Open Access

On multivariate higher order Lyapunov-type inequalities

Journal of Inequalities and Applications20142014:503

https://doi.org/10.1186/1029-242X-2014-503

Received: 30 September 2014

Accepted: 1 December 2014

Published: 12 December 2014

Abstract

In this paper, by using the best Sobolev constant method, we obtain some new Lyapunov-type inequalities for a class of even-order partial differential equations; the results of this paper are new which generalize and improve some early results in the literature.

Keywords

Lyapunov-type inequality even-order partial differential equations Sobolev constant

1 Introduction

It is well known that the Lyapunov inequality for the second-order linear differential equation
x ( t ) + q ( t ) x ( t ) = 0
(1)
states that if q C [ a , b ] , x ( t ) is a nonzero solution of (1) such that x ( a ) = x ( b ) = 0 , then the following inequality holds:
a b | q ( t ) | d t > 4 b a
(2)

and the constant 4 is sharp.

There have been many proofs and generalizations as well as improvements on this inequality. For example, the authors in [13] generalized the Lyapunov-type inequality to the partial differential equations or systems.

First let us recall some background and notations which are introduced in [1, 2].

Let A be a spherical shell R N for N > 1 , i.e. A = B ( 0 , R 2 ) B ( 0 , R 1 ) ¯ for 0 < R 1 < R 2 , where B ( 0 , R ) = { x R N : x < R } for R > 0 and is the Euclidean norm. Denote S N 1 = { x R N : x = 1 } , the unit sphere in R N with surface area
ω N = 2 π N 2 Γ ( N 2 ) , i.e.  S N 1 d ω = 2 π N 2 Γ ( N 2 ) ,
(3)
where Γ ( ) is the gamma function. Then every x R N { 0 } has a unique representation of the form x = r ω , where r = x > 0 and ω = x r S N 1 . Therefore, for any f C ( A ¯ ) , we have
A f ( x ) d x = S N 1 ( R 1 R 2 f ( r ω ) r N 1 d r ) d ω .

In [1], Aktaş obtained the following results.

Theorem A If f C 2 n ( A ¯ ) is a nonzero solution of the following even-order partial differential equation:
2 n f ( x ) r 2 n + q ( x ) f ( x ) = 0 , x A ,
(4)
where n N and q C ( A ¯ ) , with the boundary conditions
2 i f r 2 i ( B ( 0 , R 1 ) ) = 2 i f r 2 i ( B ( 0 , R 2 ) ) = 0 , i = 0 , 1 , 2 , , n 1 ,
(5)
then the following inequality holds:
A | q ( x ) | d x > 2 3 n 1 ( R 2 R 1 ) 2 n 1 2 π N 2 Γ ( N 2 ) R 1 N 1 .
(6)
Theorem B If f C 2 n ( A ¯ ) is a nonzero solution of (4) with the boundary conditions
i f r i ( B ( 0 , R 1 ) ) = i f r i ( B ( 0 , R 2 ) ) = 0 , i = 0 , 1 , 2 , , n 1 ,
(7)
then the following inequality holds:
A | q ( x ) | d x > 4 2 n 1 ( 2 n 1 ) [ ( n 1 ) ! ] 2 ( R 2 R 1 ) 2 n 1 2 π N 2 Γ ( N 2 ) R 1 N 1 .
(8)

In this paper, we generalize Theorem A and Theorem B to a more general class of even order partial differential equations. Moreover, as we shall see by the end of this paper, Theorem 1 improves Theorem A significantly.

2 Main results

Let us consider the following even-order partial differential equation:
2 n y ( x ) r 2 n + k = 0 n p k ( x ) k y ( x ) r k = 0 ,
(9)

where y ( x ) C 2 n ( A ¯ ) , p k ( x ) C ( A ¯ ) , k = 0 , 1 , 2 , , n , and x R N .

The main results of this paper are the following theorems.

Theorem 1 If y ( x ) is a nonzero solution of (9) satisfying boundary conditions (5), then the following inequality holds:
1 < ( 2 2 n 1 ) ζ ( 2 n ) ( R 2 R 1 ) 2 n 1 Γ ( N 2 ) 2 2 n π 2 n + N 2 R 1 N 1 ( A p n 2 ( x ) d x ) 1 2 + k = 0 n 1 ( R 2 R 1 ) 2 n k 1 Γ ( N 2 ) ( 2 π ) 2 n k R 1 N 1 π N 2 ( 2 2 n 1 ) ( 2 2 ( n k ) 1 ) ζ ( 2 n ) ζ ( 2 ( n k ) ) × A | p k ( x ) | d x ,
(10)

where ζ ( s ) = n = 1 + 1 n s is the Riemann zeta function.

Theorem 2 If y ( x ) is a nonzero solution of (9) satisfying boundary conditions (7), then the following inequality holds:
1 < 1 ( n 1 ) ! 2 2 n 1 ( R 2 R 1 ) 2 n 1 Γ ( N 2 ) ( 2 n 1 ) R 1 N 1 2 π N 2 ( A p n 2 ( x ) d x ) 1 2 + k = 0 n 1 ( R 2 R 1 ) 2 n k 1 Γ ( N 2 ) ( 2 n 1 ) ( 2 n 2 k 1 ) ( n 1 ) ! ( n k 1 ) ! 4 2 n k 1 R 1 N 1 2 π N 2 A | p k ( x ) | d x .
(11)

3 Proofs of theorems

For the proofs of Theorem 1 and Theorem 2, let us consider first the following ordinary even-order linear ordinary differential equation:
x ( 2 n ) ( t ) + k = 0 n p k ( t ) x ( k ) ( t ) = 0 ,
(12)

where p k ( t ) C [ a , b ] , k = 0 , 1 , 2 , , n .

Proposition 3 If (12) has a nonzero solution x ( t ) satisfying the following boundary value conditions:
x ( 2 i ) ( a ) = x ( 2 i ) ( b ) = 0 , i = 0 , 1 , 2 , , n 1 ,
(13)
then the following inequality holds:
1 < ( 2 2 n 1 ) ζ ( 2 n ) ( b a ) 2 n 1 2 2 n 1 π 2 n ( a b p n 2 ( t ) d t ) 1 2 + k = 0 n 1 ( b a ) 2 n k 1 ( 2 2 n 1 ) ( 2 2 n 2 k 1 ) ζ ( 2 n ) ζ ( 2 ( n k ) ) 2 2 n k 1 π 2 n k a b | p k ( t ) | d t ,
(14)

where ζ ( s ) is the Riemann zeta function: ζ ( s ) = k = 1 + 1 k s , s > 1 .

Proposition 4 If (12) has a nonzero solution x ( t ) satisfying the following boundary value conditions:
x ( i ) ( a ) = x ( i ) ( b ) = 0 , i = 0 , 1 , 2 , , n 1 ,
(15)
then we have the following inequality:
1 < 1 ( n 1 ) ! 2 2 n 1 ( b a ) 2 n 1 ( 2 n 1 ) ( a b p n 2 ( t ) d t ) 1 2 + k = 0 n 1 ( b a ) 2 n k 1 ( n 1 ) ! ( n k 1 ) ! 4 2 n k 1 ( 2 n 1 ) ( 2 n 2 k 1 ) a b | p k ( t ) | d t .
(16)

In order to prove the above propositions, we need the following lemmas.

Lemma 5 ([[4], Proposition 2.1])

Let M N and
H C = { u | u ( M ) L 2 ( a , b ) , u ( 2 k ) ( a ) = u ( 2 k ) ( b ) = 0 , 0 k [ ( M 1 ) / 2 ] } .
Then there exists a positive constant C such that, for any u H C , the Sobolev inequality
( sup a t b | u ( t ) | ) 2 C a b | u ( M ) ( t ) | 2 d t
holds. Moreover, the best constant C = C ( M ) is as follows:
C ( M ) = ( 2 2 M 1 ) ζ ( 2 M ) ( b a ) 2 M 1 2 2 M 1 π 2 M .

Lemma 6 ([[5], Theorem 1.2 and Corollary 1.3])

Let M N and
H D = { u | u ( M ) L 2 ( a , b ) , u ( k ) ( a ) = u ( k ) ( b ) = 0 , 0 k M 1 } .
Then there exists a positive constant D such that for any u H D , the Sobolev inequality
( sup a t b | u ( t ) | ) 2 D a b | u ( M ) ( t ) | 2 d t
holds. Moreover, the best constant D = D ( M ) is as follows:
D ( M ) = ( b a ) 2 M 1 ( 2 M 1 ) [ ( M 1 ) ! ] 2 4 2 M 1 .
(17)
We give the first seven values of ζ ( 2 n ) , C ( n ) , and D ( n ) in Table 1.
Table 1

The first seven values of ζ ( 2 n ) , C ( n ) and D ( n )

n

1

2

3

4

5

6

7

ζ(2n)

π 2 6

π 4 90

π 6 945

π 8 9 , 450

π 10 93 , 555

691 π 12 638 , 512 , 875

2 π 14 18 , 243 , 225

C(n)

b a 4

( b a ) 3 48

( b a ) 5 480

17 ( b a ) 7 80 , 640

31 ( b a ) 9 1 , 451 , 520

691 ( b a ) 11 9 , 123 , 840

( 2 14 1 ) ( b a ) 13 2 13 × 18 , 243 , 225

D(n)

b a 4

( b a ) 3 192

( b a ) 5 20 , 480

( b a ) 7 4 , 128 , 768

( b a ) 9 1 , 358 , 954 , 496

( b a ) 11 664 , 377 , 753 , 600

( b a ) 13 13 ( 6 ! ) 2 4 13

Since the proof of Proposition 4 is similar to that of Proposition 3, we give only the proof of Proposition 3 below.

Proof of Proposition 3 Multiplying both sides of (12) by x ( t ) and integrating from a to b by parts and using the boundary value condition (13), we can obtain
a b x ( 2 n ) ( t ) x ( t ) d t = ( 1 ) n a b ( x ( n ) ( t ) ) 2 d t = k = 0 n a b p k ( t ) x ( k ) ( t ) x ( t ) d t .
This yields
a b ( x ( n ) ( t ) ) 2 d t k = 0 n a b | p k ( t ) | | x ( k ) ( t ) x ( t ) | d t = a b | p n ( t ) | | x ( n ) ( t ) x ( t ) | d t + k = 0 n 1 a b | p k ( t ) | | x ( k ) ( t ) x ( t ) | d t .
(18)
Now, by using Lemma 5, we get for any t [ a , b ] , k = 1 , 2 , , n 1 ,
| x ( t ) | C ( n ) ( a b ( x ( n ) ( t ) ) 2 d t ) 1 2
(19)
and
| x ( k ) ( t ) | C ( n k ) ( a b ( x ( n ) ( t ) ) 2 d t ) 1 2 .
(20)
Substituting (19) and (20) into (18), we obtain
a b ( x ( n ) ( t ) ) 2 d t C ( n ) a b | p n ( t ) | | x ( n ) ( t ) | d t ( a b ( x ( n ) ( t ) ) 2 d t ) 1 2 + k = 0 n 1 C ( n ) C ( n k ) a b | p k ( t ) | d t a b ( x ( n ) ( t ) ) 2 d t .
(21)
Now by applying Hölder’s inequality, we get
a b | p n ( t ) x ( n ) ( t ) | d t ( a b p n 2 ( t ) d t ) 1 2 ( a b ( x ( n ) ( t ) ) 2 d t ) 1 2 .
(22)
Substituting (22) into (21) and by using the fact that x ( t ) is not a constant function, we obtain the following strict inequality:
a b ( x ( n ) ( t ) ) 2 d t < C ( n ) ( a b p n 2 ( t ) d t ) 1 2 a b ( x ( n ) ( t ) ) 2 d t + k = 0 n 1 C ( n ) C ( n k ) a b | p k ( t ) | d t a b ( x ( n ) ( t ) ) 2 d t .
(23)
Dividing both sides of (23) by a b ( x ( n ) ( t ) ) 2 d t , which can be proved to be positive by using the boundary value condition (13) and the assumption that x ( t ) 0 , we obtain
1 < C ( n ) ( a b p n 2 ( t ) d t ) 1 2 + k = 0 n 1 C ( n ) C ( n k ) a b | p k ( t ) | d t .

This is equivalent to (14). Thus we finished the proof of Proposition 3. □

Lemma 7 For any f C ( A ) , we have
A | f ( x ) | d x R 1 N 1 2 π N 2 Γ ( N 2 ) R 1 R 2 | f ( r ω ) | d r .
(24)
Proof Similar to the proofs given in [1] and [2], we have
R 1 R 2 | f ( r ω ) | d r = R 1 R 2 r 1 N r N 1 | f ( r ω ) | d r ( R 1 R 2 r N 1 | f ( r ω ) | d r ) R 1 1 N ,
which implies that
A | f ( x ) | d x = S N 1 ( R 1 R 2 r N 1 | f ( r ω ) | d r ) d ω S N 1 ( R 1 N 1 R 1 R 2 | f ( r ω ) | d r ) d ω = ( R 1 R 2 | f ( r ω ) | d r ) R 1 N 1 2 π N 2 Γ ( N 2 ) .

 □

Proof of Theorem 1 It follows from (14) and Lemma 7 that for any fixed ω S N 1 , we have
1 < ( 2 2 n 1 ) ζ ( 2 n ) ( R 2 R 1 ) 2 n 1 2 2 n 1 π 2 n ( R 1 R 2 p n 2 ( r ω ) d r ) 1 2 + k = 0 n 1 ( R 2 R 1 ) 2 n k 1 ( 2 2 n 1 ) ( 2 2 n 2 k 1 ) ζ ( 2 n ) ζ ( 2 ( n k ) ) 2 2 n k 1 π 2 n k a b | p k ( r ω ) | d r ( 2 2 n 1 ) ζ ( 2 n ) ( R 2 R 1 ) 2 n 1 Γ ( N 2 ) 2 2 n π 2 n + N 2 R 1 N 1 ( A p n 2 ( x ) d x ) 1 2 + k = 0 n 1 ( R 2 R 1 ) 2 n k 1 Γ ( N 2 ) ( 2 π ) 2 n k R 1 N 1 π N 2 ( 2 2 n 1 ) ( 2 2 ( n k ) 1 ) ζ ( 2 n ) ζ ( 2 ( n k ) ) A | p k ( x ) | d x ,

which is (10). This finishes the proof of Theorem 1. □

The proof of Theorem 2 is similar to that of Theorem 1, so we omit it for simplicity.

Let us compare Theorem 1 and Theorem 2 with Theorem A and Theorem B. It is evident that Theorem 2 is a natural generalization of Theorem B. If we let p n ( x ) = p n 1 ( x ) = = p 1 ( x ) 0 , p 0 ( x ) = q ( x ) , x A , then (10) reduces to the following inequality:
A | q ( x ) | d x > 2 2 n 1 π 2 n ( 2 2 n 1 ) ζ ( 2 n ) ( R 2 R 1 ) 2 n 1 2 π N 2 Γ ( N 2 ) R 1 N 1 .
(25)
Let us compare the right sides of inequalities (6) and (25): if we denote δ n = 2 2 n 1 π 2 n ( 2 2 n 1 ) ζ ( 2 n ) 2 3 n 1 , then we have
δ n = π 2 n 2 n ( 2 2 n 1 ) ζ ( 2 n ) > π 2 n 2 3 n ζ ( 2 n ) = ( π 2 8 ) n 1 ζ ( 2 n ) , as  n ,
since ζ ( 2 n ) 1 as n . Table 2 gives the first eight values of δ n .
Table 2

The first eight values of δ n

n

1

2

3

4

5

6

7

8

δ n

1

1.50

1.42

2.32

2.86

3.53

4.35

5.37

From Table 2 we see that δ n increases very quickly, so Theorem 1 improves Theorem A significantly even in the special case of (4).

Declarations

Acknowledgements

The authors thank the anonymous referees for their valuable suggestions and comments on the original manuscript.

Authors’ Affiliations

(1)
College of Science, Hebei University of Engineering

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Copyright

© Ji and Fan; licensee Springer. 2014

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