Weakly singular Gronwall inequalities and applications to fractional differential equations

Accepted Manuscript Weakly singular Gronwall inequalities and applications to fractional differential equations

J.R.L. Webb

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S0022-247X(18)30939-9 https://doi.org/10.1016/j.jmaa.2018.11.004 YJMAA 22690

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Journal of Mathematical Analysis and Applications

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17 August 2017

Please cite this article in press as: J.R.L. Webb, Weakly singular Gronwall inequalities and applications to fractional differential equations, J. Math. Anal. Appl. (2018), https://doi.org/10.1016/j.jmaa.2018.11.004

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Weakly singular Gronwall inequalities and applications to fractional differential equations J. R. L. Webb School of Mathematics and Statistics, University of Glasgow, Glasgow G12 8SQ, UK

Abstract We obtain some new Gronwall type inequalities which are applicable to some weakly singular Volterra integral equations similar to ones first studied by D. Henry. The main interest is that we consider cases with a double singularity and we obtain explicit L∞ bounds rather than L1 bounds. Furthermore our bounds involve the exponential function and not the Mittag-Leffler function as in some previous works. We give applications to some Volterra integral equations with a doubly singular kernel that arise from Caputo fractional differential equations where, as opposed to previous papers, we have a singularity in the nonlinearity. Keywords: Gronwall inequality, Volterra integral equation, weakly singular kernel, fractional differential equation. 2000 MSC: 26D10, 34A08, 34A12, 45D05

1. Introduction The Gronwall inequality is a well-known tool in the study of differential equations, Volterra integral equations, and evolution equations, including some cases where there are weak singularities, for the weakly singular case see, for example, the lecture notes [11, 12]. The inequality is often used to establish a priori bounds which are used in proving global existence, uniqueness, and stability results. There are a number of versions of the result, we refer to the book [19] for some history and further details on ‘classical’ versions. One well known ‘classical’ version reads as follows. Theorem 1.1. Suppose that φ ∈ L∞ + [0, T ] satisfies the inequality  φ(t) ≤ c0 (t) +

t 0

c1 (s)φ(s) ds for a.e. t ∈ [0, T ],

(1.1)

where c1 ∈ L1+ [0, T ], and c0 ∈ L∞ + [0, T ] is non-decreasing, [Subscript + denotes Email address: [email protected] (J. R. L. Webb)

Preprint submitted to J. Math. Anal. Appl.

November 5, 2018

functions that are non-negative almost everywhere (a.e.)] Then φ(t) ≤ c0 (t) exp





t 0

c1 (s) ds

for a.e. t ∈ [0, T ].

(1.2)

If c0 is not non-decreasing a similar result holds if we replace c0 (t) in the conclusion by c∗0 (t), where, for an L∞ function u, the non-decreasing function u∗ is defined by u∗ (t) = ess sups∈[0,t] u(s). For weakly singular Gronwall type inequalities the pioneering work was by Henry [12] who proved, by an iterative process, some L1 bounds given by series related to the Mittag-Leffler function. For a similar inequality involving an integral with a doubly singular kernel we shall prove some L∞ type bounds which involve the exponential function. For example we show that if a, b, β > 0, γ ≥ 0 and β + γ < 1 and u ∈ L∞ satisfies the inequality  t (t − s)−β s−γ u(s) ds, for a.e. t ∈ [0, T ], u(t) ≤ a + b 0

then it follows that u(t) ≤ b1 exp(b2 t1−γ ), for a.e. t ∈ [0, T ] where bi are explicit constants given by the data a, b, β, γ, T . In particular there is a constant C = C(b, β, γ, T ) such that u(t) ≤ aC for a.e. t ∈ [0, T ]. Secondly for the case when u can be singular with tα u(t) an L∞ function for some α ∈ (0, 1) and satisfies u(t) ≤ at−α + c



t 0

(t − s)−β s−γ u(s) ds, for a.e. t ∈ (0, T ],

we prove that there is an explicit constant C = C(c, β, γ, T ) such that u(t) ≤ at−α C for a.e. t ∈ (0, T ]. These bounds are precisely what are required when studying initial value problems for Caputo and Riemann-Liouville fractional differential equations, as we discuss in section 4. One version of a weakly singular result of Henry is the following, cf. [12, Exercise 4*]. Theorem 1.2. Let a, b, α, β be non-negative constants with α < 1, β < 1. Suppose that u ∈ L1 [0, T ] satisfies u(t) ≤ at−α + b



t 0

(t − s)−β u(s) ds, for a.e. t ∈ (0, T ].

Then there is a constant C(b, β, T ) such that u(t) ≤

at−α C(b, β, T ) for a.e. t ∈ (0, T ]. 1−α

2

(1.3)

Henry [12, Lemma 7.1.2] also has a result for a doubly singular case with an extra term of the form sγ−1 as part of the integrand when it is assumed that u(t)tγ−1 in an L1 function and β > 0, γ > 0, and β + γ > 1 and u satisfies the inequality  t (t − s)β−1 sγ−1 u(s) ds, for a.e. t ∈ (0, T ]. (1.4) u(t) ≤ a + b 0

  In this case his conclusion is u(t) ≤ aEβ,γ bΓ(β)1/(β+γ−1) t where Eβ,γ (z) is given by an infinite series related to the two-parameter Mittag-Leffler function. Kong and Ding [14] use the same method as [12] to extend this result, replacing a by atα−1 and allowing b to be a continuous nondecreasing function. For the following simpler inequality, it was shown by Henry as part of [12, Lemma 7.1.1], and by Diethelm and Ford [9, Lemma 4.3], or [8, Lemma 6.19], that if a continuous function u satisfies  t b (t − s)q−1 u(s) ds, (1.5) u(t) ≤ a + Γ(q) 0 for some real number q > 0 then u(t) ≤ aEq (btq ),

(1.6)

∞ zk is the one-parameter Mittag-Leffler function. A where Eq (z) = k=0 Γ(kq+1) similar more general result is proved by Tisdell [24] starting with the inequality u(t) ≤ a +

1 Γ(q)



t 0

(t − s)q−1 (bu(s) + c) ds,

Henry uses an iterative method, while Diethelm and Ford and Tisdell essentially use a comparison principle since the given bound is the explicit solution of the equation when equality holds in (1.5), thus this bound is optimal. Neither paper [9, 24] handles the case with the extra singular term such as sγ−1 . In this paper we study the doubly singular case for L∞ functions and obtain bounds in terms of the exponential function. Some results of this type for a special case are proved in Haraux, see [11, Lemma 6, p.33] and [11, Lemma 10, p.112]. The method of proof we employ is similar in idea to that of Haraux to reduce to the classical Gronwall inequality of Theorem 1.1, but with extra difficulties. For a special case, using some properties of the Mittag-Leffler function, we show that our bound and the estimate then obtained from (1.6) are asymptotically of the same type and we make some numerical comparisons. Medved [17, Theorem 4] also proves several Gronwall type results, including one for the case of (1.4) by an argument based on H¨older’s inequality and has a bound involving the exponential function of certain integrated terms. The second part of our paper treats Volterra integral equations that arise from fractional differential equations of Caputo and Riemann-Liouville type which have been extensively studied in recent years. We deal with a case of 3

continuous functions whose Caputo derivative is allowed to have a singularity at t = 0 which, as far as we are aware, is the first time this has been done. We prove an equivalence between the initial value problem of fractional order α ∈ (0, 1) and a Volterra integral equation. By applying our Gronwall inequality we prove a global existence theorem, continuous dependence on initial data, and a uniqueness result for the Volterra integral equation. The main new features in our work are that: (1) We study Gronwall type inequalities with a double singularity. (2) We obtain explicit L∞ bounds, rather than L1 bounds. (3) Our bounds involve the well known exponential function rather than series or the Mittag-Leffler function. (4) Theorems 3.2 and 3.9 have novel proofs. (5) We prove equivalence between Caputo fractional differential equations with a singular nonlinearity and a Volterra integral equation with a doubly singular kernel. (6) We prove global existence for such Volterra integral equations which gives new results for Caputo and Riemann-Liouville fractional differential equations. 2. Preliminaries 1 We shall make extensive use of the Beta function B(p, q) := 0 (1−s)p−1 sq−1 ds which is well defined for p > 0, q > 0 and it is well known that B(p, q) = ∞ Γ(p)Γ(q) where Γ is the Gamma function, Γ(p) := 0 sp−1 exp(−s) ds which is Γ(p + q) well defined for p > 0, and is an extension of the factorial function, Γ(n+1) = n! for n ∈ N. The following simple lemma is classical. Since it is useful to us we sketch the short proof for completeness. Lemma 2.1. Let 0 ≤ t1 < t and 0 < β, γ < 1. Then for t > t1 we have  t (t − s)−β (s − t1 )−γ ds = (t − t1 )1−β−γ B(1 − β, 1 − γ).

(2.1)

t1

Proof. Change the variable of integration from s to σ where s = t1 + σ(t − t1 ) and the integral becomes  1  −β  −γ (1 − σ)(t − t1 ) σ(t − t1 ) (t − t1 ) dσ 0

=(t − t1 )1−β−γ



1 0

(1 − σ)−β σ −γ dσ

=(t − t1 )1−β−γ B(1 − β, 1 − γ).

4

3. Singular Gronwall inequalities with L∞ bounds We are interested in inequalities of the type studied and used by Henry [12] and Haraux [11]. Henry considers the following inequality. Lemma 3.1. [12, Lemma 7.1.1] Let 0 < β < 1, a ∈ L1+ and let u ∈ L1+ satisfy the inequality  t u(t) ≤ a(t) + b (t − s)β−1 u(s) ds, for a.e. t ∈ (0, T ]. 0

Then the following bound holds:  t ∞ (bΓ(β))n (t − s)nβ−1 a(s) ds, for a.e. t ∈ (0, T ]. u(t) ≤ a(t) + 0 n=1 Γ(nβ) Henry’s method is an iterative process. This result must be interpreted with care since a(t) can have singularities, this is not an L∞ bound in general. It was shown in [25] by the same iterative method as in [12] that b can be replaced by a non-negative, nondecreasing continuous function in this result. We note that the method used in Corollary 3.3 below reduces the case of a function b to the case when b is a constant. Haraux studies a similar inequality [11, Lemma 6, p.33] for the special case β = 1/2 but for L∞ + functions. The proof is by a reduction to the classical Gronwall inequality, a method attributed to Pazy [23], and the bound involves the exponential function rather than the above infinite series which is related to the Mittag-Leffler function. Medved [17] has some results also with exponential function bounds proved by different methods. A similar inequality is used to give existence of global solutions to some semilinear evolution equations in [18]. Our first result generalizes that of Haraux in two ways, firstly we replace 1/2 by β ∈ (0, 1), secondly, and importantly, we have an extra singular term in the integrand; it gives a variant of another of Henry’s results [12, Lemma 7.1.2] which is for L1 functions and the bound there is given by an infinite series. Theorem 3.2. Let a ≥ 0 and b > 0 be constants and suppose that β > 0, γ ≥ 0 and β + γ < 1. Suppose that u ∈ L∞ + [0, T ] satisfies the inequality  t u(t) ≤ a + b (t − s)−β s−γ u(s) ds, for a.e. t ∈ [0, T ]. (3.1) 0

1/(1−β−γ)  , and let We write B0 := B(1 − β, 1 − γ). For r > 0 let tr := bBr 0 r0 := bB0 T 1−β−γ so that tr ≤ T for r ≤ r0 . Then, if r ≤ r0 and also r < 1 we have   a b t−β r (3.2) exp t1−γ , for a.e. t ∈ [0, T ]. u(t) ≤ 1−r (1 − r)(1 − γ) Moreover we always have, for r1 = β/(1 − γ), u(t) ≤

  b t−β a(1 − γ) r1 exp t1−γ , for a.e. t ∈ [0, T ]. 1−β−γ (1 − β − γ) 5

(3.3)

In particular, there is an explicit constant C(b, β, γ, T ) such that u(t) ≤ aC(b, β, γ, T ) for a.e. t ∈ [0, T ]. Moreover, the exponent constant b t−β r1 /(1−β −γ) is ‘optimal’ in the sense that it is the smallest possible for admissible choices of r. t Proof. Let v(t) := a + b 0 (t − s)−β s−γ u(s) ds, then v is continuous, u(t) ≤ v(t) for a.e t ∈ [0, T ] and  t  t −β −γ v(t) = a + b (t − s) s u(s) ds ≤ a + b (t − s)−β s−γ v(s) ds, for t ∈ [0, T ]. 0

0

Clearly it suffices to prove the inequalities (3.2), (3.3) with v replacing u, that is, we may assume u itself is continuous. Choose r ≤ r0 with r < 1, hence tr ≤ T . Let t ∈ (0, T ] and consider an arbitrary τ ∈ (0, t]. If τ ≤ tr we have, using Lemma 2.1,  τ  τ −β −γ (τ − s) s u(s) ds ≤ a + b (τ − s)−β s−γ u∗ (t) ds u(τ ) ≤ a + b (3.4) 0 0 ∗ ∗ = a + bτ 1−β−γ u∗ (t)B0 ≤ a + bB0 t1−β−γ u (t) = a + ru (t). r Next, if tr ≤ τ ≤ t we have,   τ −tr (τ − s)−β bs−γ u(s) ds + b u(τ ) ≤ a + b 0

τ τ −tr

(τ − s)−β s−γ u(s) ds.

−γ ≤ (s−τ +tr )−γ Using the inequality (τ −s)−β ≤ t−β r in the first integral and s in the second integral together with Lemma 2.1 we obtain  τ −tr  τ −β −γ s u(s) ds + b (τ − s)−β (s − τ + tr )−γ u∗ (t) ds u(τ ) ≤ a + btr

≤ a + bt−β r



0

t 0

τ −tr

s−γ u∗ (s) ds + bB0 t1−β−γ u∗ (t), r

that is, u(τ ) ≤ a + bt−β r



t

s−γ u∗ (s) ds + ru∗ (t).

(3.5)

0

From (3.4) we see that (3.5) holds for t ∈ (0, T ] and all τ ∈ [0, t]. Hence, taking the supremum for τ ∈ [0, t], since r < 1 we obtain  t (1 − r)u∗ (t) ≤ a + bt−β s−γ u∗ (s) ds, r 0

that is, u∗ (t) ≤

b −β a + t 1−r 1−r r



t

s−γ u∗ (s) ds.

0

Applying the classical Gronwall inequality of Theorem 1.1 we obtain u(t) ≤ u∗ (t) ≤

  a b t−β r exp t1−γ , for a.e. t ∈ [0, T ]. 1−r (1 − r)(1 − γ) 6

(3.6)

Now we will show that the choice r = r1 := β/(1 − γ) is ‘optimal’ for the inequality (3.6) in the sense that the term b t−β r /(1 − r)(1 − γ) is as small as possible. The set of allowed r is the interval (0, r0 ]. Define a function f by  1 β/(1−β−γ) . f (r) := (1 − r)tβr = (1 − r)rβ/(1−β−γ) bB0 Differentiating f we see that f  (r) = 0 for r = β/(1 − γ) = r1 and this is a maximum of f ; and we have (1 − r1 )(1 − γ) = 1 − β − γ. Note, however, that the choice r = r1 is only valid if r1 ≤ r0 , that is bB0 T 1−β−γ ≥ β/(1 − γ). But, if bB0 T 1−β−γ < β/(1 − γ) we have, for an arbitrary τ ≤ t ≤ T ,  τ (τ − s)−β s−γ u(s) ds u(τ ) ≤ a + b 0

= a + bB0 τ 1−β−γ u∗ (t)

(3.7)

≤ a + bB0 T 1−β−γ u∗ (t) < a + β/(1 − γ)u∗ (t). a(1 − γ) for 1−β−γ all t ∈ [0, T ], that is, the conclusion also holds in this case, in fact a stronger version. Taking the supremum in (3.7) for τ ∈ [0, t] gives u(t) ≤ u∗ (t) ≤

We can easily allow a, b to be bounded functions instead of constants. In that case the result is as follows, the main interest would be for non-decreasing functions. Corollary 3.3. Let β > 0, γ ≥ 0 and β +γ < 1 and let a, b ∈ L∞ + [0, T ]. Suppose [0, T ] satisfies the inequality that u ∈ L∞ +  t u(t) ≤ a(t) + b(t) (t − s)−β s−γ u(s) ds, for a.e. t ∈ [0, T ]. (3.8) 0

Then we have u(t) ≤

  b∗ (t) t−β a∗ (t)(1 − γ) r1 exp t1−γ , for a.e. t ∈ [0, T ], 1−β−γ (1 − β − γ)

(3.9)

Proof. For any t1 ∈ (0, T ] the following inequality holds:  t ∗ ∗ (t − s)−β s−γ u(s) ds, for a.e. t ∈ [0, t1 ]. u(t) ≤ a (t1 ) + b (t1 ) 0

By Theorem 3.2 this gives the conclusion   b∗ (t ) t−β a∗ (t1 )(1 − γ) 1 r1 exp t1−γ , for a.e. t ∈ [0, t1 ]. u(t) ≤ 1−β−γ (1 − β − γ)

(3.10)

Thus (3.10) holds with u∗ (t1 ) on the left and t replaced by t1 on the right. Since t1 is arbitrary in (0, T ], this proves the result. 7

Example 3.4. Suppose that u ∈ L∞ + [0, T ] satisfies  t u(t) ≤ a + b (t − s)−1/2 u(s) ds, for a.e. t ∈ [0, T ]. 0

Then we have

u(t) ≤ 2a exp(8b2 t), for a.e. t ∈ [0, T ],

In fact, taking β = 1/2, γ = 0 in Theorem 3.2 we obtain tr1 = 1/(16b2 ), and therefore the exponent multiplying t is 2 b t−β r1 /(1 − β) = (b)(4b)/(1/2) = 8b

as claimed. Remark 3.5. Example 3.4 is the result given by Haraux in [11, Lemma 6, p.33] where the proof is attributed to Pazy [23]. Our proof of Theorem 3.2 uses the basic idea of that one but has extra technical difficulties. We deal with a more general case and in addition we show that the choice of r1 = β = 1/2 in Example 3.4 is not only simple but also has an optimal property. Remark 3.6. For the situation of Example 3.4 the inequality obtained by Henry, Diethelm and Ford, and Tisdell, gives the bound u(t) ≤ aE 12 (bt1/2 Γ( 12 )).   z Using a known property of E 12 , namely, E 12 (z) = exp(z 2 ) 1+ π2 0 exp(−s2 ) ds see for example [13, (1.8.6)], or [8, Example 4.1 (c)] (beware of the misprint), it is easily shown that   u(t) ≤ aE 12 bt1/2 Γ( 12 ) ≤ 2a exp(πb2 t), which shows that this optimal bound is indeed better than u(t) ≤ 2a exp(8b2 t) but of the same type. Remark 3.7. For γ = 0 and β ∈ (0, 1) the estimate (3.3) is of the form u(t) ≤

a exp(a1 (β)b1/(1−β) t) 1−β

(3.11)

where a1 (β) = 1/(β (β(1 − β))1/(1−β) ). For this case the bound given by Henry, Diethelm and Ford, and Tisdell is u(t) ≤ aE1−β (bΓ(1 − β)t1−β ). The following asymptotic formula is known, see for example Proposition 3.6 and Corollary 3.7 of [10], there is a constant M such that |Eα z| ≤

1 exp(z 1/α ) + M/(1 + |z|). α

(3.12)

Therefore asymptotically (t large) the Henry, Diethelm and Ford, and Tisdell bound can be essentially estimated by a exp([bΓ(1 − β)]1/(1−β) t). u(t) ≤ (3.13) 1−β 8

This indicates that our results are of the right form and are good for γ = 0, this gives us confidence that our results are similarly good also for the general case when γ = 0 when using a bound given by a series such as in Lemma 3.1 seems less clear than the estimate (3.3). For the case γ = 0 we note that the constant inside the exponent blows up as β → 1 as is to be expected. As β → 0 it gives exponent b, which is the usual Gronwall inequality. We give some numerical comparisons between the constants in the exponents that come from the estimates (3.11) and (3.13). β 1/4 1/3 1/2 2/3 3/4 7/8

(3.11) constant ≈ 2.32955 b4/3 ≈ 3.18198 b3/2 8 b2 3 243 b /4 = 60.75 b3 ≈ 606.8148 b4 ≈ 42723175 b8

(3.13) constant ≈ 1.31133 b4/3 ≈ 1.57574 b3/2 π b2 ≈ 19.22597 b3 ≈ 172.79227 b4 ≈ 10379537 b8

Note that the constants coming from (3.11) give inequalities that are valid for all t ∈ [0, T ] while the constants from (3.13) only give asymptotic inequalities, except for β = 1/2 where there is the known equality as given earlier in Remark 3.6. We now discuss the case when u may be singular. The first result is deduced from Theorem 3.2 Theorem 3.8. Let a, b ≥ 0 and c > 0 be constants. Let 0 < α, β < 1, γ ≥ 0 with α + β + γ < 1. Suppose that u(t)tα ∈ L∞ + [0, T ] and u satisfies the inequality u(t) ≤ at−α + b + c



t 0

(t − s)−β s−γ u(s) ds, for a.e. t ∈ (0, T ].

(3.14)

Then we have   c t−β 1−γ−α r2 (at−α + b) exp t1−γ , for a.e. t ∈ (0, T ], 1−α−β−γ 1−α−β−γ (3.15) β where r2 := 1−α−γ . Thus there is an explicit constant C = C(b, c, α, β, γ, T ) such that u(t) ≤ a t−α C for a.e. t ∈ (0, T ]. u(t) ≤

Proof. Write v(t) = tα u(t) so that v ∈ L∞ and v satisfies the inequality  t v(t) ≤ a + btα + ctα (t − s)−β s−γ s−α v(s) ds. 0

This is the situation of Corollary 3.3 with γ replaced by α + γ. Hence we obtain v(t) ≤

  t−β 1−γ−α r2 (a + btα ) exp ctα t1−γ−α . 1−α−β−γ 1−α−β−γ 9

Thus we have u(t) ≤

  ct−β 1−γ−α r2 (at−α + b) exp t1−γ . 1−α−β−γ 1−α−β−γ

The second singular case we now discuss seems to be new, Exercise 3 in [12] is similar but does not give information on the terms with negative powers of t. The result appears similar to Theorem 3.8 but it has weaker hypotheses and requires a different argument; the conclusion is precise but has a more complicated appearance. Theorem 3.9. Let a, b ≥ 0 and c > 0 be constants. Let 0 < α, β, γ < 1 with α + γ < 1 and β + γ < 1. Suppose that u(t)tα ∈ L∞ + [0, T ] and u satisfies  t (t − s)−β s−γ u(s) ds, for a.e. t ∈ (0, T ]. (3.16) u(t) ≤ at−α + b + c 0

Then we have, for a.e. t ∈ (0, T ], u(t) ≤ at−α + acB1 t−α+1−β−γ + ac2 B1 B2 t−α+2(1−β−γ) + . . .  c t−β    r1 + b + acm B1 B2 . . . Bm t−α+m(1−β−γ) exp t1−γ . 1−β−γ

(3.17)

where m is the smallest positive integer such that m(1 − β − γ) − α ≥ 0, r1 = β/(1 − γ), and for n ∈ N, Bn := B(1 − β, 1 − α − γ + (n − 1)(1 − β − γ)). In particular, there is an explicit constant C = C(b, c, β, γ, T ) such that u(t) ≤ at−α C for a.e. t ∈ [0, T ]. Proof. Let w0 (t) = u(t) and let M0 be such that w0 (t)tα ≤ M0 for a.e. t ∈ [0, T ]. Define w1 by  t

w1 (t) := b + c so that

0

(t − s)−β s−γ w0 (s) ds,

w0 (t) ≤ at−α + w1 (t), for a.e. t ∈ (0, T ].

(3.18)

Firstly, using Lemma 2.1 we obtain  t w1 (t) ≤ b + c (t − s)−β s−γ M0 s−α ds 0

≤ b + cM0 t1−β−α−γ B(1 − β, 1 − α − γ) = b + cM0 t1−β−α−γ B1 . Thus w1 ∈ L∞ [0, T ] if α + β + γ ≤ 1, and tα+β+γ−1 w1 (t) is an L∞ function when α + β + γ > 1. When α + β + γ ≤ 1, w1 satisfies the inequality  t w1 (t) ≤ b + c (t − s)−β s−γ (as−α + w1 (s)) ds 0 (3.19)  t 1−α−β−γ −β −γ = b + act B1 + c (t − s) s w1 (s) ds. 0

10

Now we can apply Corollary 3.3 to get    w1 (t) ≤ b + act1−α−β−γ B1 exp

 c t−β r1 t1−γ , for a.e. t ∈ (0, T ]. 1−β−γ

Thus, for the case α + β + γ ≤ 1, from (3.18) we obtain,  c t−β    r1 t1−γ , for a.e. t ∈ (0, T ]. u(t) ≤ at−α + b + act1−α−β−γ B1 exp 1−β−γ (3.20) If α + β + γ > 1 we define w2 by  t (t − s)−β s−γ w1 (s) ds, w2 (t) := b + c 0

and we have t

α+β+γ−1

w1 (t) ≤ M1 and so by Lemma 2.1 w2 (t) ≤ b + cM1 t−α+2(1−β−γ) B2 .

Thus, if 2β + 2γ + α ≤ 2 then w2 is an L∞ function, and if 2β + 2γ + α > 2 then t2β+2γ+α−2 w2 is an L∞ function. When 2β + 2γ + α ≤ 2, that is −α + 2(1 − β − γ) ≥ 0, from (3.19) we see that w1 (t) ≤ acB1 t−α+1−β−γ + w2 (t). Therefore w2 satisfies the inequality  t (t − s)−β s−γ w1 (s) ds w2 (t) ≤ b + c 0  t ≤b+c (t − s)−β (acB1 s−α+1−β−γ + w2 (s)) ds, 0  t 2 = b + ac B1 B2 t−α+2(1−β−γ) + c (t − s)−β s−γ w2 (s)) ds.

(3.21)

(3.22)

0

Since −α + 2(1 − β) ≥ 0 and w2 ∈ L∞ we can apply Corollary 3.3 to get  c t−β    r1 w2 (t) ≤ b + ac2 B1 B2 t−α+2(1−β−γ) exp t1−γ , for a.e. t ∈ [0, T ]. 1−β−γ From (3.18), (3.21) this gives, for the case −α + 2(1 − β − γ) ≥ 0, for a.e. t ∈ (0, T ],  c t−β r1 t1−γ . 1−β−γ (3.23) Note that we ‘gain’ 1 − β − γ at each step and the process can be continued for a finite number of steps until the power of t becomes non-negative. Let m be the smallest positive integer such that m(1 − β − γ) − α ≥ 0. Then we obtain    u(t) ≤ at−α +acB1 t−α+1−β−γ + b+ac2 B1 B2 t−α+2(1−β−γ) exp

u(t) ≤ at−α + acB1 t−α+1−β−γ + ac2 B1 B2 t−α+2(1−β−γ) + . . .  c t−β    r1 t1−γ . + b + acm B1 B2 . . . Bm t−α+m(1−β−γ) exp 1−β−γ 11

(3.24)

Remark 3.10. The special case α = β = 1/2 and γ = 0 is proved in [11, Lemma 10, p.112] by reduction to the previous case, but there the proof terminates at the first step. The conclusion of Theorem 3.9, but not the method, is related to Theorem 2.7 of Kong and Ding [14] which has b = 0, c = c(t) a nonnegative, nondecreasing function and an L1 type hypothesis on u, the method is an iteration process as in lemma 7.1.2 of Henry [12] and the bound is a series, they give a complete proof; this extends Exercise 3 of [12] where c is a constant. 4. Volterra integral equations and fractional differential equations The existence and uniqueness of solutions of boundary value problems for fractional order differential equations of various types have been considered by very many authors in recent years. We only reference papers that are closely related to our work here. We study an initial value problem for a fractional order differential equation which is related to a Volterra integral equation. Volterra integral equation with convolution kernels have been well studied. They arise in many applications such as the modelling of problems in nonlinear diffusion and percolation theory, for example [22]; the physically interesting solutions are often the positive ones. The most studied fractional differential equations involve either RiemannLiouville or Caputo fractional derivatives, whose definitions are recalled below. The Caputo derivative is often preferred in applications since a well posed problem for a Caputo type fractional differential equation involves the initial values of the function and its integer order derivatives, which have a clear physical meaning, for example the initial value problem (IVP) of a Caputo fractional α u(t) = f (t, u(t)) with u(0) = u0 differential equation of order α ∈ (0, 1) is DC and is posed for functions (at least) continuous. The corresponding RiemannLiouville problem should prescribe the initial condition as limt→0+ t1−α u(t) = u0 and be posed for functions with singularities at 0, see for example the discussion in [1]. We first recall some definitions of fractional integrals and derivatives. We base the discussion on the text by Diethelm [8]. Since we always work on an interval [0, T ] we give the definitions for that case and employ a simpler notation than is often used. Definition 4.1. The Riemann-Liouville (R-L) fractional integral of order α ∈ (0, 1) of a function f ∈ L1 [0, T ] is defined to be  t 1 α I (f )(t) := (t − s)α−1 f (s) ds. Γ(α) 0 This integral is a convolution of L1 functions so I α f is defined as an L1 function, so I α (u)(t) is finite for a.e. t. If α = 1 it is the usual integration operator which we denote I. These fractional integral operators satisfy a semigroup property as follows. 12

Lemma 4.2. [8, Theorem 2.2] Let α, β > 0 and f ∈ L1 [0, T ]. Then I α I β (f )(t) = I α+β (f )(t) for a.e. t. Let D denote the usual differentiation operator. The R-L fractional derivative of order α ∈ (0, 1) is defined as follows. Definition 4.3. For α ∈ (0, 1) the Riemann-Liouville fractional derivative D α u of u where I 1−α (u) is absolutely continuous (AC) is Dα u(t) := D I 1−α (u)(t), a.e. t. When I 1−α (u) is absolutely continuous the R-L derivative Dα u(t) exists a.e.. We believe this condition is necessary, and should be stated explicitly, when discussing equations involving the R-L fractional derivative via solutions of the related integral equation, see the arguments in Theorem 4.6. Using lemma 4.2 shows that the R-L derivative Dα is the left inverse of I α . Lemma 4.4. [8, Theorem 2.14] Let 0 < α < 1. Then, for every f ∈ L1 , Dα I α (f )(t) = f (t) for almost every t. The Caputo fractional derivative is defined with the derivative and fractional integral taken in the reverse order to that of the R-L derivative. α (u) of u ∈ Definition 4.5. For α ∈ (0, 1) the Caputo fractional derivative DC AC is defined for a.e. t by α u(t) := I 1−α (Du)(t). DC

For u ∈ AC, Du ∈ L1 and so I 1−α (Du) is defined as an L1 function. There is a connection between the R-L and Caputo derivatives for functions with some regularity, for example it can be shown that for absolutely continuous functions u α u(t), for a.e. t. (4.1) Dα (u(t) − u(0)) = DC A proof can be given in a few lines, it is a special case of the result given for fractional derivatives of all orders in [8, Theorem 3.1]. Equation (4.1) is often used as the definition of Caputo derivative, for example in [8], that is define the modified Caputo derivative by D∗α u(t) = Dα (u(t) − u(0)). We do not α of have an equivalence between a Caputo fractional differential equation DC order α ∈ (0, 1) and an integral equation, but we have an equivalence using the definition D∗α of Caputo fractional derivative see Theorem 4.6 below. Solutions for a Caputo fractional initial value problem of order α ∈ (0, 1), α u(t) = f (t, u(t)) with u(0) = u0 , are often studied via fixed points in the DC space C[0, T ] of the Volterra integral equation  t 1 (t − s)α−1 f (s, u(s)) ds u(t) = u0 + Γ(α) 0 and it has been assumed previously that f is continuous, see for example [3]. α However, when u ∈ AC, DC u(t) can have singularities at t = 0, for example, for 13

α u(t) := tρ we have DC u(t) = tρ−α Γ(1 + ρ)/Γ(1 + ρ − α) which is an L1 function when ρ > α − 1 but can be singular at t = 0; it is continuous if ρ ≥ α. Therefore it makes sense to allow the Caputo derivative to be singular at t = 0. In the following result we prove an equivalence between these two problems. The case when there is no singular term, s−γ , is essentially well known, for example, Diethelm [8, Lemma 6.2].

Theorem 4.6. Let f be continuous on [0, T ] × R, let 0 < α < 1 and let 0 ≤ γ < α. If a function u ∈ AC satisfies the Caputo fractional initial value problem α DC u(t) = t−γ f (t, u(t)), for a.e. t ∈ (0, T ], u(0) = u0 ,

then u satisfies the Volterra integral equation  t 1 u(t) = u0 + (t − s)α−1 s−γ f (s, u(s)) ds, t ∈ [0, T ]. Γ(α) 0

(4.2)

(4.3)

Secondly, if u ∈ C[0, T ] satisfies (4.3) then I α−1 (u − u0 ) ∈ AC and u satisfies D∗α (u)(t) = t−γ f (t, u(t)), for a.e. t ∈ (0, T ], u(0) = u0 .

(4.4)

Thirdly, if u ∈ C[0, T ] and I α−1 (u − u0 ) ∈ AC, and u satisfies (4.4), then u satisfies (4.3). α Proof. Suppose that u ∈ AC and DC u(t) = t−γ f (t, u(t)), for a.e. t ∈ (0, T ]. Let −γ g(t) := t f (t, u(t)), t ∈ (0, T ], and note that g ∈ L1 since γ ≤ α < 1. Then, by the definition of Caputo derivative, we have I 1−α (Du) = g where Du ∈ L1 since u ∈ AC. This yields I α I 1−α (Du) = I α g, hence, by Lemma 4.2, I(Du) = I α g, that is u(t) − u(0) = I α g since u ∈ AC. Secondly, suppose that u is continuous and u(t) − u0 = I α g. Since we have t (t − s)α−1 s−γ ds = tα−γ B(α, 1 − γ) with α > γ, and f (s, u(s)) is bounded, 0 taking the limit as t → 0+ in (4.3) gives u(0) = u0 . Since g ∈ L1 by Lemma 4.4 we may apply the R-L derivative Dα to obtain Dα (u(t) − u(0)) = g(t) for a.e. t, that is (4.4). Moreover, I 1−α (u − u0 ) = I 1−α I α g = Ig ∈ AC since g ∈ L1 . Thirdly, suppose that u ∈ C[0, T ] satisfies I 1−α (u − u0 ) ∈ AC and (4.4). Then we have

DI 1−α (u − u0 ) = g(t), a.e. where g(t) := t−γ f (t, u(t)), thus g ∈ L1 . We assert that I 1−α (u − u0 )(0) = 0. In fact, for t > 0, we have  t 1 1−α (u − u0 )(t) = (t − s)−α (u(s) − u0 ) ds, I Γ(1 − α) 0 where u − u0 ∈ C[0, T ] so is bounded, say |u(s) − u0 | ≤ M . Thus  t M 1−α (u − u0 )(t)| ≤ (t − s)−α ds |I Γ(1 − α) 0 M t1−α → 0 as t → 0+, = Γ(2 − α) 14

(4.5)

hence by continuity I 1−α (u − u0 )(0) = 0. Moreover, since I α−1 (u − u0 ) ∈ AC we can integrate (4.5) to give I 1−α (u − u0 )(t) = I 1−α (u − u0 )(0) + Ig(t) = Ig(t). α

α

(4.6)

α

Now we apply I to both sides to get I(u − u0 ) = I Ig = I I g and by differentiation we deduce that u − u0 = I α g. Remark 4.7. It has often been asserted, imprecisely, that (4.2) is equivalent to (4.3) but it seems that solutions u of (4.3) have never been shown to be absolutely continuous when f is at best continuous. Some of these issues are discussed in the paper [16] where some positive results for boundary value problems β for 1 < β < 2 are obtained under involving the Caputo fractional derivative DC a Lipschitz condition on f . It is important to have I 1−α (u − u0 ) ∈ AC in the third part of Theorem 4.6 otherwise the integration in (4.6) is not valid. When the term t−γ is absent and (4.4) is satisfied for t ∈ [0, T ] then D(I 1−α (u − u0 )) is continuous so I 1−α (u − u0 ) ∈ C 1 and therefore is AC. It is quite natural to assume I 1−α (u − u0 ) ∈ AC since for existence of D∗α (u)(t) it is required that I 1−α (u − u0 ) be differentiable a.e., but also necessary for the equivalence as we have shown. We will now discuss solutions of the Volterra integral equation (4.3) which can be called a mild solution of the Caputo fractional equation (4.2) or a solution of the modified Caputo fractional equation (4.4). As far as we are aware the case γ = 0 is new. We will employ fixed point index theory (details may be found in many texts, for example Deimling [6]) to prove a result for the existence of a non-negative solution of the integral equation  t 1 (t − s)α−1 s−γ f (s, u(s)) ds u(t) = u0 + Γ(α) 0 when 0 ≤ γ < α and f ≥ 0 is continuous. The hypothesis imposed on the nonlinear term f for the existence result concerns only the behaviour for u large. Under a stronger Lipschitz condition we also obtain continuous dependence on the initial data and a uniqueness result. The proof uses an equicontinuity result which requires extra care because of the singular term s−γ . Let P := {u ∈ C[0, T ] : u(t) ≥ 0, t ∈ [0, T ]} be the usual cone of nonnegative continuous functions. Theorem 4.8. Let f : [0, T ] × [0, ∞) → [0, ∞) be continuous, let 0 < α < 1, 0 ≤ γ < α, and let u0 ∈ P . Suppose there is a constant M > 0 such that f (t, u) ≤ M (1 + u) for all t ∈ [0, T ] and u ≥ 0. Then the integral operator  t 1 N u(t) := u0 + (t − s)α−1 s−γ f (s, u(s)) ds Γ(α) 0 has a fixed point in P . If, in addition, there exists L > 0 such that f satisfies the Lipschitz condition |f (t, u) − f (t, v)| ≤ L|u − v|, for all t ∈ [0, T ] and all u, v ≥ 0, 15

then fixed points depend continuously on the initial data and in particular, for a given u0 there is a unique fixed point. Proof. We will show that N : P → P and is completely continuous, that is N is continuous and maps bounded subsets of P into relatively compact subsets of P . Supposing temporarily that this has been shown we will show that there is a bounded open set U containing 0 such that N u = λu for all u ∈ ∂U ∩ P and all λ ≥ 1. This will prove the fixed point index iP (N, U ∩ P ) = 1 which proves that N has a fixed point in U . We begin with finding a suitable open set U . In fact, if there exists λ ≥ 1 and u = 0 such that λu(t) = N u(t) then u(t) ≤ λu(t) = u0 +

1 Γ(α)



t 0

1 M ≤ u0 + Γ(α)

(t − s)α−1 s−γ f (s, u(s)) ds



t 0

(t − s)α−1 s−γ (1 + u(s)) ds

1 1 ≤ u0 + M B(α, 1 − γ)T α−γ + M Γ(α) Γ(α)



t 0

(t − s)α−1 s−γ u(s) ds.

Since 1 − α + γ < 1, by Theorem 3.2 and continuity of u, there is a constant C(u0 , α, γ, M, T ), independent of u, such that u(t) ≤ C for all t ∈ [0, T ]. Choose R > C, and let UR , be the open ball of radius R centred at 0. Thus we have shown that N u = λu for all u ∈ ∂UR ∩ P and all λ ≥ 1. We turn to proving N is compact. This will be achieved by showing that a theorem applies. N (U R ) is bounded and equicontinuous so the Ascoli-Arzel` Firstly, we show N (U R ) is bounded. We have u ≤ R so u(t) ≤ R for all t ∈ [0, T ]. As f is uniformly continuous on [0, T ] × [0, R], there exists MR < ∞ such that f (t, u) ≤ MR for all t ∈ [0, T ], u ∈ [0, R]. Thus for u ∈ U R we have 1 N u(t) = u0 + Γ(α)



t 0

(t − s)α−1 s−γ f (s, u(s)) ds

 t 1 ≤ u0 + MR (t − s)α−1 s−γ ds Γ(α) 0 1 1 MR tα−γ B(α, 1 − γ) ≤ u0 + MR T α−γ B(α, 1 − γ). = u0 + Γ(α) Γ(α)

This proves that the set N (U R ) is bounded. Secondly we prove that N (U R ) is an equicontinuous family, that is, for ε > 0, there exists δ = δ(ε) > 0 such that |t1 − t2 | < δ implies |N u(t1 ) − N u(t2 )| < ε for all u ∈ U R . Without loss of generality suppose that 0 ≤ t2 < t1 ≤ T . Then we have f (t, u(t)) ≤ MR for all

16

u ∈ U R and all t ∈ [0, T ], and



Γ(α)|N u(t1 ) − N u(t2 )| =

t1

(t1 − s)α−1 s−γ f (s, u(s)) ds  t2



(t2 − s)α−1 s−γ f (s, u(s)) ds

−

 ≤

0

t2

0

|(t1 − s)α−1 − (t2 − s)α−1 |s−γ MR ds 0  t1 + (t1 − s)α−1 s−γ MR ds. t2

1

Since the second term has an L integrand, that term is < ε/3 for |t1 − t2 | < δ1 . η t For 0 < η < t2 , we write the first term as 0 + η 2 . By integrability we can fix η η so small that 0 |(t1 − s)α−1 − (t2 − s)α−1 |s−γ MR ds < ε/3. The final part is estimated by the following calculations.  t2 |(t1 − s)α−1 − (t2 − s)α−1 |s−γ MR ds η

≤



t2 η −γ

|(t1 − s)α−1 − (t2 − s)α−1 |η −γ MR ds

η

MR |(t1 − η)α − (t1 − t2 )α − (t2 − η)α | α η −γ MR 2|t1 − t2 |α . ≤ α This term is then < ε/3 for |t1 − t2 | < δ2 and the equicontinuity is shown. The proof of equicontinuity shows that each N u is continuous so N (U R ) ⊂ P . We now show that N is continuous, that is un → u (that is un (s) → u(s) uniformly in s) implies N un → N u. Since {un } is bounded, say un ≤ R and f is uniformly continuous on [0, T ] × [0, R], for ε > 0 there exists n0 = n0 (ε) ∈ N such that |f (s, un (s)) − f (s, u(s))| < ε for all n > n0 and all s ∈ [0, T ]. Then we have, for every t ∈ [0, T ],  t 1 |N un (t) − N u(t)| ≤ (t − s)α−1 s−γ |f (s, un (s)) − f (s, u(s))| ds Γ(α) 0  t ε < (t − s)α−1 s−γ ds, for n > n0 , Γ(α) 0 ε T α−γ B(α, 1 − γ), for n > n0 . ≤ Γ(α) =

This proves that N un → N u. This completes the proof for existence of a solution. Now suppose that f satisfies the Lipschitz condition and let u, v be fixed points with u(0) = u0 , v(0) = v0 . Then we have  t   1 u(t) − v(t) = u0 − v0 + (t − s)α−1 s−γ f (s, u(s)) − f (s, v(s) ds. Γ(α) 0 17

This gives |u(t) − v(t)| ≤ |u0 − v0 | +

1 Γ(α)

By Theorem 3.2 we obtain, with r3 = |u(t) − v(t)| ≤



t 0

(t − s)α−1 s−γ L|u(s) − v(s)| ds.

1−α 1−γ ,

  L tα−1 |u0 − v0 |(1 − γ) r3 exp t1−γ , for t ∈ [0, T ]. α−γ Γ(α)(α − γ)

This proves the continuous dependence on initial data and taking u0 = v0 proves uniqueness. Remark 4.9. By some similar arguments, without imposing a sign restriction, replacing P with C[0, T ] and using degree theory, with practically the same proof a result for existence (plus continuous dependence on initial data, and uniqueness) follows in the same way if it is supposed that |f (t, u)| ≤ M (1 + |u|) for t ∈ [0, T ] and u ∈ R (plus a Lipschitz condition). Remark 4.10. A local existence theorem for fractional equations in the special case γ = 0 is given in [8, Theorem 6.1] when f is continuous. A global existence result is proved in [8, Corollary 6.3] when it is assumed that γ = 0 and f is continuous and there exist constants c1 > 0, c2 > 0, 0 ≤ μ < 1 such that f (t, u) ≤ c1 + c2 |u|μ but that result does not allow μ = 1. Since, for 0 < μ < 1, |u|μ ≤ 1 + |u| our result includes that one and covers the case μ = 1. Under a Lipschitz condition [8, Theorem 6.8] proves an existence and uniqueness result by a very different argument. Tisdell [24] studies the existence of solutions to the higher order initial value problem. In one case he uses an approximate solution approach using the Gronwall inequality to get a priori bounds, in another he uses the contraction mapping principle with a weighted norm with weight determined by a Mittag-Leffler function, and in another he uses Leray-Schauder degree. Kosmatov [15] studies the solvability of integral equations associated with α u(t) = f (t, Dβ u(t)) when the nonlinear term f is the initial value problems DC assumed to be continuously differentiable and depends on the fractional derivative of lower order. The study of that case was continued in [7]. The Volterra integral equation  t (t − s)α−1 f (u(s)) ds (4.7) u(t) = 0

for f continuous has been well studied, for some recent contributions see [3] and references therein. When f (0) = 0, u = 0 is a solution but necessary and sufficient conditions have been given in [20] for nontrivial solutions to also exist. There can be finite time blow up, see [21]. Note that the equation we study can be written  t 1 (t − s)α−1 s−γ fˆ(s, u(s)) ds u(t) = Γ(α) 0 18

where fˆ(s, u) = f (s, u) +

u0 sα−γ . Γ(1 − α)

The paper [5] discusses existence and non-existence for (4.7) and points out that, for α > 0 and 0 < p < 1 and with the non-Lipschitz nonlinearity f (u) = up , (4.7) has the positive solution u(t) = tα/(1−p) [B(α, 1 + αp/(1 − p))]1/(1−p) which exhibits non-uniqueness explicitly. For the R-L derivative case, it is shown in [1, Theorem 6.2] that if u(t) and h(t, u(t)) are continuous on (0, T ] and belong to L1 [0, 1], if limt→0+ t1−α u(t) exists then u satisfies the initial value problem, with 0 < α < 1, D α u(t) = h(t, u(t)), t ∈ (0, T ], lim t1−α u(t) = u0 , t→0+

if and only if it satisfies the Volterra integral equation  t 1 u(t) = u0 tα−1 + (t − s)α−1 h(s, u(s)) ds, t ∈ (0, T ]. Γ(α) 0 The appropriate space to work in is the following weighted space. Let w(t) := t1−α and let Xw denote the space of continuous functions u : (0, T ] → R for which u w := sup0
· w is a Banach space. The IVP Dα u(t) = h(t, u(t)), lim t1−α u(t) = u0 , t→0+

for the R-L derivative can be studied as fixed points of the operator  t 1 (t − s)α−1 h(s, u(s)) ds, N u(t) := u0 tα−1 + Γ(α) 0 in the space Xw under suitable conditions on h. We give one result. Theorem 4.11. Let f : [0, T ] × [0, ∞) → [0, ∞) be continuous, let 0 < α < 1 and 0 ≤ γ < α and let h(t, u) := t−γ f (t, u) for t ∈ (0, T ]. Suppose u0 ≥ 0 and there is a constant M > 0 such that f (t, u) ≤ M (1 + u) for all t ∈ [0, T ] and u ≥ 0. Then the integral operator  t 1 (t − s)α−1 h(s, u(s)) ds, N u(t) := u0 tα−1 + Γ(α) 0 has a non-negative fixed point in Xw . If, in addition, f satisfies the Lipschitz condition |f (t, u) − f (t, v)| ≤ L|u − v| for some L > 0 for all t ∈ [0, T ] and all u, v ≥ 0, then the fixed point is unique. Moreover, for ‘initial conditions’ u0 , v 0 the solutions satisfy |u(t) − v(t)|t1−α ≤ |u0 − v 0 |C(α, γ, L, T ), for 0 < t ≤ T, where C can be determined explicitly from (3.17). 19

Proof. The proof is similar to that of Theorem 4.8, one step is to prove t1−α N u(t) is an equicontinuous family when u is in a bounded subset of Xw . The methods are similar to previously with some extra care with the terms involving t1−α . Therefore we only show that there is a bounded set U in Xw such that N u = λu for all u ∈ ∂U . If there exists λ ≥ 1 and u = 0 such that λu(t) = N u(t) then  t 1 (t − s)α−1 s−γ f (s, u(s)) ds Γ(α) 0  t 1 0 α−1 ≤u t M + (t − s)α−1 s−γ (1 + u(s)) ds Γ(α) 0  t 1 1 0 α−1 α−γ ≤u t M B(α, 1 − γ)T M + + (t − s)α−1 s−γ u(s) ds. Γ(α) Γ(α) 0

u(t) ≤ λu(t) = u0 tα−1 +

By Theorem 3.9 there is a constant C(u0 , M, α, γ, T ) such that u(t) ≤ tα−1 C for t ∈ (0, T ]. This proves u w is uniformly bounded. Taking UR the ball in Xw with R > C we have shown that N u = λu for all u ∈ ∂UR . By applying fixed point index theory we obtain a solution. Uniqueness and ‘continuous dependence’ follow as previously. In [4, Theorem 3.1] the authors prove a local existence theorem under weaker conditions on h. The question of whether local solutions can be continued is discussed in [2]. Our result is for global existence without going via local existence. Acknowledgement I thank the referee for pointing out some useful references and for valuable comments which led to improvements of the paper. References [1] L.C. Becker, T.A. Burton and I.K. Purnaras, Complementary equations: A fractional differential equation and a Volterra integral equation, Electron. J. Qual. Theory Differ. Equ. 12 (2015), 1–24. [2] L. C. Becker, T. A. Burton and I. K. Purnaras, An inversion of a fractional differential equation and fixed points, Nonlinear Dyn. Syst. Theory, 15 (2015), 242–271. [3] L. C. Becker, T. A. Burton and I. K. Purnaras, Integral and fractional equations, positive solutions, and Schaefer’s fixed point theorem. Opuscula Math. 36 (2016), 431–458. [4] L.C. Becker, T.A. Burton and I.K. Purnaras, Existence of solutions of nonlinear fractional differential equations of Riemann-Liouville type, J. Fract. Calc. Appl., 7(2) (2016), 20–39. 20

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