On [p,q],φ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$[p,q]_{,\varphi }$$\end{document}-Order and Complex Differential Equations

The fast growing solutions of the following linear differential equation (∗)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$(*)$$\end{document} is investigated by using a more general scale [p,q],φ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${[p,q]_{,\varphi }}$$\end{document}-order, f(k)+Ak-1(z)f(k-1)+···+A0(z)f=0,(∗)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} f^{(k)}+A_{k-1}(z)f^{(k-1)}+\cdot \cdot \cdot +A_0(z)f=0,\qquad (*) \end{aligned}$$\end{document}where Ai(z)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$A_i(z)$$\end{document} are entire functions in the complex plane, i=0,1,…,k-1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$i=0,1,\ldots ,k-1$$\end{document}. The growth relationships between entire coefficients and solutions of the equation (∗)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$(*)$$\end{document} is found by using the concepts of [p,q],φ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${[p,q]_{,\varphi }}$$\end{document}-order and [p,q],φ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${[p,q]_{,\varphi }}$$\end{document}-type, which extend and improve some previous results.


Introduction and Main Results
We assume that the reader is familiar with the fundamental results and the standard notation of Nevanlinna theory in the complex plane ℂ , see [8,15] for more details.Considering the linear differential equation Journal of Nonlinear Mathematical Physics (2023) 30:  where A 0 (z), … , A k−1 (z) are entire functions in ℂ and k(≥ 2) is integer.Usually, order and hyper order are used to study the growth of solutions of Eq. (1.1), for example, see [7,10,14,15,18,19,20,21] and therein references.For the fast growing entire function, the iterated order is defined to measure their growing.It is well-known that Kinnunen firstly used the idea of iterated order to study the fast growing of solutions of Eq. (1.1) in [13].Since then, the iterated order of solutions of Eq. (1.1) is very interesting topic, many results concerning iterated order of solutions of Eq. (1.1) have been obtained, for example [3,9] and therein references.To estimate precisely the fast growing of entire functions, the concept of [p, q]-order is defined in [12].From then, many results concerning [p, q]-order of solutions of Eq. (1.1) have been found by different researchers, for example [16,17] and theirin references.
In [4], Chyzhykov and Semochko have pointed out that the definition of [p, q]order have weaknesses is that it do not cover arbitrary growth, and given Examples 1.4 and 1.7 in [4] to show the case.And the same time, they given more general growth scale of meromorphic function as follows.
Recently, Belaïdi defined the concept of -type of meromorphic functions which is used to study the growth of solutions of Eq. (1.1), and the following Theorem 1.4 is obtained.

Definition 2 ([2]
) Let be an increasing unbounded function on [1, +∞) , and f be a meromorphic function with i (f ) ∈ (0, +∞), i = 0, 1 .The -types of f are defined by If f is an entire function, then the -types of f are defined by Theorem 1.4 ([2]) Let ∈ Φ , A 0 (z), … , A k−1 (z) be entire functions.Assume that and Then all nontrivial solutions f of Eq. (1.1) Motivated to the [p, q]-order of meromorphic function.We introduce the concepts of [p, q] , -order and [p, q] , -type, where p ≥ q ≥ 1 .For all r ∈ (0, +∞) , exp 1 r = e r , exp n+1 r = exp(exp n r) and log 1 r = log r and log n+1 r = log(log n r) , n ∈ N .We also denote exp 0 r = r = log 0 r , exp −1 r = log 1 r .The [p, q] , -order and [p, q] , -type are defined as follows, respectively.Definition 3 Let be an increasing unbounded function on [1, +∞) , and f be a meromorphic function.The [p, q] , -orders of f are defined by .

3
Journal of Nonlinear Mathematical Physics (2023) 30:932-955 If f is an entire function, then the [p, q] , -orders of f are defined by Definition 4 Let be an increasing unbounded function on [1, +∞) , and f be a meromorphic function with i [p,q], (f ) ∈ (0, +∞), i = 0, 1 .The [p, q] , -types of f are defined by If f is an entire function with ρi [p,q], (f ) ∈ (0, +∞), i = 0, 1 , then the [p, q] , -types of f are defined by The following two examples show that [p, q] , -order is indeed superior to -order when studying the same fast growth functions.
Example 1 It follows from [5] that exp 4 ( (log r) ) is convex in log r .Then there exists an entire function f that satisfies where ,  > 0.
Example 2 It follows from [5] that exp 2 ( (log r) ) is convex in log r .Then there exists an entire function f that satisfies where ,  > 0.

Properties of [p, q] ,' -order
In [4], Chyzhykov and Semochko defined the class of positive unbounded increasing function on [1, +∞) by Φ such that (e t ) is slowly growing, i. e., First, we recall properties of functions from the class Φ.
2) is equivalent to the definition of the class Φ.
Next, we obtain some basic properties of [p, q] , -order by using standard method.
Proposition 2.2 Let ∈ Φ , and f be an entire function.Then Proof First, we prove that this is true when j = 1 , and it can be proved for the case of j = 0 by using similar reason as the case of j = 1.
According to the monotonicity of function and the following inequality we get that Next, by (2.3) and choose R = kr, k > 1 , we have In fact, by the properties of function , Hence,

It is implies that
Therefore, this is completely proved.◻ Proposition 2.3 Let ∈ Φ , and let f , f 1 , f 2 be three meromorphic functions.Then the following statements hold.
Proof (i) We prove that this is true when j = 1 , and similarly it can be proved for the case of . Without loss of generality, suppose that a ≤ b < +∞ .Now by the definition of 1 [p,q], -order, for any  > 0 and suffi- ciently large r, It follows from the properties of Nevanlinna characteristic functions that log q kr log q kr log q r , r → +∞.

It is implies that
The properties (ii), (iii) and (iv) can be proved by using similar way as in the proof of the case (i).
Proof Obviously, we can easily conclude that this is true by Proposition 2.3.◻ Proposition 2.5 Let ∈ Φ , and f 1 , f 2 be two meromorphic functions.Then the fol- lowing statements hold. (2.4) Proof We just prove the case of j = 1 , and the case of j = 0 is very similar.(i) By the definition of the 1 [p,q], -type, for any given  > 0 , there exists a sequence r n which tending to infinity and N 1 ∈ Z + , such that for n > N 1 , On the other hand, there exists N 2 ∈ Z + , such that for n > N 2 , Obviously, By the properties of and n > N , we have . By the monotonicity of , we have And then . By using similar discussion as in the proof above, we obtain easily that (2.5) 1 3 Journal of Nonlinear Mathematical Physics (2023) 30:932-955 So, 1 [p,q], (f 2 ) = 1 [p,q], (f 1 f 2 ).(ii) By (2.5), we have Hence, by the monotonicity of , Without loss of generality, suppose  1  [p,q], (f 1 ) <  1 [p,q], (f 2 ) .Then, by (2.6) and . (iii) is proved by using similar reason as in the proof of (i) and (ii).◻ The following Corollary can be obtain from (i) and (ii) of Proposition 2.5.
Proposition 2.7 Let ∈ Φ , and f be a meromorphic function.Then Proof Set 1 [p,q], (f ) = .From the definition of 1 [p,q], -order, for any  > 0 , there exists r 0 > 1 , such that for all r ≥ r 0 , . By the Lemma of logarithmic deriva- tive (p.34 in [8]), we have (2.6) (2.7) where E ⊂ [0, +∞) is of finite linear measure.By Lemma 3.2 in Sect. 3 and for all sufficiently large r, On the other hand, we prove the inequality 1 [p,q], (f ) ≤ 1 [p,q], (f � ) .The definition of 1 [p,q], (f � ) = implies that for any given above  > 0 , there exists r 1 > 1 , such that for all r > r 1 , By the properties of and we can get that By the monotonicity of , we get

Auxiliary Results
In the proof of Theorems 1.5 and 1.6, the classical reduced order method is adopted for Eq.(1.1), which aims to find the estimation of m(r, A j )(j = 0, … , k − 1) by using the estimation of m(r, f (k) f )(k ≥ 1) .The following lemma is an estimation of m(r, f (k) f ).
Proof Let k = 1 .The definition of 1 [p,q], -order implies that for any  > 0 , there exists r 0 > 1 , such that for all r > r 0 , It follows from (3.1) and the lemma of logarithmic derivative that where E ⊂ (0, +∞) is of finite linear measure.Now, we assume that for some k ∈ ℕ, It follows from (3.2) and (3.3) that m r, f (k+1) Thus,

◻
The following lemma is needed to prove Theorems 1.5 and 1.6.
Wiman-Valiron theory is needed in proving our results, which can be found [15].Let f (z) = +∞ ∑ n=0 a n z n be an entire function.Then are called the maximal term and the central index of f, respectively.The following estimation of the radius r of the polynomial P(z) is used in the proof of Theorem 1.5.Lemma 3.4 ([15, p.10]) Let P(z) = a n z n + a n−1 z n−1 + ⋯ + a 0 be a polynomial, where a n ≠ 0 .Then all zero of P(z) lie in the discs D(0, r) of radius We need the following two lemmas to get estimations of T(r, f) and m(r, f), which is used in proving Theorems 1.6 and 1.8.
Proof The definition of 0 [p,q], -order implies that there exists a sequence (R j ) +∞ j=1 satisfying From the equality above, for any ∈ (0, 0 − ) , there exists an integer j 1 such that for j ≥ j 1 , Since  <  0 −  , there exists an integer j 2 such that for j ≥ j 2 , It follows from this inequality and (3.4) that for j ≥ j 3 = max j 1 , j 2 and for any r ∈ [R j , (1 1 3 Journal of Nonlinear Mathematical (2023) 30:932-955

◻
We can also prove the following result by using similar reason as in the proof of Lemma 3.5.

Lemma 3.7 ([9]
) Let f be a solution of Eq. (1.1), and let 1 ≤  < +∞ .Then for all 0 < r < R , where 0 < R < +∞, where C > 0 is constant which depends on and the initial value of f in a point z 0 , where A j ≠ 0 for some j = 0, … , k − 1 , and where The following logarithmic derivative estimation was found in [6] from Gundersen.

Proof Set
By the definition of ρ0 [p,q], (A j ) , for any  > 0 and for sufficiently large r, By Lemma 3. The classical way of reducing the order is adopted for Eq.(1.1) in proofs of Theorems 1.5 and 1.6, and T(r, A j )(j = 0, 1, … , k − 1) is estimated by T(r, f (k) f )(k ≥ 1) in reducing the order.
To state our proving concisely, let E represents the finite logarithmic measure, I represents the infinite logarithmic measure and F represents the finite linear measure in the proofs of Theorems 1.5-1.8.Next we start prove our results by using the similar way as in the proofs of Theorems 1.1-1.4.

T(r, A
This implies that [p,q], ≤ [p,q], .◻ Proof of Theorem 1.6 By the assumption there exist two numbers 1 and such that 0 [p,q], (f i ) <  , this contradict with Theorem 1.5.Hence, m < k − 1 .Applying the order reduction procedure as in the proof of Theorem 1.5.We use the notation 0 instead of f, and A 0,0 , … , A 0,k−1 instead of A 0 , … , A k−1 .On the general reduction step, we obtain an equation of the form where and the functions determine at each reduction step a solution base of (4.9) in terms of the preceding solution base.We may express (1.1) and the mth reduction steps by the following Table .The rows correspond to (4.9) for 0 , … , m , i.e., the first row corresponds to (1.1), and columns from k to 0 give the coefficients of these equations, while the last column lists those solutions with  1 [p,q], (f ) < .
Journal of Nonlinear Mathematical Physics (2023) 30:932-955 By Lemma 3.1 and (4.10), we see that in the second row, corresponding to the first reduction step, m(r, A 1 Similarly, in each reduction step (4.10) implies that when l = m + 1 − j, … , k − (j + 1) , i.e., for all coefficients to the left from the bold- face coefficient A j,m−j , while for j = 1, … , m, In particular, Applying Lemma 3.

Conclusions
We define new measure [p, q] , -order to describe the growing of meromorphic func- tion, and the new measure is used to study the growth of solutions of complex differential equations.