# On Subnormal Solutions of Periodic Differential Equations

Authors: Zong-Xuan Chen; Kwang Ho Shon
Journal: Abstract and Applied Analysis
 (2010)

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## Abstract

We investigate the existence and the form of subnormal solutions of higher-order linear periodic differential equations, and precisely estimate the growth of all solutions.


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## Body

## 1. Introduction and Results

In this paper we use standard notations from the value distribution theory (see [1–3]). In addition, we denote the order of growth of f(z) by σ(f) and also use the notation σ2(f) to denote the hyperorder of f(z), which is defined as (1.1)σ2(f)=lim⁡¯r→∞log⁡log⁡T(r,f)log⁡r.Consider the second-order homogeneous linear periodic differential equation(1.2)f′′+P(ez)f′+Q(ez)f=0,
where P(z) and Q(z) are polynomials in z, but both are not constants. It is well known that every solution f of (1.2) is an entire function.Supposef(≢0) is a solution of (1.2), and if f satisfies the condition (1.3)lim⁡¯r→∞log⁡T(r,f)r=0,
then we say that f is a nontrivial subnormal solution of (1.2). For convenience, we also say that f≡0 is a subnormal solution of (1.2) (see [4, 5]).It clearly follows that(i)
if
(1.4)0<lim⁡¯r→∞log⁡T(r,f)r<∞,
then σ2(f)=1;(ii)
ifσ2(f)<1, then (1.3) holds.Wittich [5] investigated the subnormal solution of (1.2), which gives the form of all subnormal solutions of (1.2) in the following theorem.Theorem A (see [5]).
Iff≢0 is a subnormal solution of (1.2), then  f must have the form
(1.5)f(z)=ecz(h0+h1ez+⋯+hmemz),
where m≥0 is an integer and c,h0,…,hm are constants with h0≠0 and hm≠0.Gundersen and Steinbart [4] refined Theorem A and got the following theorem.Theorem B (see [4]).
Under the assumption of TheoremA, the following statements hold.
(i)
IfdegP>degQ and Q≢0, then any subnormal solution f(≢0) of (1.2) must have the form
(1.6)f(z)=∑k=0mhke-kz,
where m≥1 is an integer and h0,h1,…,hm are constants with h0≠0 and hm≠0.(ii)
If Q≡0 and degP≥1, then any subnormal solution of (1.2) must be a constant.(iii)
IfdegP<degQ, then the only subnormal solution of (1.2) is f≡0.In [6], Chen and Shon proved that supposing that (1.7)P(ez)=an(z)enz+⋯+a1(z)ez,Q(ez)=bs(z)esz+⋯+b1(z)ez,
where an(z),…,a1(z),bs(z),…,b1(z) are polynomials satisfying an(z)bs(z)≢0, if n≠s, then every solution f(≢0) of (1.2) satisfies σ2(f)=1.In [6], the condition “all constant terms of P(ez) and Q(ez) are equal to zero” plays an important role in the growth of solutions of (1.2). This makes us consider that the condition may be applied to higher-order differential equations.Gundersen and Steinbart [4] consider a subnormal solution of higher-order linear nonhomogeneous differential equation (1.8)f(k)+Pk-1(ez)f(k-1)+⋯+P0(ez)f=Q1(ez)+Q2(e-z),
where Qd(z)(d=1,2), Pj(z)(j=0,…,k-1) are polynomials in z and obtain the following theorem.Theorem C.
Suppose that, in (1.8), one has k≥2 and
(1.9)deg⁡P0>degPj
for all 1≤j≤k-1. Then any subnormal solution f of (1.8) must have the form
(1.10)f(z)=S1(ez)+S2(e-z),
where S1(z) and S2(z) are polynomials in z.From the proof of TheoremC, we see that the condition (1.9) of Theorem C guarantees that the corresponding homogeneous differential equation of (1.8) (1.11)f(k)+Pk-1(ez)f(k-1)+⋯+P0(ez)f=0
has no nontrivial subnormal solution.Thus, a natural question is whether or not (1.11) has a nontrivial subnormal solution if the condition (1.9) is replaced by the condition “there exists some s satisfying degPs>degPj(j≠s)”.Examples1.1 and 1.2 show that if degPs>degPj(j≠s), (1.11) may have a nontrivial subnormal solution.Example 1.1.
The equation(1.12)f′′′-(14e3z+2e2z)f′′+12ezf′+(ez+8)f=0
has a subnormal solution f=e-2z+1.Example 1.2.
The equation(1.13)f′′′+(e3z+4)f′′+(e4z+ez)f′+(3ez-9)f=0
has a subnormal solution f=e-3z+1.Thus, a natural question is what conditions will guarantee that (1.11) has no nontrivial subnormal solution under the condition degPs>degPj(j≠s).In Theorem1.3, we answer this question. We conclude that if all constant terms of Pj are equal to zero under the conditions degPs>degPj(j≠s) and P0≢0, then (1.11) has no nontrivial subnormal solution, and we also prove that all solutions of (1.11) satisfy σ2(f)=1.Examples1.1 and 1.2 show that the condition “all constant terms of Pj are equal to zero” cannot be deleted in Theorem 1.3.In this paper, we firstly investigate the existence of subnormal solutions. It is an important problem in theory of periodic differential equations.Theorem1.4 generalizes the result of Theorem C, shows that (1.8) has at most one nontrivial subnormal solution, gives the form of subnormal solution of (1.8), and proves that all other solutions f of (1.8) satisfy σ2(f)=1.Theorem1.6 refines Theorem C.Our method for obtaining the proof is totally different from the method applied in [4, 5].Theorem 1.3.
LetPj(z)(j=0,…,k-1) be polynomials in z such that all constant terms of Pj are equal to zero and degPj=mj, that is,
(1.14)Pj(ez)=ajmjemjz+aj(mj-1)e(mj-1)z+⋯+aj1ez,
where ajmj,aj(mj-1),…,aj1 are constants and ajmj≠0;mj≥1 are integers. Suppose that there exists ms(s∈{0,…,k-1}) satisfying
(1.15)ms>max{mj:j=0,…,s-1,s+1,…,k-1}=m.
Then, one has the following properties.(i)
IfP0≢0, then (1.11) has no nontrivial subnormal solution and every solution of (1.11) is of hyper order σ2(f)=1.(ii)
IfP0≡⋯≡Pd-1≡0 and Pd≢0(d<s), then any polynomials with degree ≤d-1 are subnormal solutions of (1.11) and all other solutions f of (1.11) satisfy σ2(f)=1.Considering proof of theorems, if the setez=ζ, then (1.8) (or (1.11)) becomes an equation with rational coefficients, but the equation with rational coefficients may have nonmeromorphic solution. For example, the equation (1.16)zf′′+zf′-2f=0
has a solution f=exp{1/z}. This shows that we cannot use the transformation ez=ζ to prove that every solution of (1.11) is of σ2(f)=1.Theorem 1.4.
LetPj(ez)(j=0,…,k-1)  satisfy (1.14) and (1.15). Let Q1(z) and Q2(z) be polynomials in z. If P0≢0,  then(i)
(1.8) possesses at most one nontrivial subnormal solution f0, and f0 is of the form (1.10), where S1(z) and S2(z) are polynomials in z;(ii)
all other solutionsf of (1.8) satisfy σ2(f)=1 except the possible subnormal solution in (i).Example1.5 shows the existence of subnormal solution in Theorem 1.4.Example 1.5.
The equation(1.17)f′′′-2e2zf′′-ezf′+ezf=-2e3z-e-z+2
has a subnormal solution f=ez+e-z+1.Theorem 1.6.
Under the assumption of Theorem C, the following statements hold.(i)
Equation (1.11) has no nontrivial subnormal solution, and all solutions of (1.11) satisfy σ2(f)=1.(ii)
Equation (1.8) has at most one nontrivial subnormal solution f0, and f0 is of the form (1.10); all other solutions f of (1.8) satisfy σ2(f)=1.


## 2. Lemmas for the Proofs of Theorems

Lemma 2.1 (see [7, 8]).
Letfj(z)(j=1,…,n)(n≥2) be meromorphic functions, and let  gj(z)(j=1,…,n) be entire functions and satisfy
(i)
∑j=1nfj(z)egj(z)≡0;(ii)
when1≤j<k≤n, then gj(z)-gk(z) is not a constant;(iii)
when1≤j≤n, 1≤h<k≤n, then(2.1)T(r,fj)=o{T(r,egh-gk)}(r→∞,r∉E),
where E⊂(1,∞) is of finite linear measure or logarithmic measure.
Thenfj(z)≡0(j=1,…,n).Lemma 2.2.
LetPj,  mj,  ms,  and m satisfy the hypotheses of Theorem 1.3.(i)
If P0≢0, then (1.11) has no nonzero polynomial solution.(ii)
IfP0≡⋯≡Pd-1≡0 and Pd≢0(d<s), then all polynomials with degree ≤d-1 are solutions of (1.11), and any polynomial with degree ≥d is not solution of (1.11).Proof.
(i) Firstly, byP0≢0, we see that all nonzero constants cannot be a solution of (1.11). Now suppose that f0=bnzn+⋯+b1z+b0(n≥1,bn,…,b0 are constants, bn≠0) is a solution of (1.11). If n≥s, then f0(s)≢0. Substituting f0 into (1.11) and taking z=r, we conclude that
(2.2)|asms|emsr|bn|n(n-1)⋯(n-s+1)rn-s(1-o(1))≤|-Ps(ez)f0(s)(z)|≤|f0(k)(z)|+|Pk-1(ez)f0(k-1)(z)|+⋯+|Ps+1(ez)f0(s+1)(z)|+|Ps-1(ez)f0(s-1)(z)|+⋯+|P0(ez)f0(z)|≤Mrnemr(1+o(1)),
where M(>0) is some constant. Since ms>m, we see that (2.2) is a contradiction. If n<s, then
(2.3)Pn(ez)f0(n)(z)+⋯+P0(ez)f0(z)=0.
Set max{degPj:j=0,…,n}=h. If degPj=mj<h, then we can rewrite
(2.4)Pj(ez)=ajhehz+⋯+aj(mj+1)e(mj+1)z+ajmjemjz+⋯+aj1ez(j=0,…,n),
where ajh=⋯=aj(mj+1)=0. Thus, we conclude by (2.3) and (2.4) that
(2.5)(anhf0(n)+a(n-1)hf0(n-1)+⋯+a0hf0)ehz+⋯+(anjf0(n)+a(n-1)jf0(n-1)+⋯+a0jf0)ejz+⋯+(an1f0(n)+a(n-1)1f0(n-1)+⋯+a01f0)ez=0.
Set
(2.6)Qj(z)=anjf0(n)+a(n-1)jf0(n-1)+⋯+a0jf0(j=1,…,h).
Since f0 is the polynomial, we see that
(2.7)m(r,Qj)=o{m(r,e(α-β)z}(1≤β<α≤h).
By Lemma 2.1, (2.5)–(2.7), we conclude that
(2.8)Q1(z)≡Q2(z)≡⋯≡Qh(z)≡0.
Since degf0>degf0′>⋯>degf0(n), by (2.6) and (2.8), we see that
(2.9)a00=a01=⋯=a0h=0.
Since h≥m0=degP0, we have P0≡0. This contradicts our assumption that P0≢0.
(ii) SinceP0≡⋯≡Pd-1≡0 and Pd≢0(d<s), clearly all polynomials with degree ≤d-1 are solutions of (1.11). By Pd≢0 and (i), we see that f(d) cannot be a nonzero polynomial, and hence f cannot be a polynomial with degree (≥d).Lemma 2.3 (see [9]).
Letf be a transcendental meromorphic function with σ(f)=σ<∞, and let H={(k1,j1),(k2,j2),…,(kq,jq)} be a finite set of distinct pairs of integers that satisfy ki>ji≥0, for  i=1,…,q, and let ε>0 be a given constant. Then, there exists a set E⊂[-π/2,3π/2) that has linear measure zero, such that if ψ∈[-π/2,3π/2)∖E, then there is a constant R0=R0(ψ)>1 such that for all z satisfying argz=ψ and |z|≥R0 and for all (k,j)∈H, one has
(2.10)|f(k)(z)f(j)(z)|≤|z|(k-j)(σ-1+ε).Lemma 2.4 (see [10]).
Letf(z) be an entire function and suppose that |f(k)(z)| is unbounded on some ray argz=θ. Then there exists an infinite sequence of points zn=rneiθ(n=1,2,…),  where rn→∞,  such that f(k)(zn)→∞ and
(2.11)|f(j)(zn)f(k)(zn)|≤|zn|k-j(1+o(1))(j=0,…,k-1).Lemma 2.5 (see [11]).
Letf(z) be an entire function with σ(f)=σ<∞. Let there exists a set E⊂[0,2π) that has linear measure zero, such that for any ray argz=θ0∈[0,2π)∖E,|f(reiθ0)|≤Mrk(M=M(θ0)>0 is a constant, and k(>0) is a constant independent of θ0). Then f(z) is a polynomial with degf≤k.Lemma2.6 can be obtained from [12, Theorem 4] or [2, Theorem 7.3].Lemma 2.6.
LetA0,…,Ak-1 be entire functions of finite order. If f(z)  is a solution of equation
(2.12)f(k)+Ak-1f(k-1)+⋯+A0f=0,
then σ2(f)≤max{σ(Aj):j=0,…,k-1}.Lemma 2.7 (see [13]).
Letg(z) be an entire function of infinite order with the hyperorder σ2(g)=σ,  and let ν(r) be the central index of g.  Then
(2.13)lim⁡¯r→∞log⁡log⁡ν(r)log⁡r=σ2(g)=σ.Lemma 2.8 (see [6]).
Letf(z) be an entire function of infinite order with σ2(f)=α(0≤α<∞), and let a set E⊂[1,∞) have finite logarithmic measure. Then there exists {zk=rkeiθk} such that |f(zk)|=M(rk,f),  θk∈[-π/2,3π/2),  limk→∞θk=θ0∈[-π/2,3π/2),rk∉E,rk→∞ such that(i)
ifσ2(f)=α(0<α<∞), then for any given ε1(0<ε1<α),
(2.14)exp⁡{rkα-ε1}<ν(rk)<exp⁡{rkα+ε1};(ii)
ifσ(f)=∞ and σ2(f)=0, then for any given ε2(0<ε2<1/2) and for any large M(>0), one has as rk sufficiently large(2.15)rkM<ν(rk)<exp⁡{rkε2}.Lemma 2.9 (see [9]).
Letf be a transcendental meromorphic function, and let α>1 be a given constant. Then there exist a set E⊂(1,∞) with finite logarithmic measure and a constant B>0 that depends only on α and i,j(i<j(i,j∈ℕ)), such that for all z satisfying |z|=r∉[0,1]∪E, one has
(2.16)|f(j)(z)f(i)(z)|≤B(T(αr,f)r(log⁡αr)log⁡T(αr,f))j-i.Remark 2.10.
From the proof of Lemma2.9 (i.e., Theorem 3 in [9]), we can see that exceptional set E satisfies that if an and bm(n,m=1,2,…) denote all zeros and poles of f, respectively, O(an) and O(bm) denote sufficiently small neighborhoods of an and bm, respectively, then
(2.17)E={|z|:z∈(⋃n=1+∞O(an))⋃(⋃m=1+∞O(bm))}.
Hence, if f(z) is a transcendental entire function and z is a point such that it satisfies that |f(z)| is sufficiently large, then (2.16) holds.Lemma 2.11 (see [4]).
Consider annth-order linear differential equation of the form
(2.18)P0(ez,e-z)f(n)+P1(ez,e-z)f(n-1)+⋯+Pn(ez,e-z)f=Pn+1(ez,e-z),
where each Pj(z,w) is a polynomial in z and w with P0(z,w)≢0. Suppose that f=ϕ(z) is an entire subnormal solution of (2.18), that is, an entire solution of (2.18) that also satisfies (1.3). If ϕ is periodic with period 2πi, then
(2.19)ϕ(z)=S1(ez)+S2(e-z),
where S1(z) and S2(z) are polynomials in z.


## 3. Proof of Theorem1.3

(i) Suppose thatP0≢0 and f(≢0) are the solution of (1.11). Then f is an entire function. By Lemma 2.2 (i), we see that f is transcendental.Step 1.
We prove thatσ(f)=∞. Suppose to the contrary that σ(f)=σ<∞. By Lemma 2.3, we know that for any given ɛ>0, there exists a set E⊂[-π/2,3π/2) of linear measure zero, such that if ψ∈[-π/2,3π/2)∖E, then there is a constant R0=R0(ψ)>1 such that for all z satisfying argz=ψ and |z|=r>R0, we have
(3.1)|f(j)(z)f(s)(z)|≤r(σ-1+ε)(j-s),j=s+1,…,k.
Now we take a rayargz=θ∈(-π/2,π/2)∖E, then cosθ>0. We assert that |f(s)(reiθ)| is bounded on the ray argz=θ. If |f(s)(reiθ)| is unbounded on the ray argz=θ, then by Lemma 2.4, there exists an infinite sequence of points zt=rteiθ(t=1,2,…) such that, as rt→∞, f′(zt)→∞ and
(3.2)|f(i)(zt)f(s)(zt)|≤rts-i(1+o(1))(i=0,…,s-1).
By (1.11), we get that
(3.3)-Ps(ezt)=f(k)(zt)f(s)(zt)+Pk-1(ezt)f(k-1)(zt)f(s)(zt)+⋯+Ps+1(ezt)f(s+1)(zt)f(s)(zt)+Ps-1(ezt)f(s-1)(zt)f(s)(zt)+⋯+P0(ezt)f(zt)f(s)(zt).
Since cosθ>0 and (1.14), we have
(3.4)|Ps(ezt)|=|asms|emsrtcos⁡θ(1+o(1)),|Pj(ezt)|≤Memrtcos⁡θ(1+o(1))(j=0,…,s-1,s+1,…,k-1),
where M(>0) is some constant. Substituting (3.1), (3.2), and (3.4) into (3.3), we get that
(3.5)|asms|emsrtcos⁡θ(1+o(1))≤kMrt(σ+1)kemrtcos⁡θ(1+o(1)).
By ms>m, we know that when rt→∞, (3.5) is a contradiction. So,
(3.6)|f(reiθ)|≤M1rs
on the ray argz=θ∈(-π/2,π/2)∖E, where M1(>0) is some constant.
Now we take a rayargz=θ∈(π/2,3π/2)∖E. Then, cosθ<0. If |f(k)(reiθ)| is unbounded on the ray argz=θ, then by Lemma 2.4, there exists an infinite sequence of points zt′=rt′eiθ(t=1,2,…) such that, as rt′→∞, f(k)(zt′)→∞ and
(3.7)|f(i)(zt′)f(k)(zt′)|≤(rt′)k-i(1+o(1))(i=0,…,k-1).
By (1.11), we get that
(3.8)|-1|≤|Pk-1(ezt′)f(k-1)(zt′)f(k)(zt′)|+⋯+|P0(ezt′)f(zt′)f(k)(zt′)|.
Since cosθ<0 and (3.7), for j=0,…,k-1 as rt′→∞,
(3.9)|Pj(ezt′)f(j)(zt′)f(k)(zt′)|≤(|ajmj|emjrt′cos⁡θ+⋯+|aj1|ert′cos⁡θ)(rt′)k-j(1+o(1))⟶0.
By (3.8) and (3.9), we get that 1≤0; this is a contradiction. So,
(3.10)|f(reiθ)|≤M1rk
on the ray argz=θ∈(π/2,3π/2)∖E.By Lemma2.5, (3.6), and (3.10), we know that f(z) is a polynomial, which contradicts the above assertion that f(z) is transcendental. Therefore σ(f)=∞.Step 2.
We prove that (1.11) has no nontrivial subnormal solution. Now suppose that (1.11) has a nontrivial subnormal solution f0, and we will deduce a contradiction. By the conclusion in Step 1, f0 satisfies (1.3) and and σ(f0)=∞. By Lemma 2.6, we see that σ2(f)≤1. Set σ2(f)=α≤1. By Lemma 2.9, we see that there exist a subset E1⊂(1,∞) having finite logarithmic measure and a constant B>0 such that for all z satisfying |z|=r∉[0,1]∪E1, we have
(3.11)|f0(j)(z)f0(z)|≤B[T(2r,f0)]k+1(j=1,…,k).
From the Wiman-Valiron theory (see [2, page 51]), there is a set E2⊂(1,∞) having finite logarithmic measure, so we can choose z satisfying |z|=r∉[0,1]∪E2 and |f0(z)|=M(r,f0). Thus, we get
(3.12)f0(j)(z)f0(z)=(ν(r)z)j(1+o(1)),j=1,…,k,
where ν(r) is the central index of f0(z).
By Lemma2.8, we see that there exists a sequence {zn=rneiθn} such that |f(zn)|=M(rn,f), θn∈[-π/2,3π/2), limθn=θ0∈[-π/2,3π/2), rn∉[0,1]∪E1∪E2, rn→∞, and if α>0, then by (2.14), we see that for any given ε1(0<ε1<α), and for sufficiently large rn,
(3.13)exp⁡{rnα-ε1}<ν(rn)<exp⁡{rnα+ε1},
and if α=0, then by σ(f)=∞ and (2.15), we see that for any given ε2(0<ε2<1/2) and for any sufficiently large M2>2k+3, as rn is sufficiently large,
(3.14)rnM2<ν(rn)<exp⁡{rnε2}.Sinceθ0 may belong to (-π/2,π/2), (π/2,3π/2), or {-π/2,π/2}, we divide this proof into three cases to prove.Case 1.
Suppose thatθ0∈(-π/2,π/2), then cosθ0>0. If we take δ=(1/4)(π/2-|θ0|), then [θ0-δ,θ0+δ]⊂(-π/2,π/2). By θn→θ0, we see that there is a constant N(>0) such that, as n>N, θn∈[θ0-δ,θ0+δ], and 0<cos(|θ0|+δ)≤cosθn. By (3.11), we see that for any given ε3 satisfying 0<ε3<(1/(4(k+1)))cos(|θ0|+δ),
(3.15)[T(2rn,f0)]k+1≤eε3(k+1)2rn≤e(1/2)cos⁡(|θ0|+δ)rn≤e(1/2)cos⁡θnrn
holds for n>N. By (3.11), (3.12), and (3.15), we see that
(3.16)|f0(k-s)(zn)f0(zn)|=(ν(rn)rn)k-s(1+o(1))≤B[T(2rn,f0)]k+1≤Be(1/2)cos⁡θnrn.
By (1.11), we get
(3.17)-f0(s)(zn)f0(zn)(asmsemszn+⋯+as1ezn)=f0(k)(zn)f0(zn)+Pk-1(ezn)f0(k-1)(zn)f0(zn)+⋯+Ps+1(ezn)f0(s+1)(zn)f0(zn)Ps-1(ezn)f0(s-1)(zn)f0(zn)+⋯+P1(ezn)f0′(zn)f0(zn)+P0(ezn).
Since cosθn>0 and (1.14), we get that
(3.18)|asmsemszn+⋯+as1ezn|=|asms|emsrncos⁡θn(1+o(1)),|Pj(ezn)|≤M3emrncos⁡θn(j=0,…,s-1,s+1,…k-1),
where M3(>0) is some constant. Substituting (3.12) and (3.18) into (3.17), we deduce that for sufficiently large rn,
(3.19)(ν(rn)rn)s|asms|emsrncos⁡θn(1+o(1))≤(ν(rn)rn)k(1+o(1))+M3emrncos⁡θn∑j=0,j≠sk-1(ν(rn)rn)j(1+o(1)).
From (3.13) or (3.14), we have
(3.20)ν(rn)>rnM2>rn2k+3>rn.
By (3.16), (3.19), and (3.20), we get
(3.21)(ν(rn)rn)s|asms|emsrncos⁡θn(1+o(1))≤kM3emrncos⁡θn(ν(rn)rn)k(1+o(1)),(3.22)|asms|e(ms-m)rncos⁡θn(1+o(1))≤kM3Be(1/2)rncos⁡θn.
Since ms-m≥1>1/2, we see that (3.22) is a contradiction.Case 2.
Suppose thatθ0∈(π/2,3π/2). By cosθ0<0 and θn→θ0, we see that for sufficiently large n, cosθn<0. By (1.11), we get
(3.23)-f0(k)(zn)f0(zn)=Pk-1(ezn)f0(k-1)(zn)f0(zn)+⋯+P1(ezn)f0′(zn)f0(zn)+P0(ezn).
Since cosθn<0 and (1.14), we get that for j=0,…,k-1(3.24)|Pj(ezn)|≤|ajmj|emjrncos⁡θn+⋯+|aj1|erncos⁡θn≤Cj,
where Cj(j=0,…,k-1) are constants. By (3.12), (3.20), (3.23), and (3.24), we get that
(3.25)(ν(rn)rn)k(1+o(1))=|f0(k)(zn)f0(zn)|≤[Ck-1+⋯+C0](ν(rn)rn)k-1(1+o(1)).
Thus, we have
(3.26)ν(rn)(1+o(1))≤[Ck-1+⋯+C0]rn(1+o(1)).
By (3.20), we see that (3.26) is also a contradiction.Case 3.
Suppose thatθ0=π/2 or θ0=-π/2. Since the proof of θ0=-π/2 is the same as the proof of θ0=π/2, we only prove the case that θ0=π/2. Since θn→θ0, for any given ε4(0<ε4<1/10), we see that there is an integer N(>0), as n>N, θn∈[π/2-ε4,π/2+ε4], and
(3.27)zn=rneiθn∈Ω¯={z:π2-ε4≤argz≤π2+ε4}.
By Lemma2.9, there exist a subset E3⊂(1,∞) having finite logarithmic measure and a constant B>0 such that for all z satisfying |z|=r∉[0,1]∪E3, we have
(3.28)|f0(d)(z)f0(s)(z)|≤B[T(2r,f0(s))]k+1(d=s+1,…,k).Now we consider the growth off0(reiθ) on a ray argz=θ∈Ω¯∖{π/2}. If θ∈[π/2-ε4,π/2), then cosθ>0. By (1.3), for any given ε5 satisfying 0<ε5<(1/(4(k+1)))cosθ, (3.29)[T(2r,f0(s))]k+1≤eε5(k+1)2r≤e(1/2)cos⁡θr.
If |f0(s)(reiθ)| is unbounded on the ray argz=θ, then by Lemma 2.4, there exists a sequence {yj=Rjeiθ} such that, as Rj→∞, f0(s)(yj)→∞ and (3.30)|f0(i)(yj)f0(s)(yj)|≤Rjs-i(1+o(1))(i=0,…,s-1).
By Remark 2.10 and f0(s)(yj)→∞, we know that yj satisfies (3.28). By (3.28) and (3.29), we see that for sufficiently large j, (3.31)|f0(d)(yj)f0(s)(yj)|≤B[T(2Rj,f0(s))]k+1≤e(1/2)cos⁡θRj(d=s+1,…,k).
By (1.11), (3.18), (3.30), and (3.31), we deduce that (3.32)|asms|emsRjcos⁡θ(1+o(1))=|-Ps(eyj)|≤kM3Be(m+(1/2))Rjcos⁡θRjs(1+o(1)).
Since ms>m+1/2, we know that when Rj→∞, (3.32) is a contradiction. Hence (3.33)|f0(reiθ)|≤Mrs
on the ray argz=θ∈[π/2-ε4,π/2).Ifθ∈(π/2,π/2+ε4], then cosθ<0. We assert that |f0(k)(reiθ)| is bounded on the ray argz=θ. If |f0(k)(reiθ)| is unbounded on the ray argz=θ, then by Lemma 2.4, there exists a sequence {yj*=Rj*eiθ} such that, as Rj*→∞, f0(k)(yj*)→∞ and (3.34)|f0(i)(yj*)f0(k)(yj*)|≤(Rj*)k-i(1+o(1))(i=0,…,k-1).
Since cosθ<0, for fixed t∈{0,1,…,k-1}, we deduce that as Rj*→∞(3.35)|Pt(eyj*)|(Rj*)k≤|atmt|emtRj*cos⁡θ(Rj*)k+⋯+|at1|eRj*cos⁡θ(Rj*)k⟶0.
By (1.11), (3.34), and (3.35), we deduce that as Rj*→∞(3.36)1≤|Pk-1(eyj*)f0(k-1)(yj*)f0(k)(yj*)|+⋯+|P0(eyj*)f0(yj*)f0(k)(yj*)|≤|Pk-1(eyj*)|(Rj*)k(1+o(1))+⋯+|P0(eyj*)|(Rj*)k(1+o(1))⟶0.
This, (3.36) is a contradiction. Hence (3.37)|f0(reiθ)|≤Mrk
on the ray argz=θ∈(π/2,π/2+ε4].By (3.33) and (3.37), we see that |f0(reiθ)| satisfies (3.38)|f0(reiθ)|≤Mrk
on the ray argz=θ∈Ω¯∖{π/2}.However, sincef0(z) is of infinite order and {zn=rneiθn} satisfies |f0(zn)|=M(rn,f0), we see that for any large N(>k), as n is sufficiently large (3.39)|f0(zn)|=|f0(rneiθn)|≥exp⁡{rnN}.
Since zn∈Ω¯, by (3.38) and (3.39), we see that for sufficiently large n(3.40)θn=π2.
Thus cosθn=0 and for sufficiently large n(3.41)|Pj(ezn)|=|ajmjemjzn+⋯+aj1ezn|≤C(j=0,…,k-1),
where C(>0) is some constant. By (1.11) and (3.12), we get that (3.42)-(ν(rn)zn)k(1+o(1))=Pk-1(ezn)(ν(rn)zn)k-1(1+o(1))+⋯+P1(ezn)ν(rn)zn(1+o(1))+P0(ezn).
By (3.20), (3.41) and (3.42), we get that (3.43)ν(rn)≤kCrnk.
By (3.13) (or (3.14)), we see that (3.43) is a contradiction. Hence (1.11) has no nontrivial subnormal solution.Step 3.
We prove that all solutions of (1.11) satisfy σ2(f)=1. If there is a solution f1 satisfying σ2(f1)<1, then f1 satisfies (1.3), that is, f1 is subnormal, but this contradicts the conclusion in Step 2. Hence every solution f satisfies σ2(f)≥1. By this and σ2(f)≤1, we get σ2(f)=1. Theorem 1.3(i) is thus proved.(ii) SinceP0≡⋯≡Pd-1≡0 and Pd≢0, we clearly see that all polynomials with degree ≤d-1 are subnormal solutions of (1.11). By (i), we see that every f(d) satisfies σ2(f(d))=1 or f(d)≡0. Hence σ2(f)=1 or f is a polynomial with degree ≤d-1.


## 4. Proof of Theorem1.4

Suppose thatf1 and f2(≢f1) are nontrivial subnormal solutions of (1.8), then f1-f2(≢0) is a subnormal solution of the corresponding homogeneous equation (1.11) of (1.8). This contradicts the assertion of Theorem 1.3(i). Hence (1.8) possesses at most one nontrivial subnormal solution.Now suppose thatf0 is a nontrivial subnormal solution of (1.8), then f0(z+2πi) is also nontrivial subnormal solution, so, f0(z)=f0(z+2πi) by the above assertion. Thus, by Lemma 2.11, we see that f0 satisfies (1.10).By Theorem1.3(i), we see that all solutions of the corresponding homogeneous equation (1.11) of (1.8) are of σ2(f)=1. By variation of parameters, we see that all solutions of (1.8) satisfy σ2(f)≤1. If σ2(f)<1, then f clearly satisfies (1.3); that is, f is subnormal. Hence all other solutions f of (1.8) satisfy σ2(f)=1 except at most one nontrivial subnormal solution.


## 5. Proof of Theorem1.6

(i) By Lemma2.6 and σ(Pj)=1(j=0,…,k-1), we see that σ2(f)≤1. By Lemma 2.9, we see that there exist a subset E⊂(1,∞) having finite logarithmic measure and a constant B>0 such that for all z satisfying |z|=r∉[0,1]∪E, (5.1)|f(j)(z)f(z)|≤B[T(2r,f)]k+1(j=1,…,k).Takingz=r, by (1.11) and (5.1), we deduce that (5.2)|a0m0|em0r(1+o(1))=|-P0(eyj)|≤kB[T(2r,f)]k+1Memr(1+o(1)).
Since m0>m, by (5.2), we get that σ2(f)≥1. Hence σ2(f)=1.(ii) Using a similar method as in the proof of Theorem1.4, we can prove (ii).



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*Source: 170762-2011-01-02.xml*