Multiple Positive Solutions for Nonlinear Semipositone Fractional Differential Equations

Abstract

We present some new multiplicity of positive solutions results for nonlinear semipositone fractional boundary value problem 𝐷^𝛼₀₊𝑢(𝑡)=𝑝(𝑡)𝑓(𝑡,𝑢(𝑡))−𝑞(𝑡),0<𝑡<1,𝑢(0)=𝑢(1)=𝑢′(1)=0, where 2<𝛼≤3 is a real number and 𝐷^𝛼₀₊ is the standard Riemann-Liouville differentiation. One example is also given to illustrate the main result.

1. Introduction

This paper is mainly concerned with the multiplicity of positive solutions of nonlinear fractional differential equation boundary value problem (BVP for short)𝐷^𝛼₀₊𝑢(𝑡)(𝑡)𝑓((𝑡(𝑡)𝑢(0)(1)(1)=𝑝𝑡,𝑢))−𝑞,0<𝑡<1,=𝑢=𝑢′=0, (1.1) where 2<𝛼≤3 is a real number and 𝐷^𝛼₀₊ is the standard Riemann-Liouville differentiation, and 𝑓,𝑝,𝑞 is a given function satisfying some assumptions that will be specified later.

In the last few years, fractional differential equations (in short FDEs) have been studied extensively the motivation for those works stems from both the development of the theory of fractional calculus itself and the applications of such constructions in various sciences such as physics, mechanics, chemistry, and engineering. For an extensive collection of such results, we refer the readers to the monographs by Kilbas et al. [1], Miller and Ross [2], Oldham and Spanier [3], Podlubny [4], and Samko et al. [5].

Some basic theory for the initial value problems of FDE involving the Riemann-Liouville differential operator has been discussed by Lakshmikantham and Vatsala [6–8], Babakhani and Daftardar-Gejji [9–11], and Bai [12], and others. Also, there are some papers that deal with the existence and multiplicity of solutions (or positive solution) for nonlinear FDE of BVPs by using techniques of nonlinear analysis (fixed point theorems, Leray-Schauders theory, topological degree theory, etc.), see [13–22] and the references therein.

Bai and Lü [15] studied the following two-point boundary value problem of FDEs𝐷^𝑞₀₊𝑢(𝑡)+𝑓(𝑡,𝑢(𝑡))=0,𝑢(0)=𝑢(1)=0,0<𝑡<1,1<𝑞≤2, (1.2) where 𝐷^𝑞₀₊ is the standard Riemann-Liouville fractional derivative. They obtained the existence of positive solutions by means of the Guo-Krasnosel’skii fixed point theorem and Leggett-Williams fixed point theorem.

Zhang [22] considered the existence and multiplicity of positive solutions for the nonlinear fractional boundary value problem^𝑐𝐷^𝑞₀₊𝑢(𝑡)=𝑓(𝑡,𝑢(𝑡)),0<𝑡<1,𝑢(0)+𝑢^′(0)=0,𝑢(1)+𝑢^′(1)=0, (1.3) where 1<𝑞≤2 is a real number, 𝑓∶[0,1]×[0,+∞)→[0,+∞), and ^𝑐𝐷^𝑞₀₊ is the standard Caputo’s fractional derivative. The author obtained the existence and multiplicity results of positive solutions by means of the Guo-Krasnosel’skii fixed point theorem.

From the above works, we can see the fact that although the fractional boundary value problems have been investigated by some authors to the best of our knowledge, there have been few papers that deal with the boundary value problem (1.1) for nonlinear fractional differential equation. Motivated by all the works above, in this paper we discuss the boundary value problem (1.1), using the Guo-Krasnosel’skii fixed point theorem, and we give some new existence of multiple positive solutions criteria for boundary value problem (1.1).

The paper is organized as follows. In Section 2, we give some preliminary results that will be used in the proof of the main results. In Section 3, we establish the existence of multiple positive solutions for boundary value problem (1.1) by the Guo-Krasnosel’skii fixed point theorem. In the end, we illustrate a simple use of the main result.

2. Preliminaries and Lemmas

For the convenience of the reader, we present here the necessary definitions from fractional calculus theory. These definitions can be found in the recent literature such as [1, 4, 15].

Definition 2.1 (see [1, 4]). The Riemann-Liouville fractional integral of order 𝛼(𝛼>0) of a function 𝑓∶(0,+∞)→ℝ is given by 𝐼^𝛼₀₊𝑓(𝑡)=^𝑡₀(𝑡−𝑠)^𝛼−1Γ(𝛼)𝑓(𝑠)𝑑𝑠, (2.1) provided that the right side is pointwise defined on (0,+∞), where Γ is the gamma function.

Definition 2.2 (see [1, 4]). The Riemann-Liouville fractional derivative of order 𝛼(𝛼>0) of a continuous function 𝑓∶(0,+∞)→ℝ is given by 𝐷^𝛼₀₊𝑓(𝑡)=1Γ()𝑑𝑛−𝛼𝑑𝑡^𝑛𝑡₀(𝑡−𝑠)^{𝑛−𝛼+1}𝑓(𝑠)𝑑𝑠, (2.2) provided that the right side is pointwise defined on (0,+∞), where 𝑛=[𝛼]+1 and [𝛼] denotes the integer part of 𝛼.

Lemma 2.3 (see [15]). Let 𝛼>0. If one assumes 𝑢∈𝐶(0,1)∩𝐿(0,1), then fractional differential equation 𝐷^𝛼𝑢(𝑡)=0 (2.3) has 𝑢(𝑡)=𝐶₁𝑡^𝛼−1+𝐶₂𝑡^𝛼−2+⋯+𝐶_𝑁𝑡^𝛼−𝑁,𝐶_𝑖∈ℝ,𝑖=1,2,…,𝑁, (2.4) as unique solutions, where 𝑁 is the smallest integer greater than or equal to 𝛼.

Lemma 2.4 (see [15]). Assume that ℎ∈𝐶(0,1)∩𝐿(0,1) with a fractional derivative of order 𝛼>0 that belongs to 𝐶(0,1)∩𝐿(0,1). Then 𝐼^𝛼𝐷^𝛼ℎ(𝑡)=ℎ(𝑡)+𝐶₁𝑡^𝛼−1+𝐶₂𝑡^𝛼−2+⋯+𝐶_𝑁𝑡^𝛼−𝑁, (2.5) for some 𝐶_𝑖∈ℝ, 𝑖=1,2,…,𝑁, where 𝑁 is the smallest integer greater than or equal to 𝛼.

In the following, we present Green’s function of the fractional differential equation boundary value problem.

Lemma 2.5. Let ℎ∈𝐶[0,1] and 2<𝛼≤3, then the unique solution of 𝐷^𝛼𝑢(𝑡)(𝑡)𝑢(0)(1)+ℎ=0,0<𝑡<1,=𝑢=𝑢^′(1)=0 (2.6) is given by 𝑢(𝑡)=¹₀𝐺(𝑡,𝑠)ℎ(𝑠)𝑑𝑠, (2.7) where 𝐺(𝑡,𝑠) is Green’s function given by 𝐺()=1𝑡,𝑠Γ(𝛼)(1−𝑠)^𝛼−1𝑡^𝛼−1−(𝑡−𝑠)^𝛼−1(),if0≤𝑠≤𝑡≤1,1−𝑠^𝛼−1𝑡^𝛼−1,if0≤𝑡≤𝑠≤1. (2.8)

The following properties of Green’s function form the basis of our main work in this paper.

Lemma 2.6. The function 𝐺(𝑡,𝑠) defined by (2.8) possesses the following properties:(i)𝐺(𝑡,𝑠)=𝐺(1−𝑠,1−𝑡) for 𝑡,𝑠∈(0,1);(ii)𝑡^𝛼−1(1−𝑡)𝑠(1−𝑠)^𝛼−1≤𝐺(𝑡,𝑠)Γ(𝛼)≤(𝛼−1)𝑠(1−𝑠)^𝛼−1 for 𝑡, 𝑠∈(0,1);(iii)𝑡^𝛼−1(1−𝑡)𝑠(1−𝑠)^𝛼−1≤𝐺(𝑡,𝑠)Γ(𝛼)≤(𝛼−1)𝑡^𝛼−1(1−𝑡) for 𝑡, 𝑠∈(0,1);(iv)𝐺(𝑡,𝑠)>0 for 𝑡,𝑠∈(0,1).

The following Krasnosel’skii’s fixed point theorem will play a major role in our next analysis.

Lemma 2.7 (see [23]). Let 𝑋 be a Banach space, and let 𝑃⊂𝑋 be a cone in 𝑋. Assume Ω₁,Ω₂ are open subsets of 𝑋 with 0∈Ω₁⊂Ω₁⊂Ω₂, and let 𝒜∶𝑃→𝑃 be a completely continuous operator such that either(i)‖𝒜𝑢‖≤‖𝑢‖,𝑢∈𝑃∩𝜕Ω₁, ‖𝒜𝑢‖≥‖𝑢‖,𝑢∈𝑃∩𝜕Ω₂, or(ii)‖𝒜𝑢‖≥‖𝑢‖,𝑢∈𝑃∩𝜕Ω₁, ‖𝒜𝑢‖≤‖𝑢‖,𝑢∈𝑃∩𝜕Ω₂.Then 𝒜 has a fixed point in 𝑃∩(Ω₂⧵Ω₁).

3. Main Results

In this section, we establish some new existence results for the fractional differential equation (1.1). Given 𝑎∈𝐿¹(0,1), we write a 𝑎≻0, if 𝑎≥0 for 𝑡∈[0,1], and it is positive in a set of positive measure.

Let us list the following assumptions:

(H1) 𝑓∶[0,1]×[0,∞)→[0,∞) is continuous, 𝑝,𝑞≻0;

(H2) there exists 𝜃∈(0,1/2), such that_𝜃^1−𝜃𝑝(𝑠)𝑠(1−𝑠)^𝛼−1𝑑𝑠>0. (3.1)

In view of Lemmas 2.5 and 2.6, we obtain the following.

Lemma 3.1. Let 𝑞∈𝐿¹[0,1] with 𝑞>0 on (0,1), and 𝛾(𝑡) is the unique solution of 𝐷^𝛼₀₊𝑢(𝑡)(𝑡)𝑢(0)(1)=𝑞,0<𝑡<1,=𝑢=𝑢^′(1)=0, (3.2) Then 0≤𝛾(𝑡)≤𝛼−1Γ(𝛼)‖𝑞‖₁𝑡^𝛼−1(1−𝑡)∶=𝒞𝑡^𝛼−1(1−𝑡),𝑓𝑜𝑟𝑡∈[0,1], (3.3) where  𝒞=(𝛼−1/Γ(𝛼))‖𝑞‖₁, ‖𝑞‖₁=∫¹₀|𝑞(𝑡)|𝑑𝑡.

Next, we consider𝐷^𝛼₀₊𝑢(𝑡)(𝑡)𝑔(𝑡)(𝑡𝑢(0)(1)=𝑝𝑡,𝑢−𝛾),0<𝑡<1,=𝑢=𝑢^′(1)=0, (3.4) where𝑔()=𝑓()𝑓()𝑡,𝑢𝑡,𝑢,if𝑢≥0𝑡,0,if𝑢<0, (3.5) Then (3.4) is equivalent to the following integral equation:𝑢(𝑡)=¹₀𝐺(𝑡,𝑠)𝑝(𝑠)𝑔𝑠,𝑢(𝑠)−𝛾(𝑠)𝑑𝑠. (3.6)

Lemma 3.2. Let 𝑢(𝑡)≥𝛾(𝑡) for 𝑡∈[0,1], and 𝑢(𝑡) is positive solution of the problem (3.4). Then 𝑢(𝑡)−𝛾(𝑡) is positive solution of the problem (1.1).

Proof. In fact, let 𝑥(𝑡)=𝑢(𝑡)−𝛾(𝑡). Then 𝑥(𝑡)≥0 and 𝑢(𝑡)=𝑥(𝑡)+𝛾(𝑡). Since 𝑢(𝑡) is positive solution of the problem (3.4), we have 𝐷^𝛼₀₊𝑥(𝑡)(𝑡(𝑡)𝑔((𝑡(0)=(1)=+𝛾)=𝑝𝑡,𝑥)),0<𝑡<1,𝑥+𝛾𝑥+𝛾𝑥+𝛾^′(1)=0. (3.7) So 𝐷^𝛼₀₊𝑥(𝑡)(𝑡)𝑔((𝑡(𝑡)𝑥(0)(1)=𝑝𝑡,𝑥))−𝑞,0<𝑡<1,=𝑥=𝑥^′(1)=0. (3.8)
For our constructions, we will consider the Banach space 𝐸=𝐶[0,1] equipped with standard norm ‖𝑢‖=max_0≤𝑡≤1|𝑢(𝑡)|,𝑢∈𝐸.
Define a cone 𝐾 by (𝑡)≥𝑡𝐾=𝑢∈𝐸∶𝑢^𝛼−1(1−𝑡)[](𝛼−1‖𝑢‖,∀𝑡∈0,1,𝛼∈2,3]. (3.9)
Let the operator 𝒜∶𝐾→𝐸 be defined by the formula (𝒜𝑥)(𝑡)∶=¹₀𝐺(𝑡,𝑠)𝑝(𝑠)𝑔𝑠,𝑥(𝑠)−𝛾(𝑠)𝑑𝑠,0≤𝑡≤1,𝑢∈𝐾. (3.10)

Lemma 3.3. Assume that (H1) holds. Then 𝒜(𝐾)⊂𝐾.

Proof. Notice from (3.10) and Lemma 2.6 that, for 𝑥∈𝐾, 𝒜𝑥(𝑡)≥0 on [0,1] and ‖‖≤1𝒜𝑥Γ(𝛼)¹₀(𝛼−1)𝑠(1−𝑠)^𝛼−1𝑝(𝑠)𝑔𝑠,𝑥(𝑠)−𝛾(𝑠)𝑑𝑠. (3.11) On the other hand, we have (𝑡)≥1𝒜𝑥Γ(𝛼)¹₀𝑡^𝛼−1(1−𝑡)𝑠(1−𝑠)^𝛼−1𝑝(𝑠)𝑔(𝑠)(𝑠)≥𝑡𝑠,𝑥−𝛾𝑑𝑠_𝛼−1(1−𝑡)(𝛼−1)Γ(𝛼)‖𝒜𝑥‖. (3.12) Thus we have 𝒜(𝐾)⊂𝐾. The proof is finished.

It is standard that 𝒜∶𝐾→𝐾 is continuous and completely continuous.

For convenience, we introduce the following notations: 𝑁=((𝛼−1)‖𝑝‖₁/Γ(𝛼))max_0≤𝑠≤1𝑠(1−𝑠)^𝛼−1, 𝑁=𝜎∫_𝜃^1−𝜃(𝑠(1−𝑠)^𝛼−1/Γ(𝛼))𝑝(𝑠)𝑑𝑠, 𝜎=min_{𝜃≤𝑡≤1−𝜃}(1−𝑡)𝑡^𝛼−1.

Theorem 3.4. Assume that (H1) and (H2) are satisfied. Also suppose the following conditions are satisfied:(A1) there exists a constant 𝑅₁>(𝛼−1)𝒞 such that 𝑁𝑓(𝑡,𝑢)≤𝑅₁ for all (𝑡,𝑢)∈[0,1]×[0,𝑅₁];(A2) there exists a constant 𝑅₂>2𝑅₁ such that 𝑁𝑓(𝑡,𝑢)>𝑅₂ for all (𝑡,𝑢)∈[0,1]×[𝜎𝑅₂,𝑅₂];(A3)lim_𝑢→+∞max_0≤𝑡≤1(𝑓(𝑡,𝑢)/𝑢)=0. Then the problem (1.1) has at least two positive solutions.

Proof. To show that (1.1) has at least two positive solutions, we will assume the problem (3.4) has at least two positive solutions 𝑥₁ and 𝑥₂ with 𝑅₁≤‖𝑥₁‖<𝑅₂<‖𝑥₂‖≤𝑅₃.
We now show ‖𝒜𝑥‖≤‖𝑥‖,for𝑥∈𝐾∩𝜕Ω₁, (3.13)
To see this, let Ω₁={𝑥∈𝐾∣‖𝑥‖<𝑅₁}, then for 𝑥∈𝐾∩𝜕Ω₁, 𝑡∈[0,1], by Lemma 3.1 and (A1), we have 𝑥(𝑡)−𝛾(𝑡)≤𝑥(𝑡)≤‖𝑥‖=𝑅₁,𝑥(𝑡)(𝑡)≥𝑡−𝛾^𝛼−1(1−𝑡)𝑅𝛼−1¹−𝒞𝑡^𝛼−1()≥𝑅1−𝑡¹𝑡𝛼−1−𝒞^𝛼−1(1−𝑡)≥0. (3.14) Thus, we see, from Lemma 2.6 and (A1), that ‖𝒜𝑥‖=max_0≤𝑡≤1(𝒜𝑢)(𝑡)=max_0≤𝑡≤1¹₀𝐺()𝑝(𝑠)𝑔(𝑠)(𝑠)≤()𝑡,𝑠𝑠,𝑥−𝛾𝑑𝑠𝛼−1Γ(𝛼)¹₀𝑠(1−𝑠)^𝛼−1𝑝(𝑠)𝑓(𝑠)(𝑠)≤()𝑠,𝑥−𝛾𝑑𝑠𝛼−1Γ(𝛼)¹₀𝑅₁𝑁𝑠(1−𝑠)^𝛼−1𝑝(𝑠)𝑑𝑠≤𝑅₁, (3.15) from which we see that ‖𝒜𝑥‖≤‖𝑥‖, for 𝑥∈𝐾∩𝜕Ω₁.
Next we now show ‖𝒜𝑥‖≥‖𝑥‖,for𝑥∈𝐾∩𝜕Ω₂. (3.16) To see this, let Ω₂={𝑥∈𝐾|‖𝑥‖<𝑅₂}; then, for 𝑥∈𝐾∩𝜕Ω₂, 𝑡∈[0,1], by 𝑅₂>2𝑅₁, we have 𝑥(𝑡)(𝑡)≥𝑡−𝛾^𝛼−1(1−𝑡)𝑅𝛼−1²−𝒞𝑡^𝛼−1()≥𝑡1−𝑡^𝛼−1(1−𝑡)2()𝑅𝛼−1². (3.17)
For 𝑥∈𝜕Ω₂; 𝑡∈[𝜃,1−𝜃]𝜎, then, it follows from (3.17) that 𝑅₂≤𝑡^𝛼−1(1−𝑡)2()𝑅𝛼−1²≤𝑥(𝑡)−𝛾(𝑡)≤𝑅₂. (3.18) In view of (A2), (3.17) and Lemma 2.6, we have that for all 𝑥∈𝜕Ω₂, 𝑡∈[𝜃,1−𝜃]𝜎‖𝒜𝑢‖≥¹₀𝐺(𝑡,𝑠)𝑝(𝑠)𝑔𝑠,𝑥(𝑠)−𝛾(𝑠)𝑑𝑠≥𝑡^𝛼−1(1−𝑡)¹₀𝑠(1−𝑠)^𝛼−1Γ(𝛼)𝑝(𝑠)𝑓𝑠,𝑥(𝑠)−𝛾(𝑠)𝑑𝑠>𝑡^𝛼−1(1−𝑡)_𝜃^1−𝜃𝑠(1−𝑠)^𝛼−1𝑅₂Γ(𝛼)𝑁𝑝(𝑠)𝑑𝑠≥𝜎_𝜃^1−𝜃𝑠(1−𝑠)^𝛼−1𝑅₂Γ(𝛼)𝑁𝑝(𝑠)𝑑𝑠=𝑅₂, (3.19) from which we see that ‖𝒜𝑥‖>‖𝑥‖, for 𝑥∈𝐾∩𝜕Ω₂.
On the other hand, let 𝜀>0, where 𝜀(𝛼−1)Γ(𝛼)max_0≤𝑡≤1𝑡(1−𝑡)^𝛼−1𝑝₁≤1. (3.20)
Supposing that (A3) holds, one can find 𝑁>𝑅₂>0, so that 𝑓(𝑡,𝑢)≤𝜀𝑢,∀𝑡∈[0,1],𝑢≥𝑁. (3.21) Setting 𝑅₃=(𝛼−1)max_0≤𝑡≤1𝑡(1−𝑡)^𝛼−1‖𝑝‖₁max_{(𝑡,𝑢)∈[0,1]×[0,𝑁]}𝑓(𝑡,𝑢)Γ(𝛼)−𝜀(𝛼−1)max_0≤𝑡≤1𝑡(1−𝑡)^𝛼−1‖𝑝‖₁+𝑁, (3.22) then 𝑅₃>𝑁>𝑅₂, and so ‖𝒜𝑢‖=max_0≤𝑡≤1¹₀𝐺()𝑝(𝑠)𝑔(𝑠)(𝑠)≤𝑡,𝑠𝑠,𝑥−𝛾𝑑𝑠₁₀(𝛼−1)𝑠(1−𝑠)^𝛼−1Γ(𝛼)𝑝(𝑠)max_{()∈[]×[]𝑠,𝑢0,10,𝑁}𝑓()+𝑠,𝑢𝑑𝑠₁₀(𝛼−1)𝑠(1−𝑠)^𝛼−1Γ(𝛼)𝑝(𝑠)𝜀𝑥(𝑠)−𝛾(𝑠)𝑑𝑠≤𝑅₃, (3.23) from which we see that ‖𝒜𝑥‖≤‖𝑥‖, for 𝑥∈𝐾∩𝜕Ω₃.
In view of Lemma 2.7, the problem (3.4) has at least two positive solutions 𝑥₁ and 𝑥₂ with 𝑅₁≤‖𝑥₁‖<𝑅₂<‖𝑥₂‖≤𝑅₃. Since 𝑅₂>𝑅₁>(𝛼−1)𝒞, we have 𝑥₁(𝑡)(𝑡)≥𝑡−𝛾^𝛼−1(1−𝑡)𝑅𝛼−1¹−𝒞𝑡^𝛼−1()≥𝑅1−𝑡¹𝑡𝛼−1−𝒞^𝛼−1()𝑥1−𝑡≥0,₂(𝑡)(𝑡)≥𝑡−𝛾^𝛼−1(1−𝑡)𝑅𝛼−1²−𝒞𝑡^𝛼−1()≥𝑅1−𝑡²𝑡𝛼−1−𝒞^𝛼−1(1−𝑡)≥0. (3.24) Therefore 𝑥₁,𝑥₂ are solutions of the problem (1.1). This completes the proof.

Theorem 3.5. Suppose that (H1), (H2) are satisfied. Furthermore assume that(A4) there exists a constant 𝑅₁>2(𝛼−1)𝒞 such that 𝑁𝑓(𝑡,𝑢)≥𝑅₁ for all (𝑡,𝑢)∈[0,1]×[𝜎𝑅₁,𝑅₁];(A5) there exists a constant 𝑅₂>𝑚𝑎𝑥{𝑅₁,(𝑅₁/𝑁)𝑁} such that 𝑁𝑓(𝑡,𝑢)<𝑅₂ for all (𝑡,𝑢)∈[0,1]×[0,𝑅₂];(A6)lim_𝑢→+∞min_{𝜃≤𝑡≤1−𝜃}(𝑓(𝑡,𝑢)/𝑢)=+∞. Then the problem (1.1) has at least two positive solutions.

4. An Example

As an application of the main results, we consider𝐷^5/2𝑦(𝑡)𝑦−1=𝑓√𝑡𝑦(0),0<𝑡<1,=𝑦^′(0)=𝑦(1)=0, (4.1) Set𝑓𝑦=−2(𝑦−7)²+1100,if0≤𝑦≤7,−2𝑦−7+1100,if7≤𝑦≤450,𝑦−450²+214,if𝑦≥450, (4.2) Then we have 𝒞=(𝛼−1)/Γ(𝛼)‖𝑞‖₁≈2.25676,𝑁=((𝛼−1)‖𝑞‖₁/Γ(𝛼))max_0≤𝑠≤1𝑠(1−𝑠)^𝛼−1≈0.4, letting, 𝜃=1/4, then 𝜎=min_{𝜃≤𝑡≤1−𝜃}(1−𝑡)𝑡^𝛼−1≈0.09375,𝑁=𝜎∫^3/4_1/4(𝑠(1−𝑠)^𝛼−1/Γ(𝛼))𝑑𝑠≈0.008, choosing 𝑅₁=7, 𝑅₂=450, then 𝑅₁>2(𝛼−1)𝒞=6.77, 𝑅₂>max{𝑅₁,(𝑅₁/𝑁)𝑁}=max{7,350}=350; therefore, we have 𝑁𝑓(𝑢)=0.008[−2(𝑦−7)²+1100]≥8.156>𝑅₁,(𝑡,𝑢)→[1/4,3/4]×[0.65625,7], 𝑁𝑓(𝑢)=0.4[−2(𝑦−7)²+1100]≤440<𝑅₂,(𝑡,𝑢)→[0,1]×[0,7], 𝑁𝑓(𝑢)=0.4[−2(𝑦−7)+1100]≤440<𝑅₂,(𝑡,𝑢)→[0,1]×[7,450], and lim_𝑦→+∞(𝑓(𝑦)/𝑦)=lim_𝑦→+∞(((𝑦−450)²+214)/𝑦)=+∞.

It is clear that 𝑓∶[0,1]×[0,∞)→[0,∞) is continuous. Since all the conditions of Theorem 3.5 are satisfied, the problem (4.1) has at least two positive solutions.

Acknowledgments

This work supported by the Nature Science Foundation of China under the Contact no. 10901075 and the Key Project of Chinese Ministry of Education (210226).