A New Fractional Integral Inequality with Singularity and Its Application

Abstract

We prove an integral inequality with singularity, which complements some known results. This inequality enables us to study the dependence of the solution on the initial condition to a fractional differential equation in the weighted space.

1. Introduction

Integral inequalities provide an excellent tool for the properties of solutions to differential equations, such as boundedness, existence, uniqueness, and stability (e.g., see [1–10]). For this reason, the study of integral inequalities has been emphasized by many authors. For example, in 1919, Gronwall in [11] proved a remarkable inequality which can be described by the following.

Suppose that 𝑥(𝑡) satisfies the relation 𝑥(𝑡)≤ℎ(𝑡)+^𝑡_𝑡0𝑘(𝑠)𝑥(𝑠)𝑑𝑠,𝑡₀≤𝑡<𝑇, (1.1) where all the functions involved are continuous on the interval [𝑡₀,𝑇), 𝑇≤∞, and 𝑘(𝑡)≥0. Consider 𝑥(𝑡)≤ℎ(𝑡)+^𝑡_𝑡0ℎ(𝑠)𝑘(𝑠)exp^𝑡_𝑠𝑘(𝜏)𝑑𝜏𝑑𝑠,𝑡₀≤𝑡<𝑇. (1.2) The inequality has attracted and continues to attract considerable attention in the literature. In 2007, Ye et al. [12] reported an integral inequality with singular kernel. The inequality can be stated as follows.

If 𝛽>0, 𝑎(𝑡) is a nonnegative and locally integrable on 0≤𝑡<𝑇, 𝑔(𝑡) is a nonnegative, nondecreasing continuous function on 0≤𝑡<𝑇, and 𝑔(𝑡)≤𝑀, where 𝑇≤∞, 𝑀 is a positive constant. Further suppose that 𝑢(𝑡) is nonnegative and locally integrable on 0≤𝑡<𝑇 with 𝑢(𝑡)≤𝑎(𝑡)+𝑔(𝑡)^𝑡₀(𝑡−𝑠)^𝛽−1𝑢(𝑠)𝑑𝑠,0≤𝑡<𝑇. (1.3) Then 𝑢(𝑡)≤𝑎(𝑡)+^∞_𝑛=1𝑔(𝑡)Γ𝛽^𝑛Γ𝑛𝛽^𝑡₀(𝑡−𝑠)^𝑛𝛽−1𝑎(𝑠)𝑑𝑠,0≤𝑡<𝑇. (1.4) Besides the above-mentioned inequalities, there are still many inequalities (e.g., see [13–15]).

But in the analysis of the dependence of the solution on the initial condition of a fractional differential equation in the weighted space, the bounds provided by the existing inequalities are not adequate. So it is natural and necessary to seek new inequality in order to obtain our desired results. In this paper, we present a new integral inequality, and then apply our inequality to investigate the dependence of the solution on the initial condition of a fractional differential equations in the weighted space.

2. An Integral Inequality

In this section, our main aim is to establish an integral inequality with singularity. Before proceeding, we give some useful definitions and lemmas.

Definition 2.1 (see [14, 16]). The gamma function is defined by Γ(𝑧)=∫^∞₀𝑒^−𝑡𝑡^𝑧−1𝑑𝑡,𝑧>0.

Definition 2.2 (see [14, 16]). The beta function is defined by 𝐵(𝑧,𝑤)=∫¹₀(1−𝑡)^𝑧−1𝑡^𝑤−1𝑑𝑡,𝑧,𝑤>0.

The beta function is connected with gamma function by the following relation [3, 14]: 𝐵()=Γ(𝑧)Γ(𝑤)𝑧,𝑤Γ(𝑧+𝑤),𝑧,𝑤>0. (2.1)

Lemma 2.3 (see [14]). Let 𝑧>0,  𝑎,𝑏∈𝑅. Then the quotient expansion of two gamma functions at infinity can be represented as follows: Γ(𝑧+𝑎)Γ(𝑧+𝑏)=𝑧^𝑎−𝑏11+𝑂𝑧,𝑧→∞. (2.2)

Lemma 2.4. Let 𝑧>0,  𝑎,𝑏∈𝑅. Then one has Γ(𝑧+𝑎)Γ()𝑧𝑧+𝑏=𝑂^𝑎−𝑏,𝑧⟶∞. (2.3)

Proof. By Lemma 2.3, we have lim_𝑧→∞(Γ(𝑧+𝑎)/Γ(𝑧+𝑏))/𝑧^𝑎−𝑏=lim_𝑧→∞(1+𝑂(1/𝑧))=1, which proves that Γ(𝑧+𝑎)/Γ(𝑧+𝑏)=𝑂(𝑧^𝑎−𝑏) as 𝑧→∞. The proof of this lemma is completed.

Based on Lemma 2.4, we can define a function.

Definition 2.5. Let 𝑏>𝑎>0,  𝜌>0. Then the following definition: 𝐹_{𝜌,𝑎,𝑏}(𝑧)∶=^∞_𝑘=0𝑐_𝑘𝑧^𝑘,𝑧∈𝑅 (2.4) is well defined, where 𝑐₀ is a positive constant, and 𝑐_𝑘+1=(Γ(𝑘𝜌+𝑎)/Γ(𝑘𝜌+𝑏))𝑐_𝑘.

Proof. We only need to show that the series in (2.4) is uniformly convergent for 𝑧∈𝑅. By Lemma 2.4, we know that 𝑐_𝑘+1/𝑐_𝑘=Γ(𝑘𝜌+𝑎)/Γ(𝑘𝜌+𝑏)=𝑂((𝑘𝜌)^𝑎−𝑏) as 𝑘→∞. Since 𝑏>𝑎>0,  𝑐_𝑘+1/𝑐_𝑘→0 as 𝑘→∞. This implies that the series in (2.4) is uniformly convergent for 𝑧∈𝑅. It follows that the definition is well defined.

Lemma 2.6. Let 𝑧,𝑤>0,  𝑡,𝑠∈𝑅 and 𝑡≠𝑠. Then one has ^𝑡_𝑠(𝑡−𝜏)^𝑧−1(𝜏−𝑠)^𝑤−1𝑑𝜏=(𝑡−𝑠)^{𝑧+𝑤−1}Γ(𝑧)Γ(𝑤)Γ().𝑧+𝑤 (2.5)

Proof. Making the substitution 𝜏=𝑠+𝜉(𝑡−𝑠) and combining the relation (2.1), we obtain ^𝑡_𝑠(𝑡−𝜏)^𝑧−1(𝜏−𝑠)^𝑤−1𝑑𝜏=(𝑡−𝑠)^{𝑧+𝑤−11}₀(1−𝜉)^𝑧−1𝜉^𝑤−1=()𝑑𝜉𝑡−𝑠^{𝑧+𝑤−1}𝐵(𝑧,𝑤)=(𝑡−𝑠)^{𝑧+𝑤−1}Γ(𝑧)Γ(𝑤)Γ().𝑧+𝑤 (2.6) The proof of this lemma is completed.

Now we can state the integral inequality.

Theorem 2.7. Let 𝛼,𝛽,𝛾>0,  𝛿=𝛼+𝛾−1>0,  𝜈=𝛽+𝛾−1>0,  𝑎>0, and let 𝑏(𝑡) be a nonnegative, nondecreasing continuous function on 0≤𝑡<𝑇,  𝑏(𝑡)≤𝑀, where 𝑇≤∞, 𝑀 is a positive constant. Further suppose that 𝑢(𝑡) is nonnegative and 𝑡^𝛾−1𝑢(𝑡) is locally integrable on 0≤𝑡<𝑇 with 𝑢(𝑡)≤𝑎𝑡^𝛼−1+𝑏(𝑡)^𝑡₀(𝑡−𝑠)^𝛽−1𝑠^𝛾−1𝑢(𝑠)𝑑𝑠,0≤𝑡<𝑇. (2.7) Then one has 𝑢(𝑡)≤𝑎𝑡^𝛼−1𝐹_{𝜈,𝛿,𝛿+𝛽}Γ𝛽𝑏(𝑡)𝑡^𝛽,0≤𝑡<𝑇. (2.8)

Proof. For convenience, we define an operator (ℛ𝑢)(𝑡)=𝑏(𝑡)^𝑡₀(𝑡−𝑠)^𝛽−1𝑠^𝛾−1𝑢(𝑠)𝑑𝑠. (2.9) Then (2.7) can be rewritten in the form 𝑢(𝑡)≤𝑎𝑡^𝛼−1+(ℛ𝑢)(𝑡). (2.10) Since 𝑏(𝑡) and 𝑢(𝑡) are nonnegative, it is easy to induce that 𝑢(𝑡)≤^𝑛_𝑘=0ℛ^𝑘𝑎𝑡^𝛼−1(𝑡)+ℛ^𝑛+1𝑢(𝑡),𝑛∈𝑁. (2.11) Let us prove that the following relation ℛ^𝑛𝑢(𝑡)≤𝑏(𝑡)Γ𝛽𝑏(𝑡)^{𝑛−1𝑛−1}_𝑖=1Γ(𝑖𝜈)Γ𝑖𝜈+𝛽^𝑡₀(𝑡−𝑠)^{𝑛𝜈−𝛾}𝑠^𝛾−1𝑢(𝑠)Γ𝛽𝑏(𝑡)𝑑𝑠,0<𝛾<1,_𝑛𝑡^{(𝑛−1)(𝛾−1)}Γ𝑛𝛽^𝑡₀(𝑡−𝑠)^𝑛𝛽−1𝑠^𝛾−1𝑢(𝑠)𝑑𝑠,𝛾≥1, (2.12) holds for any 𝑛∈𝑁⁺, where ∏⁰_𝑖=11=1, and (ℛ^𝑛𝑢)(𝑡)→0 as 𝑛→∞ for each 𝑡 in 0≤𝑡<𝑇.
Obviously, inequality (2.12) is valid for 𝑛=1, due to ∏⁰_𝑖=11=1. Suppose that the inequality is satisfied for any fixed 𝑛∈𝑁⁺. Let us verify that it is also satisfied for 𝑛+1. We first prove the case 0<𝛾<1. According to the induction hypothesis and Lemma 2.6, we have ℛ^𝑛+1𝑢(𝑡)=𝑏(𝑡)^𝑡₀(𝑡−𝑠)^𝛽−1𝑠^𝛾−1ℛ^𝑛𝑢(𝑠)𝑑𝑠≤𝑏²(𝑡)Γ𝛽𝑏(𝑡)^{𝑛−1𝑛−1}_𝑖=1Γ(𝑖𝜈)Γ𝑖𝜈+𝛽^𝑡₀(𝑡−𝑠)^𝛽−1𝑠^𝛾−1𝑠₀(𝑠−𝜏)^{𝑛𝜈−𝛾}𝜏^𝛾−1𝑢(𝜏)𝑑𝜏𝑑𝑠=𝑏²(𝑡)Γ𝛽𝑏(𝑡)^{𝑛−1𝑛−1}_𝑖=1Γ(𝑖𝜈)Γ𝑖𝜈+𝛽^𝑡₀𝜏^𝛾−1𝑢(𝜏)𝑑𝜏^𝑡_𝜏(𝑡−𝑠)^𝛽−1𝑠^𝛾−1(𝑠−𝜏)^{𝑛𝜈−𝛾}𝑑𝑠≤𝑏²(𝑡)Γ𝛽𝑏(𝑡)^{𝑛−1𝑛−1}_𝑖=1Γ(𝑖𝜈)Γ𝑖𝜈+𝛽^𝑡₀𝜏^𝛾−1𝑢(𝜏)𝑑𝜏^𝑡_𝜏(𝑡−𝑠)^𝛽−1(𝑠−𝜏)^𝑛𝜈−1𝑑𝑠=𝑏²(𝑡)Γ𝛽𝑏(𝑡)^{𝑛−1𝑛−1}_𝑖=1Γ(𝑖𝜈)Γ𝑖𝜈+𝛽^𝑡₀𝜏^𝛾−1𝑢(𝜏)(𝑡−𝜏)^{𝑛𝜈+𝛽−1}Γ𝛽Γ(𝑛𝜈)Γ(𝑡)Γ𝛽𝑏(𝑡)𝑛𝜈+𝛽𝑑𝜏=𝑏^𝑛𝑛_𝑖=1Γ(𝑖𝜈)Γ𝑖𝜈+𝛽^𝑡₀(𝑡−𝜏)^{(𝑛+1)𝜈−𝛾}𝜏^𝛾−1𝑢(𝜏)𝑑𝜏, (2.13) which is estimated with the help of 𝑠^𝛾−1≤(𝑠−𝜏)^𝛾−1,0≤𝜏≤𝑠,0<𝛾<1. (2.14) So, for the case 0<𝛾<1, inequality (2.12) is true for any 𝑛∈𝑁⁺. Now we prove the case 𝛾≥1. Similarly, according to the induction hypothesis and Lemma 2.6, we get ℛ^𝑛+1𝑢(𝑡)=𝑏(𝑡)^𝑡₀(𝑡−𝑠)^𝛽−1𝑠^𝛾−1ℛ^𝑛𝑢(𝑠)(𝑡)𝑑𝑠≤𝑏^𝑡₀(𝑡−𝑠)^𝛽−1𝑠^𝛾−1Γ𝛽𝑏(𝑠)^𝑛𝑠^{(𝑛−1)(𝛾−1)}Γ𝑛𝛽^𝑠₀(𝑠−𝜏)^𝑛𝛽−1𝜏^𝛾−1𝑢(𝜏)(𝑡)Γ𝛽𝑏(𝑡)𝑑𝜏𝑑𝑠=𝑏^𝑛Γ𝑛𝛽^𝑡₀𝜏^𝛾−1𝑢(𝜏)𝑑𝜏^𝑡_𝜏(𝑡−𝑠)^𝛽−1𝑠^{𝑛(𝛾−1)}(𝑠−𝜏)^𝑛𝛽−1(𝑡)Γ𝛽𝑏(𝑡)𝑑𝑠≤𝑏^𝑛𝑡^{𝑛(𝛾−1)}Γ𝑛𝛽^𝑡₀𝜏^𝛾−1𝑢(𝜏)𝑑𝜏^𝑡_𝜏(𝑡−𝑠)^𝛽−1(𝑠−𝜏)^𝑛𝛽−1(𝑡)Γ𝛽𝑏(𝑡)𝑑𝑠=𝑏^𝑛𝑡^{𝑛(𝛾−1)}Γ𝑛𝛽^𝑡₀𝜏^𝛾−1𝑢(𝜏)(𝑡−𝜏)^{𝑛𝛽+𝛽−1}Γ𝛽Γ𝑛𝛽Γ=Γ𝛽𝑏(𝑡)𝑛𝛽+𝛽𝑑𝜏_𝑛+1𝑡^{𝑛(𝛾−1)}Γ(𝑛+1)𝛽^𝑡₀𝜏^𝛾−1𝑢(𝜏)(𝑡−𝜏)^{(𝑛+1)𝛽−1}𝑑𝜏, (2.15) which is calculated with the help of 𝑠^{𝑛(𝛾−1)}≤𝑡^{𝑛(𝛾−1)},0≤𝑠≤𝑡,𝛾≥1,𝑛∈𝑁⁺. (2.16) So, for the case 𝛾≥1, inequality (2.12) is true for any 𝑛∈𝑁⁺. Based on this analysis, we conclude that inequality (2.12) holds for any 𝑛∈𝑁⁺.
Next, we show that (ℛ^𝑛𝑢)(𝑡)→0 as 𝑛→∞. Now, we go back to inequality (2.12). For the case 0<𝛾<1, we denote 𝐾_𝑛(𝑡,𝑠)=𝐵_𝑛(𝑡−𝑠)^{𝑛𝜈−𝛾}, where 𝐵_𝑛=𝑏(𝑡)(Γ(𝛽)𝑏(𝑡))^𝑛−1∏^𝑛−1_𝑖=1(Γ(𝑖𝜈)/Γ(𝑖𝜈+𝛽)). Note that 𝐵₁(𝑡),𝐵=𝑏^𝑛+1𝐵_𝑛𝛽𝑏(𝑡)Γ()=Γ𝑛𝜈Γ.𝑛𝜈+𝛽 (2.17) Since 𝑏(𝑡)≤𝑀, by Lemma 2.4, we obtain 𝐵_𝑛+1/𝐵_𝑛→0 as 𝑛→∞. This implies that 𝐾_𝑛(𝑡,𝑠)→0 as 𝑛→∞. It follows that (𝑅^𝑛𝑢)(𝑡)→0 as 𝑛→∞ for the case 0<𝛾<1. For the case 𝛾≥1, we denote 𝐾_𝑛(𝑡,𝑠)=𝐵_𝑛(𝑡−𝑠)^𝑛𝛽−1, where 𝐵_𝑛=(Γ(𝛽)𝑏(𝑡))^𝑛𝑡^{(𝑛−1)(𝛾−1)}/Γ(𝑛𝛽). Note that 𝐵₁=𝑏(𝑡),𝐵_𝑛+1𝐵_𝑛=Γ𝛽𝑏(𝑡)𝑡^𝛾−1Γ𝑛𝛽Γ𝑛𝛽+𝛽. (2.18) Using the same arguments as above, we know that 𝐾_𝑛(𝑡,𝑠)→0 as 𝑛→∞. It follows that (ℛ^𝑛𝑢)(𝑡)→0 as 𝑛→∞ for the case 𝛾≥1. So, it has (ℛ^𝑛𝑢)(𝑡)→0 as 𝑛→∞ for the two cases 0<𝛾<1 and 𝛾≥1. This, together with (2.11), leads to 𝑢(𝑡)≤∑^∞_𝑘=0(ℛ^𝑘𝑎𝑡^𝛼−1)(𝑡).
Finally, we show that ℛ^𝑘𝑎𝑡^𝛼−1(𝑡)≤𝑎Γ𝛽𝑏(𝑡)^𝑘𝑐_𝑘𝑡^𝛼−1𝑡^𝑘𝜈,𝑘∈𝑁, (2.19) where 𝑐₀=1,  𝑐_𝑘=∏^𝑘−1_𝑖=0(Γ(𝑖𝜈+𝛿)/Γ(𝑖𝜈+𝛿+𝛽)),  𝑘∈𝑁⁺.
Obviously, inequality (2.19) is true for 𝑘=0. Suppose that the inequality is satisfied for any fixed 𝑘∈𝑁. Let us verify that it is also satisfied for 𝑘+1. According to the induction hypothesis and Lemma 2.6, we obtain ℛ^𝑘+1𝑎𝑡^𝛼−1(𝑡)≤𝑏(𝑡)^𝑡₀(𝑡−𝑠)^𝛽−1𝑠^𝛾−1ℛ^𝑘𝑎𝑠^𝛼−1(𝑠)Γ𝛽𝑏(𝑡)𝑑𝑠≤𝑎^𝑘𝑐_𝑘𝑏(𝑡)^𝑡₀(𝑡−𝑠)^𝛽−1𝑠^{𝑘𝜈+𝛿−1}Γ𝛽𝑏(𝑡)𝑑𝑠=𝑎^𝑘𝑐_𝑘𝑏(𝑡)𝑡^{𝑘𝜈+𝛿+𝛽−1}Γ𝛽Γ(𝑘𝜈+𝛿)ΓΓ𝛽𝑏(𝑡)𝑘𝜈+𝛿+𝛽=𝑎^𝑘+1𝑐_𝑘+1𝑡^𝛼−1𝑡^(𝑘+1)𝜈. (2.20) This proves that inequality (2.19) is satisfied for any 𝑘∈𝑁. In other words, we have proved that 𝑢(𝑡)≤^∞_𝑘=0𝑎Γ𝛽𝑏(𝑡)^𝑘𝑐_𝑘𝑡^𝛼−1𝑡^𝑘𝜈, (2.21) where 𝑐₀=1, 𝑐_𝑘=∏^𝑘−1_𝑖=0(Γ(𝑖𝜈+𝛿)/Γ(𝑖𝜈+𝛿+𝛽)), 𝑘∈𝑁⁺. By virtue of Definition 2.5, we can arrive at inequality (2.8) and the proof of this theorem is completed.

For the case 𝑏(𝑡)≡𝑏>0 in Theorem 2.7, we can obtain the following corollary, which can be found in [17].

Corollary 2.8. Let 𝛼,𝛽,𝛾>0,  𝛿=𝛼+𝛾−1>0,  𝜈=𝛽+𝛾−1>0,  𝑎,𝑏>0. And suppose that 𝑢(𝑡) is nonnegative and 𝑡^𝛾−1𝑢(𝑡) is locally integrable on 0≤𝑡<𝑇 (𝑇≤∞) with 𝑢(𝑡)≤𝑎𝑡^𝛼−1+𝑏^𝑡₀(𝑡−𝑠)^𝛽−1𝑠^𝛾−1𝑢(𝑠)𝑑𝑠,0≤𝑡<𝑇. (2.22) Then one has 𝑢(𝑡)≤𝑎𝑡^𝛼−1𝐹_{𝜈,𝛿,𝛿+𝛽}Γ𝛽𝑏𝑡^𝛽,0≤𝑡<𝑇. (2.23)

For 𝛼=𝛾=1 in Theorem 2.7, we can arrive at the following corollary, which can be found in [12].

Corollary 2.9. Let 𝛽,  𝑎>0,  𝑏(𝑡) be a nonnegative, nondecreasing continuous function on 0≤𝑡<𝑇, 𝑏(𝑡)≤𝑀, where 𝑇≤∞, 𝑀 is a positive constant. And suppose that 𝑢(𝑡) is nonnegative and locally integrable on 0≤𝑡<𝑇 with 𝑢(𝑡)≤𝑎+𝑏(𝑡)^𝑡₀(𝑡−𝑠)^𝛽−1𝑢(𝑠)𝑑𝑠,0≤𝑡<𝑇. (2.24) Then one has 𝑢(𝑡)≤𝑎𝐸_𝛽Γ𝛽𝑏(𝑡)𝑡^𝛽,0≤𝑡<𝑇. (2.25)

3. Application

In this section, we will apply our established result to study the dependence of the solution on the initial condition of a fractional differential equation with the Riemann-Liouville derivative.

For the reader's convenience, we first recall several definitions of the Reimann-Liouville integral and derivative. From now on, we assume that 𝑇 is a finite positive constant, that is, 𝑇≠∞.

Definition 3.1 (see [14, 16]). Let 0<𝑝<1. The Riemann-Liouville integral of order 𝑝 is defined by 𝐼^𝑝₀₊𝑥(𝑡)=1Γ𝑝^𝑡₀(𝑡−𝑠)^𝑝−1𝑥(𝑠)𝑑𝑠,0≤𝑡≤𝑇. (3.1)

Definition 3.2 (see [14, 16]). Let 0<𝑝<1. The Riemann-Liouville derivative of order 𝑝 is defined by 𝐷^𝑝₀₊𝑥(𝑡)=1Γ𝑑1−𝑝𝑑𝑡^𝑡₀(𝑡−𝑠)^−𝑝𝑥(𝑠)𝑑𝑠,0≤𝑡≤𝑇. (3.2)

Now we consider the following initial value problem of the form 𝐷^𝑝₀₊𝑥(𝑡)=𝑓(𝑡,𝑥(𝑡)),lim_𝑡→0+𝐼₀^1−𝑝₊𝑥(𝑡)=𝑥₀,0<𝑝<1,0<𝑡≤𝑇,𝑥₀∈𝑅. (3.3)

With regard to problem (3.3), the existence and uniqueness of the solution can be found in the book by Kilbas et al. [14]. For the completeness of this paper, we give the existence and uniqueness of the solution to (3.3) in the weighted space 𝐶_1−𝑝([0,𝑇]). The space 𝐶_1−𝑝([0,𝑇]) consists of all functions 𝑥∈𝐶((0,𝑇]) such that 𝑡^1−𝑝𝑥(𝑡)∈𝐶([0,𝑇]), which turns out to be a Banach space when endowed with the norm |𝑥|_1−𝑝=max_{0≤𝑡≤𝑇}|𝑡^1−𝑝𝑥(𝑡)|.

Theorem 3.3 (see [14]). Let 0<𝑝<1, and 𝑓(𝑡,𝑥)∶(0,𝑇]×𝑅→𝑅 be a function such that for any 𝑥∈𝑅, 𝑓(𝑡,𝑥)∈𝐶_1−𝑝([0,𝑇]). Further assume that for any 𝑡∈(0,𝑇],  𝑥,𝑦∈𝑅, the following inequality 𝑓(𝑡,𝑥)−𝑓𝑡,𝑦≤𝐿𝑥−𝑦 (3.4) holds, where 𝐿>0 is a constant. Then there exists a unique solution 𝑥(𝑡) to problem (3.3) in the space 𝐶_1−𝑝([0,𝑇]).

Theorem 3.4. Let 0<𝑝<1, and 𝑓(𝑡,𝑥)∶(0,𝑇]×𝑅→𝑅 be a function such that for any 𝑥∈𝑅, 𝑓(𝑡,𝑥)∈𝐶_1−𝑝([0,𝑇]). Further assume that for any 𝑡∈(0,𝑇], 𝑥,𝑦∈𝑅, the following inequality 𝑓(𝑡,𝑥)−𝑓𝑡,𝑦≤𝐿𝑥−𝑦 (3.5) holds, where 𝐿>0 is a constant. Assume that 𝑥 and 𝑦 are the solutions of problem (3.3) and 𝐷^𝑝₀₊𝑦(𝑡)=𝑓𝑡,𝑦(𝑡),lim_𝑡→0+𝐼₀^1−𝑝₊𝑦(𝑡)=𝑦₀,0<𝑡≤𝑇,𝑦₀∈𝑅, (3.6) respectively. Then, for 0≤𝑡≤𝑇, one has 𝑡^1−𝑝𝑥(𝑡)(𝑡≤𝑥−𝑦)⁰−𝑦₀Γ𝑝𝐹_{2𝑝−1,𝑝,2𝑝}(),1𝐿𝑡2𝑥<𝑝<1,0≤𝑡≤𝑇,₀−𝑦₀Γ𝑝𝐹_{2𝑝+𝑞−1,𝑝+𝑞,2𝑝+𝑞}𝐿^∗𝑡1,0<𝑝≤2𝑥,0≤𝑡<1,₀−𝑦₀Γ𝑝𝐹_{2𝑝+𝑞′−1,𝑝+𝑞′,2𝑝+𝑞′}𝐿^′𝑡1,0<𝑝≤2,1≤𝑡≤𝑇, (3.7) where 𝑞,𝑞^′,𝐿^∗,𝐿^′ are positive constants such that 1−2𝑝<𝑞<log^𝐿/𝐿∗_𝑡1,0<𝑝≤2𝑞,0<𝑡<1,_′>max1−2𝑝,log^𝐿/𝐿′_𝑡1,0<𝑝≤2,1≤𝑡≤𝑇. (3.8)

Proof. The proof is rather technical. We first prove the case 1/2<𝑝<1 and 0≤𝑡≤𝑇. Since 𝑥(𝑡) and 𝑦(𝑡) are the solutions of (3.3) and (3.6), we have 𝑥(𝑡)=𝑥₀𝑡^𝑝−1Γ𝑝+1Γ𝑝^𝑡₀(𝑡−𝑠)^𝑝−1𝑓(𝑠,𝑥(𝑠))𝑑𝑠, (3.9)𝑦(𝑡)=𝑦₀𝑡^𝑝−1Γ𝑝+1Γ𝑝^𝑡₀(𝑡−𝑠)^𝑝−1𝑓𝑠,𝑦(𝑠)𝑑𝑠. (3.10) Subtracting (3.10) from (3.9) and using the Lipschitz condition (3.5), we obtain 𝑥(𝑡)(𝑡≤𝑥−𝑦)⁰−𝑦₀𝑡^𝑝−1Γ𝑝+𝐿Γ𝑝^𝑡₀(𝑡−𝑠)^𝑝−1𝑥(𝑠)−𝑦(𝑠)𝑑𝑠. (3.11) Taking into account that 𝑥(𝑡),𝑦(𝑡)∈𝐶_1−𝑝([0,𝑇]), we multiply at both sides of inequality (3.11) by 𝑡^1−𝑝 to get 𝑡^1−𝑝𝑥(𝑡)(𝑡≤𝑥−𝑦)⁰−𝑦₀Γ𝑝+𝐿𝑡^1−𝑝Γ𝑝^𝑡₀(𝑡−𝑠)^𝑝−1𝑠^𝑝−1𝑠^1−𝑝𝑥(𝑠)−𝑦(𝑠)𝑑𝑠. (3.12) Denote 𝑢(𝑡)=𝑡^1−𝑝|𝑥(𝑡)−𝑦(𝑡)|. Then, (3.12) can be written as 𝑢(𝑡)≤𝑥₀−𝑦₀Γ𝑝+𝐿𝑡^1−𝑝Γ𝑝^𝑡₀(𝑡−𝑠)^𝑝−1𝑠^𝑝−1𝑢(𝑠)𝑑𝑠. (3.13) Putting 𝑎=|𝑥₀−𝑦₀|/Γ(𝑝), 𝑏(𝑡)=𝐿𝑡^1−𝑝/Γ(𝑝), 𝛼=1, 𝛽=𝑝, 𝛾=𝑝, we see that 𝛼,𝛽,𝛾>0, 𝛿=𝛼+𝛾−1=𝑝>0, 𝜈=𝛽+𝛾−1=2𝑝−1>0, 𝑎>0, and 𝑏(𝑡) is a nondecreasing continuous function due to 𝑝,𝐿>0. So, applying Theorem 2.7 to (3.13), we obtain 𝑢(𝑡)≤𝑥₀−𝑦₀Γ𝑝𝐹_{2𝑝−1,𝑝,2𝑝}(),1𝐿𝑡2<𝑝<1,0≤𝑡≤𝑇. (3.14)
Next, we prove the case 0<𝑝≤1/2 and 0≤𝑡<1. Notice that the Lipschitz condition (3.5) holds for each 𝑡 in 𝑡∈(0,𝑇]. Since 𝑡>0 and 𝐿>0, we can always choose two positive constants 𝑞, 𝐿^∗ such that 1−2𝑝<𝑞<log^𝐿/𝐿∗_𝑡,0<𝑡<1. (3.15) Condition (3.15) means that 0≤1−2𝑝<𝑞 and 𝐿<𝐿^∗𝑡^𝑞. That is to say, if the Lipschitz condition (3.5) holds for each 𝑡 in 𝑡∈(0,𝑇], then we can always choose two positive constants 𝑞,  𝐿^∗ such that the following condition 0≤1−2𝑝<𝑞,𝑓(𝑡,𝑢)−𝑓(𝑡,𝑣)≤𝐿^∗𝑡^𝑞|𝑢−𝑣| (3.16) holds for each 𝑡 in 𝑡∈(0,𝑇].
Subtracting (3.10) from (3.9) and using condition (3.16), we obtain 𝑥(𝑡)(𝑡≤𝑥−𝑦)⁰−𝑦₀𝑡^𝑝−1Γ𝑝+𝐿^∗Γ𝑝^𝑡₀(𝑡−𝑠)^𝑝−1𝑠^𝑞𝑥(𝑠)−𝑦(𝑠)𝑑𝑠. (3.17) Multiplying 𝑡^1−𝑝 on both sides of (3.17), we get 𝑢(𝑡)≤𝑥₀−𝑦₀Γ𝑝+𝐿^∗𝑡^1−𝑝Γ𝑝^𝑡₀(𝑡−𝑠)^𝑝−1𝑠^{𝑝+𝑞−1}𝑢(𝑠)𝑑𝑠, (3.18) where 𝑢(𝑡) is defined as before. Now, putting 𝑎=|𝑥₀−𝑦₀|/Γ(𝑝), 𝑏(𝑡)=𝐿^∗𝑡^1−𝑝/Γ(𝑝), 𝛼=1, 𝛽=𝑝, 𝛾=𝑝+𝑞, we see that 𝛼,𝛽,𝛾>0, 𝛿=𝛼+𝛾−1=𝑝+𝑞>0, 𝜈=𝛽+𝛾−1=2𝑝+𝑞−1=𝑞−(1−2𝑝)>0, 𝑎>0, 𝑏(𝑡) is a nondecreasing continuous function due to 𝑝,𝐿^∗>0. So, applying Theorem 2.7 to (3.18), we have 𝑢(𝑡)≤𝑥₀−𝑦₀Γ𝑝𝐹_{2𝑝+𝑞−1,𝑝+𝑞,2𝑝+𝑞}𝐿^∗𝑡1,0<𝑝≤2,0≤𝑡<1. (3.19)
Finally, we prove the case 0<𝑝≤1/2 and 1≤𝑡≤𝑇. Since 𝑡>0 and 𝐿>0, we can always choose two positive constants 𝑞^′, 𝐿^′ such that 𝑞^′>max1−2𝑝,log^𝐿/𝐿′_𝑡,1≤𝑡≤𝑇. (3.20) Condition (3.20) means that 0≤1−2𝑝<𝑞^′ and 𝐿<𝐿^′𝑡^𝑞. Using the same arguments as above, we can obtain that 𝑢(𝑡)≤𝑥₀−𝑦₀Γ𝑝𝐹_{2𝑝+𝑞′−1,𝑝+𝑞′,2𝑝+𝑞′}𝐿^′𝑡1,0<𝑝≤2,1≤𝑡≤𝑇, (3.21) where 𝑢(𝑡) is defined as before. So the conclusion of this theorem is true.

From the proof of Theorem 3.4, we can see that the integral inequality in Theorem 2.7 is very useful. This demonstrates that our investigation is meaningful.

Acknowledgmets

This work was supported by the National Natural Science Foundation of China (11071192) and the International Science and Technology (S&T) Cooperation Program of China (2010DFA14700).