**2. Preliminaries**

In this section, we briefly show some necessary definitions and results which will be used in our main results.

**Definition 1.** *[24] Let <sup>f</sup>* : [0, <sup>+</sup>∞) <sup>→</sup> <sup>R</sup> *and <sup>t</sup>* <sup>&</sup>gt; <sup>0</sup>*. The conformable fractional derivative of order* <sup>0</sup> <sup>&</sup>lt; *<sup>α</sup>* <sup>≤</sup> <sup>1</sup> *is defined by*

$$D\_{\alpha}f(t) = \lim\_{\rho \to 0} \frac{f(t + \rho t^{1-\alpha}) - f(t)}{\rho}.$$

*for t* > 0 *and the conformable fractional derivative at* 0 *is defined as D<sup>α</sup> f*(0) = lim *<sup>t</sup>*→0<sup>+</sup>(*D<sup>α</sup> <sup>f</sup>*)(*t*)*. If <sup>f</sup> is differentiable then D<sup>α</sup> f*(*t*) = *t* <sup>1</sup>−*<sup>α</sup> f* (*t*)*.*

**Definition 2.** *[24] Let <sup>α</sup>* <sup>∈</sup> (0, 1]*. The conformable fractional integral of a function <sup>f</sup>* : [0, <sup>+</sup>∞) <sup>→</sup> <sup>R</sup> *of order α is denoted by I<sup>α</sup> f*(*t*) *and is defined as*

$$I\_{\alpha}f(t) = \int\_{0}^{t} s^{\alpha - 1} f(s) ds.$$

**Lemma 1.** *[25] Let f* : (0, <sup>+</sup>∞) <sup>→</sup> <sup>R</sup> *be differentiable and* <sup>0</sup> <sup>&</sup>lt; *<sup>α</sup>* <sup>≤</sup> <sup>1</sup>*. Then, for all t* <sup>&</sup>gt; <sup>0</sup> *we have*

$$I\_a D\_a f(t) = f(t) - f(0).$$

**Lemma 2.** *[24] Let <sup>α</sup>* <sup>∈</sup> (0, 1]*, l*1, *<sup>l</sup>*2, *<sup>q</sup>*, *<sup>k</sup>* <sup>∈</sup> <sup>R</sup>*, and the functions f , h be <sup>α</sup>-differentiable on* [0, <sup>+</sup>∞)*. Then*

*(i) Dαk* = 0 *for all constant functions f*(*t*) = *k; (ii) Dα*(*l*<sup>1</sup> *f* + *l*<sup>2</sup> *f*) = *l*1*D<sup>α</sup> f*(*t*) + *l*2*Dαh*(*t*)*; (iii) Dαt <sup>q</sup>* = *qtq*−*α;*

*(iv) Dα*(*f h*) = *f*(*t*)*Dαh*(*t*) + *h*(*t*)*D<sup>α</sup> f*(*t*)*; (v) Dα*( *<sup>f</sup> <sup>h</sup>* ) = *hD<sup>α</sup> <sup>f</sup>*−*f Dα<sup>h</sup> <sup>h</sup>*<sup>2</sup> *when h*(*t*) = <sup>0</sup>*.*

**Theorem 1.** *[24] (Mean value theorem) Let* [*a*, *<sup>b</sup>*] <sup>⊂</sup> [0, <sup>+</sup>∞)*, and let f* : [0, <sup>+</sup>∞) <sup>→</sup> <sup>R</sup>*. Suppose that*


*Then there exists a constant <sup>ξ</sup>* <sup>∈</sup> (*a*, *<sup>b</sup>*)*, such that D<sup>α</sup> <sup>f</sup>*(*ξ*) = *<sup>f</sup>*(*b*)−*f*(*a*) <sup>1</sup> *<sup>α</sup> <sup>b</sup>α*<sup>−</sup> <sup>1</sup> *<sup>α</sup> <sup>a</sup><sup>α</sup> .*

**Definition 3.** *A function u* <sup>∈</sup> *<sup>C</sup>*([0, 1], <sup>R</sup>) *is known as a lower solution of (1), if it satisfies*

$$D\_{\alpha}u(t) \le f(t, u(t)), t \in [0, 1]. \tag{2}$$

$$u(0) \le \int\_{0}^{1} u(t)d\mu(t). \tag{3}$$

*If inequalities (2), (3) are reversed, then u is an upper solution of problem (1).*

Next, we present the following existence and uniqueness results for linear equations.

**Lemma 3.** *Let* <sup>0</sup> <sup>&</sup>lt; *<sup>α</sup>* <sup>≤</sup> <sup>1</sup>*, <sup>a</sup>* <sup>∈</sup> <sup>R</sup> *and <sup>M</sup>*, *<sup>N</sup>* <sup>∈</sup> *<sup>C</sup>*([0, 1], <sup>R</sup>)*. Then linear fractional differential equation involving integral boundary problem:*

$$\begin{cases} \begin{aligned} D\_t u(t) &= -M(t)u(t) + N(t), \ t \in [0, 1], \\ u(0) &= \int\_0^1 u(t) d\mu(t) + a \end{aligned} \end{cases} \tag{4}$$

*has a unique solution provided <sup>α</sup>* <sup>=</sup> <sup>1</sup> <sup>−</sup> - 1 <sup>0</sup> *<sup>e</sup>*<sup>−</sup> *t* <sup>0</sup> *<sup>s</sup>α*−1*M*(*s*)*dsdμ*(*t*) <sup>=</sup> <sup>0</sup>*.*

**Proof.** Multiplying both sides of the first equation of the problem (4) by *e t* <sup>0</sup> *<sup>s</sup>α*−1*M*(*s*)*ds* and using Lemma 2, we can get

$$e^{\int\_0^t s^{a-1} M(s)ds} D\_a u(t) + M(t)u(t)e^{\int\_0^t s^{a-1} M(s)ds} = N(t)e^{\int\_0^t s^{a-1} M(s)ds}.$$

In other words, by means of the product rule (item (*iv*) of Lemma 2), the above equality turns to

$$D\_a[e^{\int\_0^t s^{a-1} M(s)ds} u(t)] = N(t)e^{\int\_0^t s^{a-1} M(s)ds}.\tag{5}$$

Applying the conformable fractional integral of order *α* to both side of (5), we have

$$\begin{array}{rcl} e^{\int\_0^t s^{a-1} M(s)ds} u(t) - u(0) &=& I\_a[N(t)e^{\int\_0^t s^{a-1} M(s)ds}] \\ &=& \int\_0^t s^{a-1} N(s)e^{\int\_0^s \tau^{a-1} M(\tau)d\tau} ds. \end{array}$$

Then

$$u(t) = e^{-\int\_0^t s^{a-1} M(s)ds} [u(0) + \int\_0^t s^{a-1} N(s)e^{\int\_0^s \tau^{a-1} M(\tau)d\tau} ds].\tag{6}$$

From the boundary condition of (4), we have

$$\begin{aligned} & \quad \left(1 - \int\_0^1 e^{-\int\_0^t s^{a-1} M(s) ds} d\mu(t)\right) u(0) \\ &= \int\_0^1 e^{-\int\_0^t s^{a-1} M(s) ds} \int\_0^t s^{a-1} N(s) e^{\int\_0^s \tau^{a-1} M(\tau) d\tau} ds d\mu(t) + a. \end{aligned}$$

On account of condition *<sup>α</sup>* = 0, then

$$\mu(0) = \frac{\int\_0^1 e^{-\int\_0^t s^{a-1} M(s)ds} \int\_0^t s^{a-1} N(s) e^{\int\_0^s \tau^{a-1} M(\tau)d\tau} ds d\mu(t) + a}{1 - \int\_0^1 e^{-\int\_0^t s^{a-1} M(s)ds} d\mu(t)}$$

,

thus problem (4) has a unique solution. The proof is finished.

In the next Lemma, we discuss comparison results for the linear problem which play a key role in the proof of the main result.

**Lemma 4.** *Let* <sup>0</sup> <sup>&</sup>lt; *<sup>α</sup>* <sup>≤</sup> <sup>1</sup>*. Suppose that M*, *<sup>u</sup>* <sup>∈</sup> *<sup>C</sup>*([0, 1], <sup>R</sup>) *satisfies*

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

*Then u*(*t*) ≤ 0 *on* [0, 1] *provided <sup>α</sup>* > 0*.*

**Proof.** Let *<sup>N</sup>*(*t*) = *<sup>D</sup>αu*(*t*) + *<sup>M</sup>*(*t*)*u*(*t*) and *<sup>a</sup>* <sup>=</sup> *<sup>u</sup>*(0) <sup>−</sup> - 1 <sup>0</sup> *u*(*t*)*dμ*(*t*), we know that *N*(*t*) ≤ 0, *a* ≤ 0 and

$$\begin{cases} \begin{aligned} D\_a u(t) &= -M(t)u(t) + N(t), \; t \in [0, 1], \\ u(0) &= \int\_0^1 u(t) d\mu(t) + a \end{aligned} \end{cases}$$

Using *<sup>α</sup>* > 0, we have

$$\mu(0) = \frac{\int\_0^1 \mathfrak{e}^{-\int\_0^t s^{a-1} M(s)ds} \int\_0^t s^{a-1} N(s) \mathfrak{e}^{\int\_0^s \tau^{a-1} M(\tau)d\tau} ds d\mu(t) + a}{1 - \int\_0^1 \mathfrak{e}^{-\int\_0^t s^{a-1} M(s)ds} d\mu(t)} \le 0.$$

Then by (6), we can conclude that

$$
\mu(t) \le e^{-\int\_0^t s^{n-1} M(s) ds} \mu(0) \le 0.
$$

The proof is complete.
