AAAAbstract and Applied Analysis1687-04091085-3375Hindawi Publishing Corporation36392710.1155/2009/363927363927Research ArticleGrowth of Solutions of Nonhomogeneous Linear Differential EquationsWangJun1LaineIlpo2EloePaul1School of Mathematics ScienceFudan UniversityShanghai 200433Chinafudan.edu.cn2Department of MathematicsUniversity of JoensuuFI-80101 JoensuuFinlandjoensuu.fi200904032009200905112008170220092009Copyright © 2009This 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.

This paper is devoted to studying growth of solutions of linear differential equations of type f(k)+Ak1(z)f(k1)++A1(z)f+A0(z)f=H(z) where Aj(j=0,,k1) and H are entire functions of finite order.

1. Introduction and Main Results

We assume that the reader is familiar with the usual notations and basic results of the Nevanlinna theory . 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, andρ(f)=limsuprlogT(r,f)logr.

We now recall some previous results concerning nonhomogeneous linear differential equations of typef(k)+Ak1(z)f(k1)++A1(z)f+A0(z)f=H(z),where Aj(j=0,1,,k1) and A00, H0 are entire functions of finite-order, k2. In the case that the coefficients Aj(j=0,1,,k1) are polynomials, growth properties of solutions of (1.2) have been extensively studied, see, for example, . 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 equationf(k)+Ak1(z)f(k1)++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 whenmaxjd{ρ(Aj),ρ(H)}<ρ(Ad)12, Hellerstein et al.  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, . As for the special case of k=2, Wang and Laine studied equations of typef+A1(z)eazf+A0(z)ebzf=H(z),where A10,A00,H are entire functions of order less than one, and the complex numbers a,b satisfy ab0. They proved that every nontrivial solution of (1.5) is of infinite-order if ab, see . 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 function f(z)=ez satisfies the equationf(k)+f(k1)++f+ezf+Q(z)f=(k1+Q(z))ez+1,where Q(z) can be any entire function. Choosing Q(z)=1k 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 function f(z)=ez2 satisfies the equationf+ezf+ezf+e2zf=(8z3+12z+4z2ez+2ez+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 . We next prove two results of (1.2), which generalize Theorems 2 and 4 in  and Theorem 1.1 in .

Theorem 1.3.

Suppose that Aj(z)=hj(z)ePj(z)(j=0,,k1) where Pj(z)=ajnzn++aj0 are polynomials with degree n1, 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,,k1) are distinct complex numbers, then every solution of (1.2) is of infinite-order.

Theorem 1.4.

Suppose that Aj(z)=hj(z)ePj(z)(j=0,,k1) where Pj(z)=ajnzn++aj0 are polynomials with degree n1, 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 0s<lk1, θs,θl[0,2π), θsθl, hshl0, and for all js,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 when Aj(z)=hjeajz+gj where hj,gj(j=0,,k1) are polynomials, Chen considered the growth of solutions of (1.3) with some additional conditions imposed upon on aj, see . Our last results generalizes his result and [7, Theorem 1.3].

Theorem 1.5.

Suppose that Aj(z)=hj(z)ePj(z)+gj(z)(j=0,,k1) where Pj(z)=ajnzn++aj0 are polynomials with degree n1, 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 0s<lk1 such that for js,l, ajn=djeiφ(dj0) or ajn=djeiφ(dj0), and max{dj,js,l}=d<min{ds,dl}. If hshl0, then every transcendental solution of (1.2) is of infinite-order.

Remark 1.6.

Under the assumptions of Theorem 1.4, 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 F0, respectively, with Lemma 2.2, if F0.

Remark 1.7.

In the preceding three theorems, if ρ(f)=, then we also have λ(f)= for the exponent of convergence of the zero-sequence of f. Indeed, rewriting (1.2) in the form1f=1H(f(k)f+Ak1f(k1)f++A0),we havem(r,1f)m(r,1H)+j=0k1m(r,Aj)+j=0k1m(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 LemmasLemma 2.1 (see [<xref ref-type="bibr" rid="B15">10</xref>]).

Suppose that f1(z),f2(z),,fn(z)(n2) are meromorphic functions and g1(z),g2(z),,gn(z) are entire functions satisfying the following conditions:

j=1nfj(z)egj(z)0,

gj(z)gk(z) are not constants for 1j<kn,

for 1jn,1h<kn,T(r,fj)=o{T(r,eghgk)},(r,rE),

where E is a set with finite linear measure.

Then fj0(j=1,2,,n).

Lemma 2.2 (see [<xref ref-type="bibr" rid="B15">10</xref>]).

Suppose that f1(z),f2(z),,fn(z)(n2) are linearly independent meromorphic functions satisfying the following identity:j=1nfj1.Then for 1jn, one has 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),S(r)=o(max1kn{T(r,fk)}),(r,rE),E is a set with finite linear measure.

Lemma 2.3 (see [<xref ref-type="bibr" rid="B1">11</xref>, <xref ref-type="bibr" rid="B12">12</xref>]).

Suppose that P(z)=(α+iβ)zn+ (α,β are real numbers, |α|+|β|0) is a polynomial with degree n1, 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)(H1H2), there is R>0 such that for |z|=r>R, one has

if δ(P,θ)>0, thenexp{(1ε)δ(P,θ)rn}<|g(reiθ)|<exp{(1+ε)δ(P,θ)rn};

if δ(P,θ)<0, thenexp{(1+ε)δ(P,θ)rn}<|g(reiθ)|<exp{(1ε)δ(P,θ)rn},

where H2={θ[0,2π);δ(P,θ)=0}.

Lemma 2.4 (see [<xref ref-type="bibr" rid="B4">13</xref>]).

Let f(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,0j<k, one has |f(k)(z)f(j)(z)||z|(kj)(ρ1+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,0j<k, the inequality (2.7) holds.

Lemma 2.5.

Let f(z) be an entire function and suppose thatG(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|f(j)(zn)f(k)(zn)|1(kj)!(1+o(1))rnkj,j=0,,k1,as n.

Proof.

The first assertion is trivial. DenotingM(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). SinceG(zn)as n, we immediately see that|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.

Let f(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 that σ<ρ. Since E has linear measure zero, we may choose θj[0,2π)E such that 0θ1<θ2<<θn+1=2π, andmax{θj+1θj,1jn}πρ+1.We first treat the sectorH1:={zθ1argzθ2},definingϕ(z)=f(z)exp{beiθ0zσ},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 ρπ/(θ2θ1)1. Therefore,0>arg(eiθ0zσ)=arg(eiθ0rσeiσθ1)=σ(θ1θ2)2π2+(θ2θ1)2on the ray argz=θ1, and, respectively,0<arg(eiθ0zσ)=arg(eiθ0rσeiσθ2)=σ(θ2θ1)2π2(θ2θ1)2on the ray argz=θ2. Hence, we may now fix b>0 so thatbcos(π2(θ2θ1)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|f(z)||exp{beiθ0zσ}|exp{brσ}on H1. Repeating the same reasoning for all the sectors Hj={zθjargzθj+1} where θj are determined in (2.13), the assertion immediately follows.

3. Proof of Theorem <xref ref-type="statement" rid="thm1.1">1.3</xref>

Suppose, contrary to the assertion, that f is a solution of (1.2) with ρ(f)=ρ<, then nρ. Indeed, if f(k)=H, we may apply Lemma 2.1 to conclude that hsf(s)0 for some s, 0sk1 such that hs0. 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 nρ since the exponential functions ePj(j=0,1,,k1) are linearly independent.

By Lemma 2.3, there is a set E[0,2π) of linear measure such that whenever θ[0,2π)E, then δ(Pj,θ)0 for all 0jk1 and δ(PjPi,θ)0 for all i,j with 0i<jk1. 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|f(j)(z)f(i)(z)||z|kρ,0i<jk.Since ajn are distinct complex numbers, then for any fixed θ[0,2π)E, there exists exactly one s{0,,k1} such thatδ(Ps,θ)=δ:=max{δ(Pj,θ)j=0,,k1}.Denoting δ1=max{δ(Pj,θ)js}, then δ1<δ and δ0. We now discuss two cases separately.

Case 1.

Assume first that δ>0. By Lemma 2.3, for any given ε with 0<3ε<min{(δδ1)/δ,nρ(H)}, we have|As(reiθ)|exp{(1ε)δrn},|Aj(reiθ)|exp{(1+ε)δ1rn},for js, provided that r is sufficiently large. We now proceed to show thatlog+|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 thatlog+|f(s)(zm)|rmρ(H)+ε,|f(j)(zm)f(s)(zm)|(1+o(1))rmsj,(j=0,,s1).From (3.5) and the definition of order, it is easy to see that|H(zm)f(s)(zm)|0,for m is large enough. From (1.2), we obtain|As(zm)||f(k)(zm)f(s)(zm)|++|As+1(zm)||f(s+1)(zm)f(s)(zm)|+|As1(zm)||f(s1)(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 thatexp{(1ε1)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|f(z)|(1+o(1))rs|f(s)(z)|(1+o(1))Mrserρ(H)+εMerρ(H)+2εon the ray argz=θ.

Case 2.

Suppose now that δ<0. From (1.2), we get1=Ak1f(k1)f(k)++Ajf(j)f(k)++A0ff(k)Hf(k).Again by Lemma 2.3, for any given ε with 0<3ε<min{1,nρ(H)}, we have|Aj(reiθ)|exp{(1ε)δrn},(j=0,1,,k1),for r sufficiently large. As in Case 1, we prove thatlog+|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|f(j)(zm)f(k)(zm)|rmkj(1+o(1)),(j=0,,k1),|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|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 Theorem <xref ref-type="statement" rid="thm1.2">1.4</xref>

Suppose that f is a transcendental solution of (1.2) with ρ(f)=ρ<.

If f(k)H and ρ<n, it follows from (1.2) thatf(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 θsθl and θs,θl[0,2π), we conclude that djuPl(z)djvPs(z) are polynomials of degree n. Indeed, if djualn=djvasn, we have0<djudjv|alnasn|=ei(θsθl)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)hlf(s)hs0. Since hshl0, f has to be a polynomial of degree less than s, then H0, a contradiction.

Therefore, we may proceed under the assumption that f(k)H. By Lemma 2.2, if f(k)H, then nρ since the exponential functions ePl,ePs,edjuPl(u=1,2,,p) and edjvPs(v=1,2,,q) are linearly independent.

Since θsθl, 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, andδ(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 that δ2>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 that δ2<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 Lemma 2.6, we obtain a contradiction, so ρ(f)=.

5. Proof of Theorem <xref ref-type="statement" rid="thm1.3">1.5</xref>

Contrary to the assertion, suppose that f is a transcendental solution of (1.2) of finite-order. If ρ<n, then it follows from (1.2) thatf(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, dju0(u=1,,p) are distinct, and djv0(v=1,,q) are also distinct. Similarly as in the proof of Theorem 1.4, we may assume that nρ. Since σ=max{ρ(gj)(j=0,,k1)}<n, we havemax{|gj(z)|(j=0,,k1),|H(z)|}exp{rσ+ε}for any ε with 0<3ε<nσ, 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<(dld)dl, we obtain for |z| sufficiently large that|Al(reiθ)|exp{(1ε)dlcos(φ+nθ)rn}.For all other coefficients Aj(js), considering the hypothesis of ajn, we have|Aj(reiθ)|exp{(1+ε)dcos(φ+nθ)rn},when r is large enough. It follows from (1.2) thatAl=f(k)f(s)++Al+1f(l+1)f(l)+Al1f(l1)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 thatlog+|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)=.

Acknowledgments

The authors wish to express their thanks to the referee for his valuable suggestions and comments. The authors have been partially supported by the National Natural Science Foundation of China (Grant No. 10871047) and the Academy of Finland (Grant no. 210245 and no. 124954).

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