# Relations between Solutions of Differential Equations and Small Functions

Authors: Wei Liu; Zong-Xuan Chen
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We investigate relations between solutions, their derivatives of differential equationf(k)+Ak−1f(k−1)+⋯+A1f’+A0f=0, and functions of small growth, where Aj(j=0,1,…,k−1) are entire functions
of finite order. By these relations, we see that every transcendental solution and its
derivative of above equation have infinitely many fixed points.


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## Body

## 1. Introduction and Results

In this paper, we use the standard notations of the Nevanlinna’s value distribution theory ([1–3]). We use λ(f) and λ¯(f) to denote exponents of convergence of the zero sequence and the sequence of distinct zeros of a meromorphic function f(z), and σ(f) to denote the order of growth of f(z).In 2000, Chen [4] considered fixed points of solutions of second-order linear differential equations and obtained precise estimation of the number of fixed points of solutions. Recently, a number of papers (including [5–11]) considered relations between solutions, their derivatives of some differential equations, and functions of small growth.In 2006, Chen and Shon [7] proved the following theorem.Theorem A.
LetAj(z)(≢0)(j=0,1) be entire functions of σ(Aj)<1,a,b be complex constants such that ab≠0 and arga≠argb or a=cb(o<c<1). Let φ(z)(≢0) be an entire function of finite order. Then, every solution f(≢0) of the equation
(1.1)f′′+A1eazf′+A0ebzf=0,
satisfies
(1.2)λ¯(f-φ)=λ¯(f′-φ)=λ¯(f′′-φ)=∞.In 2010, Xu and Yi [11] proved the following theorem.Theorem B.
LetAj(z)(≢0)(j=0,1) be entire functions of σ(Aj)<1,a,b be complex constants such that ab≠0 and a/b∉{1,2}. Let φ(z)(≢0) be an entire function of σ(φ)<1. Then, every solution f(≢0) of (1.1) satisfies (1.2).In [5, 6, 8–10], authors considered similar problems in Theorems A and B. For relations between solutions, their derivatives of some differential equations, and functions of small growth, particularly, relations between derivatives and functions of small growth are difficult problems. Such problems on higher-order differential equations are more difficult.In this paper, we consider the higher-order differential equation(1.3)f(k)+Ak-1f(k-1)+⋯+A1f′+A0f=0,
and prove the following results.Theorem 1.1.
LetAj(j=0,1,…,k-1) be entire functions of finite order, not all identically equal to zero, such that if Aj≢0, then λ(Aj)<σ(Aj); if i≠j, then σ(Ai/Aj)=max{σ(Ai),σ(Aj)}. Suppose that φ(z) is a finite-order transcendental entire function. Then, every transcendental solution f of (1.3) satisfies λ¯(f-φ)=σ(f)=∞. Furthermore, if λ(φ)<λ(A0), then every solution f(≢0) of (1.3) satisfies λ¯(f′-φ)=σ(f)=∞.Theorem 1.2.
LetAj(j=0,1,…,k-1) satisfy conditions of Theorem 1.1 and A0≢0. Suppose that H is a nonzero polynomial. Then, every solution f(≢0) of (1.3) satisfies λ¯(f′-H)=λ¯(f-H)=σ(f)=∞.Corollary 1.3.
LetAj(j=0,1,…,k-1) satisfy all conditions of Theorem 1.2. Then, every solution f(≢0) and its derivative of (1.3) have infinitely many fixed points.To prove Theorems1.1 and 1.2, we use a new method. Our method is different from methods before (including methods applied in [4–13]) which cannot be applied to prove our Theorems 1.1 and 1.2.


## 2. Auxiliary Lemmas

Lemma 2.1 (see [12]).
LetAj(j=0,1,…,k-1) be entire functions of finite order, not all identically zero. Suppose that if Aj≢0, then λ(Aj)<σ(Aj); if i≠j, then σ(Ai/Aj)=max{σ(Ai),σ(Aj)}. Then, every transcendental solution f of (1.3) satisfies σ(f)=∞. Furthermore, according to the order of A0,A1,…,Ak-1, if Aj is the first coefficient satisfying Aj≢0, then (1.3) may at most have polynomial solutions of degree ≤j-1, and all other solutions are of infinite order. If A0≢0, then every nonzero solution f of (1.3) has infinite order.Lemma 2.2 (see [13]).
LetAj(j=0,1,…,k-1),F(≢0) be meromorphic functions of finite order. Then, every meromorphic solution of
(2.1)f(k)+Ak-1f(k-1)+⋯+A1f′+A0f=F,
satisfies λ¯(f)=λ(f)=σ(f).Lemma 2.3 (see [14]).
Letf be a transcendental meromorphic function of σ(f)=σ<∞. Let H={(k1,j1),(k2,j2),…,(kq,jq)} be a finite set of distinct pairs of integers that satisfy ki>ji⩾0 for i=1,2,…,q. Also, let ε>0 be a given constant. Then,
(i) there exists a set E⊂[0,2π) of linear measure zero such that, if ψ∈[0,2π)∖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, we have
(2.2)|f(k)(z)f(j)(z)|⩽|z|(k-j)(σ-1+ε),
(ii) there exists a set E⊂(1,∞) of finite logarithmic measure, such that, for all z satisfying |z|∉E∪[0,1] and for all (k,j)∈H, we have
(2.3)|f(k)(z)f(j)(z)|⩽|z|(k-j)(σ-1+ε),
(iii) there exists a set E⊂(0,∞) of finite linear measure such that, for all z satisfying |z|∉E and for all (k,j)∈H, we have
(2.4)|f(k)(z)f(j)(z)|⩽|z|(k-j)(σ+ε).Lemma 2.4 (see [7]).
Letg(z) be a meromorphic function of σ(g)=β<∞. Then, for any given ε>0, there is a set E⊂[0,2π) that has linear measure zero, such that, if ψ∈[0,2π)∖E, there is a constant R=R(ψ)>1 such that, for all z satisfying argz=ψ and |z|=r≥R, we have
(2.5)exp⁡{-rβ+ε}≤|g(z)|≤exp⁡{rβ+ε}.Lemma 2.5 (see [12, 15]).
Suppose thatP(z)=(α+iβ)zn+⋯  be a polynomial with degree n≥1, where α,β are real numbers satisfying |α|+|β|≠0. Let ω(z)≢0 be an entire function with σ(ω)<n. Set g=ωeP,z=reiθ,δ(P,θ)=αcosnθ-βsinnθ. Then, for any given ε(0<ε<1), there exists a set H1⊂[0,2π) of linear measure zero such that, for θ∈[0,2π)∖(H1∪H2), there is a constant R>0 such that, for |z|=r>R, we have(i)
ifδ(P,θ)>0, then
(2.6)exp⁡⁡{(1-ε)δ(P,θ)rn}≤|g(reiθ)|≤exp⁡⁡{(1+ε)δ(P,θ)rn},(ii)
ifδ(P,θ)<0, then
(2.7)exp⁡⁡{(1+ε)δ(P,θ)rn}≤|g(reiθ)|≤exp⁡⁡{(1-ε)δ(P,θ)rn},where H2={θ∈[0,2π);δ(P,θ)=0} is a finite set.


## 3. Proof

Proof of Theorem1.1.
Suppose thatf(z) is a transcendental solution of (1.3). By Lemma 2.1, we know that σ(f)=∞. Set g0=f-φ. Then, σ(g0)=σ(f)=∞ and λ¯(g0)=λ¯(f-φ). Substituting f=g0+φ into (1.3), we obtain
(3.1)g0(k)+Ak-1g0(k-1)+⋯+A1g0′+A0g0=-[φ(k)+Ak-1φ(k-1)+⋯+A1φ′+A0φ].
Since all transcendental solutions of (1.3) have infinite order and φ is a transcendental entire function of finite order, we see that φ(k)+Ak-1φ(k-1)+⋯+A1φ′+A0φ≢0. So that, by Lemma 2.2, we obtain λ¯(g0)=σ(g0)=∞, that is, λ¯(f-φ)=σ(f)=∞.
Now suppose thatλ(φ)<λ(A0). Thus, A0≢0. In what follows, we prove that λ¯(f′-φ)=σ(f)=∞.
Setg1=f′-φ. Then, σ(g1)=σ(f′)=σ(f)=∞ and λ¯(g1)=λ¯(f′-φ). Differentiating both sides of (1.3), we obtain
(3.2)f(k+1)+Ak-1f(k)+(Ak-1′+Ak-2)f(k-1)+⋯+(A1′+A0)f′+A0′f=0.
By (1.3), we obtain
(3.3)f=-1A0[f(k)+Ak-1f(k-1)+⋯+A1f′].
Substituting (3.3) into (3.2), we deduce that
(3.4)f(k+1)+(Ak-1-A0′A0)f(k)+(Ak-1′+Ak-2-A0′A0Ak-1)f(k-1)+⋯+(A1′+A0-A0′A0A1)f′=0.
Substituting f′=g1+φ,f′′=g1′+φ′,…,f(k+1)=g1(k)+φ(k) into (3.4), we obtain
(3.5)g1(k)+(Ak-1-A0′A0)g1(k-1)+(Ak-1′+Ak-2-A0′A0Ak-1)g1(k-2)+⋯+(A1′+A0-A0′A0A1)g1=h,
where
(3.6)-h=φ(k)+(Ak-1-A0′A0)φ(k-1)+(Ak-1′+Ak-2-A0′A0Ak-1)φ(k-2)+⋯+(A1′+A0-A0′A0A1)φ.
Since whenAj≢0,λ(Aj)<σ(Aj), by Hadamard-Borel theorem, we know that Aj(z)=hj(z)ePj(z) where hj(z) is nonzero entire function, Pj(z) is a nonzero polynomial, such that σ(hj)=λ(Aj)<σ(Aj)=degPj. By Aj(z)=hj(z)ePj(z), we obtain
(3.7)Aj′(z)Aj(z)=hj′(z)hj(z)+Pj′(z),Aj′(z)=(hj′(z)+Pj′(z)hj(z))ePj(z).
Next we proveh≢0. Suppose to the contrary h≡0. Then,
(3.8)φ(k)+(Ak-1-A0′A0)φ(k-1)+(Ak-1′+Ak-2-A0′A0Ak-1)φ(k-2)+⋯+(A1′+A0-A0′A0A1)φ=0.
Dividing φ into both sides of (3.8) and substituting (3.7) into (3.8), we obtain
(3.9)Bk-1(z)ePk-1(z)+Bk-2(z)ePk-2(z)+⋯+B1(z)eP1(z)+B0(z)eP0(z)+B(z)=0,
where
(3.10)B0=h0,B1=h1φ′φ+(h1′+h1P1′-A0′A0h1),B2=h2φ′′φ+(h2′+h2P2′-A0′A0h2)φ′φ,⋮Bj=hjφ(j)φ+(hj′+hjPj′-A0′A0hj)φ(j-1)φ,⋮Bk-1=hk-1φ(k-1)φ+(hk-1′+hk-1Pk-1′-A0′A0hk-1)φ(k-2)φ,B=φ(k)φ-A0′A0φ(k-1)φ.
Sinceλ(h0)=λ(A0)>λ(φ), we see that B0=h0≢0. It is obviously that not all Bk-1,Bk-2,…,B0 are equal to zero. Without loss of generality, we may suppose that all Bj(j=0,1,…,k-1) are not identically zero. In fact, if there exists some Bj≡0, we can remove it and rewrite the subscript of each function in (3.9).
Sinceσ(φ)<+∞ and σ(A0)<+∞, by Lemma 2.3, there exists a set E1⊂[0,2π) of linear measure zero, such that, if θ∈[0,2π)∖E1, there is a contant R=R(θ)>1, such that, for all z satisfying argz=θ and |z|≥R, we have
(3.11)|φ(j)(z)φ(z)|≤|z|j⋅σ(φ)(j=1,2,…,k-1),|A0′(z)A0(z)|≤|z|σ(A0).
By (3.10) and (3.11), we obtain
(3.12)|B(z)|≤|φ(k)(z)φ(z)|+|A0′(z)A0(z)||φ(k-1)(z)φ(z)|≤|z|kσ(φ)+|z|σ(A0)|z|kσ(φ)≤2r2kσ,
where σ=max{σ(φ),σ(A0)}.
Sinceφ,A0 are entire functions of finite order, then we obtain
(3.13)m(r,A0′A0)=O(log⁡⁡r),m(r,φ(j)φ)=O(log⁡⁡r)(j=1,…,k-1).
By (3.10), (3.13), for sufficiently large r, we obtain
(3.14)m(r,Bj)≤3m(r,hj)+m(r,φ(j)φ)+m(r,hj′)+m(r,Pj′)+m(r,A0′A0)+m(r,φ(j-1)φ)+O(1)≤4T(r,hj)+O(log⁡⁡r)(j=2,…,k-1),N(r,Bj)≤N(r,1φ)+N(r,1A0)(j=1,…,k-1).
By (3.14), we obtain
(3.15)T(r,Bj)=m(r,Bj)+N(r,Bj)≤4T(r,hj)+N(r,1φ)+N(r,1A0)+O(log⁡⁡r)(j=2,…,k-1).
Since σ(hj)=λ(Aj),λ(φ)<λ(A0), by (3.15), we obtain
(3.16)σ(Bj)≤max⁡⁡{λ(φ),λ(A0),σ(hj)}=max⁡⁡{λ(A0),λ(Aj)}(j=2,…,k-1).
Using the same method as above, we obtain
(3.17)σ(B1)≤max⁡⁡{λ(A0),λ(A1)}.
Clearly,
(3.18)σ(B0)=σ(h0)=λ(A0).
By (3.16)–(3.18), we obtain
(3.19)σ(Bs)≤max⁡⁡{λ(Aj)∣0≤j≤k-1}(s=0,1,…,k-1).
Set
(3.20)d=max⁡⁡{deg⁡Pj∣j=0,1,…,k-1},d̃=max⁡⁡{deg⁡Pj,λ(Ai)∣i=0,…,k-1,deg⁡Pj<d,j∈{0,1,…,k-1}}.
According to definitions of d and d̃ and (3.19), we obtain d̃<d and
(3.21)σ(Bs)≤max⁡⁡{λ(Aj)∣0≤j≤k-1}≤d̃(s=0,1,…,k-1).
Next, we discuss functionsBjePj(j=0,1,2…,k-1). We divide them into two cases:
(3.22)I={BjePj∣deg⁡Pj<d,j∈{0,1,…,k-1}},II={BjePj∣deg⁡Pj=d,j∈{0,1,…,k-1}}.
Firstly, we considerBjePj∈I. By the definition of d̃ and (3.21), for any BjePj∈I, we have
(3.23)σ(BjePj)≤d̃.
By Lemma 2.4, for any given ε1(0<ε1<d-d̃), there is a set E2 of linear measure zero, such that, if θ∈[0,2π)∖E2, there is a constant R=R(θ)>1 such that, for all z satisfying argz=θ and |z|=r≥R, we have
(3.24)|Bj(z)ePj(z)|≤exp⁡{rd̃+ε1}.
As r→∞, we have rd̃+ε1/rd→0, that is, rd̃+ε1≤ε1rd. Then, inequality (3.24) can be rewritten as form
(3.25)|Bj(z)ePj(z)|≤exp⁡{ε1rd}.
Secondly, we considerBjePj∈II. By the definition of II, for every BjePj, we have degPj=d. By (3.21), we obtain σ(Bj)≤d̃<d. So, for any BjePj∈II, we have σ(Bj)<d=σ(Pj). By Lemma 2.5, there is a set E3⊂[0,2π) which has the linear measure zero, such that, for any given ε2(0<ε2<1), and we have that for all z satisfying argz=θ∈[0,2π)∖E3 and |z|=r≥R, if δ(Pj,θ)>0, then
(3.26)exp⁡{(1-ε)δ(Pj,θ)rd}≤|BjePj|≤exp⁡{(1+ε)δ(Pj,θ)rd},
if δ(Pj,θ)<0, then
(3.27)exp⁡{(1+ε)δ(Pj,θ0)rd}≤|BjePj|≤exp⁡{(1-ε)δ(Pj,θ0)rd}.
Now, we further considerBjePj(j=0,1,…,k-1). Take a fixed polynomial Ps∈II. Thus, degPs=d. Set
(3.28)E={θ∈[0,2π)∣δ(Ps,θ)>0},E4={θ∈[0,2π)∣δ(Pi-Pj,θ)=0,0≤i<j≤k-1},⋃{θ∈[0,2π)∣δ(Pj,θ)=0,j=0,1,…,k-1}.
Clearly, the linear measure of E∖(E1∪E2∪E3∪E4) is greater than zero. Now, we take ray argz=θ0∈E∖(E1∪E2∪E3∪E4), then δ(Ps,θ0)>0. When j≠s, we have δ(Pj,θ0)≠0; when i<j and degPi=degPj, we have δ(Pi,θ0)≠δ(Pj,θ0). Set
(3.29)δ=max⁡⁡{δ(Pj,θ0)∣BjePj∈II}.
It is clearly δ>0. Since δ(Pi,θ0)≠δ(Pj,θ0)(i<j and degPi=degPj), there exists a unique integer t(0≤t≤k-1), such that δ(Pt,θ0)=δ. Suppose that Ps satisfies δ(Ps,θ0)=δ. On the ray argz=θ0, we have that
(3.30)|Bs(z)ePs(z)|≥exp⁡{(1-ε)δrd}.
Set
(3.31)δ̃=max⁡⁡{δ(Pj,θ0)∣Bjepj∈II∖{BsePs}}.
Thus, δ̃<δ. For any BjejP∈II∖{BsePs}, by (3.26) and (3.27), we see that, on the ray argz=θ0, if δ(Pj,θ0)>0, we have
(3.32)|BjePj|≤exp⁡{(1+ε)δ(Pj,θ0)rd},
if δ(Pj,θ0)<0, we have
(3.33)|Bj(z)ePj(z)|≤exp⁡{(1-ε)δ(Pj,θ0)rd}<1.
Hence, if BjejP∈II∖{BsePs}, then we have
(3.34)|Bj(z)ePj(z)|≤exp⁡{(1+ε)δ(Pj,θ0)rd}+1≤exp⁡{(1+ε)δ̃rd}+1.
Hence, (3.9) can be rewritten as form
(3.35)Bs(z)ePs(z)=Σj≠sBj(z)ePj(z)+B(z).
By (3.12), (3.25), (3.30), (3.34), and (3.35), for above ε, set
(3.36)ε=12min⁡⁡{δ1+δ,δ-δ̃δ+δ̃,ε1,ε2},
then, for all z satisfying argz=θ0 and sufficiently large r, we have
(3.37)exp⁡{(1-ε)δrd}≤|Bs(z)ePs(z)|≤Σj≠s|Bj(z)ePj(z)|+|B(z)|≤O(1)exp⁡{εrd}+O(1)exp⁡{(1+ε)δ̃rd}+O(1)+2r2kσ.
Thus, we obtain 1≤0. This is a contradiction which shows h≢0.
Sinceh≢0 and σ(g1)=∞, by Lemma 2.2 and (3.5), we obtain λ¯(g1)=σ(g1)=∞, that is, λ¯(f′-φ)=σ(f)=∞.
Thus, Theorem1.1 is proved.Proof of Theorem1.2.
Using the same method as in the proof of Theorem1.1, we can prove Theorem 1.2.



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