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On a first-order differential system with initial and nonlocal boundary conditions

  • Le Thi Phuong Ngoc and Nguyen Thanh Long EMAIL logo
Published/Copyright: August 1, 2022
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Abstract

This paper is devoted to the existence of solutions and the multiplicity of positive solutions of an initial-boundary value problem for a nonlinear first-order differential system with nonlocal conditions. The main tool is the fixed-point theorem in which we construct the novel representation of the associated Green’s functions with useful properties and define a cone in the Banach space suitably. Some examples are also given to demonstrate the validity of the main results.

MSC 2010: 34B07; 34B10; 34B18; 34B27

1 Introduction

In this paper, we consider the following nonlinear system

(1.1) u ( t ) = f ( t , u ( t ) , v ( t ) ) , t ( 0 , T ) , v ( t ) = g ( t , u ( t ) , v ( t ) ) , t ( 0 , T ) ,

associated with the initial and nonlocal boundary conditions

(1.2) u ( 0 ) = u 0 , v ( 0 ) = j = 1 N μ j v ( T j ) + 0 T h ( t ) v ( t ) d t ,

where f , g : [ 0 , T ] × R n × R n R n , h : [ 0 , T ] R are given functions and u 0 R n ; μ j , T j ( j = 1 , N ¯ ) are given constants, with 0 < T 1 < T 2 < < T N = T .

Nonlocal initial and boundary value problems with integral boundary conditions or multipoint boundary value problems for differential equations have been extensively studied due to their numerous applications in various fields of science and applied mathematics; see [1,2, 3,4,5, 6,7,8, 9,10,11, 12,13,14, 15,16,17, 18,19,20, 21,22] and the references given therein.

In [4], Bolojan-Nica et al. proved the solvability of a system of N first-order differential equations of the form

(1.3) x 1 ( t ) = f 1 ( t , x 1 ( t ) , , x N ( t ) ) , x N ( t ) = f N ( t , x 1 ( t ) , , x N ( t ) ) , t [ 0 , 1 ] ,

subject to the coupled nonlocal conditions

(1.4) x 1 ( 0 ) = j = 1 N α 1 j [ x j ] , x N ( 0 ) = j = 1 N α N j [ x j ] ,

where α i j : C ( [ 0 , 1 ] ) R were linear continuous functionals. As noted in [4], α i j could be written in a form involving Stieltjes integrals, namely,

(1.5) α i j [ v ] = 0 1 ν ( s ) d A i j ( s ) , v C ( [ 0 , 1 ] ) ,

for some functions of bounded variation A i j ( i , j = 1 , , n ). On the other hand, the formula (1.5) covered, as special cases, initial conditions of the type:

α i j [ v ] = k = 1 m a k v ( t k ) ,

where a k R and 0 t 1 < t 2 < < t m 1 were given points. When all the functionals α i j had discrete expressions, (1.4) gave a multipoint condition. The approach there was based on the fixed-point theorems of Perov, Schauder, and Schaefer and on a vector method for treating systems, which used matrices with spectral radius less than one.

In [3], by extending the use of techniques in [4] to nonlocal conditions given by nonlinear functionals, O. Bolojan et al. proved the existence results of solutions to an initial value problem for a nonlinear first-order differential system subject to nonlinear nonlocal initial conditions of the form

(1.6) x ( t ) = f 1 ( t , x ( t ) , y ( t ) ) , y ( t ) = f 2 ( t , x ( t ) , y ( t ) ) , a.e. on [ 0 , 1 ] , x ( 0 ) = α [ x , y ] , y ( 0 ) = β [ x , y ] ,

where f 1 , f 2 : [ 0 , 1 ] × R 2 R were L 1 -Carathéodory functions, α , β : C ( [ 0 , 1 ] ) × C ( [ 0 , 1 ] ) R were nonlinear continuous functionals, and the solution ( x , y ) was sought in W 1 , 1 ( 0 , 1 ; R 2 ) . That problem was studied by using the fixed-point principles by Perov, Schauder, and Leray-Schauder, together with the technique that used convergent matrices and vector norms.

In [16], by applying the Banach’s fixed-point theorem and the Schaefer’s fixed-point theorem, Mardanov et al. proved the existence and uniqueness theorems for the system of ordinary differential equations with three-point boundary conditions as follows:

(1.7) y = f ( t , y ) , t ( 0 , T ) , A y ( 0 ) + B y ( t 1 ) + C y ( T ) = d ,

where A , B , and C were constant square matrices of order n such that det ( A + B + C ) 0 , f : [ 0 , T ] × R n R n was a given function, d R n was a given vector, t 1 satisfied the condition of 0 < t 1 < T , and y : [ 0 , T ] R n was unknown.

In [9], Han considered the second-order three-point boundary value problem in the form

(1.8) x ( t ) = f ( t , x ( t ) ) , t ( 0 , 1 ) , x ( 0 ) = 0 , x ( η ) = x ( 1 ) ,

with η ( 0 , 1 ) . By means of the fixed-point theorem in cones, the existence and multiplicity of positive solutions were proved.

In [20], Truong et al. studied the following m -point boundary value problem:

(1.9) x ( t ) = f ( t , x ( t ) ) , t ( 0 , 1 ) , x ( 0 ) = 0 , x ( 1 ) = j = 1 m 2 α j x ( η j ) ,

where m 3 , η j ( 0 , 1 ) , and α j 0 , for all j = 1 , m 2 ¯ such that j = 1 m 2 α j < 1 . By applying well-known Guo-Krasnoselskii’s fixed-point theorem and applying the monotone iterative technique, the results obtained in [20] were the existence and multiplicity of positive solutions. Furthermore, the compactness of the set of positive solutions was proved.

In [5], Boucherif applied the fixed-point theorem in a cone to study the existence of positive solutions for the problem given by

(1.10) x ( t ) = f ( t , x ( t ) ) , t ( 0 , 1 ) , x ( 0 ) c x ( 0 ) = 0 1 g 0 ( s ) x ( s ) d s , x ( 1 ) d x ( 1 ) = 0 1 g 1 ( s ) x ( s ) d s ,

where f : [ 0 , 1 ] × R R was continuous, g 0 , g 1 : [ 0 , 1 ] [ 0 , + ) were continuous and positive, and c and d were nonnegative real parameters.

In [1], Agarwal et al. formulated existence results for solutions to discrete equations that approximate three-point boundary value problems for second-order ordinary differential equations. The proofs of these results were finished based on extending the notion of discrete compatibility, which was a degree-based relationship between the given boundary conditions and the lower or upper solutions chosen, to three-point boundary conditions. On the other hand, the invariance of the degree under the homotopy of the degree theory was also applied in the aforementioned proofs.

In [11], Henderson and Luca investigated the following multipoint boundary value problem for the system of nonlinear higher-order ordinary differential equations of the type

(1.11) u ( n ) ( t ) = f ( t , v ( t ) ) , t ( 0 , T ) , n N , n 2 , v ( m ) ( t ) = g ( t , u ( t ) ) , t ( 0 , T ) , m N , m 2 ,

with the multipoint boundary conditions

(1.12) u ( 0 ) = u ( 0 ) = = u ( n 2 ) ( 0 ) = 0 , v ( 0 ) = v ( 0 ) = = v ( m 2 ) ( 0 ) = 0 , u ( T ) = i = 1 p 2 a i u ( ξ i ) , p N , p 3 , v ( T ) = i = 1 q 2 b i v ( η i ) , q N , q 3 .

Under sufficient assumptions on f and g , the authors proved the existence and multiplicity of positive solutions of the aforementioned problem by applying the fixed-point index theory.

Motivated by the aforementioned works, by applying standard fixed-point theorems, we study the initial and multipoint boundary problems (1.1) and (1.2).

This paper is organized as follows. In Section 2, we present some preliminaries. Here, the Green function is established for Prob. (1.1) and (1.2) such that the considered problem is reduced to the equivalent integral system. Section 3 is devoted to the existence and uniqueness of solutions based on the fixed-point theorems of Banach and Krasnoselskii. In Section 4, we prove the existence of positive solutions by using the Guo-Krasnoselskii’s fixed-point theorem in a cone. In Section 5, we observe some multiplicity results for positive solutions. Finally, a remark is also given in Section 6 for a system of multiple differential equations. Furthermore, various examples (Examples 3.1, 3.2, and 4.1) are also presented to demonstrate the validity of the main results. The methods used for proofs are standard, but their applications in this framework of the system (1.1) and (1.2) with nonlocal conditions are new, where we construct the novel representation of the Green’s functions with useful properties and define a cone suitably. We note more that properties of the associated Green’s function are of fundamental importance in many of our arguments, and based on these properties, we establish sufficient conditions for which solutions/positive solutions exist.

2 Preliminaries

Throughout this paper, x i denotes the ith component of x R n , x 1 = x 1 + + x n denotes the norm of x = ( x 1 , , x n ) R n , C ( [ 0 , T ] ; R n ) denotes the Banach space of all continuous functions u : [ 0 , T ] R n endowed with the usual maximum norm u C ( [ 0 , T ] ; R n ) = max 0 t T u ( t ) 1 ; C ( [ 0 , T ] ) C ( [ 0 , T ] ; R ) , and L 1 ( 0 , T ) denotes the set of all integrable functions from ( 0 , T ) to R . Denote the Banach space of continuously differentiable functions u : [ 0 , T ] R n by C 1 ( [ 0 , T ] ; R n ) endowed with the usual norm as follows:

u C 1 ( [ 0 , T ] ; R n ) = u C ( [ 0 , T ] ; R n ) + u C ( [ 0 , T ] ; R n ) .

In addition, for all x = ( x 1 , , x n ) R n , we say x 0 (or x R + n ) if and only if x i 0 , for all i = 1 , , n ; x > 0 if and only if x 0 , and x 0 ; and we say x y if and only if x y 0 , for all x , y R n . It is clear to see that if one has x y 0 then x 1 y 1 ; therefore, x 1 = x y 1 + y 1 . Moreover, if u L 1 ( 0 , T ; R n ) , u ( t ) 0 , t [ 0 , T ] , and then, we have

0 T u ( t ) d t 1 = 0 T u 1 ( t ) d t , , 0 T u n ( t ) d t 1 = 0 T u 1 ( t ) d t + + 0 T u n ( t ) d t = 0 T u ( t ) 1 d t .

Now, for the convenience of the reader, we recall the following fixed-point theorems.

Theorem 2.1

(Krasnosellskii) [19]. Let M be a nonempty bounded closed convex subset of a Banach space X . Suppose that U : M X is a contraction and C : M X is a compact operator such that

U ( x ) + C ( y ) M , x , y M .

Then, U + C has a fixed-point in M .

Theorem 2.2

(Guo-Krasnoselskii) [8]. Let X be a Banach space and let K X be a cone. Assume that Ω 1 , Ω 2 are two open bounded subsets of X with 0 Ω 1 , Ω 1 ¯ Ω 2 and let P : K ( Ω 2 ¯ Ω 1 ) K be a completely continuous operator satisfying one of the following conditions:

  1. P u u , u K Ω 1 and P u u , u K Ω 2 ;

  2. or

  3. P u u , u K Ω 1 and P u u , u K Ω 2 .

Then, the operator P has a fixed-point in K ( Ω 2 ¯ Ω 1 ) .

Next, we construct an expression for solutions of Prob. (1.1) and (1.2), with f , g C ( [ 0 , T ] × R n × R n ; R n ) , where C ( [ 0 , T ] × R n × R n ; R n ) is the space of all continuous functions mapping from [ 0 , T ] × R n × R n to R n . For details, in the following, we shall construct the Green function for Prob. (1.1) and (1.2) such that the considered problem is reduced to an equivalent integral system.

For each α , β > 0 , we denote f α ( t , u , v ) = f ( t , u , v ) + α u and g β ( t , u , v ) = g ( t , u , v ) + β v .

Lemma 2.3

Assume that f , g C ( [ 0 , T ] × R n × R n ; R n ) , h C ( [ 0 , T ] ) and σ 0 , with σ 1 0 T e β t h ( t ) d t j = 1 N μ j e β T j . The pair of functions ( u , v ) C ( [ 0 , T ] ; R n ) × C ( [ 0 , T ] ; R n ) is a solution of Prob. (1.1) and (1.2) if and only if ( u , v ) is a solution of the following integral equations system:

(2.1) u ( t ) = u 0 e α t + 0 t e α ( t s ) f α ( s , u ( s ) , v ( s ) ) d s , v ( t ) = 0 t e β ( t s ) g β ( s , u ( s ) , v ( s ) ) d s + e β t σ 0 T s T e β ( τ s ) h ( τ ) d τ g β ( s , u ( s ) , v ( s ) ) d s + e β t σ j = 1 N μ j 0 T j e β ( T j s ) g β ( s , u ( s ) , v ( s ) ) d s .

Proof

Let ( u , v ) C ( [ 0 , T ] ; R n ) × C ( [ 0 , T ] ; R n ) is a solution of Prob. (1.1) and (1.2). It is obvious that ( u , v ) C 1 ( [ 0 , T ] ; R n ) × C 1 ( [ 0 , T ] ; R n ) and ( u , v ) satisfies (1.1) and (1.2). For each α , β > 0 , the system (1.1) can be transformed into an equivalent form as follows:

(2.2) u + α u = f α ( t , u , v ) , t ( 0 , T ) , v + β v = g β ( t , u , v ) , t ( 0 , T ) .

Multiplying (2.2) 1 and (2.2) 2 by e α t and e β t , respectively, and integrating from 0 to t , we obtain

(2.3) u ( t ) = u 0 e α t + 0 t e α ( t s ) f α ( s , u ( s ) , v ( s ) ) d s , v ( t ) = v ( 0 ) e β t + 0 t e β ( t s ) g β ( s , u ( s ) , v ( s ) ) d s .

Multiplying (2.2) 2 by e β t and taking an integration from 0 to T j , we have

(2.4) v ( T j ) e β T j v ( 0 ) = 0 T j e β ( T j s ) g β ( s , u ( s ) , v ( s ) ) d s .

Combining (1.2) 2 , (2.3), and (2.4), we infer that ( u ( t ) , v ( t ) ) satisfies the system (2.1), so ( u , v ) is a solution of the nonlinear integral system (2.1).

Otherwise, let ( u , v ) C ( [ 0 , T ] ; R n ) × C ( [ 0 , T ] ; R n ) is a solution of the nonlinear integral equations (2.1). It is not difficult to prove that ( u , v ) C 1 ( [ 0 , T ] ; R n ) × C 1 ( [ 0 , T ] ; R n ) and ( u , v ) satisfies Prob. (1.1) and (1.2). Lemma 2.3 is completely proved.□

The second integral equation of (2.1) can be written in the following form:

(2.5) v ( t ) = 0 T G ( t , s ) g β ( s , u ( s ) , v ( s ) ) d s ,

where G ( t , s ) is the Green’s function for Prob. (1.1) and (1.2), and this function is defined as follows:

(2.6) G ( t , s ) = e β ( t s ) , 0 s t T , 0 , 0 t s T , + e β ( t s ) σ s T e β τ h ( τ ) d τ + e β ( t s ) σ j = 1 N μ j e β T j , 0 s T 1 , j = k N μ j e β T j , T k 1 < s T k , μ N e β T T N 1 < s T .

We have a property of the Green’s function G ( t , s ) as follows.

Lemma 2.4

Suppose that h ( t ) 0 t [ 0 , T ] , μ j 0 , for j = 1 , N 1 ¯ , μ N > 0 , such that

σ 1 0 T e β t h ( t ) d t j = 1 N μ j e β T j > 0 .

Then,

(2.7) μ N e β T σ e β ( t s ) G ( t , s ) 1 σ e β ( t s ) , t , s [ 0 , T ] .

Moreover, there exist positive constants g ¯ 0 , g ¯ 1 such that

(2.8) g ¯ 0 G ( t , s ) g ¯ 1 , t , s [ 0 , T ] .

Proof

By direct computations, we have

(2.9) G ( t , s ) e β ( t s ) σ μ N e β T = μ N e β T σ e β ( t s )

and

(2.10) G ( t , s ) 1 + 1 σ 0 T e β τ h ( τ ) d τ + j = 1 N μ j e β T j e β ( t s ) = 1 σ e β ( t s ) .

We note that e β T e β ( t s ) e β T , for all t , s [ 0 , T ] , and we obtain (2.8) with g ¯ 0 = μ N e 2 β T σ , g ¯ 1 = e β T σ . Lemma 2.4 is completely proved.□

If the sign of μ j and h ( t ) can not be determined, we note more that

(2.11) G ( t , s ) G max , t , s [ 0 , T ] ,

where

(2.12) G max = 1 + 1 σ 0 T e β τ h ( τ ) d τ + j = 1 N μ j e β T j e β T .

3 Existence and uniqueness

In this section, we shall prove two existence results (Theorems 3.1 and 3.5) for Prob. (1.1) and (1.2), where h C ( [ 0 , T ] ) and f , g C ( [ 0 , T ] × R n × R n ; R n ) . Based on the Banach’s fixed-point theorem, Theorem 3.1 gives a sufficient condition for the unique existence of a solution. Under weaker conditions, Theorem 3.5 gives a sufficient condition for the existence, and this is proved by using the Krasnoselskii’s fixed-point theorem.

We consider the Banach space X = C ( [ 0 , T ] ; R n ) × C ( [ 0 , T ] ; R n ) equipped with the norm

(3.1) ( u , v ) X = u C ( [ 0 , T ] ; R n ) + v C ( [ 0 , T ] ; R n ) ,

and we define an operator P : X X as follows:

P : X X , ( u , v ) ( P 1 ( u , v ) , P 2 ( u , v ) ) ,

where

(3.2) P 1 ( u , v ) ( t ) = u 0 e α t + 0 t e α ( t s ) f α ( s , u ( s ) , v ( s ) ) d s , P 2 ( u , v ) ( t ) = 0 T G ( t , s ) g β ( s , u ( s ) , v ( s ) ) d s .

We make the following assumptions:

  1. h C ( [ 0 , T ] ) ;

  2. 0 < 0 T h ( t ) d t + j = 1 N μ j < 1 ;

  3. There exists a positive function L f L 1 ( 0 , T ) such that

    (3.3) f ( t , u , v ) f ( t , u ¯ , v ¯ ) 1 L f ( t ) ( u u ¯ 1 + v v ¯ 1 ) ,

    for all ( t , u , v ) , ( t , u ¯ , v ¯ ) [ 0 , T ] × R n × R n ;

  4. There exists a positive function L g L 1 ( 0 , T ) such that

    (3.4) g ( t , u , v ) g ( t , u ¯ , v ¯ ) 1 L g ( t ) ( u u ¯ 1 + v v ¯ 1 ) ,

    for all ( t , u , v ) , ( t , u ¯ , v ¯ ) [ 0 , T ] × R n × R n .

Theorem 3.1

Suppose that ( H 1 ) ( H 4 ) are satisfied. In addition, assume that there exist two positive constants α and β small enough such that

(3.5) L = α T + L f L 1 ( 0 , T ) + ( β T + L g L 1 ( 0 , T ) ) g ¯ 1 < 1 .

Then, Prob. (1.1) and (1.2) have a unique solution.

Proof

First, we put f T = max 0 t T f ( t , 0 , 0 ) 1 , g T = max 0 t T g ( t , 0 , 0 ) 1 and choose R > 0 large enough such that

(3.6) R > u 0 1 + T ( f T + g ¯ 1 g T ) 1 L .

Next, we will finish the proof of this theorem through a process with two steps as follows.

Step 1. Let B R = { ( u , v ) X : ( u , v ) X R } . We show that P ( B R ) B R .

Indeed, for ( u , v ) B R and for all t [ 0 , T ] , we have the following estimates:

(3.7) P 1 ( u , v ) ( t ) 1 u 0 1 + 0 t f α ( s , u ( s ) , v ( s ) ) f α ( s , 0 , 0 ) 1 d s + 0 t f α ( s , 0 , 0 ) 1 d s u 0 1 + R ( α T + L f L 1 ( 0 , T ) ) + T f T ,

and

(3.8) P 2 ( u , v ) ( t ) 1 g ¯ 1 0 T g β ( s , u ( s ) , v ( s ) ) g β ( s , 0 , 0 ) 1 d s + 0 T g β ( s , 0 , 0 ) 1 d s g ¯ 1 [ R ( β T + L g L 1 ( 0 , T ) ) + T g T ] .

Combining (3.7) and (3.8) and the choice of R as in (3.6), we infer that P ( B R ) B R , and it means that operator P : B R B R is defined.

Step 2. We prove that the operator P is a contraction mapping.

Indeed, let ( u , v ) and ( u ¯ , v ¯ ) be arbitrary elements in B R . We have

(3.9) P 1 ( u , v ) ( t ) P 1 ( u ¯ , v ¯ ) ( t ) 1 0 t f α ( s , u ( s ) , v ( s ) ) f α ( s , u ¯ ( s ) , v ¯ ( s ) ) 1 d s ( α T + L f L 1 ( 0 , T ) ) ( u , v ) ( u ¯ , v ¯ ) X

and

(3.10) P 2 ( u , v ) ( t ) P 2 ( u ¯ , v ¯ ) ( t ) 1 g ¯ 1 0 T g β ( s , u ( s ) , v ( s ) ) g β ( s , u ¯ ( s ) , v ¯ ( s ) ) 1 d s g ¯ 1 ( β T + L g L 1 ( 0 , T ) ) ( u , v ) ( u ¯ , v ¯ ) X .

It follows from (3.9) and (3.10) and the assumption in Theorem 3.1 that P : B R B R is a contraction mapping. Applying the Banach’s fixed-point theorem, we verify that Prob. (1.1) and (1.2) have a unique solution ( u , v ) . Theorem 3.1 is proved.□

In what follows, under weaker conditions, the second result is given without the Lipschitzian condition on g as in ( H 3 ) . The main tool is the Krasnoselskii’s fixed-point theorem. We make the assumption ( H 4 ) as below.

( H ¯ 4 ) g : [ 0 , T ] × R n × R n R n is a continuous function, and there exist two positive functions g ˆ 1 , g ˆ 2 L 1 ( 0 , T ) such that

(3.11) g ( t , u , v ) 1 g ˆ 1 ( t ) ( u 1 + v 1 ) + g ˆ 2 ( t ) , ( t , u , v ) [ 0 , T ] × R n × R n .

We now define two operators U , C : X X as follows:

(3.12) U : X X , ( u , v ) ( P 1 ( u , v ) , 0 ) ,

with

(3.13) P 1 ( u , v ) ( t ) = u 0 e α t + 0 t e α ( t s ) f α ( s , u ( s ) , v ( s ) ) d s ,

and

(3.14) C : X X , ( u , v ) ( 0 , P 2 ) ,

with

(3.15) P 2 ( u , v ) ( t ) = 0 T G ( t , s ) g β ( s , u ( s ) , v ( s ) ) d s .

It is easy to check that P = U + C .

Lemma 3.2

Let ( H 1 )–( H 3 ) and ( H ¯ 4 ) hold. In addition, suppose that there exist two positive constants α and β small enough such that

(3.16) L 1 = α T + L f L 1 ( 0 , T ) + g ¯ 1 ( β T + g ˆ 1 L 1 ( 0 , T ) ) < 1 .

Then, there exists a positive constant R > 0 such that

(3.17) U ( u , v ) + C ( u ¯ , v ¯ ) B R

for all ( u , v ) , ( u ¯ , v ¯ ) B R = { ( u , v ) X : ( u , v ) X R } .

Proof

Let ( u , v ) be an arbitrary element in B R . We have the following estimate:

(3.18) P 1 ( u , v ) ( t ) 1 u 0 1 + 0 t f α ( s , u ( s ) , v ( s ) ) 1 d s u 0 1 + T f T + R ( α T + L f L 1 ( 0 , T ) ) .

Besides, we also have an estimate for P 2 ( u , v ) as follows:

(3.19) P 2 ( u , v ) ( t ) g ¯ 1 0 T g β ( s , u ( s ) , v ( s ) ) 1 d s g ¯ 1 [ R ( β T + g 1 L 1 ( 0 , T ) ) + g ˆ 2 L 1 ( 0 , T ) ] .

Choosing R > 0 large enough such that

(3.20) R > u 0 1 + T f T + g ¯ 1 g ˆ 2 L 1 ( 0 , T ) 1 L 1 .

Combining (3.18)–(3.20) and doing some direct calculations, we have (3.17). Lemma 3.2 is proved.□

Lemma 3.3

Suppose that the conditions in Lemma 3.2 are satisfied. Then, the operator U : X X is a contraction.

Proof of Lemma 3.3

Let ( u , v ) and ( u ¯ , v ¯ ) be arbitrary elements in X . We have

(3.21) P 1 ( u , v ) ( t ) P 1 ( u ¯ , v ¯ ) ( t ) 1 0 t e α ( t s ) f α ( s , u ( s ) , v ( s ) ) f α ( s , u ¯ ( s ) , v ¯ ( s ) ) 1 d s ( α T + L f L 1 ( 0 , T ) ) ( u , v ) ( u ¯ , v ¯ ) X .

By the assumption α T + L f L 1 ( 0 , T ) L 1 < 1 , we infer that P 1 : X C ( [ 0 , T ] ; R n ) is a contraction mapping, so is the operator U = ( P 1 , 0 ) : X X . Lemma 3.3 is proved.□

Lemma 3.4

Suppose that the conditions in Lemma 3.2 are satisfied. Then, the operator C : B R X is continuous and compact.

Proof

Step 1: P 2 is continuous. Let { ( u n , v n ) } B R and ( u , v ) B R such that

(3.22) ( u n , v n ) ( u , v ) X 0 , as n + .

By the continuity of g β and the Lebesgue’s dominated convergence theorem, we obtain

(3.23) 0 T g β ( t , u n ( t ) , v n ( t ) ) g ( t , u ( t ) , v ( t ) ) 1 d t , as n + .

By using (3.23), we infer that

(3.24) sup 0 t T P 2 ( u n , v n ) ( t ) P 2 ( u , v ) ( t ) 1 g ¯ 1 0 T g β ( t , u n ( t ) , v n ( t ) ) g β ( t , u ( t ) , v ( t ) ) 1 d t 0 , as n + .

Step 2: P 2 ( B R ) is relatively compact. It follows from the continuity of g β that there exists M R > 0 such that g β ( t , u ( t ) , v ( t ) ) 1 M R for all ( u , v ) B R , t [ 0 , T ] . Hence, the set P 2 ( B R ) is bounded in C ( [ 0 , T ] ; R n ) .

Taking arbitrary ( u , v ) B R and t 1 , t 2 [ 0 , T ] , t 2 < t 1 , we obtain

(3.25) P 2 ( u , v ) ( t 1 ) P 2 ( u , v ) ( t 2 ) 1 M R t 1 t 2 + T e β T M R 1 + 1 σ 0 T e β t h ( t ) d t + j = 1 N μ j e β T j e β t 1 e β t 2 M R t 1 t 2 + T e β T M R 1 0 T e β t h ( t ) d t j = 1 N μ j e β T j e β t 1 e β t 2 ,

it leads to P 2 ( B R ) which is equicontinuous. Therefore, the set P 2 ( B R ) is relatively compact in C ( [ 0 , T ] ; R n ) due to the Arzelà-Ascoli’s theorem. Lemma 3.4 is proved.□

Theorem 3.5

Suppose that the conditions in Lemma 3.2 are satisfied. Then, Prob. (1.1) and (1.2) have a solution.

Proof

Combining Lemmas 3.2–3.4 and applying the Krasnoselskii’s fixed-point theorem, it is clear to see that P = U + C has a fixed-point. Theorem 3.5 is proved.□

Example 3.6

We consider the following problem:

(3.26) u ( t ) = δ 1 ( u ( t ) + v ( t ) ) e t + u ( t ) 1 + v ( t ) 1 , t ( 0 , 1 ) , v ( t ) = δ 2 ( u ( t ) v ( t ) ) e t + u ( t ) 1 + v ( t ) 1 , t ( 0 , 1 ) , u ( 0 ) = u 0 R 2 , v ( 0 ) = 1 4 v ( 1 / 2 ) + 1 8 v ( 1 ) + 1 4 0 1 e t v ( t ) d t ,

where 2 ( 1 e 1 ) δ 1 + 8 e δ 2 3 e + 2 < 1 , u ( t ) = ( u 1 ( t ) , u 2 ( t ) ) , v ( t ) = ( v 1 ( t ) , v 2 ( t ) ) , u ( t ) 1 = u 1 ( t ) + u 2 ( t ) , v ( t ) 1 = v 1 ( t ) + v 2 ( t ) .

We have f ( t , u , v ) = δ 1 ( u + v ) e t + u 1 + v 1 , g ( t , u , v ) = δ 2 ( u v ) e t + u 1 + v 1 , h ( t ) = 1 4 e t , N = 2 , μ 1 = 1 4 , T 1 = 1 2 , μ 2 = 1 8 , T 2 = T = 1 .

Hence,

0 < 0 1 h ( t ) d t + μ 1 + μ 2 = 1 4 ( 1 e 1 ) + 1 4 + 1 8 = 5 8 1 4 e < 1 ; σ = 1 0 T e β t h ( t ) d t μ 1 e β T 1 μ 2 e β T 2 = 1 1 e β 1 4 ( β + 1 ) 1 4 e β 2 1 8 e β ; g ¯ 1 = e β T σ = e β 1 1 e β 1 4 ( β + 1 ) 1 4 e β 2 1 8 e β 8 e 3 e + 2 , as β 0 + .

It is easy to see that assumptions ( H 1 ) ( H 4 ) are satisfied with L f ( t ) = 2 δ 1 e t , L g ( t ) = 2 δ 2 e t .

On the other hand,

L = α + L f L 1 ( 0 , 1 ) + ( β T + L g L 1 ( 0 , 1 ) ) g ¯ 1 = α + 2 δ 1 ( 1 e 1 ) + ( β + 2 δ 2 ( 1 e 1 ) ) g ¯ 1 2 ( 1 e 1 ) δ 1 + 8 e δ 2 3 e + 2 < 1 , as α 0 + , β 0 + .

Hence, we can choose α > 0 , β > 0 small enough such that L < 1 .

Then, the conditions of Theorem 3.1 are satisfied. Thus, we deduce that Prob. (3.26) has a unique solution.

Example 3.7

Let us consider the following system:

(3.27) u ( t ) = δ 1 u ( t ) e t + u ( t ) 1 + v ( t ) 1 , t ( 0 , 1 ) , v ( t ) = δ 2 e t ( v ( t ) + u ( t ) sin 2 ( v 1 ( t ) 3 ) ) , t ( 0 , 1 ) , u ( 0 ) = u 0 R 2 , v ( 0 ) = 1 4 v ( 1 / 4 ) 1 8 v ( 1 ) + 1 16 0 1 e t v ( t ) d t ,

where 2 ( 1 e 1 ) δ 1 + 8 e δ 2 1 + 13 e < 1 .

We have f ( t , u , v ) = δ 1 u e t + u 1 + v 1 , g ( t , u , v ) = δ 2 e t ( v + u sin 2 ( v 1 3 ) ) , h ( t ) = 1 16 e t , N = 2 , μ 1 = 1 4 , T 1 = 1 4 , μ 2 = 1 8 , T 2 = T = 1 .

Obviously, assumptions ( H 1 ) ( H 3 ) and ( H ¯ 4 ) are satisfied with L f ( t ) = 2 δ 1 e t , g ˆ 1 ( t ) = δ 2 e t , g ˆ 2 ( t ) = 0 .

Moreover,

0 < 0 1 h ( t ) d t + μ 1 + μ 2 = 1 16 ( 1 e 1 ) + 1 4 + 1 8 = 7 16 1 16 e < 1 ; σ = 1 0 1 e β t h ( t ) d t μ 1 e β T 1 μ 2 e β T 2 = 1 1 e β 1 16 ( β + 1 ) 1 4 e β 4 + 1 8 e β ; g ¯ 1 = e β T σ = e β 1 1 e β 1 16 ( β + 1 ) 1 4 e β 4 + 1 8 e β 16 e 1 + 13 e , as β 0 + ; L 1 = α + L f L 1 ( 0 , 1 ) + ( β + g ˆ 1 L 1 ( 0 , 1 ) ) g ¯ 1 = α + 2 δ 1 ( 1 e 1 ) + ( β + δ 2 ( 1 e 1 ) ) g ¯ 1 2 ( 1 e 1 ) δ 1 + 8 e δ 2 1 + 13 e < 1 , as α 0 + , β 0 + .

Therefore, we can choose α > 0 , β > 0 small enough such that L 1 < 1 .

Applying Theorem 3.5, we also verify that the system (3.27) has a solution.

4 Positive solutions

The main purpose of this section is to prove the existence of positive solutions for Prob. (1.1) and (1.2), in which f , g C ( [ 0 , T ] × R n × R n ; R n ) . The main tool is the Guo-Krasnoselskii’s fixed-point theorem in a cone. For the sake of simplicity, we consider the case u 0 = 0 . Then, based on the preliminaries, the integral system (2.1) can be written as follows:

(4.1) u ( t ) = 0 t e α ( t s ) f α ( s , u ( s ) , v ( s ) ) d s , v ( t ) = 0 T G ( t , s ) g β ( s , u ( s ) , v ( s ) ) d s .

We make the following assumptions:

( H ˜ 1 ) h C ( [ 0 , T ] ) , h ( t ) 0 , t [ 0 , T ] ;

( H ˜ 2 ) { μ i } satisfies ( H 2 ) and μ j 0 , j = 1 , N 1 ¯ , μ N > 0 ;

( H ˜ 3 ) f : [ 0 , T ] × R n × R n R n is a continuous function, and there exists a positive constant α such that

f ( t , u , v ) α u , for all ( t , u , v ) [ 0 , T ] × R + n × R + n ;

( H ˜ 4 ) g : [ 0 , T ] × R n × R n R n is a continuous function, and there exists a positive constant β such that

g ( t , u , v ) β v , for all ( t , u , v ) [ 0 , T ] × R + n × R + n .

We have the following lemmas.

Lemma 4.1

Suppose that ( H ˜ 1 ) ( H ˜ 4 ) are satisfied. Then, for each ( u , v ) X such that u ( t ) , v ( t ) 0 , t [ 0 , T ] , we have P 1 ( u , v ) ( t ) , P 2 ( u , v ) ( t ) 0 , for all t [ 0 , T ] .

Lemma 4.2

There exist a constant γ ( 0 , 1 ) and a function Φ : [ 0 , T ] [ 0 , + ) such that

(4.2) γ Φ ( s ) G ( t , s ) Φ ( s ) , s , t [ 0 , T ] .

Proof

By (2.7), we obtain that

(4.3) G ( t , s ) 1 σ e β ( t s ) 1 σ e β s 1 + 1 σ e β s Φ ( s ) , t , s [ 0 , T ]

and

(4.4) G ( t , s ) μ N e β T σ e β ( t s ) μ N e 2 β T σ e β s , t , s [ 0 , T ] .

On the other hand, if we choose γ > 0 defined as follows:

(4.5) γ = μ N e 2 β T 1 + σ μ N j = 1 N μ j < 1 ,

and then we immediately obtain that G ( t , s ) γ Φ ( s ) for all t , s [ 0 , T ] .

Lemma 4.2 is proved.□

We define the set K in X as follows:

(4.6) K = { ( u , v ) X : u ( t ) 0 , v ( t ) 0 , v ( t ) 1 γ ( u , v ) X , t [ 0 , T ] } .

Obviously, K is a cone. Indeed, since K X , ( 0 , 0 ) K and by the continuity of the norms, it is clear to see that K is a nonempty, closed subset of the Banach space X ; furthermore, we have

(i) K + K K :

If we have ( u , v ) X , ( u ¯ , v ¯ ) X such that

u ( t ) 0 , u ¯ ( t ) 0 , v ( t ) 0 , v ¯ ( t ) 0 , v ( t ) 1 γ ( u , v ) X , v ¯ ( t ) 1 γ ( u ¯ , v ¯ ) X , t [ 0 , T ] ,

and then we have

u ( t ) + u ¯ ( t ) 0 , v ( t ) + v ¯ ( t ) 0 , t [ 0 , T ]

and

v ( t ) + v ¯ ( t ) 1 = v ( t ) 1 + v ¯ ( t ) 1 γ ( ( u , v ) X + ( u ¯ , v ¯ ) X ) , t [ 0 , T ] ,

where ( u , v ) X = u C ( [ 0 , T ] ; R n ) + v C ( [ 0 , T ] ; R n ) , ( u ¯ , v ¯ ) X = u ¯ C ( [ 0 , T ] ; R n ) + v ¯ C ( [ 0 , T ] ; R n ) .

Therefore, v ( t ) + v ¯ ( t ) 1 γ ( u + u ¯ C ( [ 0 , T ] ; R n ) + v + v ¯ C ( [ 0 , T ] ; R n ) ) = γ ( u + u ¯ , v + v ¯ ) X .

It means that ( u , v ) K , ( u ¯ , v ¯ ) K ( u , v ) + ( u ¯ , v ¯ ) K .

(ii) λ K K for all λ 0 :

Let ( u , v ) K and λ 0 , by u ( t ) 0 , v ( t ) 1 γ ( u , v ) X , t [ 0 , T ] , we deduce that

λ u ( t ) 0 , t [ 0 , T ] ,

and

λ v ( t ) 1 = λ v ( t ) 1 λ γ ( u , v ) X = γ λ ( u , v ) X = γ ( λ u , λ v ) X , t [ 0 , T ] .

Thus, λ ( u , v ) K .

(iii) K ( K ) = { ( 0 , 0 ) } :

It is obvious that ( 0 , 0 ) K ( K ) .

On the other hand, if ( u , v ) K ( K ) , then u ( t ) 0 , v ( t ) 0 , u ( t ) 0 , v ( t ) 0 , t [ 0 , T ] , so ( u , v ) = ( 0 , 0 ) .

Lemma 4.3

Suppose that the following conditions are fulfilled

  1. α β ;

  2. f α ( t , u , v ) g β ( t , u , v ) for all ( t , u , v ) [ 0 , T ] × R + n × R + n .

Then, P : K K .

Proof

Let ( u , v ) be an arbitrary element in K . We have

(4.7) P ( u , v ) X = max 0 t T 0 t e α ( t s ) f α ( s , u ( s ) , v ( s ) ) 1 d s + max 0 t T 0 T G ( t , s ) g β ( s , u ( s ) , v ( s ) ) 1 d s 0 T e α s f α ( s , u ( s ) , v ( s ) ) 1 d s + 0 T 1 σ e β s g β ( s , u ( s ) , v ( s ) ) 1 d s 0 T Φ ( s ) g β ( s , u ( s ) , v ( s ) ) 1 d s .

On the other hand, we also have

(4.8) P 2 ( u , v ) ( t ) 1 = 0 T G ( t , s ) g β ( s , u ( s ) , v ( s ) ) 1 d s γ 0 T Φ ( s ) g β ( s , u ( s ) , v ( s ) ) 1 d s .

It follows from (4.7) and (4.8) that P 2 ( u , v ) ( t ) 1 γ P ( u , v ) X , so P : K K .

Lemma 4.3 is proved.□

Theorem 4.4

Suppose that ( H ˜ 1 ) ( H ˜ 4 ) and conditions in Lemma 4.3 are satisfied. In addition, the following assertions are fulfilled

  1. There is a constant 0 < θ T 1 / 2 such that

    (4.9) f α ( t , u , v ) 1 θ ( u 1 + v 1 ) , ( t , u , v ) [ 0 , T ] × R n × R n .

  2. There exist two positive constants r , R , r < R , such that

    (4.10) g β ( t , u , v ) 1 r 2 T g ¯ 1 , ( t , u , v ) [ 0 , T ] × R + n × R + n , u 1 r , γ r v 1 r , g β ( t , u , v ) 1 R T g ¯ 0 , ( t , u , v ) [ 0 , T ] × R + n × R + n , u 1 R , γ R v 1 R ,

    or

    (4.11) g β ( t , u , v ) 1 r T g ¯ 0 , ( t , u , v ) [ 0 , T ] × R + n × R + n , u 1 r , γ r v 1 r , g β ( t , u , v ) 1 R 2 T g ¯ 1 , ( t , u , v ) [ 0 , T ] × R + n × R + n , u 1 R , γ R v 1 R .

    Then, the boundary value problem (1.1) and (1.2) has a solution ( u , v ) with u ( t ) 0 , v ( t ) 0 , for all t [ 0 , T ] .

Proof

By using similar calculations as shown in Lemma 3.4, we obtain that operator P is completely continuous.

Let us consider two bounded sets as follows:

(4.12) Ω r = { ( u , v ) X : ( u , v ) X < r } , Ω R = { ( u , v ) X : ( u , v ) X < R } .

It is easy to see that Ω r and Ω R are open subsets of X with 0 Ω r and Ω r ¯ Ω R . We shall consider two cases.

Case 1. The (4.10) is true.

Take an arbitrary element ( u , v ) K with ( u , v ) X = r . We have the following estimates:

(4.13) P 1 ( u , v ) ( t ) 1 θ 0 t ( u ( s ) 1 + v ( s ) 1 ) d s T θ ( u , v ) X

and

(4.14) P 2 ( u , v ) ( t ) 1 g ¯ 1 0 T g β ( s , u ( s ) , v ( s ) ) 1 d s r 2 = 1 2 ( u , v ) X .

It follows from (4.13), (4.14), and (4.10) ( 1 ) that

(4.15) P ( u , v ) X ( u , v ) X , ( u , v ) K Ω r .

On the other hand, for each ( u , v ) K Ω R , we have

(4.16) P ( u , v ) X P 2 ( u , v ) ( t ) 1 = 0 T G ( t , s ) g β ( s , u ( s ) , v ( s ) ) 1 d s g ¯ 0 0 T g β ( s , u ( s ) , v ( s ) ) 1 d s R = ( u , v ) X .

Combining (4.15) and (4.16) and applying the first part of Theorem 2.2, we deduce that there exists ( u , v ) K ( Ω R ¯ Ω r ) such that P ( u , v ) = ( u , v ) . It means that problems (1.1) and (1.2), with u 0 = 0 , have positive solutions.

Case 2. The (4.11) is true.

By using the same method as in Case 1, by applying the second part of Theorem 2.2, we obtain the similar result.

Theorem 4.4 is proved.□

Remark 4.5

To show the existence of positive solutions of Prob. (1.1) and (1.2) with u 0 > 0 , we put u ¯ ( t ) = u ( t ) u 0 . Then, the pair of functions ( u ¯ , v ) is the solution of the following problem:

(4.17) u ¯ = f ¯ ( t , u ¯ , v ) = f ( t , u ¯ + u 0 , v ) , t ( 0 , T ) , v = g ¯ ( t , u ¯ , v ) = g ( t , u ¯ + u 0 , v ) , t ( 0 , T ) , u ¯ ( 0 ) = 0 , v ( 0 ) = j = 1 N μ j v ( T j ) + 0 T h ( t ) v ( t ) d t ,

where f ¯ ( t , u ¯ , v ) = f ( t , u ¯ + u 0 , v ) , g ¯ ( t , u ¯ , v ) = g ( t , u ¯ + u 0 , v ) .

Applying results in Theorem 4.4 for system (4.17), we can obtain the existence of a solution ( u , v ) such that u ( t ) u 0 , v ( t ) 0 for all t [ 0 , T ] .

We will provide an example for Theorem 4.4.

Example 4.6

Let g ¯ 0 , g ¯ 1 be denoted as in Lemma 2.4, γ be defined by (4.5) such that r < γ R . Let us consider the nonlinear first-order ordinary differential system as follows:

(4.18) u ( t ) = f ( u ( t ) , v ( t ) ) , t ( 0 , T ) , v ( t ) = g ( u ( t ) , v ( t ) ) , t ( 0 , T ) , u ( 0 ) = u 0 , v ( 0 ) = j = 1 N μ j v ( T j ) + 0 T h ( t ) v ( t ) d t ,

where h and { μ j ) satisfy hypotheses ( H ˜ 1 ) and ( H ˜ 2 ) , respectively, and

(4.19) f ( u , v ) = α u + δ g ( u , v ) + β v + α v 1 sin 2 ( v 1 5 ) , v 2 1 + v 2 3 ,

g ( u , v ) β v = c 1 u 1 v 1 1 + u 1 2 + v 1 2 m , u , v R 2 , v 1 r , γ R v 1 γ R r c 1 r u 1 1 + r 2 + u 1 2 m + v 1 r γ R r c 2 u + v 1 γ R v , u , v R 2 , r v 1 γ R , c 2 ( u + v ) , u , v R 2 , v 1 γ R ,

for m R 2 , m > 0 , 0 < c 1 < 1 T g ¯ 1 m 1 , c 2 1 γ T g ¯ 0 .

Now, we verify that ( H ˜ 3 ) and ( H ˜ 4 ) are satisfied.

It is clear to see that g C ( R + 2 × R + 2 ; R 2 ) and for all ( u , v ) R + 2 × R + 2 , we have

  1. u , v R + 2 , v 1 r : g ( u , v ) β v = c 1 u 1 v 1 1 + u 1 2 + v 1 2 m 0 ;

  2. u , v R + 2 , r v 1 γ R :

    g ( u , v ) β v = γ R v 1 γ R r c 1 r u 1 1 + r 2 + u 1 2 m + v 1 r γ R r c 2 u + v 1 γ R v = 1 v 1 r γ R r c 1 r u 1 1 + r 2 + u 1 2 m + v 1 r γ R r c 2 u + v 1 γ R v = ( 1 λ ) c 1 r u 1 1 + r 2 + u 1 2 m + λ c 2 u + v 1 γ R v 0 ,

    with λ = v 1 r γ R r [ 0 , 1 ] ;

  3. u , v R + 2 , v 1 γ R : g ( u , v ) β v = c 2 ( u + v ) 0 .

Thus g β ( u , v ) = g ( u , v ) β v + 2 β v 0 , ( u , v ) R + 2 × R + 2 . It implies that ( H ˜ 4 ) holds.

On the other hand, f satisfies ( H ˜ 3 ) . Indeed, by

f ( u , v ) = α u + δ g ( u , v ) + β v + α v 1 sin 2 ( v 1 5 ) , v 2 1 + v 2 3 ,

we have f C ( R + 2 × R + 2 , R 2 ) and for all ( u , v ) R + 2 × R + 2 ,

f ( u , v ) + α u = δ g ( u , v ) + β v + α v 1 sin 2 ( v 1 5 ) , v 2 1 + v 2 3 δ g β ( u , v ) 0 .

Next, the conditions in Lemma 4.3 are satisfied. We need to prove that if α β , then

f ( u , v ) + α u g ( u , v ) + β v , ( u , v ) R + 2 × R + 2 .

We have

f α ( u , v ) = δ g ( u , v ) + β v + α v 1 sin 2 ( v 1 5 ) , v 2 1 + v 2 3 δ [ g ( u , v ) + β v + α v ] = δ g ( u , v ) + 1 + α β β v δ 1 + α β g β ( u , v ) g β ( u , v ) , ( u , v ) R + 2 × R + 2 ,

with δ > 0 small enough, such that 0 < δ 1 + α β 1 .

Finally, (4.9) and (4.10) are true.

Indeed, f satisfies condition (4.9), i.e.,

θ 0 , 1 2 T : f α ( u , v ) 1 = f ( u , v ) + α u 1 θ ( u 1 + v 1 ) , u , v R 2 .

For all ( u , v ) R 2 × R 2 , we have

  1. ( u , v ) R 2 × R 2 , v 1 r :

    g ( u , v ) β v 1 = c 1 u 1 v 1 1 + u 1 2 + v 1 2 m 1 c 1 m 1 2 v 1 c 1 m 1 2 ( u 1 + v 1 ) ;

  2. ( u , v ) R 2 × R 2 , r v 1 γ R : with λ = v 1 r γ R r [ 0 , 1 ] , we obtain

    g ( u , v ) β v 1 = ( 1 λ ) c 1 r u 1 1 + r 2 + u 1 2 m + λ c 2 u + v 1 γ R v 1 ( 1 λ ) c 1 r u 1 1 + r 2 + u 1 2 m 1 + λ c 2 u 1 + v 1 γ R v 1 c 1 r m 1 1 + r 2 u 1 + c 2 ( u 1 + v 1 ) c 1 m 1 2 u 1 + c 2 ( u 1 + v 1 ) c 1 m 1 2 + c 2 ( u 1 + v 1 ) ;

  3. ( u , v ) R 2 × R 2 , v 1 γ R : g ( u , v ) β v 1 = c 2 ( u + v ) 1 c 2 ( u 1 + v 1 ) .

It implies from (i)–(iii) that

g ( u , v ) β v 1 c 1 m 1 2 + c 2 ( u 1 + v 1 ) , ( u , v ) R 2 × R 2 .

Hence,

f ( u , v ) + α u 1 = δ g ( u , v ) + β v + α v 1 sin 2 ( v 1 5 ) , v 2 1 + v 2 3 1 δ [ g ( u , v ) β v 1 + ( 2 β + α ) v 1 ] δ c 1 m 1 2 + c 2 ( u 1 + v 1 ) + ( 2 β + α ) v 1 δ c 1 m 1 2 + c 2 + 2 β + α ( u 1 + v 1 ) θ ( u 1 + v 1 ) , u , v R 2 ,

where θ = δ c 1 m 1 2 + c 2 + 2 β + α 1 2 T , with δ > 0 small enough.

The function g β satisfies condition (4.10) because

  1. u , v R + 2 , u 1 r , γ r v 1 r :

    g β ( u , v ) 1 g ( u , v ) β v 1 + 2 β v 1 = c 1 u 1 v 1 1 + u 1 2 + v 1 2 m 1 + 2 β v 1 c 1 m 1 2 v 1 + 2 β v 1 c 1 m 1 2 + 2 β r r 2 T g ¯ 1 ,

    since c 1 m 1 2 < 1 2 T g ¯ 1 , c 1 m 1 2 + 2 β 1 2 T g ¯ 1 with β > 0 small enough.

  2. u , v R + 2 , u 1 R , γ R v 1 R :

    g β ( u , v ) 1 = g ( u , v ) β v + 2 β v 1 = c 2 ( u + v ) + 2 β v 1 c 2 v 1 1 γ T g ¯ 0 γ R = R T g ¯ 0 .

We deduce that assumptions and conditions in Theorem 4.4 are satisfied, so we also verify that the system (4.18) has a positive solution.

5 Multiplicity of positive solutions

In this section, we will show that problems (1.1) and (1.2) can have two distinct solutions or even finitely many distinct solutions. The multiplicity of positive solutions depends strongly on the nonlinear term in (1.1). For the sake of simplicity, we just consider the case u 0 = 0 .

First, to prove the multiplicity result, we assume that there exists R 1 < γ R 2 < γ 2 R 3 such that for j = 1 , 2 ¯

( G ¯ 1 ) g β ( t , u , v ) 1 R j 2 T g ¯ 1 for all ( t , u , v ) [ 0 , T ] × R + n × R + n , u 1 R j , γ R j v 1 R j ;

( G ̲ 1 ) g β ( t , u , v ) 1 R j + 1 T g ¯ 0 for all ( t , u , v ) [ 0 , T ] × R + n × R + n , u 1 R j + 1 , γ R j + 1 v 1 R j + 1 .

Theorem 5.1

Assume that ( H ˜ 1 ) ( H ˜ 3 ) , (4.9) and ( G ¯ 1 ) ( G ̲ 1 ) are satisfied. Then, the boundary value problems (1.1)–(1.2) have two solutions ( u 1 , v 1 ) and ( u 2 , v 2 ) such that

(5.1) R 1 < ( u 1 , v 1 ) X R 2 < ( u 2 , v 2 ) X R 3 .

Proof

We denote the sets

(5.2) Ω j = { ( u , v ) X : ( u , v ) X < R j } , j = 1 , 3 ¯ .

For ( u , v ) K Ω 1 , we have

u ( t ) 1 ( u , v ) X = R 1 ,

(5.3) γ R 1 = γ ( u , v ) X v ( t ) 1 ( u , v ) X = R 1 .

It follows from (5.3) and ( G ¯ 1 ) that

(5.4) g β ( t , u , v ) 1 R 1 2 T g ¯ 1 .

Combining (4.9) and (5.4), we obtain the following estimate:

P ( u , v ) X = max 0 t T 0 t f α ( s , u ( s ) , v ( s ) ) 1 d s + max 0 t T 0 T G ( t , s ) g β ( s , u ( s ) , v ( s ) ) 1 d s ,

so

(5.5) P ( u , v ) X 1 2 ( u , v ) X + R 1 2 = ( u , v ) X .

If ( u , v ) K Ω 2 , we have

u ( t ) 1 ( u , v ) X = R 2

and

(5.6) γ R 2 = γ ( u , v ) X v ( t ) 1 ( u , v ) = R 2 .

It follows from (5.6) and assumption ( G ̲ 1 ) that

(5.7) P ( u , v ) X ( u , v ) X .

Applying the Guo-Krasnoselskii’s fixed-point theorem, we verify that there exists a pair of functions ( u 1 , v 1 ) K ( Ω 2 ¯ Ω 1 ) such that P ( u 1 , v 1 ) = ( u 1 , v 1 ) .

By using the similar calculations as given in the previous section, we also deduce that there exists a pair of functions ( u 2 , v 2 ) K ( Ω 3 ¯ Ω 2 ) , which is a fixed-point of the operator P .

Theorem 5.1 is proved.□

Next, we shall generalize results obtained in Theorem 5.1 to have the existence of finitely many distinct solutions. For this purpose, we assume that there exists { R j } 1 j p such that R j 1 < γ R j . We make the following assumptions:

( G ¯ p ) g β ( t , u , v ) 1 R j 2 T g ¯ 1 for all ( t , u , v ) [ 0 , T ] × R + n × R + n , u 1 R j , γ R j v 1 R j , j = 1 , p 1 ¯ ,

( G ̲ p ) g β ( t , u , v ) 1 R j + 1 T g ¯ 0 for all ( t , u , v ) [ 0 , T ] × R + n × R + n , u 1 R j + 1 , γ R j + 1 v 1 R j + 1 , j = 1 , p 1 ¯ ; or

( G ̲ p ) g β ( t , u , v ) 1 R j T g ¯ 0 for all ( t , u , v ) [ 0 , T ] × R + n × R + n , u 1 R j , γ R j v 1 R j , j = 1 , p 1 ¯ ,

( G ¯ p ) g β ( t , u , v ) 1 R j + 1 2 T g ¯ 1 for all ( t , u , v ) [ 0 , T ] × R + n × R + n , u 1 R j + 1 , γ R j + 1 v 1 R j + 1 , j = 1 , p 1 ¯ .

Theorem 5.2

Assume that ( H ˜ 1 ) ( H ˜ 3 ) , (4.9), and ( G ¯ p ) ( G ̲ p ) (or ( G ̲ p ) ( G ¯ p ) ) are satisfied. Then, the boundary value problems (1.1) and (1.2) have at least p 1 solutions ( u j , v j ) , j = 1 , p 1 ¯ such that

(5.8) R j < ( u , v ) X R j + 1 , j = 1 , p 1 ¯ .

Finally, assume that we have a positive sequence { R j } such that R j R j + 1 < γ < 1 such that for each j N ,

( G ¯ ) g β ( t , u , v ) 1 R j 2 T g ¯ 1 for all ( t , u , v ) [ 0 , T ] × R + n × R + n , u 1 R j , γ R j v 1 R j ,

( G ̲ ) g β ( t , u , v ) 1 R j + 1 T g ¯ 0 for all ( t , u , v ) [ 0 , T ] × R + n × R + n , u 1 R j + 1 , γ R j + 1 v 1 R j + 1 ; or

( G ̲ ) g β ( t , u , v ) 1 R j T g ¯ 0 for all ( t , u , v ) [ 0 , T ] × R + n × R + n , u 1 R j , γ R j v 1 R j ,

( G ¯ ) g β ( t , u , v ) 1 R j + 1 2 T g ¯ 1 for all ( t , u , v ) [ 0 , T ] × R + n × R + n , u 1 R j + 1 , γ R j + 1 v 1 R j + 1 .

Then, we have the following theorem.

Theorem 5.3

Assume that ( H ˜ 1 ) ( H ˜ 3 ) , (4.9) and ( G ¯ ) ( G ̲ ) (or ( G ̲ ) ( G ¯ ) ) are satisfied. Then, the boundary value problems (1.1) and (1.2) have infinitely many solutions { ( u j , v j ) } , j N such that

(5.9) R j < ( u , v ) X R j + 1 , j N .

6 A remark

We remark that the methods used in the previous sections can be applied again to obtain the same results as mentioned earlier for the following problems:

Problem 1.

(6.1) u = f ( t , u , v ) , t ( 0 , T ) , v = g ( t , u , v ) , t ( 0 , T ) ,

associated with the initial and nonlocal boundary conditions:

(6.2) u ( 0 ) = u 0 , v ( 0 ) = j = 1 N B j v ( T j ) + 0 T H ( t ) v ( t ) d t ,

where f , g : [ 0 , T ] × R n × R n R n , H : [ 0 , T ] M n are given functions and u 0 R n , B j M n ( j = 1 , N ¯ ), 0 < T 1 < T 2 < < T N = T are given constants with 0 < j = 1 N B j M n + 0 T H ( t ) M n d t < 1 , M n is the set of real matrices of order n .

Problem 2.

(6.3) u k ( t ) = f k ( t , u 1 , , u m , u m + 1 , , u m + n ) , t ( 0 , T ) , k = 1 , m + n ¯ ,

asscociated with the initial and multipoint conditions

(6.4) u k ( 0 ) = u 0 k , k = 1 , m ¯ , u k ( 0 ) = j = 1 N k μ k j u k ( T k j ) + 0 T h k ( t ) u k ( t ) d t , k = m + 1 , m + n ¯ ,

where f k : [ 0 , T ] × R m + n R , h k : [ 0 , T ] R ( k = 1 , m + n ¯ ) are given functions and u 0 k ( k = 1 , m ¯ ), 0 < T k 1 < T k 2 < < T k N k = T ( k = m + 1 , m + n ¯ ), μ k j ( j = 1 , N k ¯ ) are given constants, with max m + 1 k m + n j = 1 N k μ k j + 0 T h k ( t ) d t 1 .

Acknowledgements

The authors wish to express their sincere thanks to the referees and the editor for the valuable comments and suggestions for the improvement of the paper.

  1. Author contributions: All authors have accepted responsibility for the entire content of this manuscript and approved its submission.

  2. Conflict of interest: No potential conflict of interest was reported by the authors.

  3. Data availability statement: Data sharing is not applicable to this article as no datasets were generated or analysed during the current study.

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Received: 2021-07-16
Revised: 2022-02-17
Accepted: 2022-04-04
Published Online: 2022-08-01

© 2022 Le Thi Phuong Ngoc and Nguyen Thanh Long, published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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