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{SECT 0 {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 60 "      Code for finding a Greens function g(x,t) wh
ich solves" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 28 "                            " }{XPPEDIT 18 0 "a2(x)*diff(g(x,t)
,x$2)+a1(x)*diff(g(x,t),x)+a0(x)*g(x,t)=delta(x-t)" "/,(*&-%#a2G6#%\"x
G\"\"\"-%%diffG6$-%\"gG6$F(%\"tG-%\"$G6$F(\"\"#F)F)*&-%#a1G6#F(F)-F+6$
-F.6$F(F0F(F)F)*&-%#a0G6#F(F)-F.6$F(F0F)F)-%&deltaG6#,&F(F)F0!\"\"" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 93 "      sat
isfying prescribed boundary conditions. Below we define the differenti
al operator L:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "a2:=x->x^2:a1:=x->2*x:a0:=x-
>0:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 69 "L:=u->a2(x)*diff(u,x
$2)+a1(x)*diff(u,x)+a0(x)*u:L(g(x,t))=delta(x-t);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#/,&*&%\"xG\"\"#-%%diffG6$-F)6$-%\"gG6$F&%\"tGF&F&\"\"\"
F1*&F&F1F+F1F'-%&deltaG6#,&F&F1F0!\"\"" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 73 "     The distributional e
quation g(x,t) we will solve is as shown above. " }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 77 "      Next we need to pre
scribe boundary conditions for g(x,t) at x=a and x=b" }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "a:=1:b:=2
:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 72 "B1:=u->subs(x=a,u):B2:
=u->subs(x=b,diff(u,x)):B1(g(x,t))=0;B2(g(x,t))=0;" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#/-%\"gG6$\"\"\"%\"tG\"\"!" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#/-%%diffG6$-%\"gG6$\"\"#%\"tGF*\"\"!" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 105 "     The boundary conditions  are shown above \+
(Note: Maple does not display the evaluation of derivatives" }}{PARA 
0 "" 0 "" {TEXT -1 42 "    of g(x,t) at x=a and x=b very nicely)." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 52 "    To find g(x,t) we de
fine it piecewise as follows" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 53 "                    g(x,t)    = Gp(x,t)        \+
   x>t" }}{PARA 0 "" 0 "" {TEXT -1 55 "                               \+
    Gm(x,t)          x<t" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 23 "    where we must have " }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 39 "                     L(Gp) = 0 \+
     x>t" }}{PARA 0 "" 0 "" {TEXT -1 38 "                     L(Gm) = \+
0     x<t" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
50 "    The resulting general forms for Gp and Gm are:" }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "Gp:=su
bs(\{_C1=C1p,_C2=C2p\},rhs(dsolve(L(g(x))=0,g(x))));" }}{PARA 11 "" 1 
"" {XPPMATH 20 "6#>%#GpG,&%$C1pG\"\"\"*&%$C2pGF'%\"xG!\"\"F'" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "Gm:=subs(\{_C1=C1m,_C2=C2m\}
,rhs(dsolve(L(g(x))=0,g(x))));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#G
mG,&%$C1mG\"\"\"*&%$C2mGF'%\"xG!\"\"F'" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 87 "     where the 4 constant
s C1p,C2p,C1m,C2m  really depend on t. We find these constants" }}
{PARA 0 "" 0 "" {TEXT -1 90 "     by requiring g(x,t) to satisfy the b
oundary conditions, a continuity condition at x=t" }}{PARA 0 "" 0 "" 
{TEXT -1 61 "     and a jump condition in the derivative of g(x,t) at \+
x=t:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 46 "BCOND1:=eval(B1(Gm))=0;BCOND2:=eval(B2(Gp))=0;" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'BCOND1G/,&%$C1mG\"\"\"%$C2mGF(\"\"!
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'BCOND2G/,$%$C2pG#!\"\"\"\"%\"\"
!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "CONTY:=subs(x=t,Gp-Gm)
=0;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&CONTYG/,*%$C1pG\"\"\"*&%$C2p
GF(%\"tG!\"\"F(%$C1mGF,*&%$C2mGF(F+F,F,\"\"!" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 46 "JUMP:=subs(x=t,diff(Gp,x)-diff(Gm,x))=1/a2(t);" 
}}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%JUMPG/,&*&%$C2pG\"\"\"%\"tG!\"#!
\"\"*&%$C2mGF)F*F+F)*$F*F+" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 56 "    Solving these 4 equations for the 4 c
onstants gives:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 63 "Constants:=solve(\{BCOND1,BCOND2,CONTY,JUMP\},
\{C1p,C2p,C1m,C2m\});" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%*ConstantsG
<&/%$C2mG\"\"\"/%$C1mG!\"\"/%$C1pG,$*&,&%\"tGF(F+F(F(F1F+F+/%$C2pG\"\"
!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 33 "    Thus, for x>t, g(x,t) equals " }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "Gplus:=si
mplify(subs(Constants,Gp));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&Gplu
sG,$*&,&%\"tG\"\"\"!\"\"F)F)F(F*F*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 30 "    and for x<t, g(x,t) equals" }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "Gminus:=simplify(subs(C
onstants,Gm));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'GminusG,$*&,&%\"x
G\"\"\"!\"\"F)F)F(F*F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" 
}}}}{MARK "24 0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 }