# New Results for Generalized Gronwall Inequalities and Their Applications

Authors: Shi-you Lin
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

In succession to our earlier work, we further provide some new generalized Gronwall inequalities and apply these inequalities to the study of qualitative estimations of solutions to certain fractional differential equations.


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## Body

## 1. Introduction

It is well known that the Gronwall inequality contributes significantly to research on many differential and integral equations. An increasing number of generalizations of this inequality have been made in recent years to derive qualitative properties of solutions to various fractional differential equations. One remarkable result was obtained by Ye et al. in 2007 and is presented below.Theorem 1 (see [1, Theorem 1]).
For anyt∈[0,T),
(1)u(t)≤a(t)+b(t)∫0t‍(t-s)β-1u(s)ds,
where all functions are nonnegative and continuous. The constant β>0. b is a bounded and monotonically increasing function of [0,T); then
(2)u(t)≤a(t)+∫0t‍[∑n=1∞‍(b(t)Γ(β))nΓ(nβ)(t-s)nβ-1a(s)]ds,hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhht∈[0,T).This result continuously extends and improves and has been widely used in many studies (e.g., see [1–8]). Among them, our previous research [5] provides the following generalizations to deal with complex fractional differential equations.Theorem 2 (see [5, Theorem 1.4]).
For anyt∈[0,T),
(3)u(t)≤a(t)+∑i=1n‍bi(t)∫0t‍(t-s)βi-1u(s)ds,
where all functions are nonnegative and continuous. The constants βi>0. bi(i=1,2,…,n) are bounded and monotonically increasing functions of [0,T). Then
(4)u(t)≤a(t)11111+∑k=1∞‍(∑1′,2′,…,k′=1n‍∏i=1k[bi′(t)Γ(βi′)]Γ(∑i=1kβi′)1111111111111111111×∫0t‍(t-s)∑i=1kβi′-1a(s)ds∑1′,2′,…,k′=1n‍),11111111111111111111111111111111111t∈[0,T).Theorem 3 (see [5, Theorem 1.5]).
For anyt∈[1,T),
(5)u(t)≤a(t)+∑i=1n‍bi(t)∫1t‍(ln⁡ts)βi-1u(s)sds,
where all functions are nonnegative and continuous. The constants βi>0. bi(i=1,2,…,n) are bounded and monotonically increasing functions of [1,T). Then
(6)u(t)≤a(t)+∑k=1∞(∑1′,2′,…,k′=1n‍∏i=1k[bi′(t)Γ(βi′)]Γ(∑i=1kβi′)‍111111111111111×∫1t‍[(ln⁡ts)∑i=1kβi′-1a(s)]dss∑1′,2′,…,k′=1n‍),111111111111111111111111111111111111t∈[1,T).In this paper, we aim to discuss further issues by using the aforementioned conclusions and suitable analytical techniques according to the above facts. We will establish several new classes of generalized Gronwall inequalities in the next section. In the last section, we will qualitatively analyze certain fractional differential equations to highlight the applications of the inequalities.


## 2. Main Results and Proofs

We introduce the useful Young’s inequality withϵ>0 (see [9, page 622]); that is, for any a,b>0 and 1<p, q<+∞, 1/p+1/q=1,
(7)ab≤ϵap+C(ϵ)bq,
where C(ϵ)=(ϵp)-q/pq-1.The first main result and its proof procedure are presented as follows.Theorem 4.
For anyt∈[0,T),
(8)u(t)≤a(t)+∑i=1n‍bi(t)∫0t‍(t-s)βi-1ci(s)uλi(s)ds,
where all functions are nonnegative and continuous. For any i∈{1,2,…,n}, the constants βi>0 and 0<λi<1, and the function bi(t) is bounded and monotonically increasing on [0,T). Then
(9)u(t)≤a~(t)+∑k=1∞‍(∑1′,2′,…,k′=1n‍∏i=1k[b~i′(t)Γ(βi′)]Γ(∑i=1kβi′)11111111111111111×∫0t‍(t-s)∑i=1kβi′-1a~(s)ds∑1′,2′,…,k′=1n‍),111111111111111111111111111111111t∈[0,T),
where
(10)a~(t)=a(t)+∑i=1n‍(1-λi)(λiɛ)λi/(1-λi)bi(t)hhhhhhhh×∫0t‍(t-s)βi-1[ci(s)]1/(1-λi)ds;b~i(t)=ɛ·bi(t).
Here ɛ is an arbitrary given positive number.Proof.
By Young’s inequality,(11)ci(s)uλi(s)≤ɛ·[uλi(s)]1/λi+(λiɛ)λi/(1-λi)·(1-λi)·[ci(s)]1/(1-λi)=ɛ·u(s)+(λiɛ)λi/(1-λi)·(1-λi)·[ci(s)]1/(1-λi),
which implies that, for any t∈[0,T),
(12)u(t)≤a(t)+∑i=1n‍bi(t)∫0t‍(t-s)βi-1ci(s)uλi(s)ds≤a(t)+∑i=1n‍bi(t)1111111111×∫0t‍(t-s)βi-1{+(λiɛ)λi/(1-λi)·(1-λi)·[ci(s)]1/(1-λi)ɛ·u(s)1111111111111111111111111+(λiɛ)λi/(1-λi)·(1-λi)1111111111111111111111111·[ci(s)]1/(1-λi)(λiɛ)λi/(1-λi)}ds={a(t)+∑i=1n‍(1-λi)(λiɛ)λi/(1-λi)bi(t)111111111111×∫0t‍(t-s)βi-1[ci(s)]1/(1-λi)ds∑i=1n‍}+∑i=1n‍ɛ·bi(t)∫0t‍(t-s)βi-1u(s)ds=a~(t)+∑i=1n‍b~i(t)∫0t‍(t-s)βi-1u(s)ds.
We immediately derived estimation (9) by using Theorem 2. This estimation completes the proof of Theorem 4.Our second result can be proved in the same manner by applying Young’s inequality and Theorem3. The proof procedure is similar to that of Theorem 4 and is omitted here.Theorem 5.
For anyt∈[1,T),
(13)u(t)≤a(t)+∑i=1n‍bi(t)∫1t‍(ln⁡ts)βi-1ci(s)uλi(s)sds,
where all functions are nonnegative and continuous. For any i∈{1,2,…,n}, the constants βi>0 and 0<λi<1, and the function bi(t) is bounded and monotonically increasing on [1,T). Then
(14)u(t)≤a¯(t)+∑k=1∞‍(∑1′,2′,…,k′=1n‍∏i=1k[b¯i′(t)Γ(βi′)]Γ(∑i=1kβi′)111111111111111111×∫1t‍(ln⁡ts)∑i=1kβi′-1a¯(s)sds∑1′,2′,…,k′=1n‍),11111111111111111111111111111111111t∈[1,T),
where
(15)a¯(t)=a(t)+∑i=1n‍(1-λi)(λiɛ)λi/(1-λi)bi(t)1111111111×∫1t‍(ln⁡ts)βi-1[ci(s)]1/(1-λi)dss;b¯i(t)=ɛ·bi(t).
Here ɛ is an arbitrary given positive number.Evidently, the following two corollaries are directly obtained by using Theorems4 and 5 when choosing βi=1 for any i∈{1,2,…,n}.Corollary 6.
For anyt∈[0,T),
(16)u(t)≤a(t)+∑i=1n‍bi(t)∫0t‍ci(s)uλi(s)ds,
where all functions are nonnegative and continuous. For any i∈{1,2,…,n}, the constant 0<λi<1, and the function bi(t) is bounded and monotonically increasing on [0,T). Then
(17)u(t)≤a~(t)+∑k=1∞‍(∑1′,2′,…,k′=1n‍∏i=1kb~i′(t)k!111111111111111111×∫0t‍(t-s)k-1a~(s)ds∑1′,2′,…,k′=1n‍∏i=1kb~i′k!),111111111111111111111111111111t∈[0,T),
where
(18)a~(t)=a(t)+∑i=1n‍(1-λi)(λiɛ)λi/(1-λi)bi(t)11111111111111×∫0t‍[ci(s)]1/(1-λi)ds;b~i(t)=ɛ·bi(t).
Here ɛ is an arbitrary given positive number.Corollary 7.
For anyt∈[1,T),
(19)u(t)≤a(t)+∑i=1n‍bi(t)∫1t‍ci(s)uλi(s)sds,
where all functions are nonnegative and continuous. For any i∈{1,2,…,n}, the constant 0<λi<1, and the function bi(t) is bounded and monotonically increasing on [1,T). Then
(20)u(t)≤a¯(t)+∑k=1∞‍(∑1′,2′,…,k′=1n‍∏i=1kb¯i′(t)k!1111111111111111111×∫1t‍[(ln⁡ts)k-1a¯(s)]dss∑1′,2′,…,k′=1n‍∏i=1kb¯i′(t)k!),1111111111111111111111111111111111t∈[1,T),
where
(21)a¯(t)=a(t)+∑i=1n‍(1-λi)(λiɛ)λi/(1-λi)bi(t)111111111×∫1t‍[ci(s)]1/(1-λi)dss;b¯i(t)=ɛ·bi(t).
Here ɛ is an arbitrary given positive number.Our next task is estimating the nonnegative and continuous functionu(t), which satisfies, for any t∈[0,T),
(22)u(t)≤a(t)+b(t)∫0t‍c(s)u(s)ds+∑i=1n‍bi(t)∫0t‍(t-s)βi-1ci(s)uλi(s)ds,
where all the functions are nonnegative and continuous. For any i∈{1,2,…,n}, the constants βi>0 and 0<λi<1, and the functions b(t) and bi(t) are bounded and monotonically increasing on [0,T). Put
(23)Mi(t)=max⁡0≤s≤t{bi(s)c(s)}.
Suppose that
(24)A(t)=a(t)+b(t)∫0t‍c(s)e∫st‍c(w)b(w)dw·a(s)ds;Bi(t)=bi(t)+t·b(t)·Mi(t)βie∫0t‍c(w)b(w)dw,i=1,2,…,n.
Obviously, Bi(t)≥0(i=1,2,…,n) are bounded and monotonically increasing functions. Moreover, we assumed that
(25)A~(t)=A(t)+∑i=1n‍(1-λi)(λiɛ)λi/(1-λi)Bi(t)111111111×∫0t‍(t-s)βi-1[ci(s)]1/(1-λi)ds;B~i(t)=ɛBi(t),i=1,2,…,n.
Here ɛ is an arbitrary given positive number. The third main result is given as follows.Theorem 8.
For anyt∈[0,T), the nonnegative and continuous function u(t) satisfies the inequality (22); then
(26)u(t)≤A~(t)+∑k=1∞‍(∑1′,2′,…,k′=1n‍∏i=1k[B~i′(t)Γ(βi′)]Γ(∑i=1kβi′)1111111111111111×∫0t‍(t-s)∑i=1kβi′-1A~(s)ds∑1′,2′,…,k′=1n‍),111111111111111111111111111111111t∈[0,T),
where the expressions A~(t) and B~i(t) are described in (25).Proof.
Suppose that(27)I(t)=a(t)+∑i=1n‍bi(t)∫0t‍(t-s)βi-1ci(s)uλi(s)ds.
Then, (22) transforms into the following form:
(28)u(t)≤I(t)+b(t)∫0t‍c(s)u(s)ds.
Therefore,
(29)c(t)u(t)≤c(t)I(t)+c(t)b(t)∫0t‍c(s)u(s)ds.
Letting u1(t)=c(t)u(t), I1(t)=c(t)I(t), and J1(t)=c(t)b(t) obtains
(30)u1(t)≤I1(t)+J1(t)∫0t‍u1(s)ds.
By using the classical Gronwall inequality (see [10, page 15]), we have
(31)u1(t)≤I1(t)+J1(t)∫0t‍I1(s)e∫st‍J1(w)dwds;
that is,
(32)u(t)≤I(t)+b(t)∫0t‍c(s)I(s)e∫st‍c(w)b(w)dwds=[a(t)+b(t)∫0t‍c(s)a(s)e∫st‍c(w)b(w)dwds]+∑i=1n‍bi(t)∫0t‍(t-s)βi-1ci(s)uλi(s)ds+b(t)∑i=1n‍∫0t‍∫0s‍c(s)e∫st‍c(w)b(w)dw111111111111111111×bi(s)(s-v)βi-1ci(v)uλi(v)dvds,
given that
(33)∑i=1n‍∫0t‍∫0s‍bi(s)c(s)e∫st‍c(w)b(w)dw11111111×(s-v)βi-1ci(v)uλi(v)dvds=∑i=1n‍∫0t‍∫vt‍bi(s)c(s)e∫st‍c(w)b(w)dw11111111×(s-v)βi-1ci(v)uλi(v)dsdv≤e∫0t‍c(w)b(w)dw×∑i=1n‍Mi(t)∫0t‍∫vt‍(s-v)βi-1ci(v)uλi(v)dsdv=e∫0t‍c(w)b(w)dw∑i=1n‍1βiMi(t)111111111111111×∫0t‍(t-v)βi-1(t-v)ci(v)uλi(v)dv≤e∫0t‍c(w)b(w)dw∑i=1n‍1βitMi(t)111111111111111×∫0t‍(t-v)βi-1ci(v)uλi(v)dv,
where the function Mi(t) is defined in (23). Combining (24), (32), and (33) yields the following inequality:
(34)u(t)≤A(t)+∑i=1n‍Bi(t)∫0t‍(t-s)βi-1ci(s)uλi(s)ds.
The estimation (26) is obtained according to Theorem 4. This process completes the proof of Theorem 8.In the same manner, the final result in this section can be obtained by applying the conclusion of Theorem5.Theorem 9.
For anyt∈[1,T),
(35)u(t)≤a(t)+b(t)∫1t‍c(s)u(s)dss+∑i=1n‍bi(t)∫1t‍(ln⁡ts)βi-1ci(s)uλi(s)dss,
where all functions are nonnegative and continuous. For any i∈{1,2,…,n}, the constants βi>0 and 0<λi<1, and the functions b(t) and bi(t) are bounded and monotonically increasing on [1,T). Then
(36)u(t)≤A¯(t)+∑k=1∞‍(∑1′,2′,…,k′=1n‍∏i=1k[B¯i′(t)Γ(βi′)]Γ(∑i=1kβi′)111111111111111×∫1t‍(ln⁡ts)∑i=1kβi′-1A¯(s)sds∑1′,2′,…,k′=1n‍),11111111111111111111111111sssss11t∈[1,T),
where
(37)A¯(t)=A′(t)+∑i=1n‍(1-λi)(λiɛ)λi/(1-λi)Bi′(t)111111111×∫1t‍(ln⁡ts)βi-1[ci(s)]1/(1-λi)dss;B¯i(t)=ɛ·Bi′(t).
Here
(38)A′(t)=a(t)+b(t)∫1t‍c(s)e∫st‍c(w)b(w)(dw/w)·a(s)dss;Bi′(t)=bi(t)+ln⁡t·b(t)·Mi′(t)βie∫1t‍c(w)b(w)(dw/w),1111111111111111111111i=1,2,…,n.
Also Mi′(t)=max⁡1≤s≤t{bi(s)c(s)}, and ɛ is an arbitrary given positive number.The proof procedure of Theorem9 is relatively similar to that of Theorem 8. Hence, the procedure will not be presented in this paper.


## 3. Applications

In this section, we apply the main results in Section2 to provide qualitative conclusions for solutions of certain fractional differential equations. First, the definitions and some properties of the Riemann-Liouville fractional derivative and integral need to be recalled.Definition 10 (see [11–16]).
For any0<β<1 and a continuous function w, the βth Riemann-Liouville type fractional order derivative DRβw and the corresponding fractional integral operator IRβw are defined by
(39)IRβw(t)=1Γ(β)∫0t‍(t-s)β-1w(s)ds;DRβw(t)=1Γ(1-β)ddt∫0t‍(t-s)-βw(s)ds.Lemma 11 (see [13, Lemma 2.2]).
For any0<β<1 and a continuous function w,
(40)IRβDRβw(t)=w(t)+k·tβ-1,
where k is a certain constant in R.Lemma 12 (see [17, Page 14]).
For anyα,β>0 and a continuous function w,
(41)IRαIRβw(t)=IRα+βw(t);IRαtβ=Γ(β+1)Γ(α+β+1)tα+β.Given the aforementioned preliminary knowledge, we consider the following initial value problem:(42)∑i=1n‍DRβn+1-βi[ci(t)uλi(t)]+DRβn+1[uλ0(t)]=f(t,u(t));{∑i=1n‍IR1+βi-βn+1[ci(t)uλi(t)]+IR1-βn+1[uλ0(t)]}|t=0=δ,
where all functions are continuous. Moreover ci(t)>0 and the constants λi,βi>0 (i=1,2,…,n). Consider λ0>0, δ∈R, and 1>βn+1>max⁡1≤i≤n{βi}.Theorem 13.
Suppose that, for anyt∈[0,T] and y,z∈R,
(43)|f(t,y)-f(t,z)|≤cn+1(t)|yλn+1-zλn+1|,
where the function cn+1(t)>0 is continuous and the constant λn+1∈(0,1).(i)
Ifλ0>max⁡1≤i≤n+1{λi}, then for any solution u(t) of the problem (42),
(44)|u(t)|≤{∑k=1∞‍(∑1′,2′,…,k′=1n+1‍ɛkΓ(∑i=1kβi′)A~0(t)hhh+∑k=1∞‍(∑1′,2′,…,k′=1n+1‍ɛkΓ(∑i=1kβi′)1111111111111111×∫0t(t-s)∑i=1kβi′-1‍A~0(s)ds∑k=1∞‍(∑1′,2′,…,k′=1n+1‍ɛkΓ(∑i=1kβi′))}1/λ0,1111111111111ssss1111111111111111111111t∈[0,T],
where
(45)A~0(t)=|δ|tβn+1-1Γ(βn+1)+1Γ(βn+1)∫0t‍(t-s)βn+1-1|f(s,0)|ds+∑i=1n+1‍1Γ(βi)(1-λiλ0)(λiɛλ0)λi/(λ0-λi)111111111111×∫0t‍(t-s)βi-1[ci(s)]λ0/(λ0-λi)ds.
Here ɛ is an arbitrary given positive number.(ii)
Ifλ0≤min⁡1≤i≤n+1⁡{λi}, then the continuous solution of problem (42) is unique.Proof.
Since1>βn+1>max⁡1≤i≤n{βi}, we obtain the following by using Lemmas 11 and 12:
(46)IRβn+1f(t,u(t))=IRβn+1DRβn+1[uλ0(t)]+∑i=1n‍IRβn+1DRβn+1-βi[ci(t)uλi(t)]=IRβn+1DRβn+1[uλ0(t)]+∑i=1n‍IRβi(IRβn+1-βiDRβn+1-βi[ci(t)uλi(t)])=uλ0(t)+kn+1tβn+1-1+∑i=1n‍IRβi(ci(t)uλi(t)+kitβn+1-βi-1)=uλ0(t)+∑i=1n‍IRβi[ci(t)uλi(t)]+tβn+1-1Γ(βn+1){kn+1Γ(βn+1)+∑i=1n‍kiΓ(βn+1-βi)},
where ki, i=1,2,…,n+1, are some constants. Therefore,
(47)0={∫0t‍f(s,u(s))ds}|t=0={IR1-βn+1IRβn+1f(t,u(t))}|t=0={∑i=1n‍IR1+βi-βn+1[ci(t)uλi(t)]+IR1-βn+1[uλ0(t)]∑i=1n‍}|t=0+kn+1Γ(βn+1)+∑i=1n‍kiΓ(βn+1-βi)=δ+kn+1Γ(βn+1)+∑i=1n‍kiΓ(βn+1-βi);
that is,
(48)kn+1Γ(βn+1)+∑i=1n‍kiΓ(βn+1-βi)=-δ.
Integrating this equality into (46) obtains
(49)uλ0(t)=IRβn+1f(t,u(t))-∑i=1n‍IRβi[ci(t)uλi(t)]+δtβn+1-1Γ(βn+1)=1Γ(βn+1)∫0t‍(t-s)βn+1-1f(s,u(s))ds+δtβn+1-1Γ(βn+1)-∑i=1n‍1Γ(βi)∫0t‍(t-s)βi-1ci(s)uλi(s)ds.
Let U(t)=uλ0(t), and applying (49), given the fact that
(50)|f(s,u(s))|≤|f(s,0)|+|f(s,u(s))-f(s,0)|≤|f(s,0)|+cn+1(s)|u(s)|λn+1,
obtains the following:
(51)|U(t)|=|u(t)|λ0≤1Γ(βn+1)∫0t‍(t-s)βn+1-1|f(s,u(s))|ds+|δ|tβn+1-1Γ(βn+1)+∑i=1n‍1Γ(βi)∫0t‍(t-s)βi-1ci(s)|u(s)|λids≤1Γ(βn+1)∫0t‍(t-s)βn+1-1|f(s,0)|ds+|δ|tβn+1-1Γ(βn+1)+∑i=1n+1‍1Γ(βi)∫0t‍(t-s)βi-1ci(s)|u(s)|λids={1Γ(βn+1)∫0t‍(t-s)βn+1-1|f(s,0)|ds+|δ|tβn+1-1Γ(βn+1)}+∑i=1n+1‍1Γ(βi)∫0t‍(t-s)βi-1ci(s)|U(s)|λi/λ0ds.
If λ0>max⁡1≤i≤n+1⁡{λi}, then, for any i∈{1,2,…,n+1}, 0<λi/λ0<1. According to Theorem 4, for any t∈[0,T],
(52)|U(t)|≤A~0(t)11111111+∑k=1∞‍(∑1′,2′,…,k′=1n+1‍ɛkΓ(∑i=1kβi′)111111111111111111111×∫0t‍(t-s)∑i=1kβi′-1A~0(s)ds∑1′,2′,…,k′=1n+1‍),
where the expression of A~0(t) is shown in (45). Hence, the conclusion of (i) is derived.
In proving (ii), we assume that problem (42) has two continuous solutions u and v. Combining with the fact that ci(t)∈C[0,T] for any 1≤i≤n+1 and the boundedness of the continuous function on a closed interval, there exists a finite number M which satisfies that, for any t∈[0,T],
(53)M>max⁡{|u(t)|,|v(t)|,max⁡1≤i≤n+1|ci(t)|}.
Cauchy’s mean value theorem provides
(54)|uλi(t)-vλi(t)|=|uλ0(t)-vλ0(t)|·|λi·ξiλi-1λ0·ξiλ0-1|=λiλ0|uλ0(t)-vλ0(t)|·|ξi|λi-λ0,
where ξi, i=1,2,…,n+1, are the numbers between u(t) and v(t). The following estimation is deduced by applying (53) and the hypothesis of λ0≤min⁡1≤i≤n+1{λi} in (ii):
(55)|uλi(t)-vλi(t)|≤λiλ0|uλ0(t)-vλ0(t)|·Mλi-λ0,
holds for any t∈[0,T] and i=1,2,…,n+1. Therefore, (49), (53), and (55) obtain
(56)|uλ0(t)-vλ0(t)|=|1Γ(βn+1)∫0t‍(t-s)βn+1-11111111122111×[f(s,u(s))-f(s,v(s))]ds-∑i=1n‍1Γ(βi)∫0t‍(t-s)βi-1ci(s)111111111111111×[uλi(s)-vλi(s)]ds1Γ(βn+1)|≤∑i=1n+1‍1Γ(βi)∫0t‍(t-s)βi-11111111111111×ci(s)|uλi(s)-vλi(s)|ds≤∑i=1n+1‍λiM1+λi-λ0λ0Γ(βi)∫0t‍(t-s)βi-1111111111111111111×|uλ0(s)-vλ0(s)|ds.
According to Theorem 2,
(57)|uλ0(t)-vλ0(t)|≤0,
which means that
(58)u(t)=v(t),t∈[0,T].
This completes the proof of (ii).Moreover, we can also address the following initial value problem with the Hadamard type fractional derivative:(59)∑i=1n‍DHαn+1-αi[di(t)uγi(t)]+DHαn+1[uγ0(t)]=g(t,u(t));{∑i=1n‍IH1+αi-αn+1[di(t)uγi(t)]+IH1-αn+1[uγ0(t)]}|t=1=η,
where all functions are continuous. di(t)>0 and the constants γi,αi>0 (i=1,2,…,n). Also γ0>0, η∈R, and 1>αn+1>max⁡1≤i≤n{αi}. For any α∈(0,1) and a continuous function w, the operators DHα and IHα are presented below (see [18, page 110]):
(60)DHαw(t)=1Γ(1-α)(tddt)∫1t‍(ln⁡ts)-αw(s)sds;IHαw(t)=1Γ(α)∫1t‍(ln⁡ts)α-1w(s)sds.
We are able to deduce the following result by Theorems 3 and 5.Theorem 14.
Suppose that, for anyt∈[1,T] and y,z∈R, |g(t,y)-g(t,z)|≤dn+1(t)|yγn+1-zγn+1|, where the function dn+1(t)>0 is continuous and the constant γn+1∈(0,1).(i)
Ifγ0>max⁡1≤i≤n+1{γi}, then, for any solution u(t) of problem (59),
(61)|u(t)|≤{A¯0(t)+∑k=1∞‍(∑1′,2′,…,k′=1n+1‍ɛkΓ(∑i=1kαi′)1111111222222111111111×∫1t‍(ln⁡ts)∑i=1kA¯0(s)dss∑1′,2′,…,k′=1n+1‍)∑k=1∞‍(∑1′,2′,…,k′=1n+1‍ɛkΓ(∑i=1kαi′)}1/γ0,11111111122222211111111111121111111111111t∈[1,T],
where
(62)A¯0(t)=|η|(ln⁡t)αn+1-1Γ(αn+1)+1Γ(αn+1)∫1t‍(ln⁡ts)αn+1-1|g(s,0)|dss+∑i=1n+1‍1Γ(αi)(1-γiγ0)(γiɛγ0)γi/(γ0-γi)111111111×∫1t‍(ln⁡ts)αi-1[di(s)]γ0/(γ0-γi)dss.
Here ɛ is an arbitrary given positive number.(ii)
Ifγ0≤min⁡1≤i≤n+1{γi}, then the continuous solution of problem (59) is unique.The proof procedure is similar to that of Theorem13. Thus, the procedure is omitted here.



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