# A Generalized Henry-Type Integral Inequality and Application to Dependence on Orders and Known Functions for a Fractional Differential Equation

Authors: Jun Zhou
Journal: Journal of Applied Mathematics
 (2014)

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## Abstract

We discuss on integrable solutions for a generalized Henry-type integral inequality in which weak singularity and delays are involved. Not requiring continuity or differentiability for some given functions, we use a modified iteration argument to give an estimate of the unknown function in terms of the multiple Mittag-Leffler function. We apply the result to give
continuous dependence of solutions on initial data, derivative orders, and known functions for a fractional differential equation.


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## Body

## 1. Introduction

Since Gronwall [1] and Bellman [2] discussed the integral inequalities
(1)0≤u(t)≤∫αt‍(a+bu(s))ds,(2)u(t)≤a+∫αt‍f(s)u(s)ds,
respectively, there have been made many generalizations, some of which were applied to existence, uniqueness, boundedness, and stability of solutions and invariant manifolds for differential equations and integral equations. In 1956 Bihari [3] discussed a nonlinear version of the integral inequality
(3)u(t)≤a+∫0t‍f(s)w(u(s))ds,t≥0,
where the given function w is continuous, nondecreasing, and positive definite (i.e., w(u)≥0 for all u and w(u)=0 if and only if u=0) on [0,∞). In 2000 this result was generalized by Lipovan [4] to the delay case
(4)u(t)≤a+∫t0t‍f(s)w(u(s))ds+∫b(t0)b(t)‍g(s)w(u(s))ds,t0≤t<t1,
where b is a continuously differentiable and nondecreasing function from [t0,t1) to [t0,t1) such that b(t)≤t. In 2005 Agarwal et al. [5] investigated the integral inequality of a finite sum
(5)u(t)≤a(t)+∑i=1n‍∫bi(t0)bi(t)‍fi(t,s)wi(u(s))ds,t0≤t<t1,
where the monotonicity of a is not required but wi+1 has stronger monotonicity than wi for each i. In recent years some new generalizations were given in, for example, [6–8]. Many results on integral inequalities can be found in Pachpatte’s monograph (see [9]).Among various types of integral inequalities, integral inequalities with weakly singular kernels are important tools in the discussion of reaction-diffusion equations and fractional differential equations. As shown in [10], the integral ∫t0tK(t,s) is singular on the line s=t; it is referred to beweakly singularif it is singular and ∫t0t|f(t,s)|ds<∞ for all t0≤t<T≤∞. In fractional differential equations we need to consider the following Riemann-Liouville derivative operator and integral operator (see [11])
(6)Dt0αf(x)∶=1Γ(1-α)ddx×∫t0x‍(x-t)-αf(t)dt,0<α<1,It0βf(x):=1Γ(β)∫t0x‍(x-t)β-1f(t)dt,β>0,
in which the singular kernels (x-t)-α and (x-t)β-1 are included, respectively. Another example can be seen from the Cauchy problem of the evolution equation
(7)x˙+A(t)x=f(t),x(t0)=x0,
in a Banach space X, where A is a sectorial operator and f is locally Hölder continuous with Hölder index 0<θ<1. From the variation-constant formula (see [12]), we encounter the following:
(8)∥Ae-A(t-s)∥=O((t-s)-1),∥f(t)-f(s)∥=O((t-s)θ),
and therefore the singular kernel (t-s)θ-1 is the estimation of its solution. For this reason the integral inequality with a weakly singular kernel
(9)u(t)≤a(t)+b∫0t‍(t-s)β-1u(s)ds,t≥0,
where b≥0, 0<β<1 and both a and u are nonnegative and locally integrable, was considered in Henry’s book [12] in 1981 and the estimate
(10)u(t)≤a(t)+∫0t{∑n=1∞(bΓ(β))nΓ(nβ)(t-s)nβ-1a(s)}ds,t≥0,
where Γ is the Gamma function, was given by an iteration approach. Following Henry’s idea, in 1994 Sano and Kunimatsu [13] extended Henry’s result to a more general integral inequality
(11)u(t)≤c1+c2tβ-1+c3∫0t‍u(s)ds+c4∫0t‍(t-s)β-1u(s)ds,t≥0,
where c1,c2,c3,c4≥0, 0<β<1. Inequality (9) was also extended by Ye et al. [14] in 2007 by replacing the constant b with a nonnegative, nondecreasing, bounded, and continuous function g(t).Another idea for inequalities with a weakly singular kernel was introduced by Medved [15] in 1997 for the following Henry-Bihari type integral inequality:
(12)u(t)≤a(t)+∫0t‍(t-s)β-1F(s)w(u(s))ds,t≥0,
where 0<β<1. This inequality is of Bihari’s form (3) with a weakly singular kernel. He applied the well-known Hölder inequality to separate the unknown u from the singular kernel, that is,
(13)u(t)≤a(t)+∫0t‍(t-s)β-1esF(s)e-sw(u(s))ds≤a(t)+(∫0t‍(t-s)p(β-1)epsds)1/p×(∫0t‍F(s)qe-qsw(u(s))qds)1/q,
where p,q>1 are certain constants such that 1/p+1/q=1, so that the inserted exponential factor es makes the singular integral ∫0t‍(t-s)p(β-1)epsds convergent as s→t and the inequality is reduced to the classic Bihari’s form (3). In 2002 Ma and Yang [16] improved Medved’s method and gave an estimation to the Volterra-type integral inequality with a more general form of weakly singular kernel
(14)u(t)≤a(t)+b(t)∫0t‍(tσ-sσ)μ-1sτ-1g(s)u(s)ds+c(t)∫0t‍(tα-sα)β-1sγ-1f(s)w(u(s))ds,kt≥0,
where α,β,γ,σ,μ,τ≥0. Recently Ma and Pečarić [17] also employed the separation approach to discuss another weakly singular integral inequality
(15)up(t)≤a(t)+b(t)∫0t‍(tα-sα)β-1sγ-1f(s)uq(s)ds,mmmmmmmmmmmmmmmmmmmmmmmmt≥0,
under the condition p≥q≥0.In this paper we investigate the following integral inequality of finite sum:(16)u(t)≤a(t)+∑i=1m‍∫bi(t0)bi(t)‍(t-fi(s))βihi(t,s)u(s)ds,mmmmmmmmmmmmmmmmmmmmmmit≥0,
in which weakly singular kernels and delays are involved. Not requiring continuity of a or differentiability of bi, we give an estimate for locally integrable u in terms of the multiple Mittag-Leffler function (see [18]). We prefer Henry’s iteration approach in our proof because the approach does not reduce the problem to the classic one so that no more continuity and differentiability are required. Our result generalizes the works made in [14–17] in some sense because of the more general form (16). Finally, we apply the result to give continuous dependence of solutions on initial data, derivative orders, and known functions for a fractional differential equation.


## 2. Main Results

For constantst0,  T such that t0<T≤+∞, consider inequality (16), where a,bi,fi and hi are given nonnegative functions and satisfy the following hypotheses:(H1)
a is a locally integrable function on [t0,T), that is, Lebesgue integrable on every compact subset of [t0,T), and the integrations in (16) are bounded by replacing u with a in (16);(H2)
everyhi,  i=1,…,m, is continuous on [t0,T)×[t0,T);(H3)
everybi:[t0,T)→[t0,T), i=1,…,m, is continuous and strictly increasing such that bi(t)≤t for all t∈[t0,T);(H4)
everyfi, i=1,…,m, is continuously differentiable on [t0,T) such that fi′(t)>0 and t≤fi(t)≤bi-1(t) for all t∈[t0,T).Theorem 1.
Suppose that (H1)–(H4) hold and βi>-1, i=1,…,m. Then, every nonnegative and locally integrable function u(t) which satisfies (16) on [t0,T) and the integrations in (16) bounded has the estimate
(17)u(t)≤a(t)+∑n=1∞‍{∑1≤k1,…,kn≤m∏i=1n‍Γ(βki+1)h~ki(t)Γ(∑i=1n‍βki+n)·∫bk1(t0)bk1(t)‍(t-fk1(s))∑i=1n‍βki+n-1×fk1′(s)a(s)ds∑1≤k1,…,kn≤m∏i=1n‍Γ(βki+1)h~ki(t)Γ(∑i=1n‍βki+n)},t∈[t0,T),
where h~i(t)=max⁡(s,τ)∈[t0,t]×[t0,t]hi(s,τ)/fi′(τ), i=1,…,m.We leave the proof of this theorem to next section. In what follows, we express the estimate of series form in terms of the multiple Mittag-Leffler function (see [18]).The original Mittag-Leffler function(18)Eα(z)∶=∑k=0∞‍zkΓ(kα+1),z∈C,α>0,
was proposed as an extension of the exponential function by Mittag-Leffler ([19]) in 1903. An extension of two parameters
(19)Eα,β(z)∶=∑k=0∞‍zkΓ(kα+β),α,β,z∈C,R(α),R(β)>0,
was proposed by Wiman [20] in 1905. Later, two Mittag-Leffler functions with three parameters were given separately by Prabhakar [21] and Kilbas and Saigo [22]. In 1996 Hadid and Luchko [18] generalized the function into the multiple form
(20)E(α1,…,αn),β(z1,…,zn)∶=∑k=0∞∑l1+⋯+ln=kl1,…,ln≥0‍(k;l1,…,ln)∏i=1n‍ziliΓ(∑i=1n‍liαi+β),
where αi,β,zi∈C,R(αi), R(β)>0,  i=1,…,n and
(21)(k;l1,…,ln)∶=k!l1!×⋯×ln!.
These generalized Mittag-Leffler functions have been treated as significant special functions since they played an important role in computing fractional calculus and solving fractional differential and integral equations modeled in physics, chemistry, biology, engineering, and applied sciences (see monographs [11, 23]). We have the following inequality:
(22)E(α1,…,αn),1(z1,…,zn)≤∏i=1n‍Eαi(zi),mmmmmiαi,zi∈R+,i=1,…,n.
In fact, we can show that
(23)Γ(∑i=1m‍ai+1)≥∏i=1m‍Γ(ai+1),ai>0,i=1,…,m.
By Euler’s definition on Gamma function (see [24]), we have
(24)Γ(ai+1)=1ai+1∏n=1∞‍(1+(1/n))ai+11+((ai+1)/n),Γ(∑i=1m‍ai+1)=1∑i=1m‍ai+1∏n=1∞‍(1+(1/n))∑i=1m‍ai+11+(∑i=1m‍(ai+1)/n).
It follows that
(25)Γ(∑i=1m‍ai+1)∏i=1m‍Γ(ai+1)=∏i=1m‍(ai+1)∑i=1m‍ai+1∏n=1∞‍∏i=1m‍(1+((ai+1)/n))(1+(1/n))m-1(1+(∑i=1m‍(ai+1)/n)).
It is easy to know that ∏i=1m‍(ai+1)≥∑i=1m‍ai+1. Then, we can prove by induction that
(26)∏i=1m‍(1+ai+1n)≥(1+1n)m-1(1+∑i=1m‍ai+1n).
Obviously, (26) is true for m=1. Assume that (26) is true for m=l. We get
(27)∏i=1l+1‍(1+ai+1n)=(1+1n)l-1(1+∑i=1l‍ai+1n)×(1+al+1+1n)≥(1+1n)l(1+∑i=1l+1‍ai+1n),
since
(28)(1+∑i=1l‍ai+1n)(1+al+1+1n)≥(1+1n)(1+∑i=1l+1‍ai+1n),
implying that (26) holds for m=l+1. Thus, from (26) we see that (23) holds. By (23), we have
(29)E(α1,…,αn),1(z1,…,zn)=∑m=0∞∑1≤k1,…,km≤n‍∏i=1m‍zkiΓ(∑i=1m‍αki+1)≤∑k1=0∞‍⋯∑kn=0∞‍∏i=1n‍ziki∏i=1n‍Γ(kiαi+1)=∏i=1n‍Eαi(zi).
Therefore, (22) is proved.Corollary 2.
Suppose that the hypotheses of Theorem1 hold and additionally that a(t) is continuous on [t0,T). Then,
(30)u(t)≤a~(t)E(β1+1,…,βm+1),1×(h~1(t)Γ(β1+1)(t-t0)β1+1,…,h~m(t)Γ(βm+1)(t-t0)βm+1)≤a~(t)∏i=1m‍Eβi+1(h~i(t)Γ(βi+1)(t-t0)βi+1),
where a~(t)=max⁡t0≤τ≤ta(τ) and Eα(z) and E(α1,…,αn),1 are defined in (18) and (20).Proof.
Starting from (17), we have
(31)u(t)≤a~(t){1+∑n=1∞{∑1≤k1,…,kn≤m‍∏i=1n‍Γ(βki+1)h~ki(t)Γ(∑i=1n‍βki+n)×∫bk1(t0)bk1(t)‍(t-fk1(s))∑i=1n‍βki+n-1×fk1′(s)ds∑1≤k1,…,kn≤m‍∏i=1n‍Γ(βki+1)h~ki(t)Γ(∑i=1n‍βki+n)}}≤a~(t){1+∑n=1∞‍(∑1≤k1,…,kn≤m‍∏i=1n‍Γ(βki+1)h~ki(t)Γ(∑i=1n‍βki+n+1)×(t-t0)∑i=1n‍(βki+1)∑1≤k1,…,kn≤m‍∏i=1n‍Γ(βki+1)h~ki(t)Γ(∑i=1n‍βki+n+1))}=a~(t)∑n=0∞∑1≤k1,…,kn≤m(∏i=1n‍Γ(βki+1)h~ki(t)×(t-t0)βki+1∏i=1n‍)×(Γ(∑i=1n‍βki+n+1))-1=a~(t)×∑k=0∞∑k1+⋯+km=kk1,…,km≥0(∏i=1m‍(k;k1,…,km)‍ccccccccccccccccc×∏i=1m((t-t0)βi+1Γ(βi+1)h~i(t)‍ccccccccccccccccc×(t-t0)βi+1)ki∏i=1m(Γ(βi+1)h~i(t)‍)×(Γ(∑i=1m‍ki(βi+1)+1))-1=a~(t)E(β1+1,…,βm+1),1×(h~1(t)Γ(β1+1)(t-t0)β1+1,…,h~m(t)Γ(βm+1)(t-t0)βm+1).
It follows from (22) that
(32)a~(t)E(β1+1,…,βm+1),1×(h~1(t)Γ(β1+1)(t-t0)β1+1,…,h~m(t)Γ(βm+1)(t-t0)βm+1)≤a~(t)∏i=1m‍Eβi+1(h~i(t)Γ(βi+1)(t-t0)βi+1).
The corollary is proved.


## 3. Proof of Theorem

LetLloc⁡,w1[t0,T) consist of all locally integrable nonnegative functions ϕ on [t0,T) such that ∑i=1m‍∫bi(t0)bi(t)‍(t-fi(s))βihi(t,s)ϕ(s)ds<∞ for arbitrarily given t∈[t0,T). We can verify that Lloc⁡,w1[t0,T) is a linear space, due to the linearity of integration. Define a linear operator B on Lloc⁡,w1[t0,T)as
(33)Bϕ(t)∶=∑i=1m‍∫bi(t0)bi(t)‍(t-fi(s))βihi(t,s)ϕ(s)ds,t∈[t0,T),
where ϕ∈Lloc⁡,w1[t0,T). We claim that B is self-mapping on Lloc⁡,w1[t0,T). In fact, for all ϕ∈Lloc⁡,w1[t0,T) and given t∈[t0,T), ∑i=1m‍∫bi(t0)bi(t)‍(t-fi(s))βihi(t,s)ϕ(s)ds<∞, implying that Bϕ∈C+[t0,T), the set of all continuous and nonnegative functions on [t0,T). We know that C+[t0,T)⊂Lloc⁡1[t0,T). For all ψ∈C+[t0,T), from the continuity of ψ and hi, we know that they are all locally bounded. We also know that (t-fi(s))βi are integrable on s∈[bi(t0),bi(t)] by (H3), (H4) and βi>-1. It follows that ∑i=1m‍∫bi(t0)bi(t)‍(t-fi(s))βihi(t,s)ψ(s)ds<∞, implying that C+[t0,T)⊂Lloc⁡,w1[t0,T). Then, inequality (16) can be simplified as u(t)≤a(t)+Bu(t), from which we can prove by induction that
(34)u(t)≤∑l=0n-1‍Bla(t)+Bnu(t).
We claim that for all integers n≥1,
(35)Bnu(t)≤∑1≤k1,…,kn≤m‍∏i=1n‍Γ(βki+1)h~ki(t)Γ(∑i=1n‍βki+n)·∫bkn(t0)bkn(t)‍(t-fkn(s))∑i=1n‍βki+n-1fkn′(s)u(s)ds,
where h~ki are well-defined on [t0,T) since hki and fki′ are continuous and fki′(t)>0 on [t0,T). In fact, (35) is true for n=1. Assume that (35) is true for n=l. Then
(36)Bl+1u(t)=Bl(Bu(t))≤∑1≤k1,…,kl≤m‍∏i=1l‍Γ(βki+1)h~ki(t)Γ(∑i=1l‍βki+l)×∫bkl(t0)bkl(t)‍(t-fkl(s))∑i=1l‍βki+l-1fkl′(s)·∑kl+1=1m‍{∫bkl+1(t0)bkl+1(s)‍hkl+1(s,τ)×(s-fkl+1(τ))βkl+1×u(τ)dτ∫bkl+1(t0)bkl+1(s)‍}ds.
It follows that
(37)Bl+1u(t)≤∑1≤k1,…,kl≤m‍∏i=1l‍Γ(βki+1)h~ki(t)Γ(∑i=1l‍βki+l)×∑kl+1=1m‍h~kl+1(t)Φ(t),
where
(38)Φ(t)∶=∫bkl(t0)bkl(t)‍(t-fkl(s))∑i=1l‍βki+l-1fkl′(s)φ(s)ds,(39)φ(s)∶=∫bkl+1(t0)bkl+1(s)‍(s-fkl+1(τ))βkl+1fkl+1′(τ)u(τ)dτ.
Note that fki(bki(t))≤t by (H4). It implies that s-fkl+1(τ)≥0 for all τ∈[bkl+1(t0),bkl+1(s)] by the monotonicity of fkl+1. It follows from (39) that φ is a nondecreasing function. Hence, φ(fkl-1(t))≤φ(t) by (H4). On the other hand, from (H3) and (H4) we see that bi(t0)=fi(t0)=t0, implying that fkl(bkl(t0))=t0. With the change of variables ξ=fkl(s), it follows from (38) that
(40)Φ(t)=∫fkl(bkl(t0))fkl(bkl(t))‍(t-ξ)∑i=1l‍βki+l-1φ(fkl-1(ξ))dξ≤∫t0t‍(t-ξ)∑i=1l‍βki+l-1φ(fkl-1(ξ))dξ≤∫t0t‍(t-ξ)∑i=1l‍βki+l-1φ(ξ)dξ.
Letting s denote ξ and substituting (39) in (40), we get
(41)Φ(t)≤∫t0t‍∫bkl+1(t0)bkl+1(s)‍(t-s)∑i=1l‍βki+l-1(s-fkl+1(τ))βkl+1×fkl+1′(τ)u(τ)dτds=∫bkl+1(t0)bkl+1(t)‍∫bkl+1-1(τ)t‍(t-s)∑i=1l‍βki+l-1×(s-fkl+1(τ))βkl+1ds×fkl+1′(τ)u(τ)dτ,
by interchanging the order of integration. In (41) we observe that
(42)∫bkl+1-1(τ)t‍(t-s)∑i=1l‍βki+l-1(s-fkl+1(τ))βkl+1ds=(t-fkl+1(τ))∑i=1l+1‍βki+l×∫bkl+1-1(τ)t‍(t-st-fkl+1(τ))∑i=1l‍βki+l-1×(s-fkl+1(τ)t-fkl+1(τ))βkl+1d(s-fkl+1(τ)t-fkl+1(τ))=(t-fkl+1(τ))∑i=1l+1‍βki+l×∫z11‍(1-z)∑i=1l‍βki+l-1zβkl+1dz≤(t-fkl+1(τ))∑i=1l+1‍βki+l×∫01‍(1-z)∑i=1l‍βki+l-1zβkl+1dz=(t-fkl+1(τ))∑i=1l+1‍βki+lB(∑i=1l‍βki+l,βkl+1+1)=Γ(∑i=1l‍βki+l)Γ(βkl+1+1)Γ(∑i=1l+1‍βki+l+1)×(t-fkl+1(τ))∑i=1l+1‍βki+l,
where z:={s-fkl+1(τ)}/{t-fkl+1(τ)} and z1:={bkl+1-1(τ)-fkl+1(τ)}/{t-fkl+1(τ)}≥0 by (H4). Thus, from (37) we see that inequality (35) holds for n=l+1 and the claimed (35) is proved.For everyϕ(t)∈Lloc⁡,w1[t0,T) and each n=0,1,…,∞, define a sequence of linear operators An:Lloc⁡,w1[t0,T)→C[t0,T)by
(43)Anϕ(t)∶=∑1≤k1,…,kn≤m‍∏i=1n‍Γ(βki+1)h~ki(t)Γ(∑i=1n‍βki+n)·∫bkn(t0)bkn(t)‍(t-fkn(s))∑i=1n‍βki+n-1×fkn′(s)ϕ(s)ds,
where ∫bkn(t0)bkn(t)‍(t-fkn(s))∑i=1n‍βki+n-1fkn′(s)ϕ(s)ds<∞ since ∫bkn(t0)bkn(t)‍(t-fkn(s))βknϕ(s)ds<∞ from the supposition of Theorem 1 and (t-fkn(s))∑i=1n-1‍(βki+1)fkn′(s) are continuous on [bkn(t0),bkn(t)). It implies from (35) that
(44)Bnϕ(t)≤Anϕ(t).
In what follows we will prove that for all ϕ∈Lloc⁡,w1[t0,T) and arbitrarily given t∈[t0,T), ∑n=1∞‍Anϕ(t) converges. Define
(45)Ψa(x)∶=Γ(x)Γ(x+a),
where x∈(0,∞) and a>0 is a constant. By Euler’s definition on Gamma function (see [24]), Γ(x)=(1/x)∏n=1∞‍(1+(1/n))x/(1+(x/n)), we know that
(46)Ψa(x)=(1+ax)∏n=1∞‍1(1+(1/n))a(1+an+x),
and therefore Ψa is strictly decreasing on (0,∞). Thus, for all integers 1≤ki, kj≤m, 1≤i, j≤n and δ≥0 satisfying that -1<βkj-δ<0, we have Ψa(βkj+1)≤Ψa(βkj-δ+1), where a:=(∑i=1n‍βki)+n-βkj-1, implying that
(47)Γ(βkj+1)Γ(∑i=1n‍βki+n)≤Γ(βkj-δ+1)Γ(∑i=1n‍βki-δ+n).
Multiplying the above inequality by (∏i=1n‍Γ(βki+1))/Γ(βkj+1), we get that
(48)∏i=1n‍Γ(βki+1)Γ(∑i=1n‍βki+n)≤∏i=1j-1‍Γ(βki+1)Γ(βkj-δ+1)∏i=j+1n‍Γ(βki+1)Γ(∑i=1n‍βki-δ+n).
Let β*:=min⁡1≤i≤mβi, β*:=max⁡1≤i≤mβi and c:=(β*+1)/(β*+1)≥1. Then (48) implies that
(49)∏i=1n‍Γ(βki+1)Γ(∑i=1n‍βki+n)≤(Γ(β*+1))nΓ(n(β*+1)).
When t∈(t0+1,T), we have
(50)Anϕ(t)=∑1≤k1,…,kn≤m‍∏i=1n‍Γ(βki+1)h~ki(t)Γ(∑i=1n‍βki+n)×∫bkn(t0)bkn(t)‍(t-fkn(s))∑i=1n‍βki+n-1fkn′(s)ϕ(s)ds=∑1≤k1,…,kn≤m‍∏i=1n‍Γ(βki+1)h~ki(t)Γ(∑i=1n‍βki+n)×{∫bkn(t0)fkn-1(t-1)‍(t-fkn(s))βknϕ(s)×(t-fkn(s))∑i=1n-1‍βki+n-1fkn′(s)ds+∫fkn-1(t-1)bkn(t)‍(t-fkn(s))βknϕ(s)×(t-fkn(s))∑i=1n-1‍βki+n-1fkn′(s)ds∫bkn(t0)fkn-1(t-1)‍}≤mn-1(max⁡i=1,…,m{h~i(t)})n(Γ(β*+1))nΓ(n(β*+1))×∑i=1m‍{∫bi(t0)fi-1(t-1)‍(t-fi(s))βiϕ(s)×(t-fi(s))(n-1)c(β*+1)fi′(s)ds+∫fi-1(t-1)bi(t)‍(t-fi(s))βiϕ(s)×(t-fi(s))(n-1)(β*+1)fi′(s)ds∫bi(t0)fi-1(t-1)‍∫bi(t0)fi-1(t-1)‍}≤(mmax⁡i=1,…,m{h~i(t)})n(Γ(β*+1))nΓ(n(β*+1))×{∫fj-1(t-1)bj(t)‍(t-t0)(n-1)c(β*+1)×∫bj(t0)fj-1(t-1)‍(t-fj(s))βjϕ(s)fj′(s)ds+∫fj-1(t-1)bj(t)‍(t-fj(s))βjϕ(s)fj′(s)ds}≤(mmax⁡i=1,…,m{h~i(t)})n(Γ(β*+1))nΓ(n(β*+1))×(t-t0)(n-1)c(β*+1)×∫bj(t0)bj(t)‍(t-fj(s))βjϕ(s)fj′(s)ds,
where
(51)∫bj(t0)bj(t)‍(t-fj(s))βjϕ(s)fj′(s)ds=max⁡i=1,…,m⁡{∫bi(t0)bi(t)‍(t-fi(s))βiϕ(s)fi′(s)ds}<∞.Let(52)cn(t)∶=(mmax⁡i=1,…,m{h~i(t)})n(Γ(β*+1))nΓ(n(β*+1))×(t-t0)(n-1)c(β*+1)×∫bj(t0)bj(t)‍(t-fj(s))βjϕ(s)fj′(s)ds.
Then ∑n=1∞‍cn(t) is convergent on (t0+1,T) by the ratio test ([25, pages 66-67]) because
(53)lim⁡n→+∞cn+1(t)cn(t)=lim⁡n→+∞mmax⁡i=1,…,m{h~i(t)}×Γ(β*+1)Γ(n(β*+1))Γ((n+1)(β*+1))(t-t0)c(β*+1)=lim⁡n→+∞B(β*+1,n(β*+1))m×max⁡i=1,…,m{h~i(t)}(t-t0)c(β*+1)=lim⁡n→+∞Γ(β*+1)(n(β*+1))β*+1m×max⁡i=1,…,m{h~i(t)}(t-t0)c(β*+1)=0,
where we note that mmax⁡i=1,…,m{h~i(t)}(t-t0)c(β*+1) is a continuous function on [t0,T) and Γ(β*+1)/(n(β*+1))β*+1 is a Stirling’s approximation of B(β*+1,n(β*+1)) as n→+∞, as known in [26, pages 59]. It implies that ∑n=1∞‍Anϕ(t) is convergent on (t0+1,T). The case of t∈[t0,t0+1] can be proved similarly. Then, ∑n=1∞‍Anϕ(t) is convergent for ϕ∈Lloc⁡,w1[t0,T) and arbitrarily given t∈[t0,T). Passing to the limit as n→+∞ in (34), by (44) we get
(54)u(t)≤∑n=0∞‍Ana(t)=a(t)+∑n=1∞‍{∑1≤k1,…,kn≤m‍∏i=1n‍Γ(βki+1)h~ki(t)Γ(∑i=1n‍βki+n)×∫bkn(t0)bkn(t)‍(t-fkn(s))∑i=1n‍βki+n-1×fkn′(s)a(s)ds∑1≤k1,…,kn≤m‍∏i=1n‍Γ(βki+1)h~ki(t)Γ(∑i=1n‍βki+n)}.
Since ki’s are chosen arbitrarily, by interchanging kn with k1 in (54), we obtain the estimate (17) and complete the proof of the theorem.


## 4. Application to Dependence

Recently, increasing interest was given to fractional differential equations (see monographs [11, 23]). In this section we consider the Cauchy problem of the general fractional differential equation
(55)Dt01x(t)=∑i=1m‍Dt0αifi(t,x(bi(t))),(56)x(t0)=a,
where 0≤αi<1, t0≤t≤T<+∞, Dt0αi is defined as in the Introduction, particularly, Dt01 denotes d/dt and Dt00 denotes identity map, bi:[t0,T]→[t0,T] and fi:[t0,T]×R→R, i=1,…,m. This system includes the system
(57)D01x(t)+pD0αx(t)+D00x(t)=q(t),x(0+)=a,
as a special case, which was considered in [27] and corresponds to the Basset problem when α=1/2, a classical problem in fluid dynamics concerning the unsteady motion of a particle accelerating in a viscous fluid under the action of gravity (see [28]). We will give continuous dependence for solutions of (55) associated with the initial condition (56) on the derivative orders αi’s, the initial data a, and the functions fi and bi.Proposition 3.
Suppose that eachbi∈C1([t0,T],[t0,T]) is strictly increasing such that bi(t)≤t and each fi∈C([t0,T]×[a-ρ,a+ρ],R) satisfies that |fi(t,x1)-fi(t,x2)|≤Li|x1-x2|, i=1,…,m, where T>0, ρ>0, and Li>0 are constant. Then, the Cauchy problem (55)-(56) has a unique solution on [t0,t0+h] for a certain constant 0<h<T and the solution depends continuously on a, αi’s, fi’s, and bi’s for all t∈[t0,t0+h].Proof.
The Cauchy problem (55)-(56) is equivalent to the integral equation
(58)x(t)=a+∑i=1m‍∫t0t‍1Γ(1-αi)d(∫t0s‍(s-τ)-αifi(τ,x(bi(τ)))dτ)=a+∑i=1m‍1Γ(1-αi)∫t0t‍(t-s)-αifi(s,x(bi(s)))ds.
Define a sequence {φn(t)} such that φ0(t)≡a and
(59)φn(t)=a+∑i=1m‍1Γ(1-αi)×∫t0t‍(t-s)-αifi(s,φn-1(bi(s)))ds,ccccccccccccccccccccccccn=1,2,…,
where t∈[t0,t0+h]. We first claim that all φn(t) are well-defined and continuous in [t0,t0+h], such that |φn(t)-a|≤ρ. In fact, it is true for n=0. Suppose that it is true for some n. Then, φn+1 is also well defined by (59) and continuous in [t0,t0+h]. Moreover,
(60)|φn+1(t)-a|≤∑i=1m‍1Γ(1-αi)∫t0t‍(t-s)-αi|fi(s,φn(bi(s)))|ds≤∑i=1m‍Mh1-αiΓ(2-αi)≤ρ,
where
(61)M≔max⁡(t,x)∈[t0,T]×[a-ρ,a+ρ]i=1,…,m|fi(t,x)|,h≔min⁡⁡{T-t0,σ},
and σ is a positive constant such that
(62)∑i=1m‍Mσ1-αiΓ(2-αi)=ρ.
The existence of σ is guaranteed by the intermediate theorem for continuous functions. Thus, the claim is proved by induction for all n.
The convergence of the sequence{φn(t)} is equivalent to the convergence of the series φ0(t)+∑l=1∞‍(φl(t)-φl-1(t)). We claim that
(63)|φl(t)-φl-1(t)|≤∑1≤k1,…,kl≤m‍M∏i=2l‍LkiΓ(1+∑i=1l‍(1-αki))(t-t0)∑i=1l‍(1-αki),
for all t∈[t0,t0+h] and all integers l. It can be checked easily for l=1. Suppose that it is true for some l. By changing t into bkl+1(s) in (63), we get from bkl+1(s)≤s that
(64)|φl(bkl+1(s))-φl-1(bkl+1(s))|≤∑1≤k1,…,kl≤m‍M∏i=2l‍LkiΓ(1+∑i=1l‍(1-αki))×(bkl+1(s)-t0)∑i=1l‍(1-αki)≤∑1≤k1,…,kl≤m‍M∏i=2l‍LkiΓ(1+∑i=1l‍(1-αki))×(s-t0)∑i=1l‍(1-αki).
Combining with (59), we have
(65)|φl+1(t)-φl(t)|≤∑kl+1=1m‍1Γ(1-αkl+1)×∫t0t‍(t-s)-αkl+1×|fkl+1(s,φl(bkl+1(s)))-fkl+1(s,φl-1(bkl+1(s)))|ds≤∑kl+1=1m‍Lkl+1Γ(1-αkl+1)×∫t0t‍(t-s)-αkl+1|φl(bkl+1(s))-φl-1(bkl+1(s))|ds≤∑1≤k1,…,kl+1≤m‍M∏i=2l+1‍LkiΓ(1+∑i=1l‍(1-αki))Γ(1-αkl+1)×∫t0t‍(t-s)-αkl+1(s-t0)∑i=1l‍(1-αki)ds=∑1≤k1,…,kl+1≤m‍(M(∏i=2l+1‍Lki)×B(1+∑i=1l‍(1-αki),1-αkl+1)×(Γ(1+∑i=1l‍(1-αki))×Γ(1-αkl+1)∑i=1l‍)-1)×(t-t0)∑i=1l+1‍(1-αki)=∑1≤k1,…,kl+1≤m‍M∏i=2l+1‍LkiΓ(1+∑i=1l+1‍(1-αki))(t-t0)∑i=1l+1‍(1-αki).
Thus, (63) holds for l+1 and the claim is proved for all l by induction. From (63), for t≤t0+h we have
(66)|φl(t)-φl-1(t)|≤∑1≤k1,…,kl≤m‍Mh∑i=1l‍(1-αki)∏i=2l‍LkiΓ(1+∑i=1l‍(1-αki))≤mlMhl(1-α*)Ll-1Γ(1+l(1-α*)),
where L:=max⁡1≤i≤mLi, α*:=max⁡1≤i≤mαi, and α*:=min⁡1≤i≤mαi. By the ratio test ([25, pp. 66-67]), we get that ∑l=1∞‍mlMhl(1-α*)Ll-1/Γ(1+l(1-α*)) is convergent, implying that the series φ0(t)+∑l=1∞‍(φl(t)-φl-1(t)) and therefore the sequence {φn(t)} are uniformly convergent in [t0,t0+h]. Let φ(t):=lim⁡n→∞φn(t), which is well-defined and continuous in [t0,t0+h] such that |φ(t)-a|≤ρ. One can check that φ is a continuous solution of the Cauchy problem (55)-(56) in [t0,t0+h].
Next, we prove the continuous dependence of the solution. Consider the Cauchy problem(67)Dt01x(t)=∑i=1m‍Dt0α~if~i(t,x(b~i(t))),(68)x(t0)=a~,
where 0≤α~i<1, t0≤t≤T<+∞, b~i∈C1([t0,T],[t0,T]) is strictly increasing such that bi(t)≤t, and f~i∈C([t0,T]×[a~-ρ,a~+ρ],R) is Lipschitzian in the second variable with the Lipschitz constant L~i, i=1,…,m. As above, we similarly obtain a solution ψ of the Cauchy problem (67)-(68). Similar to (58), we have
(69)φ(t)=a+∑i=1m‍1Γ(1-αi)∫t0t‍(t-s)-αifi(s,φ(bi(s)))ds,ψ(t)=a~+∑i=1m‍1Γ(1-α~i)∫t0t‍(t-s)-α~if~i(s,ψ(b~i(s)))ds.
It follows that
(70)|φ(t)-ψ(t)|≤|a-a~|+∑i=1m‍|1Γ(1-αi)×∫t0t‍(t-s)-αifi(s,φ(bi(s)))ds-1Γ(1-α~i)×∫t0t‍(t-s)-α~if~i(s,ψ(b~i(s)))ds|≤Ξ(t)+∑i=1m{|1Γ(1-αi)×∫t0t‍(t-s)-αifi(s,φ(bi(s)))ds-1Γ(1-αi)×∫t0t‍(t-s)-αifi(s,ψ(bi(s)))ds1Γ(1-αi)|}≤Ξ(t)+∑i=1m‍LiΓ(1-αi)×∫bi(t0)bi(t)‍(t-bi-1(s))-αi|φ(s)-ψ(s)|d(bi-1(s)),
where
(71)Ξ(t)≔|a-a~|+∑i=1m‍{|1Γ(1-αi)∫t0t‍×∫t0t‍(t-s)-αifi(s,ψ(bi(s)))ds-1Γ(1-α~i)×∫t0t‍(t-s)-α~ifi(s,ψ(bi(s)))ds|+|1Γ(1-α~i)∫t0t‍(t-s)-α~ifi(s,ψ(bi(s)))ds-1Γ(1-α~i)∫t0t‍(t-s)-α~if~i(s,ψ(bi(s)))ds|+|1Γ(1-α~i)∫t0t‍(t-s)-α~if~i(s,ψ(bi(s)))ds-1Γ(1-α~i)∫t0t‍(t-s)-α~if~i(s,ψ(b~i(s)))ds|}≤|a-a~|+∑i=1m‍{|(t-t0)1-αiΓ(2-αi)-(t-t0)1-α~iΓ(2-α~i)|∥fi∥+(t-t0)1-α~iΓ(2-α~i)∥fi-f~i∥+L~iΓ(1-α~i)∫t0t‍(t-s)-α~i×|ψ(bi(s))-ψ(b~i(s))|ds(t-t0)1-αiΓ(2-αi)},∥fi∥:=max⁡(t,x)∈[t0,T]×[a-ρ,a+ρ]|fi(t,x)|, and ∥fi-f~i∥:=max⁡(t,x)∈[t0,T]×[a-ρ,a+ρ]|fi(t,x)-f~i(t,x)|. Thus, applying the inequality given in Corollary 2 to (70), we obtain that
(72)|φ(t)-ψ(t)|≤Ξ~(t)∏i=1m‍E1-αi(Li(t-t0)1-αi),mmmmmmmmmmmmmmk∀t∈[t0,t0+h],
where Ξ~(t):=max⁡t0≤τ≤tΞ(τ).
Note from (71) that Ξ~(t) contains |(t-t0)1-αi/Γ(2-αi)-(t-t0)1-α~i/Γ(2-α~i)| and |ψ(bi(s))-ψ(b~i(s))|, which will be estimated with ∥bi-b~i∥:=max⁡t0≤τ≤t0+h|bi(τ)-b~i(τ)| and |αi-α~i|, respectively. In (71) we have ∥fi∥≤M in (61). Since (t-t0)1-αi and Γ(2-αi) are both C1 functions in αi∈[0,1], and Γ(2-αi)>0 for all αi∈[0,1], we see that (t-t0)1-αi/Γ(2-αi) is C1 in αi∈[0,1]. By the Lagrange mean value theorem (see [25, page 108]), function (t-t0)1-αi/Γ(2-αi) satisfies the Lipschitz condition uniformly with respect to t∈[t0,t0+h]; that is, there exists a constant N>0 such that
(73)|(t-t0)1-αiΓ(2-αi)-(t-t0)1-α~iΓ(2-α~i)|≤N|αi-α~i|,∀αi,α~i∈[0,1].
On the other hand, from the equivalent integral equation of (67)
(74)ψ(t)=ψ(t0)+∑j=1m‍1Γ(1-α~j)∫t0t‍(t-s)-α~jf~j(s,ψ(b~j(s)))ds,
we get
(75)|ψ(t)-ψ(t0)|≤∑j=1m‍M~Γ(2-α~j)|t-t0|1-α~j,
where
(76)M~:=max⁡(t,x)∈[t0,T]×[a-ρ,a+ρ]j=1,…,m|f~j(t,x)|.
Choosing t=bi(s) and t0=b~i(s) in (74), we obtain
(77)|ψ(bi(s))-ψ(b~i(s))|≤∑j=1m‍M~Γ(2-α~j)|bi(s)-b~i(s)|1-α~j≤∑j=1m‍M~Γ(2-α~j)∥bi-b~i∥1-α~j.
Summarily, from (71), (73) and (75) we get
(78)Ξ(t)≤|a-a~|+∑i=1m‍{MN|αi-α~i|+(t-t0)1-α~iΓ(2-α~i)∥fi-f~i∥+L~iΓ(1-α~i)∫t0t‍(t-s)-α~i∑j=1m‍M~Γ(2-α~j)×∥bi-b~i∥1-α~jds(t-t0)1-α~iΓ(2-α~i)}≤|a-a~|+∑i=1m‍{∥bi-b~i∥1-α~jΓ(2-α~j)MN|αi-α~i|+h1-α~iΓ(2-α~i)∥fi-f~i∥+M~L~ih1-α~iΓ(2-α~i)∑j=1m‍∥bi-b~i∥1-α~jΓ(2-α~j)}.
It follows from (72) that
(79)∥φ-ψ∥≤∏i=1m‍E1-αi(Lih1-αi)|a-a~|+MN∏i=1m‍E1-αi(Lih1-αi)∑i=1m‍|αi-α~i|+∏i=1m‍E1-αi(Lih1-αi)∑i=1m‍h1-α~iΓ(2-α~i)∥fi-f~i∥+M~∏i=1m‍E1-αi(Lih1-αi)×∑1≤i,j≤m‍L~ih1-α~iΓ(2-α~i)Γ(2-α~j)∥bi-b~i∥1-α~j;
that is, the solution φ of the Cauchy problem (55)-(56) depends continuously on a, αi, fi and bi.
As a complement, we note from (79) that φ≡ψ in [t0,t0+h] if a=a~, αi=α~i, fi≡f~i, and bi≡b~i. This implies the uniqueness of the solution.



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