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Advanced Nonlinear Studies

Editor-in-Chief: Ahmad, Shair


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Volume 16, Issue 1

Issues

Estimates for Eigenvalues of Poly-Harmonic Operators

Guoxin Wei / Lingzhong Zeng
Published Online: 2016-02-01 | DOI: https://doi.org/10.1515/ans-2015-5020

Abstract

In this paper, we study the eigenvalues of the poly-Laplacian with arbitrary order on a bounded domain in an n-dimensional Euclidean space and we obtain a lower bound which generalizes the results due to Cheng and Wei [5] and gives an improvement of the results due to Cheng, Qi and Wei [3].

Keywords: Eigenvalue Problem; Lower Bound for Eigenvalues; Poly-Laplacian with Arbitrary Order

MSC 2010: 35P15

1 Introduction

Let Ω be a bounded domain with piecewise smooth boundary Ω in an n-dimensional Euclidean space n. Let λi be the i-th eigenvalue of the Dirichlet eigenvalue problem of the poly-Laplacian with arbitrary order

{(-Δ)lu=λuin Ω,u=uν==l-1uνl-1=0on Ω,(1.1)

where l is a positive integer, Δ is the Laplacian in n, and ν denotes the outward unit normal vector field of the boundary Ω. It is well known that the spectrum of this eigenvalue problem is real and discrete, that is,

0<λ1λ2λ3+,

where each λi has finite multiplicity and is repeated according to its multiplicity. Let V(Ω) denote the volume of Ω and let Bn denote the volume of the unit ball in n.

When l=1, the eigenvalue problem (1.1) is called a fixed membrane problem. In this case, one has Weyl’s asymptotic formula

λk4π2(BnV(Ω))2nk2n,k+.

From the above asymptotic formula, one can derive the formula

1ki=1kλinn+24π2(BnV(Ω))2nk2n,k+.(1.2)

Pólya [11] proved that

λk4π2(BnV(Ω))2nk2n,k=1,2,,

if Ω is a tiling domain in n . Furthermore, he proposed the following conjecture.

(Pólya)

If Ω is a bounded domain in n, then the k-th eigenvalue λk of the fixed membrane problem satisfies

λk4π2(BnV(Ω))2nk2n,k=1,2,.

Berezin [2] and Lieb [8] gave a partial solution for the conjecture of Pólya. In particular, Li and Yau [7] proved that

1ki=1kλinn+24π2(BnV(Ω))2nk2n,k=1,2,.(1.3)

The formula (1.2) shows that the result of Li and Yau is sharp in the sense of average. From (1.3) one can infer that

λknn+24π2(BnV(Ω))2nk2n,k=1,2,,

which gives a partial solution for the conjecture of Pólya with a factor nn+2. Recently, Melas [9] has improved (1.3) to the estimate

1ki=1kλinn+24π2(BnV(Ω))2nk2n+124(n+2)V(Ω)I(Ω),k=1,2,,

where

I(Ω)=minanΩ|x-a|2𝑑x

is called the moment of inertia of Ω.

When l=2, the eigenvalue problem (1.1) is called a clamped plate problem. For the eigenvalues of the clamped plate problem, Agmon [1] and Pleijel [10] obtained

λk16π4(BnV(Ω))4nk4n,k+.(1.4)

From (1.4) one can obtain

1ki=1kλinn+416π4(BnV(Ω))4nk4n,k+.(1.5)

Furthermore, Levine and Protter [6] proved that the eigenvalues of the clamped plate problem satisfy the inequality

1ki=1kλinn+416π4(BnV(Ω))4nk4n.(1.6)

The formula (1.5) shows that the coefficient of k4n is the best possible constant. By adding to its right-hand side two terms of lower order in k, Cheng and Wei [4] obtained the estimate

1ki=1kλinn+416π4(BnV(Ω))4nk4n+(n+212n(n+4)-11152n2(n+4))4π2(BnV(Ω))2nnn+2V(Ω)I(Ω)k2n+(1576n(n+4)-127648n2(n+2)(n+4))(V(Ω)I(Ω))2,(1.7)

which is an improvement of (1.6). Very recently, Cheng and Wei [5] have improved the estimate (1.7) to the estimate

1ki=1kλinn+416π4(BnV(Ω))4nk4n+n+212n(n+4)4π2(BnV(Ω))2nnn+2V(Ω)I(Ω)k2n+(n+2)21152n(n+4)2(V(Ω)I(Ω))2.

When l is arbitrary, Levine and Protter [6] proved the estimate

1ki=1kλinn+2lπ2l(BnV(Ω))2lnk2ln,k=1,2,,(1.8)

which implies that

λknn+2lπ2l(BnV(Ω))2lnk2ln,k=1,2,.

By adding l terms of lower order of k2ln to its right-hand side, Cheng, Qi and Wei [3] obtained a sharper result than (1.8), that is,

1ki=1kλinn+2l(2π)2l(BnV(Ω))2lnk2ln+n(n+2l)p=1ll+1-p(24)pn(n+2p-2)(2π)2(l-p)(BnV(Ω))2(l-p)n(V(Ω)I(Ω))pk2(l-p)n.(1.9)

In this paper, we study the eigenvalues of the Dirichlet eigenvalue problem (1.1) of a Laplacian with arbitrary order and prove the following result.

Let Ω be a bounded domain in an n-dimensional Euclidean space n. Assume that l2 and λi is the i-th eigenvalue of the eigenvalue problem (1.1). Then, the eigenvalues satisfy

1kj=1kλjnn+2l(2π)2l(BnV(Ω))2lnk2ln+l24(n+2l)(2π)2(l-1)(BnV(Ω))2(l-1)nV(Ω)I(Ω)k2(l-1)n+l(n+2(l-1))22304n(n+2l)2(2π)2(l-2)(BnV(Ω))2(l-2)n(V(Ω)I(Ω))2k2(l-2)n.(1.10)

When l=2, Theorem 1.2 reduces to the result of Cheng and Wei [5].

When l2, we give an important improvement of the result in (1.9) due to Cheng, Qi and Wei [3], since the inequality (1.10) is sharper than (1.9). We will give a proof of this fact in Section 2.

2 Proofs of Theorem 1.2 and Remark 1.4

In this section, we give the proof of Theorem 1.2. At first, we need the following key lemma which will play an important role in the proof of Theorem 1.2. Its proof will be given in Appendix A.

Let b2 be a positive real number and μ>0. If ψ:[0,+)[0,+) is a decreasing function such that

-μψ(s)0

and

A:=0sb-1ψ(s)𝑑s>0,

then, for any positive integer l2, we have

0sb+2l-1ψ(s)𝑑s1b+2l(bA)b+2lbψ(0)-2lb+l6b(b+2l)μ2(bA)b+2(l-1)bψ(0)2b-2l+2b+l(b+2(l-1))2144b2(b+2l)2μ4(bA)b+2l-4bψ(0)4b-2l+4b.

Proof of Theorem 1.2.

In this proof, we will use the same notation as in [3]. Let φ^j(z) be the Fourier transform of the trial function φj(x) defined as

φj(x)={uj(x),xΩ,0,xnΩ,

where uj(x) is an orthonormal eigenfunction corresponding to the eigenvalue λj, f(z):=j=1k|φ^j(z)|2, and f is the symmetric decreasing rearrangement of f. We assume that A=knBn, b=n, and ψ(s)=ϕ(s), where ϕ:[0,+)[0,(2π)-nV(Ω)] is a non-increasing function of |x| and ϕ(x) is defined by ϕ(|x|):=f(x). Then, from Lemma 2.1 we can obtain that

j=1kλjnBn0sn+2l-1ϕ(s)𝑑snBn(kBn)n+2lnn+2lϕ(0)-2ln+lBn(kBn)n+2(l-1)n6(n+2l)μ2ϕ(0)2n-2l+2n+l(n+2(l-1))2Bn(kBn)n+2l-4n144n(n+2l)2μ4ϕ(0)4n-2l+4n.(2.1)

Now, we define a function ξ(t) as

ξ(t)=nBnn+2l(kBn)n+2lnt-2ln+lBn6(n+2l)μ2(kBn)n+2(l-1)nt2n-2l+2n+l(n+2(l-1))2Bn144n(n+2l)2μ4(kBn)n+2l-4nt4n-2l+4n.(2.2)

Here, we assume that ln+1. The other cases (i.e., n+1<l<2(n+1), l2(n+1)) can be discussed by using a similar method. After differentiating (2.2) with respect to the variable t, we derive

ξ(t)=Bnn+2l(kBn)n+2lnt-2ln-1[-2l+l(2n-2l+2)6nμ2(kBn)-2nt2n+2n+l(4n-2l+4)(n+2(l-1))2144n2(n+2l)μ4(kBn)-4nt4n+4n].(2.3)

Putting

ζ(t)=ξ(t)n+2lBn(kBn)-n+2lnt2ln+1

and noticing that μ(2π)-nBn-1nV(Ω)n+1n, from (2.3) we have

ζ(t)=-2l+l(2n-2l+2)6nμ2(kBn)-2nt2n+2n+l(4n-2l+4)(n+2(l-1))2144n2(n+2l)μ4(kBn)-4nt4n+4n-2l+l(2n-2l+2)6n(2π)-2nBn-2nV(Ω)2(n+1)n(kBn)-2nt2n+2n+l(4n-2l+4)(n+2(l-1))2144n2(n+2l)(2π)-4nBn-4nV(Ω)4(n+1)n(kBn)-4nt4n+4n.(2.4)

Since the right-hand side of (2.4) is an increasing function of t, if the right-hand side of (2.4) is not larger than 0 at t=(2π)-nV(Ω), that is,

ζ(t)-2l+l(2n-2l+2)6nk-2nBn4n(2π)2+l(4n-2l+4)(n+2(l-1))2144n2(n+2l)k-4nBn8n(2π)40,(2.5)

we can claim from (2.5) that ξ(t)0 on (0,(2π)-nV(Ω)]. If ξ(t)0, then ξ(t) is a decreasing function on (0,(2π)-nV(Ω)]. In fact, by a direct calculation, we can obtain

ζ(t)-2l+l(2n-2l+2)6n+l(4n-2l+4)(n+2(l-1))2144n2(n+2l)0,

since

Bn4n(2π)2<1.

On the other hand, since 0<ϕ(0)(2π)-nV(Ω) and the right-hand side of (2.1) is ξ(ϕ(0)), which is a decreasing function of ϕ(0) on (0,(2π)-nV(Ω)], we can replace ϕ(0) by (2π)-nV(Ω) in (2.1), which gives the inequality

1kj=1kλjnn+2l(2π)2l(BnV(Ω))2lnk2ln+l24(n+2l)(2π)2(l-1)(BnV(Ω))2(l-1)nV(Ω)I(Ω)k2(l-1)n+l(n+2(l-1))22304n(n+2l)2(2π)2(l-2)(BnV(Ω))2(l-2)n(V(Ω)I(Ω))2k2(l-2)n.

This completes the proof of the theorem. ∎

Next, we will prove that the inequality (1.10) is sharper than (1.9).

Proof of Remark 1.4.

Under the same assumptions as in Lemma 2.1, letting b=n and A=knBn, from

μ(2π)-nBn-1nV(Ω)n+1n

we obtain that

(bA)-2bψ(0)2+2bμ2(2π)-2n-2V(Ω)2+2n(2π)-2nBn-2nV(Ω)2(n+1)n(kBn)-2n=(2π)-2(Bn)-4nk-2n<(2π)-2(Bn)-4n<1

and then we have

1b+2lp=2l(l+1-p)6pb(b+2p-2)μ2p(bA)b+2(l-p)bψ(0)2pb-2(l-p)b<1b+2lp=2l(l+1-p)6pb(b+2p-2)μ4(bA)b+2l-4bψ(0)4b-2l+4b<l-136(b+2l)b(b+2)μ4p=016p(b+2)p(bA)b+2l-4bψ(0)4b-2l+4b=l-16b(b+2l)(6(b+2)-1)μ4(bA)b+2l-4bψ(0)4b-2l+4b.(2.6)

By a direct calculation, we derive

l(6(b+2)-1)(b+2(l-1))2>24b(b+2l)(l-1)>0.

In fact,

l(6(b+2)-1)(b+2(l-1))2-24b(b+2l)(l-1)=4b(l-1)(6b(l-1)-l)+l(6b+11)(b2+4(l-1)2)>24b2(l-1)2+4bl(l-1)(6(l-1)-1)>0,

that is,

24b(b+2l)(l-1)l(6(b+2)-1)(b+2(l-1))2<1.(2.7)

Therefore, from (2.6) and (2.7) we get

1b+2lp=2l(l+1-p)6pb(b+2p-2)μ2p(bA)b+2(l-p)bψ(0)2pb-2(l-p)b<l-16b(b+2l)(6(b+2)-1)μ4(bA)b+2l-4bψ(0)4b-2l+4b=24b(b+2l)(l-1)l(6(b+2)-1)(b+2(l-1))2×l(b+2(l-1))2144b2(b+2l)2μ4(bA)b+2l-4bψ(0)4b-2l+4b<l(b+2(l-1))2144b2(b+2l)2μ4(bA)b+2l-4bψ(0)4b-2l+4b.(2.8)

Taking

b=n,A=knBn,ψ(0)=(2π)-nV(Ω),μ=2(2π)-nV(Ω)I(Ω)(2.9)

and substituting (2.9) into (2.8), we have

nn+2l(2π)2l(BnV(Ω))2lnk2ln+l24(n+2l)(2π)2(l-1)(BnV(Ω))2(l-1)nV(Ω)I(Ω)k2(l-1)n+l(n+2(l-1))22304n(n+2l)2(2π)2(l-2)(BnV(Ω))2(l-2)n(V(Ω)I(Ω))2k2(l-2)n>nn+2l(2π)2l(BnV(Ω))2lnk2ln+n(n+2l)p=1ll+1-p(24)pn(n+2p-2)(2π)2(l-p)(BnV(Ω))2(l-p)n(V(Ω)I(Ω))pk2(l-p)n.

This completes the proof of the remark. ∎

A Proof of Lemma 2.1

In this section, we give the proof of Lemma 2.1.

Proof of Lemma 2.1.

Letting

ϱ(t)=ψ(ψ(0)μt)ψ(0),

we have ϱ(0)=1 and -1ϱ(t)0. Without loss of generality, we can assume that

ψ(0)=1andμ=1.

Define

Dl:=0sb+2l-1ψ(s)𝑑s.

One can assume that Dl<, otherwise there is nothing to prove. Since Dl<, we can conclude that

limssb+2l-1ψ(s)=0.

Putting h(s)=-ψ(s) for s0, we get

0h(s)1and0h(s)𝑑s=ψ(0)=1.

By making use of integration by parts, one has

0sbh(s)𝑑s=b0sb-1ψ(s)𝑑s=bA

and

0sb+2lh(s)𝑑s(b+2l)Dl,

since ψ(s)>0. By the same assertion as in [9], one can infer that there exists an ϵ0 such that

ϵϵ+1sb𝑑s=0sbh(s)𝑑s=bA(A.1)

and

ϵϵ+1sb+2l𝑑s0sb+2lh(s)𝑑s(b+2l)Dl.(A.2)

Letting

Θ(s)=bsb+2l-(b+2l)τ2lsb+2lτb+2l-2lτb+2(l-1)(s-τ)2,

we can prove that Θ(s)0. By integrating the function Θ(s) from ϵ to ϵ+1, from (A.1) and (A.2) we deduce, for any τ>0, that

b(b+2l)Dl-(b+2l)τ2lbA+2lτb+2ll6τb+2(l-1).(A.3)

Defining

f(τ):=(b+2l)τ2lbA-2lτb+2l+l6τb+2(l-1),

for any τ>0, we can obtain from (A.3) that

Dl=0sb+2l-1ψ(s)𝑑sf(τ)b(b+2l).

Then, taking

τ=(bA)1b(1+b+2(l-1)12(b+2l)(bA)-2b)1b,

we have

f(τ)=(bA)b+2lb(b-l(b+2(l-1))6(b+2l)(bA)-2b)(1+b+2(l-1)12(b+2l)(bA)-2b)2lb+l6(bA)b+2(l-1)b(1+b+2(l-1)12(b+2l)(bA)-2b)b+2(l-1)b.

Next, we consider the following four cases.

Case 1. b2l. For t>0, from Taylor’s formula we have

(1+t)2lb1+2lbt+2l(2l-b)2b2t2+2l(2l-b)(2l-2b)6b3t3+2l(2l-b)(2l-2b)(2l-3b)24b4t4

and

(1+t)b+2(l-1)b1+2(l-1)+bbt+(2(l-1)+b)(l-1)b2t2+(2(l-1)-b)(l-1)(2(l-1)+b)3b3t3.

Putting

t=b+2(l-1)12(b+2l)(bA)-2b,

we have from

(bA)2b1(b+1)2b13>14

that t<13 and b-2lt>4l3>0 (see also [5]). Then, we obtain

(b-l(b+2(l-1))6(b+2l)(bA)-2b)(1+b+2(l-1)12(b+2l)(bA)-2b)2lb=(b-2lt)(1+t)2lb(b-2lt)(1+2lbt+2l(2l-b)2b2t2+2l(2l-b)(2l-2b)6b3t3+2l(2l-b)(2l-2b)(2l-3b)24b4t4)b-l(2l+b)b(b+2(l-1)12(b+2l)(bA)-2b)2-(2l-b)(8l2+4lb)6b2(b+2(l-1)12(b+2l)(bA)-2b)3   -(2l-b)(2l-2b)(12l2+6lb)24b3(b+2(l-1)12(b+2l)(bA)-2b)4

and also

(1+b+2(l-1)12(b+2l)(bA)-2b)b+2(l-1)b=(1+t)b+2(l-1)b1+2(l-1)+bb(b+2(l-1)12(b+2l)(bA)-2b)+(2(l-1)+b)(l-1)b2(b+2(l-1)12(b+2l)(bA)-2b)2+(2(l-1)-b)(l-1)(2(l-1)+b)3b3(b+2(l-1)12(b+2l)(bA)-2b)3.

Therefore, we have

f(τ)=(b+2l)τ2lbA-2lτb+2l+l6τb+2(l-1)(bA)b+2lb[b-l(2l+b)b(b+2(l-1)12(b+2l)(bA)-2b)2-(2l-b)(8l2+4lb)6b2(b+2(l-1)12(b+2l)(bA)-2b)3-(2l-b)(2l-2b)(12l2+6lb)24b3(b+2(l-1)12(b+2l)(bA)-2b)4]+l6(bA)b+2(l-1)b[1+2(l-1)+bb(b+2(l-1)12(b+2l)(bA)-2b)+(2(l-1)+b)(l-1)b2(b+2(l-1)12(b+2l)(bA)-2b)2+(2(l-1)-b)(l-1)(2(l-1)+b)3b3(b+2(l-1)12(b+2l)(bA)-2b)3]=b(bA)b+2lb+l6(bA)b+2(l-1)b+l(b+2(l-1))2144b(b+2l)(bA)b+2l-4b+η1,

where

η1=2l(l+b-3)(b+2l)3b2(b+2(l-1)12(b+2l))3(bA)b+2l-6b+l(b+2(l-1))(4(l-1)(2(l-1)-b)-3(2l-b)(l-b))72b3(b+2(l-1)12(b+2l))3(bA)b+2l-8b.

Since

(bA)2b1(b+1)2b13>14

and b2l, we have

η12l(l+b-3)(b+2l)12b2(b+2(l-1)12(b+2l))3(bA)b+2l-8b+l(b+2(l-1))(4(l-1)(2(l-1)-b)-3(2l-b)(l-b))72b3(b+2(l-1)12(b+2l))3(bA)b+2l-8b=l(9b3+(35l-26)b2+(36l2-90l)b+(4l3-36l2+48l-16))72b3(b+2(l-1)12(b+2l))3(bA)b+2l-8bl(72l3+(70l2-52l)b+(36l2-90l)b-36l2)72b3(b+2(l-1)12(b+2l))3(bA)b+2l-8bl((72l3-36l2)+(140l-52l)b+(72l-90l)b)72b3(b+2(l-1)12(b+2l))3(bA)b+2l-8b0,

which implies

f(τ)b(bA)b+2lb+l6(bA)b+2(l-1)b+l(b+2(l-1))2144b(b+2l)(bA)b+2l-4b.

Case 2. 2l-2b<2l. By using Taylor’s formula, for t>0, we obtain the inequalities

(1+t)2lb1+2lbt+2l(2l-b)2b2t2+2l(2l-b)(2l-2b)6b3t3

and

(1+t)b+2(l-1)b1+2(l-1)+bbt+(2(l-1)+b)(l-1)b2t2+(2(l-1)+b)(l-1)(2(l-1)-b)3b3t3.

Putting

t=b+2(l-1)12(b+2l)(bA)-2b,

we have b-2lt>l3>0,

(b-l(b+2(l-1))6(b+2l)(bA)-2b)(1+b+2(l-1)12(b+2l)(bA)-2b)2lb=(b-2lt)(1+t)2lb(b-2lt)(1+2lbt+2l(2l-b)2b2t2+2l(2l-b)(2l-2b)6b3t3)=b-l(b+2l)b(b+2(l-1)12(b+2l)(bA)-2b)2-(2l-b)(8l2+4lb)6b2(b+2(l-1)12(b+2l)(bA)-2b)3   -4l2(2l-b)(2l-2b)6b3(b+2(l-1)12(b+2l)(bA)-2b)4,

and

(1+b+2(l-1)12(b+2l)(bA)-2b)b+2(l-1)b=(1+t)b+2(l-1)b1+2(l-1)+bb(b+2(l-1)12(b+2l)(bA)-2b)+(2(l-1)+b)(l-1)b2(b+2(l-1)12(b+2l)(bA)-2b)2+(2(l-1)+b)(l-1)(2(l-1)-b)3b3(b+2(l-1)12(b+2l)(bA)-2b)3.

Furthermore, using the same method as in Case 1, we deduce that

f(τ)=(b+2l)τ2lbA-2lτb+2l+l6τb+2(l-1)(bA)b+2lb[b-l(b+2l)b(b+2(l-1)12(b+2l)(bA)-2b)2-(2l-b)(8l2+4lb)6b2(b+2(l-1)12(b+2l)(bA)-2b)3-4l2(2l-b)(2l-2b)6b3(b+2(l-1)12(b+2l)(bA)-2b)4]+l6(bA)b+2(l-1)b[1+2(l-1)+bb(b+2(l-1)12(b+2l)(bA)-2b)+(2(l-1)+b)(l-1)b2(b+2(l-1)12(b+2l)(bA)-2b)2+(2(l-1)+b)(l-1)(2(l-1)-b)3b3(b+2(l-1)12(b+2l)(bA)-2b)3]=b(bA)b+2lb+l6(bA)b+2(l-1)b+l(b+2(l-1))2144b(b+2l)(bA)b+2l-4b+η2,

where

η2=2l(l+b-3)(b+2l)3b2(b+2(l-1)12(b+2l))3(bA)b+2l-6b+l(2(l-1)+b)((l-1)(2(l-1)-b)(b+2l)-l(2l-b)(2l-2b))18b3(b+2l)(b+2(l-1)12(b+2l))3(bA)b+2l-8b2l(l+b-3)(b+2l)9b2(b+2(l-1)12(b+2l))3(bA)b+2l-8b+l(2(l-1)+b)(l-1)(2(l-1)-b)18b3(b+2(l-1)12(b+2l))3(bA)b+2l-8b=(4bl(l+b-3)(b+2l)18b3+l(2(l-1)+b)(l-1)(2(l-1)-b)18b3)(b+2(l-1)12(b+2l))3(bA)b+2l-8b(4bl(l+b-3)(b+2l)18b3+lb(b+2l)(2(l-1)-b)18b3)(b+2(l-1)12(b+2l))3(bA)b+2l-8bbl(b+2l)(6l+3b-14)18b3(b+2(l-1)12(b+2l))3(bA)b+2l-8b0,

since

(bA)2b1(b+1)2b13.

Therefore, we have

f(τ)b(bA)b+2lb+l6(bA)b+2(l-1)b+l(b+2(l-1))2144b(b+2l)(bA)b+2l-4b.

Case 3. lb<2l-2. By using Taylor’s formula, for t>0, we have

(1+t)2lb1+2lbt+l(2l-b)b2t2+l(2l-b)(2l-2b)3b3t3

and

(1+t)b+2(l-1)b1+2(l-1)+bbt+(2(l-1)+b)(l-1)b2t2.

Putting

t=b+2(l-1)12(b+2l)(bA)-2b>0,

we have b-2lt>0,

(b-l(b+2(l-1))6(b+2l)(bA)-2b)(1+b+2(l-1)12(b+2l)(bA)-2b)2lb=(b-2lt)(1+t)2lb(b-2lt)(1+2lbt+l(2l-b)b2t2+l(2l-b)(2l-2b)3b3t3)=b-l(b+2l)b(b+2(l-1)12(b+2l)(bA)-2b)2-(2l-b)(4l2+2lb)3b2(b+2(l-1)12(b+2l)(bA)-2b)3   -4l2(2l-b)(l-b)3b3(b+2(l-1)12(b+2l)(bA)-2b)4,

and

(1+b+2(l-1)12(b+2l)(bA)-2b)b+2(l-1)b=(1+t)b+2(l-1)b1+2(l-1)+bbb+2(l-1)12(b+2l)(bA)-2b+(2(l-1)+b)(l-1)b2(b+2(l-1)12(b+2l)(bA)-2b)2.

By the same argument as in Case 2, we can deduce that

f(τ)=(b+2l)τ2lbA-2lτb+2l+l6τb+2(l-1)(bA)b+2lb[b-l(b+2l)b(b+2(l-1)12(b+2l)(bA)-2b)2-(2l-b)(4l2+2lb)3b2(b+2(l-1)12(b+2l)(bA)-2b)3-4l2(2l-b)(l-b)3b3(b+2(l-1)12(b+2l)(bA)-2b)4]+l6(bA)b+2(l-1)b[1+2(l-1)+bbb+2(l-1)12(b+2l)(bA)-2b+(2(l-1)+b)(l-1)b2(b+2(l-1)12(b+2l)(bA)-2b)2]=b(bA)b+2lb+l6(bA)b+2(l-1)b+l(b+2(l-1))2144b(b+2l)(bA)b+2l-4b+η3,

where

η3=2l(l+b-3)(b+2l)3b2(b+2(l-1)12(b+2l))3(bA)b+2l-6b-4l2(2l-b)(l-b)3b3(b+2(l-1)12(b+2l))4(bA)b+2l-8b0.

Therefore, we have

f(τ)b(bA)b+2lb+l6(bA)b+2(l-1)b+l(b+2(l-1))2144b(b+2l)(bA)b+2l-4b.

Case 4. 2b<l. Since 2b<l, there exists a positive integer k such that 2k-12lb<k and, for t>0, we have

(1+t)2lb1+2lbt+12!2lb(2lb-1)t2+13!2lb(2lb-1)(2lb-2)t3++1(k+1)!2lb(2lb-1)(2lb-k)tk+1=1+p=0k{1(p+1)!q=0p(2lb-q)}tp+1

and

(1+t)b+2lb1+b+2lbt+12!b+2lb2lbt2+13!b+2lb2lb(2lb-1)t3++1(k+1)!b+2lb2lb(2lb-1)(2lb-(k-1))tk+1=1+p=0k{1(p+1)!q=0p(2lb-q+1)}tp+1,

and also

(1+t)b+2(l-1)b1+2(l-1)+bbt+12!(2(l-1)+b)b2(l-1)bt2+13!(2(l-1)+b)b2(l-1)b(2(l-1)b-1)t3++1k!(2(l-1)+b)b2(l-1)b(2(l-1)b-(k-2))tk-|1(k+1)!(2(l-1)+b)b2(l-1)b(2(l-1)b-(k-1))|tk+1=1+p=0k-1{1(p+1)!q=0p(2(l-1)b-q+1)}tp+1-|1(k+1)!q=0k(2(l-1)b-q+1)|tk+1.

Putting

t=b+2(l-1)12(b+2l)(bA)-2b

and f(τ)=(bA)b+2lbh(τ), where

h(τ)=(b+2l)(1+t)2lb-2l(1+t)b+2lb+16(bA)-2b(1+t)b+2(l-1)b,

then, for 2b<l, we have

h(τ)(b+2l){1+p=0k[1(p+1)!q=0p(2lb-q)]tp+1}-2l{1+p=0k[1(p+1)!q=0p(2lb-q+1)]tp+1}+l6(bA)-2b{1+p=0k-1[1(p+1)!q=0p(2(l-1)b-q+1)]tp+1-|1(k+1)!q=0k(2(l-1)b-q+1)|tk+1}=b+l6(bA)-2b+p=1k{b+2l(p+1)!2lb[q=1p(2lb-q)-q=1p(2lb-q+1)]}tp+1+p=0k-1{l(bA)-2b6(p+1)!q=0p(2(l-1)b-q+1)}tp+1-|l(bA)-2b6(k+1)!q=0k(2(l-1)b-q+1)|tk+1=b+l6(bA)-2b-p=1k{p2l(b+2l)b(p+1)!q=1p-1(2lb-q)}tp+1+p=1k{l(bA)-2b6p!q=0p-1(2(l-1)b-q+1)}tp-|l(bA)-2b6(k+1)!q=0k(2(l-1)b-q+1)|tk+1=b+l6(bA)-2b-p=1k{pbp(p+1)!q=0p(2l-(q-1)b)}(b+2(l-1)12(b+2l)(bA)-2b)p+1+p=1k{l(bA)-2b6bpp!q=0p-1(2(l-1)-(q-1)b)}(b+2(l-1)12(b+2l)(bA)-2b)p-|l(bA)-2b6bk+1(k+1)!q=0k(2(l-1)-(q-1)b)|(b+2(l-1)12(b+2l)(bA)-2b)k+1.

Furthermore, we have

f(τ)b(bA)b+2lb+l6(bA)b+2(l-1)b+l(b+2(l-1))2144b(b+2l)(bA)b+2l-4b-p=2k{pbp(p+1)!q=0p(2l-(q-1)b)}(b+2(l-1)12(b+2l))p+1(bA)b+2l-2p-2b+p=2k{l6bpp!q=0p-1(2(l-1)-(q-1)b)}(b+2(l-1)12(b+2l))p(bA)b+2l-2p-2b-|l6bk+1(k+1)!q=0k(2(l-1)-(q-1)b)|(b+2(l-1)12(b+2l))k+1(bA)b+2l-2k-4b=b(bA)b+2lb+l6(bA)b+2(l-1)b+l(b+2(l-1))2144b(b+2l)(bA)b+2l-4b+η4,

where

η4=p=2k{(b+2(l-1))2l12bpp![q=1p-1(2(l-1)-(q-1)b)-pp+1q=1p-1(2l-qb)]}(b+2(l-1)12(b+2l))p(bA)b+2l-2p-2b-|l6bk+1(k+1)!q=0k(2(l-1)-(q-1)b)|(b+2(l-1)12(b+2l))k+1(bA)b+2l-2k-4b.

From k-22(l-1)b<k we have

k-2-ik+1-i2(l-1)b-ik+1-i<k-ik+1-i.(A.4)

Then, from (A.4) we have

|2(l-1)b-ik+1-i|1,i=0,1,2,,k-1.

Note that

2(l-1)-(q-1)b2l-qb0,p=2,3,,k,

and

q=1p-1(2(l-1)-(q-1)b)-pp+1q=1p-1(2l-qb)q=1p-1(2(l-1)-(q-1)b)-q=1p-1(2l-qb)0.

Therefore, we obtain

p=2k{(b+2(l-1))2l12bpp![q=1p-1(2(l-1)-(q-1)b)-pp+1q=1p-1(2l-qb)]}(b+2(l-1)12(b+2l))p(bA)b+2l-2p-2b(b+2(l-1))2l24b2(2(l-1)-2(2l-b)3)(b+2(l-1)12(b+2l))2(bA)b+2l-6b.

From

(bA)2b1(b+1)2b13

we have

η4(b+2(l-1))2l24b2(2(l-1)-2(2l-b)3)(b+2(l-1)12(b+2l))2(bA)b+2l-6b-|l6bk+1(k+1)!q=0k(2(l-1)-(q-1)b)|(b+2(l-1)12(b+2l))k+1(bA)b+2l-2k-4b=l(b+2(l-1))12b2{(2(l-1)-2(2l-b)3)-2b|1bk(k+1)!q=1k(2(l-1)-(q-1)b)|(b+2(l-1)12(b+2l))k-1(bA)-2k+2b}(b+2(l-1)12(b+2l))2(bA)b+2l-6bl(b+2(l-1))12b2{2l+2b-63-2b|q=0k-1(2(l-1)b-qk+1-q)|(14)k-1}(b+2(l-1)12(b+2l))2(bA)b+2l-6bl(b+2(l-1))12b2(2l+2b-63-b8)(b+2(l-1)12(b+2l))2(bA)b+2l-6b0,

which implies that

f(τ)b(bA)b+2lb+l6(bA)b+2(l-1)b+l(b+2(l-1))2144b(b+2l)(bA)b+2l-4b.

This completes the proof of the lemma. ∎

The authors wish to express their gratitude to Professor Qing-Ming Cheng for his continuous encouragement and his enthusiastic help. The authors would also like to thank the referees for their useful comments and suggestions.

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About the article

Received: 2014-06-28

Revised: 2014-11-11

Accepted: 2014-11-16

Published Online: 2016-02-01

Published in Print: 2016-02-01


Funding Source: National Natural Science Foundation of China

Award identifier / Grant number: 11371150

Award identifier / Grant number: 11401268

The first author acknowledges the support of the NSFC (Grant No. 11371150) and of the project Pearl River New Star of Guangzhou (Grant No. 2012J2200028). The second author acknowledges the support of the NSFC (Grant No. 11401268).


Citation Information: Advanced Nonlinear Studies, Volume 16, Issue 1, Pages 31–44, ISSN (Online) 2169-0375, ISSN (Print) 1536-1365, DOI: https://doi.org/10.1515/ans-2015-5020.

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