# On Generalized Growth of Analytic Functions Solutions of Linear Homogeneous Partial Differential Equation of Second Order

Authors: Devendra Kumar
Journal: Journal of Complex Analysis
 (2017)

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## Abstract

The generalized growth parameters of analytic functions solutions of linear homogeneous partial differential equations of second order have been studied. Moreover, coefficients characterizations of generalized order and generalized type of the solutions represented in convergent series of Laguerre polynomials have been obtained.


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## Body

## 1. Introduction

Hu and Yang [1, 2] studied the behavior of meromorphic solutions of the following homogeneous linear partial differential equations of the second order: (1)t2∂2u∂t2-z2∂2u∂z2+2t+2∂u∂t-2z∂u∂z=0,t2∂2u∂t2-z2∂2u∂z2+t∂u∂t-z∂u∂z+t2u=0.They showed that these solutions are closely related to Bessel functions and Bessel polynomials for (t,z)∈C2. Several authors such as Berenstein and Li [3], Hu and Yang [4], Hu and Li [5], Li [6], and Li and Saleeby [7] have investigated the global solutions of some first-order partial differential equations. McCoy [8] and Kumar [9] studied the approximation of pseudoanalytic functions on the disk and obtained some coefficient and Bernstein-type growth theorems. Kapoor and Nautiyal [10] and Kumar and Basu [11] characterized the order and type of solutions (not necessarily entire solutions) of certain linear partial differential equations in terms of rates of decay of approximation errors in various norms. These solutions are related to Jacobi polynomials. Wang et al. [12] obtained the growth parameters order and type of entire function solutions of the partial differential equation (2)t∂2u∂t2+δ+1-t∂u∂t+z∂u∂z=0for a real δ>0. These solutions are related to Laguerre polynomials. In this paper, our aim is to characterize the generalized growth parameters order and type of solutions of (2) which are analytic on bidisk of finite radius R. The generalized growth parameters have been studied by several authors such as Šeremeta [13], Shah [14], Srivastava and Kumar [15], and Vakarchuk and Zhir [16, 17]. Wang et al. [12] proved that the PDE (2) has an entire solution u=f(t,z) on C2, if and only if u=f(t,z) has a series expansion f(t,z)=∑n=0∞anLn(δ,t)zn. Here (3)Lnδ,t=∑k=0nn+δn-k-tkk!are the Laguerre polynomials. Bernstein theorem identifies a real analytic function on the closed unit disk as the restriction of an analytic function defined on an open disk of radius R>1 by computing R from the sequence of minimal errors generated from optimal polynomials approximates. The disk DR of maximum radius on which analytic function f(t,z) exists is denoted by f(t,z)∈A(DR×DR). If f is an entire function, it has no singularities in the finite positive C2 plane and write f∈A(C2). Let α and β be two positive, strictly increasing, and differentiable functions from (0,∞)→(0,∞), which satisfies the following conditions for every γ>0: (4)limx→∞⁡αγxαx=1,limx→∞⁡β1+xωxβx=1,limx→∞⁡ωx=0,limx→∞⁡dβ-1γαxdlog⁡x≤b,αxβ-1γαx=1+0xαx,x⟶∞;here d(u) denotes the differential of u. Now we define the generalized order and generalized type of an entire f(t,z)∈A(C2) by (5)ρα,β=lim⁡supr→∞⁡αlog⁡Mr,r,fβlog⁡r,Tα,β=lim⁡supr→∞⁡αMr,r,fβrρα,β,where (6)Mr,r,f=supz,t∈D¯R⁡ft,z,r<R.In view of the concept introduced by MacLane [18] to the measures of order and type for an analytic function on a disk z<R, we normalize above definitions relative to the boundary under the transformation r→R/(R-r). Thus an analytic function f(t,z)∈A(DR×DR) with radial limits is said to be of generalized regular growth (ρo(α,β),To(α,β)) if it satisfies (7)ρoα,β=lim⁡supr→R⁡αlog⁡Mr,r,fβR/R-r,Toα,β=lim⁡supr→R⁡αMr,r,fβR/R-rρoα,β,1<r<R,where ρo(α,β) is referred to as the (α,β)-order of f(t,z) provided that 0<ρo(α,β)<∞ and To(α,β) is referred to as the (α,β)-type.Example A.
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## 2. Auxiliary and Main Results

First we prove the following lemma.Lemma 1.
Letf(t,z)=∑n=0∞anLn(δ,t)zn and f∈A(DR×DR). For every 1<r<R, we set (14)M¯r,r,f=supn∈N⁡anLnδ,t·rn,r>0,ρ¯oα,β=lim⁡supr→Rαlog⁡M¯r,r,fβR/R-r;then (15)ρ¯oα,β≤μα,β,ρoα,β≤ρ¯oα,β.Proof.
From (13), for r sufficiently close to R, we have (16)log⁡nδΓδ+1anRn≤nβ-1αn/μ¯,μ¯=μ+εor (17)log⁡nδΓδ+1anRn≤nlog⁡rR+nβ-1αn/μ¯.Using the result (18)limx→∞⁡dβ-1cαxdlog⁡x≤blog⁡1+ξ=1+oξ,ξ⟶0,with the method of Calculus that, for every r>1 and μ>0, the maximum of the function (19)xlog⁡rR+xβ-1αx/μ¯is reached at (20)x=α-1μ¯β1-dlog⁡β-1αx/μ/dlog⁡xlog⁡R/r,we get (21)x=1+o1α-1μ¯βRR-r,r⟶∞.Now we have (22)log⁡nδΓδ+1anrn≤Coα-1μ¯βRR-r,log⁡Mr,r,f≤Coα-1μ¯βRR-r.Using the property of α, we get (23)αlog⁡M¯r,r,fβR/R-r≤μ¯.Applying the limit supremum as r→R, we obtain (24)ρ¯oα,β≤μα,β.Now consider (25)ft,z=∑n=0∞anLnδ,0zn;putting r=r·Rr/R in the above, we get (26)Mr,r,f≤∑n=0∞anLnδ,tr·RnrRn,rR<1,or(27)Mr,r,f≤∑n=0∞sup⁡anLnδ,tr·RnrRn,or (28)Mr,r,f≤M¯r′,r′,f∑n=0∞rRn≤M¯r′,r′,f11-r/R,r′=r·R,log⁡Mr,r,f≤log⁡M¯r′,r′,f-log⁡1-rR,or (29)αlog⁡Mr,r,fβR/R-r≤αlog⁡M¯r′,r′,f-log⁡1-r/RβR/R-r·R·βR/R-r·RβR/R-r.Proceeding to limits, we obtain (30)ρoα,β≤ρ¯oα,β.Combining (24) and (30), we get (31)ρoα,β≤μα,β.Theorem 2.
Letf(t,z)=∑n=0∞anLn(δ,t)zn such that (32)lim⁡supn→∞⁡αnβn/log⁡nδ/Γδ+1anRn=μα,β<∞;then f is the restriction of analytic function in (DR×DR)(R>1) and its (α,β)-order ρ(α,β)=μ(α,β).Proof.
Sincef(0,z)=∑n=0∞anLn(δ,0)zn is an analytic function in DR, we have (33)lim⁡supn→∞⁡anLnδ,01/n=1R,R>1.Using Ln(δ,0)=nδ/Γ(δ+1), we get (34)lim⁡supn→∞⁡an1/n=1R,which is necessary and sufficient condition for f∈A(DR×DR). So, for every 1<r<R, the series ∑n=0∞anLn(δ,t)zn is convergent in DR×DR when ∑n=0∞anLn(δ,t)zn is analytic in DR×DR.
Now we have to prove thatμ(α,β) is the (α,β)-order of f(t,z).
In order to complete the proof by Lemma1, it is only to show that ρ(α,β)≥μ(α,β). In view of the definition of ρ(α,β), we have, for every δ>0, that there exists 1<rε<R such that, for every rε<r<R, (35)log⁡Mr,r,f≤α-1ρα,β+εβRR-r.Using Cauchy’s estimate of analytic functions, we have (36)∂nf∂zn0,0≤nr-nMr,r,fwith the coefficients formula of the Taylor expansion,(37)∂nf∂zn0,0=anLnδ,0n!;we get M(r,r,f)≥anLnδ,0rn. Since Lnδ,0≥1, it gives (38)anrn≤Mr,r,f.Now using (35) in (38), we get (39)log⁡annδΓδ+1Rn≤-nlog⁡rR+δlog⁡n-log⁡Γδ+1+α-1ρα,β+εβRR-r.The minimum value of right-hand side is estimated at (40)r=R1-1β-11/ρ+εαn/β-1αn/ρ+ε,ρ=ρα,β.Using the properties of functions α and β, (41)αxβ-1cαx≃1+o1αx;for c>0, x→∞, and the properties of logarithm, we get (42)log⁡annδΓδ+1Rn≤C1nβ-1αn/ρ+ε,where C1 is a constant. Hence, (43)βC1nlog⁡annδ/Γδ+1Rn≥αnρ+ε.Proceeding to limit supremum as n→∞, we obtain (44)μα,β≤ρα,β.Hence the proof is completed.Example B.
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Letf(t,z)=∑n=0∞anLn(δ,t)zn. For every 1<r<R, (46)σ¯1α,β=lim⁡supr→Rαlog⁡M¯r,r,fβR/R-rρα,β;then (47)σα,β≤σ¯1α,β.Proof.
Following the same reasoning as in the proof of Lemma1, we obtain (48)αlog⁡Mr,r,fβR/R-rρα,β≤αlog⁡M¯r′,r′,f-log⁡1-r/RβR/R-r·Rρα,β·βR/R-r·Rρα,ββR/R-rρα,β.Applying the limit supremum as r→R, we get (49)σα,β≤σ¯1α,β.Theorem 4.
Letf(t,z)=∑n=0∞anLn(δ,t)zn be of finite generalized (α,β)-order ρ(α,β) and (50)Tα,β=lim⁡supn→∞⁡αnβn/lognδ/Γδ+1anRnρα,β<∞.Then f is the restriction of an analytic function in (DR×DR)(R>1) and its (α,β)-type σ(α,β)=T(α,β).Proof.
We have proven in Theorem2 that f is the restriction of an analytic function in (DR×DR). Now, in order to complete the proof, first we shall prove that σ(α,β)≤T(α,β). In view of the definition of T, for every ε>0, there exists n≥nε:(51)αn≤T+εβnlog⁡nδ/Γδ+1anRnρα,β,T≡Tα,β<∞,log⁡annδΓδ+1Rn≤nβ-1αn/T¯1/ρ,T¯=T+ε,since (52)log⁡annδΓδ+1rn≤nlog⁡rR+log⁡annδΓδ+1Rn.From inequality (51), we get (53)log⁡annδΓδ+1rn≤nlog⁡rR+nβ-1αn/T¯1/ρ.For every 1<r<R and r sufficiently close to R, we put (54)φx,r=xlog⁡rR+xβ-1αx/T¯1/ρ.The maximum value of φ(x,r) is reached at (55)x=xr=α-1T¯β1-dlog⁡β-1αx/T¯1/ρ/dlog⁡xlog⁡R/rρ.Using the relation (56)log⁡rR=log⁡r-RR+1~r-RR(as (r-R)/R→0) and (57)dlog⁡β-1αx/T¯1/ρdlog⁡x≤ko,where ko is a positive constant, it gives (58)xr=1+o1α-1T¯βRR-rρ.Now, from relation (53), we have (59)log⁡annδΓδ+1rn≤sup⁡φx,ror (60)log⁡annδΓδ+1rn≤1+o1α-1T¯βR/R-rρR/R-r.Since R/(R-r)>1, it gives (61)log⁡annδΓδ+1rn≤k1α-1T¯βRR-rρ.Then (62)log⁡Mr,r,f≤k1α-1T¯βRR-rρ,or (63)αlog⁡Mr,r,fβR/R-rρ≤T¯,or (64)σα,β≤T.The inequality obviously holds for T=∞. Now we shall prove that σ(α,β)≥T(α,β). Let σ(α,β)<∞. In view of definition of σ(α,β), we have, for every ε>0, that there exist 1<rε<R, such that, for every r>rε(R>r>rε>1), (65)log⁡Mr,r,f≤α-1σ¯βRR-rρ,σ¯=σ+ε.Now, using (51), we get (66)log⁡annδΓδ+1Rn≤-nlog⁡rR+δlog⁡n-log⁡Γδ+1+α-1σ¯βRR-rρ,or (67)log⁡annδ/Γδ+1Rnn≤-log⁡rR+1nδlog⁡n-log⁡Γδ+1+α-1σ¯βRR-rρ,or (68)1nlog⁡annδΓδ+1Rn≤1+o1log⁡Rr,as n→∞. Now, using log⁡(R/r)~(R-r)/r as r→R, we have (69)1nlog⁡annδΓδ+1Rn≤1+o1log⁡Rr-1.Set (70)rR=β-1αn/σ¯1/ρ1+β-1αn/σ¯1/ρ;in above inequality, we get (71)1nlog⁡annδΓδ+1Rn≤1β-1αn/σ¯1/ρor (72)αnβn/log⁡annδ/Γδ+1Rn≤σ¯=σ+ε.Applying the limit supremum as n→∞, we obtain (73)σα,β≥Tα,β.The result is obviously true for σ(α,β)=∞. This completes the proof.Example C.
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