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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 112 "This worksheet finds the \+
eigenvalues and eigenfunctions for the general problem of cooling of a
 disk of radius a" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "DE := \+
r*diff(y(r),r,r)+diff(y(r),r)+q^2*r*y(r)=0;" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 16 "dsolve(DE,y(r));" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 71 "plot([BesselJ(0,x),BesselY(0,x)],x=0..18,color=[blue,
red],thickness=2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "z1 :=
 fsolve(BesselJ(0,z)=0,z=2..3);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 34 "z := evalf(BesselJZeros(0,1..10));" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 93 "plot([BesselJ(0,x),sqrt(2/(Pi*x))*cos(x-Pi/4)],x=0.
.18,-0.5..1,color=[blue,red],thickness=1);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 59 "We define a general inner
 product of two functions u and v." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "IP := (r,u,v) -> evalf(int(
u*v*r,r=0..R));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 170 "We define the eigenfunctions, with n as an independ
ent variable. We also define the inner product of two eigenfunctions a
nd the time-dependent factor of the solution u_n." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "y := (n,r) \+
-> BesselJ(0,z[n]*r/R);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "
IPy := (m,n) -> IP(BesselJ(0,z[m]*r/R),BesselJ(0,z[n]*r/R));" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "g := (n,t) -> exp(-z[n]^2*k*
t/R^2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 49 "We verify that our eigenfunctions are orthogonal." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 41 "IP(r,y(1,r),y(2,r));\nIP(r,y(1,r),y(2,r));" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 75 "We choose values f
or R and k and a function f(r) for the initial condition." }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "k:=1
;   R:=1;   f:=r->1-r;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 36 "We compute the Fourier coefficients." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 69 "for n from 1 to 10 do c[n]:=IP(r,f(r),y(n,r))/IP(r,y(n,r),y(n,r)
) od;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 49 "The first 9 terms will give a good approximation." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 45 "u := (r,t) -> sum(c[m]*g(m,t)*y(m,r),m=1..9);" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 75 "The eigenfuncti
on y1(x) indicates the shape of the graph after a long time." }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "p
lot(y(1,r),r=0..R);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 98 "The above graph indicates that the maximum valu
e of u with respect to r at large times is at r=0. " }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "plot(u(0,
t),t=0..1,0..1);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 75 "The above graph suggests that a time interval of 0
..0.6 will be sufficient." }}{PARA 0 "" 0 "" {TEXT -1 92 "Here we see \+
a plot of temperature profiles, each at a time twice that of the previ
ous graph." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 147 "plot([u(x,0),u(x,0.0125),u(x,0.025),u(x,0.05),u(x,
0.1),u(x,0.2),u(x,0.4)],x=0..1,0..1, color=[green,blue,black,maroon,re
d,orange,tan],thickness=2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "29" 0 }
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