Hindawi Publishing Corporation Journal of Inequalities and Applications Volume 2007, Article ID 24385, 18 pages doi:10.1155/2007/24385 Research Article New Integral Inequalities for Iterated Integrals with Applications Ravi P. Agarwal, Cheon Seoung Ryoo, and Young-Ho Kim Received 20 September 2007; Accepted 15 November 2007 Recommended by Sever S. Dragomir Some new nonlinear retarded integral inequalities of Gronwall type are established. These inequalities can be used as basic tools in the study of certain classes of integrodifferential equations. Copyright © 2007 Ravi P. Agarwal et al. 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 The Gronwall-type integr al inequalities provide necessary tools in the study of the theory of differential equations, integral equations, and inequalities of various types. Some such inequalities can be found in the works of Agarwal, Deng et al. [1]. The result has been used in the study of global existence of solutions of a retarded di fferential equations and esti- mation of solution of function differential equation, Cheung [2]. The result has been used in the study of certain initial boundary value problem for hyperbolic partial differential equations, Cheung and Ma [3]. The result has been used in the study of global existence of solutions for a partial differential equations, Pachpatte [4–9]. The results have been ap- plied in the study of certain properties of solutions for the integrodifferential equations, partial integrodifferential equations, retarded Volterra-Fredholm integral equations, re- tarded nonself–adjoint hyperbolic partial differential equations, Ye et al. [10]. The result has been used in the study of the Riemann-Liouville fractional integral equations, Zhao and Meng [11]. The result has been used in the study of integral equations. During the past few years, several authors (see [12–19] and some of the references cited therein) have established many other very useful Gronwall—like integral inequalities. Recently, in [16] a new interesting Gronwall—like integral inequality involving iterated integrals has been established. 2 Journal of Inequalities and Applications Theorem 1.1. Let u(t) be nonnegative continuous function in J = [α, β] and let a(t) be positive nondecreasing continuous function in J, and let f i (t,s), i = 1, ,n, be nonne gative continuous functions for α ≤ s ≤ t ≤ β which are nondecreasing in t for fixed s ∈ J.If u(t) ≤ a(t)+ t α f 1 t,t 1 t 1 α f 2 t 1 ,t 2 ··· t n−1 α f n t n−1 ,t n u p t n dt n ··· dt 1 (1.1) for t ∈ J,wherep ≥ 0, p =1, is a constant. Then u(t) ≤ Y 1 (t,t),whereY 1 (T,t) can be suc- cessively determined from the formulas Y n (T,t) = exp t α n−1 i=1 f i (T,s)ds × a q (T)+q t α f n (T,s)exp − q s α n −1 i=1 f i (T,τ)dτ ds 1/q (1.2) for t ∈ [α,β 1 ),withq = 1 − p and β 1 is chosen so that the expression between [···] is posi- tive in the subinterval [α,β 1 ), and Y k (T,t) = E k (T,t) a(T)+ t α f k (T,s) Y k+1 (T,s) E k (T,s) ds , E k (T,t) = exp t α k−1 i=1 f i (T,τ) − f k (T,τ) dτ , (1.3) for k = n − 1, ,1,α ≤ t ≤ T ≤ β. The main aim of the present paper is to establish some nonlinear retarded inequalities, which extend the above theorem and other results appeared in [16]. We will also illustrate the usefulness of our results. 2. Gronwall-type inequalities First we int roduce some notation, R denotes the set of real numbers and R + = [0,∞), J = [α,β]isthegivensubsetofR. Denote by C i (M,N)theclassofalli-times continuously differentiable functions defined on the set M to the set N for i = 1,2, ,andC 0 (M,N) = C(M,N). Theorem 2.1. Let u(t) and a(t) be nonnegative continuous functions in J = [α, β] with a(t) nondecreasing in J,andlet f i (t,s), i = 1, ,n, be nonnegative continuous functions for α ≤ s ≤ t ≤ β which are nondec reasing in t for fixed s ∈ J.Supposethatφ ∈ C 1 (J,J) is nondecreasing w ith φ(t) ≤ t on J, g(u) is a nondecreasing continuous function for u ∈ R + with g(u) > 0 for u>0,andϕ ∈ C(R + ,R + ) is an increasing function with ϕ(∞) =∞.If ϕ u(t) ≤ a(t)+ φ(t) φ(α) f 1 t,t 1 φ(t 1 ) φ(α) f 2 t 1 ,t 2 ··· φ(t n−1 ) φ(α) f n t n−1 ,t n g u t n dt n ··· dt 1 (2.1) Ravi P. Agarwal et al. 3 for t ∈ [α,β],thenfort ∈ [α, T 1 ], u(t) ≤ ϕ −1 G −1 G a(t) + n i=1 φ(t) φ(α) f i (t,s)ds , (2.2) where G(r) = r r 0 ds s + g(s) , r ≥ r 0 > 0, (2.3) G −1 denotes the inverse function of G,andT 1 ∈J is chosen so that (G(a(t))+ n i =1 φ(t) φ(α) f i (t, s)ds) ∈ Dom(G −1 ). Proof. Let us first assume that a(t) > 0. Fix T ∈ (α,β]. For α ≤ t ≤ T,weobtainfrom(2.1) ϕ u(t) ≤ a(T)+ φ(t) φ(α) f 1 T,t 1 φ(t 1 ) φ(α) f 2 T,t 2 ··· φ(t n−1 ) φ(α) f n T,t n g u t n dt n ··· dt 1 . (2.4) Now we introduce the functions m 1 (t)= a(T)+ φ(t) φ(α) f 1 T,t 1 φ(t 1 ) φ(α) f 2 T,t 2 ··· φ(t n−1 ) φ(α) f n T,t n g u t n dt n ··· dt 1 , m k (t) = m k−1 (t)+ φ(t) φ(α) f k T,t k φ(t k ) φ(α) f k+1 T,t k+1 ··· × φ(t n−1 ) φ(α) f n T,t n g m k−1 t n dt n ··· dt k , (2.5) for t ∈ [α,T]andk = 2, ,n.Thenwehavem k (α) = a(T)fork = 1, , n,andm 1 (t) ≤ m 2 (t) ≤ ··· ≤ m n (t), t ∈ [α,T]. From the inequality (2.4), we obtain u(t) ≤ ϕ −1 (m 1 (t)) or u(t) ≤ ϕ −1 (m n (t)), t ∈ [α,T]. Moreover the function m 1 (t) is nondecreasing. Differ- entiating m 1 (t), we get m 1 (t) = f 1 T,φ(t) φ(φ(t)) φ(α) f 2 T,t 2 ··· φ(t n−1 ) φ(α) f n T,t n g u t n dt n ··· dt 2 φ (t), ≤ − f 1 T,φ(t) φ (t)m 1 (t)+ f 1 T,φ(t) φ (t)m 2 (t) . (2.6) Thus, induction with respect to k gives m k (t) ≤ k−1 i=1 f i T,φ(t) − f k T,φ(t) φ (t)m k (t)+ f k T,φ(t) φ (t)m k+1 (t), (2.7) 4 Journal of Inequalities and Applications for t ∈ [α,T], k = 1,2, ,n − 1. From the definition of the function m n (t) and inequality (2.7), we have m n (t) = m n−1 (t)+ f n T,φ(t) g m n−1 φ(t) φ (t) ≤ n−2 i=1 f i T,φ(t) m n−1 (t)+ f n−1 T,φ(t) m n (t)+ f n T,φ(t) g m n (t) φ (t) ≤ n−1 i=1 f i T,φ(t) m n (t)+ f n T,φ(t) g m n (t) φ (t) ≤ n i=1 f i T,φ(t) φ (t) m n (t)+g m n (t) . (2.8) That is, m n (t) m n (t)+g m n (t) ≤ n i=1 f i T,φ(t) φ (t). (2.9) Tak ing t = s in (2.9) and then integrating it from α to any t ∈ [α,β], changing the variable and using the definition of the function G,wefind G m n (t) ≤ G m n (α) + n i=1 φ(t) φ(α) f i (T,s)ds, (2.10) or m n (t) ≤ G −1 G m n (α) + n i=1 φ(t) φ(α) f i (T,s)ds (2.11) for α ≤ t ≤ T ≤ β.Now,acombinationofu(t) ≤ ϕ −1 (m n (t)) and the last inequality gives the required inequality in (2.2)forT = t.Ifa(t) = 0, we replace a(t)bysomeε>0and subsequently let ε →0. This completes the proof. For the special case g(u) = u p (p>0 is a constant), Theorem 2.1 gives the following retarded integral inequality for iterated integrals. Corollary 2.2. Let u(t), a(t) , f i (t,s), φ(t),andϕ(u) be as in Theorem 2.1.Andletp>0 be a constant. Suppose that ϕ u(t) ≤ a(t)+ φ(t) φ(α) f 1 t,t 1 φ(t 1 ) φ(α) f 2 t 1 ,t 2 ··· φ(t n−1 ) φ(α) f n t n−1 ,t n u p t n dt n ··· dt 1 (2.12) for any t ∈ [α,β].Then,foranyt ∈ [α, T 1 ], u(t) ≤ ϕ −1 G −1 1 G 1 a(t) + n i=1 φ(t) φ(α) f i (t,s)ds , (2.13) Ravi P. Agarwal et al. 5 where G 1 (r) = r r 0 ds s + s p , r ≥ r 0 > 0, (2.14) G −1 1 denotes the inverse function of G 1 ,andT 1 ∈J is chosen so that (G 1 (a(t))+ n i =1 φ(t) φ(α) f i (t, s)ds) ∈ Dom (G −1 1 ). Remark 2.3. (i) When ϕ(u) = u and g(u) = u,formTheorem 2.1, we derive the following retarded integr al inequality: u(t) ≤ a(t)exp 2 n i=1 φ(t) φ(α) f i (t,s)ds . (2.15) (ii) When ϕ(u) = u,inTheorem 2.1, we obtain the following retarded integral in- equality: u(t) ≤ G −1 G a(t) + n i=1 φ(t) φ(α) f i (t,s)ds . (2.16) (iii) When ϕ(u) = u p (p>0 is a constant) in Theorem 2.1, we have the following retarded integr al inequality: u(t) ≤ G −1 G a(t) + n i=1 φ(t) φ(α) f i (t,s)ds 1/p . (2.17) Now we introduce the following notation. For α<β,letJ i ={(t 1 ,t 2 , ,t i ) ∈ R i : α ≤ t i ≤ ··· ≤ t 1 ≤ β} for i = 1, ,n. Theorem 2.4. Let u(t) and a(t) be nonnegative continuous functions in J = [α,β] with a(t) nondecreasing in J,andletp i (t), i = 1, ,n, be nonnegative continuous functions for α ≤ t ≤ β.Supposethatφ ∈ C 1 (J,J) is nondecreasing with φ( t) ≤ t on J, g(u) is a nondecreasing continuous function for u ∈ R + with g(u) > 0 for u>0,andϕ ∈ C(R + ,R + ) is an increasing function with ϕ( ∞) =∞. If ϕ u(t) ≤ a(t)+ φ(t) φ(α) p 1 t 1 g u t 1 dt 1 + n i=2 φ(t) φ(α) p 1 t 1 φ(t 1 ) φ(α) p 2 t 2 ··· φ(t i−2 ) φ(α) p i−1 t i−1 × φ(t i−1 ) φ(α) p i t i g u t i dt i dt i−1 ··· dt 2 dt 1 , (2.18) for any t ∈ J, then u(t) ≤ ϕ −1 G −1 G(a(t + F(t) (2.19) 6 Journal of Inequalities and Applications for t ∈ [α,T 2 ],whereT 2 ∈ I is chosen so that (G(a(t)) + F(t)) ∈ Dom (G −1 ), G(r) = r r 0 ds g ϕ −1 (s) , r ≥ r 0 > 0, (2.20) G −1 denotes the inverse function of G,and F(t) = φ(t) φ(α) p 1 t 1 dt 1 + n i=2 φ(t) φ(α) p 1 t 1 φ(t 1 ) φ(α) p 2 t 2 ··· φ(t i−2 ) φ(α) p i−1 t i−1 × φ(t i−1 ) φ(α) p i t i dt i dt i−1 ··· dt 2 dt 1 , (2.21) for any t ∈ I. Proof. Let the function a(t) be positive. Define a function v(t) by the right side of (2.18). Clearly, v(t) is nondecreasing continuous, u(t) ≤ ϕ −1 (v(t)) for t ∈ I and v(α) = a(α). Differentiating v(t) and rew riting, we have v (t) − a (t) φ (t)p 1 φ(t) − g u φ(t) ≤ v 1 (t), (2.22) where v 1 (t) = φ(t) φ(α) p 2 t 2 g u t 2 dt 2 + n i=3 φ(t) φ(α) p 2 t 2 φ(t 2 ) φ(α) p 3 t 3 ··· φ(t i−2 ) φ(α) p i−1 t i−1 × φ(t i−1 ) φ(α) p i t i g u t i dt i dt i−1 ··· dt 3 dt 2 . (2.23) Now differentiating v 1 (t) and rewriting, we get v 1 (t) φ (t)p 2 φ(t) − g u φ(t) ≤ v 2 (t), (2.24) where v 2 (t)= φ(t) φ(α) p 3 t 3 g u t 3 dt 3 + n i=4 φ(t) φ(α) p 3 t 3 φ(t 3 ) φ(α) p 4 t 4 ··· φ(t i−2 ) φ(α) p i−1 t i−1 × φ(t i−1 ) φ(α) p i t i g u t i dt i dt i−1 ··· dt 4 dt 3 . (2.25) Ravi P. Agarwal et al. 7 Continuing in this way, we obtain v n−2 (t) φ (t)p n−1 φ(t) − g u φ(t) ≤ v n−1 (t), (2.26) where v n−1 (t) = φ(t) φ(α) p n t n g u t n dt n . (2.27) From the definition of v n−1 (t) and the inequality u(t) ≤ ϕ −1 (v(t)), we find v n−1 (t) g ϕ −1 v(t) ≤ φ (t)p n φ(t) . (2.28) Integrating the inequality (2.28), we get t α v n−1 (s) g ϕ −1 v(s) ds ≤ φ(t) φ(α) p n (s)ds. (2.29) Now integrating by parts the left–hand side of (2.29), we obtain t α v n−1 (s) g ϕ −1 v(s) ds = v n−1 (t) g ϕ −1 v(t) + t α v n−1 g ϕ −1 (v) g 2 ϕ −1 (v) v ϕ ϕ −1 (v) ds ≥ v n−1 (t) g ϕ −1 v(t) . (2.30) From the i nequalities (2.29)and(2.30), we have v n−1 (t) g ϕ −1 v(t) ≤ φ(t) φ(α) p n (s)ds. (2.31) Next from the inequality (2.26), we observe that v n−2 (t) ≤ φ (t)p n−1 φ(t) g u φ(t) + φ (t)p n−1 φ(t) v n−1 (t), (2.32) Thus, it follows that v n−2 (t) g ϕ −1 v(t) ≤ φ (t)p n−1 φ(t) g u φ(t) g ϕ −1 v(t) + φ (t)p n−1 φ(t) v n−1 (t) g ϕ −1 v(t) ≤ φ (t)p n−1 φ(t) + φ (t)p n−1 φ(t) v n−1 (t) g ϕ −1 v(t) . (2.33) Using the same procedure from (2.29)to(2.31) to the inequality (2.33), we get v n−2 (t) g ϕ −1 v(t) ≤ φ(t) φ(α) p n−1 t 1 dt 1 + φ(t) φ(α) p n−1 t 1 v n−1 t 1 g ϕ −1 v t 1 dt 1 . (2.34) 8 Journal of Inequalities and Applications Now combining the inequalities (2.31)and(2.34), we find v n−2 (t) g ϕ −1 v(t) ≤ φ(t) φ(α) p n−1 t 1 dt 1 + φ(t) φ(α) p n−1 t 1 φ(t 1 ) φ(α) p n t 2 dt 2 dt 1 . (2.35) Proceeding in this way we arrive at v 1 (t) g ϕ −1 v(t) ≤ φ(t) φ(α) p 2 t 1 dt 1 + ··· + φ(t) φ(α) p 2 t 1 φ(t 1 ) φ(α) p 3 t 2 ··· φ(t n−2 ) φ(α) p n t n−1 dt n−1 ··· dt 2 dt 1 . (2.36) On the other hand, from the inequality (2.22), we have v (t) − a (t) ≤ φ (t)p 1 φ(t) g u φ(t) + φ (t)p 1 φ(t) v 1 (t), (2.37) or v (t) − a (t) g ϕ −1 v(t) ≤ φ (t)p 1 φ(t) g u φ(t) g ϕ −1 v(t) + φ (t)p 1 φ(t) v 1 (t) g ϕ −1 v(t) ≤ φ (t)p 1 φ(t) + φ (t)p 1 φ(t) v 1 (t) g ϕ −1 v(t) , (2.38) that is, v (t) g ϕ −1 v(t) − a (t) g ϕ −1 a(t) ≤ φ (t)p 1 φ(t) + φ (t)p 1 φ(t) v 1 (t) g ϕ −1 v(t) . (2.39) Setting t = t 1 and integrating from α to t, and using the definition of G,weobtain G v(t) ≤ G a(t) + φ(t) φ(α) p 1 t 1 dt 1 + φ(t) φ(α) p 1 t 1 v 1 t 1 g ϕ −1 v(t 1 ) dt 1 . (2.40) Consequently, using (2.36) to the inequality (2.40), we get v(t) ≤ G −1 G a(t) + F(t) , (2.41) where the function F(t)isdefinedin(2.21). Now, the desired inequality in (2.24)fol- lows by the inequalit y u(t) ≤ ϕ −1 (v(t)). If a(t) = 0, we replace a(t)bysomeε>0and subsequently let ε →0. This completes the proof. Ravi P. Agarwal et al. 9 For the special case ϕ(u) = u p (p>1 is a constant), Theorem 2.4 gives the fol lowing retarded integral inequality for iterated integrals. Corollary 2.5. Let u(t), a(t), p i (t), φ(t),andg(u) be as in Theorem 2.4.Andletp>0 be a constant. If u p (t) ≤ a(t)+ φ(t) φ(α) p 1 t 1 g u t 1 dt 1 + n i=2 φ(t) φ(α) p 1 t 1 φ(t 1 ) φ(α) p 2 t 2 ··· φ(t i−2 ) φ(α) p i−1 t i−1 × φ(t i−1 ) φ(α) p i t i g u t i dt i dt i−1 ··· dt 2 dt 1 , (2.42) for any t ∈ J, then u(t) ≤ G −1 G a(t) + F(t) 1/p (2.43) for t ∈ [α,T 3 ],whereT 3 ∈ I is chosen so that (G 1 (a(t)) + F(t)) ∈ Dom (G −1 1 ), G 1 (r) = r r 0 ds g v 1/p (s) , r ≥ r 0 > 0, (2.44) G −1 denotes the inverse function of G,andthefunctionF(t) is defined in (2.21)foranyt ∈ I. Theorem 2.6. Let u(t) and a(t) be nonnegative continuous functions in J = [α,β] with a(t) nondecreasing in J,andlet f i (t) and p i (t), i = 1, ,n, be nonnegative continuous functions for α ≤ t ≤ β.Supposethatφ ∈ C 1 (J,J) is nondec reasing with φ(t) ≤ t on J, g(u) is a non- decreasing continuous function for u ∈ R + with g(u) > 0 for u>0,andϕ ∈ C(R + ,R + ) is an increasing function with ϕ( ∞) =∞.If ϕ u(t) ≤ a(t)+ φ(t) φ(α) p 1 t 1 f 1 t 1 u t 1 g u t 1 dt 1 + n i=2 φ(t) φ(α) p 1 t 1 φ(t 1 ) φ(α) p 2 t 2 ··· φ(t i−2 ) φ(α) p i−1 t i−1 × φ(t i−1 ) φ(α) p i t i f i t i u t i g u t i dt i dt i−1 ··· dt 2 dt 1 , (2.45) for any t ∈ J, then u(t) ≤ ϕ −1 Φ −1 G −1 2 G 2 Φ a(t) + F 1 (t) (2.46) 10 Journal of Inequalities and Applications for t ∈ [α,T 4 ],whereT 4 ∈ I is chosen so that (G 2 [Φ(a(t))] + F 1 (t)) ∈ Dom (G −1 2 ), [G −1 2 (G 2 [Φ(a(t))] + F 1 (t))] ∈ Dom (Φ −1 ), G 2 (r) = r r 0 ds g ϕ −1 Φ −1 (s) , r ≥ r 0 > 0, Φ(r) = r r 0 ds ϕ −1 (s) , r ≥ r 0 > 0, (2.47) G −1 2 denotes the inverse function of G 2 ,and F 1 (t) = φ(t) φ(α) p 1 t 1 f 1 (t)dt 1 + n i=2 φ(t) φ(α) p 1 t 1 φ(t 1 ) φ(α) p 2 t 2 × ··· φ(t i−2 ) φ(α) p i−1 t i−1 × φ(t i−1 ) φ(α) p i t i f i t i dt i dt i−1 ··· dt 2 dt 1 , (2.48) for any t ∈ I. Proof. Let the function a(t) be positive. Define a function w(t) by the right side of (2.45). Clearly, w(t) is a nondecreasing continuous function, u(t) ≤ ϕ −1 (w(t)) for t ∈ I and w(α) = a(α). Differentiating w(t) and rewriting, we have w (t) − a (t) φ (t)p 1 φ(t) − f 1 φ(t) u φ(t) g u φ(t) ≤ w 1 (t), (2.49) where w 1 (t) = φ(t) φ(α) p 2 t 2 f 2 t 2 u t 2 g u t 2 dt 2 + n i=3 φ(t) φ(α) p 2 t 2 φ(t 2 ) φ(α) p 3 t 3 ··· φ(t i−2 ) φ(α) p i−1 t i−1 × φ(t i−1 ) φ(α) p i t i f i t i u t i g u t i dt i dt i−1 ··· dt 3 dt 2 . (2.50) Now differentiating v 1 (t) and rewriting, we get w 1 (t) φ (t)p 2 φ(t) − f 2 φ(t) u φ(t) g u φ(t) ≤ w 2 (t), (2.51) [...]... Kim, “On some integral inequalities with iterated integrals, ” Journal of the Korean Mathematical Society, vol 43, no 3, pp 563–578, 2006 [14] S S Dragomir and Y.-H Kim, “On certain new integral inequalities and their applications,” Journal of Inequalities in Pure and Applied Mathematics, vol 3, no 4, article 65, pp 1–8, 2002 [15] S S Dragomir and Y.-H Kim, “Some integral inequalities for functions... certain new Gronwall-Ou-Iang type integral inequalities in two variables and their applications,” Journal of Inequalities and Applications, vol 2005, no 4, pp 347–361, 2005 [4] B G Pachpatte, Inequalities for Differential and Integral Equations, vol 197 of Mathematics in Science and Engineering, Academic Press, San Diego, Calif, USA, 1998 [5] B G Pachpatte, “Explicit bounds on certain integral inequalities, ”... [6] B G Pachpatte, “On some retarded integral inequalities and applications,” Journal of Inequalities in Pure and Applied Mathematics, vol 3, no 2, article 18, pp 1–7, 2002 [7] B G Pachpatte, “On a certain retarded integral inequality and applications,” Journal of Inequalities in Pure and Applied Mathematics, vol 5, no 1, article 19, pp 1–9, 2004 [8] B G Pachpatte, Inequalities applicable to certain... some Gronwall type inequalities for a system integral equation,” Bulletin of the Korean Mathematical Society, vol 42, no 4, pp 789–805, 2005 [17] O Lipovan, “A retarded integral inequality and its applications,” Journal of Mathematical Analysis and Applications, vol 285, no 2, pp 436–443, 2003 [18] Q.-H Ma and J Peˇ ari´ , “On certtain new nonlinear retared integral inequalities for functions c c in... Pachpatte, Inequalities applicable to certain partial differential equations,” Journal of Inequalities in Pure and Applied Mathematics, vol 5, no 2, article 27, pp 1–12, 2004 [9] B G Pachpatte, “On some new nonlinear retarded integral inequalities, ” Journal of Inequalities in Pure and Applied Mathematics, vol 5, no 3, article 80, pp 1–8, 2004 [10] H Ye, J Gao, and Y Ding, “A generalized Gronwall inequality... Applications, vol 328, no 2, pp 1075–1081, 2007 [11] X Zhao and F Meng, “On some advanced integral inequalities and their applications,” Journal of Inequalities in Pure and Applied Mathematics, vol 6, no 3, article 60, pp 8 pages, 2005 18 Journal of Inequalities and Applications [12] D Ba˘nov and P Simeonov, Integral Inequalities and Applications, vol 57 of Mathematics and Its ı Applications, Kluwer Academic,... Ou-Iang type-retarded integral inequality with iterated integrals Corollary 2.7 Let u, a, fi , pi , φ, and g(u) be as in Theorem 2.6 and let p > 1 be a constant If φ(t) u p (t) ≤ a(t) + n + φ(α) p1 t1 f1 t1 u t1 g u t1 dt1 φ(t) φ(t1 ) p1 t1 i=2 φ(α) φ(α) ··· p2 t2 × φ(ti−2 ) φ(α) pi−1 ti−1 φ(ti−1 ) φ(α) pi ti fi ti u ti g u ti dti dti−1 · · · dt2 dt1 , (2.76) for any t ∈ J, then u(t) ≤ G−1 G3 a(p−1)/... from the inequalities (2.72) and (2.74), we get k(t) ≤ G−1 G2 Φ a(T) 2 + F1 (t) , (2.75) where the function F1 (t) is defined in (2.46) In particular, for T = t, we find that the desired inequality (2.46) follows by the inequalities u(t) ≤ ϕ−1 (w(t)) and w(t) ≤ Φ−1 (k(t)) This completes the proof When ϕ(u) = u p (p > 1 is a constant) in Theorem 2.6, we get the following Ou-Iang type-retarded integral. .. = η2 + τ(t2 ), for 1 η1 , η2 ∈ I, and B Mb1 η1 ,M 2 b1 η1 b2 η2 = φ(η) φ(α) φ(η) Mb1 η1 dη1+ φ(η1 ) φ(α) φ(α) M 2 b2 η1 b3 η2 dη2 dη1 , (3.4) where φ(γ) = γ − τ(γ) for γ ∈ I Proof It is easy to see that the solution x(η) of the problem (3.1) satisfies the equivalent integral equation x p (η) = x p (α) + η α F t1 ,x t1 − τ t1 , t1 −τ(t1 ) α G t2 ,x t2 − τ t2 dt2 dt1 (3.5) 16 Journal of Inequalities and... ∈ I for i = 1,2,3 Now when ϕ(u) = u and f1 = f4 = · · · = fn = 0, a suitable application of the inequality given in Theorem 2.6 to (3.12) yields the desired result References [1] R P Agarwal, S Deng, and W Zhang, “Generalization of a retarded Gronwall-like inequality and its applications,” Applied Mathematics and Computation, vol 165, no 3, pp 599–612, 2005 [2] W S Cheung, “Some new nonlinear inequalities . Corporation Journal of Inequalities and Applications Volume 2007, Article ID 24385, 18 pages doi:10.1155/2007/24385 Research Article New Integral Inequalities for Iterated Integrals with Applications Ravi. useful Gronwall—like integral inequalities. Recently, in [16] a new interesting Gronwall—like integral inequality involving iterated integrals has been established. 2 Journal of Inequalities and. some integral inequalities w ith iterated integrals, ” Journal of the Korean Mathematical Society, vol. 43, no. 3, pp. 563–578, 2006. [14] S. S. Dragomir and Y H. Kim, “On certain new integral inequalities