**2. Preliminaries**

Let us first outline some preliminary concepts of fractional calculus [5].

**Definition 1.** *The Caputo derivative for a function h* ∈ *ACn*[*<sup>a</sup>*, *b*] *of fractional order q* ∈ (*n* − 1, *<sup>n</sup>*], *n* ∈ N, *existing almost everywhere on* [*a*, *b*], *is defined by*

$$\, ^cD^q h(t) = \frac{1}{\Gamma(n-q)} \int\_a^t (t-u)^{n-q-1} h^{(n)}(u) du, \quad t \in [a,b].$$

**Definition 2.** *The Riemann–Liouville fractional integral for a function h* ∈ *<sup>L</sup>*1[*<sup>a</sup>*, *b*] *of order r* > 0*, which exists almost everywhere on* [*a*, *b*], *is defined by*

$$I^r h(t) = \frac{1}{\Gamma(r)} \int\_a^t \frac{h(u)}{(t-u)^{1-r}} du, \ t \in [a, b].$$

**Lemma 1.** *For m* − 1 < *r* ≤ *m, the general solution of the fractional differential equation cDrx*(*t*) = 0 *can be written as*

$$\mathbf{x}(t) = b\_0 + b\_1 t + b\_2 t^2 + \dots + b\_{m-1} t^{m-1},\tag{2}$$

*where bi* ∈ R, *i* = 0, 1, 2, . . . , *m* − 1.

> It follows by Lemma 1 that

$$I^{r \in} D^r \mathbf{x}(t) = \mathbf{x}(t) + b\_0 + b\_1 t + b\_2 t^2 + \dots + b\_{m-1} t^{m-1},\tag{3}$$

for some *bi* ∈ R, *i* = 0, 1, 2, . . . , *m* − 1, are arbitrary constants.

The following lemma deals with the linear version of the problem (1).

**Lemma 2.** *For a given function h* ∈ *C*([0, 1], R) *the unique solution of the following boundary value problem*

$$\begin{cases} \,^cD^q \mathbf{x}(t) = h(t), \; 0 < t < 1, \\\ x(0) = a, \; \mathbf{x}'(0) = 0, \; \mathbf{x}''(0) = 0, \dots, \mathbf{x}^{(m-2)}(0) = 0, \\\ ax(1) + \beta \mathbf{x}'(1) = \gamma\_1 \int\_0^\zeta \mathbf{x}(s) ds + \sum\_{j=1}^p a\_j \, \mathbf{x}(\eta\_j) + \gamma\_2 \int\_\zeta^1 \mathbf{x}(s) ds, \end{cases} \tag{4}$$

*is given by*

$$\begin{split} \mathbf{x}(t) &= \int\_{0}^{t} \frac{(t-s)^{q-1}}{\Gamma(q)} h(s)ds + a - \frac{t^{m-1}}{\Lambda\_{1}} \left[ a \int\_{0}^{1} \frac{(1-s)^{q-1}}{\Gamma(q)} h(s)ds + \beta \int\_{0}^{1} \frac{(1-s)^{q-2}}{\Gamma(q-1)} h(s)ds \right] \\ &- \gamma\_{1} \int\_{0}^{\tilde{\zeta}} \int\_{0}^{s} \frac{(s-u)^{q-1}}{\Gamma(q)} h(u)duds - \sum\_{j=1}^{p} a\_{j} \int\_{0}^{\eta\_{j}} \frac{(\eta\_{j}-s)^{q-1}}{\Gamma(q)} h(s)ds \\ &- \gamma\_{2} \int\_{\tilde{\zeta}}^{1} \int\_{0}^{s} \frac{(s-u)^{q-1}}{\Gamma(q)} h(u)duds + \Lambda\_{2} \right], \end{split} \tag{5}$$

*where it is assumed that*

$$
\Lambda\_1 = \left( a + \beta (m - 1) - \gamma\_1 \left( \frac{\zeta^m}{m} \right) - \sum\_{j=1}^p a\_j \eta\_j^{m-1} - \gamma\_2 \left( \frac{1 - \zeta^m}{m} \right) \right) \neq 0,\tag{6}
$$

$$
\Lambda\_2 = a \left[ a - \gamma\_1 \mathbb{J} - \sum\_{j=1}^p a\_j - \gamma\_2 (1 - \mathbb{J}) \right]. \tag{7}
$$

**Proof.** Using (3), we can write the general solution of the fractional differential equation in (4) as

$$\mathbf{x}(t) = \int\_0^t \frac{(t-s)^{q-1}}{\Gamma(q)} h(s)ds - c\_0 - c\_1t - c\_2t^2 - \dots - c\_{m-1}t^{m-1},\tag{8}$$

where *c*0, *c*1, *c*2,..., *cm*−1 ∈ R are arbitrary constants. From (8), we have

$$\mathbf{x}'(t) = \int\_0^t \frac{(t-s)^{q-2}}{\Gamma(q-1)} h(s)ds - c\_1 - 2c\_2t - \dots - (m-1)c\_{m-1}t^{m-2},$$

$$\mathbf{x}''(t) = \int\_0^t \frac{(t-s)^{q-3}}{\Gamma(q-2)} h(s)ds - 2c\_2 - \dots - (m-1)(m-2)c\_{m-1}t^{m-3},\tag{9}$$

$$\vdots$$

*<sup>x</sup>*(*s*)*ds*, we ge<sup>t</sup>

Applying the conditions *x*(0) = *a*, *x* (0) = 0, . . . , *x*(*<sup>m</sup>*−<sup>2</sup>)(0) = 0 in (8), it is found that *c*0 = <sup>−</sup>*a*, *c*1 = 0, . . . , *cm*−2 = 0. Then (8) becomes

$$\mathbf{x}(t) = \int\_0^t \frac{(t-s)^{q-1}}{\Gamma(q)} h(s)ds + a - c\_{m-1}t^{m-1},\tag{10}$$

and

+ *γ*2

$$\mathbf{x}'(t) = \int\_0^t \frac{(t-s)^{q-2}}{\Gamma(q-1)} h(s)ds - (m-1)\mathbf{c}\_{m-1}t^{m-2}.\tag{11}$$

Combining (10) and (11) with the condition *αx*(1) + *βx* (1) = *γ*1 *ζ*0 *x*(*s*)*ds* + *p*∑*j*=1 *αj <sup>x</sup>*(*ηj*) 1

$$\begin{split} &\quad \alpha \left[ \int\_{0}^{1} \frac{(1-s)^{q-1}}{\Gamma(q)} h(s) ds + a - c\_{m-1} \right] + \beta \left[ \int\_{0}^{1} \frac{(1-s)^{q-2}}{\Gamma(q-1)} h(s) ds - (m-1)c\_{m-1} \right] \\ &= \quad \gamma\_{1} \int\_{0}^{\zeta} \left[ \int\_{0}^{s} \frac{(s-u)^{q-1}}{\Gamma(q)} h(u) du + a - c\_{m-1} s^{m-1} \right] ds + \sum\_{j=1}^{p} a\_{j} \left[ \int\_{0}^{\eta\_{j}} \frac{(\eta\_{j}-s)^{q-1}}{\Gamma(q)} h(s) ds \right] \\ &\quad + a - c\_{m-1} \eta\_{j}^{m-1} \Big] + \gamma\_{2} \int\_{\zeta}^{1} \left[ \int\_{0}^{s} \frac{(s-u)^{q-1}}{\Gamma(q)} h(u) du + a - c\_{m-1} s^{m-1} \right] ds, \end{split}$$

which, together with (6) and (7), yields

$$\begin{split} c\_{m-1} &= \quad \frac{1}{\Lambda\_1} \Big[ \mathfrak{a} \int\_0^1 \frac{(1-s)^{q-1}}{\Gamma(q)} h(s) ds + \beta \int\_0^1 \frac{(1-s)^{q-2}}{\Gamma(q-1)} h(s) ds \\ &- \gamma\_1 \int\_0^\zeta \int\_0^s \frac{(s-u)^{q-1}}{\Gamma(q)} h(u) du ds - \sum\_{j=1}^p a\_j \int\_0^{\eta\_j} \frac{(\eta\_j - s)^{q-1}}{\Gamma(q)} h(s) ds \\ &- \gamma\_2 \int\_\xi^1 \int\_0^s \frac{(s-u)^{q-1}}{\Gamma(q)} h(u) du ds + a\kappa - a\gamma\_1 \zeta - a \sum\_{j=1}^p a\_j - a\gamma\_2 (1-\xi) \Big]. \end{split}$$

Substituting the value of *cm*−1 in (10), we obtain

$$\begin{split} &\quad x(t) \\ &= \quad \int\_{0}^{t} \frac{(t-s)^{q-1}}{\Gamma(q)} h(s) ds + a - \frac{t^{m-1}}{\Lambda\_{1}} \Big[ a \int\_{0}^{1} \frac{(1-s)^{q-1}}{\Gamma(q)} h(s) ds + \beta \int\_{0}^{1} \frac{(1-s)^{q-2}}{\Gamma(q-1)} h(s) ds \\ &\quad - \gamma\_{1} \int\_{0}^{\tilde{\zeta}} \int\_{0}^{s} \frac{(s-u)^{q-1}}{\Gamma(q)} h(u) du ds - \sum\_{j=1}^{p} a\_{j} \int\_{0}^{\eta\_{j}} \frac{(\eta\_{j}-s)^{q-1}}{\Gamma(q)} h(s) ds \\ &\quad - \gamma\_{2} \int\_{\tilde{\zeta}}^{1} \int\_{0}^{s} \frac{(s-u)^{q-1}}{\Gamma(q)} h(u) du ds + \Lambda\_{2} \Big]. \end{split}$$

We can obtain the converse of this lemma by direct computation. This finishes the proof.

Using Lemma 2, we can transform problem (1) into a fixed point problem as *x* = F*<sup>x</sup>*, where the operator F : C→C is defined by

$$\mathcal{F}\mathbf{x}(t) \;=\int\_0^t \frac{(t-s)^{q-1}}{\Gamma(q)} f(s, \mathbf{x}(s)) ds - \sum\_{i=1}^k \int\_0^t \frac{(t-s)^{q+p\_i-1}}{\Gamma(q+p\_i)} g\_i(s, \mathbf{x}(s)) ds + a\_i$$

$$\begin{split} & -\frac{t^{m-1}}{\Gamma\_{0}} \Big[ \begin{split} & \int\_{0}^{1} \left( \frac{(1-s)^{q-1}}{\Gamma(q)} f(s, \mathbf{x}(s)) - \sum\_{i=1}^{k} \frac{(1-s)^{q+p\_{i}-1}}{\Gamma(q+p\_{i})} g\_{i}(s, \mathbf{x}(s)) \right) ds \\ & + \beta \int\_{0}^{1} \left( \frac{(1-s)^{q-2}}{\Gamma(q-1)} f(s, \mathbf{x}(s)) - \sum\_{i=1}^{k} \frac{(1-s)^{q+p\_{i}-2}}{\Gamma(q+p\_{i}-1)} g\_{i}(s, \mathbf{x}(s)) \right) ds \\ & - \gamma\_{1} \int\_{0}^{\zeta} \int\_{0}^{s} \frac{(s-u)^{q-1}}{\Gamma(q)} f(u, \mathbf{x}(u)) \, du \, ds \\ & + \gamma\_{1} \sum\_{i=1}^{k} \int\_{0}^{\zeta} \int\_{0}^{s} \frac{(s-u)^{q-1}}{\Gamma(q)} \int\_{0}^{u} \frac{(u-w)^{p\_{i}-1}}{\Gamma(p\_{i})} g\_{i}(w, \mathbf{x}(w)) dw \, du \, ds \\ & - \sum\_{j=1}^{p} x\_{j} \left( \int\_{0}^{\eta\_{i}} \frac{(\eta\_{j}-s)^{q-1}}{\Gamma(q)} f(s, \mathbf{x}(s)) ds - \sum\_{i=1}^{k} \int\_{0}^{\eta\_{i}} \frac{(\eta\_{j}-s)^{q+p\_{i}-1}}{\Gamma(q+p\_{i})} g\_{i}(s, \mathbf{x}(s)) ds \right) \\ & - \gamma\_{2} \int\_{\zeta}^{1} \int\_{0}^{s} \frac{(s-u)^{q-1}}{\Gamma(q$$

Here C, represents the Banach space of all continuous functions *x* : [0, 1] → R equipped with the norm *x* = sup*<sup>t</sup>*∈[0,1] |*x*(*t*)|.

By a solution of (1), we mean a function *x* ∈ C of class C*m*[0, 1] satisfying the nonlocal integro-multipoint boundary value problem (1).

For computational convenience, we set

$$\begin{aligned} \Omega &= \begin{bmatrix} \frac{1}{\Gamma(q+1)} + \frac{1}{|\Lambda\_1|} \left( \frac{|a|}{\Gamma(q+1)} + \frac{|\beta|}{\Gamma(q)} + \frac{|\gamma\_1|\xi^{q+1}}{\Gamma(q+2)} + \frac{\sum\_{j=1}^p |a\_j| \eta\_j^q}{\Gamma(q+1)} \right) \\ &+ \frac{|\gamma\_2|(|1-\xi^{q+1}| )}{\Gamma(q+2)} \end{bmatrix}, &\leq \\ \begin{bmatrix} \frac{1}{\Gamma(q+1)} & \frac{1}{\Gamma(q+2)} & |\beta| & |\gamma\_2|\gamma^q + p\_1 + 1 \end{bmatrix} \end{aligned} \tag{13}$$

$$\Omega\_i = \begin{cases} \frac{1}{\Gamma(q+p\_i+1)} + \frac{1}{|\Lambda\_1|} \left( \frac{|a|}{\Gamma(q+p\_i+1)} + \frac{|\beta|}{\Gamma(q+p\_i)} + \frac{|\gamma\_1| \zeta^{q+p\_i+1}}{\Gamma(q+p\_i+2)} \right) \\ + \frac{\nu}{\Gamma(q+p\_i+1)} + \frac{|\gamma\_2|(1-\xi^{q+p\_i+1})}{\Gamma(q+p\_i+2)} \end{cases}, \quad i=1,2,\ldots,k. \tag{14}$$

#### **3. Existence and Uniqueness Results**

In the following theorem, we make use of Banach's fixed point theorem.

**Theorem 1.** *Let f* , *gi* : [0, 1] × R → R *be continuous functions and let there exist constants L*, *Li* > 0*,* (*i* = 1, . . . , *k*) *such that:*

(*<sup>A</sup>*1) | *f*(*<sup>t</sup>*, *x*) − *f*(*<sup>t</sup>*, *y*)| ≤ *L* |*x* − *y*|*, and* |*gi*(*<sup>t</sup>*, *x*) − *gi*(*<sup>t</sup>*, *y*)| ≤ *Li* |*x* − *y*| *for all t* ∈ [0, 1]*, x*, *y* ∈ R*.*

*Then, the boundary value problem (1) has a unique solution on* [0, 1] *if*

$$L\Omega + \sum\_{i=1}^{k} L\_i \Omega\_i \,<\, 1\,,\tag{15}$$

*where* Ω, Ω*i*, *i* = 1, 2, . . . , *k are given by (13) and (14), respectively.*

**Proof.** The proof will be given in two steps. **Step 1.** We show that F*Br* ⊂ *Br*, where *Br* = {*x* ∈ C : *x* ≤ *r*} with *r* ≥ *M* Ω + *k* ∑ *i*=1 *Mi*Ω*i* + |*a*| + (Λ2/Λ1)4 1 − (*L*Ω + *k* ∑ *i*=1 *Li*Ω*i*) and *M*, *Mi* are positive numbers such that *M* = sup *<sup>t</sup>*∈[0,1] | *f*(*<sup>t</sup>*, 0)| and *Mi* = sup *<sup>t</sup>*∈[0,1] |*gi*(*<sup>t</sup>*, 0)| , *i* = 1, 2, . . . , *k*. For*x*∈*Br*and*t*∈[0,1],itfollowsby(*<sup>A</sup>*1)that

$$|f(t, \mathbf{x}(t))| \le |f(t, \mathbf{x}(t)) - f(t, 0)| + |f(t, 0)| \le L \|\mathbf{x}\| + M \le Lr + M. \tag{16}$$

In a similar manner, we have |*gi*(*<sup>t</sup>*, *x*(*t*))| ≤ *Li r* + *Mi*, *i* = 1, 2, . . . , *k*. Then

F*x* ≤ sup *<sup>t</sup>*∈[0,1] - *t* 0 (*t* − *s*)*<sup>q</sup>*−<sup>1</sup> <sup>Γ</sup>(*q*) | *f*(*<sup>s</sup>*, *x*(*s*))|*ds* + *k* ∑ *i*=1 *t* 0 (*t* − *s*)*q*+*pi*−<sup>1</sup> Γ (*q* + *pi*) |*gi*(*<sup>s</sup>*, *x*(*s*))|*ds* + |*a*| +*t <sup>m</sup>*−1 |<sup>Λ</sup>1| & |*α*| 1 0 (1 − *s*)*<sup>q</sup>*−<sup>1</sup> <sup>Γ</sup>(*q*) | *f*(*<sup>s</sup>*, *x*(*s*))| + *k* ∑ *i*=1 (1 − *s*)*q*+*pi*−<sup>1</sup> Γ (*q* + *pi*) |*gi*(*<sup>s</sup>*, *x*(*s*))| *ds* +|*β*| 1 0 (1 − *s*)*<sup>q</sup>*−<sup>2</sup> <sup>Γ</sup>(*q* − 1) | *f*(*<sup>s</sup>*, *x*(*s*))| + *k* ∑ *i*=1 (1 − *s*)*q*+*pi*−<sup>2</sup> <sup>Γ</sup>(*q* + *pi* − 1) |*gi*(*<sup>s</sup>*, *x*(*s*))| *ds* <sup>+</sup>|*<sup>γ</sup>*1| *ζ* 0 *s* 0 (*s* − *u*)*<sup>q</sup>*−<sup>1</sup> <sup>Γ</sup>(*q*) | *f*(*<sup>u</sup>*, *x*(*u*))| *du ds* <sup>+</sup>|*<sup>γ</sup>*1| *k* ∑ *i*=1 *ζ* 0 *s* 0 (*s* − *u*)*<sup>q</sup>*−<sup>1</sup> <sup>Γ</sup>(*q*) *u* 0 (*u* − *w*)*pi*−<sup>1</sup> Γ (*pi*) |*gi*(*<sup>w</sup>*, *x*(*w*))| *dw du ds* + *p* ∑ *j*=1 |*<sup>α</sup>j*| *ηj* 0 (*ηj* − *s*)*<sup>q</sup>*−<sup>1</sup> <sup>Γ</sup>(*q*) | *f*(*<sup>s</sup>*, *x*(*s*))|*ds* + *k* ∑ *i*=1 *ηj* 0 (*ηj* − *s*)*q*+*pi*−<sup>1</sup> <sup>Γ</sup>(*q* + *pi*) |*gi*(*<sup>s</sup>*, *x*(*s*))|*ds* <sup>+</sup>|*<sup>γ</sup>*2| 1 *ξ s* 0 (*s* − *u*)*<sup>q</sup>*−<sup>1</sup> <sup>Γ</sup>(*q*) | *f*(*<sup>u</sup>*, *x*(*u*))| *du ds* <sup>+</sup>|*<sup>γ</sup>*2| *k* ∑ *i*=1 1 *ξ s* 0 (*s* − *u*)*<sup>q</sup>*−<sup>1</sup> <sup>Γ</sup>(*q*) *u* 0 (*u* − *w*)*pi*−<sup>1</sup> Γ (*pi*) |*gi*(*<sup>w</sup>*, *x*(*w*))| *dw du ds* + <sup>|</sup><sup>Λ</sup>2<sup>|</sup>' ≤ (*Lr* + *M*) ⎡ ⎢ ⎢ ⎢ ⎣ sup *<sup>t</sup>*∈[0,1] ⎧ ⎪⎪⎪⎨ ⎪⎪⎪⎩ *t q* <sup>Γ</sup>(*q* + 1) + *t <sup>m</sup>*−1 |<sup>Λ</sup>1| ⎛ ⎜⎜⎜⎝ |*α*| <sup>Γ</sup>(*q* + 1) + |*β*| <sup>Γ</sup>(*q*) + |*<sup>γ</sup>*1|*ζ<sup>q</sup>*+<sup>1</sup> <sup>Γ</sup>(*q* + 2) + *p* ∑ *j*=1 |*<sup>α</sup>j*|*η<sup>q</sup> j* <sup>Γ</sup>(*q* + 1) +|*<sup>γ</sup>*2|(<sup>1</sup> − *ξq*+<sup>1</sup>) <sup>Γ</sup>(*q* + 2) ! + *k* ∑ *i*=1 (*Lir* + *Mi*) *t q*+*pi* <sup>Γ</sup>(*q* + *pi* + 1) + *t <sup>m</sup>*−1 |<sup>Λ</sup>1| |*α*| <sup>Γ</sup>(*q* + *pi* + 1) + |*β*| <sup>Γ</sup>(*q* + *pi*) + |*<sup>γ</sup>*1|*ζ<sup>q</sup>*+*pi*+<sup>1</sup> <sup>Γ</sup>(*q* + *pi* + 2) + *p* ∑ *j*=1 |*<sup>α</sup>i*|*ηq*+*pi j* <sup>Γ</sup>(*q* + *pi* + 1) + |*<sup>γ</sup>*2|(<sup>1</sup> − *ξq*+*pi*+<sup>1</sup>) <sup>Γ</sup>(*q* + *pi* + 2) ⎞ ⎟⎟⎟⎠ ⎤ ⎥ ⎥ ⎥ ⎦ <sup>+</sup>|*a*| + *t <sup>m</sup>*−<sup>1</sup>|<sup>Λ</sup>2| |<sup>Λ</sup>1| , = (*Lr* + *M*)Ω + *k* ∑ *i*=1 (*Lir* + *Mi*)<sup>Ω</sup>*i* + |*a*| + Λ2 Λ1 = *L*Ω + *k* ∑ *i*=1*Li*Ω*i r* + *M* Ω + *k* ∑ *i*=1*Mi*Ω*i* + |*a*| + Λ2 Λ1 ≤ *r*,

which shows that F*Br* ⊂ *Br*.

> **Step 2.** We show that F is a contraction. For *x*, *y* ∈ C and for each *t* ∈ [0, 1], we obtain

F*x* − F*y* ≤ sup *<sup>t</sup>*∈[0,1] + *t* 0 (*t* − *s*)*<sup>q</sup>*−<sup>1</sup> <sup>Γ</sup>(*q*) | *f*(*<sup>s</sup>*, *x*(*s*)) − *f*(*<sup>s</sup>*, *y*(*s*))|*ds* + *k* ∑ *i*=1 *t* 0 (*t* − *s*)*q*+*pi*−<sup>1</sup> Γ (*q* + *pi*) |*gi*(*<sup>s</sup>*, *x*(*s*)) − *gi*(*<sup>s</sup>*, *y*(*s*))|*ds* +*t <sup>m</sup>*−1 |<sup>Λ</sup>1| |*α*| 1 0 (1 − *s*)*<sup>q</sup>*−<sup>1</sup> <sup>Γ</sup>(*q*) | *f*(*<sup>s</sup>*, *x*(*s*)) − *f*(*<sup>s</sup>*, *y*(*s*))| + *k* ∑ *i*=1 (1 − *s*)*q*+*pi*−<sup>1</sup> <sup>Γ</sup>(*q* + *pi*) |*gi*(*<sup>s</sup>*, *x*(*s*)) − *gi*(*<sup>s</sup>*, *y*(*s*))| *ds* +|*β*| 1 0 (1 − *s*)*<sup>q</sup>*−<sup>2</sup> <sup>Γ</sup>(*q* − 1) | *f*(*<sup>s</sup>*, *x*(*s*)) − *f*(*<sup>s</sup>*, *y*(*s*))| + *k* ∑ *i*=1 (1 − *s*)*q*+*pi*−<sup>2</sup> <sup>Γ</sup>(*q* + *pi* − 1) |*gi*(*<sup>s</sup>*, *x*(*s*)) − *gi*(*<sup>s</sup>*, *y*(*s*))| *ds* <sup>+</sup>|*<sup>γ</sup>*1| *ζ* 0 *s* 0 (*s* − *u*)*<sup>q</sup>*−<sup>1</sup> <sup>Γ</sup>(*q*) | *f*(*<sup>u</sup>*, *x*(*u*)) − *f*(*<sup>u</sup>*, *y*(*u*))| *du ds* <sup>+</sup>|*<sup>γ</sup>*1| *k* ∑ *i*=1 *ζ* 0 *s* 0 (*s* − *u*)*<sup>q</sup>*−<sup>1</sup> <sup>Γ</sup>(*q*) *u* 0 (*u* − *w*)*pi*−<sup>1</sup> Γ (*pi*) |*gi*(*<sup>w</sup>*, *x*(*w*)) − *gi*(*<sup>w</sup>*, *y*(*w*))|*dwduds* + *p* ∑ *j*=1 |*<sup>α</sup>j*| *ηj* 0 (*ηj* − *s*)*<sup>q</sup>*−<sup>1</sup> <sup>Γ</sup>(*q*) | *f*(*<sup>s</sup>*, *x*(*s*)) − *f*(*<sup>s</sup>*, *y*(*s*))|*ds* + *k* ∑ *i*=1 *ηi* 0 (*ηi* − *s*)*q*+*pi*−<sup>1</sup> <sup>Γ</sup>(*q* + *pi*) |*gi*(*<sup>s</sup>*, *x*(*s*)) − *gi*(*<sup>x</sup>*, *y*(*s*))|*ds* <sup>+</sup>|*<sup>γ</sup>*2| 1 *ξ s* 0 (*s* − *u*)*<sup>q</sup>*−<sup>1</sup> <sup>Γ</sup>(*q*) | *f*(*<sup>u</sup>*, *x*(*u*)) − *f*(*<sup>u</sup>*, *y*(*u*))| *du ds* <sup>+</sup>|*<sup>γ</sup>*2| *k* ∑ *i*=1 1 *ξ s* 0 (*s* − *u*)*<sup>q</sup>*−<sup>1</sup> <sup>Γ</sup>(*q*) *u* 0 (*u* − *w*)*pi*−<sup>1</sup> Γ (*pi*) |*gi*(*<sup>w</sup>*, *x*(*w*)) − *gi*(*<sup>w</sup>*, *y*(*w*))| *dw du ds*' ≤ *L* ⎡ ⎢ ⎢ ⎢ ⎣ sup *<sup>t</sup>*∈[0,1] ⎧ ⎪⎪⎪⎨ ⎪⎪⎪⎩ *t q* <sup>Γ</sup>(*q* + 1) + *t <sup>m</sup>*−1 |<sup>Λ</sup>1| ⎛ ⎜⎜⎜⎝ |*α*| <sup>Γ</sup>(*q* + 1) + |*β*| <sup>Γ</sup>(*q*) + |*<sup>γ</sup>*1|*ζ<sup>q</sup>*+<sup>1</sup> <sup>Γ</sup>(*q* + 2) + *p* ∑ *j*=1 |*<sup>α</sup>j*|*η<sup>q</sup> j* <sup>Γ</sup>(*q* + 1) +|*<sup>γ</sup>*2|(<sup>1</sup> − *ξq*+<sup>1</sup>) <sup>Γ</sup>(*q* + 2) ,! *x* − *y* + *k* ∑ *i*=1 *Li t q*+*pi* <sup>Γ</sup>(*q* + *pi* + 1) + *t <sup>m</sup>*−1 |<sup>Λ</sup>1| |*α*| <sup>Γ</sup>(*q* + *pi* + 1) + |*β*| <sup>Γ</sup>(*q* + *pi*) + |*<sup>γ</sup>*1|*ζ<sup>q</sup>*+*pi*+<sup>1</sup> <sup>Γ</sup>(*q* + *pi* + 2) + *p* ∑ *j*=1 |*<sup>α</sup>j*|*ηq*+*pi j* <sup>Γ</sup>(*q* + *pi* + 1) + |*<sup>γ</sup>*2|(<sup>1</sup> − *ξq*+*pi*+<sup>1</sup>) <sup>Γ</sup>(*q* + *pi* + 2) ⎞ ⎟⎟⎟⎠ ⎤ ⎥ ⎥ ⎥ ⎦ *x* − *y* ≤ *L*Ω + *k* ∑ *i*=1*Li*Ω*i x* − *y*,

which, by the condition (15), implies that F is a contraction. Thus the conclusion of the Banach contraction mapping principle applies and hence the operator F has a unique fixed point. Therefore, there exists a unique solution for the boundary value problem (1) on [0, 1].

Next, we prove an existence result for the boundary value problem (1), which relies on Krasnosel'ski ˘ i's fixed point theorem [46].

**Theorem 2.** *Let f* , *gi* : [0, 1] × R → R*,* (*i* = 1, ... , *k*) *be continuous functions satisfying the condition* (*<sup>A</sup>*1)*. In addition, we assume that:*

(*<sup>A</sup>*2) | *f*(*<sup>t</sup>*, *x*)| ≤ *μ*(*t*)*,* |*gi*(*<sup>t</sup>*, *x*)| ≤ *μi*(*t*)*, for all* (*t*, *x*) ∈ [0, 1] × R, *μ*, *μi* ∈ *C*([0, 1], <sup>R</sup>+)*.*

*Then, the boundary value problem (1) has at least one solution on* [0, 1]*, provided that*

$$L\left[\Omega - \left(\frac{1}{\Gamma(q+1)}\right)\right] + \sum\_{i=1}^{k} L\_i \left[\Omega\_i - \left(\frac{1}{\Gamma(q+p\_i+1)}\right)\right] < 1,\tag{17}$$

*where* Ω, Ω*i*, *i* = 1, 2, . . . , *k are given by (13) and (14), respectively.*

**Proof.** Consider *Bρ* = {*x* ∈ C : *x* ≤ *ρ*}, *μ* = sup *<sup>t</sup>*∈[0,1] |*μ*(*t*)|, *μi* = sup *<sup>t</sup>*∈[0,1] |*μi*(*t*)|, *i* = 1, 2, ... , *k* with *ρ* ≥ *μ*<sup>Ω</sup> + ∑*ki*=<sup>1</sup> *μi*<sup>Ω</sup>*i* + |*a*| + |<sup>Λ</sup>2|/|<sup>Λ</sup>1|. Then, we define the operators Φ and Ψ on *Bρ* as

<sup>Φ</sup>*x*(*t*) = *t* 0 (*t* − *s*)*<sup>q</sup>*−<sup>1</sup> <sup>Γ</sup>(*q*) *f*(*<sup>s</sup>*, *x*(*s*))*ds* + *k*∑*i*=1 *t* 0 (*t* − *s*)*q*+*pi*−<sup>1</sup> Γ (*q* + *pi*) *gi*(*<sup>s</sup>*, *<sup>x</sup>*(*s*))*ds*, *t* ∈ [0, 1], <sup>Ψ</sup>*x*(*t*) = *a* − *t<sup>m</sup>*−<sup>1</sup> Λ1 &*α* 10 (1 − *s*)*<sup>q</sup>*−<sup>1</sup> <sup>Γ</sup>(*q*) *f*(*<sup>s</sup>*, *x*(*s*)) − *k*∑*i*=1 (1 − *s*)*q*+*pi*−<sup>1</sup> Γ (*q* + *pi*) *gi*(*<sup>s</sup>*, *x*(*s*)) *ds* +*β* 10 (1 − *s*)*<sup>q</sup>*−<sup>2</sup> <sup>Γ</sup>(*q* − 1) *f*(*<sup>s</sup>*, *x*(*s*)) − *k*∑*i*=1 (1 − *s*)*q*+*pi*−<sup>2</sup> <sup>Γ</sup>(*q* + *pi* − 1) *gi*(*<sup>s</sup>*, *x*(*s*)) *ds* −*γ*1 *ζ*0 *s* 0 (*s* − *u*)*<sup>q</sup>*−<sup>1</sup> <sup>Γ</sup>(*q*) *f*(*<sup>u</sup>*, *x*(*u*)) *du ds* +*γ*1 *k* ∑ *i*=1 *ζ*0 *s* 0 (*s* − *u*)*<sup>q</sup>*−<sup>1</sup> <sup>Γ</sup>(*q*) *u*0 (*u* − *w*)*pi*−<sup>1</sup> Γ (*pi*) *gi*(*<sup>w</sup>*, *x*(*w*))*dw du ds* − *p* ∑ *j*=1 *αj ηj* 0 (*ηj* − *s*) <sup>Γ</sup>(*q*) *f*(*<sup>s</sup>*, *x*(*s*))*ds* − *k*∑*i*=1 *ηj* 0 (*ηj* − *s*)*q*+*pi*−<sup>1</sup> <sup>Γ</sup>(*q* + *pi*) *gi*(*<sup>s</sup>*, *x*(*s*))*ds* −*γ*2 1*ξ s* 0 (*s* − *u*)*<sup>q</sup>*−<sup>1</sup> <sup>Γ</sup>(*q*) *f*(*<sup>u</sup>*, *x*(*u*)) *du ds* +*γ*2 *k* ∑ *i*=1 1*ξ s* 0 (*s* − *u*)*<sup>q</sup>*−<sup>1</sup> <sup>Γ</sup>(*q*) *u*0 (*u* − *w*)*pi*−<sup>1</sup> Γ (*pi*) *gi*(*<sup>w</sup>*, *x*(*w*))*dwduds* + Λ2' , *t* ∈ [0, 1].

We complete the proof in three steps.

**Step 1.** We show that Φ*x* + Ψ*y* ∈ *<sup>B</sup>ρ*. For *x*, *y* ∈ *<sup>B</sup>ρ*, we find that

$$\begin{split} & \quad \|\Phi x + \Psi y\| \\ & \leq \sup\_{t \in [0,1]} \left\{ \int\_{0}^{t} \frac{(t-s)^{q-1}}{\Gamma(q)} |f(s,x(s))| ds + \sum\_{i=1}^{k} \int\_{0}^{t} \frac{(t-s)^{q+p\_{i}-1}}{\Gamma(q+p\_{i})} |g\_{i}(s,x(s))| ds + |a| \right\} \\ & \quad + \frac{t^{m-1}}{|\Lambda\_{1}|} \left[ |a| \int\_{0}^{1} \left( \frac{(1-s)^{q-1}}{\Gamma(q)} |f(s,y(s))| + \sum\_{i=1}^{k} \frac{(1-s)^{q+p\_{i}-1}}{\Gamma(q+p\_{i})} |g\_{i}(s,y(s))| \right) ds \right. \\ & \left. + |\beta| \int\_{0}^{1} \left( \frac{(1-s)^{q-2}}{\Gamma(q-1)} |f(s,y(s))| + \sum\_{i=1}^{k} \frac{(1-s)^{q+p\_{i}-2}}{\Gamma(q+p\_{i}-1)} |g\_{i}(s,y(s))| \right) ds \\ & \quad + |\gamma\_{1}| \int\_{0}^{\zeta} \int\_{0}^{s} \frac{(s-u)^{q-1}}{\Gamma(q)} |f(u,y(u))| \, du \, ds \end{split}$$

<sup>+</sup>|*<sup>γ</sup>*1| *k* ∑ *i*=1 *ζ*0 *s* 0 (*s* − *u*)*<sup>q</sup>*−<sup>1</sup> <sup>Γ</sup>(*q*) *u*0 (*u* − *w*)*pi*−<sup>1</sup> Γ (*pi*) |*gi*(*<sup>w</sup>*, *y*(*w*))| *dw du ds* + *p* ∑ *j*=1 |*<sup>α</sup>j*| *ηj* 0 (*ηj* − *s*)*<sup>q</sup>*−<sup>1</sup> <sup>Γ</sup>(*q*) | *f*(*<sup>s</sup>*, *y*(*s*))|*ds* + *k*∑*i*=1 *ηj* 0 (*ηj* − *s*)*q*+*pi*−<sup>1</sup> <sup>Γ</sup>(*q* + *pi*) |*gi*(*<sup>s</sup>*, *y*(*s*))|*ds* <sup>+</sup>|*<sup>γ</sup>*2| 1*ξ s* 0 (*s* − *u*)*<sup>q</sup>*−<sup>1</sup> <sup>Γ</sup>(*q*) | *f*(*<sup>u</sup>*, *y*(*u*))| *du ds* <sup>+</sup>|*<sup>γ</sup>*2| *k* ∑ *i*=1 1*ξ s* 0 (*s* − *u*)*<sup>q</sup>*−<sup>1</sup> <sup>Γ</sup>(*q*) *u*0 (*u* − *w*)*pi*−<sup>1</sup> Γ (*pi*) |*gi*(*<sup>w</sup>*, *y*(*w*))| *dw du ds* + <sup>|</sup><sup>Λ</sup>2<sup>|</sup>' ≤ *μ* ⎡ ⎢⎢⎢⎣ sup *<sup>t</sup>*∈[0,1] ⎧⎪⎪⎪⎨⎪⎪⎪⎩ *tq* <sup>Γ</sup>(*q* + 1) + *t<sup>m</sup>*−<sup>1</sup> |<sup>Λ</sup>1| ⎛⎜⎜⎜⎝ |*α*| <sup>Γ</sup>(*q* + 1) + |*β*| <sup>Γ</sup>(*q*) + |*<sup>γ</sup>*1|*ζ<sup>q</sup>*+<sup>1</sup> <sup>Γ</sup>(*q* + 2) + *p*∑*j*=1 |*<sup>α</sup>j*|*ηqj* <sup>Γ</sup>(*q* + 1) +|*<sup>γ</sup>*2|(<sup>1</sup> − *ξq*+<sup>1</sup>) <sup>Γ</sup>(*q* + 2) ,! + *k*∑*i*=1 *μi tq*+*pi* <sup>Γ</sup>(*q* + *pi* + 1) + *t<sup>m</sup>*−<sup>1</sup> |<sup>Λ</sup>1| |*α*| <sup>Γ</sup>(*q* + *pi* + 1) + |*β*| <sup>Γ</sup>(*q* + *pi*) + |*<sup>γ</sup>*1|*ζ<sup>q</sup>*+*pi*+<sup>1</sup> <sup>Γ</sup>(*q* + *pi* + 2) + *p* ∑ *j*=1 |*<sup>α</sup>j*|*ηq*+*pi j* <sup>Γ</sup>(*q* + *pi* + 1) + |*<sup>γ</sup>*2|(<sup>1</sup> − *ξq*+*pi*+<sup>1</sup>) <sup>Γ</sup>(*q* + *pi* + 2) ⎞⎟⎟⎟⎠⎤⎥⎥⎥⎦ + |*a*| + Λ2 Λ1 =*μ*<sup>Ω</sup> + *k* ∑ *i*=1 *μi*<sup>Ω</sup>*i* + |*a*| + Λ2 Λ1 ≤ *ρ*.

Thus, Φ*x* + Ψ*y* ∈ *<sup>B</sup>ρ*.

**Step 2.** We show that Ψ is a contraction mapping. For that, let *x*, *y* ∈ C. Then, for each *t* ∈ [0, 1], we have

<sup>Ψ</sup>*x* − <sup>Ψ</sup>*y* ≤sup *<sup>t</sup>*∈[0,1] +*t<sup>m</sup>*−<sup>1</sup> <sup>|</sup><sup>Λ</sup>1<sup>|</sup> <sup>|</sup>*α*<sup>|</sup> 10 (1 − *s*)*<sup>q</sup>*−<sup>1</sup> <sup>Γ</sup>(*q*) | *f*(*<sup>s</sup>*, *x*(*s*)) − *f*(*<sup>s</sup>*, *y*(*s*))| + *k* ∑ *i*=1 (1 − *s*)*q*+*pi*−<sup>1</sup> <sup>Γ</sup>(*q* + *pi*) |*gi*(*<sup>s</sup>*, *x*(*s*)) − *gi*(*<sup>s</sup>*, *y*(*s*))| *ds* +|*β*| 10 (1 − *s*)*<sup>q</sup>*−<sup>2</sup> <sup>Γ</sup>(*q* − 1) | *f*(*<sup>s</sup>*, *x*(*s*)) − *f*(*<sup>s</sup>*, *y*(*s*))| + *k* ∑ *i*=1 (1 − *s*)*q*+*pi*−<sup>2</sup> <sup>Γ</sup>(*q* + *pi* − 1)|*gi*(*<sup>s</sup>*, *x*(*s*)) − *gi*(*<sup>s</sup>*, *y*(*s*))| *ds* <sup>+</sup>|*<sup>γ</sup>*1| *ζ*0 *s* 0 (*s* − *u*)*<sup>q</sup>*−<sup>1</sup> <sup>Γ</sup>(*q*) | *f*(*<sup>u</sup>*, *x*(*u*)) − *f*(*<sup>u</sup>*, *y*(*u*))| *du ds* <sup>+</sup>|*<sup>γ</sup>*1| *k* ∑ *i*=1 *ζ*0 *s* 0 (*s* − *u*)*<sup>q</sup>*−<sup>1</sup> <sup>Γ</sup>(*q*) *u*0 (*u* − *w*)*pi*−<sup>1</sup> Γ (*pi*) |*gi*(*<sup>w</sup>*, *x*(*w*)) − *gi*(*<sup>w</sup>*, *y*(*w*))| *dw du ds* + *p* ∑ *j*=1 |*<sup>α</sup>j*| *ηj* 0 (*ηj* − *s*)*<sup>q</sup>*−<sup>1</sup> <sup>Γ</sup>(*q*) | *f*(*<sup>s</sup>*, *x*(*s*)) − *f*(*<sup>s</sup>*, *y*(*s*))|*ds* + *k* ∑ *i*=1 *ηj* 0 (*ηj* − *s*)*q*+*pi*−<sup>1</sup> <sup>Γ</sup>(*q* + *pi*) |*gi*(*<sup>s</sup>*, *x*(*s*)) − *gi*(*<sup>x</sup>*, *y*(*s*))|*ds* <sup>+</sup>|*<sup>γ</sup>*2| 1*ξ s* 0 (*s* − *u*)*<sup>q</sup>*−<sup>1</sup> <sup>Γ</sup>(*q*) | *f*(*<sup>u</sup>*, *x*(*u*)) − *f*(*<sup>u</sup>*, *y*(*u*))| *du ds*

<sup>+</sup>|*<sup>γ</sup>*2| *k* ∑ *i*=1 1*ξ s* 0 (*s* − *u*)*<sup>q</sup>*−<sup>1</sup> <sup>Γ</sup>(*q*) *u*0 (*u* − *w*)*pi*−<sup>1</sup> Γ (*pi*) |*gi*(*<sup>w</sup>*, *x*(*w*)) − *gi*(*<sup>w</sup>*, *y*(*w*))| *dw du ds*' ≤ *L* ⎡ ⎢⎢⎢⎣ sup *<sup>t</sup>*∈[0,1] ⎧⎪⎪⎪⎨⎪⎪⎪⎩*tm*−1 |<sup>Λ</sup>1| ⎛⎜⎜⎜⎝ |*α*| <sup>Γ</sup>(*q* + 1) + |*β*| <sup>Γ</sup>(*q*) + |*<sup>γ</sup>*1|*ζ<sup>q</sup>*+<sup>1</sup> <sup>Γ</sup>(*q* + 2) + *p*∑*j*=1 |*<sup>α</sup>j*|*ηqj* <sup>Γ</sup>(*q* + 1) + |*<sup>γ</sup>*2|(<sup>1</sup> − *ξq*+<sup>1</sup>) <sup>Γ</sup>(*q* + 2) ⎞⎟⎟⎟⎠⎫⎪⎪⎪⎬⎪⎪⎪⎭⎤⎥⎥⎥⎦ ×*x* − *y* + *k* ∑ *i*=1 *Li <sup>t</sup><sup>m</sup>*−<sup>1</sup> |<sup>Λ</sup>1| |*α*| <sup>Γ</sup>(*q* + *pi* + 1) + |*β*| <sup>Γ</sup>(*q* + *pi*) + |*<sup>γ</sup>*1|*ζ<sup>q</sup>*+*pi*+<sup>1</sup> <sup>Γ</sup>(*q* + *pi* + 2) + *p* ∑ *j*=1 |*<sup>α</sup>j*|*ηq*+*pi j* <sup>Γ</sup>(*q* + *pi* + 1) + |*<sup>γ</sup>*2|(<sup>1</sup> − *ξq*+*pi*+<sup>1</sup>) <sup>Γ</sup>(*q* + *pi* + 2) ⎞⎟⎟⎟⎠⎤⎥⎥⎥⎦ *x* − *y* ≤ *L* Ω − 1 <sup>Γ</sup>(*q* + 1)! + *k*∑*i*=1 *Li* <sup>Ω</sup>*i* − 1 <sup>Γ</sup>(*q* + *pi* + 1)! *x* − *y*,

which is a contraction by the condition (17).

**Step 3.** We show that Φ is compact and continuous.

(i) Observe that the continuity of the operator Φ follows from that of *f* and *gi*, *i* = 1, . . . , *k*.

(ii) Φ is uniformly bounded on *Bρ* as:

$$\begin{split} \|\Phi\mathbf{x}\| &\leq \sup\_{t\in[0,1]} \left\{ \int\_{0}^{t} \frac{(t-s)^{q-1}}{\Gamma(q)} |f(s,\mathbf{x}(s))| ds + \sum\_{i=1}^{k} \int\_{0}^{t} \frac{(t-s)^{q+p\_{i}-1}}{\Gamma(q+p\_{i})} |g\_{i}(s,\mathbf{x}(s))| ds \right\} \\ &\leq \|\mu\| \sup\_{t\in[0,1]} \left\{ \int\_{0}^{t} \frac{(t-s)^{q-1}}{\Gamma(q)} ds \right\} + \|\mu\_{i}\| \sup\_{t\in[0,1]} \left\{ \sum\_{i=1}^{k} \int\_{0}^{t} \frac{(t-s)^{q+p\_{i}-1}}{\Gamma(q+p\_{i})} ds \right\} \\ &\leq \frac{||\mu||}{\Gamma(q+1)} + \sum\_{i=1}^{k} \frac{||\mu\_{i}||}{\Gamma(q+p\_{i}+1)}. \end{split}$$

(iii) Φ is equicontinuous.

Let us set max (*<sup>t</sup>*,*<sup>x</sup>*)∈[0,1]×*B<sup>ρ</sup>* | *f*(*<sup>t</sup>*, *x*)| = *f* / and max (*<sup>t</sup>*,*<sup>x</sup>*)∈[0,1]×*B<sup>ρ</sup>* |*gi*(*<sup>t</sup>*, *x*)| = *g*/*i*, *i* = 1, 2, ... , *m*. Then, for *t*1, *t*2 ∈ [0, 1], *t*1 > *t*2, we have


$$\begin{aligned} &+\frac{\hat{\mathcal{G}}\_{i}}{\Gamma(q+p\_{i}+1)} \left\{ |t\_{2}^{q+p\_{i}}-t\_{1}^{q+p\_{i}}|+|(t\_{1}-t\_{2})^{q+p\_{i}}|+|(t\_{1}-t\_{2})^{q+p\_{i}}| \right\} \\ &\leq \quad \frac{\hat{f}}{\Gamma(q+1)} \left| 2\left(t\_{1}-t\_{2}\right)^{q}+t\_{1}^{q}-t\_{2}^{q}\right|+\frac{\hat{\mathcal{G}}\_{i}}{\Gamma(q+p\_{i}+1)} |2\left(t\_{1}-t\_{2}\right)^{q+p\_{i}}-t\_{1}^{q+p\_{i}}+t\_{2}^{q+p\_{i}}| \right\} \end{aligned}$$

which tends to zero, independent of *x*, as *t*1 − *t*2 → 0. So, Φ is equicontinuous. Hence, we deduce by the Arzelá–Ascoli Theorem that Φ is compact on *Br*. So, the hypothesis (Steps 1–3) of Krasnosel'ski ˘ i's fixed point theorem [46] holds true. Consequently, there exists at least one solution for the boundary value problem (1) on [0, 1]. The proof is completed.
