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John Shackell Canad. Math. Bull. Vol. 41 (2), 1998 pp. 214±224 ON A PROBLEM OF RUBEL CONCERNING THE SET OF FUNCTIONS SATISFYING ALL THE ALGEBRAIC DIFFERENTIAL EQUATIONS SATISFIED BY A GIVEN FUNCTION JOHN SHACKELL ABSTRACT. For two functions f and g, de®ne g − f to mean that g satis®es every algebraic differential equation over the constants satis®ed by f . The order − was introduced in one of a set of problems on algebraic differential equations given by the late Lee Rubel. Here we characterise the set of g such that g − f,whenfis a given Liouvillian function. 1. Introduction. One tiny part of the legacy to Mathematics of the late Lee Rubel is the following problem, which appears as part of Problem 22 in [2]. For two functions g and f , we de®ne g − f to mean that g satis®es every algebraic differential equation (over C) which f satis®es. Discuss the order − P ; in particular, do this for the case when f is an exponential polynomial, n ïkx k=1 ake . In order to discuss the order − when more general functions are involved, it is necessary to say something about the domains of de®nition of the functions to be considered. It is generally too restrictive to require g to have the same domain of de®nition as f .On the other hand, one would at least want a non-empty open subset of C on which both functions are de®ned. In fact for the functions we shall be considering, we shall generally be able to take that subset to be dense. We shall also want to use a topology on various sets of functions. Since the functions concerned will be Liouvillian, most natural choices of topology are likely to give the same answers. We shall use uniform convergence of the functions and their derivatives on compact subsets. Although a description of the order − is of interest for its own sake, there are also applications to asymptotics. If one searches for a series solution to a non-linear ordinary differential equation in terms of base functions fen(x)Ò n =1Ò2Òg,whereen denote the n-times iterated exponential function, one would like to boundÁ the possible n that j might occur. Suppose we have a solution of the form f (x)+ en(x) g(x)wheref is a ÒÒ Ù polynomial in e1(Áx) en1(x), j 0andggives the tail of the series. If we do not j have f (x)+ e (x) g(x) − f there is a differential polynomial P which vanishes at f but n Á Á j j not at f (x)+ en(x) g(x). Then Phf (x)+ en(x) g(x)itends to zero approximately as a ®xed negative power of en(x). If n is too large compared with the order of the differential Received by the editors June 20, 1996. AMS subject classi®cation: 34A34, 12H05. c Canadian Mathematical Society 1998. 214 Downloaded from https://www.cambridge.org/core. 30 Sep 2021 at 19:26:57, subject to the Cambridge Core terms of use. FUNCTIONS SATISFYING ALGEBRAIC DIFFERENTIAL EQUATIONS 215 equation, one can proveÁ that this is impossible. Under suitable conditions, one can also j prove that f (x)+ en(x) g(x)−f is impossible, thereby bounding possible values of n. This has been done within the context of nested expansions in [3]. We begin Section 2 with a purely elementary consideration of the special case in which f is an exponential polynomial. Here we are able to give a completely explicit characterisation of the set of g for which g − f . Then we re-frame our results in terminology which is closer to differential algebra. This leads into Section 3, where we consider the case when f is an arbitrary Liouvillian function. Here we characterise the set of g for which g − f as the closure of a set of explicitly given transformations of f . Our argument in this section has some similarities with that of Section 5 of [3]. A major part of the work for this paper was done while the author was visiting the University of Limoges during the summer of 1995. The author would like to thank that University and its Mathematics staff for their splendid hospitality during this period. Special thanks are due to Dominique Duval for having arranged the invitation, and also for some useful comments during an informal seminar on the problem treated here. P n ïkx 2. Exponential polynomials. The ®rst thing to be said is that if f = k=1 ake P ï − n kx ÒÒ 2C and g f ,theng= k=1 Ake for someQ A1 An . This is because f satis®es h i f n Û ï g the linear differential equation L y = k=1(d dx k) y =0.Herewehaveused the notation Lhyi to indicate that L is a polynomial in y and its derivatives, where y isP an indeterminate. However in many cases, it is not suf®cient that g be of the form n ïkx k=1 Ake . We suppose that the ïis are all different and that no ai is zero. We write Q[ï1ÒÒïn] for the Q-linear space generated by ï1ÒÒïn,andletdbe its dimension. Now d is ï ï also equal to the degree of transcendence of C(e 1xÒÒe nx) over C, and hence f cannot satisfy an algebraic differential equation of order less than d. P n ïkx Let g be of the form k=1 Ake , and suppose ®rst that d = n.Thenwehavea linear differential equation of order d satis®ed by f , namely Lhyi =0.Iffsatis®es another algebraic differential equation of order n,sayPhyi=0,thenLmust divide P. Ò For otherwise the resultant, resy(n) (P L), would be a non-zero differential polynomial of order less than n annulled by f . Hence g satis®es every differential equation of order n satis®ed by f .Letm½nand suppose inductively that g satis®es every differential equation of order m satis®ed by f .LetQhyi= 0 be a differential equation of order m +1 satis®ed by f . On differentiating m n +1 times the equation Lhf i = 0, we obtain a linear expression for f (m+1) in terms of f ÒÒf(m),sayf(m+1) = X(f ÒÒf(m)); note that g also satis®es this equation. On substituting X(yÒÒy(m))fory(m+1) in Qhyi = 0, we get an equation of order m satis®ed by f , and therefore by g. When we replace X(gÒÒg(m)) where it occurs in this last equation by g(m+1), we see that also Qhgi = 0. Thus g satis®es every equation of order m + 1 which is satis®ed by f , and by induction this holds for P ï − n kx all m. Hence when d = n,wehavethatg f if and only if g = k=1 Ake with A1ÒÒAn 2C. It may be noted that we have in effect used the order on differential polynomials given in [1]. Downloaded from https://www.cambridge.org/core. 30 Sep 2021 at 19:26:57, subject to the Cambridge Core terms of use. 216 JOHN SHACKELL Now we consider the case when d Ú n. By rearranging as necessary, we may suppose that ï1ÒÒïd are linearly independent, and that ïd+1ÒÒïn are Q-linear combinations of ï1ÒÒïd. Then for i = d +1ÒÒnwe have Xd ï jï Ò (1) i = ci j j=1 1ÒÒ d 2Q with ci ci . On differentiating the equation Xn ïkx f = ake k=1 ï ï 1xÒÒ nx n 1 times,Q we obtain n linear equations for a1e ane . The determinant of the system is iÚj(ïj ïi), which is non-zero. Hence we may obtain linear expressions for ï x (n1) each ake k in terms of f ÒÒf ;say ïkx (n1) (2) ake =Rk(fÒÒf )Ò k =1ÒÒn On combining these with (1) and taking suitable powers to remove roots, we obtain, for j =1ÒÒnd, Yd j Yd j b ç b ç j i j i ÒÒ (n 1) Ò (3) ad+j ai = Rd+j Ri = Sj(f f ) i=1 i=1 2N çj j C where each bj , each i is an integer equal to cibj,andSj is a rational function over . P ï − n kx If g f then the equations (3) must also be satis®ed by g. However if g = k=1 Ake with all the Aks non-zero, then applying to g the same process of elimination that was applied to f yields ïkx (n1) Ake = Rk(gÒÒg ) On substituting into (3), we obtain Yd j Yd j b ç b ç j i j i Ò (4) ad+j ai = Ad+j Ai i=1 i=1 for j =1ÒÒnd.SotheAk must satisfy these equations in order that g − f . It is easy to see that this holds even when some of the Aks are zero provided that negative powers of Aks are removed from (4) by cross multiplication. In fact these conditions are also suf®cient. P ï n kx ï ÒÒï THEOREM 1. Let f = k=1 ake . Suppose that 1 d are linearly independent Q ï ÒÒï 1ÒÒ d 2Q − over and that d+1 n satisfy (1) with ci ci .Theng f if and only P ï n kx ÒÒ if g = k=1 Ake , with A1 An satisfying the relations obtained from (4) by cross multiplying to remove negative powers. Downloaded from https://www.cambridge.org/core. 30 Sep 2021 at 19:26:57, subject to the Cambridge Core terms of use.
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