Webb, J.

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Наукова електронна бібліотека періодичних видань НАН України (Vernadsky National Library of Ukraine)

An extension of Gronwall’s inequalityc© Jeffrey Webb1 IntroductionThe Gronwall inequality is a well-known tool in the study of differential equations,Volterra integral equations, and evolution equations [2]. It is often used to establisha priori bounds which are used in proving global existence, uniqueness, and stabilityresults. There are a number of versions of the result, a well-known version was firstproved by Bellman according to the book [3]. One classical version reads as follows.Theorem 1.1. Suppose that φ(t) ≤ c0(t) +∫ t0c1(s)φ(s) ds for a.e. t ∈ [0, T ], wherec0 is non-negative and non-decreasing, c1 ∈ L1+, and φ ∈ L∞+ . [Subscript + denotesfunctions that are ≥ 0 a.e.] Thenφ(t) ≤ c0(t) exp(∫ t0c1(s) ds)a.e. . (1.1)Another somewhat different inequality is the following, apparently first given byLiang Ou-Iang [5], and also given in [2],(both authors consider continuous functions).Theorem 1.2. Suppose that φ2(t) ≤ c20 + 2∫ t0h(s)φ(s) ds for a.e. t ∈ [0, T ], where c0is a constant, h ∈ L1+, and φ ∈ L∞+ . Then, for a.e. t ∈ [0, T ],φ(t) ≤ c0 +∫ t0h(s) ds.If h ∈ L2, by using the inequality hφ ≤ h2/2 + φ2/2 this case could be reduced tothe classical case but would give a different type of conclusion.We have discovered a ‘nonlinear’ version of Gronwall’s inequality that contains bothof these results and many other inequalities of Gronwall type given in the literature,see for example [3].To describe the type of results we prove consider the following simple case. Supposethat a non-negative L∞ function φ satisfies a.e. the inequalityφ2(t) ≤ c20 +∫ t02c1(s)φ(s) + 2c2(s)φ2(s) ds,where ci are non-negative L1 functions. Then we prove thatφ(t) ≤ [c0 + ∫ t0 c1(s) ds] exp(∫ t0 c2) a.e..This result contains both of the results just cited and also a special case of a resultof Dafermos [1], see below. We are interested in more general inequalities and inobtaining explicit L∞ bounds for φ which depend only on c0 and integrals of the givenfunctions ci. Also we consider the case when the inequalities hold a.e. since this is acommonly occurring situation.2 Extended Gronwall inequalityWe give a general result which contains Theorems 1.1 and 1.2 and many other resultsas well as giving new ones. To make the formula a little easier we insert some constants.Theorem 2.1. Let k ≥ 1 be an integer. Suppose thatφk(t) ≤ ck0(t) +∫ t0kc1(s)φ(s) + kc2(s)φ2(s) + · · ·+ kck(s)φk(s) ds, for a.e. t ∈ [0, T ],(2.1)where c0 ≥ 0 is non-decreasing, φ ∈ L∞+ , and ci ∈ L1+ for i ≥ 1. Then,φ(t) ≤ Wk(c0, c1, . . . , ck−1) exp(∫ t0ck)a.e. (2.2)where the function Wk is defined recursively byW1(c0) =c0Wk+1(c0, c1, . . . , ck) =Wk([ck0 + k∫ t0c1]1/k, c2, . . . , ck), k ≥ 1.We use in the proof the classical Gronwall inequality quoted above. For complete-ness, and because we use similar arguments without further explanation, we give aproof here.Proof of Theorem 1.1. Fix τ in [0, T ] and let w(t) := c0(τ) +∫ t0c1(s)φ(s) ds. Then wis absolutely continuous (AC), φ(t) ≤ w(t) for a.e. t ∈ [0, τ ] andw′(t) = c1(t)φ(t) ≤ c1(t)w(t) for a.e. t ∈ [0, τ ].Write C1(t) :=∫ t0c1(s) ds. The function w exp(−C1) is AC because C1 is AC, exp isLipschitz on bounded intervals, so the composition exp(−C1) is AC, and a product ofAC functions is also AC. Therefore, for a.e. t ∈ [0, τ ],[w(t) exp(−C1(t))]′ ≤ 0, (2.3)and hence, for all t ∈ [0, τ ],w(t) exp(−C1(t)) ≤ w(0).Since w(0) = c0(τ) this gives w(t) ≤ c0(τ) exp(C1(t)) for all t ∈ [0, τ ]. In particular,w(τ) ≤ c0(τ) exp(C1(τ)) and, as τ is arbitrary, (1.1) holds.Proof of Theorem 2.1. The proof is by induction. As before we replace c0(t) by c0(τ)and reduce the proof to the case where c0 is a constant which we suppose to be positive(otherwise add ε > 0). The case k = 1 is the classical result proved above. Supposethe inequality holds for an integer k and consider the case for k + 1. Thus supposethat for a.e. t ∈ [0, T ],φk+1(t) ≤ ck+10 +∫ t0(k+1)c1(s)φ(s)+(k+1)c2(s)φ2(s)+ · · ·+(k+1)ck+1(s)φk+1(s) ds.Write, for t ∈ [0, T ],uk+1(t) := ck+10 +∫ t0(k+1)c1(s)φ(s)+(k+1)c2(s)φ2(s)+ · · ·+(k+1)ck+1(s)φk+1(s) ds.Since uk+1 is AC and u ≥ c0 > 0, it follows that u is AC. Therefore u′ exists a.e. andfor a.e. t ∈ [0, T ],uku′ =c1φ+ c2φ2 + · · ·+ ck+1φk+1≤c1u+ c2u2 + · · ·+ ck+1uk+1.Henceuk−1u′ ≤ c1 + c2u+ · · ·+ ck+1uk, a.e. .So we have for all t ∈ [0, T ],uk(t)− uk(0) ≤∫ t0kc1 + kc2u+ · · ·+ kck+1uk,that is,uk(t) ≤ [ck0 + ∫ t0kc1]+∫ t0kc2u+ · · ·+ kck+1uk.This shows, by the result for the case k, that for a.e. t ∈ [0, T ],u(t) ≤ Wk([ck0 +∫ t0kc1]1/k, c2, . . . , ck) exp(∫ t0ck+1).As u is continuous this holds for all t ∈ [0, T ] and therefore holds almost everywherefor φ. This completes the inductive step.The special case k = 2 includes the classical case and the result quoted in Theo-rem 1.2 since we may take c2 = 0. It includes a special case of a result of Dafermos[1] who shows that ifφ2(t) ≤M2φ2(0) +∫ t02Ng(s)φ(s) + 2γφ2(s) dswhere N,M, γ are non-negative constants and φ ∈ L∞+ , g ∈ L1+ thenφ(t) ≤ (Mφ(0) +N ∫ t0g(s) ds)exp(γt).Pachpatte [4] has an extension of this special case of Dafermos’s result for continuousfunctions when the integral is replaced by a suitable multiple integral, which hasapplications to certain higher order ODE’s. In Dafermos, [1], an extra term is includedwith γ replaced by γ + βt in the first inequality. We shall use the same idea to giveanother result below which will extend Dafermos’s result and allow some t dependenceon the functions ci. We first consider some special cases of the inequality proved inTheorem 2.1.Example 2.2. If u ≤ a0 +∫ t0a1(s)um/k(s) ds where m < k are positive integers, anda1 ∈ L1+, u ∈ L∞+ , thenu ≤ [a1−m/k0 + ∫ t0(1−m/k)a1(s) ds] 11−m/k .As usual we may suppose that a0 > 0. Write v = u1/k so that the inequality readsvk ≤ a0 +∫ t0a1(s)vm(s) ds.This is now a special case of Theorem 2.1 so we can writev ≤ Wk(a1/k0 , 0, . . . , a1/k, 0, . . . , 0),where a1/k occurs in the (m+ 1)-st place. Thus,v ≤Wk−m+1(a1/k0 , a1/k, 0, . . . , 0)=Wk−m([ak−mk0 +∫ t0(k −m)a1(s)/k ds] 1k−m , 0, . . . , 0).Since Wr(c0, 0, . . . , 0) = c0 for every r, this establishes the result.By an approximation argument this result shows the following: ifu(t) ≤ a0 +∫ t0a1(s)uα(s) ds, for a.e. t ∈ [0, T ],where 0 < α < 1, thenu(t) ≤ [a1−α0 + ∫ t0(1− α)a1(s) ds] 11−α , a.e..This last inequality is well-known and is usually shown via a comparison theorem: Ifh is continuous and u(t) ≤ u0+∫ t0h(s, u(s)) ds then u(t) ≤ v(t) where v is the solutionof the differential equation v′ = h(t, v), v(0) = u0; [“the faster runner goes further”].Example 2.3. Given the inequalityu ≤ a0 +∫ t0a1um/k + a2uj/kwe can immediately write down an inequality for u, namelyu1/k ≤ Wk(a1/k0 , 0, . . . , 0, a1/k, 0, . . . , 0, a2/k, 0, . . . , 0),where a1/k (respectively a2/k) occurs in the (m+ 1) (resp. (j + 1)) place. Hence, asabove, we obtainu1/k ≤ Wk−m−(j−m)({[ak−mk0 +∫ t0(k −m)a1(s)/k ds] k−jk−m + k−jk∫ t0a2(s) ds}, 0, . . . , 0),that is,u ≤ {[a1−mk0 + ∫ t0(1−m/k)a1(s) ds] 1−j/k1−m/k + (1− j/k)∫ t0a2(s) ds}1/(1−j/k).By an approximation argument this gives the following result. The inequalityu ≤ a0 +∫ t0a1uα + a2uβ,where 0 < α < β < 1, impliesu ≤ {[a1−α0 + ∫ t0(1− α)a1(s) ds] 1−β1−α + (1− β)∫ t0a2(s) ds}1/(1−β).It is not clear whether the comparison result applies to this situation, becausepossibly noncontinuous functions are involved. Also one encounters the problem ofsolving the differential equation v′ = a1vα + a2vβ where ai need not be constants.For example, in the simple case when we haveu(t) ≤ a0 +∫ t0a1(s)u1/3 + a2(s)u2/3 dswe obtain the estimate u ≤ w := W3(a1/30 , a1/3, a2/3)3 which can be computed directlyto bew(t) =([a2/30 + 2A1(t)/3]1/2+ A2(t)/3)3,where Ai(t) =∫ t0ai(s) ds. To apply the comparison result we have to take a0 > 0or else find a maximal (or minimal) solution because of lack of uniqueness. A smallcalculation shows that w′ ≥ a1w1/3+a2w2/3 so that w(t) ≥ v(t) where v is the solutionof v′ = a1v1/3 + a2v2/3, v(0) = a0. In other words the estimate we obtain is not asgood as could (in theory) be obtained from the comparison result. However, even witha1 and a2 constant, the solution v in this case is found only in implicit form, so theestimate obtained from the comparison result is of less practical use than our explicitestimate.Example 2.4. The inequalityu(t) ≤ a0 +∫ t0a1(s)um/k(s) + a2(s)u(s) dshas been studied by Perov when m/k is replaced by α, 0 ≤ α < 1. [In fact he has aresult also for α > 1.] We can immediately write down an inequality for u, namelyu1/k ≤ Wk(a1/k0 , 0, . . . , 0, a1/k, 0, . . . , 0) exp(∫ t0a2(s)/k ds).where a1/k is in the (m+ 1)-place. As in example 2.2 this givesu1/k ≤ [a1−m/k0 + ∫ t0(1−m/k)a1(s) ds] 1k−m exp(∫ t0a2(s)/k ds).Hence we obtainu ≤ [a1−m/k0 + ∫ t0(1−m/k)a1(s) ds] 11−m/k exp(∫ t0a2(s) ds).By a simple approximation argument we can then obtain: the inequalityu(t) ≤ a0 +∫ t0a1(s)uα(s) + a2(s)u(s) dsimplies thatu ≤ [a1−α0 + ∫ t0(1− α)a1(s) ds] 11−α exp(∫ t0a2(s) ds).This is a minor change from Perov’s result.Remark 2.5. We could, of course, write down other consequences of the theorem butsuch inequalities await application. Another ‘nonlinear’ Gronwall inequality is thatattributed to Bihari but apparently proved earlier by Lasalle [3].If u(t) ≤ a+ b ∫ t0k(s)g(u(s)) ds where g is nondecreasing, thenu(t) ≤ G−1[G(a) + b∫ t0k(s) ds]where G(s) =∫ s0dξg(ξ). This can, in theory, be applied to some of the situations weinclude, but the problems of calculating G and G−1 do not appeal in general. Alsosuch inequalities do not fit our aim of giving explicit bounds in terms of the givendata.We now give a further extension which allows for noninteger powers.Theorem 2.6. Let m ≥ 1 be an integer and p ≥ m be a real number. Suppose thatφp(t) ≤ cp0(t) +∫ t0pc1(s)φp−m(s) + pc2(s)φp−m+1(s) + · · ·+ pcm+1(s)φp(s) ds, (2.4)for a.e. t ∈ [0, T ], where c0 ≥ 0 is non-decreasing, φ ∈ L∞+ , and ci ∈ L1+, i ≥ 1. Then,φ(t) ≤ Wm+1(c0, c1, c2, . . . , cm) exp(∫ t0cm+1), a.e. . (2.5)Proof. As usual we may and do suppose that c0 is a positive constant and we omit‘almost everywhere’ in the following. Writeup(t) = cp0 +∫ t0pc1(s)φp−m(s) + pc2(s)φp−m+1(s) + · · ·+ pcm+1(s)φp(s) ds.Then we obtainup−1u′ ≤ c1up−m + · · ·+ cm+1upso thatum−1u′ ≤ c1 + c2u+ · · ·+ cm+1um.Hence um ≤ cm0 +∫ t0mc1 +∫ t0mc2u+ · · ·+mcm+1um, which givesu(t) ≤ Wm([cm0 +∫ t0mc1]1/m, c2, . . . , cm) exp(∫ t0cm+1),and yields the conclusion.Remark 2.7. When p−m = j/k it is possible to reduce the inequality in the hypoth-esis of Theorem 2.6 to a special case of the one in Theorem 2.1. For example, giventhe inequalityu5/2(t) ≤ c5/20 +∫ t052c1(s)u1/2(s) + 52c2(s)u3/2(s) + 52c3(s)u5/2(s) dswe can say at once from Theorem 2.6 thatu ≤ W3(c0, c1, c2) exp(∫ t0c3(s) ds).However writing v = u1/2 the given inequality may be writtenv5(t) ≤ (c1/20 )5 +∫ t05(c1/2)v + 5(c2/2)v3 + 5(c3/2)v5.Theorem 2.1 then gives the conclusionv ≤ W5(c1/20 , c1/2, 0, c2/2, 0) exp(∫ t0c3(s)/2 ds).A small calculation checks thatW5(c1/20 , c1/2, 0, c2/2, 0) = W3(c0, c1, c2)1/2so the apparent two conclusions are but one. However, using Theorem 2.6 is simplerhere.3 A further extensionWe now give a result of the same type where the functions ci are allowed to dependon t. The extra idea used is that employed by Dafermos [1] in proving the following:φ2(t) ≤M2φ2(0) +∫ t02Ng(s)φ(s) + (2γ + 4βt)φ2(s) dswhere N,M, γ are non-negative constants and φ ∈ L∞+ , g ∈ L1+ thenφ(t) ≤ (Mφ(0) +N ∫ t0g(s) ds)exp(αt+ βt2),where α = γ + β/γ.We will prove the following result.Theorem 3.1. Let k ≥ 1 be an integer. Suppose that for i = 1, 2, . . . , k, and fors, t ∈ [0, T ], we have 0 ≤ gi(s, t) ≤ h(s) where h is an L1 function, and that ∂gi/∂texists and satisfies 0 ≤ ∂gi(s, t)/∂t ≤ Agi(s, t) for some non-negative constant A.Suppose thatφk(t) ≤ ck0(t) + k∫ t0g1(s, t)φ(s) + · · ·+ gk(s, t)φk(s) ds, for a.e. t ∈ [0, T ], (3.1)where c0 ≥ 0 is non-decreasing, φ ∈ L∞+ . Then,φ(t) ≤ Wk(c0(t), g1(t, t), . . . , gk−1(t, t)) exp(∫ t0[gk(s, s) + A/k] ds). (3.2)Proof. Letuk(t) = ck0 +∫ t0kg1(s, t)φ(s) + kg2(s, t)φ2(s) + · · ·+ kgk(s, t)φk(s) ds,where we suppose that c0 is a positive constant. The hypotheses made allow us todifferentiate under the integral sign to obtainuk−1u′ =k∑i=1gi(t, t)φi(t) +∫ t0k∑i=1∂gi(s, t)∂tφi(s) ds≤k∑i=1gi(t, t)ui(t) +∫ t0k∑i=1Agi(s, t)φi(s) ds≤k−1∑i=1gi(t, t)ui(t) + [A/k + gk(t, t)]uk(t).This yields the inequalityuk(t) ≤ uk(0) +∫ t0k−1∑i=1kgi(s, s)ui(s) + [A+ kgk(s, s)]uk(s) ds.Theorem 2.1 now applies to give the conclusionφ(t) ≤ u(t) ≤ Wk(c0, g1(t, t), . . . , gk−1(t, t)) exp(∫ t0[A/k + gk(s, s)] ds).This includes Dafermos’s result for we may take g1(s, t) = Ng(s) and g2(s, t) =γ + 2βt. Then ∂g1/∂t = 0 and ∂g2(s, t)/∂t = 2β ≤ A(γ + 2βt) for A = 2β/γ.References[1] Dafermos C.M., The second law of thermodynamics and stability, Arch. RationalMech. Anal., 70 (1979), 167–179.[2] Haraux, A., Nonlinear evolution equations-global behaviour of solutions, LectureNotes in Mathematics 841, Springer-Verlag 1981.[3] Mitrinovic´, D. S., Pec˘aric´, J. E., and Fink, A.M., Inequalities involving functionsand their integrals and derivatives, Kluwer Academic Publishers, Dordrecht /Boston / London, 1991.[4] Pachpatte, B.G., On some fundamental integral inequalities arising in the theoryof differential equations, Chinese J. Math. (Taiwan, R.O.C.), 22 (1994), 261-273.[5] Liang Ou-Iang, The boundedness of solutions of linear differential equations y′′+A(t)y = 0, Shuxue Jinzhan 3 (1957), 409–415.Department of Mathematics, University of Glasgow,Glasgow, G12 8QW, Scotland, U. K.Phone: +44 141 330 5181FAX: +44 141 330 4111E-mail: J.Webb@maths.gla.ac.uk

An extension of Gronwall's inequality

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An extension of Gronwall's inequality

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Наукова електронна бібліотека періодичних видань НАН України (Vernadsky National Library of Ukraine)