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Impulsive differential and impulsive integral inequalities with integral jump conditions

Abstract

In this article, we establish some impulsive differential and impulsive integral inequalities for integral jump conditions. The new jump conditions for impulse effects are related to the integral conditions of the past state. Two examples are given to illustrate the advantage of our results.

2010 Mathematics Subject Classification: 34A37; 34A40.

1 Introduction

In [1], Lakshmikantham et al. developed a famous impulsive differential inequality given as Theorem A below.

Lakshmikantham et al. assume that 0 ≤ t0 < t1 < t2 <, limk→∞t k = ∞, R+ = [0, +∞) and I R. They define PC(R+, I) = {u: R+I; u(t) is continuous for tt k , and u(0+), u ( t k - ) , and u ( t k + ) exist, and u ( t k ) = u ( t k ) , k = 1 , 2 , } and PC1(R+,I) = {u PC(R+, I): u'(t) is continuous everywhere for tt k , and u'(0+), u ( t k + ) and u ( t k - ) exist, and u ( t k ) = u ( t k ) , k = 1 , 2 , } .

Theorem A. Assume that

(H0) the sequence {t k } satisfies 0 ≤ t0 < t1 < t2 < , limk→∞t k = ∞;

(H1) m PC1[R+, R] and m(t) is left-continuous at t k , k = 1, 2,...;

(H2) for k = 1, 2,..., t ≥ t0,

m ( t ) p ( t ) m ( t ) + q ( t ) , t t k ,
(1.1)
m ( t k + ) d k m ( t k ) + b k ,
(1.2)

where q, p C[R+, R], d k ≥ 0 and b k are constants.

Then,

m ( t ) m ( t 0 ) t 0 < t k < t d k e t 0 t p ( s ) d s + t 0 < t k < t t k < t j < t d j e t k t p ( s ) d s b k + t 0 t s < t k < t d k e s t p ( σ ) d σ q ( s ) d s , t t 0 .
(1.3)

Impulsive differential and impulsive integral inequalities play an important role in the study of the theory of impulsive differential equations (see [14]). In recent years, many authors have used impulsive (differential or integral) inequalities to investigate properties of solutions of various impulsive problems, such as existence, uniqueness, boundedness, stability, asymptotic behavior, and oscillation etc. (see, for example [539]). There are many good results on the impulsive differential and impulsive integral inequalities (see for example [4048]). However, most of these articles deal with jump conditions at impulse point t k depending on the left-hand limit m(t k ) or a time-delay value, m(t k -τ), τ > 0. Our main goal is to extend the theory of impulsive differential and impulsive integral inequalities to include integral jump conditions.

In the present article, we will show that Theorem A can be generalized to obtain differential inequalities for integral jump conditions by replacing the inequality in (1.2) by the inequality in (1.4).

m ( t k + ) d k m ( t k ) + c k t k - τ k t k - σ k m ( s ) d s + b k , k = 1 , 2 , ,
(1.4)

where 0 ≤ σ k τ k t k - tk-1. We note that if c k = 0 for all k = 1, 2,..., then condition (1.4) reduces to condition (1.2). If d k = 0, c k ≠ 0 and 0 ≤ σ k < τ k t k -tk-1, k = 1, 2,..., then condition (1.4) means that the bound of the jump condition at t k is a functional of past states on the interval (t k - τ k , t k - σ k ] before the impulse point t k . Moreover, we establish some new impulsive integral inequalities with integral jump conditions.

At the end of this article, we will show some applications of our results to prove a maximum principle and the boundedness of solutions for impulsive problems.

2 Main results

Denote l = max{k: tt k , k = 1, 2,...}. Now we are in the position to state and prove our results.

Theorem 2.1. Let (H 0 ) and (H1) hold. Suppose that p, q C[R+, R] and for k = 1, 2,..., tt0,

m ( t ) p ( t ) m ( t ) + q ( t ) , t t k ,
(2.1)
m ( t k + ) d k m ( t k ) + c k t k - τ k t k - σ k m ( s ) d s + b k ,
(2.2)

where c k , d k ≥ 0, 0 ≤ σ k τ k t k - tk-1and b k are constants.

Then,

m ( t ) m ( t 0 ) t 0 < t k < t d k e t k - 1 t k p ( ξ ) d ξ + c k t k - τ k t k - σ k e t k - 1 s p ( ξ ) d ξ d s + t 0 < t k < t t k < t j < t d j e t j - 1 t j p ( ξ ) d ξ + c j t j - τ j t j - σ j e t j - 1 s p ( ξ ) d ξ d s × d k t k - 1 t k q ( s ) e s t k p ( ξ ) d ξ d s + c k t k - τ k t k - σ k t k - 1 s q ( r ) e r s p ( ξ ) d ξ d r d s + b k e t l t p ( ξ ) d ξ + t l t q ( s ) e s t p ( ξ ) d ξ d s , t t 0 .
(2.3)

Proof. From (2.1) we have that

d d t m ( t ) e - t 0 t p ( ξ ) d ξ q ( t ) e - t 0 t p ( ξ ) d ξ ,
(2.4)

for t [t0, t1]. Integrating (2.4) from t0 to t for t [t0, t1], we get

m ( t ) m ( t 0 ) e t 0 t p ( ξ ) d ξ + t 0 t q ( s ) e s t p ( ξ ) d ξ d s .
(2.5)

Hence (2.3) is valid on [t0, t1]. Assume that (2.3) holds for t [t0, t n ] for some integer n > 1. Then, for t [t n , tn+1], it follow from (2.1) and (2.5) that

m ( t ) m ( t n + ) e t n t p ( ξ ) d ξ + t n t q ( s ) e s t p ( ξ ) d ξ d s .
(2.6)

Now using (2.2) and (2.6), we have

m ( t ) d n m ( t n ) + c n t n - τ n t n - σ n m ( s ) d s + b n e t n t p ( ξ ) d ξ + t n t q ( s ) e s t p ( ξ ) d ξ d s .
(2.7)

By the principle of mathematical induction, (2.7) can be expressed as

m ( t ) { d n ( { m ( t 0 ) t 0 < t k < t n ( d k e t k 1 t k p ( ξ ) d ξ + c k t k τ k t k σ k e t k 1 s p ( ξ ) d ξ d s ) + t 0 < t k < t n [ t k < t j < t n ( d j e t j 1 t j p ( ξ ) d ξ + c j t j τ j t j σ j e t j 1 s p ( ξ ) d ξ d s ) × ( d k t k 1 t k q ( s ) e s t k p ( ξ ) d ξ d s + c k t k τ k t k σ k t k 1 s q ( r ) e r s p ( ξ ) d ξ d r d s + b k ) ] } e t n 1 t n p ( ξ ) d ξ + t n 1 t n q ( s ) e s t n p ( ξ ) d ξ d s ) + c n t n τ n t n σ n { { m ( t 0 ) t 0 < t k < s ( d k e t k 1 t k p ( ξ ) d ξ + c k t k τ k t k σ k e t k 1 v p ( ξ ) d ξ d v ) + t 0 < t k < s [ t k < t j < s ( d j e t j 1 t j p ( ξ ) d ξ + c j t j τ j t j σ j e t j 1 v p ( ξ ) d ξ d v ) ( d k t k 1 t k q ( v ) e v t k p ( ξ ) d ξ d v + c k t k τ k t k σ k t k 1 v q ( r ) e r v p ( ξ ) d ξ d r d v + b k ) ] } e t n 1 s p ( ξ ) d ξ + t n 1 s q ( v ) e v s p ( ξ ) d ξ d v } d s + b n } e t n t p ( ξ ) d ξ + t n t q ( s ) e s t p ( ξ ) d ξ d s .
(2.8)

Set

E k = d k e t k - 1 t k p ( ξ ) d ξ + c k t k - τ k t k - σ k e t k - 1 s p ( ξ ) d ξ d s
(2.9)
G k = d k t k - 1 t k q ( s ) e s t k p ( ξ ) d ξ d s + c k t k - τ k t k - σ k t k - 1 s q ( r ) e r s p ( ξ ) d ξ d r d s + b k .
(2.10)

Substituting (2.9), (2.10) into (2.8), we get that for t [t n , tn+1]

m ( t ) { d n ( { m ( t 0 ) t 0 < t k < t n E k + t 0 < t k < t n [ t k < t j < t n E j G k ] } e t n 1 t n p ( ξ ) d ξ + t n 1 t n q ( s ) e s t n p ( ξ ) d ξ d s ) + c n t n τ n t n σ n { { m ( t 0 ) t 0 < t k < s E k + t 0 < t k < s [ t k < t j < s E j G k ] } e t n 1 s p ( ξ ) d ξ + t n 1 s q ( v ) e v s p ( ξ ) d ξ d v } d s + b n } e t n t p ( ξ ) d ξ + t n t q ( s ) e s t p ( ξ ) d ξ d s = { ( m ( t 0 ) t 0 < t k < t n E k + t 0 < t k < t n [ t k < t j < t n E j G k ] ) d n e t n 1 t n p ( ξ ) d ξ + d n t n 1 t n q ( s ) e s t n p ( ξ ) d ξ d s + ( ( m ( t 0 ) ) t 0 < t k < t n E k + t 0 < t k < t n [ t k < t j < t n E j G k ] ) c n t n τ n t n σ n e t n 1 s p ( ξ ) d ξ d s + c n t n τ n t n σ n t n 1 s q ( v ) e v s p ( ξ ) d ξ d v d s + b n } e t n t p ( ξ ) d ξ + t n t q ( s ) e s t p ( ξ ) d ξ d s = { ( m ( t 0 ) t 0 < t k < t n E k + t 0 < t k < t n [ t k < t j < t n E j G k ] ) E n + G n } e t n t p ( ξ ) d ξ d s + t n t q ( s ) e s t p ( ξ ) d ξ d s = { m ( t 0 ) t 0 < t k < t E k + t 0 < t k < t [ t k < t j < t E j G k ] } e t n t p ( ξ ) d ξ d s + t n t q ( s ) e s t p ( ξ ) d ξ d s .

Hence,

m ( t ) m ( t 0 ) t 0 < t k < t d k e t k - 1 t k p ( ξ ) d ξ + c k t k - τ k t k - σ k e t k - 1 s p ( ξ ) d ξ d s + t 0 < t k < t t k < t j < t d j e t j - 1 t j p ( ξ ) d ξ + c j t j - τ j t j - σ j e t j - 1 s p ( ξ ) d ξ d s × d k t k - 1 t k q ( s ) e s t k p ( ξ ) d ξ d s + c k t k - τ k t k - σ k t k - 1 s q ( r ) e r s p ( ξ ) d ξ d r d s + b k e t n t p ( ξ ) d ξ + t n t q ( s ) e s t p ( ξ ) d ξ d s ,

for t n ttn+1. Therefore, the estimate (2.3) holds for t0ttn+1. This completes the proof.

Remark 2.2. If c k = 0 for all k = 1, 2,..., then Theorem 2.1 reduces to Theorem A.

Corollary 2.3. Let (H0) and (H1) hold. Suppose that p, q C[R+, R] and for k = 1, 2,..., tt0,

m ( t ) p ( t ) m ( t ) + q ( t ) , t t k ,
(2.11)
m ( t k + ) c k t k - τ k t k - σ k m ( s ) d s + b k ,
(2.12)

where c k ≥ 0, 0 ≤ σ k τ k t k - tk-1and b k are constants.

Then,

m ( t ) m ( t 0 ) t 0 < t k < t c k t k - τ k t k - σ k e t k - 1 s p ( ξ ) d ξ d s + t 0 < t k < t t k < t j < t c j t j - τ j t j - σ j e t j - 1 s p ( ξ ) d ξ d s × c k t k - τ k t k - σ k t k - 1 s q ( r ) e r s p ( ξ ) d ξ d r d s + b k e t l t p ( ξ ) d ξ + t l t q ( s ) e s t p ( ξ ) d ξ d s , t t 0 .
(2.13)

The following corollary will be used in our examples. For convenience, we set

A k = c k p ( e - p σ k - e - p τ k ) ,
(2.14)
B k = c k p e p ( t k - σ k ) t k - 1 t k - σ k q ( r ) e - p r d r - e p ( t k - τ k ) t k - 1 t k - τ k q ( r ) e - p r d r - t k - τ k t k - σ k q ( r ) d r + b k .
(2.15)

Corollary 2.4. Let (H0) and (H1) hold. Suppose that q C[R+, R], and for k = 1, 2,..., tt0,

m ( t ) p m ( t ) + q ( t ) , t t k ,
(2.16)
m ( t k + ) c k t k - τ k t k - σ k m ( s ) d s + b k ,
(2.17)

where p ≠ 0, c k ≥ 0, 0 ≤ σ k τ k t k - tk-1and b k are constants.

Then,

m ( t ) m ( t 0 ) t 0 < t k < t A k e p ( t - t 0 ) + t 0 < t k < t t k < t j < t A j B k e p ( t - t k ) + t l t q ( s ) e p ( t - s ) d s ,
(2.18)

for tt0where A k , B k are defined by (2.14), (2.15), respectively.

Proof. By using Corollary 2.3 and reversing the order of double integration, we have the required result.

Corollary 2.5. Let (H0) and (H1) hold. Suppose that q C[R+, R], and for k = 1, 2,...,tt0,

m ( t ) q ( t ) , t t k ,
(2.19)
Δ m ( t k ) c k t k - τ k t k - σ k m ( s ) d s + b k ,
(2.20)

where Δ m ( t k ) = m ( t k + ) - m ( t k ) , c k ≥ 0, 0 ≤ σ k τ k t k - tk-1and b k are constants.

Then,

m ( t ) m ( t 0 ) t 0 < t k < t [ 1 + c k ( τ k - σ k ) ] + t 0 < t k < t t k < t j < t 1 + c j ( τ j - σ j ) × [ 1 + c k ( τ k - σ k ) ] t k - 1 t k - τ k q ( s ) d s + t k - τ k t k - σ k [ 1 + c k ( t k - σ k - s ) ] q ( s ) d s + t k - σ k t k q ( s ) d s + b k + t l t q ( s ) d s , t t 0 .
(2.21)

Proof. By setting p(t) ≡ 0 and d k = 1(k = 1, 2,...) in Theorem 2.1 and reversing the order of double integration, we have the required result.

Next, we give an application of Theorem 2.1 to the determination of a bound for the solutions of impulsive integral inequalities with integral jump conditions.

Theorem 2.6. Assume that (H0) and (H1) hold. Suppose that p C[R+, R+] and for k = 1, 2,...

m ( t ) C + t 0 t p ( s ) m ( s ) d s + t 0 < t k < t β k m ( t k ) + t 0 < t k < t α k t k - τ k t k - σ k m ( s ) d s , t t 0 ,
(2.22)

where α k , β k ≥ 0, 0 ≤ σ k τ k t k - tk-1and C are constants. Then

m ( t ) C t 0 < t k < t ( 1 + β k ) e t k - 1 t k p ( ξ ) d ξ + α k t k - τ k t k - σ k e t k - 1 s p ( ξ ) d ξ d s e t l t p ( ξ ) d ξ , t t 0 .
(2.23)

Proof. Defining a function v(t) by the right side of (2.22), we have

v ( t ) = p ( t ) m ( t ) , t t k , v ( t 0 ) = C , v ( t k + ) = v ( t k ) + β k m ( t k ) + α k t k - τ k t k - σ k m ( s ) d s .

Since m(t) ≤ v(t), we get

v ( t ) p ( t ) v ( t ) , t t k , v ( t 0 ) = C , v ( t k + ) ( 1 + β k ) v ( t k ) + α k t k - τ k t k - σ k v ( s ) d s .

Applying Theorem 2.1, we obtain

v ( t ) C t 0 < t k < t ( 1 + β k ) e t k - 1 t k p ( ξ ) d ξ + α k t k - τ k t k - σ k e t k - 1 s p ( ξ ) d ξ d s e t l t p ( ξ ) d ξ , t t 0 ,

which results in (2.23).

Theorem 2.7 . Assume that (H0) and (H1) hold. Suppose that p C[R+, R+], h PC[R+, R] and for k = 1, 2,...

m ( t ) h ( t ) + t 0 t p ( s ) m ( s ) d s + t 0 < t k < t β k m ( t k ) + t 0 < t k < t α k t k - τ k t k - σ k m ( s ) d s , t t 0 ,
(2.24)

where α k , β k ≥ 0 and 0 ≤ σ k τ k t k - tk-1are constants.

Then,

m ( t ) h ( t ) + t 0 < t k < t t k < t j < t ( 1 + β j ) e t j - 1 t j p ( ξ ) d ξ + α j t j - τ j t j - σ j e t j - 1 s p ( ξ ) d ξ d s × ( 1 + β k ) t k - 1 t k p ( s ) h ( s ) e s t k p ( ξ ) d ξ d s + α k t k - τ k t k - σ k t k - 1 s p ( r ) h ( r ) e r s p ( ξ ) d ξ d r d s + β k h ( t k ) + α k t k - τ k t k - σ k h ( s ) d s e t l t p ( ξ ) d ξ + t l t p ( s ) h ( s ) e s t p ( ξ ) d ξ d s , t t 0 .
(2.25)

Proof. Setting

v ( t ) = t 0 t p ( s ) m ( s ) d s + t 0 < t k < t β k m ( t k ) + t 0 < t k < t α k t k - τ k t k - σ k m ( s ) d s ,

and from the fact that m(t) ≤ h(t) + v(t), we obtain

v ( t ) p ( t ) v ( t ) + p ( t ) h ( t ) , t t k , v ( t 0 ) = 0 , v ( t k + ) ( 1 + β k ) v ( t k ) + α k t k - τ k t k - σ k v ( s ) d s + β k h ( t k ) + α k t k - τ k t k - σ k h ( s ) d s .

Using Theorem 2.1 together with m(t) ≤ h(t) + v(t), we then obtain the estimate (2.25).

Remark 2.8. If α k = 0 for all k = 1, 2,..., then Theorem 2.6 and Theorem 2.7 are reduced to the Theorems 1.5.1 and 1.5.2 in [1], respectively.

3 Some examples

In this section, two applications of impulsive differential and impulsive integral inequalities with integral jump conditions are given.

Corollary 3.1. Assume that u PC1[J, R] satisfies

u ( t ) - M u ( t ) + a ( t ) 0 , t t k , t J = [ 0 , T ] , u ( t k + ) c k t k - τ k t k - σ k u ( s ) d s , k = 1 , , n , u ( 0 ) = u ( T ) + λ ,
(3.1)

where M > 0, a C[R+, R+], 0 < t1 < t2 < < t n < T. c k ≥ 0, 0 ≤ σ k τ k t k - tk-1, k = 1, 2,..., n.

Suppose in addition that

(D1) k = 1 n c k M ( e - σ k M - e - τ k M ) < e - M T ,

(D2)

e ( t k - τ k ) M t k - 1 t k - τ k a ( r ) e - M r d r + t k - τ k t k - σ k a ( r ) d r e ( t k - σ k ) M t k - 1 t k - σ k a ( r ) e - M r d r , k = 1 , 2 , , n ,

(D3) λ t n T a ( s ) e M ( T - s ) ds.

Then u(t) ≤ 0 for t [0, T].

Proof. By Corollary 2.4 for t [0, T] we can write that

u ( t ) u ( 0 ) t 0 < t k < t A ¯ k e M t + t 0 < t k < t t k < t j < t A ¯ j B ¯ k e M ( t - t k ) - t l t a ( s ) e M ( t - s ) d s ,

where

A ¯ k = c k M ( e - σ k M - e - τ k M ) 0 ,

and

B ¯ k = c k M ( e ( t k τ k ) M t k 1 t k τ k a ( r ) e M r d r + t k τ k t k σ k a ( r ) d r e ( t k σ k ) M t k 1 t k σ k a ( r ) e M r d r ) , k = 1 , 2 , , n .

Condition (D2) implies that B ¯ k 0 for k = 1, 2,..., n. Then, it is sufficient to show that u(0) ≤ 0. For t = T we have

u ( T ) u ( 0 ) k = 1 n A ¯ k e M T + t 0 < t k < T t k < t j < T A ¯ j B ¯ k e M ( T - t k ) - t n T a ( s ) e M ( T - s ) d s .

By the conditions (D1) and (D3), we see that

u ( 0 ) 1 - k = 1 n A ¯ k e M T λ + t 0 < t k < T t k < t j < T A ¯ j B ¯ k e M ( T - t k ) - t n T a ( s ) e M ( T - s ) d s 0 ,

which implies that u(0) ≤ 0.

Corollary 3.2. Let v PC1[R+, R] such that

v ( t ) = f ( t , v ( t ) ) , t t k , t [ t 0 , ) , Δ v ( t k ) = I k t k - τ k t k - σ k v ( s ) d s , k = 1 , 2 , , v ( t 0 ) = v 0 ,
(3.2)

where f C(R × R, R), I k C(R, R), 0 ≤ t0 < t1 < t2 < , limk→∞t k = ∞, 0 ≤ σ k τ k t k - tk-1, k = 1, 2,.... Assume that

(D4) there exists a constant N > 0, such that

f ( t , v ( t ) ) N v ( t ) for t t 0 ,

(D5) there exist constants L k ≥ 0 such that

I k ( x ) L k x , x R , k = 1 , 2 , .

Then the following inequality is valid

v ( t ) v 0 t 0 < t k < t 1 + L k N ( e - σ k N - e - τ k N ) e ( t - t 0 ) N , t t 0 .
(3.3)

Proof. The solution v(t) of problem (3.2) satisfies the equation

v ( t ) = v ( t 0 ) + t 0 t f ( s , v ( s ) ) d s + t 0 < t k < t I k t k - τ k t k - σ k v ( s ) d s .

From the hypothesis (D4), (D5) it follows for tt0 that

v ( t ) v 0 + t 0 t f ( s , v ( s ) ) d s + t 0 < t k < t I k t k - τ k t k - σ k v ( s ) d s v 0 + t 0 t N v ( s ) d s + t 0 < t k < t L k t k - τ k t k - σ k v ( s ) d s .

Hence Theorem 2.6 yields the estimate

v ( t ) v 0 t 0 < t k < t e ( t k - t k - 1 ) N + L k t k - τ k t k - σ k e ( s - t k - 1 ) N d s e ( t - t l ) N .

Therefore, the inequality (3.3) holds for tt0 and the proof is complete.

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Acknowledgements

The authors thank the referees for several useful remarks and interesting comments. This research was supported by the Centre of Excellence in Mathematics, Thailand.

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Thiramanus, P., Tariboon, J. Impulsive differential and impulsive integral inequalities with integral jump conditions. J Inequal Appl 2012, 25 (2012). https://doi.org/10.1186/1029-242X-2012-25

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