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Lyapunov-type inequalities for 2M th order equations under clamped-free boundary conditions

  • Kohtaro Watanabe1Email author,
  • Kazuo Takemura2,
  • Yoshinori Kametaka3,
  • Atsushi Nagai2 and
  • Hiroyuki Yamagishi4
Journal of Inequalities and Applications20122012:242

https://doi.org/10.1186/1029-242X-2012-242

Received: 29 May 2012

Accepted: 9 October 2012

Published: 23 October 2012

Abstract

This paper generalizes the well-known Lyapunov-type inequalities for second-order linear differential equations to certain 2M th order linear differential equations

( 1 ) M u ( 2 M ) ( x ) r ( x ) u ( x ) = 0 ( s x s )

under clamped-free boundary conditions. The usage of the best constant of some kind of a Sobolev inequality helps clarify the process for obtaining the result.

Keywords

Green FunctionClassical SolutionLinear Differential EquationSobolev InequalityDistributional Sense

1 Introduction

Let us consider the second-order linear differential equation
{ u ( x ) + r ( x ) u ( x ) = 0 ( s x s ) , u ( ± s ) = 0 ,
(1)
where . It is well known that the Lyapunov inequality
s s r + ( x ) d x > 2 s
(2)
gives a necessary condition for the existence of non-trivial classical solutions of (1), where r + ( x ) = ( r ( x ) + | r ( x ) | ) / 2 . There are various extensions and applications for the above result; see, for example, surveys of Brown and Hinton [1] for relations to other fields and Tiryaki [2] for recent developments. Extensions to higher-order equations
{ u ( n ) ( x ) + r ( x ) u ( x ) = 0 ( s x s ) , Boundary Conditions ,
(3)

will be one important aspect. The first result for the high-order equation (3) is due to Levin [3], which states without proof:

Theorem A Let n = 2 M , and a non-trivial solution of (3) satisfies the clamped boundary condition, u ( i ) ( ± s ) = 0 ( i = 0 , 1 , , M 1 ). Then it holds that
s s r + ( x ) d x > 2 2 M 1 ( 2 M 1 ) { ( M 1 ) ! } 2 s 2 M 1 .
Later, Das and Vatsala [4] gave the proof and extended the result by constructing the Green function. Other interesting developments for higher-order equations are seen in [59]. For example, as shown in Yang [8], Lyapunov-type inequalities can be obtained under the following conditions:
{ ( a ) n = 2 M + 1 , u ( i ) ( ± s ) = 0 ( i = 0 , , M 1 ) , u ( 2 M ) ( d ) = 0 ( s < d < s )  (see also [10]) ( b ) n 2 , u ( s ) = u ( t 2 ) = = u ( t n 1 ) = u ( s ) = 0 , where,  s = t 1 < t 2 < < t n 1 < t n = s  (see also [10]) ( c ) n = 2 M , u ( 2 i ) ( ± s ) = 0 ( i = 0 , , M 1 )  (see also [5]) ( d ) n 2 , u ( i ) ( s ) = 0 ( i = 0 , , k 1 ) , u ( j ) ( s ) = 0 ( j = 0 , , n k 1 ) where  k  runs over the range  ( 1 k n ) .
Here we note for the condition (c), very recently Çakmak [11], He and Tang [12], He and Zang [13] and [14] improved and extended the results of [5] and [8]. This paper considers the necessary condition for the existence of a non-trivial solution of the 2M th order linear differential equation
( 1 ) M u ( 2 M ) ( x ) r ( x ) u ( x ) = 0 ( s x s )
(4)

under yet another boundary condition:

Clamped-free boundary condition
u ( i ) ( s ) = 0 , u ( M + i ) ( s ) = 0 ( i = 0 , , M 1 ) .

The main result is as follows.

Theorem 1 Suppose a non-trivial solution u of (4) exists under the clamped-free boundary condition, then it holds
s s r + ( x ) d x > { ( M 1 ) ! } 2 ( 2 M 1 ) ( 2 s ) 2 M 1 .
(5)

Moreover, the estimate is sharp in the sense that there exists a function r ( x ) , and for this r ( x ) , the solution u of (4) exits such that the right-hand side is arbitrarily close to the left-hand side.

The result is obtained using Takemura [[15], Theorem 1], which computes the best constant of some kind of a Sobolev inequality. In Section 4, we give a concise proof for an L p extension of Theorem 1 of [15].

2 Proof of Theorem 1

Now, let us introduce the following L p -type Sobolev inequality:
( sup s x s | u ( m ) ( x ) | ) p C s s | u ( M ) ( x ) | p d x ,
(6)
where u belongs to
W ( M , p ) : = { u | u ( M ) L p ( s , s ) , u ( i ) ( s ) = 0 ( i = 0 , , M 1 ) } ,

1 < p , m runs over the range 0 m M 1 , and u ( i ) is the i th derivative of u in a distributional sense. We denote by C C F ( M , m , p ) the best constant of the above Sobolev inequality (6). Here, we note that in [15], Takemura obtained the best constant for p = 2 , m = 0 by constructing the Green function of the clamped-free boundary value problem. Although, for the proof of Theorem 1, we simply need the value C C F ( M , 0 , 2 ) , we would like to compute C C F ( M , m , p ) for general p and m since the proof presented in Section 4 does not depend on special values of p and m and quite simplifies the proof of Theorem 1 of [15]. Now, we have the following propositions.

Proposition 1 The best constant of (6) is
C C F ( M , m , p ) = 1 { ( M m 1 ) ! } p ( ( p 1 ) ( 2 s ) p ( M m ) 1 p 1 p ( M m ) 1 ) p 1 ,
(7)
and it is attained by
u ( x ) = s x ( x t ) M 1 ( M 1 ) ! { ( s t ) M 1 m ( M 1 m ) ! } p 1 d t .
(8)
Proposition 2 Suppose a C 2 M [ s , s ] solution of (4) with the clamped-free boundary condition exists, then it holds that
s s r + ( x ) d x > 1 C C F ( M , 0 , 2 ) .
(9)

Moreover, the estimate is sharp.

Proof of Theorem 1 Clearly, Theorem 1 is obtained from Propositions 1 and 2. □

Thus, all we have to do is to show Propositions 1 and 2. Before proceeding with the proof of these propositions, we would like to show a corollary obtained from Proposition 1.

Corollary 1 Suppose a non-trivial solution u of the non-linear equation
( 1 ) M u ( x ) u ( 2 M ) ( x ) r ( x ) ( u ( m ) ( x ) ) 2 = 0
(10)
exists under the clamped-free boundary condition, where m satisfies ( 1 m M 1 ), then it holds
s s r + ( x ) d x > { ( M 1 m ) ! } 2 ( 2 ( M m ) 1 ) ( 2 s ) 2 ( M m ) 1 .
(11)

The following are the examples of Theorem 1 and Corollary 1.

Example 1 The following example corresponds to the case M = 1 and r ( x ) = 6 / ( 11 s 2 + 2 s x + x 2 ) of (4) with the clamped-free boundary condition
{ u ( x ) + 6 11 s 2 + 2 s x + x 2 u ( x ) = 0 , u ( s ) = u ( s ) = 0 .
It is easy to see that u ( x ) = ( s + x ) ( 11 s 2 2 s x x 2 ) is the solution of the above equation. Moreover, it holds that
s s r + ( x ) d x = s s 6 11 s 2 + 2 s x + x 2 d x = 3 log ( 2 + 3 ) 2 s > 1 C C F ( 1 , 0 , 2 ) = 1 2 s .
Example 2 The following example corresponds to the case M = 2 , m = 1 and r ( x ) = ( 3 ( 17 s 2 6 s x + x 2 ) ) / ( 2 ( 7 s 2 4 s x + x 2 ) 2 ) of (10) with the clamped-free boundary condition
{ u ( x ) u ( 4 ) ( x ) 3 ( 17 s 2 6 s x + x 2 ) 2 ( 7 s 2 4 s x + x 2 ) 2 ( u ( x ) ) 2 = 0 , u ( s ) = u ( s ) = u ( s ) = u ( s ) = 0 .
It is easy to see that u ( x ) = ( s + x ) 2 ( 17 s 2 6 s x + x 2 ) is the solution of the above equation. Moreover, it holds that
s s r + ( x ) d x = s s 3 ( 17 s 2 6 s x + x 2 ) 2 ( 7 s 2 4 s x + x 2 ) 2 d x = 3 + 2 3 π 12 s > 1 C C F ( 2 , 1 , 2 ) = 1 2 s .

3 Proof of Proposition 2

Assuming Proposition 1, we first prove Proposition 2.

Proof of Proposition 2 Let u be a solution of equation (4). Since u satisfies the clamped-free boundary condition, multiplying (4) by u and integrating it over [ s , s ] , we have
s s ( u ( M ) ( x ) ) 2 d x = s s ( 1 ) M u ( x ) u ( 2 M ) ( x ) d x = s s r ( x ) ( u ( x ) ) 2 d x ( sup s x s | u ( x ) | ) 2 s s r + ( x ) d x C ( M , 0 , 2 ) s s ( u ( M ) ( x ) ) 2 d x s s r + ( x ) d x .
(12)
Here, if u ( M ) 0 , then there exists ( i = 0 , , M 1 ) such that u ( x ) = i = 0 M 1 a i x i . Since u satisfies the clamped boundary condition at x = s , we have u 0 . This contradicts the assumption that u is a non-trivial solution of (4). So, canceling u ( M ) 2 2 , we obtain
s s r + ( x ) d x 1 C C F ( M , 0 , 2 ) .
(13)
Next, we show that the inequality (13) is strict. To see this, we note that in (12), if the equality holds for the first inequality, then u is a constant. But, again from the clamped boundary condition at x = s , we have u 0 . Thus, the inequality is strict. Finally, we see (5) is sharp. For this purpose, let us define the functional
J ( ϕ ) : = s s | ϕ ( M ) | 2 d x s s r ˜ | ϕ | 2 d x ( ϕ W ( M , 2 ) , ϕ 0 ) ,
where is defined later. By the standard argument of the variational method, J has the minimizer u W ( M , 2 ) (see, for example, [[16], Lemma 3]), i.e.,
λ 1 : = min ϕ W ( M , 2 ) , ϕ 0 J ( ϕ ) = J ( u ) .
Hence, it satisfies the Euler-Lagrange equation (as a classical solution by the regularity argument)
( 1 ) M u ( 2 M ) ( x ) = λ 1 r ˜ ( x ) u ( x ) ( s x s ) .
(14)
Further, it holds that
λ 1 = min ϕ W ( M , 2 ) , ϕ 0 s s | ϕ ( M ) | 2 d x s s r ˜ | ϕ | 2 d x > s s | ϕ ( M ) | 2 d x ( sup s x s | ϕ | ) 2 s s r ˜ d x 1 C C F ( M , 0 , 2 ) s s r ˜ d x .
(15)
Here, let us fix r ˜ as
r ˜ ( x ) : = { x s δ + 1 ( s δ < x s ) , 0 ( s x s δ ) .
For such r ˜ , let us substitute ϕ = u (of Proposition 1) into (15). It is easy to see that u takes its maximum at x = s , hence by taking δ sufficiently small, we see that the right-hand side of (15) can be arbitrarily close to the left-hand side, i.e., for a small positive ϵ 1 , λ 1 can be written as
λ 1 = 1 C C F ( M , 0 , 2 ) s s r ˜ d x + ϵ 1 .
(16)
Putting r = λ 1 r ˜ , we see from (14) a solution u of
( 1 ) M u ( 2 M ) ( x ) = r ( x ) u ( x ) ( s x s )
(17)
exists, and from (16) r satisfies
s s r ( x ) d x = 1 C ( M , 0 , 2 ) + ϵ 1 s s r ˜ d x = 1 C ( M , 0 , 2 ) + ϵ 2 .
(18)

Hence, (5) is sharp. □

Proof of Corollary 1 Integrating equation (10), we have
s s ( u ( M ) ( x ) ) 2 d x = s s r ( x ) ( u ( m ) ( x ) ) 2 d x ( sup s x s | u ( m ) ( x ) | ) 2 s s r + ( x ) d x C ( M , m , 2 ) s s ( u ( M ) ( x ) ) 2 d x s s r + ( x ) d x .
(19)
As in the proof of Proposition 2, by canceling u ( M ) 2 2 , we have
s s r + ( x ) d x 1 C C F ( M , m , 2 ) .

Next, we show that the inequality (11) is strict. To see this, we note that in (19), the equality holds for the first inequality if and only if u ( m ) is a constant. Hence, from the clamped boundary condition at x = s , we have u ( m ) 0 . So, there exists ( i = 0 , , m 1 ) such that u ( x ) = i = 0 m 1 a i x i . But, again from the clamped boundary condition at x = s , we have u 0 . Thus, inequality (11) is strict. □

4 Proof of Proposition 1

We prepare the following lemmas for the proof of Proposition 1.

Lemma 1 Suppose there exists a function u W ( M , p ) which attains the best constant C ( M , n , p ) of (6), then it holds that
max s x s | u ( m ) ( x ) | = | u ( m ) ( s ) | .
Proof Suppose it holds that
max s x s | u ( m ) ( x ) | = | u ( m ) ( a ) | ,
where a s . Further, let us define
u ˜ ( x ) : = { 0 ( s x a ) , u ( x + a s ) ( a x s ) .
Then it holds u ˜ W ( M , p ) and
max s x s | u ˜ ( m ) ( x ) | = max s x s | u ( m ) ( x ) | = | u ( m ) ( a ) |
and u ˜ ( M ) L p ( s , s ) < u ( M ) L p ( s , s ) . Hence,
C ( M , m , p ) = ( max s x s | u ( m ) ( x ) | ) p u ( M ) L p ( s , s ) < ( max s x s | u ˜ ( m ) ( x ) | ) p u ˜ ( M ) L p ( s , s ) .

This contradicts the assumption that C ( M , m , p ) is the best constant of (6). □

Lemma 2 Let
H m ( x ) : = ( 1 ) M 1 m ( x s ) M 1 m ( M 1 m ) ! ,
then for u W ( M , p ) it holds that
u ( m ) ( s ) = s s u ( M ) ( x ) H m ( x ) d x .
(20)

Proof Integrating by parts, we obtain the result. □

Proof of Proposition 1 From Lemma 2, we see that if the function attains the best constant C ( M , m , p ) , it belongs to W ( M , m , p ) W ( M , p ) :
W ( M , m , p ) = { u W ( M , p ) | max s x s | u ( m ) ( x ) | = | u ( m ) ( s ) | } .
Let u W ( M , m , p ) . Then applying Hölder’s inequality to (20), we have
max s x s | u ( m ) ( x ) | = | u ( m ) ( s ) | H m L q ( s , s ) u ( M ) L p ( s , s ) ,
(21)
where q satisfies 1 / p + 1 / q = 1 . Hence, if there exists the function u W ( M , m , p ) which attains the equality of (21), it holds that C ( M , m , p ) = H m L q ( s , s ) p . On the contrary, we see that the equality holds for (21) if and only if u satisfies
u ( M ) ( x ) = ( sgn H m ( x ) ) | H m ( x ) | q 1 .
(22)
It is easy to see that
u ( x ) = s x ( x t ) m 1 ( M 1 ) ! ( sgn H m ( t ) ) | H m ( t ) | q 1 d t = s x ( x t ) m 1 ( M 1 ) ! { ( s t ) M 1 m ( M 1 m ) ! } 1 p 1 d t
satisfies (22) and belongs to W ( M , m , p ) . Thus, we have shown C ( M , m , p ) = H m L q ( s , s ) p . Now, we compute H m L q ( s , s ) p . It is
H m L q ( s , s ) p = 1 { ( M 1 m ) ! } p { s s ( s x ) q ( M 1 m ) d x } p q = 1 { ( M 1 m ) ! } p { ( p 1 ) ( 2 s ) p ( M m ) 1 p 1 p ( M m ) 1 } p 1 .

This completes the proof. □

Declarations

Authors’ Affiliations

(1)
Department of Computer Science, National Defense Academy, Yokosuka, Japan
(2)
Liberal Arts and Basic Sciences, College of Industrial Technology, Nihon University, Narashino, Japan
(3)
Graduate School of Mathematical Sciences, Faculty of Engineering Science, Osaka University, Toyonaka, Japan
(4)
Tokyo Metropolitan College of Industrial Technology, Shinagawa, Tokyo, Japan

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© Watanabe et al.; licensee Springer 2012

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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