Hindawi Publishing Corporation Journal of Inequalities and Applications Volume 2008, Article ID 312395, 8 pages doi:10.1155/2008/312395 ResearchArticleOnSomeNewImpulsiveIntegral Inequalities Jianli Li Department of Mathematics, Hunan Normal University, Changsha, Hunan 410081, China Correspondence should be addressed to Jianli Li, ljianli@sina.com Received 4 June 2008; Accepted 21 July 2008 Recommended by Wing-Sum Cheung We establish somenewimpulsiveintegral inequalities related to certain integral inequalities arising in the theory of differential equalities. The inequalities obtained here can be used as handy tools in the theory of some classes of impulsive differential and integral equations. Copyright q 2008 Jianli Li. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. 1. Introduction Differential and integral inequalities play a fundamental role in global existence,uniqueness, stability, and other properties of the solutions of various nonlinear differential equations; see 1–4. A great deal of attention has been given to differential and integral inequalities; see 1, 2, 5–8 and the references given therein. Motivated by the results in 1, 5, 7,the main purpose of this paper is to establish somenewimpulsiveintegral inequalities similar to Bihari’s inequalities. Let 0 ≤ t 0 <t 1 <t 2 < ··· , lim k→∞ t k ∞, R 0, ∞,andI ⊂ R, then we introduce the following spaces of function: PCR ,I{u : R → I, u is continuous for t / t k , u0 ,ut k ,andut − k exist, and ut − k ut k ,k 1, 2, }, PC 1 R ,I{u ∈ PCR ,I : u is continuously differentiable for t / t k , u 0 ,u t k , and u t − k exist, and u t − k u t k ,k 1, 2, }. To prove our main results, we need the following result see 1, Theorem 1.4.1. Lemma 1.1. Assume that A 0 the sequence {t k } satisfies 0 ≤ t 0 <t 1 <t 2 < ···,withlim k→∞ t k ∞; A 1 m ∈ PC 1 R , R and mt is left-continuous at t k , k 1, 2, ; A 2 for k 1, 2, , t≥ t 0 , m t ≤ ptmtqt,t / t k , m t k ≤ d k mt k b k , 1.1 where q, p ∈ PCR , R,d k ≥ 0, and b k are constants. 2 Journal of Inequalities and Applications Then, mt ≤ mt 0 t 0 <t k <t d k exp t t 0 psds t t 0 s<t k <t d k exp t s pσdσ qsds t 0 <t k <t t k <t j <t d j exp t t k psds b k ,t≥ t 0 . 1.2 2. Main results In this section, we will state and prove our results. Theorem 2.1. Let u, f ∈ PCR , R , b k ≥ 1, and c ≥ 0 be constants. If u 2 t ≤ c 2 2 t 0 fsusds 0<t k <t b 2 k − 1 u 2 t k , 2.1 for t ∈ R ,then ut ≤ c 0<t k <t b k t 0 0<t k <t b k fsds, 2.2 for t ∈ R . Proof. Define a function zt by ztc ε 2 2 t 0 fsusds 0<t k <t b 2 k − 1 u 2 t k , 2.3 where ε>0 is an arbitrary small constant. For t / t k ,differentiating 2.3 and then using the fact that ut ≤ zt, we have z t2ftut ≤ 2ft zt, 2.4 and so d zt dt z t 2 zt ≤ ft. 2.5 For t t k , we have zt k − zt k b 2 k − 1u 2 t k ≤ b 2 k − 1zt k ;thuszt k ≤ b 2 k zt k .Let ztxt; it follows that x t ≤ ft,t / t k ,t≥ 0, x t k ≤ b k xt k ,k 1, 2 2.6 From Lemma 1.1,weobtain xt ≤ x0 0<t k <t b k t 0 s<t k <t b k fsds ≤ c ε 0<t k <t b k t 0 s<t k <t b k fsds. 2.7 Now by using the fact that ut ≤ ztxt in 2.7 and then letting ε → 0, we get the desired inequality in 2.2. This proof is complete. Jianli Li 3 Theorem 2.2. Let u, f ∈ PCR , R and b k ≥ 1 be constants, and let c be a nonnegative constant. If u 2 t ≤ c 2 2 t 0 fsu 2 shsus ds 0<t k <t b 2 k − 1 u 2 t k , 2.8 for t ∈ R ,then ut ≤ c 0<t k <t b k exp t 0 fsds t 0 s<t k <t b k exp t s fτdτ hsds, 2.9 for t ∈ R . Proof. This proof is similar to that of Theorem 2.1; thus we omit the details here. Theorem 2.3. Let u, f, g, h ∈ PCR , R ,c≥ 0, and b k ≥ 1 be constants. If u 2 t ≤ c 2 2 t 0 fsus us s 0 gτuτdτ hsus ds 0<t k <t b 2 k − 1 u 2 t k , 2.10 for t ∈ R ,then ut ≤ c 0<t k <t b k t 0 s<t k <t b k fsashsds, 2.11 for t ∈ R ,where atc 0<t k <t b k exp t 0 fτgτdτ t 0 s<t k <t b k exp t s fτgτdτ hsds. 2.12 Proof. Let ε>0 be an arbitrary small constant, and define a function zt by ztc ε 2 2 t 0 fsus us s 0 gτuτdτ hsus ds 0<t k <t b 2 k − 1 u 2 t k . 2.13 Let ztxt; similar to the proof of Theorem 2.1, we have x t ≤ ft xt t 0 gsxsds ht,t / t k , x t k ≤ b k xt k ,k 1, 2, 2.14 Set vtxt t 0 gsxsds; then vt ≥ xt,andsofrom2.14 we get that x t ≤ ftvt ht.Thus,fort / t k , v tx tgtxt ≤ ftvthtgtxt ≤ ftgtvtht, 2.15 4 Journal of Inequalities and Applications and for t t k , v t k − vt k x t k − xt k ≤ b k − 1xt k ≤ b k − 1vt k , 2.16 and so vt k ≤ b k vt k .ByLemma 1.1, we have vt≤ cε 0<t k <t b k exp t 0 fτgτdτ t 0 s<t k <t b k exp t s fτgτdτ hsds. 2.17 Let ε → 0, then we obtain vt ≤ at, 2.18 where at is defined in 2.12. Substituting 2.18 into 2.14, we have x t ≤ ftatht,t / t k , x t k ≤ b k xt k ,k 1, 2, 2.19 Applying Lemma 1.1 again, we obtain xt ≤ c ε 0<t k <t b k t 0 s<t k <t b k fsashsds. 2.20 Now using ut ≤ xt and letting ε → 0, we get the desired inequality in 2.11. Theorem 2.4. Let u, f, g, h ∈ PCR , R ,c≥ 0, and b k ≥ 1 be constants. If u 2 t ≤ c 2 2 t 0 fsus s 0 gτuτdτ hsus ds 0<t k <t b 2 k − 1u 2 t k , 2.21 for t ∈ R ,then ut ≤ c 0<t k <t b k exp t 0 fs s 0 gτdτ ds t 0 s<t k <t b k exp t s fτ τ 0 gωdω dτ hsds, 2.22 for t ∈ R . Proof. Set ztc ε 2 2 t 0 fsus s 0 gτuτdτ hsus ds 0<t k <t b 2 k − 1 u 2 t k , 2.23 where ε is an arbitrary small constant; then zt is nondecreasing. Let xt zt, then it follows for t / t k that x t ≤ ft t 0 gsxsds ht ≤ ft t 0 gsds xtht2.24 Jianli Li 5 since xt is nondecreasing. Also, for t t k , we have xt k ≤ b k xt k . Applying Lemma 1.1, we obtain xt ≤ c εc 0<t k <t b k exp t 0 fs s 0 gτdτ ds t 0 s<t k <t b k exp t s fτ τ 0 gωdω dτ hsds. 2.25 Now by using the fact that ut ≤ xt in 2.25 and letting ε → 0, we get the inequality 2.22. Remark 2.5. If b k ≡ 1, then 2.1, 2.8, 2.10,and2.21 have no impulses. In this case, it is clear that Theorems 2.2-2.3 improve the corresponding results of 5, Theorem 1. Theorem 2.6. Let u, f ∈ PCR , R ,ht, s ∈ CR 2 , R ,for0 ≤ s ≤ t<∞,c≥ 0,b k ≥ 1, and p>1 be constants. Let g ∈ PCR , R be a nondecreasing function with gu > 0,foru>0, and gλu ≥ μλgu,forλ>0,u∈ R;hereμλ > 0,forλ>0.If u p t ≤ c t 0 fsgus s 0 hs, σguσdσ ds 0<t k <t b k − 1u p t k , 2.26 for t ∈ R , then for 0 ≤ t<T, ut ≤ G −1 G c 0<t k <t b k t 0 s<t k <t b k μ b 1/p k psds 1/p , 2.27 where ptft t 0 ht, σdσ, 2.28 Gr r r 0 ds g s 1/p for r ≥ r 0 > 0, 2.29 T sup t ≥ 0: G c 0<t k <t b k t 0 s<t k <t b k μ b 1/p k psds ∈ dom G −1 . 2.30 Proof. We first assume that c>0 and define a function zt by the right-hand side of 2.26. Then, zt > 0,z0c, ut ≤ zt 1/p ,andzt is nondecreasing. For t / t k , z tftgut t 0 ht, σguσdσ ≤ ftg zt 1/p t 0 ht, σg zσ 1/p dσ ≤ g zt 1/p ft t 0 ht, σdσ , 2.31 6 Journal of Inequalities and Applications and for t t k ,zt k ≤ b k zt k .Ast ∈ 0,t 1 ,from2.31 we have Gzt − Gz0 zt z0 ds g s 1/p ≤ t 0 psds, 2.32 and so zt ≤ G −1 Gc t 0 psds . 2.33 Now assume that for 0 ≤ t ≤ t n , we have zt ≤ G −1 G c 0<t k <t b k t 0 0<t k <t b k μ b 1/p k psds . 2.34 Then, for t ∈ t n ,t n1 , it follows from 2.32 that Gzt ≤ Gzt n t t n psds.Usingzt k ≤ b k zt k , we arrive at Gzt ≤ Gb n zt n t t n psds. 2.35 From the supposition of g,weseethat Gλu − Gλv λu 0 ds g s 1/p − λv 0 ds g s 1/p ≤ λ μ λ 1/p Gu − Gv, for u ≥ v, λ > 0. 2.36 If Gzt n ≤ Gc n−1 k1 b k , then Gzt ≤ Gb n zt n t t n psds ≤ G c n k1 b k t 0 s<t k <t b k μ b 1/p k psds. 2.37 Otherwise, we have Gb n zt n − G c 0<t k <t b k ≤ b n μ b 1/p n Gzt n − G c n−1 k1 b k . 2.38 This implies, by induction hypothesis, that Gb n zt n − G c 0<t k <t b k ≤ b n μ b 1/p n t n 0 s<t k <t n b k μ b 1/p k psds t n 0 s<t k <t b k μ b 1/p k psds. 2.39 Thus, 2.35 and 2.39 yield, for 0 <t≤ t n1 , Gzt ≤ G c 0<t k <t b k t 0 s<t k <t b k μ b 1/p k psds, 2.40 and so zt ≤ G −1 G c 0<t k <t b k t 0 s<t k <t b k μ b 1/p k psds . 2.41 Using 2.41 in ut ≤ zt 1/p , we have the required inequality in 2.27. If c is nonnegative, we carry out the above procedure with c ε instead of c, where ε>0 is an arbitrary small constant, and by letting ε → 0, we obtain 2.27. The proof is complete. Jianli Li 7 Remark 2.7. If ∞ r 0 ds/gs 1/p ∞, then G∞∞ and the inequality in 2.27 is true for t ∈ R . An interesting and useful special version of Theorem 2.6 is given in what follows. Corollary 2.8. Let u, f, h, c, p, and b k be as in Theorem 2.6.If u p t ≤ c t 0 fsus s 0 hs, σuσdσ ds 0<t k <t b k − 1u p t k , 2.42 for t ∈ R ,then ut ≤ c 0<t k <t b k p−1/p p − 1 p t 0 s<t k <t b p−1/p k psds p/p−1 , 2.43 for t ∈ R ,wherept is defined by 2.28. Proof. Let guu in Theorem 2.6. Then, 2.26 reduces to 2.42 and Gr p p − 1 r p−1/p − r p−1/p 0 , G −1 r p − 1 p r r p−1/p 0 p/p−1 . 2.44 Consequently, by Theorem 2.6, we have ut ≤ c 0<t k <t b k p−1/p p − 1 p t 0 s<t k <t b p−1/p k psds p/p−1 . 2.45 This proof is complete. 3. Application Example 3.1. Consider the integrodifferential equations x t − F t, xt, t 0 Kt, s, xsds ht, x t k b k xt k ,k 1, 2, , x0x 0 , 3.1 where 0 t 0 <t 1 <t 2 < ··· with lim k→∞ t k ∞; h : R → R and K : R 2 × R → R are continuous; F : R × R 2 → R is continuous at t / t k ; lim t→t k Ft, ·, · and lim t→t − k Ft, ·, · exist and lim t→t − k Ft, ·, ·Ft, ·, ·; b k are constants with |b k |≥1 k 1, 2, . Here, we assume that the solution xt of 3.1 exists on R . Multiplying both sides of 3.1 by xt and then integrating them from 0 to t,weobtain x 2 tx 2 0 2 t 0 xsF s, xs, s 0 Ks, τ, xτdτ hsxs ds 0<t k <t b 2 k − 1 x 2 t k . 3.2 8 Journal of Inequalities and Applications We assume that |Kt, s, xs|≤ftgs|xs|, |Ft, xt,v|≤ft|xt| |v|, 3.3 where f, g ∈ CR , R .From3.2 and 3.3,weobtain |xt| 2 ≤|x 0 | 2 2 t 0 fs|xs| |xs| s 0 gτ|xτ|dτ |hs||xs| ds 0<t k <t |b k | 2 −1 |xt k | 2 . 3.4 Now applying Theorem 2.3, we have |xt|≤|x 0 | 0<t k <t |b k | t 0 s<t k <t |b k | fsashsds, 3.5 where at|x 0 | 0<t k <t |b k | exp t 0 fτgτdτ t 0 s<t k <t |b k | exp t s fτgτdτ hsds, 3.6 for all t ∈ R . The inequality 3.5 gives the bound on the solution xt of 3.1. Acknowledgments This work is supported by the National Natural Science Foundation of China Grants nos. 10571050 and 60671066. The project is supported by Scientific Research Fund of Hunan Provincial Education Department 07B041 and Program for Young Excellent Talents at Hunan Normal University. References 1 V. 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