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{SECT 0 {PARA 256 "" 0 "" {TEXT -1 0 "" }{TEXT 256 53 "A String in a V
iscous Medium: Separation of Variables" }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 59 "In Module 25, we performed a change o
f variables to the PDE" }}{PARA 0 "" 0 "" {TEXT -1 29 " \+
" }{XPPEDIT 18 0 "diff(w,`$`(x,2))-k*diff(w,t)-g = diff(
w,`$`(t,2));" "6#/,(-%%diffG6$%\"wG-%\"$G6$%\"xG\"\"#\"\"\"*&%\"kGF.-F
&6$F(%\"tGF.!\"\"%\"gGF4-F&6$F(-F*6$F3F-" }}{PARA 0 "" 0 "" {TEXT -1
0 "" }}{PARA 0 "" 0 "" {TEXT -1 65 " \+
w(t, 0) = 0 = w(t, L)" }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}
{PARA 0 "" 0 "" {TEXT -1 59 " \+
w( 0, x) = F(x)" }}{PARA 0 "" 0 "" {TEXT -1 44 " \+
" }{XPPEDIT 18 0 "w[t]" "6#&%\"wG6#%\"tG" }
{TEXT -1 17 " ( 0 , x) = G(x)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 278 "to arrive at a different PDE to be solve
d. If you stepped out into the hall from your workplace, stopped the f
irst person you saw, and asked them how to solve the original PDE, the
y would respond something such as, \"Separate variables. Isn't that ho
w you do all those problem?\" " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 280 "I say to you: \"Surely. Separation of va
riables is one of the main ideas of this course.\" So, abandon with me
the change of variable from Lecture 25 and from Module 25, and let's \+
solve the equation like the person we might stop in the hall. Let's us
e the techniques of this course." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 73 "To be specific, we will get solutions tha
t we can graph for the equation:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 26 " " }{XPPEDIT 18
0 "diff(w,`$`(x,2))-diff(w,t)/5-32 = diff(w,`$`(t,2));" "6#/,(-%%diffG
6$%\"wG-%\"$G6$%\"xG\"\"#\"\"\"*&-F&6$F(%\"tGF.\"\"&!\"\"F4\"#KF4-F&6$
F(-F*6$F2F-" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT
-1 64 " w(t, 0) = 0 = w(t, \+
" }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 1 ")" }}{PARA 0 "" 0 ""
{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 61 " \+
w( 0, x) = sin(x)" }}{PARA 0 "" 0 "" {TEXT -1 44 "
" }{XPPEDIT 18 0 "w[t]" "6
#&%\"wG6#%\"tG" }{TEXT -1 14 " ( 0 , x) = 0." }}{PARA 0 "" 0 "" {TEXT
-1 157 "Recognize that this is a non-homogeneous problem. The -32 term
makes it thus. So, we must separate out the steady state solution and
the transient solution. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 100 "The steady state solution S(x) is independent of \+
time. Thus, that solution will satisfy the equation" }}{PARA 0 "" 0 "
" {TEXT -1 17 " " }{XPPEDIT 18 0 "diff(S(x),`$`(x,2));
" "6#-%%diffG6$-%\"SG6#%\"xG-%\"$G6$F)\"\"#" }{TEXT -1 30 " - 32 = 0, \+
with S(0) = 0 = S(" }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 2 ")." }
}{PARA 0 "" 0 "" {TEXT -1 58 "We solve this second order ordinary diff
erential equation." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart
;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "dsolve(\{diff(S(x),x,x
)=32,S(0)=0,S(Pi)=0\},S(x));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1
0 21 "S:=x->16*x^2-16*Pi*x;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0
0 "" }}}{PARA 0 "" 0 "" {TEXT -1 42 "We find that the steady state sol
ution is " }{XPPEDIT 18 0 "S(x) = 16*x^2-16*Pi*x;" "6#/-%\"SG6#%\"xG,&
*&\"#;\"\"\"*$F'\"\"#F+F+*(F*F+%#PiGF+F'F+!\"\"" }{TEXT -1 1 "." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 63 "Next we s
olve the transient equation. The transient equation is" }}{PARA 0 ""
0 "" {TEXT -1 24 " " }{XPPEDIT 18 0 "diff(w,`$`
(x,2))-diff(w,t)/5 = diff(w,`$`(t,2));" "6#/,&-%%diffG6$%\"wG-%\"$G6$%
\"xG\"\"#\"\"\"*&-F&6$F(%\"tGF.\"\"&!\"\"F4-F&6$F(-F*6$F2F-" }}{PARA
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 63 " \+
w(t, 0) = 0 = w(t, " }{XPPEDIT 18 0 "Pi;
" "6#%#PiG" }{TEXT -1 1 ")" }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA
0 "" 0 "" {TEXT -1 68 " w( \+
0, x) = sin(x) - S(x)" }}{PARA 0 "" 0 "" {TEXT -1 44 " \+
" }{XPPEDIT 18 0 "w[t]" "6#&%\"wG6#%\"tG"
}{TEXT -1 14 " ( 0 , x) = 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA
0 "" 0 "" {TEXT -1 119 "How do we solve this transient equation? Just \+
like the hall walker said: we separate variables. This leads to equati
ons" }}{PARA 0 "" 0 "" {TEXT -1 44 " X''/X = (T'' +
T '/5 )/T." }}{PARA 0 "" 0 "" {TEXT -1 128 "In the usual manner, we ar
rive at two ordinary differential equations, the solutions of which le
ad to the solutions for the PDE:" }}{PARA 0 "" 0 "" {TEXT -1 19 " \+
X '' = " }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT -1 20 " \+
X, with X(0) = X( " }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 6 " ) = \+
0" }}{PARA 0 "" 0 "" {TEXT -1 3 "and" }}{PARA 0 "" 0 "" {TEXT -1 26 " \+
T '' + T'/5 = " }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }
{TEXT -1 3 " T." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 56 "We've seen the X equation enough to know all solutions: \+
" }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT -1 3 " = " }{XPPEDIT
18 0 "-n^2;" "6#,$*$%\"nG\"\"#!\"\"" }{TEXT -1 5 " and " }{XPPEDIT 18
0 "X[n];" "6#&%\"XG6#%\"nG" }{TEXT -1 10 "(x) = sin(" }{XPPEDIT 18 0 "
n*Pi*x/L;" "6#**%\"nG\"\"\"%#PiGF%%\"xGF%%\"LG!\"\"" }{TEXT -1 10 ") =
sin(n " }{XPPEDIT 18 0 "x;" "6#%\"xG" }{TEXT -1 24 "). Check this by \+
asking:" }}{PARA 0 "" 0 "" {TEXT -1 7 "(1) Is " }{XPPEDIT 18 0 "X[n];
" "6#&%\"XG6#%\"nG" }{TEXT -1 5 "'' = " }{XPPEDIT 18 0 "lambda;" "6#%'
lambdaG" }{TEXT -1 2 " " }{XPPEDIT 18 0 "X[n];" "6#&%\"XG6#%\"nG" }
{TEXT -1 2 " ?" }}{PARA 0 "" 0 "" {TEXT -1 7 "(2) Is " }{XPPEDIT 18 0
"X[n];" "6#&%\"XG6#%\"nG" }{TEXT -1 15 "(0) = 0 ? Is " }{XPPEDIT 18
0 "X[n];" "6#&%\"XG6#%\"nG" }{TEXT -1 1 "(" }{XPPEDIT 18 0 "Pi;" "6#%#
PiG" }{TEXT -1 7 ") = 0 ?" }}{PARA 0 "" 0 "" {TEXT -1 41 "The answer t
o the three questions is yes." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 76 "For each n, we solve the T equation. Be s
ure we know what the T equation is:" }}{PARA 0 "" 0 "" {TEXT -1 38 " \+
T '' + T'/5 = " }{XPPEDIT 18 0 "-n^2;" "6#,$*$%
\"nG\"\"#!\"\"" }{TEXT -1 3 " T." }}{PARA 0 "" 0 "" {TEXT -1 141 "Sinc
e this is a second order equation in T, we expect two constants. These
will be constants that we will determine by some Fourier process. " }
}{PARA 0 "" 0 "" {TEXT -1 94 " How many terms shall we do? Let's c
all the number of terms we choose to do by the name N." }}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "N:=5;" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 89 "for n from 1 to N do\n dsolve(diff(T(t),t,t)+diff(T
(t),t)/5=-n^2*T(t),T(t));\nod;\nn:='n':" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 224 "It would be nice to
have a good formula for those trig terms. The question is: how can I \+
write that multiple of t inside the sine or cosine function as a funct
ion of n. I'll solve the equation again without saying what n is." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "dsolve(diff(T(t),t,t)+diff(T
(t),t)/5=-n^2*T(t),T(t));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0
"" }}}{PARA 0 "" 0 "" {TEXT -1 133 "It is clear what the terms are now
, isn't it? Using the Euler identity relating exponential and trig fun
ctions, each T is of the form" }}{PARA 0 "" 0 "" {TEXT -1 13 " \+
" }{XPPEDIT 18 0 "T[n](t) = A[n]*exp(-t/10)*cos(t*sqrt(100*n^2-1)
/10)+B[n]*exp(-t/10)*sin(t*sqrt(100*n^2-1)/10);" "6#/-&%\"TG6#%\"nG6#%
\"tG,&*(&%\"AG6#F(\"\"\"-%$expG6#,$*&F*F0\"#5!\"\"F7F0-%$cosG6#*(F*F0-
%%sqrtG6#,&*&\"$+\"F0*$F(\"\"#F0F0F0F7F0F6F7F0F0*(&%\"BG6#F(F0-F26#,$*
&F*F0F6F7F7F0-%$sinG6#*(F*F0-F=6#,&*&FAF0*$F(FCF0F0F0F7F0F6F7F0F0" }
{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 68 "I can now make the gener
al solution for the PDE. I hope you can too." }}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 139 "u:=(t,x)->sum(A[n
]*exp(-t/10)*cos(t*sqrt(100*n^2-1)/10)*sin(n*x),\n n=1..N)+sum(B[n]*e
xp(-t/10)*sin(t*sqrt(100*n^2-1)/10)*sin(n*x),n=1..N);" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 162 "W
e check that this is a general solution. Ask: does it satisfy the PDE \+
and the boundary conditions? (The initial conditions will determine th
e values of A and B.)" }}{PARA 0 "" 0 "" {TEXT -1 20 "Boundary conditi
ons:" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "u(t,0); u(t,Pi);" }}
}{PARA 0 "" 0 "" {TEXT -1 8 "The PDE:" }}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 61 "simplify(diff(u(t,x),x,x)-diff(u(t,x),t)/5-diff(u(t,x
),t,t));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 ""
0 "" {TEXT -1 57 "It remains to determine the A's and B's. Observe wha
t is " }{XPPEDIT 18 0 "u(0,x);" "6#-%\"uG6$\"\"!%\"xG" }{TEXT -1 5 " a
nd " }{XPPEDIT 18 0 "u[t](0,x);" "6#-&%\"uG6#%\"tG6$\"\"!%\"xG" }
{TEXT -1 1 "." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "u(0,x);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "D[1](u)(0,x);" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 101 "DerivTerm:=collect(collect(collect
(collect(collect(%,sin(x)),sin(2*x)),sin(3*x)),sin(4*x)),sin(5*x));" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT
-1 193 "It seems clear that we can get the A's through a simple Fourie
r Series. Having the A's we can get the B's. We do that here. This is \+
the first time in this problem to use the initial conditions." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "f:=x->sin(x)-S(x);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 85 "for n from 1 to N do\n A[n
]:=int(f(x)*sin(n*x),x=0..Pi)/int(sin(n*x)^2,x=0..Pi);\nod;" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "g:=x->0;" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 15 "g(x)=DerivTerm;" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 63 "for n from 1 to N do\n B[n]:=A[n]/sqrt(100*n^2-1)
;\nod;\nn:='n';" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{PARA 0 "" 0 "" {TEXT -1 105 "We now have the transient solution compl
etely determined. We create the solution to the original problem:" }}
{PARA 0 "" 0 "" {TEXT -1 53 " w(t,x) = u(
t,x) + S(x)." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "w:=(t,x)->u(
t,x)+S(x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "
" 0 "" {TEXT -1 97 "It seems appropriate to check that this is really \+
the solution. We check the boundary conditions." }}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 16 "w(t,0); w(t,Pi);" }}}{PARA 0 "" 0 "" {TEXT -1
17 "We check the PDE." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "sim
plify(diff(w(t,x),x,x)-diff(w(t,x),t)/5-32-diff(w(t,x),t,t));" }}}
{PARA 0 "" 0 "" {TEXT -1 107 "We check the initial conditions. For eac
h one, we draw graphs to see how close we are to the initial value." }
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "plot([sin(x),w(0,x)],x=0..P
i,color=[black,red]);" }}}{PARA 0 "" 0 "" {TEXT -1 96 "We draw a graph
to see how close the initial velocity is to zero. Expect a small, err
atic graph." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "plot(eval(sub
s(t=0,diff(w(t,x),t))),x=0..Pi);" }}}{PARA 0 "" 0 "" {TEXT -1 261 "We \+
now show an animation. Can you predict the movement? It should be a vi
brating string; the first derivative term is a retarding force so the \+
oscillations should decrease in size; and the downward force should pu
ll the string down to that hanging, steady state" }}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 45 "with(plots):\nanimate(w(t,x),x=0..Pi,t=0..15);
" }{TEXT -1 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 221 " Remember why we did this problem: The Lecture 25 se
emed to take emphasis away from one of the main ideas of the course: M
any problems encountered in this course can be solved straight away by
separation of variables." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "
" }}}{EXCHG }{EXCHG }{PARA 0 "" 0 "" {TEXT -1 1 " " }}}{MARK "0 1" 53
}{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }