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{SECT 0 {PARA 3 "" 0 "" {TEXT -1 49 "Module 42: Infinite Strings And T
he Wave Equation" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 25 "Problem 1: Se
nding a wave" }}{PARA 0 "" 0 "" {TEXT -1 147 "We have done a problem s
uch as the following but using d'Alembert's techniques. We re-visit th
is problem with the techniques of Laplace Transforms." }}{PARA 0 "" 0 
"" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 8 "        " }{XPPEDIT 18 
0 "diff(u,`$`(t,2)) = 9*diff(u,`$`(x,2));" "6#/-%%diffG6$%\"uG-%\"$G6$
%\"tG\"\"#*&\"\"*\"\"\"-F%6$F'-F)6$%\"xGF,F/" }{TEXT -1 22 "  for x > \+
0 and t > 0," }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 20 "          u(0, x) = " }{XPPEDIT 18 0 "diff(u,t)" "6#-%%diffG6$%
\"uG%\"tG" }{TEXT -1 11 "(0, x) = 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}}{PARA 0 "" 0 "" {TEXT -1 33 "          u(t, 0) = sin(t), u(t, " }
{XPPEDIT 18 0 "infinity" "6#%)infinityG" }{TEXT -1 14 ")  is bounded.
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 37 "restart; with(inttrans): assume(x>0);" }}}{PARA 0 "" 0 "" 
{TEXT -1 44 "We define the partial differential equation." }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "PDE:= diff(U(t,x),t,t)=9*diff(U(t,x
),x,x);" }}}{PARA 0 "" 0 "" {TEXT -1 41 "We take the Laplace Transform
 of the PDE." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "Lpde:=laplac
e(PDE,t,s);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 
"" 0 "" {TEXT -1 49 "This calculation asks for the initial conditions.
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "subs(\{U(0,x)=0,D[1](U)(
0,x)=0\},Lpde);" }}}{PARA 0 "" 0 "" {TEXT -1 96 "We now make an ordina
ry differential equation by defining v(x) to be the Laplace transform \+
of U." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "ODE:=subs(laplace(U
(t,x),t,s)=v(x),%);" }}}{PARA 0 "" 0 "" {TEXT -1 105 "The solution for
 the differential equation will need the value of v(0). Then, the equa
tion can be solved." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "q:=un
apply(laplace(sin(t),t,s),s):\nq(s);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 39 "dsolve(\{ODE,v(0)=q(s),D(v)(0)=b\},v(x));" }}}{PARA 
0 "" 0 "" {TEXT -1 32 "I collect the exponential terms." }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "sol:=collect(collect(rhs(%),exp(s*x
/3)),exp(-s*x/3));" }}}{PARA 0 "" 0 "" {TEXT -1 121 "We need the coeff
icient of the term which goes unbounded to be zero. The question is ho
w to choose b to make that happen." }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 26 "C2:=coeff(sol,exp(s*x/3));" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 17 "B:=solve(C2=0,b);" }}}{PARA 0 "" 0 "" {TEXT -1 80 
"Using this B, we solve the equation again. We should not get any unbo
unded term." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "sol2:=dsolve(
\{ODE,v(0)=q(s),D(v)(0)=B\},v(x),method=laplace);" }}}{PARA 0 "" 0 "" 
{TEXT -1 58 "We have now only to compute the inverse transform of v(x)
." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "invlaplace(rhs(sol2),s,
t);" }}}{PARA 0 "" 0 "" {TEXT -1 15 "This defines u." }}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 20 "u:=unapply(%,(t,x));" }}}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }{TEXT -1 43 "We will make a plot. Give it a little time
." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "plot3d(u(t,x),x=0..10,t
=0..4,axes=NORMAL,orientation=[-135,45]);" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 53 "plots[animate](u(t,x),x=0.1..30,t=0..10,color=BLACK
);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 31 "Problem 2: Sendin
g a brief wave" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 68 "This is essentially the same problem. There is only one \+
\"bump\" sent." }}{PARA 0 "" 0 "" {TEXT -1 8 "        " }{XPPEDIT 18 
0 "diff(u,`$`(t,2)) = 9*diff(u,`$`(x,2));" "6#/-%%diffG6$%\"uG-%\"$G6$
%\"tG\"\"#*&\"\"*\"\"\"-F%6$F'-F)6$%\"xGF,F/" }{TEXT -1 22 "  for x > \+
0 and t > 0," }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 20 "          u(0, x) = " }{XPPEDIT 18 0 "diff(u,t)" "6#-%%diffG6$%
\"uG%\"tG" }{TEXT -1 11 "(0, x) = 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}}{PARA 0 "" 0 "" {TEXT -1 56 "          u(t, 0) = (1 - Heaviside(t - \+
))  sin(t), u(t, " }{XPPEDIT 18 0 "infinity" "6#%)infinityG" }{TEXT 
-1 6 ") = 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 37 "restart; with(inttrans): assume(x>0);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "pde:= diff(U(t,x),t,t)-9*diff(U(t,x
),x,x)=0;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "laplace(pde,t,
s);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "subs(\{U(0,x)=0,D[1]
(U)(0,x)=0\},%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "ODE:=su
bs(laplace(U(t,x),t,s)=v(x),%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 60 "q:=unapply(laplace((1-Heaviside(t-Pi))*sin(t),t,s),s):\nq(s);
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "sol:=dsolve(\{ODE,v(0)=
q(s),D(v)(0)=b\},v(x));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "
expand(sol);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "sol:=collec
t(collect(rhs(%),exp(s*x/3)),exp(-s*x/3));" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 16 "op(1,op(2,sol));" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 16 "C2:=simplify(%);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 17 "B:=solve(C2=0,b);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 77 "sol2:=dsolve(\{ODE,\n       v(0)=q(s),D(v)(0)=B\},\n \+
       v(x),method=laplace);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 20 "simplify(rhs(sol2));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
18 "invlaplace(%,s,t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "u
:=unapply(%,(t,x));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "plot
3d(u(t,x),x=0..10,t=0..4,axes=NORMAL,orientation=[-135,45]);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "plots[animate](u(t,x),x=0.1.
.30,t=0..10,color=BLACK);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 
"" }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 49 "Problem 3: A Long String F
alling Under its Weight" }}{PARA 0 "" 0 "" {TEXT -1 202 "This problem \+
is different. It is a model for a string falling due to gravity. The t
echniques of Laplace Transforms seems more appropriate for this proble
m than the techniques of separation of variables." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 29 "We use the transform pack
age." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart:" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "with(inttrans); assume(x>0);" }}}
{PARA 0 "" 0 "" {TEXT -1 81 "We define the partial differential equati
on. We add accelleration due to gravity." }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 48 "pde:= diff(U(t,x),t,t)-c^2*diff(U(t,x),x,x)+g=0;" }}}
{PARA 0 "" 0 "" {TEXT -1 44 "We compute the Laplace Transform of the P
DE." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "laplace(pde,t,s);" }}
}{PARA 0 "" 0 "" {TEXT -1 129 "This computation asks for the value of \+
u and the velocity of u at t = 0. We are assuming the string was at re
st at the beginning." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "subs
(\{U(0,x)=0,D[1](U)(0,x)=0\},%);" }}}{PARA 0 "" 0 "" {TEXT -1 61 "As e
xpected, the Laplace Transform of a PDE generated an ODE." }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "ODE:=subs(laplace(U(t,x),t,s)=v(x),
%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "dsolve(\{ODE,v(0)=0,
D(v)(0)=b\},v(x));" }}}{PARA 0 "" 0 "" {TEXT -1 56 "We prefer to have \+
the solutions in term of exponentials." }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 20 "convert(rhs(%),exp);" }}}{PARA 0 "" 0 "" {TEXT -1 38 
"We need the coefficient of exp(s x/c)." }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 43 "collect(collect(%,exp(s*x/c)),exp(-s*x/c));" }}}
{PARA 0 "" 0 "" {TEXT -1 77 "Is it clear that this coefficient is zero
 with b as defined in the next line?" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 14 "B:=-g/(c*s^2);" }}}{PARA 0 "" 0 "" {TEXT -1 52 "Now, \+
resolve the ODE with this value for the v '(0)." }}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 42 "sol2:=dsolve(\{ODE,v(0)=0,D(v)(0)=B\},v(x));" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "sol2:=expand(convert(sol2
,exp));" }}}{PARA 0 "" 0 "" {TEXT -1 67 "Take g = 32 and c = 1. Then, \+
v(x) is the transform of the solution." }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 54 "g:=32: c:=1;\ninvlaplace(rhs(sol2),s,t); subs(x='x',%
);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "u:=unapply(%,(t,x));
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "plot3d(u(t,x),x=0.1..10
,t=0..4,axes=NORMAL,orientation=[-75,35]);" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 41 "plots[animate](u(t,x),x=0.1..30,t=0..10);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT 256 11 "Assignment:" }{TEXT -1 113 " V
erify that d'ALembert's method gives the same solution for Problems 1 \+
and 2. Hint: review Worksheet 26, part 4." }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}}{MARK "0 0" 49 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }
{PAGENUMBERS 0 1 2 33 1 1 }