{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 Comment" 2 18 "" 0 1 0 0 0 0 0 0 0 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 "" 0 1 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 "" -1 261 "" 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 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 273 "" 0 1 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 "" 1 24 0 0 0 0 0 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 277 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 278 "" 1 18 0 0 0 0 0 0 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 }{CSTYLE "" -1 280 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 281 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 282 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 283 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 284 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 285 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 286 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 287 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 288 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 289 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 290 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 291 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 292 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 293 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 294 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 295 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 296 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 297 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 298 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 299 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 300 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 301 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 302 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 303 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 304 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 305 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 306 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 307 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 308 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 309 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 310 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 311 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 312 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 313 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 314 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 315 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 316 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 317 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 318 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 319 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 320 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 321 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 322 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 323 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 324 "" 1 18 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 325 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 326 "" 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 "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 77 "Decomposin g a 4'th order linear differential equation as a symmetric product." } }{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 118 "by:\n \+ Mark van Hoeij\n Florida State University\n hoe ij@math.fsu.edu\n May 2001.\n " }}{PARA 0 "" 0 "" {TEXT -1 10 "Let L in " }{TEXT 258 1 "C" }{TEXT -1 4 "(x)[" } {XPPEDIT 18 0 "d/dx;" "6#*&%\"dG\"\"\"%#dxG!\"\"" }{TEXT -1 109 "] be \+ a linear homogeneous differential operator with rational functions as \+ coefficients. Denote V(L) = \{y in " }{XPPEDIT 18 0 "Omega;" "6#%&Omeg aG" }{TEXT -1 20 " | L(y) = 0\} as the " }{TEXT 257 14 "solution space " }{TEXT -1 13 " of L. Here " }{XPPEDIT 18 0 "Omega" "6#%&OmegaG" } {TEXT -1 39 " is a differential field that contains " }{TEXT 259 1 "C " }{TEXT -1 69 "(x) as well as n linearly independent solutions of L, \+ where n is the " }{TEXT 326 5 "order" }{TEXT -1 51 " of L (the highest derivative in L). We could take " }{XPPEDIT 18 0 "Omega" "6#%&OmegaG " }{TEXT -1 201 " as the field of fractions of the analytic functions \+ at x=p, where p is chosen as a point that will be a regular point for \+ all operators L under consideration. Then the dimension of V(L) is n=o rder(L)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 38 "If L1 and L2 are operators, then the " }{TEXT 261 18 "symmetric p roduct " }{TEXT -1 54 "of L1,L2 is defined as the monic operator L suc h that:" }}{PARA 0 "" 0 "" {TEXT -1 31 " V(L) = SPAN\{ y1*y 2 " }{TEXT 323 0 "" }{TEXT 324 1 "|" }{TEXT -1 31 " y1 in V(L1), y2 in V(L2) \}." }}{PARA 0 "" 0 "" {TEXT -1 37 "It is known that such op erator L in " }{TEXT 262 1 "C" }{TEXT -1 78 "(x)[d/dx] exists and is \+ unique. If n1, n2, n are the orders of L1, L2, L then:" }}{PARA 0 "" 0 "" {TEXT -1 43 " n1 + n2 - 1 <= n <= n1 * n2" }} {PARA 0 "" 0 "" {TEXT -1 74 "Here is an example how the symmetric prod uct L can be computed with maple:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT 284 10 "Example A:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 102 "restart:\nwith(DEtools):\n_Envdiffopdomain := [Dx, x ]:\n # remark: this means that d/dx is denoted by Dx\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "L1 := Dx^2+2/x*Dx+x;" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#>%#L1G,(*$)%#DxG\"\"#\"\"\"F**&*&F)F*F(F*F*%\"xG !\"\"F*F-F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "L2 := Dx^2 - x;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#L2G,&*$)%#DxG\"\"#\"\"\"F*% \"xG!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "L := symmetric _product(L1,L2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"LG,**$)%#DxG\" \"%\"\"\"F**&*&\"\"$F*)F(F-F*F*%\"xG!\"\"F**&*&F-F*)F(\"\"#F*F**$)F/F4 F*F0F0*&F)F*F6F*F*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 120 "The equati ons corresponding to L1 and L2 are given by the command diffop2de. If \+ y1(x) and y2(x) satisfy these equations:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "diffop2de(L1,y1(x))=0; diffop2de(L2,y2(x))=0;" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#/,(*&%\"xG\"\"\"-%#y1G6#F&F'F'*&*&\"\" #F'-%%diffG6$F(F&F'F'F&!\"\"F'-F/6$F(-%\"$G6$F&F-F'\"\"!" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#/,&*&%\"xG\"\"\"-%#y2G6#F&F'!\"\"-%%diffG6$F(-% \"$G6$F&\"\"#F'\"\"!" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 76 "then y(x) := y1(x)*y2(x) will satisfy the equation of the symmetric product:" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "diffop2de(L,y(x))=0;" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#/,**&)%\"xG\"\"#\"\"\"-%\"yG6#F'F)\"\" %*&*&\"\"$F)-%%diffG6$F*-%\"$G6$F'F(F)F)*$F&F)!\"\"F8*&*&F0F)-F26$F*-F 56$F'F0F)F)F'F8F)-F26$F*-F56$F'F-F)\"\"!" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 41 "For brevity, write symmetric_product as: " }{TEXT 263 4 " sp. " }{TEXT -1 22 "It is well known that:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 9 " (1) " }{TEXT 266 2 "sp" } {TEXT -1 68 "(Dx - d, Dx^n + a*Dx^(n-1) + ... ) = Dx^n + (a-n*d)*Dx^ (n-1) + ..." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 150 "where the dots refer to terms with lower order derivatives. I f y is non-zero solution of Dx-d then 1/y is a solution of Dx+d, and s o it follows that:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 8 " (2) " }{TEXT 269 2 "sp" }{TEXT -1 3 "( " }{TEXT 264 2 "sp" }{TEXT -1 14 "(Dx+d, L1), " }{TEXT 265 2 "sp" }{TEXT -1 18 "( Dx-d, L2) ) = " }{TEXT 267 3 " sp" }{TEXT -1 10 "( L1, L2 )" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 38 "for any o perators L1, L2 and any d in " }{TEXT 268 1 "C" }{TEXT -1 9 "(x). If " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 13 " (3) L = " }{TEXT 271 2 "sp" }{TEXT -1 41 "( Dx^2 + a1*Dx + b1, Dx^2 + a2*Dx + b2 )" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 24 "for some a1,a2,b1,b2 in " }{TEXT 325 1 "C" }{TEXT -1 89 " (x), then by taking d=(a2-a1)/4 in equation (2) it follows that there \+ exist a, B1, B2 in " }{TEXT 270 1 "C" }{TEXT -1 13 "(x) such that" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 12 " (4) L = " }{TEXT 272 3 "sp(" }{TEXT -1 38 " Dx^2 + a*Dx + B1, Dx^2 + a*Dx \+ + B2 )" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 144 "Note that if a,B1,B2 satisfies (4) then so does a,B2,B1. We wish \+ to reduce the number of possibilities. For this purpose we will rewrit e (4) as:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 11 " (5) L = " }{TEXT 286 2 "sp" }{TEXT -1 12 "(L1, L2) = " }{TEXT 273 3 "sp(" }{TEXT -1 56 " Dx^2 + a*Dx + b + sqrt(c), Dx^2 + a*Dx + b - sqrt(c) )" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 69 "where a=(a1+a2)/2, b=(B1+B2)/2, and c=((B1-B2)/2)^2 are elemen ts of " }{TEXT 274 1 "C" }{TEXT -1 5 "(x). " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 276 13 "Main problem:" }}{PARA 0 "" 0 "" {TEXT -1 88 "The problem to be solved in this document is the follo wing: Given a monic operator L in " }{TEXT 310 1 "C" }{TEXT -1 53 "(x) [d/dx] of order 4, decide if there exist a,b,c in " }{TEXT 275 1 "C" } {TEXT -1 57 "(x) such that equation (5) holds. If so, find such a,b,c. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 277 12 "Appl ication:" }}{PARA 0 "" 0 "" {TEXT -1 198 "If we can find such a,b,c, t hen solving L has been reduced to solving two second order operators L 1, L2. A basis of V(L) is then obtained by multiplying the elements of the bases of V(L1) and V(L2)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT 278 30 "Easy subproblem: The case a=0." }}{PARA 0 "" 0 "" {TEXT -1 91 "We will first calculate formulas for finding b, c under the simplifying assumption that a=0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 11 " (6) L = " }{TEXT 287 2 "sp" }{TEXT -1 11 "(L1, L2) = " }{TEXT 279 3 "sp(" }{TEXT -1 43 " Dx^2 + b \+ + sqrt(c), Dx^2 + b - sqrt(c) )" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "L:=symmetric_product(Dx^2+b( x)+sqrt(c(x)), Dx^2+b(x)-sqrt(c(x)));" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%\"LG,,*$)%#DxG\"\"%\"\"\"F**&#F*\"\"#F**&*&-%%diffG6$-%\"cG6#%\"x GF6F*)F(\"\"$F*F*F3!\"\"F*F9*(F)F*-%\"bGF5F*)F(F-F*F**&*(F-F*,&*&F0F*F ;F*F**(F8F*-F16$F;F6F*F3F*F9F*F(F*F*F3F9F9*&,(*&F0F*FCF*F9*&F)F*)F3F-F *F**(F-F*-F16$F;-%\"$G6$F6F-F*F3F*F*F*F3F9F*" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 111 "Now we can write the coefficient of Dx^3 as:\n (7) \+ C = -1/2 * c'/c\nso that we can substitute the following" }}{PARA 0 "" 0 "" {TEXT -1 22 " (7') c' = -2*C*c" }}{PARA 0 "" 0 "" {TEXT -1 7 "into L:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "coll ect(subs( diff(c(x),x) = -2*C(x)*c(x),L),Dx,normal);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,0*$)%#DxG\"\"%\"\"\"F(*&-%\"CG6#%\"xGF()F&\"\"$F(F( *(F'F(-%\"bGF,F()F&\"\"#F(F(*&,&*&F*F(F1F(F'*&\"\"'F(-%%diffG6$F1F-F(F (F(F&F(F(*(F4F(F*F(F:F(F(*&F'F(-%\"cGF,F(F(*&F4F(-F;6$F1-%\"$G6$F-F4F( F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 84 "So, if L is of the form (6) , then the values of C(x) and b(x) are easily determined:" }}{PARA 0 " " 0 "" {TEXT -1 49 "(8) C(x) = coeff(L,Dx,3) = coefficient of Dx^3 " }}{PARA 0 "" 0 "" {TEXT -1 33 "(9) b(x) = 1/4 * coeff(L,Dx,2)" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 55 "At this p oint, we can use the coefficient of Dx^1 as a " }{TEXT 280 11 "first c heck" }{TEXT -1 261 ": If this coefficient is not equal to 4*C(x)*b(x) +6*diff(b(x),x) then L can not be of the form (6) for any functions b, c in any differential field. Assume that the coefficient of Dx^1 passe s the check. Then we need to calculate c(x). It is clear how to find i t:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 82 " \+ c(x) = 1/4*( coeff(L,Dx,0) - 2*C(x)*diff(b(x),x) - 2*diff(b( x),x,x) )" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 29 "Then equation (7) gives us a " }{TEXT 281 13 "second check:" } {TEXT -1 214 " test if C(x) equals -1/2*diff(c(x),x)/c(x). Again, if \+ this test fails, then L can not have form (6) for any b,c. It is easy \+ to see that if L does have form (6), then it will pass both tests and \+ b,c will be found." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 135 "We will now illustrate the a=0 subproblem by two example s. In the first example, L is of the form (6). In the second example i t is not." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 282 10 "Example 1." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "L := \+ Dx^4-2/x*Dx^3+4*Dx^2-8/x*Dx+4*x^4;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6# >%\"LG,,*$)%#DxG\"\"%\"\"\"F**&*&\"\"#F*)F(\"\"$F*F*%\"xG!\"\"F1*&F)F* )F(F-F*F**&*&\"\")F*F(F*F*F0F1F1*&F)F*)F0F)F*F*" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 40 "C:=coeff(L,Dx,3); b:=1/4*coeff(L,Dx,2);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"CG,$*&\"\"\"F'%\"xG!\"\"!\"#" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"bG\"\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "4*C*b+6*diff(b,x) = coeff(L,Dx,1);" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#/,$*&\"\"\"F&%\"xG!\"\"!\")F$" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 110 "Now the first check is that this equation hold s, i.e. that lhs-rhs (LeftHandSide minus RightHandSide) equals 0" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "first_check := normal(lhs(%) -rhs(%));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%,first_checkG\"\"!" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "c := 1/4*( coeff(L,Dx,0) - 2 *C*diff(b,x) - 2*diff(b,x,x) );" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>% \"cG*$)%\"xG\"\"%\"\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 " C = -1/2*diff(c,x)/c;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/,$*&\"\"\"F& %\"xG!\"\"!\"#F$" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "second_ check := normal(lhs(%)-rhs(%));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%- second_checkG\"\"!" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 74 "Both checks pass, therefore L is the symmetric product of L1 and L2 below:" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "sqc := sqrt(c, symbolic);" } }{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$sqcG*$)%\"xG\"\"#\"\"\"" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "L1:=Dx^2 + b + sqc; L2:=Dx^2 + b - sqc;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#L1G,(*$)%#DxG\"\"#\" \"\"F*F*F**$)%\"xGF)F*F*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#L2G,(*$ )%#DxG\"\"#\"\"\"F*F*F**$)%\"xGF)F*!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "symmetric_product(L1,L2); # equals L" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,,*$)%#DxG\"\"%\"\"\"F(*&*&\"\"#F()F&\"\"$F(F(%\"x G!\"\"F/*&F'F()F&F+F(F(*&*&\"\")F(F&F(F(F.F/F/*&F'F()F.F'F(F(" }}} {EXCHG {PARA 0 "" 0 "" {TEXT 283 35 "Example 2 (same L as in example A )." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "L := Dx^4+3/x*Dx^3-3/ x^2*Dx^2+4*x^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"LG,**$)%#DxG\" \"%\"\"\"F**&*&\"\"$F*)F(F-F*F*%\"xG!\"\"F**&*&F-F*)F(\"\"#F*F**$)F/F4 F*F0F0*&F)F*F6F*F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "C:=co eff(L,Dx,3): b:=1/4*coeff(L,Dx,2):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "4*C*b+6*diff(b,x) = coeff(L,Dx,1):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "first_check := normal(lhs(%)-rhs(%));" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%,first_checkG\"\"!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "c:=1/4 * ( coeff(L,Dx,0) - 2*C*diff(b,x) \+ - 2*diff(b,x,x) ):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "C = - 1/2*diff(c,x)/c:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "second_ check := normal(lhs(%) - rhs(%));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#> %-second_checkG,$*&\"\"\"F'%\"xG!\"\"\"\"%" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 198 "The second check fails. So in example 2, there exist no \+ b,c for which equation (6) hold. However, this operator is the same as in example A. And so there do exist a,b,c such that equation (5) hold s." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 112 "Th ere is a well known trick to eliminate the coefficient of Dx^(n-1) whe re n=order(L). In Maple this is done by:" }}{PARA 0 "" 0 "" {TEXT -1 2 " " }{TEXT 285 2 "sp" }{TEXT -1 25 "(Dx - coeff(L,Dx,n)/n, L)" }} {PARA 0 "" 0 "" {TEXT -1 394 "In many algorithms for linear differenti al equations, the input is first \"normalized\" with this trick. The t wo second order operators L1, L2 in equation (5) are normalized precis ely when a=0. We could normalize L1,L2 if we knew what L1,L2 are, but \+ we only know L. We may hope that normalizing L would have the same eff ect, however, one easily finds out that that is not so by trying an ex ample:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 288 39 "Example 2a: normalize L from example 2." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "n:=4:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "L:=symmetric_product(Dx - coeff(L,Dx,n-1)/n, L);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"LG,**$)%#DxG\"\"%\"\"\"F**&#\"#:\"\")F**&*$)F(\"\"# F*F**$)%\"xGF2F*!\"\"F*F6*&*&#F-F.F*F(F*F**$)F5\"\"$F*F6F**&*&#F*\"$c# F*,&*$)F5\"\"'F*\"%C5\"$:$F6F*F**$)F5F)F*F6F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "C:=coeff(L,Dx,3): b:=1/4*coeff(L,Dx,2):" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "4*C*b+6*diff(b,x) = coeff(L, Dx,1):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "first_check := no rmal(lhs(%)-rhs(%));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%,first_check G,$*&\"\"\"F'*$)%\"xG\"\"$F'!\"\"#\"#:\"\"%" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 73 "The normalized L already fails the first check, so it i s not of form (6)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 316 13 "Remark: sp" }{TEXT -1 8 "(Dx-a, " } {TEXT 311 2 "sp" }{TEXT -1 13 "(L1,L2)) = " }{TEXT 312 2 "sp" } {TEXT -1 2 "( " }{TEXT 313 2 "sp" }{TEXT -1 15 "(Dx-a/2, L1), " } {TEXT 314 2 "sp" }{TEXT -1 14 "(Dx-a/2, L2) )" }}{PARA 0 "" 0 "" {TEXT -1 61 "So if L is of form (5), and if we knew the value of a, th en " }{TEXT 315 2 "sp" }{TEXT -1 58 "(Dx-a,L) is of form (6) and then the problem becomes easy." }}{PARA 0 "" 0 "" {TEXT 317 11 "Corollary: " }{TEXT -1 68 " Finding a (if it exists) is equivalent to solving th e main problem." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 76 "The corollary could be used in the following way: Apply a transformation L:=" }{TEXT 318 2 "sp" }{TEXT -1 545 "(Dx-a(x), L) wh ere a(x) is an undetermined function. Then compute first_check and sec ond_check. Both of these are non-linear differential equations in a(x) . We could then try to simplify these equations and hope to find a li near differential equation. From equation (8) we see that we could als o work with the Dx^(n-1) coefficient by using C(x) or c(x) (see equat ion (7)). Taking c(x) makes the equations much easier. This way, after determining equation first_check, we obtain a linear equation by redu cing it to an equation for c(x)^(-1/4)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 319 12 "Proposition:" }}{PARA 0 "" 0 "" {TEXT -1 48 "Let L = Dx^4 + A4*Dx^3 + A3*Dx^2 + A2*Dx + A1." }} {PARA 0 "" 0 "" {TEXT -1 66 "If L is of form (5), then c^(-1/4) is a \+ solution of the operator:" }}{PARA 0 "" 0 "" {TEXT -1 44 " L3 = 20* Dx^3 + (8*A3-12*A4'-3*A4^2)*Dx" }}{PARA 0 "" 0 "" {TEXT -1 54 " \+ + 12*A3'-8*A2+4*A3*A4-10*A4''-A4^3-9*A4*A4' " }}{PARA 0 "" 0 "" {TEXT 320 6 "Proof:" }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "restart: with(DEtools): _Envdiffopdomain:=[Dx,x]:" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 90 "L := symmetric_product(Dx^2 + a(x)*Dx + b(x)+sqrt(c(x)), Dx^2 +a(x)*Dx + b(x)-sqrt(c(x))):" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 62 "A[4]:=coeff(L,Dx,3): A[3]:=c oeff(L,Dx,2): A[2]:=coeff(L,Dx,1):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 133 "L3:=20*Dx^3 + (8*A[3]-12*diff(A[4],x)-3*A[4]^2)*Dx + 12*diff(A[3],x)-8*A[2]+4*A[3]*A[4]-10*diff(A[4],x,x)-A[4]^3-9*A[4]*di ff(A[4],x):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "sol:=c(x)^(- 1/4): diffop2de(L3,y(x)): subs(y(x)=sol,%):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "normal(%); # if 0 then the proof is completed." } }{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"!" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 75 "Note: If L is normalized (i.e. if A4=0) then we can use t he shorter formula" }}{PARA 0 "" 0 "" {TEXT -1 35 " L3 = 5/2*Dx^3+A 3*Dx+3/2*A3'-A2." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 65 "The following proposition shows how the formula for L3 wa s found:" }}{PARA 0 "" 0 "" {TEXT 290 12 "Proposition:" }}{PARA 0 "" 0 "" {TEXT -1 48 " i is a non-zero solution of L3 if and only if " } {TEXT 289 2 "sp" }{TEXT -1 55 "(Dx - 1/4 * A4 + 1/2 * i'/i, L) satisf ies first_check." }}{PARA 0 "" 0 "" {TEXT 291 7 "Proof: " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "L:=Dx^4+A4(x)*Dx^3+A3(x)*Dx^2+A2(x) *Dx+A1(x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"LG,,*$)%#DxG\"\"%\" \"\"F**&-%#A4G6#%\"xGF*)F(\"\"$F*F**&-%#A3GF.F*)F(\"\"#F*F**&-%#A2GF.F *F(F*F*-%#A1GF.F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "L:=sym metric_product(Dx-A4(x)/4 + diff(i(x),x)/i(x)/2, L):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "C:=coeff(L,Dx,3): b:=1/4*coeff(L,Dx,2): " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "4*C*b+6*diff(b,x) = coe ff(L,Dx,1):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "first_check \+ := primpart(numer(normal(lhs(%)-rhs(%))));" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%,first_checkG,6*&)-%#A4G6#%\"xG\"\"#\"\"\"-%%diffG6$- %\"iGF*F+F-!\"$*(\"\")F--%#A3GF*F-F.F-F-*(\"#7F--F/6$F(F+F-F.F-!\"\"** \"\"*F-F(F-F1F-F:F-F<*(F9F--F/6$F6F+F-F1F-F-*(\"#5F--F/6$F(-%\"$G6$F+F ,F-F1F-F<*&\"#?F--F/6$F1-FG6$F+\"\"$F-F-**\"\"%F-F6F-F1F-F(F-F-*&)F(FO F-F1F-F<*(F5F--%#A2GF*F-F1F-F<" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "de2diffop(first_check, i(x)); # = L3" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#,2*$)%#DxG\"\"$\"\"\"\"#?*&,(*$)-%#A4G6#%\"xG\"\"#F(!\" $*&\"\")F(-%#A3GF0F(F(*&\"#7F(-%%diffG6$F.F1F(!\"\"F(F&F(F(*(\"\"*F(F. F(F:F(F=*&F9F(-F;6$F6F1F(F(*$)F.F'F(F=*&F5F(-%#A2GF0F(F=*&\"#5F(-F;6$F .-%\"$G6$F1F2F(F=*(\"\"%F(F6F(F.F(F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 38 "This leads to the following algorithm:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 294 6 "Input:" }{TEXT -1 6 " L i n " }{TEXT 292 1 "C" }{TEXT -1 9 "(x)[d/dx]" }}{PARA 0 "" 0 "" {TEXT 295 7 "Output:" }{TEXT -1 10 " a,b,c in " }{TEXT 296 1 "C" }{TEXT -1 54 "(x) such that equation (5) holds, if such a,b,c exist." }}{PARA 0 "" 0 "" {TEXT 297 7 "Step 1:" }{TEXT -1 36 " Compute L3 with the form ula above." }}{PARA 0 "" 0 "" {TEXT 298 7 "Step 2:" }{TEXT -1 53 " Co mpute all solutions of L3 whose 4'th power is in " }{TEXT 293 1 "C" } {TEXT -1 30 "(x), so L3(i)=0 and i^4 is in " }{TEXT 300 1 "C" }{TEXT -1 4 "(x)." }}{PARA 0 "" 0 "" {TEXT 299 8 "Step 3: " }{TEXT -1 62 " De termine the set of all i'/i for all non-zero i from step 2." }}{PARA 0 "" 0 "" {TEXT 304 9 "Step 4: " }{TEXT -1 28 "For each such i'/i, te st if " }{TEXT 305 3 " sp" }{TEXT -1 55 "(Dx - 1/4 * A4 + 1/2 * i'/i, \+ L) satisfies second_check." }}{PARA 0 "" 0 "" {TEXT 306 9 "Step 5: " }{TEXT -1 40 "If so, then we can find L1,L2 for which " }{TEXT 307 2 " sp" }{TEXT -1 74 "(Dx - 1/4 * A4 + 1/2 * i'/i, L) meets condition (6). Replace L1 and L2 by " }{TEXT 308 2 "sp" }{TEXT -1 37 "(Dx + 1/8 * A4 - 1/4 * i'/i, L1) and " }{TEXT 309 2 "sp" }{TEXT -1 49 "(Dx + 1/8 * A 4 - 1/4 * i'/i, L2). Then (5) holds." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 301 8 "Remarks:" }}{PARA 0 "" 0 "" {TEXT -1 64 "Step 2: recall that c^(-1/4) must be a solution of L3, and c in " }{TEXT 321 1 "C" }{TEXT -1 26 "(x), hence i^4 must be in " }{TEXT 322 1 "C" }{TEXT -1 4 "(x)." }}{PARA 0 "" 0 "" {TEXT -1 48 "Step 3: As an \+ algebraic set, this set is either:" }}{PARA 0 "" 0 "" {TEXT -1 48 " c ase 1: A finite set with 0, 1, 2 or 3 points." }}{PARA 0 "" 0 "" {TEXT -1 97 " case 2: A projective line, so the i'/i are parametrized by homogeneous parameters (s:t) in P^1(" }{TEXT 302 1 "C" }{TEXT -1 2 ")." }}{PARA 0 "" 0 "" {TEXT -1 62 " case 3: A disjoint union of a \+ projective line and one point." }}{PARA 0 "" 0 "" {TEXT -1 79 " case \+ 4: A projective plane, then the i'/i are parametrized by (s:t:u) in P^ 2(" }{TEXT 303 1 "C" }{TEXT -1 2 ")." }}{PARA 0 "" 0 "" {TEXT -1 190 " Step 4: i'/i depends on parameters in cases 2, 3, and 4, which means w e need to translate second_check into homogeneous polynomial equations for s,t (in cases 2, 3) or for s,t,u (in case 4)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 649 "The algorithm is impleme nted in dsolve in Maple7. Cases 2, 3, 4 have not been fully implemente d (it will consider only 2 points on the line or 3 points on the plane ). The reason is that if L is reducible then it will be treated by DFa ctor, and if L is irreducible then cases 2, 3, 4 are rare. However, in order to have a complete procedure these cases must be implemented. I t is easy to do so, and the efficiency would not be bad because no Gro ebner basis is needed to solve the polynomial equations because the nu mber of homogeneous variables is <= 3, so they can be solved with resu ltants and gcd's. To view the code, type the following in Maple7:" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 105 "interface(verboseproc=2): # to display Maple procedures\n prin t(`dsolve/diffeq/higherorder/is_sympr_o2`);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#R6%%\"AG%\"xG%\"VG6(%\"vG%\"iG%\"jG%#BxG%\"wG%#DFG6#%\\ pCopyright~(c)~2000~Waterloo~Maple~Inc.~All~rights~reserved.~~Author:~ M.~van~HoeijG6\"C%>8$-%$mapG6$%+NormalizerG7&,.-%%diffG6$&9$6#\"\"$9%# FA\"\"&*&#\"\"#FD\"\"\"&F?6#FGFH!\"\"*(#FHFDFHF>FH&F?6#\"\"%FHFH*&#FHF GFH-F<6%FNFBFBFHFK*&#FH\"#?FH*$)FNFAFHFHFK*&#\"\"*FWFH*&FNFH-F<6$FNFBF HFHFK,(F>#FGFD*&#FAFDFHFhnFHFK*&#FAFWFH*$)FNFGFHFHFK\"\"!FH@$-%%typeG6 $F?-%%listG6#-%(ratpolyG6$%)anythingGFBC$>F4-%0DEtools/ExpsolsG6*F4Fbo FB%(use~IntG%*no~algextG%(radicalG%&denomGFP?&8%-%%sortG6$F4R6$F*F+F16 $%)operatorG%&arrowGF1-%&evalbG6#2-%'lengthG6#F?-Ffq6#FBF1F1F1%%trueGC $>8'-F86#,&FN#FHFP*&#FHFGFH*&-F<6$FhpFBFHFhpFKFHFK@$0F]rFboC&-%2ODEtoo ls/userinfoG6&FG%'dsolveG%9Multiplying~solutions~byG-%$expG6#-%$IntG6$ F]rFB>F4-%*traperrorG6#-&%(DEtoolsG6#.%2symmetric_productG6%,&8)FHF]rF K-%$addG6$*&&F?6#,&8&FHFHFHFH)FatFitFH/Fit;FboFP7$FatFB@$/F4%*lasterro rGO%&falseG@$-%Mdsolve/diffeq/higherorder/is_sympr_o2_no_DF1G6%7#-%$se qG6$-%&coeffG6%F4FatFitF[uFB.8(C$>9&7$-%#opG6#F_v,$F]rFKOFjqFbuF1F1F1 " }}}}{MARK "9 12 2" 5 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 } {PAGENUMBERS 0 1 2 33 1 1 }