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Infinite growth of solutions of second order complex differential equation

Guowei Zhang
Published Online: 2018-11-03 | DOI: https://doi.org/10.1515/math-2018-0103

Abstract

In this paper we study the growth of solutions of second order differential equation f′′ + A(z)f′ + B(z)f = 0. Under certain hypotheses, the non-trivial solution of this equation is of infinite order.

MSC 2010: 30D20; 30D35; 34M05

1 Introduction and main results

In this article, we assume the reader is familiar with standard notations and basic results of Nevanlinna theory in the complex plane ℂ see [1, 2]. Nevanlinna theory is an important tool to studying the complex differential equations, and there appears many results in this areas recent years. In this paper, the order of an entire function f is defined as

$ρ(f)=lim supr→+∞log+⁡T(r,f)log⁡r=lim supr→+∞log+log+⁡M(r,f)log⁡r,$

where log+ x = max{logx,0} and M(r, f) denotes the maximum modulus of f on the circle |z| = r.

Our main purpose is to consider the second order linear differential equation

$f″+A(z)f′+B(z)f=0,$(1)

where A(z) and B(z) are entire functions. It’s well known that all solutions of (1) are entire functions. If B(z) is transcendental and f1,f2 are two linearly independent solutions of this equation, then at least one of f1,f2 is of infinite order, see . What conditions on A(z) and B(z) can guarantee that every solution f≢0 of equation (1) is of infinite order? There has been many results on this subject. For example, we collect some results and give the following theorems:

Theorem 1.1

Let A(z) and B(z) be nonconstant entire functions, satisfying any one of the following additional hypotheses:

1. ρ(A) < ρ(B), see ;

2. A(z) is a polynomial and B(z) is transcendental, see ;

3. $\begin{array}{}\rho \left(B\right)<\rho \left(A\right)\le \frac{1}{2}\end{array}$, see ,

then every solution f(≢0) of equation (1) has infinite order.

Theorem 1.2

Let A(z)=ez and B(z) be nonconstant entire functions, satisfying any one of the following additional hypotheses:

1. B(z) is a nonconstant polynomial, see ;

2. B(z) is transcendental entire function with order ρ(B) ≠ 1, see ;

3. B(z) = h(z)edz, where h(z)(≢ 0) is an entire function with ρ(h) < 1, d(≠ 1) is a nonzero complex constant, see ,

then every solution f(≢ 0) of equation (1) has infinite order.

In the above theorem we can see that A(z) could not get the value zero which means zero is a deficient value of A(z). In general, if A(z) has a finite deficient value, we have the following collection theorem.

Theorem 1.3

Let A(z) be an finite order entire function with a finite deficient value and B(z) be a transcendental entire function, satisfying any one of the following additional hypotheses:

1. μ(B) < 1/2, see ;

2. T(r,B) ∼ logM(r, B) as r→∞ outside a set of finite logarithmic measure, see [10, Lemma 2.7],

then every solution f( 0) of equation (1) has infinite order.

In this paper, we continue to study the above in order to find conditions which A(z), B(z)should satisfy to ensure that nontrivial solution of (1) has infinite order. Similar as in Theorem 1.2, we also assume zero is a Picard exceptional value of A(z), and the first main result is as follows.

Theorem 1.4

Suppose A(z) = ep(z), where p(z) is a nonconstant polynomial and B(z) is a transcendental entire function with nonzero finite order. If

$1ρ(A)+1ρ(B)>2,$

then every solution f(≢ 0) of equation

$f″+ep(z)f′+B(z)f=0$(2)

is of infinite order.

Remark 1.1

It’s clear that f(z) = ez satisfies f′′ + ezf′−(ez + 1)f = 0, here ρ(A) = ρ(B) = 1. Therefore, the inequality in the above theorem is sharp.

Remark 1.2

Simple calculation shows that the infinite order function f(z) = exp{ez} satisfies f′′ + ezf′−(e2z + ez + 1)f = 0. But ρ(A) and ρ(B) do not satisfy the inequality. Thus, the converse problem of Theorem 1.4 is not true.

Remark 1.3

As in Theorem 1.3, if A(z) has a finite deficient value, does the conclusion still hold?

Remark 1.4

We adopt the method of Rossi  to prove this theorem. This method was also used by Cao  to affirm Brück conjecture provided the hyper-order of f(z) is equal to 1/2.

Remark 1.5

By the idea of reviewer, the conclusion of this theorem can also be obtained by a relevant result of [4, Theorem 7]Gundersen, but the proof in this paper is totally different from it.

In the following, we consider the case that B(z) is a nonconstant polynomial and p(z) is transcendental entire. Rewrite (1) as

$ep(z)=f″f′+B(z)ff′.$(3)

Observe the order of both sides, we can deduce that the solution has infinite order. For the remaining case that p(z) and B(z) are both nonconstant polynomials, we have the following result.

Theorem 1.5

Suppose A(z) = ep(z), where p(z) is a polynomial with degree n ≥ 2 and B(z) = Q(z) is also a nonconstant polynomial with degree m. If m + 2 > 2n and nm + 2, then every solution f(≢ 0) of equation

$f″+ep(z)f′+Q(z)f=0$(4)

is of infinite order.

Remark 1.6

The case when the degree of p(z) is one could be proved by a minor modification of the proof of Theorem 2 in , here the degree of Q(z) can be arbitrary positive integer.

2 Preliminary lemmas

Lemma 2.1. ()

Let f be a transcendental meromorphic function. Let α > 1 be a constant, and k, j be integers satisfying k > j ≥ 0. Then the following two statements hold:

1. There exists a set E1⊂(1, + ∞) which has finite logarithmic measure, and a constant K > 0, such that for all z satisfying |z| = rE1 ∪ [0,1], we have

$f(k)(z)f(j)(z)≤KT(αr,f)r(log⁡r)αlog⁡T(αr,f)k−j.$(5)

2. There exists a set E2 ⊂ [0, 2π) which has zero linear measure, such that if θ ∈ [0, 2π) ∖ E2, then there is a constant R = R(θ) > 0 such that (5) holds for all z satisfying argz = θ and |z| ≥ R.

The following Lemma is proved in  by using [14, Theorem III.68]. We need some notations to state it. Let D be a region in ℂ. For each $\begin{array}{}r\in {\mathbb{R}}^{+}\phantom{\rule{thinmathspace}{0ex}}\text{set}\phantom{\rule{thinmathspace}{0ex}}{\theta }_{D}^{\ast }\left(r\right)={\theta }^{\ast }\left(r\right)=+\mathrm{\infty }\end{array}$ if the entire circle |z| = r lies in D. Otherwise, let $\begin{array}{}{\theta }_{D}^{\ast }\left(r\right)={\theta }^{\ast }\left(r\right)\end{array}$ be the measure of all θ in [0, 2π) such that reD. As usual, we define the order ρ(u) of a function u subharmonic in the plane as

$ρ(u):=lim supR→+∞log⁡M(r,u)log⁡r,$

where M(r, u) is the maximum modulus of subharmonic function u on a circle of radius r.

Lemma 2.2()

Let u be a subharmonic function inand let D be an open component of {z : u(z) > 0}. Then

$ρ(u)≥lim supR→+∞πlog⁡R∫1RdttθD∗(t).$(6)

Furthermore, given ε > 0, define $\begin{array}{}F=\left\{r:{\theta }_{D}^{\ast }\le \epsilon \pi \right\}\end{array}$. Then

$lim supR→+∞1log⁡R∫F∩[1,R]dtt≤ερ(u).$(7)

Lemma 2.3 ([15, 11])

Let l1(t) > 0, l2(t) > 0(tt0) be two measurable functions on (0, + ∞) with l1(t) + l2(t) ≤ (2 + ε)π, where ε > 0. If G⊆(0, + ∞) is any measurable set and

$π∫Gdttl1(t)≤α∫Gdtt,α≥12,$(8)

then

$π∫Gdttl2(t)≥α(2+ε)α−1∫Gdtt.$(9)

Lemma 2.4.()

Let S be the strip

$z=x+iy,x≥x0,|y|≤4.$

Suppose that in S

$Q(z)=anzn+O(|z|n−2),$

where n is positive integer and an > 0. Then there exists a path Γ tending to infinity in Ssuch that all solutions of y′′ + Q(z)y = 0 tend to zero on Γ.

Lemma 2.5 ()

Suppose that A(z) is analytic in a sector containing the ray Γ : re and that as r→∞, A(re) = O(rn) for some n ≥ 0. Then all solutions of y′′ + A(z)y = 0 satisfy

$log+⁡|y(reiθ)|=O(r(n+2)/2)$

on Γ.

We recall the definition of an R-set; for reference, see . Set B(zn,rn) = {z:|zzn| < rn}. If $\begin{array}{}\sum _{n=1}^{\mathrm{\infty }}{r}_{n}<\mathrm{\infty }\end{array}$ and zn → ∞, then $\begin{array}{}{\cup }_{n=1}^{\mathrm{\infty }}B\left({z}_{n},{r}_{n}\right)\end{array}$ is called an R-set. Clearly, the set $\begin{array}{}\left\{|z|:z\in {\cup }_{n=1}^{\mathrm{\infty }}B\left({z}_{n},{r}_{n}\right)\right\}\end{array}$ is of finite linear measure. Moreover, by [1, Lemma 5.9], the set of angles θ for which the ray re meets infinitely many discs of a given R-set has linear measure zero.

Lemma 2.6 ([1, Proposition 5.12])

Let f be a meromorphic function of finite order. Then there exists N = N(f) > 0 such that

$f′(z)f(z)=O(rN)$(10)

holds outside of an R-set.

In the next Lemma, let p(z) = (α + )zn+ ⋯ + a0 be a polynomial with α, β real, and denote δ(p, θ) := αcosβsin .

Lemma 2.7 ([1, Lemma 5.14])

Let p(z) be a polynomial of degree n ≥ 1, and consider the exponent function A(z) := ep(z) on a ray re. Then we have

1. If δ(p, θ) > 0, there exists an r(θ) such that log|A(re)| is increasing on [r(θ), + ∞) and

$|A(reiθ)|≥exp12δ(p,θ)rn$(11)

holds here;

2. If δ(p, θ) < 0, there exists an r(θ) such that log|A(re)| is decreasing on [r(θ), + ∞) and

$|A(reiθ)|≤exp12δ(p,θ)rn$(12)

holds here.

Lemma 2.8. ([16, Theorem 7.3])

Phragmén-Lindelöf Theorem Let f(z) be an analytic function of z = re, regular in a region D between rays making an angle π/α at the origin and on the straight lines themselves. Suppose that |f(z)| ≤ M on the lines and as r → ∞, f(z) = O(erβ), where β < α uniformly. Then |f(z)| ≤ M throughout D.

Remark 2.1

If f(z) → a as z → ∞ along a straight line, f(z) → b as z → ∞ along another straight line, and f(z) is analytic and bounded in the angle between, then a = b and f(z) → a uniformly in the angle. The straight lines may be replaced by curves approaching ∞.

3 Proofs of theorems

Proofs of Theorem 1.4

We assume that ρ(f) = ρ < +∞, and would obtain the assertion by reduction to contradiction. By Lemma 2.1 and definition of the growth order, there exist constants K > 0, β > 1 and C = C(ε) (depending on ε) such that

$f(j)(z)f(z)≤KT(βr,f)log⁡r(log⁡r)βlog⁡T(βr,f)j≤rC,j=1,2$(13)

holds for all r > r0 = R(θ) and θJ(r), where J(r) is a set with zero linear measure. We may say m(J(r)) ≤ επ where ε( > 0) is given arbitrarily small. Fix ε > 0 and let N be an integer such that N > C = C(ε), and

$log⁡M(2,B)(14)

Since B(z) is transcendental there exists z0,|z0| > 2, such that log|B(z0)| > Nlog|z0|. Let D1 be the component of the set {z:log|B(z)|−Nlog|z| > 0} containing z0. Clearly D1 is open and since (14) holds, log|B(z)|−Nlog|z| is subharmonic in D1 and identically zero on ∂D1. Thus, if we define

$log⁡|B(z)|−Nlog⁡|z|,z∈D1,0,z∈C∖D1,$(15)

we have that u(z) is subharmonic in ℂ with

$ρ(u)≤ρ(B).$(16)

Let D2 be an unbounded component of the set {z : log|ep(z)| > 0}, such that if we define

$v(z)=log⁡|e−p(z)|,z∈D2,0,z∈C∖D2,$

then v(z) is subharmonic in ℂ with

$ρ(v)=ρ(A).$

Moreover, define D3:={re : θJ(r)}. For the above given ε, if (D1D2) ∖ D3contains an unbounded sequence {rneiθn}, then we get from the above discussions that

$rnN<|B(rneiθn)|≤f″(rneiθn))f(rneiθn)+ep(rneiθn)f′(rneiθn))f(rneiθn)≤2rnC,$(17)

and this clearly contradicts N > C for n large enough. Thus for arbitrary ε, we may assume that (D1D2) ∖ D3 is bounded, this implies that for rr1r0,

$Kr:={θ:reiθ∈D1∩D2}⊆J(r),$

Obviously, we have

$m(Kr)≤επ.$(18)

(We remark here that the proof of Theorem 1.4 would now follow easily from (6) and Lemma 2.3 if D1 and D2 were disjoint. As we shall see, (6), (13) and (18) imply that these sets are "essentially" disjoint. Define

$2π,ifθDj∗(t)=∞,θDj∗(t),otherwise,$(19)

for j = 1,2. Since B(z) is transcendental, it follows D1 and D2 are unbounded open sets. Then we get l1(t) > 0,l2(t) > 0 for t sufficiently large, and

$l1(t)+l2(t)≤2π+επ.$(20)

Set

$α:=lim supR→∞πlog⁡R∫1Rdttl1(t).$(21)

By (21) and the fact l1(t) ≤ 2π, we have

$α≥12log⁡R∫1Rdtt=12.$

Thus

$π∫1Rdttl1(t)≤αlog⁡R=α∫1Rdtt.$

Therefor, from Lemma 2.3 we obtain

$π∫1Rdttl2(t)≥α(2+ε)α−1∫1Rdtt=α(2+ε)α−1log⁡R,$

and thus,

$lim supR→+∞πlog⁡R∫1Rdttl2(t)≥α(2+ε)α−1.$(22)

Define the set

$Bj:={r:θDj∗(r)=+∞}$(23)

for j = 1, 2. If rB1 and rr1, then $\begin{array}{}{\theta }_{{D}_{2}}^{\ast }\left(r\right)\le \epsilon \pi \end{array}$ by (20). Thus $\begin{array}{}{B}_{1}\subseteq \left\{r:{\theta }_{{D}_{2}}^{\ast }\left(r\right)\le \epsilon \pi \right\}\end{array}$. By Lemma 2.2 we have

$lim supR→∞1log⁡R∫B1∩[1,R]dtt≤ερ(ep(z))=ερ(e−p(z)).$(24)

The last equality follows by the first Nevanlinna theorem. Let $\begin{array}{}\stackrel{~}{{B}_{j}}={R}^{+}\setminus {B}_{j},j=1,2\end{array}$. Then (6), (21) and (24) give

$ρ(u)≥lim supR→∞πlog⁡R∫1RdttθD1∗(t)=lim supR→∞πlog⁡R∫B1~∩[1,R]dttθD1∗(t)=lim supR→∞1log⁡Rπ∫1Rdttl1(t)−12∫B1∩[1,R]dtt≥α−ερ(ep(z))2,$(25)

which together with (1) and A(z) = ep(z) shows

$ρ(B)≥α−ερ(A)2.$(26)

Applying the similar arguments as above to B2, if rB2 and rr1, then $\begin{array}{}{\theta }_{{D}_{1}\left(r\right)}^{\ast }\le \epsilon \pi \end{array}$. Thus $\begin{array}{}{B}_{2}\subseteq \left\{r:{\theta }_{{D}_{1}\left(r\right)}^{\ast }\le \epsilon \pi \right\}\end{array}$. Then we obtain also from Lemma 2.2 that

$lim supR→∞1log⁡R∫B2∩[1,R]dtt≤ερ(u).$(27)

Combining (6), (22) with (27) we obtain

$ρ(A)=ρ(e−p(z))≥lim supR→∞πlog⁡R∫1RdttθD2∗(t)=lim supR→∞πlog⁡R∫B2~∩[1,R]dttθD2∗(t)=lim supR→∞1log⁡Rπ∫1Rdttl2(t)−12∫B2∩[1,R]dtt≥α(2+ε)α−1−ερ(u)2.$(28)

Since $\begin{array}{}\frac{\alpha }{\left(2+\epsilon \right)\alpha -1}\end{array}$ is a monotone decreasing function of α, inequalities (16), (26) and (28) give

$ρ(A)≥ρ(B)+ερ(A)2(2+ε)(ρ(B)+ερ(A)2)−1−ερ(B)2.$(29)

Note that ε is positive arbitrary small and ρ(A), ρ(B) are finite, we obtain

$ρ(A)≥ρ(B)2ρ(B)−1.$

That is,

$1ρ(A)+1ρ(B)≤2,$

which contradicts the assumption. Thus, every solution f(≢ 0) of equation (2) is of infinite order. □

Proofs of Theorem 1.5

We assume that (4) has a solution f(z) with finite order. Set

$f=yexp−12∫0zep(z)dz,$(30)

equation (4) can be transformed into

$y″+Q(z)−14e2p(z)−12ep(z)p′(z)y=0.$(31)

By a translation we may assume that

$Q(z)=amzm+am−2zm−2+⋯,m>2.$(32)

We define the critical ray for Q(z) as those ray rej for which

$θj=−arg⁡am+2jπm+2,$(33)

where j = 0, 1, 2⋯, m + 1, and note that the substitution z = xej transforms equation (31) into

$d2ydx2+(Q1(x)+P1(x))y=0,$(34)

where

$Q1(x)=α1xm+O(xm−2),α1>0$

and

$P1(x)=e2iθj−14e2p(xeiθj)−12ep(xeiθj)p′(xeiθj).$

For the polynomial p(z) with degree n, set p(z) = (α + )zn + pn−1(z) with α, β real, and denote δ(p, θ) := αcos βsin . The rays

$arg⁡z=θk=arctan⁡αβ+kπn,k=0,1,2,⋯,2n−1$

satisfying δ(p, θk) = 0 can split the complex domain into 2n equal angular domains. Without loss of generality, denote these angle domains as

$Ω+:=z=reiθ:0(35)

i = 0,1, ⋯ , n−1, where δ(p, θ) > 0 on Ω+ and δ(p, θ) < 0 on Ω. By Lemma 2.7, we obtain

$|P1(x)|≤|e2p(xeiθj)|+|ep(xeiθj)||p′(xeiθj)|≤exp⁡{δ(p,θ)xn}+exp12δ(p,θ)xnO(xn−1)→0$(36)

for xej ∈ Ω as x → ∞, then by Lemma 2.4 and (34), for any critical line argz = θj lying in Ω there exists a path Γθj tending to infinity, such that argzθj on Γθj while y(z) → 0 there. Moreover, by

$exp−12∫0zep(z)dz≤exp12∫0zep(z)dz≤exp12rexp12δ(p,θ)rn→1$(37)

for z ∈ Ω as r → ∞, together with (2) we have f(z) → 0 along Γθjtending to infinity.

Setting V = f′/f, equation (4) can be written as

$V′+V2+ep(z)V+Q(z)=0.$(38)

By Lemma 2.6, we have

$|V′|+|V|2=O(|z|N)$(39)

outside an R-set U, where N is a positive constant. Moreover, if z = re ∈ Ω+ is such that the ray arg z = ϕ meets only finitely many discs of U we see that V = o(|z|−2) as z tends to infinity on this ray and hence f tends to a finite, nonzero limit. Applying this reasoning to a set of ϕ outside a set of zero measure we deduce by the Phragmén-Lindelöf principle that without loss of generality, for any small enough given positive ε,

$f(reiθ)→1$(40)

as r → ∞ with

$z=reiθ∈Ωε+:=z=reiθ:0(41)

For any z = re ∈ Ω, we have that δ(p, θ) < 0, and by Lemma 2.7 we have

$Q(z)−14e2p(z)−12ep(z)p′(z)≤|Q(z)|+|e2p(z)|+|ep(z)||p′(z)|≤O(rm)+exp⁡{δ(p,θ)rn}+exp12δ(p,θ)rnO(rn−1)≤O(rm)$(42)

for sufficiently large r. Applying Lemma 2.5 to (31) and together with (42), y(z) satisfies

$log+⁡|y(reiθ)|=O(rm+22)$(43)

as r → ∞ for any z = re ∈ Ω. From (30) and (37), we have

$log+⁡|f(reiθ)|=O(rm+22)$(44)

as r → ∞ for any z = re ∈ Ω.

On the rays arg z = θk such that δ(p, θk) = 0, we have |ep(z)| = |epn−1(z)|. Consider the two cases δ(pn−1, θk) > 0 or δ(pn−1, θk) < 0, by the same method above, we get f(z) → 1 or $\begin{array}{}{\mathrm{log}}^{+}|f\left(z\right)|=O\left({r}^{\frac{m+2}{2}}\right)\end{array}$, respectively, on the ray arg z = θk. If δ(pn−1, θk) = 0 also, repeating these arguments again. Finally, we deduce that either f(z) → 1 or $\begin{array}{}{\mathrm{log}}^{+}|f\left(z\right)|=O\left({r}^{\frac{m+2}{2}}\right)\end{array}$ on the rays arg z = θk, k = 0,1, ⋯ ,2n − 1. Thus, (30), (40), (44) and the fact ε is arbitrary imply that, by the Phragmén-Lindelöf principle,

$σ(f)≤m+22.$(45)

We claim that $\begin{array}{}\frac{\left(2i+1\right)\pi }{n}\left(i=0,1,\cdots ,n-1\right)\end{array}$ are critical rays for Q(z). For otherwise there exists a critical θj for Q(z) in

$(2i+1)πn<θj<(2i+1)πn+2πm+2(i=0,1,⋯,n−1)$

because m + 2 > 2n. This implies the existence of an unbounded domain of angular measure at most $\begin{array}{}\frac{2\pi }{m+2}+\epsilon \end{array}$, bounded by a path on which f(z) → 0 and a ray on which f(z) → 1. By the remark following Lemma 2.7 implies that $\begin{array}{}\sigma \left(f\right)>\frac{m+2}{2}\end{array}$, contradicting (45). Then there exists a positive integer k satisfying $\begin{array}{}\frac{2\pi }{n}=k\frac{2\pi }{m+2}\end{array}$, that is, m + 2 = kn, which contradicts nm + 2. Thus, we complete the proof. □

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Accepted: 2018-09-14

Published Online: 2018-11-03

FundingThis work was supported by the key scientific research project for higher education institutions of Henan Province, China (no. 18A110002) and training program for young backbone teachers of colleges and universities in Henan Province, China (no. 2017GGJS126).

AbbreviationsNot applicable.

Availability of data and materialsNot applicable.

Ethics approval and consent to participateNot applicable.

Competing interestsThe author declares that he has no competing interests.

Author’s informationSchool of Mathematics and Statistics, Anyang Normal University, Anyang, China.

Author’s contributionsThe author read and approved the final manuscript.

Citation Information: Open Mathematics, Volume 16, Issue 1, Pages 1233–1242, ISSN (Online) 2391-5455,

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