*3.1. Existence Result Via Schaefer'S Fixed Point Theorem*

This subsection will provide the proof of the existence results of Equation (5) using Schaefer's fixed point theorem.

**Theorem 2** ([61])**.** *Let* A : X→X *be a completely continuous operator. Suppose that the set* E(A) = {*p* ∈ X : *p* = A*p*, *f or some* ∈ [0, 1]} *is bounded, then* A *has a fixed point.*

Thus we need the following assumptions:

(*A*1) Let *<sup>f</sup>* : J × <sup>R</sup><sup>3</sup> <sup>→</sup> <sup>R</sup> be a function such that *<sup>f</sup>* ∈ C1−*q*;*φ*[<sup>J</sup> , <sup>R</sup>] for any *<sup>z</sup>* ∈ C1−*q*;*φ*[<sup>J</sup> , <sup>R</sup>]. (*A*2) There exist *<sup>k</sup>*, *<sup>l</sup>*, *<sup>m</sup>*, *<sup>n</sup>* ∈ C1−*q*;*φ*[<sup>J</sup> , <sup>R</sup>] with *<sup>k</sup>*<sup>∗</sup> <sup>=</sup> sup *t*∈J |*k*(*t*)| < 1 such that

$$|f(t, u, v, w)| \le k(t) + l(t)|x| + m(t)|y| + n(t)|z|, \quad t \in \mathcal{J}, \quad u, v, w \in \mathbb{R}.$$

**Theorem 3.** *Let* 0 < *r* < 1*,* 0 ≤ *p* ≤ 1 *and q* = *r* + *p* − *rp. Suppose that the assumptions* (*A*1) *and* (*A*2) *are satisfied. Then there exist at least one solution of the problem* (5) *in the space* <sup>C</sup>*r*,*<sup>p</sup>* <sup>1</sup>−*q*;*φ*[<sup>J</sup> , <sup>R</sup>].

**Proof.** Define the operator *<sup>F</sup>* : <sup>C</sup>1−*q*;*φ*[<sup>J</sup> , <sup>R</sup>] → C1−*q*;*φ*[<sup>J</sup> , <sup>R</sup>] by

$$\begin{split} \phi(Fz)(t) &= \frac{\delta \Gamma(\rho+q)}{\Gamma(q)\Gamma(\rho+r)} (\phi(t) - \phi(0))^{q-1} \sum\_{i=1}^{m} b\_i \int\_{0^+}^{\tilde{\varsigma}\_i} \phi'(s) (\phi(\tilde{\varsigma}\_i) - \phi(s))^{\rho+r-1} T\_z(s) ds \\ &+ \frac{1}{\Gamma(r)} \int\_{0^+}^{t} \phi'(s) (\phi(t) - \phi(s))^{r-1} T\_z(s) ds, \end{split} \tag{28}$$

then, clearly the operator *F* is well-defined. The proof is given in the following steps: Step 1: the operator *<sup>F</sup>* is continuous. Let *zn* be a sequence such that *zn* <sup>→</sup> *<sup>z</sup>* in <sup>C</sup>1−*q*,*φ*[<sup>J</sup> , <sup>R</sup>]. Then for each *t* ∈ J , we have

$$\begin{split} & \left| \langle (Fz\_n)(t) - (Fz)(t) \rangle (\phi(t) - \phi(0))^{1-q} \right| \\ & \quad \le \frac{|\delta| \Gamma(\rho+q)}{\Gamma(q) \Gamma(\rho+r)} \sum\_{i=1}^{m} b\_i \int\_{0^+}^{\mathbb{S}\_i} \phi'(s) (\phi(\xi\_i) - \phi(s))^{\rho+r-1} |T\_{z\_n}(s) - T\_z(s)| ds \\ & \quad + \frac{1}{\Gamma(r)} (\phi(t) - \phi(0))^{1-q} \int\_{0^+}^{\mathbb{S}\_i} \phi'(s) (\phi(t) - \phi(s))^{r-1} |T\_{z\_n}(s) - T\_z(s)| ds \\ & \quad \le \frac{|\delta| \Gamma(\rho+q)}{\Gamma(q) \Gamma(\rho+r)} \mathcal{B}(q, \rho+r) \sum\_{i=1}^m b\_i (\phi(\xi\_i) - \phi(s))^{\rho+r+q-1} \|T\_{z\_n}(\cdot) - T\_z(\cdot)\|\_{\mathcal{L}\_{1-q\theta}} \\ & \quad + \frac{\mathcal{B}(q,r)}{\Gamma(r)} (\phi(T) - \phi(0))^r \|T\_{z\_n}(\cdot) - T\_z(\cdot)\|\_{\mathcal{L}\_{1-q\theta}} \\ & \quad \le \left[ \frac{|\delta| \Gamma(\rho+q)}{\Gamma(q) \Gamma(\rho+r)} \mathcal{B}(q, \rho+r) \sum\_{i=1}^m b\_i (\phi(\xi\_i) - \phi(s))^{\rho+r+q-1} \\ & \quad + \frac{\mathcal{B}(q,r)}{\Gamma(r)} (\phi(T) - \phi(0))^r \right] \|T\_{z\_n}(\cdot) - T\_z(\cdot)\|\_{\mathcal{L}\_{1-q\theta}}. \end{split} \tag{29}$$

Since *f* is continuous, this implies that *Tz* is also continuous. Therefore, we have

$$||T\_{z\_n} - T\_z||\_{\mathcal{C}\_{1-q;\phi}} \to 0, \quad \text{as} \quad n \to \infty.$$

Step 2: *<sup>F</sup>* maps bounded sets into bounded sets in <sup>C</sup>1−*q*;*φ*[<sup>J</sup> , <sup>R</sup>].

Indeed, it suffices to show that for any *κ* > 0, there exist a *μ* > 0 such that for any *<sup>z</sup>* <sup>∈</sup> **<sup>B</sup>***<sup>κ</sup>* <sup>=</sup> {*<sup>z</sup>* ∈ C1−*q*;*φ*[<sup>J</sup> , <sup>R</sup>] : *<sup>z</sup>*<sup>≤</sup> *<sup>κ</sup>*}, thus we have *<sup>F</sup>*(*z*) C<sup>1</sup>−*q*;*<sup>φ</sup>* <sup>≤</sup> *<sup>μ</sup>*.

For simplicity, we put

$$E\_1 = \frac{|\boldsymbol{\delta}|\Gamma(\rho+q)}{\Gamma(q)\Gamma(\rho+r)}\sum\_{i=1}^m b\_i \int\_{0^+}^{\xi\_i} \phi'(\mathbf{s}) (\phi(\xi\_i) - \phi(\mathbf{s}))^{\rho+r-1} |T\_z(\mathbf{s})| ds \tag{30}$$

and

$$E\_2 = \frac{1}{\Gamma(r)} (\phi(t) - \phi(0))^{1-q} \int\_{0^+}^t \phi'(s) (\phi(t) - \phi(s))^{r-1} |T\_z(s)| ds. \tag{31}$$

It follows from assumption (*A*2) that

$$\begin{split} |T\_z(t)| &= |f(t, z(t), z(\gamma t), T\_z(t))| \\ &\le k(t) + l(t)|z| + m(t)|z| + n(t)|T\_z(t)| \\ &\le \frac{k^\* + (l^\* + m^\*)|z(t)|}{1 - n^\*}. \end{split} \tag{32}$$

Thus, in view of Equations (30)–(32), we get

$$\begin{split} E\_1 &\leq \quad \frac{|\delta|\Gamma(\rho+q)}{\Gamma(q)(1-n^\*)} \sum\_{i=1}^m b\_i \left( \frac{k^\*}{\Gamma(\rho+r+1)} (\phi(\xi\_i) - \phi(0))^{\rho+r} \\ &\quad + (l^\*+m^\*) \frac{(\phi(\xi\_i) - \phi(0))^{\rho+r+q-1}}{\Gamma(\rho+r)} \mathcal{B}(q,\rho+r) ||z||\_{\mathcal{C}\_{1-q\phi}} \right) \\ E\_2 &\leq \quad \frac{1}{(1-n^\*)} \left( \frac{k^\*}{\Gamma(r+1)} (\phi(T) - \phi(0))^{\rho+r-q+1} \\ &\quad + \frac{(l^\*+m^\*)\mathcal{B}(q,r)}{\Gamma(r)} (\phi(T) - \phi(0))^r ||z||\_{\mathcal{C}\_{1-q\phi}} \right). \end{split}$$

This implies that,

$$\begin{split} |(F(z)(t)((\phi(t)-\phi(0))^{q-1})| \\ &\leq \frac{k^\*}{(1-n^\*)} \left[ \frac{|\delta|\Gamma(\rho+q)}{\Gamma(q)\Gamma(\rho+r+1)} \sum\_{i=1}^m b\_i(\phi(\xi\_i)-\phi(0))^{\rho+r} \\ &\quad + \frac{k^\*}{\Gamma(r+1)} (\phi(T)-\phi(0))^{\rho+r-q+1} \right] \\ &\quad + \frac{(l^\*+m^\*)}{(1-n^\*)} \left[ \frac{|\delta|\Gamma(\rho+q)}{\Gamma(q)\Gamma(\rho+r)} \mathcal{B}(q,\rho+r) \sum\_{i=1}^m b\_i(\phi(\xi\_i)-\phi(0))^{\rho+r+q-1} \\ &\quad + \frac{\mathcal{B}(q,r)}{\Gamma(r)} (\phi(T)-\phi(0))^r \right] \|z\|\_{\mathcal{L}\_{1-\psi\delta}} \\ &=: \mu. \end{split} \tag{33}$$

Step 3: *<sup>F</sup>* maps bounded sets into equicontinuous set of <sup>C</sup>1−*q*;*φ*[<sup>J</sup> , <sup>R</sup>]. Let *<sup>t</sup>*1, *<sup>t</sup>*<sup>2</sup> ∈ J such that *<sup>t</sup>*<sup>1</sup> <sup>≥</sup> *<sup>t</sup>*<sup>2</sup> and *<sup>B</sup><sup>κ</sup>* be a bounded set of <sup>C</sup>1−*q*;*φ*[<sup>J</sup> , <sup>R</sup>] as defined in Step 2. Let *<sup>z</sup>* <sup>∈</sup> *<sup>B</sup>κ*, then

<sup>|</sup>((*φ*(*t*1) <sup>−</sup> *<sup>φ</sup>*(*a*))*q*−1)(*Fz*)(*t*1) <sup>−</sup> ((*φ*(*t*2) <sup>−</sup> *<sup>φ</sup>*(0))*q*−1)(*Fz*)(*t*2)<sup>|</sup> ≤ 1 Γ(*r*) (*φ*(*t*1) <sup>−</sup> *<sup>φ</sup>*(0))1−*<sup>q</sup> t*<sup>1</sup> <sup>0</sup><sup>+</sup> *<sup>φ</sup>* (*s*)(*φ*(*t*1) <sup>−</sup> *<sup>φ</sup>*(*s*))*r*−1*Tz*(*s*)*ds* <sup>−</sup> <sup>1</sup> Γ(*r*) (*φ*(*t*2) <sup>−</sup> *<sup>φ</sup>*(0))1−*<sup>q</sup> t*<sup>2</sup> <sup>0</sup><sup>+</sup> *<sup>φ</sup>* (*s*)(*φ*(*t*2) <sup>−</sup> *<sup>φ</sup>*(*s*))*r*−1*Tz*(*s*)*ds* ≤ 1 Γ(*r*) *t*<sup>1</sup> <sup>0</sup><sup>+</sup> *<sup>φ</sup>* (*s*) ((*φ*(*t*1) <sup>−</sup> *<sup>φ</sup>*(0))*q*−1)(*φ*(*t*1) <sup>−</sup> *<sup>φ</sup>*(*s*))*r*−<sup>1</sup> <sup>−</sup> ((*φ*(*t*2) <sup>−</sup> *<sup>φ</sup>*(0))*q*−1)(*φ*(*t*2) <sup>−</sup> *<sup>φ</sup>*(*s*))*r*−1*Tz*(*s*)*ds* + (*φ*(*t*2) <sup>−</sup> *<sup>φ</sup>*(0))*q*−<sup>1</sup> Γ(*r*) *t*<sup>2</sup> *t*1 *φ* (*s*)(*φ*(*t*1) <sup>−</sup> *<sup>φ</sup>*(0))*r*−1*Tz*(*s*)*ds* → 0, *as t*<sup>1</sup> → *t*2. (34)

Thus, steps 1–3, together with the Arzela–Ascoli theorem, show that the operator *F* is completely continuous.

Step 4: a priori bounds.

It is enough to show that the set *<sup>χ</sup>* <sup>=</sup> {*<sup>z</sup>* ∈ C1−*q*;*φ*[<sup>J</sup> , <sup>R</sup>] : *<sup>z</sup>* <sup>=</sup> *<sup>σ</sup>*(*Fz*), 0 <sup>&</sup>lt; *<sup>σ</sup>* <sup>&</sup>lt; <sup>1</sup>} is bounded. Now, let *z* ∈ *χ*, *z* = *σ*(*Fz*) for some 0 < *σ* < 1. Thus for each *t* ∈ J , we obtain

$$\begin{split} z(t) &= \sigma \left[ \frac{\delta \Gamma(\rho+q)}{\Gamma(q)\Gamma(\rho+r)} (\phi(t) - \phi(0))^{q-1} \sum\_{i=1}^{\mathrm{nr}} b\_i \int\_{0^+}^{\xi\_i} \phi'(s) (\phi(\xi\_i) - \phi(s))^{\rho+r-1} T\_z(s) ds \\ &+ \frac{1}{\Gamma(r)} \int\_{0^+}^t \phi'(s) (\phi(t) - \phi(s))^{r-1} T\_z(s) ds \right]. \end{split}$$

It follows from assumption (*A*2), that for every *t* ∈ J ,

$$\begin{split} |z(t)(\boldsymbol{\phi}(t) - \boldsymbol{\phi}(0))^{1-q}| &\leq |(Fz)(t)(\boldsymbol{\phi}(t) - \boldsymbol{\phi}(0))^{1-q}| \\ &\leq \frac{k^\*}{(1-n^\*)} \left[ \frac{|\boldsymbol{\delta}|\Gamma(\boldsymbol{\rho}+q)}{\Gamma(q)\Gamma(\boldsymbol{\rho}+r+1)} \sum\_{i=1}^m b\_i(\boldsymbol{\phi}(\xi\_i) - \boldsymbol{\phi}(0))^{\boldsymbol{\rho}+r} \\ &\quad + \frac{k^\*}{\Gamma(r+1)} (\boldsymbol{\phi}(T) - \boldsymbol{\phi}(0))^{\boldsymbol{\rho}+r-q+1} \right] \\ &\quad + \frac{(l^\*+m^\*)}{(1-n^\*)} \left[ \frac{|\boldsymbol{\delta}|\Gamma(\boldsymbol{\rho}+q)}{\Gamma(q)\Gamma(\boldsymbol{\rho}+r)} \mathcal{B}(q,\boldsymbol{\rho}+r) \sum\_{i=1}^m b\_i(\boldsymbol{\phi}(\xi\_i) - \boldsymbol{\phi}(0))^{\boldsymbol{\rho}+r+q-1} \right] \\ &\quad + \frac{\mathcal{B}(q,r)}{\Gamma(r)} (\boldsymbol{\phi}(T) - \boldsymbol{\phi}(0))^{r} \Big] \|z\|\_{c\_{1-q\tilde{\rho}}} \\ &< \infty. \end{split} \tag{35}$$

This shows that the set *χ* is bounded. Hence, by the Schaefer's fixed point theorem, problem (5) has at least one solution.
