# On the Complex Oscillation of Meromorphic Solutions of Nonhomogeneous Linear Differential Equations with Meromorphic Coefficients

Authors: Chuang-Xin Chen; Ning Cui; Zong-Xuan Chen
Journal: Discrete Dynamics in Nature and Society
 (2020)

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## Abstract

In this paper, we study the higher order differential equationfk+Bf=H, where B is a rational function, having a pole at ∞ of order n>0, and H≡0 is a meromorphic function with finite order, and obtain some properties related to the order and zeros of its meromorphic solutions.


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## Body

## 1. Introduction and Results

In this paper, a meromorphic function means a function that is meromorphic in the whole complex planeℂ. We assume that the reader is familiar with the fundamental results and the standard notations of Nevanlinna’s value distribution theory of meromorphic functions (see [1, 2]) and use the same notations as in [3]. In addition, we use the notations λf and λ¯f to denote the exponents of convergence of the zeros and distinct zeros of a meromorphic function fz, respectively, and σf to denote the growth order of f.In 1993, Chen [4] obtained the following theorems.Theorem 1.
LetA be a rational function with n-th order pole at ∞, and let E be a transcendental meromorphic function with σE=β<n+k/k. If all solutions of the differential equation(1)fk+Af=E,k≥2,are meromorphic, then every solution fz of (1) satisfies(2)λ¯f=λf=σf=n+kk,λ1f=λ1E,with at most one exceptional solution f0 satisfying β≤σf0<n+k/k.Theorem 2.
LetA be a rational function with n-th order pole at ∞, and let E be a transcendental meromorphic function with n+k/k<σE=β<∞. If all solutions fz of (1) are meromorphic, then(a)
σf=β,λ1/f=λ1/E.(b)
Ifβ=λE>λ1/E, then λf=β.(c)
Ifβ>maxλE,λ1/E, then fz satisfies(3)λ¯f=λf=σf=β,with at most one exceptional solution f0 satisfying σf0<β.We all know from the fundamental theory of complex differential equation that all solutions of linear differential equation with entire coefficients are entire functions, but a solution of linear differential equation with meromorphic coefficients is not always a meromorphic function. For example,fz=exp1/z is a solution of the equation f″−1/z4+2/z3f=0, but fz is not a meromorphic function. Hence, the condition “all solutions fz of (1) are meromorphic” in Theorems 1 and 2 is very rigorous. A natural question to ask is whether the condition “all solutions fz of (1) are meromorphic” in Theorems 1 and 2 can be omitted? In this paper, we consider the above problem and give a positive answer by proving the following theorems.Theorem 3.
LetB be a rational function with n-th order pole at ∞, and let H≡0 be a meromorphic function with σH=β. If the differential equation(4)fk+Bf=H,has a meromorphic solution, then(a)
Ifβ<n+k/k, then all meromorphic solutions fz of (4) satisfy(5)λ¯f=λf=σf=n+kk,with at most one exceptional solutionf0 satisfying σf0=σH<n+k/k.(b)
Ifβ=n+k/k, then all meromorphic solutions fz of (4) satisfy σf=n+k/k and(6)maxλf,λ1f≥maxλH,λ1H.Theorem 4.
LetB be a rational function with n-th order pole at ∞, and let H≡0 be a meromorphic function with n+k/k<σH=β<∞. If differential equation (4) has meromorphic solution, then(a)
σf=β.(b)
Ifβ=λH>λ1/H, then λf=β.(c)
Ifβ>maxλH,λ1/H, then all meromorphic solutions fz of (4) satisfy(7)λ¯f=λf=σf=β,with at most one exceptional solution f0 satisfying(8)maxλf0,λ1f0≥maxλH,λ1H.


## 2. Lemmas for the Proof of Theorems

Lemma 1 (see [5]).
LetH be a meromorphic function and σH=β<∞. Then, for any ε>0, there exists a set E1⊂1,∞ that has finite linear measure and finite logarithmic measure, such that(9)Hz≤exprβ+ε,holds for all z satisfying z=r∉E1∪0,1.Lemma 2 (see [5]).
Letgz be a transcendental entire function and σg=α<∞. Then, there exists a set E2⊂1,∞ that has infinite logarithmic measure, such that(10)limr∈E2r⟶∞loglogMr,glogr=limr∈E2r⟶∞logνgrlogr=α,where νgr is the central index of gz.Lemma 3 (see [4]).
LetB be a rational function with n-th order pole at ∞. If fz≡0 is a meromorphic solution of the differential equation(11)fk+Bf=0,then σf=n+k/k.Lemma 4 (see [4]).
LetE≡0 be a meromorphic function, and let bk−jj=1,2,…,k be some rational functions. If fz is a meromorphic solution of the equation(12)fk+bk−1fk−1+⋯+b0f=E,satisfying σE<σf<∞, then λ¯f=λf=σf.Lemma 5 (see [6]).
Letu1z be a meromorphic function with σu1=β<∞, and let ε>0 be a given constant. Then, there exists a set E3⊂1,∞ that has finite logarithmic measure, such that(13)u1kzu1jz≤zk−jβ−1+ε,j=0,1,…,k=j+1,j+2,…,holds for all z satisfying z=r∉E3∪0,1.Lemma 6.
Letb0,b1,…,bk−1,H≡0 be some meromorphic functions and σH=β<∞. For b0,b1,…,bk−1, there exists a set E4⊂1,∞ that has finite logarithmic measure, such that for all z satisfying z=r∉E4∪0,1, we have(14)bjz≤zcj,j=0,1,…,k−1,where cjj=0,1,…,k−1 are positive constants. If there is an entire function gz satisfying the equation(15)gk+bk−1gk−1+⋯+b0g=H,then σg<∞.Proof.
Suppose thatσg=∞, μr is the maximum term of gz when z=r, and νr is the central index of gz.
Using a similar method as in the proof of Lemma2 in [5], we know that there exists a subset E2⊂1,∞ with infinite logarithmic measure, such that(16)limr∈E2r⟶∞logνr/2logr/2=∞.
Sinceνr is a step function of r, without loss of generality, we suppose that 0=t0<t1<t2<⋯ are discontinuity points of νr and μt=aνttνt has a fixed central index νt=m when t∈tj,tj+1. Hence, for t∈tj,tj+1, we have(17)μ′r=mamtm−1=μtνtt,outside of finite points in 0,r. Since μt is a continuous function, for r>2, we have(18)logμr−logμ1=∫1rμ′tμtdt=∫1rνttdt>∫r/2rνttdt≥νr2log2.
By Cauchy’s inequality, we haveμr≤Mr,g. From the inequality above, we can get that(19)νr2log2≤logMr,g−logμ1.
Choosing a sufficiently largeα such that(20)α>maxc0,c1,…,ck−1,β+k+2.
By (16)–(19), we can see that for sufficiently large r, r∈E2, we have(21)νr≥νr2≥r2α=d2rα,(22)Mr,g>d3expd4rα,where d2,d3,andd4 are the positive constants.
From Wiman–Valiron theory (see [7]), we can choose z=r,gz=Mr,g, such that we have(23)gjzgz=νrzj1+o1,j=1,2,…,k,r∉E4,where E4 is of finite logarithmic measure.
By Lemma1, we can see that there exists a set E1⊂1,∞ of finite logarithmic measure, such that for all z satisfying z=r∈1,∞−E1,r⟶∞, we have(24)Hz≤exprβ+1/2.
Now, we can choose sufficiently largez=r∈E2−E1∪E3∪E4,gz=Mr,g, E2−E1∪E3∪E4 is of infinite logarithmic measure; by (15) and (23), we have(25)νrzk1+o1+bk−1νrzk−11+o1+⋯+b0=Hzgz,that is,(26)νr1+o1+bk−1·z1+o1+bk−2·z2ν1+o1+⋯+b0·zkνk−1=Hgzkνk−1.
Byα>β, (20)–(22), and (24), we have(27)Hgzkνk−1≤exprβ+1/2·rkd3expd4rα·d2rαk−1⟶0,r⟶∞,where z=r∈E2−E1∪E3∪E4,gz=Mr,g. For j=1,2,…,k−1, we have(28)bj−1zk−j+1νk−j≤rcj−1+k−j+1d2rαk−j⟶0,holds for z=r∈E2−E1∪E3∪E4,r⟶∞. Hence, (29)Hgzkνk−1−bk−1·z1+o1−bk−2·z2ν1+o1−⋯−b0·zkνk−1=Obk−1·z=Orck−1+1=orck−1+k+1.
On the other hand, by (20) and (21), we can see that for sufficiently large r, r∈E2, we have(30)νr1+o1>d22rα>rck−1+k+1.
By (29) and (30), we can know that (26) implies a contradiction. Hence, σg<∞.Lemma 7.
LetB be a rational function with n-th order pole at ∞, and let H≡0 be a meromorphic function with σH=β. If fz is a meromorphic solution of (4), then(a)
Ifβ<n+k/k, then all meromorphic solutions fz of (4) satisfy σf=n+k/k, with at most one exceptional solution f0 satisfying σf0=β.(b)
Ifβ≥n+k/k, then σf=β.Proof.
To our aim, we will consider the following three cases.Case 1. Suppose that Hz=B0zexpPz, where B0z is a rational function and Pz is a polynomial and then degP=β. By Lemma 3 in [4], we can know that the conclusion holds.Case 2. Suppose that Hz is a meromorphic function with infinitely many poles. Then, by (4), we have σf≥β.(a)
First, we will prove that ifσf=α>β, then σf=n+k/k.Suppose thatz′ is a pole of fz of order m1≥1 and BandH are all analytic at z′. Then, z′ must be a pole of fk+Bf of order m1+k, which contradicts the fact that H is analytic at z′. Hence, all poles of f come from poles of B and H. Since B has only finitely many poles, we can see that all poles of f come from poles of H with finitely many exceptions. Suppose that z″ is a pole of H of order m2 and B is analytic at z″, then z″ is a pole of fz of order m2−k. Thus, we can know that fz has infinitely many poles. From(31)nr,f=nr,H−kn¯r,H+O1≤nr,H+O1,nr,H=nr,f+kn¯r,f+O1≤k+1nr,f+O1,we can see thatfz and H have the same exponent of convergence of poles, that is,(32)λ1f=λ1H≤β.By Hadamard factorization theory, we can setfz=gz/zk2uz=gz/u1z, where k2 is a nonnegative integer, gz is an entire function, and u1z=zk2·uz, where uz is the canonical product formed with the nonzero poles zj:j=1,2,⋯;zj=rj,0<r1≤r2≤⋯ of fz. λu1=σu1=λ1/f≤β. By σf=α>β, we have σg=σf=α.By Lemma1, we can choose ε, such that(33)0<k+1ε<minα−β,n+kk−β,holds, and there exists a setE1⊂1,+∞ of finite logarithmic measure, such that for all z satisfying z=r∉0,1∪E1,r⟶∞, we have(34)1u1z≤exprβ+ε.By (32), we can know that fz and H have the same exponent of convergence of poles. From the above proofs, we can see that except for finitely many exceptions, fz and H have the same poles with the difference of their orders being k; hence, the above set E1 concerning fz, for H, we still have for all z satisfying z=r∉0,1∪E1,r⟶∞,(35)Hz≤exprβ+ε,holds. Substitutingfz=gz/u1z into (4), we get(36)gkg+bk−1gk−1g+⋯+b0+B=u1Hg,wherebk−j are some differential polynomials, with constant coefficients, in u1′/u1,u1″/u1,…,u1j/u1. By σu1≤β and Lemma 5, we can see that there exists a set E3⊂1,+∞ with finite logarithmic measure, such that for all z satisfying z=r∉0,1∪E3,r⟶∞, we have(37)u1jzu1z≤zjβ−1+ε,j=1,2,…,k.By (33), we have kβ−1+ε<n. Hence, by (37), we can easily obtain(38)bk−j≤czjβ−1+ε,j=1,2,…,k,b0+B=dzn1+o1,wherecandd are two nonzero constants and z=r∈1,∞−E3,r⟶∞. Thus, by Lemma 6, (36) and (38), we have σg=α<∞.By Lemma2 and σg=α<∞, we can see that there exists a set E2⊂1,+∞ with infinite logarithmic measure, such that(39)limr∈E2r⟶∞loglogMr,glogr=limr∈E2r⟶∞logνrlogr=α,whereνr is the central index of gz.Choosingz=r,gz=Mr,g, from Wiman–Valiron theory, we can know that (23) holds outside a set E4 with finite logarithmic measure. For sufficiently large r, by (39), we have(40)Mr,g≥exprα−ε,wherez=r∈E2−E1∪E3∪E4,gz=Mr,g. Since u1 is entire, σu1≤β, combining (35), (40), and β+2ε≤β+k+1ε<α, we have(41)u1z·Hzgz=u1z·HzMr,g≤exp2rβ+ε−rα−ε⟶0,r⟶∞,wherez=r∈E2−E1∪E3∪E4,gz=Mr,g.Choosingz such that z=r∈E2−E1∪E3∪E4,gz=Mr,g, since E2−E1∪E3∪E4 is of infinite logarithmic measure, by (23), (36), (38), and (41), we can see that for r⟶∞, we have(42)νrzk1+o1+Orβ−1+ενrzk−11+o1+⋯+drn1+o1=u1zHzgz=o1.By (39), we can see that for r∈E2−E1∪E3∪E4,r⟶∞, we have(43)νr=rα+o1.By (43), we can see that the degree of terms of the left-hand side of (42) in r is(44)kα+o1−1,α+o1−1k−1+β−1+ε,…,α+o1−1+k−1β−1+ε,n.By (33), we have β−1+ε<α+o1−1, which implies that(45)kα+o1−1>α+o1−1k−1+β−1+ε>α+o1−1k−2+2β−1+ε>⋯>α+o1−1+k−1β−1+ε.Thus, comparing the degree of terms of both sides of (42) in r, we obtain kα−1=n, which implies α=n+k/k, that is, σf=σg=n+k/k.Now, we will prove that all meromorphic solutionsfz of (4) satisfy σf=n+k/k, with at most one exceptional solution f0 satisfying σf0=β.Suppose thatf0 and f1f1≡f0 are two meromorphic solutions of (4) and satisfy σf0=σf1=β. Then, σf0−f1<n+k/k. Since f0−f1 is a meromorphic solution of homogeneous equation (11) corresponding to (4), by Lemma 3, we have σf0−f1=n+k/k, which is a contradiction. Hence, equation (4) has at most one exceptional solution f0 satisfying σf0=β, and all other meromorphic solutions are satisfying σf=n+k/k.(b)
Sinceσf≥β≥n+k/k, we will just prove that σf=α>β is false.Suppose thatα>β, and set fz=gz/u1z, where gzandu1z have the same meanings as in (a). Then, using a similar method as in the proof of (a), we can see that for any given ε0<k+1ε<α−β, (34)–(37) hold. By β≥n+k/k and (37), we can see that for z=r∈1,∞−E3,r⟶∞, we have(46)bk−j≤czjβ−1+ε,j=1,2,…,k,b0+B=dzn1+o1,wherecandd are the two nonzero constants.By using (36) and (46) in conjunction with Lemma 6, we have σg=α<∞. Continuing using a similar method as in the proof of (a), we can see (39)–(43) hold. Hence, for z=r∈E2−E1∪E3∪E4, we can see that in the left-hand side of (42), only one term νr/zk1+o1 has the highest degree kα−1 in r. It is impossible. Hence, σf=β.Case 3. Suppose that Hz is a meromorphic function with finite many poles and infinite many zeros. Then, we can use a similar method as in the proof of Case 2 to obtain our conclusions.
Using a similar method as in the proof of (a) in Lemma7, we can obtain the following lemma.Lemma 8.
Letβ>1 be a positive integer, and let Bk−j be some rational functions with nk−j-th order pole at ∞, where nk−j=jβ−1j=1,2,…,k. U≡0 is a meromorphic function, and σU<β. If the differential equation(47)hk+Bk−1hk−1+⋯+B0h=U,has meromorphic solution, then all meromorphic solutions hz of (47) satisfy σh=β, with at most one exceptional solution h0 satisfying σh0=σU.Lemma 9.
LetBk−jj=1,2,…,k,Q be rational functions and Bk−j has a pole of order nk−j>0 at ∞, and P is a polynomial with degP=β. If fz is a meromorphic solution of the equation(48)fk+Bk−1fk−1+⋯+B0f=QeP,then σf<∞.Proof.
Without loss of generality, we suppose thatfz is transcendental. Otherwise, fz is a rational function, and we have σf<∞.
Differentiating both sides of (48), we have(49)fk+1+Bk−1−Q′Q−P′fk−Bk−1′+Bk−2−Bk−1Q′Q+P′fk−1+⋯+B1′+B0−B1Q′Q+P′f′+B0′−B0Q′Q+P′f=0,that is,(50)fk+1+∑j=0kBk−j′+Bk−j−1−Bk−jQ′Q+P′fk−j=0,where Bk=1andB−1=0.
Since the poles offz just appear at the poles of Bk−jj=1,2,…,k,Q, fz has finitely many poles. Using f1 to denote the sum of the major part of all poles of f, f2=f−f1 is a transcendental entire function and σf=σf2. Substituting f=f1+f2 into (50), we have(51)f2k+1+∑j=0kBk−j′+Bk−j−1−Bk−jQ′Q+P′f2k−j=−f1k+1+∑j=0kBk−j′+Bk−j−1−Bk−jQ′Q+P′f1k−j.
Since the right-hand side of (51) is a rational function, its order is finite.
Suppose thatBk−j′=ak−jznk−j1+o1 and P′=bzβ−11+o1, where ak−jandb are some nonzero constants. Then, (52)Bk−j′+Bk−j−1−Bk−jQ′Q+P′=nk−jak−j·znk−j−11+o1+ak−j−1·znk−j−11+o1−bak−j·znk−j+β−11+o1,hence, there must exist some positive constants ck−jj=0,1,…,k such that(53)Bk−j′+Bk−j−1−Bk−jQ′Q+P′≤zck−j,j=0,1,…,k.
Applying Lemma6 on (51) and (53), we have σf2<∞. Hence, σf<∞.Lemma 10.
LetBk−jj=1,2,…,k be some rational functions with nk−j-th order pole at ∞, and let H≡0 be a meromorphic function with σH=β<∞. If the equation(54)fk+Bk−1fk−1+⋯+B0f=H,has meromorphic solution f, then(55)λ1f=λ1H,λ¯1f=λ¯1H,maxλf,λ1f≥maxλH,λ1H.Proof.
SinceBk−jj=1,2,…,k have only finitely many poles, we can see that if z0 is not a pole of Bk−jj=1,2,…,k, then z0 is an α-th order pole of f if and only if z0 is an α+k-th order pole of H; hence, λ¯1/f=λ¯1/H. Since(56)α+k2k=α2k+12≤α,we have(57)12knr,H+O1≤nr,f≤nr,H+O1,12kNr,H+Ologr≤Nr,f≤Nr,H+Ologr.
Thus,(58)λ1f=λ1H.
Now, we will prove that every meromorphic solutionf of (54) satisfies σf<∞. So, we will consider the following three cases:Case 1. Suppose that Hz=QzexpPz, where Qz is a rational function and Pz is a polynomial. Then, degP=β. Thus, by Lemma 9, we have σf<∞.Case 2. Suppose that Hz is a meromorphic function with infinitely many poles. Then, by (54), we have σf≥β. If σf=β, then σf<∞. Without loss of generality, we suppose that σf=α>β.Setfz=gz/u1z, where gzandu1z have the same meanings as in the proof of (a) of Lemma 7. Using a similar method as in the proof of (a) of Lemma 7, we can get σu1≤β and σg=σf=α. Substituting fz=gz/u1z into (54), we have(59)gkg+bk−1gk−1g+⋯+b0=u1Hg,wherebk−j are some differential polynomials, with constant coefficients, in u1′/u1,u1″/u1,…,u1j/u1, and Bk−j,…,Bk−1. Using a similar method as in the proof of (a) of Lemma 7, we know (37) holds obviously. Hence, we can easily obtain(60)bk−j≤czjβ−1+ε+n′,j=1,2,…,k,wherec is a nonzero constant, ε satisfies 0<ε<α−β, and n′=maxn0,n1,…,nk−1. By Lemma 6 and (59) and (60), we have σg=α<∞. Hence, σf<∞.Case 3. Suppose that Hz is a meromorphic function with finitely many poles and infinitely many zeros. Then, using a similar method as in the proof of Case 2, we can get σf<∞.
Thus, we can writefandH in the following form:(61)f=zm1E1Q1eP1,H=zm2E2Q2eP2,where m1andm2 are the integers, E1andE2 are the canonical product formed with the nonzero zeros of f and H, Q1andQ2 are the canonical product formed with the nonzero poles of f and H, and P1andP2 are the polynomials such that degP1≤σf and degP2≤σH. Substituting (61) into (54), we have(62)FE1,Q1=zm2E2Q2eP2−P1,where F is a rational function in E1, Q1, and its derivative, with constant coefficients. Comparing the growth order of both sides of (62) and noting that E2andQ2 are canonical products, we have(63)maxσE1,σQ1≥σF=σzm2E2Q2eP2−P1=maxσE2,σQ2,degP2−P1≥maxσE2,σQ2,that is,(64)maxλf,λ1f≥maxλH,λ1H.


## 3. Proofs of Theorems

The proof of Theorem3.
(a)
By Lemma7, we can see that all meromorphic solutions fz of equation (4) are satisfying σf=n+k/k, with at most one exceptional meromorphic solution f0 satisfying σf0=σH=β. From Lemma 4, we can know that meromorphic solutions fz of (4) with σf=n+k/k satisfy λ¯f=λf=σf=n+k/k.(b)
By Lemmas7 and 10, we get the required result.The proof of Theorem4.
(a)
By Lemma7, we have σf=β.(b)
Ifβ=λH>λ1/H, then, by Lemma 10, we have λf=β.(c)
Ifβ>maxλH,λ1/H, set H=UeP, where U=zkV1/V2, k is an integer and V1 and V2 are the canonical product (or polynomial) formed with the nonzero zeros and nonzero poles of H, respectively. σU=maxλH,λ1/H<β, and Pz is a polynomial with degPz=β. Setting f=h·eP, where h is a meromorphic function. Thus, f and h have the same zeros and poles. Substituting f=h·ePandH=UeP into (4), we have(65)hk+bk−1hk−1+⋯+b0h=U,where(66)bk−1=kP′,bk−j=CkjP′j+Hj−1P′,j=2,…,k−1,b0=CkkP′k+Hk−1P′+B,where Ckj are the binomial coefficients and Hj−1P′ are the polynomials in P′ and its derivatives, with constant coefficients and having degree j−1. It is easy to see the derivative of Hj−1P′ with respect to z having the same form with Hj−1P′. Since β>n+k/k, we know degbk−j=jβ−1j=1,2,…,k. By σU<β and Lemma 8, we can see that all meromorphic solutions hz of (65) satisfy σf=n+k/k, with at most one exceptional meromorphic solution h0 satisfying σh0=σU. By Lemma 4, we have λ¯h=λh=σh=β. From Lemma 10, we can know that h0 satisfies(67)σh0≥maxλh0,λ1h0≥maxλH,λ1H.
Hence, (4) has at most one exceptional meromorphic solution f0=h0eP satisfying(68)maxλf0,λ1f0≥maxλH,λ1H,and all other meromorphic solutions f=heP satisfy λ¯f=λf=λh=β.


## 4. Conclusion

Our paper investigates the nonhomogeneous linear differential equationfk+Bf=H, where B is a rational function, having a pole at ∞ of order n>0, and H≡0 is a meromorphic function with finite order, and obtains some properties related to the order and zeros of its meromorphic solutions.



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