Convergence of solutions for functional integro-differential equations with nonlinear boundary conditions

Abstract

This paper is concerned with the convergence of solutions for a class of functional integro-differential equations with nonlinear boundary conditions. New comparison principles are obtained. By using the comparison principles and quasilinearization method, we present two monotone iterative sequences uniformly and monotonically converging to the unique solution with rate of order 2. Meanwhile, an example is given to demonstrate applications of the result reported.

Introduction

Integro-differential equations are widely used in many fields such as control theory, biology, and mechanics, and the qualitative theory of integro-differential equations creates an important branch of nonlinear analysis; see, for instance, the monographs [2, 6, 7] and the papers [5, 10–12, 16, 24, 26, 31, 33, 34]. For the results of existence of solutions and existence of extremal solutions for such equations under different boundary conditions, we refer the reader to the monographs by Guo et al. [13] and Lakshmikantham and Rama Mohana Rao [23], the related literature for integro-differential equations [1, 4, 8, 15, 27, 30], and for functional integro-differential equations [3, 14, 17, 20, 21, 28, 35, 36], and the references cited therein.

Recently, various classes of differential/difference equations with nonlinear boundary conditions have attracted extensive attention of researchers. For instance, Franco et al. [9] discussed the existence conditions of solutions for first-order differential equations with nonlinear boundary conditions; Jankowski [19] obtained the existence conditions of first-order advanced differential equations with nonlinear boundary conditions; Mahdavi [29] investigated the nonlinear boundary value problems involving abstract Volterra operators; Wang et al. [36] presented the existence conditions of extreme solutions for first-order functional difference equations with nonlinear boundary conditions; Wang et al. [37] proved uniform convergence approximate solutions for second-order functional differential equations with periodic boundary conditions; Wang [38] and Wang and Tian [39] established the existence conditions of extreme solutions for causal differential equations and impulsive differential equations with causal operators, respectively. However, we noticed that the previous studies mostly focused on the existence of solutions and extremal solutions as well as the uniform convergence approximate solutions via the method of upper and lower solutions coupled with the monotone iterative technique; see [22, 31]. There are few results of rapid convergence for integro-differential equations with nonlinear boundary conditions. From the perspective of application, the convergence rate of the solution is both important and meaningful. In [18], by using the quasilinearization method [25], Jankowski obtained the quadratic approximation of solutions for differential equations with nonlinear boundary conditions. In [32], Sun et al. presented quadratic approximation of solutions for boundary value problems with nonlocal boundary conditions. Inspired by [18, 25, 32], in this paper, we consider the following functional integro-differential equation with nonlinear boundary conditions:

$$ \left \{ \textstyle\begin{array}{ll} x'(t)=f(t, x(t), x(\theta(t)), (Sx)(t)), \quad t\in J, \\ 0=g(x(0), x(T)), \end{array}\displaystyle \right . $$

(1.1)

where $f\in C(J\times R^{3}, R)$, $g\in C(R\times R, R)$, $J=[0, T]$, $\theta\in C(J, R_{+})$, $\theta(t)\leq t$, $(Sx)(t)=\int^{t}_{0}k(t, s)x(s)\,ds$, $k\in C(D, R_{+})$, $D=\{(t, s)\in J\times J: 0\leq s\leq t, t\in J\}$, $k_{0}=\max_{(t, s)\in D}k(t, s)$.

The aim of this paper is to investigate the problem of the convergence of solutions for Eq. (1.1). By employing the comparison principles and the quasilinearization method, we obtain two monotone sequences of iterates converging uniformly and quadratically to the unique solution of the problem. Meanwhile, an example is given to demonstrate applications of the result established. Equation (1.1) contains many special types. In addition, the nonlinear boundary conditions of Eq. (1.1) contain a lot of special types. For instance, Eq. (1.1) can be reduced to initial value problems for $g(x(0), x(T))=x(0)-c$, that is, $x(0)=c$; Eq. (1.1) reduces to anti-periodic boundary value problems for $g(x(0), x(T))=x(0)+x(T)$, that is, $x(0)=-x(T)$.

Preliminaries

We introduce the following definitions and lemmas which are used throughout this paper.

Definition 2.1

We say that a function $\alpha\in{C^{1}(J,R)}$ is a lower solution of Eq. (1.1) if

$$ \left \{ \textstyle\begin{array}{ll} \alpha'(t)\leq f(t, \alpha(t), \alpha(\theta(t)), (S\alpha)(t)), \quad t\in J, \\ g(\alpha(0), \alpha(T))\leq0. \end{array}\displaystyle \right . $$

Definition 2.2

We say that a function $\beta\in{C^{1}(J,R)}$ is an upper solution of Eq. (1.1) if

$$ \left \{ \textstyle\begin{array}{ll} \beta'(t)\geq f(t, \beta(t), \beta(\theta(t)), (S\beta)(t)), \quad t\in J, \\ g(\beta(0), \beta(T))\geq0. \end{array}\displaystyle \right . $$

Definition 2.3

We say that the functions $\alpha, \beta\in{C^{1}(J,R)}$ are coupled quasisolutions of Eq. (1.1) if

$$ \left \{ \textstyle\begin{array}{ll} \alpha'(t)= f(t, \alpha(t), \alpha(\theta(t)), (S\alpha)(t)), \quad t\in J, \\ g(\alpha(0), \beta(T))=0, \end{array}\displaystyle \right . $$

or

$$ \left \{ \textstyle\begin{array}{ll} \beta'(t)= f(t,\beta(t),\beta(\theta(t)),(S\beta)(t)), \quad t\in J, \\ g(\beta(0),\alpha(T))=0. \end{array}\displaystyle \right . $$

Definition 2.4

We say that the functions $\alpha, \beta\in{C^{1}(J,R)}$ are coupled lower solution and upper solution of Eq. (1.1) if

$$ \left \{ \textstyle\begin{array}{ll} \alpha'(t)\leq f(t, \alpha(t), \alpha(\theta(t)), (S\alpha)(t)), \quad t\in J, \\ g(\alpha(0), \beta(T))\leq0, \end{array}\displaystyle \right . $$

(2.1)

and

$$ \left \{ \textstyle\begin{array}{ll} \beta'(t)\geq f(t, \beta(t), \beta(\theta(t)), (S\beta)(t)), \quad t\in J, \\ g(\beta(0), \alpha(T))\geq0, \end{array}\displaystyle \right . $$

(2.2)

respectively.

Lemma 2.1

Assume that the following condition holds.

$(H_{2.1})$:

There exist integrable functions $h_{i}(t)<0$, $i=1, 2, 3$such that

$$ \int^{T}_{0} \bigl\{ h_{1}(t)+h_{2}(t)+ k_{0} T h_{3}(t) \bigr\} \,dt\geq-1. $$

(2.3)

If there exists a function $p\in C^{1}(J, R)$such that

$$ \textstyle\begin{cases} p'(t)\leq h_{1}(t)p(t)+h_{2}(t)p(\theta(t) )+h_{3}(t) (Sp) (t), \quad t\in J, \\ p(0) \leq0, \end{cases} $$

(2.4)

then $p(t)\leq0$.

Proof

Suppose, to the contrary, that there exists a $t^{*}\in(0, T]$ such that $p(t^{*})>0$. Let $t_{*}\in[0, t^{*}]$ be such that $p(t_{*})=\inf p(t)=-b$, $b\geq0$. By virtue of (2.4), we have

$$ p'(t)\leq-b \bigl[h_{1}(t)+h_{2}(t)+k_{0}Th_{3}(t) \bigr]. $$

Integrating both sides of the above inequality, we get

$$ p \bigl(t^{*} \bigr)- p(t_{*})\leq-b \int^{t^{*}}_{t_{*}} \bigl\{ h_{1}(t)+h_{2}(t)+k_{0}Th_{3}(t) \bigr\} \,dt, $$

and so

$$ 0< p \bigl(t^{*} \bigr)\leq p(t_{*})-b \int^{t^{*}}_{t_{*}} \bigl\{ h_{1}(t)+h_{2}(t)+k_{0}Th_{3}(t) \bigr\} \,dt. $$

Furthermore, we obtain

$$ b< -b \int^{T}_{0} \bigl\{ h_{1}(t)+h_{2}(t)+k_{0}Th_{3}(t) \bigr\} \,dt, $$

(2.5)

which contradicts (2.3), and thus $p(t)\leq0$. This completes the proof. □

Lemma 2.2

Assume that condition $(H_{2.1})$is satisfied. If there exist functions $p\in C^{1}(J, R)$and $q\in C^{1}(J, R)$such that

$$ \textstyle\begin{cases} p'(t)\leq h_{1}(t)p(t)+h_{2}(t)p (\theta(t) )+h_{3}(t) (Sp) (t), &p(0)\leq A q(T), t\in J, \\ q'(t) \leq h_{1}(t)q(t)+h_{2}(t) q ( \theta(t) )+h_{3}(t) (Sq) (t), & q(0)\leq A p(T), t\in J, \end{cases} $$

(2.6)

where $0 < A\leq1$is a constant, then $p(t)\leq0$and $q(t)\leq0$, $t\in J$.

Proof

Without loss of generality, we prove one case of $p(t)\leq0$. Suppose that the conclusion is not true. We consider the following two cases where $p(0)\leq0$ and $p(0)>0$, respectively.

Case 1. Let $p(0)\leq0$. As in the proof of Lemma 2.1, we arrive at (2.5), which contradicts (2.3).

Case 2. Let $p(0)>0$. There are two cases where for all $t\in J$, $p(t)>0$ and there exist t̄, $\underline{t}$ such that $p(\underline{t})\leq0$, $p(\bar{t})>0$, respectively.

Case 2.1. When $p(t)>0$, $t\in J$, if $q(0)\leq0$, Lemma 2.1 implies that $q(T)\leq0$, then we have $p(0)\leq A q(T)\leq0$, which is a contradiction.

If $q(0)>0$, by $0< p(0)\leq A q(T)$, then $q(T)>0$. Hence, there are two cases where for all $t\in J$, $q(t)>0$ and there exist t̃, t̂ such that $q(\tilde{t})\leq0$, $q(\hat{t})>0$, respectively.

Case 2.1.1. First, when $q(t)>0$ for all $t\in J$, condition (2.6) implies that $q'(t)<0$, and so q is decreasing. The inequalities $p(t)>0$ and (2.6) yield $p'(t)<0$, and hence p is decreasing and $A^{2}q(0)> A^{2}q(T)> Ap(0)> Ap(T)> q(0)$, which is a contradiction.

Case 2.1.2. Then there exist t̃, t̂ such that $q(\tilde{t})\leq0$, $q(\hat{t})>0$, and we obtain $q(\tilde{t}_{*})=\inf q(t)=-b$, where $\tilde{t}_{*}\in(0, T)$, $b\geq0$. Condition (2.6) implies that

$$ q'(t)\leq-b \bigl[h_{1}(t)+h_{2}(t)+k_{0}Th_{3}(t) \bigr]. $$

Integrating the above inequality from $\tilde{t}_{*}$ to T, we deduce that

$$ 0< q(T)\leq q(\tilde{t}_{*})-b \int^{T}_{\tilde{t}_{*}} \bigl\{ h_{1}(t)+h_{2}(t)+k_{0}Th_{3}(t) \bigr\} \,dt, $$

and thus

$$ b< -b \int^{T}_{0} \bigl\{ h_{1}(t)+h_{2}(t)+k_{0}Th_{3}(t) \bigr\} \,dt, $$

which is a contradiction.

Case 2.2. Next, we consider the following case, there exist t̄, $\underline{t}$ such that $p(\underline{t})\leq0$, $p(\bar{t})>0$. If $q(0)\leq0$, Lemma 2.1 yields $q(t)\leq0$, then $p(0)\leq A q(T)\leq0$, it is a contradiction.

If $q(0)> 0$ and $0< p(0)\leq A q(T)$, then $q(T)>0$, and so $q(\bar{t}_{*})=\inf q(t)=-b$, where $\bar{t}_{*}\in(0, T)$, $b\geq0$. Similarly to the proof of Case 2.1.2, we can get a contradiction. Therefore, we conclude that $p(t)\leq0$.

Similarly, the case of $q(t)\leq0$ can be proved. The proof is complete. □

Remark 2.1

When $p(T)=q(T)$, Lemma 2.2 is also valid.

Lemma 2.3

Assume that condition $(H_{2.1})$is satisfied and

$$ \begin{aligned}[b]f \bigl(t,\bar{\gamma},\bar{\gamma}(\theta),T\bar{\gamma} \bigr) -f \bigl(t,\gamma,\gamma(\theta),T\gamma \bigr)\geq{}& h_{1}(t) [\bar{ \gamma}- \gamma] +h_{2}(t) \bigl[\bar{\gamma}(\theta)-\gamma(\theta) \bigr]\\&+h_{3}(t) \bigl[(S\bar{\gamma }) (t)-(S\gamma) (t) \bigr],\end{aligned} $$

(2.7)

where $\gamma(t)\leq\bar{\gamma}(t)$, $\gamma(\theta)\leq\bar{\gamma}(\theta)$, $(S\gamma)(t)\leq(S\bar{\gamma})(t)$, $t\in J$, $h_{i}(t)<0$, $i=1, 2, 3$.

$(H_{2.2})$:

There exist two constants $0< N_{2}< N_{1}$such that

$$ g(\bar{\alpha},\beta)-g(\alpha,\bar{\beta})\leq N_{1}[ \bar{\alpha }-\alpha]-N_{2}[\bar{\beta}-\beta], $$

(2.8)

where $\alpha(0)\leq\bar{\alpha}(0)$, $\beta(T)\leq\bar{\beta}(T)$.

Moreover, let $u, v\in C^{1}(J, R)$be coupled lower and upper solutions of problem (1.1). If $y, z\in C^{1}(J, R)$and

$$\begin{aligned}& \textstyle\begin{cases} y'(t)=f (t, u(t), u (\theta(t) ), (Su) (t) )+h_{1}(t) [y(t)-u(t) ]+h_{2}(t) [y(\theta)-u(\theta) ] \\ \phantom{y'(t)=}{} +h_{3}(t) [(Sy) (t)-(Su) (t) ], \quad t\in J, \\ 0=g (u(0),v(T) )+N_{1} [y(0)-u(0) ]+N_{2} [z(T)-v(T) ], \end{cases}\displaystyle \end{aligned}$$

(2.9)

$$\begin{aligned}& \textstyle\begin{cases} z'(t)=f (t, v(t), v (\theta(t) ), (Sv) (t) )+h_{1}(t) [z(t)-v(t) ]+h_{2}(t) [z(\theta)-v(\theta) ] \\ \phantom{z'(t)=}{} +h_{3}(t) [(Sz) (t)-(Sv) (t) ], \quad t\in J, \\ 0=g (v(0), u(T) )+N_{1} [z(0)-v(0) ]+N_{2} [y(T)-u(T) ], \end{cases}\displaystyle \end{aligned}$$

(2.10)

then $u(t)\leq y(t)\leq z(t)\leq v(t)$, andy, zare coupled lower and upper solutions of (1.1), respectively.

Proof

First, we need to prove that the inequalities $u(t)\leq y(t)$ and $z(t)\leq v(t)$ hold. Set

$$ p(t)=u(t)- y(t) \quad \text{and} \quad q(t)=z(t)- v(t). $$

This yields

$$ \begin{aligned} p'(t)\leq{}& f \bigl(t, u(t), u \bigl( \theta(t) \bigr), (Su) (t) \bigr)-f \bigl(t, u(t), u \bigl(\theta(t) \bigr), (Su) (t) \bigr) \\ &-h_{1}(t) \bigl[y(t)-u(t) \bigr]-h_{2}(t) \bigl[y( \theta)-u( \theta ) \bigr]-h_{3}(t) \bigl[(Sy) (t)-(Su) (t) \bigr] \\ ={}&h_{1}(t)p(t)+h_{2}(t)p(\theta)+h_{3}(t) (Sp) (t). \end{aligned} $$

Condition (2.9) implies that $0=g(u(0), v(T))-N_{1}p(0)+N_{2}q(T)$. By virtue of (2.1), we obtain $p(0)\leq({N_{2}}/{N_{1}})q(T)$.

Similarly, we can conclude that

$$ \textstyle\begin{cases} q'(t) \leq h_{1}(t)q(t)+h_{2}(t)q( \theta)+h_{3}(t) (Sq) (t), \quad t\in J, \\ q(0)\leq\frac{N_{2}}{N_{1}}p(T). \end{cases} $$

It follows from Lemma 2.2 that $p(t)\leq0$, $q(t)\leq0$, that is, $u(t)\leq y(t)$, $z(t)\leq v(t)$.

Now, we prove that $y(t)\leq z(t)$. Letting $m(t)=y(t)- z(t)$, we have

$$\begin{aligned} m'(t)\leq{}&f \bigl(t, u(t), u \bigl(\theta(t) \bigr), (Su) (t) \bigr)-f \bigl(t, v(t), v \bigl(\theta(t) \bigr), (Sv) (t) \bigr) \\ &+h_{1}(t) \bigl[y(t)-u(t) \bigr]+h_{2}(t) \bigl[y( \theta)-u( \theta ) \bigr]+h_{3}(t) \bigl[(Sy) (t)-(Su) (t) \bigr] \\ &-h_{1}(t) \bigl[z(t)-v(t) \bigr]-h_{2}(t) \bigl[z( \theta)-v( \theta ) \bigr]-h_{3}(t) \bigl[(Sz) (t)-(Sv) (t) \bigr] \\ \leq{}& {-}h_{1}(t) \bigl[v(t)-u(t) \bigr]-h_{2}(t) \bigl[v( \theta)-u(\theta ) \bigr]-h_{3}(t) \bigl[(Sv) (t)-(Su) (t) \bigr] \\ &+h_{1}(t) \bigl[v(t)-u(t)+m(t) \bigr]+h_{2}(t) \bigl[v( \theta)-u(\theta)+m(\theta) \bigr] \\ &+h_{3}(t) \bigl[(Sv) (t)-(Su) (t)+(Sm) (t) \bigr] \\ ={}&h_{1}(t)m(t)+h_{2}(t)m(\theta)+h_{3}(t) (Sm) (t). \end{aligned}$$

In view of (2.9) and (2.10), we arrive at

$$\begin{aligned} 0={}&g \bigl(u(0), v(T) \bigr)-g \bigl(v(0),u(T) \bigr)+N_{1} \bigl[y(0)-u(0) \bigr]+N_{2} \bigl[z(T)-v(T) \bigr] \\ &-N_{1} \bigl[z(0)-v(0) \bigr]-N_{2} \bigl[y(T)-u(T) \bigr] \\ \geq{}&{-}N_{1} \bigl[v(0)-u(0) \bigr]+N_{2} \bigl[v(T)-u(T) \bigr]+N_{1} \bigl[v(0)-u(0)+m(0) \bigr] \\ &+N_{2} \bigl[m(T)-v(T)+u(T) \bigr] \\ ={}& N_{1}m(0)-N_{2}m(T), \end{aligned}$$

which finally gives $m(0)\leq({N_{2}}/{N_{1}}) m(T)$. It follows now from Lemma 2.2 that $m(t)\leq0$.

Now, we need to prove that y and z are coupled lower and upper solutions of Eq. (1.1). In fact, by (2.7) and (2.8), we get

$$\begin{aligned} y'(t)={}&f \bigl(t, u(t), u \bigl(\theta(t) \bigr), (Su) (t) \bigr)-f \bigl(t, y(t), y \bigl(\theta(t) \bigr), (Sy) (t) \bigr) \\ &+f \bigl(t, y(t),y \bigl(\theta(t) \bigr), (Sy) (t) \bigr)+h_{1}(t) \bigl[y(t)-u(t) \bigr] \\ &+h_{2}(t) \bigl[y(\theta)-u(\theta) \bigr]+h_{3}(t) \bigl[(Sy) (t)-(Su) (t) \bigr] \\ \leq{}&f \bigl(t, y(t), y \bigl(\theta(t) \bigr), (Sy) (t) \bigr) \end{aligned}$$

and

$$\begin{aligned} g \bigl(y(0), z(T) \bigr)-g \bigl(u(0), v(T) \bigr) \leq{}& N_{1} \bigl[y(0)-u(0) \bigr]-N_{2} \bigl[v(T)-z(T) \bigr] \\ \leq{}& g \bigl(u(0), v(T) \bigr)+ N_{1} \bigl[y(0)-u(0) \bigr]-N_{2} \bigl[v(T)-z(T) \bigr] \\ ={}&0. \end{aligned}$$

Similarly, we deduce that $z'(t)\geq f(t, z(t), z(\theta(t)), (Sz)(t))$ and $g(z(0), y(T))\geq0$. This proves that y and z are coupled lower and upper solutions of Eq. (1.1). □

Main result

In this section, the quadratic convergence of successive approximation sequences is proved by the quasilinearization method.

Theorem 3.1

Set $\varOmega=\{(t, u)\in J\times R: y_{0}(t)\leq u(t)\leq z_{0}(t)\}$, $\varOmega_{1}=[y_{0}(0),z_{0}(0)]$, and $\varOmega _{2}=[y_{0}(T),z_{0}(T)]$. Assume that the following conditions hold.

$(A_{3.1})$:

$y_{0}$, $z_{0}$are coupled lower and upper solutions of Eq. (1.1), and $y_{0}(t)\leq z_{0}(t)$onJ;

$(A_{3.2})$:

$f_{x}\in C(\varOmega, R)$, $g_{x}, g_{y}\in C(\varOmega_{1}\times\varOmega_{2}, R)$, $f_{x}<0$, $f_{y}<0$, $f_{z}<0$, $0<{g_{y}}<{g_{x}}<1$;

$(A_{3.3})$:

$f_{xx}, f_{xy}, f_{yy}\in C(\varOmega, R)$, $g_{xx}, g_{xy}, g_{yy}\in C(\varOmega_{1}\times\varOmega_{2}, R)$, $f_{xx}\geq0$, $f_{xy}\geq0$, $f_{xz}\geq0$, $f_{yy}\geq0$, $f_{zz}\geq 0$, $f_{yz}\geq0$, $g_{xx}\leq0$, $g_{xy}\leq0$, $g_{yy}\leq0$.

If

$$\begin{aligned}[b] &\int^{T}_{0} \bigl\{ f_{x} \bigl(t, y_{0}, y_{0}(\theta), (Sy_{0}) (t) \bigr) +f_{y} \bigl(t, y_{0}, y_{0}(\theta), (Sy_{0}) (t) \bigr)+k_{0}Tf_{z} \bigl(t, y_{0}, y_{0}(\theta), (Sy_{0}) (t) \bigr) \bigr\} \,dt\hspace{-24pt}\\&\quad\geq-1,\end{aligned} $$

(3.1)

then there exist the monotone sequences $\{y_{n}(t)\}$and $\{z_{n}(t)\}$converging uniformly to the unique solutionxof Eq. (1.1) and the convergence is quadratic, that is,

$$ \begin{gathered} \max_{t\in J} \bigl\vert x(t)-y_{n+1}(t) \bigr\vert \leq d_{1} \max _{t\in J} \bigl\vert x(t)-y_{n}(t) \bigr\vert ^{2}+d_{2} \max_{t\in J} \bigl\vert x(t)-z_{n}(t) \bigr\vert ^{2}, \\ \max_{t\in J} \bigl\vert x(t)-z_{n+1}(t) \bigr\vert \leq d_{3} \max_{t\in J} \bigl\vert x(t)-y_{n}(t) \bigr\vert ^{2}+d_{4} \max _{t\in J} \bigl\vert x(t)-z_{n}(t) \bigr\vert ^{2}, \end{gathered} $$

where the coefficients $d_{1}$, $d_{2}$, $d_{3}$, and $d_{4}$are nonnegative constants.

Proof

Consider the following problems:

$$\begin{aligned}& \textstyle\begin{cases} y_{n+1}'(t)=f (t, y_{n}(t), y_{n} (\theta(t) ), (Sy_{n}) (t) ) +f_{x} (t, y_{n}(t), y_{n} (\theta(t) ), (Sy_{n}) (t) ) [y_{n+1}(t) \\ \phantom{y_{n+1}'(t)=}{}-y_{n}(t) ]+f_{y} (t, y_{n}(t), y_{n} (\theta(t) ), (Sy_{n}) (t) ) [y_{n+1}(\theta)-y_{n}(\theta) ] \\ \phantom{y_{n+1}'(t)=}{}+f_{z} (t, y_{n}(t), y_{n} (\theta(t) ), (Sy_{n}) (t) ) [(Sy_{n+1}) (t)-(Sy_{n}) (t) ], \quad t\in J, \\ 0=g (y_{n}(0), z_{n}(T) )+g_{x} (y_{n}(0), y_{n}(T) ) [y_{n+1}(0)-y_{n}(0) ] \\ \phantom{0=}{}+g_{y} (y_{n}(0), z_{n}(T) ) [z_{n+1}(T)-z_{n}(T) ], \end{cases}\displaystyle \end{aligned}$$

(3.2)

$$\begin{aligned}& \textstyle\begin{cases} z_{n+1}'(t)=f (t, z_{n}(t), z_{n} (\theta(t) ), (Sz_{n}) (t) )+f_{x} (t, y_{n}(t), y_{n} (\theta(t) ), (Sy_{n}) (t) ) [z_{n+1}(t) \\ \phantom{z_{n+1}'(t)=}{}-z_{n}(t) ]+f_{y} (t, y_{n}(t), y_{n} (\theta(t) ), (Sy_{n}) (t) ) [z_{n+1}(\theta)-z_{n}(\theta) ] \\ \phantom{z_{n+1}'(t)=}{}+f_{z} (t, y_{n}(t), y_{n} (\theta(t) ), (Sy_{n}) (t) ) [(Sz_{n+1}) (t)-(Sz_{n}) (t) ], \quad t\in J, \\ 0=g (z_{n}(0), y_{n}(T) )+g_{x} (y_{n}(0), y_{n}(T) ) [z_{n+1}(0)-z_{n}(0) ] \\ \phantom{0=}{}+g_{y} (y_{n}(0), z_{n}(T) ) [y_{n+1}(T)-y_{n}(T) ], \end{cases}\displaystyle \end{aligned}$$

(3.3)

in which $n=0, 1, \dots$. By the mean value theorem, we conclude that

$$ \begin{aligned} g(\bar{\alpha}, \beta)-g(\alpha, \bar{\beta})&=g(\bar{ \alpha}, \beta) -g(\alpha, \beta)+g(\alpha, \beta)-g(\alpha, \bar{\beta}) \\ &\leq g_{x}(\delta_{1}, \beta)[\bar{\alpha}- \alpha]-g_{y}(\alpha, \delta_{2})[\bar{\beta}-\beta], \end{aligned} $$

where $\alpha(0)\leq\delta_{1}\leq\bar{\alpha}(0)$, $\beta(T)\leq\delta _{2}\leq\bar{\beta}(T)$. Note that

$$\begin{aligned} &f \bigl(t,\bar{\gamma}(t), \bar{\gamma}(\theta), (S\bar{\gamma}) (t) \bigr)-f \bigl(t, \gamma(t), \gamma(\theta), (S\gamma) (t) \bigr) \\ &\quad=f_{x} \bigl(t, \delta_{3}, \bar{\gamma}(\theta), (S\bar{ \gamma}) (t) \bigr) \bigl[\bar {\gamma}(t)-\gamma(t) \bigr] +f_{y} \bigl(t, \gamma(t), \delta_{4}, (S\bar{\gamma}) (t) \bigr) \bigl[\bar{ \gamma }( \theta)-\gamma(\theta) \bigr] \\ &\quad\quad+f_{z} \bigl(t, \gamma(t), \gamma(\theta), \delta_{5} \bigr) \bigl[(S\bar{\gamma }) (t)-(S\gamma) (t) \bigr] \\ &\quad\geq f_{x} \bigl(t, \gamma(t), \gamma(\theta), (S\gamma) (t) \bigr) \bigl[\bar{\gamma }(t)-\gamma(t) \bigr] +f_{y} \bigl(t, \gamma(t), \gamma(\theta), (S\gamma) (t) \bigr) \bigl[\bar{\gamma}(\theta )-\gamma(\theta) \bigr] \\ &\quad\quad{}+f_{z} \bigl(t, \gamma(t), \gamma(\theta), (S\gamma) (t) \bigr) \bigl[(S\bar{\gamma }) (t)-(S\gamma) (t) \bigr], \end{aligned}$$

where $\gamma(t) \leq\delta_{3}\leq\bar{\gamma}(t)$, $\gamma(\theta)\leq\delta_{4}\leq\bar{\gamma}(\theta)$, $(S\gamma)(t)\leq\delta_{5}\leq(S\bar{\gamma})(t)$, and

$$\begin{aligned} & \int^{T}_{0}\bigl\{ f_{x} \bigl(t, z_{n}, z_{n}(\theta), (Sz_{n}) (t) \bigr) +f_{y} \bigl(t, z_{n}, z_{n}(\theta), (Sz_{n}) (t) \bigr)+k_{0}Tf_{z} \bigl(t, z_{n}, z_{n}(\theta), (Sz_{n}) (t) \bigr) \bigr\} \,dt \\ &\quad\geq \int^{T}_{0} \bigl\{ f_{x} \bigl(t, y_{n}, y_{n}(\theta), (Sy_{n}) (t) \bigr) +f_{y} \bigl(t, y_{n}, y_{n}(\theta), (Sy_{n}) (t) \bigr) \\ &\quad\quad+k_{0}Tf_{z} \bigl(t, y_{n}, y_{n}(\theta), (Sy_{n}) (t) \bigr) \bigr\} \,dt \\ &\quad \geq \int^{T}_{0} \bigl\{ f_{x} \bigl(t, y_{0}, y_{0}(\theta), (Sy_{0}) (t) \bigr) +f_{y} \bigl(t, y_{0}, y_{0}(\theta), (Sy_{0}) (t) \bigr) \\ &\quad\quad+k_{0}Tf_{z} \bigl(t, y_{0}, y_{0}(\theta), (Sy_{0}) (t) \bigr) \bigr\} \,dt \\ &\quad\geq-1. \end{aligned}$$

Using Lemma 2.3 and mathematical induction, we can deduce that

$$ y_{0}(t)\leq y_{1}(t)\leq\cdots\leq y_{n}(t) \leq z_{n}(t)\leq\cdots\leq z_{1}(t)\leq z_{0}(t), \quad n=0, 1, \ldots , t\in J. $$

Thus, the sequences $\{y_{n}\}$ and $\{z_{n}\}$ are uniformly bounded and equicontinuous on J.

$$\begin{aligned} \bigl\vert y_{n}(t)-y_{n}(s) \bigr\vert ={}& \biggl\vert y_{n}(0)+ \int_{0}^{t} \bigl\{ f \bigl(\varphi, y_{n}(\varphi), y_{n} \bigl(\theta(\varphi) \bigr), (Sy_{n}) (\varphi) \bigr) \\ &+f_{x} \bigl(\varphi, y_{n}(\varphi), y_{n} \bigl(\theta(\varphi) \bigr), (Sy_{n}) (\varphi) \bigr) \bigl[y_{n+1}(\varphi) \\ &-y_{n}(\varphi) \bigr]+f_{y} \bigl(\varphi, y_{n}(\varphi), y_{n} \bigl(\theta(\varphi ) \bigr), (Sy_{n}) (\varphi) \bigr) \bigl[y_{n+1}(\theta)-y_{n}( \theta) \bigr] \\ &+f_{z} \bigl(\varphi, y_{n}(\varphi), y_{n} \bigl(\theta(\varphi) \bigr), (Sy_{n}) (\varphi) \bigr) \bigl[(Sy_{n+1}) (\varphi)-(Sy_{n}) (\varphi) \bigr] \bigr\} \,d\varphi \\ &- \biggl\{ y_{n}(0)+ \int_{0}^{s} \bigl\{ f \bigl(\varphi, y_{n}(\varphi), y_{n} \bigl(\theta(\varphi) \bigr), (Sy_{n}) (\varphi) \bigr) \\ &+f_{x} \bigl(\varphi, y_{n}(\varphi), y_{n} \bigl(\theta(\varphi) \bigr), (Sy_{n}) (\varphi) \bigr) \bigl[y_{n+1}(\varphi) \\ &-y_{n}(\varphi) \bigr]+f_{y} \bigl(\varphi, y_{n}(\varphi), y_{n} \bigl(\theta(\varphi ) \bigr), (Sy_{n}) (\varphi) \bigr) \bigl[y_{n+1}(\theta)-y_{n}( \theta) \bigr] \\ &+f_{z} \bigl(\varphi, y_{n}(\varphi), y_{n} \bigl(\theta(\varphi) \bigr), (Sy_{n}) (\varphi) \bigr) \bigl[(Sy_{n+1}) (\varphi)-(Sy_{n}) (\varphi) \bigr] \bigr\} \,d\varphi \biggr\} \biggr\vert \\ ={}& \biggl\vert \int_{s}^{t} \bigl\{ f \bigl(\varphi, y_{n}(\varphi), y_{n} \bigl(\theta(\varphi ) \bigr), (Sy_{n}) (\varphi) \bigr) \\ &+f_{x} \bigl(\varphi, y_{n}(\varphi), y_{n} \bigl(\theta(\varphi) \bigr), (Sy_{n}) (\varphi) \bigr) \bigl[y_{n+1}(\varphi) \\ &-y_{n}(\varphi) \bigr]+f_{y} \bigl(\varphi, y_{n}(\varphi), y_{n} \bigl(\theta(\varphi ) \bigr), (Sy_{n}) (\varphi) \bigr) \bigl[y_{n+1}(\theta)-y_{n}( \theta) \bigr] \\ &+f_{z} \bigl(\varphi, y_{n}(\varphi), y_{n} \bigl(\theta(\varphi) \bigr), (Sy_{n}) (\varphi) \bigr) \bigl[(Sy_{n+1}) (\varphi)-(Sy_{n}) (\varphi) \bigr] \bigr\} \,d\varphi \biggr\vert \\ \leq{}& \int^{t}_{s} \bigl\vert f \bigl(\varphi, y_{n}(\varphi), y_{n} \bigl(\theta(\varphi) \bigr), (Sy_{n}) (\varphi) \bigr) \\ &+f_{x} \bigl(\varphi, y_{n}(\varphi), y_{n} \bigl(\theta(\varphi) \bigr), (Sy_{n}) (\varphi) \bigr) \bigl[y_{n+1}(\varphi) \\ &-y_{n}(\varphi) \bigr]+f_{y} \bigl(\varphi, y_{n}(\varphi), y_{n} \bigl(\theta(\varphi ) \bigr), (Sy_{n}) (\varphi) \bigr) \bigl[y_{n+1}(\theta)-y_{n}( \theta) \bigr] \\ &+f_{z} \bigl(\varphi, y_{n}(\varphi), y_{n} \bigl(\theta(\varphi) \bigr), (Sy_{n}) (\varphi) \bigr) \bigl[(Sy_{n+1}) (\varphi)-(Sy_{n}) (\varphi) \bigr] \bigr\vert \,d\varphi \\ \leq{}&M \vert t-s \vert . \end{aligned}$$

By virtue of Arzelà–Ascoli theorem, there exist the subsequences $\{y_{n_{k}}\}$ and $\{z_{n_{k}}\}$ converging uniformly on J to some continuous functions y and z, respectively, and

$$\begin{aligned}& \textstyle\begin{cases} y_{n_{k+1}}'(t)=f (t, y_{n_{k}}(t), y_{n_{k}} (\theta(t) ), (Sy_{n_{k}}) (t) ) +f_{x} (t, y_{n_{k}}(t), y_{n_{k}} ( \theta(t) ), (Sy_{n_{k}}) (t) ) [y_{n_{k+1}}(t) \\ \phantom{y_{n_{k+1}}'(t)=}{}-y_{n_{k}}(t) ]+f_{y} (t, y_{n_{k}}(t), y_{n_{k}} (\theta(t) ), (Sy_{n_{k}}) (t) ) [y_{n_{k+1}}(\theta)-y_{n_{k}}(\theta) ] \\ \phantom{y_{n_{k+1}}'(t)=}{}+f_{z} (t, y_{n_{k}}(t), y_{n_{k}} (\theta(t) ), (Sy_{n_{k}}) (t) ) [(Sy_{n_{k+1}}) (t)-(Sy_{n_{k}}) (t) ], \quad t\in J, \\ 0=g (y_{n_{k}}(0), z_{n_{k}}(T) )+g_{x} (y_{n_{k}}(0), y_{n_{k}}(T) ) [y_{n_{k+1}}(0)-y_{n_{k}}(0) ] \\ \phantom{0=}{}+g_{y} (y_{n_{k}}(0), z_{n_{k}}(T) ) [z_{n_{k+1}}(T)-z_{n_{k}}(T) ], \end{cases}\displaystyle \\& \textstyle\begin{cases} z_{n_{k+1}}'(t)=f (t, z_{n_{k}}(t), z_{n_{k}} (\theta(t) ), (Sz_{n_{k}}) (t) )+f_{x} (t, y_{n_{k}}(t), y_{n_{k}} ( \theta(t) ), (Sy_{n_{k}}) (t) ) [z_{n_{k+1}}(t) \\ \phantom{z_{n_{k+1}}'(t)=}{}-z_{n_{k}}(t) ]+f_{y} (t, y_{n_{k}}(t), y_{n_{k}} (\theta(t) ), (Sy_{n_{k}}) (t) ) [z_{n_{k+1}}(\theta)-z_{n_{k}}(\theta) ] \\ \phantom{z_{n_{k+1}}'(t)=}{}+f_{z} (t, y_{n_{k}}(t), y_{n_{k}} (\theta(t) ), (Sy_{n_{k}}) (t) ) [(Sz_{n_{k+1}}) (t)-(Sz_{n_{k}}) (t) ], \quad t\in J, \\ 0=g (z_{n_{k}}(0), y_{n_{k}}(T) )+g_{x} (y_{n_{k}}(0), y_{n_{k}}(T) ) [z_{n_{k+1}}(0)-z_{n_{k}}(0) ] \\ \phantom{0=}{}+g_{y} (y_{n_{k}}(0), z_{n_{k}}(T) ) [y_{n_{k+1}}(T)-y_{n_{k}}(T) ], \end{cases}\displaystyle \end{aligned}$$

when $n_{k}\rightarrow\infty$, y and z satisfy the equations

$$ \textstyle\begin{cases} y'(t)=f (t, y(t), y (\theta(t) ), (Sy) (t) ), \quad t\in J, \\ 0=g (y(0), z(T) ), \end{cases} $$

and

$$ \textstyle\begin{cases} z'(t) =f (t, z(t), z ( \theta(t) ), (Sz) (t) ), \quad t\in J, \\ 0=g (z(0), y(T) ). \end{cases} $$

Thus, $y, z\in C^{1}(J, R)$ are coupled solutions of Eq. (1.1).

Now, we prove that $y=z$ is a unique solution of Eq. (1.1). Clearly, $y(t)\leq z(t)$. Let $p(t)=z(t)-y(t)$. Then

$$\begin{aligned} p'(t)={}& f \bigl(t, z(t), z \bigl(\theta(t) \bigr), (Sz) (t) \bigr)-f \bigl(t, y(t), y \bigl(\theta(t) \bigr), (Sy) (t) \bigr) \\ ={}& f_{x} \bigl(t, \xi_{1}, z \bigl(\theta(t) \bigr), (Sz) (t) \bigr)p(t)+f_{y} \bigl(t, y(t), \xi _{2}, (Sz) (t) \bigr)p( \theta) \\ &+f_{z} \bigl(t, y(t), y \bigl(\theta(t) \bigr), \xi_{3} \bigr) (Sp) (t), \end{aligned}$$

where $y(t)\leq\xi_{1}\leq z(t)$, $y(\theta(t))\leq\xi_{2}\leq z(\theta (t))$, $(Sy)(t)\leq\xi_{3}\leq(Sz)(t)$, and

$$\begin{aligned} g \bigl(z(0), y(T) \bigr)-g \bigl(y(0), z(T) \bigr) &\leq N_{1} \bigl[z(0)-y(0) \bigr]-N_{2} \bigl[z(T)-y(T) \bigr] \\ &= N_{1} p(0)-N_{2} p(T). \end{aligned}$$

In view of (2.1), we get $p(0)\leq({N_{2}}/{N_{1}})p(T)$. An application of Lemma 2.2 yields $p(t)\leq0$, that is, $z(t)\leq y(t)$. Hence, we have $y(t)= z(t)$.

Let $x\in[y_{0}, z_{0}]$ be any solution of Eq. (1.1). It is not difficult to prove that $y_{n}(t)\leq x(t)\leq z_{n}(t)$. Letting $n\rightarrow\infty$, then $y(t)= z(t)=x(t)$, it means that $\{y_{n_{k}}\}$ and $\{z_{n_{k}}\}$ converge to the unique solution x of Eq. (1.1).

Finally, we prove the quadratic convergence of $\{y_{n}\}$ and $\{ z_{n}\}$ to x. Let $p_{n+1}(t)= x(t)-y_{n+1}(t)\geq0$ and $q_{n+1}(t)= z_{n+1}(t)-x(t)\geq0$. Then

$$\begin{aligned} p_{n+1}'(t) ={}&f_{x} \bigl(t, \rho_{1}, x \bigl(\theta(t) \bigr), (Sx) (t) \bigr)p_{n}(t)+f_{y} \bigl(t, y_{n}(t), \rho_{2}, (Sx) (t) \bigr) p_{n}( \theta) \\ &+f_{z} \bigl(t, y_{n}(t), y_{n} \bigl( \theta(t) \bigr), \rho_{3} \bigr) (Sp_{n}) (t) \\ &-f_{x} \bigl(t, y_{n}(t), y_{n} \bigl( \theta(t) \bigr), (Sy_{n}) (t) \bigr)p_{n}(t) \\ &-f_{y} \bigl(t, y_{n}(t), y_{n} \bigl( \theta(t) \bigr), (Sy_{n}) (t) \bigr)p_{n}(\theta) \\ &-f_{z} \bigl(t, y_{n}(t), y_{n} \bigl( \theta(t) \bigr), (Sy_{n}) (t) \bigr) (Sp_{n}) (t) \\ &+f_{x} \bigl(t, y_{n}(t), y_{n} \bigl( \theta(t) \bigr), (Sy_{n}) (t) \bigr)p_{n+1}(t) \\ &+f_{y} \bigl(t, y_{n}(t), y_{n} \bigl( \theta(t) \bigr), (Sy_{n}) (t) \bigr) p_{n+1}(\theta) \\ &+f_{z} \bigl(t, y_{n}(t), y_{n} \bigl( \theta(t) \bigr), (Sy_{n}) (t) \bigr) (Sp_{n+1}) (t) \\ \leq{}& f_{xx} \bigl(t, \rho_{4}, x \bigl(\theta(t) \bigr), (Sx) (t) \bigr)p_{n}^{2}(t) \\ &+f_{xy} \bigl(t, y_{n}(t), \rho_{5}, (Sx) (t) \bigr)p_{n}(t)p_{n}(\theta) \\ &+f_{xz} \bigl(t, y_{n}(t), y_{n} \bigl( \theta(t) \bigr), \rho_{6} \bigr)p_{n}(t) (Sp_{n}) (t) \\ &+f_{yy} \bigl(t, y_{n}(t), \rho_{7}, (Sx) (t) \bigr)p_{n}^{2}(\theta) \\ &+f_{yz} \bigl(t, y_{n}(t), y_{n} \bigl( \theta(t) \bigr), \rho_{8} \bigr) p_{n}(\theta ) (Sp_{n}) (t) \\ &+f_{zz} \bigl(t, y_{n}(t), y_{n} \bigl( \theta(t) \bigr), (Sy_{n}) (t) \bigr) (Sp_{n})^{2}(t) \\ &+f_{x} \bigl(t, y_{n}(t), y_{n} \bigl( \theta(t) \bigr), (Sy_{n}) (t) \bigr)p_{n+1}(t) \\ &+f_{y} \bigl(t, y_{n}(t), y_{n} \bigl( \theta(t) \bigr), (Sy_{n}) (t) \bigr) p_{n+1}(\theta) \\ &+f_{z} \bigl(t, y_{n}(t), y_{n} \bigl( \theta(t) \bigr), (Sy_{n}) (t) \bigr) (Sp_{n+1}) (t) \\ \leq{}& f_{xx} \bigl(t, \rho_{4}, x \bigl(\theta(t) \bigr), (Sx) (t) \bigr)p_{n}^{2}(t) + \frac{1}{2}f_{xy} \bigl(t, y_{n}(t), \rho_{5}, (Sx) (t) \bigr) \bigl[ p_{n}^{2}(t) \\ &+ p_{n}^{2}(\theta) \bigr]+ \frac{1}{2}f_{xz} \bigl(t, y_{n}(t), y_{n} \bigl(\theta(t) \bigr), \rho_{6} \bigr) \bigl[p_{n}^{2}(t)+(Sp_{n})^{2}(t) \bigr] \\ &+f_{yy} \bigl(t, y_{n}(t), \rho_{7}, (Sx) (t) \bigr)p_{n}^{2}(\theta)+\frac {1}{2}f_{yz} \bigl(t, y_{n}(t), y_{n} \bigl(\theta(t) \bigr), \rho_{8} \bigr) \bigl[p_{n}^{2}(\theta ) \\ &+(Sp_{n})^{2}(t) \bigr]+f_{zz} \bigl(t, y_{n}(t), y_{n} \bigl(\theta(t) \bigr), (Sy_{n}) (t) \bigr) (Sp_{n})^{2}(t) \\ \leq{}& \biggl\{ f_{xx} \bigl(t, \rho_{4}, x \bigl(\theta(t) \bigr), (Sx) (t) \bigr)+ \frac {1}{2}f_{xy} \bigl(t,y_{n}(t), \rho_{5},(Sx) (t) \bigr) \\ &+ \frac{1}{2}f_{xz} \bigl(t,y_{n}(t),y_{n} \bigl(\theta(t) \bigr),\rho_{6} \bigr) \biggr\} p_{n}^{2}(t)+ \biggl\{ f_{yy} \bigl(t, y_{n}(t), \rho_{7}, (Sx) (t) \bigr) \\ &+\frac{1}{2}f_{xy} \bigl(t, y_{n}(t), \rho_{5}, (Sx) (t) \bigr)+\frac {1}{2}f_{yz} \bigl(t, y_{n}(t), y_{n} \bigl(\theta(t) \bigr), \rho_{8} \bigr) \biggr\} p_{n}^{2}(\theta) \\ &+ \biggl\{ f_{zz} \bigl(t, y_{n}(t), y_{n} \bigl( \theta(t) \bigr), (Sy_{n}) (t) \bigr) +\frac {1}{2}f_{xz} \bigl(t, y_{n}(t), y_{n} \bigl(\theta(t) \bigr), \rho_{6} \bigr) \\ &+\frac{1}{2}f_{yz} \bigl(t, y_{n}(t), y_{n} \bigl(\theta(t) \bigr), \rho_{8} \bigr) \biggr\} (Sp_{n})^{2}(t) \\ \leq{}& D_{0}p_{n}^{2}(t)+D_{1}p_{n}^{2}( \theta)+D_{2}(Sp_{n})^{2}(t), \end{aligned}$$

where $y_{n}(t)\leq\rho_{4}\leq\rho_{1}\leq x(t)$, $y_{n}(\theta)\leq \rho_{5}\leq x(\theta)$, $(Sy_{n})(t)\leq\rho_{3}, \rho_{6}, \rho_{8}\leq(Sx)(t)$, $y_{n}(\theta)\leq\rho_{7}\leq\rho_{2}\leq x(\theta)$. Hence, we conclude that

$$ \begin{aligned}[b] p_{n+1}(t)&\leq p_{n+1}(0)+ \int^{t}_{0} \bigl\{ D_{0}p_{n}^{2}(s)+D_{1}p_{n}^{2}( \theta)+D_{2}(Sp_{n})^{2}(s) \bigr\} \,ds \\ &\leq p_{n+1}(0)+ \int^{T}_{0} \bigl\{ D_{0}p_{n}^{2}(s)+D_{1}p_{n}^{2}( \theta)+D_{2}(Sp_{n})^{2}(s) \bigr\} \,ds \\ &\leq p_{n+1}(0)+C_{0}\max_{t\in J} p_{n}^{2}(t), \end{aligned} $$

(3.4)

where $C_{0}=T[D_{0}+D_{1}+D_{2}T^{2}k_{0}^{2}]$. Moreover, we obtain

$$\begin{aligned} 0={}&g \bigl(x(0), x(T) \bigr)-g \bigl(y_{n}(0), z_{n}(T) \bigr)-g_{x} \bigl(y_{n}(0), y_{n}(T) \bigr) \bigl[-p_{n+1}(0)+p_{n}(0) \bigr] \\ &-g_{y} \bigl(y_{n}(0), z_{n}(T) \bigr) \bigl[q_{n+1}(T)-q_{n}(T) \bigr] \\ ={}&g_{x} \bigl(\delta_{1}, x(T) \bigr)p_{n}(0)-g_{y} \bigl(y_{n}(0), \delta _{2} \bigr)q_{n}(T)-g_{x} \bigl(y_{n}(0), y_{n}(T) \bigr) \bigl[-p_{n+1}(0) \\ &+p_{n}(0) \bigr]-g_{y} \bigl(y_{n}(0), z_{n}(T) \bigr) \bigl[q_{n+1}(T)-q_{n}(T) \bigr] \end{aligned}$$

and

$$\begin{aligned} &g_{x} \bigl(y_{n}(0), y_{n}(T) \bigr)p_{n+1}(0)\\ &\quad= \bigl[g_{y} \bigl(y_{n}(0), \delta _{2} \bigr)-g_{y} \bigl(y_{n}(0), z_{n}(T) \bigr) \bigr]q_{n}(T) \\ &\quad\quad+ \bigl[g_{x} \bigl(y_{n}(0), y_{n}(T) \bigr)-g_{x} \bigl(\delta_{1}, x(T) \bigr) \bigr]p_{n}(0)+g_{y} \bigl(y_{n}(0), z_{n}(T) \bigr)q_{n+1}(T) \\ &\quad=-g_{yy} \bigl(y_{n}(0), \delta_{3} \bigr)q_{n}^{2}(T)-g_{xx} \bigl( \delta_{4}, x(T) \bigr)p_{n}^{2}(0)-g_{xy} \bigl(y_{n}(0), \delta_{5} \bigr)p_{n}(T)p_{n}(0) \\ &\qquad+g_{y} \bigl(y_{n}(0), z_{n}(T) \bigr)q_{n+1}(T), \end{aligned}$$

where $y_{n}(0)\leq\delta_{4}\leq\delta_{1}\leq x(0)$, $x(T)\leq\delta _{2}\leq\delta_{3}\leq z_{n}(T)$. Therefore, we deduce that

$$ \begin{aligned}[b] p_{n+1}(0)&\leq B_{0}q_{n}^{2}(T)+B_{1}p_{n}^{2}(0)+B_{2}p_{n}^{2}(T)+g_{y} \bigl(y_{n}(0), z_{n}(T) \bigr)q_{n+1}(T) \\ &\leq B_{0}\max_{t\in J} q_{n}^{2}(t)+B_{1} \max_{t\in J} p_{n}^{2}(t)+B_{2} \max_{t\in J} p_{n}^{2}(t)+ \frac{g_{y}(y_{n}(0), z_{n}(T))}{g_{x}(y_{n}(0), y_{n}(T))}q_{n+1}(T) \\ &\leq C_{1}\max_{t\in J} p_{n}^{2}(t)+C_{2} \max_{t\in J} q_{n}^{2}(t)+C_{1}^{0}q_{n+1}(T), \end{aligned} $$

(3.5)

where

$$\begin{gathered} B_{0}=-\frac{g_{yy}(y_{n}(0),\delta_{3})}{g_{x}(y_{n}(0), y_{n}(T))}, \qquad B_{1}=-\frac{g_{xx}(\delta_{4}, x(T))}{g_{x}(y_{n}(0), y_{n}(T))}, \\ B_{2}=-\frac{1}{2}\frac{g_{xy}(y_{n}(0), \delta_{5})}{g_{x}(y_{n}(0), y_{n}(T))}, \qquad C_{1}=B_{1}+B_{2}, \qquad C_{2}=B_{0}, \qquad C_{1}^{0}= \frac{g_{y}(y_{n}(0), z_{n}(T))}{g_{x}(y_{n}(0), y_{n}(T))}.\end{gathered} $$

In a similar way, we can arrive at

$$\begin{aligned} q_{n+1}'(t) \leq{}& f_{xx} \bigl(t, \xi_{4}, z_{n} \bigl(\theta(t) \bigr), (Sz_{n}) (t) \bigr)q_{n}(t) \bigl(q_{n}(t)+p_{n}(t) \bigr) \\ &+f_{xy} \bigl(t, y_{n}(t), \xi_{5}, (Sz_{n}) (t) \bigr)q_{n}(t) \bigl(q_{n}(\theta )+p_{n}(\theta) \bigr) \\ &+f_{xz} \bigl(t, y_{n}(t), y_{n} \bigl( \theta(t) \bigr), \xi _{6} \bigr)q_{n}(t) \bigl[(Sq_{n}) (t)+(Sp_{n}) (t) \bigr] \\ &+f_{yx} \bigl(t, \xi_{7}, \xi_{2}, (Sz_{n}) (t) \bigr)q_{n}(\theta)p_{n}(t) \\ &+f_{yy} \bigl(t, y_{n}(t), \xi_{8}, (Sz_{n}) (t) \bigr)q_{n}(\theta) \bigl[p_{n}( \theta )+q_{n}(\theta) \bigr] \\ &+f_{yz} \bigl(t, y_{n}(t), y_{n} \bigl( \theta(t) \bigr), \xi_{9} \bigr)q_{n}(\theta ) \bigl[(Sp_{n}) (t)+(Sq_{n}) (t) \bigr] \\ &+f_{zx} \bigl(t, \xi_{10}, x \bigl(\theta(t) \bigr), \xi_{3} \bigr)p_{n}(t) (Sq_{n}) (t) \\ &+f_{zy} \bigl(t, y_{n}(t), \xi_{11}, \xi_{3} \bigr)p_{n}(\theta) (Sq_{n}) (t) \\ &+f_{zz} \bigl(t, y_{n}(t), y_{n} \bigl( \theta(t) \bigr), \xi _{12} \bigr) \bigl[(Sp_{n}) (t)+(Sq_{n}) (t) \bigr](Sq_{n}) (t) \\ \leq{}& \biggl\{ \frac{3}{2}f_{xx} \bigl(t, \xi_{4}, z_{n} \bigl(\theta(t) \bigr), (Sz_{n}) (t) \bigr)+f_{xy} \bigl(t, y_{n}(t), \xi_{5}, (Sz_{n}) (t) \bigr) \\ &+f_{xz} \bigl(t, y_{n}(t), y_{n} \bigl( \theta(t) \bigr), \xi_{6} \bigr) \biggr\} q_{n}^{2}(t)+ \frac {1}{2} \bigl\{ f_{xx} \bigl(t, \xi_{4}, z_{n} \bigl(\theta(t) \bigr), (Sz_{n}) (t) \bigr) \\ &+f_{yx} \bigl(t, \xi_{7}, \xi_{2}, (Sz_{n}) (t) \bigr)+f_{zx} \bigl(t, \xi_{10}, x \bigl(\theta (t) \bigr), \xi_{3} \bigr) \bigr\} p_{n}^{2}(t) \\ &+ \biggl\{ \frac{1}{2}f_{xy} \bigl(t, y_{n}(t), \xi_{5}, (Sz_{n}) (t) \bigr)+\frac {1}{2}f_{yx} \bigl(t, \xi_{7}, \xi_{2}, (Sz_{n}) (t) \bigr) \\ &+\frac{3}{2}f_{yy} \bigl(t,y_{n}(t), \xi_{8}, (Sz_{n}) (t) \bigr)+f_{yx} \bigl(t, \xi _{7}, \xi_{2}, (Sz_{n}) (t) \bigr) \biggr\} q_{n}^{2}(\theta) \\ &+\frac{1}{2} \bigl\{ f_{xy} \bigl(t, y_{n}(t), \xi_{5}, (Sz_{n}) (t) \bigr)+f_{yy} \bigl(t, y_{n}(t), \xi_{8}, (Sz_{n}) (t) \bigr) \\ &+f_{zy} \bigl(t, y_{n}(t), \xi_{11}, \xi_{3} \bigr) \bigr\} p_{n}^{2}(\theta) + \frac{1}{2} \bigl\{ f_{xz} \bigl(t, y_{n}(t), y_{n} \bigl(\theta(t) \bigr), \xi_{6} \bigr) \\ &+f_{yz} \bigl(t, y_{n}(t), y_{n} \bigl( \theta(t) \bigr), \xi_{9} \bigr)+f_{zx} \bigl(t, \xi_{10}, x \bigl(\theta(t) \bigr), \xi_{3} \bigr) \\ &+f_{zy} \bigl(t, y_{n}(t), \xi_{11}, \xi_{3} \bigr)+3f_{zz} \bigl(t, y_{n}(t), y_{n} \bigl(\theta(t) \bigr), \xi_{12} \bigr) \bigr\} (Sq_{n})^{2}(t) \\ &+\frac{1}{2} \bigl\{ f_{xz} \bigl(t, y_{n}(t), y_{n} \bigl(\theta(t) \bigr), \xi_{6} \bigr) +f_{yz} \bigl(t, y_{n}(t), y_{n} \bigl(\theta(t) \bigr), \xi_{9} \bigr) \\ & +f_{zz} \bigl(t, y_{n}(t), y_{n} \bigl( \theta(t) \bigr), \xi_{12} \bigr) \bigr\} (Sp_{n})^{2}(t) \\ \leq{}& D_{3}q_{n}^{2}(t)+D_{4}p_{n}^{2}(t)+D_{5}q_{n}^{2}( \theta )+D_{6}p_{n}^{2}(\theta)+D_{7}(Sq_{n})^{2}(t) +D_{8}(Sp_{n})^{2}(t) \end{aligned}$$

and

$$ \begin{aligned} [b]q_{n+1}(t)\leq{}& q_{n+1}(0)+ \int^{T}_{0} \bigl\{ D_{3}q_{n}^{2}(s)+D_{4}p_{n}^{2}(s)+D_{5}q_{n}^{2}( \theta )+D_{6}p_{n}^{2}(\theta) \\ &+D_{7}(Sq_{n})^{2}(s)+D_{8}(Sp_{n})^{2}(s) \bigr\} \,ds \\ \leq{}& q_{n+1}(0)+\max_{t\in J} p_{n}^{2}(t)T \bigl[D_{4}+D_{6}+D_{8}T^{2}k_{0}^{2} \bigr]+\max_{t\in J} q_{n}^{2}(t)T \bigl[D_{3} \\ &+D_{5}+D_{7}T^{2}k_{0}^{2} \bigr] \\ ={}&q_{n+1}(0)+C_{3}\max_{t\in J} p_{n}^{2}(t)+C_{4}\max_{t\in J} q_{n}^{2}(t), \end{aligned} $$

(3.6)

where $C_{3}=T[D_{4}+D_{6}+D_{8}T^{2}k_{0}^{2}]$, $C_{4}=T[D_{3}+D_{5}+D_{7}T^{2}k_{0}^{2}]$, $y_{n}(t)\leq\xi_{4}\leq\xi_{1}$, $y_{n}(\theta)\leq\xi_{5}\leq z_{n}(\theta)$, $x(t)\leq\xi_{1}\leq z_{n}(t)$, $(Sy_{n})(t)\leq\xi_{6},\xi_{9}\leq (Sz_{n})(t)$, $x(\theta)\leq\xi_{2}\leq z_{n}(\theta)$, $y_{n}(t)\leq\xi_{7},\xi _{10}\leq x(t)$, $(Sx)(t)\leq\xi_{3}\leq(Sz_{n})(t)$, $y_{n}(\theta)\leq\xi_{8}\leq\xi_{2}$, $y_{n}(\theta)\leq\xi_{11}\leq x(\theta)$, $(Sy_{n})(t)\leq\xi _{12}\leq\xi_{3}$. Meanwhile, we have

$$\begin{aligned} 0={}&{-}g \bigl(x(0), x(T) \bigr)+g \bigl(z_{n}(0), y_{n}(T) \bigr)+g_{x} \bigl(y_{n}(0), y_{n}(T) \bigr) \bigl[q_{n+1}(0)-q_{n}(0) \bigr] \\ &+g_{y} \bigl(y_{n}(0), z_{n}(T) \bigr) \bigl[-p_{n+1}(T)+p_{n}(T) \bigr] \\ ={}&g_{x}\bigl(\alpha_{1}, y_{n}(T) \bigr)q_{n}(0)-g_{y} \bigl(x(0), \alpha_{2} \bigr)p_{n}(T) \\ &+g_{x} \bigl(y_{n}(0), y_{n}(T) \bigr) \bigl[q_{n+1}(0)-q_{n}(0) \bigr]+g_{y} \bigl(y_{n}(0), z_{n}(T) \bigr) \bigl[-p_{n+1}(T)+p_{n}(T) \bigr] \end{aligned}$$

and

$$\begin{aligned} &g_{x} \bigl(y_{n}(0), y_{n}(T) \bigr)q_{n+1}(0)\\ &\quad\leq-g_{xx} \bigl(\alpha_{3}, y_{n}(T) \bigr)q_{n}(0) \bigl[q_{n}(0)+p_{n}(0) \bigr] \\ &\qquad-g_{yy} \bigl(y(0), \alpha_{5} \bigr)p_{n}(T) \bigl[q_{n}(T)+p_{n}(T) \bigr]+g_{yx} \bigl(\alpha _{4}, \alpha_{2}(T) \bigr)p_{n}(T)p_{n}(0) \\ &\qquad+g_{y} \bigl(y_{n}(0), z_{n}(T) \bigr)p_{n+1}(T) \\ &\quad\leq-g_{xx} \bigl(\alpha_{3}, y_{n}(T) \bigr)q_{n}(0) \bigl[q_{n}(0)+p_{n}(0) \bigr]-g_{yy} \bigl(y(0), \alpha_{5} \bigr) p_{n}(T) \bigl[q_{n}(T)+p_{n}(T) \bigr] \\ &\qquad+g_{y} \bigl(y_{n}(0), z_{n}(T) \bigr)p_{n+1}(T). \end{aligned}$$

Hence, we conclude that

$$ \begin{aligned}[b] q_{n+1}(0) \leq{}& B_{3}q_{n}^{2}(T)+B_{4}q_{n}^{2}(0)+B_{5}p_{n}^{2}(0)+B_{6}p_{n}^{2}(T)+g_{y} \bigl(y_{n}(0), z_{n}(T) \bigr)p_{n+1}(T) \\ \leq{}& B_{3}\max_{t\in J} q_{n}^{2}(t)+B_{4} \max_{t\in J} q_{n}^{2}(t)+B_{5} \max_{t\in J} p_{n}^{2}(t) +B_{6} \max_{t\in J} p_{n}^{2}(t) \\ &+\frac{g_{y}(y_{n}(0), z_{n}(T))}{g_{x}(y_{n}(0), y_{n}(T))}p_{n+1}(T) \\ \leq{}& C_{5}\max_{t\in J} p_{n}^{2}(t)+C_{6} \max_{t\in J} q_{n}^{2}(t)+C_{2}^{0}p_{n+1}(T), \end{aligned} $$

(3.7)

where

$$\begin{gathered} B_{3}=-\frac{1}{2}\frac{g_{yy}(y(0), \alpha_{5})}{g_{x}(y_{n}(0), y_{n}(T))}, \qquad B_{4}=- \frac{3}{2}\frac{g_{xx}(\alpha_{3}, y_{n}(T))}{g_{x}(y_{n}(0), y_{n}(T))}, \\ B_{5}=-\frac{1}{2} \frac{g_{xx}(\alpha_{3}, y_{n}(T))}{g_{x}(y_{n}(0), y_{n}(T))}, \qquad B_{6}=-\frac{3}{2}\frac{g_{yy}(y(0), \alpha_{5})}{g_{x}(y_{n}(0), y_{n}(T))}, \\ C_{5}=B_{5}+B_{6}, \qquad C_{6}=B_{3}+B_{4}, \qquad C_{2}^{0}=\frac {g_{y}(y_{n}(0), z_{n}(T))}{g_{x}(y_{n}(0), y_{n}(T))}.\end{gathered} $$

It follows from (3.4)–(3.7) that

$$ \begin{gathered} p_{n+1}(0)\leq C^{1}_{1}p_{n}^{2}(t)+C^{1}_{2}q_{n}^{2}(t), \\ q_{n+1}(0)\leq C^{1}_{3}p_{n}^{2}(t)+C^{1}_{4}q_{n}^{2}(t), \end{gathered} $$

(3.8)

where

$$\begin{gathered} C^{1}_{1}=\frac {1}{(1-C^{0}_{2})(1-C^{0}_{1})} \bigl[C^{0}_{1}C^{0}_{2}C_{0}+C^{0}_{1}C_{5}+C^{0}_{1}C_{3}+C_{1} \bigr], \\ C^{1}_{2}=\frac {1}{(1-C^{0}_{2})(1-C^{0}_{1})} \bigl[C^{0}_{1}C_{2}+C^{0}_{1}C_{4}+C_{6} \bigr], \\ C^{1}_{3}=C_{5}+C^{1}_{2}C^{0}_{2}+C_{0}C^{0}_{2}, \qquad C^{1}_{4}=C_{6}+C^{1}_{2}C^{0}_{2}.\end{gathered} $$

By virtue of (3.4), (3.6), and (3.8), we see that

$$ \begin{gathered} \max_{t\in J} \bigl\vert x(t)-y_{n+1}(t) \bigr\vert \leq d_{1} \max _{t\in J} \bigl\vert x(t)-y_{n}(t) \bigr\vert ^{2}+d_{2} \max_{t\in J} \bigl\vert x(t)-z_{n}(t) \bigr\vert ^{2}, \\ \max_{t\in J} \bigl\vert x(t)-z_{n+1}(t) \bigr\vert \leq d_{3} \max_{t\in J} \bigl\vert x(t)-y_{n}(t) \bigr\vert ^{2}+d_{4} \max _{t\in J} \bigl\vert x(t)-z_{n}(t) \bigr\vert ^{2}, \end{gathered} $$

where $d_{1}=C^{1}_{1}+C_{0}$, $d_{2}=C^{1}_{2}$, $d_{3}=C^{1}_{3}+C_{3}$, $d_{4}=C^{1}_{4}+C_{4}$. This completes the proof. □

Example

To illustrate the validity of the theoretical result obtained in the previous section, we give the following example.

Example 4.1

Consider the boundary value problem

$$ \textstyle\begin{cases} x'(t)= \frac{1}{10}x^{2}(t)-\frac{1}{4}x (\frac{t}{2} )-\frac {1}{80} \int_{0}^{t}x(s)\,ds-\frac{1}{10}x(t), \quad t \in[0, 1], \\ \frac{1}{2}x(0)-\frac{1}{12}x^{2}(1)+ \frac{1}{12}x(1)+\frac{1}{4}=0, \end{cases} $$

(4.1)

where

$$ \begin{gathered} f \bigl(t, x(t), x \bigl(\theta(t) \bigr), (Sx) (t) \bigr)=\frac{1}{10}x^{2}(t)-\frac{1}{4}x \biggl( \frac{t}{2} \biggr) -\frac{1}{80} \int_{0}^{t}x(s)\,ds-\frac{1}{10}x(t), \quad t \in[0, 1], \\ g \bigl(x(0), x(1) \bigr)=\frac{1}{2}x(0)-\frac{1}{12}x^{2}(1)+ \frac {1}{12}x(1)+\frac{1}{4}. \end{gathered} $$

Letting $y_{0}(t)=-1$, $z_{0}(t)=0$, $t\in[0, 1]$, then $y_{0}(t)< z_{0}(t)$ and

$$ \begin{gathered} f \bigl(t, y_{0}(t), y_{0} \bigl( \theta(t) \bigr), (Sy_{0}) (t) \bigr)\ge y_{0}'(t), \quad t\in [0, 1], \qquad g \bigl(y_{0}(0), z_{0}(1) \bigr)=- \frac{1}{4}, \\ f \bigl(t, z_{0}(t), z_{0} \bigl(\theta(t) \bigr), (Sz_{0}) (t) \bigr)=0=z_{0}'(t), \quad t\in [0, 1], \qquad g \bigl(z_{0}(0), y_{0}(1) \bigr)= \frac{1}{12}. \end{gathered} $$

Thus, $y_{0}$ and $z_{0}$ are coupled lower and upper solutions of Eq. (4.1), and $y_{0}(t)\leq x(t)\leq z_{0}(t)$, that is, $-1\leq x(t)\leq0$. Moreover, by Eq. (4.1), we have

$$\begin{aligned} &\int^{1}_{0} \bigl\{ f_{x} \bigl(t, y_{0}, y_{0}(\theta), (Sy_{0}) (t) \bigr) +f_{y} \bigl(t, y_{0}, y_{0}(\theta), (Sy_{0}) (t) \bigr)+f_{z} \bigl(t, y_{0}, y_{0}(\theta), (Sy_{0}) (t) \bigr)k_{0}T \bigr\} \,dt \\ &\quad=-\frac {9}{16} \geq-1. \end{aligned}$$

It is not difficult to verify that all conditions of Theorem 3.1 are satisfied. Therefore, there exist the monotone sequences $\{y_{n}(t)\}$ and $\{z_{n}(t)\}$ converging uniformly to the unique solution x of equation (4.1) and the convergence is quadratic.