Electronic Journal of Differential Equations, Vol. 2021 (2021), No. 80, pp. 1–22. ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu A FRACTIONAL GRONWALL INEQUALITY AND THE ASYMPTOTIC BEHAVIOUR OF GLOBAL SOLUTIONS OF CAPUTO FRACTIONAL PROBLEMS JEFFREY R. L. WEBB Abstract. We study the asymptotic behaviour of global solutions of some nonlinear integral equations related to some Caputo fractional initial value problems. We consider problems of fractional order between 0 and 1 and of order between 1 and 2, each in two cases: when the nonlinearity depends only on the function, and when the nonlinearity also depends on fractional derivatives of lower order. Our main tool is a new Gronwall inequality for integrals with singular kernels, which we prove here, and a related boundedness property of a fractional integral of an L1[0,∞) function. 1. Introduction We investigate the asymptotic behaviour of global solutions of problems of the form u(t) = u0 + Iαg(t, u(t)) := u0 + 1 Γ(α) ∫ t 0 (t− s)α−1g(s, u(s)) ds (1.1) where 0 < α < 1, Iαg is the Riemann-Liouville (R-L) fractional integral of g, and also in the case when g depends on lower order Caputo fractional derivatives, g(t, u(t), Dγ ∗u(t)); precise definitions are given later in the paper. We also study u(t) = u0 +B1t β + 1 Γ(α+ β) ∫ t 0 (t− s)α+β−1(Gu)(t)) ds, (1.2) where (Gu)(t) = g(t, u(t), Dγ ∗u(t)) for 1 < α+ β < 2 and 0 ≤ γ ≤ β. The function g will be of the form φ(t)f(t, u,Dγ ∗u(t)) where f is continuous and φ ∈ L1[0,∞) or dominated by such a term. These problems are motivated by the study of initial value problems for Caputo fractional differential equations. Under certain conditions solution of (1.1) are also solutions of Dα ∗ u(t) = g(t, u(t)), for a.e. t > 0, u(0) = u0. (1.3) Similarly, solutions of Dα ∗ (Dβ ∗u)(t) = g(t, u(t), Dγ ∗u(t)), u(0) = u0, Dβ ∗u(0) = B1Γ(β + 1), (1.4) are closely related to solutions of (1.2). 2010 Mathematics Subject Classification. 34A08, 34A12, 26A33, 26D10. Key words and phrases. Fractional derivatives; asymptotic behaviour; Gronwall inequality; weakly singular kernel. c©2021 Texas State University. Submitted July 23, 2021. Published September 20, 2021. 1 2 J. R. L. WEBB EJDE-2021/80 For problem (1.1) the kernel (t− s)α−1 has a singularity when t = s, whereas in problem (1.2) the term (t − s)α+β−1 is not singular. For the singular case, (1.1), some special inequality is needed, we will use a new Gronwall inequality suitable for this purpose. For the case (1.2) when f does not depend on Dγ ∗u(t), that is γ = 0, we can use the classical Gronwall inequality, but when there is explicit dependence on Dγ ∗u(t) we again need the new Gronwall inequality to handle some fractional derivatives. This paper was motivated by some previous papers, mainly those by Medveď and Posṕı̌sil [14] and more recently by Kassim and Tatar [11]. 2. Known Gronwall inequalities We state one classical version of the Gronwall inequality that we shall use, which was proved by Bellman [1]. This, and many other versions, may be found in several books, for example [5, 15]. Theorem 2.1. Suppose that u ∈ C+[0, T ] satisfies u(t) ≤ a(t) + ∫ t 0 φ(s)u(s) ds for t ∈ [0, T ], where a is non-negative, continuous and non-decreasing, and φ ∈ L1 +[0, T ]. Then u(t) ≤ a(t) exp (∫ t 0 φ(s) ds ) for t ∈ [0, T ]. (2.1) If also a is uniformly bounded for all t > 0 and φ ∈ L1[0,∞) then u is uniformly bounded for all t > 0. Here, L1 +[0, T ] denotes the Lebesgue integrable functions u defined on a finite interval [0, T ] with u(t) ≥ 0 a.e., similarly C+[0, T ] denotes the non-negative con- tinuous functions. For Gronwall type inequalities with singular kernels the pioneering work was done by Henry [8]. A version of one result is as follows. Theorem 2.2. Let a, g be non-decreasing and g(t) ≤ C for all t ∈ [0, T ] and let 0 < β < 1. If x ∈ L∞+ [0, T ] satisfies the inequality x(t) ≤ a(t) + g(t) ∫ t 0 (t− s)β−1x(s)ds, t ∈ [0, T ], then x(t) ≤ a(t)Eβ [g(t)Γ(β)(tβ)] where Eβ is the Mittag-Leffler function. The Mittag-Leffler function is an entire function of z ∈ C defined by a power series Eβ(z) = ∑∞ k=0 zk Γ(βk+1) . The special case E1(z) is the exponential function. Further information can be found in many places, for example in the books [3, 17]. The original result in [8] was proved by an iteration argument. Another proof was given by Dixon and McKee [4, Theorem 3.1] when g is a constant. It is simple to replace a constant by a non-decreasing function as in Corollary 5.4 below. Such a generalization of Henry’s work was done by Ye, Gao and Ding [22] but using a modification of Henry’s original proof. A different proof is given by Tisdell [19]. In a recent paper [20] we proved a new result for such an inequality by a different method and we obtained bounds in terms of the much better known exponential function. It is this result we will extend to enable us to handle our problems. We do not know of a result similar to this new one involving Mittag-Leffler functions. EJDE-2021/ 80 ASYMPTOTIC BEHAVIOUR 3 There are other results with exponential bounds. For example the following integral inequality is a special case of Medveď [12, Theorem 2]. If f ≥ 0 is continuous and u ≥ 0 is continuous and satisfies u(t) ≤ a+ ∫ t 0 (t− s)β−1f(s)u(s) ds, t ∈ [0, T ], then under certain hypotheses on f , u(t) is bounded on [0, T ] by a exp(H(t) + t) for an explicit function H. (He also considers Bihari type inequalities.) The proofs involve applications of various inequalities such as Hölder’s inequality. Further inequalities using the idea from [12] have been obtained by Zhu [23, 24]. These types of inequality are quite different from the one we will prove. 3. Definitions of fractional integrals and derivatives All functions we consider will be real-valued and defined on an interval [0, T ] or on [0,∞). The space of continuous functions defined on [0, T ] is denoted C[0, T ] and is endowed with the norm ‖u‖∞ := supt∈[0,T ] |u(t)|. AC[0, T ] denotes the space of Absolutely Continuous functions on [0, T ]. L1[0, T ] denotes the space of Lebesgue integrable functions u, that is ∫ T 0 |u(s)| ds < ∞, and L1[0,∞) are those functions such that ∫∞ 0 |u(s)| ds <∞. A subscript + will denote the non-negative functions in the corresponding space. For η > −1 we define the space denoted Cη = Cη[0, T ] by Cη[0, T ] := {u ∈ C(0, T ] such that lim t→0+ t−ηu(t) exists}, then u ∈ Cη if and only if u(t) = tηv(t) for some function v ∈ C[0, T ] and we define ‖u‖η := ‖v‖∞. The spaces of functions with singularity at t = 0 are C−η where η > 0. The space C0 coincides with the space C0 = C[0, T ]. Clearly, for η > 0 the space Cη is a subspace of C[0, T ]. The Gamma and Beta functions occur frequently so we recall them here. The Gamma function is, for p > 0, given by Γ(p) := ∫ ∞ 0 sp−1 exp(−s) ds (3.1) which is an improper Riemann integral but is well defined as a Lebesgue integral, and is an extension of the factorial function: Γ(n + 1) = n! for n ∈ N. The Beta function is defined by B(p, q) := ∫ 1 0 (1− s)p−1sq−1 ds (3.2) which is a well defined Lebesgue integral for p > 0, q > 0 and it is well known, and proved in many calculus texts, that B(p, q) = Γ(p)Γ(q) Γ(p+ q) . The Riemann-Liouville (R-L) fractional integral Iαu is ‘defined’ informally by: Iαu(t) := 1 Γ(α) ∫ t 0 (t− s)α−1u(s) ds, provided the integral exists. This does not specify to which space of functions u belongs, and leaves open whether the integral is to exist for all t, or for all nonzero t, or for almost every (a.e.) t. The most interesting case is when 0 < α < 1 then the integrand has a singularity. A precise definition for integrable functions is the following. 4 J. R. L. WEBB EJDE-2021/80 Definition 3.1. The Riemann-Liouville (R-L) fractional integral of order α > 0 of a function u ∈ L1[0, T ] is defined for a.e. t by Iαu(t) := 1 Γ(α) ∫ t 0 (t− s)α−1u(s) ds. The integral Iαu is the convolution of the L1 functions h, u, where h(t) = tα−1/Γ(α), so, by well known results on convolutions, Iαu is defined as an L1 function, in particular Iαu(t) is finite for a.e. t. When α = 1 this is the usual inte- gration operator which we denote I. We set I0u = u. If α ∈ (0, 1) and f ∈ Lp[0, T ] with p > 1/α then Iαf belongs to a Hölder space hence is defined at all points and is continuous, a result of Hardy and Littlewood [7, Theorem 12]. The Riemann-Liouville (R-L) fractional derivative of order α ∈ (0, 1) is defined as follows. We write D for the usual derivative operator, that is, Du = u′. Definition 3.2. For α ∈ (0, 1) and u ∈ L1[0, T ] the R-L fractional derivative Dαu is defined when I1−αu ∈ AC[0, T ] by Dαu(t) := D I1−αu(t), a.e. t ∈ [0, T ]. The condition I1−αu ∈ AC is often not stated in published work, but it is necessary if we want to relate solutions of R-L fractional differential equations with solutions of a Volterra integral equation. It is not enough to assume that I1−αu is differentiable for a.e. t. This has been noted in the monograph [17], see [17, Definition 2.4] and the related comments in the ‘Notes to §2.6’. The Caputo fractional derivative is frequently defined with the derivative and fractional integral taken in the reverse order to that of the R-L derivative. Definition 3.3. For α ∈ (0, 1) and u ∈ AC[0, T ] the Caputo fractional derivative Dα Cu is defined for a.e. t by Dα Cu(t) := I1−αDu(t). For u ∈ AC, we have Du ∈ L1 and so Dα Cu = I1−α(Du) is defined as an L1 function. However, this definition has a severe disadvantage. It is often claimed that for 0 < α < 1 and f continuous Dα Cu(t) = f(t), u(0) = u0 is equivalent to u(t) = u0 + Iαf. However, ‘solution’ means different things on each side of the equation. Usually solution of the integral equation is a function in C[0, T ], and it is never shown that such a function is in AC[0, T ] for the very good reason that it is false in general, so the derivative Dα Cu(t) is not shown to exist. Again this was proved by Hardy and Littlewood [7]. For 0 < α < 1, Iα maps C−η[0, T ] into Cα−η[0, T ], Iα maps C[0, T ] into C[0, T ] and maps AC[0, T ] into AC[0, T ] but does not map C[0, T ] into AC[0, T ] and does not map C1[0, T ] into C1[0, T ]. Details can be found the paper [21] and its addendum. To get an equivalence it is necessary to use the definition of Caputo differential operator as is used in Diethelm’s book [3]. Definition 3.4. The Caputo differential operator of order α ∈ (0, 1) is defined by Dα ∗ u := Dα(u − u(0)) = D(I1−α(u − u(0))) whenever this R-L derivative exists, that is when u(0) exists and I1−αu ∈ AC. EJDE-2021/ 80 ASYMPTOTIC BEHAVIOUR 5 The two definitions of Dα Cu and Dα ∗ u coincide when u ∈ AC. For 1 < β < 2, the R-L derivative is Dβu = D2I2−βu if D(I2−βu) ∈ AC, and the Caputo derivative is Dβ ∗u(t) := D2 ( I2−β(u− u(0)− tu′(0)) ) provided that u(0), u′(0) exist and D(I2−βu) ∈ AC. Higher order derivatives are defined similarly, for details see for example the books [3, 17]. The main advantages of the Caputo derivative over the R-L derivative are that Dα ∗ (c) = 0 (any α > 0) when c is a constant function, whereas the R-L derivative of a constant has a singularity at zero, and initial value problems for the Caputo derivative are well posed when initial values are prescribed on the function and its ordinary derivatives, fractional integrals and derivatives should be prescribed in the R-L case. An equivalence that can be proved is as follows, in fact a somewhat more general result holds for f ∈ C−η for 0 ≤ η < α, see [21, Theorem 5.1]. Theorem 3.5. Let f be continuous on [0, T ] × R, let 0 < α < 1. If u ∈ C[0, T ] satisfies u(t) = u0 + Iαf then I1−α(u − u0) ∈ AC, Dα ∗ u exists a.e. and satisfies Dα ∗ u(t) = f(t) a.e., u(0) = u0. Conversely, if u ∈ C[0, T ], I1−α(u− u0) ∈ AC and u satisfies Dα ∗ u(t) = f(t) a.e., u(0) = u0, then u satisfies u(t) = u0 + Iαf . This means that Caputo fractional equations can be studied in the space C[0, T ], but the corresponding R-L fractional equations for α ∈ (0, 1) should be studied in a space that allows functions to be singular at 0 such as Cα−1. 4. Some properties of fractional integrals It is well known that fractional integral operators satisfy a semigroup property as follows. Lemma 4.1. Let α, β > 0 and u ∈ L1[0, T ]. Then IαIβ(u)(t) = Iα+β(u)(t) for each t for which Iα+β |u|(t) exists (finite), that is a.e. t ∈ [0, T ]. If u is continuous, or if u ∈ C−η and α+ β ≥ η, this holds for all t ∈ [0, T ]. If u ∈ L1 and α+ β ≥ 1 equality again holds for all t ∈ [0, T ]. The proof uses Fubini’s theorem, and all but the part concerning C−η is usually sketched in the texts, for example [3, Theorem 2.2], [17, Eq.(2.21)]. A detailed proof of Lemma 4.1 is given in [21]. We first prove some properties of fractional integrals that we will use, the first one is probably known. Proposition 4.2. Let f ∈ L1[0, T ] and α ≥ β > 0. Then Γ(α)(Iα|f |)(t) tα ≤ Γ(β)(Iβ |f |)(t)) tβ for a.e. t ∈ (0, T ]. (4.1) Proof. Clearly it suffices to give the proof for f ∈ L1 +[0, T ] and omit absolute value signs. By standard properties, each fractional integral exists as an L1 function, hence exists for a.e. t. For a.e. t we have Γ(α)Iαf(t) = ∫ t 0 (t− s)α−1f(s) ds = ∫ t 0 tα−1(1− s/t)α−1f(s) ds 6 J. R. L. WEBB EJDE-2021/80 ≤ ∫ t 0 tα−1(1− s/t)β−1f(s) ds = ∫ t 0 tα−β(t− s)β−1f(s) ds = tα−βΓ(β)Iβf(t). � Corollary 4.3. Let f ∈ L1[0, T ], γ > 0, and β > 0. Then Iγ+β |f |(t)) ≤ tγ Γ(β) Γ(γ + β) Iβ |f |(t), for a.e. t ∈ [0, T ]. (4.2) Proof. Replace α by γ + β in Proposition 4.2. � Corollary 4.4. Let f ∈ L1[0, T ] and β > 0. Then I1+β |f |(t) ≤ t β Iβ |f |(t), for a.e. t ∈ [0, T ]. (4.3) This corollary follows from Corollary 4.3 with γ = 1 noting that Γ(β) Γ(1+β) = 1 β . If the only information is that f ∈ L1[0, T ] then Corollary 4.4 is optimal as the following example shows. Example 4.5. Let f(t) := tp for p > −1, t ∈ [0, T ]. It is known, and easy to show, that Iβf(t) = Γ(1 + p) Γ(1 + β + p) tβ+p. Then, by the semigroup property of fractional integration Lemma 4.1, we obtain, for every t ∈ [0, T ], I1+βf(t) = I(Iβf)(t) = Γ(1 + p) Γ(2 + β + p) t1+β+p. Hence, I1+βf(t) = 1 1 + β + p t Iβf(t) for t > 0 and the constant approaches 1/β as p→ −1. We now consider some results given by Kassim-Tatar [11]. Proposition 4.6. Let ρ > 1 and let F be a function such that F ′ exists a.e. and F ′ ∈ L1([0,∞), that is, ∫∞ 0 |F ′(t)| dt exists and is finite. Then we have (a) lim t→∞ IρF ′(t) tρ−1 = 1 Γ(ρ) ∫ ∞ 0 F ′(t)dt. (4.4) (b) If, in addition, F ∈ AC[0, T ] for every T > 0 then lim t→∞ IρF ′(t) tρ−1 = 1 Γ(ρ) lim t→∞ (F (t)− F (0)). (4.5) Moreover in this case lim t→∞ Iρ−1F (t) tρ−1 = 1 Γ(ρ) lim t→∞ F (t). (4.6) EJDE-2021/ 80 ASYMPTOTIC BEHAVIOUR 7 Proof. (a) Let χJ denote the characteristic function of an interval J . Then we have IρF ′(t) tρ−1 = 1 Γ(ρ) ∫ t 0 (1− s/t)ρ−1F ′(s) ds = 1 Γ(ρ) ∫ ∞ 0 χ[0,t](s)(1− s/t)ρ−1F ′(s) ds For each t, since ρ > 1, s 7→ |χ[0,t](s)(1 − s/t)ρ−1F ′(s)| is dominated by the inte- grable function |F ′(s)|, so the limit as t→∞ exists by the dominated convergence theorem, and taking the limit proves (4.4). (b) As F ∈ AC[0, T ] for every T > 0, we have F (t)−F (0) = ∫ t 0 F ′(s)ds for every t > 0. Since the right hand side in this equation has a limit as t→∞, the limit of the left hand side exists and equals that limit, so using (a), we see that (4.5) holds. For the last part, we note that IρF ′(t) = Iρ−1IF ′(t), for all t, by the semigroup property of fractional integral, as given in Lemma 4.1. Since F ∈ AC[0, T ] we have IF ′(t) = F (t) − F (0) and, by a simple calculation, Iα(F (0))(t) = tα Γ(1+α)F (0) for any α > 0. Therefore, applying (4.5) we obtain 1 Γ(ρ) lim t→∞ (F (t)− F (0)) = lim t→∞ IρF ′(t) tρ−1 = lim t→∞ (Iρ−1F (t) tρ−1 − 1 Γ(ρ) F (0) ) , that is 1 Γ(ρ) lim t→∞ F (t) = lim t→∞ Iρ−1F (t) tρ−1 . � Remark 4.7. The above results hold if we have f ∈ L1[0,∞) and take F (t) :=∫ t 0 f(s)ds for then F ∈ AC[0, T ] and F ′ = f a.e.. However, F ′ ∈ L1 does not imply F ∈ AC as shown by the well-known Lebesgue’s singular function F (also known as the Cantor-Vitali function, or Devil’s staircase). Remark 4.8. Result (4.4) is stated as [11, Lemma 2.10 ] with ρ = 1 + α for α > 0, for the case f ∈ L1 as in above Remark 4.7, and was previously proved in [10, Lemma 7] by a more detailed version of the proof we gave. The result (4.6) is essentially the same result as claimed in [11, Lemma 2.11] and proved in [10, Lemma 18] which is stated when f(t) = Dα Cx(t) for a function x ∈ AC, but those lemmas statements and proof have omitted the necessary L1[0,∞) hypotheses. Moreover, our different proof is shorter. In part (a) it is necessary that F ′ ∈ L1[0,∞) otherwise both sides can be infinite, but the result does not have a weaker meaning in that case, it can happen that I1+αF ′(t)/tα∫ t 0 F ′(s) ds/Γ(1 + α) 6→ 1, as t→∞. Example 4.9. Let 0 < α < 1 and let F (t) = tγ/γ with γ > 0 so that F ∈ AC and F ′(t) = tγ−1 for t > 0 and F ′ ∈ L1[0, T ] for every T > 0 but F ′ /∈ L1[0,∞). Then the numerator is I1+αF ′(t) tα = 1 tαΓ(1 + α) ∫ t 0 (t− s)αsγ−1 ds = 1 tαΓ(1 + α) tα+γB(1 + α, γ) 8 J. R. L. WEBB EJDE-2021/80 = tγ Γ(γ) Γ(1 + α+ γ) . The denominator is 1 Γ(1 + α) ∫ t 0 sγ−1 ds = 1 Γ(1 + α) tγ γ . The ratio numerator denominator = Γ(γ) Γ(1 + α+ γ) γΓ(1 + α) = Γ(1 + γ)Γ(1 + α) Γ(1 + α+ γ) , and this is not equal to 1, except in the excluded cases γ = 0 or α = 0. 5. New Gronwall type of inequality To discuss the asymptotic behaviour of solutions of some fractional integral equa- tions of order less than 1, or higher order with nonlinearities depending on fractional derivatives of order less than 1, it is useful to have an inequality of Gronwall type appropriate for discussing fractional integrals with singular kernels. We will prove the following inequality which has similar conclusions to those of the classical Gronwall inequality, with extra restrictions necessary. We believe this result to be new. Theorem 5.1. Let a > 0, b > 0, 0 < β < 1 and let φ be non-increasing, φ ∈ L1 +[0, T ] for all T > 0. Suppose that u ∈ C+[0,∞) satisfies the inequality u(t) ≤ a+ b ∫ t 0 (t− s)−βφ(s)u(s) ds, for t > 0, (5.1) If there exists r ∈ (0, 1) and tr > 0 such that bΓ(1−β)(I1−βφ)(t) ≤ r for 0 ≤ t ≤ tr, then u(t) ≤ a 1− r exp ( b tβr (1− r) ∫ t 0 φ(s) ds ) , for every t > 0. (5.2) Moreover, if, in addition, φ ∈ L1 +[0,∞) then u(t) ≤ a 1− r exp ( b tβr (1− r) ∫ ∞ 0 φ(s) ds ) , (5.3) so that u(t) is uniformly bounded. Proof. We will use the notation u∗(t) := maxs∈[0,t] u(s). Let t > 0 and let τ ∈ (0, t] be arbitrary. For τ ≤ tr we have u(τ) ≤ a+ b ∫ τ 0 (τ − s)−βφ(s)u(s) ds ≤ a+ bu∗(t) ∫ τ 0 (τ − s)−βφ(s) ds, = a+ bΓ(1− β) ( I1−βφ ) (τ)u∗(t) ≤ a+ ru∗(t). (5.4) Now we consider the case when τ > tr. We have u(τ) ≤ a+ b ∫ τ−tr 0 (τ − s)−βφ(s)u(s) ds+ b ∫ τ τ−tr (τ − s)−βφ(s)u(s) ds. In the first integral we use the fact that τ −s ≥ tr so that (τ −s)−β ≤ t−βr , while in the second integral we use s ≥ s−(τ− tr) ≥ 0 and the fact that φ is non-increasing. EJDE-2021/ 80 ASYMPTOTIC BEHAVIOUR 9 This gives u(τ) ≤ a+ b ∫ τ−tr 0 t−βr φ(s)u∗(s) ds+ bu∗(t) ∫ τ τ−tr (τ − s)−βφ(s− (τ − tr)) ds (5.5) In the second integral we now let σ = s− (τ − tr) and it becomes∫ tr 0 (tr − σ)−βφ(σ)dσ = Γ(1− β)(I1−βφ)(tr) ≤ r/b. Thus (5.5) gives u(τ) ≤ a+ b ∫ τ−tr 0 t−βr φ(s)u∗(s) ds+ ru∗(t) ≤ a+ b ∫ τ 0 t−βr φ(s)u∗(s) ds+ ru∗(t). (5.6) As (5.4) holds, (5.6) holds for all τ ∈ (0, t], and taking the supτ∈(0,t], we obtain u∗(t) ≤ a+ b ∫ t 0 t−βr φ(s)u∗(s) ds+ ru∗(t). Hence we have u∗(t) ≤ a 1− r + b 1− r ∫ t 0 t−βr φ(s)u∗(s) ds, and this holds for all t > 0. This is now a classical Gronwall inequality, Theorem 2.1, and we can immediately deduce that u(t) ≤ u∗(t) ≤ a 1− r exp ( b 1− r t−βr ∫ t 0 φ(s)ds ) , for all t > 0. The last assertion is now obvious. � Remark 5.2. The proof of Theorem 5.1 is a modification of the proof of Theorem 3.2 in [20] which considered the Gronwall inequality on a finite interval [0, T ] for the special case where φ(t) = t−η for η < 1− β. We allow φ to be possibly singular at 0 with essentially the same type of singularity. The proof in [20] was based on the proof in Haraux [6, Lemma 6, p.33], which had β = 1/2, he attributed the method to Pazy [16]. In the paper [20] it was possible to show that r = β/(1 − η) (when φ(t) = t−η) had an optimal property; nothing similar seems possible in the general case here. Clearly tr should be chosen as large as possible to give a better estimate in (5.2). A simpler case is when r1 = 1/2 is an allowed value, the conclusion is then u(t) ≤ 2a exp ( 2 b t−βr1 ∫ t 0 φ(s)ds ) . (5.7) Remark 5.3. The hypotheses that there exist r ∈ (0, 1) and tr > 0 such that bΓ(1− β)(I1−βφ)(t) ≤ r for 0 ≤ t ≤ tr, can be replaced by the hypothesis (I1−βφ)(t) → 0 as t → 0+ and then we may choose any r ∈ (0, 1). The property (I1−βφ)(t)→ 0 as t→ 0+ holds in the following two cases: (a) φ ∈ Lp[0, τ ] for some p > 1/(1− β) and some τ > 0. (b) φ ∈ C−η[0, τ ] for some η < 1− β and some τ > 0. 10 J. R. L. WEBB EJDE-2021/80 In case (a), a result of Hardy and Littlewood [7] shows that if p > 1/(1− β), then I1−β maps Lp[0, τ ] into a Hölder space C0,1−β−1/p (hence I1−βφ is continuous) and also that I1−βφ(t) → 0 as t → 0+. Further information may be found in [21, Proposition 3.2 (3)]. Case (b) is a special case of case (a) but more can be easily proved in this case. It is shown in [21, Proposition 3.2 (5)] that I1−β maps C−η[0, τ ] into C1−β−η[0, τ ] ⊂ C[0, τ ] and that I1−βφ(t) → 0 as t → 0+. In fact, if φ(t) = t−ηv(t) where v is continuous, then I1−βφ(t) = 1 Γ(1− β) ∫ t 0 (t− s)−βs−ηv(s) ds = 1 Γ(1− β) t1−β−η ∫ 1 0 (1− σ)−βσ−ηv(tσ) dσ, which proves the result since ∫ 1 0 (1− σ)−βσ−η dσ = B(1− β, 1− η). Of course in this remark τ can be as small as we wish. In specific cases, tr in Theorem 5.1 can be determined explicitly. The constants may be replaced by non-decreasing functions as we now show by a simple method. Corollary 5.4. Suppose a, b are positive non-decreasing functions, 0 < β < 1 and u ∈ C+[0,∞) satisfies the inequality u(t) ≤ a(t) + b(t) ∫ t 0 (t− s)−βφ(s)u(s) ds, for t > 0, where r, tr, φ are as in Theorem 5.1. Then we have u(t) ≤ a(t) 1− r exp ( b(t) 1− r t−βr ∫ t 0 φ(s)ds ) , for every t > 0 In particular, if r1 = 1/2 is an allowed value then using tr1 , we obtain u(t) ≤ 2a(t) exp (2b(t) tβr1 ∫ t 0 φ(s) ds ) , for every t > 0. Proof. In the proof of Theorem 5.1, for a fixed t̄ > 0, first replace the inequality by u(t) ≤ a(t̄) + b(t̄) ∫ t 0 (t− s)−βφ(s)u(s) ds, for t ≤ t̄. Applying the result proved, and noting that the final inequality holds for t = t̄, gives the conclusion since t̄ is arbitrary. � The following fact will be important in the following discussions. It seems to be a new result concerning properties of fractional integrals for it was shown by Hardy and Littlewood [7, §3.5,(iv)] that, for p > 1, I1/p does not map Lp into L∞. The non-increasing property of the function is important for our proof and prevents φ having any spikes. Theorem 5.5. Let 0 < α < 1 and suppose that φ is non-increasing, φ ∈ L1 +(0,∞) and there exist t1 > 0 and a constant M > 0 such that (Iαφ)(t) ≤M for 0 ≤ t ≤ t1. Then Iαφ(t) is uniformly bounded for all t > 0. EJDE-2021/ 80 ASYMPTOTIC BEHAVIOUR 11 Proof. Take t1 ∈ (0, 1) such that Iαφ(t) ≤M for every t ≤ t1. For t > t1 we argue as in Theorem 5.1 and write Iαφ(t) = 1 Γ(α) ∫ t−t1 0 (t− s)α−1φ(s) ds+ 1 Γ(α) ∫ t t−t1 (t− s)α−1φ(s) ds ≤ 1 Γ(α) ∫ t−t1 0 tα−1 1 φ(s) ds+ 1 Γ(α) ∫ t t−t1 (t− s)α−1φ(s− (t− t1)) ds ≤ tα−1 1 Γ(α) ∫ ∞ 0 φ(s) ds+ 1 Γ(α) ∫ t1 0 (t1 − σ)α−1φ(σ) dσ ≤ tα−1 1 Γ(α) ∫ ∞ 0 φ(s) ds+M =: C0. Thus Iαφ(t) ≤ C0 for every t > 0. � Remark 5.6. We could assume φ ∈ C−η[0, t2] for some η < α and some t2 > 0, or that φ ∈ Lp[0, t2] for some p > 1/(1− α), which imply that limt→0+ I αφ(t)→ 0 as in Remark 5.3. The hypotheses do not imply that Iαφ ∈ L1[0,∞) as the following example shows. Example 5.7. Let f(t) = t−γ 1+t for 0 < γ < 1. Then f ∈ L1[0,∞) and for 0 < γ < α < 1 we have Iαf(t) → 0 as t → 0+, Iαf is uniformly bounded but Iαf /∈ L1[0,∞). Clearly f ∈ L1[0,∞), in fact, ∫∞ 0 f(t) dt = π sin(γπ) (Beta function). Firstly, we have Iαf(t) = 1 Γ(α) ∫ t 0 (t− s)α−1 s −γ 1 + s ds ≤ 1 Γ(α) ∫ t 0 (t− s)α−1s−γ ds = tα−γ Γ(α) B(α, 1− γ). (5.8) This shows that Iαf(t)→ 0 as t→ 0+. Secondly we will show that Iαf is uniformly bounded. For t ≤ 1 the above shows that Iαf(t) ≤ C1. For t > 1 we write Iαf(t) ≤ C1 + 1 Γ(α) ∫ t 1 (t− s)α−1 s −γ 1 + s ds ≤ C1 + 1 Γ(α) ∫ t 1 (t− s)α−1 1 1 + s ds ≤ C1 + 1 Γ(α) ∫ t 0 (t− s)α−1 1 1 + s ds ≤ C2. The integral here, according to Maple, involves hypergeometric functions in general. For the special case when α = 1/2 we have∫ t 0 (t− s)−1/2 1 1 + s ds = 2 tanh−1 ( √t√ t+ 1 ) 1√ t+ 1 , 12 J. R. L. WEBB EJDE-2021/80 with the inverse of the tanh function; this is bounded by 4/3. Thirdly, we have Iαf(t) = 1 Γ(α) ∫ t 0 (t− s)α−1 s −γ 1 + s ds ≥ 1 Γ(α) ∫ t 0 (t− s)α−1 t −γ 1 + t ds = tα−γ Γ(α+ 1)(1 + t) , which shows that Iαf /∈ L1[0,∞) since α− γ > 0. Remark 5.8. It is a basic fact that Iα maps L1[0, T ] into L1[0, T ] for every finite T > 0 but that does not hold for T = ∞; the above example is an explicit proof of this fact. 6. Asymptotic behaviour for equations of order α < 1 In this section we investigate the asymptotic behaviour of solutions of the integral equation corresponding to solutions of the initial value problem Dα ∗ u(t) = φ(t)f(t, u(t), Dγ ∗u(t)), for a.e. t > 0, u(0) = u0, (6.1) when 0 < γ < α < 1. 6.1. When f does not depend on fractional derivatives. We first consider the simpler case Dα ∗ u(t) = φ(t)f(t, u(t)), for a.e. t > 0, u(0) = u0, (6.2) and will study the asymptotic behaviour of solutions of the corresponding integral equation u(t) = u0 + Iα(φ(t)f(t, u(t))) = u0 + 1 Γ(α) ∫ t 0 (t− s)α−1φ(s)f(s, u(s)) ds. (6.3) with f continuous. These problems are essentially equivalent on a finite interval as in Theorem 3.5. Thus we will study solutions of (6.3) in the space C[0, T ] that are assumed to exist for an arbitrary T > 0 and consider the behaviour of any solution as T →∞. Remark 6.1. When |f(t, u)| ≤ C(1 + |u|) problems such as (6.3) have global solutions, that is solutions that exist on intervals [0, T ] for all T > 0, such solutions will often grow exponentially so to have a linear or polynomial decay φ(t) will need some suitable smallness property for large t. Global existence for this problem can be proved exactly as in [20, Theorem 4.8] under the assumptions we use below by using the estimates from the fractional Gronwall inequality of Theorem 5.1. The following remark can be used to reduce the number of hypotheses that are stated in the following results, but we have not necessarily done this in every case. Remark 6.2. (1) If φ(s) and spφ(s) belong to L1[0,∞) for some p < 1 then srφ(s) ∈ L1[0,∞) for each r ∈ [p, 1]. This follows from the fact that, for each s > 0, sr = exp(r ln(s)) is a convex function of r, and every r ∈ [p, 1] can be written r = (1−λ)p+λ for some λ ∈ [0, 1], hence, by the convexity, sr ≤ (1−λ)sp+λs. EJDE-2021/ 80 ASYMPTOTIC BEHAVIOUR 13 (2) If sqφ(s) is non-increasing for some q > 0 then srφ(s) is also non-increasing for each r ∈ [0, q]. (3) If δ > 0 and Iδφ(t) is bounded on a neighbourhood of 0 then also Iδ(tpφ(t)) is bounded on a neighbourhood of 0 for any p > 0. (4) Recall Proposition 4.2; if α > β then Iαφ(t) ≤ Ctα−βIβφ(t), thus Iβφ bounded near 0 implies Iαφ(t)→ 0 as t→ 0. Our result for this problem is the following. Theorem 6.3. Let 0 < α < 1, φ ∈ L1 +[0,∞), and let φ be non-increasing and suppose that Iαφ(t) → 0 as t → 0. Let f be continuous and satisfy |f(t, u)| ≤ C(1 + |u|) for a constant C > 0, all t ∈ [0,∞) and all u ∈ R. If u is a global solution of (6.3), then |u| is uniformly bounded on [0,∞). Proof. Since Cφ satisfies the same assumptions as φ, without loss of generality we can and do take C = 1. We note that Iαφ(t) ≤ C0 for every t > 0 by Theorem 5.5 and then we have |u(t)| ≤ |u0|+ 1 Γ(α) ∫ t 0 (t− s)α−1φ(s)|f(s, u(s))| ds ≤ |u0|+ 1 Γ(α) ∫ t 0 (t− s)α−1φ(s)(1 + |u(s)|) ds ≤ |u0|+ C0 + 1 Γ(α) ∫ t 0 (t− s)α−1φ(s)|u(s)|) ds. (6.4) By the fractional Gronwall inequality (5.7), for t1 chosen so that Γ(α)(Iαφ)(t) < 1/2 for t ≤ t1, we obtain u(t) ≤ 2(|u0|+ C0) exp ( 2tα−1 1 ∫ t 0 φ(s) ds ) , for every t > 0, (6.5) hence |u(t)| ≤ 2(|u0|+C0) exp ( 2tα−1 1 ∫∞ 0 φ(s) ds ) , that is |u| is uniformly bounded since φ ∈ L1[0,∞). � Remark 6.4. The result applies to the slightly more general case of a function f1 replacing φf where we suppose that |f1(t, u(t))| ≤ φ(t)|f(t, u(t))|, with the given hypotheses, and no changes in the proof. 6.2. When f depends on fractional derivatives. We now turn to the much trickier case Dα ∗ u(t) = φ(t)f(t, u(t), Dγ ∗u(t)), for a.e. t > 0, u(0) = u0, (6.6) when 0 < γ < α < 1 and f is continuous. If u is a sufficiently regular solution of (6.6) then u is a solution of the integral equation u(t) = u0 + Iαφ(t)f(t, u(t), Dγ ∗u(t)) = u0 + 1 Γ(α) ∫ t 0 (t− s)α−1φ(s)f(s, u(s), Dγ ∗u(s)) ds. (6.7) However, the two problems are not necessarily equivalent. Equivalence depends on how ‘solution’ of each problem is defined. It is not entirely clear in what space we should seek solutions of (6.6) and of (6.7). A solution in C[0, T ] of (6.7) is 14 J. R. L. WEBB EJDE-2021/80 not enough since it is necessary that Dγ ∗u exists at least a.e. for the problem to have a meaning. Solutions in C1[0, T ] would certainly be sufficient and (6.6) would make sense, but this requires knowledge of u′ which formally would be given by Iα−1φ(t)f(t, u(t), Dγ ∗u(t)), unfortunately that expression might not have a finite value for every t. An appropriate subspace of continuous functions is X := {u ∈ C[0, T ], Dγ ∗u ∈ C[0, T ]} endowed with the norm ‖u‖X := ‖u‖∞ + ‖Dγ ∗u‖∞. Thus u ∈ X is equivalent to u ∈ C[0, T ] and I1−γ(u− u0) ∈ C1[0, T ]. Definition 6.5. For a continuous function f we say that u ∈ X is a solution of (6.6), on an interval [0, T ], if Dα ∗ u exists and equals φf a.e. We say that u ∈ X is a solution of (6.7) if u satisfies (6.7) for all t ∈ [0, T ]. A global solution is one that exists on [0, T ] for every T > 0. We now show the equivalence of the two problems. Proposition 6.6. Let f be continuous. Then u ∈ X is a solution of (6.6) if and only if u ∈ X is a solution of (6.7). Proof. Firstly suppose that u ∈ X is a solution of (6.7). Then F (t) := φ(t)f(t, u(t), Dγ ∗u(t)) ∈ L1[0, T ] so that I1−α(u− u0)(t) = I1−αIαF (t) = IF (t) where IF ∈ AC hence Dα ∗ u(t) = F (t) exists a.e., that is u is a solution of (6.6). Conversely, if u ∈ X is a solution of (6.6) then Dα ∗ u(t) = F (t) a.e., that is D(I1−α(u − u0)(t) = F (t) a.e. and I1−α(u − u0) ∈ AC[0, T ]. Thus we have I1−α(u−u0)(t) = IF (t) (note that I1−α(u−u0)(0) = 0 by Remark 5.3). Applying Iα, and using the semigroup property of fractional integrals, we obtain I(u− u0)(t) = IαI1−α(u− u0)(t) = IαIF = IIαF (t), for every t. The left hand side is a C1 function, the right hand side is in AC. Therefore, the derivatives exist a.e. and u(t) − u0 = IαF (t) for all a.e. t, so u is a solution of (6.6). � It is useful to know that X is Banach space, that is, is complete. This is a known result proved for the Riemann-Liouville derivative in Su [18], but we include a proof for completeness. Proposition 6.7. (X, ‖ · ‖) is a Banach space. Proof. Let {un} be a Cauchy sequence in X, then {un} and {Dγ ∗un} are Cauchy sequences in C[0, T ] hence are uniformly convergent, say un → u and Dγ ∗un → v uniformly on [0, T ] with u and v continuous. We have to prove that Dγ ∗u exists and v = Dγ ∗u. Let un,0 denote the constant function with value un(0). As I1−γ is a bounded linear operator from C[0, T ] to itself, we have I1−γ(un − un,0) → I1−γ(u − u0) uniformly on [0, T ]. By definition, Dγ ∗un = D(I1−γ(un − un,0)) so that I1−γ(un − un,0) ∈ C1 for un ∈ X. Therefore we have, for every n, I1−γ(un − un,0)(t) = I1−γ(un − un,0)(0) + ∫ t 0 D(I1−γ(un − un,0))(s) ds. EJDE-2021/ 80 ASYMPTOTIC BEHAVIOUR 15 Passing to the limit on both sides gives I1−γ(u− u0)(t) = I1−γ(u− u0)(0) + ∫ t 0 v(s) ds. This proves that v(t) = D(I1−γ(u− u0))(t) = Dγ ∗u(t), as required. � We now assume that a global solution of (6.7) exists and we discuss its asymptotic behaviour. In doing this we establish some a priori bounds which could be used to prove an existence result. Theorem 6.8. Let 0 < γ < α < 1 and let φ be non-increasing and (Iα−γφ)(t)→ 0 as t→ 0+. Suppose also that φ(s) and sγφ(s) are in L1[0,∞). Let f be continuous and satisfy |f(t, u, p)| ≤ C(1 + |u| + |p|) for all t ∈ [0,∞) and all u, p ∈ R. If u is a global solution of (6.7), then |u| and |Dγ ∗u| are uniformly bounded on [0,∞). Proof. Without loss of generality we take C = 1. We first note that if u is a global solution of (6.7), that is, u ∈ X and u(t) = u0 + Iα ( φ(t)f(t, u(t), Dγ ∗u(t)) ) , then from Proposition 4.2 we obtain |u− u0| = |Iα(φf)| ≤ Iα|φf | ≤ tγΓ(α− γ)Iα−γ |φf |, hence |u| ≤ |u0|+ tγΓ(α− γ)Iα−γ |φf |. Also we have, by definition, and the semigroup property that Dγ ∗u = D(I1−γ)(u− u0) = D(I1−γIα)φf = DIIα−γφf = Iα−γφf, and hence |Dγ ∗u(t)| ≤ Iα−γ |φf | a.e. We now have Iα−γ |φf | = 1 Γ(α− γ) ∫ t 0 (t− s)α−γ−1|φ(s)f(s, u(s), Dγ ∗u(s))| ds ≤ 1 Γ(α− γ) ∫ t 0 (t− s)α−γ−1φ(s)(1 + |u(s)|+ |Dγ ∗u(s))| ds ≤ 1 Γ(α− γ) ∫ t 0 (t− s)α−γ−1φ(s) ( 1 + |u0|+ sγΓ(α− γ)Iα−γ |φf |+ Iα−γ |φf | ) ds. Write v(t) := Iα−γ |φf |(t), then v satisfies the inequality v(t) ≤ 1 Γ(α− γ) ∫ t 0 (t− s)α−γ−1φ(s) ( 1 + |u0|+ sγΓ(α− γ)v(s) + v(s) ) ds, = (1 + |u0|)(Iα−γφ)(t) + 1 Γ(α− γ) ∫ t 0 (t− s)α−γ−1φ(s) ( Γ(α− γ)sγ + 1) ) v(s) ds Now we note that Iα−γφ is bounded on [0,∞) by Theorem 5.5, say Iα−γφ(t) ≤M for all t ≥ 0. We then have v(t) ≤M(1 + |u0|) + ∫ t 0 (t− s)α−γ−1φ(s) ( sγ + 1 Γ(α− γ) ) v(s) ds. 16 J. R. L. WEBB EJDE-2021/80 By our assumption, φ(s)(sγ + 1 Γ(α−γ) ) ∈ L1[0,∞), and (Iα−γφ)(t) → 0 as t → 0+ gives Iα−γ(tγφ(t))→ 0 also, so we may apply the fractional Gronwall inequality of Theorem 5.1 to deduce that, for every t > 0, v(t) ≤ 2M(1 + |u0|) exp ( 2 t1+γ−α 1 ∫ t 0 φ(s) ( sγ + 1 Γ(α− γ) ) ds ) , ≤ 2M(1 + |u0|) exp ( 2 t1+γ−α 1 ∫ ∞ 0 φ(s) ( sγ + 1 Γ(α− γ) ) ds ) := M1. (6.8) Since |Dγ ∗u(t)| ≤ Iα−γ |φf | = v(t) this has proved that |Dγ ∗u(t)| ≤M1. Next we have u(t) = u0 + Iαφf thus |u(t)| ≤ |u0|+ Iα|φf |. This gives |u(t) ≤ |u0|+ 1 Γ(α) ∫ t 0 (t− s)α−1|φ(s)f(s, u(s), Dγ ∗u(s))| ds ≤ |u0|+ 1 Γ(α) ∫ t 0 (t− s)α−1φ(s)(1 + |u(s)|+M1) ds = |u0|+ (1 +M1)Iαφ(t) + ∫ t 0 (t− s)α−1φ(s)|u(s)| ds. (6.9) By Theorem 5.5, Iαφ(t) is uniformly bounded, so that |u(t)| ≤M2 + ∫ t 0 (t− s)α−1φ(s)|u(s)| ds. By the fractional Gronwall inequality of Theorem 5.1, we deduce that |u(t)| ≤M3 for all t > 0. This completes the proof. � Remark 6.9. Of course the result applies to the case of a function f1 replacing φf where we suppose that |f1(t, u(t), Dγ ∗u(t))| ≤ φ(t)|f(t, u(t), Dγ ∗u(t))|, with the given hypotheses. The apparently more general case when |f(t, u, p)| ≤ a+ b|u|+ c|p| is really the same since a+ b|u|+ c|p| ≤ max{a, b, c}(1 + |u|+ |p|). Remark 6.10. Medveď and Posṕı̌sil [14, Theorem 1] studied this problem under different growth assumptions on f , not necessarily linear but with some integrability conditions. They use Hölder’s inequality and the Bihari inequality to prove that |u(t)|/tγ and |Dγ ∗u(t)| are bounded and the dominated convergence theorem to show that |u(t)|/tγ has a limit L as t→∞. In our case we have L = 0. It is not easy to make a comparison of these results. This problem was also recently studied by Kassim-Tatar [11, Theorem 6.2] and they proved boundedness of |u| and |Dγ ∗u| assuming f satisfies a multiplicative type inequality which appears more restrictive than our sum inequality. Their example 6.3 also follows from our result, where they essentially have φ(t) = t−1/3 exp(−t), which is smaller than necessary, but it is an example and optimal conditions are unknown. 7. Higher order equation with non derivative dependent nonlinearity We will investigate the asymptotic behaviour of global solutions of the integral equation u(t) = u0 + a(t)u1 + Iα+βf(t, u(t)), t > 0, (7.1) EJDE-2021/ 80 ASYMPTOTIC BEHAVIOUR 17 where a is continuous, u0, u1 are constants, 0 < α, β ≤ 1 with 1 < α+ β < 2; here the φ is essentially included in f term. By a solution we mean that u ∈ C[0, T ] for all T > 0 and satisfies equation (7.1) on [0, T ]; a global solution is one which exists on [0, T ] for all T > 0. The motivation for studying this form is that for a(t) = t this arises from seeking solutions of the IVP Dα+β ∗ (t) = f(t, u(t)), u(0) = u0, u′(0) = u1, (7.2) and for a(t) = tβ with 0 < β < 1 it arises from the similar problem Dα ∗ (Dβ ∗u)(t) = f(t, u(t)), u(0) = u0, D β ∗ (0) = u1Γ(β + 1). (7.3) Note: Dα+β ∗ u is not the same as Dα ∗ (Dβ ∗u). For example, if α+β < 1, the problem Dα+β ∗ (t) = f(t) requires only one initial condition u(0) = u0 to be well posed, whereas Dα ∗ (Dβ ∗u)(t) = f(t) requires two initial conditions; see also discussion and examples in Diethelm [3]. The special case of the problem (7.3) when α = 1 was recently studied by Kassim- Tatar [11, Theorem 4.1] where a special inequality was used. We will show that the more general problem (7.1) can be tackled by a single method which involves using the classical Gronwall inequality Theorem 2.1. Clearly the asymptotic behaviour of a solution u of (7.1) depends in an essential way on the behaviour of Iα+βf(t, u(t)) as t→∞. Theorem 7.1. Let a ∈ C[0,∞). Suppose that u is continuous and is a global solution of the equation u(t) = u0 + a(t)u1 + Iα+βf(t, u(t)), t > 0, (7.4) where u0, u1 are constants and 0 < α, β ≤ 1 with 1 < α + β < 2. Write v(t) := Iα+βf(t, u(t)). If f satisfies the growth assumption |f(t, u)| ≤ φ(t)(1 + |u|), (7.5) where φ ∈ L1[0,∞), a, φ ∈ L1[0,∞), and tα+β−1φ(t) ∈ L1[0,∞), then there is a constant C > 0 such that |v(t)|/tα+β−1 ≤ C. Moreover, v(t)/tα+β−1 → L as t → ∞ and therefore u(t) − ( u0 + a(t)u1 + Ltα+β−1 ) → 0 as t → ∞, where L = 1 Γ(α+β) ∫∞ 0 f(s, u(s)) ds. Remark 7.2. We do not need any non-increasing property of φ in this result. Proof of Theorem 7.1. We have, since α+ β > 1, |v(t)| tα+β−1 ≤ 1 Γ(α+ β) ∫ t 0 (1− s/t)α+β−1|f(s, u(s))| ds ≤ 1 Γ(α+ β) ∫ t 0 |f(s, u(s))| ds ≤ 1 Γ(α+ β) ∫ t 0 φ(s)(1 + |u0|+ |a(s)u1|+ |v(s)|)) ds ≤ 1 Γ(α+ β) ∫ ∞ 0 (1 + |u0|)φ(s) + |u1a(s)|φ(s) ds + 1 Γ(α+ β) ∫ t 0 sα+β−1φ(s) |v(s)| sα+β−1 ds 18 J. R. L. WEBB EJDE-2021/80 ≤ C1 + 1 Γ(α+ β) ∫ t 0 sα+β−1φ(s) |v(s)| sα+β−1 ds, using (7.5). By the classical Gronwall inequality of Theorem 2.1, this gives |v(t)| tα+β−1 ≤ C1 exp ( 1 Γ(α+ β) ∫ t 0 sα+β−1φ(s)ds ) ≤ C1 exp ( 1 Γ(α+ β) ∫ ∞ 0 sα+β−1φ(s)ds ) ≤ C. From above we have |f(s, u(s))| ≤ φ(s)(1 + |u0|+ |a(s)u1 + |v(s)|) ≤ φ(s)(1 + |u0|+ |a(s)u1|+ Csα+β−1), so f(s, u(s)) is an L1[0,∞) function, and hence, by the dominated convergence theorem, lim t→∞ v(t) tα+β−1 = 1 Γ(α+ β) ∫ ∞ 0 f(s, u(s)) ds =: L. As u(t) = u0 +a(t)u1 +v(t) this gives u(t) ∼ u0 +a(t)u1 +Ltα+β−1 asymptotically. � Remark 7.3. If f does not depend on u then we can apply Proposition 4.6(a) to immediately deduce that if f ∈ L1[0,∞) then u ∼ u0 + a(t)u1 + ∫∞ 0 f(s)ds Γ(α+ β) tα+β−1, which confirms that our result is giving a correct answer. The main requirement of our proof is to prove f(s, u(s)) is an L1[0,∞) function. Corollary 7.4. (1) For a(t) = t, we have u(t) ∼ u0+t u1 when φ ∈ L1[0,∞), tφ(t) ∈ L1[0,∞) since tα+β−1 ≤ t for t ≥ 1. (2) For a(t) = tβ we have u(t) ∼ u0 + C2t β since α + β − 1 ≤ β, provided that φ ∈ L1[0,∞), tβφ(t) ∈ L1[0,∞). (3) For a(t) = tγ and γ ∈ (0, 1), we have, under the corresponding conditions on φ, u(t) ∼ u0 + u1t γ when α + β − 1 < γ, and we have u(t) ∼ u0 + Ltα+β−1 when α+ β − 1 > γ. Case (1) is essentially studied in [13] when β = 1 with the same conclusion under different multiplicative type assumptions using some ideas from his own paper [12]. Case (2) with α = 1 is the case studied in [11]. Their Theorem 4.1 proves Dβ Cu(t) and u(t) tβ have limits at t→∞ under some different multiplicative type hypotheses. Using the inequality |xy| ≤ |x|p/p+ |y|p′/p′, p > 1, 1/p+1/p′ = 1, their hypotheses seem to imply ours; in [11, Example 4.2] they have f(t, u) = exp(−t)ur where 0 < r ≤ 1 so that |f(t, u)| ≤ φ(t)(1 + |u|) for φ(t) = exp(−t). 8. Higher order equation with derivative dependence We now consider the problem u(t) = u0 + Iβb1 + Iα+βf(t, u(t), Dγ ∗u(t)), (8.1) for 0 < γ ≤ β ≤ 1, 0 < α ≤ α + β − γ < 1 and α + β > 1. We do not impose an ordering between α and β. Here Iβb1 = b1t β/Γ(β + 1) but it is convenient to often EJDE-2021/ 80 ASYMPTOTIC BEHAVIOUR 19 write Iβb1. This is the integral equation version of the initial value problem of the sequential fractional differential problem Dα ∗ ( Dβ ∗u(t) ) = f(t, u(t), Dγ ∗u(t)), a.e. u(0) = u0, D β ∗u(0) = b1. (8.2) The special case α = 1 is studied in [11], but the general sequential problem here is not studied in either [11] or [14]. Writing F (t) := f(t, u(t), Dγ ∗u(t)), informally we can ‘solve’ (8.2) as follows: Dα ∗ ( Dβ ∗u(t) ) = F (t), (8.3) =⇒ Dβ ∗u(t) = (Dβ ∗u)(0) + IαF = b1 + IαF (8.4) =⇒ u(t) = u0 + Iβb1 + Iα+βF = u0 + b1 Γ(β + 1) tβ + Iα+βF. (8.5) Equation (8.1) should be studied in a space where at least u and Dβ ∗u are continuous but we will not study existence of solutions here. We suppose that a global solution u exists, that is u and Dβ ∗u exist and are continuous on [0, T ] for all T > 0 and (8.4) and (8.5) are satisfied. We investigate the asymptotic behaviour of u and fractional derivatives, under suitable conditions on f similar to the ones used previously. We require a fact about the relationship between Dγ ∗u(t) and other terms in (8.1) as follows. Dγ ∗u(t) = Dγ(u− u0)(t) = D ( I1−γ(Iβb1 + Iα+βF ) ) (t) = Iβ−γb1 + Iα+β−γF (t). (8.6) Theorem 8.1. Suppose that there exists φ such that |f(t, u, p)| ≤ φ(t)(1 + |u|+ |p|), for all t ≥ 0, u, p ∈ R, (8.7) where φ(s) and sβφ(s) are non-increasing, Iα+β−γφ(t) ≤M for t near 0, and φ(s), sγφ(s), sα+β−γφ(s) are all L1[0,∞) functions. Then, for a global solution u(t) of (8.4) and (8.5), there is a constant L such that u(t)− ( u0 + b1 Γ(β + 1) tβ + Ltα+β−1 ) → 0 as t→∞. In fact, writing F (t) := f(t, u(t), Dγ ∗u(t)), L is given by L = limt→∞ Iα+βF (t) tα+β−1 . Proof. Let u satisfy (8.4) and (8.5) and let F (t) := f(t, u(t), Dγ ∗u(t)) and ρ := α+ β − γ, so ρ < 1 by assumption. We note that Iρ|F |(t) ≤ 1 Γ(ρ) ∫ t 0 (t− s)ρ−1|f(s, u(s), Dγ ∗u(s))| ds ≤ 1 Γ(ρ) ∫ t 0 (t− s)ρ−1φ(s) ( 1 + |u(s)|+ |Dγ ∗u(s)| ) ds ≤ 1 Γ(ρ) ∫ t 0 (t− s)ρ−1φ(s) ( 1 + |u0|+ |Iβb1|+ Iα+β |F |+ |Iβ−γb1|+ Iρ|F | ) ds, (8.8) where we have used (8.4) and (8.6). Since φ(s), sβφ(s) and sβ−γφ(s) are non- increasing L1[0,∞) functions, and Iρφ(t) is assumed to be bounded for t near 0, 20 J. R. L. WEBB EJDE-2021/80 the terms ∫ t 0 (t − s)ρ−1φ(s) ( 1 + |u0|) ds and ∫ t 0 (t − s)ρ−1φ(s) ( |Iβb1| + |Iβ−γb1| ) ds are uniformly bounded by Theorem 5.5. By Proposition 4.2 we have Iα+β |F |(s) ≤ Γ(α+ β − γ) Γ(α+ β) sγIα+β−γ |F |(s) = Γ(ρ) Γ(α+ β) sγIρ|F |(s) (8.9) so from (8.8) we obtain Iρ|F |(t) ≤ C1 + 1 Γ(ρ) ∫ t 0 (t− s)ρ−1Φ(s)Iρ|F |(s) ds, (8.10) where Φ(s) = φ(s) ( 1 + sγ Γ(ρ) Γ(α+β) ) which is an L1[0,∞) non-increasing function by our hypotheses. Since 0 < ρ < 1 we may apply the fractional Gronwall inequality Theorem 5.1 to give Iρ|F | is uniformly bounded, by M1 say, which proves that |Dγ ∗u(t)| ≤ |Iβ−γb1|+M1. Now we consider Iα+β |F | which is related to u by (8.5). Since α+β > 1 this is not the case of singular kernel as in Theorem 5.1. We argue similarly to Theorem 7.1. We consider Iα+β |F (t)|/tα+β−1 and we have Iα+β |F |(t) tα+β−1 = 1 Γ(α+ β) ∫ t 0 (1− s/t)α+β−1|f(s, u(s), Dγ ∗u(s))| ds, ≤ 1 Γ(α+ β) ∫ t 0 φ(s) ( 1 + |u0|+ |Iβb1|+ Iα+β |F |(s) + |Iβ−γb1|+M1 ) ds ≤ C2 + 1 Γ(α+ β) ∫ t 0 φ(s)Iα+β |F |(s) ds, ≤ C2 + 1 Γ(α+ β) ∫ t 0 φ(s)sα+β−1 I α+β |F |(s) sα+β−1 ds, (8.11) where C2 comes from using Theorem 5.5 similarly to above, since φ(s), sβφ(s), sβ−γ ∈ L1[0,∞) and are non-increasing. This is a classical Gronwall inequality and by the hypothesis sα+β−1φ(s) ∈ L1[0,∞) we deduce that Iα+β |F |(t) tα+β−1 is uniformly bounded for all t > 0, by M2 say. Now we note that Iα+βF (t) tα+β−1 = 1 Γ(α+ β) ∫ t 0 (1− s/t)α+β−1F (s) ds and, when F ≥ 0, the right side is a non-decreasing function of t since α + β ≥ 1. Write F = F+−F−, the positive and negative parts, so that F+ +F− = |F |. Each of Iα+βF±(t) tα+β−1 is a non-decreasing function of t, bounded above by Iα+β |F |(t) tα+β−1 ≤ M2, hence each has a limit as t → ∞, say L±. Thus Iα+βF (t) tα+β−1 → L := L+ − L− as t → ∞. (This is simpler than applying the Dominated Convergence Theorem which also works.) In total we have, for L = 1 Γ(α+β) ∫∞ 0 F (s) ds, u(t)− ( u0 + b1 Γ(β + 1) tβ + Ltα+β−1 ) → 0 as t→∞. (8.12) � Remark 8.2. The asymptotic behaviour does not depend explicitly on γ but the hypotheses on φ have some γ dependence. The term with tβ dominates in (8.12) as EJDE-2021/ 80 ASYMPTOTIC BEHAVIOUR 21 t → ∞ since α ≤ 1 so we have u(t) ∼ b1 Γ(β+1) t β when α < 1, but (8.12) is a more precise conclusion. Remark 8.3. A similar problem with f = f(t, u(t), u′(t), Dβ ∗u(t)) is studied in [14, Theorem 2] under some hypotheses and it is shown, for that case, that, u(t)/t and x′(t) have a limit as t → ∞. The result uses Hölder’s inequality and the Bihari inequality. Also higher order equations are studied in this paper. The special case of α = 1 and 0 < γ < β < 1 is studied in [11, Theorem 5.2] under different hypotheses and they conclude that u(t)/tβ has a limit as t→∞, which is the same as our conclusion. 9. Conclusion We have shown that, by using a new fractional Gronwall inequality (in terms of exponential functions) and the classical Gronwall inequality, we can tackle many problems concerning the asymptotic behaviour of solutions of equations involving fractional integrals with nonlinearities possibly depending on fractional derivatives, under realistic and reasonable hypotheses. We do not need a different type of inequality for each problem. References [1] R. Bellman, The stability of solutions of linear differential equations, Duke Math. J., 10 (1943), 643–647. [2] E. Brestovanská, M. Medveď, Asymptotic behavior of solutions to second-order differential equations with fractional derivative perturbations, Electron. J. Differential Equations Vol. 2014 (2014), No. 201, 10 pp. [3] K. 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Equ. 2020, Paper No. 5, 16 pp. Jeffrey R. L. Webb School of Mathematics and Statistics, University of Glasgow, Glasgow G12 8SQ, UK Email address: jeffrey.webb@glasgow.ac.uk 1. Introduction 2. Known Gronwall inequalities 3. Definitions of fractional integrals and derivatives 4. Some properties of fractional integrals 5. New Gronwall type of inequality 6. Asymptotic behaviour for equations of order <1 6.1. When f does not depend on fractional derivatives 6.2. When f depends on fractional derivatives 7. Higher order equation with non derivative dependent nonlinearity 8. Higher order equation with derivative dependence 9. Conclusion References