NONLINEAR LOEWY FACTORIZABLE ALGEBRAIC ODES AND HAYMAN S CONJECTURE

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1 ISRAEL JOURNAL OF MATHEMATICS 00 (XXXX, 1 37 DOI: /s NONLINEAR LOEWY FACTORIZABLE ALGEBRAIC ODES AND HAYMAN S CONJECTURE BY Tuen-Wai Ng Department of Mathematics, The University of Hong Kong Pofulam Road, Hong Kong ntw@maths.hu.h AND Cheng-Fa Wu Institute for Advanced Study, Shenzhen University Shenzhen, PR China cfwu@szu.edu.cn ABSTRACT In this paper, we introduce certain n-th order nonlinear Loewy factorizable algebraic ordinary differential equations for the first time and study the growth of their meromorphic solutions in terms of the Nevanlinna characteristic function. It is shown that for generic cases all their meromorphic solutions are elliptic functions or their degenerations and hence their order of growth are at most two. Moreover, for the second order factorizable algebraic ODEs, all the meromorphic solutions of them (except for one case are found explicitly. This allows us to show that a conjecture proposed by Hayman in 1996 holds for these second order ODEs. Dedicated to Professor Walter Hayman on the occasion of his 90th birthday Supported by PROCORE - France/Hong Kong joint research grant, F-HK39/11T and RGC grant HKU P. Supported by NSFC grant (No and RGC grant HKU P. 1

2 2 T.W. NG AND C.F. WU Isr. J. Math. 1. Introduction One important aspect of the studies of complex differential equations is to investigate the growth of their solutions which are meromorphic on the whole complex plane. A well nown problem in this direction is the following conjecture proposed by Hayman in [21] (see also [27, p. 344]. It is also referred as the classical conjecture in [6]. Hayman s conjecture for algebraic ODEs : If f is a meromorphic solution of P (z, f, f,, f (n = 0, where P is a polynomial in all its arguments, then there exist a, b, c R + such that (1 T (r, f < a exp n 1 (br c, 0 r <, where T (r, f is the Nevanlinna characteristic of f(z and exp l (x is the l times iterated exponential, i.e., exp 0 (x = x, exp 1 (x = e x, exp l (x = exp{exp l 1 (x}. Note that the conjecture is due to Ban [2] for the case n = 2. Also, it has been listed as an open problem by Eremeno [3, p. 491] and Rubel [31, p. 662] for the case of entire solutions. This conjecture is closely related to a false conjecture due to E. Borel on the growth of real-valued solutions. In [5], Borel proved that any real-valued solution defined on the interval (x 0, of the first-order algebraic ODE is dominated by exp 2 (x for all sufficiently large x (improvements of this result were later made by Lindelöf [28] and Hardy [20]. In the same paper, Borel dealt with higher-order ODEs as well and showed that all such solutions of n-th order algebraic ODEs are eventually dominated by exp n+1 (x. However, it was later pointed out by Fowler [16], Vijayaraghavan [35] etc. that Borel s proof in the higher-order case was incorrect. The counter-examples constructed by Vijayaraghavan etc. [35, 4] demonstrate that second-order algebraic ODEs may possess real-valued solutions dominating any given increasing function for a sequence of points tending to. Hayman s conjecture is true when n = 1 by a result of Gol dberg [18] while it is still open for any n 2. For general n N, it was proved by Eremeno, Liao and Ng [15] that the conjecture is true for the ODE P (f (n, f = 0, where P

3 Vol. 00, XXXX FACTORIZABLE ODES AND HAYMAN S CONJECTURE 3 C[x, y] is a non-constant polynomial. In fact, they proved that any meromorphic solution with at least one pole must be an elliptic function or its degenerations. For some other partial results of this conjecture, we refer the readers to [2, 11, 17, 19, 21] and the references therein. Since Hayman s conjecture seems to be out of reach currently, we introduce and study the following factorizable n-th order algebraic ODE (2 [D f n (u] [D f 2 (u][d f 1 (u](u α = 0, where u = u(z, D = d dz, α C and f i C[x] (i = 1, 2,..., n, n N. As we will see later, the study of ODE (2 is motivated by the consideration of Lowey decomposition (Theorem 1.2 to linear ODEs (see [32, 33] and the references therein and it also covers some interesting well-nown ODEs. Another reason to consider it is that the particular meromorphic solutions of (2 can have a tower structure because a solution of [D f (u] [D f 1 (u](u α = 0 will also be a solution of [D f +1 (u][d f (u] [D f 1 (u](u α = 0 for = 1,..., n 1 and it seems that one can get meromorphic solutions which grow faster and faster and eventually produce solutions which show that the estimate in (1 is sharp. This is at least the case when n = 2 and all f i are linear polynomials (see Remar A special case for the ODE (2 is that all the f i are constants, for which the equation (2 becomes linear and thus Hayman s conjecture holds. On the other hand, according to the following proposition, any linear ODE with constant coefficients can be rewritten in the form (2. Proposition 1.1: For any n N, the linear ODE (3 u (n (z + c n 1 u (n 1 (z + + c 0 = 0, c i C, i = 1, 2,..., n can be decomposed into the form (4 [D b n ] [D b 2 ][D b 1 ](u α = 0, D = d dz, for some α, b C. Proposition 1.1 is connected to a special case of Loewy decomposition (Theorem 1.2 and Corollary 1.3. To state them, we recall some terminologies. A differential operator L of order n is defined by (5 L := D n + r n 1 D n r 1 D + r 0

4 4 T.W. NG AND C.F. WU Isr. J. Math. where the coefficients r i, i = 1,..., n, are rational functions over Q, i.e., r i Q(z. L is called reducible if it can be represented as the product of two operators L 1 and L 2, i.e., L = L 1 L 2, both of order lower than n. In this case, L 1 is called the exact quotient of L by L 2, and L 2 is called the right factor of L. Otherwise, the operator L is called irreducible. The number of irreducible factors of L in any two decompositions into irreducible factors is the same and any two such decompositions are lined by a permutation of the irreducible factors (see Proposition 1.1 of [32]. It follows that there are only finitely many irreducible right factors of L. For any two operators L 1 and L 2, the least common left multiple denoted by Lclm( L 1, L 2 is the operator of lowest order such that both L 1 and L 2 divide it from the right. An operator which can be represented as Lclm of irreducible operators is called completely reducible. Given L, consider all its irreducible right factors. Let L (d1 1 be the Lclm of all these irreducible right factors and by construction L (d1 1 is the unique completely reducible right factors of maximal order d 1. Factoring L (d1 1 out from L and repeating the same procedure to the remaining left factor of L, we have the following. Theorem 1.2: ([29, 32, p. 4] Let L be an operator of order n as defined in (5, then it can be uniquely written as the product of completely reducible factors L (d of maximal order d over Q(z of the form where d d m = n. L = L (dm m L (dm 1 m 1 L (d1 1, Corollary 1.3 ([33]: Each factor L (d, = 1, 2,..., m, in Theorem 1.2 can be expressed as L (d = Lclm(l (e1 j 1, l (e2 j 2,... l (e j, where e e = d and each l (ei j i, i = 1,...,, is an irreducible operator of order e i over Q(z. The decomposition obtained in Theorem 1.2 is called the Loewy decomposition of L and it has been generalized to linear partial differential operators [34]. However, as far as we now, there is no similar study on Loewy decomposition for nonlinear ODEs. Therefore, we try to study nonlinear ODE of type (6 below and we call ODE (2 nonlinear Loewy factorizable algebraic ODE.

5 Vol. 00, XXXX FACTORIZABLE ODES AND HAYMAN S CONJECTURE 5 Among the non-linear cases of the equation (2, the simplest case is perhaps the one with deg(f j 1, i = 1, 2,..., n, which we will study in this paper. In this case, we may assume f i = a i u + b i, where a i, b i C, i = 1, 2,..., n. Then the equation (2 reduces to (6 [D (a n u + b n ] [D (a 2 u + b 2 ][D (a 1 u + b 1 ](u α = 0, D = d dz, and our main results are as follows. Theorem 1.4: For all n N and a = (a 1, a 2,..., a n C n \S, where S is the union of at most countably many hypersurfaces in C n, all meromorphic solutions (if they exist of the ODE (6 belong to the class W, which consists of elliptic functions and their degenerations. Hence, for any generic a C n, Hayman s conjecture is true for (6. The proof of Theorem 1.4 is based on a long and careful application of Painlevé analysis as well as a simple application of Wiman-Valiron theory [26, p. 51]. We expect that this general method can also be used to show that for other types of non-linear algebraic ODEs with constant coefficients, a generic choice of the coefficients will mae the corresponding ODE has all meromorphic solutions (if exist in the class W. If n = 1, then the equation (2 is a particular Riccati equation and its meromorphic solutions can be easily derived, which are given by α + b 1ce (αa1+b1z u(z = a 1 ce (αa1+b1z 1, αa 1 + b 1 0, (7 c arbitrary. 1 α a 1 z c, αa 1 + b 1 = 0, Meanwhile, we can see from above that the Hayman s conjecture is sharp for n = 1. Then the first non-trivial case for (2 is n = 2, which has been studied in [12], and we will show that Hayman s conjecture is true for n = 2 apart from an exceptional case. Theorem 1.5: Consider the ordinary differential equation (8 [D f 2 (u][d f 1 (u](u α = 0, where u = u(z, D = d dz, α C and f i(u = a i u+b i, a i, b i C, i = 1, 2. If either a 1 a 2 = 0 or 2 4a 1 a 2 N\{1, 2, 3, 4, 6}, then (7 is a particular meromorphic

6 6 T.W. NG AND C.F. WU Isr. J. Math. solution of the equation (8 and all other meromorphic solutions of (8 are given in Table 1, 2 and 3 in the Appendix. Remar 1.6: The ODE (8 reduces to the traveling wave reduction of the KPP equation [24] under certain choice of parameters. Remar 1.7: For n 2, the meromorphic solutions of (8 given in Table 1, 2 and 3 are particular solutions of the ODE (6 as well. Remar 1.8: After normalization and expansion, the following case of equation (8 remains unsolved ( 2 j u +(j 4uu (b 1 +b 2 u +u 2 u + b 1 (2u + b 2 = 0, j N\{1, 2, 3, 4, 6}, for which only particular meromorphic solutions have been found but not all of them. Remar 1.9: We note that equation (8 (which is equivalent to equation (17 is a special case of the equation (G in [22, p. 326]. In Ince s boo [22], a classification of all equations of the form (G such that all their solutions have no movable critical points is given. There, except for a few simple cases, no explicit solutions have been given while here we are interested in constructing all meromorphic solutions of (8 or (17. We would also lie to emphasize that in the proof of Theorem 1.5, Subcase A1 and Subcase A4 correspond to the canonical form VI in [22, p. 334] whereas Subcase A2 and Subcase A3 correspond to the canonical form X in [22, p. 334]. No explicit solutions have been given for either of these two forms in [22]. The readers can find these explicit solutions in Table 2. Table 1-3 may loo scary but we thin that they are of sufficient interest to applied mathematicians, physicists and engineers who are interested in explicit solutions of non-linear ODEs. Theorem 1.10: With the same assumption on a 1, a 2 as given in Theorem 1.5, Hayman s conjecture holds for the equation (8 and it is sharp in certain cases.

7 Vol. 00, XXXX FACTORIZABLE ODES AND HAYMAN S CONJECTURE 7 Remar 1.11: From the Appendix, we see that the equation (8 may have meromorphic solutions of the form u 1 (z = q i q 2 u 2 (z = αa 1 b 1 β e 2a 1 a 1 u 3 (z = α e q i q λ z (e qi q λ z ζ 0 ; g 2, 0 2b1 c 0 e b1z tanh ( 1 2 (e q i q λ z ζ 0 ; g 2, 0 + q, g 2 C, b 2 z 2 (c 1 J ν (ζ + c 2 Y ν (ζ, or (c 1 J ν (ζ + c 2 Y ν (ζ ( 2c0 e b1z + c 1 a 2 under some constraints on the parameters, and it will be shown in the proof of Theorem 1.10 that for u i, i = 1, 2, 3, Hayman s conjecture is sharp for n = 2. Here, J ν (ζ and Y ν (ζ are Bessel functions of the first and second inds respectively. Finally, Remar 1.11 shows that the ODE (8 may have meromorphic solutions outside the class W. 2. Proof of Proposition 1.1 Proof. We claim that for a fixed n N, the characteristic equation of (4 is given by Π n m=1(z b m = 0. We prove this by induction. Let A n = [D b n ] [D b 2 ][D b 1 ](u α, then the claim holds obviously for n = 1. Assume it is true for n =, then if n = + 1, as A +1 = da dz b +1A, the characteristic equation of the linear ODE A +1 = 0 is 0 = zπ m=1(z b m b +1 Π m=1(z b m = Π +1 m=1 (z b m. Next, for the equation (3, one may express its characteristic equation as 0 = z n + c n 1 z n c 0 = Π n m=1(z d m, d m C. Consequently, one can rewrite the equation (3 as the form of (4 by choosing α, b m C, 1 m n such that b m = d m and ( 1 n+1 απ n m=1b m = c 0 for c 1 = ( 1 n Π n m=1b m 0, otherwise α will be taen as an arbitrary constant.

8 8 T.W. NG AND C.F. WU Isr. J. Math. Remar 2.1: It is clear from the proof that the decomposition (4 for the equation (3 is unique up to a permutation of the b m s. 3. Proof of Theorem 1.4 We first introduce some terminologies and notations. Let I = (i 0, i 1,..., i n, i N {0}, 0 n and H(y, y,, y (n = I Λ c I y i0 (y i1 (y (n in, y = y(z, c I C\{0}. If y = z p, p N, then H(y, y,, y (n = I Λ C I z α I, where C I C, α I = i 0 p + i 1 (p i n (p n. Next, let A be the set of those negative integers p such that min I Λ α I is attained by at least two I s. Note that if A =, then H(y, y,, y (n = 0 has no meromorphic solutions with at least one pole. Suppose A, then for each p A, denote by Λ = {I Λ α I = min I Λ α I } and we define the dominant terms for each p A to be Ê = I Λ c I y i0 (y i1 (y (n in. Suppose u(z = + n=0 u nz n+p (u 0 0, p N with a pole at z = 0 is a meromorphic solution of H(y, y,, y (n = 0. Then if we plug y = u(z into H, we will get an expression of the form E = + j=0 E jz j+q = 0, E j C. Since y = u(z is a solution of H = 0, we must have E j = 0, for all j N {0}. Note that E 0 = E 0 (u 0 ; p is a polynomial in u 0 with coefficients depending on p. On the other hand, for j = 1, 2,..., we can express E j as: (9 E j P (u 0 ; ju j + Q j ({u l l < j}, where P (u 0 ; j is a polynomial in j determined by u 0 and Q j is a polynomial in j with coefficients in u l (l < j. In fact, it is nown that [13] (see also [8, p. 15] (10 P (u 0 ; j = lim z 0 z j q Ê (u 0 z p z j+p, where Ê (u is defined by Ê Ê(u + λv (uv := lim Ê(u. λ 0 λ

9 Vol. 00, XXXX FACTORIZABLE ODES AND HAYMAN S CONJECTURE 9 In order to have E j = 0 for all j N, we must have for each j, either 1 u j is uniquely determined by P (u 0 ; j and Q j, or 2 both P (u 0 ; j and Q j vanish, otherwise there is no meromorphic function satisfying H(y, y,, y (n = 0. Therefore if the polynomial P (u 0 ; j in j does not have any nonnegative integer root, then each u j is uniquely determined by P (u 0 ; j and Q j. Definition 3.1: The zeros of P (u 0 ; j are defined to be the Fuchs indices of the equation H(y, y,, y (n = 0 and the indicial equation is defined as P (u 0 ; j = 0. From the above definition and (10, one sees that the Fuchs indices of an ODE are determined by its dominant terms and the values of u 0. Therefore, to compute the Fuchs indices of the ODE (6, we have to find its dominant terms, denoted by Ên, and u 0. We will see that any terms involving b i s will not be included in the dominant terms when all the a i s are non-zero. Therefore, it would be useful to first loo at (11 D n = D n (u(z := [D a n u] [D a 2 u][d a 1 u]u. Then we may express D n as (12 D n = I Ω c I u i0 (u i1 (u (n in = u (n + ( 1 n Π n =1a u n+1 +, where c I C, i κ N {0}, κ = 0, 1,..., n, and we have Lemma 3.2: For any (i 0, i 1,..., i n Ω in (12, we have i 0 + 2i (n + 1i n = n + 1.

10 10 T.W. NG AND C.F. WU Isr. J. Math. Proof. We prove by induction. It is obvious for n = 1. Now suppose i 0 + 2i (n + 1i n = n + 1, then (13 aaad n+1 = [D a n+1 u]d n = dd n dz a n+1ud n [ n ] = c i0,i 1,...,i n i u i0 (u i1 (u (n u(+1 in u ( (i 0,i 1,...,i n =0 a n+1 c i0,i 1,...,i n u i0+1 (u i1 (u (n in = (i 0,i 1,...,i n (j 0,j 1,...,j n,j n+1 d j0,j 1,...,j n,j n+1 u j0 (u j1 (u (n jn (u (n+1 jn+1. Here (j 0, j 1,..., j n, j n+1 = (i 0 +1, i 1,..., i n, 0 or (i 0, i 1,..., i 1, i +1 +1,..., and in both cases we have j 0 + 2j (n + 1j n + (n + 2j n+1 = n + 2. Lemma 3.3: Suppose u(z = r=0 u rz r+p (u 0 0, p N is a meromorphic solution of the ODE (6 with all the a i 0, then for any n N (i p = 1. (ii The dominant terms E ˆ n of the equation (6 satisfies E ˆ n = D n and hence P (u 0 ; j does not depend on } b i s for j N. (iii u 0 { 1a1, 2a2,..., nan. Proof. (i For a fixed n N, we prove by contradiction. First rewrite the ODE (6 as (14 D n + I Ω c Iu i0 (u i1 (u (n in = 0, then by Lemma 3.2, one can see that i 0 + 2i (n + 1i n < n + 1 for any I = (i 0, i 1,..., i n Ω. Assume now p 2, then for any term u i0 (u i1 (u (n in in (14 with (i 0, i 1,..., i n (l, 0,..., 0, 0 l n + 1, according to Lemma

11 Vol. 00, XXXX FACTORIZABLE ODES AND HAYMAN S CONJECTURE , its order at z = 0 is = > i 0 p + i 1 (p i n (p n ( n n i (p + 1 ( + 1i =0 =0 ( n i (p + 1 (n + 1 =0 ( n ( + 1i (p + 1 (n + 1 =0 p(n + 1. As the order of ( 1 n Π n =1 a u n+1 at z = 0 is p(n + 1, which is lower than that of any other term in (14, it cannot be balanced unless u 0 = 0. Consequently, we must have p = 1. (ii As p = 1, we now that the order at z = 0 of each term u i0 (u i1 (u (n in in (14 is no less than (n + 1. Therefore E ˆ n consists of all terms with order (n + 1 at z = 0, and thus E ˆ n = D n. (iii To compute u 0, without loss of generality, we may assume u(z = u0 z. We then prove by induction. } It is obvious for n = 1. Suppose u 0 { 1a1, 2a2,..., nan for an n N and we consider the n + 1 case. If u 0 = for some 1 n, then since a Ên = D n we have by direct checing that and hence ( u0 D n+1 = z ( u0 D n = 0, z ( ddn dz (u0 ( ddn dz z (u0 = 0, z a +1 u 0 z D n For u 0 = n + 1 a n+1, from (13 we now that ( u0 = 0. z

12 12 T.W. NG AND C.F. WU Isr. J. Math. ( u0 D n+1 z [ n ] = c i0,i 1,...,i n i u i0 (u i1 (u (n u(+1 in a u ( n+1 ud n ( u 0 z (i 0,i 1,...,i n =0 [ ] = c i0,i n 1,...,i n ( + 1i u i0 (u i1 (u (n in a n+1 ud n ( u 0 z z (i 0,i 1,...,i n =0 = 1 n ( + 1i c i0,i z 1,...,i n u i0 (u i1 (u (n in a n+1 ud n ( u 0 z =0 (i 0,i 1,...,i n ( = 1 n ( u0 ( u0 ( + 1i D n a n+1 ud n z z z =0 ( = n + 1 ( u0 a n+1 u D n z z = 0. On the other hand, it is easy to chec that zn+2 ( D u0 n+1 u z is a polynomial in u 0 of degree n + 1 with coefficients depending only on a i, 1 0 ( i n + 1, and hence the set of nonzero roots u 0 of z n+2 D u0 { n+1 z = 0 is 1, 2,..., n + 1 }. a 1 a 2 a n+1 ( Now we denote by R n (u 0 = z n+1 D u0 n z, then Rn (u 0 is a polynomial } of degree n + 1 in u 0 with the set of zeros {0, 1a1, 2a2,..., nan. Let the indicial equation of D n = 0 be P n (u 0 ; j = 0. From Lemma 3.3, we now that P n (u 0 ; j = 0 is also the indicial equation of (6 when all the a i s are non-zero. Let the indicial equation of dd n dz Proposition 3.4: For any n N, = 0 be P n (u 0; j = 0, then we have 1 P n (u 0 ; j = P n (u 0 ; j(j n 1; 2 P n+1 (u 0 ; j = P n (u 0 ; j(j n 1 a n+1 u 0 a n+1 R n (u 0 ;

13 Vol. 00, XXXX FACTORIZABLE ODES AND HAYMAN S CONJECTURE 13 3 If u 0 = a, where 1 n, then P n+1 (u 0 ; j = 0 P n (u 0 ; j = 0 or j = n + 1 a n+1 a. If u 0 = n + 1 a n+1, then P n+1 (u 0 ; j = jp n (u 0 ; j a n+1 R n (u 0. Remar 3.5: If we choose a 1 = a 2 = 1, then P 1 ( 1, j = j + 1 and P 2 ( 1, j = j 2 1. Proof. We set v = u 0 ( 1! for = 0, 1, 2,..., then we have v +1 = ( + 1v and from (10 and Ên = D n = I Ω c Iu i0 (u i1 (u (n in, P n (u 0 ; j = I c I [ n =1 ] Π n i α=1vα iα (j 1(j 2 (j + i 0 v i0 1 0 v i1 1 v vin n Since dd n dz = I c I [ n =0 i u i0 (u i1 (u (n u(+1 in u ( ], we have from (10, P n (u 0 ; j = I n c I =1 i n m=1 m,+1 +v i0 1 0 v i1 1 vi 1 i m v i0 0 vi1 1 vi 1 v i (i 1v i0 0 vi1 1 vi 2 +1 vn in v i +1+1 v i vm im 1 +1 vn in (j 1 (j v in n (j 1 (j m +(i v i0 0 vi1 1 vi 1 v i vin n (j 1 (j (j 1 +i 0 ( n m=2 i m v i0 1 0 v i1+1 +(i 0 1v i0 2 0 v i1+1 1 vm im 1 1 vm im 1 v in n (j 1 (j m v in n + (i 1 + 1v i0 1 0 v i1 1 vin n (j 1 ].

14 14 T.W. NG AND C.F. WU Isr. J. Math. Now we compute each term in the above equality, for 1 n, [ c I I = I n m=1 m,+1 +v i0 1 0 v i1 1 vi 1 i m v i0 0 vi1 1 vi 1 v i (i 1v i0 0 vi1 1 vi 2 +1 vn in v i +1+1 v i vm im 1 +1 vn in (j 1 (j v in n (j 1 (j m +(i v i0 0 vi1 1 vi 1 v i vin n (j 1 (j (j 1 [ n c I ( + 1 i m v i0 0 vi v i vim 1 m vn in (j 1 (j m m=1 m,+1 ( + 1v i0 1 0 v i v i vin n ( + 1(i 1v i0 0 vi 1 v i vin n (j 1 (j ] ( + 1(i v i0 0 vi v i vn in (j 1 (j (j 1 { = ( + 1 P n (u 0 ; j + [ c I v i0 0 vi 1 v i vin n (j 1 (j I ]} +v i0 0 vi v i vn in (j 1 (j (j 1 { = ( + 1 P n (u 0 ; j + [ c I v i0 0 vi 1 v i vin n (j 1 (j I 1 ] } + 1 vi0 0 vi 1 v i vin n (j 1 (j (j 1 = ( + 1[P n (u 0 ; j I c I j + 1 vi0 0 vi 1 v i vin n (j 1 (j ] ] = ( + 1P n (u 0 ; j + I c I jv i0 0 vi 1 v i vin n (j 1 (j. Similarly, for = 0, [ c I I n m=2 i m v i0 1 0 v i1+1 +(i 0 1v i0 2 0 v i1+1 = P n (u 0 ; j + I 1 vm im 1 1 vm im 1 v in n c I jv i0 1 0 v i1 1 vin n. v in n (j 1 (j m ] + (i 1 + 1v i0 1 0 v i1 1 vin n (j 1

15 Vol. 00, XXXX FACTORIZABLE ODES AND HAYMAN S CONJECTURE 15 Therefore n P n (u 0 ; j = ( + 1i P n (u 0 ; j + jp n (u 0 ; j =0 = P n (u 0 ; j(j n 1. As D n+1 = dd n dz a n+1ud n, one can easily deduce that P n+1 (u 0 ; j = P n (u 0 ; j a n+1 u 0 P n (u 0 ; j a n+1 R n (u 0 = P n (u 0 ; j(j n 1 a n+1 u 0 a n+1 R n (u 0. Finally, 3 can be obtained by directly substituting the values of u 0 into the equality in 2. Remar 3.6: The above method can be used to get a similar relation between the indicial equations of H(y, y,, y (n = 0 and dh = 0, where y = y(z dz and H is a polynomial in y and its derivatives. Proof of Theorem 1.4. Let L i = {(a 1, a 2,..., a n C n a i = 0}, i = 1, 2..., n. Then for any a C n \ ( n i=1 L i, one can see immediately that the equation (6 has only one top degree term ( 1 n Π n =1 a u n+1 and thus by Wiman-Valiron theory, we conclude that it does not admit any transcendental entire solution. On the other hand, it is easy to see that the equation (6 does not have any nonconstant polynomial solution. Therefore any meromorphic solution of the ODE (6 has at least one pole on C. Then by Lemma 3.3, any meromorphic solution of (6 with a pole at z = z 0 C can be expressed as u(z = r=0 u r(z z 0 r 1, u 0 0. Now for any 1 n, j N {0}, we define H,j := {(a 1, a 2,..., a n C n \ ( n i=1l i a n P n (u 0 ; j = 0, u 0 = a }, where P n (u 0 ; j is the indicial equation of (6. From Proposition 3.4-3, one can easily chec by induction that H,j C n \ ( n i=1 L i and thus it defines a hypersurface in C n. Next, we define ( n S := L i H,j, i=1 1 n, j N {0}

16 16 T.W. NG AND C.F. WU Isr. J. Math. then according to the method used in [10, 14], for any a C n \S, since the equation (6 does not have any nonnegative integer Fuchs index for any u 0 { z = 1, 2,..., n}, all meromorphic solutions of the equation (6 belong to the class W. 4. Proof of Theorem 1.5 We first recall some lemmas that will be needed. Lemma 4.1: [7, p. 210] Let f and g be two transcendental entire functions. Then log M(r, f g T (r, f g lim =, lim =, r log M(r, f r T (r, f log M(r, f g T (r, f g lim =, lim =. r log M(r, g r T (r, g Lemma 4.2: The equation (15 w (z + cw (z 6 λ (w(z e 1 (w(z e 2 = 0, λ 0 has non-constant meromorphic solutions if and only if c ( c 2 λ + 25e 1 25e 2 ( c 2 λ 25e e 2 = 0 and they are given respectively as follows 1 if c = 0, then the general solution to the equation (15 is w 1 (z = λ (z z 0 ; g 2, g (e 1 + e 2, where g 2 = 3(e1 e22 λ and z 2 0, g 3 are arbitrary. 2 if c 2 λ = 25(e i e j 0, i, j {1, 2}, then the general solution to the equation (15 is w 2 (z = (e i e j e 2c 5 z (e c 5 z ζ 0 ; 0, g 3 + e j, where ζ 0, g 3 C are arbitrary, see [1, 25]. Lemma 4.3: [26, p. 53] All solutions of the linear differential equation (16 L(f := f (n + α n 1 (zf (n α 0 (zf = 0 with entire coefficients α 0 (z,..., α n (z, are entire functions.

17 Vol. 00, XXXX FACTORIZABLE ODES AND HAYMAN S CONJECTURE 17 Proof of Theorem 1.5. Expanding (8 gives (17 u +u (αa 1 + ( 2a 1 a 2 u b 1 b 2 +(u α (a 1 u + b 1 (a 2 u + b 2 = 0. Before presenting the complete analysis of meromorphic solutions of the ODE (8, we give a simple observation to derive some of its particular meromorphic solutions. Let G(z = [D a 1 u b 1 ](u α, then we have [D a 2 u b 2 ]G(z = 0 from which one can solve for G(z = βe a 2udz e b2z, β C. If β = 0, from the Riccati equation G(z = [D a 1 u b 1 ](u α = 0, we are able to obtain the particular meromorphic solution (7 of the equation (8. For β 0, in order to characterize all the meromorphic solutions of (8, we distinguish the following cases according to the values of a i and b i, i = 1, 2. (I a 1 a 2 0, 2a 1 + a 2 0, (II a 1 a 2 0, 2a 1 + a 2 = 0, (III a 1 = 0, a 2 0, b 1 0, (IV a 1 0, a 2 = 0, b 2 0, (V other cases. (I a 1 a 2 0, 2a 1 + a 2 0. For the convenience of applications, we first compare equation (17 with the following second order ODE (18 1 ( 2 d 2 (u α 1 (u α 2 (u α 3 + (u bu + u = 0, 0, 2 d One can see immediately that the ODE (17 is a special case of (18 and further calculations imply that the ODE (18 can be written in the form of (8 if and only if the coefficients involved in (18 satisfy either one of the following two conditions [(αi + α j ( + d + 2α ( d 4b] = 0, [(αi + α j ( d + 2α ( + d 4b] = 0, where i, j, {1, 2, 3} are distinct and the product is taen over all the permutations of (123. We now come bac to the ODE (8. Suppose u(z is a meromorphic solution of the ODE (8 with a pole at z = z 0, W.L.O.G., we may assume z 0 = 0 then u(z = + j=p u jz j, p N, u p 0. Substituting the series expansion of u(z

18 18 T.W. NG AND C.F. WU Isr. J. Math. into (8 gives p = 1, u 1 = 1 a 1 or 2 a 2 and the Fuchs indices j = 1, 2 a2 a 1, u 1 = 1 a 1, j = 1, 2 4a1 a 2, u 1 = 2 a 2. We denote by j 1 = 2 a2 a 1, j 2 = 2 4a1 a 2, then we have (19 j 1 = j 2 = 0, or 2 j j 2 = 1. For G(z 0, we let H(z = e a 2udz which satisfies H (z = a 2 u(zh(z and (20 [D a 1 u b 1 ](u α = βe b2z H(z, β 0, hence, u(z is meromorphic if and only if H(z is meromorphic. By the substitution of u = 1 H into (20, we have a 2 H (21 e b2z a 2 2βH 3 + αa 2 2b 1 H 2 + a 2 HH + (αa 1 a 2 a 2 b 1 HH (a 1 + a 2 (H 2 = 0. If we let H(z = e b2z h(z, which implies u(z = and u(z is meromorphic if and only if so is h(z, then the ODE (21 reduces to h (z a 2h(z b2 a 2 (22 h 2 (a 2 b 1 a 1 b 2 (αa 2 + b 2 + hh (a 1 (αa 2 + 2b 2 a 2 b 1 a 2 2βh 3 + a 2 hh (a 1 + a 2 (h 2 = 0, β 0. It is obvious that the ODE (22 does not have any polynomial solutions. By Wiman-Valiron theory [26, p. 51], we can conclude that (22 does not have any transcendental entire solutions. Hence, each meromorphic solution of the ODE (22 should have at least one pole on C. Next suppose h(z is a meromorphic solution of (22, W.L.O.G, we assume that it has a pole at z = 0 and h(z = + j=p h jz j, p N, h p 0. Now the following cases are distinguished. (A If both of j 1 and j 2 are integers, then by solving the Diophantine equation (19, we have three choices j 1 = j 2 = 0 a 2 = 2a 1, {j 1, j 2 } = {3, 6} a 2 = a 1 or a 2 = 4a 1, {j 1, j 2 } = {1, 2} a 2 = a 1 or a 2 = 4a 1.

19 Vol. 00, XXXX FACTORIZABLE ODES AND HAYMAN S CONJECTURE 19 Subcase A0. If a 2 = 2a 1, then the ODE (22 reduces to (23 (2b 1 b 2 h 2 (2αa 1 + b 2 + 2hh (αa 1 b 1 + b 2 4a 1 βh 3 + 2hh 3 (h 2 = 0, β 0. One can see that there does not exist any negative integer p with h p 0 such that h(z = + j=p h jz j satisfies (23 and therefore in this case all the meromorphic solutions of (8 are those given in (7. Subcase A1. For a 2 = a 1, the ODE (22 reduces to (24 h(z (αa 1 b 1 + 2b 2 h (z (b 2 b 1 h(z 2 (αa 1 + b 2 a 1 βh(z 3 +h(zh (z 2h (z 2 = 0. Let h(z = 1, then the ODE (24 reduces to a linear ODE v(z v + ( αa 1 + b 1 2b 2 v + v (b 1 b 2 (αa 1 + b 2 a 1 β = 0, with solutions c 2 e ( αa1 b2z + c 1 e (b1 b2z a + 1β (b, 2 1 b 2(αa 1+b 2 i=1 (αa 1 + b i (b 1 b 2 0 βb 1 α(b 1 b 2 + e (b1 b2z (c 2 2 z + c 1, b 1 + αa 1 = 0, (b 1 b 2 (αa 1 + b 2 0; v(z = c 2 c1e (αa 1 +b 2 z +a 1βz αa 1+b 2, b 1 = b 2, αa 1 + b 2 0; c 1e z(αa 1 +b 1 +a 1(αc 2+βz+b 1c 2 αa 1+b 1, b 1 b 2, αa 1 + b 2 = 0; 1 2 a 1βz 2 + c 2 z + c 1, b 1 = b 2 = αa 1. After substitution, we obtain the follow meromorphic solutions of the ODE (8 (b 1 b 2(αa 1+b 2(αa 1c 2 b 1c 1e z(αa 1 +b 1 a 1βb 2e z(αa 1 +b 2, a 1(a 1(α(b 1 b 2(c 1e z(αa 1 +b 1 +c 2+βe z(αa 1 +b 2 +(b 1 b 2b 2(c 1e z(αa 1 +b 1 +c 2 (b 1 + αa 1 (b 1 b 2 (αa 1 + b 2 0; α(α(b 1 b 2 2 e b 1 z (b 1(c 2z+c 1+c 2+βb 1b 2e b 2 z, b 1(α(b 1 b 2 2 e b 1 z (c 2z+c 1+βb 1e b 2 z u(z = b 1 + αa 1 = 0, (b 1 b 2 (αa 1 + b 2 0; e z(αa 1 +b 2 (b 2 2 c2 a1(β+b2(βz αc2+αa1c1 a 1(e z(αa 1 +b 2 (a 1(βz αc 2 b 2c 2+c 1 a 1(αa 1(αc 2+βz β+αb 1c 2 b 1c 1e z(αa 1 +b 1 a 1(c 1e z(αa 1 +b 1 +a 1(αc 2+βz+b 1c 2 2a 1(αc 1+z(β+αc 2+αa 2 1 βz2 +2c 2 a 1(a 1βz 2 2(c 2z+c 1, b 1 = b 2 = αa 1, where β, c 1, c 2 C are arbitrary., b 1 = b 2, αa 1 + b 2 0;, b 1 b 2, αa 1 + b 2 = 0;

20 20 T.W. NG AND C.F. WU Isr. J. Math. Remar 4.4: In general, the above u(z does not belong to the class W. For b 1 = b 2 = α = 0, β = 1, a 1 = a 2 = 1, we recover the general solution u(z = 1 z a + 1 z b, a + b = 2c 2, ab = 2c 1, c 1, c 2 C to the ODE u + 3uu + u 3 = 0 [30, 9]. Remar 4.5: It seems that we have three arbitrary constants β, c 1 and c 2 to a second order ODE, but actually the arbitrariness of β can be absorbed into c 1 and c 2. Subcase A2. For the case a 2 = a 1, the ODE (22 reduces to the Fisher equation (25 h (z ( αa 1 + b 1 + 2b 2 h (z (b 1 + b 2 h(z (αa 1 b 2 +a 1 βh(z 2 = 0. According to Lemma 4.2, the necessary condition for the existence of meromorphic solutions of the ODE (25 is c ( c 2 λ + 25e 1 25e 2 ( c 2 λ 25e e 2 = 0, where c = αa 1 b 1 2b 2, λ = 6 a 1 β, e 1 = 0, e 2 = (b 1 + b 2 (αa 1 b 2. a 1 β (i If c = 0, then the general solution to the equation (25 is h(z = λ (z z 0 ; g 2, g (e 1 + e 2, where g 2 = 3(e1 e22 λ and z 2 0, g 3 are arbitrary. (ii For c 2 λ = 25(e i e j 0, i, j {1, 2}, then the general solution to the equation (25 is h(z = (e i e j e 2c 5 z (e c 5 z ζ 0 ; 0, g 3 + e j, After substitution, we obtain the following meromorphic solutions of the ODE (8 (i if a 2 = a 1, c = 0, u(z = 12 (z z 0 ; g 2, g 3 a 1 [(b 2 αa (z z 0 ; g 2, g 3 ] + b 2 a 1,

21 Vol. 00, XXXX FACTORIZABLE ODES AND HAYMAN S CONJECTURE 21 where g 2 = 1 12 (b 2 αa 1 4 and z 0, g 3 are arbitrary. (ii for a 2 = a 1, c 2 λ = 25(e i e j 0, i, j {1, 2}, ( u(z = b c (e i e j 2e cz a 1 a 1 [5 (e i e j e cz 5 where ζ 0, g 3 are arbitrary. (e 1 5 (cz ζ 0 ; 0, g 3 + (e 1 5 (cz ζ 0 ; 0, g 3 (e 15 (cz ζ 0 ; 0, g 3 ], + 5e j e 3cz 5 Remar 4.6: The above solutions may degenerate to rational functions in exponential or rational functions due to the degeneration of. Subcase A3. If a 2 = 4a 1, set u(z = w(z + α, then the ODE (8 becomes (26 [D + 4a 1 w b 2][D a 1 w b 1]w = 0, where b 1 = b 1 + αa 1, b 2 = b 2 4αa 1. The compatibility conditions for the existence of meromorphic solutions of (26 are b 2 = 2b 1, 2b 1, or 6b 1. Note that, under the compatibility conditions, the equation (26 can be factorized into another form (27 [D + α 1 w β 2 ][D α 1 w β 1 ](w α 0 = 0. Here α 1 β 1 β 2 α 0 = 2a 1 2b 1 b 1 0, 2a 1 b 1 2b 1 0 or 2a 1 3b 1 0 b 1 a 1 which correspond to b 2 = 2b 1, 2b 1, or 6b 1 respectively. Remar 4.7: The ODE (8 admits more than one distinct factorizations only for some specific cases including (26 with the compatibility conditions satisfied. Since the equation (27 shares the same form as (8 with a 2 = a 1, one can obtain its meromorphic solutions given below by using the results of Subcase A2.

22 22 T.W. NG AND C.F. WU Isr. J. Math. For b 2 = 2b 1, 12 (z z 0 ; g 2, g 3 u(z = 2a 1 ((αa 1 + b (z z 0 ; g 2, g 3 b 1 αa 1, 2a 1 where g 2 = 1 12 (b 1 + αa 1 4 and z 0, g 3 are arbitrary. For b 2 = 2b 1 0, [ u(z = c (e 2 e 1 2e cz 5 2a 1 [5 (e 2 e 1 e cz 5 b 1 a 1, (e 1 5 (cz ζ 0 ; 0, g 3 + (e 1 5 (cz ζ 0 ; 0, g 3 ] (e 15 (cz ζ 0 ; 0, g 3 ] + 5e 1 e 3cz 5 where c = 5 (αa 1 + b 1 0, e 1 = 0, e 2 = 3 (αa 1 + b 1 2, and β a 1 β 0, ζ 0, g 3 are arbitrary. For b 2 = 6b 1 0, [ ] c (e 2 e 1 2e cz 5 (e 1 5 (cz ζ 0 ; 0, g 3 + (e 1 5 (cz ζ 0 ; 0, g 3 u(z = α ], 2a 1 [5 (e 2 e 1 e cz 5 (e 15 (cz ζ 0 ; 0, g 3 + 5e 1 e 3cz 5 where c = 5 (αa 1 + b 1 0, e 1 = 0, e 2 = 3 (αa 1 + b 1 2, and β 0, ζ 0, g 3 a 1 β are arbitrary. Remar 4.8: The above solutions may degenerate to rational functions in exponential or rational functions due to the degeneration of. Subcase A4. If a 2 = 4a 1, then the ODE (22 reduces to 2h(z (2αa 1 2b 1 + b 2 h (z + (4b 1 b 2 h(z 2 (4αa 1 + b 2 16a 1 βh(z 3 +4h(zh (z 5h (z 2 = 0, with Fuchs indices j = 1, 1 and the compatibility condition (28 2αa 1 2b 1 + b 2 = 0. Again, we mae the change of variables u u + α, b 1 b 1, b 2 b 2, where b 1 = b 1 + αa 1, b 2 = b 2 + 4αa 1, then with the compatibility condition b 2 = 2b 1 satisfied, the ODE (21 reduces to 16a 1 βe 2b 1 z H(z 3 4b 1H(zH (z + 4H(zH (z 5H (z 2 = 0, β 0.

23 Vol. 00, XXXX FACTORIZABLE ODES AND HAYMAN S CONJECTURE 23 (29 (30 For b 1 0, performing the transformation H(z = v(ζ, ζ = e b 1 z gives 16a 1 β (b 1 2 v3 4vv + 5v 2 = 0, = d dζ. Upon integration of (29, we have ( c 0 v 5 + (v 2 16a 2 1β b 2 v 3 = 0, c 0 arbitrary, 1 which has the general solution (b 1 2 4a v(ζ = 1β(ζ+c 1, c 2 0 = 0, 256c 0(b 1 4 (256a 1β+c 0[b 1 (ζ c1]2, c 2 0 0, For b 2 = 2b 1 = 0, the ODE (8 reduces to [D 4a 1 u][d a 1 u]u = 0, c 0, c 1 arbitrary. which admits another factorization [D α 1 u][d α 1 u]u = 0, α 1 = 2a 1. The above ODE belongs to Subcase A1, and hence we can obtain the following solution of the ODE (30 1 u(z = 2a 1 (z c 0 1 2a 1 (z c 1, c 0, c 1 arbitrary. After substitution, with (28 satisfied, we obtain the meromorphic solutions of the ODE (8 1 α 2a 1 1(z c 0 2a, αa 1(z c b 1 = 0, u(z = α (αa1+b1ez(αa 1 +b 1, c 2a 1(e z(αa 1 +b 1 +c 1 0 = 0, αa 1 + b 1 0, α c0(αa1+b13 e z(αa 1 +b 1 (e z(αa 1 +b 1 c 1 a 1(256a 1β+c 0(αa 1+b 1 2 (e z(αa 1 +b 1 c 1 2, c 0 0, αa 1 + b 1 0, where c 0, c 1 are arbitrary. Again, it seems that we have three arbitrary constants to a second order ODE, but actually the arbitrariness of β can be absorbed into c 1 and c 2. (B If only one of j 1 and j 2 is an integer, then we should have a 2 ±a 1, ±4a 1 or 2a 1. For the ODE (22, one can chec that p = 2, h 1 = 2(a 2 2a 1 a 2 2 β and its Fuchs indices are j = 1, 2 4a1 a 2.

24 24 T.W. NG AND C.F. WU Isr. J. Math. Subcase B1. If 2 4a1 a 2 N {0} and 2 a2 a 1 Z, i.e., 2 4a 1 N\{1, 2, 3, 4, 6}, then the method in [10, 14] fails and so in this case meromorphic solutions are unfortunately not yet nown. Subcase B2. If 2 4a1 a 2 N {0, 2} (a 2 a 1, then h belongs to the class W and so is u. Using the argument used in [10, 14], one can find all the meromorphic solutions of (22 2(a 2 2a 1(αa 1+b 1 2 e 2(z z 0 (αa 1 +b 1 a 2 2 β(e(z z 0 (αa 1 +b 1 1 2, b 2 = 2αa 1 αa 2 + 2b 1, αa 1 + b 1 0; 2(a 2 2a 1(αa 1+b 1 2 a 2 2 β(e(z z 0 (αa 1 +b 1 1, b 2 2 = 2αa2 1 2a1b1+a2b1 a 1, αa 1 + b 1 0; h(z = (2a 1 a 2(αa 1+b 1 2 u(z = 2a 2 1 β (e a 2 (z z 0 (αa 1 +b 1 4a 1 e a 2 (z z 0 (αa 1 +b 1 4a 1 2 2(a 2 2a 1, b a 2 1 = αa 1, b 2 = αa 2. 2 β(z z02 a 2, b 2 = a2b1 αa1a2 2a 1, αa 1 + b 1 0; After substitution, we obtain the meromorphic solutions of the ODE (8 in Subcase B2 2αa 1+αa 2 2b 1 a 2 2(αa 1+b 1 a 2(e (z z 0 (αa 1 +b 1 1 b1 a2(z z0(αa1+b1 αa1+b1e 2a 1 a 2 (z z 0 (αa 1 +b 1 a 1 (e 2a 1 2(αa 1+b 1 a 2(e (z z 0 (αa 1 +b 1 1, b 2 = 2αa 1 αa 2 + 2b 1, αa 1 + b 1 0, 2 a b2 2(z z 0 a 2, b 1 = αa 1, b 2 = αa 2. a 1, b 2 = 2αa2 1 2a1b1+a2b1 a 1, αa 1 + b 1 0, 1, b 2 = a2b1 αa1a2 2a 1, αa 1 + b 1 0, where z 0 is arbitrary. (C If neither j 1 nor j 2 is an integer, then we conclude that all meromorphic solutions of the ODE (8 belong to the class W and they can be found by using the same method as that in Subcase B2. (II a 1 a 2 0, a 2 = 2a 1. We consider a more general ODE which includes the equation (17 and hence equation (8 (31 u (z + cu (z 2 λ 2 (u(z q 1 (u(z q 2 (u(z q 3 = 0,

25 Vol. 00, XXXX FACTORIZABLE ODES AND HAYMAN S CONJECTURE 25 where λ( 0, c, q 1, q 2, q 3 C, for which the Fuchs indices are 1, 4 and the compatibility conditions are (32 c (cλ + q 1 + q 2 2q 3 (cλ + q 1 2q 2 + q 3 (cλ 2q 1 + q 2 + q 3 = 0, if u 1 = λ, c (cλ + 2q 1 q 2 q 3 (cλ q 1 + 2q 2 q 3 (cλ q 1 q 2 + 2q 3, if u 1 = λ. Now we compare the ODE (31 with the equation (8. One can chec that the compatibility conditions (32 hold if and only if c = 0 or, c = αa 1 b 1 b 2 0, λ 2 = 1 a 2 1 {q 1, q 2, q 3 } = {α, b 1, a 1 b 2 2a 1 }. In other words, the compatibility conditions (32 hold if and only if c = 0 or the equation (31 can be factorized into the form (8. If c = 0, then the ODE (31 reduces to a first order Briot-Bouquet differential equation through multiplying it by u and performing an integration. Therefore all its meromorphic solutions belong to the class W and can be found by using the method introduced in [10, 14]. For c 0 and assuming (32 from now on, due to the symmetry in (32 and the fact that u(z has at least one pole in C, it suffices to consider the case c = q 1 + 2q 2 q 3 λ 0 and one choice for a i, b i, i = 1, 2 and α is a 1 = 1 λ, a 2 = 2 λ, b 1 = q 3 λ, b 2 = 2q 2 λ, α = q 1. Define G(z, H(z, h(z in the same way as in the case 1, that is u = λ H 2 H, H(z = e b2z h(z. Then we have u is meromorphic H is meromorphic h is meromorphic. For β 0, using the same argument as in the case 1, we have (33 2q 3 HH + 4q 1q 3 λ H2 4βe 2q2 λ z H 3 + 2λHH 2q 1 HH λh 2 = 0, and (34 2h ((q 1 + q 3 2q 2 h λh + 4 λ (q 3 q 2 (q 1 q 2 h 2 λh 2 4βh 3 = 0. Suppose h(z is a meromorphic solution of (34, W.L.O.G, we assume that it has a pole at z = 0 and h(z = + j=p h jz j, p N, h p 0 then one can

26 26 T.W. NG AND C.F. WU Isr. J. Math. chec that p = 2 and the Fuchs indices of the ODE (34 are 1, 4 with the compatibility condition (35 (q 1 + q 2 2q 3 (2q 1 q 2 q 3 (q 1 2q 2 + q 3 = 0, which implies q 3 = 1 2 (q 1 + q 2 or q 1 = 1 2 (q 2 + q 3 since c = 2q 2 q 1 q 3 λ Then by the substitution of (35, the ODE (34 reduces to λ 2 h (z 2 + λh(z (2λh (z + 3 (q 2 q 1 h (z +2 (q 2 q 1 2 h(z 2 4βλh(z 3 = 0, q 3 = 1 2 (q 1 + q 2, or λ 2 h (z 2 + λh(z (2λh (z + 3 (q 2 q 3 h (z +2 (q 2 q 3 2 h(z 2 4βλh(z 3 = 0, q 1 = 1 2 (q 2 + q Next, it suffices to consider the case q 3 = 1 2 (q 1 + q 2 due to the symmetry in the above two equations. By the translation against the dependent variable u, we may further assume q 3 = 0 which implies q 1 + q 2 = 0. Let us come bac to equation (33, which by the substitution of (35 with q 1 = q 2 0, q 3 = 0 reduces to λh (z 2 + 2H(z (λh (z + q 2 H (z 4βH(z 3 e 2q 2 z λ = 0. Performing the transformation H(z = v(ζ, ζ = e q 2 λ z gives (36 Upon integration of (36, we have which has the general solution where g 2 = Cβλ 2q2 2, g 3 = 0, C, ζ 0 C. 2vv (v 2 4βλ q2 2 v 3 = 0, = d dζ. (v 2 2βλ q2 2 v 3 + Cv = 0, C C v(ζ = 2q2 2 βλ (ζ ζ 0; g 2, g 3,

27 Vol. 00, XXXX FACTORIZABLE ODES AND HAYMAN S CONJECTURE 27 Finally, for c = q 1 + 2q 2 q 3 0 and q 3 = 1 λ 2 (q 1 + q 2, which implies c = 2q 1 q 2 q 3, we obtain the meromorphic solution of the ODE (31 (which λ meanwhile is the general solution u(z = q 2 q 3 2 e q 2 q 3 λ z (e q2 q3 λ z ζ 0 ; g 2, 0 (e q 2 q 3 λ z ζ 0 ; g 2, 0 + q 3, ζ 0, g 2 C. (III a 1 = 0, a 2 b 1 0. We first loo at the entire solution u of the equation (8, for which we now that (37 βe (a 2u+b 2dz = u b 1 (u α, β C. For β = 0, the entire solutions are given by α + ce b1z, b 1 0, u(z = c arbitrary. c, b 1 = 0, For β 0, we claim that the equation (37 does not have any nonconstant entire solution. Otherwise, suppose u is a transcendental entire solution of (37 as one can chec immediately that (37 does not admit any nonconstant polynomial solution. Let U(z = e (a 2u+b 2dz, then U is transcendental entire and the equation (37 becomes It implies that βe U = 1 a 2 (U b 1 U + b 1 b 2 + αb 1 a 2. T (r, e U = O(T (r, U, for all r (0, + outside a possible exceptional set with finite linear measure, which contradicts Lemma 4.1. Next, we consider meromorphic solutions of (8 with at least one pole on C. In this case, the ODE (17 reduces to (38 u + u ( a 2 u b 1 b 2 + b 1 (u α (a 2 u + b 2 = 0, which has the Fuchs indices j = 1, 2 with compatibility condition for the existence of meromorphic solutions: αa 2 2b 1 + b 2 = 0.

28 28 T.W. NG AND C.F. WU Isr. J. Math. We mae the change of variables u u + α, b 1 b 1, b 2 b 2, where b 1 = b 1 + αa 1, b 2 = b 2 + 4αa 1, then with b 2 = 2b 1 0, the ODE (21 reduces to a 2 βe 2b1z H(z 3 b 1 H(zH (z + H(zH (z H (z 2 = 0, β 0. Performing the transformation H(z = v(ζ, ζ = e b1z gives (39 Upon integration of (39, we have (v 2 2 a 2 β b 2 v 3 vv + v 2 = 0, = d 1 dζ. a 2β b 2 v 3 + c 0 v 2 = 0, c 0 arbitrary, 1 which has the general solution 4b 2 1 a v(ζ = 2β(ζ c 1, c 2 0 = 0, b2 1 2c 2 0 a 2β cosh(, c 2c 0 0, 0ζ+c 1+1 c 1 arbitrary. After substitution, with (4 satisfied, we obtain the meromorphic solution of the ODE (38 α 2b1eb 1 z u(z =, c a 2(e b 1 z c 1 0 = 0, 2b1c 0e α b 1 z tanh( 1 2( 2c 0e b 1 z +c 1 a 2, c 0 0, where c 0, c 1 are arbitrary. Remar 4.9: If a 1 = 0, the particular solution (7 is entire. (IV a 1 b 2 0, a 2 = 0. Upon the translation u = w + α and integration, the ODE (8 reduces to a Riccati equation dw (40 dz a 1w 2 b 1w βe b2z = 0, where β C, b 1 = b 1 + a 1 α. It suffices to consider the case β 0, otherwise the meromorphic solutions are given by (7. Denote by w = 1 v, then the a 1 v equation (40 is transformed to d 2 v dz 2 dv (41 b 1 dz + a 1βe b2z v = 0. According to Lemma 4.3, all solutions of the ODE (41 are entire functions and thus all the solutions of the ODE (40 are meromorphic functions.

29 Vol. 00, XXXX FACTORIZABLE ODES AND HAYMAN S CONJECTURE 29 To find the general (meromorphic solution of the ODE (41, we set v(z = e b 1 z 2 f (ζ, ζ = 2 a 1 β b 2 e b2z 2. With the new variables, the equation (41 is transformed to the Bessel equation which has the general solution ζ 2 d2 f dζ 2 + ζ df dζ + (ζ2 ν 2 f = 0, ν = b 1 b 2, f(ζ = c 1 J ν (ζ + c 2 Y ν (ζ, where c 1, c 2 C are arbitrary, J ν (ζ and Y ν (ζ are Bessel functions of the first second inds respectively. Consequently, for a 1 b 2 0, a 2 = 0, the general solution of (8 which is meromorphic is given by where ν = αa1+b1 b 2 u(z = αa 1 b 1 2a 1 b 2 z β e 2 (c 1 J ν (ζ + c 2 Y ν (ζ, a 1 (c 1 J ν (ζ + c 2 Y ν (ζ, ζ = 2 a 1 β e b2z 2 and β, c 1, c 2 C are arbitrary. b 2 Remar 4.10: Although J ν (ζ and Y ν (ζ as functions of ζ are not entire in general, J ν (e z and Y ν (e z as functions of z are entire for any ν C. (V For other cases, the nonconstant meromorphic solutions given below of the ODE (8 can be easily derived c 1 e b1z + c 2 e b2z + α, a 1 = a 2 = 0, b 1 b 2, c 1 e b1z + c 2 ze b1z + α, a 1 = a 2 = 0, b 1 = b 2, u(z = c 1 cot(c 2 c 1 z 2 b 2 a 2, a 1 = b 1 = 0, a 2 0, c 1 cot(c 2 c 1 z 2 +αa 1 b 1 2a 1, a 1 0, a 2 = b 2 = 0, where c 1, c 2 C are arbitrary. Remar 4.11: The above solutions may degenerate to rational functions due to the degeneration of c cot(cz as c approaches 0. Thus, the proof of Theorem 1.5 is completed.

30 30 T.W. NG AND C.F. WU Isr. J. Math. 5. Proof of Theorem 1.10 We first recall a lemma and some terminologies that will be needed. For more details, see [23]. Lemma 5.1: ([26, p. 5] Let g : (0, + R and h : (0, + R be monotone increasing functions such that g(r h(r outside of an exceptional set F with finite linear measure. Then, for any α > 1, there exists r 0 > 0 such that g(r < h(αr holds for all r r 0. The iterated order of a meromorphic function is defined by ρ j (f := lim sup r log j T (r, f, log r where log 1 (r := log r, log j (r := log log j 1 (r. The finiteness degree of growth i(f of a meromorphic function f is defined as 0, for f rational, i(f := min{j N; ρ j (f < }, for f transcendental, otherwise. For the differential operator L(f defined by (16 with entire coefficients, we define δ(l := max{i(f; L(f = 0}, γ j (L := max{ρ j (f; L(f = 0}, j N, p(l := max{i(α j ; j = 0, 1,..., n 1}. Another notation is defined for 0 < p := p(l <, Then we have κ(l := max{ρ p (α j ; j = 0, 1,..., n 1}. Theorem 5.2: [23] If 0 < p <, then δ(l = p + 1 and γ p+1 (L = κ(l. Moreover, if α j is the last one in the sequence of coefficients α 0,..., α n 1 such that i(α j = p, then the differential equation L(f = 0 possesses at most j linearly independent solutions f such that i(f p.

31 Vol. 00, XXXX FACTORIZABLE ODES AND HAYMAN S CONJECTURE 31 Proof of Theorem It has been shown in [11] that for any C, α 1, β 1 > 0 such that max{t (r, (e z, T (r, (e z } < α 1 exp(β 1 r, 0 r <, and the same method also gives that for any 1, 2 C, α 2, β 2 > 0 such that T (r, exp{ 1 e 2z } < α 2 exp(β 2 r, 0 r <. On the other hand, for any function f in the class W, we have T (r, f = O(r 2 or o(r 2. Then, according to the following properties of T (r, f n i=1 T (r, f n m i m j=1 g T (r, f i + T (r, g j + O(1, j i=1 j=1 T (r, fg T (r, f + T (r, g, we only need to consider case IV, because for other cases all the meromorphic fi h i solutions can be expressed as u(z =, where f i, h i, g j, y j belong to either gj y j the class W or {exp{ 1 e 2z }, (e 3z, (e 4z i C, i = 1, 2, 3, 4}. For case IV, to obtain an upper bound for T (r, u, we only need to estimate T (r, v because T (r, u T (r, w + O(1 T (r, v + T (r, v + O(1 3T (r, v + S(r, v (3 + εt (r, v, ε > 0 for all r (0, + outside a possible exceptional set E (0, + with finite linear measure. For any v satisfying (41 with p = 1, α 0 (z = a 1 βe b2z, α 1 (z = b 1, Theorem 5.2 implies that v is of infinite order, δ(l = 2 and γ 2 (L = κ(l = 1. As a consequence, ρ 2 (v 1 and T (r, v e r for r > 0 sufficiently large. Thus we have T (r, u (3 + εe r, ε > 0 for all r (0, + \E. By Lemma 5.1, we conclude that T (r, u (3+εe αr, ε > 0, α > 1, for all sufficiently large r.

32 32 T.W. NG AND C.F. WU Isr. J. Math. To show the sharpness of Hayman s conjecture, we consider the following three types of solutions of (8 u 1 (z = q i q 2 u 2 (z = αa 1 b 1 β e 2a 1 a 1 u 3 (z = α e q i q λ z (e qi q λ z ζ 0 ; g 2, 0 2b1 c 0 e b1z tanh ( 1 2 (e q i q λ z ζ 0 ; g 2, 0 + q, g 2 C, b 2 z 2 (c 1 J ν (ζ + c 2 Y Here, we choose ν = 1 2, ζ = e b 2 z 2 for which ( u 2 (z = αa 1 b 1 + b 2e b 2 z 2 c 1 cot a 1 4a 1 c 2 cot ν (ζ, c 1, c 2 C, (c 1 J ν (ζ + c 2 Y ν (ζ ( 2c0 e b1z + c 1 a 2, c 0, c 1 C. (e b 2 z 2 (e b 2 z 2 + c 2 c 1 Then one can apply the same argument as that in [11] to get lower bounds for the Nevanlinna counting functions N(r, u i, i = 1, 2, 3, namely, there exist positive α i, β i, γ i, i = 1, 2, 3 such that Thus, the proof is completed. α i exp{β i r γi } N(r, u i T (r, u i.. Acnowledgement We would lie to than A. E. Eremeno and Y. M. Chiang for helpful discussions on the historical bacground of Hayman s conjecture and Robert Conte and the referee for their very valuable comments.

33 Vol. 00, XXXX FACTORIZABLE ODES AND HAYMAN S CONJECTURE 33 Appendix Table 1. Meromorphic Solutions of ODE (8. In the following, z0, ζ0, c0, c1, c2 C are arbitrary. a1a2 0, 2a1 + a a1 a2 2 4a1 a2 2 a2 N {0, 2} N {0, 2}, Nonconstant meromorphic solutions other than (7 Constraints on the parameters u(z = 2αa1+αa2 2b1 a2 b2 = 2αa1 αa2 + 2b1, αa1 + b1 0 u(z = 2(αa1+b1 a2(e (z z 0 (αa 1 +b 1 1 b1 a2(z z0(αa1+b1 u(z = αa1+b1e 2a 1 a1e u(z = 2 a1 Z Not yet nown 2(αa1+b1 a2(e (z z 0 (αa 1 +b 1 1 a1 b2 = 2αa2 1 2a1b1+a2b1 a1 a 2 (z z 0 (αa 1 +b 1 b2 = a2b1 αa1a2 2a1 2a 1 a1 a2(z z0 b2 a2, αa1 + b1 0, αa1 + b1 0 b1 = αa1, b2 = αa2

34 34 T.W. NG AND C.F. WU Isr. J. Math. Table 2. Meromorphic Solutions of ODE (8. In the following, z0, ζ0, c0, c1, c2 C are arbitrary. Nonconstant meromorphic solutions other than (7 Constraints on the parameters a2 = 2a1 Nil 2 4a1 = 0 a2 (b1 b2(αa1+b2(αa1c2 b1c1e u(z = z(αa 1 +b 1 a1βb2e z(αa 1 +b 2 a1(a1(α(b1 b2(c1e z(αa 1 +b 1 +c2+βe z(αa 1 +b 2 +(b1 b2b2(c1e z(αa 1 +b 1 (b1 + αa1 (b1 b2 (αa1 + b2 0 +c2 2 4a1 a2 a1a2 0, 2a1 + a2 0, 2 a2 Z a1 and N {0, 2}, a2 = a1 2 4a1 a2 = 2 a2 = a1 a2 = 4a1 a2 = 4a1 u(z = α ( α (b1 b2 2 e b1z (b1 (c2z + c1 + c2 + βb1b2e b2z b1 (α (b1 b2 2 e b1z (c2z + c1 + βb1e b2z b1 + αa1 = 0, (b1 b2 (αa1 + b2 0 ( u(z = ez(αa1+b2 b 2 2c2 a1 (β + b2 (βz αc2 + αa1c1 ( a1 e z(αa1+b2 b1 = b2, αa1 + b2 0 (a1 (βz αc2 b2c2 + c1 u(z = a1 (αa1 (αc2 + βz β + αb1c2 b1c1ez(αa1+b1 ( a1 c1e z(αa1+b1 b1 b2, αa1 + b2 = 0 + a1 (αc2 + βz + b1c2 u(z = 2a1 (αc1 + z (β + αc2 + αa2 1βz 2 + 2c2 a1 (a1βz 2 2 (c2z + c1 u(z = u(z = b2 u(z = 12 (z z0; g2, g3 a1[(b2 αa (z z0; g2, g3] + b2 a1 + c(ei ej a1 ( [ 2e cz ( 5 e 1 5 (cz ζ0;0,g3 + ( e 1 5 (cz ζ0;0,g3 5(ei eje cz ( 5 e 1 5 (cz ζ0;0,g3 +5eje 3cz ], g3 C 5 12 (z z0; g2, g3 2a1 ((αa1 + b (z z0; g2, g3 a1 b1 αa1 2a1 b1 = b2 = αa1 a1 b1 2b2 = 0, g2 = 1 12 (b2 αa1 4, g3 C c 2 λ = 25(ei ej 0, i, j {1, 2}, c = αa1 b1 2b2, λ = 6 a1β, e1 = 0, e2 = (b1 + b2 (αa1 b2 /(a1β b2 = 2 (αa1 b1, g2 = (b1 + αa1 4 /12, g3 C [ u(z = c(e2 e1 2e cz ( 5 e 1 5 (cz ζ0;0,g3 + ( e 1 ] 5 (cz ζ0;0,g3 [ 2a1 5(e2 e1e cz ( 5 e 1 5 (cz ζ0;0,g3 +5e1e 3cz ] b1 b2 = 2 (3αa1 + b1, c = 5 (αa1 + b1 0, 5 a1 e1 = 0, e2 = 3 (αa1 + b1 2 /(a1β, β 0, g3 C [ u(z = α c(e2 e1 2e cz ( 5 e 1 5 (cz ζ0;0,g3 + ( e 1 ] 5 (cz ζ0;0,g3 b2 = 2 (αa1 + 3b1, c = 5 (αa1 + b1 0, [ 2a1 5(e2 e1e cz ( 5 e 1 5 (cz ζ0;0,g3 +5e1e 3cz ] 5 e1 = 0, e2 = 3 (αa1 + b1 2 /(a1β, β 0, g3 C 1 u(z = α 2a1(z c0 1 αa1 + b1 = 0 2a1(z c1 (αa1 + b1 ez(αa1+b1 u(z = α ( 2a1 e z(αa1+b1 c0 = 0, αa1 + b1 0 + c1 u(z = α c0 (αa1 + b1 3 e z(αa1+b1 ( e z(αa1+b1 c1 a1 ( 256a1β + c0 (αa1 + b1 2 ( e z(αa1+b1 c1 2 c0 0, αa1 + b1 0

35 Vol. 00, XXXX FACTORIZABLE ODES AND HAYMAN S CONJECTURE 35 Table 3. Meromorphic Solutions of ODE (8. In the following, z0, ζ0, c0, c1, c2 C are arbitrary. a1a2 0, 2a1 + a2 = 0 and set λ 2 = 1 a 2 1 c = αa1 b1 b2, {q1, q2, q3} = {α, b 1 a1 {i, j, } = {1, 2, 3} b2 2a1 }, c = 0 c 0 Nonconstant meromorphic solutions other than (7 Constraints on the parameters 3λ u(z = 2 (z z0[(qj q(z z0±3λ] + q j qi = qj u(z = ±λm1 cot [m1 (z z0] (q 1 + q2 + q3 q = q i + qj 1, m1 = 2 2λ (q i qj 0 1 (q j qi(q qi 0, 2λ u(z = λm2 (cot [m2 (z z0] cot [m2 (z z0 a] + h u(z = λ (a (z z0 (a + h, h C, u(z = q je qj (z z0 ±λ qe q (z z 0 ±λ u(z = q i q a1 0, a2 = 0, b2 0 u(z = αa1 b1 2a1 a1 0, a2 = 0, b2 = 0 u(z = a1 = 0, a2 0, b1 0 e q j (z z 0 ±λ e q (z z 0 ±λ 2 e q i q β a1 h = qi, m2 = ± m2 cot m2a = q j + q 2qi 3λ ( (a = 1 6λ 3h 2 2hs s2, (a = 1 λ (h q 1 (h q2 (h q3, 3 ( g2 = 1 3λ 3h 4 + 4s 1h 3 6s2h s3h + s s1s3, [ ( ( ( g3 = 1 27λ 3 s s2 h 4 4s 1 s 2 1 3s2 h 3 + 6s 2 s 2 1 3s2 12s3 ( s 2 1 3s2 h s s1s2s3 27s 2 3], s1 = q1 + q2 + q3, s2 = q2q3 + q3q1 + q1q2, s3 = q1q2q3. c = 2q i qj q ±λ λ z (e qi q λ z ζ 0; g2, 0 (e q i q λ z ζ 0; g2, 0 + q g2 C, c = 2q i qj q λ e b 2 z c c1 cot(c2 1z 2 +αa1 b1 2a1 u(z = α 2b1eb 1 z u(z = α a1 = 0, a2 0, b1 = 0 u(z = a2(e b 1 z c1 2 (c1j ν (ζ + c2y ν (ζ (c1jν (ζ + c2yν (ζ 2b1c0e b 1 z tanh( 1 2( 2c0e b 1 z +c1 c c1 cot(c2 1z 2 b2 a1 = 0, a2 = 0, b1 b2 u(z = c1e b1z + c2e b2z + α a1 = 0, a2 = 0, b1 = b2 u(z = c1e b1z + c2ze b1z + α a2 a2 ν = αa1+b1 b2, ζ = 2 a1β b2 0 = q i + 2qj q λ c0 = 0, b2 = αa2 + 2b1 c0 0, b2 = αa2 + 2b1 e b 2 z 2, β, c1, c2 C 0 h 2,,