**3. Main Results**

In this section, we prove the existence of extremal solutions for conformable fractional differential equations involving integral boundary condition. For convenience, we list some assumptions.

(*H*1): *<sup>f</sup>* : [0, 1] <sup>×</sup> <sup>R</sup> <sup>→</sup> <sup>R</sup> is continuous.

(*H*2): Assume that *v*0, *w*<sup>0</sup> ∈ *E* = *C*[0, 1] is lower and upper solution of problem (1), and *v*0(*t*) ≤ *w*0(*t*).

(*H*3): There exists a function *M* ∈ *E* with *<sup>α</sup>* > 0 which satisfies

$$f(t, \mathbf{x}) - f(t, \overline{\mathbf{x}}) \le M(t)(\overline{\mathbf{x}} - \mathbf{x}),$$

for *v*0(*t*) ≤ *x* ≤ *x* ≤ *w*0(*t*).

**Theorem 2.** *Assume that* (*H*1)*,* (*H*2)*,* (*H*3) *hold. Then there exist monotone iterative sequences* {*vn*}<sup>∞</sup> *<sup>n</sup>*=0, {*wn*}<sup>∞</sup> *<sup>n</sup>*=<sup>0</sup> ⊂ *E such that*

$$\lim\_{n \to \infty} v\_n = v, \lim\_{n \to \infty} w\_n = w$$

*uniformly on* [0, 1]*, and v*, *w are the extremal solutions of problem (1) in the sector* [*v*0, *w*0] = {*g* ∈ *E* : *v*0(*t*) ≤ *g*(*t*) ≤ *w*0(*t*), 0 ≤ *t* ≤ 1}*.*

**Proof.** For all *vn*, *wn* ∈ *E*, let

$$\begin{cases} D\_a v\_{n+1}(t) = f(t, v\_n(t)) - M(t)(v\_{n+1}(t) - v\_n(t)), & t \in [0, 1], \\ D\_a w\_{n+1}(t) = f(t, w\_n(t)) - M(t)(w\_{n+1}(t) - w\_n(t)), & t \in [0, 1], \\ v\_{n+1}(0) = \int\_0^1 v\_{n+1}(t) d\mu(t), \; w\_{n+1}(0) = \int\_0^1 w\_{n+1}(t) d\mu(t). \end{cases} \tag{7}$$

Thus, the iterative sequences {*vn*} and {*wn*} can be constructed by Lemma 3.

Firstly, we shall prove that

*vn* ≤ *vn*+<sup>1</sup> ≤ *wn*+<sup>1</sup> ≤ *wn*, *n* = 0, 1, 2, . . . .

Let *p* = *v*<sup>0</sup> − *v*1. According to (7) and Definition 3, we have

$$\begin{cases} D\_u p(t) = D\_u v\_0(t) - D\_u v\_1(t) \le f(t, v\_0(t)) - f(t, v\_0(t)) + M(t)(v\_1(t) - v\_0(t)), & t \in [0, 1], \\ p(0) \le \int\_0^1 v\_0(t) d\mu(t) - \int\_0^1 v\_1(t) d\mu(t), & \end{cases}$$

i.e.,

$$\begin{cases} \quad D\_t p(t) \le -M(t) p(t), & t \in [0, 1], \\\quad p(0) \le \int\_0^1 p(t) d\mu(t). \end{cases}$$

Therefore, by Lemma 4, we have *v*0(*t*) ≤ *v*1(*t*). Similarly, we can prove that *w*1(*t*) ≤ *w*0(*t*), *t* ∈ [0, 1]. Now, let *r* = *v*<sup>1</sup> − *w*1, according to (7) and (*H*3), we have

$$\begin{cases} D\_a r(t) &= f(t, v\_0(t)) - f(t, w\_0(t)) - M(t)(v\_1(t) - v\_0(t) - w\_1(t) + w\_0(t)) \\ &\le M(t)(w\_0(t) - v\_0(t)) - M(t)(v\_1(t) - v\_0(t) - w\_1(t) + w\_0(t)) \\ &= -M(t)r(t), \\\ r(0) &= \int\_0^1 r(t)d\mu(t). \end{cases}$$

By Lemma 4, we have *v*1(*t*) ≤ *w*1(*t*), *t* ∈ [0, 1].

Secondly, we show that *v*1, *w*<sup>1</sup> are lower and upper solutions of (1), respectively.

$$\begin{cases} D\_a v\_1(t) &= f(t, v\_0(t)) - M(t)(v\_1(t) - v\_0(t)) - f(t, v\_1(t)) + f(t, v\_1(t)) \\ &\le M(t)(v\_1(t) - v\_0(t)) - M(t)(v\_1(t) - v\_0(t)) + f(t, v\_1(t)) \\ &= f(t, v\_1(t)), \\ v\_1(0) &= \int\_0^1 v\_1(t) d\mu(t). \end{cases}$$

According to (*H*3) and Definition 3, we deduce that *v*<sup>1</sup> is a lower solution of (1). Similarly, *w*<sup>1</sup> is a upper solutions of (1). By the above arguments and mathematical induction, it is clear that

$$w\_0 \le \dots \le v\_n \le v\_{n+1} \le w\_{n+1} \le w\_n \le \dots \le w\_0, \ n = 0, 1, 2, \dots \tag{8}$$

Thirdly, we show that lim*n*→<sup>∞</sup> *vn* <sup>=</sup> *<sup>v</sup>*, lim*n*→<sup>∞</sup> *wn* <sup>=</sup> *<sup>w</sup>*. Hence, we need to conclude that *vn*, *wn* are uniformly bounded and equicontinuous on [0, 1]. Obviously, the uniform boundedness of sequences *vn*, *wn* follows from (8). Thus, there exists *L* > 0 such that

$$|f(t, v\_n(t)) - M(t)(v\_{n+1}(t) - v\_n(t))| \le L$$

and

$$|f(t, w\_n(t)) - M(t)(w\_{n+1}(t) - w\_n(t))| \le L.$$

Using Theorem 1, we get

$$\begin{aligned} \left| v\_n(t\_1) - v\_n(t\_2) \right| &= \frac{1}{a} |D\_a v\_n(\xi)| |t\_1^a - t\_2^a| \\ &= \frac{1}{a} |f(\xi, v\_{n-1}(\xi)) - M(\xi)(v\_n(\xi) - v\_{n-1}(\xi))| |t\_1^a - t\_2^a|. \end{aligned}$$

Therefore, {*vn*} are equicontinuous. Similarly, we obtain that {*wn*} are equicontinuous too. By Arzela-Ascoli Theorems, we conclude that {*vn*}, {*wn*} have subsequences {*vnk*}, {*wnk*} such that {*vnk*} → *v*, and {*wnk*} → *w* when *k* → ∞. This together with the monotonicity of sequences {*vn*} and {*wn*} implies

$$\lim\_{n \to \infty} v\_n(t) = v(t), \lim\_{n \to \infty} w\_n(t) = w(t)$$

uniformly on [0, 1]. Please note that the sequence {*vn*} satisfies

$$\begin{cases} v\_n(t) = e^{-\int\_0^t s^{n-1} M(s) ds} [v\_{n-1}(0) + Rv\_{n-1}(t)], & t \in [0, 1], \\ v\_n(0) = \int\_0^1 v\_n(t) d\mu(t), & n = 1, 2, \dots, \end{cases} \tag{9}$$

where

$$Rv\_{n-1}(t) = \int\_0^t s^{a-1} [f(t, v\_{n-1}(s)) + M(s)v\_{n-1}(s)] e^{\int\_0^s \tau^{a-1} M(\tau)d\tau} ds.$$

Let *n* → ∞ in (9) . We have

$$\begin{cases} \boldsymbol{v}(t) = \boldsymbol{\varepsilon}^{-\int\_{0}^{t} \boldsymbol{s}^{a-1} M(\boldsymbol{s}) d\boldsymbol{s}} [\boldsymbol{v}(0) + \boldsymbol{R} \boldsymbol{v}(t)], & t \in [0, 1], \\\ \boldsymbol{v}(0) = \int\_{0}^{1} \boldsymbol{v}(t) d\boldsymbol{\mu}(t). \end{cases}$$

This shows that *v* is a solution of the nonlinear problem (1). Similarly, we obtain *w* is a solution of the nonlinear problem (1) too. And

$$w\_0(t) \le v(t) \le w(t) \le w\_0(t), \ t \in [0, 1].$$

Finally, we are going to prove that *v*, *w* are minimal and maximal solutions of (1) in the sector [*v*0, *w*0]. In the following, we show this using induction arguments. Suppose that *g*(*t*) is any solution of (1) in the [*v*0, *w*0] that is

$$w\_0(t) \le \mathbf{g}(t) \le w\_0(t), \ t \in [0, 1].$$

Assume that *vn*(*t*) ≤ *g*(*t*) ≤ *wn*(*t*) hold. Let *p*(*t*) = *vn*+1(*t*) − *g*(*t*), we have

$$\begin{cases} D\_h p(t) &= \, \_D \boldsymbol{v}\_{n+1}(t) - \boldsymbol{D}\_h \boldsymbol{g}(t) \\ &= \, \_f (t, \boldsymbol{v}\_n(t)) - \boldsymbol{M}(t)(\boldsymbol{v}\_{n+1}(t) - \boldsymbol{v}\_n(t)) - f(t, \boldsymbol{g}(t)) \\ &\leq \, \_M (t)(\boldsymbol{g}(t) - \boldsymbol{v}\_n(t)) - \boldsymbol{M}(t)(\boldsymbol{v}\_{n+1}(t) - \boldsymbol{v}\_n(t)) \\ &= \, \_ -\boldsymbol{M}(t)p(t) \\ \boldsymbol{p}(0) &= \int\_0^1 p(t) d\mu(t) .\end{cases}$$

Then, by Lemma 4, we have *vn*+1(*t*) ≤ *g*(*t*), *t* ∈ [0, 1]. By similar method, we can show that *g*(*t*) ≤ *wn*+1(*t*), *t* ∈ [0, 1]. Therefore,

$$w\_n \le g \le w\_{n'} \quad n = 1, 2, \dots$$

By taking *n* → ∞ in the above inequalities, we get that *v* ≤ *g* ≤ *w*. That is *v*, *w* are extremal solutions of problem (1) in [*v*0, *w*0]. Thus, the proof is finished.

**Example 1.** *Consider the following nonlinear problem:*

$$\begin{cases} \,^1D\_{\frac{1}{2}}\mathbf{x}(t) = -\frac{2}{9}(1+\mathbf{x}(t))^3 + 9\sin\frac{\mathbf{x}^2(t)}{4}, & t \in [0, 1], \\\\ \mathbf{x}(0) = \frac{1}{3}\mathbf{x}(\frac{1}{4}) + \frac{1}{6}\mathbf{x}(\frac{1}{2}). \end{cases} \tag{10}$$

*Let*

$$\mu(t) = \begin{cases} \ 0, & t \in [0, \frac{1}{4}), \\\ 1 \\\ \frac{1}{3}, & t \in [\frac{1}{4}, \frac{1}{2}), \\\ 1 \\\ \frac{1}{2}, & t \in [\frac{1}{2}, 1]. \end{cases}$$

*Obviously, α* = <sup>1</sup> <sup>2</sup> *, f*(*t*, *<sup>x</sup>*) = <sup>−</sup><sup>2</sup> <sup>9</sup> (<sup>1</sup> <sup>+</sup> *<sup>x</sup>*)<sup>3</sup> <sup>+</sup> 9 sin *<sup>x</sup>*<sup>2</sup> <sup>4</sup> *and*

$$\int\_0^1 \mathbf{x}(t)d\mu(t) = \frac{1}{3}\mathbf{x}(\frac{1}{4}) + \frac{1}{6}\mathbf{x}(\frac{1}{2})\dots$$

*We can get*

$$\int\_0^1 d\mu(t) = \frac{1}{2}.$$

*Take*

$$w\_0(t) = -2, \ w\_0(t) = 0,$$

*then,*

$$\begin{cases} \, \, \_2\upsilon\_0(t) = 0 < \frac{2}{9} + 9\sin 1 = f(t, \upsilon\_0(t)), \\\\ \, \, \upsilon\_0(0) = -2 < -1 = \int\_0^1 \upsilon\_0(t) d\mu(t), \end{cases}$$

*and*

$$\begin{cases} \,^\flat D\_{\frac{1}{2}} w\_0(t) = 0 > -\frac{2}{9} = f(t, w\_0(t)), \\\,^\flat w\_0(0) = 0 = \int\_0^1 w\_0(t) d\mu(t). \end{cases}$$

*Then v*0, *w*<sup>0</sup> *are lower and upper solutions of (10). When M*(*t*) = 1*, it is easy to verify that assumption* (*H*3) *holds. In addition,*

$$\int\_0^1 e^{-\int\_0^t s^{a-1} M(s) ds} d\mu(t) = \int\_0^1 e^{-t\frac{1}{2}} d\mu(t) < \int\_0^1 d\mu(t) = \frac{1}{2} < 1.$$

*By Theorem 2, problem (10) has an extremal iterative solution in* [*v*0, *w*0]*.*
