Existence of Positive Solutions for Multiterm Fractional Differential Equations of Finite Delay with Polynomial Coefficients

Abstract

Existence of positive solutions has been studied by A. Babakhani and V. Daftardar-Gejji (2003) in case of multiterm nonautonomous fractional differential equations with constant coefficients. In the present paper we discuss existence of positive solutions in case of multiterm fractional differential equations of finite delay with polynomial coefficients.

1. Introduction

In last 30 years, the theory of ordinary differential equations of fractional order has become a new important branch (see, e.g., [1–5] and the references therein). Numerous applications of such equations have been presented [3–10]. Existence of positive solution of fractional ordinary differential equations has been well investigated for fractional functional differential equations [1, 6, 11–14]. Ye et al. [6] have addressed the question of existence of positive solutions for the nonlinear fractional functional differential equation 𝐷^𝛼[𝑥(𝑡)−𝑥(0)]=𝑥(𝑡)𝑓(𝑡,𝑥_𝑡),𝑡∈(0,𝑇],𝑥(𝑡)=𝜙(𝑡)≥0,𝑡∈[−𝑤,0], (1.1) by using the sub- and supersolution method, where 0<𝛼<1, 𝐷^𝛼 is the standard Riemann-Liouville fractional derivative, 𝜙∈𝐶 and 𝑓∶[0,𝑇]×𝐶→ℝ⁺ is continuous, as usual, 𝐶=𝐶([−𝑤,0],ℝ⁺) is the space of continuous function from [−𝑤,0] to ℝ⁺, 𝑤>0, equipped with the sup norm: ‖𝜙‖=max_{−𝑤≤Θ≤0}|𝜙(𝑡)|, (1.2) and 𝑥_𝑡 denotes the function in 𝐶 defined by 𝑥_𝑡(𝜃)=𝑥(𝑡+𝜃),−𝑤≤𝜃≤0. (1.3) They require that the nonlinearity 𝑓(𝑡,𝑥_𝑡) is increasing in 𝑥_𝑡 for each 𝑡∈[0,𝑇].

As a pursuit of this in the present paper, we deal with the existence of positive solutions in the case of multiterm differential equations with polynomial coefficients of the fractional type: ℒ(𝐷)[𝑥(𝑡)−𝑥(0)]=𝑓(𝑡,𝑥_𝑡),𝑡∈(0,𝑇],𝑥(𝑡)=𝜙(𝑡)≥0,𝑡∈[−𝑤,0], (1.4) where ℒ(𝐷)=𝐷^𝛼𝑛−^𝑛−1_𝑗=1𝑝_𝑗(𝑡)𝐷^{𝛼𝑛−𝑗},0<𝛼₁<⋯<𝛼_𝑛<1,𝑝_𝑗(𝑡)=^𝑁𝑗_𝑘=0𝑎_𝑗𝑘𝑡^𝑘,𝑝_𝑗^(2𝑚)(𝑡)≥0,𝑝_𝑗^(2𝑚+1)𝑁(𝑡)≤0,𝑚=0,1,…,[^𝑗2],𝑗=1,2,…,𝑛−1, (1.5) and 𝐷^𝛼𝑗 is the standard Riemann-Liouville fractional derivative, 𝑇>0, 𝑤>0, 𝜙∈𝐶=𝐶([−𝑤,0],ℝ⁺) and 𝑓∶𝐼×𝐶→ℝ⁺ is a given continuous function, 𝐼=[0,𝑇].

2. Preliminaties

Let 𝐸 be a real Banach space with a cone 𝐾. 𝐾 introduces a partial order ≤ in 𝐸 in the following manner [13]: 𝑥≤𝑦if𝑦−𝑥∈𝐾. (2.1)

Definition 2.1 (see [15]). For 𝑥,𝑦∈𝐸 the order interval ⟨𝑥,𝑦⟩ is defined as ⟨𝑥,𝑦⟩={𝑧∈𝐸∶𝑥≤𝑧≤𝑦}. (2.2)

Definition 2.2 (see [15]). A cone 𝐾 is called normal, if there exists a positive constant 𝑟 such that 𝑓,𝑔∈𝐾 and 𝜗≺𝑓≺𝑔 implies ‖𝑓‖≤𝑟‖𝑔‖, where 𝜗 denotes the zero element of 𝐾.

Definition 2.3 (see [16, 17]). Let 𝑓∶[𝑎,𝑏]→ℝ, and 𝑓∈𝐿¹[𝑎,𝑏]. The left-sided Riemann-Liouville fractional integral of 𝑓 of order 𝛼 is defined as 𝐼^𝛼_𝑎1𝑓(𝑥)=Γ(𝛼)^𝑥_𝑎(𝑥−𝑡)^𝛼−1𝑓(𝑡)𝑑𝑡,𝛼>0,𝑥∈[𝑎,𝑏]. (2.3)

Definition 2.4 (see [16, 17]). The left-sided Riemann-Liouville fractional derivative of a function 𝑓∶[𝑎,𝑏]→ℝ is defined as 𝐷^𝛼_𝑎𝑓(𝑥)=𝐷^𝑚[𝐼_𝑎^𝑚−𝛼𝑓(𝑥)],𝑥∈[𝑎,𝑏], (2.4) where 𝑚=[𝛼]+1, 𝐷^𝑚=𝑑^𝑚/𝑑𝑡^𝑚. We denote 𝐷^𝛼₀ by 𝐷^𝛼 and 𝐼^𝛼₀ by 𝐼^𝛼. If the fractional derivative 𝐷^𝛼_𝑎𝑓(𝑥) is integrable, then [16, page 71] 𝐼^𝛼_𝑎(𝐷^𝛽_𝑎𝑓(𝑥))=𝐼_𝑎^𝛼−𝛽𝑓(𝑥)−[𝐼_𝑎^1−𝛽𝑓(𝑥)]_𝑥=𝑎𝑥^𝛼−1Γ(𝛼),0<𝛽≤𝛼<1. (2.5)
If 𝑓 is continuous on [𝑎,𝑏], then [𝐼_𝑎^1−𝛽𝑓(𝑥)]_𝑥=𝑎=0 and (2.5) reduces to 𝐼^𝛼_𝑎(𝐷^𝛽_𝑎𝑓(𝑥))=𝐼_𝑎^𝛼−𝛽𝑓(𝑥),0<𝛽≤𝛼<1. (2.6)

Proposition 2.5. Let 𝑦 be continuous on [0,𝑇], 𝑇>0 and let 𝑛 be a nonnegative integer, then 𝐼^𝛼(𝑡^𝑛𝑥(𝑡))=^𝑛_𝑘=0𝑘−𝛼[𝐷^𝑘𝑡^𝑛][𝐼^𝛼+𝑘𝑥(𝑡)]=^𝑛_𝑘=0𝑘−𝛼𝑛!𝑡^𝑛−𝑘𝐼(𝑛−𝑘)!^𝛼+𝑘𝑥(𝑡), (2.7) where 𝑘−𝛼=(−1)^𝑘Γ(𝛼+1)𝑘!Γ(𝛼)=(−1)^𝑘𝛼𝑘=Γ(1−𝛼)Γ(𝑘+1)Γ(1−𝛼−𝑘). (2.8)

The proof of the above proposition can be found in [17, page 53].

Corollary 2.6. Let 𝑥∈𝐶[0,𝑇], 𝑇>0 and 𝑝_𝑗(𝑡)=∑^𝑁𝑗_𝑘=0𝑎_𝑗𝑘𝑡^𝑘, 𝑁_𝑗∈ℕ∪{0}, 𝑗=1,2,…,𝑛, then 𝐼^𝛼(^𝑛_𝑗=1𝑝_𝑗(𝑡)𝑥(𝑡))=^𝑛𝑁_𝑗=1^𝑗𝑘_𝑘=0𝑟=0𝑎_𝑗𝑘𝑟−𝛼𝑘!𝑡^𝑘−𝑟(𝑘−𝑟)![𝐼^𝛼+𝑟𝑥(𝑡)]. (2.9)

Theorem 2.7 (see [10]). Let 𝐸 be a Banach space with 𝐶⊆𝐸 closed and convex. Assume that 𝑈 is a relatively open subset of 𝐶 with 0∈𝑈 and 𝐹∶𝑈→𝐶 is a continuous and compact map. Then either(1)𝐹 has a fixed point in 𝑈, or(2)there exists 𝑢∈𝜕𝑈 and 𝜆∈(0,1) with 𝑢=𝜆𝐹(𝑢).

3. Existence of Positive Solution

In this section, we discuss the existence of positive solutions for (1.4). Using (2.5), (2.6), and Corollary 2.6, (1.4) is equivalent to the integral equation 𝑥(𝑡)=𝑥(0)+ℐ[𝑥(𝑡)−𝑥(0)]+𝐼^𝛼𝑛𝑓(𝑡,𝑥_𝑡),𝑡∈(0,𝑇],𝜙(𝑡),𝑡∈[−𝑤,0], (3.1) where ℐ=^𝑛−1𝑁_𝑗=1^𝑗𝑘_𝑘=0𝑟=0𝑎_𝑗𝑘−𝛼_𝑛𝑟𝑘!𝑡^𝑘−𝑟𝐼(𝑘−𝑟)!^{𝛼𝑛−𝛼𝑛−𝑗+𝑟}. (3.2) Let 𝑦(⋅)∶[−𝑤,𝑇]→[0,+∞) be the function defined by 𝑦(𝑡)=𝜙(0),𝑡∈𝐼,𝜙(𝑡)≥0,𝑡∈[−𝑤,0], (3.3) then 𝑦₀=𝜙, for each 𝑧∈𝐶(𝐼,ℝ) with 𝑧(0)=0, we denote by 𝑧 the function define by 𝑧(𝑡)=𝑧(𝑡),𝑡∈𝐼,0,𝑡∈[−𝑤,0]. (3.4) We can decompose 𝑥(⋅) as 𝑥(𝑡)=𝑧(𝑡)+𝑦(𝑡), 𝑡∈[−𝑤,𝑇], which implies 𝑥_𝑡=𝑧_𝑡+𝑦_𝑡, for 𝑡∈𝐼. Therefore, (3.1) is equivalent to the integral equation 𝑧(𝑡)=ℐ𝑧(𝑡)+𝐼^𝛼𝑛𝑓(𝑡,𝑧_𝑡+𝑦_𝑡),𝑡∈𝐼, (3.5) where ℐ is defined (3.2). Set 𝐴₀={𝑧∈𝐶(𝐼,ℝ)∶𝑧₀=0} and let ‖𝑧‖_𝑇 be the seminorm in 𝐴₀ defined by ‖𝑧‖_𝑇=‖𝑧₀‖+‖𝑧‖=‖𝑧‖=∶sup{|𝑧(𝑡)|∶𝑡∈𝐼},𝑧∈𝐴₀, (3.6) and 𝐴₀ is a Banach space with norm ‖⋅‖_𝑇. Let 𝐾 be a cone of 𝐴₀, 𝐾={𝑧∈𝐴₀;𝑧(𝑡)≥0,𝑡∈𝐼} and 𝐾^∗={𝑥∈𝐶([−𝑤,𝑇],ℝ⁺);𝑥(𝑡)=𝜙(𝑡)≥0,𝑡∈[−𝑤,0]}. (3.7) Define the operator 𝐹∶𝐾→𝐾 by 𝐹𝑧(𝑡)=ℐ𝑧(𝑡)+𝐼^𝛼𝑛𝑓(𝑡,𝑧_𝑡+𝑦_𝑡),𝑡∈𝐼. (3.8)

Theorem 3.1. Suppose that the following conditions hold:(1)there exist 𝑝,𝑞∈𝐶(𝐼,ℝ⁺) such that 𝑓(𝑡,𝑥_𝑡)≤𝑝(𝑡)+𝑞(𝑡)‖𝑥_𝑡‖, for 𝑡∈𝐼, 𝑥_𝑡∈𝐶, and ‖𝐼^𝛼𝑛𝑝‖=sup_{𝑡∈[0,𝑇]}𝐼^𝛼𝑛𝑝(𝑡)<∞, ‖𝐼^𝛼𝑛𝑞‖=sup_{𝑡∈[0,𝑇]}𝐼^𝛼𝑛𝑞(𝑡)<∞,(2)1−ℐ(𝑇)−‖𝐼^𝛼𝑛𝑞‖>0, where ℐ(𝑇)=^𝑛−1𝑁_𝑗=1^𝑗𝑘_𝑘=0𝑟=0|𝑎_𝑗𝑘−𝛼_𝑛𝑟|𝑘!𝑇^{𝛼𝑛−𝛼𝑛−𝑗+𝑘}(𝑘−𝑟)!Γ(𝛼_𝑛−𝛼_𝑛−𝑗+𝑟). (3.9)Then (1.4) has at least a positive solution 𝑥^∗∈𝐾^∗, satisfying ‖𝑥^∗‖≤max{‖𝜙‖,ℎ}, where ℎ=𝜆‖𝜙‖‖𝐼^𝛼𝑛𝑞‖+𝜆‖𝐼^𝛼𝑛𝑝‖1−𝜆ℐ(𝑇)−𝜆‖𝐼^𝛼𝑛𝑞‖+1. (3.10)

Proof. We will show that the operator 𝐹∶𝐾→𝐾 is continuous and completely continuous.Step 1. The operator 𝐹∶𝐾→𝐾 is continuous in view of the continuity of 𝑓.Step 2. 𝐹 maps bounded sets into bounded sets in 𝐾.
Let 𝐺⊂𝐾 be bounded; that is, there exists a positive constant 𝑙 such that ‖𝑧‖_𝑇≤𝑙, for all 𝑧∈𝐺. For each 𝑧∈𝐺, we have |𝐹𝑧(𝑡)|≤^𝑛−1𝑁_𝑗=1^𝑗𝑘_𝑘=0𝑟=0|𝑎_𝑗𝑘−𝛼_𝑛𝑟|𝑘!𝑡^𝑘−𝑟𝐼(𝑘−𝑟)!^{𝛼𝑛−𝛼𝑛−𝑗+𝑟}|𝑧(𝑡)|+𝐼^𝛼𝑛𝑓(𝑡,𝑧_𝑡+𝑦_𝑡)≤^𝑛−1𝑁_𝑗=1^𝑗𝑘_𝑘=0𝑟=0|𝑎_𝑗𝑘−𝛼_𝑛𝑟|𝑙𝑘!𝑡^{𝛼𝑛−𝛼𝑛−𝑗+𝑘}(𝑘−𝑟)!Γ(𝛼_𝑛−𝛼_𝑛−𝑗+𝑟)+𝐼^𝛼𝑛𝑓(𝑡,𝑧_𝑡+𝑦_𝑡)≤𝑙ℐ(𝑇)+𝐼^𝛼𝑛{𝑝(𝑡)+𝑞(𝑡)‖𝑧_𝑡+𝑦_𝑡‖}, (3.11) where ℐ(𝑇) is defined in (3.9). It follows that ‖𝐹𝑧‖_𝑇≤𝑙ℐ(𝑇)+‖𝐼^𝛼𝑛𝑝‖+𝑙‖𝐼^𝛼𝑛𝑞‖+‖𝜙(𝑡)‖‖𝐼^𝛼𝑛𝑞‖. (3.12) Hence 𝐹𝐺 is bounded.
Step 3. 𝐹 maps bounded sets into equicontinuous sets of 𝐾.
We will show that 𝐹𝐺 is equicontinuous. For each 𝑧∈𝐺, 𝑡₁,𝑡₂∈𝐼 and 𝑡₁<𝑡₂, then for given 𝜖>0, choose 𝛿=min{[𝜖𝐶(𝑗,𝑘,𝑟)4]^{1/(𝛼𝑛−𝛼𝑛−𝑗+𝑟)},[𝜖Γ(𝛼_𝑛+1)]4(‖𝑝‖+‖𝑞‖(𝑙+‖𝜙‖))^1/𝛼𝑛}, (3.13) where 𝑗=1,2,…,𝑛−1, 𝑘=0,1,…,𝑁_𝑗, 𝑟=0,1,…,𝑘, 𝐶(𝑗,𝑘,𝑟)=(𝑘−𝑟)!∑^𝑛−1_𝑖=1(𝑁_𝑖+1)(𝑁_𝑖×+2)Γ(𝛼_𝑛−𝛼_𝑛−𝑗+𝑟+1)|𝑎_𝑗𝑘(^−𝛼𝑛_𝑟)|𝑙𝜂𝑘! (3.14) and 𝜂=max{1,𝑇^𝑁𝑗,𝑗=1,2,…,𝑛−1}. If |𝑡₁−𝑡₂|<𝛿, |𝐹𝑧(𝑡₁)−𝐹𝑧(𝑡₂)|=|^𝑛−1𝑁_𝑗=1^𝑗𝑘_𝑘=0𝑟=0|𝑎_𝑗𝑘(^−𝛼𝑛_𝑟)|𝑘!𝑡₁^𝑘−𝑟(𝑘−𝑟)!Γ(𝛼_𝑛−𝛼_𝑛−𝑗+𝑟)^𝑡1₀(𝑡₁−𝑠)^{𝛼𝑛−𝛼𝑛−𝑗+𝑟−1}−𝑧(𝑠)𝑑𝑠_{𝑛−1𝑁𝑗=1𝑗𝑘𝑘=0𝑟=0}|𝑎_𝑗𝑘(^−𝛼𝑛_𝑟)|𝑘!𝑡₂^𝑘−𝑟(𝑘−𝑟)!Γ(𝛼_𝑛−𝛼_𝑛−𝑗+𝑟)^𝑡2₀(𝑡₂−𝑠)^{𝛼𝑛−𝛼𝑛−𝑗+𝑟−1}+1𝑧(𝑠)𝑑𝑠|Γ(𝛼_𝑛)|^𝑡1₀(𝑡₁−𝑠)^𝛼𝑛−1𝑓(𝑠,𝑧_𝑠+𝑦_𝑠)𝑑𝑠−^𝑡2₀(𝑡₂−𝑠)^𝛼𝑛−1𝑓(𝑠,𝑧_𝑠+𝑦_𝑠)𝑑𝑠|≤|^𝑛−1𝑁_𝑗=1^𝑗𝑘_𝑘=0𝑟=0|𝑎_𝑗𝑘(^−𝛼𝑛_𝑟)|𝑘!𝑡₂^𝑘−𝑟(𝑘−𝑟)!Γ(𝛼_𝑛−𝛼_𝑛−𝑗+𝑟)^𝑡1₀(𝑡₂−𝑠)^{𝛼𝑛−𝛼𝑛−𝑗+𝑟−1}−𝑧(𝑠)𝑑𝑠_{𝑛−1𝑁𝑗=1𝑗𝑘𝑘=0𝑟=0}|𝑎_𝑗𝑘(^−𝛼𝑛_𝑟)|𝑘!𝑡₁^𝑘−𝑟(𝑘−𝑟)!Γ(𝛼_𝑛−𝛼_𝑛−𝑗+𝑟)^𝑡1₀(𝑡₁−𝑠)^{𝛼𝑛−𝛼𝑛−𝑗+𝑟−1}+𝑧(𝑠)𝑑𝑠|_{𝑛−1𝑁𝑗=1𝑗𝑘𝑘=0𝑟=0}|𝑎_𝑗𝑘(^−𝛼𝑛_𝑟)|𝑘!𝑡₂^𝑘−𝑟(𝑘−𝑟)!Γ(𝛼_𝑛−𝛼_𝑛−𝑗+𝑟)^𝑡2_𝑡1(𝑡₂−𝑠)^{𝛼𝑛−𝛼𝑛−𝑗+𝑟−1}+|𝑧(𝑠)|𝑑𝑠‖𝑓‖_∞Γ(𝛼_𝑛){^𝑡1₀[(𝑡₂−𝑠)^𝛼𝑛−1−(𝑡₁−𝑠)^𝛼𝑛−1]𝑑𝑠+^𝑡2_𝑡1(𝑡₂−𝑠)^𝛼𝑛−1≤𝑑𝑠}_{𝑛−1𝑁𝑗=1𝑗𝑘𝑘=0𝑟=0}|𝑎_𝑗𝑘(^−𝛼𝑛_𝑟)|𝑙𝑘!𝑇^𝑘−𝑟(𝑘−𝑟)!Γ(𝛼_𝑛−𝛼_𝑛−𝑗×+𝑟)_𝑡20{(𝑡₂−𝑠)^{𝛼𝑛−𝛼𝑛−𝑗+𝑟−1}−(𝑡₁−𝑠)^{𝛼𝑛−𝛼𝑛−𝑗+𝑟−1}+}𝑑𝑠‖𝑝‖+‖𝑞‖(𝑙+‖𝜙‖)Γ(𝛼_𝑛){^𝑡1₀[(𝑡₂−𝑠)^𝛼𝑛−1−(𝑡₁−𝑠)^𝛼𝑛−1]𝑑𝑠+^𝑡2_𝑡1(𝑡₂−𝑠)^𝛼𝑛−1≤𝑑𝑠}_{𝑛−1𝑁𝑗=1𝑗𝑘𝑘=0𝑟=0}|𝑎_𝑗𝑘(^−𝛼𝑛_𝑟)|2𝑙𝑘!𝜂(𝑡₂−𝑡₁)^{𝛼𝑛−𝛼𝑛−𝑗+𝑟}(𝑘−𝑟)!Γ(𝛼_𝑛−𝛼_𝑛−𝑗++𝑟+1)2(‖𝑝‖+‖𝑞‖(𝑙+‖𝜙‖))(𝑡₂−𝑡₁)^𝛼𝑛Γ(𝛼_𝑛=+1)_{𝑛−1𝑁𝑗=1𝑗𝑘𝑘=0𝑟=0}|𝑎_𝑗𝑘(^−𝛼𝑛_𝑟)|2𝑙𝑘!𝜂𝛿^{𝛼𝑛−𝛼𝑛−𝑗+𝑟}(𝑘−𝑟)!Γ(𝛼_𝑛−𝛼_𝑛−𝑗++𝑟+1)2(‖𝑝‖+‖𝑞‖(𝑙+‖𝜙‖))𝛿^𝛼𝑛Γ(𝛼_𝑛≤𝜖+1)2+𝜖2=𝜖. (3.15)(1)(2)(3)(4)(5)(6)(7)(8)(9)(10)(11)(12)(13) Hence 𝐹𝐺 is equicontiuous. The Arzela-Ascoli theorem implies that 𝐹(𝐺) is compact and 𝐹∶𝐾→𝐾 is continuous and completely continuous.
Step 4. We now show that there exists an open set 𝑈⊆𝐾 with 𝑧≠𝜆𝐹(𝑧) for 𝜆∈(0,1) and 𝑧∈𝜕𝑈. Let 𝑧∈𝐾 be any solution of 𝑧=𝜆𝐹𝑧, 𝜆∈(0,1), where 𝐹 is given by (3.8); since 𝐹∶𝐾→𝐾 is continuous and completely continuous, we have 𝑧(𝑡)=𝜆𝐹𝑧(𝑡)≤𝜆^𝑛−1𝑁_𝑗=1^𝑗𝑘_𝑘=0𝑟=0|𝑎_𝑗𝑘−𝛼_𝑛𝑟|𝑘!𝑡^𝑘−𝑟(𝑘−𝑟)!Γ(𝛼_𝑛−𝛼_𝑛−𝑗×+𝑟)_𝑡0(𝑡−𝑠)^{𝛼𝑛−𝛼𝑛−𝑗+𝑟−1}𝜆𝑧(𝑠)𝑑𝑠+Γ(𝛼_𝑛)^𝑡₀(𝑡−𝑠)^𝛼𝑛−1𝑓(𝑠,𝑧_𝑠+𝑦_𝑠)𝑑𝑠≤𝜆^𝑛−1𝑁_𝑗=1^𝑗𝑘_𝑘=0𝑟=0|𝑎_𝑗𝑘−𝛼_𝑛𝑟|‖𝑧‖𝑘!𝑡^𝑘−𝑟(𝑘−𝑟)!Γ(𝛼_𝑛−𝛼_𝑛−𝑗+𝑟)^𝑡₀(𝑡−𝑠)^{𝛼𝑛−𝛼𝑛−𝑗+𝑟−1}+𝜆𝑑𝑠Γ(𝛼_𝑛)^𝑡₀(𝑡−𝑠)^𝛼𝑛−1[𝑝(𝑠)+𝑞(𝑠)‖𝑧_𝑠+𝑦_𝑠‖]𝑑𝑠≤𝜆{‖𝑧‖ℐ(𝑇)+‖𝑧‖𝐼^𝛼𝑛𝑞(𝑡)+‖𝜙‖𝐼^𝛼𝑛𝑞(𝑡)+𝐼^𝛼𝑛𝑝(𝑡)}. (3.16) So ‖𝑧‖(1−𝜆ℐ(𝑇)−𝜆‖𝐼^𝛼𝑛𝑝‖)≤𝜆‖𝜙‖‖𝐼^𝛼𝑛𝑞‖+𝜆‖𝐼^𝛼𝑛𝑝‖. (3.17) Now, by (3.10) and (3.17), we know that any solution 𝑧 of (3.8) satisfies ‖𝑧‖≠ℎ; let 𝑈={𝑧∈𝐾;‖𝑧‖<ℎ}. (3.18) Therefore, Theorem 2.7 guarantees that (3.1) has at least a positive solution 𝑧∈𝑈. Hence, (1.4) has at least a positive solution 𝑥^∗∈𝐾^∗, satisfying ‖𝑥^∗‖≤max{‖𝜙‖,ℎ} and the proof is complete.

Note that we can complete the above mentioned procedure by using only the continuity of 𝑓 without condition (1), but with our procedure and details of condition (1) in Theorem 3.1 answers all the questions exist in the following remark.

Remark 3.2. When 𝑓 is continuous on (0,𝑇]×𝐶, lim_𝑡→0⁺𝑓(𝑡,⋅)=+∞, (i.e., 𝑓 is singular at 𝑡=0) in (1.4). Suppose ∃𝜎∈(0,𝛼_𝑛], such that 𝑡^𝜎𝑓(𝑡,𝑥_𝑡) is a continuous function on [0,𝑇]×𝐶, then 𝐼^𝛼𝑛𝑓(𝑡,𝑥_𝑡)=𝐼^𝛼𝑛𝑡^−𝜎𝑡^𝜎𝑓(𝑡,𝑥_𝑡) is continuous on 𝐼×𝐶 by Lemma 2.1 in [12, page 613]. We also obtain results about the existence to (1.4) by using a nonlinear alternative of Leray-Schauder type. The proof is similar to that of Theorem 3.1 as long as we let(1)𝑡^𝜎𝑓(𝑡,𝑥_𝑡)≤𝑝(𝑡)+𝑞(𝑡)‖𝑥_𝑡‖, for 𝑡∈𝐼, 𝑥_𝑡∈𝐶, and ‖𝐼^𝛼𝑛𝑡^−𝜎𝑝‖<∞, ‖𝐼^𝛼𝑛𝑡^−𝜎𝑞‖<∞, (2)1−ℐ(𝑇)−‖𝐼^𝛼𝑛𝑡^−𝜎𝑞‖>0, then (1.4) has at least a positive solution 𝑥^∗∈𝐾^∗, satisfying ‖𝑥^∗‖≤max{‖𝜙‖,ℎ}, where ℎ=𝜆‖𝐼^𝛼𝑛𝑡^−𝜎𝑝‖+𝜆‖𝜙‖‖𝐼^𝛼𝑛𝑡^−𝜎𝑞‖1−𝜆ℐ(𝑇)−𝜆‖𝐼^𝛼𝑛𝑡^−𝜎𝑞‖+1. (3.19)

4. Unique Existence of Solution

In this section, we will give uniqueness of positive solution to (1.4).

Theorem 4.1. Let 𝑓∶𝐼×𝐶→ℝ⁺ be continuous and 𝜆∈𝐿¹([0,𝑇],ℝ⁺) with ‖𝐼^𝛼𝑛𝜆‖<∞. Further assume(i)|𝑓(𝑡,𝑢_𝑡+𝑦_𝑡)−𝑓(𝑡,𝑣_𝑡+𝑦_𝑡)|≤𝜆(𝑡)‖𝑢_𝑡−𝑣_𝑡‖, for all 𝑢,𝑣∈𝐾, 𝑡∈[0,𝑇], (ii)ℐ(𝑇)+‖𝐼^𝛼𝑛𝜆‖<1.Then (1.4) has unique solution which is positive, where ℐ(𝑇) is given in (3.9).

Proof. Let 𝑢,𝑣∈𝐾. Then we obtain |𝐹𝑢(𝑡)−𝐹𝑣(𝑡)|≤^𝑛−1𝑁_𝑗=1^𝑗𝑘_𝑘=0𝑟=0|𝑎_𝑗𝑘(^−𝛼𝑛_𝑟)|𝑘!𝑡^𝑘−𝑟𝐼(𝑘−𝑟)!^{𝛼𝑛−𝛼𝑛−𝑗+𝑟}|𝑢(𝑡)−𝑣(𝑡)|+𝐼^𝛼𝑛|𝑓(𝑡,𝑢_𝑡+𝑦+𝑡)−𝑓(𝑡,𝑣_𝑡+𝑦_𝑡)|≤‖𝑢−𝑣‖_𝑇{^𝑛−1𝑁_𝑗=1^𝑗𝑘_𝑘=0𝑟=0|𝑎_𝑗𝑘(^−𝛼𝑛_𝑟)|𝑘!𝑡^{𝛼𝑛−𝛼𝑛−𝑗+𝑘}(𝑘−𝑟)!Γ(𝛼_𝑛−𝛼_𝑛−𝑗+𝑟+1)+𝐼^𝛼𝑛𝜆(𝑡)}≤‖𝑢−𝑣‖_𝑇{^𝑛−1𝑁_𝑗=1^𝑗𝑘_𝑘=0𝑟=0|𝑎_𝑗𝑘(^−𝛼𝑛_𝑟)|𝑘!𝑇^{𝛼𝑛−𝛼𝑛−𝑗+𝑘}(𝑘−𝑟)!Γ(𝛼_𝑛−𝛼_𝑛−𝑗+𝑟+1)+𝐼^𝛼𝑛𝜆(𝑡)}, (4.1) where 𝐹 is given in (3.8). Hence, ‖𝐹𝑢−𝐹𝑣‖_𝑇≤(ℐ(𝑇)+‖𝐼^𝛼𝑛𝜆‖)‖𝑢−𝑣‖_𝑇. (4.2) In view of Banach fixed point theorem 𝐹 has unique fixed point in 𝐾, which is the unique positive solution of (2.7) and (1.4) has a unique positive solution in 𝐾^∗.

Remark 4.2. When 𝜆(𝑡)=𝐿>0, then condition (i) reduces to the Lipschitz condition.

Example 4.3. Let 𝜆(𝑡)=𝐿>0 and 𝑓(𝑡,𝑥_𝑡)=𝐿𝑥_𝑡+𝑒^𝑡=𝐿𝑥(𝑡−𝜔)+𝑒^𝑡, 𝜔>0. Consider the equation (𝐷^1/2−𝑎𝑡²𝐷^1/4−𝑏𝑡𝐷^1/6−𝑐𝐷^1/8)𝑥=𝐿𝑥(𝑡−𝑤)+𝑒^𝑡,𝑡∈(0,64],𝑥(𝑡)=0,𝑡∈[−𝑤,0]. (4.3) Then (4.3) is equivalent to the integral equation, 𝑥(𝑡)=^3𝑁_𝑗=1^𝑗𝑘_𝑘=0𝑟=0𝑎_𝑗𝑘−12𝑟𝑘!𝑡^𝑘−𝑟𝐼(𝑘−𝑟)!^{1/2−𝛼3−𝑗+𝑟}𝑥(𝑡)+𝐼^1/2(𝐿𝑥(𝑡−𝜔)+𝑒^𝑡). (4.4) Here 𝛼₃=1/2, 𝑝₁(𝑡)=∑²_𝑘=0𝑎_1𝑘𝑡^𝑘=𝑎𝑡², then 𝑁₁=2, 𝑎₁₀=𝑎₁₁=0, 𝑎₁₂=𝑎, 𝑝₂(𝑡)=∑¹_𝑘=0𝑎_2𝑘𝑡^𝑘=𝑏𝑡, then 𝑁₂=1, 𝑎₂₀=0,𝑎₂₁=𝑏 and 𝑝₃(𝑡)=∑⁰_𝑘=0𝑎_3𝑘𝑡^𝑘=𝑐, then 𝑁₃=0, 𝑎₃₀=𝑐. Hence 𝑥(𝑡)=𝑎₁₀−120𝐼^1/2−1/4𝑥+𝑎₁₁[−120𝑡𝐼^1/2−1/4−1𝑥+21𝐼^1/2−1/4+1]+𝑎₁₂[−120𝑡²𝐼^1/2−1/4−1𝑥+221𝑡𝐼^1/2−1/4+1−1𝑥+222𝐼^1/2−1/4+2𝑥]+𝑎₂₀−120𝐼^1/2−1/6𝑥+𝑎₂₁[−120𝑡𝐼^1/2−1/6−1𝑥+21𝐼^1/2−1/6+1𝑥]+𝑎₃₀−120𝐼^1/2−1/8𝑥+𝐿𝐼^1/2𝑥(𝑡−𝜔)+𝐼^1/2𝑒^𝑡. (4.5) In view of (2.8) and that Γ(1/2)=√𝜋, Γ(−1/2)=−2√𝜋 and Γ(−3/4)=4√𝜋/3 we obtain 𝑥(𝑡)=𝑎[𝑡²𝐼^1/4𝑥(𝑡)−𝑡𝐼^5/43𝑥(𝑡)+4𝐼^9/4𝑥(𝑡)]+𝑏[(1+𝑡)𝐼^1/31𝑥(𝑡)−2𝐼^4/3𝑥(𝑡)]+𝑐𝐼^3/8+𝐿𝐼^1/2𝑥(𝑡−𝜔)+𝐼^1/2𝑒^𝑡. (4.6) If |𝑎|≤3/5, |𝑏|≤2/5, |𝑐|≤1/5, 0<𝐿<4/5 in the above equation satisfy the conditions required in Theorem 4.1, the iterated sequence is 𝑥₁(𝑡)=𝐼^1/2𝑒^𝑡=𝑡^1/2𝐸_1,3/2𝑥(𝑡),₂(𝑡)=[𝑎(𝑡²𝐼^1/4−𝑡𝐼^5/4+34𝐼^9/4)+𝑏((1+𝑡)𝐼^1/3−12𝐼^4/3)+𝑐𝐼^3/8+𝐿𝐼^1/2]𝑥₁(𝑡)+𝑥₁𝑥(𝑡),_𝑛+1(𝑡)=^𝑛_𝑘=0[𝑎(𝑡²𝐼^1/4−𝑡𝐼^5/4+34𝐼^9/4)+𝑏((1+𝑡)𝐼^1/3−12𝐼^4/3)+𝑐𝐼^3/8+𝐿𝐼^1/2]^𝑛−𝑘𝑥₁(𝑡), (4.7) for 𝑛=1,2,3,…, where 𝐼^𝛼𝑥₁=𝑡^𝛼+1/2𝐸_1,𝛼+3/2(𝑡), 𝛼>0, 𝑥(𝑡)=lim_𝑛→∞𝑥_𝑛(𝑡) is the unique solution, which may not be positive, where 𝐸_𝛼,𝛽(𝑡)=∑^∞_𝑘=0(𝑡^𝑘/Γ(𝛼𝑘+𝛽)) is Mittag-Leffler function.

Acknowledgment

The authors thank the referee’s efforts for their remarks and Professor Ferhan Merdivenci Atici, Western Kentucky University, USA, for her regular contact.