The purpose of this section is concerning to establish the required theory for existence of at least one solutions to the BVP (1).
Lemma 3.1
For
\(\omega \in L^{1}(\mathfrak{J}, \mathbf{R})\), the solution of the linear BVP of FDEs
$$ \begin{aligned} & {}^{C}\mathbf{D}^{q} z(t)+\omega (t)=0;\quad t\in \mathfrak{J}, q \in (1, 2], \\ & u(t)\vert _{t=0}=\phi (z),\qquad z(1)= \delta{}^{C} \mathbf{D}^{p} z(\eta ), \end{aligned} $$
(3)
is given as
$$z(t)=\phi (z) (1td)+ \int_{0}^{1} \mathcal{H}(t,\tau )\omega (\tau )\,d\tau , $$
where
\(\mathcal{H}(t,s)\)
is the Green’s function given by
$$ \mathcal{H}(t,\tau )= \textstyle\begin{cases} \frac{td (1\tau )^{q1}}{\Gamma (q)} \frac{1(t\tau )^{q1}}{\Gamma (q)}\frac{\delta t d(\eta \tau )^{qp1}}{\Gamma (qp)};\quad 0\leq \tau \leq t \leq \eta \leq 1, \\ \frac{td(1\tau )^{q1}}{\Gamma (q)}\frac{\delta t d(\eta \tau )^{qp1}}{ \Gamma (qp)};\quad 0\leq t\leq \tau \leq \eta \leq 1, \\ \frac{td(1\tau )^{q1}}{\Gamma (q)}\frac{(t\tau )^{q1}}{\Gamma (q)} ;\quad 0\leq \eta \leq \tau \leq t \leq 1, \\ \frac{td(1\tau )^{q1}}{\Gamma (q)};\quad 0\leq \eta \leq t\leq \tau \leq 1. \end{cases} $$
(4)
Proof
Consider equation (3) with the associated given boundary conditions; applying \(\mathbf{I}^{q}\) on \({}^{C}\mathbf{D}^{q} z(t)= \omega (t)\) and thanks to Theorem 2.3, we have
$$ z(t)=\mathbf{I}^{q} (t)+e_{0}+e_{1}t $$
(5)
for some \(e_{0},e_{1}\in \mathbf{R}\). From the nonlocal condition \(z(t)_{t=0}=\phi (z)\) implies that \(e_{0}=\phi (z)\) and \(z(1)= \delta {}^{C}\mathbf{D}^{p} u(\eta )\) yields \(e_{1}=d [ \mathbf{I}^{q} \omega (1)\delta \mathbf{I}^{qp}\omega (\eta )\phi (z) ] \), where \(d=\frac{\Gamma (2p)}{\Gamma (2p)\delta \eta^{1p}}>1\). It implies that
$$ z(t)=\mathbf{I}^{q} \omega (t)+td \mathbf{I}^{q} \omega (1)\delta td \mathbf{I}^{qp} \omega ( \eta )+(1td)\phi (z). $$
(6)
Thus we get a solution z in the form
$$z(t)=\phi (z) (1td)+ \int_{0}^{1} \mathcal{H}(t,\tau )\omega (\tau )\,d\tau , $$
where the kernel \(\mathcal{H}(t,\tau )\) is the Green’s function and is given as (4). □
Lemma 3.2
A function
\(z\in \mathbf{Z}\)
will be the solution of the fractional integral equation (6) if and only if
z
is a solution of (1).
Proof
The proof is obvious. □
To derive formally the required results as regards the data dependence and existence of at least one of solutions to the proposed BVP (1), we state the following hypotheses:
 (\(A_{1}\)):

For arbitrary \(u,v\in \mathbf{Z}\) and if there exists a constant \(\mathcal{K}_{\phi } \in [0,1)\), then one has
$$\bigl\lvert \phi (z)\phi (\bar{z})\bigr\rvert \leq \mathcal{K}_{\phi } \lVert z \bar{z} \rVert_{\mathbf{Z}}; $$
 (\(A_{2}\)):

for constants \(\mathcal{C}_{\phi }\), \(q_{1} \in [0,1)\) and \(\mathcal{M}_{\phi }>0\), we have the following growth condition:
$$\bigl\lvert \phi (z)\bigr\rvert \leq \mathcal{C}_{\phi } \lVert z \rVert_{\mathbf{Z}} ^{q_{1}}+\mathcal{M}_{\phi },\quad \mbox{for each } z\in \mathbf{Z}; $$
 (\(A_{3}\)):

in the same fashion, for constants \(\mathcal{C}_{ \theta }\), \(q_{2} \in [0,1)\) and \(\mathcal{M}_{\theta }>0\), we have the following growth condition:
$$\bigl\lvert \theta (t,z(t))\bigr\rvert \leq \mathcal{C}_{\theta } \lVert z \rVert_{\mathbf{Z}}^{q_{2}}+\mathcal{M}_{\theta }. $$
To show that equation (6) has at least one solution \(z\in \mathbf{Z}\) based on assumptions (\(A_{1}\))(\(A_{3}\)), we define the operators by
$$\mathcal{F}:\mathbf{Z}\rightarrow \mathbf{Z} $$
as follows:
$$(\mathcal{F}z) (t)=(1td)\phi (z),\quad d>1 $$
and \(\mathcal{G}:\mathbf{Z}\rightarrow \mathbf{Z}\) is defined by
$$\begin{aligned}& \begin{aligned} (\mathcal{G}z) (t) ={}&\frac{td}{\Gamma (q)} \int_{0}^{1}(1s)^{q1} \theta \bigl(\tau , z(\tau )\bigr)\,d\tau\frac{1}{\Gamma (q)} \int_{0}^{t}(t\tau )^{q1}\theta \bigl(\tau , z( \tau )\bigr)\,d\tau \\ &{} \frac{1}{\Gamma (qp)} \int_{0}^{\eta }(\eta \tau )^{qp1} \theta \bigl(\tau , z(\tau )\bigr)\,d\tau , \end{aligned} \\& \mathbf{T}:\mathbf{Z}\rightarrow \mathbf{Z}, \qquad \mathbf{T}z=\mathcal{F}z+ \mathcal{G}z. \end{aligned}$$
Obviously, the operator T is well defined because the function θ is continuous. So, we can write (6) as an operator equation given by
$$ z=\mathbf{T}z=\mathcal{F}z+\mathcal{G}z. $$
(7)
Here by the existence of a solution to equation (1), we mean the existence of a fixed point for operator T as defined afore and satisfying equation (7).
Lemma 3.3
The operator
\(\mathcal{F}: \mathbf{Z}\rightarrow \mathbf{Z}\)
is Lipschitz and consequently
βLipschitz with constant
\(\mathcal{K}_{\mathcal{F}}<1\). Moreover, the operator
\(\mathcal{F}\)
satisfies the following growth condition:
$$ \lVert \mathcal{F}z \rVert \leq \mathcal{Q} \lVert z \rVert_{ \mathbf{Z}}^{q_{1}}+\mathcal{M}_{\phi },\quad \textit{for every } z \in \mathbf{Z}. $$
(8)
Proof
For \(\mathcal{F}\) to be Lipschitz, we consider \(\lvert \mathcal{F}z(t) \mathcal{F}\bar{z}(t) \rvert \), and apply assumptions (\(A_{1}\)) and (\(A_{2}\)), we have
$$\begin{aligned} \bigl\lvert \mathcal{F}z(t)\mathcal{F}\bar{z}(t) \bigr\rvert =& \bigl\lvert (1t d) \bigl(\phi \bigl(z(t)\bigr)\phi \bigl(\bar{z}(t)\bigr) \bigr) \bigr\rvert \\ =& \bigl\lvert (1t d)\bigr\rvert \big\rvert \bigl(\phi \bigl(z(t)\bigr)\phi \bigl(\bar{z}(t)\bigr) \bigr) \big\rvert \\ \leq & \bigl\lvert (1t d)\bigr\rvert \mathcal{K}_{\phi } \lVert z \bar{z} \rVert_{\mathbf{Z}} \\ \leq & \mathcal{K}_{\mathcal{F}} \lVert z\bar{z} \rVert_{\mathbf{Z}},\quad \mbox{where } \mathcal{K}_{\mathcal{F}}=\bigl\lvert (1t d)\bigr\rvert \mathcal{K}_{\phi }< 1. \end{aligned}$$
Hence, we get
$$\lVert \mathcal{F}z\mathcal{F}\bar{z} \rVert_{\mathbf{Z}} \leq \mathcal{K}_{\mathcal{F}} \lVert z\bar{z} \rVert_{\mathbf{Z}},\quad \mbox{for every } z\in \mathbf{Z}. $$
Thanks to proposition 2.4, \(\mathcal{F}\) is also βLipschitz with the same coefficient \(K_{\mathcal{F}}\).
To derive the growth condition, we consider \((\mathcal{F}z)(t)=(1td) \phi (z)\), and applying assumption (\(A_{2}\)), we get
$$\begin{aligned} \lVert \mathcal{F}z \rVert \leq \mathcal{Q} \lVert z \rVert_{ \mathbf{Z}}^{q_{1}}+ \mathcal{M}_{\phi }, \quad \mbox{for every } z \in \mathbf{Z}, \end{aligned}$$
where \(\mathcal{Q}=\vert d\vert C_{\phi }\). □
Lemma 3.4
The operator
\(\mathcal{G}:\mathbf{Z}\rightarrow \mathbf{Z}\)
is continuous. Moreover, it also satisfies the growth condition as
$$ \lVert \mathcal{G}z \rVert_{\mathbf{Z}} \leq \frac{2d+1}{\Gamma (qp+1)} \bigl( \mathcal{C}_{\theta } \lVert z \rVert_{\mathbf{Z}}^{q_{2}}+ \mathcal{M}_{\theta } \bigr) , $$
(9)
for every
\(z\in \mathbf{Z}\).
Proof
Let \(\{z_{m} \}\) be a sequence in the bounded set \(\bar{\mathbf{B}}= \{\lVert z\rVert_{\mathbf{Z}} \leq \kappa : z \in \mathbf{Z}\}\) such that \(z_{m} \rightarrow z\) as \(m\rightarrow \infty \) in \(\bar{ \mathbf{B}}\). We need to show that \(\lVert \mathcal{G}z_{m}  \mathcal{G}z \rVert_{\mathbf{Z}}\rightarrow 0\) as \(m\rightarrow \infty \). Since θ is continuous and \(z_{m}\rightarrow z\), therefore, \(\theta (\tau , z_{m}(\tau ))\rightarrow \theta (\tau , z( \tau ))\) as \(m\rightarrow 0\). Now consider
$$\begin{aligned}& \bigl\lvert (\mathcal{G}z_{m}) (t) (\mathcal{G}z) (t) \bigr\rvert \\& \quad \leq \frac{t d}{\Gamma (q)} \int_{0}^{1}(1\tau )^{q1}\bigl\lvert \theta \bigl(\tau ,z_{m}( \tau )\bigr) \theta \bigl(\tau , z(\tau ) \bigr)\bigr\rvert \,d\tau \\& \qquad {}+\frac{t \delta d}{\Gamma (qp)} \int_{0}^{\eta }(\eta \tau )^{qp1} \bigl\lvert \theta \bigl(\tau ,z_{m}(\tau )\bigr) \theta \bigl(\tau , z( \tau )\bigr)\bigr\rvert \,d \tau \\& \qquad {}+\frac{1}{\Gamma (q)} \int_{0}^{t}(t\tau )^{q1}\bigl\lvert \theta \bigl( \tau , z_{m}(\tau )\bigr) \theta \bigl(\tau , z(\tau ) \bigr)\bigr\rvert \,d\tau . \end{aligned}$$
In view of assumption (\(A_{3}\)) and thanks to the Lebesgue dominated convergence theorem, one has
$$\bigl\lVert (\mathcal{G}z_{m}) (t) (\mathcal{G}z) (t) \bigr\rVert _{\mathbf{Z}} \rightarrow 0\quad \mbox{as } m\rightarrow \infty , $$
which shows that \(\mathcal{G}\) is continuous. To obtain the growth condition for the nonlinear operator \(\mathcal{G}\), consider
$$\begin{aligned} \bigl\lvert (\mathcal{G}z) (t)\bigr\rvert =& \biggl\lvert \frac{t d}{\Gamma (q)} \int_{0}^{1}(1\tau )^{q1}\theta \bigl( \tau , z(\tau )\bigr)\,d\tau \\ &{} \frac{t \delta d}{\Gamma (qp)} \int_{0}^{\eta }(\eta \tau )^{qp1}\theta \bigl( \tau , z(\tau )\bigr)\,d\tau \\ &{} \frac{1}{\Gamma (q)} \int_{0}^{t}(t\tau )^{q1}\theta \bigl( \tau , z(\tau )\bigr)\,d\tau \biggr\rvert \\ \leq& \frac{2d+1}{\Gamma (qp+1)} \bigl(\Vert z\Vert _{\mathbf{Z}} ^{q_{2}}+\mathcal{M}_{\theta } \bigr). \end{aligned}$$
Applying assumption (\(A_{3}\)), we obtain the condition (9). □
Lemma 3.5
The operator
\(\mathcal{G}:\mathbf{Z}\rightarrow \mathbf{Z}\)
is compact. Consequently, \(\mathcal{G}\)
is
βLipschitz with zero constant.
Proof
To prove the required result, we take a bounded set \(\mathbf{D} \subset \bar{B}\subseteq \mathbf{Z}\). Let \(\{z_{m}\}\) be a sequence on \(\mathbf{D}\subset \bar{\mathbf{B}}\), then from (9) for every \(z_{m}\in \mathbf{D}\), we have
$$\lVert \mathcal{G}z_{m} \rVert_{\mathbf{Z}} \leq \frac{2d+1}{\Gamma (qp+1)} \bigl(\mathcal{C}_{\theta } \lVert z_{m} \rVert_{\mathbf{Z}}^{q_{2}}+ \mathcal{M}_{\theta }\bigr), $$
which implies that \(\mathcal{G}(\mathbf{D})\) is bounded in Z. Next, we will show that \(\{\mathcal{G}z_{m}\}\) is equicontinuous. For this purpose, let \(t_{1}< t_{2} \in (0,1)\), and using these relations \(\delta \eta^{qp}<1\), \(\frac{1}{\Gamma (q+1)}<\frac{1}{ \Gamma (qp+1)}\), we have
$$\begin{aligned} \bigl\lvert (\mathcal{G}z_{m}) (t_{1})(\mathcal{G}z) (t_{2}) \bigr\rvert \leq & \frac{(t_{2}t_{1})d}{\Gamma (q)} \int_{0}^{1}(1\tau )^{q1}\bigl\lvert \theta \bigl(\tau , z(\tau )\bigr)\bigr\rvert \,d\tau \\ &{}+ \frac{(t_{2}t_{1}) \delta d}{\Gamma (qp)} \int_{0}^{\eta }( \eta \tau )^{qp1}\bigl\lvert \theta \bigl(\tau , z(\tau )\bigr)\bigr\rvert \,d\tau \\ &{}+\frac{1}{\Gamma (q)} \int_{0}^{t_{1}} \bigl( (t_{1}\tau )^{q1}(t _{2}\tau )^{q1} \bigr) \bigl\lvert \theta \bigl(\tau , z(\tau )\bigr)\bigr\rvert \,d\tau \\ &{} \int_{t_{1}}^{t_{2}}(t_{2}\tau )^{q1} \bigl\lvert \theta \bigl(\tau , z( \tau )\bigr)\bigr\rvert \,d\tau , \end{aligned}$$
which on simplification takes the form
$$ \bigl\lvert (\mathcal{G}z_{m}) (t_{1}) ( \mathcal{G}z) (t_{2}) \bigr\rvert \leq \frac{( \mathcal{C}_{\theta }\kappa^{q_{2}}+\mathcal{M}_{\theta })}{\Gamma (qp+1)} \bigl[ (t_{2}t_{1})2d+\bigl(t_{2}^{q}t_{1}^{q} \bigr)2(t_{2}t_{1})^{q} \bigr] . $$
(10)
The right hand side of the inequality (10) goes to zero as \(t_{2}\rightarrow t_{1}\). Thus, \(\{\mathcal{G}z_{m}\}\) is equicontinuous and also \(\mathcal{G}(\mathbf{D})\) is relatively compact in Z by using the ArzeláAscoli theorem. Furthermore, in view of Proposition 2.3, the nonlinear operator \(\mathcal{G}\) is βLipschitz with constant zero. □
From now on, we will prove our main results.
Theorem 3.1
Under the hypotheses (\(A_{1}\))(\(A_{3}\)) equation (1) has at least one solution
\(u\in \mathbf{Z}\). Also, the set of solutions for (1) is bounded in
Z.
Proof
Thank to Proposition 2.2, the operator T is a strict βcontraction with constant \(\mathcal{K}_{\phi }\). Consider the set
$$\mathbf{S}_{0}=\bigl\{ z\in \mathbf{Z}: \mbox{there exists } \lambda \in [0,1] \mbox{ such that } z=\lambda \mathbf{T}z\bigr\} . $$
We need to show that \(\mathbf{S}_{0}\subset \mathbf{Z}\) is bounded. For this purpose, consider
$$\lvert z \rvert =\lvert \lambda \mathbf{T}z \rvert =\lambda \lvert \mathbf{T}z \rvert \leq \lambda \bigl(\lVert Fz \rVert_{\mathbf{Z}}+ \lVert \mathcal{G}z \rVert_{\mathbf{Z}} \bigr), $$
using (8) and (9), we have
$$\lvert z \rvert \leq \lambda \biggl[ \mathcal{Q} \lVert z \rVert_{\mathbf{Z}}^{q_{1}}+ \mathcal{M}_{\phi }+ \frac{2d+1}{\Gamma (qp+1)} \bigl(\Vert z\Vert _{\mathbf{Z}}^{q_{2}}+ \mathcal{M}_{\theta } \bigr) \biggr] , $$
which implies using \(\lambda <1\) that
$$ \lVert z \rVert_{\mathbf{Z}} \leq \biggl[ \mathcal{Q} \lVert z \rVert_{\mathbf{Z}}^{q_{1}}+\mathcal{M}_{\phi }+ \frac{2d+1}{\Gamma (qp+1)} \bigl(\Vert z\Vert _{\mathbf{Z}}^{q_{2}}+ \mathcal{M}_{\theta } \bigr) \biggr] . $$
(11)
Hence, (11) and \(q_{1}, q_{2}\in (0, 1)\) shows that \(\mathbf{S}_{0}\) is bounded in Z. If it is not bounded then assume that \(\Vert z\Vert _{\mathbf{Z}}=\rho \) and consider that \(\rho \rightarrow \infty \). Then from (11), we have
$$ 1 \leq \frac{ [ \mathcal{Q} \lVert z \rVert_{\mathbf{Z}}^{q_{1}}+ \mathcal{M}_{\phi }+\frac{2d+1}{\Gamma (qp+1)} (\Vert z\Vert _{ \mathbf{Z}}^{q_{2}}+\mathcal{M}_{\theta } ) ] }{\rho }, $$
(12)
where due to assumption
$$\rho \rightarrow \infty \quad \mbox{yields } 1\leq 0. $$
This is impossible, so we take \(\mathbf{S}_{0}\) to be bounded.
Therefore, we conclude that the operator T has at least one fixed point and the set of fixed points is bounded in Z. □
We make the following assumption for discussion of data dependence of solutions:
 (\(A_{4}\)):

There exist constants \(\mathcal{L}_{\theta }>0\), \(\lambda \in [0,1\frac{1}{r}]\) for some \(r\in (0,1\frac{1}{1q})\) such that
$$\bigl\lvert \theta (t,z)\theta (t,\bar{z}) \bigr\rvert \leq \mathcal{L}_{ \theta }\lvert z\bar{z} \rvert^{\lambda }, \quad \mbox{for each } t\in \mathfrak{J}, \mbox{and for all } z, \bar{z} \in \mathbf{R}. $$
Theorem 3.2
Assuming that (\(A_{1}\))(\(A_{6}\)) hold, let
\(z(t) \bar{z}(t)\)
be the solutions of (FDE) (1) with associated boundary conditions. Then there exists a constant
\(\mathcal{M}^{\ast }>0\)
such that
$$\bigl\lvert z(t)\bar{z}(t) \bigr\rvert \leq \mathcal{M}^{\ast } \biggl( \frac{1}{(p(1q)+1)} \biggr)^{\frac{1}{p}}. $$
Proof
Consider \(\lvert z(t)\bar{z}(t) \rvert \), and thanks to (\(A_{1}\)), (\(A_{3}\)), and (\(A_{4}\)), we get
$$\begin{aligned} \bigl\lvert z(t)\bar{z}(t) \bigr\rvert \leq& \mathcal{K}_{\phi } \bigl\lvert z(t) \bar{z}(t) \bigr\rvert + \frac{\mathcal{L}_{\theta }}{\Gamma (q)} \int_{0} ^{t}(t\tau )^{q1}\bigl\lvert z( \tau )\bar{z}(\tau ) \bigr\rvert ^{\lambda }\,d\tau \\ &{} + \frac{ d\mathcal{L}_{\theta }}{\Gamma (q)} \int_{0}^{1}(1 \tau )^{q1}\bigl\lvert z( \tau )\bar{z}(\tau ) \bigr\rvert ^{\lambda }\,d\tau \\ &{} +\frac{\delta \mathcal{L}_{\theta } d}{\Gamma (qp)} \int _{0}^{\eta }(\eta \tau )^{qp1} \bigl\lvert z(\tau )\bar{z}(\tau ) \bigr\rvert ^{\lambda }\,d\tau . \end{aligned}$$
Upon further simplification and using (2), we get
$$\begin{aligned} \bigl\lvert z(t)\bar{z}(t) \bigr\rvert \leq& \frac{1}{1\mathcal{K}_{\phi }} \biggl[ \frac{\mathcal{L}_{\theta }}{\Gamma (q)} \int_{0}^{t}(t\tau )^{q1} \bigl\lvert z( \tau )\bar{z}(\tau ) \bigr\rvert ^{\lambda }\,d\tau \biggr] \\ &{} +\frac{1}{1\mathcal{K}_{\phi }} \biggl[\frac{d \mathcal{L}_{\theta }}{\Gamma (q)} \int_{0}^{1}(1\tau )^{q1}\bigl\lvert z( \tau )\bar{z}(\tau ) \bigr\rvert ^{\lambda }\,d\tau \\ &{} +\frac{\delta \mathcal{L}_{\theta } d}{\Gamma (qp)} \int _{0}^{\eta }(\eta \tau )^{qp1} \bigl\lvert z(\tau )\bar{z}(\tau ) \bigr\rvert ^{\lambda }\,d\tau \biggr]. \end{aligned}$$
Hence using (2) and Theorem 2.9, we obtain
$$\lvert z\bar{z} \rvert \leq \mathcal{M}^{\ast }, $$
where \(\mathcal{M}^{\ast }= e^{\mathcal{M}}\) and
$$\begin{aligned} M =&\frac{1}{1\mathcal{K}_{\phi }} \biggl[ \frac{\mathcal{L}_{\theta }}{ \Gamma (q)} \int_{0}^{t}(t\tau )^{q1}\bigl\lvert z( \tau )\bar{z}(\tau ) \bigr\rvert ^{\lambda }\,d\tau \biggr] \\ &{}+\frac{1}{1\mathcal{K}_{\phi }} \biggl[ \frac{d \mathcal{L}_{\theta }}{ \Gamma (q)} \int_{0}^{1}(1\tau )^{q1}\bigl\lvert z( \tau )\bar{z}(\tau ) \bigr\rvert ^{\lambda }\,d\tau \\ &{} +\frac{\delta \mathcal{L}_{\theta } d}{\Gamma (qp)} \int_{0}^{\eta }(\eta \tau )^{qp1} \bigl\lvert z(\tau )\bar{z}( \tau ) \bigr\rvert ^{\lambda }\,d\tau \biggr] . \end{aligned}$$
□
Now, rewrite assumption (\(A_{4}\)) as follows.
 (\(A_{5}\)):

For a \(\mathcal{L}_{\theta }>0\), the following relation holds:
$$\bigl\lvert \theta (t,z)\theta (t,\bar{z}) \bigr\rvert \leq \mathcal{L}_{ \theta }\lvert z\bar{z }\rvert ,\quad \mbox{for each } t\in \mathfrak{J}, \mbox{and for each } z, \bar{z} \in \mathbf{R}. $$
Theorem 3.3
Assume that the hypotheses (\(A_{1}\))(\(A_{5}\)) hold, then FDE (1) has a unique solution
\(z\in \mathbf{Z}\)
if
\(\frac{ \mathcal{M}^{\ast }}{1\mathcal{K}_{\phi }}<1\).
Proof
As we investigated in Theorem 3.1
\(z(t)\in \mathbf{Z}\) is a solution of (1). Let \(\overline{z}(t)\) be another solution of (1). Then, repeating the same procedure as in Theorem 3.2 and using assumptions (\(A_{1}\)), (\(A_{3}\)) and (\(A_{5}\)), we obtain
$$\begin{aligned} \bigl\lvert \mathbf{T}z(t)\mathbf{T}\bar{z}(t) \bigr\rvert \leq& \frac{1}{1 \mathcal{K}_{\phi }} \biggl[ \mathcal{K}_{\phi } \lVert z\bar{z} \rVert_{\mathbf{Z}}+\frac{\mathcal{L}_{\theta }}{\Gamma (q)} \int_{0} ^{t}(t\tau )^{q1}\bigl\lvert z( \tau )\bar{z}(\tau ) \bigr\rvert ^{\lambda }\,d\tau \biggr] \\ &{} +\frac{1}{1\mathcal{K}_{\phi }} \biggl[\frac{ \mathcal{L} _{\theta }t d}{\Gamma q} \int_{0}^{1}(1\tau )^{q1}\bigl\lvert z_{1}( \tau )z_{2}(\tau ) \bigr\rvert ^{\lambda }\,d\tau \\ &{} +\frac{\delta d\mathcal{L}_{\theta }}{\Gamma (qp)} \int _{0}^{\eta }(\eta s)^{qp1} \bigl\lvert z(\tau )\bar{z}(\tau ) \bigr\rvert ^{\lambda }\,d\tau \biggr], \end{aligned}$$
using the inequality in Theorem 2.9, we get
$$\lVert \mathbf{T}z\mathbf{T}\bar{z} \rVert_{\mathbf{Z}} \leq \frac{ \mathcal{M}^{\ast }}{1\mathcal{K}_{\phi }}\lVert z\bar{z} \rVert_{\mathbf{Z}},\quad t\in \mathfrak{J}, $$
which produces the uniqueness of z. □