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# Generalized monotone iterative method for nonlinear boundary value problems with causal operators

## Abstract

This paper discusses nonlinear boundary value problems for causal differential equations where the right-hand side is the sum of two monotone functions. We develop the monotone iterative technique and establish the existence results of the extremal solutions. The results obtained include several special cases and extend previous results; two examples satisfying the assumptions are also presented.

MSC: 34B15, 39B12.

## 1 Introduction

As we all know, the monotone iterative technique is an effective and a flexible method, and it provides a useful mechanism to prove existence results for nonlinear differential equations, for detail see for monographs , papers –, and the references therein. The basic idea of this method is that using the upper and lower solutions as an initial iteration, one can construct monotone sequences from a corresponding linear equations, and these sequences converge monotonically to the minimal and maximal solutions of the nonlinear equations.

In 2004, West  developed this method, considered the generalized monotone iterative method for initial value problems, obtained the existence of extremal solutions for differential equations where the forcing function is the sum of two monotone functions, one of which is monotone non-decreasing and the other is non-increasing.

Recently, this method has been extended to causal differential equations. Its theory has the powerful quality of unifying ordinary differential equations, integro differential equations, differential equations with finite or infinite delay, Volterra integral equations and neutral equations. We refer to the monograph by Lakshmikantham  and papers , .

In 2009, Lakshmikantham discussed in  the following problem with causal operators: $x ′ (t)=(Px)(t)+(Qx)(t)$, $x(0)= x 0$, where $t∈J=[0,T]$, $P,Q:E→E=C(J,R)$ are causal operators. However, we notice that the results are only valid for initial value problems. Motivated by the above excellent work, we extend the notion of casual operators to nonlinear boundary value problems and develop the monotone iterative technique.

In this paper, we deal with the following causal differential equation:

${ u ′ ( t ) = ( Q u ) ( t ) + ( S u ) ( t ) , t ∈ J = [ 0 , T ] , g ( u ( 0 ) , u ( T ) ) = 0 ,$
(1.1)

where $Q,S:E→E=C(J,R)$ are causal operators.

Note that the nonlinear boundary value problem (1.1) reduce to periodic boundary value problems for $g(u(0),u(T))=u(0)−u(T)$, initial value problems for $g(u(0),u(T))=u(0)− u 0$ which has been studied in  and other general conditions such as $g(u(0),u(T))=h(u(0))−u(T)$. Thus problem (1.1) can be regarded as a generalization of the boundary value problems mentioned above.

The rest of this paper is organized as follows. In Section 2, we develop the monotone technique for (1.1); four theorems and several special cases are given. In Section 3, we give two examples to illustrate the results obtained. Finally, a brief summary is given in Section 4.

## 2 Main results

Let $J=[0,T]$, $Q,S:E→E=C(J,R)$. In order to prove general results, we need the following definitions.

### Definition 2.1

For (1.1), the functions $y 0 , z 0 ∈ C 1 [J,R]$ are said to be

1. (1)

natural lower and upper solutions if

${ y 0 ′ ( t ) ≤ ( Q y 0 ) ( t ) + ( S y 0 ) ( t ) , g ( y 0 ( 0 ) , y 0 ( T ) ) ≤ 0 , z 0 ′ ( t ) ≥ ( Q z 0 ) ( t ) + ( S z 0 ) ( t ) , g ( z 0 ( 0 ) , z 0 ( T ) ) ≥ 0 ;$
1. (2)

coupled lower and upper solutions of type I, if

${ y 0 ′ ( t ) ≤ ( Q y 0 ) ( t ) + ( S z 0 ) ( t ) , g ( y 0 ( 0 ) , y 0 ( T ) ) ≤ 0 , z 0 ′ ( t ) ≥ ( Q z 0 ) ( t ) + ( S y 0 ) ( t ) , g ( z 0 ( 0 ) , z 0 ( T ) ) ≥ 0 ;$
1. (3)

coupled lower and upper solutions of type II, if

${ y 0 ′ ( t ) ≤ ( Q z 0 ) ( t ) + ( S y 0 ) ( t ) , g ( y 0 ( 0 ) , y 0 ( T ) ) ≤ 0 , z 0 ′ ( t ) ≥ ( Q y 0 ) ( t ) + ( S z 0 ) ( t ) , g ( z 0 ( 0 ) , z 0 ( T ) ) ≥ 0 ;$
1. (4)

coupled lower and upper solutions of type III, if

${ y 0 ′ ( t ) ≤ ( Q z 0 ) ( t ) + ( S z 0 ) ( t ) , g ( y 0 ( 0 ) , y 0 ( T ) ) ≤ 0 , z 0 ′ ( t ) ≥ ( Q y 0 ) ( t ) + ( S y 0 ) ( t ) , g ( z 0 ( 0 ) , z 0 ( T ) ) ≥ 0 .$

If we set $y 0 (t)≤ z 0 (t)$, $t∈J$, $(Qu)$ is non-decreasing and $(Su)$ is non-increasing, then the natural lower and upper solutions and the coupled lower and upper solutions of type III satisfy type II. Thus, we only need to consider the case of the coupled lower and upper solutions of type I and II for (1.1).

### Definition 2.2

Relative to the causal differential equation (1.1):

1. (1)

A function $U∈ C 1 [J,R]$ is said to be a natural solution if it satisfies (1.1).

2. (2)

$U,V∈ C 1 [J,R]$ are said to be coupled solutions of type I, if

${ U ′ ( t ) = ( Q U ) ( t ) + ( S V ) ( t ) , g ( U ( 0 ) , U ( T ) ) = 0 , V ′ ( t ) = ( Q V ) ( t ) + ( S U ) ( t ) , g ( V ( 0 ) , V ( T ) ) = 0 .$
1. (3)

U, V are said to be coupled solutions of type II, if

${ U ′ ( t ) = ( Q V ) ( t ) + ( S U ) ( t ) , g ( U ( 0 ) , U ( T ) ) = 0 , V ′ ( t ) = ( Q U ) ( t ) + ( S V ) ( t ) , g ( V ( 0 ) , V ( T ) ) = 0 .$
1. (4)

U, V are said to be coupled solutions of type III, if

${ U ′ ( t ) = ( Q V ) ( t ) + ( S V ) ( t ) , g ( U ( 0 ) , U ( T ) ) = 0 , V ′ ( t ) = ( Q U ) ( t ) + ( S U ) ( t ) , g ( V ( 0 ) , V ( T ) ) = 0 .$

### Definition 2.3

Coupled solutions $ρ,r∈ C 1 [J,R]$, are said to be couple minimal and maximal solutions of (1.1), if for any coupled solutions U, V, we have $ρ≤U,V≤r$.

### Theorem 2.1

We suppose that the following hypotheses hold:

H1: $y 0 , z 0 ∈ C 1 (J,R)$are the coupled lower and upper solutions of type I for (1.1) with$y 0 (t)≤ z 0 (t)$on J;

H2: the operators Q, S in (1.1) are such that$Q,S:E→E$, $(Qu)$is non-decreasing in u and$(Su)$is non-increasing in u;

H3: the function$g(u,v)∈C( R 2 ,R)$is non-increasing in the second variable and there exists a constant$M>0$such that

$g( u 1 ,v)−g( u 2 ,v)≤M( u 1 − u 2 ),$

for$y 0 (0)≤ u 2 ≤ u 1 ≤ z 0 (0)$, $y 0 (T)≤v≤ z 0 (T)$.

Then there exist two monotone sequences ${ y n (t)}$, ${ z n (t)}$ such that $lim n → ∞ y n (t)=ρ(t)$, $lim n → ∞ z n (t)=r(t)$ uniformly and monotonically on J and that ρ, r are coupled minimal and maximal solutions of type I for (1.1). Furthermore, if u is any natural solution of (1.1) such that $y 0 ≤u≤ z 0$ on J, then $ρ≤u≤r$ on J.

### Proof

We consider the following linear problem:

${ y n + 1 ′ ( t ) = ( Q y n ) ( t ) + ( S z n ) ( t ) , y n + 1 ( 0 ) = y n ( 0 ) − 1 M g ( y n ( 0 ) , y n ( T ) ) ,$
(2.1)
${ z n + 1 ′ ( t ) = ( Q z n ) ( t ) + ( S y n ) ( t ) , z n + 1 ( 0 ) = z n ( 0 ) − 1 M g ( z n ( 0 ) , z n ( T ) ) .$
(2.2)

This is an adequate definition since by general results on the initial value problem of causal differential equations  the existence and uniqueness of solution for (2.1) and (2.2) are guaranteed.

First, we show that $y 0 ≤ y 1 ≤ z 1 ≤ z 0$, putting $n=0$ in (2.1) and setting $p= y 0 − y 1$, we acquire

$p ( 0 ) = y 0 ( 0 ) − y 0 ( 0 ) + 1 M g ( y 0 ( 0 ) , y 0 ( T ) ) ≤ 0 , p ′ = y 0 ′ − y 1 ′ ≤ ( Q y 0 ) ( t ) + ( S z 0 ) ( t ) − ( Q y 0 ) ( t ) − ( S z 0 ) ( t ) = 0 .$

It follows that $p(t)≤0$ on J, which implies $y 0 ≤ y 1$ on J. Similarly, we may obtain $z 1 ≤ z 0$ on J.

Next, take $p= y 1 − z 1$, then from the hypotheses H2, H3, and the fact $y 0 ≤ z 0$, one attains

$p ( 0 ) = y 1 ( 0 ) − z 1 ( 0 ) = y 0 ( 0 ) − z 0 ( 0 ) + 1 M ( g ( z 0 ( 0 ) , z 0 ( T ) ) − g ( y 0 ( 0 ) , y 0 ( T ) ) ) ≤ 0 , p ′ = y 1 ′ − z 1 ′ = ( Q y 0 ) ( t ) + ( S z 0 ) ( t ) − ( Q z 0 ) ( t ) − ( S y 0 ) ( t ) ≤ 0 .$

This implies that $p(t)≤0$ on J, and $y 1 ≤ z 1$.

In the following, we shall show that $y 1$, $z 1$ are the coupled lower and upper solutions of type I for (1.1). Following H2 and $y 0 ≤ y 1$, $z 1 ≤ z 0$, we obtain

$y 1 ′ (t)=(Q y 0 )(t)+(S z 0 )(t)≤(Q y 1 )(t)+(S z 1 )(t),$
(2.3)

and by means of the facts that $y 0 ≤ y 1$, $z 1 ≤ z 0$, H1, and H3, we have

$g ( y 1 ( 0 ) , y 1 ( T ) ) = g ( y 1 ( 0 ) , y 1 ( T ) ) − g ( y 0 ( 0 ) , y 0 ( T ) ) − M y 1 ( 0 ) + M y 0 ( 0 ) ≤ M ( y 1 ( 0 ) − y 0 ( 0 ) ) − M ( y 1 ( 0 ) − y 0 ( 0 ) ) = 0 .$
(2.4)

Similarly, we can get

$z 1 ′ (t)≥(Q z 1 )(t)+(S y 1 )(t),g ( z 1 ( 0 ) , z 1 ( T ) ) ≥0,$
(2.5)

from (2.3)-(2.5), we show that $y 1$, $z 1$ are coupled lower and upper solutions of type I for (1.1).

Now employing the mathematical induction, assume that, for some integer $k>1$,

We need to show that

For this purpose, let $p= y k − y k + 1$ and use H2, H3; we note that

$p ( 0 ) = y k ( 0 ) − y k + 1 ( 0 ) = y k ( 0 ) − y k ( 0 ) + 1 M g ( y k ( 0 ) , y k ( T ) ) ≤ 0 , p ′ = y k ′ − y k + 1 ′ = ( Q y k − 1 ) ( t ) + ( S z k − 1 ) ( t ) − ( Q y k ) ( t ) − ( S z k ) ( t ) ≤ 0 .$

This implies $y k ≤ y k + 1$ on J. Similarly, we can prove that $z k + 1 ≤ z k$ on J by using (2.2) and H2, H3. To prove $y k + 1 ≤ z k + 1$, set $p= y k + 1 − z k + 1$, then by using H2, H3, and the fact that $y k ≤ z k$, we get

$p ( 0 ) = y k + 1 ( 0 ) − z k + 1 ( 0 ) = y k ( 0 ) − z k ( 0 ) + 1 M ( g ( z k ( 0 ) , z k ( T ) ) − g ( y k ( 0 ) , y k ( T ) ) ) ≤ 0 , p ′ = y k + 1 ′ − z k + 1 ′ = ( Q y k ) ( t ) + ( S z k ) ( t ) − ( Q z k ) ( t ) − ( S y k ) ( t ) ≤ 0 .$

That yields $p(t)≤0$, i.e.$y k + 1 ≤ z k + 1$. From the above discussion, we have

$y 0 ≤ y 1 ≤⋯≤ y n ≤ z n ≤⋯≤ z 1 ≤ z 0 .$

Obviously, the constructed sequences ${ y n }$, ${ z n }$ are equicontinuous and uniform bounded. Thus, by the Ascoli-Arzela theorem, we have ${ y n }→ρ$, ${ z n }→r$ on J. Since the sequences ${ y n }$, ${ z n }$ are monotone, the entire sequences converge uniformly and monotonically to ρ, r on J, respectively.

Using the definition of (2.1), (2.2), and passing to the limit when $n→∞$, we obtain the result that ρ, r are coupled solutions of type I for (1.1).

It remains to show that ρ, r are coupled minimal and maximal solutions of type I for (1.1). Let $u 1 , u 2 ∈[ y 0 , z 0 ]$ be any coupled solutions of type I for (1.1). Assume that there exists a positive integer k such that $y k ≤ u 1 , u 2 ≤ z k$ on J. Then, putting $p= y k + 1 − u 1$, and employing H2 and H3, we arrive at

$p ( 0 ) = y k + 1 ( 0 ) − u 1 ( 0 ) = y k ( 0 ) − u 1 ( 0 ) + 1 M ( g ( u 1 ( 0 ) , u 1 ( T ) ) − g ( y k ( 0 ) , y k ( T ) ) ) ≤ 0 , p ′ = y k + 1 ′ − u 1 ′ = ( Q y k ) ( t ) + ( S z k ) ( t ) − ( Q u 1 ) ( t ) − ( S u 2 ) ( t ) ≤ 0 .$

That implies $p(t)≤0$, which proves $y k + 1 ≤ u 1$ on J. Using similar arguments we can conclude $y k + 1 ≤ u 1 , u 2 ≤ z k + 1$ on J. Since $y 0 ≤ u 1 , u 2 ≤ z 0$, by the principle of induction, $y n ≤ u 1 , u 2 ≤ z n$ holds for all n. Taking the limit as $n→∞$, we have $ρ≤ u 1 , u 2 ≤r$ on J proving ρ, r are coupled minimal and maximal solutions of type I for (1.1). Since any natural solution u of (1.1) can be considered as $(u,u)$ coupled solutions of type I, we also have $ρ≤u≤r$ on J. This completes the proof. □

### Theorem 2.2

Let the hypotheses of Theorem  2.1hold. Then, for any natural solution u of (1.1) with$y 0 ≤u≤ z 0$, there exist alternating sequences${ y 2 n , z 2 n + 1 }→ρ$, ${ z 2 n , y 2 n + 1 }→r$uniformly on J with$y 0 ≤ z 1 ≤⋯≤ y 2 n ≤ z 2 n + 1 ≤u≤ y 2 n + 1 ≤ z 2 n ≤⋯≤ y 1 ≤ z 0$. Here ρ, r are the coupled minimal and maximal solutions of type I for (1.1). Also, $ρ≤u≤r$on J.

### Proof

Consider the following initial problem:

$y n + 1 ′ ( t ) = ( Q z n ) ( t ) + ( S y n ) ( t ) , y n + 1 ( 0 ) = z n ( 0 ) − 1 M g ( z n ( 0 ) , z n ( T ) ) , z n + 1 ′ ( t ) = ( Q y n ) ( t ) + ( S z n ) ( t ) , z n + 1 ( 0 ) = y n ( 0 ) − 1 M g ( y n ( 0 ) , y n ( T ) ) . }$
(2.6)

First we show that $y 0 ≤ y 1$ and $z 1 ≤ z 0$.

Set $p= y 0 − y 1$, following (2.6) and the hypotheses of Theorem 2.1, we obtain

$p ( 0 ) = y 0 ( 0 ) − y 1 ( 0 ) ≤ y 0 ( 0 ) − z 0 ( 0 ) + 1 M g ( z 0 ( 0 ) , z 0 ( T ) ) ≤ 1 M g ( y 0 ( 0 ) , y 0 ( T ) ) ≤ 0 , p ′ = y 0 ′ − y 1 ′ = ( Q y 0 ) ( t ) + ( S z 0 ) ( t ) − ( Q z 0 ) ( t ) − ( S y 0 ) ( t ) ≤ 0 .$

Thus we prove $p(t)≤0$, which gives $y 0 ≤ y 1$ on J. Similarly, we can show $z 1 ≤ z 0$.

Now we wish to prove that

$y 0 ≤ z 1 ≤ y 2 ≤ z 3 ≤u≤ y 3 ≤ z 2 ≤ y 1 ≤ z 0 .$
(2.7)

Setting $p=u− y 1$, we get

$p ′ = u ′ − y 1 ′ =(Qu)(t)+(Su)(t)−(Q z 0 )(t)−(S y 0 )(t)≤0,$

using the monotone nature of the operators Q, S, and the fact $y 0 ≤u≤ z 0$ on J, u being any natural solution of (1.1). Also, by using H3, we have

$p ( 0 ) = u ( 0 ) − y 1 ( 0 ) = u ( 0 ) − z 0 ( 0 ) + 1 M ( g ( z 0 ( 0 ) , z 0 ( T ) ) − g ( u ( 0 ) , u ( T ) ) ) ≤ 0 .$

Hence $p(t)≤0$ on J, i.e., $u≤ y 1$ on J. A similar argument yields $z 1 ≤u$. In order to avoid repetition, we can prove each of the following: $y 2 ≤u$, $u≤ z 2$, $u≤ y 3$, and $z 3 ≤u$.

Now we shall show that $y 0 ≤ z 1 ≤ y 2 ≤ z 3$ and $y 3 ≤ z 2 ≤ y 1 ≤ z 0$. Take $p= y 0 − z 1$, and one attains

$p ′ = y 0 ′ − z 1 ′ ≤ ( Q y 0 ) ( t ) + ( S z 0 ) ( t ) − ( Q y 0 ) ( t ) − ( S z 0 ) ( t ) = 0 , p ( 0 ) = y 0 ( 0 ) − z 1 ( 0 ) = y 0 ( 0 ) − y 0 ( 0 ) + 1 M g ( y 0 ( 0 ) , y 0 ( T ) ) ≤ 0 .$

Then $p(t)≤0$ on J, thus $y 0 ≤ z 1$. Similarly, we can obtain $y 1 ≤ z 0$ on J.

Next, take $p= z 1 − y 2$, by H2 and H3, we derive

$p ′ = z 1 ′ − y 2 ′ = ( Q y 0 ) ( t ) + ( S z 0 ) ( t ) − ( Q z 1 ) ( t ) − ( S y 1 ) ( t ) ≤ 0 , p ( 0 ) = z 1 ( 0 ) − y 2 ( 0 ) = y 0 ( 0 ) − z 1 ( 0 ) + 1 M ( g ( z 1 ( 0 ) , z 1 ( T ) ) − g ( y 0 ( 0 ) , y 0 ( T ) ) ) ≤ 0 .$

This implies that $p(t)≤0$ on J, i.e., $z 1 ≤ y 2$ on J. Using similar arguments we can show $y 2 ≤ z 3$, $y 3 ≤ z 2$, $z 2 ≤ y 1$. Combining all these arguments, we now have the desired relations (2.7).

Now, suppose there exists an integer $k≥2$ such that $z 2 k − 1 ≤ y 2 k ≤ z 2 k + 1 ≤u≤ y 2 k + 1 ≤ z 2 k ≤ y 2 k − 1$ on J. Thus, we need to show

Setting $p= z 2 k + 1 − y 2 k + 2$, then for $t∈J$ and utilizing the hypotheses $y 2 k ≤ z 2 k + 1$ and H2, H3, we may get

$p ( 0 ) = z 2 k + 1 ( 0 ) − y 2 k + 2 ( 0 ) = y 2 k ( 0 ) − z 2 k + 1 ( 0 ) + 1 M ( g ( z 2 k + 1 ( 0 ) , z 2 k + 1 ( T ) ) − g ( y 2 k ( 0 ) , y 2 k ( T ) ) ) ≤ 0 , p ′ = z 2 k + 1 ′ − y 2 k + 2 ′ = ( Q y 2 k ) ( t ) + ( S z 2 k ) ( t ) − ( Q z 2 k + 1 ) ( t ) − ( S y 2 k + 1 ) ( t ) ≤ 0 .$

This implies that $p≤0$ and $z 2 k + 1 ≤ y 2 k + 2$. Similarly, we obtain $z 2 k + 2 ≤ y 2 k + 1$, $y 2 k + 2 ≤ z 2 k + 3$, and $y 2 k + 3 ≤ z 2 k + 2$.

To prove $y 2 k + 2 ≤u$, by using H2, H3, and the inequalities $z 2 k + 1 ≤u$, consider the relations

$p ′ = y 2 k + 2 ′ − u ′ =(Q z 2 k + 1 )(t)+(S y 2 k + 1 )(t)−(Qu)(t)−(Su)(t)≤0$

and

$p ( 0 ) = y 2 k + 2 ( 0 ) − u ( 0 ) = z 2 k + 1 ( 0 ) − u ( 0 ) + 1 M ( g ( u ( 0 ) , u ( T ) ) − g ( z 2 k + 1 ( 0 ) , z 2 k + 1 ( T ) ) ) ≤ 0 ,$

one attains $p≤0$ on J, i.e.$y 2 k + 2 ≤u$. Hence, as before, we can conclude that $u≤ z 2 k + 2$, $z 2 k + 3 ≤u$, and $u≤ y 2 k + 3$ on J. Now, with the principle of induction, we have

$y 0 ≤ z 1 ≤⋯≤ y 2 n ≤ z 2 n + 1 ≤u≤ y 2 n + 1 ≤ z 2 n ≤⋯≤ y 1 ≤ z 0 .$

By employing a reasoning similar to that of Theorem 2.1, we get the sequences ${ y 2 n , z 2 n + 1 }$, ${ z 2 n , y 2 n + 1 }$ which converge uniformly and monotonically to ρ, r on J, respectively. Thus, ρ, r are coupled solutions of type I for (1.1).

Finally, to prove that ρ, r are coupled minimal and maximal solutions of (1.1), let $u 1 , u 2 ∈[ y 0 , z 0 ]$ be any coupled solutions of type I for (1.1). Similar to the proof of the above, if $z 2 k − 1 ≤ y 2 k ≤ z 2 k + 1 ≤ u 1 , u 2 ≤ y 2 k + 1 ≤ z 2 k ≤ y 2 k − 1$ for some positive integer k, we can easily see that $z 2 k + 1 ≤ y 2 k + 2 ≤ z 2 k + 3 ≤ u 1 , u 2 ≤ y 2 k + 3 ≤ z 2 k + 2 ≤ y 2 k + 1$ on J, by the induction, one has $y 2 n ≤ z 2 n + 1 ≤ u 1 , u 2 ≤ y 2 n + 1 ≤ z 2 n$ holds on J for all n. Taking the limit as $n→∞$, we have $ρ≤ u 1 , u 2 ≤r$ on J proving ρ, r are coupled minimal and maximal solutions of type I for (1.1). Since we have already shown that $y 2 n ≤ z 2 n + 1 ≤u≤ y 2 n + 1 ≤ z 2 n$ holds on J for all n. Now taking the limit as $n→∞$, we get $ρ≤u≤r$ on J. This completes the proof. □

### Corollary 2.1

Under the hypotheses of Theorem  2.1, let$(Su)≡0$, then$y 0$, $z 0$are natural lower and upper solutions of (1.1), and we can get the following results:

1. (i)

There exist two monotone sequences ${ y n }$, ${ z n }$ such that $y 0 ≤ y 1 ≤⋯≤ y n ≤ z n ≤⋯≤ z 1 ≤ z 0$ which converge uniformly to the minimal and maximal solutions of (1.1), respectively.

2. (ii)

There exist alternating sequences ${ y 2 n , z 2 n + 1 }$, ${ z 2 n , y 2 n + 1 }$ such that $y 0 ≤ z 1 ≤⋯≤ y 2 n ≤ z 2 n + 1 ≤ y 2 n + 1 ≤ z 2 n ≤⋯≤ y 1 ≤ z 0$ which converge uniformly to the minimal and maximal solutions of (1.1), respectively.

### Corollary 2.2

If$(Su)≡0$and$(Qu)$is not non-decreasing, then$( Q ˜ u)=(Qu)+Lu$is non-decreasing for some$L>0$, and we can consider the following problem:

$u ′ (t)=(Qu)(t)=( Q ˜ u)(t)−Lu(t),g ( u ( 0 ) , u ( T ) ) =0,$

we see that it can be seen as (1.1) with$(Qu)$replaced by$Q ˜ u$and$(Su)$replaced by$−Lu$. Thus we get the same conclusions as for Theorem  2.1and Theorem  2.2.

### Corollary 2.3

Under the hypotheses of Theorem  2.1, let$(Qu)≡0$, if u is any natural solution of (1.1) such that$y 0 ≤u≤ z 0$on J, then we have

1. (i)

there exist the monotone sequences ${ y n }$, ${ z n }$ converging to ρ and r, where $(ρ,r)$ are coupled minimal and maximal solutions of (1.1), respectively, satisfying

${ ρ ′ ( t ) = ( S r ) ( t ) , g ( ρ ( 0 ) , ρ ( T ) ) = 0 , r ′ ( t ) = ( S ρ ) ( t ) , g ( r ( 0 ) , r ( T ) ) = 0 ,$

for$t∈J$and$ρ≤u≤r$;

1. (ii)

there exist alternating sequences ${ y 2 n , z 2 n + 1 }$ converging to ρ, ${ z 2 n , y 2 n + 1 }$ converging to r, where $(ρ,r)$ are coupled minimal and maximal solutions of (1.1), respectively, satisfying

${ ρ ′ ( t ) = ( S r ) ( t ) , g ( ρ ( 0 ) , ρ ( T ) ) = 0 , r ′ ( t ) = ( S ρ ) ( t ) , g ( r ( 0 ) , r ( T ) ) = 0 ,$

for$t∈J$and$ρ≤u≤r$.

### Corollary 2.4

If$(Qu)≡0$and$(Su)$is not non-increasing, then$( S ˜ u)=(Su)−Nu$is non-increasing for some$N>0$, we can consider the following problem:

$u ′ (t)=(Su)(t)=( S ˜ u)(t)+Nu(t),g ( u ( 0 ) , u ( T ) ) =0,$

which is the same as (1.1) with$(Qu)$replaced by Nu and$(Su)$replaced by$( S ˜ u)$. Hence the conclusions of Theorem  2.1and Theorem  2.2remain valid.

### Corollary 2.5

Suppose$(Qu)$is non-decreasing but$(Su)$is not non-increasing, then$( Q ˜ u)=(Qu)+Nu$is non-decreasing and$( S ˜ u)=(Su)−Nu$is non-increasing for some$N>0$, we can consider the following problem:

$u ′ (t)=(Qu)+(Su)(t)=( Q ˜ u)(t)+( S ˜ u)(t),g ( u ( 0 ) , u ( T ) ) =0,$

which is the same as (1.1) with$(Qu)$replaced by$( Q ˜ u)$and$(Su)$replaced by$( S ˜ u)$and the conclusions of Theorem  2.1and Theorem  2.2hold.

### Corollary 2.6

If$(Qu)$is not non-decreasing but$(Su)$is non-increasing, then$( Q ˜ u)=(Qu)+Lu$is non-decreasing and$( S ˜ u)=(Su)−Lu$is non-increasing for some$L>0$, we can consider the following problem:

$u ′ (t)=(Qu)+(Su)(t)=( Q ˜ u)(t)+( S ˜ u)(t),g ( u ( 0 ) , u ( T ) ) =0,$

which is the same as (1.1) with$(Qu)$replaced by$( Q ˜ u)$and$(Su)$replaced by$( S ˜ u)$and the conclusions of Theorem  2.1and Theorem  2.2hold.

### Corollary 2.7

If$(Qu)$is not non-decreasing and$(Su)$is not non-increasing, then for some$L>0$, such that$( Q ˜ u)=(Qu)+Lu$is non-decreasing and$( S ˜ u)=(Su)−Lu$is non-increasing, we can get the conclusions of Theorem  2.1and Theorem  2.1with$(Qu)$replaced by$( Q ˜ u)$and$(Su)$replaced by$( S ˜ u)$.

We can always construct coupled upper and lower solutions of type II as in this paper. To avoid repetition, we will merely state the next two theorems without proof since it follows along the same lines as Theorems 2.1 and 2.2.

### Theorem 2.3

Assume conditions H2, H3of Theorem  2.1hold, let$y 0 , z 0 ∈ C 1 (J,R)$be the coupled lower and upper solutions of type II with$y 0 (t)≤ z 0 (t)$on J, we have the iterates${ y n }$, ${ z n }$satisfying

$y 0 ≤ y 1 ≤⋯≤ y n ≤ z n ≤ z n − 1 ≤⋯≤ z 1 ≤ z 0 ,$

provided$y 0 ≤ y 1$and$z 1 ≤ z 0$on J, where the iterates are developed by

$y n + 1 ′ ( t ) = ( Q y n ) ( t ) + ( S z n ) ( t ) , y n + 1 ( 0 ) = y n ( 0 ) − 1 M g ( y n ( 0 ) , y n ( T ) ) , z n + 1 ′ ( t ) = ( Q z n ) ( t ) + ( S y n ) ( t ) , z n + 1 ( 0 ) = z n ( 0 ) − 1 M g ( z n ( 0 ) , z n ( T ) ) .$

Moreover, the monotone sequences${ y n }$, ${ z n }$converge uniformly to ρ and r, respectively, where$(ρ,r)$are coupled minimal and maximal solutions of type II for (1.1), they satisfy the coupled system

${ ρ ′ ( t ) = ( Q ρ ) ( t ) + ( S r ) ( t ) , g ( ρ ( 0 ) , ρ ( T ) ) = 0 , r ′ ( t ) = ( Q r ) ( t ) + ( S ρ ) ( t ) , g ( r ( 0 ) , r ( T ) ) = 0 ,$

for$t∈J$and if u is any natural solution of (1.1) such that$y 0 ≤u≤ z 0$on J, then$ρ≤u≤r$on J.

### Theorem 2.4

Assume the hypotheses H2, H3of Theorem  2.1to hold, and let$y 0 , z 0 ∈ C 1 (J,R)$be the coupled lower and upper solutions of type II with$y 0 (t)≤ z 0 (t)$on J, then for any natural solution u of (1.1) with$y 0 ≤u≤ z 0$on J, we get the alternating sequences${ y 2 n , z 2 n + 1 }$, ${ z 2 n , y 2 n + 1 }$satisfying

$y 0 ≤ z 1 ≤⋯≤ y 2 n ≤ z 2 n + 1 ≤u≤ y 2 n + 1 ≤ z 2 n ≤⋯≤ y 1 ≤ z 0 ,$

provided$y 0 ≤ y 1$and$z 1 ≤ z 0$on J, for every$n≥1$, where the iterative schemes are constructed by

$y n + 1 ′ ( t ) = ( Q z n ) ( t ) + ( S y n ) ( t ) , y n + 1 ( 0 ) = z n ( 0 ) − 1 M g ( z n ( 0 ) , z n ( T ) ) , z n + 1 ′ ( t ) = ( Q y n ) ( t ) + ( S z n ) ( t ) , z n + 1 ( 0 ) = y n ( 0 ) − 1 M g ( y n ( 0 ) , y n ( T ) ) .$

Moreover, the monotone sequences${ y 2 n , z 2 n + 1 }$converge to ρ and${ z 2 n , y 2 n + 1 }$converge to r on J, where$(ρ,r)$are coupled minimal and maximal solutions of type I for (1.1), respectively, satisfying the coupled system

${ ρ ′ ( t ) = ( Q ρ ) ( t ) + ( S r ) ( t ) , g ( ρ ( 0 ) , ρ ( T ) ) = 0 , r ′ ( t ) = ( Q r ) ( t ) + ( S ρ ) ( t ) , g ( r ( 0 ) , r ( T ) ) = 0 ,$

for$t∈J$and$ρ≤u≤r$on J.

## 3 Examples

In this section, we give two simple but illustrative examples, thereby validating the proposed theorems.

### Example 3.1

Consider the following problem:

${ u ′ ( t ) = ( sin t ) u ( t ) + 1 15 e − u ( t ) , t ∈ J = [ 0 , 1 ] , g ( u ( 0 ) , u ( 1 ) ) = u ( 0 ) − 1 100 u ( 1 ) − 1 20 e = 0 .$
(3.1)

We construct a pair of coupled upper and lower solutions of type I for (3.1),

$y 0 (t)=0, z 0 (t)=t+ 1 10 ,t∈J.$

Obviously, $y 0 (t)≤ z 0 (t)$, and

${ y 0 ′ ( t ) = 0 ≤ ( sin t ) y 0 ( t ) + 1 15 e − z 0 ( t ) = 1 15 e − t − 1 10 , g ( y 0 ( 0 ) , y 0 ( 1 ) ) = − 1 20 e < 0 , { z 0 ′ ( t ) = 1 ≥ ( sin t ) z 0 ( t ) + 1 15 e − y 0 ( t ) = ( sin t ) ( t + 1 10 ) + 1 15 , g ( z 0 ( 0 ) , z 0 ( 1 ) ) = 1 10 − 1 100 ( 1 + 1 10 ) − 1 20 e > 0 .$

It proves that $y 0 (t)$, $z 0 (t)$ are coupled lower and upper solutions of problem (3.1). Then assumptions H1, H2, and H3 hold with $M=1$. By Theorem 2.1, we obtain the existence of monotone sequences that approximate the extremal solutions of (3.1) in the sector $[0,t+ 1 10 ]$. By Theorem 2.2, we obtain the existence of alternating sequences that also converge to the extremal solutions.

### Example 3.2

Consider the following problem:

${ u ′ ( t ) = u 3 ( t ) − t 2 u ( t ) , t ∈ J = [ 0 , 1 ] , g ( u ( 0 ) , u ( 1 ) ) = u ( 0 ) − u 2 ( 1 ) + 1 4 = 0 .$
(3.2)

Put $y 0 (t)=−1$, $z 0 (t)=1$, $t∈J$. Obviously, $y 0 (t)≤ z 0 (t)$, then

${ y 0 ′ ( t ) = 0 ≤ z 0 3 ( t ) − t 2 y 0 ( t ) = 1 + t 2 , g ( y 0 ( 0 ) , y 0 ( 1 ) ) = − 1 − ( − 1 ) 2 + 1 / 4 < 0 , { z 0 ′ ( t ) = 0 ≥ y 0 3 ( t ) − t 2 z 0 ( t ) = − 1 − t 2 , g ( z 0 ( 0 ) , z 0 ( 1 ) ) = 1 / 4 > 0 .$

Functions $y 0 (t)$, $z 0 (t)$ are coupled lower and upper solutions of type II for problem (3.2). It is easy to see that H1, H2, and H3 hold with $M=1$. Consider the function $y 1 (t)=− t 2$, $z 1 (t)= t 2$ satisfy $y 0 ≤ y 1$, $z 1 ≤ z 0$, respectively. Therefore, applying Theorem 2.3, by using (2.1) and (2.2), we obtain the existence of monotone sequences that approximate the extremal solutions of (3.2) in the sector $[−1,1]$. From Theorem 2.4, we obtain the existence of alternating sequences that also converge to the extremal solutions.

## 4 Conclusions

In this paper, we have developed monotone iterative method for casual differential equations with nonlinear boundary conditions. The method is based on the new concepts of lower and upper solutions (resp. coupled lower and upper solutions). We have constructed monotone sequences and alternating sequences from a corresponding linear equations. It was proven that these sequences converge uniformly to the coupled minimal and maximal solutions of the problems.

The condition on the function $g(u,v)$ was necessary to carry out the proofs. A future research direction would be to consider this method for nonlinear conditions of $B(u(0),u)=0$ or $B(u,u(T))=0$. Also, since the theory of causal differential equations has recently gained more attention, much work can be done on its theoretical research.

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## Acknowledgements

The authors would like to thank the reviewers and the editors for their valuable suggestions and comments. This work was supported by the NNSF of China (No. 61073121), the Fundamental Research Funds for the Central Universities (No. 13ZD19), and the Higher School Science Research of Hebei Province of China (No. Z2013038).

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