# A Linear Homogeneous Partial Differential Equation with Entire Solutions Represented by Laguerre Polynomials

Authors: Xin-Li Wang; Feng-Li Zhang; Pei-Chu Hu
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We study a homogeneous partial differential equation and get its entire solutions represented in convergent series of Laguerre polynomials. Moreover, the formulae of the order and type of the solutions are established.


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## Body

## 1. Introduction and Main Results

The existence and behavior of global meromorphic solutions of homogeneous linear partial differential equations of the second order(1.1)a0∂2u∂t2+2a1∂2u∂t∂z+a2∂2u∂z2+a3∂u∂t+a4∂u∂z+a6u=0,
where ak=ak(t,z) are polynomials for (t,z)∈ℂ2, have been studied by Hu and Yang [1]. Specially, in [1, 2], they have studied the following cases of  (1.1)(1.2)t2∂2u∂t2-z2∂2u∂z2+(2t+2)∂u∂t-2z∂u∂z=0,(1.3)t2∂2u∂t2-z2∂2u∂z2+t∂u∂t-z∂u∂z+t2u=0
and showed that the solutions of (1.2) and (1.3) are closely related to Bessel functions and Bessel polynomials, respectively. Hu and Li [3] studied meromorphic solutions of homogeneous linear partial differential equations of the second order in two independent complex variables:(1.4)(1-t2)∂2u∂t2+z2∂2u∂z2-{α-β+(α+β+2)t}∂u∂t+(α+β+2)z∂u∂z=0,
where α,β∈ℂ. Equation (1.4) has a lot of entire solutions on ℂ2 represented by Jacobian polynomials. Global solutions of some first-order partial differential equations (or system) were studied by Berenstein and Li [4], Hu and Yang [5], Hu and Li [6], Li [7], Li and Saleeby [8], and so on.In this paper, we concentrate on the following partial differential equation (PDE)(1.5)t∂2u∂t2+(α+1-t)∂u∂t+z∂u∂z=0
for a real α>0. We will characterize the entire solutions of (1.5), which are related to Laguerre polynomials. Further, the formulae of the order and type of the solutions are obtained.It is well known that the Laguerre polynomials are defined by(1.6)Ln(α,t)=∑k=0n(n+αn-k)(-t)kk!,
which are solutions of the following ordinary differential equations (ODE):(1.7)td2ωdt2+(α+1-t)dωdt+nω=0.
Moreover, Hu [9] pointed out that the generating function of Ln(α,t)(1.8)F(α,t,z)=(1-z)-α-1e-tz/(1-z)=∑n=0∞Ln(α,t)zn
is a solution of the PDE (1.5). Based on the methods from Hu and Yang [2], we get the following results.Theorem 1.1.
The partial differential equation (1.5) has an entire solution u=f(t,z) on ℂ2, if and only if u=f(t,z) has a series expansion
(1.9)f(t,z)=∑n=0∞cnLn(α,t)zn
such that
(1.10)limsup⁡n→∞|cn|n=0.Iff(t,z) is an entire function on ℂ2, set
(1.11)M(r,f)=max⁡|t|≤r,|z|≤r|f(t,z)|,
we define its order by
(1.12)ord(f)=limsup⁡r→∞log⁡+log⁡+M(r,f)log⁡r,
where(1.13)log⁡+x={log⁡x,ifx≥1,0,ifx<1.Theorem 1.2.
Iff(t,z) is defined by (1.9) and (1.10), then
(1.14)ρ≤ord(f)≤max⁡(1,ρ),
where
(1.15)ρ=limsup⁡n→∞log⁡nlog⁡⁡(1/|cn|n).Valiron [10] showed that each entire solution of a homogeneous linear ODE with polynomial coefficients was of finite order. By studying (1.2) and (1.3), Hu and Yang showed that Valiron’s theorem was not true for general partial differential equations. Here by using Theorems 1.1 and 1.2, we can construct entire solution of (1.5) with arbitrary order ρ(ρ≥1).If0<λ=ord(f)<∞, we define the type of f by(1.16)typ(f)=limsup⁡r→∞log⁡+M(r,f)rλ.Theorem 1.3.
Iff(t,z) is defined by (1.9) and (1.10), and 1<λ=ord(f)<∞, then the type σ=typ(f) satisfies
(1.17)eσλ=limsup⁡n→∞n|cn|λn.Lindelöf-Pringsheim theorem [11] gave the expression of order and type for one complex variable entire function, and for two variable entire function the formulae of order and type were obtained by Bose and Sharma in [12]. Hu and Yang [2] established an analogue of Lindelöf-Pringsheim theorem for the entire solution of PDE (1.2). But from Theorems 1.2 and 1.3, we find that the analogue theorem for the entire solution of (1.5) is different from the results due to Hu and Yang.


## 2. An Estimate of Laguerre Polynomials

Before we prove our theorems, we give an upper bound ofLn(α,t), which will play an important role in this paper. The following asymptotic properties of Ln(α,t) can be found in [13]:(a)(2.1)Ln(α,t)=12πet/2(-t)-α/2-1/4nα/2-1/4e2-nt(1+O(n-1/2))(n⟶∞)
holds for t in the complex plane cut along the positive real semiaxis; thus, for |t|≤r, we obtain that
(2.2)|Ln(α,t)|≤nα/2-1/4er/2r-α/2-1/4e2nr
holds when n is large enough.(b)(2.3)Ln(α,t)nα/2=et/2t-α/2Jα(2nt)+O(n-3/4)(n⟶∞)
holds uniformly on compact subsets of (0,+∞), where Jα is the Bessel function and
(2.4)Jα(2nt)=2πΓ(α+1/2)(nt)α∫0π/2cos⁡(ntcos⁡x)sin⁡2αxdx,
combining with (2.3), for |t|≤r we can deduce that
(2.5)|Ln(α,t)|≤πnαπΓ(α+1/2)et/2≤πnαπΓ(α+1/2)er/2≤nα/2-1/4er/2r-α/2-1/4e2nr
holds when n is large enough. Then (2.2) and (2.5) imply
(2.6)M(r,Ln)≤nα/2-1/4er/2r-α/2-1/4e2nr,
where
(2.7)M(r,Ln)=max⁡|t|≤r|Ln(α,t)|.


## 3. Proof of Theorem1.1

Assuming thatu=f(t,z) is an entire solution on ℂ2 satisfying (1.5), we have Taylor expansion(3.1)f(t,z)=∑n=0∞wn(t)n!zn,
where(3.2)wn(t)=∂nf∂zn(t,0).
Hence wn(t) is an entire solution of (1.7).By the method of Frobenius (see [14]), we can get a second independent solution Xn(α,t) of (1.7) which is(3.3)Xn(α,t)=qLn(α,t)log⁡t+∑i=0∞piti,
where q(≠0),pi are constants.So there existcn and bn satisfying(3.4)wn(t)=n!cnLn(α,t)+bnXn(α,t).
Because of the singularity of Xn(α,t) at t=0, we obtain bn=0. That shows(3.5)f(t,z)=∑n=0∞cnLn(α,t)zn.Now we need to estimate the terms ofcn. Since(3.6)f(0,z)=∑n=0∞cnLn(α,0)zn
is an entire function, we have(3.7)limsup⁡n→∞|cn|Ln(α,0)n=0.
Since(3.8)Ln(α,0)=(n+αn)≈nαΓ(α+1),
we easily get(3.9)limsup⁡n→∞|cn|n=0.Conversely, the relations (1.7), (1.9), and (1.10) imply that(3.10)t∂2f∂t2+(α+1-t)∂f∂t+z∂f∂z=∑n=0∞cn{td2Ln(α,t)dt2+(α+1-t)dLn(α,t)dt+nLn(α,t)}zn=0
holds for all (t,z)∈ℂ2. Since (2.6) implies(3.11)limsup⁡n→∞M(r,Ln)n≤1,
we have(3.12)limsup⁡n→∞|cnLn(α,t)|n≤limsup⁡n→∞|cn|M(r,Ln)n≤limsup⁡n→∞|cn|n.
Combining (1.10), (3.10) with (3.12), we can get that u=f(t,z) is obviously an entire solution of (1.5) on ℂ2.


## 4. Proof of Theorem1.2

Firstly, we proveρ≤ord(f). If ρ=0, the result is trivial. Now we assume 0<ρ≤∞ and prove ord(f)≥k1 for any 0<k1<ρ. The relation (1.15) implies that there exists a sequence nj→∞ such that(4.1)njlog⁡nj≥k1log⁡1|cnj|.
By using Cauchy’s inequality of holomorphic functions, we have(4.2)|∂nf∂zn(0,0)|≤n!r-nM(r,f),
together with the formula of the coefficients of the Taylor expansion(4.3)∂nf∂zn(0,0)=cnLn(α,0)n!,
we obtain M(r,f)≥|cnLn(α,0)|rn. Since |Ln(α,0)|≥1, we have(4.4)|cn|rn≤M(r,f),
then(4.5)log⁡M(r,f)≥log⁡|cn|+nlog⁡r≥nj(log⁡r-1k1log⁡nj).
Putting rj=(enj)1/k1, we have(4.6)log⁡M(rj,f)≥rjk1ek1,
which means ord(f)≥k1. Then we can get ord(f)≥ρ.Next, we will proveord(f)≤max(1,ρ). Set ρ′=max(1,ρ). The result is easy for ρ′=∞; then we assume ρ′<∞. For any ɛ>0, (1.15) implies that there exists n0>0, when n>n0, we have(4.7)|cn|<n-n(1+2ɛ)/ρ′(1+3ɛ)=n-n(1+2ɛ)/k2,
where k2=ρ′(1+3ɛ)>1. For any α≥0, there exists n1>n0 such that when n>n1,
(4.8)n(α/2-1/4)<nn(ɛ/k2),
combining with (2.6) and (4.7), we get
(4.9)M(r,f)≤∑n=0∞|cn|M(r,Ln)rn≤Cr2n1+C∑n>n1∞|cn|nα/2-1/4er/2e2nrrn≤Cr2n1+Cer/2∑n>n1∞n-(1+ɛ)n/k2e2nrrn,
where C is a constant but not necessary to be the same every time.Setm1(r)=((1/ɛ)(2r/logr))2, which means that e2nr<rɛn for n>m1(r). Further set m2(r)=(2r)k2, which yields that when n≥m2(r), (4.10)(n-1/k2r)(1+ɛ)n≤(12)(1+ɛ)n<(12)n.
Obviously, we can choose r0>0 such that m2(r)>m1(r) for r>r0. Then(4.11)∑n>m2(r)n-(1+ɛ)n/k2e2nrrn≤∑n>m2(r)∞n-(1+ɛ)n/k2r(1+ɛ)n≤∑n>m2(r)∞(12)n≤1.
We also have(4.12)∑n1<n≤m2(r)n-(1+ɛ)n/k2e2nrrn≤∑n1<n≤m2(r)n-(1+ɛ)n/k2e2m2(r)rrm2(r)≤∑n1<n≤m2(r)n-(1+ɛ)n/k2rm2(r)rɛm2(r)≤r(1+ɛ)m2(r)∑n1<n≤m2(r)n-(1+ɛ)n/k2≤Cr(1+ɛ)m2(r)=Cr(1+ɛ)(2r)k2.
Therefore, when |t|=r≥r0, we have(4.13)M(r,f)≤Cr2n1+Cer/2+Cer/2r(1+ɛ)(2r)k2,
which means ord(f)≤k2. Hence ord(f)≤ρ′=max(1,ρ) follows by letting ɛ→0.


## 5. Proof of Theorem1.3

Set(5.1)κ=limsup⁡n→∞n|cn|λn.
At first, we prove eλσ≥κ. The result is trivial for κ=0, we assume 0<κ≤∞ and take ɛ with 0<ɛ<κ, set(5.2)k3={κ-ɛ,ifκ<∞,1ɛ,ifκ=∞.
Equation (5.1) implies that there exists a sequence nj→∞ satisfying(5.3)|cnj|>(k3nj)nj/λ,
combining with (4.4), we can deduce that(5.4)M(r,f)≥|cnj|rnj≥(k3nj)nj/λrnj=(k3njrλ)nj/λ.
Taking rjλ=enj/k3, we get M(rj,f)>ek3rjλ/eλ, which yields σ≥k3/eλ, so eλσ≥κ.Next, we proveeλσ≤κ. We may assume κ<∞. Equation (5.1) implies that for any ɛ>0, there exists n0>0, such that when n>n0,(5.5)|cn|<(κ+(ɛ/2)n)n/λ.
For any α≥0, we choose n1(>n0) such that when n>n1,(5.6)nα/2-1/4<(κ+ɛκ+(ɛ/2))n/λ,
combining with (2.6), we have(5.7)M(r,f)≤∑n=0∞|cn|M(r,Ln)rn≤Cr2n1+C∑n>n1∞|cn|nα/2-1/4er/2e2nrrn≤Cr2n1+Cer/2∑n>n1∞(κ+ɛn)n/λe2nrrn.Setm3(r)=16rλ2, when n>m3(r), we deduce e2nr<en/2λ. Set m4(r)=2(κ+ɛ)rλ, it is obvious that (κ+ɛ)rλ/n<1/2 for n>m4(r). Since λ>1, there exists r1 such that when r>r1,(5.8)m4(r)=2(κ+ɛ)rλ>16rλ2=m3(r).
Then(5.9)∑n>m4(r)∞(κ+ɛn)n/λe2nrrn=∑n>m4(r)∞((κ+ɛ)rλn)n/λe2nr≤∑n>m4(r)∞(e4)n/2λ≤C.
We note that for a>0, b>0, maxx>0(a/x)x/b=ea/eb, then we have(5.10)(κ+ɛn)n/λrn=((κ+ɛ)rλn)n/λ≤e(κ+ɛ)rλ/eλ.
This shows(5.11)∑n1<n≤m4(r)(κ+ɛn)n/λe2nrrn≤m4(r)e2m4(r)re(κ+ɛ)rλ/eλ≤2(κ+ɛ)rλe22(κ+ɛ)rλ+1e(κ+ɛ)rλ/eλ.Therefore when|t|=r≥r1,(5.12)M(r,f)≤Cer/2(κ+ɛ)rλe22(κ+ɛ)rλ+1e(κ+ɛ)rλ/eλ+Cr2n1+Cer/2.
Together with λ>1 and the definition of type, we can get σ≤(κ+ɛ)/eλ, which yields eλσ≤κ by letting ɛ→0.



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*Source: 609862-2012-05-20.xml*