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{SECT 0 {PARA 256 "" 0 "" {TEXT -1 50 "Module 23: d'Alembert's Solutio
n for Wave Equation" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 64 "Hereafter we acknowledge that d'Alembert's solution has
 the form" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
22 "            u(t, x) = " }{XPPEDIT 18 0 "psi(x+c*t);" "6#-%$psiG6#,
&%\"xG\"\"\"*&%\"cGF(%\"tGF(F(" }{TEXT -1 3 " + " }{XPPEDIT 18 0 "phi(
x-c*t);" "6#-%$phiG6#,&%\"xG\"\"\"*&%\"cGF(%\"tGF(!\"\"" }{TEXT -1 1 "
." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 120 "Not
e that we have not used any information about boundary conditions to a
rrive at this result. to get information about " }{XPPEDIT 18 0 "psi;
" "6#%$psiG" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "phi;" "6#%$phiG" }
{TEXT -1 119 "  we used information about the initial conditions. We r
eview the situation with the assumption that u(0,x) = f(x) and " }
{XPPEDIT 18 0 "u[t];" "6#&%\"uG6#%\"tG" }{TEXT -1 15 " (0, x) = g(x).
" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 41 " A general solution for the w
ave equation" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 99 "     We use the ideas of the previous module to construct a gen
eral solution for the wave equation." }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 8 "restart:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 
"f:='f': g:='g': u:='u':" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 87 
"u:=(f,g,t,x,c)->(f(x+c*t)+f(x-c*t))/2\n                 +int(g(s),s=x
-c*t..x+c*t)/(2*c);" }}}{PARA 0 "" 0 "" {TEXT -1 91 "     To illustrat
e this general solution, we take a particular example. We will take c \+
= 1." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "f:=x->4*cos(4*x);\ng
:=x->x^2-1/3;" }}}{PARA 0 "" 0 "" {TEXT -1 52 "     Here is an evaluat
ion of this general solution." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 13 "u(f,g,t,x,1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "with
(plots):\nanimate(u(f,g,t,x,1),x=-2..2,t=0..2);" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 
{PARA 3 "" 0 "" {TEXT -1 21 "Half Infinite Strings" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 137 "This solution is for all
 number x. It is as though we had a string that was infinite in both d
irectons. That is, this solution holds for " }{XPPEDIT 18 0 "-infinity
 < x;" "6#2,$%)infinityG!\"\"%\"xG" }{TEXT -1 5 " and " }{XPPEDIT 18 
0 "x < infinity;" "6#2%\"xG%)infinityG" }{TEXT -1 1 "." }}{PARA 0 "" 
0 "" {TEXT -1 80 "     Our next idea is to suppose that we solve the e
quation on the interval [0, " }{XPPEDIT 18 0 "infinity;" "6#%)infinity
G" }{TEXT -1 144 " ). We begin by imposing only one boundary condition
. We suppose that x > 0. With this assumption, we have a boundary cond
ition at the left end." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 32 "Boundary Condition: u(t, 0) = 0." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 39 "Initial Conditions: u(0, \+
x) = f(x) and " }{XPPEDIT 18 0 "u[t];" "6#&%\"uG6#%\"tG" }{TEXT -1 24 
"(0, x) = g(x) for x > 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 79 "     With no initial conditions or boundary condit
ions, we have concluded that " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 32 "                      u(t, x) = " }
{XPPEDIT 18 0 "psi(x+c*t);" "6#-%$psiG6#,&%\"xG\"\"\"*&%\"cGF(%\"tGF(F
(" }{TEXT -1 3 " + " }{XPPEDIT 18 0 "phi(x-c*t);" "6#-%$phiG6#,&%\"xG
\"\"\"*&%\"cGF(%\"tGF(!\"\"" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 52 "as t increases. With initial co
nditions, we got that" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 
"" {TEXT -1 31 "                     u(t, x) = " }{XPPEDIT 18 0 "(f(x+
c*t)+f(x-c*t))/2;" "6#*&,&-%\"fG6#,&%\"xG\"\"\"*&%\"cGF*%\"tGF*F*F*-F&
6#,&F)F**&F,F*F-F*!\"\"F*F*\"\"#F2" }{TEXT -1 3 " + " }{XPPEDIT 18 0 "
(G(x+c*t)-G(x-c*t))/2;" "6#*&,&-%\"GG6#,&%\"xG\"\"\"*&%\"cGF*%\"tGF*F*
F*-F&6#,&F)F**&F,F*F-F*!\"\"F2F*\"\"#F2" }{TEXT -1 1 " " }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 34 "where G is an antide
rivative of g." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 69 "     Now, we are supposing only that x > 0. We must do ex
tensions of " }{XPPEDIT 18 0 "psi;" "6#%$psiG" }{TEXT -1 5 " and " }
{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 77 " for as t increases for \+
c t will exceed x and we will need the definition of " }{XPPEDIT 18 0 
"phi;" "6#%$phiG" }{TEXT -1 96 " for negative values.  Happily, there \+
is information available through the boundary conditions. " }}{PARA 0 
"" 0 "" {TEXT -1 35 "     The first boundary conditon is" }}{PARA 0 "
" 0 "" {TEXT -1 24 "          0 = u(t, 0) = " }{XPPEDIT 18 0 "psi;" "6
#%$psiG" }{TEXT -1 11 " ( c t ) + " }{XPPEDIT 18 0 "phi;" "6#%$phiG" }
{TEXT -1 12 " (- c t ) = " }{XPPEDIT 18 0 "(f(c*t)+G(c*t)+C)/2;" "6#*&
,(-%\"fG6#*&%\"cG\"\"\"%\"tGF*F*-%\"GG6#*&F)F*F+F*F*%\"CGF*F*\"\"#!\"
\"" }{TEXT -1 6 "  +   " }{XPPEDIT 18 0 "(f(-c*t)-G(-c*t)-C)/2;" "6#*&
,(-%\"fG6#,$*&%\"cG\"\"\"%\"tGF+!\"\"F+-%\"GG6#,$*&F*F+F,F+F-F-%\"CGF-
F+\"\"#F-" }{TEXT -1 3 "  ." }}{PARA 0 "" 0 "" {TEXT -1 6 " Thus," }}
{PARA 0 "" 0 "" {TEXT -1 65 "               0 = f( c t ) + f( - c t ) \+
+ G( c t ) - G( - c t )." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 62 "Since this holds for all f and for all g, then it \+
must be that" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 77 "          f( c t ) = - f( - c t )   and  G( c t ) = G( - c t ) \+
for all t > 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 73 "Thus, f should have an odd extension and G should have an
 even extension." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "
" 0 "" {TEXT -1 30 "Making even and odd extensions" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 245 "It's clear that in order
 to illustrate these ideas, we are going to need to be able to make ev
en and odd extensions for functions. Here is a procedure. You must exe
cute these two parts inorder to be able to work what the rest of this \+
work sheet." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 80 "ef:=proc(f,x)
     \n      if x>0 then f(x);\n      else f(-x);\n      fi;\n     end
;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 77 "of:=proc(f,x) \n      \+
if x>0 then f(x);\n      else -f(-x);\n      fi;\n     end;" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT 
-1 57 "Having defined this part we proceed with the development." }}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 46 "C
omparing Infinite and Half-Infinite Problems." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 281 "It is wise to constrast \+
this \"half\" infinite string with the infinite string of the previous
 module. Let's do one example twice: first we let x range over all num
bers and use just the original d'Alembert solution, and then we let x \+
only be positive and use the odd extension of f. " }}{PARA 0 "" 0 "" 
{TEXT -1 20 "     Here's f and g." }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 59 "f:=x->-(Heaviside(2*Pi-x)-Heaviside(Pi-x))*sin(x);\ng
:=x->0;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "plot(f(x),x=0..1
0);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 121 "
Now, we solve the PDE with initial conditions and no boundary conditio
ns. we use the d'Alembert solution, and take c = 1." }}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 28 "u:=(t,x)->(f(x+t)+f(x-t))/2;" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "plot3d(u(t,x),x=-10..10,t=0..5,orie
ntation=[-70,25]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "with(
plots):\nanimate(u(t,x),x=-10..10,t=0..5);" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 81 "Next, we solve the
 PDE with initial conditions and boundary conditon u(t, 0) = 0." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 34 "u:=(t,x)->(of(f,x+t)+of(f,x-t))/2;" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "plot3d('u(t,x)',x=0..4*Pi,t=0..10,o
rientation=[-90,10]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 74 "wi
th(plots):\nanimate('u(t,x)',x=0..4*Pi,t=0..3*Pi,numpoints=75,color=RE
D);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 
1 {PARA 3 "" 0 "" {TEXT -1 14 "Finite Strings" }}{PARA 0 "" 0 "" 
{TEXT -1 204 "     If we have a finite string, we have at end point at
 x = 0 and at x = L. We have used information at x = 0. We have not us
ed the other end point. At that end point, we can get in a similar man
ner that" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
71 "        0 = f( L + c t ) + f( L  - c t ) + G( L + c t ) - G( L - c
 t )," }}{PARA 0 "" 0 "" {TEXT -1 3 "and" }}{PARA 0 "" 0 "" {TEXT -1 
89 "          f( L + c t ) = - f( L - c t )   and  G( L + c t ) = G( L
 - c t ) for all t > 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 57 "The oddness of f and evenness of f and G changes the
se to" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 91 "          f( L + c t ) = f(-  L + c t
 )   and  G( L + c t ) = G( - L + c t ) for all t > 0." }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 106 "That is, f and G ha
ve period 2 L. Now we know how to make the extensions of f and of G so
 we can define u." }}{PARA 0 "" 0 "" {TEXT -1 116 "     We work two an
 easy example. Then, we will do a more interesting one. Take c = 1 her
e. For the first example,  " }{XPPEDIT 18 0 "f(x) <> 0;" "6#0-%\"fG6#%
\"xG\"\"!" }{TEXT -1 66 " and g(x) = 0. Let us choose f(x) = sin( x ) \+
on the interval [ 0, " }{XPPEDIT 18 0 "pi;" "6#%#piG" }{TEXT -1 50 " ]
. This function is already odd and has period 2 " }{XPPEDIT 18 0 "PI;
" "6#%#PIG" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 100 "     We have found how to construct a general \+
solution for the wave equation on the interval [0, L]." }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart:" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 35 "f:=x->sin(x); g:=x->0; L:=Pi; c:=1;" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "u:=(t,x)->(f(x+c*t)+f(x-c*t))/2;" }
}}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 21 "Check \+
the conditions:" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "u(t,0);\n
u(t,L);\nu(0,x);\nsimplify(subs(t=0,diff(u(t,x),t)));" }}}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "with(pl
ots):\nanimate(u(t,x),x=0..2*Pi,t=0..2*Pi,color=RED);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "plot3d(u(t,x),x=0..L,t=0..2*Pi,axes
=NORMAL,orientation=[-155,65]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 52 "Even periodi
c extensions and odd periodic extensions" }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 295 "In order to be able to graph solu
tions for the finite string, we need to be able to make a different ex
tensions of functions from simply even or odd. We need to make odd and
 even,  2 L  periodic extensions of functions of functions defined on \+
the interval [0, L]. Here is a procedure to do this." }}{PARA 0 "" 0 "
" {TEXT -1 53 "     First, we make an even, 2 L  periodic extension." 
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart:" }}}{PARA 0 "" 0 "
" {TEXT -1 109 "First we make the extension on the interval [0, L] awa
re that the goal is to get an even, periodic extension." }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 113 "pfe:= proc(f,x)\n     global L;\n
\011\011if 0 <= x and x < L then f(x)\n\011\011elif L <= x and x <= 2*
L then f(2*L-x)\n\011\011fi end;" }}}{PARA 0 "" 0 "" {TEXT -1 57 "Here
 is where we make extend the function to all numbers." }}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 43 "epf:= (f,x)->pfe(f,frac(abs(x)/(2*L))*2*L
);" }}}{PARA 0 "" 0 "" {TEXT -1 270 "We test this setup. We define an \+
interval L and a function f and draw the graph. Be aware that we want \+
an even, 2 L extension. Look at the following to see that on the inter
val [0, L], we have the graph of L, that the graph is even, and that t
he function has period 2 L." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
14 "L:=2;\nf:=x->x;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "plot
('epf(f,x)',x=-3*L..3*L);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
15 "L:='L':\nf:='f':" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 47 " Next, w
e make an odd, 2 L  periodic extension." }}{PARA 0 "" 0 "" {TEXT -1 
108 "First we make the extension on the interval [0, L] aware that the
 goal is to get an odd, periodic extension." }}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 114 "pfo:= proc(f,x)\n     global L;\n\011\011if 0 <= x
 and x < L then f(x)\n\011\011elif L <= x and x <= 2*L then -f(2*L-x)
\n\011\011fi end;" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 57 "Here is where we make extend the function to all numbers.
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "opf:= (f,x)->sign(x)*pfo
(f,frac(abs(x)/(2*L))*2*L);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 290 "As befoe, we test this setup. We define an int
erval L and a function f and draw the graph. This time, be aware that \+
we want an even, 2 L extension. Look at the following to see that on t
he interval [0, L], we have the graph of L, that the graph is odd, and
 that the function has period 2 L." }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 14 "L:=2;\nf:=x->x;" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 29 "plot('opf(f,x)',x=-3*L..3*L);" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 0 "" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "
" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 
13 "Illustrations" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }{TEXT 256 9 "Examples:" }{TEXT -1 65 " Take c = 1 and L
, f, and g are specified:\n1.  L = 2, f(x) = sin(" }{XPPEDIT 18 0 "pi;
" "6#%#piG" }{TEXT -1 47 "x), g(x) = 0.\n2.  L = 2, f(x) = 0, g(x) =  \+
sin(" }{XPPEDIT 18 0 "pi" "6#%#piG" }{TEXT -1 11 "x).\n3.  L =" }
{XPPEDIT 18 0 "pi" "6#%#piG" }{TEXT -1 9 ", f(x) = " }{XPPEDIT 18 0 "p
i" "6#%#piG" }{TEXT -1 8 "/2 - |x-" }{XPPEDIT 18 0 "pi" "6#%#piG" }
{TEXT -1 188 "/2|, g(x) = 0.\n4.  L = 4, f(x) = (x-1)*(Heaviside(x-1)-
Heaviside(x-2)) + \n\011\011\011(3-x)*(Heaviside(x-2)-Heaviside(x-3))
\n\011\011g(x) = 0.\n5.  L = 4, f(x) = 0, and g(x) = Heaviside(x-3)-He
aviside(x-4)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 98 "In order to work these example, we need to have the functions e
pf and opf from the section above. " }}{PARA 0 "" 0 "" {TEXT -1 18 "He
re is example 1." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "L:=2;\nf
:=x->sin(Pi*x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "plot('op
f(f,x)',x=-2*L..2*L);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "u:
=(t,x)->(opf(f,x+t)+opf(f,x-t))/2;" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 67 "plot3d('u(t,x)',x=0..L,t=0..2*L,axes=NORMAL,orientati
on=[-135,45]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "with(plot
s):\nanimate('u(t,x)',x=0..L,t=0..L);" }}}{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 18 "Here is example 2." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "L:=2;\ng:=x->sin(Pi*x);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "int(g(x),x);\nG:=unapply(%,x
);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "plot('epf(G,x)',x=-2*
L..2*L);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "u:=(t,x)->(epf(
G,x+t)-epf(G,x-t))/2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "pl
ot3d('u(t,x)',x=0..L,t=0..2*L,axes=NORMAL,orientation=[-135,45]);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "with(plots):\nanimate('u(t,x
)',x=0..L,t=0..L);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 257 11 "Assignme
nt:" }{TEXT -1 40 " Work the remaining Illustrations above." }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 
{PARA 3 "" 0 "" {TEXT -1 13 "Note In Proof" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 137 "Get periodic extensions of fun
ctions is an interesting problem in the context of these notes. Here i
s another method from the literature." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 3 "" 0 "" {TEXT -1 44 "Generating periodic extensions of fun
ctions." }}{PARA 0 "" 0 "" {TEXT -1 201 "Suppose that f is defined on \+
the interval [a,b], in MapleTech Vol. 3, NO.3, Monagan and Lopez provi
ded the following method to output a function which extends the defini
tion of f to the real line using" }}{PARA 0 "" 0 "" {TEXT -1 23 "     \+
  f(a + k (b-a) + " }{XPPEDIT 18 0 "delta" "6#%&deltaG" }{TEXT -1 10 "
) = f(a + " }{XPPEDIT 18 0 "delta" "6#%&deltaG" }{TEXT -1 1 ")" }}
{PARA 0 "" 0 "" {TEXT -1 14 "for integer k." }}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 165 "PeriodicExtender:=proc(f,d::range)\nsubs( \{'F' = \+
f, 'L'=lhs(d),\n        'D'=rhs(d)-lhs(d)\},\nproc(x::algebraic) local
 y;\n    y:=floor((x-L)/D);\n    F(x-y*D);\nend)\nend;" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "sw:=PeriodicExtender(signum,-1..1);
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "plot('sw(x)','x'=-4..4)
;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "g:=x->piecewise(x<0,-x
,x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "st:=PeriodicExtende
r(g,-1..1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "plot('st(x)'
,'x'=-3..3);" }}}{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 "" }}}{MARK "0 0" 0 }
{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }