# On the Iterated Exponent of Convergence of Solutions of Linear Differential Equations with Entire and Meromorphic Coefficients

Authors: Rabab Bouabdelli; Benharrat Belaïdi
Journal: Journal of Mathematics
 (2013)

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## Abstract

We investigate the zeros of the difference of the derivative of solutions of the higher-order linear differential equationsf(k)+Ak-1(z)f(k-1)+⋯+A1(z)f′+A0(z)f=0 and small functions, where A0(z),…,Ak-1(z) are entire or meromorphic functions of finite iterated p order.


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## Body

## 1. Introduction and Main Results

In this paper, a meromorphic function will mean meromorphic in the whole complex plane. We shall use the standard notations in Nevanlinna value distribution of meromorphic functions [1, 2] such as T(r,f), N(r,f), m(r,f). For the definition of the iterated order of a meromorphic function, we use the same definition as in [3], [4, p. 317], [2, p. 129]. For all r∈ℝ, we define exp1r∶=er  and expp+1r∶=exp(exppr),  p∈ℕ. We also define for all r sufficiently large log1r∶=logr and logp+1r∶=log(logpr),p∈ℕ. Moreover, we denote by exp0r∶=r, log0r∶=r, log-1r∶=exp1r and exp-1r∶=log1r.Definition 11 (see [2, 3]).
Letf be a meromorphic function. Then the iterated p order ρp(f) of f is defined by(1)ρp(f)=limsupr→+∞logpT(r,f)logr,(p⩾1isaninteger).
For p=1, this notation is called order and for p=2 hyperorder. If f is an entire function, then the iterated p order ρp(f) of f is defined by
(2)ρp(f)=limsupr→+∞logpT(r,f)logr=limsupr→+∞logp+1M(r,f)logr,(p⩾1isaninteger),
where M(r,f)=max|z|=r|f(z)|.Definition 2 (see [3]).
The finiteness degree of the order of a meromorphic functionf is defined by
(3)i(f)={0forfrational,min{j∈ℕ:ρj(f)<+∞},forftranscendentalforwhichsomej∈ℕwithρj(f)<+∞exists,+∞,forfwithρj(f)=+∞∀j∈ℕ.Definition 3 (see [3]).
The iterated convergence exponent of the sequence of zeros of a meromorphic functionf(z) is defined by
(4)λp(f)=limsupr→+∞logpN(r,1/f)logr,(p⩾1isaninteger),
where N(r,1/f) is the counting function of zeros of f(z) in {z:|z|<r}. Similarly, the iterated convergence exponent of the sequence of distinct zeros of f(z) is defined by
(5)λ¯p(f)=limsupr→+∞logpN¯(r,1/f)logr,(p⩾1isaninteger),
where N¯(r,1/f) is the counting function of distinct zeros of f(z) in {z:|z|<r}.Definition 4 (see [3]).
The finiteness degree of the iterated convergence exponent of the sequence of zeros of a meromorphic functionf(z) is defined by
(6)iλ(f)={0,ifn(r,1f)=O(logr),min{j∈ℕ:λj(f)<∞},ifλj(f)<∞forsomej∈ℕ,∞,ifλj(f)=∞∀j∈ℕ.Remark 5.
Similarly, we can define the finiteness degreeiλ¯(f) of λ¯p(f).Definition 6 (see [5]).
Letf be an entire function. Then the iterated p type of f, with iterated p order 0<ρp(f)<∞ is defined by
(7)τp(f)=limsupr→+∞logpM(r,f)rρp(f),(p⩾1isaninteger),
where M(r,f)=max|z|=r|f(z)|.
We define the linear measure of a setE⊂[0,+∞) by m(E)=∫0+∞‍χE(t)dt and the logarithmic measure of a set F⊂(1,+∞) by lm(F)=∫1+∞‍(χF(t)/t)dt, where χH is the characteristic function of a set H.Recently, Xu, Tu, and Zheng have investigated the relationship between the small functions and the derivative of solutions of higher-order linear differential equation:(8)f(k)+Ak-1(z)f(k-1)+⋯+A1(z)f′+A0(z)f=0,
where A0(z),…,Ak-1(z) are entire or meromorphic functions and have obtained the following results.Theorem A (see [6]).
LetAj(z)  (j=0,1,…,k-1) be entire functions with finite order and satisfy one of the following conditions:(i)
max{ρ(Aj):j=1,2,…,k-1}<ρ(A0)<∞.(ii)
0<ρ(Ak-1)=⋯=ρ(A1)=ρ(A0)<∞ and max{τ(Aj):j=1,2,…,k-1}=τ1<τ(A0)=τ. Then for every solution f≢0 of (8) and for any entire function φ(z)≢0 satisfying ρ2(φ)<ρ(A0), we have
(9)λ¯2(f(i)-φ)=ρ2(f)=ρ(A0),(i=0,1,…).Theorem B (see [6]).
LetAj(z)  (j=1,…,k-1) be polynomials and A0(z) be a transcendental entire function. Then for every solution f≢0 of (8) and for any entire function φ(z) of finite order, we have(i)
λ¯(f-φ)=λ(f-φ)=ρ(f)=∞,(ii)
λ¯(f(i)-φ)=λ(f(i)-φ)=ρ(f(i)-φ)=∞,
(i⩾1,i∈ℕ).When the coefficients are meromorphic functions, they have proved the following theorem.Theorem C (see [6]).
LetAj(z)  (j=0,1,…,k-1) be meromorphic functions satisfying max{ρ(Aj):j=1,2,…,k-1}<ρ(A0) and δ(∞,A0)>0. Then for every meromorphic solution f≢0 of (8) and for any meromorphic function φ(z)≢0 satisfying ρ2(φ)<ρ(A0), we have
(10)λ¯2(f(i)-φ)=λ2(f(i)-φ)⩾ρ(A0),(i=0,1,…),wheref(0)=f.In this paper, we improve and extend the above results by considering the iterated order, and we obtain the following theorems. For some closely related to applications of this paper, see the papers of Gupta et al. [7, 8].Theorem 7.
LetAj(z)  (j=0,1,…,k-1) be entire functions of finite iterated order with i(A0)=p(0<p<∞) and satisfy one of the following conditions:(i)
max{ρp(Aj):j=1,2,…,k-1}<ρp(A0).(ii)
max{ρp(Aj):j=1,2,…,k-1}⩽ρp(A0)=ρ(0<ρ<∞) and max{τp(Aj):ρp(Aj)=ρp(A0)}<τp(A0)=τ(0<τ<∞). Then for every solution f≢0 of (8) and for any entire function φ(z)≢0 satisfying ρp+1(φ)<ρp(A0), we have
(11)λ¯p+1(f(i)-φ)=λp+1(f(i)-φ)=ρp+1(f)=ρp(A0),(i=0,1,…).Theorem 8.
LetAj(z)  (j=1,2,…,k-1) be polynomials and A0(z) be a transcendental entire function with 0<ρp(A0)<∞. Then for every solution f≢0 of (8) and for any entire function φ(z) of finite iterated order ρp(φ)<∞, we have(i)
λ¯p(f-φ)=λp(f-φ)=ρp(f)=∞.(ii)
λ¯p(f(i)-φ)=λp(f(i)-φ)=ρp(f(i)-φ)=∞,i⩾1,i∈ℕ.Theorem 9.
LetAj(z)  (j=0,1,…,k-1) be meromorphic functions of finite iterated order with i(A0)=p(0<p<∞) satisfying max{ρp(Aj):j=1,2,…,k-1}<ρp(A0) and δ(∞,A0)>0. Then for every meromorphic solution f≢0 of (8) whose poles are of uniformly bounded multiplicity and for any meromorphic function φ(z)≢0 satisfying ρp+1(φ)<ρp(A0), we have
(12)λ¯p+1(f(i)-φ)=λp+1(f(i)-φ)=ρp(A0),(i=0,1,…).Corollary 10.
Under the assumptions of Theorem7, if φ(z)=z, then, for every solution f≢0 of (8), we have
(13)λ¯p+1(f(i)-z)=λp+1(f(i)-z)=ρp+1(f)=ρp(A0),(i=0,1,…).Corollary 11.
Under the assumptions of Theorem8, if φ(z)=z, then, for every solution f≢0 of (8), we have(i)
λ¯p(f-z)=λp(f-z)=ρp(f)=∞.(ii)
λ¯p(f(i)-z)=λp(f(i)-z)=ρp(f(i)-z)=∞,
i⩾1,
i∈ℕ.Corollary 12.
Under the assumptions of Theorem9, if φ(z)=z, then for every meromorphic solution f≢0 of (8) whose poles are of uniformly bounded multiplicity, we have
(14)λ¯p+1(f(i)-z)=λp+1(f(i)-z)=ρp(A0),(i=0,1,…).


## 2. Auxiliary Lemmas

In order to prove our theorems, we need the following lemmas.Lemma 13 (see [6]).
Assume thatf≢0 is a solution of (8), set gi=f(i)-φ, and then gi satisfies the equation
(15)gi(k)+Uk-1igi(k-1)+⋯+U0igi=-[φ(k)+Uk-1iφ(k-1)+⋯+U0iφ],
where
(16)Uji=Uj+1i-1′+Uji-1-U0i-1′U0i-1Uj+1i-1,Uj0=Aj,j=0,1,2,…,k-1,Uki-1≡1,(i⩾1,i∈ℕ).Lemma 14 (see [3]).
Letf be a meromorphic function for which i(f)=p⩾1 and ρp(f)=ρ, and let k⩾1 be an integer. Then for any ε>0,
(17)m(r,f(k)f)=O(expp-2rρ+ε)
outside of a possible exceptional set E1 of finite linear measure.Lemma 15 (see [1]).
Letf be a transcendental meromorphic function and k⩾1 be an integer. Then
(18)m(r,f(k)f)=O(log(rT(r,f)))
outside of a possible exceptional set E2 of finite linear measure, and if f is of finite order of growth, then
(19)m(r,f(k)f)=O(logr).Lemma 16 (see [9]).
Letφ:[0,+∞)→ℝ and ψ:[0,+∞)→ℝ be monotone nondecreasing functions such that φ(r)⩽ψ(r) for all r∉E3∪[0,1], where E3⊂(1,+∞) is a set of finite logarithmic measure. Let α>1 be a given constant. Then there exists an r0=r0(α)>0 such that φ(r)⩽ψ(αr) for all r>r0.Lemma 17 (see [10]).
Letf be a meromorphic solution of (8), assuming that not all coefficients Aj are constants. Given a real constant γ>1, and denoting T(r)=∑j=0k-1T(r,Aj), we have
(20)logm(r,f)<T(r){(logr)logT(r)}γ,ifp=0,logm(r,f)<r2p+γ-1T(r){logT(r)}γ,ifp>0
outside of an exceptional set Ep with ∫Eptp-1dt<+∞.Remark 18.
We note that in the above lemma,p=1 corresponds to Euclidean measure and p=0 to logarithmic measure.Lemma 19 (see [11]).
Letf(z) be a meromorphic function of finite iterated order satisfying i(f)=p. Then there exists a set E4⊂(1,+∞) with infinite logarithmic measure such that for all r∈E4, we have
(21)limr→∞logpT(r,f)logr=ρ.Lemma 20.
LetA0(z),A1(z),…,Ak-1(z) be entire functions of finite iterated order with i(A0)=p(0<p<∞) and satisfy max{ρp(Aj):j=1,2,…,k-1}=ρ1<ρp(A0) and set
(22)Uj1=Aj+1′+Aj-A0′A0Aj+1,Uji=Uj+1i-1′+Uji-1-U0i-1′U0i-1Uj+1i-1,
where j=0,1,…,k-1;  Ak≡1,  Uki-1≡1 and i⩾1,  i∈ℕ. Then there exists a set E5 with infinite logarithmic measure such that for all r∈E5(23)limr→∞logpm(r,U0i)logr=ρp(A0)>limsupr→∞max1⩽j⩽k-1{logpm(r,Uji)}logr=ρ1.Proof.
We prove this lemma by using mathematical induction. First, wheni=1, we have Uj1=Aj+1′+Aj-(A0′/A0)Aj+1,  j=0,1,…,k-1 and Ak≡1.
Whenj=0, that is, U01=A1′+A0-(A0′/A0)A1. Then we have
(24)m(r,U01)⩽m(r,A1)+m(r,A0)+m(r,A1′A1)+m(r,A0′A0)+O(1).
From -A0=A1′-U01-(A0′/A0)A1, we have
(25)m(r,A0)⩽m(r,A1)+m(r,U01)+m(r,A1′A1)+m(r,A0′A0)+O(1).
When j≠0, from the definitions of Uj1, we have
(26)m(r,Uj1)⩽m(r,Aj+1)+m(r,Aj)+m(r,Aj+1′Aj+1)+m(r,A0′A0)+O(1),j=1,2,…,k-1. Since Aj(z) are entire functions with max{ρp(Aj):j=1,2,…,k-1}<ρp(A0)<∞, then by Lemma 19, there exists a set E4⊂(1,+∞) with infinite logarithmic measure such that for all r∈E4(27)m(r,Aj+1)=o(m(r,A0)),(j=1,2,…,k-1).
By this, (26), and Lemma 14, we get for all r∈E4(28)max1⩽j⩽k-1{m(r,Uj1)}⩽max1⩽j⩽k-1{m(r,Aj)+o(m(r,A0))+O(expp-2rβ)+O(1)(expp-2rβ)}
for some constant β<∞ outside of a possible exceptional set E1 of finite linear measure. From (24), (25), and (28) we get for all r∈E4∖E1(29)limr→∞logpm(r,U01)logr=ρp(A0)>ρ1=limsupr→∞max1⩽j⩽k-1{logpm(r,Aj)}logr⩾limsupr→∞max1⩽j⩽k-1{logpm(r,Uj1)}logr.
Now, suppose that (23) holds, for i⩽n,n∈ℕ; that is, there is a set E5 with infinite logarithmic measure such that for all r∈E5(30)limsupr→∞logpm(r,U0n)logr=ρp(A0)>limsupr→∞max1⩽j⩽k-1{logpm(r,Ujn)}logr=ρ1.
Next, we prove that (23) holds for i=n+1. Since i=n+1, then we have
(31)Ujn+1=Uj+1n′+Ujn-U0n′U0nUj+1n,
where j=0,1,…,k-1;  Ukn≡1. When j=0, we have U0n+1=U1n′+U0n-(U0n′/U0n)U1n. Then we obtain that
(32)m(r,U0n+1)⩽m(r,U0n)+m(r,U1n)+m(r,U0n′U0n)+m(r,U1n′U1n)+O(1).
Since -U0n=U1n′-U0n+1-(U0n′/U0n)U1n, then we have
(33)m(r,U0n)⩽m(r,U0n+1)+m(r,U1n)+m(r,U0n′U0n)+m(r,U1n′U1n)+O(1).
When j≠0, from the definitions of Ujn+1,  j=1,2,…,k-1 and Ukn≡1, we have
(34)m(r,Ujn+1)⩽m(r,Uj+1n)+m(r,Ujn)+m(r,Uj+1n′Uj+1n)+m(r,U0n′U0n)+O(1).
From (30)–(34), there exists a set E5 with infinite logarithmic measure such that for all r∈E5(35)limr→∞logpm(r,U0n+1)logr=limr→∞logpm(r,U0n)logr=ρp(A0)>ρ1=limsupr→∞max1⩽j⩽k-1{logpm(r,Ujn)}logr=limsupr→∞max1⩽j⩽k-1{logpm(r,Ujn+1)}logr.
Thus, the proof of Lemma 20 is completed.Lemma 21.
LetA0,A1,…,Ak-1 be meromorphic functions of finite iterated order. If
(36)limsupr→∞max1⩽j⩽k-1{logpm(r,Aj)}logr=β1
and there exists a set E6 with infinite logarithmic measure such that
(37)limr→∞logpm(r,A0)logr=β2>β1
holds for all r∈E6, then every meromorphic solution f≢0 of (8) satisfies ρp(f)=∞ and ρp+1(f)⩾β2.Proof.
Suppose thatf≢0 is a meromorphic solution of (8). If ρp(f)=ρ<∞, then from (8) we have
(38)m(r,A0)⩽∑j=1k‍m(r,f(j)f)+∑j=1k-1‍m(r,Aj)+O(1).
By using Lemma 14, we obtain
(39)m(r,A0)⩽O(expp-2rρ+ε)+∑j=1k-1‍m(r,Aj)+O(1)
outside of a possible exceptional set E1 of finite linear measure. By the hypothesis of Lemma 21, there exists a set E6 having infinite logarithmic measure such that for all |z|=r∈E6∖E1, we have for any ε(0<2ε<β2-β1)(40)expp-1{rβ2-ε}⩽O(expp-2rρ+ε)+(k-1)expp-1{rβ1+ε}
and so by Lemma 16, we obtain
(41)β2⩽β1,
and from this we obtain a contradiction. Hence ρp(f)=∞.
Now, assume thatf≢0 is a meromorphic solution of (8) with ρp(f)=∞. By (8) we have
(42)m(r,A0)⩽∑j=1k‍m(r,f(j)f)+∑j=1k-1‍m(r,Aj)+O(1).
By Lemma 15 and (42), we have
(43)m(r,A0)⩽O{logrT(r,f)}+∑j=1k-1‍m(r,Aj),r∉E2,
where E2⊂[1,+∞) is a set having finite linear measure. By the hypothesis of Lemma 21, there exists a set E6 having infinite logarithmic measure such that, for all |z|=r∈E6∖E2, we have by using (43)
(44)expp-1{rβ2-ε}⩽O{logrT(r,f)}+(k-1)expp-1{rβ1+ε},
where 0<2ε<β2-β1. By (44) and Lemma 16, we have
(45)ρp+1(f)⩾β2.Lemma 22 (see [12, Theorem 3]).
Letf(z) be a transcendental meromorphic function, and let α>1 be a given constant. Then there exists a set E7⊂(1,∞) with finite logarithmic measure and a constant B>0 that depends only on α and i,j(0⩽i<j⩽k), such that for all z satisfying |z|=r∉[0,1]∪E7, we have
(46)|f(j)(z)f(i)(z)|⩽B{T(αr,f)r(logαr)logT(αr,f)}j-i.From the above lemma, we obtain the following result.Lemma 23.
Letf be a transcendental meromorphic function with finite iterated order ρp(f)=ρ<+∞,  H={(k1,j1),(k2,j2),…,(kq,jq)} be a finite set of distinct pairs of integers satisfying ki>ji⩾0  (i=1,…,q), and let ε>0 be a given constant. Then, there exists a set E8⊂(1,+∞) with finite logarithmic measure, such that, for all z satisfying |z|∉E8∪[0,1] and for all (k,j)∈H, we have
(47)|f(k)(z)f(j)(z)|⩽{expp-1{rρ-1+ε}}k-j.Proof.
The definition(48)ρp(f)=limsupr→+∞logpT(r,f)logr
implies that for any given ε1>0 there exists R>1 such that for all r>R we have
(49)T(r,f)<expp-1{rρ+ε1}.
Combining (49) with Lemma 22, for α>0, there exists a set E8=[1,R]∪E7 that has finite logarithmic measure and a constant B>0, such that if |z|∉E8∪[0,1], we obtain
(50)|f(k)(z)f(j)(z)|⩽B{expp-1{(αr)ρ+ε1}r(logαr)expp-2{(αr)ρ+ε1}}k-j.
Hence, there exists a constant ε>1+ε1, such that if |z|∉E8∪[0,1], we have
(51)|f(k)(z)f(j)(z)|⩽{expp-1{rρ-1+ε}}k-j.Lemma 24 (see [5]).
Letf(z) be an entire function of iterated p order 0<ρp(f)<+∞ and iterated p type 0<τp(f)<∞. Then, for every given β<τp(f), there exists a set E9⊂[1,+∞) that has infinite logarithmic measure, such that, for all r∈E9, we have
(52)logpM(r,f)>βrρp(f).Lemma 25.
LetA0(z),A1(z),…,Ak-1(z) be entire functions with finite iterated order and satisfy max{ρp(Aj):j=1,2,…,k-1}⩽ρp(A0)=ρ2(0<ρ2<∞), and max{τp(Aj):ρp(Aj)=ρp(A0)}=τ1<τp(A0)=τ(0<τ<∞), and let Uj1,  Uji be stated as in Lemma 20. Then, for any given ε(0<2ε<τ-τ1), there exists a set E10 with infinite logarithmic measure such that
(53)|Uji|⩽expp{(τ1+ε)rρ2},|U0i|⩾expp{(τ-ε)rρ2},
where i⩾1,i∈ℕ and j=1,2,…,k-1.Proof.
We prove this lemma by using mathematical induction.(i)
We first prove thatUji  (j=0,1,…,k-1) satisfy (53) when i=1. From the definition of Uj1=Aj+1′+Aj-(A0′/A0)Aj+1(j≠0) and U01=A1′+A0-(A0′/A0)A1, we have
(54)|U01|⩾|A0|-|A1|(|A1′A1|+|A0′A0|),|Uj1|⩽|Aj|+|Aj+1|(|Aj+1′Aj+1|+|A0′A0|),j=1,2,…,k-1,Ak≡1.
From Lemmas 23, 24, and (54), for any ε(0<4ε<τ-τ1), there exists a set E10 with infinite logarithmic measure such that
(55)|U01|⩾expp{(τ-ε4)rρ2}-2expp{(τ1+ε4)rρ2}expp-1{rM}⩾expp{(τ-ε2)rρ2},|Uj1|⩽expp{(τ1+ε4)rρ2}+2expp{(τ1+ε4)rρ2}expp-1{rM}⩽expp{(τ1+ε2)rρ2},j≠0,
where M>0 is a constant, not necessarily the same at each occurrence.(ii)
Next, we show thatUji  (j=0,1,…,k-1) satisfy (53) when i=2. From
(56)U02=U11′+U01-U01′U01U11,Uj2=Uj+11′+Uj1-U01′U01Uj+11,j=1,2,…,k-1,Uk1≡1,
we have
(57)|U02|⩾|U01|-|U11|(|U11′U11|+|U01′U01|),|Uj2|⩽|Uj1|+|Uj+11|(|Uj+11′Uj+11|+|U01′U01|),j=1,2,…,k-1.By the conclusions of (i) and Lemma 23, (55)–(57), we obtain for all |z|=r∈E10(58)|U02|⩾expp{(τ-ε2)rρ2}-2expp{(τ1+ε2)rρ2}expp-1{rM}⩾expp{(τ-ε)rρ2},|Uj2|⩽expp{(τ1+ε2)rρ2}+2expp{(τ1+ε2)rρ2}expp-1{rM}⩽expp{(τ1+ε)rρ2},j≠0.(iii)
Now, suppose that (53) holds for i⩽n,  n∈ℕ; that is, for any given ε(0<4ε<τ-τ1), there exists a set E10 with infinite logarithmic measure such that
(59)|Uji|⩽expp{(τ1+ε)rρ2},|U0i|⩾expp{(τ-ε)rρ2},i⩽n,j=1,2,…,k-1.From U0n+1=U1n′+U0n-(U0n′/U0n)U1n and Ujn+1=Uj+1n′+Ujn-(U0n′/U0n)Uj+1n  (j=1,…,k-1),  Ukn≡1, we have
(60)|U0n+1|⩾|U0n|-|U1n|(|U1n′U1n|+|U0n′U0n|),|Ujn+1|⩽|Ujn|+|Uj+1n|(|Uj+1n′Uj+1n|+|U0n′U0n|),j=1,2,…,k-1.
Then, from Lemma 23 and (59)-(60), for all |z|=r∈E10,
(61)|Ujn+1|⩽expp{(τ1+ε)rρ2}+2expp{(τ1+ε)rρ2}expp-1{rM}⩽expp{(τ1+2ε)rρ2}
for j≠0, and
(62)|U0n+1|⩾expp{(τ-ε)rρ2}-2expp{(τ1+ε)rρ2}expp-1{rM}⩾expp{(τ-2ε)rρ2}.
Thus, the proof of Lemma 25 is complete.Lemma 26.
LetA0,A1,…,Ak-1 be meromorphic functions of finite iterated order with i(A0)=p(0<p<∞) such that max{ρp(Aj):j=1,2,…,k-1}=ρ4<ρp(A0)=ρ3 and δ=δ(∞,A0)=liminfr→∞(m(r,A0))/(T(r,f))>0. Then every meromorphic solution f≢0 of (8) satisfies ρp+1(f)⩾ρp(A0)=ρ3.Proof.
Assume thatf(z)≢0 is a meromorphic solution of (8). By (8) we have
(63)m(r,A0)⩽∑j=1k‍m(r,f(j)f)+∑j=1k-1‍m(r,Aj)+O(1).
By Lemma 15, we obtain
(64)m(r,A0)⩽O{logrT(r,f)}+∑j=1k-1‍m(r,Aj),r∉E2,
where E2⊂[1,+∞) is a set having finite linear measure. By Lemma 19, there exists a set E4 having infinite logarithmic measure such that, for all |z|=r∈E4, we have
(65)limr→∞logpT(r,A0)logr=ρ3.
Since δ(∞,A0)>0, then for any given ε(0<2ε<min{δ,ρ3-ρ4}) and for all r∈E4, by (65), we have
(66)m(r,A0)⩾(δ-ε)T(r,A0)⩾(δ-ε)expp-1{rρ3-ε}.
From (64) and (66), we have for any given ε(0<2ε<min{δ,ρ3-ρ4}) and for all |z|=r∈E4∖E2(67)(δ-ε)expp-1{rρ3-ε}⩽O{logrT(r,f)}+(k-1)expp-1{rρ4+ε}.
By (67) and Lemma 16, we obtain
(68)ρp+1(f)⩾ρ3.Lemma 27.
LetA0,A1,…,Ak-1 be meromorphic functions of finite iterated order. If there exist positive constants ρ5,α,β(0<α<β) and a set E11 with infinite logarithmic measure such that
(69)|A0(z)|⩾expp{βrρ5},max{|Aj(z)|:j=1,2,…,k-1}⩽expp{αrρ5}
hold for all |z|=r∈E11, then every meromorphic solution f≢0 of (8) satisfies ρp+1(f)⩾ρ5.Proof.
Assume thatf≢0 is a meromorphic solution of (8). By (8), we obtain
(70)|A0|⩽|f(k)f|+∑j=1k-1‍|Aj||f(j)f|.
By Lemma 22, there exists a set E7 having a finite logarithmic measure such that, for all |z|=r∉E7, we have
(71)|f(j)(z)f(z)|⩽B[T(2r,f)]k+1,j=1,2,…,k,
where B>0 is a constant. Substituting (69) and (71) into (70), we obtain for all |z|=r∈E11-E7(72)expp{βrρ5}⩽Bk[T(2r,f)]k+1expp{αrρ5}.
Since 0<α<β, then from (72), Definitions 1, 2, and Lemma 16, we can deduce that i(f)⩾p+1 and
(73)ρp+1(f)⩾ρ5.Lemma 28 (see [13]).
Letp⩾1 be an integer, and let f(z) be a meromorphic solution of the differential equation
(74)f(k)+Ak-1f(k-1)+⋯+A0f=F,
where A0,A1,…,Ak-1 and F≢0 are meromorphic functions:(i)
ifmax{ρp(F),ρp(Aj)(j=0,1,2,…,k-1)}<ρp(f)=ρ⩽∞, then  λ¯p(f)=λp(f)=ρp(f),(ii)
ifmax{ρp+1(F),ρp+1(Aj)(j=0,1,2,…,k-1)}<ρp+1(f)=ρ<∞,  then  λ¯p+1(f)=λp+1(f)=ρp+1(f).Lemma 29 (see [3]).
LetA0(z),…,Ak-1(z) be entire functions such that i(A0)=p(0<p<∞). If either max{i(Aj): j=1,2,…,k-1}<p or max{ρp(Aj):j=1,2,…,k-1}<ρp(A0), then every solution f≢0 of (8) satisfies i(f)=p+1 and ρp+1(f)=ρp(A0).Lemma 30 (see [5]).
LetA0,A1,…,Ak-1 be entire functions of finite iterated order, and let i(A0)=p(0<p<∞). Assume that max{ρp(Aj):j=1,2,…,k-1}⩽ρp(A0)=ρ(0<ρ<∞) and max{τp(Aj):ρp(Aj)=ρp(A0)}<τp(A0)=τ(0<τ<∞). Then every solution f≢0 of (8) satisfies i(f)=p+1 and ρp+1(f)=ρp(A0)=ρ.


## 3. Proof of the Theorems

Proof of Theorem7.
We consider two cases.
Case  1. Suppose that max{ρp(Aj):j=1,2,…,k-1}<ρp(A0)<∞.
(i) First, we prove thatλ¯p+1(f-φ)=λp+1(f-φ)=ρp+1(f)=ρp(A0). Assume that f≢0 is a solution of (8). From Lemma 29, we have ρp+1(f)=ρp(A0). Set g=f-φ. Since ρp+1(φ)<ρp(A0), then ρp+1(g)=ρp+1(f)=ρp(A0) and λ¯p+1(g)=λ¯p+1(f-φ),λp+1(g)=λp+1(f-φ). By substituting f=g+φ into (8), we get that g satisfies the following equation:
(75)g(k)+Ak-1g(k-1)+⋯+A0g=-[φ(k)+Ak-1φ(k-1)+⋯+A0φ].
Set F=φ(k)+Ak-1φ(k-1)+⋯+A0φ. If F≡0, then, from Lemma 29, we have ρp+1(φ)=ρp(A0), which is a contradiction. Hence F≢0. From the assumptions of Theorem 7, we get
(76)max{ρp+1(F),ρp+1(Aj)(j=0,1,2,…,k-1)}<ρp+1(g)=ρp(A0).
From Lemma 28  (ii), we have λ¯p+1(f-φ)=λp+1(f-φ)=ρp+1(f)=ρp(A0).
(ii) Second, we prove thatλ¯p+1(f′-φ)=λp+1(f′-φ)=ρp+1(f)=ρp(A0). Set g1=f′-φ, then ρp+1(g1)=ρp+1(f)=ρp(A0). By Lemma 13, we get that g1 satisfies the (15). Set F1=φ(k)+Uk-11φ(k-1)+⋯+U01φ, where Uj1  (j=0,1,…,k-1) are stated as in Lemma 13. If F1≡0, then, from Lemmas 20 and 21, we have ρp+1(φ)⩾ρp(A0), a contradiction with ρp+1(φ)<ρp(A0). Hence F1≢0. By Lemma 28 (ii), we have
(77)λ¯p+1(f′-φ)=λp+1(f′-φ)=ρp+1(f)=ρp(A0).
(iii) Now, we prove thatλ¯p+1(f′′-φ)=λp+1(f′′-φ)=ρp+1(f)=ρp(A0). Set g2=f′′-φ, then ρp+1(g2)=ρp+1(f)=ρp(A0). By Lemma 13, we get that g2 satisfies (15). Set F2=φ(k)+Uk-12φ(k-1)+⋯+U02φ, where Uj2  (j=0,1,…,k-1) are stated as in Lemma 13. If F2≡0, then, from Lemmas 20 and 21, we have ρp+1(φ)⩾ρp(A0), a contradiction with ρp+1(φ)<ρp(A0). Hence F2≢0. By Lemma 28  (ii), we have
(78)λ¯p+1(f′′-φ)=λp+1(f′′-φ)=ρp+1(f)=ρp(A0).
(iv) Setg3=f′′′-φ, and then ρp+1(g3)=ρp+1(f)=ρp(A0). From Lemmas 13, 20, 21, and 28  (ii), using the same argument as in Case 1  (iii), we can get λ¯p+1(f′′′-φ)=λp+1(f′′′-φ)=ρp+1(f)=ρp(A0).
(v) We prove thatλ¯p+1(f(i)-φ)=λp+1(f(i)-φ)=ρp+1(f)=ρp(A0),  i>3,  i∈ℕ. Set gi=f(i)-φ, and then ρp+1(gi)=ρp+1(f)=ρp(A0). By Lemma 13, we get that gi satisfies (15). Set Fi=φ(k)+Uk-1iφ(k-1)+⋯+U0iφ, where Uji  (j=0,1,…,k-1,i>3,i∈ℕ) are stated as in Lemma 13. If Fi≡0, then from Lemmas 20 and 21, we have ρp+1(φ)⩾ρp(A0), a contradiction with ρp+1(φ)<ρp(A0). Hence Fi≢0. By Lemma 28  (ii), we have λ¯p+1(f(i)-φ)=λp+1(f(i)-φ)=ρp+1(f)=ρp(A0)  (i>3,i∈ℕ).
Case
2. Suppose that max{ρp(Aj):j=1,2,…,k-1}⩽ρp(A0)=ρ(0<ρ<∞) and max{τp(Aj):ρp(Aj)=ρp(A0)}<τp(A0)=τ(0<τ<∞).
(i) First, we prove thatλ¯p+1(f-φ)=λp+1(f-φ)=ρp+1(f)=ρp(A0). Assume that f≢0 is a solution of (8). From Lemma 30, we have ρp+1(f)=ρp(A0). Set g=f-φ. Since φ≢0 is an entire function satisfying ρp+1(φ)<ρp(A0), then ρp+1(g)=ρp+1(f-φ)=ρp(A0) and λ¯p+1(g)=λ¯p+1(f-φ),  λp+1(g)=λp+1(f-φ). By substituting f=g+φ into (8), we get that g satisfies (75). Set F=φ(k)+Ak-1φ(k-1)+⋯+A0φ. If F≡0, then, from Lemma 30, we have ρp+1(φ)=ρp(A0), which is a contradiction. Hence F≢0. From the assumptions of Theorem 7, we get
(79)max{ρp+1(F),ρp+1(Aj)(j=0,1,2,…,k-1)}<ρp+1(g)=ρp(A0).
From Lemma 28 (ii), we have λ¯p+1(f-φ)=λp+1(f-φ)=ρp+1(f)=ρp(A0).
(ii) Now, we prove thatλ¯p+1(f′-φ)=λp+1(f′-φ)=ρp+1(f)=ρp(A0). Set g1=f′-φ. Since φ≢0 is an entire function satisfying ρp+1(φ)<ρp(A0), then ρp+1(g1)=ρp+1(f)=ρp(A0). By Lemma 13, we get that g1 satisfies the (15). If F1≡0, then, from Lemmas 25 and 27, we have ρp+1(φ)⩾ρp(A0), a contradiction with ρp+1(φ)<ρp(A0). Hence F1≢0. By Lemma 28 (ii), we have λ¯p+1(f′-φ)=λp+1(f′-φ)=ρp+1(f)=ρp(A0). Similar to the arguments as in Case 1 (iii)–(v) and by using Lemmas 13, 25, and 27, we can get λ¯p+1(f(i)-φ)=λp+1(f(i)-φ)=ρp+1(f)=ρp(A0),  i>1, i∈ℕ. Thus, the proof of Theorem 7 is completed.Proof of Theorem8.
SinceAj(z)  (j=1,2,…,k-1) are polynomials and A0 is a transcendental entire function with 0<ρp(A0)<∞; that is, Aj(z)  (j=0,1,2,…,k-1) satisfy the conditions of Theorem 7 (i). By using the same argument as in Theorem 7 and Lemma 28 (i), we can get the conclusions of Theorem 8 easily.Proof of Theorem9.
Assume thatf≢0 is a meromorphic solution of (8). By (8) we get that the poles of f(z) can only occur at the poles of A0,A1,…,Ak-1. Note that the multiplicities of poles of f are uniformly bounded, and thus we have
(80)N(r,f)⩽M1N¯(r,f)⩽M1∑j=0k-1‍N¯(r,Aj)⩽Mmax{N(r,Aj):j=0,1,…,k-1},
where M1 and M are some suitable positive constants. This gives
(81)T(r,f)=m(r,f)+O(max{N(r,Aj):j=0,1,…,k-1}).
Applying now (81) with Lemma 17, we obtain
(82)ρp+1(f)⩽ρp(A0).
According to the conditions of Theorem 9, we can easily get the conclusions of Theorem 9 by using the similar argument as in Theorem 7, the estimation (82), and Lemmas 26 and 28 (ii).



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