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Hindawi Publishing Corporation Journal of Inequalities and Applications Volume 2008, Article ID 312395, 8 pages doi:10.1155/2008/312395 Research Article On Some New Impulsive Integral 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 some new impulsive integral 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 some new impulsive integral 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: PCR  ,I{u : R  → I, u is continuous for t /  t k , u0  ,ut  k ,andut − k  exist, and ut − k ut k ,k 1, 2, }, PC 1 R  ,I{u ∈ PCR  ,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 mt is left-continuous at t k , k  1, 2, ; A 2  for k  1, 2, , t≥ t 0 , m  t ≤ ptmtqt,t /  t k , m  t  k  ≤ d k mt k b k , 1.1 where q, p ∈ PCR  , R,d k ≥ 0, and b k are constants. 2 Journal of Inequalities and Applications Then, mt ≤ mt 0   t 0 <t k <t d k exp   t t 0 psds    t t 0  s<t k <t d k exp   t s pσdσ  qsds   t 0 <t k <t   t k <t j <t d j  exp   t t k psds  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 ∈ PCR  , R  , b k ≥ 1, and c ≥ 0 be constants. If u 2 t ≤ c 2  2  t 0 fsusds   0<t k <t  b 2 k − 1  u 2 t k , 2.1 for t ∈ R  ,then ut ≤ c   0<t k <t b k    t 0   0<t k <t b k  fsds, 2.2 for t ∈ R  . Proof. Define a function zt by ztc  ε 2  2  t 0 fsusds   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 ut ≤  zt, we have z  t2ftut ≤ 2ft  zt, 2.4 and so d   zt  dt  z  t 2  zt ≤ ft. 2.5 For t  t k , we have zt  k  − zt k b 2 k − 1u 2 t k  ≤ b 2 k − 1zt k ;thuszt  k  ≤ b 2 k zt k .Let  ztxt; it follows that x  t ≤ ft,t /  t k ,t≥ 0, x  t  k  ≤ b k xt k ,k 1, 2 2.6 From Lemma 1.1,weobtain xt ≤ x0   0<t k <t b k    t 0   s<t k <t b k  fsds ≤ c  ε   0<t k <t b k    t 0   s<t k <t b k  fsds. 2.7 Now by using the fact that ut ≤  ztxt 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 ∈ PCR  , R   and b k ≥ 1 be constants, and let c be a nonnegative constant. If u 2 t ≤ c 2  2  t 0  fsu 2 shsus  ds   0<t k <t  b 2 k − 1  u 2 t k , 2.8 for t ∈ R  ,then ut ≤ c   0<t k <t b k  exp   t 0 fsds    t 0   s<t k <t b k  exp   t s fτdτ  hsds, 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 ∈ PCR  , R  ,c≥ 0, and b k ≥ 1 be constants. If u 2 t ≤ c 2  2  t 0  fsus  us  s 0 gτuτdτ   hsus  ds   0<t k <t  b 2 k − 1  u 2 t k , 2.10 for t ∈ R  ,then ut ≤ c   0<t k <t b k    t 0   s<t k <t b k  fsashsds, 2.11 for t ∈ R  ,where atc   0<t k <t b k  exp   t 0 fτgτdτ    t 0   s<t k <t b k  exp   t s fτgτdτ  hsds. 2.12 Proof. Let ε>0 be an arbitrary small constant, and define a function zt by ztc  ε 2  2  t 0  fsus  us  s 0 gτuτdτ   hsus  ds   0<t k <t  b 2 k − 1  u 2 t k . 2.13 Let  ztxt; similar to the proof of Theorem 2.1, we have x  t ≤ ft  xt  t 0 gsxsds   ht,t /  t k , x  t  k  ≤ b k xt k ,k 1, 2, 2.14 Set vtxt  t 0 gsxsds; then vt ≥ xt,andsofrom2.14 we get that x  t ≤ ftvt ht.Thus,fort /  t k , v  tx  tgtxt ≤ ftvthtgtxt ≤ ftgtvtht, 2.15 4 Journal of Inequalities and Applications and for t  t k , v  t  k  − vt k x  t  k  − xt k  ≤ b k − 1xt k  ≤ b k − 1vt k , 2.16 and so vt  k  ≤ b k vt k .ByLemma 1.1, we have vt≤ cε   0<t k <t b k  exp   t 0 fτgτdτ    t 0   s<t k <t b k  exp   t s fτgτdτ  hsds. 2.17 Let ε → 0, then we obtain vt ≤ at, 2.18 where at is defined in 2.12. Substituting 2.18 into 2.14, we have x  t ≤ ftatht,t /  t k , x  t  k  ≤ b k xt k ,k 1, 2, 2.19 Applying Lemma 1.1 again, we obtain xt ≤ c  ε   0<t k <t b k    t 0   s<t k <t b k  fsashsds. 2.20 Now using ut ≤ xt and letting ε → 0, we get the desired inequality in 2.11. Theorem 2.4. Let u, f, g, h ∈ PCR  , R  ,c≥ 0, and b k ≥ 1 be constants. If u 2 t ≤ c 2  2  t 0  fsus   s 0 gτuτdτ   hsus  ds   0<t k <t b 2 k − 1u 2 t k , 2.21 for t ∈ R  ,then ut ≤ c   0<t k <t b k  exp   t 0 fs   s 0 gτdτ  ds    t 0   s<t k <t b k  exp   t s fτ   τ 0 gωdω  dτ  hsds, 2.22 for t ∈ R  . Proof. Set ztc  ε 2  2  t 0  fsus   s 0 gτuτdτ   hsus  ds   0<t k <t  b 2 k − 1  u 2 t k , 2.23 where ε is an arbitrary small constant; then zt is nondecreasing. Let xt  zt, then it follows for t /  t k that x  t ≤ ft  t 0 gsxsds  ht ≤  ft  t 0 gsds  xtht2.24 Jianli Li 5 since xt is nondecreasing. Also, for t  t k , we have xt  k  ≤ b k xt k . Applying Lemma 1.1, we obtain xt ≤ c  εc   0<t k <t b k  exp   t 0 fs   s 0 gτdτ  ds    t 0   s<t k <t b k  exp   t s fτ   τ 0 gωdω  dτ  hsds. 2.25 Now by using the fact that ut ≤ xt 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,and2.21 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 ∈ PCR  , R  ,ht, s ∈ CR 2  , R  ,for0 ≤ s ≤ t<∞,c≥ 0,b k ≥ 1, and p>1 be constants. Let g ∈ PCR  , R   be a nondecreasing function with gu > 0,foru>0, and gλu ≥ μλgu,forλ>0,u∈ R;hereμλ > 0,forλ>0.If u p t ≤ c   t 0  fsgus   s 0 hs, σguσdσ  ds   0<t k <t b k − 1u p t k , 2.26 for t ∈ R  , then for 0 ≤ t<T, ut ≤  G −1  G  c  0<t k <t b k    t 0  s<t k <t b k μ  b 1/p k  psds  1/p , 2.27 where ptft  t 0 ht, σdσ, 2.28 Gr  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  psds  ∈ dom G −1  . 2.30 Proof. We first assume that c>0 and define a function zt by the right-hand side of 2.26. Then, zt > 0,z0c, ut ≤ zt 1/p ,andzt is nondecreasing. For t /  t k , z  tftgut   t 0 ht, σguσdσ ≤ ftg  zt 1/p    t 0 ht, σg  zσ 1/p  dσ ≤ g  zt 1/p   ft  t 0 ht, σdσ  , 2.31 6 Journal of Inequalities and Applications and for t  t k ,zt  k  ≤ b k zt k .Ast ∈ 0,t 1 ,from2.31 we have Gzt − Gz0   zt z0 ds g  s 1/p  ≤  t 0 psds, 2.32 and so zt ≤ G −1  Gc  t 0 psds  . 2.33 Now assume that for 0 ≤ t ≤ t n , we have zt ≤ G −1  G  c  0<t k <t b k    t 0  0<t k <t b k μ  b 1/p k  psds  . 2.34 Then, for t ∈ t n ,t n1 , it follows from 2.32  that Gzt ≤ Gzt  n    t t n psds.Usingzt  k  ≤ b k zt k , we arrive at Gzt ≤ Gb n zt n    t t n psds. 2.35 From the supposition of g,weseethat Gλu − Gλv  λu 0 ds g  s 1/p  −  λv 0 ds g  s 1/p  ≤ λ μ  λ 1/p  Gu − Gv, for u ≥ v, λ > 0. 2.36 If Gzt n  ≤ Gc  n−1 k1 b k , then Gzt ≤ Gb n zt n    t t n psds ≤ G  c n  k1 b k    t 0  s<t k <t b k μ  b 1/p k  psds. 2.37 Otherwise, we have Gb n zt n  − G  c  0<t k <t b k  ≤ b n μ  b 1/p n   Gzt n  − G  c n−1  k1 b k  . 2.38 This implies, by induction hypothesis, that Gb n zt 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  psds   t n 0  s<t k <t b k μ  b 1/p k  psds. 2.39 Thus, 2.35 and 2.39  yield, for 0 <t≤ t n1 , Gzt ≤ G  c  0<t k <t b k    t 0  s<t k <t b k μ  b 1/p k  psds, 2.40 and so zt ≤ G −1  G  c  0<t k <t b k    t 0  s<t k <t b k μ  b 1/p k  psds  . 2.41 Using 2.41 in ut ≤ zt 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/gs 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  fsus  s 0 hs, σuσdσ  ds   0<t k <t b k − 1u p t k , 2.42 for t ∈ R  ,then ut ≤  c  0<t k <t b k  p−1/p  p − 1 p  t 0  s<t k <t b p−1/p k psds  p/p−1 , 2.43 for t ∈ R  ,wherept is defined by 2.28. Proof. Let guu in Theorem 2.6. Then, 2.26 reduces to 2.42 and Gr 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 ut ≤  c  0<t k <t b k  p−1/p  p − 1 p  t 0  s<t k <t b p−1/p k psds  p/p−1 . 2.45 This proof is complete. 3. Application Example 3.1. Consider the integrodifferential equations x  t − F  t, xt,  t 0 Kt, s, xsds   ht, x  t  k   b k xt k ,k 1, 2, , x0x 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 Ft, ·, · and lim t→t − k Ft, ·, · exist and lim t→t − k Ft, ·, ·Ft, ·, ·; b k are constants with |b k |≥1 k  1, 2, . Here, we assume that the solution xt of 3.1 exists on R  . Multiplying both sides of 3.1 by xt and then integrating them from 0 to t,weobtain x 2 tx 2 0  2  t 0  xsF  s, xs,  s 0 Ks, τ, xτdτ   hsxs  ds   0<t k <t  b 2 k − 1  x 2 t k . 3.2 8 Journal of Inequalities and Applications We assume that |Kt, s, xs|≤ftgs|xs|, |Ft, xt,v|≤ft|xt|  |v|, 3.3 where f, g ∈ CR  , R  .From3.2 and 3.3,weobtain |xt| 2 ≤|x 0 | 2 2  t 0  fs|xs|  |xs|  s 0 gτ|xτ|dτ  |hs||xs|  ds  0<t k <t  |b k | 2 −1  |xt k | 2 . 3.4 Now applying Theorem 2.3, we have |xt|≤|x 0 |   0<t k <t |b k |    t 0   s<t k <t |b k |  fsashsds, 3.5 where at|x 0 |   0<t k <t |b k |  exp   t 0 fτgτdτ    t 0   s<t k <t |b k |  exp   t s fτgτdτ  hsds, 3.6 for all t ∈ R  . The inequality 3.5  gives the bound on the solution xt of 3.1. 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Hindawi Publishing Corporation Journal of Inequalities and Applications Volume 2008, Article ID 312395, 8 pages doi:10.1155/2008/312395 Research Article On Some New Impulsive Integral Inequalities Jianli. Applications, vol. 5, no. 4, pp. 649–657, 2002. 8 N E. Tatar, “An impulsive nonlinear singular version of the Gronwall-Bihari inequality,” Journal of Inequalities and Applications, vol. 2006, Article. 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