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On multivariate higher order Lyapunov-type inequalities

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.

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 ) | dt> 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 [1–3] 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 Ï€ N 2 Γ ( N 2 ) ,i.e.  ∫ S N − 1 dω= 2 Ï€ 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=rω, where r=∥x∥>0 and ω= x r ∈ S N − 1 . Therefore, for any f∈C( A ¯ ), we have

∫ A f(x)dx= ∫ S N − 1 ( ∫ R 1 R 2 f ( r ω ) r N − 1 d r ) dω.

In [1], AktaÅŸ 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 ) | dx> 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 ) | dx> 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 dt

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 dt

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(2n), C(n), and D(n) in Table 1.

Table 1 The first seven values of ζ(2n) , C(n) and D(n)

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)dt= ( − 1 ) n ∫ a b ( x ( n ) ( t ) ) 2 dt=− ∑ k = 0 n ∫ a b p k (t) x ( k ) (t)x(t)dt.

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 ) | dt≤ ( ∫ 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 dt, 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 ) | dt.

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 ) | dx≥ R 1 N − 1 2 π N 2 Γ ( N 2 ) ∫ R 1 R 2 | f ( r ω ) | dr.
(24)

Proof Similar to the proofs given in [1] and [2], we have

∫ R 1 R 2 | f ( r ω ) | dr= ∫ R 1 R 2 r 1 − N r N − 1 | f ( r ω ) | dr≤ ( ∫ 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 ) | dx> 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 = Ï€ 2 n 2 n ( 2 2 n − 1 ) ζ ( 2 n ) > Ï€ 2 n 2 3 n ζ ( 2 n ) = ( Ï€ 2 8 ) n 1 ζ ( 2 n ) →∞,as n→∞,

since ζ(2n)→1 as n→∞. Table 2 gives the first eight values of δ n .

Table 2 The first eight values of δ n

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).

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

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Ji, T., Fan, J. On multivariate higher order Lyapunov-type inequalities. J Inequal Appl 2014, 503 (2014). https://doi.org/10.1186/1029-242X-2014-503

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