2. F ; =(,1)F,1; +F,1;,1 is satised by thestirlig ubers of the rst kid ([1], p. 824). 3. F ; = F,1; + F,1;,1 is satised by the Stirlig ubers of the se

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1 O First-Order Two-Diesioal Liear Hoogeeous Partial Dierece Equatios G. Neil Have y Ditri A. Gusev z Abstract Aalysis of algoriths occasioally requires solvig of rst-order two-diesioal liear hoogeeous partial dierece equatios. We survey solutios to special cases of the liear recurrece equatio (a + b + c)f ; = (d + e + f)f,1; +(g + h + i)f,1;,1 i ters of kow fuctios ad establish equivaleces betwee usolved cases. The article also reviews solutio techiques used to siplify recurreces ad establish equivaleces betwee the. 1 Itroductio The stadard for for the recurrece equatio we will be cosiderig is x(; )F ; = y(; )F,1; + z(; )F,1;,1 : Recurreces of this for arise i probles related to aalysis of algoriths ad cobiatorics (see, for exaple, [7], Sectios ad 8.2). We are especially iterested i the case whe x(; ) a + b + c, y(; ) d + e + f, z(; ) g + h + i, i.e., (a + b + c)f ; =(d + e + f)f,1; +(g + h + i)f,1;,1 ; where a, b, c, d, e, f, g, h, adi are real ubers. Our goal is to deterie which fors of this recurrece have solutios i ters of kow fuctios. If we classify these fors accordig to which of the coeciets a, b, d, e, g, ad h are o-zero, there are 64 distict cases to cosider. The values of the costat ters c, f, adi have little eect o whether a give for has a solutio. Redeig variables or shiftig idices will usually eliiate the fro the recurrece. Tables of solutios i Sectio 3 cotai particular solutios that satisfy the correspodig equatios. Fidig solutios that also t particular boudary coditios is a separate proble. It is discussed i [6] ad related to a well-kow siilar proble for partial dieretial equatios cosidered i [2, 3]ad ay other sources. For siplicity's sake we also igore the proble of dig the radii of covergece for the solutios we d. Iterested readers are referred to [9]. 2 Notatio ad Caoical Recurreces I this sectio, we review the otatio for several well-kow fuctios ad reid the reader of recurreces these fuctios satisfy. We alsoprovide refereces to sources where oe ay read ore about these fuctios (how to copute the, their properties, etc.). I fact, there are ay ore sources, because the fuctios uder cosideratio are extreely coo. 1. F ; = F,1; + F,1;,1 is satised by thebioial coeciets, (see [7], Sectio 3.2, ad [1], pp. 822{823). This work was partially supported by NSF Grats CCR ad CCR yautospect Ic., 4750 Veture Drive, A Arbor, MI (have@cs.idiaa.edu). z Idiaa Uiversity, Coputer Sciece Departet, Lidley Hall, Blooigto, IN (diguse@cs.idiaa.edu). 1

2 2. F ; =(,1)F,1; +F,1;,1 is satised by thestirlig ubers of the rst kid ([1], p. 824). 3. F ; = F,1; + F,1;,1 is satised by the Stirlig ubers of the secod kid ([1], pp. 824{825). 4. F ; = F,1; +(, +1)F,1;,1 is satised by the Euleria ubers ([4], p.35). 5. F ; = ( +, 1)F,1; + F,1;,1 is satised by the Lah ubers (,1)! (,1)!, (see [5] ad [7], p. 353). 6. Riorda's recurrece F ; = F,1; +( +, 1)F,1;,1 has the solutio (+)!!(,)!2 ([8], p. 85). 7. Kuer's equatio z d2 w +(b, z) dw, aw =0hastwo idepedet solutios deoted by M(a; b; z) dz 2 dz ad U(a; b; z) ([1], p. 504). Recurrece (,1)F ; =(,)F,1; +(,1)F,1;,1 is satised by M(; ; z) for all z. Recurrece cf ; =(,1)F,1;, (,1)F,1;,1 is satised by M(; ; c). cf ; =(,, 1)F,1; + F,1;,1 is satised by U(; ; c). P 8. Gauss hypergeoetric fuctio F (a; b; c; z)=,(c) 1,(a+),(b+)z,(a),(b) =0 has the followig properties,(c+)! ([1], p. 556{558). a(,1)f ; =(+,k,1)f,1; +(k,)f,1;,1 is satised by F (; ; k;1,a). a(, k, 1)F ; =(a, 1)(, 1)F,1; +(, 1)F,1;,1 is satised by F (; k; ; a). 3 Tables of Solutios The 64 fors of the geeral recurrece are divided ito 17 cases with solutios i ters of kow fuctios ad 47 cases with o such kow solutios. The 47 cases with o kow solutio ca be further divided ito classes of cases solvable i ters of oe aother but ot i ters of kow fuctios. Usig the techiques of substitutio, perutatio, ad idex shiftig described i Sectio 4, it is possible to exhibit 14 such equivalece classes aog the 47 usolved cases. Table 3.1 presets geeral case solutios ad equivaleces. The cases with o kow geeral solutio are soeties solvable if certai relatios are itroduced betwee the values of the coeciets a, b, c, d, e, f, g, h, adi. Soe special case solutios are give i Table Solutio Techiques The techiques of substitutio, perutatio, ad idex shiftig were used i the preparatio of the above tables with oe of two purposes: rst, to trasfor a recurrece ito a for with a kow solutio; secod, to trasfor a ukow recurrece ito aother ukow recurrece ad so exhibit the equivalece betwee the. 4.1 Siplicatio by Substitutio I the case that x(; ) = x() we ay siplify the recurrece by akig the substitutio F ; = p()h ; where p is yet to be deteried ([7], Sectio 8.2.2): F ; = p()h ; ; x()p()h ; = y(; )p(, 1)H,1; + z(; )p(, 1)H,1;,1 ; p() x() p(, 1) H ; = y(; )H,1; + z(; )H,1;,1 : 2

3 Table 3.1 Case x(; ) y(; ) z(; ) Solutio (or Equivalece) f i 0 c f i c f f h,( + i=h +1) 1 c f h + i c f g f + i=g +1 2 c f g + i f c, 3 c f g + h + i special cases e i 4 c e + f i c e + f=e e h,( + i=h +1) 5 c e + f h + i c e + f=e 6 c e + f g + i? 7 c e + f g + h + i special case d i + f=d+1 8 c d + f i c d d h,( + i=h +1) + f=d+1 9 c d + f h + i c d 10 c d + f g + i special case 11 c d + f g + h + i? 12 c d + e + f i Case 3 13 c d + e + f h + i Case 3 14 c d + e + f g + i Case c d + e + f g + h + i? (,i) f +1,, 1 16 b + c f i f b +1 + c=b f +1 (,h),( + i=h +1),, 1 17 b + c f h + i f b +1 + c=b f +1 g 18 b + c f g + i, + c=b, i=g, 2 b +1 f + c=b 19 b + c f g + h + i? 20 b + c e + f i Case b + c e + f h + i Case b + c e + f g + i? 23 b + c e + f g + h + i? 24 b + c d + f i Case 6 25 b + c d + f h + i Case 6 26 b + c d + f g + i Case b + c d + f g + h + i? 28 b + c d + e + f i Case b + c d + e + f h + i Case b + c d + e + f g + i? 31 b + c d + e + f g + h + i? 3

4 Table 3.1 (cotiued) Case x(; ) y(; ) z(; ) Solutio (or Equivalece) f i 32 a + c f i a f,( + c=a +1) f h,( + i=h +1) 33 a + c f h + i a f,( + c=a +1) g f + i=g a + c f g + i f a,( + c=a +1), 35 a + c f g + h + i Case 3 e i 36 a + c e + f i a e,( + c=a +1) + f=e e h,( + i=h +1) 37 a + c e + f h + i a e,( + c=a +1) + f=e 38 a + c e + f g + i Case 6 39 a + c e + f g + h + i Case 7 d i + f=d+1 40 a + c d + f i a d,( + c=a +1) d h,( + i=h +1) + f=d+1 41 a + c d + f h + i a d,( + c=a +1) 42 a + c d + f g + i Case a + c d + f g+ h + i Case a + c d + e + f i Case 3 45 a + c d + e + f h + i Case 3 46 a + c d + e + f g + i Case a + c d + e + f g + h + i Case a + b + c f i Case 3 49 a + b + c f h + i Case 3 50 a + b + c f g + i Case a + b + c f g + h + i Case a + b + c e + f i Case a + b + c e + f h + i Case a + b + c e + f g + i Case a + b + c e + f g + h + i? 56 a + b + c d + f i Case 7 57 a + b + c d + f h + i Case 7 58 a + b + c d + f g + i Case a + b + c d + f g + h + i Case a + b + c d + e + f i Case a + b + c d + e + f h + i Case a + b + c d + e + f g + i Case a + b + c d + e + f g + h + i? 4

5 Table 3.2 Case x(; ) y(; ) z(; ) Solutio 3a c f g + g + i 3b c f 2g, g + i 7a c e + f g, g + i 10a c d + f k(d + f) Solvig the secodary recurrece for p x()p() =p(, 1) results i the siplicatio H ; = y(; )H,1; + z(; )H,1;,1 : f g ( + + i=g +1)! c f ( + i=g, 1)!(,, i=g +3)!2 +i=g,1 f g!, 1 c f (, + i=g +1)!, + i=g g e +1 + f=e+ i=g, 1 e c +1 + f=e d k,( + f=d+1) c Likewise, i the case that z(; ) = z() we ay siplify the recurrece by akig the substitutio F ; = q()h ; (where q is yet to be deteried): F ; = q()h ; ; x(; )q()h ; = y(; )q()h,1; + z()q(, 1)H,1;,1 ; x(; )H ; = y(; )H,1; + z() Solvig the secodary recurrece for q q() =z()q(, 1) results i the siplicatio x(; )H ; = y(; )H,1; + H,1;,1 : q(, 1) H,1;,1 : q() Case 0 fro Table 3.1 is a good exaple of a siple recurrece that ca be solved with this techique. 4.2 Siplicatio by Shiftig Idices Stadard techiques ay be used to shift the idices of the recurrece equatio. This ay result i sipli- catio, as show i the followig exaple. Solve Dee The F ; =(, 4)F,1; + F,1;,1 : G ;,4 def = F ; : G ;,4 =(, 4)G,1;,4 + G,1;,5 ; G ; = G,1; + G,1;,1 : This is the failiar recurrece for Stirlig ubers of the secod kid. Hece, G ; = ; F ; = :, 4 5

6 4.3 Siplicatio by Perutatio The geeral recurrece x(; )H,;, = y(; )H,;, + z(; )H,;, ay be trasfored ito a recurrece i the stadard for x 0 ( 0 ; 0 )F 0 ; 0 = y0 ( 0 ; 0 )F 0,1; 0 + z0 ( 0 ; 0 )F 0,1; 0,1 by a appropriate coordiate traslatio ad rotatio. First, traslate the idices to yield H,;, G ; x(; )G ; = y(; )G +,;+, + z(; )G +,;+, : Secod, followig [7], rotate the coordiates by replacig ad i x, y, ad z accordig to def = (, )(, ), (, )(, ); x 0 ( 0 ; 0 )=x((, ) 0, (, ) 0 ; (, ) 0 +(, ) 0 ); y 0 ( 0 ; 0 )=x((, ) 0, (, ) 0 ; (, ) 0 +(, ) 0 ); z 0 ( 0 ; 0 )=x((, ) 0, (, ) 0 ; (, ) 0 +(, ) 0 ): (If = 0 the ters of the recurrece are coliear ad the recurrece ay be solved as a oe-diesioal recurrece.) This rotatio yields a ew recurrece i stadard for i ters of the variables 0 ad 0. If this ew recurrece ca be solved, the the solutio to the origial recurrece ca be writte dow by eas of backward substitutio for 0 ad 0 accordig to G ; = F (,)+(,) ; (,)+(,) ; H ; = G +;+ = F (,)(+)+(,)(+) ; (,)(+)+(,)(+) : Makig use of the above techique we ay perute the order of the coeciets i this recurrece accordig to the followig table: Table 4.1 Forward Substitutio Backward Substitutio Perutatio =x =y =z = = = 0 = 0 (1)(2)(3) x y z 0 0 (12)(3) y x,z 0, 0 0,, 1 (13)(2) z,y x, 0, 0,, 1,, 1 (1)(23) x z y 0 0, 0, (123) z x,y 0, 0, 0,, 1, (132) y,z x, 0 0, 0,, 1,, 1 Here the forward substitutios are used to trasfor oe recurrece i stadard for ito aother recurrece i stadard for i ters of the variables 0 ad 0. This secod recurrece ay thebesolved. Backward substitutio ito this solutio for 0 ad 0 i ters of the origial variables ad yields a solutio to the origial recurrece. To illustrate the eect of perutatio (123) o the geeral recurrece x(; )H ; = y(; )H,1; + z(; )H,1;,1 ; 6

7 we ay perute accordig to the forward substitutios to yield: z( 0, 0 ;, 0 )F 0 ; 0 = x(0, 0 ;, 0 )F 0,1; 0, y(0, 0 ; 0 )F 0,1; 0,1: The solutio to the origial recurrece is the obtaied fro the solutio to this recurrece by backward substitutio for 0 ad 0 H ; = F,,1;, : To see the perutatio techique i actio, cosider the followig exaple. Solve H ; = H,1; +(, 1)H,1;,1 : Perute accordig to (12)(3) to yield F 0 ; 0 = 0 F 0,1; 0, (0, 0, 1)F 0,1; 0,1; F 0 ; 0 = 0 F 0,1; 0 +(0, 0 +1)F 0,1; 0,1: This is the failiar Euleria recurrece, which leads to the solutio 0 F 0 ; 0 = 0 : Backward substitutio yields the solutio of the origial recurrece,, 1 F ; = : Ackowledget. The authors thak Paul W. Purdo, Jr. for his helpful suggestios o presetatio of the paper. Refereces [1] Milto Abraowitz ad Iree A. Stegu (eds.), Hadbook of Matheatical Fuctios, Dover Publicatios, New York (1965). [2] Paul W. Berg, Jaes L. McGregor, Eleetary Partial Dieretial Equatios, Holde-Day, Sa Fracisco (1966). [3] P. R. Garabedia, Partial Dieretial Equatios, Chelsea Publishig Copay, New York, secod editio (1986). [4] Doald E. Kuth, The Art of Coputer Prograig, Vol. 3, Addiso-Wesley, Readig, Mass. (1973). [5] I. Lah, Eie eue Art vo Zahle, ihre Eigeschafte ud Awedug i der Matheatische Statistik, Mitteilugsbl. Math. Statist. 7 (1955), pp. 203{212. [6] Roald E. Mickes, Dierece Equatios, Va Nostrad Reihold Co., New York (1987). [7] Paul W. Purdo, Jr. ad Cytia A. Brow, The Aalysis of Algoriths, Holt, Riehart ad Wisto, New York (1985). [8] Joh Riorda, A Itroductio to Cobiatorial Aalysis, Joh Wiley & Sos, New York (1958). [9] A. N. Sharkovsky, Y. L. Maistreko, E. Y. Roaeko, Dierece Equatios ad Their Applicatios, Kluwer Acadeic Publishers, Dordrecht (1993). 7