The hyper order and fixed points of solutions of a class of linear differential equations

Author(s): Nour el imane Khadidja CHERIET1, Benharrat BELAÏDI2
1Department of Mathematics, Ibnou-Khaldoun University, Tiaret-Algeria.
2Department of Mathematics, Laboratory of Pure and Applied Mathematics, University of Mostaganem (UMAB), B. P. 227 Mostaganem-Algeria.
Copyright © Nour el imane Khadidja CHERIET, Benharrat BELAÏDI. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

In this paper, we precise the hyper order of solutions for a class of higher order linear differential equations and investigate the exponents of convergence of the fixed points of solutions and their first derivatives for the second order case. These results generalize those of Nan Li and Lianzhong Yang and of Chen and Shon.

Keywords: Linear differential equations; Hyper order; Fixed points.

1. Introduction

In this paper, we use standard notations from the value distribution theory of meromorphic functions (see [1,2,3]). We suppose that f is a meromorphic function in whole complex plane C. In addition, we denote the order of growth of f by ρ(f), and use the notation ρ2(f) to denote the hyper-order of f, defined by ρ2(f)=limsupr+loglogT(r,f)logr, where T(r,f) is the Nevanlinna characteristic function of f.

To give the precise estimate of fixed points, we denote the exponent of convergence of fixed points by τ(f), which is defined by

τ(f)=λ(fz)=limsupr+logN(r,1fz)logr and the hyper-exponent of convergence of fixed points and distinct fixed points are denoted by τ2(f) and τ¯2(f) and are defined by τ2(f)=λ2(fz)=limsupr+loglogN(r,1fz)logr, and τ¯2(f)=λ¯2(fz)=limsupr+loglogN¯(r,1fz)logr, respectively, where N(r,1fz) and N¯(r,1fz) are respectively the integrated counting function of fixed points and distinct fixed points of f. We denote the exponent of convergence of zeros (distinct zeros) of f by λ(f) (λ¯(f)) and the hyper-exponent of convergence of zeros (distinct zeros) of f by λ2(f) (λ¯2(f)).

Consider the second-order homogeneous linear differential equation

f+P(ez)f+Q(ez)f=0,
(1)
where P(w) and Q(w) are not constants polynomials in w=ez (zC). It’s well-known that every solution of Equation (1) is entire.

Suppose f0 is a solution of (1). If f satisfies the condition

limsupr+logT(r,f)r=0, then we say that f is a nontrivial subnormal solution of (1), and if f satisfies the condition [4], limsupr+logT(r,f)rn=0, then we say that f is a nontrivial n-subnormal solution of (1). In [5], Wittich investigated the subnormal solution of (1), and obtained the form of all subnormal solutions in the following theorem:

Theorem 1. [5] If f0 is a subnormal solution of (1), then f must have the form f(z)=ecz(a0+a1ez++amemz), where m0 is an integer and c, a0, a1,…,am are constants with a0am0.

Gundersen and Steinbert [6] refined Theorem 1 and got the following theorem:

Theorem 2. [6] Under the assumption of Theorem 1, the following statements hold:

  • (i) If degP>degQ and Q0, then any subnormal solution f0 of (7) must have the form f(z)=k=0mhkekz, where m1 is an integer and h0, h1, …, hm are constants with h00 and hm0.
  • (ii) If degP1 and Q0, then any subnormal solution of Equation (7) must be constant.
  • (iii) If degP<degQ, then the only subnormal solution of (7) is f0.

Chen and Shon [7] investigated more general equation than (7), and got the following theorem: Set
aj(z)=ajdjzdj+aj(dj1)zdj1++aj1z+aj0, (j=0,,n),
(2)
bk(z)=bkmkzmk+bk(mk1)zmk1++bk1z+bk0, (k=0,,s),
(3)
where dj0 (j=0,,n), mk0 (k=0,,s) are integers, ajdj,,aj0; bkmk,,bk0 are complex constants such that ajdj0, bkmk0.

Theorem 3. [7] Let an(z),,a1(z), a0(z), bs(z),,b1(z), b0(z) be polynomials and satisfy (2) and (3), and an(z)bs(z)0. Suppose that P(ez)=an(z)enz++a1(z)ez+a0(z), Q(ez)=bs(z)esz++b1(z)ez+b0(z). If n<s, then every solution f (0) of equation f+P(ez)f+Q(ez)f=0 satisfies ρ2(f)=1.

Many authors investigated the growth of solutions and the existence of subnormal solutions for some class of higher order linear differential equations (see [4,7,8,9,10,11,12,13]). For the higher-order linear homogeneous differential equation

f(k)+Pk1(ez)f(k1)++P0(ez)f=0,
(4)
where Pj(ez) (j=0,,k1) are polynomials in z, Yang and Li [11] generalized the result of Theorem 2 to the higher order and obtained the following results: Set
ajmj(z)=ajmjdjmjzdjmj+ajmj(djmj1)zdjmj1++ajmj1z+ajmj0,
(5)
where djmj0 (j=0,,k1) are integers, ajmjdjmj,,ajmj0 are complex constants, ajmjdjmj0.

Theorem 4. [11] Let ajmj(z) be polynomials and satisfy (5). Suppose that Pj(ez)=ajmj(z)emjz++aj1(z)ez, where ajmj(z)0. If there exists an integer s (s{0,,k1}) satisfying ms>max{mj:j=0,,s1,s+1,,k1}=m, then every solution f0 of Equation (4) satisfies ρ2(f)=1 if one of the following condition holds:

  • (i) s=0 or 1.
  • (ii) s2 and dega0j(z)>degaij(z) (i0).

Theorem 5. [11] Under the assumption of Theorem 4, if zP0(ez)+P1(ez) 0, then we have every solution f0 of Equation (4) satisfies τ2(f)=τ¯2(f)=ρ2(f)=1.

In particular, they also investigated the exponents of convergence of the fixed points of solutions and their first derivatives for a second order Equation (1) and obtained the following theorem:

Theorem 6. [11] Let an(z),…, a1(z), bs(z),…, b1(z) be polynomials and satisfy (2) and (3), and an(z)bs(z)0. Suppose that P(ez)=an(z)enz++a1(z)ez, Q(ez)=bs(z)esz++b1(z)ez. If ns, then every solution f (0) of Equation (1) satisfy λ(fz)=λ(fz)=ρ(f)= and λ2(fz)=λ2(fz)=ρ2(f)=1.

Thus, it is natural to ask what will happen if we change exp{z} in the coefficients of (4) into exp{A(z)}? In this paper, we consider the above problem to Theorems 3, 4, 5 and 6, we obtain the following results: We set

A(z)=cnzn+cn1zn1++c1z+c0, where n1 is an integer and c0,,cn are complex constants such that Recn>0, throughout the rest of this paper.

Theorem 7. Let ajmj(z) be polynomials and satisfy (5). Suppose that

Pj(eA(z))=ajmj(z)emjA(z)++aj1(z)eA(z),
(6)
where ajmj(z)0. If there exists an integer s (s{0,,k1}) satisfying ms>max{mj:j=0,,s1,s+1,,k1}=m, then every solution f0 of equation
f(k)+Pk1(eA(z))f(k1)++P0(eA(z))f=0
(7)
satisfies ρ(f)= and ρ2(f)=n if one of the following condition holds:
  • (i) s=0 or 1.
  • (ii) s2 and dega0j(z)>degaij(z) (i0).

Example 1. Let f=eez2 be a solution of the equation f(4)2zez2f(3)12z2ez2f24z3ez2f[24z2e3z2+(96z2+12)e2z2+(16z4+48z2+12)ez2]f=0. Set P3(eA(z))=a3,1(z)eA(z)=2zez2,P2(eA(z))=a2,1(z)eA(z)=12z2ez2,P1(eA(z))=a1,1(z)eA(z)=24z3ez2,P0(eA(z))=a0,3(z)e3A(z)+a0,2(z)e2A(z)+a0,1(z)eA(z)=24z2e3z2(96z2+12)e2z2(16z4+48z2+12)ez2. We remark that s=0 and m0=3>max{mj:j=1,2,3}=m=1. Obviously, the conditions of Theorem 7 are satisfied, we see that ρ(f)= and ρ2(f)=n=2.

Remark 1. Very recently, Li et al., [4] have investigated n subnormal solutions of the Equation (7) with Pj(eA(z))=ajmjemjA(z)++aj1eA(z) (j=0,,k1), where ajmj,,aj1 (j=0,,k1) are complex constants instead of polynomials and obtained some results concerning their growth.

Corollary 1. Under the assumption of Theorem 7, if zP0(eA(z))+P1(eA(z))0, then we have every solution f0 of Equation (4) satisfies τ(f)=τ¯(f)=ρ(f)= and τ2(f)=τ¯2(f)=ρ2(f)=n.

In particular, we also investigate the exponents of convergence of the fixed points of solutions and their first derivatives for a second order equation
f+P(eA(z))f+Q(eA(z))f=0,
(8)
and we obtain the following theorems:

Theorem 8. Let ap(z),…, a1(z), bs(z),…, b1(z) be polynomials and satisfy (2) and (3), and ap(z)bs(z)0. Suppose that P(eA(z))=ap(z)epA(z)++a1(z)eA(z), Q(eA(z))=bs(z)esA(z)++b1(z)eA(z). If ps, then every solution f (0) of Equation (8) satisfies λ(fz)=λ(fz)=ρ(f)= and λ2(fz)=λ2(fz)=ρ2(f)=n.

Example 2. Let f=eez2 be a solution of the equation f3zez2f+[2z2e2z2(4z2+2)ez2]f=0. Set P(eA(z))=a1(z)eA(z)=3zez2,Q(eA(z))=b2(z)e2A(z)+b1(z)eA(z)=2z2e2z2(4z2+2)ez2. It is clear that the conditions of Theorem 8 are satisfied with p=1s=2, we see that λ(eez2z)=λ(2zez2eez2z)=ρ(f)= and λ2(eez2z)=λ2(2zez2eez2z)=ρ2(f)=n=2.

Remark 2. If p=s, then the conclusions of Theorem 8 does not hold. For instance, consider the following equation

f+((z4+2iz)e2(1+5i)z3+2z+(z2+(2i)z)e(1+5i)z3+z)f((z3+2i)e2(1+5i)z3+2z+(z+2i)e(1+5i)z3+z)f=0.
(9)
We can easily see that (9) has solution f(z)=z which satisfies ρ(f)=0 and ρ2(f)=0n=3.  In this example, we have p=s=2, A(z)=(1+5i)z3+z, a2(z)=z4+2iz, a1(z)=z2+(2i)z, b2(z)=(z3+2i) and b1(z)=(z+2i).

Theorem 9. Let ap(z),,a1(z),a0(z),bs(z),,b1(z),b0(z) be polynomials and satisfy (2) and (3), and ap(z)bs(z)0. Suppose that P(eA(z))=ap(z)epA(z)++a1(z)eA(z)+a0(z), Q(eA(z))=bs(z)esA(z)++b1(z)eA(z)+b0(z). If p<s, then every solution f (0) of equation

f+P(eA(z))f+Q(eA(z))f=0
(10)
satisfies ρ(f)= and ρ2(f)=n.

Example 3. Let f=ezeez be a solution of the equation f+(ez+13)f+[(e2e1)e2(z+1)ez+1+2]f=0. Set P(eA(z))=a1(z)eA(z)+a0(z)=ez+13,Q(eA(z))=b2(z)e2A(z)+b1(z)eA(z)+b0(z)=(e2e1)e2(z+1)ez+1+2. It is clear that the conditions of Theorem 9 are satisfied with p=1<s=2, here we have ρ(f)= and ρ2(f)=n=1.

Remark 3. If ps, then the conclusions of Theorem 9 does not hold. For instance, consider the following equation

f((2z2+3z)e(1i)z2+2z+i+iz3z2+(1+i)z)f+((2z+3)e(1i)z2+2z+i+iz2z+1+i)f=0.
(11)
It is easy to see that (11) has solution f(z)=z which satisfies ρ(f)=0 and ρ2(f)=0n=2. In this example, we have p=s=1, A(z)=(1i)z2+2z+i, a1(z)=(2z2+3z), a0(z)=(iz3z2+(1+i)z), b1(z)=2z+3 and b0(z)=iz2z+1+i.

Remark 4. Setting cn=1, cn1==c0=0 and n=1, in Theorem 7, Corollary 1, Theorem 8 and Theorem 9, we obtain Theorem 4, Theorem 5, Theorem 6 and Theorem 3 respectively.

2. Auxiliary Lemmas

Recall that A(z)=cnzn+cn1zn1++c0,,cl=αleiθl, z=reiθ, Recn>0, we set δl(A,θ)=Re(cl(eiθ)l)=αlcos(θl+lθ), and Hl,0={θ[0,2π):δl(A,θ)=0}, Hl,+={θ[0,2π):δl(A,θ)>0}, Hl,={θ[0,2π):δl(A,θ)<0}, for l=1,,n, throughout the rest of this paper. Obviously, if δn(A,θ)0, as r+, we get
|eA(z)|=eδn(A,θ)rn++δ1(A,θ)r+Rec0=eδn(A,θ)rn(1+o(1)).
(12)

Lemma 1. [3] Let fj(z) (j=1,,n) (n2) be meromorphic functions, gj(z) (j=1,,n) be entire functions, and satisfy

  • (i) j=1negj(z)0;
  • (ii) when 1jkn, then gi(z)gk(z) is not a constant;
  • (iii) when 1jn, 1hkn,
then T(r,fj)=o{T(r,eghgk)}(r+,rE), where E(1,) is of finite linear measure or logarithmic measure. Also, fj(z)0 (j=1,,n).

Lemma 2. Let A(z), Pj(eA(z)), mj, ms, m and aij(z) satisfy the hypotheses of Theorem 7, then Equation (7) has no constant polynomial solution.

Proof. Suppose that f0(z)=blzl++b1z+b0 (l1) is a nonconstant polynomial solution of (7), where bl0,,b0 are complex constants.

If ls, then f(s)0. Taking z=r, we have

|eA(z)|=|eA(r)|=|ecnrn+cn1rn1++c0|=eRecnrn+Recn1rn1++Rec0=eRecnrn(1+o(1)).
(13)
Substituting f0 into (7) and using (13), we conclude that
|asmsdsmsrdsmsemsRecnrn(1+o(1))|bll(l1)(ls+1)|rls(1+o(1))|Ps(eA(r))f0(s)(r)||f(k)(r)|+|Pk1(eA(r))f0(k1)(r)|++|Ps+1(eA(r))f0(s+1)(r)|+|Ps1(eA(r))f0(s1)(r)|++|P0(eA(r))f0(r)|M0rdemRecnrn(1+o(1))(1+o(1)),
(14)
where d=max{djmj:j=0,,s1,s+1,,k1} and M0>0 is some constant. Since ms>m, we see that (3) is a contradiction. Obviously, when s=0 or 1, we can get that the Equation (7) has nonconstant polynomial solution from the above process. If l<s, then
Pl(eA(z))f0(l)(z)++P0(eA(z))f0(z)=0.
(15)
Set max{mj:j=0,,l}=h. If mj<h, then we can rewrite Pj(eA(z))=ajh(z)ehA(z)++aj(mj+1)(z)e(mj+1)A(z)+ajmj(z)emjA(z)++aj1(z)eA(z) for j=0,,l, where ajh(z)==aj(mj+1)(z)=0. Thus, we conclude by (15) that
[alh(z)f0(l)+a(l1)h(z)f0(l1)++a0h(z)f0]ehA(z)++[alj(z)f0(l)+a(l1)j(z)f0(l1)++a0j(z)f0]ejA(z)++[al1(z)f0(l)+a(l1)1(z)f0(l1)++a01(z)f0]eA(z)=0.
(16)
Set
Qj(z)=alj(z)f0(l)+a(l1)j(z)f0(l1)++a0jf0(j=1,,h).
(17)
Since f0 and aij(z) are polynomials, we see that
m(r,Qj)=o{m(r,e(αβ)A(z))},(1β<αh).
(18)
By Lemma 1 and (16)-(18), we conclude that
Q1(z)Q2(z)Qh(z)0.
(19)
Since degf0>degf0>>degf0(l) and dega0j(z)>degaij(z) (i0), so by (16) and (19), we get a contradiction.

Lemma 3. [14,15] Let f(z) be an entire function and suppose that |f(k)(z)| is unbounded on some ray argz=θ. Then, there exists an infinite sequence of points zm=rmeiθ (m=1,2,), where rm+ such that f(k)(zm) and |f(j)(zm)f(k)(zm)||zm|kj(1+o(1))(j=0,,k1).

Lemma 4. [16] Let f(z) be a transcendental meromorphic function of finite order ρ. Let Γ={(k1,j1),(k2,j2),,(km,jm)} denote a set of distinct pairs of integers satisfying ki>ji0 (i=1,2,,m) and let ε>0 be a given constant. Then, there exists a set E1[0,2π) that has linear measure zero such that if θ[0,2π)E1, then there is a constant R1=R1(θ)>1 such that for all z satisfying argz=θ and |z|R1 and for all (k,j)Γ, we have |f(k)(z)f(j)(z)||z|(kj)(ρ1+ε).

Lemma 5. [17] Let f(z) be an entire function with ρ(f)=ρ0 is a constant and k>0 is a constant independent of θ0). Then f(z) is a polynomial with degfk.

Lemma 6. [16] Let f be a transcendental meromorphic function, and α>1 be a given constant. Then, there exists a set E3(1,) with finite logarithmic measure and a constant C>0 that depends only on α and i, j (i,jN), such that for all z satisfying |z|=rE3[0,1],

|f(j)(z)f(i)(z)|C(T(αr,f)r(logαr)logT(αr,f))ji.
(20)

Remark 5. From the proof of Lemma 6 ([16, Theorem 3]), we can see that the exceptional set E4 equals {|z|:z(n=1+O(an))}, where an(n=1,2,) denote all zeros and poles of f(i), and O(an) denote sufficiently small neighborhoods of an. Hence, if f(z) is a transcendental entire function and z is a point that satisfies |f(z)| to be sufficiently large, then the point zE4 satisfies (20). For details, see , [9, Remark 2.10].

Lemma 7. [10,18] Let A0, , Ak1 be entire functions of finite order. If f(z) is a solution of equation f(k)+Ak1f(k1)++A0f=0, then ρ2(f)max{ρ(Aj):j=0,,k1}.

Lemma 8. [19] Let g(z) be an entire function of infinite order with the hyper-order ρ2(g)=ρ, and let ν(r) be the central index of g. Then, limsupr+loglogν(r)logr=ρ2(g)=ρ.

Lemma 9. [7] Let f(z) be an entire function that satisfies ρ(f)=ρ(n<ρ<); or ρ(f)= and ρ2=0; or ρ2=α(0<α<), and a set E5[1,) has a finite logarithmic measure. Then, there exists a sequence {zk=rkeiθk} such that |f(zk)|=M(rk,f), θk[0,2π), limkθk=θ0[0,2π), rkE5, and rk, such that

  • (i) if ρ(f)=ρ (n<ρ<), then for any given ε1(0<ε1<ρn2), rkρε1<ν(rk)<rkρ+ε1;
  • (ii) if ρ(f)= and ρ2(f)=0, then for any given ε2(0<ε20), we have, as rk is sufficiently large, rkM<ν(rk)<exp{rkε2};
  • (iii) if ρ2(f)=α(0<α<), then for any given ε3(0<ε3<α), exp{rkαε3}<ν(rk)<exp{rkα+ε3}.

Lemma 10. [20] Let g be a non-constant entire function, and let 0<δ<1. There exists a set E6[1,) of finite logarithmic measure with the following property. For r[1,)E6, the central index ν(r) of g satisfies ν(r)(logM(r,g))1+δ.

Lemma 11. [21,22] Let A0, …, Ak1, F0 be finite order meromorphic functions. If f is a meromorphic solution of the equation f(k)+Ak1f(k1)++A0f=F, with ρ(f)=+ and ρ2(f)=ρ, then f satisfies λ¯(f)=λ(f)=ρ(f)= and λ¯2(f)=λ2(f)=ρ2(f)=ρ.

Lemma 12. [14] Let φ:[0,+)R and ψ:[0,+)R be monotone non-decreasing functions such that φ(r)ψ(r) for all rE7[0,1], where E7(1,+) is a set of finite logarithmic measure. Let γ>1 be a given constant. Then there exists a r1=r1(γ)>0 such that φ(r)ψ(γr) for all r>r1.

3. Proofs of the results

Proof of Theorem 7. Suppose that f0 is a solution of (7), then f is an entire function. By Lemma 2, we see that f is transcendental.

First step. We prove that ρ(f)=.

Suppose, to the contrary, that ρ(f)=ρ0, there exists a set E1[0,2π) with linear measure zero, such that if θ[0,2π)E1, then there exists a constant R1=R1(θ)>1, such that for all z satisfying argz=θ and |z|=r>R1, we have

|f(j)(z)f(s)(z)|r(ρ1+ε)(js)j=s+1,,k.
(21)

Case 1. Take a ray argz=θHn,+E1, then δn(A,θ)>0. We assume that |f(s)(reiθ)| is bounded on the ray argz=θ. If |f(s)(reiθ)| is unbounded on the ray argz=θ, then by Lemma 3, there exists a sequence {zt=rteiθ} such that as rt+, f(s)(zt) and

|f(i)(zt)f(s)(zt)|rtsi(1+o(1))2rts,i=0,,s1.
(22)
By (7), we get
|Ps(eA(zt))||f(k)(zt)f(s)(zt)|+j=0jsk1|Pj(eA(zt))||f(j)(z)f(s)(z)|.
(23)
For rt+, we have
|Ps(eA(zt))|=|asms(zt)emsA(zt)++as1(zt)eA(zt)||asms(zt)emsA(zt)||as(ms1)(zt)ems1A(zt)++as1(zt)eA(zt)||asms(zt)emsA(zt)|[|as(ms1)(zt)ems1A(zt)|++|as1(zt)eA(zt)|]=|asmsdsms|rtdsmsemsδn(A,θ)rtn(1+o(1))(1+o(1))[|as(ms1)ds(ms1)|rtds(ms1)e(ms1)δn(A,θ)rtn(1+o(1))(1+o(1))++|as1ds1|rtds1eδn(A,θ)rtn(1+o(1))(1+o(1))] 12|asmsdsms|rtdsmsemsδn(A,θ)rtn(1+o(1))(1+o(1)),
(24)
and
|Pj(eA(zt))|=|ajmj(zt)emjA(zt)++aj1(zt)eA(zt)||ajmjdjmj|rtdjmjemjδn(A,θ)rtn(1+o(1))(1+o(1))++|ajmj1|rtdj1eδn(A,θ)rtn(1+o(1))(1+o(1))2|ajmjdjmj|rtdemδn(A,θ)rtn(1+o(1))(1+o(1)), (js),
(25)
where d=max{djmj:j=0,,s1,s+1,,k1}. Substituting (21), (22), (24), (25) into (23), we obtain that for sufficiently large rt
12|asmsdsms|rtdsmsemsδn(A,θ)rtn(1+o(1))(1+o(1))C0rtd+kρemδn(A,θ)(1+o(1))rtn(1+o(1)),
(26)
where C0>0 is a constant. From (26), we can get a contradiction by ms>m and δn(A,θ)>0, so
|f(reiθ)|MrsM1rk,M1>0,
(27)
on the ray argz=θHn,+E1.

Case 2. Now, we take a ray argz=θHn,, then δn(A,θ)<0. If |f(k)(reiθ)| is unbounded on the ray argz=θ, then by Lemma 3, there exists a sequence {zt=rteiθ} such that as rt+, f(s)(zt) and

|f(i)(zt)f(k)(zt)|rtki(1+o(1))2rtk,i=0,,k1.
(28)
By (7), we get
1=Pk1(eA(zt))f(k1)(zt)f(k)(zt)++P0(eA(zt))f(zt)f(k)(zt).
(29)
For rt+, we have
|Pj(eA(zt))|=|ajmj(zt)emjA(zt)++aj1(zt)eA(zt)||ajmjdjmj|rdjmjemjδn(A,θ)rtn(1+o(1))(1+o(1))++|ajmj1|rdj1eδn(A,θ)rtn(1+o(1))(1+o(1))2|ajmj1|rdeδn(A,θ)rtn(1+o(1))(1+o(1)) (j=0,,k1).
(30)
Substituting (28) and (30) into (29), we obtain that for sufficiently large rt
1C1rtk+deδn(A,θ)rtn(1+o(1))(1+o(1)),C1>0.
(31)
Since δn(A,θ)<0, when rt+, by (31), we get 10, this is a contradiction. Hence
|f(reiθ)|M2rk,M2>0,
(32)
on the ray argz=θHn,E1. From Lemma 5, (27) and (32), we know that f(z) is a polynomial, which contradicts the assertion that f(z) is transcendental. Therefore, ρ(f)=.

Step 2. We prove that ρ2(f)=n. By Lemma 7 and ρ(Pj(eA(z)))=n (j=0,,k1), we see that ρ2(f)max{ρ(Pj(eA(z)))}=n.

Now, we suppose that there exists a solution f0 satisfies ρ2(f0)=α<n. Then we have

limsupr+logT(r,f0)rn=0.
(33)
By Lemma 6, we see that there exists a subset E3(1,) having finite logarithmic measure such that for all z satisfying |z|=rE3[0,1],
|f0(j)(z)f0(z)|C[T(2r,f0)]k+1,j=1,,k,
(34)
where C(>0) is some constant. From the Wiman-Valiron theory, there is a set E8(1,) having finite logarithmic measure, such that we can choose a z satisfying |z|=r[0,1]E8 and |f0(z)|=M(r,f0), then we get
f0(j)(z)f0(z)=(ν(r)z)j(1+o(1)),j=1,,k1,
(35)
where ν(r) is the central index of f0(z). By Lemma 9, we see that there exists a sequence {zt=rteiθt} such that |f0(zt)|=M(rt,f0), θt[0,2π), with rt[0,1]E5E8, rt+ and for any sufficiently large M3(>2k+3)
ν(rt)>rtM3>rt.
(36)

Case 1. Suppose θ0Hn,+. Since δn(A,θ)=αncos(θn+nθ) is a continuous function of θ, by θtθ0 we get limtδn(A,θt)=δn(A,θ0)>0. Therefore, there exists a constant N(>0) such that as t>N,

δn(A,θt)12δn(A,θ0)>0. By (33), for any given ε1(0<ε1N,
[T(2rt,f0)]k+1eε1(k+1)(2rt)ne12δn(A,θt)rtn.
(37)
By (34), (35) and (37), we have
(ν(rt)rt)ks(1+o(1))=|f0(ks)(zt)f0(zt)|C[T(2rt,f0)]k+1Ce12δn(A,θ0)rtn.
(38)
By (7), we get
f0(s)(zt)f0(zt)Ps(eA(zt))=f0(k)(zt)f0(zt)+j=0,jsk1Pj(eA(zt))f0(j)(zt)f0(zt).
(39)
Substituting (24), (25) and (35) into (39), we get for sufficiently large rt,
(ν(rt)rt)s12|asmsdsms|rtdsmsemsδn(A,θt)rtn(1+o(1))(1+o(1))(ν(rt)rt)k(1+o(1))+j=0,jsk12|ajmjdjmj|rtdemδn(A,θ)rtn(1+o(1))(ν(rt)rt)j(1+o(1)).
(40)
By (36), (38) and (40), we get |asmsdsms|rtdsmsemsδn(A,θt)rtn(1+o(1))(1+o(1))2(ν(rt)rt)ks(1+o(1))+j=0,jsk14|ajmjdjmj|rtdemδn(A,θ)rtn(1+o(1))(ν(rt)rt)js(1+o(1))2(ν(rt)rt)ks(1+o(1))+j=0s14|ajmjdjmj|rtdemδn(A,θ)rtn(1+o(1))(ν(rt)rt)js(1+o(1))+j=s+1k14|ajmjdjmj|rtdemδn(A,θ)rtn(1+o(1))(ν(rt)rt)js(1+o(1)) C2rtdemδn(A,θ)rtn(1+o(1))(ν(rt)rt)ks(1+o(1)), where C2>0 is a constant. From this inequality and (38), it follows that
|asmsdsms|rtdsmse(msm)δn(A,θt)rtn(1+o(1))(1+o(1))C2rtd(ν(rt)rt)ks(1+o(1))CC2|ajmjdjmj|rtde12δn(A,θ0)rtn.
(41)
Since msm1>12 and δ(A,θt)12δn(A,θ0)>0, we see that (41) is a contradiction.

Case 2. Suppose θ0Hn,. Since δn(A,θ) is a continuous function of θ, by θtθ0 we get limtδn(A,θt)=δn(A,θ0)0) such that as t>N,

δn(A,θt)12δn(A,θ0)<0. By (7), we can write
emsA(zt)f0(k)(zt)f0(zt)=emsA(zt)Pk1(eA(zt))f(k1)(zt)f0(zt)++emsA(zt)P0(eA(zt)).
(42)
From (6) and δn(A,θt)<0, we get
|emsA(zt)Pj(eA(zt))|=|emsA(zt)(ajmj(zt)emjA(zt)++aj1(zt)eA(zt))|=|ajmj(zt)e(msmj)A(zt)++aj1(zt)e(ms1)A(zt)|C3rtdj1e(ms1)δ(A,θt)rtn(1+o(1))(1+o(1)),
(43)
where C3>0 is a constant. Substituting (35) and (43) into (42), we get
emsδ(A,θt)rtn(1+o(1))ν(rt)C4rtd+ke(ms1)δ(A,θt)rtn(1+o(1))(1+o(1)),
(44)
where C4>0 is a constant. By substituting (36) into (44), we have
rtM3emsδ(A,θt)rtn(1+o(1))C4rtd+ke(ms1)δ(A,θt)rtn(1+o(1))(1+o(1)).
(45)
Since δ(A,θt) 12δn(A,θ0)<0, we see (45) is also a contradiction.

Case 3. Suppose θ0Hn,0. Since θtθ0, for any given ε2 (0<ε20), such that as t>N, θt[θ0ε2,θ0+ε2], and

zt=rteiθtΩ¯={z:θ0ε2argzθ0+ε2}. By Lemma 6, we se that there exist a subset E3(1,) having logarithmic measure lmE30 such that for all z satisfying |z|=rE3[0,1],
|f0(i)(z)f0(s)(z)|C[T(2r,f0(s))]ks+1,i=s+1,,k,
(46)
Now, we consider the growth of f0(reiθ) on a ray argz=θΩ¯{θ0}. By the properties of cosine function, we suppose without loss of generality that δn(A,θ)>0 for θ[θ0ε2,θ0) and δn(A,θ)<0 for θ(θ0,θ0+ε2].

Subcase 3.1 For a fixed θ[θ0ε2,θ0), we have δn(A,θ)>0. Since ρ2(f0)<n, we get that f0 satisfies (33). From T(r,f0(s))<(s+1)T(r,f0)+S(r,f0), where S(r,f)=o(T(r,f)), as r+ outside of a possible exceptional set of finite logarithmic measure, we get that f0(s) also satisfies (33). So for any given ε2 satisfying 0<ε2<12n+1(ks+1)δn(A,θ), we have

[T(2rt,f0(s))]ks+1eε2(ks+1)(2rt)ne12δn(A,θ0)rtn.
(47)
We assert that |f0(s)(reiθ)| is bounded on the ray argz=θ[θ0ε2,θ0). If |f(s)(reiθ)| is unbounded on the ray argz=θ, then, by Lemma 3, there exists a sequence {yj=Rjeiθ} such that as Rj, f0(s)(yj) and
|f0(i)(yj)f0(s)(yj)|Rjsi(1+o(1))2Rjs,i=0,,s1.
(48)
By Remark 5 and f0(s)(yj), we know that |yj|=RjE4. By (46) and (47), we have for sufficiently large j,
|f0(j)(yj)f0(s)(yj)|C[T(2Rj,f0(s))]ks+1Ce12δn(A,θ0)Rjn,j=s+1,,k.
(49)
Substituting (24), (25), (48) and (49) into (23)
12|asmsdsms|Rjdsmsemsδn(A,θ)Rjn(1+o(1))(1+o(1))=|Ps(eA(yj))||f(k)(yj)f(s)(yj)|+j=0jsk1|Pj(eA(yj))||f(j)(yj)f(s)(yj)|=|f(k)(yj)f(s)(yj)|+j=0s1|Pj(eA(yj))||f(j)(yj)f(s)(yj)|+j=s+1k1|Pj(eA(yj))||f(j)(yj)f(s)(yj)|Ce12δn(A,θ0)Rjn+j=0s14|ajmjdjmj|Rjdemδn(A,θ)Rjn(1+o(1))Rjs(1+o(1))+j=s+1k12|ajmjdjmj|Rjdemδn(A,θ)Rjn(1+o(1))Ce12δn(A,θ0)RjnC5Rjde(12+m)δn(A,θ)Rjn,
(50)
where C5>0 is a constant, which yields a contradiction by msm1>12 and δn(A,θ)>0. Hence |f0(s)(reiθ)| is bounded on the ray argz=θ, so
|f0(reiθ)|M4rs,M4>0,
(51)
on the ray argz=θ[θ0ε4,θ0).

Subcase 3.2 For a fixed θ(θ0,θ0+ε2], we have δn(A,θ)<0. Using a reasoning similar to that in Subcase 3.1, we obtain

|f0(reiθ)|M5rk,M5>0,
(52)
on the ray argz=θ(θ0,θ0+ε4]. By (51) and (52), we see that on the ray argz=θΩ¯{θ0},
|f0(reiθ)|M5rk,M5>0.
(53)
But since ρ(f0(reiθ))= and {zt=rteiθt} satisfies |f0(zt)|=M(rt,f0), we see that, for any large M6(>k), as t is sufficiently large,
|f0(zt)|=|f0(zt)|=|f0(zt)|=|f0(rteiθt)|exp{rtM6}.
(54)
Since ztΩ¯, by (53) and (54), we see that θt=θ0 as t. Therefore, δn(A,θt)=0 as t. Thus, for sufficiently large t,
|Pj(ezt)|=|ajmj(zt)emjA(zt)+ajmj1(zt)emj1A(zt)++aj1(zt)eA(zt)||ajmj(zt)|+|ajmj1(zt)|++|aj1(zt)|C6rd,
(55)
where j=0,,k1 and C6>0 is a constant. By (7), (35) and (55), we get that |(ν(rt)zt)k(1+o(1))|=|f0(k)(zt)f0(zt)|C7rd(ν(rt)zt)k1(1+o(1)), i.e.,
ν(rt)(1+o(1))C7rd+1(1+o(1)),
(56)
where C7>0 is a constant. Substituting (36) into (56), we obtain also a contradiction. So we have ρ2(f)=n.

Proof of Corollary 1. From Theorem 7, we get ρ(f)= and ρ2(f)=n. Let g=fz, then f=g+z. Substituting it into (7), we have g(k)+Pk1(eA(z))g(k1)++P0(eA(z))g=zP0(eA(z))P1(eA(z)). Since zP0(eA(z))P1(eA(z))0, from Lemma 11, ρ(g)= and ρ2(g)=n we conclude λ¯(g)=λ(g)=ρ(g)= and λ¯2(g)=λ2(g)=ρ2(g)=n. So τ¯(f)=τ(f)=ρ(f)= and τ¯2(f)=τ2(f)=ρ2(f)=n.

Proof of Theorem 8. From Theorem 7, we get ρ(f)= and ρ2(f)=n.

  • (i) Let g=fz, then f=g+z. Substituting it into (8), we have g+P(eA(z))g+Q(eA(z))g=P(eA(z))zQ(eA(z)). Since ps, we get P(eA(z))Q(eA(z))z0. From Lemma 11, we obtain λ(g)=ρ(g)=ρ(f)= and λ2(g)=ρ2(g)=ρ2(f)=n. So λ(fz)= and λ2(fz)=n.
  • (ii) Differentiating both sides of (8), we get that
    f+P(eA(z))f+[(P(eA(z)))+Q(eA(z))]f+(Q(eA(z)))f=0.
    (57)
    By (8), we have
    f=f+P(eA(z))fQ(eA(z)).
    (58)
    Substituting (58) into (57), we get
    f+[(P(eA(z)))(Q(eA(z)))Q(eA(z))]f+[(P(eA(z)))+Q(eA(z))(Q(eA(z)))Q(eA(z))P(eA(z))]f=0.
    (59)
    Let g=fz, then f=g+z, f=g+1, f=g. Substituting these into (59), we get that
    g+[(P(eA(z)))(Q(eA(z)))Q(eA(z))]g+[(P(eA(z)))+Q(eA(z))(Q(eA(z)))Q(eA(z))P(eA(z))]g=P(eA(z))+(Q(eA(z)))Q(eA(z))[(P(eA(z)))+Q(eA(z))(Q(eA(z)))Q(eA(z))P(eA(z))]z=h(z).
    (60)
    Next, we prove that h(z)0. If h(z)0, then P(eA(z))+(Q(eA(z)))Q(eA(z))[(P(eA(z)))+Q(eA(z))(Q(eA(z)))Q(eA(z))P(eA(z))]z. Since Q(z)0, we have
    (Q(eA(z)))(Q(eA(z)))2zP(eA(z))Q(eA(z))+[(P(eA(z)))Q(eA(z))(Q(eA(z)))P(eA(z))]z.
    (61)
    Suppose p>s. By taking z=r, we have P(eA(r))=ap(r)epA(r)++a1(r)eA(r),andQ(eA(r))=bs(r)esA(r)++b1(r)eA(r). We get (P(eA(r)))=j=1p(aj(r)+jA(r)aj(r))ejA(r)=(ap(r)+pA(r)ap(r))epA(r)++(a1(r)+A(r)a1(r))eA(r) and (Q(eA(r)))=j=1s(bj(r)+jA(r)bj(r))ejA(r)=(bs(r)+sA(r)bs(r))esA(r)++(b1(r)+A(r)b1(r))eA(r). So, we obtain |P(eA(r))Q(eA(r))+(P(eA(r)))Q(eA(r))r(Q(eA(r)))P(eA(r))r|=|ap(r)bs(r)+(ps)rA(r)ap(r)bs(r)+(ap(r)bs(r)ap(r)bs(r))r|e(p+s)Recnrn(1+o(1))(1+o(1)). Since ap(r), bs(r) and A(r) are polynomials and p>s, we get deg((ps)rA(r)ap(r)bs(r))>deg[ap(r)bs(r)+(ap(r)bs(r)ap(r)bs(r))r]. So, we have |(ps)rA(r)ap(r)bs(r)+ap(r)bs(r)+(ap(r)bs(r)ap(r)bs(r))r|=Mrd1(1+o(1))0, where M>0 and d1>0 are some constants. It follows that |P(eA(r))Q(eA(r))+(P(eA(r)))Q(eA(r))r(Q(eA(r)))P(eA(r))r|=Mrd1e(p+s)Recnrn(1+o(1))(1+o(1)). From (61), we have Mrd1e(p+s)Recnrn(1+o(1))(1+o(1))=|P(eA(r))Q(eA(r))+(P(eA(r)))Q(eA(r))r(Q(eA(r)))P(eA(r))r|=|(Q(eA(r)))(Q(eA(r)))2r|M1rd2e2sRecnrn(1+o(1))(1+o(1)), where M1>0 and d2>0 are some constants, which is a contradiction. So we have h(z)0. If \(p0,\) {d3>0, }M3>0 and d4>0 are some constants. This is a contradiction. So, we obtain h(z)0. Hence, if ps we have h(z)0. From Lemma 11, we get λ(g)=ρ(g)=ρ(fz)=ρ(f)= and λ2(g)=ρ2(g)=ρ2(fz)=ρ2(f)=n.

Proof of Theorem 9. Suppose that f0 is a solution of (10). Since ρ(P)=ρ(Q)=n, then by Lemma 7, we see that

ρ2(f)max{ρ(P),ρ(Q)}=n.
(62)
By Lemma 6, we se that there exist a subset E3(1,) having logarithmic measure lmE30 such that for all z satisfying |z|=rE3[0,1],
|f(j)(z)f(z)|C[T(2r,f)]j+1,j=1,2.
(63)
Taking z=r, in (2) and (3), we obtain that for sufficiently large r
|P(eA(r))|=|ap(r)epA(r)++a1(r)eA(r)+a0(r)|2|apdp|rdpepRecnrn(1+o(1))(1+o(1)),
(64)
and
|Q(eA(r))|=|bs(r)esA(r)++b1(r)eA(r)+b0(r)|12|bsms|rmsesRecnrn(1+o(1))(1+o(1)).
(65)
Substituting (63)-(65) into (10), we deduce that for all z satisfying |z|=rE3[0,1]
12|bsms|rmsesRecnrn(1+o(1))(1+o(1))|f(z)f(z)+P(eA(z))f(z)f(z)||f(z)f(z)|+|P(eA(z))||f(z)f(z)|C[T(2r,f)]3+2|apdp|rdpepRecnrn(1+o(1))C[T(2r,f)]2(1+o(1))3C|apdp|rdpepRecnrn(1+o(1))[T(2r,f)]3(1+o(1)).
(66)
By (66), we deduce that for all z satisfying |z|=rE3[0,1]
|bsms|rmsdpe(sp)Recnrn(1+o(1))(1+o(1))6C|apdp|[T(2r,f)]3(1+o(1)).
(67)
Since sp>0, by (67) and Lemma 12, we get
ρ(f)limsupr+logT(r,f)logr=+, ρ2(f)limsupr+loglogT(r,f)logr=n.
(68)
From (62) and (68) we obtain ρ(f)=+ and ρ2(f)=n.

Acknowledgments

This paper was supported by Directorate-General for Scientific Research and Technological Development(DGRSDT).

Author Contributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

Conflicts of Interest

The authors declare no conflict of interest.

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