Universal bounds for eigenvalues of a buckling problem

Qing-Ming Cheng* and Hongcang Yang**

Abstract

In this paper, we investigate an eigenvalue problem for biharmonic operator on a bounded domain in an n-dimensional Euclidean space, which is also called a buckling problem. We introduce a new method to construct ”nice” trial functions and we derive a universal inequality for higher eigenvalues of the buckling problem by making use of the trial functions. Thus, we give an affirmative answer for the problem on universal bounds for eigenvalues of the buckling problem, which was proposed by Payne, Pólya and Weinberger in

[14] and this problem has been mentioned again by Ashbaugh in [1].

1 Introduction

Let Ω ⊂ R^{n} be a bounded domain in an n-dimensional Euclidean space R^{n}. An eigenvalue problem of Dirichlet Laplacian is given by

\{ \begin{matrix}△u = −λu, & in Ω, \\ u = 0, & on ∂Ω, \end{matrix}  (1.1)

which is also called a fixed membrane problem. The spectrum of this eigenvalue problem is real and purely discrete:

0 < λ_{1} < λ_{2} ≤ ⋯ ≤ λ_{k} ≤ ⋯ → ∞.

Let λ_{k+1} be the (k + 1)-th eigenvalue of the eigenvalue problem (1.1). The universal bounds for the eigenvalue λ_{k+1} are studied by many mathematicians. The main contributions were obtained by Payne, Pólya and Weinberger [14] and Thompson [15], Hile and Protter [9] and Yang [16] (cf. Cheng and Yang [7]), that is, Payne, Pólya and Weinberger [14] and Thompson [15] proved, for k = 1, 2, ⋯,

Key words and phrases: eigenvalue, trial function, universal bound for eigenvalues, biharmonic operator and a buckling problem.

2001 Mathematics Subject Classification: 35P15, 58G25, 53C42.

* Research partially Supported by a Grant-in-Aid for Scientific Research from the Japan Society for the Promotion of Science.

** Research partially Supported by SF of CAS.

λ_{k+1} − λ_{k} ≤ \frac{4}{kn} ∑_{i=1}^{k} λ_{i}.  (1.2)

Hile and Protter [9] improved the above result of Payne, Pólya and Weinberger and Thompson to

∑_{i=1}^{k} \frac{λ_{i}}{λ_{k+1} − λ_{i}} ≥ \frac{kn}{4}.  (1.3)

Further, Yang [16] has obtained very sharp inequality:

\begin{aligned}λ_{k+1} ≤ & (1 + \frac{2}{n}) \frac{1}{k} ∑_{i=1}^{k} λ_{i} \\ & + [(\frac{2}{n} \frac{1}{k} ∑_{i=1}^{k} λ_{k})^{2} − (1 + \frac{4}{n}) \frac{1}{k} ∑_{j=1}^{k} (λ_{j} − \frac{1}{k} ∑_{i=1}^{k} λ_{i})^{2}]^{1/2} . \end{aligned}  (1.4)

Yang’s inequality (1.4) is sharper than the inequalities (1.2) and (1.3) ( see [1] and [2] for details). In order to obtain the universal bounds for the eigenvalues of the fixed membrane problem (1.1), it is important to find appropriated trial functions. For the fixed membrane problem (1.1), one chooses

φ_{i} = x^{p}u_{i} − ∑_{j=1}^{k} a_{ij}u_{j}, a_{ij} = ∫_{Ω} x^{p}u_{i}u_{j},

as trial functions by making use of the basic method of Payne, Pólya and Weinberger, where u_{j} is an orthonormal eigenfunction corresponding to the j-th eigenvalue λ_{j}, x^{p} denotes the p-th standard coordinate function of R^{n}.

On the other hand, in order to describe vibrations of a clamped plate, we need to consider an eigenvalue problem for Dirichlet biharmonic operator, so called a clamped plate problem:

\{ \begin{matrix}Δ^{2}u = & λu in Ω, \\ u|_{∂Ω} = & \frac{∂u}{∂n⃗}|_{∂Ω} = 0, \end{matrix}  (1.5)

where Δ^{2} is the biharmonic operator in R^{n} and n⃗ denotes the unit outward normal vector on the boundary ∂Ω of Ω. For this clamped plate problem (1.5), Payne, Pólya and Weinberger [14] also established the following inequality for the eigenvalues:

λ_{k+1} − λ_{k} ≤ \frac{8(n + 2)}{n^{2}} \frac{1}{k} ∑_{i=1}^{k} λ_{i}.  (1.6)

Hile and Yeh [8] obtained, as a generalization of their result,

∑_{i=1}^{k} \frac{λ_{i}^{1/2}}{λ_{k+1} − λ_{i}} ≥ \frac{n^{2}k^{3/2}}{8(n + 2)} (∑_{i=1}^{k} λ_{i})^{−1/2} .  (1.7)

They used an improved method of Hile and Protter [9]. Further, Hook [11], Chen and Qian [4], independently, proved the following inequality

\frac{n^{2}k^{2}}{8(n + 2)} ≤ [∑_{i=1} \frac{λ_{i}^{1/2}}{λ_{k+1} − λ_{i}} ] ∑_{i=1}^{k} λ_{i}^{1/2}.  (1.8)

The following problem 1 has been proposed by Ashbaugh in [1]:

Problem 1. Whether can one obtain a universal inequality for the eigenvalues of the clamped plate problem (1.5), which is similar to Yang’s universal inequality (1.4) for the eigenvalues of the fixed membrane problem (1.1)?

This problem 1 has been solved by Cheng and Yang in [6], as stated:

\begin{aligned}λ_{k+1} ≤ & [1 + \frac{4(n + 2)}{n^{2}}] \frac{1}{k} ∑_{i=1}^{k} λ_{i} \\ & + \{[\frac{4(n + 2)}{n^{2}} \frac{1}{k} ∑_{i=1}^{k} λ_{i}]^{2} − \frac{8(n + 2)}{n^{2}} \frac{1}{k} ∑_{i=1}^{k} (λ_{i} − \frac{1}{k} ∑_{j=1}^{k} λ_{j})^{2}\}^{1/2} . \end{aligned}

In order to prove the universal inequalities for the eigenvalues of the clamped plate problem (1.5), we see that

φ_{i} = x^{p}u_{i} − ∑_{j=1}^{k} a_{ij}u_{j}, a_{ij} = ∫_{Ω} x^{p}u_{i}u_{j},

are also very appropriated trial functions.

In this paper, we consider an eigenvalue problem:

\{ \begin{matrix}Δ^{2}u = & − ΛΔu in Ω, \\ u|_{∂Ω} = & \frac{∂u}{∂n⃗}|_{∂Ω} = 0, \end{matrix}  (1.9)

which is called a buckling problem. This eigenvalue problem (1.9) is used to describe the critical buckling load of a clamped plate subjected to a uniform compressive force around its boundary. It is our purpose of this paper to derive a universal inequality for the (k + 1)-th eigenvalue of the buckling problem (1.9). It is very hard to discuss this problem. The reason is that we can not find appropriated trial functions because we must use Dirichlet inner product (f, h)_{D} = ∫_{Ω} ∇f · ∇h for functions f and h. As in the fixed membrane problem and the clamped plate problem, one wants to choose functions

φ_{i} = x^{p}u_{i} − ∑_{j=1}^{k} a_{ij}u_{j}, a_{ij} = ∫_{Ω} ∇(x^{p}u_{i}) · ∇u_{j}

as trial functions. However, they are not very appropriated trial functions for the buckling problem (1.9) because the a_{ij}’s are not symmetric (cf. Ashbaugh [1] and Payne, Pólya and Weinberger [14]). Thus, the crucial point for the solution of the finding a universal bound for the eigenvalues of the buckling problem (1.9) is how to find ”nice” trial functions. In [14], Payne, Pólya and Weinberger proposed the following:

Problem 2. Whether can one obtain a universal inequality for the eigenvalues of the buckling problem (1.9), which is similar to the universal inequalities for the eigenvalues of the fixed membrane problem (1.1) or of the clamped plate problem (1.5)?

Remark 1. In [1], Ashbaugh has mentioned the problem 2 of Payne, Pólya and Weinberger again.

In this paper, first of all, we introduce, in section 2, a new method to construct trial functions for the buckling problem (1.9). Secondly, we derive a universal inequality for the (k + 1)-th eigenvalue of the buckling problem (1.9) by making use of the trial functions. Thus, we give an affirmative answer for the problem 2 of Payne, Pólya and Weinberger [14].

We shall review known results about estimates on the eigenvalues of the buckling problem (1.9). From now, we shall use Λ_{i} and u_{i} to denote eigenvalues and eigenfunctions of the buckling problem. In [14], Payne, Pólya and Weinberger proved, for n = 2,

Λ_{2} ≤ 3Λ_{1}.

For general n, one obtained

Λ_{2} ≤ (1 + \frac{4}{n})Λ_{1}.

Subsequently, Hile and Yeh [10] improved the above results. They proved

Λ_{2} ≤ \frac{n^{2} + 8n + 20}{(n + 2)^{2}}Λ_{1}.

Ashbaugh [1] has obtained

∑_{i=1}^{n} Λ_{i+1} ≤ (n + 4)Λ_{1}.  (1.10)

From our knowledge, these are the only known results about the universal bounds for the eigenvalues of the buckling problem (1.9).

In this paper, we prove the following:

Theorem. Let Λ_{i} be the i-th eigenvalue of the buckling problem (1.9). Then, we have

∑_{i=1}^{k}(Λ_{k+1} − Λ_{i})^{2} ≤ \frac{4(n + 2)}{n^{2}} ∑_{i=1}^{k}(Λ_{k+1} − Λ_{i})Λ_{i}.  (1.11)

Corollary 1. Under the assumption of the theorem, we have

\begin{matrix}Λ_{k+1} ≤ & (1 + \frac{2(n + 2)}{n^{2}}) \frac{1}{k} ∑_{i=1}^{k} Λ_{i} \\ + & [(\frac{2(n + 2)}{n^{2}} \frac{1}{k} ∑_{i=1}^{k} Λ_{i})^{2} − (1 + \frac{4(n + 2)}{n^{2}}) \frac{1}{k} ∑_{j=1}^{k} (Λ_{j} − \frac{1}{k} ∑_{i=1}^{k} Λ_{i})^{2}]^{\frac{1}{2}} . \end{matrix}

Corollary 2. Under the assumption of the theorem, we have

Λ_{k+1} − Λ_{k} ≤2 [(\frac{2(n + 2)}{n^{2}} \frac{1}{k} ∑_{i=1}^{k} Λ_{i})^{2} − (1 + \frac{4(n + 2)}{n^{2}}) \frac{1}{k} ∑_{j=1}^{k} (Λ_{j} − \frac{1}{k} ∑_{i=1}^{k} Λ_{i})^{2}]^{\frac{1}{2}} .

Remark 2. From the corollary 1, we have

Λ_{k+1} − Λ_{k} ≤ Λ_{k+1} − \frac{1}{k} ∑_{i=1}^{k} Λ_{i} ≤ \frac{4(n + 2)}{n^{2}} \frac{1}{k} ∑_{i=1}^{k} Λ_{i}.  (1.12)

Acknowledgements. We would like to express our gratitude to the referee for his/her valuable comments and suggestions.

2 The construction of trial functions

For functions f and h, we define Dirichlet inner product (f, h)_{D} of f and h by

(f, h)_{D} = ∫_{Ω} ∇f · ∇h.

Dirichlet norm of a function f is defined by

||f||_{D} = {(f, f)_{D}}^{1/2} = (∫_{Ω} ∑_{α=1}^{n} |∇_{α}f|^{2})^{1/2} .

Let u_{i} be the i-th orthonormal eigenfunction of the buckling problem (1.9) corresponding to the eigenvalue Λ_{i}, namely, u_{i} satisfies

\{ \begin{matrix}Δ^{2}u_{i} = −Λ_{i}Δu_{i} in Ω, \\ u_{i}|_{∂Ω} = \frac{∂u_{i}}{∂n⃗} |_{∂Ω} = 0 \\ (u_{i}, u_{j})_{D} = ∫_{Ω} ∇u_{i} · ∇u_{j} = δ_{ij} \\ ∥u_{i}∥_{D}^{2} = (u_{i}, u_{i})_{D} = ∫_{Ω} |∇u_{i}|^{2} = 1. \end{matrix}  (2.1)

Define H_{2}^{2}(Ω) by

H_{2}^{2}(Ω) = {f : f, ∇_{α}f, ∇_{α}∇_{β}f ∈ L^{2}(Ω), α, β = 1, …, n}.

Then, H_{2}^{2}(Ω) is a Hilbert space with norm ∥ · ∥_{2}:

∥f∥_{2} = (∫_{Ω} |f|^{2} + ∫_{Ω} |∇f|^{2} + ∑_{β,α=1}^{n}(∇_{α}∇_{β}f)^{2})^{1/2} .

Let H_{2,D}^{2}(Ω) be a subspace of H_{2}^{2}(Ω)defined as

H_{2,D}^{2}(Ω) = \{f ∈ H_{2}^{2}(Ω) : f|_{∂M} = \frac{∂}{∂n⃗}f|_{∂Ω} = 0\} .

The biharmonic operator Δ^{2} defines a self-adjoint operator acting on H_{2,D}^{2}(Ω) with discrete eigenvalues {0 < Λ_{1} < Λ_{2} ≤ ⋯ ≤ Λ_{k} ≤ ⋯ } for the buckling problem (1.9) and the eigenfunctions defined in (2.1)

{u_{i}}_{i=1}^{∞} = {u_{1}, u_{2}, ⋯, u_{k}, ⋯ }

form a complete orthonormal basis for Hilbert space H_{2,D}^{2}(Ω).

If φ ∈ H_{2,D}^{2}(Ω), (φ, u_{j})_{D} = 0, for j = 1, 2, ⋯, k, then, from the Rayleigh-Ritz inequality, we have

Λ_{k+1}∥φ∥_{D}^{2} ≤ ∫ φΔ^{2}φ = − ∫ ∇φ · ∇(Δφ).  (2.2)

We define an inner product (f, h) for vector-valued functions f = (f^{1}, f^{2}, ⋯, f^{n}) ∈ R^{n} and h = (h^{1}, h^{2}, ⋯, h^{n}) ∈ R^{n} by

(f, h) ≡ ∫_{Ω} f · h = ∫_{Ω} ∑_{α=1}^{n} f^{α}h^{α}.

The norm of f is defined by

∥f∥ = (f, f)^{1/2} = {∫_{Ω} ∑_{α=1}^{n}(f^{α})^{2}}^{1/2} .

Denote a Hilbert space H_{1}^{2}(Ω) of the vector-valued functions as

H_{1}^{2}(Ω) = {f : f^{α}, ∇_{β}f^{α} ∈ L^{2}(Ω), for α, β = 1, …, n}

with norm ∥ · ∥_{1}:

∥f∥_{1} = (∥f∥^{2} + ∫_{Ω} ∑_{α,β=1}^{n} |∇_{α}f^{β}|^{2})^{1/2} .

Let H_{1,D}^{2}(Ω) ⊂ H_{1}^{2}(Ω) be a subspace of H_{1}^{2}(Ω) spanned by the vector-valued functions {∇u_{i}}_{i=1}^{∞}, which form a complete orthonormal basis of H_{1,D}^{2}(Ω). It is easy to see that for any f ∈ H_{2,D}^{2}(Ω), ∇f ∈ H_{1,D}^{2}(Ω) and for any h ∈ H_{1,D}^{2}(Ω), there exists a function f ∈ H_{2,D}^{2}(Ω) such that h = ∇f. Let g = x^{p} for a fixed p with p = 1, 2, ⋯, n. For a vector-valued function g∇u_{i}, i = 1, …, k, we decompose it into

g∇u_{i} = f_{i} + w_{i},  (2.3)

where f_{i} is the projection of g∇u_{i} onto H_{1,D}^{2}(Ω) and w_{i} ⊥ H_{1,D}^{2}(Ω) , namely,

f_{i} ∈ H_{1,D}^{2}(Ω), f_{i} = ∇h_{i}, for some h_{i} ∈ H_{2,D}^{2}(Ω)  (2.4)

and

(w_{i}, ∇u) = ∫_{Ω} ∑_{α=1}^{n} w_{i}^{α}∇_{α}u = 0, for any u ∈ H_{2,D}^{2}(Ω).  (2.5)

Therefore, since H_{2,D}^{2}(Ω) is dense in L^{2}(Ω) and C^{1}(Ω) is dense in L^{2}(Ω), we have, for any function h ∈ C^{1}(Ω) ∩ L^{2}(Ω),

(w_{i}, ∇h) = 0.  (2.6)

Hence, from the definition of w_{i} and (2.6), we have

\{ \begin{matrix}w_{i}|_{∂Ω} = 0, \\ ∥ divw_{i}∥^{2} = 0, ( divw_{i} ≡ ∑_{α=1}^{n} ∇_{α}w_{i}^{α}). \end{matrix}  (2.7)

From (2.3) and (2.4), we can write

g∇_{α}u_{i} = f_{i}^{α} + w_{i}^{α}, α = 1, …, n,

with f_{i}^{α} = ∇_{α}h_{i} for some function h_{i} ∈ H_{2,D}^{2}(Ω). Since

∇_{β}(g∇_{α}u_{i}) − ∇_{α}(g∇_{β}u_{i}) = ∇_{β}w_{i}^{α} − ∇_{α}w_{i}^{β}  (2.8)

holds, we infer, from (2.7) and (2.8),

\begin{matrix}∥∇w_{i}∥^{2} = & ∑_{α,β=1}^{n} ∥∇_{α}w_{i}^{β}∥^{2} \\ = & \frac{1}{2} ∑_{α,β=1}^{n} ∥∇_{β}w_{i}^{α} − ∇_{α}w_{i}^{β}∥^{2} + ∥divw_{i}∥^{2} \\ = & \frac{1}{2} ∑_{α,β=1}^{n} ∥∇_{β}(g∇_{α}u_{i}) − ∇_{α}(g∇_{β}u_{i})∥^{2} \\ = & 1 − ∥∇_{p}u_{i}∥^{2}. \end{matrix}  (2.9)

Now we can construct the trial functions φ_{i} by

φ_{i} = h_{i} − ∑_{j=1}^{k} b_{ij}u_{j},  (2.10)

where

∇h_{i} = f_{i} and b_{ij} = ∫ g∇u_{i} · ∇u_{j} = b_{ji}.

It is easy to check, from the definition (2.3) of f_{i}, that φ_{i} satisfies

φ_{i}|_{∂Ω} = \frac{∂φ_{i}}{∂n⃗} |_{∂Ω} = 0 and (φ_{i}, u_{j})_{D} = (∇φ_{i}, ∇u_{j}) = 0,  (2.11)

for any j = 1, 2, ⋯, k. Hence, we know that φ_{i} is a trial function. In section 3, we shall make use of the trial function φ_{i} to prove our Theorem.

3 Proof of Theorem

In this section, we shall use the notation in the section 2. Proof of Theorem. Defining a vector-valued function v_{i} = ∇φ_{i}, we have

v_{i} = f_{i} − ∑_{j=1}^{k} b_{ij}∇u_{j}, (v_{i}, ∇u_{j}) = ∫_{Ω} ∇u_{j} · v_{i} = 0, i, j = 1, ⋯, k.  (3.1)

From (2.10), (2.11), (3.1) and the Rayleigh-Ritz inequality (2.2), we have

Λ_{k+1}∥v_{i}∥^{2} ≤ −(v_{i}, Δv_{i}).  (3.2)

From the definition of f_{i}, we have

∥g∇u_{i}∥^{2} = ∥f_{i}∥^{2} + ∥w_{i}∥^{2}  (3.3)

and, from (2.6), we infer

−(f_{i}, Δf_{i}) = − ∫_{Ω}(g∇u_{i} − w_{i})(Δ(g∇u_{i}) − Δw_{i})
= − (g∇u_{i}, Δ(g∇u_{i})) + 2(w_{i}, Δ(g∇u_{i})) − (w_{i}, Δw_{i})
= − (g∇u_{i}, Δ(g∇u_{i})) + 2(w_{i}, Δ(f_{i} + w_{i})) − (w_{i}, Δw_{i})
= − (g∇u_{i}, Δ(g∇u_{i})) − ∥∇w_{i}∥^{2}.

(3.4)

\begin{matrix}− & (g∇u_{i}, Δ(g∇u_{i})) = − ∫_{Ω} g∇u_{i} · (g∇Δu_{i} + 2∇(∇g · ∇u_{i})) \\ = & ∫_{Ω}(gu_{i}gΔ^{2}u_{i} + 2gu_{i}∇g · ∇Δu_{i})−2 ∫_{Ω} g∇u_{i} · ∇(∇g · ∇u_{i}), \end{matrix}  (3.5)
∫_{Ω} gu_{i}gΔ^{2}u_{i} = − Λ_{i} ∫_{Ω} u_{i}g^{2}Δu_{i}
=2Λ_{i} ∫_{Ω} gu_{i}∇g · ∇u_{i} + Λ_{i}||g∇u_{i}||^{2}
= − Λ_{i}||u_{i}||^{2} + Λ_{i}(||f_{i}||^{2} + ||w_{i}||^{2}).

(3.6)

Because of

−2 ∫_{Ω} g∇u_{i} · ∇(∇g · ∇u_{i}) = 2∥∇u_{i}∥^{2} + 2 ∫_{Ω} g∇u_{i} · ∇(∇g · ∇u_{i}),

we have

−2 ∫_{Ω} g∇u_{i} · ∇(∇g · ∇u_{i}) = ∥∇u_{i}∥^{2}.  (3.7)

According to

2 ∫_{Ω} gu_{i}∇g · ∇Δu_{i} = 2||∇u_{i}||^{2} + 4||∇_{p}u_{i}||^{2} − 2 ∫_{Ω} gu_{i}∇g · ∇Δu_{i},

we derive

2 ∫_{Ω} gu_{i}∇g · ∇Δu_{i} = ||∇u_{i}||^{2} + 2||∇_{p}u_{i}||^{2}.  (3.8)

From (3.5), (3.6), (3.7) and (3.8), we obtain

\begin{matrix}− (g∇u_{i}, Δ(g∇u_{i})) \\ = 2||∇u_{i}||^{2} + 2||∇_{p}u_{i}||^{2} − Λ_{i}||u_{i}||^{2} + Λ_{i}(||f_{i}||^{2} + ||w_{i}||^{2}). \end{matrix}  (3.9)

From (3.1), we have

−(v_{i}, Δv_{i}) = − (f_{i} − ∑_{j=1}^{k} b_{ij}∇u_{j}, Δf_{i} − ∑_{j=1}^{k} b_{ij}Δ∇u_{j})

= −(f_{i}, Δf_{i}) + 2 ∑_{j=1}^{k} b_{ij}(f_{i}, Δ∇u_{j}) + ∑_{j=1}^{k} Λ_{j}b_{ij}^{2}.

(3.10)

Here we used (∇u_{j}, ∇u_{ℓ}) = δ_{jℓ}. From (2.3), (2.4) and (2.5), we derive

2 ∑_{j=1}^{k} b_{ij}(f_{i}, Δ∇u_{j}) = 2 ∑_{j=1}^{k} b_{ij}(∇h_{i}, Δ∇u_{j})

= −2 ∑_{j=1}^{k} Λ_{j}b_{ij}(f_{i}, ∇u_{j}) = −2 ∑_{j=1}^{k} Λ_{j}b_{ij}^{2}.

From (3.1), we infer

||f_{i}||^{2} = ||v_{i}||^{2} + ∑_{j=1}^{k} b_{ij}^{2}.  (3.11)

Therefore, from (3.2), (3.4), (3.9), (3.10) and (3.11), we infer

Λ_{k+1}∥v_{i}∥^{2} ≤ 2 + 2∥∇_{p}u_{i}∥^{2} − ∥∇w_{i}∥^{2} + Λ_{i}(∥v_{i}∥^{2} − ∥u_{i}∥^{2} + ∥w_{i}∥^{2}) + ∑_{j=1}^{k}(Λ_{i} − Λ_{j})b_{ij}^{2}.

Thus, from (2.9), we obtain

(Λ_{k+1} − Λ_{i})∥v_{i}∥^{2} ≤ 1 + 3∥∇_{p}u_{i}∥^{2} − Λ_{i}(∥u_{i}∥^{2} − ∥w_{i}∥^{2}) + ∑_{j=1}^{k}(Λ_{i} − Λ_{j})b_{ij}^{2}. (3.12)

Since ∇(gu_{i}) = u_{i}∇g + g∇u_{i} ∈ H_{1,D}^{2}(Ω), for h_{i} = ∇(gu_{i}) − f_{i} ∈ H_{1,D}^{2}(Ω), we have

u_{i}∇g = h_{i} − w_{i}.

Hence, we obtain

∥u_{i}∥^{2} = ∥u_{i}∇g∥^{2} = ∥w_{i}∥^{2} + ∥h_{i}∥^{2}.  (3.13)

Because of (∇(∇g · ∇u_{i}), w_{i}) = 0, we infer

\begin{matrix}2∥∇_{p}u_{i}∥^{2} = & − 2 ∫ u_{i}∇g · ∇(∇g · ∇u_{i}) \\ = & − 2 ∫ h_{i} · ∇(∇g · ∇u_{i}) \\ ≤ & Λ_{i}∥h_{i}∥^{2} + \frac{1}{Λ_{i}} ∥∇_{p}∇u_{i}∥^{2}. \end{matrix}  (3.14)

Then, from (3.12), (3.13) and (3.14), we have

(Λ_{k+1} − Λ_{i})∥v_{i}∥^{2} ≤ 1 + ∥∇_{p}u_{i}∥^{2} + \frac{1}{Λ_{i}} ∥∇_{p}∇u_{i}∥^{2} + ∑_{j=1}^{k}(Λ_{i} − Λ_{j})b_{ij}^{2}.  (3.15)

According to

− 2 ∫_{Ω} g∇u_{i} · ∇(∇g · ∇u_{i})
= 2||∇_{p}u_{i}||^{2} + 2 ∫_{Ω} gΔu_{i}∇_{p}u_{i}
= 2||∇_{p}u_{i}||^{2} − 2 ∫_{Ω} u_{i}Δu_{i} − 2 ∫_{Ω} gu_{i}Δ∇_{p}u_{i}
= 2 + 2 ∫_{Ω} g∇u_{i} · ∇(∇g · ∇u_{i}),

we have,

1 = −2 ∫_{Ω} g∇u_{i} · ∇(∇g · ∇u_{i}).

On the other hand,

\begin{matrix}1 = & − 2 ∫ g∇u_{i} · ∇(∇g · ∇u_{i}) \\ = & − 2 ∫ (f_{i} + w_{i}) · ∇(∇g · ∇u_{i}) \\ = & − 2 ∫ (v_{i} + ∑_{j=1}^{k} b_{ij}∇u_{j}) · ∇(∇g · ∇u_{i}) \\ = & − 2 ∫ v_{i} · ∇(∇g · ∇u_{i}) − 2 ∑_{j=1}^{k} b_{ij}c_{ij}, \end{matrix}

where

c_{ij} = ∫ ∇(∇g · ∇u_{i}) · ∇u_{j} = −c_{ji}.

Hence, we have

(Λ_{k+1} − Λ_{i})^{2}(1 + 2 ∑_{j=1}^{k} b_{ij}c_{ij})

= (Λ_{k+1} − Λ_{i})^{2} (−2v_{i}, (∇(∇g · ∇u_{i}) − ∑_{j=1}^{k} c_{ij}∇u_{j}))

≤ A(Λ_{k+1} − Λ_{i})^{3}∥v_{i}∥^{2} + \frac{1}{A}(Λ_{k+1} − Λ_{i}) (∥∇_{p}∇u_{i}∥^{2} − ∑_{j=1}^{k} c_{ij}^{2}) .

where A is a positive constant. From (3.14), we infer

(Λ_{k+1} − Λ_{i})^{2}(1 + 2 ∑_{j=1}^{k} b_{ij}c_{ij})

≤ A(Λ_{k+1} − Λ_{i})^{2}(1 + ∥∇_{p}u_{i}∥^{2} + \frac{1}{Λ_{i}} ∥∇_{p}∇u_{i}∥^{2} + (Λ_{i} − Λ_{j})b_{ij}^{2})

+ \frac{1}{A}(Λ_{k+1} − Λ_{i}) (∥∇_{p}∇u_{i}∥^{2} − ∑_{j=1}^{k} c_{ij}^{2}) .

By taking sum for i from 1 to k and noticing that b_{ij} is symmetric and c_{ij} is antisymmetric, we have

∑_{i=1}^{k}(Λ_{k+1} − Λ_{i})^{2} − 2 ∑_{i,j=1}^{k}(Λ_{k+1} − Λ_{i})(Λ_{i} − Λ_{j})b_{ij}c_{ij}

≤A ((Λ_{k+1} − Λ_{i})^{2}(1 + ∥∇_{p}u_{i}∥^{2} + \frac{1}{Λ_{i}} ∥∇_{p}∇u_{i}∥^{2})− ∑_{i,j=1}^{k}(Λ_{k+1} − Λ_{i})(Λ_{i} − Λ_{j})^{2}b_{ij}^{2})

+ \frac{1}{A} (∑_{i=1}^{j}(Λ_{k+1} − Λ_{i})∥∇_{p}∇u_{i}∥^{2} − ∑_{i,j=1}^{k}(Λ_{k+1} − Λ_{i})c_{ij}^{2}) ,

namely,

\begin{matrix}∑_{i=1}^{k}(Λ_{k+1} − Λ_{i})^{2} ≤ & A ∑_{i=1}^{k}(Λ_{k+1} − Λ_{i})^{2} (1 + ∥∇_{p}u_{i}∥^{2} + \frac{1}{Λ_{i}} ∥∇_{p}∇u_{i}∥^{2}) \\ + \frac{1}{A} ∑_{i=1}^{k}(Λ_{k+1} − Λ_{i})∥∇_{p}∇u_{i}∥^{2}. \end{matrix}

Taking sum for p from 1 to n and noticing ∑_{p=1}^{n} ∥∇_{p}∇u_{i}∥^{2} = Λ_{i}, we have

n ∑_{i=1}^{k}(Λ_{k+1} − Λ_{i})^{2} ≤A ∑_{i=1}^{k}(Λ_{k+1} − Λ_{i})^{2}(n + 2) + \frac{1}{A} ∑_{i=1}^{k}(Λ_{k+1} − Λ_{i})Λ_{i}.

Putting A = \frac{n}{2(n+2)}, we have

∑_{i=1}^{k}(Λ_{k+1} − Λ_{i})^{2} ≤ \frac{4(n + 2)}{n^{2}} ∑_{i=1}^{k}(Λ_{k+1} − Λ_{i})Λ_{i}.

This finishes the proof of the theorem.

□

Proof of Corollary 1. From the theorem, we know that (1.11) is a quadratic inequality of Λ_{k+1}. Hence, we infer

\begin{matrix}Λ_{k+1} ≤ & (1 + \frac{2(n + 2)}{n^{2}}) \frac{1}{k} ∑_{i=1}^{k} Λ_{i} \\ + & [(\frac{2(n + 2)}{n^{2}} \frac{1}{k} ∑_{i=1}^{k} Λ_{i})^{2} − (1 + \frac{4(n + 2)}{n^{2}}) \frac{1}{k} ∑_{j=1}^{k} (Λ_{j} − \frac{1}{k} ∑_{i=1}^{k} Λ_{i})^{2}]^{\frac{1}{2}} . \end{matrix}

This completes the proof of the corollary 1.

□

From the corollary 1, it is obvious that we have

Λ_{k+1} ≤ (1 + \frac{4(n + 2)}{n^{2}}) \frac{1}{k} ∑_{i=1}^{k} Λ_{i}.

Proof of Corollary 2. From the theorem, we know that (1.11) is a quadratic inequality of Λ_{k+1}. On the other hand, since k is any positive integer, we know that Λ_{k} also satisfies the same quadratic inequality. Hence, we obtain

\begin{matrix}Λ_{k} ≥ & (1 + \frac{2(n + 2)}{n^{2}}) \frac{1}{k} ∑_{i=1}^{k} Λ_{i} \\ − & [(\frac{2(n + 2)}{n^{2}} \frac{1}{k} ∑_{i=1}^{k} Λ_{i})^{2} − (1 + \frac{4(n + 2)}{n^{2}}) \frac{1}{k} ∑_{j=1}^{k} (Λ_{j} − \frac{1}{k} ∑_{i=1}^{k} Λ_{i})^{2}]^{\frac{1}{2}} . \end{matrix}

Thus, we derive

Λ_{k+1}−Λ_{k} ≤ 2 [(\frac{2(n + 2)}{n^{2}} \frac{1}{k} ∑_{i=1}^{k} Λ_{i})^{2} − (1 + \frac{4(n + 2)}{n^{2}}) \frac{1}{k} ∑_{j=1}^{k} (Λ_{j} − \frac{1}{k} ∑_{i=1}^{k} Λ_{i})^{2}]^{\frac{1}{2}} .

Hence, the corollary 2 is true.

□

Remark 3. We conjecture that under the assumption of the theorem, the following inequality is satisfied

∑_{i=1}^{k}(Λ_{k+1} − Λ_{i})^{2} ≤ \frac{4}{n} ∑_{i=1}^{k}(Λ_{k+1} − Λ_{i})Λ_{i}.

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Qing-Ming Cheng Department of Mathematics Faculty of Science and Engineering Saga University Saga 840-8502, Japan e-mail: cheng@ms.saga-u.ac.jp

Hongcang Yang Academy of Mathematics and Systematical Sciences CAS, Beijing 100080, China e-mail: yanghc2@netease.com