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Lyapunov-type inequalities for even order difference equations
Applied Mathematics Letters 25 (2012) 1830–1834
Contents lists available at SciVerse ScienceDirect
Applied Mathematics Letters journal homepage: www.elsevier.com/locate/aml
Lyapunov-type inequalities for even order difference equations✩ Qi-Ming Zhang a,b , X.H. Tang a,∗ a
School of Mathematical Sciences and Computing Technology, Central South University, Changsha 410083, Hunan, PR China
b
College of Science, Hunan University of Technology, Zhuzhou 412000, Hunan, PR China
article
abstract
info
Article history: Received 8 September 2011 Received in revised form 12 February 2012 Accepted 12 February 2012
In this work, we establish two new Lyapunov-type inequalities for the 2k-order difference equation
△2k x(n) + (−1)k−1 q(n)x(n + 1) = 0. Applying our inequalities, we obtain the lower bounds of the eigenvalue for a related eigenvalue problem. © 2012 Elsevier Ltd. All rights reserved.
Keywords: Even order Difference equation Lyapunov-type inequality
1. Introduction In 1907, Lyapunov [1] first established the so-called Lyapunov inequality
(b − a)
b
|q(t )|dt > 4,
(1.1)
a
for where the Hill equation x′′ (t ) + q(t )x(t ) = 0
(1.2)
has a real solution x(t ) such that x(a) = x(b) = 0,
x(t ) ̸≡ 0,
t ∈ [a, b],
(1.3)
where a, b ∈ R with a < b. The constant 4 in (1.1) cannot be replaced by a larger number. Since then, many generalizations of (1.1) have been established in the literature. A thorough literature review of continuous and discrete Lyapunov-type inequalities and their applications can be found in the survey paper [2] by Cheng. For other related references, we refer the reader to the few papers [3–9]. In 1983, Cheng [10] first obtained the following Lyapunov inequality which is a discrete analogue of (1.1):
F(b − a)
b−2
q(n) ≥ 4,
(1.4)
n=a
where a, b ∈ Z and
F(m) =
2 m − 1, m
m,
if m − 1 is even,
(1.5)
if m − 1 is odd,
✩ This work was partially supported by the NNSF (No. 11171351) of China and by Hunan Provincial Natural Science Foundation of China (11JJ2005).
∗
Corresponding author. Tel.: +86 73188877331. E-mail address: [email protected] (X.H. Tang).
0893-9659/$ – see front matter © 2012 Elsevier Ltd. All rights reserved. doi:10.1016/j.aml.2012.02.031
Q.-M. Zhang, X.H. Tang / Applied Mathematics Letters 25 (2012) 1830–1834
1831
if the second-order difference equation
△2 x(n) + q(n)x(n + 1) = 0
(1.6)
has a real solution x(n) such that x(a) = x(b) = 0,
x(n) ̸≡ 0,
n ∈ Z[a, b],
(1.7)
where and in the sequel, a, b ∈ N, Z[a, b] = {a, a + 1, a + 2, . . . , b − 1, b} and Z(a, b) = {a + 1, a + 2, . . . , b − 2, b − 1}. The constant 4 in (1.4) cannot be replaced by a larger number, either. For more discrete Lyapunov-type inequalities, we refer the reader to Cheng [11–13], Clark and Hinton [14,15], Guseinov and Kaymakcalan [16], Lin and Yang [17] and Zhang and Tang [18]. Consider the following even order difference equation:
△2k x(n) + (−1)k−1 q(n)x(n + 1) = 0,
(1.8)
where k ∈ N, n ∈ Z and q(n) is a real-valued function defined on Z. When k = 1, (1.8) reduces to (1.6). In this work, we establish two discrete Lyapunov-type inequalities for (1.8) under the following boundary conditions:
△2i x(a) = △2i x(b) = 0,
i = 0, 1, . . . , k − 1; x(n) ̸≡ 0, n ∈ Z[a, b].
(1.9)
Furthermore, applying our Lyapunov-type inequalities to the following eigenvalue problem:
△2k x(n) + (−1)k−1 λq(n)x(n + 1) = 0,
△2i x(a) = △2i x(b) = 0, i = 0, 1, . . . , k − 1,
(1.10)
we can obtain two different lower bounds of the eigenvalue λ. 2. The main results In this section, we establish some new Lyapunov-type inequalities for (1.8). Lemma 2.1. Assume that x(n) is a real-valued function on Z[a, b], x(a) = x(b) = 0, x(n) ̸≡ 0 for n ∈ Z[a, b]. Then
|x(n)| ≤ b−1
b−1 (n − a) (b − n) | △2 x(s)|, b−a s=a
|x(s)| ≤
s=a
b −1 1
2 s=a
∀ n ∈ Z(a, b − 1),
[(s − a + 1)(b − s − 1)| △2 x(s)|] ≤
(2.1)
b−1 (b − a)2
8
| △2 x(s)|,
(2.2)
s=a
1/2 √ b −1 (n − a) (b − n)[2(n − a) (b − n) + 1] 2 2 |x(n)| ≤ | △ x(s)| , √ 6(b − a) s =a
∀ n ∈ Z[a, b − 1],
(2.3)
and b−1
|x(n)|2 ≤
b−1 [(b − a)2 − 1][2(b − a)2 + 7]
180
n=a
| △2 x(s)|2 .
(2.4)
s=a
Proof. Since x(a) = x(b) = 0, it is easy to verify that x(n) = −
1
b −1 [G(n, s) △2 x(s)],
(2.5)
b − a s =a
where G(n, s) =
(s + 1 − a) (b − n), (n − a) (b − s − 1),
s ≤ n − 1; n, s ∈ Z[a, b], n ≤ s; n, s ∈ Z[a, b − 1].
(2.6)
Since G(n, s) ≤ (n − a) (b − n),
n, s ∈ Z[a, b − 1],
(2.7)
it follows from (2.5) that
|x(n)| ≤
b −1 b −1 (n − a) (b − n) [G(n, s)| △2 x(s)|] ≤ | △2 x(s)|, b − a s =a b−a s=a
1
n ∈ Z(a, b − 1),
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Q.-M. Zhang, X.H. Tang / Applied Mathematics Letters 25 (2012) 1830–1834
which implies that (2.1) holds. Next, by (2.5) and (2.6), we have b−1
|x(n)| ≤
n =a
=
=
≤
b −1 b −1 [G(n, s)| △2 x(s)|]
1
b − a n=a s=a b −1
1
b − a s =a b −1 1
2 s=a
| △ x(s)| 2
b−1
G(n, s)
n=a
[(s − a + 1) (b − s − 1)| △2 x(s)|]
b−1 (b − a)2
8
| △2 x(s)|,
s=a
which implies that (2.2) holds. By a relatively simple sum computation, we have b −1 [(b − a)2 − 1][2(b − a)2 + 7] [G(n, s)]2 = . 6(b − a) s=a
(2.8)
And so, by (2.5), (2.6) and (2.8) and the Cauchy inequality, we have
2 b −1 2 △ [ G ( n , s ) x ( s )] |x(n)| = (b − a)2 s=a 2 b −1 1 2 ≤ [G(n, s)| △ x(s)|] (b − a)2 s=a 1
2
≤
=
b −1
1
(b − a)
2
[G(n, s)]2
s=a
b−1
| △2 x(s)|2
s=a
b−1 (n − a) (b − n)[2(n − a) (b − n) + 1] | △2 x(s)|2 , 6(b − a) s=a
∀ n ∈ Z[a, b − 1],
(2.9)
which implies that (2.3) holds. Summing (2.9) from a to b − 1, we have b−1
|x(n)| ≤
1
2
n =a
= Hence, (2.4) holds.
(b − a)2
b −1 b−1 b −1 2 2 2 [G(n, s)] | △ x(s)| n =a
s=a
s=a
b −1 [(b − a)2 − 1][2(b − a)2 + 7]
180
| △2 x(s)|2 .
(2.10)
s=a
Theorem 2.2. Assume that k ∈ N and q(n) is a real-valued function on Z. If (1.8) has a solution x(n) satisfying the boundary value conditions (1.9), then b−1 [|q(n)|(n − a + 1) (b − n − 1)] ≥ n =a
23(k−1)
(b − a)2k−3
.
(2.11)
Proof. Choose c ∈ Z[a, b] such that |x(c )| = maxn∈[a,b] |x(n)|. Since (1.9), it follows from Lemma 2.1 that
|x(c )| ≤
b −1 b−1 b−a (c − a) (b − c ) | △2 x(n)| ≤ | △2 x(n)|, b−a 4 n=a n =a
(2.12)
and b −1 n =a
| △2i x(n)| ≤
b−1 (b − a)2
8
n=a
| △2i+2 x(n)|,
i = 1, 2, . . . , k − 2,
(2.13)
Q.-M. Zhang, X.H. Tang / Applied Mathematics Letters 25 (2012) 1830–1834
1833
and b−1
| △2k−2 x(n)| ≤
n=a
b−1 1
2 n =a
[(n − a + 1) (b − n − 1)| △2k x(n)|].
(2.14)
From (1.8) and (2.12)–(2.14), we obtain
|x(c )| ≤ ≤
≤
b−1 b−a
4
| △2 x(n)|
n=a
b−a
(b − a)2
4
k−2 b −1
8
b−a
n =a
(b − a)2
8
8
| △2k−2 x(n)|
k−2 b −1 [(n − a + 1) (b − n − 1)| △2k x(n)|] n =a
b−1 (b − a)2k−3 = [(n − a + 1) (b − n − 1)|q(n)x(n + 1)|] 3(k−1)
2
n=a
( b − a)
2k−3
≤
23(k−1)
|x(c )|
b −1 [(n − a + 1) (b − n − 1)|q(n)|].
(2.15)
n =a
Since |x(c )| > 0, it follows from (2.15) that (2.11) holds.
Corollary 2.3. Assume that k ∈ N and q(n) is a real-valued function on Z. If (1.8) has a solution x(n) satisfying the boundary value conditions (1.9), then b−1
|q(n)| ≥
n=a
23k−1
(b − a)2k−1
.
(2.16)
Theorem 2.4. Assume that k ∈ N and q(n) is a real-valued function on Z. If (1.8) has a solution x(n) satisfying the boundary value conditions (1.9), then b−1
|q(n)|2 ≥
n=a
2k+3 32k−1 5k−1
k−1 . (b − a)[(b − a)2 + 2][(b − a)2 − 1]k−1 (b − a)2 + 27
(2.17)
Proof. Choose c ∈ Z[a, b] such that |x(c )| = maxn∈[a,b] |x(n)|. Since (1.9), it follows from Lemma 2.1 and (1.8) that
|x(c )|2 ≤ ≤
=
b−1 (c − a) (b − c )[2(c − a) (b − c ) + 1] | △2 x(s)|2 6(b − a) s=a
(b − a)[(b − a)2 + 2] 48
×
b−1 [(b − a)2 − 1]k−1 [2(b − a)2 + 7]k−1
180k−1
| △2k x(s)|2
s=a
b −1 (b − a)[(b − a)2 + 2][(b − a)2 − 1]k−1 [2(b − a)2 + 7]k−1 [|q(n)|2 |x(n + 1)|2 ] 48 × 180k−1 n =a
≤ |x(c )|2
= |x(c )|
2
b −1 (b − a)[(b − a)2 + 2][(b − a)2 − 1]k−1 [2(b − a)2 + 7]k−1 |q(n)|2 48 × 180k−1 n =a
k−1 b−1 (b − a)[(b − a)2 + 2][(b − a)2 − 1]k−1 (b − a)2 + 27 2k+3 32k−1 5k−1
Since |x(c )| > 0, it follows from (2.18) that (2.17) holds.
|q(n)|2 .
(2.18)
n =a
Now, we give an application of the obtained Lyapunov-type inequalities (2.11) and (2.17) for the eigenvalue problem (1.10). We have 23k−1
|λ| ≥ (b − a)2k−3
b−1
n=a
[|q(n)|(n − a + 1) (b − n − 1)]
,
(2.19)
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Q.-M. Zhang, X.H. Tang / Applied Mathematics Letters 25 (2012) 1830–1834
and
|λ| ≥
2(k+3)/2 3(2k−1)/2 5(k−1)/2 −1 k−1 b [(b − a)[(b − a)2 + 2][(b − a)2 − 1]k−1 (b − a)2 + 27 |q(n)|2 ]1/2
.
(2.20)
n =a
References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18]
A.M. Liapunov, Problème général de la stabilité du mouvement, Ann. Fac. Sci. Univ. Toulouse 2 (1907) 203–407. S.S. Cheng, Lyapunov inequalities for differential and difference equations, Fasc. Math. 23 (1991) 25–41. D. Cakmak, Lyapunov-type integral inequalities for certain higher order differential equations, Appl. Math. Comput. 216 (2010) 368–373. X. He, X.H. Tang, Lyapunov-type inequalities for even order differential equations, Commun Pure Appl. Anal. 11 (2) (2012) 465–473. L.Q. Jiang, Z. Zhou, Lyapunov inequality for linear Hamiltonian systems on time scales, J. Math. Anal. Appl. 310 (2005) 579–593. X. Wang, Stability criteria for linear periodic Hamiltonian systems, J. Math. Anal. Appl. 367 (2010) 329–336. X. Yang, On Liapunov-type inequality for certain higher-order differential equations, Appl. Math. Comput. 134 (2–3) (2003) 307–317. X. Yang, On inequalities of Lyapunov type, Appl. Math. Comput. 134 (2-3) (2003) 293–300. X. Yang, K. Lo, Lyapunov-type inequality for a class of even-order differential equations, Appl. Math. Comput. 215 (2010) 3884–3890. S.S. Cheng, A discrete analogue of the inequality of Lyapunov, Hokkaido Math. J. 12 (1983) 105–112. S.S. Cheng, L.Y. Hsieh, On discrete analogue of Lyapunov’s inequality: addendum, Tamkang J. Math. 20 (1989) 333–340. S.S. Cheng, A sharp condition for the ground state of difference equation, Appl. Anal. 34 (1989) 105–109. S.S. Cheng, L.Y. Hsieh, D. Chao, Discrete Lyapunov inequality conditions for partial difference equations, Hokkaido Math. J. 19 (1990) 229–239. S. Clark, D.B. Hinton, Discrete Lyapunov inequalities for linear Hamiltonian systems, Math. Inequal. Appl. 1 (1998) 201–209. S. Clark, D.B. Hinton, Discrete Lyapunov inequalities, Dynam. Systems Appl. 8 (1999) 369–380. G.Sh. Guseinov, B. Kaymakcalan, Lyapunov inequalities for discrete linear Hamiltonian systems, Comput. Math. Appl. 45 (2003) 1399–1416. S.H. Lin, G.S. Yang, On discrete analogue of Lyapunov’s inequality, Tamkang J. Math. 20 (1989) 169–186. Q. Zhang, X.H. Tang, Lyapunov inequalities and stability for discrete linear Hamiltonian system, Appl. Math. Comput. 218 (2011) 574–582.
Contents lists available at SciVerse ScienceDirect
Applied Mathematics Letters journal homepage: www.elsevier.com/locate/aml
Lyapunov-type inequalities for even order difference equations✩ Qi-Ming Zhang a,b , X.H. Tang a,∗ a
School of Mathematical Sciences and Computing Technology, Central South University, Changsha 410083, Hunan, PR China
b
College of Science, Hunan University of Technology, Zhuzhou 412000, Hunan, PR China
article
abstract
info
Article history: Received 8 September 2011 Received in revised form 12 February 2012 Accepted 12 February 2012
In this work, we establish two new Lyapunov-type inequalities for the 2k-order difference equation
△2k x(n) + (−1)k−1 q(n)x(n + 1) = 0. Applying our inequalities, we obtain the lower bounds of the eigenvalue for a related eigenvalue problem. © 2012 Elsevier Ltd. All rights reserved.
Keywords: Even order Difference equation Lyapunov-type inequality
1. Introduction In 1907, Lyapunov [1] first established the so-called Lyapunov inequality
(b − a)
b
|q(t )|dt > 4,
(1.1)
a
for where the Hill equation x′′ (t ) + q(t )x(t ) = 0
(1.2)
has a real solution x(t ) such that x(a) = x(b) = 0,
x(t ) ̸≡ 0,
t ∈ [a, b],
(1.3)
where a, b ∈ R with a < b. The constant 4 in (1.1) cannot be replaced by a larger number. Since then, many generalizations of (1.1) have been established in the literature. A thorough literature review of continuous and discrete Lyapunov-type inequalities and their applications can be found in the survey paper [2] by Cheng. For other related references, we refer the reader to the few papers [3–9]. In 1983, Cheng [10] first obtained the following Lyapunov inequality which is a discrete analogue of (1.1):
F(b − a)
b−2
q(n) ≥ 4,
(1.4)
n=a
where a, b ∈ Z and
F(m) =
2 m − 1, m
m,
if m − 1 is even,
(1.5)
if m − 1 is odd,
✩ This work was partially supported by the NNSF (No. 11171351) of China and by Hunan Provincial Natural Science Foundation of China (11JJ2005).
∗
Corresponding author. Tel.: +86 73188877331. E-mail address: [email protected] (X.H. Tang).
0893-9659/$ – see front matter © 2012 Elsevier Ltd. All rights reserved. doi:10.1016/j.aml.2012.02.031
Q.-M. Zhang, X.H. Tang / Applied Mathematics Letters 25 (2012) 1830–1834
1831
if the second-order difference equation
△2 x(n) + q(n)x(n + 1) = 0
(1.6)
has a real solution x(n) such that x(a) = x(b) = 0,
x(n) ̸≡ 0,
n ∈ Z[a, b],
(1.7)
where and in the sequel, a, b ∈ N, Z[a, b] = {a, a + 1, a + 2, . . . , b − 1, b} and Z(a, b) = {a + 1, a + 2, . . . , b − 2, b − 1}. The constant 4 in (1.4) cannot be replaced by a larger number, either. For more discrete Lyapunov-type inequalities, we refer the reader to Cheng [11–13], Clark and Hinton [14,15], Guseinov and Kaymakcalan [16], Lin and Yang [17] and Zhang and Tang [18]. Consider the following even order difference equation:
△2k x(n) + (−1)k−1 q(n)x(n + 1) = 0,
(1.8)
where k ∈ N, n ∈ Z and q(n) is a real-valued function defined on Z. When k = 1, (1.8) reduces to (1.6). In this work, we establish two discrete Lyapunov-type inequalities for (1.8) under the following boundary conditions:
△2i x(a) = △2i x(b) = 0,
i = 0, 1, . . . , k − 1; x(n) ̸≡ 0, n ∈ Z[a, b].
(1.9)
Furthermore, applying our Lyapunov-type inequalities to the following eigenvalue problem:
△2k x(n) + (−1)k−1 λq(n)x(n + 1) = 0,
△2i x(a) = △2i x(b) = 0, i = 0, 1, . . . , k − 1,
(1.10)
we can obtain two different lower bounds of the eigenvalue λ. 2. The main results In this section, we establish some new Lyapunov-type inequalities for (1.8). Lemma 2.1. Assume that x(n) is a real-valued function on Z[a, b], x(a) = x(b) = 0, x(n) ̸≡ 0 for n ∈ Z[a, b]. Then
|x(n)| ≤ b−1
b−1 (n − a) (b − n) | △2 x(s)|, b−a s=a
|x(s)| ≤
s=a
b −1 1
2 s=a
∀ n ∈ Z(a, b − 1),
[(s − a + 1)(b − s − 1)| △2 x(s)|] ≤
(2.1)
b−1 (b − a)2
8
| △2 x(s)|,
(2.2)
s=a
1/2 √ b −1 (n − a) (b − n)[2(n − a) (b − n) + 1] 2 2 |x(n)| ≤ | △ x(s)| , √ 6(b − a) s =a
∀ n ∈ Z[a, b − 1],
(2.3)
and b−1
|x(n)|2 ≤
b−1 [(b − a)2 − 1][2(b − a)2 + 7]
180
n=a
| △2 x(s)|2 .
(2.4)
s=a
Proof. Since x(a) = x(b) = 0, it is easy to verify that x(n) = −
1
b −1 [G(n, s) △2 x(s)],
(2.5)
b − a s =a
where G(n, s) =
(s + 1 − a) (b − n), (n − a) (b − s − 1),
s ≤ n − 1; n, s ∈ Z[a, b], n ≤ s; n, s ∈ Z[a, b − 1].
(2.6)
Since G(n, s) ≤ (n − a) (b − n),
n, s ∈ Z[a, b − 1],
(2.7)
it follows from (2.5) that
|x(n)| ≤
b −1 b −1 (n − a) (b − n) [G(n, s)| △2 x(s)|] ≤ | △2 x(s)|, b − a s =a b−a s=a
1
n ∈ Z(a, b − 1),
1832
Q.-M. Zhang, X.H. Tang / Applied Mathematics Letters 25 (2012) 1830–1834
which implies that (2.1) holds. Next, by (2.5) and (2.6), we have b−1
|x(n)| ≤
n =a
=
=
≤
b −1 b −1 [G(n, s)| △2 x(s)|]
1
b − a n=a s=a b −1
1
b − a s =a b −1 1
2 s=a
| △ x(s)| 2
b−1
G(n, s)
n=a
[(s − a + 1) (b − s − 1)| △2 x(s)|]
b−1 (b − a)2
8
| △2 x(s)|,
s=a
which implies that (2.2) holds. By a relatively simple sum computation, we have b −1 [(b − a)2 − 1][2(b − a)2 + 7] [G(n, s)]2 = . 6(b − a) s=a
(2.8)
And so, by (2.5), (2.6) and (2.8) and the Cauchy inequality, we have
2 b −1 2 △ [ G ( n , s ) x ( s )] |x(n)| = (b − a)2 s=a 2 b −1 1 2 ≤ [G(n, s)| △ x(s)|] (b − a)2 s=a 1
2
≤
=
b −1
1
(b − a)
2
[G(n, s)]2
s=a
b−1
| △2 x(s)|2
s=a
b−1 (n − a) (b − n)[2(n − a) (b − n) + 1] | △2 x(s)|2 , 6(b − a) s=a
∀ n ∈ Z[a, b − 1],
(2.9)
which implies that (2.3) holds. Summing (2.9) from a to b − 1, we have b−1
|x(n)| ≤
1
2
n =a
= Hence, (2.4) holds.
(b − a)2
b −1 b−1 b −1 2 2 2 [G(n, s)] | △ x(s)| n =a
s=a
s=a
b −1 [(b − a)2 − 1][2(b − a)2 + 7]
180
| △2 x(s)|2 .
(2.10)
s=a
Theorem 2.2. Assume that k ∈ N and q(n) is a real-valued function on Z. If (1.8) has a solution x(n) satisfying the boundary value conditions (1.9), then b−1 [|q(n)|(n − a + 1) (b − n − 1)] ≥ n =a
23(k−1)
(b − a)2k−3
.
(2.11)
Proof. Choose c ∈ Z[a, b] such that |x(c )| = maxn∈[a,b] |x(n)|. Since (1.9), it follows from Lemma 2.1 that
|x(c )| ≤
b −1 b−1 b−a (c − a) (b − c ) | △2 x(n)| ≤ | △2 x(n)|, b−a 4 n=a n =a
(2.12)
and b −1 n =a
| △2i x(n)| ≤
b−1 (b − a)2
8
n=a
| △2i+2 x(n)|,
i = 1, 2, . . . , k − 2,
(2.13)
Q.-M. Zhang, X.H. Tang / Applied Mathematics Letters 25 (2012) 1830–1834
1833
and b−1
| △2k−2 x(n)| ≤
n=a
b−1 1
2 n =a
[(n − a + 1) (b − n − 1)| △2k x(n)|].
(2.14)
From (1.8) and (2.12)–(2.14), we obtain
|x(c )| ≤ ≤
≤
b−1 b−a
4
| △2 x(n)|
n=a
b−a
(b − a)2
4
k−2 b −1
8
b−a
n =a
(b − a)2
8
8
| △2k−2 x(n)|
k−2 b −1 [(n − a + 1) (b − n − 1)| △2k x(n)|] n =a
b−1 (b − a)2k−3 = [(n − a + 1) (b − n − 1)|q(n)x(n + 1)|] 3(k−1)
2
n=a
( b − a)
2k−3
≤
23(k−1)
|x(c )|
b −1 [(n − a + 1) (b − n − 1)|q(n)|].
(2.15)
n =a
Since |x(c )| > 0, it follows from (2.15) that (2.11) holds.
Corollary 2.3. Assume that k ∈ N and q(n) is a real-valued function on Z. If (1.8) has a solution x(n) satisfying the boundary value conditions (1.9), then b−1
|q(n)| ≥
n=a
23k−1
(b − a)2k−1
.
(2.16)
Theorem 2.4. Assume that k ∈ N and q(n) is a real-valued function on Z. If (1.8) has a solution x(n) satisfying the boundary value conditions (1.9), then b−1
|q(n)|2 ≥
n=a
2k+3 32k−1 5k−1
k−1 . (b − a)[(b − a)2 + 2][(b − a)2 − 1]k−1 (b − a)2 + 27
(2.17)
Proof. Choose c ∈ Z[a, b] such that |x(c )| = maxn∈[a,b] |x(n)|. Since (1.9), it follows from Lemma 2.1 and (1.8) that
|x(c )|2 ≤ ≤
=
b−1 (c − a) (b − c )[2(c − a) (b − c ) + 1] | △2 x(s)|2 6(b − a) s=a
(b − a)[(b − a)2 + 2] 48
×
b−1 [(b − a)2 − 1]k−1 [2(b − a)2 + 7]k−1
180k−1
| △2k x(s)|2
s=a
b −1 (b − a)[(b − a)2 + 2][(b − a)2 − 1]k−1 [2(b − a)2 + 7]k−1 [|q(n)|2 |x(n + 1)|2 ] 48 × 180k−1 n =a
≤ |x(c )|2
= |x(c )|
2
b −1 (b − a)[(b − a)2 + 2][(b − a)2 − 1]k−1 [2(b − a)2 + 7]k−1 |q(n)|2 48 × 180k−1 n =a
k−1 b−1 (b − a)[(b − a)2 + 2][(b − a)2 − 1]k−1 (b − a)2 + 27 2k+3 32k−1 5k−1
Since |x(c )| > 0, it follows from (2.18) that (2.17) holds.
|q(n)|2 .
(2.18)
n =a
Now, we give an application of the obtained Lyapunov-type inequalities (2.11) and (2.17) for the eigenvalue problem (1.10). We have 23k−1
|λ| ≥ (b − a)2k−3
b−1
n=a
[|q(n)|(n − a + 1) (b − n − 1)]
,
(2.19)
1834
Q.-M. Zhang, X.H. Tang / Applied Mathematics Letters 25 (2012) 1830–1834
and
|λ| ≥
2(k+3)/2 3(2k−1)/2 5(k−1)/2 −1 k−1 b [(b − a)[(b − a)2 + 2][(b − a)2 − 1]k−1 (b − a)2 + 27 |q(n)|2 ]1/2
.
(2.20)
n =a
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