The main objective of the present paper is to establish some new Gronwall type inequalities involving iterated integrals

S. Dragomir Sever

Yeol Je Cho

Young-Ho Kim

English

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Mathematical Communications 12(2007), 63-73 63On some Gronwall type inequalities with iteratedintegralsYeol Je Cho∗, Sever S. Dragomir†and Young-Ho Kim‡Abstract. The main objective of the present paper is to establishsome new Gronwall type inequalities involving iterated integrals.Key words: Gronwall-Bellman inequality, integral inequality, iter-ated integral, nondecreasing functionAMS subject classifications: 26D15, 35A05Received March 5, 2007 Accepted March 12, 20071. IntroductionLet u : [α, α+h] → R be a continuous real-valued function satisfying the inequality0 ≤ u(t) ≤∫ tα[a + bu(s)] ds, for all t ∈ [α, α + h],where a, b are nonnegative constants. Then u(t) ≤ ahebh for t ∈ [α, α + h]. Thisresult was proved by T. H. Gronwall [8] in the year 1919, and it is a prototype forthe study of several integral inequalities of Volterra type, and also for obtainingexplicit bounds of the unknown function. Among the several publications on thissubject, the paper of Bellman [3] is very well known. It is clear that Bellman’sresult contains that of Gronwall. This is the reason why inequalities of this typewere called “Gronwall-Bellman inequalities” or “Inequalities of Gronwall type”.The Gronwall type integral inequalities provide a necessary tool for the study ofthe theory of differential equations, integral equations and inequalities of varioustypes (see Gronwall [8] and Guiliano [9]). Some applications of this result to thestudy of stability of the solution of linear and nonlinear differential equations maybe found in Bellman [3]. Some applications to existence and uniqueness theoryof differential equations may be found in Nemyckii-Stepanov [13], Bihari [4], and∗Department of Mathematics Education, The Research Institute of Natural Sciences, Col-lege of Education, Gyeongsang National University, Chinju 660-701, Republic of Korea, e-mail:yjcho@gsnu.ac.kr†School of Computer Science and Mathematics, Victoria University of Technology, POBox14428, MCMC, Melbourne, Victoria 8001, Australia, e-mail: sever@matilda.vu.edu.au‡Department of Applied Mathematics, Changwon National University, Changwon, Kyung-Nam641-773, Republic of Korea, e-mail: yhkim@sarim.changwon.ac.kr64 Y. J.Cho, S. S.Dragomir and Y.-H.KimLangenhop [10]. During the past few years several authors (see references belowand some of the references cited therein) have established several Gronwall typeintegral inequalities in two or more independent real variables. Of course, suchresults have application in the theory of partial differential equations and Volterraintegral equations.Bainov and Simeonov [1, p. 101] proved the following interesting integral in-equality involving iterated integrals:Let u(t), a(t), and b(t) be nonnegative continuous functions in J = [α, β], andsuppose thatu(t) ≤ a(t) + b(t)[∫ tαk1(t, t1)u(t1) dt1 + · · ·+∫ tα(∫ t1α· · ·(∫ tn−1αkn(t, t1, . . . , tn)u(tn) dtn). . .)dt1](1.1)for all t ∈ J, where ki(t, t1, . . . , ti) are nonnegative continuous functions in Ji+1, i =1, 2, . . . , n, which are nondecreasing in t ∈ J for all fixed (t1, . . . , ti) ∈ Ji, i =1, 2, . . . , n. Then, for all t ∈ Ju(t) ≤ a(t) + b(t)∫ tαR̂[a](t, s) exp(∫ tsR̂[b](t, τ) dτ)ds,where, for all (t, s) ∈ J2,R̂[w](t, s)) = k1(t, s)w(s) +∫ sαk2(t, s, t2)w(t2)dt2+n∑i=3∫ sα(∫ t2α· · ·(∫ ti−1αki(t, s, t2, . . . , ti)w(ti) dti)· · ·)dt2,for each continuous function w(t) in J.In this paper we consider simple inequalities involving iterated integrals in in-equality (1.1) for functions when the function u on the right-hand side of inequality(1.1) is replaced by the function up for some p, We also provide some integralinequalities involving iterated integrals.2. The main resultsIn this section we state and prove some new nonlinear integral inequalities involvingiterated integrals. Throughout the paper all functions which appear in inequalitiesare assumed to be real-valued.Before considering our first integral inequality, we need the following lemmas,which appeared in [1, p. 2, p. 38].Lemma 2.1. Let b(t) and f(t) be continuous functions for t ≥ α, let v(t) be adifferentiable function for t ≥ α, and supposev′(t) ≤ b(t)v(t) + f(t), t ≥ αOn some Gronwall type inequalities 65and v(α) ≤ v0. Then, for all t ≥ α,v(t) ≤ v0 exp(∫ tαb(s) ds)+∫ tαf(s) exp(∫ tsb(τ) dτ)ds.Lemma 2.2. Let v(t) be a positive differential function satisfying the inequalityv′(t) ≤ b(t)v(t) + k(t)vp(t), t ∈ J = [α, β],where functions b and k are continuous in J, and p ≥ 0, p = 1, is a constant. Thenv(t) ≤ exp(∫ tαb(s) ds)[vq(α) + q∫ tαk(s) exp(−q∫ sαb(τ) dτ)ds]1/qfor all t ∈ [α, β1), where β1 is chosen so that the expression between [...] is positivein the subinterval [α, β1).In the next theorems we consider some simple inequalities involving iteratedintegrals.Let α < β, and setJi = {(t1, t2, . . . , ti) ∈ Ri : α ≤ ti ≤ · · · ≤ t1 ≤ β}, i = 1, · · · , n.Theorem 2.1. Let u(t), a(t) and b(t) be nonnegative continuous functions inJ = [α, β] and let p > 1 be a constant. Suppose that a(t)b(t) is nondecreasing in J andu(t) ≤ a(t) + b(t)[∫ tαk1(t, t1)up(t1) dt1 + · · ·+∫ tα(∫ t1α· · ·(∫ tn−1αkn(t, t1, · · · , tn)up(tn) dtn)· · ·)dt1](2.1)for any t ∈ J, where ki(t, t1, . . . , ti) are nonnegative continuous functions in Ji+1for i = 1, 2, · · · , n, which are nondecreasing in t ∈ J for all fixed (t1, · · · , ti) ∈ Ji,i = 1, 2, · · · , n. Then, for any t ∈ [α, βp)u(t) ≤ a(t)[1− (p− 1)∫ tα(a(s)b(s))p−1R[bp](t, s) ds] 11−p(2.2)where, for any (t, s) ∈ J2,βp = sup{t ∈ J : (p− 1)∫ tα(a(s)b(s))p−1R[bp](t, s) ds < 1},andR[w](t, s) = k1(t, s)w(s) +∫ sαk2(t, s, t2)w(t2)dt2+n∑i=3∫ sα(∫ t2α· · ·(∫ ti−1αki(t, s, t2, · · · , ti)w(ti) dti)· · ·)dt2,66 Y. J.Cho, S. S.Dragomir and Y.-H.Kimfor each continuous function w(t) in J.Proof. For a fixed T ∈ (α, β] and α ≤ t ≤ T we haveu(t) ≤ a(t) + b(t)v(t), (2.3)wherev(t) =∫ tαk1(T, t1)up(t1) dt1 + · · ·+∫ tα(∫ t1α· · ·(∫ tn−1αkn(T, t1, · · · , tn)up(tn) dtn)· · ·)dt1.Since ∂ki∂t (T, t1, . . . , ti) = 0 for i = 1, . . . , n and t ∈ [α, T ], we havev′(t) = R[up](T, t) ≤ (R[bp](T, t))(a(t)b(t)+ v(t))p,that is,v′(t) ≤ Q(T, t)[a(T )/b(T ) + v(t)], (2.4)where Q(T, t) = (R[bp](T, t))[a(t)/b(t) + v(t)]p−1. Lemma 2.1 and (2.4) implyv(t) +a(T )b(T )≤ a(T )b(T )exp(∫ tαQ(T, s) ds), α ≤ t ≤ T.Hence, for t = T,v(t) +a(t)b(t)≤ a(t)b(t)exp(∫ tαQ(T, s) ds). (2.5)From (2.5), we successively obtain[v(t) +a(t)b(t)]p−1≤[a(t)b(t)]p−1exp(∫ tα(p− 1)Q(T, s) ds),Q(T, t) ≤ (R[bp](T, t))[a(t)b(t)]p−1exp(∫ tα(p− 1)Q(T, s) ds),Z(T, t) ≤ (p− 1)(R[bp](T, t))[a(t)b(t)]p−1exp(∫ tα(p− 1)Q(T, s) ds),where Z(T, t) = (p− 1)Q(T, t). Consequently, we haveZ(T, s) exp(−∫ sαZ(T, s) ds)≤ (p− 1)R[bp](T, s)[a(s)b(s)]p−1ordds[− exp(−∫ sαZ(T, τ) dτ)]≤ (p− 1)R[bp](T, s)[a(s)b(s)]p−1.On some Gronwall type inequalities 67Integrating this from α to t yields1− exp(−∫ tαZ(T, s) ds)≤ (p− 1)∫ tα(a(s)b(s))p−1R[bp](T, s) ds,from which we conclude thatexp(∫ tαQ(T, s) ds)≤[1− (p− 1)∫ tα(a(s)b(s))p−1R[bp](T, s) ds] 11−p.This, together with (2.3) and (2.5), impliesu(t) ≤ a(t)[1− (p− 1)∫ tα(a(s)b(s))p−1R[bp](T, s) ds] 11−p.In particular, for T = t we find (2.2). This completes the proof. ✷Theorem 2.2. Let u(t) and b(t) be nonnegative continuous functions in J =[α, β], and suppose thatu(t) ≤ b(t)[a(t) +∫ tαk1(t, t1)up(t1) dt1 + · · ·+∫ tα(∫ t1α· · ·(∫ tn−1αkn(t, t1, . . . , tn)up(tn) dtn). . .)dt1]for t ∈ J, where p ≥ 0, p = 1 be a constant, a(t) > 0 is a nondecreasing continuousfunction in t ∈ J , and ki(t, t1, . . . , ti) are nonnegative continuous functions inJi+1, i = 1, 2, . . . , n, which are nondecreasing in t ∈ J for all fixed (t1, . . . , ti) ∈Ji, i = 1, 2, . . . , n. Thenu(t) ≤ b(t)[aq(t) + q∫ tαR[bp](t, s) ds]1/q(2.6)for any t ∈ [α, β1), where q = 1− p and β1 is chosen so that the expression between[...] is positive in the subinterval [α, β1).Proof. For a fixed T ∈ (α, β] and α ≤ t ≤ T we haveu(t) ≤ b(t)v(t)≡ b(t)[a(T ) +∫ tαk1(T, t1)up(t1) dt1 + · · ·+∫ tα(∫ t1α· · ·(∫ tn−1αkn(T, t1, . . . , tn)up(tn) dtn). . .)dt1].Since v(α) = a(T ), v(t) is nondecreasing and continuous in J and ∂ki∂t (T, t1, . . . , ti) ≡0 for i = 1, . . . , n and t ∈ [α, T ], we havev′(t) = R[up](T, t) ≤ R[bpvp](T, t)≤ (R[bp](t))vp(T, t). (2.7)68 Y. J.Cho, S. S.Dragomir and Y.-H.KimLemma 2.2 and (2.7) implyv(t) ≤[aq(T ) + q∫ tαR[bp](T, s) ds,]1/qfrom which we obtainu(t) ≤ b(t)[aq(T ) + q∫ tαR[bp](T, s) ds]1/qfor α ≤ t ≤ T. In particular, for T = t we find (2.6). This completes the proof. ✷Theorem 2.3. Let u(t), a(t) and b(t) be nonnegative continuous functions inJ = [α, β]. Suppose thatu(t) ≤ a(t) + b(t)[∫ tαk1(t, t1)up(t1) dt1 + · · ·+∫ tα(∫ t1α· · ·(∫ tn−1αkn(t, t1, . . . , tn)up(tn) dtn). . .)dt1]for all t ∈ J, where 0 < p ≤ 1 be a constant, a(t)b(t) ≥ 1 is nondecreasing in J andki(t, t1, . . . , ti) are nonnegative continuous functions in Ji+1, i = 1, 2, . . . , n, whichare nondecreasing in t ∈ J for all fixed (t1, . . . , ti) ∈ Ji, i = 1, 2, . . . , n. Thenu(t) ≤ a(t) exp(∫ tαR[bp](t, τ) dτ)(2.8)for all t ∈ [α, β].Proof. For a fixed T ∈ (α, β] and α ≤ t ≤ T , we haveu(t) ≤ a(t) + b(t)w(t)≡ a(t) + b(t)[∫ tαk1(T, t1)up(t1) dt1 + · · ·+∫ tα(∫ t1α· · ·(∫ tn−1αkn(T, t1, . . . , tn)up(tn) dtn). . .)dt1].Since w(α) = 0, w(t) is nondecreasing and continuous in J and ∂ki∂t (T, t1, . . . , ti) ≡ 0for i = 1, . . . , n and t ∈ [α, T ], we havew′(t) = R[up](T, t) ≤ R[bp](T, t)(a(t)b(t)+ w(t))p≤ R[bp](T, t)(a(t)b(t)+ w(t)). (2.9)Lemma 2.1 and (2.9) implyw(t) ≤∫ tαR[bp](T, s)(a(s)b(s))exp(∫ tsR[bp](T, τ) dτ)ds,On some Gronwall type inequalities 69from which we obtainu(t) ≤ a(t) + b(t)∫ tαR[bp](T, s)(a(s)b(s))exp(∫ tsR[bp](T, τ) dτ)ds. (2.10)Indeed, (2.10) implies thatu(t) ≤ a(t)[1 +∫ tαR[bp](T, s) exp(∫ tsR[bp](T, τ) dτ)ds]= a(t) exp(∫ tαR[bp](T, τ) dτ)for α ≤ t ≤ T. In particular, for T = t we find (2.8). This completes the proof. ✷Theorem 2.4. Let u, f1, . . . , fn be nonnegative continuous functions in J =[α, β], and suppose thatu(t) ≤ a +∫ tαf1(t1)up(t1) dt1 + · · ·+∫ tαf1(t1)(∫ t1αf2(t2) · · ·(∫ tn−1αfn(tn)up(tn) dtn)· · ·)dt1 (2.11)for all t ∈ J, where a ≥ 1 and 0 < p ≤ 1 are constants. Thenu(t) ≤ aR1(t), t ∈ J, (2.12)whereRn(t) = exp(∫ tαfn(s) ds), t ∈ J,andRn(t) = 1 +∫ tαfi(t)Ri+1(s) exp(∫ sαfi(τ) dτ)dsfor all t ∈ J and i = n− 1, · · · , 1.Proof. We setu1(t) = a + L1[up](t), uj+1(t) = uj + Lj+1[up](t)for all t ∈ J and j = 1, · · · , n− 1, whereLk[up](t) =∫ tαfk(tk)up(tk) dtk + · · ·+∫ tαfk(tk)(∫ tαfk+1(tk+1) · · ·(∫ tαfn(tn) dtn)· · ·)dtkfor all t ∈ J and k = 1, · · · , n. Now, (2.11) impliesu(t) ≤ u1(t). (2.13)70 Y. J.Cho, S. S.Dragomir and Y.-H.KimTaking into account thatuk(t) ≤ uk+1(t),(Lk[up])′ = fk(up(t) + Lk+1[up]), k = 1, · · · , n− 1,and(Ln[up])′ = fn(t)up(t).We successively findu′1(t) = (L1[up](t))′ = f1[up(t) + L2[up]]≤ f1[u1(t) + L2[up]] = f1u2,u′k(t) ≤ (f1 + · · · + fk−1)uk(t) + fkuk(t), k = 2, . . . , n− 1, (2.14)u′n(t) ≤ (f1 + · · · + fn)un(t).Since uk(α) = a, k = 1, . . . , n, (2.14) gives, by successive application of Lemma 2.1,uk(t) ≤ aRk(t) exp(∫ tαk−1∑j=1fj(s) ds), k = n, n− 1, . . . , 1.For k = 1 this and (2.13) imply (2.12). ✷Remark 2.1. In the case when a ≥ 0, p = 1, the inequality given in (2.11)reduces to the inequality established earlier by Ráb in [16] (see also [1, Theorem11.6, p. 102]).Corollary 2.1. Let u, f, g be nonnegative continuous functions in J = [α, β],u0 ≥ 1 and suppose thatu(t) ≤ u0 +∫ tαf(s)[up(s) +∫ sαg(τ)up(τ) dτ]dsfor all t ∈ J, where 0 < p ≤ 1 is a constant. Thenu(t) ≤ u0[1 +∫ tαf(s) exp(∫ sα(f(τ) + g(τ)) dτ)ds]for all t ∈ J.Theorem 2.5. Let u, fi, i = 1, · · · , n be nonnegative continuous functions inJ = [α, β], and suppose thatu(t) ≤ a(t) +∫ tαf1(t1)up(t1) dt1 + · · ·+∫ tαf1(t1)(∫ t1αf2(t2) · · ·(∫ tn−1αfn(tn)up(tn) dtn)· · ·)dt1 (2.15)for all t ∈ J, where a(t) ≥ 1 is a continuous function in J and 0 < p ≤ 1 is aconstant. Thenu(t) ≤ a(t) +∫ tαf1(s)[a(s) + v2(s)] ds, (2.16)On some Gronwall type inequalities 71whereun(t) =∫ tα(f1(s) + · · ·+ fn(s))a(s) exp(∫ ts(f1(τ) + · · · + fn(τ)) dτ)dsanduk(t) =∫ tα[( k∑j=1fj(s))a(s) + fk(s)vk+1(s)]exp(∫ tsk−1∑j=1fj(τ) dτ)dsfor all t ∈ J and k = n− 1, . . . , 2.Proof. Let Lk[up](t) be defined as in Theorem 2.4 and putv1(t) = L1[up](t), vk+1(t) = vk + Lk+1[up](t)for all t ∈ J and k = 1, · · · , n− 1. Then (2.15) impliesu(t) ≤ a(t) + v1(t) (2.17)and we successively findv′1(t) = (L1[up](t))′ = f1[up(t) + L2[up]]≤ f1[a(t) + v1(t) + L2[up]] = f1[a(t) + v2(t)],v′k(t) ≤ (f1 + · · ·+ fk−1)vk(t) + (f1 + · · ·+ fk)a + fkvk+1(t), (2.18)k = 2, · · · , n− 1,v′n(t) ≤ (f1 + · · ·+ fn)vn(t) + (f1 + · · · + fn)a(t).Since vk(α) = 0, k = 1, . . . , n, solving the system (2.18) ‘backward’ and applyingLemma 2.1, we arrive atv1(t) ≤∫ tαf1(s)[a(s) + v2(s)] ds, (2.19)wherevk(t) =∫ tα[( k∑j=1fj(s))a(s) + fk(s)vk+1(s)]exp(∫ tsk−1∑j=1fj(τ) dτ)ds,for all t ∈ J , k = n− 1, . . . , 2, andun(t) =∫ tα(f1(s) + · · ·+ fn(s))a(s) exp(∫ ts(f1(τ) + · · ·+ fn(τ)) dτ)ds.The inequalities (2.17) and (2.19) imply (2.16). This completes the proof. ✷Remark 2.2. In the case when a(t) ≥ 0, p = 1, the inequality given in (2.15)reduces to the inequality established earlier by Young in [18] (see also [1, Theorem11.7, p. 103]).72 Y. J.Cho, S. S.Dragomir and Y.-H.KimCorollary 2.2. Let u, f, g, h be nonnegative continuous functions in J = [α, β].Suppose thatu(t) ≤ u0 +∫ tα(f(s)up(s) + h(s)) ds +∫ tαf(s)(∫ sαg(τ)up(τ) dτ)dsfor all t ∈ J, where u0 +∫ tαh(s) ds ≥ 1 is a continuous function in J and 0 < p ≤ 1is a constant. Thenu(t) ≤ u0 +∫ tαh(s) ds+∫ tαf(s)[u0 +∫ sαh(τ) dτ]exp(∫ sα(f(τ) + g(τ) τ)ds.Proof. Indeed, the proof follows from Theorem 2.5 with f1 = f, f2 = g anda(t) = u0 +∫ tαh(s) ds. ✷Acknowledgement. Y. J. Cho and S. S. Dragomir greatly acknowledge the fi-nancial support from the Brain Pool Program (2002) of the Korean Federation ofScience and Technology Societies. The research was performed under the “Memo-randum of Understanding” between Victoria University and Gyeongsang NationalUniversity.References[1] D.Bainov, P. Simeonov, Integral Inequalities and Applications, Kluwer Aca-demic Publishers, Dordrecht, 1992.[2] E.F. 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On some Gronwall type inequalities with iterated integrals

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