**3. Main Results**

In this section, we first adopt some techniques from Lemma 7 in order to establish an important mixed-type integral equation of problem (5). Thus, we need the following auxiliary lemma.

*Mathematics* **2020**, *8*, 94

**Lemma 8.** *Let* <sup>0</sup> <sup>&</sup>lt; *<sup>r</sup>* <sup>&</sup>lt; <sup>1</sup>*,* <sup>0</sup> <sup>≤</sup> *<sup>p</sup>* <sup>≤</sup> <sup>1</sup> *and <sup>q</sup>* <sup>=</sup> *<sup>r</sup>* <sup>+</sup> *<sup>p</sup>* <sup>−</sup> *rp. Suppose <sup>f</sup>* : J × <sup>R</sup><sup>3</sup> <sup>→</sup> <sup>R</sup> *is 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>]*. If <sup>z</sup>* ∈ C*<sup>q</sup>* <sup>1</sup>−*q*;*φ*[<sup>J</sup> , <sup>R</sup>] *then <sup>z</sup> satisfies the problem* (5) *if and only if z satisfies the mixed-type integral equation:*

$$\begin{split} z(t) &= \frac{\delta \Gamma(\rho+q)}{\Gamma(q)\Gamma(\rho+r)} (\phi(t) - \phi(0))^{q-1} \sum\_{i=1}^{m} b\_i \int\_{0^+}^{\mathbb{T}\_i} \phi'(s) (\phi(\xi\_i) - \phi(s))^{\rho+r-1} T\_z(s) ds \\ &+ \frac{1}{\Gamma(r)} \int\_{0^+}^{t} \phi'(s) (\phi(t) - \phi(s))^{r-1} T\_z(s) ds, \end{split} \tag{9}$$

*where*

$$\begin{aligned} \delta &= \frac{1}{\Gamma(\rho + q) - \sum\_{i=1}^{m} b\_i (\phi(\xi\_i) - \phi(0))^{\rho + q - 1}} \\ \text{such that} \quad \Gamma(\rho + q) &\neq \sum\_{i=1}^{m} b\_i (\phi(\xi\_i) - \phi(0))^{\rho + q - 1} .\end{aligned} \tag{10}$$

*For simplicity, we take*

$$T\_z(t) = {}^H D\_{0^+} ^{r, p; \phi} z(t) = f(t, z(t), z(\gamma t), T\_z(t)). \tag{11}$$

**Proof.** Suppose *<sup>z</sup>* ∈ C*<sup>q</sup>* <sup>1</sup>−*q*;*φ*[<sup>J</sup> , <sup>R</sup>] is a solution to the problem (5), then, we show that *<sup>z</sup>* is also a solution of (5). Indeed, from Lemma 7, we have

$$z(t) = \frac{(\phi(t) - \phi(0))^{q-1}}{\Gamma(q)} T\_{0^{+}}^{1 - q; \phi} z(0) + \frac{1}{\Gamma(r)} \int\_{0^{+}}^{t} \phi'(s) (\phi(t) - \phi(s))^{r-1} T\_{z}(s) ds. \tag{12}$$

Now, if we substitute *t* = *ξ<sup>i</sup>* and multiply both sides by *bi* in Equation (12), we obtain

$$b\_{i}z(\zeta\_{i}) = \frac{(\phi(\zeta\_{i}) - \phi(0))^{q-1}}{\Gamma(q)} b\_{i} \mathcal{Z}\_{0^{+}}^{1-q;\mathfrak{b}} z(0) + \frac{b\_{i}}{\Gamma(r)} \int\_{0^{+}}^{\overline{\zeta}\_{i}} \phi'(s) (\phi(\zeta\_{i}) - \phi(s))^{r-1} T\_{\overline{z}}(s) ds. \tag{13}$$

Next, by applying <sup>I</sup>*ρ*;*<sup>φ</sup>* <sup>0</sup> to both sides of Equation (13) and using Lemma 1 and Proposition 1, we get

$$\begin{split} T\_{0^{+}}^{\rho;\phi} b\_{i} z(\xi\_{i}) &= \frac{(\phi(\xi\_{i}) - \phi(0))^{\rho+q-1}}{\Gamma(\rho+q)} b\_{i} T\_{0}^{1-\eta;\phi} z(0) \\ &+ \frac{b\_{i}}{\Gamma(\rho+r)} \int\_{0^{+}}^{\xi\_{i}} \phi'(s) (\phi(\xi\_{i}) - \phi(s))^{\rho+r-1} T\_{z}(s) ds. \end{split} \tag{14}$$

This implies that

$$\begin{split} \sum\_{i=1}^{m} \mathcal{Z}\_{0+}^{p, \phi} b\_{i} z(\xi\_{i}) &= \frac{1}{\Gamma(\rho + q)} \left( \sum\_{i=1}^{m} b\_{i} (\phi(\xi\_{i}) - \phi(0))^{\rho + q - 1} \right) \mathcal{Z}\_{0+}^{1 - q; \phi} z(0) \\ &+ \frac{1}{\Gamma(\rho + r)} \sum\_{i=1}^{m} b\_{i} \int\_{0^{+}}^{\xi\_{i}} \phi'(s) (\phi(\xi\_{i}) - \phi(s))^{\rho + r - 1} T\_{z}(s) ds. \end{split} \tag{15}$$

Inserting the initial condition: <sup>I</sup>1−*q*;*<sup>φ</sup>* <sup>0</sup><sup>+</sup> *<sup>z</sup>*(0+) = *m* ∑ *i*=1 I*ρ*;*<sup>φ</sup>* <sup>0</sup><sup>+</sup> *biz*(*ξi*) in Equation (15) we have

$$\begin{split} \mathcal{Z}\_{0^{+}}^{1-q;\phi} z(0) &= \frac{1}{\Gamma(\rho+q)} \left( \sum\_{i=1}^{m} b\_{i} (\phi(\xi\_{i}) - \phi(0))^{\rho+q-1} \right) \mathcal{Z}\_{0^{+}}^{1-q;\phi} z(0) \\ &+ \frac{1}{\Gamma(\rho+r)} \sum\_{i=1}^{m} b\_{i} \int\_{0^{+}}^{\xi\_{i}} \phi'(s) (\phi(\xi\_{i}) - \phi(s))^{\rho+r-1} T\_{z}(s) ds, \end{split} \tag{16}$$

which implies that

$$\begin{split} \frac{1}{\Gamma(\rho+r)} \sum\_{i=1}^{m} b\_{i} \int\_{0^{+}}^{\zeta\_{i}} \phi'(s) (\phi(\xi\_{i}) - \phi(s))^{\rho+r-1} T\_{z}(s) ds \\ &= \left( 1 - \frac{1}{\Gamma(\rho+q)} \sum\_{i=1}^{m} b\_{i} (\phi(\xi\_{i}) - \phi(0))^{\rho+q-1} \right) \mathcal{Z}\_{0^{+}}^{1-q; \phi} z(0) \\ &= \frac{1}{\delta \Gamma(\rho+q)} \mathcal{Z}\_{0^{+}}^{1-q; \phi} z(0). \end{split} \tag{17}$$

Thus,

$$\mathcal{L}\_{0^{+}}^{1-q;\phi}z(0) = \frac{\delta\Gamma(\rho+q)}{\Gamma(\rho+r)}\sum\_{i=1}^{m}b\_{i}\int\_{0^{+}}^{\overline{\varsigma}\_{i}}\phi'(s)(\phi(\zeta\_{i}^{\circ})-\phi(s))^{\rho+r-1}T\_{z}(s)ds.\tag{18}$$

Hence, the result follows by putting Equation (18) in Equation (12). This implies that *z*(*t*) satisfies Equation (9).

Conversely, suppose that *<sup>z</sup>* ∈ C*<sup>q</sup>* <sup>1</sup>−*q*;*<sup>φ</sup>* satisfies the mixed-type integral Equation (9), then, we show that *<sup>z</sup>* satisfies Equation (5). Applying <sup>D</sup>*q*;*<sup>φ</sup>* <sup>0</sup><sup>+</sup> to both sides of Equation (9) and using Lemma 2 and Proposition 1, we get

$$\begin{split} \mathcal{D}\_{0+}^{p\beta} z(t) &= \mathcal{D}\_{0+}^{p\beta} \left( \frac{\delta \Gamma(\rho+q)}{\Gamma(q)\Gamma(r)} (\phi(t) - \phi(0))^{q-1} \sum\_{i=1}^{m} b\_{i} \int\_{0^{+}}^{\xi\_{i}} \phi'(s) (\phi(\xi\_{i}) - \phi(s))^{\rho+r-1} T\_{z}(s) ds \right) \\ &+ \mathcal{D}\_{0+}^{p\beta} \left( \frac{1}{\Gamma(r)} \int\_{0^{+}}^{t} \phi'(s) (\phi(t) - \phi(s))^{r-1} T\_{z}(s) ds \right) \\ &= \mathcal{D}\_{0+}^{p(1-r)\beta} f(t, z(t), z(\gamma t), \mathcal{D}\_{0+}^{r, p\beta} z(\gamma t)). \end{split} \tag{19}$$

Since <sup>D</sup>*r*,*p*;*<sup>φ</sup>* <sup>0</sup><sup>+</sup> *<sup>z</sup>* ∈ C1−*q*;*φ*[<sup>J</sup> , <sup>R</sup>], then by definition of <sup>C</sup>*<sup>q</sup>* <sup>1</sup>−*q*;*φ*[<sup>J</sup> , <sup>R</sup>] and make use of Equation (19), we have

$$\mathcal{D}\_{0^{+}}^{p(1-r);\phi}f = \mathcal{D}\mathcal{I}\_{0^{+}}^{1-p(1-r);\phi}f \in \mathcal{C}\_{1-q;\phi}[\mathcal{J},\mathbb{R}].$$

For every *<sup>f</sup>* ∈ C1−*q*;*φ*[<sup>J</sup> , <sup>R</sup>] and Lemma 3, we can see that <sup>I</sup>1−*p*(1−*r*);*<sup>φ</sup>* <sup>0</sup><sup>+</sup> *<sup>f</sup>* ∈ C1−*q*;*φ*[<sup>J</sup> , <sup>R</sup>], which implies that <sup>I</sup>1−*p*(1−*r*);*<sup>φ</sup>* <sup>0</sup><sup>+</sup> *<sup>f</sup>* ∈ C<sup>1</sup> <sup>1</sup>−*q*;*φ*[<sup>J</sup> , <sup>R</sup>] from the definition of <sup>C</sup>*<sup>n</sup> <sup>q</sup>*;*φ*[<sup>J</sup> , <sup>R</sup>]. Applying <sup>I</sup>*p*(1−*r*);*<sup>φ</sup>* 0+ on both sides of Equation (19) and using Lemma 3, we have

$$\begin{split} \mathcal{T}\_{0^{+}}^{p(1-r);\phi} \mathcal{D}\_{0^{+}}^{p\phi} z(t) &= \mathcal{T}\_{0^{+}}^{p(1-r);\phi} \mathcal{D}\_{0^{+}}^{p(1-r);\phi} T\_{z}(t) \\ &= T\_{\bar{z}}(t) - \frac{\left(\mathcal{Z}\_{0^{+}}^{1-p(1-r);\phi} \mathcal{T}\_{\bar{z}}\right)(0^{+})}{\Gamma(p(1-r))} (\phi(t) - \phi(0))^{p(r-1)-1} \\ &= T\_{\bar{z}}(t) = f(t, z(t), z(\gamma t), \mathcal{D}\_{0^{+}}^{r, p;\phi} z(\gamma t)). \end{split} \tag{20}$$

*Mathematics* **2020**, *8*, 94

Finally, we show that if *<sup>z</sup>* ∈ C*<sup>q</sup>* <sup>1</sup>−*q*[<sup>J</sup> , <sup>R</sup>] satisfies Equation (9), it also satisfies the initial condition. Thus, by applying <sup>I</sup>1−*q*;*<sup>φ</sup>* <sup>0</sup><sup>+</sup> to both sides of Equation (9) and using Lemma 1 and Proposition 1, we obtain

$$\begin{split} & T\_{0^{+}}^{1-\eta;\phi} z(t) \\ &= \mathcal{I}\_{0^{+}}^{1-\eta;\phi} \left( \frac{\delta \Gamma(\rho+q)}{\Gamma(q)\Gamma(\rho+r)} (\phi(t)-\phi(0))^{q-1} \sum\_{i=1}^{m} b\_{i} \int\_{0^{+}}^{\mathbb{T}\_{i}} \phi'(s) (\phi(\xi\_{i})-\phi(s))^{\rho+r-1} T\_{z}(s) ds \right) \\ &+ \mathcal{I}\_{0^{+}}^{1-\eta;\phi} \left( \frac{1}{\Gamma(r)} \int\_{0^{+}}^{\mathbb{T}\_{i}} \phi'(s) (\phi(t)-\phi(s))^{r-1} T\_{z}(s) ds \right) \\ &= \frac{\delta \Gamma(\rho+q)}{\Gamma(\rho+r)} \sum\_{i=1}^{m} b\_{i} \int\_{0^{+}}^{\mathbb{T}\_{i}} \phi'(s) (\phi(\xi\_{i})-\phi(s))^{\rho+r-1} T\_{z}(s) ds + \mathcal{I}\_{0^{+}}^{1-\eta(1-r);\phi} T\_{z}(t). \end{split} \tag{21}$$

Using Lemma 4 and the fact that 1 − *q* < 1 − *p*(1 − *r*), then taking limit as *t* → 0 in Equation (21) yields

$$\mathcal{Z}\_{0^{+}}^{1-q;\phi}z(0^{+}) = \frac{\delta\Gamma(\rho+q)}{\Gamma(\rho+r)}\sum\_{i=1}^{m}b\_{i}\int\_{0^{+}}^{\xi\_{i}}\phi'(s)(\phi(\xi\_{i}^{x})-\phi(s))^{\rho+r-1}T\_{z}(s)ds.\tag{22}$$

Now, substituting *t* = *ξ<sup>i</sup>* and multiplying through by *bi* in Equation (9), we get

$$\begin{split} b\_{i}z(\boldsymbol{\xi}\_{i}) &= \frac{\delta\Gamma(\rho+q)}{\Gamma(q)\Gamma(\rho+r)} b\_{i} (\boldsymbol{\phi}(\boldsymbol{\xi}\_{i}) - \boldsymbol{\phi}(0))^{q-1} \sum\_{i=1}^{m} b\_{i} \int\_{0^{+}}^{\boldsymbol{\xi}\_{i}} \boldsymbol{\phi}'(\boldsymbol{s}) (\boldsymbol{\phi}(\boldsymbol{\xi}\_{i}) - \boldsymbol{\phi}(\boldsymbol{s}))^{\rho+r-1} T\_{z}(\boldsymbol{s}) ds \\ &+ \mathcal{I}\_{0^{+}}^{1-q;\phi} \frac{b\_{i}}{\Gamma(r)} \int\_{0^{+}}^{\boldsymbol{\xi}\_{i}} \boldsymbol{\phi}'(\boldsymbol{s}) (\boldsymbol{\phi}(\boldsymbol{\xi}\_{i}) - \boldsymbol{\phi}(\boldsymbol{s}))^{r-1} T\_{z}(\boldsymbol{s}) ds. \end{split} \tag{23}$$

Applying <sup>I</sup>*ρ*;*<sup>φ</sup>* <sup>0</sup><sup>+</sup> to both sides of Equation (23), we obtain

$$\begin{split} \mathcal{Z}\_{0^{+}}^{\rho\phi} b\_{i} z(\xi\_{i}) &= \frac{\delta b\_{i} (\phi(\xi\_{i}) - \phi(0))^{\rho+q-1}}{\Gamma(\rho+r)} \sum\_{i=1}^{m} b\_{i} \int\_{0^{+}}^{\xi\_{i}} \phi'(s) (\phi(\xi\_{i}) - \phi(s))^{\rho+r-1} T\_{z}(s) ds \\ &+ \frac{b\_{i}}{\Gamma(\rho+r)} \int\_{0^{+}}^{\xi\_{i}} \phi'(s) (\phi(\xi\_{i}) - \phi(s))^{\rho+r-1} T\_{z}(s) ds, \end{split} \tag{24}$$

which implies

$$\begin{split} \sum\_{i=1}^{m} b\_{i} \mathbb{E}\_{\boldsymbol{\theta}\_{0}^{\boldsymbol{\theta}}}^{\boldsymbol{\rho},\boldsymbol{\rho}} z(\xi\_{i}) \\ &= \frac{\delta}{\Gamma(\boldsymbol{\rho}+\boldsymbol{r})} \sum\_{i=1}^{m} b\_{i} \int\_{0^{+}}^{\xi\_{i}} \phi'(s) (\phi(\xi\_{i}) - \phi(s))^{\rho+r-1} T\_{\boldsymbol{z}}(s) ds \sum\_{i=1}^{m} b\_{i} (\phi(\xi\_{i}) - \phi(0))^{\rho+q-1} \\ &+ \frac{1}{\Gamma(\boldsymbol{\rho}+\boldsymbol{r})} \sum\_{i=1}^{m} b\_{i} \int\_{0^{+}}^{\xi\_{i}} \phi'(s) (\phi(\xi\_{i}) - \phi(s))^{\rho+r-1} T\_{\boldsymbol{z}}(s) ds \\ &= \frac{1}{\Gamma(\boldsymbol{\rho}+\boldsymbol{r})} \sum\_{i=1}^{m} b\_{i} \int\_{0^{+}}^{\xi\_{i}} \phi'(s) (\phi(\xi\_{i}) - \phi(s))^{\rho+r-1} T\_{\boldsymbol{z}}(s) ds \times \\ &\left(1 + \delta \sum\_{i=1}^{m} b\_{i} (\phi(\xi\_{i}) - \phi(0))^{\rho+q-1}\right) \end{split} \tag{25}$$

and

$$T\_{0^{+}}^{1-q;\phi}z(0^{+}) = \frac{\delta\Gamma(\rho+q)}{\Gamma(\rho+r)}\sum\_{i=1}^{m}b\_{i}\int\_{0^{+}}^{\xi\_{i}}\phi'(s)(\phi(\xi\_{i}^{x})-\phi(s))^{\rho+r-1}T\_{z}(s)ds.\tag{26}$$

Therefore, in view of Equations (22) and (26), we have

$$\mathcal{Z}\_{0^{+}}^{1-q;\phi}z(0^{+}) = \sum\_{i=1}^{m} b\_{i}\mathcal{Z}\_{0^{+}}^{q;\phi}z(\zeta\_{i}).\tag{27}$$
