On Pfaff s solution of the Pfaff problem

Transcription

1 Zur Pfaff scen Lösung des Pfaff scen Probles Mat. Ann. 7 (880) On Pfaff s soluton of te Pfaff proble By A. MAYER n Lepzg Translated by D. H. Delpenc Te way tat Pfaff adopted for te ntegraton of an equaton of te for: dx + dx + + dx = 0 rests upon te repeated applcaton of one and te sae transforaton wc can be expressed by te followng proble: Deterne x x x as utually ndependent functons of new varables t tat gve dentcally: I. dx = d = N = A were N s a functon of t but te quanttes A A are functons of just and t s necessary for te entre constructon of te Pfaff etod as well as te general applcablty of te Pfaff etod tat ts proble s always soluble for an even wle n general no soluton s allowed wen s odd. Neter te one case nor te oter one as actually been proved up to now. Naely as far as I can see t wll always be assued tat te so-called frst Pfaff syste of ordnary dfferental equatons by tself already suffces to solve te proble. However ts s correct only as long as te skew deternant tat s defned fro te eleents: = does not vans (a case tat wll generally be consdered exclusvely as a rule). Te goal of te present note s to sow ts as well as to place te Pfaff etod on fr foundatons and everytng tat does not see edately necessary for ts purpose as been left asde wle n lne wt te desre to found te etod n an absolutely clear anner peraps one gt excuse te presence of entrely too uc rgor n regard to soe oter ponts naely te ones tat one ordnarly consders to be selfexplanatory. Te requreent (I) frst leads to te condton:

2 A. Mayer. On te Pfaff proble. () B = 0 = and lkewse gves te followng values for te A : () A = N. = Tese sould be free of t. Terefore equaton () and te equatons: (3) A N x + N = = = 0 are te necessary and suffcent condtons tat te proble s supposed to satsfy by way of ndependent functons x x x of te varables t. One now as: N (4) B N N = = One ten as: NB N = x x N N B + +. = = A NB N (5) N N B = =. As a result of condtons () and (3) one ust ten ave for = t : (6) N N = 0 = = and lkewse forulas (4) and (5) yeld: II. One can conversely replace te orgnal condtons () and (3) wt te condtons (6) as long as t s possble to satsfy te latter ones wtout akng N / = 0. By coparson f t ten follows fro equatons (6) tat N / = 0 ten one ust add te condton () to tese equatons. Condtons (6) owever ay be deduced n anoter spler way. Snce: =

3 A. Mayer. On te Pfaff proble. 3 one can n fact next wrte tese equatons as: N N = = = 0. Now sould x x be ndependent functons of t ten: t ± would be non-zero. Te condtons (6) tus decopose nto te followng ones: N x N = 0. = However by eans of te forula: N N x N + = t can be converted nto: x N = x x = N. If one ten sets: (7) = ten condtons (6) fnally reduce to te followng equatons: (8) N = = log N. Te present proble wll ten be solved n all stuatons by te + equatons () and (8) assung tat one can satsfy te by ndependent functons x x. Te latter s owever always te case as long as te deternant: = 0

4 A. Mayer. On te Pfaff proble. 4 of te + equatons (8) and () s zero. Terefore f = 0 ten one can always satsfy tese equatons by values of / wc are not all zero. However f () and (8) were fulflled by te assuptons: x : x : : t : log N = u : u : : u : M were u u u are not all zero ten snce t tself does not appear at all n equatons () and (8) u u u M are functons of only x x x and wen say u s non-zero te values: u (9) = u u = u wen coupled wt te value: log N M (0) = u dentcally satsfy te equatons: = 0 = = = log N.e. te + equatons tat eerge fro () and (8) wen one assues tat t = x. Equatons (9) owever defne a syste of ordnary dfferental equatons between x x and x. Terefore f: are ts coplete solutons ten te values: x = ϕ (x ) x = x x = ϕ x = ϕ n wc tey all fulfll te condton equatons of te transforaton and lkewse: ϕ ϕ ± = x ± wc are ten non-zero yeld solutons of te gven proble * ). * ) If one wses to also fnd te assocated values of te coeffcents A teselves ten one ust only after substtutng any coplete soluton copute fro (0) by a sple quadrature: M dx u N = γ e

5 A. Mayer. On te Pfaff proble. 5 Te proble s n fact always soluble as long as te deternant s zero and ndeed ts soluton wll ten be obtaned by coplete ntegraton of a syste of ordnary dfferental equatons. By contrast t possesses no soluton wen s non-vansng. Equatons () and (8) can ten be satsfed only for te values: t = t = = = log N = 0 and ts contradcts te deand tat x x x sould be ndependent functons of t. However as a skew deternant of degree + s zero wen s even and generally non-zero wen s odd. Terefore one as te teore: III. Te gven transforaton proble s soluble wen and only wen s an even nuber and ndeed n ts case ts soluton requres te coplete ntegraton of a syste of ordnary dfferental equatons. How one can construct te Pfaff etod for te ntegraton of any lnear total dfferental equaton sply and naturally fro ts teore was set down by Gauss n s announceent of te Pfaff treatse * ) n suc classc brevty and clarty tat t would be copletely superfluous to go nto t ere. By contrast te valdty n regard to te frst Pfaff syste eerges fro te foregong alone or te stateent ade by equatons (8) would stll not be clear. We tus now fx our attenton on te case: = n ore closely! In general te deternant: A = ± s non-zero ere. However f ts s te case ten equatons (8) are utually ndependent. Terefore equaton () wen t togeter wt equatons (8) defnes a syste of vansng deternant ust necessarly a ere consequence of tese equatons. In fact wen A s not zero none of te sus: and substtutng ts value for N n te forula tat arses fro (): A = N = ϕ were one can gve te arbtrary constant γ te value snce t can ave no nfluence upon te deand tat A ust be free of x and s ten reoved fro te transforaton I tself. Ts latter coputaton owever can be copletely spared wen one as ntroduced te ntal values of x x as arbtrary constants n te solutons of syste (9) usng te Jacob etod (Crelle J. 7 pp. 56). * ) Gauss Werke III pp. 3.

6 A. Mayer. On te Pfaff proble. 6 A = vans were one understands A to ean te coeffcent of te eleent n te deternant A. However one ust ten ave eac = 0. Terefore let say: A = be non-zero. If we ten take t = x ten equatons (8) becoe: and yeld: = = log N log N A = A x = so log N / as a fnte and non-zero value. Fro II equatons (8) alone terefore suffce for a soluton to ts proble n te case consdered. On te contrary f A = 0 for even ten for tat reason te skew deternant of degree : = 0 does not need to be zero. It s ten ndependent of wle for a gven te equaton A = 0 s a partal dfferental equaton of frst order for. However wen s non-zero ten not all sub-deternants of t degree vans n te deternant and te + equatons (8) and () ten reduce to only equatons but no fewer. Tey ten unquely deterne ter unknowns wc are te ratos: t : : : : log N. As long as one can ten gve values to tese unknowns tat satsfy equatons () and (8) tese are te only values tat satsfy tese equatons n te case we assued. However one obtans suc values wen one sets log N / = 0 and ten deternes te ratos of te / fro te + equatons:

7 A. Mayer. On te Pfaff proble. 7 () = 0 = = 0. = Suc a deternaton s possble. Ten snce by assupton te skew deternant A of degree = n vanses all of ts sub-deternants of degree are also = 0 and te frst equatons () ten reduce to equatons. In te case n queston ten equatons () and (8) necessarly yeld log N / = 0 and ten fro II syste (8) alone no longer suffces as a soluton of te proble but ts wll nstead frst occur by eans of equatons (). Wen llunated n ts lgt te proble does not by any eans need to be ndeternate n te case were te deternant becoes A = 0 for a gven * ). Rater suc an ndeternacy wll frst enter te pcture wen te deternant A vanses as well as all of ts sub-deternants of order or even lower order and ndeed tere always exsts an even ndeternancy n a sngular way;.e. n suc a case one can always coose an even nuber of te ratos: t : t : : arbtrarly or wat wll always be ost convenent set te equal to zero. Ten wen all of te sub-deternants of degree r n a skew deternant vans ten as Frobenus ad proved ** ) all sub-deternants of degree r also vans. Snce = n te vansng of all sub-deternants of degree p n te deternant A necessarly ples te vansng of all sub-deternants of degree p and ts ten reduces te + equatons () to p equatons suc tat one can also terefore satsfy * ) For te total dfferental equaton: f dx + f dx + + p n dx n dz = 0 e.g. te one tat s equvalent to te partal dfferental equaton of frst order tat s free of z: z p = f (x x n p p n ) p = equatons (8) and () reduce to te known n equatons: dx dx = f p dp dx = f dz dx = f n f p p = and te transforaton proble reans copletely deterned ere. ** ) Borcardt s J. 8 pp. 4.

8 A. Mayer. On te Pfaff proble. 8 te by values of / wc are not all zero after one as set a certan p of te varables x equal to arbtrary constants. If one recalls te Jacob etod for presentng te varous total dfferental equatons by te ntroducton of te ntal values fro te outset wc s wat te ntegraton of te gven equaton wll successvely coe down to n te Pfaff etod ten one easly overlooks ow ts advantage s anded down by te frst p reductons step-by-step wt eac step decreasng by two unts.