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{SECT 0 {PARA 256 "" 0 "" {TEXT 259 32 "Module 22: d'Alembert's Soluti
on" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12
"with(plots):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 ""
0 "" {TEXT -1 400 " The solution we obtained for the wave equation
in the previous module is a natural extension of the development for \+
solutions of the heat diffusion model. The mathematics is accurate, an
d the method is satisfactory in that the solutions provided for the mo
del meet with our expectations. There was no new mathematical insight,
except that we can do the process again for this different equation.
" }}{PARA 0 "" 0 "" {TEXT -1 133 " In the 18th Century, Jean Le Ro
nd d'Alembert had formulated a different way to come to this solution.
Morris Kline in his book, " }{TEXT 260 49 "Mathematical Thought from \+
Ancient to Modern Times" }{TEXT -1 531 ", reminds us that several math
ematical concepts had to evolve with time. Kline says that d'Alembert \+
accepted only analytic curves and solutions. He wanted functions that \+
could be given with equations of relations, for example, between the e
lementary algebraic and trigonometric functions. A part of the undevel
oped understanding was that an initial function need not be expressibl
e by a single analytic expression. There was also the struggle for the
concept of an infinite sum -- a series. It is easy to imagine the dif
ficulty in " }}{PARA 0 "" 0 "" {TEXT -1 52 "(1) having the idea of an \+
infinite sum of functions," }}{PARA 0 "" 0 "" {TEXT -1 91 "(2) having \+
a series of cosine functions to converge to, for example, the sine fun
ction, and" }}{PARA 0 "" 0 "" {TEXT -1 122 "(3) having the series to c
onverge to an analytic function on a finite interval, but not to the f
unction off that interval." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0
"" 0 "" {TEXT -1 325 " When a creative person is stuck on an idea,
interesting things can happen. We present d'Alembert's solution here.
If he could see what follows, d\"Alembert might not admit that the pr
esentation is the way he had thought of the solution. But, we should t
ake advantage of the fact that we stand on the shoulders of giants! "
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 34
"Section 1: Transforming Equations." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 57 "Suppose that u is a function which satis
fies the equation" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 13 "(*) " }{XPPEDIT 18 0 "diff(u,`$`(t,2))-diff(u,`$
`(x,2)) = 0;" "6#/,&-%%diffG6$%\"uG-%\"$G6$%\"tG\"\"#\"\"\"-F&6$F(-F*6
$%\"xGF-!\"\"\"\"!" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 26 " Define v(w, z) as follows" }}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 29 "v:=(w,z)->u((w-z)/2,(w+z)/2);" }{TEXT -1 0 "" }}}
{PARA 0 "" 0 "" {TEXT -1 90 "That is, t = (w - z)/2 , and x = (w + z)/
2, or, what is the same, w = t + x and z = x - t." }}{EXCHG {PARA 0 ">
" 0 "" {MPLTEXT 1 0 35 "solve(\{t=(w-z)/2,x=(z+w)/2\},\{w,z\});" }}}
{PARA 0 "" 0 "" {TEXT -1 52 "As we see, this v will satisfy a differen
t equation." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "diff(v(w,z),w
,z);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 67 "
We see that, because u satisfies (*), v defined as above satisfies" }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 27 "(**) \+
" }{XPPEDIT 18 0 "diff(v,w,z)=0" "6#/-%%diffG6%%\"v
G%\"wG%\"zG\"\"!" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 113 "Conversely, suppose that v satisfies thi
s last equation (**) and u is defined by u( t , x ) = v( (x+t), (x-t)
)." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "v:='v';" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "u:=(t,x)->v(t+x,x-t);" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "diff(u(t,x),x,x)-diff(u(t,x),t,t);
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 ""
{TEXT -1 181 "Thus, u solves equation (*) because v satisfies (**). Th
is observation enables us to provide a solution for the wave equation \+
that can be made quite independent of the Fourier Idea." }}{PARA 0 ""
0 "" {TEXT -1 76 " You see, Equation (**) says that the derivative
with respect to z of " }{XPPEDIT 18 0 "diff(v,w);" "6#-%%diffG6$%\"
vG%\"wG" }{TEXT -1 140 " is zero. We know all functions whose derivat
ive is zero. They are constant. If taking the derivative with respect \+
to z gives zero, then " }{XPPEDIT 18 0 "diff(v,w);" "6#-%%diffG6$%\"
vG%\"wG" }{TEXT -1 26 " is constant in z. We say" }}{PARA 0 "" 0 ""
{TEXT -1 12 " " }{XPPEDIT 18 0 "diff(v,w);" "6#-%%diffG6$%
\"vG%\"wG" }{TEXT -1 9 " = C(w)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 133 "Now integrate this with respect to w and
you get some function of w, plus a constant -- a constant as far as w
is concerned. We write" }}{PARA 0 "" 0 "" {TEXT -1 24 " \+
v(w, z) = " }{XPPEDIT 18 0 "psi;" "6#%$psiG" }{TEXT -1 8 "( w ) + " }
{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 6 "( z )." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 91 "To get this back in ter
ms of t and x, recall above that w =(x + t) and z = ( x - t). Thus, "
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 21 " \+
u(t, x) = " }{XPPEDIT 18 0 "psi;" "6#%$psiG" }{TEXT -1 12 "( x + t
) + " }{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 10 "( x - t )." }}
{PARA 0 "" 0 "" {TEXT -1 64 "These are the solutions d'Alembert found.
We pursue these ideas." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 ""
}}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1
43 "Section 2. Solutions for The Wave Equation." }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 274 " Take note that noth
ing has been said about initial condtions or boundary conditons for th
is wave equation. One might conceive these function as solutions of th
e wave equation for all t > 0 and all x > 0. We found that functions s
atisfying the wave equation had the form" }}{PARA 0 "" 0 "" {TEXT -1
22 " u(t, x) = " }{XPPEDIT 18 0 "psi;" "6#%$psiG" }{TEXT
-1 12 "( x + t ) + " }{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 10 "( \+
x - t )." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1
69 "Furthermore, all functions defined in that form satisfy the equati
on." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 58 "We
conceive of finding all solutions for the wave equation" }}{PARA 0 "
" 0 "" {TEXT -1 28 " " }{XPPEDIT 18 0 "diff
(u(t,x),`$`(t,2)) = diff(u(t,x),`$`(x,2));" "6#/-%%diffG6$-%\"uG6$%\"t
G%\"xG-%\"$G6$F*\"\"#-F%6$-F(6$F*F+-F-6$F+F/" }}{PARA 0 "" 0 "" {TEXT
-1 46 "as the problem for finding the nullspace of a " }{TEXT 256 13 "
wave operator" }{TEXT -1 1 ":" }}{PARA 0 "" 0 "" {TEXT -1 27 " \+
" }{XPPEDIT 18 0 "diff(u(t,x),`$`(t,2))-diff(u(t,x)
,`$`(x,2)) = 0;" "6#/,&-%%diffG6$-%\"uG6$%\"tG%\"xG-%\"$G6$F+\"\"#\"\"
\"-F&6$-F)6$F+F,-F.6$F,F0!\"\"\"\"!" }{TEXT -1 1 "." }}{PARA 0 "" 0 "
" {TEXT -1 156 "We define a wave operator in Maple. We test that funct
ions of the special form satisfy the wave equation -- they are in the \+
null space of the wave operator." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT
1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "u:=(t,x)->sin(ex
p(cos(x-t)));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "diff(u(t,x
),t,t)-diff(u(t,x),x,x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "
" }}}{PARA 0 "" 0 "" {TEXT -1 26 " We have seen that if " }{TEXT
257 1 "f" }{TEXT -1 66 " is any function depending on the combination \+
x + t or x - t, then" }{TEXT 258 2 " f" }{TEXT -1 48 " will lie in the
nullspace of the wave operator." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT
1 0 109 "f:=(t,x)->psi(x+t);\ndiff(f(t,x),t,t)-diff(f(t,x),x,x);\nf:=(
t,x)->phi(x-t);\ndiff(f(t,x),t,t)-diff(f(t,x),x,x);" }}}{EXCHG {PARA
0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 145 " Si
nce the wave equation is linear, it is not a surprise that given any t
wo twice differentiable functions of a single variable, the function"
}}{PARA 0 "" 0 "" {TEXT -1 23 " " }{XPPEDIT 18
0 "psi(x+t)+phi(x-t);" "6#,&-%$psiG6#,&%\"xG\"\"\"%\"tGF)F)-%$phiG6#,&
F(F)F*!\"\"F)" }}{PARA 0 "" 0 "" {TEXT -1 167 "is also a solution. D'A
lembert had the insight that not only is this combination a solution, \+
it is the general solution. To understand this we need to think about \+
how " }{XPPEDIT 18 0 "phi" "6#%$phiG" }{TEXT -1 5 " and " }{XPPEDIT
18 0 "psi" "6#%$psiG" }{TEXT -1 40 " could be determine for a specific
case." }}{PARA 0 "" 0 "" {TEXT -1 46 " This is the subject for th
e next section." }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "
" 0 "" {TEXT -1 52 "Section 3. Initial conditions for the wave equatio
n." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 95 " \+
Suppose that we have an infinitely long string, with initial conditi
ons of the usual form:" }}{PARA 0 "" 0 "" {TEXT -1 29 " \+
" }{XPPEDIT 18 0 "u(0,x) = f(x)" "6#/-%\"uG6$\"\"!%\"xG-
%\"fG6#F(" }{TEXT -1 4 ", " }{XPPEDIT 18 0 "u[t](0,x)=g(x)" "6#/-&%
\"uG6#%\"tG6$\"\"!%\"xG-%\"gG6#F+" }{TEXT -1 1 "." }}{PARA 0 "" 0 ""
{TEXT -1 8 "Here, - " }{XPPEDIT 18 0 "infinity" "6#%)infinityG" }
{TEXT -1 7 " < x < " }{XPPEDIT 18 0 "infinity" "6#%)infinityG" }{TEXT
-1 106 ", so that we have no boundary conditions to impose. The questi
on is: How do we find a solution of the form" }}{PARA 0 "" 0 "" {TEXT
-1 37 " u(t,x)= " }{XPPEDIT 18 0 "psi(x+c*
t)+phi(x-c*t);" "6#,&-%$psiG6#,&%\"xG\"\"\"*&%\"cGF)%\"tGF)F)F)-%$phiG
6#,&F(F)*&F+F)F,F)!\"\"F)" }}{PARA 0 "" 0 "" {TEXT -1 59 "which solves
this problem? The initial conditions are that" }}{PARA 0 "" 0 ""
{TEXT -1 27 " f(x) = u(0, x) = " }{XPPEDIT 18 0 "psi;" "6#%$p
siG" }{TEXT -1 6 "(x) + " }{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1
4 "(x) " }}{PARA 0 "" 0 "" {TEXT -1 3 "and" }}{PARA 0 "" 0 "" {TEXT
-1 17 " g(x) = " }{XPPEDIT 18 0 "u[t];" "6#&%\"uG6#%\"tG" }
{TEXT -1 11 "(0, x) = c " }{XPPEDIT 18 0 "psi;" "6#%$psiG" }{TEXT -1
10 " '(x) - c " }{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 6 " '(x)."
}}{PARA 0 "" 0 "" {TEXT -1 46 "In the second equation, get an antideri
vative:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1
23 " G(x) + C = " }{XPPEDIT 18 0 "psi;" "6#%$psiG" }{TEXT
-1 6 "(x) - " }{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 49 "(x), whe
re C is the constant of integration and " }}{PARA 0 "" 0 "" {TEXT -1
16 " G(x) = " }{XPPEDIT 18 0 "int(g(y),y = 0 .. x)/c;" "6#*&-%
$intG6$-%\"gG6#%\"yG/F*;\"\"!%\"xG\"\"\"%\"cG!\"\"" }{TEXT -1 2 " ." }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 23 "Putting \+
this value for " }{XPPEDIT 18 0 "psi;" "6#%$psiG" }{TEXT -1 6 "(x) - \+
" }{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 23 "(x) with the value fo
r " }{XPPEDIT 18 0 "psi;" "6#%$psiG" }{TEXT -1 6 "(x) + " }{XPPEDIT
18 0 "phi;" "6#%$phiG" }{TEXT -1 13 "(x) , we have" }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 5 " " }{XPPEDIT 18 0 "psi
;" "6#%$psiG" }{TEXT -1 6 "(x) = " }{XPPEDIT 18 0 "(f(x)+G(x)+C)/2;" "
6#*&,(-%\"fG6#%\"xG\"\"\"-%\"GG6#F(F)%\"CGF)F)\"\"#!\"\"" }{TEXT -1 7
" and " }{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 7 "(x) = " }
{XPPEDIT 18 0 "(f(x)-G(x)-C)/2;" "6#*&,(-%\"fG6#%\"xG\"\"\"-%\"GG6#F(!
\"\"%\"CGF-F)\"\"#F-" }{TEXT -1 3 " ." }}{PARA 0 "" 0 "" {TEXT -1 5 "
" }}{PARA 0 "" 0 "" {TEXT -1 299 "For example, if f is a function
with one bump near x = 0 and there is no initial velocity, we can ant
icipate how this initial condition will evolve with increasing time: w
e should see the bump break into the superposition of two bumps of hal
f the height, one moving to the right and one to the left." }}{PARA 0
"" 0 "" {TEXT -1 24 " Take, for example, " }{XPPEDIT 18 0 "f(x)=ex
p(-x^2)" "6#/-%\"fG6#%\"xG-%$expG6#,$*$F'\"\"#!\"\"" }{TEXT -1 125 " a
s the initial function. We construct a solution to the wave equation w
ith this initial value and watch it evolve with time." }}{EXCHG {PARA
0 "> " 0 "" {MPLTEXT 1 0 36 "f:=x->exp(-x^2);\nplot(f(x),x=-4..4);" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "animate((f(x+t)+f(x-t))/2,x
=-4..4,t=0..10);" }}}{PARA 0 "" 0 "" {TEXT -1 90 " Sometimes it is
instructive to view u as a graph in t and x, but not as an animation.
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 87 "plot3d((f(x+t)+f(x-t))/2
,x=-10..10,t=0..6,axes=NORMAL,\n orientation=[-120,65]);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 62 " We now investigate a secon
d example -- that f(x) = 0 and " }{XPPEDIT 18 0 "g(x) <> 0;" "6#0-%\"g
G6#%\"xG\"\"!" }{TEXT -1 4 ".. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 85 "We have seen that the solution for the wa
ve equation with these initial conditions is" }}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 39 " \+
" }{XPPEDIT 18 0 "u(t,x)= int(g(s),s=x-c*t..x+c*t)/(2*c)" "6#/
-%\"uG6$%\"tG%\"xG*&-%$intG6$-%\"gG6#%\"sG/F0;,&F(\"\"\"*&%\"cGF4F'F4!
\"\",&F(F4*&F6F4F'F4F4F4*&\"\"#F4F6F4F7" }{TEXT -1 1 "." }}{PARA 0 ""
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 33 " Here is an exam
ple. We take " }{XPPEDIT 18 0 "f(x) = 0 " "6#/-%\"fG6#%\"xG\"\"!" }
{TEXT -1 5 " and " }{XPPEDIT 18 0 "g(x) = x/(1+x+x^2)" "6#/-%\"gG6#%\"
xG*&F'\"\"\",(F)F)F'F)*$F'\"\"#F)!\"\"" }{TEXT -1 1 "." }}{PARA 0 ""
0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "g:=x->x
/(1+x+x^2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "u:=(t,x)->in
t(g(s),s=x-t..x+t)/2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "u(t
,x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "simplify(diff(u(t,x
),t,t)-diff(u(t,x),x,x));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7
"u(0,x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "diff(u(t,x),t):
\nsubs(t=0,%):\nsimplify(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1
0 33 "animate(u(t,x),x=-10..10,t=0..6);" }}}{PARA 0 "" 0 "" {TEXT -1
90 " Sometimes it is instructive to view u as a graph in t and x, \+
but not as an animation." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 76 "
plot3d(u(t,x),x=-10..10,t=0..6,axes=NORMAL,\n orientation=[-12
0,65]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 ""
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 97 "In the next lecture,
we examine solutions on a half infinite line and on a line of finite \+
length." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 261
11 "Assignment:" }{TEXT -1 142 " Draw the three dimensional graphs u(t
, x) for solutions of the wave equation if f(x) = sin(x) and g(x) = 0 \+
and if f(x) = 0 and g(x) = cos(x)." }}{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 }