To establish the existence of a solution for the boundary value problem, we need to make the following assumptions.
-
\((H_{4})\)
:
-
A is defined on \([0,b]_{\mathbb{N}_{0}}\), satisfying \(G_{A}(s)\geq0\) for \(s\in{[\nu-1,b+\nu-1]_{\mathbb{N}_{\nu-1}}}\), and \(0\leq {C}<1\).
-
\((H_{5})\)
:
-
\(f(\cdot,u,s): [0,b]_{\mathbb{N}_{0}}\times[0,+\infty)\times [0,+\infty)\rightarrow[0,+\infty)\) is continuous and is nonincreasing on u and s. For all \(\lambda{\in(0,1)}\), there exist two constants \(\mu_{1},\mu_{2}>0\) such that, for any \((t,u,s)\in[0,b]_{\mathbb {N}_{0}}\times[0,+\infty)\times[0,+\infty)\),
$$\begin{aligned}& f(t,\lambda{u},s)\leq{\lambda^{-\mu_{1}}}f(t,{u},s), \end{aligned}$$
(3.1)
$$\begin{aligned}& f(t,{u},\lambda{s})\leq{\lambda^{-\mu_{2}}}f(t,{u},s). \end{aligned}$$
(3.2)
Remark 3.1
Inequalities (3.1), (3.2) are equivalent to the following inequalities (3.3), (3.4), respectively:
$$\begin{aligned}& f(t,\lambda{u},s)\geq{\lambda^{-\mu_{1}}}f(t,{u},s), \quad {\forall } { \lambda}>1, \end{aligned}$$
(3.3)
$$\begin{aligned}& f(t,{u},\lambda{s})\geq{\lambda^{-\mu_{2}}}f(t,{u},s), \quad {\forall } { \lambda}>1. \end{aligned}$$
(3.4)
Now we denote
$$\xi=\frac{{l_{y}}^{-(\mu_{1}+\mu_{2})}}{\Gamma(\beta)}, \quad\quad \zeta=\frac {{l_{y}}^{\mu_{1}+\mu_{2}}}{\Gamma(\beta)} $$
and \(g(t)=f (t',{_{b+\varepsilon-1}\nabla^{-\varepsilon }}(t'')^{\underline{\nu-\varepsilon-1}},(t'' -\varepsilon)^{\underline{\nu-\varepsilon-1}} )\), \(t\in\mathbb{T}\). Then \(g(t)\in{C (\mathbb{T},\mathbb{R} )}\), for \(m\in(0,1)\), we define
$$\Vert {g}\Vert _{\frac{1}{m}}:={ \Biggl(\sum_{s=\nu-\beta -1}^{b+\nu-\beta-2}{g^{\frac{1}{m}}(s)} \Biggr)^{m}}. $$
Theorem 3.1
Suppose that
\((H_{4})\)
and
\((H_{5})\)
hold. Then there exists a constant
\(\lambda^{\ast}>0\)
such that FBVP (2.7) has at least one positive solution
\(w(t)\)
for any
\(\lambda\in{(\lambda^{\ast},+\infty)}\). Moreover, there exists a constant
\(0< l<1\)
such that
$${l(b+\nu-2-t)^{\underline{\nu-\varepsilon-1}}}\leq{w(t)}\leq {{l^{-1}}(b+ \nu-2-t)^{\underline{\nu-\varepsilon-1}}}. $$
Proof
Let \(Q=C ([\varepsilon,b+\varepsilon]_{\mathbb{N}_{\varepsilon}},\mathbb {R} )\), and define a subset P of Q as follows:
$$P= \bigl\{ y\in{Q}:\exists l\in (0,1),\text{ such that } {l(b+\nu -2-t) ^{\underline{\nu-\varepsilon-1}}} \leq{y(t)}\leq{l^{-1}(b+\nu -2-t)^{\underline{\nu-\varepsilon-1}}} \bigr\} . $$
Clearly, P is a nonempty set since \((b+\nu-2-t)^{\underline{\nu -\varepsilon-1}}\in{P}\). Now define the operator \(T_{\lambda}\) in P.
$$ T_{\lambda}y(t)={{\sum _{s=\nu-1}^{b+\nu-2}}\overline{J}(t,s)\varphi _{q} \bigl({\triangle_{\nu-\beta-1}^{-\beta}}\lambda{f \bigl(s',_{b+\varepsilon-1} \nabla^{-\varepsilon }y \bigl(s' \bigr),y \bigl(s'+ \varepsilon \bigr) \bigr)} \bigr)}, $$
(3.5)
where \(\overline{J}(t,s)\) is given by (2.4).
We assert that \(T_{\lambda}\) is well defined and \(T_{\lambda}(P)\subset {P}\).
In fact, for any \(y\in{P}\), there exists a positive number \(0< l_{y}<1\) such that
$${l_{y}(b+\nu-2-t)^{\underline{\nu-\varepsilon-1}}}\leq{y(t)} \leq {l_{y}^{-1}(b+ \nu-2-t)^{\underline{\nu-\varepsilon-1}}}, \quad t\in {[\varepsilon,\varepsilon+b]_{\mathbb{N}_{\varepsilon}}}. $$
Thus, by Lemma 2.12, condition \((H_{5})\), Hölder’s inequality and noticing \(m\in(0,1)\), we get
$$\begin{aligned} T_{\lambda}y(t) =&{{\sum_{s=\nu-1}^{b+\nu-2}} \overline{J}(t,s)\varphi _{q} \bigl({\triangle_{\nu-\beta-1}^{-\beta}} \lambda{f \bigl(s',{_{b+\varepsilon-1}\nabla^{-\varepsilon }}y \bigl(s' \bigr),y \bigl(s'+\varepsilon \bigr) \bigr)} \bigr)} \\ \leq& \lambda^{q-1}{\sum_{s=\nu-1}^{b+\nu-2}} \overline{J}(t,s)\varphi_{q} \bigl({\triangle_{\nu -\beta-1}^{-\beta}}f \bigl(s',{_{b+\varepsilon-1}\nabla^{-\varepsilon }}l_{y} \bigl(s'' \bigr)^{\underline{\nu-\varepsilon-1}},l_{y} \bigl(s''- \varepsilon \bigr)^{\underline{\nu-\varepsilon-1}} \bigr) \bigr) \\ \leq&{\lambda^{q-1}} {{\sum_{s=\nu-1}^{b+\nu-2}} \overline{J}(t,s)\varphi _{q} \Biggl(\xi{\sum _{u=\nu-\beta-1}^{s-\beta}}(s-u-1)^{\underline{\beta -1}}g(u) \Biggr)} \\ \leq&{\sum_{s=\nu-1}^{b+\nu-2}} { \lambda^{q-1}}M(b+\nu-2-t)^{\underline {\nu-\varepsilon-1}}{\xi}^{q-1} \Biggl({ \sum_{u=\nu-\beta-1}^{s-\beta }}(s-u-1)^{\underline{\beta-1}}g(u) \Biggr)^{q-1} \\ \leq&(\xi\lambda)^{q-1}{M}(b+\nu-2-\varepsilon)^{\underline{\nu -\varepsilon-1}}\Vert {g}\Vert _{\frac{1}{m}}^{q-1} \\ &{}\times {\sum_{s=\nu-1}^{b+\nu-2}} \Biggl({\sum_{u=\nu-\beta-1}^{s-\beta}} \bigl((s-u-1)^{\underline{\beta-1}} \bigr)^{\frac{1}{1-m}} \Biggr)^{(1-m)(q-1)} \\ < &+\infty, \end{aligned}$$
i.e.,
$$ T_{\lambda}y(t)< +\infty. $$
(3.6)
On the other hand, using Lemma 2.12 and Remark 3.1, we have
$$\begin{aligned} T_{\lambda}y(t) =&{{\sum_{s=\nu-1}^{b+\nu-2}} \overline{J}(t,s)\varphi _{q} \bigl({\triangle_{\nu-\beta-1}^{-\beta}} \lambda{f \bigl(s',{_{b+\varepsilon-1}\nabla^{-\varepsilon }}y \bigl(s' \bigr),y \bigl(s'+\varepsilon \bigr) \bigr)} \bigr)} \\ \geq& \lambda^{q-1}{\sum_{s=\nu-1}^{b+\nu-2}} \overline{J}(t,s)\varphi_{q} \bigl({\triangle_{\nu-\beta-1}^{-\beta}} f \bigl(s',{_{b+\varepsilon-1}\nabla^{-\varepsilon }}l_{y}^{-1} \bigl(s'' \bigr)^{\underline{\nu-\varepsilon -1}},l_{y}^{-1} \bigl(s''-\varepsilon \bigr)^{\underline{\nu-\varepsilon-1}} \bigr) \bigr) \\ \geq&{\lambda^{q-1}} {{\sum_{s=\nu-1}^{b+\nu-2}} \overline{J}(t,s)\varphi _{q} \Biggl(\zeta{\sum _{u=\nu-\beta-1}^{s-\beta}}(s-u-1)^{\underline{\beta -1}}g(u) \Biggr)} \\ \geq&(\lambda\zeta)^{q-1}{\sum_{s=\nu-1}^{b+\nu-2}}m(s) (b+\nu -2-t)^{\underline{\nu-\varepsilon-1}}\varphi_{q} \Biggl({\sum _{u=\nu-\beta -1}^{s-\beta}}(s-u-1)^{\underline{\beta-1}}g(u) \Biggr). \end{aligned}$$
Therefore
$$ T_{\lambda}y(t)\geq(\lambda\zeta)^{q-1}(b+\nu-2-t)^{\underline{\nu -\varepsilon-1}}{ \sum_{s=\nu-1}^{b+\nu-2}} {m(s)} \Biggl({\sum _{u=\nu-\beta -1}^{s-\beta}}(s-u-1)^{\underline{\beta-1}}g(u) \Biggr)^{q-1}. $$
(3.7)
Choose
$$ \begin{aligned}[b] I_{y}&=\min \Biggl\{ \frac{1}{2}, \Biggl[(\lambda\xi)^{q-1}{M} \Vert {g}\Vert _{\frac{1}{m}}^{q-1}{ \sum_{s=\nu-1}^{b+\nu-2}} \Biggl({\sum _{u=\nu-\beta-1}^{s-\beta}} \bigl((s-u-1)^{\underline{\beta-1}} \bigr)^{\frac{1}{1-m}} \Biggr)^{(1-m)(q-1)} \Biggr]^{-1}, \\ &\quad (\lambda\zeta)^{q-1}{\sum_{s=\nu-1}^{b+\nu-2}} {m(s)} \Biggl({\sum_{u=\nu -\beta-1}^{s-\beta}}(s-u-1)^{\underline{\beta-1}}g(u) \Biggr)^{q-1} \Biggr\} . \end{aligned} $$
(3.8)
Then it follows from (3.6), (3.7) and (3.8) that
$${I_{y}(b+\nu-2-t)^{\underline{\nu-\varepsilon-1}}}\leq{T_{\lambda }y}(t) \leq{I_{y}^{-1}(b+\nu-2-t)^{\underline{\nu-\varepsilon-1}}}. $$
Next we shall devote our attention to finding the upper and lower solutions of FBVP (2.7). Let
$$e(t)={\sum_{s=\nu-1}^{b+\nu-2}} {\overline{J}(t,s)} \varphi_{q} \bigl({\triangle_{\nu-\beta-1}^{-\beta}}g(s) \bigr). $$
By Lemma 2.12, we have
$$e(t)\geq{(b+\nu-2-t)^{\underline{\nu-\varepsilon-1}}} {\sum_{s=\nu -1}^{b+\nu-2}} {m(s)}\varphi_{q} \bigl({\triangle_{\nu-\beta-1}^{-\beta }}g(s) \bigr), \quad \forall{t\in{[\varepsilon,b+\varepsilon]_{\mathbb {N}_{\varepsilon}}}}, $$
and consequently there exists a constant \(\lambda_{1}\geq{1}\) such that
$$ \lambda_{1}e(t)\geq(b+\nu-2-t)^{\underline{\nu-\varepsilon-1}}. $$
(3.9)
Thus, for any \(\lambda>\lambda_{1}\), by \((H_{5})\) and similar to (3.6), we have
$$\begin{aligned}& {\sum_{s=\nu-1}^{b+\nu-2}} \overline{J}(t,s) \varphi_{q} \bigl({\triangle _{\nu-\beta-1}^{-\beta}} {f \bigl(s',{_{b+\varepsilon-1}\nabla^{-\varepsilon}}\lambda {e} \bigl(s' \bigr),\lambda{e} \bigl(s'+\varepsilon \bigr) \bigr)} \bigr) \\& \quad \leq{\sum_{s=\nu-1}^{b+\nu-2}}\overline{J}(t,s) \varphi_{q} \bigl({\triangle_{\nu-\beta-1}^{-\beta}} {f \bigl(s',{_{b+\varepsilon-1}\nabla^{-\varepsilon}}\lambda _{1}{e} \bigl(s' \bigr),\lambda_{1}{e} \bigl(s'+\varepsilon \bigr) \bigr)} \bigr) \\& \quad \leq {\sum_{s=\nu-1}^{b+\nu-2}}\overline{J}(t,s) \varphi_{q} \Biggl[{\sum_{u=\nu-\beta-1}^{s-\beta}} \biggl(\frac{(s-u-1)^{\underline{\beta -1}}}{\Gamma(\beta)} {f \bigl(u', \bigl(_{b+\varepsilon-1} \nabla^{-\varepsilon } \bigl(u'' \bigr)^{\underline{\nu-\varepsilon-1}} \bigr), \bigl(u''-\varepsilon \bigr)^{\underline{\nu-\varepsilon-1}} \bigr)} \biggr) \Biggr] \\& \quad ={{\sum_{s=\nu-1}^{b+\nu-2}}\overline{J}(t,s) \varphi_{q} \Biggl[{\sum_{u=\nu-\beta-1}^{s-\beta}} \biggl({\frac{(s-u-1)^{\underline{\beta -1}}}{\Gamma(\beta)}}g(u) \biggr) \Biggr]} \\& \quad < +\infty, \end{aligned}$$
and
$$\begin{aligned} e(t) \leq&{M(b+\nu-2-t)^{\underline{\nu-\varepsilon-1}}} {\sum_{s=\nu -1}^{b+\nu-2}} \varphi_{q} \Biggl({\sum_{u=\nu-\beta-1}^{s-\beta}} {\frac {1}{\Gamma(\beta)}}(s-u-1)^{\underline{\beta-1}}g(u) \Biggr)< +\infty. \end{aligned}$$
Now let
$$\begin{aligned} {\rho}&={M(b+\nu-2-\varepsilon)^{\underline{\nu-\varepsilon-1}}} \biggl({\frac{1}{\Gamma(\beta)}} \biggr)^{q-1}{\Vert {g}\Vert _{\frac {1}{m}}^{q-1}} {\sum _{s=\nu-1}^{b+\nu-2}} \Biggl({\sum _{u=\nu-\beta -1}^{s-\beta}} \bigl((s-u-1)^{\underline{\beta-1}} \bigr)^{\frac {1}{1-m}} \Biggr)^{(1-m)(q-1)} \\ &\quad{} +1. \end{aligned} $$
Take
$$\begin{aligned} \lambda^{\ast}&=\max \Biggl\{ {\lambda_{1}^{\frac{1}{q-1}}}, \\ &\quad \Biggl[\rho ^{-(\mu_{1}+\mu_{2})(q-1)}{\sum_{s=\nu-1}^{b+\nu-2}}m(s) \varphi_{q} \bigl({\triangle_{\nu-\beta-1}^{-\beta}} {f \bigl(s',{_{b+\varepsilon -1}\nabla^{-\varepsilon}}1,1 \bigr)} \bigr) \Biggr]^{\frac{1}{[(\mu_{1}+\mu _{2})(q-1)-1](q-1)}} \Biggr\} . \end{aligned} $$
Then by Lemma 2.12, (3.3) and (3.4), for \({\forall{t}} \in {[\varepsilon,b+\varepsilon]_{\mathbb{N}_{\varepsilon}}}\), we can get
$$\begin{aligned} +\infty >&\sum_{s=\nu-1}^{b+\nu-2} \overline{J}(t,s){\varphi _{q}} \bigl({\triangle_{\nu-\beta-1}^{-\beta}} \lambda^{\ast}{f \bigl(s',{_{b+\varepsilon-1} \nabla^{-\varepsilon}} \bigl({\lambda^{\ast }} \bigr)^{q-1}e \bigl(s' \bigr), \bigl({\lambda^{\ast}} \bigr)^{q-1}e \bigl(s'+\varepsilon \bigr) \bigr)} \bigr) \\ \geq& (b+\nu-2-t)^{\underline{\nu-\varepsilon-1}} \bigl(\lambda^{\ast } \bigr)^{[1-(\mu_{1}+\mu_{2})(q-1)](q-1)} \\ &{}\times {{\sum_{s=\nu-1}^{b+\nu -2}}m(s)} {\varphi_{q}} { \bigl({\triangle_{\nu-\beta-1}^{-\beta}} {f \bigl(s',{_{b+\varepsilon-1}\nabla^{-\varepsilon }}e \bigl(s' \bigr),e \bigl(s'+\varepsilon \bigr) \bigr)} \bigr)} \\ \geq&{(b+\nu-2-t)^{\underline{\nu-\varepsilon-1}} \bigl(\lambda^{\ast } \bigr)^{[1-(\mu_{1}+\mu_{2})(q-1)](q-1)}{{\sum_{s=\nu-1}^{b+\nu-2}}m(s)}} {\varphi_{q} \bigl({\triangle_{\nu-\beta-1}^{-\beta}} {f \bigl(s',{_{b+\varepsilon-1}\nabla^{-\varepsilon}} {\rho},{\rho} \bigr)} \bigr)} \\ \geq& (b+\nu-2-t)^{\underline{\nu-\varepsilon-1}} \bigl(\lambda^{\ast } \bigr)^{[1-(\mu_{1}+\mu_{2})(q-1)](q-1)}{\rho^{-(\mu_{1}+\mu _{2})(q-1)}} \\ &{}\times {{\sum_{s=\nu-1}^{b+\nu-2}}m(s)} {\varphi_{q} \bigl({\triangle_{\nu-\beta-1}^{-\beta}} {f \bigl(s',{_{b+\varepsilon-1}\nabla^{-\varepsilon}}1,1 \bigr)} \bigr)} \\ \geq&(b+\nu-2-t)^{\underline{\nu-\varepsilon-1}}. \end{aligned}$$
That is to say,
$$ \begin{aligned}[b] &{\sum_{s=\nu-1}^{b+\nu-2}}\overline{J}(t,s){ \varphi_{q}} \bigl({\triangle _{\nu-\beta-1}^{-\beta}} \lambda^{\ast}{f \bigl(s',{_{b+\varepsilon -1} \nabla^{-\varepsilon}} \bigl({\lambda^{\ast}} \bigr)^{q-1}e \bigl(s' \bigr), \bigl({\lambda ^{\ast}} \bigr)^{q-1}e \bigl(s'+\varepsilon \bigr) \bigr)} \bigr) \\ &\quad \geq(b+ \nu-2-t)^{\underline{\nu-\varepsilon-1}}. \end{aligned} $$
(3.10)
Let
$$ \phi(t)= \bigl({\lambda^{\ast}} \bigr)^{q-1}e(t)=T_{\lambda^{\ast}} \bigl((b+\nu -2-t)^{\underline{\nu-\varepsilon-1}} \bigr),\quad\quad \psi(t)=T_{\lambda^{\ast }} \bigl(\phi(t) \bigr). $$
(3.11)
It follows from (3.9) and (3.10) that for any \({{t}\in{[\varepsilon,b+\varepsilon]_{\mathbb{N}_{\varepsilon}}}}\),
$$ \textstyle\begin{cases} \phi(t)=\sum_{s=\nu-1}^{b+\nu-2}\overline{J}(t,s){\varphi _{q}}(\triangle_{\nu-\beta-1}^{-\beta}\lambda^{\ast}g(s))\geq\lambda _{1}e(t)\geq(b+\nu-2-t)^{\underline{\nu-\varepsilon-1}},\\ \psi(t)=\sum_{s=\nu-1}^{b+\nu-2}\overline{J}(t,s){\varphi _{q}}({\triangle_{\nu-\beta-1}^{-\beta}}\lambda^{\ast }{f(s',{_{b+\varepsilon-1}\nabla^{-\varepsilon}}({\lambda^{\ast }})^{q-1}e(s'),({\lambda^{\ast}})^{q-1}e(s'+\varepsilon))}). \end{cases} $$
(3.12)
Moreover, by (3.11) and (3.12), we know
$$ \textstyle\begin{cases} \phi(b+\varepsilon)=0, \quad\quad [_{b+\varepsilon-1}\nabla^{\nu-\varepsilon}\phi (t)]_{\nu-2}=0,\\ [_{b+\varepsilon-1}\nabla^{-\varepsilon}\phi(t)]_{-1}=\sum_{t=0}^{b-1} (_{b+\varepsilon-1}\nabla^{-\varepsilon}\phi(t))A(t),\\ \psi(b+\varepsilon)=0,\quad\quad [_{b+\varepsilon-1}\nabla^{\nu-\varepsilon }\psi(t)]_{\nu-2}=0,\\ [_{b+\varepsilon-1}\nabla^{-\varepsilon}\psi(t)]_{-1}=\sum_{t=0}^{b-1} (_{b+\varepsilon-1}\nabla^{-\varepsilon}\psi(t))A(t). \end{cases} $$
(3.13)
Proceeding as in (3.6)-(3.8), we get that \(\phi(t),\psi(t)\in P\). By (3.10), we have
$$ \psi(t)=(T_{\lambda^{\ast}}\phi) (t)\geq(b+\nu-2-t)^{\underline{\nu -\varepsilon-1}}, $$
(3.14)
which implies
$$ \begin{aligned}[b] \psi(t)&=(T_{\lambda^{\ast}}\phi) (t) \\ &=\sum _{s=\nu-1}^{b+\nu-2}\overline{J}(t,s) {\varphi_{q}} \bigl({\triangle_{\nu-\beta-1}^{-\beta}}\lambda^{\ast }{f \bigl(s',{_{b+\varepsilon-1}\nabla^{-\varepsilon}} \bigl({ \lambda^{\ast }} \bigr)^{q-1}e \bigl(s' \bigr), \bigl({ \lambda^{\ast}} \bigr)^{q-1}e \bigl(s'+ \varepsilon \bigr) \bigr)} \bigr) \\ &\leq\sum_{s=\nu-1}^{b+\nu-2}\overline{J}(t,s){ \varphi_{q}} \bigl({\triangle_{\nu-\beta-1}^{-\beta}} \lambda^{\ast}g(s) \bigr) =\phi(t), \quad \forall t\in[\varepsilon,b+ \varepsilon]_{\mathbb{N}_{\varepsilon}}. \end{aligned} $$
(3.15)
Thus, taking account of f being nonincreasing, and by (3.11), (3.14) and (3.15), we have
$$\begin{aligned}& \triangle_{\nu-2}^{\beta} \bigl( \varphi_{p} \bigl(_{b+\varepsilon-1}\nabla^{\nu -\varepsilon}\psi \bigr) \bigr) (t)+ \lambda^{\ast}f \bigl(t',{_{b+\varepsilon-1} \nabla^{-\varepsilon}}\psi \bigl(t' \bigr),\psi \bigl(t'+ \varepsilon \bigr) \bigr) \\& \quad =\triangle_{\nu-2}^{\beta} \bigl(\varphi_{p} \bigl(_{b+\varepsilon-1}\nabla^{\nu -\varepsilon}(T_{\lambda^{\ast}}\phi) \bigr) \bigr) (t) +\lambda^{\ast}f \bigl(t',{_{b+\varepsilon-1} \nabla^{-\varepsilon}}\psi \bigl(t' \bigr),\psi \bigl(t'+ \varepsilon \bigr) \bigr) \\& \quad \geq\triangle_{\nu-2}^{\beta} \bigl(\varphi_{p} \bigl(_{b+\varepsilon-1}\nabla ^{\nu-\varepsilon}(T_{\lambda^{\ast}}\phi) \bigr) \bigr) (t) +\lambda^{\ast}f \bigl(t',{_{b+\varepsilon-1} \nabla^{-\varepsilon}}\phi \bigl(t' \bigr),\phi \bigl(t'+ \varepsilon \bigr) \bigr) \\& \quad =-\lambda^{\ast}f \bigl(t',{_{b+\varepsilon-1} \nabla^{-\varepsilon}}\phi \bigl(t' \bigr),\phi \bigl(t'+ \varepsilon \bigr) \bigr) + \lambda^{\ast}f \bigl(t',{_{b+\varepsilon-1} \nabla^{-\varepsilon}}\phi \bigl(t' \bigr),\phi \bigl(t'+ \varepsilon \bigr) \bigr) \\& \quad =0, \end{aligned}$$
(3.16)
$$\begin{aligned}& \triangle_{\nu-2}^{\beta} \bigl( \varphi_{p} \bigl(_{b+\varepsilon-1}\nabla^{\nu -\varepsilon}\phi \bigr) \bigr) (t)+ \lambda^{\ast}f \bigl(t',{_{b+\varepsilon-1} \nabla^{-\varepsilon}}\phi \bigl(t' \bigr),\phi \bigl(t'+ \varepsilon \bigr) \bigr) \\& \quad =\triangle_{\nu-2}^{\beta} \bigl(\varphi_{p} \bigl(_{b+\varepsilon-1}\nabla^{\nu -\varepsilon} \bigl(T_{\lambda^{\ast}} \bigl((b+ \nu-2-t)^{\underline{\nu-\varepsilon -1}} \bigr) \bigr) \bigr) \bigr) +\lambda^{\ast}f \bigl(t',{_{b+\varepsilon-1}\nabla^{-\varepsilon}}\phi \bigl(t' \bigr),\phi \bigl(t'+\varepsilon \bigr) \bigr) \\& \quad =-\lambda^{\ast}f \bigl(t',{_{b+\varepsilon-1} \nabla^{\nu-\varepsilon }} \bigl(t'' \bigr)^{\underline{\nu-\varepsilon-1}} , \bigl(t''-\varepsilon \bigr)^{\underline{\nu-\varepsilon-1}} \bigr) + \lambda^{\ast}f \bigl(t',{_{b+\varepsilon-1} \nabla^{-\varepsilon}}\phi \bigl(t' \bigr),\phi \bigl(t'+ \varepsilon \bigr) \bigr) \\& \quad \leq-\lambda^{\ast}f \bigl(t',{_{b+\varepsilon-1} \nabla^{\nu-\varepsilon }} \bigl(t'' \bigr)^{\underline{\nu-\varepsilon-1}} , \bigl(t''-\varepsilon \bigr)^{\underline{\nu-\varepsilon-1}} \bigr) \\& \quad\quad {} + \lambda^{\ast}f \bigl(t',{_{b+\varepsilon-1} \nabla^{\nu-\varepsilon}} \bigl(t'' \bigr)^{\underline {\nu-\varepsilon-1}} , \bigl(t''-\varepsilon \bigr)^{\underline{\nu-\varepsilon-1}} \bigr) \\& \quad =0. \end{aligned}$$
(3.17)
It follows from (3.13) and (3.15)-(3.17) that \(\psi(t),\phi(t)\) are upper and lower solutions of FBVP (2.7) and \(\psi(t),\phi(t)\in P\).
Now we define a function
$$ F \bigl(t',{_{b+\varepsilon-1} \nabla^{-\varepsilon }}y \bigl(t' \bigr),y \bigl(t' \bigr) \bigr) = \textstyle\begin{cases} f (t',{_{b+\varepsilon-1}\nabla^{-\varepsilon}}\psi(t'),\psi (t'+\varepsilon) ), & y< \psi(t),\\ f (t',{_{b+\varepsilon-1}\nabla^{-\varepsilon }}y(t'),y(t'+\varepsilon) ), & \psi(t)\leq y\leq\phi(t),\\ f (t',{_{b+\varepsilon-1}\nabla^{-\varepsilon}}\phi(t'),\phi (t'+\varepsilon) ), & y>\phi(t). \end{cases} $$
(3.18)
It then follows from \((H_{5})\) and (3.18) that \(F(t,u,s):[0,b]_{\mathbb {N}_{0}}\times [0,+\infty)\times[0,+\infty)\longrightarrow[0,+\infty)\) is continuous.
We now show that the FBVP
$$ \textstyle\begin{cases} \triangle_{\nu-2}^{\beta}(\varphi_{p}(_{b+\varepsilon-1}\nabla^{\nu -\varepsilon}y(t)))=-\lambda^{\ast}F(t',{_{b+\varepsilon-1}\nabla ^{-\varepsilon}}y(t'),y(t'+\varepsilon)), \\\quad t\in[\varepsilon ,b+\varepsilon]_{\mathbb{N}_{\varepsilon}},\\ y(b+\varepsilon)=0,\quad\quad [_{b+\varepsilon-1}\nabla^{\nu-\varepsilon}y(t)]_{\nu -2}=0,\\ [_{b+\varepsilon-1}\nabla^{-\varepsilon}y(t)]_{-1}= \sum_{t=0}^{b-1}(_{b+\varepsilon-1}\nabla^{-\varepsilon}y(t))A(t) \end{cases} $$
(3.19)
has a positive solution.
Define the operator \(D_{\lambda^{\ast}}\) by
$$ \begin{aligned}[b] &D_{\lambda^{\ast}}{y(t)}=\sum_{s=\nu-1}^{b+\nu-2} \overline{J}(t,s)\varphi _{q} \bigl(\triangle_{\nu-\beta-1}^{-\beta}{ \lambda^{\ast}}F \bigl(s',{_{b+\varepsilon-1} \nabla^{-\varepsilon }}y \bigl(s' \bigr),y \bigl(s'+ \varepsilon \bigr) \bigr) \bigr),\\ &\quad t\in[\varepsilon,b+\varepsilon ]_{\mathbb{N}_{\varepsilon}}. \end{aligned} $$
(3.20)
Then \(D_{\lambda^{\ast}}:C ([\varepsilon,b+\varepsilon]_{\mathbb {N}_{\varepsilon}},\mathbb{R} )\rightarrow C ([\varepsilon,b+\varepsilon]_{\mathbb{N}_{\varepsilon}},\mathbb {R} )\), and a fixed point of the operator \(D_{\lambda^{\ast}}\) is a solution of FBVP (3.19).
On the other hand, from the definition of F and the fact that the function f is nonincreasing on the second and third variable, we obtain
$$\begin{aligned} f \bigl(t',{_{b+\varepsilon-1}\nabla^{-\varepsilon}}\phi \bigl(t' \bigr),\phi \bigl(t'+\varepsilon \bigr) \bigr) &\leq F \bigl(t',{_{b+\varepsilon-1}\nabla^{-\varepsilon }}y \bigl(t' \bigr),y \bigl(t'+\varepsilon \bigr) \bigr) \\ & \leq f \bigl(t',{_{b+\varepsilon-1}\nabla^{-\varepsilon}}\psi \bigl(t' \bigr),\psi \bigl(t'+\varepsilon \bigr) \bigr), \end{aligned} $$
provided that \(\psi(t)\leq y(t)\leq\phi(t)\);
$$F \bigl(t',{_{b+\varepsilon-1}\nabla^{-\varepsilon }}y \bigl(t' \bigr),y \bigl(t'+\varepsilon \bigr) \bigr) = f \bigl(t',{_{b+\varepsilon-1}\nabla^{-\varepsilon}}\psi \bigl(t' \bigr),\psi \bigl(t'+\varepsilon \bigr) \bigr) $$
provided that \(y(t)< \psi(t)\);
$$F \bigl(t',{_{b+\varepsilon-1}\nabla^{-\varepsilon }}y \bigl(t' \bigr),y \bigl(t'+\varepsilon \bigr) \bigr) = f \bigl(t',{_{b+\varepsilon-1}\nabla^{-\varepsilon}}\phi \bigl(t' \bigr),\phi \bigl(t'+\varepsilon \bigr) \bigr) $$
provided that \(y(t)>\phi(t)\). So we have
$$ \begin{aligned}[b] f \bigl(t',{_{b+\varepsilon-1}\nabla^{-\varepsilon}}\phi \bigl(t' \bigr),\phi \bigl(t'+\varepsilon \bigr) \bigr) &\leq F \bigl(t',{_{b+\varepsilon-1}\nabla^{-\varepsilon }}y \bigl(t' \bigr),y \bigl(t'+\varepsilon \bigr) \bigr) \\ & \leq f \bigl(t',{_{b+\varepsilon-1}\nabla^{-\varepsilon}}\psi \bigl(t' \bigr),\psi \bigl(t'+\varepsilon \bigr) \bigr). \end{aligned} $$
(3.21)
Furthermore, by (3.13), (3.14) and (3.21), we have
$$ \begin{aligned}[b] f \bigl(t',{_{b+\varepsilon-1} \nabla^{-\varepsilon}}\phi \bigl(t' \bigr),\phi \bigl(t'+ \varepsilon \bigr) \bigr) &\leq f \bigl(t',{_{b+\varepsilon-1} \nabla^{-\varepsilon}}\psi \bigl(t' \bigr),\psi \bigl(t'+ \varepsilon \bigr) \bigr) \\ &\leq f \bigl(t',{_{b+\varepsilon-1}\nabla^{-\varepsilon }} \bigl(t'' \bigr)^{\underline{\nu-\varepsilon-1}}, \bigl(t''-\varepsilon \bigr)^{\underline{\nu-\varepsilon-1}} \bigr) \\ &= g(t). \end{aligned} $$
(3.22)
It follows from Lemma 2.12 and (3.22) that for any \(y\in P\),
$$\begin{aligned} D_{\lambda^{\ast}}y(t) =&\sum_{s=\nu-1}^{b+\nu-2} \overline{J}(t,s) \varphi_{q} \bigl(\triangle_{\nu-\beta-1}^{-\beta} \lambda^{\ast}F \bigl(s', {_{b+\varepsilon-1} \nabla^{-\varepsilon}}y \bigl(s' \bigr),y \bigl(s'+ \varepsilon \bigr) \bigr) \bigr) \\ \leq& \biggl(\frac{{\lambda^{\ast}}}{\Gamma(\beta)} \biggr)^{q-1}M(b+\nu -2-t)^{\underline{\nu-\varepsilon-1}}\sum_{s=\nu-1}^{b+\nu -2} \Biggl( \sum_{u=\nu-\beta-1}^{s-\beta}(s-u-1)^{\underline{\beta -1}}g(u) \Biggr)^{q-1} \\ \leq& \biggl(\frac{{\lambda^{\ast}}}{\Gamma(\beta)} \biggr)^{q-1}M(b+\nu -2)^{\underline{\nu-\varepsilon-1}}\Vert g\Vert _{\frac {1}{m}}^{q-1}\sum _{s=\nu-1}^{b+\nu-2} \\ &{}\times\Biggl(\sum_{u=\nu -\beta-1}^{s-\beta} \bigl((s-u-1)^{\underline{\beta-1}} \bigr)^{\frac {1}{1-m}} \Biggr)^{(1-m)(q-1)} \\ < &+\infty, \end{aligned}$$
(3.23)
namely the operator \(D_{\lambda^{\ast}}\) is uniformly bounded.
Next, let \(\Omega\subset P\) be bounded. Since the right side of (3.20) is finite sum, we can prove that \(D()\) is equicontinuous. By the Arzela-Ascoli theorem, we have \(D_{\lambda^{\ast}}:P\rightarrow P \) is completely continuous. Moreover, (3.23) implies that \(D_{\lambda^{\ast}}\) satisfies the conditions of Lemma 2.13. Thus, by using the Schauder fixed point theorem, \(D_{\lambda^{\ast}}\) has at least one fixed point w such that \(w=D_{\lambda^{\ast}}w\).
Now we prove
$$\psi(t)\leq w(t)\leq\phi(t), \quad t\in[\varepsilon,b+\varepsilon]_{\mathbb {N}_{\varepsilon}}. $$
Since w is a fixed point of \(D_{\lambda^{\ast}}\), we have
$$ \begin{gathered} w(b+\varepsilon)=0, \quad\quad \bigl[_{b+\varepsilon-1}\nabla^{\nu-\varepsilon}w(t) \bigr]_{\nu-2}=0, \\ \bigl[_{b+\varepsilon-1}\nabla^{-\varepsilon}w(t) \bigr]_{-1}=\sum_{t=0}^{b-1} \bigl(_{b+\varepsilon-1}\nabla^{-\varepsilon}w(t) \bigr)A(t). \end{gathered} $$
(3.24)
From (3.11), (3.22) and noticing that w is a fixed point of \(D_{\lambda^{\ast}}\), we also have
$$\begin{aligned} &\triangle_{\nu-2}^{\beta} \bigl( \varphi_{p} \bigl(_{b+\varepsilon-1}\nabla^{\nu -\varepsilon}\phi(t) \bigr) \bigr)-\triangle_{\nu-2}^{\beta} \bigl(\varphi _{p} \bigl(_{b+\varepsilon-1}\nabla^{\nu-\varepsilon}w(t) \bigr) \bigr) \\ &\quad =-\lambda^{\ast}f \bigl(t',{_{b+\varepsilon-1} \nabla^{-\varepsilon }} \bigl(t'' \bigr)^{\underline{\nu-\varepsilon-1}}, \bigl(t''-\varepsilon \bigr)^{\underline{\nu-\varepsilon-1}} \bigr) + \lambda^{\ast}F \bigl(t',{_{b+\varepsilon-1} \nabla^{-\varepsilon }}w \bigl(t' \bigr),w \bigl(t'+ \varepsilon \bigr) \bigr) \\ &\quad =-\lambda^{\ast}f \bigl(t',{_{b+\varepsilon-1} \nabla^{-\varepsilon }} \bigl(t'' \bigr)^{\underline{\nu-\varepsilon-1}}, \bigl(t''-\varepsilon \bigr)^{\underline{\nu-\varepsilon-1}} \bigr) + \lambda^{\ast}f \bigl(t',{_{b+\varepsilon-1} \nabla^{-\varepsilon}}\psi \bigl(t' \bigr),\psi \bigl(t'+ \varepsilon \bigr) \bigr) \\ &\quad \leq-\lambda^{\ast}f \bigl(t',{_{b+\varepsilon-1} \nabla^{-\varepsilon }} \bigl(t'' \bigr)^{\underline{\nu-\varepsilon-1}}, \bigl(t''-\varepsilon \bigr)^{\underline{\nu-\varepsilon-1}} \bigr) \\ &\quad\quad{} + \lambda^{\ast}f \bigl(t',{_{b+\varepsilon-1} \nabla^{-\varepsilon }} \bigl(t'' \bigr)^{\underline{\nu-\varepsilon-1}}, \bigl(t''-\varepsilon \bigr)^{\underline{\nu-\varepsilon-1}} \bigr) \\ &\quad =0. \end{aligned} $$
Let
$$z(t)=\varphi_{p} \bigl(_{b+\varepsilon-1}\nabla^{\nu-\varepsilon}\phi(t) \bigr)-\varphi_{p} \bigl(_{b+\varepsilon-1}\nabla^{\nu-\varepsilon}w(t) \bigr). $$
Then
$$\begin{aligned} &\triangle_{\nu-2}^{\beta}z(t)=\triangle_{\nu-2}^{\beta} \bigl(\varphi _{p} \bigl(_{b+\varepsilon-1}\nabla^{\nu-\varepsilon}\phi(t) \bigr) \bigr)-\triangle _{\nu-2}^{\beta} \bigl( \varphi_{p} \bigl(_{b+\varepsilon-1}\nabla^{\nu -\varepsilon}w(t) \bigr) \bigr)\leq 0, \quad t\in[ \varepsilon,b+\varepsilon]_{\mathbb{N}_{\varepsilon}}, \\ &z(\nu-2)=\varphi_{p} \bigl(_{b+\varepsilon-1}\nabla^{\nu-\varepsilon} \phi (\nu-2) \bigr)-\varphi_{p} \bigl(_{b+\varepsilon-1} \nabla^{\nu-\varepsilon }w(\nu-2) \bigr)=0. \end{aligned}$$
Moreover, we have \(z(t)\leq0\), i.e., \(\varphi_{p} (_{b+\varepsilon-1}\nabla^{\nu-\varepsilon}\phi(t) ) -\varphi_{p} (_{b+\varepsilon-1}\nabla^{\nu-\varepsilon}w(t) )\leq 0\).
In fact, if we denote that
$$\triangle_{\nu-2}^{\beta}z(t)=-\eta(t)\leq0, $$
according to Lemma 2.6, we have
$$\begin{aligned}& z(t)=-\triangle_{\nu-\beta-1}^{-\beta}\eta(t)+K_{1}(t-\nu+ \beta +1)^{\underline{\beta-1}}, \\& z(\nu-2)=0. \end{aligned}$$
In view of \(K_{1}=0\), hence \(z(t)\leq0\).
Noticing that \(\varphi_{p}\) is monotone increasing and \(_{b+\varepsilon-1}\nabla^{\nu-\varepsilon}\) is a linear operator, we have
$$_{b+\varepsilon-1}\nabla^{\nu-\varepsilon}(\phi-w) (t)\leq 0. $$
It follows from Remark 2.2 and (3.24) that
Thus we have \(w(t)\leq\phi(t)\) for \(t\in[\varepsilon,b+\varepsilon]_{\mathbb{N}_{\varepsilon}}\). In the same way, we also have \(w(t)\geq\psi(t)\) for \(t\in[\varepsilon,b+\varepsilon]_{\mathbb{N}_{\varepsilon}}\), so
$$ \psi(t)\leq w(t)\leq\phi(t), \quad t\in[\varepsilon,b+\varepsilon]_{\mathbb {N}_{\varepsilon}}. $$
(3.25)
Consequently,
$$F \bigl(t',{_{b+\varepsilon-1}\nabla^{-\varepsilon }}w \bigl(t' \bigr),w \bigl(t'+\varepsilon \bigr) \bigr)=f \bigl(t',{_{b+\varepsilon -1}\nabla^{-\varepsilon}}w \bigl(t' \bigr),w \bigl(t'+\varepsilon \bigr) \bigr),\quad t\in [\varepsilon,b+\varepsilon]_{\mathbb{N}_{\varepsilon}}. $$
Hence \(w(t)\) is a positive solution of FBVP (3.19), i.e., \(y(t)=_{b+\varepsilon-1}\nabla^{-\varepsilon}w(t)\) is a positive solution of problem (1.1) and (1.2).
Finally, by (3.25) and \(\phi,\psi\in P\), we have
$$l_{\psi}(b+\nu-2-t)^{\underline{\nu-\varepsilon-1}}\leq\psi(t) \leq w(t)\leq\phi(t)\leq l_{\phi}^{-1}(b+\nu-2-t)^{\underline{\nu-\varepsilon-1}}. $$
Let \(l_{y}=\min\{l_{\psi},l_{\phi}\}\), then
$$l_{y}(b+\nu-2-t)^{\underline{\nu-\varepsilon-1}}\leq\psi(t)\leq w(t)\leq \phi(t)\leq l_{y}^{-1}(b+\nu-2-t)^{\underline{\nu-\varepsilon-1}}. $$
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