{VERSION 4 0 "IBM INTEL LINUX22" "4.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 1 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 1 }{CSTYLE "2D Output" 2 20 "" 0 1 0 0 255 1 0 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 256 "" 1 24 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 257 "" 1 18 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 258 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 259 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 260 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 261 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 262 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 263 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 264 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 265 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 266 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 267 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 268 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 269 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 270 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 271 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 272 "" 1 24 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 273 "" 1 24 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 274 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 275 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 276 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 277 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 278 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 279 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Text Output " -1 2 1 {CSTYLE "" -1 -1 "Courier" 1 10 0 0 255 1 0 0 0 0 0 1 3 0 3 0 }1 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 2 6 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 2 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 } {PSTYLE "Maple Output" 0 11 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 3 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 11 12 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }1 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 256 0 "" }{TEXT 257 50 "Finding Exact solutions of Differential Equations." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 2 "A " } {TEXT 258 11 "non-linear " }{TEXT -1 49 "(it contains a product y*y) d ifferential equation" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "res tart: eq := diff(y(x),x)-y(x)^2+y(x)*sin(x)-cos(x) = 0;" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#>%#eqG/,*-%%diffG6$-%\"yG6#%\"xGF-\"\"\"*$)F*\" \"#F.!\"\"*&F*F.-%$sinGF,F.F.-%$cosGF,F2\"\"!" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 9 "Solution:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "dsolve(eq);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"yG6#%\"xG,&-%$sin GF&\"\"\"*&-%$expG6#,$-%$cosGF&!\"\"F+,&%$_C1GF+-%$IntG6$F-F'F+F3F3" } }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 2 "A " }{TEXT 259 6 "linear" }{TEXT -1 65 " (i.e. no products between y, y', y'',...) differential equatio n:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "eq := diff(y(x),x,x)+ 3/16*(x^2-x+1)/(x-1)^2/x^2*y(x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>% #eqG,&-%%diffG6$-%\"yG6#%\"xG-%\"$G6$F,\"\"#\"\"\"*&*(#\"\"$\"#;F1,(*$ )F,F0F1F1F,!\"\"F1F1F1F)F1F1*&),&F,F1F1F:F0F1F9F1F:F1" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 9 "Solution:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "dsolve(eq);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"y G6#%\"xG,&*(%$_C1G\"\"\")*&*&F'F+,&*$-%%sqrtG6#F'F+F+F+F+F+F+,&!\"\"F+ F0F+F5#F+\"\"%F+-F26#,&F'F+F+F5F+F+*(%$_C2GF+)*&*&F'F+F4F+F+F/F5F6F+F8 F+F+" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 171 "How does Maple solve suc h linear differential equations? The function y(x) depends on two arbi trary constants, so the set of all solutions is a 2-dimensional vector space." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 260 106 "The set of solutions of a linear differential equation of order n is always an n-dimensional vector space." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 317 "Sometimes we can find \"nice\" ex pressions for these solutions but not always. We can always calculate \+ the solutions as power series at a regular point. Take for example the point x=3, we can compute Taylor series of the solutions at x=3, so w e get power series in x-3. For brevity we'll use the letter T to denot e x-3." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "with(DEtools):" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "formal_sol(eq,y(x),T,x=3); " }}{PARA 12 "" 1 "" {XPPMATH 20 "6#7#7%+/%\"TG!\"\"\"\"!#\"\"(\"$%Q\" \"##!\"&\"$k)\"\"$#\"$8'\"'7\\H\"\"%#F.\"%Wh\"\"&-%\"OG6#\"\"\"\"\"'+- F&F;F;#!\"(\"%_6F0#F7\"%G " 0 "" {MPLTEXT 1 0 48 "y = Sum(c[i]*T^i,i=0..infinity); # (EQUATION_1)" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#/%\"yG-%$SumG6$*&&%\"cG6#%\"iG\"\"\")% \"TGF,F-/F,;\"\"!%)infinityG" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 13 "w here T is a " }{TEXT 270 15 "local parameter" }{TEXT -1 1 ":" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "T=x-p;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%\"TG,&%\"xG\"\"\"%\"pG!\"\"" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 301 "But at the point x=0 (i.e. T=x) at the point x=1 (i.e . T=x-1) and the point x=infinity (i.e. T=1/x) we can not write the so lutions like a convergent power series. We can only write y in that fo rm when T=x-p for all p except 0, 1, infinity (note that if p=infinity then x-p should be replaced by 1/x)." }}{PARA 0 "" 0 "" {TEXT -1 205 "In the solution of the equation you see that you'd need expansions of sqrt(x) and sqrt(x-1) and such. The series expansions of these sqrt(x ) and sqrt(x-1) look like EQUATION_1 except when p=0, 1, infinity." }} {PARA 0 "" 0 "" {TEXT -1 101 "So we can not express the solutions as p ower series at x=0, but we can do something that comes close:" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "formal_sol(eq,y(x),T,x=0);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#7$7$*&)%\"TG#\"\"$\"\"%\"\"\"+1F'F+ \"\"!#!\"\"\"\")F+#!\"*\"$G\"\"\"##!#\\\"%C5F)#!%t6\"&oF$F*#!%Ju\"'W@E \"\"&-%\"OG6#F+\"\"'F+/F'%\"xG7$*&)F'#F+F*F++1F'F+F-#!\"$F0F+#!#8F3F4# !#bF7F)#!%\\6F:F*#!%pmF=F>F?FBF+FC" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 69 "You see here that these two solutions (which form a basis) look like:" }}{PARA 0 "" 0 "" {TEXT -1 67 " T^e * (a formal power seri es in T with non-zero constant term)" }}{PARA 0 "" 0 "" {TEXT 265 154 "These two expressions are not a convergent power series (so the solut ions are not analytic at x=0) because then e would have to be a non-ne gative integer." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 97 "The numbers e, the exponents of the factor T^e in the sol ution T^e*(1+higher terms) is called an " }{TEXT 266 42 "exponent of t he equation at the point x=0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 175 "The exponents at x=0 are e=1/4 and e=3/4 . This means that the asymptotic behavior of the solutions at x=0 is \+ \"just like\" the asymptotic behavior of x^(1/4) and x^(3/4) at x=0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 50 "Lets lo ok at the exponents at x=1. Then T=x-p=x-1." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "formal_sol(eq,y(x),T,x=1,terms=3);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#7$7$*&)%\"TG#\"\"$\"\"%\"\"\"++F'F+\"\"!#F+\"\")F+ #!\"*\"$G\"\"\"#-%\"OG6#F+F)F+/F',&%\"xGF+F+!\"\"7$*&)F'#F+F*F+++F'F+F -#F)F/F+#!#8F2F3F4F)F+F7" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 54 "We se e that at x=1 the exponents are also 1/4 and 3/4." }}{PARA 0 "" 0 "" {TEXT -1 29 "Now at x=infinity. Note that:" }}{PARA 0 "" 0 "" {TEXT -1 46 " x=p <===> T=0 (if we take T=x-p)" }}{PARA 0 "" 0 "" {TEXT -1 45 " x=infinity <===> T=0 (if we take T=1/x)." }} {PARA 0 "" 0 "" {TEXT -1 102 "So to check out the point x=infinity we \+ need to write the solutions as an expansion in T, where T=1/x." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "formal_sol(eq,y(x),T,x=infin ity,terms=3);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7$7$*&++%\"TG\"\"\"\" \"!#!\"$\"\")F(#!#8\"$G\"\"\"#-%\"OG6#F(\"\"$F(*$)F'#F4\"\"%F(!\"\"/F' *&F(F(%\"xGF97$*&++F'F(F)#F9F,F(#!\"*F/F0F1F4F(*$)F'#F(F8F(F9F:" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 44 "We see that the exponents are -1/4 and -3/4." }}{PARA 0 "" 0 "" {TEXT -1 10 "Note that:" }}{PARA 0 "" 0 "" {TEXT -1 70 " A) when the exponent at x=p is negative then you hav e a pole at x=p." }}{PARA 0 "" 0 "" {TEXT -1 69 " B) when the exponent at x=p is positive then you have a root at x=p." }}{PARA 0 "" 0 "" {TEXT -1 341 " C) when you have convergent power series solutions at x =p, then all exponents are non-negative integers (but the converse is \+ not true, for example, it may happen that all exponents are non-negati ve integers but then it is still possible that you get a logarithm log (T) in the formal solution, and that's not convergent at x=p, i.e. at \+ T=0)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 267 58 "The exponents tell you the asymptotic behavior at a point." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 39 "How to solve th e differential equation?" }}{PARA 0 "" 0 "" {TEXT -1 3 "1) " }{TEXT 268 45 "Compute the exponents at all singular points." }}{PARA 0 "" 0 "" {TEXT -1 73 "--> Then we know the asymptotic behavior of the soluti ons at every point." }}{PARA 0 "" 0 "" {TEXT -1 90 " -> at regular p oints: no unusual asymptotic behavior, the solutions should be analyti c." }}{PARA 0 "" 0 "" {TEXT -1 83 " -> at singular points, the solut ion should behave as specified by the exponents." }}{PARA 0 "" 0 "" {TEXT -1 232 "For example, to solve that differential equation we firs t calculate the exponents at x=0. Once we do that, and we have the exp onents e1, e2, then we know that one solution must behave just like x^ e1 and another like x^e2 when x-->0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 3 "2) " }{TEXT 269 36 "Refine the search by linear algebra." }}{PARA 0 "" 0 "" {TEXT -1 373 "We search for soluti ons inside certain classes of functions. These exponents give big rest rictions on where to search. They may not always pinpoint exactly what the solution is (although in the above example they in fact do determ ine the solution), but often it restricts the search-space so much tha t afterwards the solutions can be pinpointed by solving linear equatio ns." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 272 45 "E xample of solving an equation via 1) and 2)." }}{PARA 0 "" 0 "" {TEXT -1 122 "We'll start with a linear differential equation with rational \+ functions as coefficients, and then we search for so-called " }{TEXT 271 22 "exponential solutions " }{TEXT -1 32 "which are solutions of t he form:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "y = P * exp(Int (r,x));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%\"yG*&%\"PG\"\"\"-%$expG6 #-%$IntG6$%\"rG%\"xGF'" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 125 "where \+ r is a rational function and P is a polynomial. Note that the previous example eq does not have solutions of this kind:" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 18 "expsols(eq, y(x));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 147 "eq:= -4/9*(x^3-13*x^2-21*x+9)*x/(x+1)^3/(x-1)^3/(x+3)*y(x)+2/9*(x^3-45*x^2- 81*x-27)/(x+3)/(x-1)^2/(x+1)^2*diff(y(x),x)+diff(diff(diff(y(x),x),x), x);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%#eqG,(*&*(,**$)%\"xG\"\"$\"\" \"F-*&\"#8F-)F+\"\"#F-!\"\"*&\"#@F-F+F-F2\"\"*F-F-F+F--%\"yG6#F+F-F-*( ),&F+F-F-F-F,F-),&F+F-F-F2F,F-,&F+F-F,F-F-F2#!\"%F5*&*(#F1F5F-,*F)F-*& \"#XF-F0F-F2*&\"#\")F-F+F-F2\"#FF2F--%%diffG6$F6F+F-F-*(F>F-)F=F1F-)F; F1F-F2F--FK6$F6-%\"$G6$F+F,F-" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 20 " Singular points are:" }}{PARA 0 "" 0 "" {TEXT -1 4 "x=-1" }}{PARA 0 " " 0 "" {TEXT -1 3 "x=1" }}{PARA 0 "" 0 "" {TEXT -1 4 "x=-3" }}{PARA 0 "" 0 "" {TEXT -1 10 "x=infinity" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 18 "The exponents are:" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 50 "gen_exp(eq,y(x),T,x=-1); exp_1 := \{seq(i[1],i =%)\};" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7%7$\"\"#/%\"TG,&%\"xG\"\"\" F*F*7$#F*\"\"$F&7$#F%F-F&" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&exp_1G <%\"\"##\"\"\"\"\"$#F&F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "gen_exp(eq,y(x),T,x=1); exp1 := \{seq(i[1],i=%)\};" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#7%7$#\"\"\"\"\"$/%\"TG,&%\"xGF&F&!\"\"7$F'F(7$#F,F'F( " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%exp1G<%\"\"$#\"\"\"F&#!\"\"F&" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "gen_exp(eq,y(x),T,x=-3); \+ exp_3 := \{seq(i[1],i=%)\};" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7#7&\" \"!\"\"\"\"\"#/%\"TG,&%\"xGF&\"\"$F&" }}{PARA 11 "" 1 "" {XPPMATH 20 " 6#>%&exp_3G<#\"\"!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "gen_e xp(eq,y(x),T,x=infinity); expi := \{seq(i[1],i=%)\};" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7%7$!\"#/%\"TG*&\"\"\"F)%\"xG!\"\"7$#F+\"\"$F&7$#F%F .F&" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%expiG<%!\"##!\"\"\"\"$#F&F) " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 159 "Note that if you have two ex ponents e1, e2 and if e2-e1 is a non-negative integer, then we may jus t as well throw e2 away, because if you have a solution like:" }} {PARA 0 "" 0 "" {TEXT -1 25 " T^e2 * power series in T" }}{PARA 0 "" 0 "" {TEXT -1 70 "and if e2 = e1 + non-negative integer then you can a lso write this as:" }}{PARA 0 "" 0 "" {TEXT -1 25 " T^e1 * power serie s in T" }}{PARA 0 "" 0 "" {TEXT -1 109 "although this latter power ser ies no longer has a non-zero constant term, but that will right now no t matter." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 52 "Now lets compare this with the candidate solution y:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 62 "y = P * exp(Int( e[-1]/(x+1) + e[1] /(x-1) + e[-3]/(x+3) , x));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%\"yG* &%\"PG\"\"\"-%$expG6#-%$IntG6$,(*&&%\"eG6#!\"\"F',&%\"xGF'F'F'F3F'*&&F 16#F'F',&F5F'F'F3F3F'*&&F16#!\"$F',&F5F'\"\"$F'F3F'F5F'" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 89 "where e[-1] e[1] and e[-3] are the expone nts of that exp(...) at the points -1, 1 and -3." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 202 "Now P is a polynomial. I ts exponent at x=-1, 1, -3 are non-negative integers. The exponent at \+ x=p is positive when P vanishes at x=p, and the exponent is simply the order of the P at the point x=p. So:" }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 40 "The exponent of y at x=-1 is the sum \+ of:" }}{PARA 0 "" 0 "" {TEXT -1 68 "*) the exponent of P at x=-1 (whic h is a non-negative integer), and:" }}{PARA 0 "" 0 "" {TEXT -1 55 "*) \+ the exponent of exp(...) at x=-1, and that is e[-1]." }}{PARA 0 "" 0 " " {TEXT -1 24 "Same goes for x=1, x=-3." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 46 "The exponent of y at x=infinity is t he sum of:" }}{PARA 0 "" 0 "" {TEXT -1 74 "*) the exponent of exp(...) at infinity, which is: -e[-1]-e[1]-e[-3], and:" }}{PARA 0 "" 0 "" {TEXT -1 57 "*) the exponent of P at infinity, which is: -degree(P,x). " }}{PARA 0 "" 0 "" {TEXT -1 216 "So the exponent of P at infinity is \+ a non-positive number. It can be either 0 when P is a constant, or neg ative when P is a non-constant polynomial (then P has a pole at x=infi nity, which implies: negative exponent)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 40 "We now have the following possibil ities:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 124 "for i in exp_1 d o for j in exp1 do for k in exp_3 do for l in expi do lprint(e[-1],e[1 ],e[-3],e[inf] = i,j,k,l) od od od od;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}{PARA 6 "" 1 "" {TEXT -1 40 "e[-1], e[1], e[-3], e[inf] = 2 , 3, 0, -2" }}{PARA 6 "" 1 "" {TEXT -1 42 "e[-1], e[1], e[-3], e[inf] \+ = 2, 3, 0, -1/3" }}{PARA 6 "" 1 "" {TEXT -1 42 "e[-1], e[1], e[-3], e[ inf] = 2, 3, 0, -2/3" }}{PARA 6 "" 1 "" {TEXT -1 42 "e[-1], e[1], e[-3 ], e[inf] = 2, 1/3, 0, -2" }}{PARA 6 "" 1 "" {TEXT -1 44 "e[-1], e[1], e[-3], e[inf] = 2, 1/3, 0, -1/3" }}{PARA 6 "" 1 "" {TEXT -1 44 "e[-1] , e[1], e[-3], e[inf] = 2, 1/3, 0, -2/3" }}{PARA 6 "" 1 "" {TEXT -1 43 "e[-1], e[1], e[-3], e[inf] = 2, -1/3, 0, -2" }}{PARA 6 "" 1 "" {TEXT -1 45 "e[-1], e[1], e[-3], e[inf] = 2, -1/3, 0, -1/3" }}{PARA 6 "" 1 "" {TEXT -1 45 "e[-1], e[1], e[-3], e[inf] = 2, -1/3, 0, -2/3" }} {PARA 6 "" 1 "" {TEXT -1 42 "e[-1], e[1], e[-3], e[inf] = 1/3, 3, 0, - 2" }}{PARA 6 "" 1 "" {TEXT -1 44 "e[-1], e[1], e[-3], e[inf] = 1/3, 3, 0, -1/3" }}{PARA 6 "" 1 "" {TEXT -1 44 "e[-1], e[1], e[-3], e[inf] = \+ 1/3, 3, 0, -2/3" }}{PARA 6 "" 1 "" {TEXT -1 44 "e[-1], e[1], e[-3], e[ inf] = 1/3, 1/3, 0, -2" }}{PARA 6 "" 1 "" {TEXT -1 46 "e[-1], e[1], e[ -3], e[inf] = 1/3, 1/3, 0, -1/3" }}{PARA 6 "" 1 "" {TEXT -1 46 "e[-1], e[1], e[-3], e[inf] = 1/3, 1/3, 0, -2/3" }}{PARA 6 "" 1 "" {TEXT -1 45 "e[-1], e[1], e[-3], e[inf] = 1/3, -1/3, 0, -2" }}{PARA 6 "" 1 "" {TEXT -1 47 "e[-1], e[1], e[-3], e[inf] = 1/3, -1/3, 0, -1/3" }}{PARA 6 "" 1 "" {TEXT -1 47 "e[-1], e[1], e[-3], e[inf] = 1/3, -1/3, 0, -2/3 " }}{PARA 6 "" 1 "" {TEXT -1 42 "e[-1], e[1], e[-3], e[inf] = 2/3, 3, \+ 0, -2" }}{PARA 6 "" 1 "" {TEXT -1 44 "e[-1], e[1], e[-3], e[inf] = 2/3 , 3, 0, -1/3" }}{PARA 6 "" 1 "" {TEXT -1 44 "e[-1], e[1], e[-3], e[inf ] = 2/3, 3, 0, -2/3" }}{PARA 6 "" 1 "" {TEXT -1 44 "e[-1], e[1], e[-3] , e[inf] = 2/3, 1/3, 0, -2" }}{PARA 6 "" 1 "" {TEXT -1 46 "e[-1], e[1] , e[-3], e[inf] = 2/3, 1/3, 0, -1/3" }}{PARA 6 "" 1 "" {TEXT -1 46 "e[ -1], e[1], e[-3], e[inf] = 2/3, 1/3, 0, -2/3" }}{PARA 6 "" 1 "" {TEXT -1 45 "e[-1], e[1], e[-3], e[inf] = 2/3, -1/3, 0, -2" }}{PARA 6 "" 1 " " {TEXT -1 47 "e[-1], e[1], e[-3], e[inf] = 2/3, -1/3, 0, -1/3" }} {PARA 6 "" 1 "" {TEXT -1 47 "e[-1], e[1], e[-3], e[inf] = 2/3, -1/3, 0 , -2/3" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 52 "For each, we can calcul ate the degree of P, because:" }}{PARA 0 "" 0 "" {TEXT -1 43 "e[infini ty] = -degree(P) - e[-1]-e[1]-e[-3]" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 138 "for i in exp_1 do for j in exp1 do for k in exp_3 do for l in expi do lprint(e[-1],e[1],e[-3],e[inf],degP = i,j,k,l,-i-j-k -l) od od od od;" }}{PARA 6 "" 1 "" {TEXT -1 50 "e[-1], e[1], e[-3], e [inf], degP = 2, 3, 0, -2, -3" }}{PARA 6 "" 1 "" {TEXT -1 55 "e[-1], e [1], e[-3], e[inf], degP = 2, 3, 0, -1/3, -14/3" }}{PARA 6 "" 1 "" {TEXT -1 55 "e[-1], e[1], e[-3], e[inf], degP = 2, 3, 0, -2/3, -13/3" }}{PARA 6 "" 1 "" {TEXT -1 54 "e[-1], e[1], e[-3], e[inf], degP = 2, 1 /3, 0, -2, -1/3" }}{PARA 6 "" 1 "" {TEXT -1 54 "e[-1], e[1], e[-3], e[ inf], degP = 2, 1/3, 0, -1/3, -2" }}{PARA 6 "" 1 "" {TEXT -1 56 "e[-1] , e[1], e[-3], e[inf], degP = 2, 1/3, 0, -2/3, -5/3" }}{PARA 6 "" 1 " " {TEXT -1 54 "e[-1], e[1], e[-3], e[inf], degP = 2, -1/3, 0, -2, 1/3 " }}{PARA 6 "" 1 "" {TEXT -1 57 "e[-1], e[1], e[-3], e[inf], degP = 2, -1/3, 0, -1/3, -4/3" }}{PARA 6 "" 1 "" {TEXT -1 55 "e[-1], e[1], e[-3 ], e[inf], degP = 2, -1/3, 0, -2/3, -1" }}{PARA 6 "" 1 "" {TEXT -1 54 "e[-1], e[1], e[-3], e[inf], degP = 1/3, 3, 0, -2, -4/3" }}{PARA 6 "" 1 "" {TEXT -1 54 "e[-1], e[1], e[-3], e[inf], degP = 1/3, 3, 0, -1/3, \+ -3" }}{PARA 6 "" 1 "" {TEXT -1 56 "e[-1], e[1], e[-3], e[inf], degP = \+ 1/3, 3, 0, -2/3, -8/3" }}{PARA 6 "" 1 "" {TEXT -1 55 "e[-1], e[1], e[- 3], e[inf], degP = 1/3, 1/3, 0, -2, 4/3" }}{PARA 6 "" 1 "" {TEXT -1 58 "e[-1], e[1], e[-3], e[inf], degP = 1/3, 1/3, 0, -1/3, -1/3" }} {PARA 6 "" 1 "" {TEXT -1 55 "e[-1], e[1], e[-3], e[inf], degP = 1/3, 1 /3, 0, -2/3, 0" }}{PARA 6 "" 1 "" {TEXT -1 54 "e[-1], e[1], e[-3], e[i nf], degP = 1/3, -1/3, 0, -2, 2" }}{PARA 6 "" 1 "" {TEXT -1 58 "e[-1], e[1], e[-3], e[inf], degP = 1/3, -1/3, 0, -1/3, 1/3" }}{PARA 6 "" 1 " " {TEXT -1 58 "e[-1], e[1], e[-3], e[inf], degP = 1/3, -1/3, 0, -2/3, \+ 2/3" }}{PARA 6 "" 1 "" {TEXT -1 54 "e[-1], e[1], e[-3], e[inf], degP = 2/3, 3, 0, -2, -5/3" }}{PARA 6 "" 1 "" {TEXT -1 57 "e[-1], e[1], e[-3 ], e[inf], degP = 2/3, 3, 0, -1/3, -10/3" }}{PARA 6 "" 1 "" {TEXT -1 54 "e[-1], e[1], e[-3], e[inf], degP = 2/3, 3, 0, -2/3, -3" }}{PARA 6 "" 1 "" {TEXT -1 53 "e[-1], e[1], e[-3], e[inf], degP = 2/3, 1/3, 0, - 2, 1" }}{PARA 6 "" 1 "" {TEXT -1 58 "e[-1], e[1], e[-3], e[inf], degP \+ = 2/3, 1/3, 0, -1/3, -2/3" }}{PARA 6 "" 1 "" {TEXT -1 58 "e[-1], e[1], e[-3], e[inf], degP = 2/3, 1/3, 0, -2/3, -1/3" }}{PARA 6 "" 1 "" {TEXT -1 56 "e[-1], e[1], e[-3], e[inf], degP = 2/3, -1/3, 0, -2, 5/3 " }}{PARA 6 "" 1 "" {TEXT -1 56 "e[-1], e[1], e[-3], e[inf], degP = 2/ 3, -1/3, 0, -1/3, 0" }}{PARA 6 "" 1 "" {TEXT -1 58 "e[-1], e[1], e[-3] , e[inf], degP = 2/3, -1/3, 0, -2/3, 1/3" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 120 "But a degree must be a non-negative integer, so whenever it is not, we throw that possibility away, and we're left with:" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 174 "for i in exp_1 do for j in \+ exp1 do for k in exp_3 do for l in expi do if type(-i-j-k-l,nonnegint) then lprint(e[-1],e[1],e[-3],e[inf],degP = i,j,k,l,-i-j-k-l) fi od od od od;" }}{PARA 6 "" 1 "" {TEXT -1 55 "e[-1], e[1], e[-3], e[inf], de gP = 1/3, 1/3, 0, -2/3, 0" }}{PARA 6 "" 1 "" {TEXT -1 54 "e[-1], e[1], e[-3], e[inf], degP = 1/3, -1/3, 0, -2, 2" }}{PARA 6 "" 1 "" {TEXT -1 53 "e[-1], e[1], e[-3], e[inf], degP = 2/3, 1/3, 0, -2, 1" }}{PARA 6 "" 1 "" {TEXT -1 56 "e[-1], e[1], e[-3], e[inf], degP = 2/3, -1/3, 0 , -1/3, 0" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 54 "Lets write down the \+ corresponding candidate solutions:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 337 "te:=0: for i in exp_1 do for j in exp1 do for k in e xp_3 do for l in expi do if type(-i-j-k-l,nonnegint) then e[-1],e[1],e [-3],e[infinity],degreeP := i,j,k,l,-i-j-k-l;\nP:=add(c[i]*x^i,i=0..de greeP); te:=te+1;\ncandidate[te]:= P*exp(Int( e[-1]/(x+1) + e[1]/(x-1) + e[-3]/(x+3) , x)); fi od od od od; for i to te do lprint(candidate[ i]) od;" }}{PARA 6 "" 1 "" {TEXT -1 36 "c[0]*exp(Int(1/3/(x+1)+1/3/(x- 1),x))" }}{PARA 6 "" 1 "" {TEXT -1 54 "(c[0]+c[1]*x+c[2]*x^2)*exp(Int( 1/3/(x+1)-1/3/(x-1),x))" }}{PARA 6 "" 1 "" {TEXT -1 45 "(c[0]+c[1]*x)* exp(Int(2/3/(x+1)+1/3/(x-1),x))" }}{PARA 6 "" 1 "" {TEXT -1 36 "c[0]*e xp(Int(2/3/(x+1)-1/3/(x-1),x))" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 132 "Now of these 4 candidates (some of which involve unknown constant s) lets determine which ones are actual solutions of the equations:" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "normal(subs(y(x)=candidate [1],eq));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"!" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 29 "so candidate[1] is a solution" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "normal(subs(y(x)=candidate[2],eq));" }} {PARA 12 "" 1 "" {XPPMATH 20 "6#,$*&*&-%$expG6#-%$IntG6$,$*&\"\"\"F.*& ,&%\"xGF.F.F.F.,&F1F.F.!\"\"F.F3#!\"#\"\"$F1F.,6*&)F1F6F.&%\"cG6#F.F.F 6*(\"$%>F.&F;6#\"\"#F.F9F.F.*(\"$w#F.F?F.)F1FAF.F.*(\"\"'F.FDF.&F;6#\" \"!F.F.*(\"#fF.F:F.FDF.F.*(\"#AF.FGF.F1F.F3*(\"#^F.F:F.F1F.F.*&\"#7F.F GF.F3*&\"#FF.F:F.F.*&\"$i\"F.F?F.F3F.F.*(,&F1F.F6F.F.)F0FAF.)F2FAF.F3# F5FS" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "factor(%/candidate[ 2]);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#,$*&,6*&)%\"xG\"\"$\"\"\"&%\"c G6#F*F*F)*(\"$%>F*&F,6#\"\"#F*F'F*F**(\"$w#F*F0F*)F(F2F*F**(\"\"'F*F5F *&F,6#\"\"!F*F**(\"#fF*F+F*F5F*F**(\"#AF*F8F*F(F*!\"\"*(\"#^F*F+F*F(F* F**&\"#7F*F8F*F?*&\"#FF*F+F*F**&\"$i\"F*F0F*F?F***,&F(F*F)F*F*),&F(F*F *F*F2F*),&F(F*F*F?F2F*,(F8F**&F+F*F(F*F**&F0F*F5F*F*F*F?#!\"#FE" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "coeffs(numer(%),x);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6&,(&%\"cG6#\"\"\"!#a*&\"$C$F'&F%6#\"\"# F'F'*&\"#CF'&F%6#\"\"!F'F',&F$!\"'*&\"$)QF'F+F'!\"\",(F+!$_&*&\"#7F'F0 F'F7*&\"$=\"F'F$F'F7,&F$!$-\"*&\"#WF'F0F'F'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "solve(\{%\});" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#< %/&%\"cG6#\"\"#\"\"!/&F&6#F)F)/&F&6#\"\"\"F)" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 32 "candidate 2 delivers no solution" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 27 "subs(y(x)=candidate[3],eq):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "normal(%/candidate[3]):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "solve(\{coeffs(numer(%),x)\});" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#<$/&%\"cG6#\"\"!F(/&F&6#\"\"\"F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 32 "candidate 3 delivers no solution" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "normal(subs(y(x)=candidate[4 ],eq));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"!" }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 25 "candidate 4 is a solution" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 43 "So the search lead to the following result:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "[cand idate[1], candidate[4]];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7$*&&%\"cG 6#\"\"!\"\"\"-%$expG6#-%$IntG6$,&*&F)F),&%\"xGF)F)F)!\"\"#F)\"\"$*&F5F ),&F3F)F)F4F4F)F3F)*&F%F)-F+6#-F.6$,&F1#\"\"#F6*&#F)F6F)*&F)F)F8F4F)F4 F3F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "simplify(value(subs (c[0]=1,%)));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7$*&),&%\"xG\"\"\"F(F (#F(\"\"$F(),&F'F(F(!\"\"F)F(*&*$)F&#\"\"#F*F(F(*$)F,#F(F*F(F-" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 39 "So we find the same solutions as M aple:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "expsols(eq,y(x)); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7$*$)*&,&%\"xG\"\"\"F)F)F),&F(F)F) !\"\"F)#F)\"\"$F)*&*&)F&#F)\"\"'F)-%%sqrtG6#F'F)F)*$-F46#F*F)F+" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 338 "Note that we find only 2 solution s, and that the solution space is 3-dimensional. So to have a basis, w e need 1 more solution. If you have n-1 linearly independent solutions of an n-dimensional solution space, then you can find that last one b y some general techniques, however, those methods do not lead to \"nic e\" answers as you can see:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "dsolve(eq);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#/-%\"yG6#%\"xG,(*&* &%$_C1G\"\"\"),&F'F,F,F,#\"\"#\"\"$F,F,*$),&F'F,F,!\"\"#F,F1F,F5F,*&%$ _C2GF,)*&F.F,F4F,#F,F1F,F,*&*&%$_C3GF,,&*&F-F,-%$IntG6$*&*&F9F,)F4F0F, F,*$),&F'F,F1F,F0F,F5F'F,F5*(-FB6$*&*&F-F,)F4#\"\"&F1F,F,*$FHF,F5F'F,F 9F,)F4F;F,F,F,F,*$)F4#F,F1F,F5F," }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 23 "Now some more examples:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 273 65 "A differential equation whose solutions involv e Bessel functions:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "eq : = diff(y(x),x,x)-y(x)/x^8;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#eqG,& -%%diffG6$-%\"yG6#%\"xG-%\"$G6$F,\"\"#\"\"\"*&F)F1*$)F,\"\")F1!\"\"F6 " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "dsolve(eq);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"yG6#%\"xG,&*(%$_C1G\"\"\"-%%sqrtG6#F'F+ -%(BesselIG6$#!\"\"\"\"',$*&F+F+*$)F'\"\"$F+F3#F3F9F+F+*(%$_C2GF+F,F+- %(BesselKG6$#F+F4F5F+F+" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 74 "Maple \+ finds this solution by \"pattern matching\". This means the following: " }}{PARA 0 "" 0 "" {TEXT -1 83 "1) Take all the standard books that c ontain lists of equations and their solutions." }}{PARA 0 "" 0 "" {TEXT -1 22 " e.g. Kamke's book." }}{PARA 0 "" 0 "" {TEXT -1 61 "So me of these equations involve parameters, like n, u, v, ..." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 198 "2) For each eq uation from the book, try to find out if there are values for the para meter such that for those values, the equation from the book is the sa me as the equation eq that we have to solve." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 171 "3) If you find parameter value s for which the pattern matches, then take the solution of the equatio n from the book, plug in those parameter values, and return the answer ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 274 29 "Str ength of pattern matching:" }}{PARA 0 "" 0 "" {TEXT -1 96 " --> Maple will solve any equation you find in any standard book. So at first si ght: very good!" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 275 29 "Weakness of pattern matching:" }}{PARA 0 "" 0 "" {TEXT -1 258 " --> After some relatively small modifications (such as differ entiating the solutions) the equation no longer looks similar to the o ne in the book. Only equations that are very close to the standard equ ation will be recognized, all others will not be solved." }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 276 26 "Solution to this \+ weakness:" }}{PARA 0 "" 0 "" {TEXT -1 364 " --> The exponents (or gene ralized exponents) modulo some equivalence (e and e+1 are equivalent) \+ remain the same under many transformations. So one has to recognize th e equation not simply by the fact that it is identical to an equation \+ in a book, but rather one has to calculate the exponents and check if \+ the exponents match the exponents of a standard equation." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "gen_exp(eq,y(x),T,x=0);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7$7$,&*&\"\"\"F'*$)%\"TG\"\"$F'!\"\"F,\"\"#F '/F*%\"xG7$,&F&F'F-F'F." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 " e1,e2 := subs(T=x,%[1][1]), subs(T=x,%[2][1]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>6$%#e1G%#e2G6$,&*&\"\"\"F**$)%\"xG\"\"$F*!\"\"F/\"\"#F *,&F)F*F0F*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 155 "This tells us tha t the asymptotic behavior at T=0 (i.e. at x=0) of the solutions of the differential equation eq is the same as the asymptotic behavior of:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "exp(Int(e1/x,x)), exp(Int (e2/x,x));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$-%$expG6#-%$IntG6$*&,&*& \"\"\"F,*$)%\"xG\"\"$F,!\"\"F1\"\"#F,F,F/F1F/-F$6#-F'6$*&,&F+F,F2F,F,F /F1F/" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 31 "This asymptotic behavior is an " }{TEXT 277 21 "essential singularity" }{TEXT -1 74 " at x=0. \+ The singularity x=0 of the differential equation is then called: " } {TEXT 278 19 "irregular singular." }}{PARA 0 "" 0 "" {TEXT -1 22 "At x =infinity we have:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "gen_e xp(eq,y(x),T,x=infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7#7%!\"\" \"\"!/%\"TG*&\"\"\"F*%\"xGF%" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 82 "I n other words: at x=infinity (i.e. at T=0 where T=1/x) the solutions b ehave like:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "T^(-1), T^0 ;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$*&\"\"\"F$%\"TG!\"\"F$" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 6 "so as:" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 21 "(1/x)^(-1), (1/x)^0 ;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%\"xG\"\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 195 "So one of t he solutions has a pole of order 1 at x=infinity, and the other has no pole, no root. So up to multiplication by a rational function T=1/x, \+ there is no singular behavior at x=infinity." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "formal_sol(eq,y(x),T,x=infinity,terms=3);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#7#7%*&+'%\"TG!\"\"\"\"!-%\"OG6#\"\"\" \"\"$F-F'F(*&+'F'F-F-F*F.F-F'F(/F'*&F-F-%\"xGF(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 211 "Note that if we allow the solutions to be multiplie d by rational functions, that is the same thing as allowing to add arb itrary integers to the exponents. So we should consider the exponents \+ modulo the integers." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 29 "Now lets modify the equation:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 3 "eq;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&-%%diffG 6$-%\"yG6#%\"xG-%\"$G6$F*\"\"#\"\"\"*&F'F/*$)F*\"\")F/!\"\"F4" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 91 "new_eq:=diff(diff(y(x),x),x) +2*(x^6+16)/(x^6+4)/x*diff(y(x),x)+(11*x^6-4)/(x^6+4)/x^8*y(x);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%'new_eqG,(-%%diffG6$-%\"yG6#%\"xG-% \"$G6$F,\"\"#\"\"\"*&*(F0F1,&*$)F,\"\"'F1F1\"#;F1F1-F'6$F)F,F1F1*&,&F5 F1\"\"%F1F1F,F1!\"\"F1*&*&,&F5\"#6F=F>F1F)F1F1*&FF1" }} }{EXCHG {PARA 0 "" 0 "" {TEXT -1 157 "For each solution y of eq, we ha ve that y'-y/(2*x) is a solution of new_eq (that's how the new equatio n was constructed, more about this construction later)." }}{PARA 0 "" 0 "" {TEXT -1 22 "Now y --> y' - y/(2*x)" }}{PARA 0 "" 0 "" {TEXT -1 122 "is a fairly simple transformation but it's already too hard for M aple to still recognize the equation by pattern matching." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "SOL_eq := rhs(dsolve(eq));" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%'SOL_eqG,&*(%$_C1G\"\"\"-%%sqrtG6#% \"xGF(-%(BesselIG6$#!\"\"\"\"',$*&F(F(*$)F,\"\"$F(F1#F1F7F(F(*(%$_C2GF (F)F(-%(BesselKG6$#F(F2F3F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "SOL_new_eq:=diff(SOL_eq,x) - SOL_eq/(2*x):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "normal(subs(y(x)=SOL_new_eq, new_eq)); # shou ld be 0" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "dsolve(new_eq); # DESol is not much of a \"so lution\"" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#/-%\"yG6#%\"xG-%&DESolG6$< #,(*&,&*$)F'\"#9\"\"\"F2*&\"\"%F2)F'\"\")F2F2F2-%%diffG6$-%#_YGF&-%\"$ G6$F'\"\"#F2F2*&,&*$)F'\"#8F2F?*&\"#KF2)F'\"\"(F2F2F2-F86$F:F'F2F2*&,& *$)F'\"\"'F2\"#6F4!\"\"F2F:F2F2<#F:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 51 "How could one recognize how to solve this equation:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "gen_exp(new_eq,y(x),T,x=0);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#7$7$,&*&\"\"\"F'*$)%\"TG\"\"$F'!\"\"F, \"\"#F,/F*%\"xG7$,&F&F'F-F,F." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 71 " Modulo the integers, this is the same as before, because before it was :" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "gen_exp(eq,y(x),T,x=0) ;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7$7$,&*&\"\"\"F'*$)%\"TG\"\"$F'! \"\"F,\"\"#F'/F*%\"xG7$,&F&F'F-F'F." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 49 "And the same is true for the other singularities:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "gen_exp(new_eq,y(x),T,x=RootOf(x^6+ 4));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7#7%\"\"!\"\"#/%\"TG,&%\"xG\" \"\"-%'RootOfG6#,&*$)%#_ZG\"\"'F+F+\"\"%F+!\"\"" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 35 "gen_exp(eq,y(x),T,x=RootOf(x^6+4));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7#7%\"\"!\"\"\"/%\"TG,&%\"xGF&-%'RootOfG6#,& *$)%#_ZG\"\"'F&F&\"\"%F&!\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 42 " Modulo the integers, that's also the same." }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 34 "gen_exp(new_eq,y(x),T,x=infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7#7%\"\"!\"\"\"/%\"TG*&F&F&%\"xG!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "gen_exp(eq,y(x),T,x=infinity);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#7#7%!\"\"\"\"!/%\"TG*&\"\"\"F*%\"xGF% " }}}{EXCHG {PARA 0 "" 0 "" {TEXT 279 48 "To solve the new_eq we have \+ to do the following:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 250 "1) Calculate the exponents modulo the integers --> thi s gives us the asymptotic behavior up to multiplication by rational fu nctions. These things are invariant under: differentiation of the sol utions, as well as multiplication by rational functions." }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 55 "2) From the expone nts, determine the standard equation." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 118 "3) Find the transformation between t he given and the standard equation, which in this example is: y --> y ' - y/(2*x)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 31 "4) Solve the standard equation." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 97 "5) Apply the transformation on the so lutions of the standard equation, and return that as answer." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 419 "Item 3) is mat hematically interesting. One can think of the solution spaces as G-mod ules where G is an algebraic group (the differential Galois group). Th en 3) consists of computing a nonzero homomorphism (as G-modules) from one solution space to the other. Maple currently has code for endomor phisms. We can also compute homomorphisms but that's not yet included \+ in the standard distribution, so we need code for that." }}}}{MARK "85 0 0" 51 }{VIEWOPTS 1 1 0 2 1 1805 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }