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∂t2z2∂2u∂z2+2t+2∂u∂t2z∂u∂z=0,t2∂2u∂t2z2∂2u∂z2+t∂u∂tz∂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 firstorder partial differential equations. McCoy [8] and Kumar [9] studied the approximation of pseudoanalytic functions on the disk and obtained some coefficient and Bernsteintype 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+1t∂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+δnktkk!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→∞dβ1γαxdlogx≤b,αxβ1γαx=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=limsupr→∞αlogMr,r,fβlogr,Tα,β=limsupr→∞αMr,r,fβrρα,β,where (6)Mr,r,f=supz,t∈D¯Rft,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/(Rr). 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=limsupr→RαlogMr,r,fβR/Rr,Toα,β=limsupr→RαMr,r,fβR/Rrρoα,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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has (α,β)order 2 and (α,β)type 1 for α(x)=logx and β(x)=x.
In a neighborhood of origin, the function f(t,z) has the local expansion (8)ft,z=∑n=0∞wntn!zn,where wn(t)=(∂nf/∂zn)(t,0) is an entire solution of the ordinary differential equation (9)td2wdt2+δ+1tdwdt+nw=0,where wn(t)zn=n!anLn(δ,t)zn. Following the method of Frobenius [19], a second independent solution of (9) can be obtained as (10)Xnδ,t=qLnδ,tlogt+∑i=0∞piti,where q≠0 and pi are constants. So there exist an and bn satisfying (11)wnt=n!anLnδ,t+bnXnδ,t.Because of the singularity of Xn(δ,t) at t=0, it leads to bn=0. Thus (12)ft,z=∑n=0∞anLnδ,tzn.Since f(0,z)=∑n=0∞anLn(δ,0)zn, we have the estimate wn0zn=n!anLnδ,0zn~n!an(nδ/Γδ+1)rn for large n. Set (13)limsupn→∞αnβn/lognδ/Γδ+1anRn=μα,β.
2. Auxiliary and Main Results
First we prove the following lemma.
Lemma 1.
Let f(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∈NanLnδ,t·rn, r>0,ρ¯oα,β=limsupr→RαlogM¯r,r,fβR/Rr;then (15)ρ¯oα,β≤μα,β,ρoα,β≤ρ¯oα,β.
Proof.
From (13), for r sufficiently close to R, we have (16)lognδΓδ+1anRn≤nβ1αn/μ¯, μ¯=μ+εor (17)lognδΓδ+1anRn≤nlogrR+nβ1αn/μ¯.Using the result (18)limx→∞dβ1cαxdlogx≤blog1+ξ=1+oξ, ξ⟶0,with the method of Calculus that, for every r>1 and μ>0, the maximum of the function (19)xlogrR+xβ1αx/μ¯is reached at (20)x=α1μ¯β1dlogβ1αx/μ/dlogxlogR/r,we get (21)x=1+o1α1μ¯βRRr, r⟶∞.Now we have (22)lognδΓδ+1anrn≤Coα1μ¯βRRr,logMr,r,f≤Coα1μ¯βRRr.Using the property of α, we get (23)αlogM¯r,r,fβR/Rr≤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∞supanLnδ,tr·RnrRn,or (28)Mr,r,f≤M¯r′,r′,f∑n=0∞rRn≤M¯r′,r′,f11r/R, r′=r·R,logMr,r,f≤logM¯r′,r′,flog1rR,or (29)αlogMr,r,fβR/Rr≤αlogM¯r′,r′,flog1r/RβR/Rr·R·βR/Rr·RβR/Rr.Proceeding to limits, we obtain (30)ρoα,β≤ρ¯oα,β.Combining (24) and (30), we get (31)ρoα,β≤μα,β.
Theorem 2.
Let f(t,z)=∑n=0∞anLn(δ,t)zn such that (32)limsupn→∞αnβn/lognδ/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.
Since f(0,z)=∑n=0∞anLn(δ,0)zn is an analytic function in DR, we have (33)limsupn→∞anLnδ,01/n=1R, R>1.Using Ln(δ,0)=nδ/Γ(δ+1), we get (34)limsupn→∞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 that μ(α,β) is the (α,β)order of f(t,z).
In order to complete the proof by Lemma 1, 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)logMr,r,f≤α1ρα,β+εβRRr.Using Cauchy’s estimate of analytic functions, we have (36)∂nf∂zn0,0≤nrnMr,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)logannδΓδ+1Rn≤nlogrR+δlognlogΓδ+1+α1ρα,β+εβRRr.The minimum value of righthand side is estimated at (40)r=R11β11/ρ+εαn/β1αn/ρ+ε, ρ=ρα,β.Using the properties of functions α and β, (41)αxβ1cαx≃1+o1αx;for c>0, x→∞, and the properties of logarithm, we get (42)logannδΓδ+1Rn≤C1nβ1αn/ρ+ε,where C1 is a constant. Hence, (43)βC1nlogannδ/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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has (α,β)order (μ/(1μ)) for α(x)=logx and β(x)=x, where l is the positive integer.
Let f(t,z)=∑n=0∞anLn(δ,t)zn be analytic function of (α,β)order ρ=ρ(α,β) and write (45)Tα,β=limsupn→∞αnβn/lognδ/Γδ+1anRnρα,β.
Lemma 3.
Let f(t,z)=∑n=0∞anLn(δ,t)zn. For every 1<r<R, (46)σ¯1α,β=limsupr→RαlogM¯r,r,fβR/Rrρα,β;then (47)σα,β≤σ¯1α,β.
Proof.
Following the same reasoning as in the proof of Lemma 1, we obtain (48)αlogMr,r,fβR/Rrρα,β≤αlogM¯r′,r′,flog1r/RβR/Rr·Rρα,β·βR/Rr·Rρα,ββR/Rrρα,β.Applying the limit supremum as r→R, we get (49)σα,β≤σ¯1α,β.
Theorem 4.
Let f(t,z)=∑n=0∞anLn(δ,t)zn be of finite generalized (α,β)order ρ(α,β) and (50)Tα,β=limsupn→∞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 Theorem 2 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+εβnlognδ/Γδ+1anRnρα,β, T≡Tα,β<∞,logannδΓδ+1Rn≤nβ1αn/T¯1/ρ, T¯=T+ε,since (52)logannδΓδ+1rn≤nlogrR+logannδΓδ+1Rn.From inequality (51), we get (53)logannδΓδ+1rn≤nlogrR+nβ1αn/T¯1/ρ.For every 1<r<R and r sufficiently close to R, we put (54)φx,r=xlogrR+xβ1αx/T¯1/ρ.The maximum value of φ(x,r) is reached at (55)x=xr=α1T¯β1dlogβ1αx/T¯1/ρ/dlogxlogR/rρ.Using the relation (56)logrR=logrRR+1~rRR(as (rR)/R→0) and (57)dlogβ1αx/T¯1/ρdlogx≤ko,where ko is a positive constant, it gives (58)xr=1+o1α1T¯βRRrρ.Now, from relation (53), we have (59)logannδΓδ+1rn≤supφx,ror (60)logannδΓδ+1rn≤1+o1α1T¯βR/RrρR/Rr.Since R/(Rr)>1, it gives (61)logannδΓδ+1rn≤k1α1T¯βRRrρ.Then (62)logMr,r,f≤k1α1T¯βRRrρ,or (63)αlogMr,r,fβR/Rrρ≤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)logMr,r,f≤α1σ¯βRRrρ, σ¯=σ+ε.Now, using (51), we get (66)logannδΓδ+1Rn≤nlogrR+δlognlogΓδ+1+α1σ¯βRRrρ,or (67)logannδ/Γδ+1Rnn≤logrR+1nδlognlogΓδ+1+α1σ¯βRRrρ,or (68)1nlogannδΓδ+1Rn≤1+o1logRr,as n→∞. Now, using log(R/r)~(Rr)/r as r→R, we have (69)1nlogannδΓδ+1Rn≤1+o1logRr1.Set (70)rR=β1αn/σ¯1/ρ1+β1αn/σ¯1/ρ;in above inequality, we get (71)1nlogannδΓδ+1Rn≤1β1αn/σ¯1/ρor (72)αnβn/logannδ/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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, where 0<γ∗<1 and 0<δ∗<∞, and has (α,β)order γ∗/(1γ∗) and (α,β)type (1γ∗)(γ∗)γ∗/(1γ∗)(δ∗)1/(1γ∗) for α(x)=logx and β(x)=x.