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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 77 " Some code to find eigenvalue equations for 2nd order different
ial operators:" }}{PARA 0 "" 0 "" {TEXT -1 69 " Specify the operator L
 and the boundary operators BC1 and BC2 below." }}{PARA 0 "" 0 "" 
{TEXT -1 2 "  " }}{PARA 0 "" 0 "" {TEXT -1 73 "Warning: Maple doesn't \+
always put things in the correct \"form\" when using" }}{PARA 0 "" 0 "
" {TEXT -1 61 "               the assume command in conjunction with d
solve." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 36 " Define the 2
nd order operator here:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "L:=U->diff(U,x$2):" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 82 "     Defi
ne the boundary conditions here using general linear boundary operator
 BC" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 49 "BC:=(U,a,b,X0)->eval(subs(x=X0,a*U+b*diff(U,x))):" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "a:=-1:b:=1:" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "BC1:=U->BC(U,1,0,a):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "BC2:=U->BC(U,1,0,b):" }}}{EXCHG 
{PARA 11 "" 1 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 79 "  The ge
neral solution of the differential equation will depend on the value o
f" }}{PARA 0 "" 0 "" {TEXT -1 72 "   lambda. Here we generate all solu
tions for lambda <0, =0, >0 and then" }}{PARA 0 "" 0 "" {TEXT -1 33 " \+
  consider eigenvalue equations " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "lambda:='lambda':assume(lamb
da<-1/4);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "UM:=rhs(dsolve
(L(u(x))=lambda*u(x),\{u(x)\}));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%
#UMG,&*&%$_C1G\"\"\"-%$sinG6#*&,$%(lambda|irG!\"\"#F(\"\"#%\"xGF(F(F(*
&%$_C2GF(-%$cosGF+F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "U
0:=rhs(dsolve(L(u(x))=-1/4*u(x),u(x)));" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%#U0G,&*&%$_C1G\"\"\"-%$cosG6#,$%\"xG#F(\"\"#F(F(*&%$_C2GF(-%$s
inGF+F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "lambda:='lambd
a':assume(lambda>-1/4);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "
UP:=rhs(dsolve(L(u(x))=lambda*u(x),\{u(x)\}));" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#UPG,&*&%$_C1G\"\"\"-%$expG6#*&%(lambda|irG#F(\"\"#%
\"xGF(F(F(*&%$_C2GF(-F*6#,$F,!\"\"F(F(" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 79 "    Now applying the boun
dary conditions to each of these solutions. The result" }}{PARA 0 "" 
0 "" {TEXT -1 84 "    is a two linear equations for (_C1,_C2) whose co
efficient matrix A is determined" }}{PARA 0 "" 0 "" {TEXT -1 66 "    b
y the code below. Eigenvalues are roots of det(A)=0. Defining" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 14 "     F= d
et(A)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 37 "
     the eigenvalue equation is F=0. " }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 91 "A:=U->array([[coeff(BC1
(U),_C1),coeff(BC1(U),_C2)],[coeff(BC2(U),_C1),coeff(BC2(U),_C2)]]):" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "with(linalg):" }}{PARA 7 
"" 1 "" {TEXT -1 32 "Warning, new definition for norm" }}{PARA 7 "" 1 
"" {TEXT -1 33 "Warning, new definition for trace" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
15 "FM:=det(A(UM));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#FMG,$*&-%$si
nG6#*$,$%(lambda|irG!\"\"#\"\"\"\"\"#F/-%$cosGF)F/!\"#" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "F0:=det(A(U0));" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#>%#F0G,$*&-%$cosG6##\"\"\"\"\"#F+-%$sinGF)F+F," }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "FP:=det(A(UP));" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#>%#FPG,&*$-%$expG6#,$*$%(lambda|irG#\"\"\"\"\"#
!\"\"F/F.*$-F(6#F+F/F0" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" 
}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 87 "  Since lambda are chosen so th
at A is singular, finding the (_C1,_C2) which are in the" }}{PARA 0 "
" 0 "" {TEXT -1 88 "  nullspace can be found by solving the first row \+
of Ac = 0. Depending on the outcome of" }}{PARA 0 "" 0 "" {TEXT -1 87 
"  the previous analysis, the following code must be changed to find a
ll eigenfunctions:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 43 "  psi  are the nonnormalized eigenfunctions" }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "A(UM)
;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'MATRIXG6#7$7$\"\"\"\"\"!7$,&-%
$cosG6#*$,$%(lambda|irG!\"\"#F(\"\"#F(*&-%$sinGF.F(F0F3F2,&F6F(*&F,F(F
0F3F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 74 "NULLA:=U->nullspac
e(array([[A(U)[1,1],A(U)[1,2]],[A(U)[1,1],A(U)[1,2]]])):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "NA:=NULLA(UM):" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 42 "psi:=subs(\{_C1=NA[1][1],_C2=NA[1][2]\},UM);
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$psiG-%$sinG6#*&,$%(lambda|irG!
\"\"#\"\"\"\"\"#%\"xGF-" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 81 "  To normalize them, the weight function \+
for the inner product must be specified:" }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 41 "   phi are the normalized eigenfun
ctions." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 8 "w:=x->1;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"wG\"\"
\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "phi:=simplify(psi/int
(psi*psi*w(x),x=0..1));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$phiG,$*(
-%$sinG6#*&,$%(lambda|irG!\"\"#\"\"\"\"\"#%\"xGF/F/,&*&-%$cosG6#*$F+F.
F/-F(F6F/F/F7F-F-F+F.!\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 
"" }}}{EXCHG {PARA 11 "" 1 "" {TEXT -1 0 "" }}}}{MARK "22 0 0" 0 }
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