Multiple positive solutions to nonlinear boundary value problems of a system for fractional differential equations.

By using Krasnoselskii's fixed point theorem, we study the existence of at least one or two positive solutions to a system of fractional boundary value problems given by -D(0+)(ν1)y1(t) = λ1a1(t)f(y1(t), y2(t)), - D(0+)(ν2)y2(t) = λ2a2(t)g(y1(t), y2(t)), where D(0+)(ν) is the standard Riemann-Liouville fractional derivative, ν1, ν2 ∈ (n - 1, n] for n > 3 and n ∈ N, subject to the boundary conditions y1((i))(0) = 0 = y ((i))(0), for 0 ≤ i ≤ n - 2, and [D(0+)(α)y1(t)] t=1 = 0 = [D(0+ (α)y2(t)] t=1, for 1 ≤ α ≤ n - 2, or y1((i))(0) = 0 = y ((i))(0), for 0 ≤ i ≤ n - 2, and [D(0+)(α)y1(t)] t=1 = ϕ1(y1), [D(0+)(α)y2(t)] t=1 = ϕ2(y2), for 1 ≤ α ≤ n - 2, ϕ1, ϕ2 ∈ C([0,1], R). Our results are new and complement previously known results. As an application, we also give an example to demonstrate our result.

1. Introduction

The purpose of this paper is to consider the existence of multiple positive solutions for the following system of nonlinear fractional differential equations: where t ∈ (0,1), ν 1, ν 2 ∈ (n − 1, n] for n > 3 and n ∈ N, and λ 1, λ 2 > 0, subject to a couple of boundary conditions. In particular, we first consider (1) subject to where f, g ∈ C([0, ∞)×[0, ∞), [0, ∞)), and a 1, a 2 ∈ C([0,1], [0, ∞)). We then consider the case in which the boundary conditions are changed to where ϕ 1, ϕ 2 ∈ C([0,1], R).

Fractional differential equations arise in many fields, such as physics, mechanics, chemistry, economics, and engineering and biological sciences; see [1-11] for example. In recent years, the study of positive solutions for fractional differential equation boundary value problems has attracted considerable attention, and fruits from research into it emerge continuously. For a small sample of such work, we refer the reader to [12-20] and the references therein. The situation of at least one positive solution has been studied in many excellent monograph; see [12–19, 21] and other references therein. In [22], by means of Schauder fixed point theorem, Su investigated the existence of one positive solution to the following boundary value problem for a coupled system of nonlinear fractional differential equations: where 1 < α, β < 2, μ, ν > 0, α − ν ≥ 1, β − μ ≥ 1.

In [21], Goodrich established the existence of one positive solution to problems (1)-(2) and (1), (3) by using Krasnoselskii's fixed point theorem. Different from the above works mentioned, in this paper we will present the existence of at least two positive solutions to problems (1)-(2) and (1), (3) by using the similar method presented in [21]. Moreover, under different conditions, we also present the existence of at least one positive solution to problems (1)-(2) and (1), (3) with λ 1 = λ 2 = 1.

2. Preliminaries

For the convenience of the reader, we present here some definitions, lemmas, and basic results that will be used in the proofs of our theorems.

Definition 1 (see [23])

Let ν > 0 with ν ∈ R. Suppose that y : [a, +∞) → R. Then the νth Riemann-Liouville fractional integral is defined to be whenever the right-hand side is defined. Similarly, with ν > 0 and ν ∈ R, we define the νth Riemann-Liouville fractional derivative to be where n ∈ N is the unique positive integer satisfying n − 1 ≤ ν < n and t > a.

Lemma 2 (see [24])

Let g ∈ C ([0,1]) be given. Then the unique solution to problem −D 0 y(t) = g(t) together with the boundary conditions y ((0) = 0 = [D 0 y(t)], where 1 ≤ α ≤ n − 2 and 0 ≤ i ≤ n − 2, is where is the Green function for this problem.

Lemma 3 (see [24])

Let G(t, s) be as given in the statement of Lemma 2. Then one finds that

G(t, s) is a continuous function on the unit square [0,1]×[0,1];

G(t, s) ≥ 0 for each (t, s)∈[0,1]×[0,1];

max⁡ G(t, s) = G(1, s), for each s ∈ [0,1].

Lemma 4 (see [24])

Let G(t, s) be as given in the statement of Lemma 2. Then there exists a constant γ ∈ (0,1) such that To prove our results, we need the following Krasnoselskii's fixed point theorem which can be seen in Guo and Lakshmikantham [25].

Lemma 5 (see [25])

Let E be a Banach space, and let P be a cone. Assume that Ω 1,  Ω 2 are open bounded subsets of E with 0 ∈ Ω 1, , and let be a completely continuous operator such that

||Tu|| ≤ ||u||, ∀u ∈ P⋂ ∂Ω 1, and ||Tu|| ≥ ||u||, ∀u ∈ P⋂ ∂Ω 2; or

||Tu|| ≥ ||u||, ∀u ∈ P⋂ ∂Ω 1, and ||Tu|| ≤ ||u||, ∀u ∈ P⋂ ∂Ω 2.

Then T has a fixed point in .

3. Main Results

In this section, we apply Lemma 5 to study problems (1)-(2) and (1), (3), and we obtain some new results on the existence of multiple positive solutions.

3.1. Problem (1)-(2) in the General Case

In our considerations, let E represent the Banach space of C([0,1]) when equipped with the usual supremum norm, ||·||. Then put X : = E × E, where X is equipped with the norm ||(y 1, y 2)|| : = ||y 1|| + ||y 2|| for (y 1, y 2) ∈ X. Observe that X is also a Banach space (see [26]). In addition, we define two operators T 1,  T 2 : X → E by where G 1(t, s) is the Green function of Lemma 2 with ν replaced by ν 1 and, likewise, G 2(t, s) is the Green function of Lemma 2 with ν replaced by ν 2. Now, we define an operator S : X → X by We claim that whenever (y 1, y 2) ∈ X is a fixed point of the operator defined in (11), it follows that y 1(t) and y 2(t) solve problems (1)-(2). That is, a pair of functions y 1, y 2 ∈ X is a solution of problems (1)-(2) if and only if y 1, y 2 is a fixed point of the operator S defined in (11) (see [26]).

In the following, we will look for fixed points of the operator S, because these fixed points coincide with solutions of problems (1)-(2). For use in the sequel, let γ 1 and γ 2 be the constants given by Lemma 4 associated, respectively, with the Green functions G 1 and G 2, and define by , and notice that .

For the sake of convenience, we set

Now we list some assumptions:such that Φ1 < λ 1, λ 2 < Φ2.

f 0, g 0 ∈ (0, +∞);

f , g ∈ (0, +∞);

there are numbers Φ1, Φ2, where

Next, we define the cone K by

Lemma 6 (see [21])

Let S be the operator defined by (11). Then S : K → K.

Lemma 7

S is a completely continuous operator.

Proof

The operator T 1 : K → E is continuous in view of nonnegativeness and continuity of G 1(t, s), f(y 1, y 2) and a 1(t).

Let Ω⊆K be bounded; that is, there exists a positive constant M > 0 such that ||(y 1, y 2)|| ≤ M, for all (y 1, y 2) ∈ Ω. Let L = max⁡0≤ | a 1(t)f(y 1(t), y 2(t))|+1; then, for (y 1, y 2) ∈ Ω, we have Hence, T 1(Ω) is bounded.

On the other hand, given ε > 0, setting δ = min⁡{(1/2)(Γ(ν 1)ε/LΦ2)1/(, εΓ(ν 1)/(ν 1 − 1)LΦ2}, then, for each (y 1, y 2) ∈ Ω, t 1, t 2 ∈ [0,1], t 1 < t 2, and t 2 − t 1 < δ, one has |T 1(y 1, y 2)(t 2) − T 1(y 1, y 2)(t 1)|<ε. That is to say, T(Ω) is equicontinuity. In fact,

In the following, we divide the proof into two cases.

Case 1. If δ ≤ t 1 < t 2 < 1, then we have where t ∈ (t 1, t 2).

Case 2. If 0 ≤ t 1 < δ, t 2 < 2δ, then we have By the means of the Arzela-Ascoli theorem, we have that T 1 is completely continuous. Similarly, T 2 is completely continuous. Consequently, S : K → K is a completely continuous operator. This completes the proof.

In [21], Goodrich established the following result.

Theorem 8 (see Theorem 3.3 in [21])

Suppose that (F 1)–(F 3) are satisfied. Then problem (1)-(2) has at least one positive solution.

From Theorem 8, the following problem is natural: whether we can obtain some conclusions or not, if f 0 = f = g 0 = g = 0  or  f 0 = f = g 0 = g = ∞? In the rest of this paper, we give some answers to this problem.

For the sake of convenience, we make some assumptions:

there exist constants ρ 1, A 1 > 0, such that

there exist constants ρ 2, A 2 > 0, such that

there are numbers Λ1, Λ2, where

such that Λ1 < λ 1,  λ 2 < Λ2;

there are numbers Λ3,  Λ4, where

such that Λ3 < λ 1,  λ 2 < Λ4.

Theorem 9

Suppose that f 0 = f = g 0 = g = ∞ and (H 1), (P 1) are satisfied. Then problem (1)-(2) has at least two positive solutions (y 1 0, y 2 0), , such that .

From Lemma 7, S is a completely continuous operator. At first, in view of f 0 = g 0 = ∞, we have f(y 1, y 2) ≥ M(y 1 + y 2), for 0 < ||(y 1, y 2)|| < r 1* < ρ 1; g(y 1, y 2) ≥ M(y 1 + y 2), for 0 < ||(y 1, y 2)|| < r 2* < ρ 1, where M satisfies M ≥ 1. Set ρ 0 : = min⁡{r 1*, r 2*}. So we define Ω by Ω : = {(y 1, y 2) ∈ X : ||(y 1, y 2)|| < ρ 0}. Then for each (y 1, y 2) ∈ K⋂∂Ω , we find that So ||T 1(y 1, y 2)|| ≥ (1/2)||(y 1, y 2)|| for (y 1, y 2) ∈ K⋂∂Ω .

Similarly, we find that ||T 2(y 1, y 2)|| ≥ (1/2)||(y 1, y 2)|| for (y 1, y 2) ∈ K⋂∂Ω . Consequently, whenever (y 1, y 2) ∈ K⋂∂Ω . Thus, S is cone expansion on K⋂∂Ω .

Next, since f = g = ∞, we have f(y 1, y 2) ≥ M 1(y 1 + y 2) for y 1 + y 2 ≥ r 1** > ρ 1;   g(y 1, y 2) ≥ M 1(y 1 + y 2) for y 1 + y 2 ≥ r 2** > ρ 1, where M 1 satisfies M 1 ≥ 1. Set ρ 10 : = max⁡{r 1**, r 2**}. Let and Ω : = {(y 1, y 2) ∈ X : ||(y 1, y 2)|| < ρ 0*}. Then (y 1, y 2) ∈ K⋂∂Ω implies So we obtain So ||T 1(y 1, y 2)|| ≥ (1/2)||(y 1, y 2)|| for (y 1, y 2) ∈ K⋂∂Ω .

Similarly, we find that ||T 2(y 1, y 2)|| ≥ (1/2)||(y 1, y 2)|| for (y 1, y 2) ∈ K⋂∂Ω .

Consequently, ||S 2(y 1, y 2)|| ≥ ||(y 1, y 2)||, whenever (y 1, y 2) ∈ K⋂∂Ω . Thus, S is cone expansion on K⋂∂Ω .

Finally, let Ω : = {(y 1, y 2) ∈ X : ||(y 1, y 2)|| < ρ 1}. For (y 1, y 2) ∈ K⋂∂Ω , from (H 1), (P 1), we have Similarly, we find that ||T 2(y 1, y 2)|| ≤ (1/2)||(y 1, y 2)|| for (y 1, y 2) ∈ K⋂∂Ω .

Consequently, ||S(y 1, y 2)|| ≤ ||(y 1, y 2)||, whenever (y 1, y 2) ∈ K⋂∂Ω . Thus, S is cone compression on K⋂∂Ω .

So, from Lemma 5, S has a fixed point and a fixed point . Both are positive solutions of BVP (1)-(2) with The proof is complete.

Theorem 10

Suppose that f 0 = f = g 0 = g = 0 and (H 2), (P 2) are satisfied. Then problem (1)-(2) has at least two positive solutions (y 1 0, y 2 0), , such that .

At first, in view of f 0 = g 0 = 0, we have f(y 1, y 2) < ε(y 1 + y 2), g(y 1, y 2) < ε(y 1 + y 2), for 0 < ||(y 1, y 2)|| ≤ ρ < ρ 2, where ε satisfies ε ≤ 1. Let Ω : = {(y 1, y 2) ∈ X : ||(y 1, y 2)|| < ρ}.

Then for each (y 1, y 2) ∈ K⋂∂Ω , we find that Like Theorem 9, we get ||S(y 1, y 2)|| ≤ ||(y 1, y 2)|| for (y 1, y 2) ∈ K⋂∂Ω .

Next, in view of f = g = 0, we have f(y 1, y 2) < ε 1(y 1 + y 2), g(y 1, y 2) < ε 1(y 1 + y 2), for y 1 + y 2 ≥ ρ′ > ρ 2, where ε 1 satisfies ε 1 ≤ 1. We consider two cases.

Case 1. Suppose that f is unbounded; there exists ρ* > ρ′ such that Since ρ* > ρ′, one has f(y 1, y 2) ≤ f(y 1*, y 2*) < ε 1(y 1* + y 2*) for 0 ≤ ||(y 1, y 2)|| ≤ ρ*. Then, for (y 1, y 2) ∈ K and ||(y 1, y 2)|| = ρ*, we obtain

Case 2. Suppose that f is bounded; there exists L 1 such that f(y 1, y 2) ≤ L 1 for all (y 1, y 2) ∈ K. Taking ρ* ≥ max⁡{2ρ 2, L 1}, for (y 1, y 2) ∈ K and ||(y 1, y 2)|| = ρ*, we obtain Hence, in either case, we always may set Ω : = {(y 1, y 2) ∈ X : ||(y 1, y 2)|| < ρ*} such that ||T 1(y 1, y 2)|| ≤ (1/2)||(y 1, y 2)|| for (y 1, y 2) ∈ K⋂∂Ω . Like Theorem 9, we get ||S(y 1, y 2)|| ≤ ||(y 1, y 2)||, for (y 1, y 2) ∈ K⋂∂Ω .

Finally, set Ω : = {(y 1, y 2) ∈ X : ||(y 1, y 2)|| < ρ 2}. Then (y 1, y 2) ∈ K⋂∂Ω implies Hence we have Consequently, ||T 1(y 1, y 2)|| ≥ (1/2)||(y 1, y 2)|| for (y 1, y 2) ∈ K⋂∂Ω . Like Theorem 9, we get ||S(y 1, y 2)|| ≥ ||(y 1, y 2)|| for (y 1, y 2) ∈ K⋂∂Ω .

So, from Lemma 5, S has a fixed point and a fixed point . Both are positive solutions of BVP (1)-(2) with which complete the proof.

3.2. Problem (1)–(3) in Case λ 1  =  λ 2  =  1

In the following, for the sake of convenience, set Assume that there exist two positive constants ρ 1 ≠ ρ 2 such that

f(y 1, y 2), g(y 1, y 2) ≤ B 1 −1 ρ 1, for 0 ≤ ||(y 1, y 2)|| ≤ ρ 1;

f(y 1, y 2), g(y 1, y 2) ≥ B 2 −1 ρ 2, for .

Theorem 11

Suppose that (H 3) and (H 4) are satisfied. Then problem (1)-(2), in the case where λ 1 = λ 2 = 1, has at least one positive solution (y 1 0, y 2 0) such that ||(y 1 0, y 2 0)|| between ρ 1 and ρ 2.

With loss of generality, we may assume that ρ 1 < ρ 2.

Let Ω : = {(y 1, y 2) ∈ X : ||(y 1, y 2)|| < ρ 1}. For (y 1, y 2) ∈ K⋂ ∂Ω , one has Like Theorem 9, we get ||S(y 1, y 2)|| ≤ ||(y 1, y 2)|| for (y 1, y 2) ∈ K⋂∂Ω .

Now, set Ω : = {(y 1, y 2) ∈ X : ||(y 1, y 2)|| < ρ 2}. Then for (y 1, y 2) ∈ K⋂∂Ω , one has Thus, we get Like Theorem 9, we get ||S(y 1, y 2)|| ≥ ||(y 1, y 2)|| for (y 1, y 2) ∈ K⋂∂Ω . Hence, from Lemma 5, we complete the proof.

Remark 12

In [21], problem (1)-(2) with λ 1 = λ 2 = 1 is not considered.

3.3. Problem (1), (3) in the General Case

Consider the following.

Lemma 13 (see [21])

A pair of functions (y 1, y 2) ∈ X is a solution of (1), (3) if and only if (y 1, y 2) is a fixed point of the operator U : X → X defined by where β 1, β 2 : [0,1]→[0,1] are defined by

Lemma 14 (see [21])

Each of β 1(t) and β 2(t) is strictly increasing in t and satisfies β 1(0) = β 2(0) = 0 and β 1(1), β 2(1)∈(0,1). Moreover, there exist constants M and M satisfying M , M ∈ (0,1) such that min⁡ β 1(t) ≥ M ||β 1|| and min⁡ β 2(t) ≥ M ||β 2||.

Let one define a new cone K 1 by where . It is obvious that γ 0 ∈ (0,1).

Lemma 15 (see [21])

U : K 1 → K 1 is a completely continuous operator.

Now, one assumes

ϕ 1(y 1) ≤ ||y 1||/4, ϕ 2(y 2) ≤ ||y 2||/4 for each (y 1, y 2) ∈ K 1;

There are numbers Λ5,  Λ6, where

such that Λ5 < λ 1,  λ 2 < Λ6.

There are numbers Λ7,  Λ8, where

such that Λ7 < λ 1,  λ 2 < Λ8.

Theorem 16

Suppose that f 0 = f = g 0 = g = ∞ and (H 1), (D 1), (P 3) are satisfied. Then problem (1), (3) has at least two positive solutions (y 1 0, y 2 0),, such that

At first, in view of f 0 = g 0 = ∞, we have f(y 1, y 2) ≥ M(y 1 + y 2), g(y 1, y 2) ≥ M(y 1 + y 2), for 0 < ||(y 1, y 2)|| ≤ ρ 0 < ρ 1, where M satisfies M ≥ 1.

Let Ω : = {(y 1, y 2) ∈ X : ||(y 1, y 2)|| < ρ 0}. Then for each (y 1, y 2) ∈ K 1⋂∂Ω , we find that So ||U 1(y 1, y 2)|| ≥ (1/2)||(y 1, y 2)|| for (y 1, y 2) ∈ K 1⋂∂Ω .

Similarly, we find that ||U 2(y 1, y 2)|| ≥ (1/2)||(y 1, y 2)|| for (y 1, y 2) ∈ K 1⋂∂Ω . Consequently, whenever (y 1, y 2) ∈ K 1⋂∂Ω . Thus, U is cone expansion on K 1⋂∂Ω .

Next, since f = g = ∞, we get f(y 1, y 2) ≥ M 1(y 1 + y 2), g(y 1, y 2) ≥ M 1(y 1 + y 2), for y 1 + y 2 ≥ ρ 10 > ρ 1, where M 1 satisfies M 1 ≥ 1. Let ρ 0* = max⁡{2ρ 1, (ρ 10/γ 0)} and Ω : = {(y 1, y 2) ∈ X : ||(y 1, y 2)|| < ρ 0*}; then, (y 1, y 2) ∈ K⋂∂Ω implies So for (y 1, y 2) ∈ K 1⋂∂Ω , we obtain That is, ||U 1(y 1, y 2)|| ≥ (1/2)||(y 1, y 2)|| for (y 1, y 2) ∈ K 1⋂∂Ω .

Similarly, we find that ||U 2(y 1, y 2)|| ≥ (1/2)||(y 1, y 2)|| for (y 1, y 2) ∈ K 1⋂∂Ω . Consequently, ||U(y 1, y 2)|| ≥ ||(y 1, y 2)||, whenever (y 1, y 2) ∈ K 1⋂∂Ω . Thus, U is cone expansion on K 1⋂∂Ω .

Finally, let Ω : = {(y 1, y 2) ∈ X : ||(y 1, y 2)|| < ρ 1}. For (y 1, y 2) ∈ K⋂∂Ω , from (H 1), (D 1), and (P 3), we have Similarly, we find that ||U 2(y 1, y 2)|| ≤ (1/2)||(y 1, y 2)|| for (y 1, y 2) ∈ K 1⋂∂Ω . Consequently, ||U(y 1, y 2)|| ≤ ||(y 1, y 2)||, whenever (y 1, y 2) ∈ K 1⋂∂Ω . Thus, U is cone compression on K 1⋂∂Ω .

So, from Lemma 5, U has a fixed point and a fixed point . Both are positive solutions of BVP (1), (3) with The proof is complete.

Theorem 17

Suppose that f 0 = f = g 0 = g = 0 and (H 2), (D 1), (P 4) are satisfied. Then problem (1), (3) has at least two positive solutions (y 1 0, y 2 0), , such that

At first, in view of f 0 = g 0 = 0, we have f(y 1, y 2) < ε(y 1 + y 2), g(y 1, y 2) < ε(y 1 + y 2) for ||(y 1, y 2)|| ≤ ρ < ρ 2, where ε satisfies ε ≤ 1. Let Ω : = {(y 1, y 2) ∈ X : ||(y 1, y 2)|| < ρ}. Then for each (y 1, y 2) ∈ K 1⋂ ∂Ω , we find that Like Theorem 16, we get ||U(y 1, y 2)|| ≤ ||(y 1, y 2)|| for (y 1, y 2) ∈ K 1⋂ ∂Ω .

Next, in view of f = g = 0, we have f(y 1, y 2) < ε 1(y 1 + y 2), g(y 1, y 2) < ε 1(y 1 + y 2), for y 1 + y 2 ≥ ρ′ > ρ 2, where ε 1 satisfies ε 1 ≤ 1. We consider two cases.

Case 1. Suppose that f is unbounded and there exists ρ* > ρ′ such that Since ρ* > ρ′, one has f(y 1, y 2) ≤ f(y 1*, y 2*) < ε 1(y 1* + y 2*) for 0 ≤ ||(y 1, y 2)|| ≤ ρ*.

Then, for (y 1, y 2) ∈ K 1 and ||(y 1, y 2)|| = ρ*, we obtain

Case 2. Suppose that f is bounded; there, exists L 1 such that f(y 1, y 2) ≤ L 1 for all (y 1, y 2) ∈ K 1. Taking ρ* ≥ max⁡{2ρ 2, L 1}, for (y 1, y 2) ∈ K 1 and ||(y 1, y 2)|| = ρ*, we obtain Hence, in either case, we always may set Ω : = {(y 1, y 2) ∈ X : ||(y 1, y 2)|| < ρ*} such that ||U 1(y 1, y 2)|| ≤ (1/2)||(y 1, y 2)|| for (y 1, y 2) ∈ K 1⋂ ∂Ω .

Like Theorem 16, we get ||U(y 1, y 2)|| ≤ ||(y 1, y 2)||, for (y 1, y 2) ∈ K 1⋂ ∂Ω .

Finally, set Ω : = {(y 1, y 2) ∈ X : ||(y 1, y 2)|| < ρ 2}. Then (y 1, y 2) ∈ K 1⋂ ∂Ω implies Hence we have So, ||U 1(y 1, y 2)|| ≥ (1/2)||(y 1, y 2)|| for (y 1, y 2) ∈ K 1⋂ ∂Ω .

Like Theorem 16, we get ||U(y 1, y 2)|| ≥ ||(y 1, y 2)|| for (y 1, y 2) ∈ K 1⋂ ∂Ω . So, from Lemma 5, U has a fixed point and a fixed point . Both are positive solutions of BVP (1), (3) with , which complete the proof.

3.4. Problem (1), (3) in Case λ 1  =  λ 2  =  1

In [21], the author obtained that problem (1), (3) with λ 1 = λ 2 = 1 having at least one positive solution. In the following, we also establish the existence of one positive solution to problem (1), (3) with λ 1 = λ 2 = 1 under different conditions.

For the sake of convenience, set Assume that there exist two positive constants ρ 1 ≠ ρ 2 such that

f(y 1, y 2), g(y 1, y 2) ≤ B 3 −1 ρ 1 for 0 ≤ ||(y 1, y 2)|| ≤ ρ 1;

f(y 1, y 2), g(y 1, y 2) ≥ B 2 −1 ρ 2 for γ 0 ρ 2 ≤ ||(y 1, y 2)|| ≤ ρ 2.

Theorem 18

Suppose that (H 5),  (H 6), and (D 1) are satisfied. Then problem (1), (3), in the case where λ 1 = λ 2 = 1, has at least one positive solution (y 1 0, y 2 0) such that ||(y 1, y 2)|| between ρ 1 and ρ 2.

With loss of generality, we may assume that ρ 1 < ρ 2. Let Ω : = {(y 1, y 2) ∈ X : ||(y 1, y 2)|| < ρ 1}. For (y 1, y 2) ∈ K 1⋂ ∂Ω , from (H 7), (D 1), one has Like Theorem 16, we get ||U(y 1, y 2)|| ≤ ||(y 1, y 2)|| for (y 1, y 2) ∈ K 1⋂ ∂Ω .

Now, set Ω : = {(y 1, y 2) ∈ X : ||(y 1, y 2)|| < ρ 2}. For (y 1, y 2) ∈ K 1⋂ ∂Ω , one has Thus, from (H 8), we get Like Theorem 16, we get ||U(y 1, y 2)|| ≥ ||(y 1, y 2)|| for (y 1, y 2) ∈ K 1⋂ ∂Ω . Hence, from Lemma 5, we complete the proof.

4. An Example

To illustrate how our main results can be used in practice, we present one example.

Example 19

Consider the following BVP, for t ∈ (0,1): subject to the boundary conditions Obviously, problem (63)–(65) fits the framework of problem (1)-(2) with In addition, we have set We can see that f, g : [0, +∞)×[0, +∞)→[0, +∞) and are continuous. The functions a 1(t) and a 2(t) are obviously nonnegative.

Now, observe that f 0 = f = g 0 = g = ∞ holds. Again set A 1 = 1/850, because f(y 1, y 2), g(y 1, y 2) is monotone increasing function for (y 1, y 2) ≥ 0, taking ρ 1 = 4; then, when ||(y 1, y 2)|| ∈ [0, ρ 1], we get which implies that (H 1) holds.

On the other hand, to calculate the admissible range of the eigenvalues λ 1, λ 2, as given by condition (P 1), observe by numerical approximation, we find that Thus, for any λ 1, λ 2 satisfying 163530 < λ 1, λ 2 < 164547.25, condition (P 1) will be satisfied.

Consequently, by Theorem 9, problem (63)–(65) has at least two positive solutions.