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{SECT 0 {PARA 256 "" 0 "" {TEXT -1 45 "Module 21: The Wave Equation in
 One Dimension" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 52 "We begin now a study of the classical wave equation." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 48 "     The \+
classical, linearized wave equation is " }}{PARA 0 "" 0 "" {TEXT -1 
39 "                                       " }{XPPEDIT 18 0 "diff(u(t,
x),`$`(t,2)) = c^2*diff(u(t,x),`$`(x,2));" "6#/-%%diffG6$-%\"uG6$%\"tG
%\"xG-%\"$G6$F*\"\"#*&%\"cGF/-F%6$-F(6$F*F+-F-6$F+F/\"\"\"" }{TEXT -1 
1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 108 "
     We classify this PDE as a special case of the more general consta
nt coefficient, second order equation:" }}{PARA 0 "" 0 "" {TEXT -1 9 "
         " }{XPPEDIT 18 0 "a*diff(u(t,x),`$`(t,2))+b*diff(u(t,x),t,x)+
c*diff(u(t,x),`$`(x,2))+d*diff(u(t,x),t)+e*diff(u(t,x),x)+f;" "6#,.*&%
\"aG\"\"\"-%%diffG6$-%\"uG6$%\"tG%\"xG-%\"$G6$F-\"\"#F&F&*&%\"bGF&-F(6
%-F+6$F-F.F-F.F&F&*&%\"cGF&-F(6$-F+6$F-F.-F06$F.F2F&F&*&%\"dGF&-F(6$-F
+6$F-F.F-F&F&*&%\"eGF&-F(6$-F+6$F-F.F.F&F&%\"fGF&" }{TEXT -1 5 " = 0.
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 256 12 "Seco
nd order" }{TEXT -1 319 " refers to the lack of derivatives of more th
an second order.We will solve the wave equation by the method of separ
ation of variables in this worksheet. We will find there are alternate
 methods for this equation. We will also see that slightly more compli
cated situations lead to the more general second order equation." }}
{PARA 0 "" 0 "" {TEXT -1 176 "     We have seen that the standard sepa
ration of variables works well for the heat equation. This method is o
ften good for linear equations. The method takes advantage of the " }
{TEXT 257 22 "superposition property" }{TEXT -1 28 ". That is, if two \+
functions " }{XPPEDIT 18 0 "u[1]" "6#&%\"uG6#\"\"\"" }{TEXT -1 5 " and
 " }{XPPEDIT 18 0 "u[2]" "6#&%\"uG6#\"\"#" }{TEXT -1 49 " are solution
s then so is any linear combination:" }}{PARA 0 "" 0 "" {TEXT -1 25 " \+
                        " }{XPPEDIT 18 0 "alpha*u[1]+beta*u[2]" "6#,&*
&%&alphaG\"\"\"&%\"uG6#F&F&F&*&%%betaGF&&F(6#\"\"#F&F&" }{TEXT -1 1 ".
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 204 "As w
ith the heat equation, the wave equation commonly comes with boundary \+
conditions and initial conditions. At this start, we take homogeneous \+
boundary conditions, assuming that x ranges between 0 and L." }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 51 "Boundary Condit
ions:  u(t, 0) = 0, and u(t, L) = 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}}{PARA 0 "" 0 "" {TEXT -1 76 "Because the equation is second order in
 t, we expect two initial conditions." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 40 "Initial Conditions:  u(0, x) = f(x) a
nd " }{XPPEDIT 18 0 "diff(u,t);" "6#-%%diffG6$%\"uG%\"tG" }{TEXT -1 
30 "(0, x) = g(x) for x in [0, L]." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 303 "The usual physical realization made for \+
this model is that of a string, held fixed at two ends, displaced init
ially by an amount f(x), and given an initial velocity g(x). This mode
l will guide our intuition and, with small displacements, gives an acc
urate impression for the vibrations of a taut string." }}{PARA 0 "" 0 
"" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 513 "     We often set c = 1 for convenience. We do this in t
his worksheet, but remove the restriction later. It is of value to con
sider the difference in the left and right side of the wave equation. \+
That difference reminds us that we are solving linear operator equatio
ns and looking for the null space of these linear operators. This exam
ple is simply a linear equation which has an infinite number of soluti
ons. We seek one which satisfies the boundary and initial conditions. \+
To give understanding, we define the " }{TEXT 258 14 "wave operator." 
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "waveop:=diff(u(t,x),t,t)-d
iff(u(t,x),x,x);" }}}{PARA 0 "" 0 "" {TEXT -1 139 "(The restriction th
at c = 1 is not a severe restriction. Equivalently if c is not one, we
 could choose a clock with time variable T = c*t.)" }}{PARA 0 "" 0 "" 
{TEXT -1 113 "     Let's test out the wave operator. We will eventuall
y try to get functions in the null space of the operator." }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "u:=(t,x)->t*sin(x)^2; waveop;" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "u:=(t,x)->sin(t-x); waveop;
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "u:='u':" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 273 " \+
    We have had experience with the method of separation of variables \+
and know that this method for solving partial differential equations i
s precisely what it says: One assumes that solutions can be written as
 products of separate functions of t and x. Thus, we make the " }
{TEXT 259 6 "ansatz" }{TEXT -1 37 " that u(t, x) is of the special for
m " }}{PARA 0 "" 0 "" {TEXT -1 26 "               T(t) X(x), " }}
{PARA 0 "" 0 "" {TEXT -1 28 "known as a product solution." }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "eq:=subs(u(t,x)=T(t)*X(x),waveop)=0
;" }}}{PARA 0 "" 0 "" {TEXT -1 55 "     This simplifies if we divide t
hrough by T(t) X(x)." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "eq/X
(x)/T(t);\nexpand(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "se
p:=(%)+(diff(X(x),x,x)/X(x)=diff(X(x),x,x)/X(x));" }}}{PARA 0 "" 0 "" 
{TEXT -1 35 "     The left side of the equation " }{TEXT 262 3 "sep" }
{TEXT -1 17 " depends only on " }{TEXT 260 1 "t" }{TEXT -1 36 " and th
e right side depends only on " }{TEXT 261 1 "x" }{TEXT -1 142 ". Thus,
 each side must be constant. We do not know the value of this constant
, yet. As in the heat equation, it will be negative; we call it -" }
{XPPEDIT 18 0 "mu^2" "6#*$%#muG\"\"#" }{TEXT -1 1 "." }}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 39 "dsolve(diff(X(x),x,x)=-mu^2*X(x),X(x));" 
}}}{PARA 0 "" 0 "" {TEXT -1 65 "In order to satisfy the boundary condi
tion at x = 0, we must have" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
16 "X:=x->sin(mu*x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "sol
ve(X(L)=0,L);" }}}{PARA 0 "" 0 "" {TEXT -1 243 "There are an infinite \+
number of solutions for this equation, they change with L and mu. Mapl
e did not pick out any solution except L = 0. We know, however where t
he sine function is zero: the sine function is zero at all integral mu
ltiples of " }{XPPEDIT 18 0 "pi;" "6#%#piG" }{TEXT -1 14 " . Thus, tak
e " }{XPPEDIT 18 0 "mu*L=n*Pi" "6#/*&%#muG\"\"\"%\"LGF&*&%\"nGF&%#PiGF
&" }{TEXT -1 15 " and solve for " }{XPPEDIT 18 0 "mu" "6#%#muG" }
{TEXT -1 1 "." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "solve(mu*L=
n*Pi,mu);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "mu:=%;" }}}
{PARA 0 "" 0 "" {TEXT -1 35 "     We now have the function X(x)." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "X(x);" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 20 "diff(X(x),x,x)/X(x);" }}}{PARA 0 "" 0 "" 
{TEXT -1 71 "     We now look at the other part of the PDE which we \"
separated off\"." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "dsolve(d
iff(T(t),t,t)=-mu^2*T(t),T(t));" }}}{PARA 0 "" 0 "" {TEXT -1 229 "    \+
 Both the sine and cosine functions give possible solutions. We have n
o reason to eliminate either of these. The general solution we find is
 a linear combination of the particular solutions we get by separating
 the equation. " }}{PARA 0 "" 0 "" {TEXT -1 56 "     The most general \+
linear combination is of the form." }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 103 "u:=(t,x)->sum((A[n]*cos(Pi*n*t/L)+B[n]*sin(Pi*n*t/L)
)*sin(n*Pi*x/L),\n                   n=1..infinity);" }}}{PARA 0 "" 0 
"" {TEXT -1 154 "     The coefficients A and B have to be determined f
rom the initial conditions in a manner that is familiar. We use the in
itial conditions. Suppose that " }{XPPEDIT 18 0 "u(0,x)=f(x)" "6#/-%\"
uG6$\"\"!%\"xG-%\"fG6#F(" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "u[t](0,x
)=g(x)" "6#/-&%\"uG6#%\"tG6$\"\"!%\"xG-%\"gG6#F+" }{TEXT -1 1 "." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "u(0,x)=f(x);\nD[1](u)(0,x)=g
(x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "
" {TEXT -1 110 "Doesn't this look like a job for Fourier Series? To sp
ell out what how to compute the coefficients recall that" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "assume(n,integer):\nint(sin(n*Pi*x/
L)^2,x=0..L);\nn:='n':" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" 
}}}{PARA 0 "" 0 "" {TEXT -1 17 "     We have that" }}{PARA 0 "" 0 "" 
{TEXT -1 25 "                         " }{XPPEDIT 18 0 "A[n];" "6#&%\"
AG6#%\"nG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "int(f(x)*sin(n*Pi*x/L),x \+
= 0 .. L)/int(sin(n*Pi*x/L)^2,x = 0 .. L);" "6#*&-%$intG6$*&-%\"fG6#%
\"xG\"\"\"-%$sinG6#**%\"nGF,%#PiGF,F+F,%\"LG!\"\"F,/F+;\"\"!F3F,-F%6$*
$-F.6#**F1F,F2F,F+F,F3F4\"\"#/F+;F7F3F4" }{TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 3 "and" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 24 "                        " }{XPPEDIT 18 0 "B[n]*n*Pi/L;
" "6#**&%\"BG6#%\"nG\"\"\"F'F(%#PiGF(%\"LG!\"\"" }{TEXT -1 3 " = " }
{XPPEDIT 18 0 "int(g(x)*sin(n*Pi*x/L),x = 0 .. L)/int(sin(n*Pi*x/L)^2,
x = 0 .. L);" "6#*&-%$intG6$*&-%\"gG6#%\"xG\"\"\"-%$sinG6#**%\"nGF,%#P
iGF,F+F,%\"LG!\"\"F,/F+;\"\"!F3F,-F%6$*$-F.6#**F1F,F2F,F+F,F3F4\"\"#/F
+;F7F3F4" }{TEXT -1 3 " = " }{TEXT -1 1 " " }{XPPEDIT 18 0 "2/L;" "6#*
&\"\"#\"\"\"%\"LG!\"\"" }{TEXT -1 1 " " }{XPPEDIT 18 0 "int(g(x)*sin(n
*Pi*x/L),x = 0 .. L);" "6#-%$intG6$*&-%\"gG6#%\"xG\"\"\"-%$sinG6#**%\"
nGF+%#PiGF+F*F+%\"LG!\"\"F+/F*;\"\"!F2" }{TEXT -1 5 " , or" }}{PARA 0 
"" 0 "" {TEXT -1 23 "                       " }{XPPEDIT 18 0 "B[n];" "
6#&%\"BG6#%\"nG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "2/(n*Pi);" "6#*&\"
\"#\"\"\"*&%\"nGF%%#PiGF%!\"\"" }{TEXT -1 1 " " }{XPPEDIT 18 0 "int(g(
x)*sin(n*Pi*x/L),x = 0 .. L);" "6#-%$intG6$*&-%\"gG6#%\"xG\"\"\"-%$sin
G6#**%\"nGF+%#PiGF+F*F+%\"LG!\"\"F+/F*;\"\"!F2" }{TEXT -1 3 "  ." }}
{PARA 0 "" 0 "" {TEXT -1 371 "Here is an example. We can animate and e
xpect to see the vibrations of the string. The following two examples \+
may give considerable insight for how the wave equation models the mot
ion of a string. In the first example, take f as below and g to be zer
o. This will mean that all the B coefficients will be zero and A coeff
icients will be determined by the Fourier quotient." }}{PARA 0 "" 0 "
" {TEXT -1 7 "       " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 96 "L:=
4;\nf:=x->(x-1)*(Heaviside(x-1)-Heaviside(x-2))+\n       (3-x)*(Heavis
ide(x-2)-Heaviside(x-3));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
18 "plot(f(x),x=0..L);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 132 "
for n from 1 to 5 do\n     A[n]:=int(f(x)*sin(n*Pi*x/L),x=0..L)/\n    \+
          int(sin(n*Pi*x/L)^2,x=0..L);\n     B[n]:=0;\nod;\nn:='n':" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 96 "u:=(t,x)->sum((A[n]*cos(Pi
*n*t/L)+B[n]*sin(Pi*n*t/L))*sin(n*Pi*x/L),\n                   n=1..5)
;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "plot3d(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..2*L);" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT 
-1 286 "In the second example, take f to be zero and g to be as below.
 The formula for f and g suggests that the string is at rest and we gi
ve the middle an initial velocity.  This will mean that all the B coef
ficients will be zero and A coefficients will be determined by the Fou
rier quotient." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "L:=4;\ng:=
x->-(Heaviside(x-1)-Heaviside(x-3));" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 41 "plot(g(x),x=0..L,discont=true,color=RED);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 106 "n:='n':\nfor n from 1 to 5 do\n   \+
  A[n]:=0;\n     B[n]:=2/(n*Pi)*int(g(x)*sin(n*Pi*x/L),x=0..L);\nod;\n
n:='n':" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 96 "u:=(t,x)->sum((A
[n]*cos(Pi*n*t/L)+B[n]*sin(Pi*n*t/L))*sin(n*Pi*x/L),\n                \+
   n=1..5);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "plot3d(u(t,x
),x=0..L,t=0..2*L,axes=NORMAL,orientation=[-135,45]);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "animate(u(t,x),x=0..L,t=0..2*L);" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 93 "Assignment: Examine the vibrati
ons of a string with these three different initial conditions." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
31 "L:=2:\nf:=x->sin(Pi*x);\ng:=x->0;" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 31 "L:=2:\nf:=x->0;\ng:=x->sin(Pi*x);" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 39 "L:=Pi;\nf:=x->Pi/2-abs(x-Pi/2);\ng:=x->0;" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}}{MARK "0 0" 45 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }
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