# Growth of Solutions of Nonhomogeneous Linear Differential Equations

Authors: Jun Wang; Ilpo Laine
Journal: Abstract and Applied Analysis
 (2009)

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## Abstract

This paper is devoted to studying growth of solutions of linear differential
equations of typef(k)+Ak−1(z)f(k−1)+⋯+A1(z)f′+A0(z)f=H(z) where Aj(j=0,…,k−1) and H are entire functions of finite order.


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## Body

## 1. Introduction and Main Results

We assume that
the reader is familiar with the usual notations and basic results of the
Nevanlinna theory [1–3]. Let now f(z) be a nonconstant meromorphic function in the
complex plane. We remark that ρ(f) will be used to denote the order of f,
and(1.1)ρ(f)=limsupr→∞logT(r,f)logr.We now recall
some previous results concerning nonhomogeneous linear differential equations
of type(1.2)f(k)+Ak−1(z)f(k−1)+⋯+A1(z)f′+A0(z)f=H(z),where Aj(j=0,1,…,k−1) and A0≢0, H≢0 are entire functions of finite-order, k≥2.
In the case that the coefficients Aj(j=0,1,…,k−1) are polynomials, growth properties of
solutions of (1.2) have been extensively studied, see, for example,  [4]. In (1.2), if p is the largest integer such that Ap is transcendental, it is well known that there
exist at most p linearly independent finite-order solutions of
the corresponding homogeneous equation(1.3)f(k)+Ak−1(z)f(k−1)+⋯+A1(z)f′+A0(z)f=0.Thus, when at least one of the
coefficients Aj is transcendental, most of the solutions of
(1.2) and (1.3) are of infinite-order. In the case when(1.4)maxj≠d{ρ(Aj),ρ(H)}<ρ(Ad)≤12,
Hellerstein et al.   [5] proved that every
transcendental solution of (1.2) is  of infinite-order. As for sectorial growth
conditions on the coefficients of (1.2) that imply that all solutions are of
infinite-order, see, for example,  [6]. As for the special case of k=2,
Wang and Laine studied equations of type(1.5)f′′+A1(z)eazf′+A0(z)ebzf=H(z),where A1≢0,A0≢0,H are entire functions of order less than one,
and the complex numbers a,b satisfy ab≢0.
They proved that every nontrivial solution of (1.5) is of infinite-order if a≢b,
see [7]. We remark
that (1.2) may indeed have solutions of finite-order as soon as ρ(H)≥max{ρ(Aj)(j=0,…,k)},
as shown by the next examples.Example 1.1.
The exponential functionf(z)=ez satisfies the equation(1.6)f(k)+f(k−1)+⋯+f′′+e−zf′+Q(z)f=(k−1+Q(z))ez+1,where Q(z) can be any entire function. Choosing Q(z)=1−k shows that (1.2) may admit a solution of
finite-order even if ρ(H)<max{ρ(Aj)(j=0,…,k)}.
On the other hand, taking Q(z)=ez,
we have the case that ρ(H)=max{ρ(Aj)(j=0,…,k)} in (1.2).Example 1.2.
The functionf(z)=ez2 satisfies the equation(1.7)f′′′+e−zf′′+ezf′+e2zf=(8z3+12z+4z2e−z+2e−z+2zez+e2z)ez2.
In this paper, we continue to consider (1.2) in the
case when ρ(H)<max{ρ(Aj)(j=0,…,k)}.
Recently, Tu and Yi investigated the growth of solutions of (1.3) when
most coefficients have the same order, see [8]. We next prove two results of (1.2), which
generalize Theorems 2 and 4 in [8] and Theorem 1.1 in [7].Theorem 1.3.
Suppose thatAj(z)=hj(z)ePj(z)(j=0,…,k−1) where Pj(z)=ajnzn+⋯+aj0 are polynomials with degree n≥1, hj(z) are entire functions of order less than n,
not all vanishing, and  H(z)≢0 is an entire function of order less than n.
If ajn(j=0,…,k−1) are distinct complex numbers, then every
solution of (1.2) is of infinite-order.Theorem 1.4.
Suppose thatAj(z)=hj(z)ePj(z)(j=0,…,k−1) where Pj(z)=ajnzn+⋯+aj0 are polynomials with degree n≥1, hj(z) and H(z)≢0 are entire functions of order less than n.
Moreover, suppose that there are two coefficients As,Al so that for asn=|asn|eiθs and aln=|aln|eiθl, where
0≤s<l≤k−1, θs,θl∈[0,2π), θs≢θl, hshl≢0, and for all j≢s,l, ajn satisfies  either ajn=djasn(0<dj<1) or ajn=djaln(0<dj<1).
Then every transcendental solution of (1.2) is of infinite-order.In the case
whenAj(z)=hjeajz+gj where hj,gj(j=0,…,k−1) are polynomials, Chen considered the growth of
solutions of (1.3) with some additional conditions imposed upon on aj,
see [9]. Our last
results generalizes his result and [7, Theorem 1.3].Theorem 1.5.
Suppose thatAj(z)=hj(z)ePj(z)+gj(z)(j=0,…,k−1) where Pj(z)=ajnzn+⋯+aj0 are polynomials with degree n≥1, hj(z),gj(z) and H(z)≢0 are entire functions of order less than n.
Moreover, suppose that there exist asn=dseiφ and aln=−dleiφ with ds>0,dl>0 and 0≤s<l≤k−1 such that for j≠s,l, ajn=djeiφ(dj≥0) or ajn=−djeiφ(dj≥0),
and max{dj,j≢s,l}=d<min{ds,dl}.
If hshl≢0,
then every transcendental solution of (1.2) is of infinite-order.Remark 1.6.
Under the
assumptions of Theorem1.4, respectively, of Theorem 1.5, polynomial solutions may
exist. However, such possible polynomial solutions must be of degree less than s.
If not, a contradiction immediately follows by combining (5.1) with Lemma 2.1,
if F≡0,
respectively, with Lemma 2.2, if F≢0.Remark 1.7.
In the
preceding three theorems, ifρ(f)=∞,
then we also have λ(f)=∞ for the exponent of convergence of the
zero-sequence of f.
Indeed, rewriting  (1.2) in the form(1.8)1f=1H(f(k)f+Ak−1f(k−1)f+⋯+A0),we have(1.9)m(r,1f)≤m(r,1H)+∑j=0k−1m(r,Aj)+∑j=0k−1m(r,f(j)f)=O(rβ)+S(r,f),for some finite β.
Therefore, N(r,1/f) must be of infinite-order.


## 2. Preliminary Lemmas

Lemma 2.1 (see [10]).
Suppose thatf1(z),f2(z),…,fn(z)(n≥2) are meromorphic functions and g1(z),g2(z),…,gn(z) are entire functions satisfying the following
conditions:
(i)
∑j=1nfj(z)egj(z)≡0,(ii)
gj(z)−gk(z) are not constants for 1≤j<k≤n,(iii)
for1≤j≤n,1≤h<k≤n,(2.1)T(r,fj)=o{T(r,egh−gk)},(r→∞,r∉E),
whereE is a set with finite linear measure. Then fj≡0(j=1,2,…,n).Lemma 2.2 (see [10]).
Suppose thatf1(z),f2(z),…,fn(z)(n≥2) are linearly independent meromorphic functions
satisfying the following identity:(2.2)∑j=1nfj≡1.Then for 1≤j≤n,
one has
(2.3)T(r,fj)≤∑j=1kN(r,1fk)+N(r,fj)+N(r,D)−∑k=1nN(r,fk)−N(r,1D)+S(r),where D is the Wronskian determinant W(f1,f2,…,fn),(2.4)S(r)=o(max1≤k≤n{T(r,fk)}),(r→∞,r∉E),E is a set with finite linear measure.Lemma 2.3 (see [11, 12]).
Suppose thatP(z)=(α+iβ)zn+⋯ (α,β are real numbers, |α|+|β|≠0) is a polynomial with degree n≥1,
and that A(z)(≢0) is an entire function with ρ(A)<n.
Set g(z)=A(z)eP(z),z=reiθ,δ(P,θ)=αcos(nθ)−βsin(nθ).
Then for any given ε>0,
there exists a set H1⊂[0,2π) of finite linear measure such that for any θ∈[0,2π)∖(H1∪H2),
there is R>0 such that for |z|=r>R,
one has
(i)
ifδ(P,θ)>0,
then(2.5)exp{(1−ε)δ(P,θ)rn}<|g(reiθ)|<exp{(1+ε)δ(P,θ)rn};(ii)
ifδ(P,θ)<0,
then(2.6)exp{(1+ε)δ(P,θ)rn}<|g(reiθ)|<exp{(1−ε)δ(P,θ)rn},
whereH2={θ∈[0,2π);δ(P,θ)=0}.Lemma 2.4 (see [13]).
Letf(z) be a transcendental meromorphic function of
finite-order ρ,
and let ε>0 be a given constant. Then there exists a set H⊂(1,∞) that has finite logarithmic measure, such that
for all z satisfying |z|∉H∪[0,1] and for all k,j,0≤j<k,
one has
(2.7)|f(k)(z)f(j)(z)|≤|z|(k−j)(ρ−1+ε).Similarly, there exists a set E⊂[0,2π) of linear measure zero such that for all z=reiθ with |z| sufficiently large and θ∈[0,2π)∖E,
and for all k,j,0≤j<k,
the inequality (2.7) holds.Lemma 2.5.
Letf(z) be an entire function and suppose that(2.8)G(z):=log+|f(k)(z)||z|ρis unbounded on some ray argz=θ with constant ρ>0.
Then there exists an infinite sequence of points zn=rneiθ(n=1,2,…), where rn→∞,
such that G(zn)→∞ and(2.9)|f(j)(zn)f(k)(zn)|≤1(k−j)!(1+o(1))rnk−j,j=0,…,k−1,as n→∞.Proof.
The first assertion is trivial.
Denoting(2.10)M(r,G,θ)=max{G(z):0≤|z|≤r,argz=θ},we may take the sequence {zn} in the first assertion so that G(zn)=M(rn,G,θ).
Since(2.11)G(zn)→∞as n→∞,
we immediately see that(2.12)|f(k)(zn)|=M(rn,f(k),θ)→∞as n→∞.
Using now the same reasoning as in the proof of [14, Lemma 4], see also [15,
Lemma 3.1], the second assertion (2.9) follows.Lemma 2.6.
Letf(z) be an entire function with ρ(f)=ρ<∞.
Suppose that there exists a set E⊂[0,2π) which has linear measure zero, such that log+|f(reiθ)|≤Mrσ for any ray argz=θ∈[0,2π)∖E,
where M is a positive constant depending on θ,
while σ is a positive constant independent of θ.
Then ρ(f)≤σ.Proof.
Clearly, we may assume thatσ<ρ.
Since E has linear measure zero, we may choose θj∈[0,2π)∖E such that 0≤θ1<θ2<⋯<θn+1=2π,
and(2.13)max{θj+1−θj,1≤j≤n}≤πρ+1.We first treat the sector(2.14)H1:={z∣θ1≤argz≤θ2},defining(2.15)ϕ(z)=f(z)exp{−be−iθ0zσ},where θ0=σ(θ1+θ2)/2 and b is a positive constant, to be determined
in what follows. Then ϕ(z) is a holomorphic inside the sector H1.
By (2.13), we have ρ≤π/(θ2−θ1)−1.
Therefore,(2.16)0>arg(e−iθ0zσ)=arg(e−iθ0rσeiσθ1)=σ(θ1−θ2)2≥−π2+(θ2−θ1)2on the ray argz=θ1,
and, respectively,(2.17)0<arg(e−iθ0zσ)=arg(e−iθ0rσeiσθ2)=σ(θ2−θ1)2≤π2−(θ2−θ1)2on the ray argz=θ2.
Hence, we may now fix b>0 so that(2.18)bcos(π2−(θ2−θ1)2)>M.By elementary computation, |ϕ(z)|≤M on the boundary of H1,
where M>0 is a bounded constant, not the same at each
occurrence. By the definition of ϕ in (2.15), it is immediate to see that ϕ is of order at most ρ.
By the Phragmén-Lindelöf theorem, we conclude that |ϕ(z)|≤M holds on the whole sector H1.
Hence(2.19)|f(z)|≤|exp{be−iθ0zσ}|≤exp{brσ}on H1.
Repeating the same reasoning for all the sectors Hj={z∣θj≤argz≤θj+1} where θj are determined in (2.13), the assertion
immediately follows.


## 3. Proof of Theorem1.3

Suppose,
contrary to the assertion, thatf is a solution of (1.2) with ρ(f)=ρ<∞,
then n≤ρ.
Indeed, if f(k)=H,
we may apply Lemma 2.1 to conclude that hsf(s)≡0 for some s, 0≤s≤k−1 such that hs≠0.
Then f has to be a polynomial of degree less than s,
so H(z)≡0,
a contradiction. Therefore, we may assume that f(k)≠H.
By Lemma 2.2, it is easy to see that n≤ρ since the exponential functions ePj(j=0,1,…,k−1) are linearly independent.By Lemma2.3, there is a set E⊂[0,2π) of linear measure such that whenever θ∈[0,2π)∖E,
then δ(Pj,θ)≠0 for all 0≤j≤k−1 and δ(Pj−Pi,θ)≠0 for all i,j with 0≤i<j≤k−1.
If, moreover, z=reiθ has r large enough, then each Aj(z) satisfies either (2.5) or (2.6). By Lemma 2.4,
we may assume, at the same time, that(3.1)|f(j)(z)f(i)(z)|≤|z|kρ,0≤i<j≤k.Since ajn are distinct complex numbers, then for any
fixed θ∈[0,2π)∖E,
there exists exactly one s∈{0,…,k−1} such that(3.2)δ(Ps,θ)=δ:=max{δ(Pj,θ)∣j=0,…,k−1}.Denoting δ1=max{δ(Pj,θ)∣j≠s},
then δ1<δ and δ≠0.
We now discuss two cases separately.Case 1.
Assume first thatδ>0.
By Lemma 2.3, for any given ε with 0<3ε<min{(δ−δ1)/δ,n−ρ(H)},
we have(3.3)|As(reiθ)|≥exp{(1−ε)δrn},|Aj(reiθ)|≤exp{(1+ε)δ1rn},for j≠s,
provided that r is sufficiently large. We now proceed to show
that(3.4)log+|f(s)(z)||z|ρ(H)+εis bounded on the ray argz=θ.
Supposing that this is not the case, then by Lemma 2.5, there is a sequence of
points zm=rmeiθ,
such that rm→∞,
and that(3.5)log+|f(s)(zm)|rmρ(H)+ε→∞,(3.6)|f(j)(zm)f(s)(zm)|≤(1+o(1))rms−j,(j=0,…,s−1).From (3.5) and the definition of
order, it is easy to see that(3.7)|H(zm)f(s)(zm)|→0,for m is large enough. From (1.2), we
obtain(3.8)|As(zm)|≤|f(k)(zm)f(s)(zm)|+⋯+|As+1(zm)||f(s+1)(zm)f(s)(zm)|+|As−1(zm)||f(s−1)(zm)f(s)(zm)|+⋯+|A0(zm)||f(zm)f(s)(zm)|+|H(zm)f(s)(zm)|.Using inequalities (3.1), (3.3), (3.6), and the limit (3.7), we conclude from the preceding inequality
that(3.9)exp{(1−ε1)δrmn}≤(k+1)exp{(1+ε1)δ1rmn}rmM,where M>0 is a bounded constant, which is a
contradiction. Therefore, log+|f(s)(z)|/|z|ρ(H)+ε is bounded, and we have |f(s)(z)|≤Mexp{rρ(H)+ε} on the ray argz=θ.
By the same reasoning as in the proof of [15, Lemma 3.1], we immediately
conclude that(3.10)|f(z)|≤(1+o(1))rs|f(s)(z)|≤(1+o(1))Mrserρ(H)+ε≤Merρ(H)+2εon the ray argz=θ.Case 2.
Suppose now thatδ<0.
From (1.2), we get(3.11)−1=Ak−1f(k−1)f(k)+⋯+Ajf(j)f(k)+⋯+A0ff(k)−Hf(k).Again by Lemma 2.3, for any
given ε with 0<3ε<min{1,n−ρ(H)},
we have(3.12)|Aj(reiθ)|≤exp{(1−ε)δrn},(j=0,1,…,k−1),for r sufficiently large. As in Case 1, we prove
that(3.13)log+|f(k)(z)||z|ρ(H)+εis bounded on the ray argz=θ.
If not, similarly as in Case 1, it follows from Lemma 2.5 that there is a
sequence of points zm=rmeiθ,
such that(3.14)|f(j)(zm)f(k)(zm)|≤rmk−j(1+o(1)),(j=0,…,k−1),|H(zm)f(k)(zm)|→0,for all m large enough. Substituting the inequalities
(3.12) and (3.14)  into (3.11), a contradiction immediately follows. Hence,
we have |f(k)(z)|≤Mexp{rρ(H)+ε} on the ray argz=θ.
This implies, as in Case 1, that(3.15)|f(z)|≤Mexp{rρ(H)+2ε}. Therefore, for
any given θ∈[0,2π)∖E, E of linear measure zero, we have got (3.15) on
the ray argz=θ,
provided that r is large enough. Then by Lemma 2.6, ρ(f)≤ρ(H)+2ε<n,
a contradiction. Hence, every transcendental solution of (1.2) must be of
infinite-order.


## 4. Proof of Theorem1.4

Suppose thatf is a transcendental solution of (1.2) with ρ(f)=ρ<∞.Iff(k)≡H and ρ<n,
it follows from (1.2) that(4.1)f(l)hlePl(z)+f(s)hsePs(z)+∑u=1pBu(z)edjuPl(z)+∑v=1qCv(z)edjvPs(z)=0,where Bu(u=1,…,p),Cv(v=1,…,q) are entire functions of order less than n.
Collecting terms of the same type together, if needed, we may assume that the
coefficients dju(u=1,…,p),
respectively, djv(v=1,…,q),
are distinct. Since θs≠θl and θs,θl∈[0,2π),
we conclude that djuPl(z)−djvPs(z) are polynomials of degree n.
Indeed, if djualn=djvasn,
we have(4.2)0<djudjv|alnasn|=ei(θs−θl)which is impossible. Similarly, Pl(z)−Ps(z),Pl(z)−djvPs(z),  and Ps(z)−djuPl(z) are also polynomials of degree n.
Therefore, applying Lemma 2.1 to (4.1), we infer that f(l)hl≡f(s)hs≡0.
Since hshl≢0, f has to be a polynomial of degree less than s,
then H≡0,
a contradiction.Therefore, we may proceed under the assumption thatf(k)≢H.
By Lemma 2.2, if f(k)≢H,
then n≤ρ since the exponential functions ePl,ePs,edjuPl(u=1,2,…,p) and edjvPs(v=1,2,…,q) are linearly independent.Sinceθs≠θl,
by Lemmas 2.3 and 2.4, there exists a set E⊂[0,2π) of linear measure zero such that whenever θ∈[0,2π)∖E then Aj(reiθ) satisfies either (2.5) or (2.6), (3.1) holds,
and(4.3)δ(Ps,θ)≠δ(Pl,θ),δ2:=max{δ(Ps,θ),δ(Pl,θ)}≠0.In what follows, we apply the
notations δ,δ1 from the proof of Theorem 1.5 as well.Case 1.
Firstly assume thatδ2>0.
Without loss of generality, we may assume that δ2=δ(Ps,θ).
From the hypothesis of ajn,
we know that δ1<δ2=δ.
Therefore, (3.3) holds by Lemma 2.3. Using the same reasoning as in
Case 1
of the proof of Theorem 1.3, we  obtain the inequality (3.15) on the
ray argz=θ.Case 2.
Finally, assume thatδ2<0.
Again by the condition on ajn,
we see that δ<0.
Then the same argument as in Case 2
of the proof of Theorem 1.3 applies, and
we again obtain (3.15).Therefore, by Lemma2.6, we obtain a contradiction, so ρ(f)=∞.


## 5. Proof of Theorem1.5

Contrary to the
assertion, suppose thatf is a transcendental solution of (1.2) of
finite-order. If ρ<n,
then it follows from (1.2) that(5.1)f(l)hlePl(z)+f(s)hsePs(z)+∑u=1pBu(z)edjuPl(z)+∑v=1qCv(z)edjvPs(z)=F(z),where Bu(u=1,…,p),Cv(v=1,…,q), and F(z) are entire functions of order less than n, dju≠0(u=1,…,p) are distinct, and djv≠0(v=1,…,q) are also distinct. Similarly as in the proof
of Theorem 1.4, we may assume that n≤ρ.
Since σ=max{ρ(gj)(j=0,…,k−1)}<n,
we have(5.2)max{|gj(z)|(j=0,…,k−1),|H(z)|}≤exp{rσ+ε}for any ε with 0<3ε<n−σ,
and for |z| sufficiently large. Since ds and dl in asn=dseiφ and aln=−dleiφ are strictly positive, the set {θ∈[0,2π),δ(Ps,θ)=δ(Pl,θ)} is of linear measure zero. Therefore, again by
Lemmas 2.3 and 2.4, there exists a set E⊂[0,2π) of linear measure zero such that for any given θ∈[0,2π)∖E, hjePj satisfies either (2.5) or (2.6), and (3.1)
holds. Moreover, δ(Ps,θ)≠δ(Pl,θ).
Without loss of generality, we may assume that δ2:=max{δ(Ps,θ),δ(Pl,θ)}=δ(Pl,θ)=−dlcos(φ+nθ), where cos(φ+nθ)<0.
Then from (2.5) and (5.2), for any ε also satisfying 0<3ε<(dl−d)∖dl,
we obtain for |z| sufficiently large that(5.3)|Al(reiθ)|≥exp{−(1−ε)dlcos(φ+nθ)rn}.For all other coefficients Aj(j≠s),
considering the hypothesis of ajn,
we have(5.4)|Aj(reiθ)|≤exp{−(1+ε)dcos(φ+nθ)rn},when r is large enough. It follows from (1.2) that(5.5)−Al=f(k)f(s)+⋯+Al+1f(l+1)f(l)+Al−1f(l−1)f(l)+⋯+A0ff(l)−Hf(l).Similarly as in Case 1
of the
proof of Theorem 1.4, and using Lemma 2.5, we may prove that(5.6)log+|f(l)(z)||z|ρ(H)+εis bounded on the ray argz=θ.
Therefore, the inequality (3.15) always holds on the ray argz=θ.
Then, by Lemma 2.6, a contradiction follows, and so ρ(f)=∞.



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