In this section, we shall establish the existence of extremal solutions of problem (1).

### Definition 3.1

Functions \(\alpha,\beta\in \Omega\) are called lower and upper solutions of (1) if

$$\left \{ \textstyle\begin{array}{@{}l} \alpha'(t)\leq(Q\alpha)(t),\quad t\in J,\\ g(\alpha(0),\alpha(T))\leq0, \end{array}\displaystyle \right . $$

and

$$\left \{ \textstyle\begin{array}{@{}l} \beta'(t)\geq(Q\beta)(t), \quad t\in J,\\ g(\beta(0),\beta(T))\geq0. \end{array}\displaystyle \right . $$

Now we state our theorems. First we discuss the existence of solutions for the following linear problem:

$$\begin{aligned} \left \{ \textstyle\begin{array}{@{}l} u'(t)=M(t)u(t)+(\mathcal{L}u)(t)+\sigma_{\eta}(t),\quad t\in J,\\ g(\eta(0),\eta(T))+M_{1}(u(0)-\eta(0))-M_{2}(u(T)-\eta(T))=0, \end{array}\displaystyle \right . \end{aligned}$$

(4)

where \(\sigma_{\eta}(t)=(Q\eta)(t)-M(t)\eta(t)-(\mathcal{L}\eta)(t)\).

### Theorem 3.1

*A function*
\(u\in\Omega\)
*is a solution of* (4) *if and only if*
*u*
*is a solution of the following integral equation*:

$$\begin{aligned} u(t)= \frac{B_{\eta}e^{\int_{0}^{t}M(s)\,ds}}{M_{1}-M_{2}e^{\int_{0}^{T}M(s)\,ds}}+ \int _{0}^{T}G(t,s) \bigl(\sigma_{\eta}(s)+( \mathcal{L}u) (s) \bigr)\,ds, \end{aligned}$$

(5)

*where*
\(B_{\eta}=-g(\eta(0),\eta(T))+M_{1}\eta(0)-M_{2}\eta(T)\), \(M\in C(\mathbb{R},\mathbb{R}^{+})\), \(M_{1}, M_{2}\)
*are constants satisfying*
\(M_{1}\neq M_{2}e^{\int_{0}^{T}M(s)\,ds}\)
*and*

$$G(t,s)= \frac{1}{M_{1}-M_{2}e^{\int_{0}^{T}M(s)\,ds}} \left \{ \textstyle\begin{array}{@{}l@{\quad}l} M_{1}e^{\int_{s}^{t}M(r)\,dr},&0\leq s< t\leq T,\\ M_{2}e^{\int_{s}^{T}M(r)\,dr}e^{\int_{s}^{t}M(r)\,dr}, &0\leq t\leq s\leq T. \end{array}\displaystyle \right . $$

### Proof

Assume \(u\in\Omega\) is a solution of (4). Set \(u(t)=v(t)e^{\int _{0}^{t}M(s)\,ds}\), we see that \(v(t)\) satisfies

$$\begin{aligned} \left \{ \textstyle\begin{array}{@{}l} v'(t)= (\sigma_{\eta}(t)+(\mathcal{L}u)(t) )e^{\int _{0}^{t}-M(s)\,ds}, \\ v(0)= \frac{B_{\eta}}{M_{1}}+ \frac {M_{2}}{M_{1}}v(T)e^{\int_{0}^{T}M(s)\,ds}. \end{array}\displaystyle \right . \end{aligned}$$

(6)

By using (6), we have

$$\begin{aligned} v(t)=v(0)+ \int_{0}^{t} \bigl(\sigma_{\eta}(s)+( \mathcal {L}u) (s) \bigr)e^{\int_{0}^{s}-M(r)\,dr}\,ds. \end{aligned}$$

(7)

If we set \(t=T\) in (7), then we get

$$\begin{aligned} v(T)=v(0)+ \int_{0}^{T} \bigl(\sigma_{\eta}(s)+( \mathcal {L}u) (s) \bigr)e^{\int_{0}^{s}-M(r)\,dr}\,ds. \end{aligned}$$

(8)

From the boundary condition \(v(T)= \frac{M_{1}v(0)-B_{\eta}}{M_{2}e^{\int_{0}^{T}M(t)\,dt}}\), we obtain

$$\begin{aligned} v(0)= \frac{B_{\eta}}{M_{1}-M_{2}e^{\int _{0}^{T}M(t)\,dt}}+ \frac{M_{2}e^{\int _{0}^{T}M(t)\,dt}}{M_{1}-M_{2}e^{\int_{0}^{T}M(t)\,dt}} \int _{0}^{T} \bigl(\sigma_{\eta}(s)+( \mathcal{L}u) (s) \bigr) e^{\int_{0}^{s}-M(r)\,dr}\,ds. \end{aligned}$$

(9)

Substituting (9) into (7) and using \(v(t)= u(t)e^{\int_{0}^{t}-M(s)\,ds}\), \(t\in J\), we have

$$\begin{aligned} u(t)={}& \frac{B_{\eta}e^{\int_{0}^{t}M(s)\,ds}}{M_{1}-M_{2}e^{\int_{0}^{T}M(t)\,dt}}+ \frac{M_{1}}{M_{1}-M_{2}e^{\int_{0}^{T}M(t)\,dt}} \int_{0}^{T} \bigl(\sigma_{\eta}(s)+( \mathcal{L}u) (s) \bigr)e^{\int_{s}^{t}M(r)\,dr}\,ds \\ &{}+ \frac{M_{2}e^{\int_{0}^{T}M(t)\,dt}}{M_{1}-M_{2}e^{\int _{0}^{T}M(t)\,dt}} \int_{0}^{T} \bigl(\sigma_{\eta}(s)+( \mathcal {L}u) (s) \bigr)e^{\int_{s}^{t}M(r)\,dr}\,ds. \end{aligned}$$

Let

$$G(t,s)=\frac{1}{M_{1}-M_{2}e^{\int_{0}^{T}M(t)\,dt}}\left \{ \textstyle\begin{array}{@{}l@{\quad}l} M_{1}e^{\int_{s}^{t}M(r)\,dr}, &0\leq s< t\leq T, \\ M_{2}e^{\int_{0}^{T}M(t)\,dt}e^{\int_{s}^{t}M(r)\,dr}, &0\leq t\leq s\leq T, \end{array}\displaystyle \right . $$

we see that *u* is a solution of (5). The proof is complete. □

In the following paper, we denote \(\xi=\Vert G(t,s)\Vert =\max \{\vert \frac {M_{1} e^{\int_{0}^{T}M(t)\,dt}}{M_{1}-M_{2}e^{\int_{0}^{T}M(t)\,dt}}\vert ,\vert \frac{M_{2} e^{\int_{0}^{T}M(t)\,dt}}{M_{1}-M_{2}e^{\int_{0}^{T}M(t)\,dt}}\vert \}\).

### Theorem 3.2

*Assume that*
\(M\in C(\mathbb {R},\mathbb{R}^{+})\), \(M_{1}\neq M_{2}e^{\int_{0}^{T}M(s)\,ds}\), *and*

$$\begin{aligned} \xi \Vert \mathcal{L}\Vert T< 1. \end{aligned}$$

(10)

*Then problem* (4) *has a unique solution*.

### Proof

By Theorem 3.1, we know that \(u\in\Omega\) is the solution of (4) if and only if *u* is a solution of the integral equation (5). Now we prove (5) has a unique solution \(u\in\Omega\). Define an operator \(F:\Omega\to\Omega\) by

$$(Fu) (t)= \frac{B_{\eta}e^{\int_{0}^{t}M(s)\,ds}}{M_{1}-M_{2}e^{\int_{0}^{T}M(s)\,ds}}+ \int _{0}^{T}G(t,s) \bigl(\sigma_{\eta}(s)+( \mathcal{L}u) (s) \bigr)\,ds. $$

For any \(u_{1}, u_{2}\in\Omega\), we have

$$\vert Fu_{1}-Fu_{2}\vert \leq \int_{0}^{T} \bigl\vert G(t,s) \bigr\vert \bigl\vert \bigl(\mathcal{L}(u_{1}-u_{2}) \bigr) (t) \bigr\vert \leq\xi T\Vert \mathcal{L}\Vert \Vert u_{1}-u_{2} \Vert . $$

Hence, \(\Vert Fu_{1}-Fu_{2}\Vert = \max_{t\in J}\vert Fu_{1}(t)-Fu_{2}(t)\vert =\tau \Vert u_{1}-u_{2}\Vert \), where

$$\tau=\xi T\Vert \mathcal{L}\Vert . $$

By (10) and the Banach contraction principle, *F* has a unique fixed point. It is clear that this fixed point is the solution of (5). The proof is complete. □

### Theorem 3.3

*Let* (3), (10) *hold and*
\(Q\in C[E,E]\). *In addition*, *we assume that*

- (H
_{1}): -
*the functions*
\(\alpha, \beta\in\Omega\)
*are lower and upper solutions of problem* (1), *respectively*, *such that*
\(\beta\leq \alpha\);

- (H
_{2}): -
\(\mathcal{L}\in C(E,E)\)
*is a positive linear operator and*
\(M\in C(\mathbb{R},\mathbb{R}^{+})\)
*such that*

$$(Qu) (t)-(Qv) (t)\leq M(t) \bigl(u(t)-v(t) \bigr)+ \bigl(\mathcal{L}(u-v) \bigr) (t),\quad \textit{for } \beta\leq v\leq u\leq\alpha; $$

- (H
_{3}): -
*there exist*
\(M_{2}\geq M_{1}>0\)
*satisfying*
\(M_{1}\neq M_{2}e^{\int _{0}^{T}M(s)\,ds}\), *and*

$$g(\bar{u},\bar{v})-g(u,v)\geq M_{1}(\bar{u}-\bar{v})-M_{2}(u-v), $$

*whenever*
\(\beta(0)\leq u(0)\leq \bar{u}(0)\leq\alpha(0),\beta(T)\leq v(T)\leq\bar{v}(T)\leq\alpha(T)\).

*Then there exist monotone sequences*
\(\{\alpha_{n}(t)\}\)
*and*
\(\{\beta_{n}(t)\}\)
*with*
\(\alpha_{0}=\alpha,\beta_{0}=\beta\), *which converge to the extremal solutions of problem* (1) *in the sector*
\([\beta,\alpha]=\{u\in C^{1}(J,\mathbb{R}):\beta(t)\leq u(t)\leq\alpha(t),t\in J\}\).

### Proof

First, we construct two sequences \(\{\alpha_{n}(t)\} \), \(\{\beta_{n}(t)\}\) which satisfy the following equations:

$$\begin{aligned} \left \{ \textstyle\begin{array}{@{}l} \alpha'_{n}(t)=M(t)\alpha_{n}(t)+(\mathcal{L}\alpha_{n})(t)+(Q\alpha _{n-1})(t)-M(t)\alpha_{n-1}(t)-(\mathcal{L}\alpha_{n-1})(t),\\ g(\alpha_{n-1}(0),\alpha_{n-1}(T))+M_{1}(\alpha_{n}(0)-\alpha _{n-1}(0))-M_{2}(\alpha_{n}(T)-\alpha_{n-1}(T))=0, \end{array}\displaystyle \right . \end{aligned}$$

(11)

and

$$\begin{aligned} \left \{ \textstyle\begin{array}{@{}l} \beta'_{n}(t)=M(t)\beta_{n}(t)+(\mathcal{L}\beta_{n})(t)+(Q\beta _{n-1})(t)-M(t)\beta_{n-1}(t)-(\mathcal{L}\beta_{n-1})(t),\\ g(\beta_{n-1}(0),\beta_{n-1}(T))+M_{1}(\beta_{n}(0)-\beta _{n-1}(0))-M_{2}(\beta_{n}(T)-\beta_{n-1}(T))=0, \end{array}\displaystyle \right . \end{aligned}$$

(12)

for \(n=1,2,\ldots\) , where \(\alpha_{0}=\alpha\), \(\beta_{0}=\beta\).

It follows from Theorem 3.2 that both (11) and (12) have a unique solution, respectively. Then we complete the proof by four steps.

*Step* 1 We prove that \(\beta_{n-1}\leq\beta_{n}\) and \(\alpha_{n}\leq \alpha_{n-1}\), \(n=1,2,\ldots\) .

Set \(m=\alpha_{1}-\alpha\), \(t\in J\). Employing (H_{1}), we have

$$\begin{aligned} m'(t)&=\alpha'_{1}(t)-\alpha'(t) \\ &\geq M(t)\alpha_{1}(t)+(\mathcal{L}\alpha_{1}) (t)+(Q \alpha) (t)-M(t)\alpha (t)-(\mathcal{L}\alpha) (t)-(Q\alpha) (t) \\ &=M(t)m(k)+(\mathcal{L}m) (t),\quad t\in J, \end{aligned}$$

and

$$\begin{aligned} m(0)&=- \frac{1}{M_{1}}g \bigl(\alpha(0),\alpha(T) \bigr)+ \frac{M_{2}}{M_{1}} \bigl(\alpha_{1}(T)-\alpha(T) \bigr)+\alpha(0)-\alpha(0) \geq \frac{M_{2}}{M_{1}}m(T). \end{aligned}$$

From Lemma 2.1, we get \(m(t)\leq0, t\in J\), so \(\alpha_{1}(t)\leq\alpha(t)\).

By mathematical induction, we obtain the sequence \(\alpha_{n}\) is a non-increasing sequence. Analogously, we can show \(\beta_{n}\) is a nondecreasing sequence.

*Step* 2 We show that \(\beta_{1}\leq\alpha_{1}\) if \(\beta\leq\alpha\).

Let \(m=\beta_{1}-\alpha_{1}\). Using (H_{1}), (H_{2}), and (H_{3}), we get

$$\begin{aligned} m'(t)={}&\beta'_{1}(t)-\alpha'_{1}(t) \\ ={}&M(t)\beta_{1}(t)+(\mathcal{L}\beta_{1}) (t)+(Q\beta) (t)-M(t)\beta (t)-(\mathcal{L}\beta) (t) \\ &{} -M(t)\alpha_{1}(t)-(\mathcal{L}\alpha_{1}) (t)-(Q\alpha ) (t)+M(t)\alpha(t)+(\mathcal{L}\alpha) (t) \\ \geq{}& M(t)m(t)+(\mathcal{L}m) (t) ,\quad t\in J, \end{aligned}$$

and

$$\begin{aligned} m(0)={}&\beta_{1}(0)-\alpha_{1}(0) \\ ={}&{-} \frac{1}{M_{1}}g \bigl(\beta(0),\beta(T) \bigr)+ \frac {M_{2}}{M_{1}} \bigl(\beta_{1}(T)-\beta(T) \bigr)+\beta(0) \\ &{} - \biggl[ \frac{1}{M_{1}}g \bigl(\alpha(0),\alpha (T) \bigr)+ \frac{M_{2}}{M_{1}} \bigl(\alpha_{1}(T)-\alpha(T) \bigr)+\alpha(0) \biggr] \\ \geq{}& \frac{M_{2}}{M_{1}}m(T). \end{aligned}$$

Then based on Lemma 2.1, we have \(m\leq0\), which implies \(\beta_{1}\leq \alpha_{1}\). By mathematical induction, we obtain \(\beta_{n}\leq\alpha_{n}\), \(n=1,2,\ldots\) .

*Step* 3 We prove that there exists a solution of problem (1) that satisfies \(\beta(t)\leq u(t)\leq\alpha(t)\) in *J*.

By the first two steps, we get

$$\begin{aligned} \beta_{0}\leq\beta_{1}\leq \beta_{2} \leq\cdots\leq\beta_{n}\leq\cdots \leq \alpha_{n}\leq\cdots \leq\alpha_{2}\leq\alpha_{1} \leq\alpha_{0}, \end{aligned}$$

(13)

and each \(\alpha_{n},\beta_{n}\) satisfies (11) and (12). It is easy to see that the sequence \(\{\beta_{n}(t)\}\) is uniformly bounded and equicontinuous, employing the Ascoli-Arzela theorem, the nondecreasing sequences \(\{\beta_{n}(t)\}\) converges pointwise to a function \(u(t)\) that satisfies \(\beta(t)\leq u(t)\leq\alpha(t)\). Therefore, there exists a solution \(u(t)\) of problem (1) that satisfies \(\beta(t)\leq u(t)\leq\alpha(t)\) in *J*.

*Step* 4 We prove that there exist extremal solutions of problem (1) in \([\beta,\alpha]\).

Apparently, from (13), there exist *ρ* and *r* such that \(\lim_{n\to\infty}\alpha_{n}(t)=\rho(t)\) and \(\lim_{n\to\infty}\beta_{n}(t)=r(t)\) uniformly on *J*. Clearly, \(\rho(t), r(t)\) satisfy problem (1). Let \(u(t)\) be any solution of (1) such that \(\beta(t)\leq u(t)\leq\alpha(t)\). Suppose that there exists a positive integer *j* such that \(\beta_{n}(t)\leq u(t)\leq\alpha _{n}(t)\). Then, setting \(m=\beta_{n+1}-u\), we have

$$\begin{aligned} m'(t)&=\Delta\beta_{n+1}(t)-u'(t) \\ &=M(t)\beta_{n+1}(t)+(\mathcal{L}\beta_{n+1}) (t)+(Q \beta_{n}) (t)-M(t)\beta _{n}(t)-(\mathcal{L} \beta_{n}) (t)-(Qy) (t) \\ &\geq M(t)m(t)+(\mathcal{L}m) (t), \quad t\in J, \end{aligned}$$

and

$$\begin{aligned} m(0)={}&\beta_{n+1}(0)-y(0)\\ ={}&{-} \frac{1}{M_{1}}g \bigl(\beta_{n}(0),\beta_{n}(T) \bigr)+ \frac {M_{2}}{M_{1}} \bigl(\beta_{n+1}(T)-\beta_{n}(T) \bigr)\\ &{}+\beta_{n}(0)-y(0)+ \frac {1}{M_{1}}g \bigl(y(0),y(T) \bigr)\\ \geq{}& \frac{M_{2}}{M_{1}}m(T). \end{aligned}$$

By Lemma 2.1, \(m\leq0\), *i.e.*, \(\beta_{n+1}\leq u\). Similarly, we can get \(u\leq\alpha_{n+1}\) on *J*. Since \(\beta_{0}(t)\leq y(t)\leq \alpha_{0}(t)\), by induction we have \(\beta_{n}\leq u\leq\alpha_{n}\), which implies \(r(t)\leq u(t)\leq\rho(t)\). The proof is complete. □