Now, after stating the above preparations, we can get our main results. First we start with a lemma.
Lemma 3.1
Let \(\pounds \in (2,3]\) and \(v(t) \in \mathfrak{A}\). Then the quantum fractional problem \(^{c}\mathcal{D}_{q}^{\pounds }l(t)=v(t)\) with boundary condition (2) has a unique solution which is obtained by
$$ l(t)=\mathcal{I}^{\pounds }_{q}v(t)+\theta \mathfrak{B}_{1}(t) \mathcal{I}^{\pounds }_{q}v(\eth )+\theta \mathfrak{B}_{2}(t) \mathcal{I}^{\pounds }_{q} \bigl[\mathfrak{I}^{\pounds }_{q}v(r)\bigr](1)+\theta \mathfrak{B}_{3}(t)\mathcal{I}^{\pounds -2}_{q}v(1){} $$
(7)
such that
$$ \theta = \biggl[\biggl(2{\boldsymbol{P}}+\frac{c}{3}\biggr) \biggl(a \eth +\frac{b}{2}-{\boldsymbol{S}}(a+b)\biggr)-c\biggl(a\eth ^{2}+ \frac{b}{3}\biggr) \biggl(\frac{1}{2}-{ \boldsymbol{S}}\biggr)\biggr]^{-1}\ne 0 $$
(8)
and
$$\begin{aligned}& \mathfrak{B}_{1}(t)=a\biggl(2{\boldsymbol{P}}+\frac{c}{3} \biggr) ({ \boldsymbol{S}}-t)+ac\biggl(\frac{1}{2}-{\boldsymbol{S}} \biggr)t^{2}, \\& \begin{aligned} \mathfrak{B}_{2}(t)={}& \biggl[c\biggl(a\eth ^{2}+ \frac{b}{3}\biggr)-(a+b) \biggl(2{ \boldsymbol{P}}+\frac{c}{3} \biggr)\biggr](t-{\boldsymbol{S}})\\ &{}+c\biggl[b\biggl({ \boldsymbol{S}}+ \frac{1}{2}\biggr)+{\boldsymbol{S}}(a+b)-a \eth -\frac{b}{2} \biggr]t^{2}, \end{aligned} \\& \mathfrak{B}_{3}(t)={\boldsymbol{P}}\biggl(a\eth ^{2}+ \frac{b}{3}\biggr) (t-{\boldsymbol{S}})+{\boldsymbol{P}}\biggl({ \boldsymbol{S}}(a+b)-a\eth -\frac{b}{2}\biggr)t^{2}. \end{aligned}$$
Proof
With regard to Lemma 2.7 the solution of \(^{c}\mathcal{D}_{q}^{\pounds }l(t)=v(t)\) is
$$ l(t)=\mathcal{I}^{\pounds }_{q}v(t)+ \mathfrak{y}_{0} + \mathfrak{y}_{1} t + \mathfrak{y}_{2} t^{2} $$
(9)
such that \(\mathfrak{y}_{0}, \mathfrak{y}_{1} \text{and} \mathfrak{y}_{2} \in \mathbb{R}\). Now, by taking derivative from \(l(t)\), we have
$$ \textstyle\begin{cases} l'(t)=\mathfrak{y}_{1}+2\mathfrak{y}_{2} t+\mathcal{I}^{\pounds -1}_{q}v(t), \\ l''(t)=2\mathfrak{y}_{2}+ \mathcal{I}^{\pounds -2}_{q}v(t), \end{cases} $$
(10)
and by exerting the boundary conditions (2) to (10) we have
$$ \textstyle\begin{cases} \mathfrak{y}_{0}+{\boldsymbol{S}}\mathfrak{y}_{1}=0, \\ (a+b)\mathfrak{y}_{0}+(a\eth +\frac{b}{2})\mathfrak{y}_{1} +(a\eth ^{2}+ \frac{b}{3})\mathfrak{y}_{2}=-a\mathcal{I}^{\pounds }_{q}v(\eth )-b \mathcal{I}^{\pounds }_{q}[\mathfrak{I}^{\pounds }_{q}v(r)]_{ (1)}, \\ c\mathfrak{y}_{0}+\frac{c}{2}\mathfrak{y}_{1}+({\boldsymbol{P}}+\frac{c}{3})\mathfrak{y}_{2}=-{\boldsymbol{P}}\mathcal{I}^{\pounds -2}_{q}v(1)-c\mathcal{I}^{\pounds }_{q}[ \mathfrak{I}^{\pounds }_{q}v(r)]_{(1)}. \end{cases} $$
Now we can compute \(\mathfrak{y}_{0}\), \(\mathfrak{y}_{1}\), \(\mathfrak{y}_{2}\) as follows:
$$ \textstyle\begin{cases} \mathfrak{y}_{0}={\boldsymbol{S}}\theta [a(2{ \boldsymbol{P}}+\frac{c}{3})\mathcal{I}^{\pounds }_{q}v( \eth )- [c(a\eth ^{2}+\frac{b}{3}) \\ \hphantom{\mathfrak{y}_{0}=}{}-(a+b)(2{\boldsymbol{P}}+\frac{c}{3}) ]\mathcal{I}^{ \pounds }_{q}[\mathfrak{I}^{\pounds }_{q}v(r)]_{(1)}-{ \boldsymbol{P}}(a\eth ^{2}+\frac{b}{3})\mathcal{I}^{ \pounds -2}_{q}v(1) ], \\ \mathfrak{y}_{1}=\theta [ [c(a\eth ^{2}+\frac{b}{3})-(a+b)(2{ \boldsymbol{P}}+\frac{c}{3}) ]\mathcal{I}^{\pounds }_{q}[ \mathfrak{I}^{\pounds }_{q}v(r)]_{(1)} \\ \hphantom{\mathfrak{y}_{1}=}{}-a(2{\boldsymbol{P}}+\frac{c}{3})\mathcal{I}^{\pounds }_{q}v( \eth )+{\boldsymbol{P}}(a\eth ^{2}+\frac{b}{3})\mathcal{I}^{ \pounds -2}_{q}v(1) ], \\ \mathfrak{y}_{2}=\theta [ac(\frac{1}{2}-{\boldsymbol{S}})\mathcal{I}^{\pounds }_{q}v(\eth )+ [c (b({ \boldsymbol{S}}+\frac{1}{2})+{\boldsymbol{S}}(a+b)-a \eth -\frac{b}{2} ) ]\mathcal{I}^{\pounds }_{q}[\mathfrak{I}^{ \pounds }_{q}v(r)]_{(1)} \\ \hphantom{\mathfrak{y}_{2}=}{} +{\boldsymbol{P}}({\boldsymbol{S}}(a+b)-a \eth -\frac{b}{2})\mathcal{I}^{\pounds -2}_{q}v(1) ]. \end{cases} $$
Now, by replacing \(\mathfrak{y}_{0}\), \(\mathfrak{y}_{1}\), \(\mathfrak{y}_{2}\) in (9), we obtain (7). □
Notation 3.2
To continue the work and for ease of understanding of the calculations performed, we introduce some symbols here. According to the definition of \(\mathfrak{B}_{1}(t)\), \(\mathfrak{B}_{2}(t)\), \(\mathfrak{B}_{3}(t)\), we have
$$\begin{aligned}& \bigl\vert \mathfrak{B}_{1}(t) \bigr\vert \le \vert a \vert \biggl(2 \vert { \boldsymbol{P}} \vert +\frac{ \vert c \vert }{3} \biggr) \vert { \boldsymbol{S}}+1 \vert + \vert c \vert \biggl\vert \frac{1}{2}+ \biggr\vert {\boldsymbol{S}} \vert \vert := \mathfrak{B}^{*}_{1}, \\& \begin{aligned} \bigl\vert \mathfrak{B}_{2}(t) \bigr\vert \le{}& \vert c \vert \biggl( \vert a \vert \eth ^{2}+\frac{ \vert b \vert }{3} \biggr)+ \vert a+b \vert \biggl(2 \vert { \boldsymbol{P}} \vert + \frac{ \vert c \vert }{3} \biggr) \bigl(1+ \vert { \boldsymbol{S}} \vert \bigr)\\ &{}+ \vert c \vert \biggl( \vert b \vert \biggl( \vert {\boldsymbol{S}} \vert +\frac{1}{2}\biggr)+ \vert {\boldsymbol{S}} \vert \vert a+b \vert + \vert a \vert \eth + \frac{ \vert b \vert }{2} \biggr):= \mathfrak{B}^{*}_{2}, \end{aligned} \\& \bigl\vert \mathfrak{B}_{3}(t) \bigr\vert \le \vert { \boldsymbol{P}} \vert \biggl( \vert a \vert \eth ^{2}+ \frac{ \vert b \vert }{3} \biggr) \bigl(1+ \vert { \boldsymbol{S}} \vert \bigr)+ \vert {\boldsymbol{P}} \vert \biggl( \vert {\boldsymbol{S}} \vert \vert a+b \vert + \vert a \vert \eth +\frac{ \vert b \vert }{2} \biggr):= \mathfrak{B}^{*}_{3}; \end{aligned}$$
moreover
$$\begin{aligned}& \bigl\vert \mathfrak{B}'_{1}(t) \bigr\vert \le \vert a \vert \biggl(2 \vert { \boldsymbol{P}} \vert + \frac{ \vert c \vert }{3} \biggr)+ \vert a \vert \vert c \vert \biggl\vert \frac{1}{2}+ \biggr\vert {\boldsymbol{S}} \vert \vert := \mathfrak{B}^{*'}_{1}, \\& \begin{aligned} \bigl\vert \mathfrak{B}'_{2}(t) \bigr\vert \le{}& \vert c \vert \biggl( \vert a \vert \eth ^{2}+ \frac{ \vert b \vert }{3} \biggr)+ \vert a+b \vert \biggl(2 \vert { \boldsymbol{P}} \vert +\frac{ \vert c \vert }{3} \biggr)\\ &{}+2 \vert c \vert \biggl( \vert b \vert \biggl( \vert {\boldsymbol{S}} \vert + \frac{1}{2}\biggr)+ \vert {\boldsymbol{S}} \vert \vert a+b \vert + \vert a \vert \eth +\frac{ \vert b \vert }{2} \biggr):=\mathfrak{B}^{*'}_{2}, \end{aligned} \\& \bigl\vert \mathfrak{B}'_{3}(t) \bigr\vert \le \vert {\boldsymbol{P}} \vert \biggl( \vert a \vert \eth ^{2}+ \frac{ \vert b \vert }{3} \biggr)+2 \vert { \boldsymbol{P}} \vert \biggl( \vert {\boldsymbol{S}} \vert \vert a+b \vert + \vert a \vert \eth + \frac{ \vert b \vert }{2} \biggr):= \mathfrak{B}^{*'}_{3}, \end{aligned}$$
also
$$\begin{aligned}& \bigl\vert \mathfrak{B}''_{1}(t) \bigr\vert \le \vert a \vert \vert c \vert \biggl\vert \frac{1}{2}+ \biggr\vert {\boldsymbol{S}} \vert \vert := \mathfrak{B}^{*''}_{1}, \\& \bigl\vert \mathfrak{B}''_{2}(t) \bigr\vert \le 2 \vert c \vert \biggl( \vert b \vert \biggl( \vert { \boldsymbol{S}} \vert +\frac{1}{2}\biggr)+ \vert {\boldsymbol{S}} \vert \vert a+b \vert + \vert a \vert \eth + \frac{ \vert b \vert }{2} \biggr):=\mathfrak{B}^{*''}_{2}, \\& \bigl\vert \mathfrak{B}''_{3}(t) \bigr\vert \le 2 \vert {\boldsymbol{P}} \vert \biggl( \vert { \boldsymbol{S}} \vert \vert a+b \vert + \vert a \vert \eth + \frac{ \vert b \vert }{2} \biggr):=\mathfrak{B}^{*''}_{3}{ .} \end{aligned}$$
Now, by applying quantum Caputo fractional derivative from order \(\hslash _{i} \in (1,2]\), \(i=1, 2\) on \(\mathfrak{B}_{1}(t)\), \(\mathfrak{B}_{2}(t)\), \(\mathfrak{B}_{3}(t)\), we get
$$\begin{aligned}& ^{c}\mathcal{D}_{q}^{\hbar _{i}} \mathfrak{B}_{1}(t)=a\biggl(2{\boldsymbol{P}}+\frac{c}{3} \biggr) \biggl[- \frac{1}{\Gamma _{q}(2-\hslash _{i})}t^{(1-\hslash _{i})} \biggr]+ac\biggl( \frac{1}{2}-{\boldsymbol{S}}\biggr) \biggl[ \frac{2}{\Gamma _{q}(3-\hslash _{i})}t^{(2-\hslash _{i})} \biggr]{ ,} \\& \begin{aligned} ^{c}\mathcal{D}_{q}^{\hbar _{i}} \mathfrak{B}_{2}(t) ={}& \biggl(c\biggl(a\eth ^{2}+ \frac{b}{3}\biggr)-(a+b) \biggl(2{\boldsymbol{P}}+\frac{c}{3} \biggr) \biggr) \biggl[\frac{1}{\Gamma _{q}(2-\hslash _{i})}t^{(1-\hslash _{i})} \biggr] \\ &{}+ c \biggl(b\biggl({\boldsymbol{S}}+\frac{1}{2}\biggr)+{ \boldsymbol{S}}(a+b)-a\eth -\frac{b}{2} \biggr) \biggl[ \frac{2}{\Gamma _{q}(3-\hslash _{i})}t^{(2-\hslash _{i})} \biggr]{ ,} \end{aligned} \\& \begin{aligned} ^{c}\mathcal{D}_{q}^{\hbar _{i}} \mathfrak{B}_{3}(t)={}&{\boldsymbol{P}}\biggl(a\eth ^{2}+ \frac{b}{3}\biggr) \biggl[ \frac{1}{\Gamma _{q}(2-\hslash _{i})}t^{(1-\hslash _{i})} \biggr]\\ &{}+{ \boldsymbol{P}} \biggl({\boldsymbol{S}}(a+b)-a \eth - \frac{b}{2} \biggr) \biggl[\frac{2}{\Gamma _{q}(3-\hslash _{i})}t^{(2- \hslash _{i})} \biggr], \end{aligned} \end{aligned}$$
from which it can be concluded
$$\begin{aligned}& \bigl\vert ^{c}\mathcal{D}_{q}^{\hbar _{i}} \mathfrak{B}_{1}(t) \bigr\vert \le \vert a \vert \biggl(2 \vert {\boldsymbol{P}} \vert + \frac{ \vert c \vert }{3} \biggr) \biggl[ \frac{1}{\Gamma _{q}(2-\hslash _{i})} \biggr]+ \vert a \vert \vert c \vert \biggl\vert \frac{1}{2}+ \biggr\vert {\boldsymbol{S}} \vert \vert \biggl[ \frac{2}{\Gamma _{q}(3-\hslash _{i})} \biggr]:=\mathfrak{B}^{**}_{1}{ ,} \\& \begin{aligned} \bigl\vert ^{c} \mathcal{D}_{q}^{\hbar _{i}}\mathfrak{B}_{2}(t) \bigr\vert \le{}& \vert c \vert \biggl( \vert a \vert \eth ^{2}+ \frac{ \vert b \vert }{3} \biggr)+ \vert a+b \vert \biggl(2 \vert { \boldsymbol{P}} \vert + \frac{ \vert c \vert }{3} \biggr) \biggl[ \frac{1}{\Gamma _{q}(2-\hslash _{i})} \biggr] \\ &{}+ \vert c \vert \biggl( \vert b \vert \biggl( \vert { \boldsymbol{S}} \vert +\frac{1}{2}\biggr)+ \vert {\boldsymbol{S}} \vert \vert a+b \vert + \vert a \vert \eth +\frac{ \vert b \vert }{2} \biggr) \biggl[ \frac{2}{\Gamma _{q}(3-\hslash _{i})} \biggr]:=\mathfrak{B}^{**}_{2}{ ,} \end{aligned} \\& \begin{aligned} \bigl\vert ^{c}\mathcal{D}_{q}^{\hbar _{i}} \mathfrak{B}_{3}(t) \bigr\vert \le {}&\vert {\boldsymbol{P}} \vert \biggl( \vert a \vert \eth ^{2}+ \frac{ \vert b \vert }{3} \biggr) \biggl[ \frac{1}{\Gamma _{q}(2-\hslash _{i})} \biggr]\\ &{}+ \vert {\boldsymbol{P}} \vert \biggl( \vert {\boldsymbol{S}} \vert \vert a+b \vert + \vert a \vert \eth + \frac{ \vert b \vert }{2} \biggr) \biggl[ \frac{2}{\Gamma _{q}(3-\hslash _{i})} \biggr]:= \mathfrak{B}^{**}_{3}. \end{aligned} \end{aligned}$$
The following conditions must be met to prove our main theorem.
(\(\mathfrak{C}_{1}\)) Given the multivalued map \(\mathcal{F}:\mathfrak{J}\times \mathbb{R}^{5}\to \mathcal{P}_{cp}( \mathbb{R})\) is integrable bounded, so that \(\mathcal{F}(\cdot,v,u,x,y,z):[0. 1]\to \mathcal{P}_{cp}(\mathbb{R})\) is measurable.
(\(\mathfrak{C}_{2}\)) For the nondecreasing (usc) map \(\psi :[0,\infty )\to [0,\infty )\), we have \(\liminf_{w \to \infty }(w - \psi (w))>0\) and \(\psi (w) < w\) for any \(w>0\).
(\(\mathfrak{C}_{3}\)) There exists \(\mho \in C(\mathfrak{J},[0,\infty ))\) such that
$$ \begin{aligned} &\mathcal{H}_{\mathfrak{d}} \bigl(\mathcal{F}(t,u_{1},u_{2},u_{3},u_{4},u_{5}), \mathcal{F}(t,v_{1},v_{2},v_{3},v_{4},v_{5}) \bigr)\\ &\quad \le \frac{1}{\boldsymbol{\Im _{1}}+\boldsymbol{\Im _{2}}+\boldsymbol{\Im _{3}}+\boldsymbol{\Im _{4_{1}}}+\boldsymbol{\Im _{4_{2}}}} \mho (t)\psi \Biggl(\sum _{i=1}^{5} \vert u_{k}-v_{k} \vert \Biggr) \end{aligned} $$
for all \(t\in \mathfrak{J}\) and \(u_{k},v_{k} \in \mathbb{R}\), \(k=1, \dots ,5\), where
$$\begin{aligned}& \boldsymbol{\Im _{1}}= \Vert \mho \Vert \biggl[ \frac{1}{\Gamma _{q}(\pounds +1)}+ \frac{ \vert \theta \vert \mathfrak{B}^{*}_{1}}{\Gamma _{q}(\pounds +1)} \eth ^{\pounds }+ \frac{ \vert \theta \vert \mathfrak{B}^{*}_{2}}{\Gamma _{q}(\pounds +2)}+ \frac{ \vert \theta \vert \mathfrak{B}^{*}_{3}}{\Gamma _{q}(\pounds -1)} \biggr], \\& \boldsymbol{\Im _{2}}= \Vert \mho \Vert \biggl[ \frac{1}{\Gamma _{q}(\pounds -1)}+ \frac{ \vert \theta \vert \mathfrak{B}^{*'}_{1}}{\Gamma _{q}(\pounds +1)} \eth ^{\pounds }+ \frac{ \vert \theta \vert \mathfrak{B}^{*'}_{2}}{\Gamma _{q}(\pounds +2)}+ \frac{ \vert \theta \vert \mathfrak{B}^{*'}_{3}}{\Gamma _{q}(\pounds -1)} \biggr], \\& \boldsymbol{\Im _{3}}= \Vert \mho \Vert \biggl[ \frac{1}{\Gamma _{q}(\pounds -2)}+ \frac{ \vert \theta \vert \mathfrak{B}^{*''}_{1}}{\Gamma _{q}(\pounds +1)} \eth ^{\pounds }+ \frac{ \vert \theta \vert \mathfrak{B}^{*''}_{2}}{\Gamma _{q}(\pounds +2)}+ \frac{ \vert \theta \vert \mathfrak{B}^{*''}_{3}}{\Gamma _{q}(\pounds -1)} \biggr], \end{aligned}$$
and for \(i=1,2\),
$$ \boldsymbol{\Im _{4_{i}}}= \Vert \mho \Vert \biggl[ \frac{1}{\Gamma _{q}(\pounds -\hslash _{i}+1)}+ \frac{ \vert \theta \vert \mathfrak{B}^{**}_{1}}{\Gamma _{q}(\pounds +1)} \eth ^{\pounds }+ \frac{ \vert \theta \vert \mathfrak{B}^{**}_{2}}{\Gamma _{q}(\pounds +2)}+ \frac{ \vert \theta \vert \mathfrak{B}^{**}_{3}}{\Gamma _{q}(\pounds -1)} \biggr]. $$
(\(\mathfrak{C}_{4}\)) Let \(\boldsymbol{\mathfrak{M}}:\mathcal{K}\to 2^{\mathcal{K}}\) be given as follows:
$$ \boldsymbol{\mathfrak{M}}(\boldsymbol{k})=\bigl\{ \mathfrak{p} \in \mathcal{K}: \boldsymbol{\exists } \mathfrak{f}\in \mathcal{S}^{*}_{ \mathcal{F},l} \mathfrak{p}(t)=\ell (t), \forall t \in \mathfrak{J} \bigr\} $$
such that
$$ \ell (t)=\mathcal{I}^{\pounds }_{q}\mathfrak{f}(t)+\theta \mathfrak{B}_{1}(t) \mathcal{I}^{\pounds }_{q} \mathfrak{f}(\eth )+\theta \mathfrak{B}_{2}(t) \mathcal{I}^{\pounds }_{q} \bigl[\mathfrak{I}^{\pounds }_{q}\mathfrak{f}(r)\bigr](1)+ \theta \mathfrak{B}_{3}(t)\mathcal{I}^{\pounds -2}_{q} \mathfrak{f}(1). $$
Theorem 3.3
Suppose that conditions (\(\mathfrak{C}_{1})-(\mathfrak{C}_{4}\)) are satisfied. If \(\boldsymbol{\mathfrak{M}}:\mathcal{K}\to 2^{\mathcal{K}}\) has the approximate endpoint property, then quantum problem (1)–(2) has a solution.
Proof
We prove that the endpoint of \(\boldsymbol{\mathfrak{M}}:\mathcal{K}\to 2^{\mathcal{K}}\) is the solution to inclusion (1)–(2). For this, we first show that \(\boldsymbol{\mathfrak{M}}(\boldsymbol{k})\) is a closed subset of \(\mathcal{K}\) for all \(\boldsymbol{k} \in \mathcal{K}\).
For \(\forall \boldsymbol{k} \in \mathcal{K}\), the map \(t \mapsto \mathcal{F}(t,l(t),l'(t),l''(t),^{c}\mathcal{D}_{q}^{ \hbar _{1}}l(t),^{c}\mathcal{D}_{q}^{\hbar _{2}}l(t))\) is measurable and closed value. So, it has measurable selection, and hence \(\mathfrak{f}\in \mathcal{S}^{*}_{\mathcal{F},l}\ne \phi \) for all \(l \in \mathcal{K}\).
Let \(\boldsymbol{k} \in \mathcal{K}\), and \(\{\mathfrak{x}_{n} \}_{n\ge 1}\) be a sequence in \(\boldsymbol{\mathfrak{M}}( \boldsymbol{k})\) such that \(\mathfrak{x}_{n} \to \mathfrak{x}\). \(\forall n \in \mathbb{N}\), choose \(\mathfrak{f}_{n} \in \mathcal{S}^{*}_{\mathcal{F},l}\), where
$$ \mathfrak{x}_{n}=\mathcal{I}^{\pounds }_{q} \mathfrak{f}_{n}(t)+\theta \mathfrak{B}_{1}(t) \mathcal{I}^{\pounds }_{q}\mathfrak{f}_{n}(\eth )+ \theta \mathfrak{B}_{2}(t)\mathcal{I}^{\pounds }_{q} \bigl[\mathfrak{I}^{ \pounds }_{q}\mathfrak{f}_{n}(r) \bigr](1)+\theta \mathfrak{B}_{3}(t) \mathcal{I}^{\pounds -2}_{q} \mathfrak{f}_{n}(1){} $$
for all \(t\in \mathfrak{J}\).
As we know, \(\mathcal{F}\) has compact values, then the sequence \(\mathfrak{f}_{n}\) has a subsequence that converges to some \(\mathfrak{f} \in L^{1}[0,1]\). We show this again with \(\mathfrak{f}_{n}\).
It is easy to check that \(\mathfrak{f} \in \mathcal{S}^{*}_{\mathcal{F},l}\) and
$$ \mathfrak{x}_{n}(t) \to \mathfrak{x}(t)=\mathcal{I}^{\pounds }_{q} \mathfrak{f}(t)+\theta \mathfrak{B}_{1}(t)\mathcal{I}^{\pounds }_{q} \mathfrak{f}(\eth )+\theta \mathfrak{B}_{2}(t)\mathcal{I}^{\pounds }_{q} \bigl[ \mathfrak{I}^{\pounds }_{q}\mathfrak{f}(r)\bigr](1)+ \theta \mathfrak{B}_{3}(t) \mathcal{I}^{\pounds -2}_{q} \mathfrak{f}(1) $$
for all \(t\in \mathfrak{J}\). Indeed, this gives that \(\mathfrak{x} \in \boldsymbol{\mathfrak{M}}(\boldsymbol{k})\), therefore \(\mathcal{K}\) has closed values. Moreover, since \(\mathcal{F}\) has compact values, then \(\boldsymbol{\mathfrak{M}}(\boldsymbol{k})\) for all \(\boldsymbol{k} \in \mathcal{K}\) is a bounded set.
Finally, we shall show that \(\mathcal{H}_{\mathfrak{d}} (\boldsymbol{\mathfrak{M}}(u), \boldsymbol{\mathfrak{M}}(v) )\le \psi (\Vert u-v\Vert )\). Let \(u,v \in \mathcal{K}\) and \(\mathfrak{p}_{1} \in \boldsymbol{\mathfrak{M}}(v)\). Choose \(\mathfrak{f}_{1}\in \mathcal{S}^{*}_{\mathcal{F},l}\) such that
$$ \mathfrak{p}_{1}(t)=\mathcal{I}^{\pounds }_{q} \mathfrak{f}_{1}(t)+ \theta \mathfrak{B}_{1}(t) \mathcal{I}^{\pounds }_{q}\mathfrak{f}_{1}( \eth )+ \theta \mathfrak{B}_{2}(t)\mathcal{I}^{\pounds }_{q} \bigl[ \mathfrak{I}^{\pounds }_{q}\mathfrak{f}_{1}(r) \bigr](1)+\theta \mathfrak{B}_{3}(t) \mathcal{I}^{\pounds -2}_{q} \mathfrak{f}_{1}(1){} $$
for almost all \(t\in \mathfrak{J}\).
But since
$$\begin{aligned}& \mathcal{H}_{\mathfrak{d}} \bigl(\mathcal{F}(t,u_{1},u_{2},u_{3},u_{4},u_{5}), \mathcal{F}(t,v_{1},v_{2},v_{3},v_{4},v_{5}) \bigr)\\& \quad \le \frac{1}{\boldsymbol{\Im _{1}}+\boldsymbol{\Im _{2}}+\boldsymbol{\Im _{3}}+\boldsymbol{\Im _{4_{1}}}+\boldsymbol{\Im _{4_{2}}}} \mho (t)\psi \Biggl(\sum _{i=1}^{5} \vert u_{k}-v_{k} \vert \Biggr){ ,} \end{aligned}$$
thus \(\exists w \in \mathcal{F}(t,l(t),l'(t),l''(t),^{c}\mathcal{D}_{q}^{ \hbar _{1}}l(t),^{c}\mathcal{D}_{q}^{\hbar _{2}}l(t))\) such that \(\forall t\in \mathfrak{J}\):
$$ \bigl\vert \mathfrak{f}_{1}(t)-w \bigr\vert \le \frac{1}{\boldsymbol{\Im _{1}}+\boldsymbol{\Im _{2}}+\boldsymbol{\Im _{3}}+\boldsymbol{\Im _{4_{1}}}+\boldsymbol{\Im _{4_{2}}}} \mho (t)\psi \Biggl(\sum_{i=1}^{5} \vert u_{k}-v_{k} \vert \Biggr). $$
Regard the set-valued map \(\boldsymbol{\mathfrak{N}}:\mathfrak{J}\to \mathcal{P}(\mathbb{R})\) by
$$ \boldsymbol{\mathfrak{N}}(t)=\Biggl\{ w\in \mathbb{R}: \bigl\vert \mathfrak{f}_{1}(t)-w \bigr\vert \le \frac{1}{\boldsymbol{\Im _{1}}+\boldsymbol{\Im _{2}}+\boldsymbol{\Im _{3}}+\boldsymbol{\Im _{4_{1}}}+\boldsymbol{\Im _{4_{2}}}} \mho (t) \psi \Biggl(\sum_{i=1}^{5} \vert u_{k}-v_{k} \vert \Biggr)\Biggr\} . $$
Since \(\frac{1}{\boldsymbol{\Im _{1}}+\boldsymbol{\Im _{2}}+\boldsymbol{\Im _{3}}+\boldsymbol{\Im _{4_{1}}}+\boldsymbol{\Im _{4_{2}}}} \mho (t)\psi (\sum_{i=1}^{5}\vert u_{k}-v_{k}\vert )\) and \(\mathfrak{f}_{1}\) are measurable, hence the set-valued map \(\boldsymbol{\mathfrak{N}}(\cdot)\cap \mathcal{F}(\cdot,l(\cdot),l'(\cdot),l''(\cdot),^{c} \mathcal{D}_{q}^{\hbar _{1}}l(\cdot),^{c}\mathcal{D}_{q}^{\hbar _{2}}l(\cdot))\) is measurable.
Choose \(\mathfrak{f}_{2}(t) \in \mathcal{F}(t,l(t),l'(t),l''(t),^{c} \mathcal{D}_{q}^{\hbar _{1}}l(t),^{c}\mathcal{D}_{q}^{\hbar _{2}}l(t))\) such that \(\forall t\in \mathfrak{J}\)
$$ \bigl\vert \mathfrak{f}_{1}(t)-\mathfrak{f}_{2}(t) \bigr\vert \le \frac{1}{\boldsymbol{\Im _{1}}+\boldsymbol{\Im _{2}}+\boldsymbol{\Im _{3}}+\boldsymbol{\Im _{4_{1}}}+\boldsymbol{\Im _{4_{2}}}} \mho (t)\psi \Biggl(\sum _{i=1}^{5} \vert u_{k}-v_{k} \vert \Biggr). $$
Now, for all \(t\in \mathfrak{J}\), let \(\mathfrak{p}_{2}\in \boldsymbol{\mathfrak{M}}(\boldsymbol{k})\) by
$$ \mathfrak{p}_{2}=\mathcal{I}^{\pounds }_{q} \mathfrak{f}_{2}(t)+\theta \mathfrak{B}_{1}(t) \mathcal{I}^{\pounds }_{q}\mathfrak{f}_{2}(\eth )+ \theta \mathfrak{B}_{2}(t)\mathcal{I}^{\pounds }_{q} \bigl[\mathfrak{I}^{ \pounds }_{q}\mathfrak{f}_{2}(r) \bigr](1)+\theta \mathfrak{B}_{3}(t) \mathcal{I}^{\pounds -2}_{q} \mathfrak{f}_{2}(1). $$
Afterwards, let \(\sup_{t \in \mathfrak{J}}\vert \mho (t)\vert =\Vert \mho \Vert \), so
$$\begin{aligned} \bigl\vert \mathfrak{p}_{1}(t)-\mathfrak{p}_{2}(t) \bigr\vert \le{} & \mathcal{I}^{ \pounds }_{q}[ \mathfrak{f}_{1}-\mathfrak{f}_{2}](t)+\theta \mathfrak{B}_{1}(t)\mathcal{I}^{\pounds }_{q}[ \mathfrak{f}_{1}- \mathfrak{f}_{2}](\eth ) \\ &{}+\theta \mathfrak{B}_{2}(t)\mathcal{I}^{\pounds }_{q} \bigl[\mathfrak{I}^{ \pounds }_{q}[\mathfrak{f}_{1}- \mathfrak{f}_{2}](r)\bigr](1) + \theta \mathfrak{B}_{3}(t) \mathcal{I}^{\pounds -2}_{q}[\mathfrak{f}_{1}- \mathfrak{f}_{2}](1) \\ \le{}& \frac{1}{\boldsymbol{\Im _{1}}+\boldsymbol{\Im _{2}}+\boldsymbol{\Im _{3}}+\boldsymbol{\Im _{4_{1}}}+\boldsymbol{\Im _{4_{2}}}} \Vert \mho \Vert \psi \bigl( \Vert u-v \Vert \bigr) \biggl[ \frac{1}{\Gamma _{q}(\pounds +1)}+ \frac{ \vert \theta \vert \mathfrak{B}^{*}_{1}}{\Gamma _{q}(\pounds +1)} \eth ^{\pounds } \\ &{}+ \frac{ \vert \theta \vert \mathfrak{B}^{*}_{2}}{\Gamma _{q}(\pounds +2)}+ \frac{ \vert \theta \vert \mathfrak{B}^{*}_{3}}{\Gamma _{q}(\pounds -1)} \biggr]= \frac{\boldsymbol{\Im _{1}}}{\boldsymbol{\Im _{1}}+\boldsymbol{\Im _{2}}+\boldsymbol{\Im _{3}}+\boldsymbol{\Im _{4_{1}}}+\boldsymbol{\Im _{4_{2}}}} \psi \bigl( \Vert u-v \Vert \bigr). \end{aligned}$$
Also,
$$\begin{aligned} \bigl\vert \mathfrak{p'}_{1}(t)- \mathfrak{p'}_{2}(t) \bigr\vert \le{}& \frac{1}{\boldsymbol{\Im _{1}}+\boldsymbol{\Im _{2}}+\boldsymbol{\Im _{3}}+\boldsymbol{\Im _{4_{1}}}+\boldsymbol{\Im _{4_{2}}}} \Vert \mho \Vert \psi \bigl( \Vert u-v \Vert \bigr) \biggl[ \frac{1}{\Gamma _{q}(\pounds -1)}+ \frac{ \vert \theta \vert \mathfrak{B}^{*'}_{1}}{\Gamma _{q}(\pounds +1)} \eth ^{\pounds } \\ &{}+ \frac{ \vert \theta \vert \mathfrak{B}^{*'}_{2}}{\Gamma _{q}(\pounds +2)}+ \frac{ \vert \theta \vert \mathfrak{B}^{*'}_{3}}{\Gamma _{q}(\pounds -1)} \biggr]= \frac{\boldsymbol{\Im _{2}}}{\boldsymbol{\Im _{1}}+\boldsymbol{\Im _{2}}+\boldsymbol{\Im _{3}}+\boldsymbol{\Im _{4_{1}}}+\boldsymbol{\Im _{4_{2}}}} \psi \bigl( \Vert u-v \Vert \bigr), \end{aligned}$$
and
$$\begin{aligned} \begin{aligned} \bigl\vert \mathfrak{p''}_{1}(t)- \mathfrak{p''}_{2}(t) \bigr\vert \le{}& \frac{1}{\boldsymbol{\Im _{1}}+\boldsymbol{\Im _{2}}+\boldsymbol{\Im _{3}}+\boldsymbol{\Im _{4_{1}}}+\boldsymbol{\Im _{4_{2}}}} \Vert \mho \Vert \psi \bigl( \Vert u-v \Vert \bigr) \biggl[ \frac{1}{\Gamma _{q}(\pounds -2)}+ \frac{ \vert \theta \vert \mathfrak{B}^{*''}_{1}}{\Gamma _{q}(\pounds +1)} \eth ^{\pounds } \\ &{}+ \frac{ \vert \theta \vert \mathfrak{B}^{*''}_{2}}{\Gamma _{q}(\pounds +2)}+ \frac{ \vert \theta \vert \mathfrak{B}^{*''}_{3}}{\Gamma _{q}(\pounds -1)} \biggr]= \frac{\boldsymbol{\Im _{3}}}{\boldsymbol{\Im _{1}}+\boldsymbol{\Im _{2}}+\boldsymbol{\Im _{3}}+\boldsymbol{\Im _{4_{1}}}+\boldsymbol{\Im _{4_{2}}}} \psi \bigl( \Vert u-v \Vert \bigr){.} \end{aligned} \end{aligned}$$
Moreover, for \(i=1,2\), we have
$$\begin{aligned} &\bigl\vert ^{c}\mathcal{D}_{q}^{\hbar _{i}} \mathfrak{f}_{1}(t)-^{c} \mathcal{D}_{q}^{\hbar _{i}} \mathfrak{f}_{2}(t) \bigr\vert \\ &\quad \le \frac{1}{\boldsymbol{\Im _{1}}+\boldsymbol{\Im _{2}}+\boldsymbol{\Im _{3}}+\boldsymbol{\Im _{4_{1}}}+\boldsymbol{\Im _{4_{2}}}} \Vert \mho \Vert \psi \bigl( \Vert u-v \Vert \bigr) \biggl[ \frac{1}{\Gamma _{q}(\pounds -\hslash _{i}+1)}+ \frac{ \vert \theta \vert \mathfrak{B}^{**}_{1}}{\Gamma _{q}(\pounds +1)} \eth ^{\pounds } \\ &\qquad {}+ \frac{ \vert \theta \vert \mathfrak{B}^{**}_{2}}{\Gamma _{q}(\pounds +2)}+ \frac{ \vert \theta \vert \mathfrak{B}^{**}_{3}}{\Gamma _{q}(\pounds -1)} \biggr]= \frac{\boldsymbol{\Im _{4_{i}}}}{\boldsymbol{\Im _{1}}+\boldsymbol{\Im _{2}}+\boldsymbol{\Im _{3}}+\boldsymbol{\Im _{4_{1}}}+\boldsymbol{\Im _{4_{2}}}} \psi \bigl( \Vert u-v \Vert \bigr). \end{aligned}$$
Finally, according to the above relations, it can be concluded that
$$\begin{aligned} \Vert \mathfrak{p}_{1}-\mathfrak{p}_{2} \Vert ={}& \sup_{t \in \mathfrak{J}} \bigl\vert \mathfrak{p}_{1}(t)- \mathfrak{p}_{2}(t) \bigr\vert + \sup_{t \in \mathfrak{J}} \bigl\vert \mathfrak{p'}_{1}(t)-\mathfrak{p'}_{2}(t) \bigr\vert +\sup_{t \in \mathfrak{J}} \bigl\vert \mathfrak{p''}_{1}(t)- \mathfrak{p''}_{2}(t) \bigr\vert \\ &{}+ \sup_{t \in \mathfrak{J}} \bigl\vert ^{c} \mathcal{D}_{q}^{\hbar _{1}} \mathfrak{f}_{1}(t)-^{c} \mathcal{D}_{q}^{\hbar _{1}}\mathfrak{f}_{2}(t) \bigr\vert +\sup_{t \in \mathfrak{J}} \bigl\vert ^{c} \mathcal{D}_{q}^{\hbar _{2}} \mathfrak{f}_{1}(t)-^{c} \mathcal{D}_{q}^{\hbar _{2}}\mathfrak{f}_{2}(t) \bigr\vert \\ \le{}& \frac{1}{\boldsymbol{\Im _{1}}+\boldsymbol{\Im _{2}}+\boldsymbol{\Im _{3}}+\boldsymbol{\Im _{4_{1}}}+\boldsymbol{\Im _{4_{2}}}} \psi \bigl( \Vert u-v \Vert \bigr) (\boldsymbol{ \Im _{1}}+\boldsymbol{\Im _{2}}+ \boldsymbol{\Im _{3}}+\boldsymbol{\Im _{4_{1}}}+ \boldsymbol{\Im _{4_{2}}})\\ ={}&\psi \bigl( \Vert u-v \Vert \bigr), \end{aligned}$$
so \(\mathcal{H}_{\mathfrak{d}} (\boldsymbol{\mathfrak{M}}(u), \boldsymbol{\mathfrak{M}}(v) )\le \psi (\Vert u-v\Vert )\) for all \(u,v \in \mathcal{K}\).
Using Lemma 2.13 and the endpoint property of \(\boldsymbol{\mathfrak{M}}\), there exists \(u^{*} \in \mathcal{K}\) such that \(\boldsymbol{\mathfrak{M}}(u^{*})=\{u^{*}\}\). Thereupon, \(u^{*}\) is a solution for quantum inclusion problem (1)–(2). □