1 40 " y' = x - 2 y \+
+ sin(t),." }}{PARA 0 "" 0 "" {TEXT -1 86 "We rewrite this Example 2 i
nto the context of our discussion by identifying the matrix" }}{PARA
0 "" 0 "" {TEXT -1 17 " A = " }{XPPEDIT 18 0 "matrix([[-2,
1], [1, -2]]);" "6#-%'matrixG6#7$7$,$\"\"#!\"\"\"\"\"7$F+,$F)F*" }
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 17 "and the function " }}
{PARA 0 "" 0 "" {TEXT -1 20 " F(t) = " }{XPPEDIT 18 0 "mat
rix([[cos(t)], [sin(t)]]);" "6#-%'matrixG6#7$7#-%$cosG6#%\"tG7#-%$sinG
6#F+" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 110 "With this identi
fication, the system of equations with a forcing function has the form
of our general problem:" }}{PARA 0 "" 0 "" {TEXT -1 26 " \+
Y' = AY + F." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT 258 9 "Example 3" }{TEXT -1 64 ": Find a periodic solution for t
he partial differential equation" }}{PARA 0 "" 0 "" {TEXT -1 19 " \+
" }{XPPEDIT 18 0 "diff(u,t);" "6#-%%diffG6$%\"uG%\"tG" }
{TEXT -1 3 " = " }{XPPEDIT 18 0 "diff(u,`$`(x,2));" "6#-%%diffG6$%\"uG
-%\"$G6$%\"xG\"\"#" }{TEXT -1 14 " + 10 sin(t) " }{XPPEDIT 18 0 "x^2*
(1-x)^2;" "6#*&%\"xG\"\"#,&\"\"\"F'F$!\"\"F%" }{TEXT -1 25 ", u(t, 0)
= 0 = u(t, 1)." }}{PARA 0 "" 0 "" {TEXT -1 251 "To realize this probl
em in the context of our equation (1.1) and to follow the paradigm we \+
will use to make solutions, we need to be a little careful about ident
ifying A. Note the character of the A in the second example: A was a m
apping of the space " }{XPPEDIT 18 0 "R^2;" "6#*$%\"RG\"\"#" }{TEXT
-1 65 " and, if t is a number, then F(t) is in the domain of A, namel
y " }{XPPEDIT 18 0 "R^2;" "6#*$%\"RG\"\"#" }{TEXT -1 171 ". For the th
ird example we take the space to be twice differentiable functions on \+
the interval [0, 1] with value zero at both ends. Note that if t is a \+
fixed number, then " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 55 " F(t, x) = 10 sin(t) \+
" }{XPPEDIT 18 0 "x^2*(1-x)^2;" "6#*&%\"xG\"\"#,&\"\"\"F'F$!\"\"F%" }
{TEXT -1 2 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 50 "as a function of x is in this space. Agreeing that" }}
{PARA 0 "" 0 "" {TEXT -1 37 " A(f) = " }
{XPPEDIT 18 0 "diff(f,`$`(x,2));" "6#-%%diffG6$%\"fG-%\"$G6$%\"xG\"\"#
" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 57 "we have that, for fix
ed t, F(t, x) is in the domain of A." }}}{SECT 1 {PARA 3 "" 0 ""
{TEXT -1 12 "What Follows" }}{PARA 0 "" 0 "" {TEXT -1 98 "In the remai
nder of this discussion, we will recall the derivation of the solution
for the problem" }}{PARA 0 "" 0 "" {TEXT -1 40 " Y' = A Y \+
+ F, with Y(0) = c." }}{PARA 0 "" 0 "" {TEXT -1 67 "This solution is o
ften called the variations of parameter solution:" }}{PARA 0 "" 0 ""
{TEXT -1 37 " Y(t) = exp(tA) c + exp(t A) " }{XPPEDIT 18 0 "in
t(exp(-sA)*F(s),s = 0 .. t);" "6#-%$intG6$*&-%$expG6#,$%#sAG!\"\"\"\"
\"-%\"FG6#%\"sGF-/F1;\"\"!%\"tG" }{TEXT -1 2 " ." }}{PARA 0 "" 0 ""
{TEXT -1 405 "The use of this formula for calculation of solutions wil
l require an understanding of the exponential function. The exponentia
l of a number can be calculated with a hand held calculator. Thus, sol
utions for the first example are easy to compute. The exponential of a
matrix A such as in the second example is not hard to compute in the \+
Maple environment. We illustrate the details for both these examples.
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 249 "The \+
exponential of the differentiable operator is more interesting. We obt
ain this exponential, develop the variations of parameter solution for
the partial differential equation with a forcing function, and find t
he periodic solution for Example 3." }}{PARA 0 "" 0 "" {TEXT -1 12 " \+
" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 45 "Section 2: A Structure \+
for Periodic Solutions" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 23 "Solutio
ns for Y'=AY + F" }}{PARA 0 "" 0 "" {TEXT -1 95 "Suppose for this sect
ion that exp(tA) can be defined for each of the examples presented in \+
the " }{TEXT 284 23 "Section 1: Introduction" }{TEXT -1 319 ". We have
stated in the introduction that this is not a severe supposition for \+
the reader in the first two examples and in this Maple environment. Th
e promise is that the exponential of the third A will be provided also
. We provide a recipe for getting a solution for differential equation
s which have this general form" }}{PARA 0 "" 0 "" {TEXT -1 39 " \+
Y' = AY + F." }}{PARA 0 "" 0 "" {TEXT -1 60 "Our d
evelopment for the variations of parameters formula is " }{TEXT 259 6
"formal" }{TEXT -1 135 " in the sense that we assume the calculus for \+
the exponential function in order to give a plausible structure for a \+
solution. That the " }{TEXT 260 8 "solution" }{TEXT -1 59 " does provi
de a solution will be verified for each example." }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 126 "Suppose that Y(t) satisf
ies the differential equation. Consider the function exp(-tA) Y(t). \+
We take the derivative formally:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 57 " ( exp(-tA) Y(t) )' = exp(-tA)Y'(t)
- exp(-tA)AY(t)" }}{PARA 0 "" 0 "" {TEXT -1 59 " \+
= exp(-tA) (Y'(t) - AY(t))" }}{PARA 0 "" 0 "" {TEXT -1 49
" = exp(-tA) F(t)." }}{PARA 0 "" 0 ""
{TEXT -1 33 "Integrate both sides from 0 to t:" }}{PARA 0 "" 0 ""
{TEXT -1 32 " exp(-tA)Y(t) - Y(0) = " }{XPPEDIT 18 0 "int(exp
(-sA)*F(s),s = 0 .. t);" "6#-%$intG6$*&-%$expG6#,$%#sAG!\"\"\"\"\"-%\"
FG6#%\"sGF-/F1;\"\"!%\"tG" }{TEXT -1 2 " ," }}{PARA 0 "" 0 "" {TEXT
-1 2 "or" }}{PARA 0 "" 0 "" {TEXT -1 52 " (2.1) Y(t) = \+
exp(tA)Y(0) + exp(tA) " }{XPPEDIT 18 0 "int(exp(-sA)*F(s),s = 0 .. t)
;" "6#-%$intG6$*&-%$expG6#,$%#sAG!\"\"\"\"\"-%\"FG6#%\"sGF-/F1;\"\"!%
\"tG" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 114 "We will use thi
s formula to compute and check solutions for each of the three example
s listed in the introduction." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}
{SECT 1 {PARA 3 "" 0 "" {TEXT -1 19 "Periodic Solutions." }}{PARA 0 "
" 0 "" {TEXT -1 192 "We are not satisfied to find solutions for the eq
uation Y' = AY + F. We want more. We suppose that F is periodic and as
k that we should be able to compute periodic solutions for the example
s. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 199 "T
o this end, take L to be a number such that F(L) = F(0). We ask: what \+
should Y(0) be in order that Y(L) = Y(0)? Using the form from equation
(2.1) which we anticipate will provide solutions, we have" }}{PARA 0
"" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 17 " \+
" }{TEXT 268 4 "Y(L)" }{TEXT -1 3 " = " }{TEXT 264 12 "exp(L A)Y(0)" }
{TEXT -1 3 " + " }{TEXT 265 7 "exp(LA)" }{TEXT -1 2 " " }{XPPEDIT 18
0 "int(`exp(-sA)`*F(s),s = 0 .. L);" "6#-%$intG6$*&%)exp(-sA)G\"\"\"-%
\"FG6#%\"sGF(/F,;\"\"!%\"LG" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT
-1 24 "If Y(L) = P = Y(0), then" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 17 " " }{TEXT 266 1 "P" }
{TEXT -1 3 " = " }{TEXT 263 8 "exp(LA)P" }{TEXT -1 3 " + " }{TEXT 262
7 "exp(LA)" }{TEXT -1 2 " " }{XPPEDIT 18 0 "int(`exp(-sA)`*F(s),s = 0
.. L);" "6#-%$intG6$*&%)exp(-sA)G\"\"\"-%\"FG6#%\"sGF(/F,;\"\"!%\"LG
" }{TEXT -1 2 " ," }}{PARA 0 "" 0 "" {TEXT -1 2 "or" }}{PARA 0 "" 0 "
" {TEXT -1 22 " (2.2) " }{TEXT 267 1 "P" }{TEXT -1 3 " \+
= " }{XPPEDIT 18 0 "(1-`exp(L A)`)^(-1);" "6#),&\"\"\"F%%)exp(L~A)G!\"
\",$F%F'" }{TEXT 261 3 "exp" }{TEXT -1 1 "(" }{TEXT 285 2 "LA" }{TEXT
-1 3 ") " }{XPPEDIT 18 0 "int(`exp(-sA)`*F(s),s = 0 .. L);" "6#-%$int
G6$*&%)exp(-sA)G\"\"\"-%\"FG6#%\"sGF(/F,;\"\"!%\"LG" }{TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 190 "This formula provides the starting point
for the computation of a periodic solution. There are some points for
concern. We need an inverse; does this inverse exist for each of the \+
problems? " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT
-1 344 "This formula purports to provide an initial value for a period
ic solution. Starting with this initial value, are the computed soluti
ons periodic? We show that the answer to this question should be yes f
rom similar analysis to the above as follows. With the supposition tha
t F is periodic with period L, let Z(t) be defined by the equation (2.
1)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 38 " \+
Z(t) = exp(tA)P + exp(tA) " }{XPPEDIT 18 0 "int(exp(-sA)*F(s)
,s = 0 .. t);" "6#-%$intG6$*&-%$expG6#,$%#sAG!\"\"\"\"\"-%\"FG6#%\"sGF
-/F1;\"\"!%\"tG" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 32 "Let X(
t) be defined as Z(t+L). " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0
"" 0 "" {TEXT -1 35 "X'(t) = Z'(t+L) = A Z(t+L) + F(t+L)" }}{PARA 0 "
" 0 "" {TEXT -1 39 " = A X(t) + F(t)." }}{PARA
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 6 "Also, " }{TEXT
286 34 "X(0) = Z(L) = exp(L A)P + exp(L A)" }{TEXT -1 2 " " }
{XPPEDIT 18 0 "int(exp(-sA)*F(s),s = 0 .. L);" "6#-%$intG6$*&-%$expG6#
,$%#sAG!\"\"\"\"\"-%\"FG6#%\"sGF-/F1;\"\"!%\"LG" }{TEXT -1 2 " ." }}
{PARA 0 "" 0 "" {TEXT -1 186 "Substitute the definition for P from abo
ve, factor, and this X(0) reduces to P. Thus, the function X(t) satisf
ies the same differential equation and the same initial value as Z(t).
Thus," }}{PARA 0 "" 0 "" {TEXT -1 37 " Z(t) = Z
(t+L)," }}{PARA 0 "" 0 "" {TEXT -1 34 "and the function Z(t) is period
ic." }}{PARA 0 "" 0 "" {TEXT -1 26 " " }}
{PARA 0 "" 0 "" {TEXT -1 66 "The examination of the examples with thes
e formulas is our intent." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 1
{PARA 3 "" 0 "" {TEXT -1 21 "Asymptotic Properties" }}{PARA 0 "" 0 ""
{TEXT -1 362 " Often in physical systems which have a periodic for
cing function, if a solution begins at some initial position different
from the one which produces the periodic solution, the values of this
solution will converge as t increases to the values of the periodic s
olution. This is true for each of the examples we examine. This observ
ation can be made precise." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0
"" 0 "" {TEXT -1 89 " We suppose that P and Q are in the domain of
A and that u and v satisfy the equation" }}{PARA 0 "" 0 "" {TEXT -1
37 " Y' = AY + F" }}{PARA 0 "" 0 "" {TEXT -1
57 "with u(0) = P and v(0) = Q. There is a number c such that" }}
{PARA 0 "" 0 "" {TEXT -1 24 " | u(t) - v(t) | " }{XPPEDIT 18 0
"` ` <= ` `;" "6#1%\"~GF$" }{TEXT -1 19 "exp(c t) | P - Q |." }}{PARA
0 "" 0 "" {TEXT -1 56 "In fact, the largest eigenvalue of A can be cho
sen as c." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1
142 " To see that the magnitude for the difference in values of u \+
and v is as suggested, one has only to use equation (2.1). There, we s
ee that" }}{PARA 0 "" 0 "" {TEXT -1 27 " | u(t) - v(t) | " }
{XPPEDIT 18 0 "` ` <= ` `;" "6#1%\"~GF$" }{TEXT -1 23 "| exp( t A) ( P
- Q) |." }}{PARA 0 "" 0 "" {TEXT -1 167 "The operator norm for the li
near operator exp(t A) will finish the explanation for the inequality.
We resist that discussion, but try to make clear what is exp( t A ).
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 406 " \+
The importance of this inequality, however, is recognized as follows.
For the three examples stated in the introduction, the number c is ne
gative. The result is that if a solution for any of the three equation
s begins at a point which does not produce the periodic solution, the \+
values of the solution quickly move toward the periodic solution. That
periodic solution is an attracting stable solution." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0
"" {TEXT -1 34 "Section 3: Construction of Exp(tA)" }}{PARA 0 "" 0 ""
{TEXT -1 165 " If c and P are numbers, then exp(c t) P is a famil
iar function known from earliest experiences with mathematics. It prov
ides a solution for Y' = c Y, Y(0) = P." }}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 44 "y:=t->exp(c*t)*P;\ndiff(y(t),t)-c*y(t);\ny(0);" }}}
{PARA 0 "" 0 "" {TEXT -1 99 "Maple would have chosen this function if \+
we had asked for a solution for the differential equation." }}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "restart:\ndsolve(\{diff(y(t),t)=c*y
(t),y(0)=P\},y(t));" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 124 "We illustrate the character of solutions for this equa
tion by choosing c = -2 and graphing solutions for several values of P
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "c:=-2;\nsol:=dsolve(\{di
ff(y(t),t)=c*y(t),y(0)=a\},y(t));" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 78 "y1:=subs(a=1/3,rhs(sol)):\ny2:=subs(a=5/3,rhs(sol)):
\ny3:=subs(a=-1/2,rhs(sol)):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1
0 55 "plot([y1,y2,y3],t=0..2,y=-1..2,color=[red,blue,black]);" }}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 387 "Consider
the Example 2. If A is an n x n matrix, then the computation of exp(t
A) is often a topic covered in a first course in differential equatio
ns, especially in the section on systems of differential equations. If
P is an n dimensional vector, then Z(t) = exp(tA)P defines a function
with values n dimensional vectors and provides a solution for the sys
tem Z'(t) = A Z(t), Z(0) = P." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1
0 50 "restart:\nwith(linalg):\nA:=matrix([[-2,1],[1,-2]]);" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "exponential(t*A);" }}}{EXCHG {PARA
0 "> " 0 "" {MPLTEXT 1 0 9 "P:=[a,b];" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 27 "evalm(exponential(t*A)&*P);" }}}{EXCHG {PARA 0 "> "
0 "" {MPLTEXT 1 0 60 "E1:=(%)[1]:\nE2:=(%%)[2]:\nx:=unapply(E1,t);\ny:
=unapply(E2,t);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 14 "We verify that" }}{PARA 0 "" 0 "" {TEXT -1 23 " \+
" }{XPPEDIT 18 0 "matrix([[diff(x,t)], [diff(y,t)]]);" "6
#-%'matrixG6#7$7#-%%diffG6$%\"xG%\"tG7#-F)6$%\"yGF," }{TEXT -1 3 " = \+
" }{XPPEDIT 18 0 "matrix([[-2, 1], [1, -2]]);" "6#-%'matrixG6#7$7$,$\"
\"#!\"\"\"\"\"7$F+,$F)F*" }{TEXT -1 1 " " }{XPPEDIT 18 0 "matrix([[x(t
)], [y(t)]]);" "6#-%'matrixG6#7$7#-%\"xG6#%\"tG7#-%\"yG6#F+" }{TEXT
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 4 "with" }}{PARA 0 "" 0 "" {TEXT
-1 31 " " }{XPPEDIT 18 0 "matrix([[x(0)]
, [y(0)]]);" "6#-%'matrixG6#7$7#-%\"xG6#\"\"!7#-%\"yG6#F+" }{TEXT -1
3 " = " }{XPPEDIT 18 0 "matrix([[a], [b]]);" "6#-%'matrixG6#7$7#%\"aG7
#%\"bG" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "simplify(diff(x(t),t)-(-2*x(t)+y(t)
));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "simplify(diff(y(t),t
)-(x(t)-2*y(t)));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "[x(0),
y(0)];" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1
166 "We illustrate the character of solutions for this equation by dra
wing four graphs for the solutions of the differential equation starti
ng at different initial values." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT
1 0 163 "y1:=subs(\{a=2,b=3\},[x(t),y(t),t=0..2]):\ny2:=subs(\{a=-2,b=
3\},[x(t),y(t),t=0..2]):\ny3:=subs(\{a=-2,b=-3\},[x(t),y(t),t=0..2]):
\ny4:=subs(\{a=2,b=-3\},[x(t),y(t),t=0..2]):" }}}{EXCHG {PARA 0 "> "
0 "" {MPLTEXT 1 0 49 "plot([y1,y2,y3,y4],color=[blue,black,red,green])
;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 60 "We turn now to Example 3.
The computation of exp(tA) in case" }}{PARA 0 "" 0 "" {TEXT -1 47 " \+
A(f) = f '', with f(0) = 0 = f(1)" }}{PARA 0 "" 0 ""
{TEXT -1 142 "may be less familiar, so we say more than we did with th
e previous two examples. First, we record the eigenvalues and eigenvec
tors for this A:" }}{PARA 0 "" 0 "" {TEXT -1 7 " " }{XPPEDIT 18
0 "lambda[n];" "6#&%'lambdaG6#%\"nG" }{TEXT -1 3 " = " }{XPPEDIT 18 0
"-n^2*Pi^2;" "6#,$*&%\"nG\"\"#%#PiGF&!\"\"" }{TEXT -1 6 " and " }
{XPPEDIT 18 0 "phi[n](x);" "6#-&%$phiG6#%\"nG6#%\"xG" }{TEXT -1 3 " = \+
" }{XPPEDIT 18 0 "sin(n*Pi*x);" "6#-%$sinG6#*(%\"nG\"\"\"%#PiGF(%\"xGF
(" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 70 "This is a complete s
et of eigenfunctions in the sense that if f is in " }{XPPEDIT 18 0 "L^
2*[0, 1];" "6#*&%\"LG\"\"#7$\"\"!\"\"\"F(" }{TEXT -1 5 " then" }}
{PARA 0 "" 0 "" {TEXT -1 22 " f(x) = " }{XPPEDIT 18 0 "S
um(f[n]*phi[n](x),n);" "6#-%$SumG6$*&&%\"fG6#%\"nG\"\"\"-&%$phiG6#F*6#
%\"xGF+F*" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 6 "where " }
{XPPEDIT 18 0 "f[n];" "6#&%\"fG6#%\"nG" }{TEXT -1 38 " is formed by th
e Fourier coefficient" }}{PARA 0 "" 0 "" {TEXT -1 14 " \+
" }{XPPEDIT 18 0 "f[n];" "6#&%\"fG6#%\"nG" }{TEXT -1 3 " = " }
{XPPEDIT 18 0 "int(f(x)*phi[n](x),x = 0 .. 1)/int(phi[n](x)^2,x = 0 ..
1);" "6#*&-%$intG6$*&-%\"fG6#%\"xG\"\"\"-&%$phiG6#%\"nG6#F+F,/F+;\"\"
!F,F,-F%6$*$-&F/6#F16#F+\"\"#/F+;F5F,!\"\"" }{TEXT -1 2 " ." }}{PARA
0 "" 0 "" {TEXT -1 43 "Convergence of the infinite sum is in norm." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 177 "We illus
trate with a graph. In this example, the convergence of the infinite s
um is uniform on [0, 1]. We will not try to make the infinite sum, but
rather use only five terms. " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1
0 30 "restart:\nf:=x->10*x^2*(1-x)^2;" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 111 "for n from 1 to 11 do\n fc[n]:=int(f(x)*sin(n*Pi*x
),x=0..1)/\n int(sin(n*Pi*x)^2,x=0..1);\nod;\nn:='n':" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "approx:=x->sum(fc[n]*sin(n*P
i*x),n=1..11);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "plot([f(x
),approx(x)],x=0..1,color=[red,blue]);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 132 "Just as we can construct the identity function from the \+
eigenvectors, so we construct A from these eigenvectors and the eigenv
alues:" }}{PARA 0 "" 0 "" {TEXT -1 25 " A(f)(x) = " }
{XPPEDIT 18 0 "Sum(-n^2*Pi^2*f[n]*phi[n](x),n);" "6#-%$SumG6$,$**%\"nG
\"\"#%#PiGF)&%\"fG6#F(\"\"\"-&%$phiG6#F(6#%\"xGF.!\"\"F(" }{TEXT -1 2
" ." }}{PARA 0 "" 0 "" {TEXT -1 27 "We illustrate with a graph." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "Afprox:=x->sum(-n^2*Pi^2*fc[
n]*sin(n*Pi*x),n=1..11);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40
"plot([Afprox(x),diff(f(x),x,x)],x=0..1);" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 15 "diff(f(x),x,x);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 454 "We are now ready to say what is exp(tA). As before, it s
hould be true that if P is in the domain of A, then Z(t) = exp(tA)P is
a function which satisfies the differential equation Z'(t) = AZ. In t
his example, for a fixed t, Z(t) is itself a function. In fact, Z(t) i
s a function of x. Instead of using the notation Z(t)(x), we follow co
nvention and write Z(t, x). Thus, what appears to be an ODE in the fun
ction space is our partial differential equation" }}{PARA 0 "" 0 ""
{TEXT -1 13 " " }{XPPEDIT 18 0 "diff(Z,t);" "6#-%%diffG6$%
\"ZG%\"tG" }{TEXT -1 5 " = " }{XPPEDIT 18 0 "diff(Z,`$`(x,2));" "6#-
%%diffG6$%\"ZG-%\"$G6$%\"xG\"\"#" }{TEXT -1 26 " , Z(t, 0) = 0 = Z(t,
1)." }}{PARA 0 "" 0 "" {TEXT -1 22 " Z(0) = f." }}{PARA
0 "" 0 "" {TEXT -1 104 "With a little knowledge of the techniques for \+
separation of variables, the way to make exp(tA) is clear:" }}{PARA 0
"" 0 "" {TEXT -1 19 " exp(tA) f = " }{XPPEDIT 18 0 "Sum(exp(-n^2
*Pi^2*t)*f[n]*phi[n](x),n);" "6#-%$SumG6$*(-%$expG6#,$*(%\"nG\"\"#%#Pi
GF-%\"tG\"\"\"!\"\"F0&%\"fG6#F,F0-&%$phiG6#F,6#%\"xGF0F," }{TEXT -1 2
" ." }}{PARA 0 "" 0 "" {TEXT -1 43 "We verify that such a sum forms a \+
solution." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "Z:=(t,x)->sum(e
xp(-n^2*Pi^2*t)*fc[n]*sin(n*Pi*x),n=1..11);" }}}{EXCHG {PARA 0 "> " 0
"" {MPLTEXT 1 0 32 "diff(Z(t,x),t)-diff(Z(t,x),x,x);" }}}{PARA 0 "" 0
"" {TEXT -1 185 "We check that Z has the right initial value by compar
ing the graphs of Z(0, x) and f(x). We off-set the graph of f(x) by 0.
001 so that it can be distinguished from the graph of Z(0, x)." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "plot([Z(0,x),f(x)+0.001],x=0
..1);" }}}{PARA 0 "" 0 "" {TEXT -1 245 "To follow the pattern of the p
revious explanations, we should draw a graph for a solution. But we al
so should expect the resulting graph to follow the pattern for the abo
ve graphs. That is, we expect solutions should collapse to zero as t g
rows." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 81 "plot3d(Z(t,x),x=0..
1,t=0..1/5,orientation=[-130,55],\n axes=NORMAL);" }}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 199 "Confiden
t that we can compute the exponential for each of the examples in Sect
ion 1, we next use this exponential to compute solutions for the diffe
rential equations with a periodic forcing function." }}{EXCHG {PARA 0
"> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT
1 {PARA 3 "" 0 "" {TEXT -1 22 "Section 4:Computations" }}{PARA 0 "" 0
"" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 184 " In this section, \+
we use the results of Section 2, together with the exponential functio
n as explored in Section 3, to find a periodic solution for the model \+
equations of Section 1." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1
{PARA 3 "" 0 "" {TEXT -1 19 "Return to Example 1" }}{PARA 0 "" 0 ""
{TEXT -1 120 " The first equation was y'(t) = - 2 y(t) + sin(t).
By Equation 2.2, the initial value for the periodic solution is" }}
{PARA 0 "" 0 "" {TEXT -1 27 " P = " }{XPPEDIT
18 0 "exp(-2*L)*int(exp(2*s)*sin(s),s = 0 .. L)/(1-exp(-2*L));" "6#*(-
%$expG6#,$*&\"\"#\"\"\"%\"LGF*!\"\"F*-%$intG6$*&-F%6#*&F)F*%\"sGF*F*-%
$sinG6#F4F*/F4;\"\"!F+F*,&F*F*-F%6#,$*&F)F*F+F*F,F,F," }{TEXT -1 0 ""
}}{PARA 0 "" 0 "" {TEXT -1 27 "where L in this example is " }{XPPEDIT
18 0 "2*pi;" "6#*&\"\"#\"\"\"%#piGF%" }{TEXT -1 40 ". We compute each
part of this formula." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "L:=2*Pi;\nint(exp(2*s)*sin(s),s=0..
L);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "expand(exp(-2*L)*%);
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "P:=simplify(expand(%/(1
-exp(-2*L))));" }}}{PARA 0 "" 0 "" {TEXT -1 63 "We now solve the diffe
rential equation with this initial value." }}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 50 "dsolve(\{diff(y(t),t)=-2*y(t)+sin(t),y(0)=P\},y(t));
" }}}{PARA 0 "" 0 "" {TEXT -1 132 "We can observe that this solution i
s periodic. We plot this solution and two others to observe the asympt
otic property of solutions." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0
55 "sol:=dsolve(\{diff(y(t),t)=-2*y(t)+sin(t),y(0)=a\},y(t));" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 72 "y1:=subs(a=1,rhs(sol)):\ny2:
=subs(a=P,rhs(sol)):\ny3:=subs(a=-1,rhs(sol)):" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 58 "plot([y1,y2,y3],t=0..2*Pi,y=-1..1,color=[blue,
red,black]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA
0 "" 0 "" {TEXT -1 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 3 " " }}{SECT 1
{PARA 3 "" 0 "" {TEXT -1 19 "Return to Example 2" }}{PARA 0 "" 0 ""
{TEXT -1 48 " The second example is a system of equations" }}
{PARA 0 "" 0 "" {TEXT -1 14 " " }{XPPEDIT 18 0 "diff(x(t)
,t) = -2*x(t)+y(t)+cos(t);" "6#/-%%diffG6$-%\"xG6#%\"tGF*,(*&\"\"#\"\"
\"-F(6#F*F.!\"\"-%\"yG6#F*F.-%$cosG6#F*F." }{TEXT -1 6 ", " }
{XPPEDIT 18 0 "diff(y(t),t) = x(t)-2*y(t)+sin(t);" "6#/-%%diffG6$-%\"y
G6#%\"tGF*,(-%\"xG6#F*\"\"\"*&\"\"#F/-F(6#F*F/!\"\"-%$sinG6#F*F/" }
{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 55 "Equation 2.2 gives the initial value for this system as"
}}{PARA 0 "" 0 "" {TEXT -1 16 " " }{TEXT 270 1 "P" }
{TEXT -1 3 " = " }{XPPEDIT 18 0 "(1-`exp(L A)`)^(-1);" "6#),&\"\"\"F%%
)exp(L~A)G!\"\",$F%F'" }{TEXT 269 3 "exp" }{TEXT -1 1 "(" }{TEXT 283
2 "LA" }{TEXT -1 3 ") " }{XPPEDIT 18 0 "int(`exp(-sA)`*F(s),s = 0 .. \+
L);" "6#-%$intG6$*&%)exp(-sA)G\"\"\"-%\"FG6#%\"sGF(/F,;\"\"!%\"LG" }
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 32 "where L in this example i
s also " }{XPPEDIT 18 0 "2*Pi;" "6#*&\"\"#\"\"\"%#PiGF%" }{TEXT -1 70
". We define A and compute the various parts to make the initial value
." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 13 "with(linalg):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT
1 0 36 "L:=2*Pi;\nA:=matrix([[-2,1],[1,-2]]);" }}}{EXCHG {PARA 0 "> "
0 "" {MPLTEXT 1 0 24 "expA:=exponential(-s*A);" }}}{PARA 0 "" 0 ""
{TEXT -1 72 "The first part is the product of exp(-s A) and F(s) = [co
s(s), sin(s) ]." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "part1:=ev
alm(expA&*vector([cos(s),sin(s)]));" }}}{PARA 0 "" 0 "" {TEXT -1 51 "T
he second part is the integral of these functions." }}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 81 "part2:=matrix([[int((part1)[1],s=0..L)],\n \+
[int((part1)[2],s=0..L)]]);" }}}{PARA 0 "" 0 "" {TEXT -1
59 "The third part is the product of exp(L A) and the integral." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "part3:=evalm(exponential(L*A
)&*part2);" }}}{PARA 0 "" 0 "" {TEXT -1 40 "Part 4 is a simplification
of the above." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 98 "part4:=mat
rix([[simplify(expand(part3[1,1]))],\n [simplify(expa
nd(part3[2,1]))]]);" }}}{PARA 0 "" 0 "" {TEXT -1 224 "We make the inve
rse. First, we define the 2 x 2 identity matrix and let part 5 be the \+
product of the inverse and what has come before. This product is best \+
simplified and separated into the two rows of its vector components."
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "id:=matrix([[1,0],[0,1]]);
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "part5:=evalm(inverse(id
-exponential(2*Pi*A))&*part4):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT
1 0 67 "r1:=simplify(expand(part5[1,1]));\nr2:=simplify(expand(part5[2
,1]));" }}}{PARA 0 "" 0 "" {TEXT -1 76 "These are the initial values f
or x and y which define the periodic solution." }}{EXCHG {PARA 0 "> "
0 "" {MPLTEXT 1 0 122 "dsolve(\{diff(x(t),t)=-2*x(t)+y(t)+cos(t),\n \+
diff(y(t),t)=x(t)-2*y(t)+sin(t),x(0)=r1,y(0)=r2\},\n \{x(
t),y(t)\});" }}}{PARA 0 "" 0 "" {TEXT -1 134 "We can observe that this
solution is periodic. We plot this solution and three others to obser
ve the asymptotic property of solutions." }}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 125 "sol:=dsolve(\{diff(x(t),t)=-2*x(t)+y(t)+cos(t),\n \+
diff(y(t),t)=x(t)-2*y(t)+sin(t),x(0)=a,y(0)=b\},\n \{x(t)
,y(t)\});" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "assign(%);" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "x(t);y(t);" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 176 "y1:=subs(\{a=1,b=2\},[x(t),y(t),t=
0..2*Pi]):\ny2:=subs(\{a=r1,b=r2\},[x(t),y(t),t=0..2*Pi]):\ny3:=subs(
\{a=-1,b=-2\},[x(t),y(t),t=0..2*Pi]):\ny4:=subs(\{a=-1,b=2\},[x(t),y(t
),t=0..2*Pi]):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "plot([y1,
y2,y3,y4],color=[black,red,green,blue]);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3
"" 0 "" {TEXT -1 19 "Return to Example 3" }}{PARA 0 "" 0 "" {TEXT -1
54 "The third example is the partial differential equation" }}{PARA 0
"" 0 "" {TEXT -1 13 " " }{XPPEDIT 18 0 "diff(u,t);" "6#-%%
diffG6$%\"uG%\"tG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "diff(u,`$`(x,2));
" "6#-%%diffG6$%\"uG-%\"$G6$%\"xG\"\"#" }{TEXT -1 13 " + 10 sin(2 " }
{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 4 " t) " }{XPPEDIT 18 0 "x^2*(
1-x)^2;" "6#*&%\"xG\"\"#,&\"\"\"F'F$!\"\"F%" }{TEXT -1 25 ", u(t, 0) \+
= 0 = u(t, 1)." }}{PARA 0 "" 0 "" {TEXT -1 113 "Just as for the other \+
examples, Equation 2.2 gives the initial value for the periodic soluti
on for this system as" }}{PARA 0 "" 0 "" {TEXT -1 9 " " }
{TEXT 272 1 "P" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "(1-`exp(L A)`)^(-1);
" "6#),&\"\"\"F%%)exp(L~A)G!\"\",$F%F'" }{TEXT 271 3 "exp" }{TEXT -1
6 "(LA) " }{XPPEDIT 18 0 "int(`exp(-sA)`*F(s),s = 0 .. L);" "6#-%$int
G6$*&%)exp(-sA)G\"\"\"-%\"FG6#%\"sGF(/F,;\"\"!%\"LG" }{TEXT -1 2 " ."
}}{PARA 0 "" 0 "" {TEXT -1 144 "What is different in this example from
the others is the techniques for computing the parts for this initial
value -- for this initial function." }}{PARA 0 "" 0 "" {TEXT -1 0 ""
}}{PARA 0 "" 0 "" {TEXT -1 166 "To make the computation of the initial
value for the periodic solution for the third example easier to under
stand, there are two useful ideas that should be recalled." }}{PARA 0
"" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 277 14 "Useful idea 1."
}{TEXT -1 4 " If " }{XPPEDIT 18 0 "phi[n];" "6#&%$phiG6#%\"nG" }{TEXT
-1 74 " is an orthogonal sequence and both A and B are linear mappings
defined by" }}{PARA 0 "" 0 "" {TEXT -1 20 " A(f) = " }
{XPPEDIT 18 0 "Sigma;" "6#%&SigmaG" }{TEXT -1 1 " " }{XPPEDIT 18 0 "la
mbda[n]*f[n]*phi[n];" "6#*(&%'lambdaG6#%\"nG\"\"\"&%\"fG6#F'F(&%$phiG6
#F'F(" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 18 "and B(g) =
" }{XPPEDIT 18 0 "Sigma;" "6#%&SigmaG" }{TEXT -1 1 " " }{XPPEDIT 18
0 "mu[n]*g[n]*phi[n];" "6#*(&%#muG6#%\"nG\"\"\"&%\"gG6#F'F(&%$phiG6#F'
F(" }{TEXT -1 2 " ," }}{PARA 0 "" 0 "" {TEXT -1 26 "for all f and g, a
nd with " }{XPPEDIT 18 0 "f[n];" "6#&%\"fG6#%\"nG" }{TEXT -1 5 " and \+
" }{XPPEDIT 18 0 "g[n];" "6#&%\"gG6#%\"nG" }{TEXT -1 55 " the Fourier
coefficients for f and g with respect to " }{XPPEDIT 18 0 "phi[n];" "
6#&%$phiG6#%\"nG" }{TEXT -1 6 ", then" }}{PARA 0 "" 0 "" {TEXT -1 25 "
A( B( f)) = " }{XPPEDIT 18 0 "Sigma;" "6#%&SigmaG" }
{TEXT -1 1 " " }{XPPEDIT 18 0 "lambda[n]*mu[n]*f[n]*phi[n];" "6#**&%'l
ambdaG6#%\"nG\"\"\"&%#muG6#F'F(&%\"fG6#F'F(&%$phiG6#F'F(" }{TEXT -1 1
"." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 278 14 "Us
eful idea 2." }{TEXT -1 32 " In the setting described above" }}{PARA
0 "" 0 "" {TEXT -1 28 " exp( tA)(f) = " }{XPPEDIT 18 0 "S
igma;" "6#%&SigmaG" }{TEXT -1 6 " exp( " }{XPPEDIT 18 0 "lambda[n];" "
6#&%'lambdaG6#%\"nG" }{TEXT -1 4 " t) " }{XPPEDIT 18 0 "f[n]*phi[n];"
"6#*&&%\"fG6#%\"nG\"\"\"&%$phiG6#F'F(" }{TEXT -1 2 " ." }}{PARA 0 ""
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 53 "To keep down the clu
tter, we write the third model as" }}{PARA 0 "" 0 "" {TEXT -1 44 " \+
Y'(t) = A Y(t) + g(t) h," }}{PARA 0 "" 0 "" {TEXT -1
54 "where we identify A as the differential operator AY = " }{XPPEDIT
18 0 "d^2;" "6#*$%\"dG\"\"#" }{TEXT -1 5 " Y / " }{XPPEDIT 18 0 "d*x^2
;" "6#*&%\"dG\"\"\"*$%\"xG\"\"#F%" }{TEXT -1 32 " with boundary condi
tions, and " }}{PARA 0 "" 0 "" {TEXT -1 12 " and define " }}{PARA 0 "
" 0 "" {TEXT -1 18 " g(t) = sin(2 " }{XPPEDIT 18 0 "Pi;" "6#%#PiG
" }{TEXT -1 15 " t), f(x) = 10 " }{XPPEDIT 18 0 "x^2*(1-x)^2;" "6#*&%
\"xG\"\"#,&\"\"\"F'F$!\"\"F%" }{TEXT -1 28 " , and F(t, x) = g(t) f(x)
. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 29 "We \+
develop a realization for " }}{PARA 0 "" 0 "" {TEXT -1 15 " \+
" }{TEXT 274 1 "P" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "(1-`exp(L A)`
)^(-1);" "6#),&\"\"\"F%%)exp(L~A)G!\"\",$F%F'" }{TEXT 273 3 "exp" }
{TEXT -1 1 "(" }{TEXT 287 2 "LA" }{TEXT -1 3 ") " }{XPPEDIT 18 0 "int
(`exp(-sA)`*g(s)*F,s = 0 .. L);" "6#-%$intG6$*(%)exp(-sA)G\"\"\"-%\"gG
6#%\"sGF(%\"FGF(/F,;\"\"!%\"LG" }{TEXT -1 2 " ." }}{PARA 0 "" 0 ""
{TEXT -1 13 "We recognize " }}{PARA 0 "" 0 "" {TEXT -1 44 " \+
" }{TEXT 275 13 " exp(-s A)" }
{TEXT -1 1 " " }{XPPEDIT 18 0 "g(s)*f;" "6#*&-%\"gG6#%\"sG\"\"\"%\"fGF
(" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 2 "as" }}{PARA 0 "" 0 ""
{TEXT -1 20 " " }{XPPEDIT 18 0 "sum(exp(-s*lambda[n
])*g(s)*f[n]*phi[n]);" "6#-%$sumG6#**-%$expG6#,$*&%\"sG\"\"\"&%'lambda
G6#%\"nGF-!\"\"F--%\"gG6#F,F-&%\"fG6#F1F-&%$phiG6#F1F-" }{TEXT -1 1 ".
" }}{PARA 0 "" 0 "" {TEXT -1 18 "We integrate this:" }}{PARA 0 "" 0 "
" {TEXT -1 25 " " }{XPPEDIT 18 0 "int(`exp(-sA
)`*g(s)*f,s = 0 .. L);" "6#-%$intG6$*(%)exp(-sA)G\"\"\"-%\"gG6#%\"sGF(
%\"fGF(/F,;\"\"!%\"LG" }{TEXT -1 6 " = " }{XPPEDIT 18 0 "sum(int(ex
p(-s*lambda[n])*g(s),s = 0 .. L)*f[n]*phi[n]);" "6#-%$sumG6#*(-%$intG6
$*&-%$expG6#,$*&%\"sG\"\"\"&%'lambdaG6#%\"nGF1!\"\"F1-%\"gG6#F0F1/F0;
\"\"!%\"LGF1&%\"fG6#F5F1&%$phiG6#F5F1" }{TEXT -1 2 " ." }}{PARA 0 ""
0 "" {TEXT -1 59 "Using the first idea again, we operate on this vecto
r with " }{TEXT 276 8 "exp(L A)" }{TEXT -1 1 ":" }}{PARA 0 "" 0 ""
{TEXT -1 9 " " }{TEXT 280 3 "exp" }{TEXT -1 1 "(" }{TEXT 279
2 "LA" }{TEXT -1 3 ") " }{XPPEDIT 18 0 "int(`exp(-sA)`*g(s)*f,s = 0 .
. L);" "6#-%$intG6$*(%)exp(-sA)G\"\"\"-%\"gG6#%\"sGF(%\"fGF(/F,;\"\"!%
\"LG" }{TEXT -1 7 " = " }{XPPEDIT 18 0 "sum(exp(lambda[n]*L)*int(e
xp(-s*lambda[n])*g(s),s = 0 .. L)*f[n]*phi[n]);" "6#-%$sumG6#**-%$expG
6#*&&%'lambdaG6#%\"nG\"\"\"%\"LGF/F/-%$intG6$*&-F(6#,$*&%\"sGF/&F,6#F.
F/!\"\"F/-%\"gG6#F9F//F9;\"\"!F0F/&%\"fG6#F.F/&%$phiG6#F.F/" }{TEXT
-1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 76 "Finally, we operate on this l
ast with the inverse indicated in Equation 2.2." }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "(1-`exp
(L A)`)^(-1);" "6#),&\"\"\"F%%)exp(L~A)G!\"\",$F%F'" }{TEXT 281 3 "exp
" }{TEXT -1 6 "(LA) " }{XPPEDIT 18 0 "int(`exp(-sA)`*g(s)*f,s = 0 .. \+
L);" "6#-%$intG6$*(%)exp(-sA)G\"\"\"-%\"gG6#%\"sGF(%\"fGF(/F,;\"\"!%\"
LG" }{TEXT -1 4 " = " }{XPPEDIT 18 0 "sum(exp(lambda[n]*L)*int(exp(-s
*lambda[n])*g(s),s = 0 .. L)/(1-exp(lambda[n]*L)));" "6#-%$sumG6#*(-%$
expG6#*&&%'lambdaG6#%\"nG\"\"\"%\"LGF/F/-%$intG6$*&-F(6#,$*&%\"sGF/&F,
6#F.F/!\"\"F/-%\"gG6#F9F//F9;\"\"!F0F/,&F/F/-F(6#*&&F,6#F.F/F0F/F " 0 "" {MPLTEXT 1 0 19 "restart:L:=1; N:=5;" }}}
{PARA 0 "" 0 "" {TEXT -1 97 " We perform each of the integrals in th
e above formula, not yet identifying what is the lambda." }}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "g:=s->sin(2*Pi*s);" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "Part1:=int(exp(-s*lambda)*g(s),s=0.
.L);" }}}{PARA 0 "" 0 "" {TEXT -1 53 " We multiply the integral by \+
the exponential term." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "Par
t2:=exp(lambda*L)*Part1;" }}}{PARA 0 "" 0 "" {TEXT -1 53 " And, we
divide this product by the inverse term." }}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 31 "Part3:=Part2/(1-exp(lambda*L));" }}}{EXCHG {PARA 0 ">
" 0 "" {MPLTEXT 1 0 22 "factor(expand(Part3));" }}}{PARA 0 "" 0 ""
{TEXT -1 71 " These calculations have enabled us to pick out the co
efficients of " }{XPPEDIT 18 0 "f[n]*phi[n];" "6#*&&%\"fG6#%\"nG\"\"\"
&%$phiG6#F'F(" }{TEXT -1 36 ". It is appropriate to identify the " }
{XPPEDIT 18 0 "lambda[n];" "6#&%'lambdaG6#%\"nG" }{TEXT -1 1 "." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "for n from 1 to N do\n c[n
]:=-2*Pi/(n^4*Pi^4+4*Pi^2);\nod;\nn:='n';" }}}{PARA 0 "" 0 "" {TEXT
-1 80 " This gives us a formula for the initial value. We record t
hat formula here." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "P:=x->s
um(c[n]*f[n]*sin(n*Pi*x),n=1..N);" }}}{PARA 0 "" 0 "" {TEXT -1 61 "It \+
should be that P(0) = 0 and P(1) = 0. This can be checked." }}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "P(0); P(1);" }}}{PARA 0 "" 0 ""
{TEXT -1 86 "This function P is the initial distribution for the perio
dic solution of the equation " }}{PARA 0 "" 0 "" {TEXT -1 41 " \+
Y '(t) = AY(t) + g(t) f." }}{PARA 0 "" 0 "" {TEXT -1 60 "We w
ill draw its graph after we define f and the resulting " }{XPPEDIT
18 0 "f[n];" "6#&%\"fG6#%\"nG" }{TEXT -1 3 "'s." }}{PARA 0 "" 0 ""
{TEXT -1 159 " There are two terms in the solution. One term is exp
(t A) P, where P is as we have just determined. Note that this term sh
ould solve Y '(t) = AY, Y(0) = P." }}{PARA 0 "" 0 "" {TEXT -1 28 "We u
se this fact as a check." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 62 "
u1:=(t,x)->sum(exp(-n^2*Pi^2*t)*c[n]*f[n]*sin(n*Pi*x),n=1..N);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "diff(u1(t,x),t)-diff(u1(t,x)
,x,x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "u1(t,0); u1(t,1);
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "u1(0,x)-P(x);" }}}
{PARA 0 "" 0 "" {TEXT -1 156 "The second term is the addition of the n
on-homogeneous part -- the forcing function. This term may be so long \+
as to be cumbersome. We suppress the printing." }}{EXCHG {PARA 0 "> "
0 "" {MPLTEXT 1 0 112 "sum(exp(-n^2*Pi^2*t)*int(exp(s*n^2*Pi^2)*g(s),s
=0..t)*\n f[n]*sin(n*Pi*x),n=1..N):\nu2:=unapply(%,(t,x)):" }
}}{PARA 0 "" 0 "" {TEXT -1 92 "This term should solve the full equatio
n, the boundary conditions, and have initial value 0." }}{EXCHG {PARA
0 "> " 0 "" {MPLTEXT 1 0 95 "simplify(diff(u2(t,x),t)-diff(u2(t,x),x,x
)-\n sin(2*Pi*t)*sum(f[n]*sin(n*Pi*x),n=1..N));" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "u2(t,0); u2(t,1);" }}}{EXCHG {PARA
0 "> " 0 "" {MPLTEXT 1 0 8 "u2(0,x);" }}}{PARA 0 "" 0 "" {TEXT -1 101
"Finally, we make u as the sum of u1 and u2. Without checking we know \+
that this is the correct answer." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT
1 0 26 "u:=(t,x)->u1(t,x)+u2(t,x);" }}}{PARA 0 "" 0 "" {TEXT -1 22 "Bu
t, we check, anyway." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 93 "simp
lify(diff(u(t,x),t)-diff(u(t,x),x,x)-\n sin(2*Pi*t)*sum(f[n]*s
in(n*Pi*x),n=1..N));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "sim
plify(u(0,x)-u(1,x));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "u(
t,0); u(t,1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA
0 "" 0 "" {TEXT -1 74 "It is time to draw graphs and see this work. Ch
oose F as in the Example 3." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0
23 "Frc:=x->10*x^2*(1-x)^2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0
118 "for n from 1 to N do\n f[n]:=int(Frc(x)*sin(n*Pi*x),x=0..L)/\n \+
int(sin(n*Pi*x)^2,x=0..L):\nod:\nn:='n';" }}}{PARA 0
"" 0 "" {TEXT -1 140 "First, we check these coefficients by comparing \+
the graph of the Fourier approximation with the graph the forcing func
tion, off set by 0.01." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "ap
prox:=x->sum(f[n]*sin(n*Pi*x),n=1..N);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 37 "plot([Frc(x),approx(x)+0.01],x=0..L);" }}}{PARA 0 ""
0 "" {TEXT -1 29 "Next, we draw the graph of P." }}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 18 "plot(P(x),x=0..1);" }}}{PARA 0 "" 0 "" {TEXT
-1 40 "We draw the graph of the surface u(t,x)." }}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 41 "plot3d(u(t,x),x=0..1,t=0..2,axes=NORMAL);" }}}
{PARA 0 "" 0 "" {TEXT -1 70 "We watch an animation comparing the forci
ng function and the solution." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1
0 64 "with(plots):\nanimate(\{u(t,x),sin(2*Pi*t)*Frc(x)\},x=0..1,t=0..
2);" }}}{PARA 0 "" 0 "" {TEXT -1 79 "We watch the term with initial va
lue zero coalesce with this periodic solution." }}{EXCHG {PARA 0 "> "
0 "" {MPLTEXT 1 0 40 "animate(\{u2(t,x),u(t,x)\},x=0..1,t=0..1);" }}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0
0 "" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{SECT 1 {PARA 3 "" 0 "" {TEXT
-1 28 "Section 5: Alternate Methods" }}{PARA 0 "" 0 "" {TEXT -1 29 "Co
nsider again the Example 3." }}{PARA 0 "" 0 "" {TEXT -1 16 " \+
" }{XPPEDIT 18 0 "diff(u,t);" "6#-%%diffG6$%\"uG%\"tG" }{TEXT
-1 3 " = " }{XPPEDIT 18 0 "diff(u,`$`(x,2));" "6#-%%diffG6$%\"uG-%\"$G
6$%\"xG\"\"#" }{TEXT -1 14 " + 10 sin(t) " }{XPPEDIT 18 0 "x^2*(1-x)^
2;" "6#*&%\"xG\"\"#,&\"\"\"F'F$!\"\"F%" }{TEXT -1 25 ", u(t, 0) = 0 =
u(t, 1)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1
27 "The method we presented in " }{TEXT 282 23 "Section 4: Computation
s" }{TEXT -1 256 " for finding periodic solutions for this equation pu
ts this problem into a setting where the structure is the same as that
for the ordinary differential equations setting. There is a satisfact
ion in making such a unity of structure between ODE's and PDE's." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 347 "The meth
od for obtaining solutions for the non-homogeneous equation and the al
ternate method for obtaining periodic solutions presented in this sect
ion takes advantage of the uniqueness for the representation of a func
tion in terms of a predetermined orthogonal basis and takes advantage \+
of the power of a computer algebra system such as this one." }}{PARA
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 474 "Here, we recal
l standard methods and techniques for the computation of solutions for
non-homogeneous partial differential equations. These methods are ill
ustrated in the text listed in the Reference. In that text, there is a
chapter for which the non-homogeneous equation is the subject. The te
chniques in the next part are the ones used. However, that book does n
ot find periodic solutions. In fact, the author is not aware of a sour
ce to illustrate finding such solutions." }}{PARA 0 "" 0 "" {TEXT -1
0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 52 "Computing solutions for th
e nonhomogeneous equation." }}{PARA 0 "" 0 "" {TEXT -1 57 "For general
ity, we rewrite the nonhomogeneous equation as" }}{PARA 0 "" 0 ""
{TEXT -1 16 " " }{XPPEDIT 18 0 "diff(u,t);" "6#-%%diffG
6$%\"uG%\"tG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "diff(u,`$`(x,2));" "6#
-%%diffG6$%\"uG-%\"$G6$%\"xG\"\"#" }{TEXT -1 5 " + " }{XPPEDIT 18 0
"F(t,x);" "6#-%\"FG6$%\"tG%\"xG" }{TEXT -1 25 ", u(t, 0) = 0 = u(t, 1
)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 38 "As \+
before, we take A to be defined on " }{XPPEDIT 18 0 "C^2;" "6#*$%\"CG
\"\"#" }{TEXT -1 10 "[0,1] with" }}{PARA 0 "" 0 "" {TEXT -1 45 " \+
A(f) = f '' and f(0) = 0 = f(1)." }}{PARA 0 "" 0 "" {TEXT -1 0
"" }}{PARA 0 "" 0 "" {TEXT -1 57 "Recall that eigenvalues and eigenfun
ctions for this A are" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0
"" {TEXT -1 17 " " }{XPPEDIT 18 0 "lambda[n];" "6#&%'l
ambdaG6#%\"nG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "-n^2*Pi^2;" "6#,$*&%
\"nG\"\"#%#PiGF&!\"\"" }{TEXT -1 9 " and " }{XPPEDIT 18 0 "phi[n];
" "6#&%$phiG6#%\"nG" }{TEXT -1 12 "(x) = sin(n " }{XPPEDIT 18 0 "Pi;"
"6#%#PiG" }{TEXT -1 4 " x)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA
0 "" 0 "" {TEXT -1 56 "Using this orthogonal family, we suppose the so
lution is" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1
25 " u(t, x) = " }{XPPEDIT 18 0 "sum(T[n](t)*sin(n*Pi*x)
,n);" "6#-%$sumG6$*&-&%\"TG6#%\"nG6#%\"tG\"\"\"-%$sinG6#*(F+F.%#PiGF.%
\"xGF.F.F+" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0
"" 0 "" {TEXT -1 63 "Substitute this into the original equation and we
find the form" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 3 " " }{XPPEDIT 18 0 "sum(diff(T[n](t),t)*sin(n*Pi*x),n);"
"6#-%$sumG6$*&-%%diffG6$-&%\"TG6#%\"nG6#%\"tGF0\"\"\"-%$sinG6#*(F.F1%#
PiGF1%\"xGF1F1F." }{TEXT -1 3 " = " }{XPPEDIT 18 0 "sum(-n^2*Pi^2*T[n]
(t)*sin(n*Pi*x),n);" "6#-%$sumG6$,$**%\"nG\"\"#%#PiGF)-&%\"TG6#F(6#%\"
tG\"\"\"-%$sinG6#*(F(F1F*F1%\"xGF1F1!\"\"F(" }{TEXT -1 3 " + " }
{XPPEDIT 18 0 "sum(g[n](t)*sin(n*Pi*x),n);" "6#-%$sumG6$*&-&%\"gG6#%\"
nG6#%\"tG\"\"\"-%$sinG6#*(F+F.%#PiGF.%\"xGF.F.F+" }{TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 4 "with" }}{PARA 0 "" 0 "" {TEXT -1 27 " \+
" }{XPPEDIT 18 0 "g[n](t);" "6#-&%\"gG6#%\"nG6#%
\"tG" }{TEXT -1 5 " = " }{XPPEDIT 18 0 "`<`*F(t,x)*`,`*sin(n*Pi*x)*`
>`/(`<`*sin(n*Pi*x)*`,`*sin(n*Pi*x)*`>`);" "6#*.%\"GF%*,F$F%-F-6#*(F0F
%F1F%F*F%F%F+F%-F-6#*(F0F%F1F%F*F%F%F2F%!\"\"" }{TEXT -1 2 " ." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 82 "Thus, we \+
seek solutions for an infinite system of numerical differential equati
ons" }}{PARA 0 "" 0 "" {TEXT -1 34 " (5.1) \+
" }{XPPEDIT 18 0 "diff(T[n](t),t) = -n^2*pi^2*T[n](t)+g[n](t);" "6#/-%
%diffG6$-&%\"TG6#%\"nG6#%\"tGF-,&*(F+\"\"#%#piGF0-&F)6#F+6#F-\"\"\"!\"
\"-&%\"gG6#F+6#F-F6" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 0 ""
}}{PARA 0 "" 0 "" {TEXT -1 148 "While this is in reality an infinite s
equence of differential equations, we construct a finite number of the
se and use these to approximate u(t, x)." }}{PARA 0 "" 0 "" {TEXT -1
0 "" }}{PARA 0 "" 0 "" {TEXT -1 117 "To get a specific solution, we ne
ed an initial value. We choose a function in the domain of A. We take \+
u(0, x) to be " }{XPPEDIT 18 0 "x*(1-x);" "6#*&%\"xG\"\"\",&F%F%F$!\"
\"F%" }{TEXT -1 19 " and F(t, x) to be " }{XPPEDIT 18 0 "sin(2*Pi*t)*x
^2*(1-x);" "6#*(-%$sinG6#*(\"\"#\"\"\"%#PiGF)%\"tGF)F)*$%\"xGF(F),&F)F
)F-!\"\"F)" }{TEXT -1 102 ". We do not expect this to lead to a period
ic solution, but we expect the solution to coalesce to one." }}{PARA
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 76 "Here is the ini
tialization. We compute only five terms of the approximation." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "N:=5;\nu0:=x->x*(1-x);\nF:=(
t,x)->sin(2*Pi*t)*x^2*(1-x);" }}}{PARA 0 "" 0 "" {TEXT -1 95 "Here is \+
the computation of solutions for five of the numerical ordinary differ
ential equations." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 268 "for n \+
from 1 to N do\n denomin:=int(sin(n*Pi*x)^2,x=0..1):\n T0:=int(u0(x)
*sin(n*Pi*x),x=0..1)/denomin:\n gn:=t->int(F(t,x)*sin(n*Pi*x),x=0..1)
/denomin:\n eq:=diff(Y(t),t)=-n^2*Pi^2*Y(t)+gn(t):\n sol:=dsolve(\{e
q,Y(0)=T0\},Y(t)):\n T[n]:=unapply(rhs(sol),t);\nod:\nn:='n';" }}}
{PARA 0 "" 0 "" {TEXT -1 61 "Here is the definition for the approximat
ion of the solution." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "u:=(
t,x)->sum(T[n](t)*sin(n*Pi*x),n=1..N);" }}}{PARA 0 "" 0 "" {TEXT -1
97 "Here is a plot of this solution. Look carefully to see the wave in
duced by the periodic solution." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT
1 0 41 "plot3d(u(t,x),x=0..1,t=0..2,axes=normal);" }}}{PARA 0 "" 0 ""
{TEXT -1 95 "Here is a check that this five term approximation satisfi
es the partial differential equation.." }}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 142 "expand(simplify(combine(diff(u(t,x),t)-diff(u(t,x),x
,x)-\n sum(2*int(F(t,x)*sin(n*Pi*x),x=0..1)*sin(n*Pi*x),n=1..N),\n \+
trig)));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 ""
}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 18 "P
eriodic Solutions" }}{PARA 0 "" 0 "" {TEXT -1 610 "We now return to th
e problem of finding a periodic solution for this heat equation with p
eriodic forcing function. As in the previous example of this section, \+
we use the Fourier expansion for the forcing function to change the pa
rtial differential equation into an infinite collection of ordinary di
fferential equations. These infinity of ordinary differential equation
s are one dimensional equations. We solve them with a yet to be determ
ined initial value. That initial value is chosen to create a periodic \+
solution. The techniques for these one dimensional equations are the s
ame techniques of Example 1 in " }{TEXT 288 23 "Section 4, Computation
s" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 243 "Here is Maple syntax to make an approximation for the so
lution. Be reminded that we do not need to specify an initial distribu
tion in this problem. Indeed, the problem is to find the initial distr
ibution which generates the periodic solution." }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 130 "First, we decided how ma
ny terms N we will use to approximation the solution and we specify th
e periodic forcing function F(t, x)." }}{PARA 0 "" 0 "" {TEXT -1 0 ""
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA
0 "> " 0 "" {MPLTEXT 1 0 38 "N:=5;\nF:=(t,x)->sin(2*Pi*t)*x^2*(1-x);"
}}}{PARA 0 "" 0 "" {TEXT -1 69 "Next, we solve the N equations 5.1 whi
ch generate periodic solutions." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT
1 0 287 "for n from 1 to N do\n denomin:=int(sin(n*Pi*x)^2,x=0..1):\n
gn:=t->int(F(t,x)*sin(n*Pi*x),x=0..1)/denomin:\n eq:=diff(Y(t),t)=-
n^2*Pi^2*Y(t)+gn(t):\n sol:=dsolve(\{eq,Y(0)=c\},Y(t)):\n YY:=unappl
y(rhs(sol),t):\n cc:=solve(YY(0)=YY(1),c):\n T[n]:=unapply(subs(c=cc
,YY(t)),t):\n od:\nn:='n';" }}}{PARA 0 "" 0 "" {TEXT -1 158 "We plot \+
these N equations. They may be so small that we need to amplify them. \+
Here, they are amplified by a power of 2 so that they can be seen in t
he graphs." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "plot([seq(2^n*
T[n](t),n=1..N)],t=0..1);" }}}{PARA 0 "" 0 "" {TEXT -1 32 "We compose \+
the solution u(t, x)." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "u:=
(t,x)->sum(T[n](t)*sin(n*Pi*x),n=1..N);" }}}{PARA 0 "" 0 "" {TEXT -1
81 "We draw the graph of the periodic solution for the partial differe
ntial equation." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "plot3d(u(
t,x),x=0..1,t=0..1,axes=normal);" }}}{PARA 0 "" 0 "" {TEXT -1 200 "As \+
a check, we draw the graph of u(0, x) and u(1, x) to see that they loo
k alike. The graphs may be so close that it is of value to off-set the
graph of one of these so that they can be distinguished." }}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "plot([u(0,x),u(1,x)+0.0001],x=0..1)
;" }}}{PARA 0 "" 0 "" {TEXT -1 49 "Finally, we check that we really ha
ve a solution." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 139 "expand(si
mplify(combine(\n diff(u(t,x),t)-diff(u(t,x),x,x)-\n sum(2*int(F(t
,x)*sin(n*Pi*x),x=0..1)*sin(n*Pi*x),\n n=1..N),trig)));" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0
"" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 258 "" 0 "" {TEXT -1 9 "REFERENC
E" }}{PARA 0 "" 0 "" {TEXT -1 20 "George A. Articolo, " }{TEXT 289 69
"Partial Differential Equations & Boundary Value Problems with Maple V
" }{TEXT -1 23 ", Academic Press, 1998." }}}{MARK "0 0" 58 }{VIEWOPTS
1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 1 1 2 33 1 1 }