Growth of meromorphic solutions of higher-order linear differential equations

In this paper, we investigate the higher-order linear differential equations with meromor- phic coefficients. We improve and extend a result of M.S. Liu and C.L. Yuan, by using the estimates for the logarithmic derivative of a transcendental meromorphic function due to Gundersen, and the extended Winman-Valiron theory which proved by J. Wang and H.X. Yi. In addition, we also consider the nonhomogeneous linear differential equations.


Introduction and main results
In this paper, we shall assume that the reader is familiar with the fundamental results and the standard notations of the Nevanlinna value distribution theory of meromorphic functions (see [11,22]).The term "meromorphic function" will mean meromorphic in the whole complex plane C.
For the second order linear differential equation where B(z) is an entire function of finite order.It is well known that each solution f of (1.1) is an entire function, and that if f 1 and f 2 are any two linearly independent solutions of (1.1), then at least one of f 1 , f 2 must have infinitely order( [12]).Hence, "most" solutions of (1.1) will have infinite order.
However, the equation (1.1) with B(z) = −(1 + e −z ) possesses a solution f = e z of finite order.
Thus a natural question is: what condition on B(z) will guarantee that every solution f ≡ 0 of (1.1) will have infinite order?Frei, Ozawa, Amemiya and Langley, and Gunderson studied the question.For the case that B(z) is a transcendental entire function, Gundersen [8] proved that if ρ(B) = 1, then for every solution f ≡ 0 of (1.1) has infinite order.
For the above question, there are many results for second order linear differential equations (see, for example [1,4,6,7,10,15]).In 2002, Z. X. Chen considered the problem and obtained the following result in [4].
Theorem 1.1.Let a, b be nonzero complex numbers and a = b, Q(z) ≡ 0 be a nonconstant polynomial or Q(z) = h(z)e bz , where h(z) is a nonzero polynomial.Then every solution f ≡ 0 of the equation has infinite order and σ 2 (f ) = 1.
In 2006, Liu and Yuan generalized Theorem 1.1 and obtained the following result.
It is natural to ask the following question: What can we say if we remove the condition In this paper, we first investigate the problem and obtain the following result.
Theorem 1.3.Let P (z) and Q(z) be a nonconstant polynomials such that 1)(h 0 ≡ 0) be meromorphic functions and σ = max{σ(h j ) : , suppose that all poles of f are of uniformly bounded multiplicity.Then every transcendental meromorphic solution f of the equation have infinite order and σ 2 (f ) = n.
Next, we continue to investigate the problem and extend Theorem 1.2.
Theorem 1.4.Let P (z) and Q(z) be a nonconstant polynomials as the above, for some complex 3) is of infinite order, and σ(h 0 ) ≤ σ 2 (f ) ≤ n.

Lemma 2.5 ([4]
, Lemma 1).Let g(z) be a meormorphic function with σ(g) = β < ∞.Then for any ε > 0, there exists a set E ⊂ (1, ∞) with lmE < ∞, such that for all z with |z| = r ∈ Applying Lemma 2.5 to 1/g(z), we can obtain that for any given ε > 0, there exists a set It is well known that the Wiman-Valiron theory (see, [14]) is an indispensable device while considering the growth of entire solution of a complex differential equation.In order to consider the growth of meromorphic function solutions of a complex differential equation, Wang and Yi [19] extended the Wiman-Valiron theory from entire functions to meromorphic functions.Here we give the special form where meromorphic function has infinite order: 19,20]).Let f (z) = g(z)/d(z) be the infinite order meromorphic function and σ 2 (f ) = σ, where g(z) and d(z) are entire function, σ(d) < ∞, there exists a sequence r j (r j → ∞) satisfying ) and j is sufficient large, we have and all poles of f are of uniformly bounded multiplicity.Then every transcendental meromorphic solution of the differential equation Obviously, the poles of f must be the poles of , note that all poles of f are of uniformly bounded multiplicity, then λ(1/f ) ≤ σ.By Hadmard factorization theorem, we know f can be written to f (z) = g(z) d(z) , where g(z) and d(z) are entire function, and . By Lemma 2.5 and Lemma 2.6, for any small ε > 0, there exists a sequence r j (r j → ∞) lim sup Substituting (2.7),(2.9)into (2.6), we obtain }. (2.10) Then by (2.8), (2.10) and for the arbitrary ε, we can obtain σ 2 (f ) ≤ σ.We complete the proof of the lemma.
Remark 3.Here we point out that the condition all poles of f are of uniformly bounded multiplicity in Theorem 1 of [3] and Theorem 1.3 of [20] was missing.Since the growth of the coefficients A j gives only an estimate for the counting function of the distinct poles of f , but not for is an infinite order meromorphic solution of the equation EJQTDE, 2009 No. 1, p. 5 3 Proofs of main results
On the other hand, by Lemma 2.7, we have σ 2 (f ) ≤ n, hence σ 2 (f ) = n, and the proof of Theorem 1.3 is completed.

Proof of Theorem 1.4
Proof.We distinguish three cases: (1) Suppose that a n = cb n with c ≥ 1, and deg(P From the equation (1.3), we obtain Hence by Lemma 2.2, from this we obtain σ(f ) = ∞ and σ 2 (f ) ≥ m.On the other hand, by Lemma 2.7, we have σ 2 (f ) ≤ n, hence m ≤ σ 2 (f ) ≤ n.
(2) We shall verify that σ 2 (f ) = n.If it is not true, then it follows from the proof of Part (1) that we shall arrive at a contradiction in the sequel.
Since σ = max{σ(h j ) : ) and k is sufficient large, we have ).By Lemma 2.4, for the above ε, there are 2n opened angles and a positive number , where j is odd.
For the above θ, if , then we may take ε sufficiently small, and there is some G j , j ∈ {0, 1, • • • , 2n − 1} such that θ 0 ∈ G j .Hence there are three cases: (i) θ 0 ∈ G j for some odd number j; (ii) θ 0 ∈ G j for some even number j; (iii) Now we split this into three cases to prove: Case (i): θ 0 ∈ G j for some odd number j.Since G j is an open set and lim k→∞ θ k = θ 0 , there is a K > 0 such that θ k ∈ G j for k > K.By Lemma 2.4, we have Since deg(P − cQ) = m ≥ 1, from (3.12), we obtain that for a sufficiently large k, where Re{P (z k ) − Q(z k )} < dr n k for a sufficiently large k.Substituting (3.10) into (1.3),we get for Thus from (3.10) and (3.12), we obtain, for a sufficiently large k, EJQTDE, 2009 No. 1, p. 8 And from (3.7), (3.10) and (3.13), we have From (3.14) we see that (3.16) is in contradiction to (3.15).
Case (ii): θ 0 ∈ G j where j is even.Since G j is an open set and lim k→∞ θ k = θ 0 , there is We may rewrite (3.14) to We may rewrite (3.14) to  (3).By using the same argument as in Theorem 2 (iv) of [13], we can prove part (3).Here we omit the detail.

Proof of Theorem 1.5
Proof.Assume f 0 is a solution of finite order of (1.4).If there exists another solution f 1 ( ≡ f 0 ) of finite order of (1.4), then σ(f 1 − f 0 ) < ∞, and f 1 − f 0 is a solution of the corresponding homogeneous differential equation (1.3).However, by Theorem 1, we get that σ(f Hence all solutions f of non-homogeneous linear differential equation (1.4), with at most one exceptional solution f 0 of finite order, satisfy σ(f ) = ∞.Now suppose that f is a solution of infinite order of (1.4), then by Lemma 2.8, we obtain In the following, we shall verify that every solution f of infinite order of (1.4) satisfy λ 2 (f ) = σ 2 (f ).
In fact, by (1.4), it is easy to see that the zeros of f occurs at the poles of h j (z)(j = 1, . . ., k − 1) or the zeros of F (z).If f has a zero at z 0 of order n, n > k, then F (z) must have a zero at z 0 of order n − k.Therefore we get by F ≡ 0 that On the other hand, (1.4) may be rewritten as follows Hence by the logarithmic derivative lemma, there exists a set E having finite linear measure such EJQTDE, 2009 No. 1, p. 10 that for all r ∈ E, we have T (r, h j ) + C log(rT (r, f )) + T (r, e P ) + T (r, e Q ) + kN (r, where C is a positive constant.Since for any ε > 0 and sufficiently large r, we have C log(rT (r, f )) ≤ 1 2 T (r, f ), T (r, F ) ≤ r σ(F )+ε , T (r, e P ) ≤ r n+ε ; T (r, e Q ) ≤ r n+ε , T (r, h j ) ≤ r σ+ε , j = 0, 1, • • • , k; so that for r ∈ E and sufficiently large r, we have T (r, f ) ≤ 2kN (r, 1 f ) + (4k + 5)r σ+ε + 4r n+ε + 2r σ(F )+ε .
Hence by Lemma 2.2, we get that σ 2 (f ) ≤ λ 2 (f ).It is obvious that λ 2 (f ) ≥ λ 2 (f ) ≥ σ 2 (f ), hence Finally, let f 0 be a solution of finite order of (1.4), then f 0 ≡ 0. Substitute it into (1.4), and rewrite it as follows It is easy to see that f 0 occurs at the poles of h j (z)(j = 1, . . ., k − 1) or the zeros of F (z).If f 0 has a zero at z 0 of order n, n > k, then F (z) must have a zero at z 0 of order n − k.Therefore we get by So by the logarithmic derivative lemma, and noting that σ(f 0 ) < +∞, we can obtain that ≤T (r, F ) + Hence σ(f 0 ) ≤ max{n, σ(F ), λ(f 0 )}, and this completes the proof of the theorem.

satisfy the hypothesis of Theorem 1 . 4 ;
F ≡ 0 be an meromorphic function of finite order.Suppose all poles of f are of uniformly bounded multiplicity and if at least one of the three statements of Theorem 1.4 hold, then all solutions f of non-homogeneous linear differential equation (1.4) with at most one exceptional solution f 0 of finite order, satisfy

k ν k g (r k ) + r k k e r σ+ε k e r σ+ε k e c− 1
≤ ν k g (r k )e r σ+2ε k .(3.24)This is in contradiction to ν g (r k ) ≥ exp{r σ−ε k }.Thus we complete the proof of Part (2) of Theorem 1.4.