a linear homogeneous partial differential equation with

Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2012, Article ID 609862, 10 pagesdoi:10.1155/2012/609862

Research ArticleA Linear Homogeneous Partial DifferentialEquation with Entire Solutions Represented byLaguerre Polynomials

Xin-Li Wang,1 Feng-Li Zhang,1 and Pei-Chu Hu2

1 College of Science, University of Shanghai for Science and Technology, Shanghai 200093, China2 Department of Mathematics, Shandong University, Shandong, Jinan 250100, China

Correspondence should be addressed to Xin-Li Wang, [email protected]

Received 11 November 2011; Revised 19 March 2012; Accepted 19 March 2012

Academic Editor: Agacik Zafer

Copyright q 2012 Xin-Li Wang et al. This is an open access article distributed under the CreativeCommons Attribution License, which permits unrestricted use, distribution, and reproduction inany medium, provided the original work is properly cited.

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

1. Introduction and Main Results

The existence and behavior of global meromorphic solutions of homogeneous linear partialdifferential equations of the second order

a0∂2u

∂t2+ 2a1

∂2u

∂t∂z+ a2

∂2u

∂z2+ a3

∂u

∂t+ a4

∂u

∂z+ a6u = 0, (1.1)

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)

t2∂2u

∂t2− z2

∂2u

∂z2+ (2t + 2)

∂u

∂t− 2z

∂u

∂z= 0, (1.2)

t2∂2u

∂t2− z2

∂2u

∂z2+ t

∂u

∂t− z

∂u

∂z+ t2u = 0 (1.3)

2 Abstract and Applied Analysis

and showed that the solutions of (1.2) and (1.3) are closely related to Bessel functionsand Bessel polynomials, respectively. Hu and Li [3] studied meromorphic solutions ofhomogeneous linear partial differential equations of the second order in two independentcomplex variables:

(1 − t2

)∂2u∂t2

+ z2∂2u

∂z2− {α − β +

(α + β + 2

)t}∂u∂t

+(α + β + 2

)z∂u

∂z= 0, (1.4)

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)

t∂2u

∂t2+ (α + 1 − t)

∂u

∂t+ z

∂u

∂z= 0 (1.5)

for a real α > 0.Wewill characterize the entire solutions of (1.5), which are related to Laguerrepolynomials. Further, the formulae of the order and type of the solutions are obtained.

It is well known that the Laguerre polynomials are defined by

Ln(α, t) =n∑

k=0

(n + αn − k

)(−t)kk!

, (1.6)

which are solutions of the following ordinary differential equations (ODE):

td2ω

dt2+ (α + 1 − t)

dω

dt+ nω = 0. (1.7)

Moreover, Hu [9] pointed out that the generating function of Ln(α, t)

F(α, t, z) = (1 − z)−α−1e−tz/(1−z) =∞∑n=0

Ln(α, t)zn (1.8)

is a solution of the PDE (1.5). Based on the methods from Hu and Yang [2], we get thefollowing 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

f(t, z) =∞∑n=0

cnLn(α, t)zn (1.9)

Abstract and Applied Analysis 3

such that

lim supn→∞

n

√|cn| = 0. (1.10)

If f(t, z) is an entire function on C2, set

M(r, f)= max

|t|≤r,|z|≤r

∣∣f(t, z)∣∣, (1.11)

we define its order by

ord(f)= lim sup

r→∞

log+log+M(r, f)

log r, (1.12)

where

log+x =

{logx, if x ≥ 1,0, if x < 1.

(1.13)

Theorem 1.2. If f(t, z) is defined by (1.9) and (1.10), then

ρ ≤ ord(f) ≤ max

(1, ρ), (1.14)

where

ρ = lim supn→∞

logn

log(1/ n√|cn|) . (1.15)

Valiron [10] showed that each entire solution of a homogeneous linear ODE withpolynomial coefficients was of finite order. By studying (1.2) and (1.3), Hu and Yang showedthat Valiron’s theorem was not true for general partial differential equations. Here by usingTheorems 1.1 and 1.2, we can construct entire solution of (1.5)with arbitrary order ρ (ρ ≥ 1).

If 0 < λ = ord(f) < ∞, we define the type of f by

typ(f)= lim sup

r→∞

log+M(r, f)

rλ. (1.16)

Theorem 1.3. If f(t, z) is defined by (1.9) and (1.10), and 1 < λ = ord(f) < ∞, then the typeσ = typ(f) satisfies

eσλ = lim supn→∞

nn

√|cn|λ. (1.17)


Lindelof-Pringsheim theorem [11] gave the expression of order and type for onecomplex variable entire function, and for two variable entire function the formulae of orderand typewere obtained by Bose and Sharma in [12]. Hu and Yang [2] established an analogueof Lindelof-Pringsheim theorem for the entire solution of PDE (1.2). But from Theorems 1.2and 1.3, we find that the analogue theorem for the entire solution of (1.5) is different from theresults due to Hu and Yang.

2. An Estimate of Laguerre Polynomials

Before we prove our theorems, we give an upper bound of Ln(α, t), which will play animportant role in this paper. The following asymptotic properties of Ln(α, t) can be foundin [13]:

(a)

Ln(α, t) =1

2√πet/2(−t)−α/2−1/4nα/2−1/4e2

√−nt(1 +O

(n−1/2

))(n −→ ∞) (2.1)

holds for t in the complex plane cut along the positive real semiaxis; thus, for |t| ≤ r, we obtainthat

|Ln(α, t)| ≤ nα/2−1/4er/2r−α/2−1/4e2√nr (2.2)

holds when n is large enough.(b)

Ln(α, t)nα/2

= et/2t−α/2Jα(2√nt)+O(n−3/4

)(n −→ ∞) (2.3)

holds uniformly on compact subsets of (0,+∞), where Jα is the Bessel function and

Jα(2√nt)=

2√πΓ(α + 1/2)

(√nt)α ∫π/2

0cos(√

nt cosx)sin2αxdx, (2.4)

combining with (2.3), for |t| ≤ r we can deduce that

|Ln(α, t)| ≤ πnα

√πΓ(α + 1/2)

et/2 ≤ πnα

√πΓ(α + 1/2)

er/2 ≤ nα/2−1/4er/2r−α/2−1/4e2√nr (2.5)

holds when n is large enough. Then (2.2) and (2.5) imply

M(r, Ln) ≤ nα/2−1/4er/2r−α/2−1/4e2√nr , (2.6)

where

M(r, Ln) = max|t|≤r

|Ln(α, t)|. (2.7)


3. Proof of Theorem 1.1

Assuming that u = f(t, z) is an entire solution on C2 satisfying (1.5), we have Taylor

expansion

f(t, z) =∞∑n=0

wn(t)n!

zn, (3.1)

where

wn(t) =∂nf

∂zn(t, 0). (3.2)

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

Xn(α, t) = qLn(α, t) log t +∞∑i=0

piti, (3.3)

where q(/= 0), pi are constants.So there exist cn and bn satisfying

wn(t) = n!cnLn(α, t) + bnXn(α, t). (3.4)

Because of the singularity of Xn(α, t) at t = 0, we obtain bn = 0. That shows

f(t, z) =∞∑n=0

cnLn(α, t)zn. (3.5)

Now we need to estimate the terms of cn. Since

f(0, z) =∞∑n=0

cnLn(α, 0)zn (3.6)

is an entire function, we have

lim supn→∞

n

√|cn|Ln(α, 0) = 0. (3.7)

Since

Ln(α, 0) =(n + αn

)≈ nα

Γ(α + 1), (3.8)

we easily get

lim supn→∞

n

√|cn| = 0. (3.9)


Conversely, the relations (1.7), (1.9), and (1.10) imply that

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

(3.10)

holds for all (t, z) ∈ C2. Since (2.6) implies

lim supn→∞

n

√M(r, Ln) ≤ 1, (3.11)

we have

lim supn→∞

n

√|cnLn(α, t)| ≤ lim sup

n→∞n

√|cn|M(r, Ln) ≤ lim sup

n→∞n

√|cn|. (3.12)

Combining (1.10), (3.10)with (3.12), we can get that u = f(t, z) is obviously an entire solutionof (1.5) on C

2.


Firstly, we prove ρ ≤ ord(f). If ρ = 0, the result is trivial. Now we assume 0 < ρ ≤ ∞ andprove ord(f) ≥ k1 for any 0 < k1 < ρ. The relation (1.15) implies that there exists a sequencenj → ∞ such that

nj lognj ≥ k1 log1∣∣∣cnj

∣∣∣. (4.1)

By using Cauchy’s inequality of holomorphic functions, we have

∣∣∣∣∂nf

∂zn(0, 0)

∣∣∣∣ ≤ n!r−nM(r, f), (4.2)

together with the formula of the coefficients of the Taylor expansion

∂nf

∂zn(0, 0) = cnLn(α, 0)n!, (4.3)

we obtain M(r, f) ≥ |cnLn(α, 0)|rn. Since |Ln(α, 0)| ≥ 1, we have

|cn|rn ≤ M(r, f), (4.4)


then

logM(r, f) ≥ log|cn| + n log r ≥ nj

(log r − 1

k1lognj

). (4.5)

Putting rj = (enj)1/k1 , we have

logM(rj , f) ≥

rk1j

ek1, (4.6)

which means ord(f) ≥ k1. Then we can get ord(f) ≥ ρ.Next, wewill prove ord(f) ≤ max(1, ρ). Set ρ′ = max(1, ρ). The result is easy for ρ′ = ∞;

then we assume ρ′ < ∞. For any ε > 0, (1.15) implies that there exists n0 > 0, when n > n0, wehave

|cn| < n−n(1+2ε)/ρ′(1+3ε) = n−n(1+2ε)/k2 , (4.7)

where k2 = ρ′(1 + 3ε) > 1. For any α ≥ 0, there exists n1 > n0 such that when n > n1,

n(α/2−1/4) < nn(ε/k2), (4.8)

combining with (2.6) and (4.7), we get

M(r, f) ≤

∞∑n=0

|cn|M(r, Ln)rn

≤ Cr2n1 + C∞∑

n>n1

|cn|nα/2−1/4er/2e2√nrrn

≤ Cr2n1 + Cer/2∞∑

n>n1

n−(1+ε)n/k2e2√nrrn,

(4.9)

where C is a constant but not necessary to be the same every time.Set m1(r) = ((1/ε)(2

√r/ log r))2, which means that e2

√nr < rεn for n > m1(r). Further

set m2(r) = (2r)k2 , which yields that when n ≥ m2(r),

(n−1/k2r

)(1+ε)n ≤(12

)(1+ε)n

<

(12

)n

. (4.10)

Obviously, we can choose r0 > 0 such that m2(r) > m1(r) for r > r0. Then

∑n>m2(r)

n−(1+ε)n/k2e2√nrrn ≤

∞∑n>m2(r)

n−(1+ε)n/k2r(1+ε)n ≤∞∑

n>m2(r)

(12

)n

≤ 1. (4.11)


We also have

∑n1<n≤m2(r)

n−(1+ε)n/k2e2√nrrn ≤

∑n1<n≤m2(r)

n−(1+ε)n/k2e2√

m2(r)rrm2(r)

≤∑

n1<n≤m2(r)

n−(1+ε)n/k2rm2(r)rεm2(r)

≤ r(1+ε)m2(r)∑

n1<n≤m2(r)

n−(1+ε)n/k2

≤ Cr(1+ε)m2(r)

= Cr(1+ε)(2r)k2.

(4.12)

Therefore, when |t| = r ≥ r0, we have

M(r, f) ≤ Cr2n1 + Cer/2 + Cer/2r(1+ε)(2r)

k2, (4.13)

which means ord(f) ≤ k2. Hence ord(f) ≤ ρ′ = max(1, ρ) follows by letting ε → 0.


Set

κ = lim supn→∞

nn

√|cn|λ. (5.1)

At first, we prove eλσ ≥ κ. The result is trivial for κ = 0, we assume 0 < κ ≤ ∞ and take εwith 0 < ε < κ, set

k3 =

⎧⎨⎩κ − ε, if κ < ∞,1ε, if κ = ∞.

(5.2)

Equation (5.1) implies that there exists a sequence nj → ∞ satisfying

∣∣∣cnj

∣∣∣ >(

k3nj

)nj/λ

, (5.3)

combining with (4.4), we can deduce that

M(r, f) ≥∣∣∣cnj

∣∣∣rnj ≥(

k3nj

)nj/λ

rnj =

(k3nj

rλ)nj/λ

. (5.4)

Taking rλj = enj/k3, we get M(rj , f) > ek3rλj /eλ, which yields σ ≥ k3/eλ, so eλσ ≥ κ.


Next, we prove eλσ ≤ κ. We may assume κ < ∞. Equation (5.1) implies that for anyε > 0, there exists n0 > 0, such that when n > n0,

|cn | <(κ + (ε/2)

n

)n/λ

. (5.5)

For any α ≥ 0, we choose n1(> n0) such that when n > n1,

nα/2−1/4 <(

κ + ε

κ + (ε/2)

)n/λ

, (5.6)

combining with (2.6), we have

M(r, f) ≤

∞∑n=0

|cn|M(r, Ln)rn

≤ Cr2n1 + C∞∑

n>n1

|cn|nα/2−1/4er/2e2√nrrn

≤ Cr2n1 + Cer/2∞∑

n>n1

(κ + ε

n

)n/λ

e2√nrrn.

(5.7)

Set m3(r) = 16rλ2, when n > m3(r), we deduce e2√nr < en/2λ. Set m4(r) = 2(κ + ε)rλ,

it is obvious that (κ + ε)rλ/n < 1/2 for n > m4(r). Since λ > 1, there exists r1 such that whenr > r1,

m4(r) = 2(κ + ε)rλ > 16rλ2 = m3(r). (5.8)

Then

∞∑n>m4(r)

(κ + ε

n

)n/λ

e2√nrrn =

∞∑n>m4(r)

((κ + ε)rλ

n

)n/λ

e2√nr ≤

∞∑n>m4(r)

(e4

)n/2λ≤ C. (5.9)

We note that for a > 0, b > 0, maxx>0(a/x)x/b = ea/eb, then we have

(κ + ε

n

)n/λ

rn =

((κ + ε)rλ

n

)n/λ

≤ e(κ+ε)rλ/eλ. (5.10)

This shows

∑n1<n≤m4(r)

(κ + ε

n

)n/λ

e2√nrrn ≤ m4(r)e2

√m4(r)re(κ+ε)r

λ/eλ

≤ 2(κ + ε)rλe2√

2(κ+ε)rλ+1e(κ+ε)rλ/eλ.

(5.11)


Therefore when |t| = r ≥ r1,

M(r, f) ≤ Cer/2(κ + ε)rλe2

√2(κ+ε)rλ+1e(κ+ε)r

λ/eλ + Cr2n1 + Cer/2. (5.12)

Together with λ > 1 and the definition of type, we can get σ ≤ (κ+ε)/eλ, which yields eλσ ≤ κby letting ε → 0.

Acknowledgments

The authors sincerely thank the reviewers for their valuable suggestions and useful com-ments that have led to the present improved version of the original paper. The first authorwas partially supported by Natural Science Foundation of China (11001057), and the thirdauthor was partially supported by Natural Science Foundation of Shandong Province.

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