Infinite growth of solutions of second order complex differential equation

Guowei Zhang

doi:10.1515/math-2018-0103

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Infinite growth of solutions of second order complex differential equation

Guowei Zhang

Published/Copyright: November 3, 2018

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$Open Mathematics$

From the journal Open Mathematics Volume 16 Issue 1

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.

Keywords: Entire function; order; Complex differential equation

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 [3]. 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:

ρ(A) < ρ(B), see [4];
A(z) is a polynomial and B(z) is transcendental, see [4];
ρ(B)<ρ(A)≤12, see [5],
then every solution f(≢0) of equation (1) has infinite order.

Theorem 1.2

Let A(z)=e^−zand B(z) be nonconstant entire functions, satisfying any one of the following additional hypotheses:

B(z) is a nonconstant polynomial, see [6];
B(z) is transcendental entire function with order ρ(B) ≠ 1, see [7];
B(z) = h(z)e^−dz, where h(z)(≢ 0) is an entire function with ρ(h) < 1, d(≠ 1) is a nonzero complex constant, see [8],
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:

μ(B) < 1/2, see [9];
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) = e^p(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) = e^z satisfies f′′ + e^−zf′−(e^−z + 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{e^z} satisfies f′′ + e^−zf′−(e^2z + e^z + 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 [11] to prove this theorem. This method was also used by Cao [121] 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) = e^p(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 n ∤ m + 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 [6], here the degree of Q(z) can be arbitrary positive integer.

2 Preliminary lemmas

Lemma 2.1. ([13])

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:

There exists a set E₁⊂(1, + ∞) which has finite logarithmic measure, and a constant K > 0, such that for all z satisfying |z| = r ∉ E₁ ∪ [0,1], we have
f(k)(z)f(j)(z)≤KT(αr,f)r(log⁡r)αlog⁡T(αr,f)k−j.(5)
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 [11] by using [14, Theorem III.68]. We need some notations to state it. Let D be a region in ℂ. For each r∈R+setθD∗(r)=θ∗(r)=+∞ if the entire circle |z| = r lies in D. Otherwise, let θD∗(r)=θ∗(r) be the measure of all θ in [0, 2π) such that re^iθ ∈ D. 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([11])

Let u be a subharmonic function in ℂ and 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, defineF={r:θD∗≤επ}. Then

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

Lemma 2.3 ([15, 11])

Let l₁(t) > 0, l₂(t) > 0(t ≥ t₀) 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.([6])

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 a_n > 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 ([6])

Suppose that A(z) is analytic in a sector containing the ray Γ : re^iθand that as r→∞, A(re^iθ) = O(r_n) 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 [1]. Set B(z_n,r_n) = {z:|z − z_n| < r_n}. If ∑n=1∞rn<∞ and z_n → ∞, then ∪n=1∞B(zn,rn) is called an R-set. Clearly, the set {|z|:z∈∪n=1∞B(zn,rn)} is of finite linear measure. Moreover, by [1, Lemma 5.9], the set of angles θ for which the ray re^iθ 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 + iβ)zⁿ+ ⋯ + a₀ be a polynomial with α, β real, and denote δ(p, θ) := αcosnθ − βsin nθ.

Lemma 2.7 ([1, Lemma 5.14])

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

If δ(p, θ) > 0, there exists an r(θ) such that log|A(re^iθ)| is increasing on [r(θ), + ∞) and
|A(reiθ)|≥exp12δ(p,θ)rn(11)
holds here;
If δ(p, θ) < 0, there exists an r(θ) such that log|A(re^iθ)| 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^iθ, 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(e^{r^β}), 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 θ ∉ J(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)<Nlog⁡2.(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|e^−p(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^iθ : θ ∈ J(r)}. For the above given ε, if (D₁∩D₂) ∖ D₃contains an unbounded sequence {r_ne^iθn}, 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 (D₁ ∩ D₂) ∖ D₃ is bounded, this implies that for r ≥ r₁ ≥ r₀,

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 r ∈ B₁ and r ≥ r₁, then θD2∗(r)≤επ by (20). Thus B1⊆{r:θD2∗(r)≤επ}. 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 Bj~=R+∖Bj,j=1,2. 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) = e^p(z) shows

ρ(B)≥α−ερ(A)2.(26)

Applying the similar arguments as above to B₂, if r ∈ B₂ and r ≥ r₁, then θD1(r)∗≤επ. Thus B2⊆{r:θD1(r)∗≤επ}. 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 α(2+ε)α−1 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 re^iθ_j for which

θj=−arg⁡am+2jπm+2,(33)

where j = 0, 1, 2⋯, m + 1, and note that the substitution z = xe^iθ_j 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 + iβ)zⁿ + p_n−1(z) with α, β real, and denote δ(p, θ) := αcos nθ − βsin nθ. The rays

arg⁡z=θk=arctan⁡αβ+kπn,k=0,1,2,⋯,2n−1

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

Ω+:=z=reiθ:0<r<+∞,2iπn<θ<(2i+1)πn,Ω−:=z=reiθ:0<r<+∞,(2i+1)πn<θ<2(i+1)πn,(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 xe^iθ_j ∈ Ω₋ as x → ∞, then by Lemma 2.4 and (34), for any critical line argz = θ_j lying in Ω⁻ there exists a path Γ_{θ_j} tending to infinity, such that argz → θ_j on Γ_{θ_j} 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 Γ_θjtending 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^iϕ ∈ 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<r<+∞,2iπn+ε<θ<(2i+1)πn−ε.(41)

For any z = re^iθ ∈ 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^iθ ∈ Ω⁻. From (30) and (37), we have

log+⁡|f(reiθ)|=O(rm+22)(44)

as r → ∞ for any z = re^iθ ∈ Ω⁻.

On the rays arg z = θ_k such that δ(p, θ_k) = 0, we have |e^p(z)| = |e^p_n−1(z)|. Consider the two cases δ(p_n−1, θ_k) > 0 or δ(p_n−1, θ_k) < 0, by the same method above, we get f(z) → 1 or log+⁡|f(z)|=O(rm+22), respectively, on the ray arg z = θ_k. If δ(p_n−1, θ_k) = 0 also, repeating these arguments again. Finally, we deduce that either f(z) → 1 or log+⁡|f(z)|=O(rm+22) on the rays arg z = θ_k, 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 (2i+1)πn(i=0,1,⋯,n−1) are critical rays for Q(z). For otherwise there exists a critical θ_j 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 2πm+2+ε, 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 σ(f)>m+22, contradicting (45). Then there exists a positive integer k satisfying 2πn=k2πm+2, that is, m + 2 = kn, which contradicts n ∤ m + 2. Thus, we complete the proof. □

Funding
This 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).
Abbreviations
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Competing interests
The author declares that he has no competing interests.
Author’s information
School of Mathematics and Statistics, Anyang Normal University, Anyang, China.
Author’s contributions
The author read and approved the final manuscript.

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Received: 2018-03-28

Accepted: 2018-09-14

Published Online: 2018-11-03

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Keywords for this article

Entire function; order; Complex differential equation

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