On a nonlocal boundary value problem for nonlinear second-order Hahn difference equation with two different $q,\omega$-derivatives

Abstract

In this paper, we study a nonlocal boundary value problem for a second-order Hahn difference equation. Our problem contains two Hahn difference operators with different numbers of q and ω. An existence and uniqueness result is proved by using the Banach fixed point theorem, and the existence of a positive solution is established by using the Krasnoselskii fixed point theorem.

1 Introduction

The quantum calculus, also known as the calculus without considering limits, deals with sets of nondifferentiable functions. There are many different types of quantum difference operators, for example, the Jackson q-difference operator, the forward (delta) difference operator, the backward (nabla) difference operator, and so on. These operators are found in many applications of mathematical areas such as orthogonal polynomials, basic hypergeometric functions, combinatorics, the calculus of variations, the theory of relativity, hypergeometric series, complex analysis, particle physics, and quantum mechanics. For some recent results and applications of the quantum calculus, see [1–9] and the references therein.

In 1949, Hahn [10] introduced the Hahn difference operator $D_{q,\omega}$,

$$D_{q,\omega}f(t)=\frac{f(qt+\omega)-f(t)}{t(q-1)+\omega},\quad t\neq\omega_{0}:= \frac{\omega}{1-q}. $$

The Hahn difference operator is generalized to two well-known difference operators, the forward difference operator and the Jackson q-difference operator. Notice that, under appropriate conditions,

$$\begin{aligned}& D_{q,\omega}f(t)=\Delta_{\omega}f(t)\quad \text{whenever }q= 1,\quad \quad D_{q,\omega}f(t)=D_{q}f(t) \quad \text{whenever }\omega= 0, \quad \text{and} \\& D_{q,\omega}f(t)=f'(t)\quad \text{whenever }q= 1,\omega \rightarrow0. \end{aligned}$$

The Hahn difference operator has been employed in many pieces of literature to construct families of orthogonal polynomials and to investigate some approximation problems; see [11–13] and the references therein.

Unfortunately, in the past, no one was interested in finding the right inverse of the Hahn difference operator. Until in 2009, Aldwoah [14, 15] (Ph.D. thesis supervised by M.H. Annaby and A.E. Hamza) defined the right inverse of $D_{q,\omega}$ in terms of both the Jackson q-integral containing the right inverse of $D_{q}$ [16] and Nörlund sum involving the right inverse of $\Delta_{\omega}$ [16].

In 2010, Malinowska and Torres [17, 18] introduced the Hahn quantum variational calculus. In 2013, Malinowska and Martins [19] studied the generalized transversality conditions for the Hahn quantum variational calculus. In the same year, Hamza et al. [20, 21] studied the theory of linear Hahn difference equations and investigated the existence and uniqueness results for the initial value problems for Hahn difference equations by using the method of successive approximations; moreover, they proved Gronwall’s and Bernoulli’s inequalities with respect to the Hahn difference operator and also established mean value theorems for this calculus.

In particular, the boundary value problem for Hahn difference equations has not been studied. The results mentioned are the motivation for this research. In this paper, we consider a nonlinear Hahn difference equation with nonlocal boundary value conditions of the form

$$ \begin{aligned} &D^{2}_{q,\omega} x(t)+f \bigl(t,x(t), D_{p,\theta}x(pt+ \theta) \bigr)=0,\quad t\in[ \omega_{0},T]_{q,\omega}, \\ &x(\omega_{0})=\varphi(x), \\ &x(T)=\lambda x(\eta),\quad\eta\in(\omega_{0},T)_{q,\omega }, \end{aligned} $$

(1.1)

where $0< q<1$, $0<\omega<T$, $\omega_{0}:=\frac{\omega}{1-q}$, $1\leq \lambda<\frac{T-\omega_{0}}{\eta-\omega_{0}}$, $p=q^{m}$, $m\in {\mathbb{N}}$, $\theta=\omega(\frac{1-p}{1-q} )$, $f: [\omega_{0},T]_{q,\omega}\times{\mathbb{R}}\times{\mathbb {R}}\rightarrow{\mathbb{R}}$ is a given function, and $\varphi: C ([\omega_{0},T]_{q,\omega},{\mathbb{R}} )\rightarrow {\mathbb{R}}$ is a given functional.

In the next section, we briefly recall some definitions and lemmas used in this research. In Section 3, we prove the existence and uniqueness of a solution to problem (1.1) by the Banach fixed point theorem. In Sections 4-5, we establish some properties of the Green function and the existence of a positive solution to problem (1.1) by using the Krasnoselskii fixed point theorem. Finally, we provide an example to illustrate our results in the last section.

The following theorem is Krasnoselskii’s fixed point theorem in a cone.

Theorem 1.1

([22])

Let E be a Banach space, and let $K\subset E$ be a cone. Let $\Omega_{1}$, $\Omega_{2}$ be open subsets of E with $0\in \Omega_{1}$, $\overline{\Omega}_{1}\subset\Omega_{2}$, and let

$$A:K\cap(\overline{\Omega}_{2}\setminus\Omega_{1}) \longrightarrow K $$

be a completely continuous operator such that

• (i) $\Vert Au\Vert \leqslant \Vert u\Vert$, $u\in K\cap \partial\Omega_{1}$, and $\Vert Au\Vert \geqslant \Vert u\Vert$, $u\in K\cap\partial \Omega_{2}$; or

• (ii) $\Vert Au\Vert \geqslant \Vert u\Vert$, $u\in K\cap \partial\Omega_{1}$, and $\Vert Au\Vert \leqslant \Vert u\Vert$, $u\in K\cap\partial \Omega_{2}$.

Then A has a fixed point in $K\cap(\overline{\Omega}_{2}\setminus\Omega_{1})$.

2 Preliminaries

We now give the notation, definitions, and lemmas used in the main results.

Definition 2.1

([10])

For $0< q<1$, $\omega>0$, and f defined on an interval $I\subseteq{\mathbb{R}}$ containing $\omega_{0}:=\frac{\omega}{1-q}$, the Hahn difference of f is defined by

$$ D_{q,\omega}f(t)=\frac{f(qt+\omega)-f(t)}{t(q-1)+\omega}\quad\text{for } t\neq \omega_{0} $$

and $D_{q,\omega}f(\omega_{0})=f'(\omega_{0})$, provided that f is differentiable at $\omega_{0}$. We call $D_{q,\omega}f$ the $q,\omega$-derivative of f and say that f is $q,\omega$-differentiable on I.

Let $a,b\in I \subseteq\mathbb{R}$ with $a<\omega_{0}<b$, and $[k]_{q}=\frac{1-q^{k}}{1-q}$, $k\in{\mathbb{N}}_{0}:={\mathbb{N}}\cup\{0\}$. We define the $q,\omega$-interval by

$$\begin{aligned} {[a,b]}_{q,\omega} :=& \bigl\{ q^{k} a+\omega[k]_{q} : k \in{\mathbb{N}}_{0} \bigr\} \cup \bigl\{ q^{k} b+ \omega[k]_{q} : k\in{\mathbb{N}}_{0} \bigr\} \cup\{ \omega_{0} \} \\ =&{[a,\omega_{0}]}_{q,\omega} \cup{[\omega_{0},b]}_{q,\omega} \\ =&{(a,b)}_{q,\omega} \cup\{a,b \} = {[a,b)}_{q,\omega} \cup\{b \} = {(a,b]}_{q,\omega} \cup\{a \}. \end{aligned}$$

Observe that, for each $s\in[a,b]_{q,\omega}$, the sequence $\{ q^{k} s+\omega[k]_{q} \}_{k=0}^{\infty}$ is uniformly convergent to $\omega_{0}$.

If f is $q,\omega$-differentiable n times on a $q,\omega$-interval $I_{q,\omega}$, then we define the higher-order derivatives by

$$D^{n}_{q,\omega}f (s ):=D_{q,\omega}D^{n-1}_{q,\omega} f (s ), $$

where $D^{0}_{q,\omega}f(s):=f(s)$, $s\in I_{q,\omega}\subset{\mathbb{R}}$.

Next, we introduce the right inverse of the operator $D_{q,\omega}$, which call the $q,\omega$-integral operator.

Definition 2.2

([14])

Let I be any closed interval of $\mathbb{R}$ containing a, b, and $\omega_{0}$. For a function $f:I\rightarrow\mathbb{R}$, we define the $q,\omega$-integral of f from a to b by

$$ \int_{a}^{b} f(t)\,d_{q,\omega}t:= \int_{\omega_{0}}^{b} f(t)\,d_{q,\omega }t- \int_{\omega_{0}}^{a} f(t)\,d_{q,\omega}t, $$

where

$$\int_{\omega_{0}}^{x} f(t)\,d_{q,\omega}t:= \bigl[x(1-q)- \omega \bigr]\sum_{k=0}^{\infty} q^{k} f \bigl(xq^{k}+\omega[k]_{q} \bigr), \quad x\in I, $$

provided that the series converges at $x=a$ and $x=b$; we say that f is $q,\omega$-integrable on $[a,b]$, and the sum on the right-hand side of the above equation is called the Jackson-Norlund sum.

Note that the actual domain of definition of f is $[a,b]_{q,\omega }\subset I$.

The following lemma is the fundamental theorem of Hahn calculus.

Lemma 2.1

([14])

Let $f:I\rightarrow\mathbb {R}$ be continuous at $\omega_{0}$. Define

$$F(x):= \int_{\omega_{0}}^{x} f(t)\,d_{q,\omega}t,\quad x\in I. $$

Then F is continuous at $\omega_{0}$. Furthermore, $D_{q,\omega _{0}}F(x)$ exists for every $x\in I$, and

$$D_{q,\omega}F(x)=f(x). $$

Conversely,

$$\int_{a}^{b} D_{q,\omega}F(t) \,d_{q,\omega}t=F(b)-F(a)\quad \textit{for all }a,b\in I. $$

Next, we give some auxiliary lemmas for simplifying calculations.

Lemma 2.2

Let $0< q<1$, $\omega>0$, and $f:I\rightarrow\mathbb{R}$ be continuous at $\omega_{0}$. Then

$$\begin{aligned} \int_{\omega_{0}}^{t} \int_{\omega_{0}}^{r} x(s) \,d_{q,\omega}s \,d_{q,\omega}r =& \int_{\omega_{0}}^{t} \int_{qs+\omega}^{t} h(s) \,d_{q,\omega}r \,d_{q,\omega}s. \end{aligned}$$

Proof

Using the definition of the $q,\omega$-integral, we have

$$\begin{aligned}& \int_{\omega_{0}}^{t} \int_{\omega_{0}}^{r} x(s)\,d_{q,\omega}s \,d_{q,\omega}r \\& \quad = \int_{\omega_{0}}^{t} \Biggl[ \bigl[r(1-q)-\omega \bigr]\sum _{k=0}^{\infty} q^{k} x \bigl(rq^{k}+\omega[k]_{q} \bigr) \Biggr] \,d_{q,\omega}r \\& \quad = \sum_{k=0}^{\infty} q^{k} \biggl[ \int_{\omega_{0}}^{t} \bigl[r(1-q)-\omega \bigr] x \bigl(rq^{k}+\omega[k]_{q} \bigr) \,d_{q,\omega }r \biggr] \\& \quad = \sum_{k=0}^{\infty} q^{k} \bigl[t(1-q)-\omega \bigr]\sum_{h=0}^{\infty} q^{h} \bigl[ \bigl(tq^{k}+\omega[k]_{q} \bigr) (1-q)-\omega \bigr] x \bigl( \bigl(tq^{k}+\omega[k]_{q} \bigr)q^{h}+\omega[h]_{q} \bigr) \\& \quad = \bigl[t(1-q)-\omega \bigr] \sum_{k=0}^{\infty} \sum_{h=0}^{\infty } q^{k+h} \bigl[ t(1-q)q^{k}-\omega q^{k} \bigr]x \biggl(tq^{k+h}+ \omega \biggl[\frac{1-q^{k}}{1-q} \biggr]q^{h}+\omega \biggl[ \frac {1-q^{h}}{1-q} \biggr] \biggr) \\& \quad = \bigl[t(1-q)-\omega \bigr] \sum_{k=0}^{\infty} \sum_{h=0}^{\infty } q^{2k+h} \bigl[ t(1-q)-\omega \bigr] x \biggl(tq^{k+h}+\omega \biggl[\frac {1-q^{k+h}}{1-q} \biggr] \biggr) \\& \quad = \bigl[t(1-q)-\omega \bigr]^{2} \sum _{k=0}^{\infty} \sum_{h=0}^{\infty} q^{2k+h} x \bigl(tq^{k+h}+\omega[k+h]_{q} \bigr) \\& \quad = \bigl[t(1-q)-\omega \bigr]^{2} \sum _{h=0}^{\infty} \bigl[ q^{h}x \bigl(tq^{h}+\omega[h]_{q} \bigr)+q^{h+2}x \bigl(tq^{h+1}+\omega[h+1]_{q} \bigr) +\cdots \bigr] \\& \quad = \bigl[t(1-q)-\omega \bigr]^{2} \bigl[x(t)+ q^{2}x (tq+\omega)+q^{4}x \bigl(tq^{2}+ \omega[2]_{q} \bigr)+\cdots \\& \quad\quad{} +qx(tq+\omega)+ q^{3}x \bigl(tq^{2}+ \omega[2]_{q} \bigr)+q^{5}x \bigl(tq^{3}+ \omega[3]_{q} \bigr)+\cdots \\& \quad\quad{} +q^{2}x \bigl(tq^{2}+\omega[2]_{q} \bigr)+ q^{4}x \bigl(tq^{3}+\omega[3]_{q} \bigr)+q^{6}x \bigl(tq^{4}+\omega[4]_{q} \bigr)+ \cdots \bigr] \\& \quad = \bigl[t(1-q)-\omega \bigr]^{2} \bigl[x(t)+ q(1+q) x (tq+\omega)+q^{2} \bigl(1+q+q^{2} \bigr) x \bigl(tq^{2}+\omega[2]_{q} \bigr)+\cdots \bigr] \\& \quad = \bigl[t(1-q)-\omega \bigr]^{2} \sum _{k=0}^{\infty} q^{k}[k+1]_{q} x \bigl(tq^{k}+ \omega[k]_{q} \bigr) \\& \quad = \int^{t}_{\omega_{0}} \bigl[t-(qs+\omega) \bigr] x(s) \,d_{q,w}s \\& \quad = \int^{t}_{\omega_{0}} \biggl[ \int^{t}_{\omega_{0}} x(s)\,d_{q,w}r - \int^{qs+\omega}_{\omega_{0}} x(s)\,d_{q,w}r \biggr] \,d_{q,w}s \\& \quad = \int_{\omega_{0}}^{t} \int_{qs+\omega}^{t} x(s)\,d_{q,\omega}r \,d_{q,\omega}s. \end{aligned}$$

 □

Lemma 2.3

Let $0< q<1$ and $\omega>0$. Then

$$\begin{aligned} \int_{\omega_{0}}^{t}\,d_{q,\omega}s=t- \omega_{0} \quad \textit{and}\quad \int_{\omega_{0}}^{t} \bigl[t-(qs+\omega) \bigr] \,d_{q,\omega}s=\frac{ (t-\omega _{0} )^{2}}{1+q}. \end{aligned}$$

Proof

Using the definition of the $q,\omega$-integral, we have

$$\begin{aligned} \int_{\omega_{0}}^{t} \,d_{q,\omega}s =& \bigl[t(1-q)- \omega \bigr]\sum_{k=0}^{\infty} q^{k}= (1-q) (t-\omega_{0} )\sum_{k=0}^{\infty} q^{k} \\ =& (1-q) (t-\omega_{0} ) \biggl[\frac{1}{1-q} \biggr]=t- \omega_{0} \end{aligned}$$

and

$$\begin{aligned} \int_{\omega_{0}}^{t} \bigl[t-(qs+\omega) \bigr] \,d_{q,\omega}s =& \bigl[t(1-q)-\omega \bigr]\sum _{k=0}^{\infty} q^{k} \bigl[t- \bigl[q \bigl(tq^{k}+\omega[k]_{q} \bigr)+\omega \bigr] \bigr] \\ =& (1-q) (t-\omega_{0} )\sum_{k=0}^{\infty} q^{k} \bigl[t \bigl(1-q^{k+1} \bigr)-\omega[k+1]_{q} \bigr] \\ =& (1-q) (t-\omega_{0} )^{2}\sum _{k=0}^{\infty} q^{k} \bigl(1-q^{k+1} \bigr) \\ =& (1-q) (t-\omega_{0} )^{2} \biggl[\frac{1}{1-q}- \frac {q}{1-q^{2}} \biggr]=\frac{ (t-\omega_{0} )^{2}}{1+q}. \end{aligned}$$

The proof is complete. □

The following lemma deals with the linear version of problem (1.1) and gives a representation of the solution.

Lemma 2.4

Let $1\leq\lambda<\frac{T-\omega_{0}}{\eta-\omega_{0}}$, $h\in C ([\omega_{0},T]_{q,\omega}, {\mathbb{R}} )$ be a given function, and $\varphi: C ([\omega_{0},T]_{q,\omega }, {\mathbb{R}} )\rightarrow{\mathbb{R}}$ be a given functional. Then the problem

$$ \begin{aligned} &D^{2}_{q,\omega} x(t)=-h(t),\quad t\in[\omega_{0},T]_{q,\omega}, \\ &x(\omega_{0})=\varphi(x),\quad \quad x(T)=\lambda x(\eta),\quad \quad \eta\in( \omega_{0},T)_{q,\omega}, \end{aligned} $$

(2.1)

has the unique solution

$$\begin{aligned} x(t) =& \varphi(x)+\frac{(t-\omega_{0})}{\Lambda} \biggl[(\lambda-1) \varphi(x)-\lambda \int_{\omega_{0}}^{\eta}\bigl[\eta-(qs+\omega) \bigr] h(s) \,d_{q,\omega}s \\ &{}+ \int_{\omega_{0}}^{T} \bigl[T-(qs+\omega) \bigr] h(s) \,d_{q,\omega}s \biggr]- \int_{\omega_{0}}^{t} \bigl[t-(qs+\omega) \bigr] h(s) \,d_{q,\omega}s, \end{aligned}$$

(2.2)

where

$$ \Lambda=(T-\omega_{0})-\lambda(\eta- \omega_{0}). $$

(2.3)

Proof

By Lemmas 2.1 and 2.2 a general solution for (2.1) can be written as

$$\begin{aligned} x(t) =&C_{1}+C_{2}(t-\omega_{0})- \int_{\omega_{0}}^{t} \int_{\omega_{0}}^{r} h(s)\,d_{q,\omega}s \,d_{q,\omega}r \\ =&C_{1}+C_{2}(t-\omega_{0})- \int_{\omega_{0}}^{t} \int_{qs+\omega}^{t} h(s) \,d_{q,\omega}r \,d_{q,\omega}s \\ =&C_{1}+C_{2}(t-\omega_{0})- \int_{\omega_{0}}^{t} \bigl[t-(qs+\omega) \bigr] h(s) \,d_{q,\omega}s \end{aligned}$$

(2.4)

for $t\in[\omega_{0},T]_{q,\omega}$.

From the conditions (2.1) we obtain

$$\begin{aligned}& C_{1}=\varphi(x), \end{aligned}$$

(2.5)

$$\begin{aligned}& C_{2}=\frac{(\lambda-1)}{\Lambda}\varphi(x)-\frac{\lambda }{\Lambda} \int_{\omega_{0}}^{\eta}\bigl[\eta-(qs+\omega) \bigr]h(s) \,d_{q,\omega }s \\& \quad\quad {}+\frac{1}{\Lambda} \int_{\omega_{0}}^{T} \bigl[T-(qs+\omega) \bigr]h(s) \,d_{q,\omega}s, \end{aligned}$$

(2.6)

where Λ is defined by (2.3).

Substituting the constants $C_{1}$, $C_{2}$ into (2.4), we obtain (2.2). □

Lemma 2.5

Problem (2.1) has the unique solution of the from

$$ x(t)= \biggl[1+\frac{(\lambda-1)(t-\omega_{0})}{\Lambda} \biggr]\varphi(x)+ \int_{\omega_{0}}^{T} G(t,qs+\omega) h(s) \,d_{q,\omega}s, $$

(2.7)

where

$$ G(t,qs+\omega)= \textstyle\begin{cases} g_{1}(t,qs+\omega), & s\in[\omega_{0},t]_{q,\omega} \cap [\omega_{0},\eta]_{q,\omega},\\ g_{2}(t,qs+\omega), & s\in[\eta,t]_{q,\omega},\\ g_{3}(t,qs+\omega), & s\in[t,\eta]_{q,\omega},\\ g_{4}(t,qs+\omega), & s\in[t,T]_{q,\omega} \cap[\eta ,T]_{q,\omega}, \end{cases} $$

(2.8)

with $g_{i}(t,s)$, $1\leq i \leq4$, defined as

$$ \begin{aligned} &g_{1}(t,qs+\omega) := \biggl[\frac{1}{\Lambda}(t- \omega_{0}) (T-\lambda\eta)-t \biggr]+(qs+\omega) \biggl[ \frac{1}{\Lambda}(t-\omega_{0}) (\lambda-1)+1 \biggr], \\ &g_{2}(t,qs+\omega) := \biggl[\frac{1}{\Lambda}(t- \omega_{0})T-t \biggr]+(qs+\omega) \biggl[1-\frac{(t-\omega_{0})}{\Lambda} \biggr], \\ &g_{3}(t,qs+\omega) := \frac{1}{\Lambda}(t-\omega_{0}) (T- \lambda\eta)+\frac{(qs+\omega )}{\Lambda}(t-\omega_{0}) (\lambda-1), \\ &g_{4}(t,qs+\omega) := \frac{1}{\Lambda}(t-\omega_{0}) \bigl[T-(qs+\omega) \bigr]. \end{aligned} $$

(2.9)

Proof

Suppose $t>\eta$. The unique solution of problem (2.1) can be written as

$$\begin{aligned} x(t) &= \biggl[1+\frac{(\lambda-1)(t-\omega_{0})}{\Lambda} \biggr]\varphi(x) \\ &\quad{} +\frac{(t-\omega_{0})}{\Lambda} \int_{\omega_{0}}^{\eta} \biggl[ \bigl[T-(qs+\omega) \bigr]- \lambda \bigl[\eta-(qs+\omega) \bigr] -\frac{\Lambda }{t-\omega_{0}} \bigl[t-(qs+\omega) \bigr] \biggr] h(s)\,d_{q,\omega}s \\ &\quad{} +\frac{(t-\omega_{0})}{\Lambda} \int_{\eta}^{t} \biggl[ \bigl[T-(qs+\omega) \bigr]- \frac{\Lambda}{t-\omega_{0}} \bigl[t-(qs+\omega) \bigr] \biggr] h(s) \,d_{q,\omega}s \\ &\quad{} +\frac{(t-\omega_{0})}{\Lambda} \int_{t}^{T} \bigl[ \bigl[T-(qs+\omega) \bigr] \bigr] h(s)\,d_{q,\omega}s \\ &= \biggl[1+\frac{(\lambda-1)(t-\omega_{0})}{\Lambda} \biggr]\varphi(x) \\ &\quad{} + \int_{\omega_{0}}^{\eta} \biggl( \biggl[\frac{1}{\Lambda}(t- \omega_{0}) (T-\lambda\eta)-t \biggr]+(qs+\omega) \biggl[ \frac{1}{\Lambda}(t-\omega_{0}) (\lambda-1)+1 \biggr] \biggr) h(s) \,d_{q,\omega}s \\ &\quad{} + \int_{\eta}^{t} \biggl( \biggl[\frac{1}{\Lambda}(t- \omega_{0})T-t \biggr]+(qs+\omega) \biggl[1-\frac{(t-\omega_{0})}{\Lambda} \biggr] \biggr) h(s)\,d_{q,\omega }s \\ &\quad{}+\frac{(t-\omega_{0})}{\Lambda} \int_{t}^{T} \bigl[T-(qs+\omega) \bigr] h(s) \,d_{q,\omega}s \\ &= \biggl[1+\frac{(\lambda-1)(t-\omega_{0})}{\Lambda} \biggr]\varphi(x)+ \int_{\omega_{0}}^{T} G(t,qs+\omega) h(s)\,d_{q,\omega}s \end{aligned}$$

and similarly for $t<\eta$. The unique solution of problem (2.1) can be written as

$$\begin{aligned} x(t) &= \biggl[1+\frac{(\lambda-1)(t-\omega_{0})}{\Lambda} \biggr]\varphi(x) \\ &\quad{}+ \int_{\omega_{0}}^{\eta} \biggl( \biggl[\frac{1}{\Lambda}(t- \omega_{0}) (T-\lambda\eta)-t \biggr]+(qs+\omega) \biggl[ \frac{1}{\Lambda}(t-\omega_{0}) (\lambda-1)+1 \biggr] \biggr) h(s) \,d_{q,\omega}s \\ &\quad{}+ \int_{\eta}^{t} \biggl( \frac{1}{\Lambda}(t- \omega_{0}) (T-\lambda\eta)+\frac{(qs+\omega )}{\Lambda}(t-\omega_{0}) ( \lambda-1) \biggr) h(s)\,d_{q,\omega }s \\ &\quad{}+ \frac{(t-\omega_{0})}{\Lambda} \int_{t}^{T} \bigl[T-(qs+\omega) \bigr] h(s) \,d_{q,\omega}s \\ &= \biggl[1+\frac{(\lambda-1)(t-\omega_{0})}{\Lambda} \biggr]\varphi(x)+ \int_{\omega_{0}}^{T} G(t,qs+\omega) h(s) \,d_{q,\omega}s. \end{aligned}$$

This completes the proof. □

3 Existence and uniqueness of a solution for problem (1.1)

In this section, we present an existence and uniqueness result for problem (1.1). Let ${\mathcal{C}}=C ([\omega_{0},T]_{q,\omega}, {\mathbb{R}} )$ be the Banach space of all continuous functions x with the norm

$$\Vert x\Vert _{\mathcal{C}}=\max \bigl\{ \Vert x\Vert ,\Vert D_{p,\theta}x \Vert \bigr\} , $$

where $\Vert x\Vert = \max_{t\in[\omega_{0},T]_{q,\omega}} \vert x(t)\vert$ and $\Vert D_{p,\theta}x \Vert = \max_{t\in[\omega_{0},T]_{q,\omega}} \vert D_{p,\theta}x(pt+\theta)\vert$. Also, define the operator ${\mathcal{F}}:{\mathcal{C}}\rightarrow{\mathcal{C}}$ by

$$\begin{aligned} ({\mathcal{F}}x) (t) =& \varphi(x)+\frac{(t-\omega _{0})}{\Lambda } \biggl[( \lambda-1) \varphi(x) \\ &{}-\lambda \int_{\omega_{0}}^{\eta}\bigl[\eta-(qs+\omega) \bigr] f \bigl(s,x(s),D_{p,\theta}x(ps+\theta) \bigr)\,d_{q,\omega}s \\ &{}+ \int_{\omega_{0}}^{T} \bigl[T-(qs+\omega) \bigr] f \bigl(s,x(s),D_{p,\theta }x(ps+\theta) \bigr)\,d_{q,\omega}s \biggr] \\ &{}- \int_{\omega_{0}}^{t} \bigl[t-(qs+\omega) \bigr] f \bigl(s,x(s),D_{p,\theta }x(ps+\theta) \bigr)\,d_{q,\omega}s, \end{aligned}$$

(3.1)

where $\Lambda\neq0$ is defined by (2.3), $p=q^{m}$, $m\in {\mathbb{N}}$, and $\theta=\omega(\frac{1-p}{1-q} )$.

Obviously, problem (1.1) has solutions if and only if the operator $\mathcal{F}$ has fixed points.

Theorem 3.1

Assume that the following conditions hold:

$(H_{1})$ :

There exist constants $\gamma_{1},\gamma_{2}>0$ such that

$$\begin{aligned} & \bigl\vert f \bigl(t,x(t),D_{p,\theta}x(pt+\theta) \bigr)-f \bigl(t,y(t),D_{p,\theta }y(pt+\theta) \bigr) \bigr\vert \\ &\quad \leq \gamma_{1} \bigl\vert x(t)-y(t) \bigr\vert + \gamma_{2} \bigl\vert D_{p,\theta }x(pt+\theta)-D_{p,\theta}y(pt+ \theta) \bigr\vert \end{aligned}$$

for all $t\in[\omega_{0},T]_{q,\omega}$ and $x, y\in\mathcal{C}$.

$(H_{2})$ :

There exists a constant $\ell>0$ such that

$$\bigl\vert \varphi(x)-\varphi(y) \bigr\vert \leq\ell \bigl\Vert x(t)-y(t) \bigr\Vert _{\mathcal{C}} $$

for each $x, y\in\mathcal{C}$.

$(H_{3})$ :

$\mathfrak{S}:=\gamma\Omega+\ell\Phi< 1$, where

$$ \begin{aligned} &\gamma = \max\{\gamma_{1}, \gamma_{2}\}, \\ &\Omega = \frac{(T-\omega_{0})}{\vert \Lambda \vert } \biggl[\frac {(T-\omega _{0})^{2}-\lambda(\eta-\omega_{0})^{2}}{(1+q)} \biggr] + \frac{(T-\omega_{0})^{2}}{1+q}, \\ &\Phi = 1+\frac{(\lambda-1)(T-\omega_{0})}{\vert \Lambda \vert }.\end{aligned} $$

(3.2)

Then problem (1.1) has a unique solution in $[\omega _{0},T]_{q,\omega}$.

Proof

Denote ${\mathcal{H}}\vert x-y\vert (t):=\vert f(t,x(t),D_{p,\theta }x(pt+\theta))-f(t,y(t),D_{p,\theta}y(pt+\theta))\vert$. Using Lemma 2.3, for all $t\in[\omega_{0},T]_{q,\omega }$ and $x,y\in{\mathcal{C}}$, we have

$$\begin{aligned} & \bigl\vert ({\mathcal{F}}x) (t)-({\mathcal{F}}y) (t) \bigr\vert \\ &\quad \leq \bigl\vert \varphi(x)-\varphi(y) \bigr\vert +\frac{(t-\omega _{0})}{\vert \Lambda \vert } \biggl\vert (\lambda-1) \bigl\vert \varphi(x)-\varphi(y) \bigr\vert \\ & \quad\quad{} -\lambda \int_{\omega_{0}}^{\eta}\bigl[\eta-(qs+\omega) \bigr] { \mathcal{H}}\vert x-y\vert (s)\,d_{q,\omega}s + \int_{\omega_{0}}^{T} \bigl[T-(qs+\omega) \bigr] { \mathcal{H}}\vert x-y\vert (s)\,d_{q,\omega}s \biggr\vert \\ & \quad\quad{}+ \int_{\omega_{0}}^{t} \bigl[t-(qs+\omega) \bigr] {\mathcal {H}}\vert x-y\vert (s)\,d_{q,\omega}s \\ &\quad \leq \ell \Vert x-y\Vert _{\mathcal{C}}+\frac{(T-\omega_{0})}{\vert \Lambda \vert } \biggl[ (\lambda-1)\ell \Vert x-y\Vert _{\mathcal{C}} + \bigl( \gamma_{1}\bigl\vert x(t)-y(t)\bigr\vert +\gamma_{2} \bigl\vert D_{p,\theta}x(pt+\theta) \\ & \quad\quad{} -D_{p,\theta}y(pt+ \theta) \bigr\vert \bigr) \biggl\vert \int_{\omega_{0}}^{T} \bigl[T-(qs+\omega) \bigr] \,d_{q,\omega}s - \lambda \int_{\omega _{0}}^{\eta}\bigl[\eta-(qs+\omega) \bigr] \,d_{q,\omega}s \biggr\vert \biggr] \\ & \quad\quad{}+ \bigl(\gamma_{1} \bigl\vert x(t)-y(t) \bigr\vert + \gamma_{2} \bigl\vert D_{p,\theta }x(pt+ \theta)-D_{p,\theta}y(pt+ \theta) \bigr\vert \bigr) \int_{\omega _{0}}^{T} \bigl[T-(qs+\omega) \bigr] \,d_{q,\omega}s \\ &\quad \leq \ell \Vert x-y\Vert _{\mathcal{C}} \biggl\{ 1+ \frac{(\lambda -1)(T-\omega_{0})}{\vert \Lambda \vert } \biggr\} \\ & \quad\quad{} + \gamma \Vert x-y\Vert _{\mathcal{C}} \biggl\{ \frac{T-\omega _{0}}{\vert \Lambda \vert } \biggl[\frac{(T-\omega_{0})^{2}-\lambda(\eta -\omega _{0})^{2}}{1+q} \biggr]+ \frac{(T-\omega_{0})^{2}}{1+q} \biggr\} \\ &\quad = \ell \Vert x-y\Vert _{\mathcal{C}}\Phi+\gamma \Vert x-y \Vert _{\mathcal {C}}\Omega \\ &\quad =\mathfrak{S} \Vert x-y\Vert _{\mathcal{C}}. \end{aligned}$$

Taking the $p,\theta$-derivative for (3.1) where $p=q^{m}$, $m\in{\mathbb{N}}$, and $\theta=\omega(\frac{1-p}{1-q} )$, we obtain

$$\begin{aligned} & \bigl\vert (D_{p,\theta}{\mathcal{F}}x ) (pt+\theta)- (D_{p,\theta}{ \mathcal{F}}y ) (pt+\theta) \bigr\vert \\ &\quad \leq \biggl\vert \frac{1}{ [-(pt+\theta)(1-p)+\theta]} \biggl\{ \frac{-(pt+\theta)(1-p)+\theta}{\Lambda} \biggl( ( \lambda-1) \bigl\vert \varphi(x)-\varphi(y) \bigr\vert \\ & \quad\quad{}- \lambda \int_{\omega_{0}}^{\eta}\bigl[\eta-(qs+\omega) \bigr] { \mathcal{H}}\vert x-y\vert (s)\,d_{q,\omega}s + \int_{\omega_{0}}^{T} \bigl[T-(qs+\omega) \bigr] { \mathcal{H}}\vert x-y\vert (s)\,d_{q,\omega}s \biggr) \\ & \quad\quad{} + \biggl( \int_{\omega_{0}}^{p(pt+\theta)+\theta} \bigl[p(pt+\theta)+\theta -(qs+\omega) \bigr] {\mathcal{H}}\vert x-y\vert (s)\,d_{q,\omega}s \\ & \quad\quad{}- \int_{\omega_{0}}^{pt+\theta} \bigl[(pt+\theta)-(qs+\omega) \bigr] { \mathcal{H}}\vert x-y\vert (s)\,d_{q,\omega}s \biggr) \biggr\} \biggr\vert \\ &\quad \leq \frac{(\lambda-1)}{\vert \Lambda \vert } \ell \Vert x-y\Vert _{\mathcal{C}} + \frac{ (\gamma_{1}\vert x(t)-y(t)\vert +\gamma_{2}\vert D_{p,\theta }x(pt+\theta)-D_{p,\theta}y(pt+\theta)\vert )}{\Lambda} \\ & \quad\quad{} \times \biggl\vert \int_{\omega_{0}}^{T} \bigl[T-(qs+\omega) \bigr] \,d_{q,\omega }s - \lambda \int_{\omega_{0}}^{\eta}\bigl[\eta-(qs+\omega) \bigr] \,d_{q,\omega}s \biggr\vert \\ &\quad\quad{}+\frac{ (\gamma_{1}\vert x(t)-y(t)\vert +\gamma_{2}\vert D_{p,\theta}x(pt+\theta)-D_{p,\theta}y(pt+\theta)\vert )}{p(1-p)(t-\omega_{0})} \\ & \quad\quad{} \times \biggl\vert \int_{\omega_{0}}^{pt+\theta} \bigl[(pt+\theta)-(qs+\omega) \bigr] \,d_{q,\omega}s- \int_{\omega_{0}}^{p^{2}t+(p+1)\theta} \bigl[p^{2}t+(p+1) \theta-(qs+\omega) \bigr]\,d_{q,\omega}s \biggr\vert \\ &\quad \leq \ell \Vert x-y\Vert _{\mathcal{C}} \biggl\{ \frac{\lambda -1}{\vert \Lambda \vert } \biggr\} + \gamma \Vert x-y\Vert _{\mathcal{C}} \biggl\{ \biggl[\frac{(T-\omega_{0})^{2}-\lambda(\eta-\omega_{0})^{2} }{(1+q)\vert \Lambda \vert } \biggr]+ \frac{p(1+p)(T-\omega _{0})}{1+q} \biggr\} \\ &\quad < \mathfrak{S} \Vert x-y\Vert _{\mathcal{C}}. \end{aligned}$$

This implies that ${\mathcal{F}}$ is a contraction. Therefore, by the Banach fixed point theorem, ${\mathcal{F}}$ has a fixed point, which is a unique solution of problem (1.1) on $t\in[\omega_{0},T]_{q,\omega}$. □

4 Properties of Green’s function for problem (1.1)

We next prove that Green’s function $G(t,s)$ in (2.8) satisfies a variety of properties that are necessary for considering the existence of a positive solution to problem (1.1). Firstly, we prove some necessary preliminary lemmas.

Theorem 4.1

([21], Mean Value Theorem)

Let $f:I\rightarrow X$ be $q,\omega$-differentiable on I. For every $s \in I$,

$$\bigl\Vert f(b)-f(a) \bigr\Vert \leq\sup_{t \in I} \bigl\Vert D_{q,\omega}f(t) \bigr\Vert (b-a) $$

for all $a,b \in \{sq^{k}+\omega[k]_{q} \}_{k=0}^{\infty}$ and $a< b$.

Theorem 4.2

Let f be $q,\omega$-differentiable on $(a,b)_{q,\omega}$ and continuous on $[a,b]_{q,\omega}$. The following statements are true:

• (i) If $D_{q,\omega}f(t)>0$ for all $t\in(a,b)_{q,\omega}$, then f is an increasing function on $[a,b]_{q,\omega}$.

• (ii) If $D_{q,\omega}f(t)<0$ for all $t\in(a,b)_{q,\omega}$, then f is a decreasing function on $[a,b]_{q,\omega}$.

• (iii) If $D_{q,\omega}f(t)=0$ for all $t\in(a,b)_{q,\omega}$, then f is a constant function on $[a,b]_{q,\omega}$.

Proof

Let $t_{1},t_{2}\in[a,b]_{q,\omega}$, $t_{1}< t_{2}$. Since f is $q,\omega$-differentiable on $(a,b)_{q,\omega}$ and continuous on $[a,b]_{q,\omega}$, we have that f is a continuous function on $(a,b)_{q,\omega}$.

By Theorem 4.1 there exists $t^{*}\in(a,b)_{q,\omega}$ such that $D_{q,\omega}f(t^{*})=\frac{\Vert f(b)-f(a)\Vert }{b-a}$.

• (i) If $D_{q,\omega}f(t)>0$ for all $t\in(a,b)_{q,\omega}$, then $D_{q,\omega}f(t^{*})>0$, which implies that

  $$f(t_{2})-f(t_{1})=(t_{2}-t_{1})D_{q,\omega}f \bigl(t^{*}\bigr)>0. $$

  So $f(t_{2})>f(t_{1})$ for all $t_{1}$, $t_{2}$, and hence f is increasing on $[a,b]_{q,\omega}$.

• (ii) If $D_{q,\omega}f(t)<0$ for all $t\in(a,b)_{q,\omega}$, then $D_{q,\omega}f(t^{*})<0$, which implies that

  $$f(t_{2})-f(t_{1})=(t_{2}-t_{1})D_{q,\omega}f \bigl(t^{*}\bigr)< 0. $$

  So $f(t_{2})< f(t_{1})$ for all $t_{1}$, $t_{2}$, and hence f is decreasing on $[a,b]_{q,\omega}$.

• (iii) If $D_{q,\omega}f(t)=0$ for all $t\in(a,b)_{q,\omega}$, then $D_{q,\omega}f(t^{*})=0$, which implies that

  $$f(t_{2})-f(t_{1})=(t_{2}-t_{1})D_{q,\omega}f \bigl(t^{*}\bigr)=0. $$

  So $f(t_{2})=f(t_{1})$ for all $t_{1}$, $t_{2}$, and hence f is constant on $[a,b]_{q,\omega}$. □

Lemma 4.1

We have that $\Lambda>0$ and $1+\frac{(\lambda-1)(t-\omega _{0})}{\Lambda}$ is positive and strictly decreasing in t for $t\in [\omega_{0},T]_{q,\omega}$. In addition,

$$\begin{aligned}& \min_{t\in[\eta,T]_{q,\omega}} \biggl[ 1+\frac{(\lambda -1)(t-\omega_{0})}{\Lambda} \biggr]= \frac{T-\eta}{\Lambda}\quad\textit{and} \\& \max_{t\in[\omega_{0},T]_{q,\omega}} \biggl[ 1+\frac {(\lambda-1)(t-\omega_{0})}{\Lambda} \biggr]= \frac{\lambda(T-\eta )}{\Lambda}. \end{aligned}$$

Proof

Considering Λ in (2.3) and $1\leq \lambda< \frac{T-\omega_{0}}{\eta-\omega_{0}}$, we obtain

$$\Lambda=(T-\omega_{0})-\lambda(\eta-\omega_{0})>(T- \omega_{0})- \biggl(\frac{T-\omega_{0}}{\eta-\omega_{0}} \biggr) (\eta- \omega_{0})=0. $$

For the proof that $1+\frac{(\lambda-1)(t-\omega_{0})}{\Lambda}>0$, $t\in[\omega_{0},T]_{q,\omega}$, it is sufficient to show that

$$\begin{aligned} \Lambda+(\lambda-1) (t-\omega_{0}) &= \bigl[(T-\omega_{0})- \lambda(\eta-\omega_{0}) \bigr]+(\lambda-1) (t-\omega_{0}) \\ &= \bigl[(T-\omega_{0})-(t-\omega_{0}) \bigr]- \lambda \bigl[(\eta-\omega_{0})- (t-\omega_{0}) \bigr] \\ &> \bigl[(T-\omega_{0})-(t-\omega_{0}) \bigr]- \frac{T-\omega_{0}}{\eta -\omega_{0}} \bigl[(\eta-\omega_{0})- (t-\omega_{0}) \bigr] \\ &= \frac{(t-\omega_{0})(T-\eta)}{\eta-\omega_{0}} \geq0. \end{aligned}$$

Next, we prove that $1+\frac{(\lambda-1)(t-\omega_{0})}{\Lambda}$ is strictly decreasing in $t\in[\omega_{0},T]_{q,\omega}$. Note that the ${q,\omega}$-derivative with respect to t for $1+\frac{(\lambda -1)(t-\omega_{0})}{\Lambda}$ is

$$\begin{aligned} {}_{t}D_{q,\omega} \biggl[ 1+\frac{(\lambda-1)(t-\omega_{0})}{\Lambda } \biggr] =& \frac{\lambda-1}{(T-\omega_{0})-\lambda(\eta-\omega _{0})}< \frac{\lambda-1}{(\eta-\omega_{0})-\lambda(\eta-\omega_{0})} \\ =&-\frac{1}{\eta-\omega_{0}} < 0. \end{aligned}$$

By Theorem 4.2 we have that $1+\frac{(\lambda-1)(t-\omega _{0})}{\Lambda}$ is strictly decreasing in $t\in[\omega_{0},T]_{q,\omega}$.

Finally, observe that

$$\begin{aligned} &\min_{t\in[\eta,T]_{q,\omega}} \biggl[ 1+\frac{(\lambda -1)(t-\omega_{0})}{\Lambda} \biggr] = \biggl[ 1+ \frac{(\lambda-1)(t-\omega_{0})}{\Lambda} \biggr]_{t=\eta}=\frac{T-\eta }{\Lambda}\quad\text{and} \\ & \max_{t\in[\omega_{0},T]_{q,\omega}} \biggl[ 1+\frac{(\lambda -1)(t-\omega_{0})}{\Lambda} \biggr] = \biggl[ 1+\frac{(\lambda-1)(t-\omega_{0})}{\Lambda} \biggr]_{t=T}=\frac{\lambda(T-\eta)}{\Lambda}. \end{aligned}$$

The proof is complete. □

Next, we show that Green’s function given in (2.8) is positive.

Lemma 4.2

Let $G(t,s)$ be Green’s function given in (2.8). Then $G(t,s)\geq0$ for each $(t,s)\in[\omega_{0},T]_{q,\omega} \times[\omega _{0},T]_{q,\omega}$.

Proof

We aim to show that $g_{i}(t,qs+\omega) >0$ for all i, $1\leq i \leq 4$, and for each admissible pair $(t,s)$.

Firstly, we consider the function $g_{4}(t,qs+\omega)=\frac{1}{\Lambda }(t-\omega_{0}) [T-(qs+\omega) ]$, $s\in[t,T]_{q,\omega}\cap [\eta,T]_{q,\omega}$. To guarantee that $g_{4}(t,qs+\omega)>0$, it suffices to show that

$$\begin{aligned} T-(qs+\omega)\geq T-(qT+\omega)>(T-\omega_{0}) (1-q)>0. \end{aligned}$$

(4.1)

Thus, we conclude that $g_{4}(t,qs+\omega)>0$ on their respective domains.

Next, we consider the function $g_{2}(t,qs+\omega)$ for $s\in[\eta ,T]_{q,\omega}$ and $t\in[\eta,T]_{q,\omega}$:

$$\begin{aligned} g_{2}(t,qs+\omega) =& \biggl(\frac{(t-\omega_{0})T}{\Lambda}-t \biggr)+(qs+\omega) \biggl(1-\frac{(t-\omega_{0})}{\Lambda} \biggr) \\ =&\frac{(t-\omega_{0})}{\Lambda} \bigl[T-(qs+\omega) \bigr]- \bigl[t-(qs+\omega) \bigr]. \end{aligned}$$

To guarantee that $g_{2}(t,qs+\omega)>0$, it suffices to show that

$$\begin{aligned} \frac{(t-\omega_{0}) [T-(qs+\omega) ]}{\Lambda [t-(qs+\omega) ]} >&\frac{(t-\omega_{0}) [(T-\omega_{0})-q(s-\omega_{0}) ]}{(T-\omega_{0}) [(t-\omega_{0})-q(s-\omega_{0}) ]} \\ =&\frac{(T-\omega_{0})-q(s-\omega_{0})}{(T-\omega_{0})- (\frac {T-\omega_{0}}{t-\omega_{0}} ) q(s-\omega_{0})} > 1. \end{aligned}$$

(4.2)

So, we conclude that $g_{2}(t,qs+\omega)>0$ on their respective domains.

We next consider the function $g_{3}(t,qs+\omega)$ for $s\in[t,\eta ]_{q,\omega}$ and $t\in[\omega_{0},\eta]_{q,\omega}$:

$$g_{3}(t,qs+\omega)= \frac{1}{\Lambda}(t-\omega_{0}) (T- \lambda\eta)+\frac{qs+\omega}{\Lambda}(t-\omega_{0}) (\lambda-1). $$

To guarantee that $g_{3}(t,qs+\omega)>0$, it suffices to show that

$$\begin{aligned}& (T-\lambda\eta)+(\lambda-1) (qs+\omega) \\& \quad = \bigl[T-(qs+\omega) \bigr]-\lambda \bigl[\eta-(qs+\omega) \bigr] \\& \quad > \bigl[T-(qs+\omega) \bigr]-\frac{(1-q)T-\omega}{(1-q)\eta -\omega} \bigl[\eta-(qs+\omega) \bigr] \\& \quad = \frac{1}{(1-q)\eta-\omega} \bigl[ \bigl[T-(qs+\omega) \bigr] \bigl((1-q)\eta- \omega \bigr)- \bigl((1-q)T-\omega \bigr) \bigl[\eta-(qs+\omega) \bigr] \bigr] \\& \quad = \frac{1}{(1-q)\eta-\omega} \bigl[ (T-\eta)q \bigl[s(1-q)-\omega \bigr] \bigr] \\& \quad > \frac{(T-\eta)q}{(1-q)\eta-\omega} \bigl[ t(1-q)-\omega \bigr] \\& \quad = \frac{(T-\eta)q}{\eta-\omega_{0}} [t-\omega_{0} ] \geq0. \end{aligned}$$

(4.3)

Hence, $g_{3}(t,qs+\omega)>0$, as claimed.

Finally, we consider the function $g_{1}(t,qs+\omega)$ for $s\in[\omega _{0},t]_{q,\omega} \cap[\omega_{0},\eta]_{q,\omega}$:

$$\begin{aligned} g_{1}(t,qs+\omega) =& \biggl[\frac{1}{\Lambda}(t- \omega_{0}) (T-\lambda\eta)-t \biggr]+(qs+\omega) \biggl[ \frac{1}{\Lambda}(t-\omega_{0}) (\lambda-1)+1 \biggr] \\ =& \frac{(t-\omega_{0})}{\Lambda} \bigl[T-(qs+\omega)-\lambda \bigl[\eta -(qs+\omega) \bigr] \bigr]-t+(qs+\omega) \\ =& \frac{(t-\omega_{0})}{\Lambda} \bigl[(T-\omega_{0})-q(s-\omega_{0})- \lambda \bigl[(\eta-\omega_{0})-q(s-\omega_{0}) \bigr] \bigr] \\ &{}- \bigl[(t-\omega_{0})-q(s-\omega_{0}) \bigr]. \end{aligned}$$

To guarantee that $g_{1}(t,qs+\omega)>0$, it suffices to show that

$$\begin{aligned} \frac{1}{\Lambda} \bigl[(T-\omega_{0})-q(s- \omega_{0})-\lambda \bigl[(\eta-\omega_{0})-q(s- \omega_{0}) \bigr] \bigr]>\frac{(t-\omega _{0})-q(s-\omega_{0})}{(t-\omega_{0})}. \end{aligned}$$

(4.4)

We observe that

$$\begin{aligned} \mathcal{I}(\lambda) :=& \frac{1}{\Lambda} \bigl[(T-\omega _{0})-q(s- \omega_{0})-\lambda \bigl[(\eta- \omega_{0})-q(s- \omega_{0}) \bigr] \bigr] \\ =&\frac{ [ (T-\omega_{0})-q(s-\omega_{0}) ] - \lambda[ (\eta-\omega_{0})-q(s-\omega_{0}) ] }{(T-\omega_{0})-\lambda(\eta -\omega_{0}) } \end{aligned}$$

(4.5)

is increasing in λ for $1<\lambda< \frac{T-\omega_{0}}{\eta -\omega_{0}}$. Note that $\mathcal{I}(\lambda)$ is increasing for λ if and only if

$$\begin{aligned} \frac{(\eta-\omega_{0}) [ (T-\omega_{0})-q(s-\omega_{0}) ] }{(T-\omega_{0}) [ (\eta-\omega_{0})-q(s-\omega_{0}) ]} =\frac{ (T-\omega _{0})-q(s-\omega_{0}) }{ (T-\omega_{0})-q (\frac {T-\omega_{0}}{\eta-\omega_{0}} ) (s-\omega_{0})}>1. \end{aligned}$$

(4.6)

Clearly, (4.6) implies that (4.4) also holds, and hence $g_{1}(t,qs+\omega)>0$.

Consequently, from this it follows that $g_{i}(t,qs+\omega)>0$ for each i, $1\leq i\leq4$. Therefore, $G(t,qs+\omega)>0$. □

Lemma 4.3

Let $G(t,s)$ be Green’s function given in (2.8). Then for given $\eta\in(\omega_{0},T)_{q,\omega}$ and $1\leq\lambda< \frac {T-\omega_{0}}{\eta-\omega_{0}}$, it follows that

$$\begin{aligned} \max_{(t,s)\in[\omega_{0},T]_{q,\omega} \times[\omega _{0},T]_{q,\omega}} G(t,qs+\omega)=G(qs+\omega,qs+ \omega). \end{aligned}$$

(4.7)

Proof

Our strategy is to consider the following two cases.

Case 1: $t<\eta$. We aim to show that ${}_{t}D_{q,\omega} g_{1}(t,s), {}_{t}D_{q,\omega} g_{2}(t,s)<0$ and ${}_{t}D_{q,\omega} g_{4}(t,s)>0$. Theorem 4.2 implies that $g_{1}$, $g_{2}$ are decreasing and $g_{4}$ is increasing in t, so $G(t,qs+\omega)\leq G(qs+\omega ,qs+\omega)$ for all $(t,s)\in[\omega_{0},T]_{q,\omega} \times [\omega_{0},T]_{q,\omega}$.

Case 2: $t>\eta$. We aim to show that ${}_{t}D_{q,\omega} g_{1}(t,s)<0$ and ${}_{t}D_{q,\omega} g_{3}(t,s) , {}_{t}D_{q,\omega} g_{4}(t,s)>0$. Theorem 4.2 implies that $g_{1}$ is decreasing and $g_{3}$, $g_{4}$ are increasing in t, so $G(t,qs+\omega)\leq G(qs+\omega ,qs+\omega)$ for all $(t,s)\in[\omega_{0},T]_{q,\omega} \times [\omega_{0},T]_{q,\omega}$.

Firstly, for $g_{4}(t,qs+\omega)$, we have that

$$\begin{aligned} {}_{t}D_{q,\omega}g_{4}(t,qs+\omega)&= {}_{t}D_{q,\omega} \biggl\{ \frac {1}{\Lambda}(t- \omega_{0}) \bigl[T-(qs+\omega) \bigr] \biggr\} \\ &= \frac{[(qt+\omega)-\omega_{0}]-[t-\omega_{0}]}{t(q-1)+\omega} \biggl[\frac{T-(qs+\omega)}{\Lambda} \biggr] \\ &= \biggl[\frac{T-(qs+\omega)}{\Lambda} \biggr]=\frac{1}{\Lambda } \bigl[(T- \omega_{0})-q(s-\omega+0) \bigr] > 0 \end{aligned}$$

(4.8)

for all $s\in[t,T]_{q,\omega}\cap[\eta,T]_{q,\omega}$ and $t\in (\omega_{0},T]_{q,\omega}$.

Later, for $g_{3}(t,qs+\omega)$, we have that

$$\begin{aligned} {}_{t}D_{q,\omega}g_{3}(t,qs+\omega) =& \frac{[(qt+\omega)-\omega _{0}]-[t-\omega_{0}]}{t(q-1)+\omega} \biggl[\frac{1}{\Lambda} \bigl( (T-\lambda\eta)-(1-\lambda) (qs+\omega) \bigr) \biggr] \\ =&\frac{1}{\Lambda} \bigl[ (T-\lambda\eta)-(1-\lambda) (qs+\omega) \bigr]. \end{aligned}$$

(4.9)

From (4.3) we obtain that ${}_{t}D_{q,\omega}g_{3}(t,qs+\omega )>0$ for all $s\in[t,\eta]_{q,\omega}$ and $t\in(\omega_{0},\eta ]_{q,\omega}$.

Next, we consider $g_{2}(t,s)$ and claim that ${}_{t}D_{q,\omega} g_{2}(t,qs+\omega)<0$ for each admissible pair $(t,s)$. To this end, noting that

$$\begin{aligned} g_{2}(t,qs+\omega)=\frac{(t-\omega_{0})}{\Lambda} \bigl[ T-(qs+\omega) \bigr]- \bigl[(t-\omega_{0})-q(s-\omega+0) \bigr], \end{aligned}$$

(4.10)

we obtain

$$\begin{aligned} {}_{t}D_{q,\omega} g_{2}(t,qs+\omega) =& \frac{[(qt+\omega)-\omega _{0}]-[t-\omega_{0}]}{t(q-1)+\omega} \biggl[\frac{1}{\Lambda} \bigl[ T-(qs+\omega) \bigr]-1 \biggr] \\ =&\frac{1}{\Lambda} \bigl( T-(qs+\omega) \bigr)-1 \\ =&\frac{ (T-\omega_{0})-q(s-\omega_{0})}{(T-\omega_{0})-\lambda(\eta -\omega_{0})}-1. \end{aligned}$$

(4.11)

So, ${}_{t}D_{q,\omega} g_{2}(t,qs+\omega)$ is nonpositive when

$$\begin{aligned} \frac{ (T-\omega_{0})-q(s-\omega_{0})}{(T-\omega_{0})-\lambda (\eta -\omega_{0})}< 1. \end{aligned}$$

(4.12)

In addition, (4.12) is true if and only if

$$\begin{aligned} \lambda< \frac{q(s-\omega_{0})}{\eta-\omega_{0}}\leq\frac {q(t-\omega _{0})}{\eta-\omega_{0}}< \frac{q(T-\omega_{0})}{\eta-\omega_{0}}< \frac {T-\omega_{0}}{\eta-\omega_{0}}. \end{aligned}$$

(4.13)

Clearly, (4.13) implies that (4.12) also holds. Hence, ${}_{t}D_{q,\omega} g_{2}(t,qs+\omega)<0$ for all $s\in[\eta ,t]_{q,\omega}$ and $t\in(\omega_{0},T]_{q,\omega}$, as desired.

Finally, to claim that ${}_{t}D_{q,\omega} g_{1}(t,qs+\omega)<0$ on its domain, we have that

$$\begin{aligned}& {}_{t}D_{q,\omega} g_{1}(t,qs+\omega) \\& \quad = {}_{t}D_{q,\omega} \biggl\{ \biggl[\frac{1}{\Lambda}(t- \omega _{0}) (T-\lambda\eta)-t \biggr]+(qs+\omega) \biggl[ \frac{1}{\Lambda }(t-\omega_{0}) (\lambda-1)+1 \biggr] \biggr\} \\& \quad = \frac{1}{\Lambda} \bigl[(T-\lambda\eta)+(\lambda-1) (qs+\omega) \bigr]-1 \\& \quad = \frac{ [ (T-\omega_{0})-q(s-\omega_{0}) ]- \lambda[ (\eta-\omega_{0})-q(s-\omega_{0}) ] }{(T-\omega_{0})-\lambda(\eta -\omega_{0})}-1 \\& \quad = \frac{q(s-\omega_{0})(\lambda-1)}{(T-\omega_{0})-\lambda(\eta -\omega_{0})} < \frac{q(s-\omega_{0})(\lambda-1)}{(\eta-\omega _{0})-\lambda(\eta-\omega_{0})} \\& \quad =-\frac{q(s-\omega_{0})}{\eta-\omega_{0}} < 0 \end{aligned}$$

(4.14)

for all $s\in[\omega_{0},t]_{q,\omega} \cap[\omega_{0},\eta ]_{q,\omega}$ and $t\in(\omega_{0},T]_{q,\omega}$.

Now, note that

$$G(\omega_{0},qs+\omega)=0=G(qs+\omega,qs+\omega) \quad \text{for all } s \in[\omega_{0},T]_{q,\omega}. $$

Consequently, this implies that

$$\max_{(t,s)\in[\omega_{0},T]_{q,\omega} \times[\omega _{0},T]_{q,\omega}} G(t,qs+\omega)=G(qs+\omega,qs+\omega). $$

Observe that $G(qs+\omega,qs+\omega)=g_{4}(qs+\omega,qs+\omega )=q(s+\omega) [T-(qs+\omega) ]$.

Thus, by the discussion in the first paragraph of this proof we deduce that (4.7) holds. The proof is complete. □

Lemma 4.4

Let $G(t,s)$ be Green’s function given in (2.8). Then it follows that

$$\begin{aligned} \min_{(t,s)\in[\eta,T]_{q,\omega}\times[\omega_{0},T]_{q,\omega}} G(t,qs+\omega) \geq& \sigma\max _{(t,s)\in[\omega_{0},T]_{q,\omega }\times[\omega_{0},T]_{q,\omega}}G(t,qs+\omega) \\ =&\sigma G(qs+\omega,qs+\omega), \end{aligned}$$

(4.15)

where σ satisfies the inequality $0<\sigma<1$, and

$$\begin{aligned} \sigma:= \min \biggl\{ \frac{\lambda(\eta-\omega_{0})}{q(T-\omega _{0})},\frac{T-\eta}{(T-\omega_{0})-q(\eta-\omega_{0})} \biggr\} . \end{aligned}$$

(4.16)

Proof

We define

$$\tilde{g}_{i}(t,qs+\omega):=\frac{g_{i}(t,qs+\omega)}{g_{k}(qs+\omega ,qs+\omega)}, $$

where $k=3$ if $i=1,3$ and $k=4$ if $i=2,4$.

For $t<\eta$, we find that

$$\begin{aligned}& \tilde{g}_{3}(t,qs+\omega) = \frac{t-\omega_{0}}{q(\eta-\omega _{0})}>\frac{\eta-\omega_{0}}{q(\eta-\omega_{0})}= \frac{1}{q}:=\sigma_{1}, \end{aligned}$$

(4.17)

$$\begin{aligned}& \tilde{g}_{4}(t,qs+\omega) = \frac{t-\omega_{0}}{q(s-\omega_{0})}>\frac {T-\omega_{0}}{q(T-\omega_{0})}= \frac{1}{q}:=\sigma_{1}. \end{aligned}$$

(4.18)

If $t>\eta$, then we consider two cases of $\tilde{g}_{1}(t,qs+\omega )$ and $\tilde{g}_{2}(t,qs+\omega)$:

$$\begin{aligned} \tilde{g}_{2}(t,qs+\omega) =& \frac{\frac{t-\omega_{0}}{\Lambda} [T-(qs+\omega) ]- [t-(qs+\omega) ]}{ \frac{qs+\omega-\omega_{0}}{\Lambda} [T-(qs+\omega) ]} \\ =&\frac{t-\omega_{0}}{q(s-\omega_{0})}-\frac{\Lambda [t-(qs+\omega) ]}{q(s-\omega_{0}) [T-(qs+\omega) ]} \\ =&\frac{q(s-\omega_{0}) [(T-t)-\lambda(\eta-\omega_{0}) ]+\lambda(\eta-\omega_{0})(t-\omega_{0})}{q(s-\omega_{0}) [(T-\omega _{0})-q(s-\omega_{0}) ]} \\ >&\frac{T-t }{(T-\omega_{0})-q(\eta-\omega_{0})}+\frac{\lambda(\eta -\omega_{0}) }{q(s-\omega_{0})} \\ >&\frac{T-t }{T-\omega_{0}}+\frac{\lambda(\eta-\omega_{0}) }{q(t-\omega_{0})} > \frac{\lambda(\eta-\omega_{0}) }{q(T-\omega_{0})} := \sigma_{2}, \end{aligned}$$

(4.19)

and

$$\begin{aligned} \tilde{g}_{1}(t,qs+\omega) =& \frac{\frac{t-\omega_{0}}{\Lambda} [(T-\lambda\eta)+(\lambda -1)(qs+\omega) ]- [t-(qs+\omega) ]}{ \frac{qs+\omega-\omega_{0}}{\Lambda} [(T-\lambda\eta)+(\lambda -1)(qs+\omega) ]} \\ =&\frac{t-\omega_{0}}{q(s-\omega_{0})}-\frac{\Lambda [t-(qs+\omega) ]}{q(s-\omega_{0}) [(T-\lambda\eta)+(\lambda -1)(qs+\omega) ]} \\ =&\frac{t-\omega_{0}}{q(s-\omega_{0})}- \biggl[\frac{(t-\omega _{0})-q(s-\omega_{0})}{q(s-\omega_{0})} \biggr] \\ &{}\times \biggl\{ \frac{(T-\omega_{0})-\lambda(\eta-\omega_{0})}{ [(T-\omega_{0})-q(s-\omega_{0}) ]-\lambda[(\eta-\omega _{0})-q(s-\omega_{0}) ]} \biggr\} \\ =:&\frac{t-\omega_{0}}{q(s-\omega_{0})}- \biggl[\frac{(t-\omega _{0})-q(s-\omega_{0})}{q(s-\omega_{0})} \biggr]\times\mathbf{\mathcal {J}}(\lambda). \end{aligned}$$

(4.20)

Observe that $\mathbf{\mathcal{J}}(\lambda)=\mathbf{\mathcal {I}}^{-1}(\lambda)$, which implies that $\mathbf{\mathcal {J}}(\lambda)$ is decreasing in λ, and we have

$$\begin{aligned} \tilde{g}_{1}(t,qs+\omega) \geq& \frac{t-\omega_{0}}{q(s-\omega_{0})}- \biggl( \frac{(t-\omega _{0})-q(s-\omega_{0})}{q(s-\omega_{0})} \biggr) \biggl[\frac{T-\omega _{0}}{(T-\omega_{0})-q(s-\omega_{0})} \biggr] \\ =&\frac{1}{q(s-\omega_{0})} \biggl[(t-\omega_{0})-(T-\omega_{0}) \biggl(\frac{(t-\omega_{0})-q(s-\omega_{0})}{(T-\omega_{0})-q(s-\omega _{0})} \biggr) \biggr] \\ =&\frac{T-t}{(T-\omega_{0})-q(s-\omega_{0})} \\ >&\frac{T-\eta}{(T-\omega_{0})-q(\eta-\omega_{0})} :=\sigma_{3}. \end{aligned}$$

(4.21)

Finally, note that since $\sigma_{1}>1$, $0<\sigma_{2}<\sigma_{1}$, and $0<\sigma_{3}<1$, it follows that

$$\begin{aligned} \sigma=\min\{\sigma_{2},\sigma_{3}\}\leq \sigma_{3}< 1. \end{aligned}$$

(4.22)

We can conclude that $\min_{t\in[\eta,T]_{q,\omega}} G(t,qs+\omega)\geq\sigma\max_{t\in [\omega_{0},T]_{q,\omega}}G(t,qs+\omega)$. □

Lemma 4.5

Let φ be a nonnegative function. Then there exists $\sigma ^{*}\in(0,1)$ such that

$$\begin{aligned} & \min_{t\in[\eta,T]_{q,\omega}} \biggl\{ \int_{\omega_{0}}^{T} G(t,qs+\omega) f \bigl(s,x(s),(D_{p,\theta})x(ps+ \theta) \bigr) \,d_{q,\omega}s \\ &\quad \quad{} + \biggl[1-\frac{(1-\lambda)(t-\omega_{0})}{\Lambda} \biggr]\varphi(x) \biggr\} \\ &\quad \geq \sigma^{*} \max_{t\in[\omega_{0},T]_{q,\omega}} \biggl\{ \int_{\omega_{0}}^{T} G(t,qs+\omega) f \bigl(s,x(s),(D_{p,\theta})x(ps+ \theta) \bigr)\,d_{q,\omega}s \\ & \quad\quad{} + \biggl[1-\frac{(1-\lambda)(t-\omega_{0})}{\Lambda} \biggr]\varphi(x) \biggr\} . \end{aligned}$$

(4.23)

Proof

Observe that by Lemma 4.4 there exists a constant $\sigma\in(0,1)$ such that

$$\begin{aligned} & \min_{t\in[\eta,T]_{q,\omega}} \int_{\omega_{0}}^{T} G(t,qs+\omega) f \bigl(s,x(s),(D_{p,\theta})x(ps+ \theta) \bigr)\,d_{q,\omega }s \\ &\quad \geq\sigma\max_{t\in[\omega_{0},T]_{q,\omega}} \int_{\omega_{0}}^{T} G(t,qs+\omega) f \bigl(s,x(s),(D_{p,\theta})x(ps+ \theta) \bigr) \,d_{q,\omega}s. \end{aligned}$$

(4.24)

Next, by Lemma 4.1 there exists a constant $\mathcal{S}>0$ such that

$$\begin{aligned}& \min_{t\in[\eta,T]_{q,\omega}} \biggl[1+\frac{(\lambda -1)(t-\omega_{0})}{\Lambda} \biggr] = \frac{T-\eta}{\Lambda}=\mathcal{S}\quad\text{and} \end{aligned}$$

(4.25)

$$\begin{aligned}& \max_{t\in[\omega_{0},T]_{q,\omega}} \biggl[1+\frac {(\lambda-1)(t-\omega_{0})}{\Lambda} \biggr]= \frac{\lambda(T-\eta )}{\Lambda} = \lambda\mathcal{S}. \end{aligned}$$

(4.26)

In particular, putting (4.25) and (4.26) together implies that by taking

$$\sigma_{0} \quad \text{such that}\quad 0< \frac{1}{\lambda}\leq \sigma_{0}< 1 $$

it follows that

$$\begin{aligned}& \min_{t\in[\eta,T]_{q,\omega}} \biggl[1+\frac{(\lambda -1)(t-\omega_{0})}{\Lambda} \biggr] \varphi(x) \\& \quad = \sigma_{0}\max_{t\in[\omega_{0},T]_{q,\omega}} \biggl[1+ \frac {(\lambda-1)(t-\omega_{0})}{\Lambda} \biggr]\varphi(x). \end{aligned}$$

(4.27)

Finally, defining

$$\begin{aligned} \sigma^{*}:=\min\{\sigma,\sigma_{0} \}\in(0,1), \end{aligned}$$

(4.28)

we obtain (4.23). This completes the proof. □

Lemma 4.6

Let G be Green’s function given in (2.8). Then

$$\begin{aligned} \int_{\omega_{0}}^{T} G(qs+\omega,qs+\omega) \,d_{q,\omega}s =& \frac {q(T-\omega_{0})}{(1+q)\Lambda} \biggl\{ T \bigl[ T+ \omega_{0} \bigl(1+q-q^{2} \bigr) \bigr] \\ & - \frac{1}{1+q+q^{2}} \bigl[ T^{2}(1+q)-\omega_{0}^{2}q \bigl(1+q^{2} \bigr) \bigr] - (T+q\omega_{0}) \biggr\} . \end{aligned}$$

Proof

Using the definition of the $q,\omega$-integral, for $s\in [\omega_{0},T]_{q,\omega}$, we obtain

$$\begin{aligned}& \int_{\omega_{0}}^{T} G(qs+\omega,qs+\omega) \,d_{q,\omega}s \\& \quad = \frac{1}{\Lambda} \int_{\omega_{0}}^{T} q(s+\omega) \bigl[T-(qs+\omega) \bigr] \,d_{q,\omega}s \\& \quad = \frac{qT}{\Lambda} \int_{\omega_{0}}^{T} (s+\omega) \,d_{q,\omega }s- \frac{q}{\Lambda} \biggl[ \int_{\omega_{0}}^{T} s(qs+\omega) \,d_{q,\omega}s+ \omega \int_{\omega_{0}}^{T} (qs+\omega) \,d_{q,\omega }s \biggr] \\& \quad = \frac{qT}{\Lambda} \bigl[T(1-q)-\omega \bigr]\sum _{k=0}^{\infty} q^{k} \bigl( Tq^{k}+ \omega[k]_{q}+\omega \bigr) \\& \quad\quad{} -\frac{q}{\Lambda} \Biggl[ \bigl[T(1-q)-\omega \bigr] \sum _{k=0}^{\infty} q^{k} \bigl[ \bigl(Tq^{k}+\omega[k]_{q} \bigr) \bigl( \bigl(Tq^{k}+ \omega[k]_{q} \bigr) q+\omega \bigr) \bigr] \\& \quad\quad{} + \omega \bigl[T(1-q)-\omega \bigr] \sum _{k=0}^{\infty} q^{k} \bigl[ \bigl(Tq^{k}+\omega[k]_{q} \bigr) q+ \omega \bigr] \Biggr] \\& \quad = \frac{q(T-\omega_{0})}{(1+q)\Lambda} \\& \quad\quad{} \times \biggl\{ T \bigl[ T+\omega_{0} \bigl(1+q-q^{2} \bigr) \bigr] - \frac{1}{1+q+q^{2}} \bigl[ T^{2}(1+q)-\omega_{0}^{2}q \bigl(1+q^{2} \bigr) \bigr] - (T+q\omega_{0}) \biggr\} . \end{aligned}$$

This completes the proof. □

5 Existence of a positive solution for problem (1.1)

In this section, we consider the existence of at least one positive solution for problem (1.1) by appealing to the Krasnoselskii fixed point theorem in a cone.

Define the cone ${\mathcal{P}}\subseteq{\mathcal{C}}$ by

$$\begin{aligned} {\mathcal{P}}:= \Bigl\{ x\in{\mathcal{C}} : x(t)\geq0, \min _{t\in [\eta,T]_{q,\omega} }x (t)\geq\sigma^{*} \bigl\Vert x(t) \bigr\Vert _{\mathcal{C}} \text{ and } \varphi(x)\geq0 \Bigr\} . \end{aligned}$$

(5.1)

Consider nonlinear equation (1.1); then x solves (1.1) if and only if x is a fixed point of the operator ${\mathcal{A}}:{\mathcal {P}}\rightarrow{\mathcal{P}}$ defined by

$$\begin{aligned} ({\mathcal{A}}x ) (t) :=& \biggl[1+\frac{(\lambda -1)(t-\omega_{0})}{\Lambda} \biggr] \varphi(x) \\ &{}+ \int_{\omega_{0}}^{T} G(t,qs+\omega) f \bigl(s,x(s),D_{p,\theta }x(ps+ \theta) \bigr)\,d_{q,\omega}s, \end{aligned}$$

(5.2)

where G is Green’s function for problem (1.1), and ${\mathcal{C}}$ is the Banach space defined in Section 3.

Lemma 5.1

Suppose that $f:[\omega_{0},T]_{q,\omega}\times[0,\infty)\times [0,\infty)\rightarrow[0,\infty)$ and $\varphi:C ([\omega _{0},T]_{q,\omega}$, $[0,\infty) )\rightarrow[0,\infty)$ are continuous. Then the operator $\mathcal{A}:{\mathcal{P}}\rightarrow{\mathcal{P}}$ is completely continuous.

Proof

Since $G(t,qs+\omega)\geq0$ for all $(t,s)\in[\omega _{0},T]_{q,\omega}\times[\omega_{0},T]_{q,\omega}$, we have ${\mathcal {A}}\geq0$ for all $x\in{\mathcal{P}}$. For a constant $L>0$, we define

$$B_{L}= \bigl\{ x\in{\mathcal{P}} : \Vert x\Vert _{\mathcal{C}}< L \bigr\} $$

and let $M= \max_{(t,x)\in{[\omega_{0},T]_{q,\omega}}\times B_{L}}\vert f (t,x(t),D_{p,\theta}x(pt+\theta) )\vert$, $N= \sup_{x\in B_{L}} \vert \varphi(x)\vert$. Then, for $x\in B_{L}$, we obtain

$$\begin{aligned}& \bigl\vert ({\mathcal{A}}x) (t)\bigr\vert \\& \quad = \biggl\vert \biggl[1+\frac{(\lambda-1)(t-\omega_{0})}{\Lambda} \biggr]\varphi(x)+ \int_{\omega_{0}}^{T} G(t,qs+\omega) f \bigl(s,x(s),D_{p,\theta}x(ps+ \theta) \bigr)\,d_{q,\omega}s \biggr\vert \\& \quad \leq \frac{N\lambda(T-\eta)}{\Lambda} +M \int_{\omega_{0}}^{T} G(qs+\omega,qs+\omega) \,d_{q,\omega}s \\& \quad = \frac{N\lambda(T-\eta)}{\Lambda} + \frac{qM(T-\omega _{0})}{(1+q)\Lambda} \\& \quad\quad{} \times \biggl\vert T \bigl[ T+\omega_{0} \bigl(1+q-q^{2} \bigr) \bigr] - \frac{1}{1+q+q^{2}} \bigl[ T^{2}(1+q)- \omega_{0}^{2}q \bigl(1+q^{2} \bigr) \bigr] - (T+q \omega_{0}) \biggr\vert \\& \quad =: \mathcal{K}. \end{aligned}$$

Similarly to the proof above and Theorem 3.1, we obtain

$$\begin{aligned}& \bigl\vert ( D_{p,\theta} {\mathcal{A}}x ) (pt+\theta) \bigr\vert \\& \quad = \biggl\vert \frac{1}{(p-1)((pt+\theta)-\omega_{0})} \biggl\{ \frac {(pt+\theta)(p-1)+\theta}{\Lambda} \\& \quad\quad{} \times \biggl[ (\lambda-1) \varphi(x) -\lambda \int_{\omega_{0}}^{\eta} G(pt+\theta,qs+\omega) f \bigl(s,x(s),D_{p,\theta}x(s) \bigr) \,d_{q,\omega}s \\ & \quad\quad{}+ \int_{\omega_{0}}^{T} G(pt+\theta,qs+\omega) f \bigl(s,x(s),D_{p,\theta}x(s) \bigr)\,d_{q,\omega}s \biggr] \\ & \quad\quad{}- \int_{\omega_{0}}^{p(pt+\theta)+\theta} \bigl[ p(pt+\theta)+\theta-(qs+ \omega) \bigr] f \bigl(s,x(s),D_{p,\theta}x(s) \bigr) \,d_{q,\omega}s \\ & \quad\quad{}+ \int_{\omega_{0}}^{pt+\theta} \bigl[(pt+\theta)-(qs+\omega) \bigr] f \bigl(s,x(s),D_{p,\theta}x(s) \bigr)\,d_{q,\omega}s \biggr\} \biggr\vert \\ & \quad < \bigl\vert ({\mathcal{A}}x ) (t)\bigr\vert . \end{aligned}$$

(5.3)

Therefore, $\Vert ({\mathcal{A}}x )(t)\Vert _{\mathcal {C}}= \mathcal{K}$, and hence ${\mathcal{A}}(B_{L})$ is uniformly bounded.

Next, we shall show that ${\mathcal{A}}(B_{L})$ is equicontinuous. For $x\in B_{L}$ and $t_{1},t_{2}\in[\omega_{0},T]_{q,\omega}$ with $t_{1}< t_{2}$, there are three cases to consider.

Case 1: If $\eta\leq t_{1}< t_{2}$, then by (2.7), letting $g_{i}(t,qs+\omega)=\frac{(t-\omega_{0})}{\Lambda} {\mathfrak {g}}_{i}(t,qs+\omega)$, we obtain

$$\begin{aligned}& \bigl\vert ({\mathcal{A}}x ) (t_{2})- ({\mathcal{A}}x ) (t_{1}) \bigr\vert \\ & \quad \leq \vert t_{2}-t_{1}\vert (\lambda-1) \frac{N}{\Lambda}+M \biggl\vert \int_{\omega_{0}}^{T} \bigl[ G(t_{2},qs+\omega) - G(t_{1},qs+\omega) \bigr]\,d_{q,\omega}s \biggr\vert \\ & \quad \leq \vert t_{2}-t_{1}\vert \frac{N(\lambda-1)}{\Lambda} +M \biggl\vert \frac{(t_{2}-t_{1})}{\Lambda} \int_{\omega_{0}}^{\eta} {\mathfrak{g}}_{1}(qs+ \omega,qs+\omega) \,d_{q,\omega}s \\ & \quad\quad{} + \biggl[ \frac{(t_{2}-\omega_{0})}{\Lambda} \int_{\eta}^{t_{2}} {\mathfrak{g}}_{2}(qs+ \omega,qs+\omega) \,d_{q,\omega}s -\frac{(t_{1}-\omega_{0})}{\Lambda} \int_{\eta}^{t_{1}} {\mathfrak{g}}_{2}(qs+ \omega,qs+\omega) \,d_{q,\omega}s \biggr] \\ & \quad\quad{}+ \biggl[ \frac{(t_{2}-\omega_{0})}{\Lambda} \int_{t_{2}}^{T} {\mathfrak{g}}_{4}(qs+ \omega,qs+\omega) \,d_{q,\omega}s -\frac{(t_{1}-\omega_{0})}{\Lambda} \int_{t_{1}}^{T} {\mathfrak{g}}_{4}(qs+ \omega,qs+\omega) \,d_{q,\omega}s \biggr] \biggr\vert \\ & \quad = \frac{N(\lambda-1)}{\Lambda} \vert t_{2}-t_{1}\vert + \frac {M}{\Lambda} \vert t_{2}-t_{1}\vert \\ & \quad\quad{} \times\biggl\vert \bigl[T-1-\lambda(\eta-1) \bigr] \int_{\omega_{0}}^{\eta }\,d_{q,\omega}s- \int_{\omega_{0}}^{\eta} \bigl[T-(qs+\omega) \bigr] \,d_{q,\omega}s \\ & \quad\quad{}+ \int_{\omega_{0}}^{T} \bigl[T-(qs+\omega) \bigr] \,d_{q,\omega}s \biggr\vert \\ & \quad = \frac{\vert t_{2}-t_{1}\vert }{\Lambda} \biggl\{ N(\lambda-1) + M \biggl\vert \bigl[T-1-\lambda(\eta-1) \bigr](\eta-\omega_{0}) \\ & \quad\quad{}+ \frac{\eta-\omega_{0}}{1+q} \bigl[(T-\omega_{0})+q(T+\eta) \bigr]+ \frac{(T-\omega_{0})^{2}}{1+q} \biggr\vert \biggr\} \\ & \quad = \frac{\vert t_{2}-t_{1}\vert }{\Lambda} {\boldsymbol{\Xi}}. \end{aligned}$$

Therefore, there exists a constant $\delta_{1}>0$ such that

$$\begin{aligned} \bigl\vert ({\mathcal{A}}x ) (t_{2})- ({\mathcal{A}}x ) (t_{1}) \bigr\vert < \frac{\epsilon}{2} \quad \text{whenever }\vert t_{2}-t_{1}\vert < \delta_{1}= \frac{\epsilon\Lambda}{2 {\boldsymbol{\Xi}}}. \end{aligned}$$

(5.4)

Case 2: If $t_{1}< t_{2}\leq\eta$, then by (2.7) we obtain

$$\begin{aligned}& \bigl\vert ({\mathcal{A}}x ) (t_{2})- ({\mathcal{A}}x ) (t_{1}) \bigr\vert \\ & \quad \leq \vert t_{2}-t_{1}\vert (\lambda-1) \frac{N}{\Lambda}+M \biggl\vert \int_{\omega_{0}}^{T} \bigl[ G(t_{2},qs+\omega) - G(t_{1},qs+\omega) \bigr]\,d_{q,\omega}s \biggr\vert \\ & \quad \leq \vert t_{2}-t_{1}\vert \frac{N(\lambda-1)}{\Lambda} +M \biggl\vert \biggl[ \frac{(t_{2}-\omega _{0})}{\Lambda} \int_{\omega_{0}}^{t_{2}} {\mathfrak{g}}_{1}(qs+ \omega,qs+\omega)\,d_{q,\omega}s \\ & \quad\quad{} -\frac{(t_{1}-\omega_{0})}{\Lambda} \int_{\omega_{0}}^{t_{1}} {\mathfrak{g}}_{1}(qs+ \omega,qs+\omega)\,d_{q,\omega}s \biggr]+ \biggl[ \frac {(t_{2}-\omega_{0})}{\Lambda} \int_{t_{2}}^{\eta} {\mathfrak{g}}_{3}(qs+ \omega,qs+\omega) \,d_{q,\omega}s \\ & \quad\quad{}- \frac{(t_{1}-\omega_{0})}{\Lambda} \int_{t_{1}}^{\eta} {\mathfrak{g}}_{3}(qs+ \omega,qs+\omega) \,d_{q,\omega}s \biggr] \biggr\vert +\frac {(t_{2}-t_{1})}{\Lambda} \int_{\eta}^{T} {\mathfrak{g}}_{4}(qs+ \omega,qs+\omega) \,d_{q,\omega}s \\ & \quad = \frac{N(\lambda-1)}{\Lambda} \vert t_{2}-t_{1}\vert +\frac {M}{\Lambda} \vert t_{2}-t_{1}\vert \\ & \quad\quad{} \times \biggl\vert \int_{\omega_{0}}^{\eta} \bigl( \bigl[T-1-\lambda(\eta-1) \bigr]- \bigl[T-(qs+\eta) \bigr] \bigr)\,d_{q,\omega}s+ \int_{\omega _{0}}^{T} \bigl[T-(qs+\omega) \bigr] \,d_{q,\omega}s \biggr\vert \\ & \quad = \frac{\vert t_{2}-t_{1}\vert }{\Lambda} \biggl\{ N(\lambda-1) + M \biggl\vert \bigl[T-1-\lambda(\eta-1) \bigr](\eta-\omega_{0}) \\ & \quad\quad{}+ \frac{\eta-\omega_{0}}{1+q} \bigl[(T-\omega_{0})+q(T+\eta) \bigr]+ \frac{(T-\omega_{0})^{2}}{1+q} \biggr\vert \biggr\} \\ & \quad = \frac{\vert t_{2}-t_{1}\vert }{\Lambda} {\boldsymbol{\Xi}}. \end{aligned}$$

Therefore, there exists a constant $\delta_{2}>0$ such that

$$\begin{aligned} \bigl\vert ({\mathcal{A}}x ) (t_{2})- ({\mathcal{A}}x ) (t_{1}) \bigr\vert < \frac{\epsilon}{2} \quad \text{whenever }\vert t_{2}-t_{1}\vert < \delta_{2}= \frac{\epsilon\Lambda}{2 {\boldsymbol{\Xi}}}. \end{aligned}$$

(5.5)

Case 3: If $t_{1}<\eta<t_{2}$ with $\vert t_{2}-t_{1}\vert <\delta=\min\{ \delta_{1},\delta_{2}\}$, then from (5.4)-(5.5) it follows that

$$\begin{aligned} \bigl\vert ({\mathcal{A}}x ) (t_{2})- ({\mathcal{A}}x ) (t_{1}) \bigr\vert \leq& \bigl\vert ({\mathcal{A}}x ) (t_{2})- ({\mathcal{A}}x ) (\eta) \bigr\vert + \bigl\vert ({\mathcal {A}}x ) (\eta)- ({\mathcal{A}}x ) (t_{1}) \bigr\vert \\ < & \frac{\epsilon}{2}+\frac{\epsilon}{2} = \epsilon. \end{aligned}$$

(5.6)

Similarly to the proof above, by (5.3) we obtain

$$\begin{aligned} \bigl\vert (D_{p,\theta}{\mathcal{A}}x ) (pt_{2}+ \theta)- (D_{p,\theta}{\mathcal{A}}x ) (pt_{1}+\theta) \bigr\vert < \bigl\vert ({\mathcal{A}}x ) (t_{2})- ({\mathcal{A}}x ) (t_{1}) \bigr\vert < \epsilon. \end{aligned}$$

(5.7)

Hence, we conclude that $\vert ({\mathcal{A}}x )(t_{2})- ({\mathcal{A}}x )(t_{1})\vert <\epsilon$ if $\vert t_{2}-t_{1}\vert <\delta=\min\{\delta_{1},\delta_{2}\}$ for $t_{2},t_{1}\in [\omega_{0},T]_{q,\omega}$, that is, ${\mathcal{A}}(B_{L})$ is equicontinuous. By the Arzelà-Ascoli theorem, $\mathcal{A}:\mathcal {C}\rightarrow\mathcal{C}$ is a completely continuous operator.

Finally, we apply Lemmas 4.2-4.4 to obtain

$$\begin{aligned} ({\mathcal{A}}x) (t)\geq0 \quad \text{for all } t\in[\omega _{0},T]_{q,\omega} \end{aligned}$$

(5.8)

and, for $f\in\mathcal{P}$,

$$\begin{aligned} \min_{t\in[\eta,T]_{q,\omega}}({\mathcal{A}}x) (t) \geq & \min _{t\in[\eta,T]_{q,\omega}} \biggl\{ \biggl[1+\frac{(\lambda-1)(t-\omega _{0})}{\Lambda} \biggr]\varphi(x) \\ &{}+ \int_{\omega_{0}}^{T} G(t,qs+\omega)f \bigl(s,x(s),D_{p,\theta }x(ps+ \theta) \bigr) \,d_{q,\omega}s \biggr\} \\ =&\sigma^{*} \biggl\{ \max_{t\in[\omega_{0},T]_{q,\omega}} \biggl[1+\frac {(\lambda-1)(t-\omega_{0})}{\Lambda} \biggr]\varphi(x) \\ &{}+ \int_{\omega_{0}}^{T} G(qs+\omega,qs+\omega) f \bigl(s,x(s),D_{p,\theta}x(ps+\theta) \bigr) \,d_{q,\omega}s \biggr\} \\ =&\sigma^{*}({\mathcal{A}}x) (qs+\omega). \end{aligned}$$

(5.9)

Hence,

$$\min_{t\in[\eta,T]_{q,\omega}}({\mathcal{A}}x) (t)\geq\sigma^{*} \Vert { \mathcal{A}}x\Vert _{\mathcal{C}},\quad \text{that is}, {\mathcal{A}} { \mathcal{P}} \subset{\mathcal{P}}. $$

Consequence, it follows that ${\mathcal{A}}: {\mathcal{P}}\rightarrow {\mathcal{P}}$ is a completely continuous operator. □

The following notation is used in the sequel:

$$\begin{aligned}& \diamondsuit := \frac{1}{ \int^{T}_{\omega_{0}}G(qs+\omega,qs+\omega )\,d_{q,\omega}s}, \\& \Theta := \frac{1}{\sigma^{*} \int^{T}_{\omega_{0}}G(qs+\omega ,qs+\omega)\,d_{q,\omega}s}, \\& \Upsilon_{1} := \frac{\Lambda}{\lambda(T-\eta)},\quad \text{and} \\& \Upsilon_{2} := \frac{\Lambda}{\sigma^{*} [\lambda (T-\eta) ]}. \end{aligned}$$

Next, we introduce some assumptions that will be helpful in the sequel.

$(A_{1})$ :

There exists a constant $r_{1}>0$ such that

$$\begin{aligned} f \bigl(t,x(t),D_{p,\theta}x(pt+\theta) \bigr)\leq\frac {1}{2} \diamondsuit r_{1} \end{aligned}$$

(5.10)

for all $t\in[\omega_{0},T]_{q,\omega}$ and $0\leq x \leq r_{1}$.

$(A_{2})$ :

There exists a constant $r_{2}>0$ with $r_{2}< r_{1}$ such that

$$\begin{aligned} f \bigl(t,x(t),D_{p,\theta}x(pt+\theta) \bigr)\geq\frac {1}{2}\Theta r_{2} \end{aligned}$$

(5.11)

for all $t\in[\omega_{0},T]_{q,\omega}$ and $\sigma^{*}r_{2} \leq x \leq r_{2}$, where $\sigma^{*}$ is defined in (4.28).

$(A_{3})$ :

There exists a constant $r_{1}>0$ such that

$$\begin{aligned} \varphi(x)\leq\frac{1}{2}\Psi_{1} r_{1} \end{aligned}$$

(5.12)

for all $x\in\mathcal{P}$ and $0\leq \Vert x\Vert _{\mathcal{C}}\leq r_{1}$.

$(A_{4})$ :

There exists a constant $r_{2}>0$ such that

$$\begin{aligned} \varphi(x)\geq\frac{1}{2}\Psi_{2} r_{2} \end{aligned}$$

(5.13)

for all $x\in\mathcal{P}$ and $\sigma^{*}r_{2}\leq \Vert x\Vert _{\mathcal {C}}\leq r_{2}$.

Now, we can prove the existence of at least one positive solution.

Theorem 5.1

Suppose that conditions $(A_{1})$-$(A_{4})$ hold. Let $f(t,x)\in C ([\eta,T]_{q,\omega}\times[0,\infty )\times[0,\infty), [0,\infty) )$ and $\varphi(x):C ([\eta ,T]_{q,\omega},[0,\infty) )\rightarrow[0,\infty)$. Then problem (1.1) has at least one positive solution, say $x^{*}$, where $r_{2}\leq \Vert x^{*}\Vert _{\mathcal{C}} \leq r_{1}$.

Proof

Set $\Psi_{1}=\{x\in C ([\omega_{0},T]_{q,\omega} ):\Vert x\Vert _{\mathcal{C}}< r_{1} \}$. Then, for $x\in{\mathcal {P}}\cap \partial\Psi_{1}$, we have

$$\begin{aligned} ({\mathcal{A}}x) (t) \leq& \max_{t\in[\omega_{0},T]_{q,\omega}} \biggl\{ \biggl[1+ \frac{(\lambda-1)(t-\omega_{0})}{\Lambda} \biggr]\varphi(x) \\ &{}+ \int_{\omega_{0}}^{T} G(t,qs+\omega) f \bigl(s,x(s),D_{p,\theta }x(ps+ \theta) \bigr) \,d_{q,\omega}s \biggr\} \\ \leq& \biggl(\frac{\lambda(T-\eta)}{\Lambda} \biggr) \frac {1}{2} \Upsilon_{1} r_{1}+ \frac{1}{2}\diamondsuit r_{1} \int_{\omega _{0}}^{T} G(qs+\omega,qs+\omega) \,d_{q,\omega}s \\ =&\frac{r_{1}}{2}+\frac{r_{1}}{2}=r_{1}. \end{aligned}$$

Since $\vert (D_{p,\theta}{\mathcal{A}}x)(t)\vert <\vert ({\mathcal{A}}x)(t)\vert \leq r_{1}$, we have

$$\begin{aligned} \Vert \mathcal{A}x\Vert _{\mathcal{C}}\leq \Vert x\Vert _{\mathcal{C}} \quad \text{whenever } x\in{\mathcal{P}}\cap\partial \Psi_{1}. \end{aligned}$$

(5.14)

Further, let $\Psi_{2}=\{x\in C ([\omega_{0},T]_{q,\omega} ):\Vert x\Vert _{\mathcal{C}}< r_{2} \}$. Then, for $x\in{\mathcal {P}}\cap \partial\Psi_{2}$, using Lemma 4.4, we find that

$$\begin{aligned} ({\mathcal{A}}x) (t) \geq& \min_{t\in[\eta,T]_{q,\omega}} \biggl\{ \biggl[1+ \frac{(\lambda-1)(t-\omega_{0})}{\Lambda} \biggr]\varphi(x) \biggr\} \\ &{}+ \int_{\omega_{0}}^{T} \min_{t\in[\eta,T]_{q,\omega}} G(t,qs+ \omega) f \bigl(s,x(s),D_{p,\theta}x(ps+\theta) \bigr) \,d_{q,\omega}s \\ \geq& \frac{1}{2}\Upsilon_{2} r_{2} \sigma^{*} \max_{t\in[\omega _{0},T]_{q,\omega}} \biggl\{ \biggl[1+\frac{(\lambda-1)(t-\omega _{0})}{\Lambda} \biggr] \varphi(x) \biggr\} \\ &{}+\frac{1}{2}\Theta r_{2}\sigma^{*} \max_{t\in[\omega_{0},T]_{q,\omega}} \int_{\omega_{0}}^{T} G(t,qs+\omega)\,d_{q,\omega}s \\ =&\frac{1}{2}\Upsilon_{2} r_{2}\sigma^{*} \biggl(\frac{\lambda(T-\eta )}{\Lambda} \biggr) + \frac{1}{2}\Theta r_{2} \sigma^{*} \int_{\omega _{0}}^{T} G(qs+\omega,qs+\omega) \,d_{q,\omega}s \\ =&\frac{r_{2}}{2}+\frac{r_{2}}{2}=r_{2}. \end{aligned}$$

Since $\vert (D_{p,\theta}{\mathcal{A}}x)(t)\vert <\vert ({\mathcal{A}}x)(t)\vert \leq r_{2}$, we have

$$\begin{aligned} \Vert \mathcal{A}x\Vert _{\mathcal{C}}\geq \Vert x\Vert _{\mathcal{C}} \quad \text{whenever } x\in{\mathcal{P}}\cap\partial \Psi_{2}. \end{aligned}$$

(5.15)

We conclude by Theorem 1.1 that the operator $\mathcal{A}$ has a fixed point. This implies that problem (1.1) has a positive solution, say $x^{*}$, where $r_{2}\leq \Vert x^{*}\Vert _{\mathcal{C}} \leq r_{1}$. □

6 Example

In this section, to illustrate our results, we consider an example.

Example

Consider the following boundary value problem for the second-order Hahn difference equation

$$ \begin{aligned} &D^{2}_{\frac{1}{3},\frac{3}{2}}x(t)+ \frac{e^{-\cos^{2}(\pi t)}}{ (t+20 )^{2}}\cdot \frac{\vert x\vert +\arctan(\cos^{2}(\pi t) ) [D_{\frac {1}{243},\frac{121}{54}}x (\frac{1}{243}t+\frac{121}{54} ) ]}{\vert x\vert +\frac{1}{e}} =0, \\ &x \biggl(\frac{9}{4} \biggr)=\sum_{i=0}^{\infty} \frac {C_{i}\vert x(t_{i})\vert }{1+\vert x(t_{i})\vert },\quad\quad t_{i}=10 \biggl(\frac{1}{3} \biggr)^{i}+\frac {3}{2}[i]_{\frac{1}{3}}, \\ &x (10 )=\frac{4}{3}x \biggl(\frac{1109}{486} \biggr), \end{aligned} $$

(6.1)

where $t\in[\frac{9}{4},10 ]_{\frac{1}{3},\frac {3}{2}}$, and $C_{i}$ are given positive constants with $\frac {1}{e^{2}}\leq\sum_{i=0}^{\infty}C_{i}\leq\frac{2}{e^{2}}$.

Set $q=\frac{1}{3}$, $\omega=\frac{3}{2}$, $\omega_{0}=\frac{9}{4}$, $p=\frac{1}{3^{5}}=\frac{1}{243}$, $\theta= \frac{3}{2} [\frac{1-(\frac{1}{3})^{5}}{1-\frac{1}{3}} ]=\frac{121}{54}$, $T=10$, $\eta=10 (\frac{1}{3} )^{5}+\frac{3}{2}[5]_{\frac {1}{3}}=\frac{1109}{486}$, $\lambda=\frac{4}{3}$, $\varphi(x)=\sum_{i=0}^{\infty}\frac{C_{i}\vert x(t_{i})\vert }{1+\vert x(t_{i})\vert }$, and

$$\begin{aligned} f \bigl(t,x(t),D_{p,\theta}x(pt+\theta) \bigr)&= \frac{e^{-\cos^{2}(\pi t)}}{ (t+20 )^{2}} \\ &\quad{} \times\frac{\vert x\vert +\arctan(\cos^{2}(\pi t) ) [D_{\frac {1}{243},\frac{121}{54}}x (\frac{1}{243}t+\frac{121}{54} ) ]}{\vert x\vert +\frac{1}{e}}. \end{aligned}$$

I. The existence and uniqueness of solution to problem ( 6.1 ). We can show that

$$1 \leq\lambda< \frac{30132}{5956} \approx5,\quad\quad \Lambda\approx7.707,\quad\quad \Omega \approx90.362,\quad \text{and}\quad \Phi\approx1.335. $$

Clearly,

$$\begin{aligned}& \bigl\vert f \bigl(t,x(t),D_{p,\theta}x(pt+\theta) \bigr)-f \bigl(t,y(t),D_{p,\theta }y(pt+\theta) \bigr) \bigr\vert \\& \quad < \frac{\vert x(t)-y(t)\vert }{(t+20)^{2}}+ \frac{\arctan (1)\vert D_{p,\theta }x(pt+\theta) - D_{p,\theta}y(pt+\theta)\vert }{(t+20)^{2}} \\& \quad \leq 0.002 \bigl\vert x(t)-y(t) \bigr\vert + 0.0016 \bigl\vert D_{p,\theta}x(pt+\theta) - D_{p,\theta}y(pt+\theta) \bigr\vert , \end{aligned}$$

so that $(H_{1})$ holds with $\gamma_{1}=0.002$, $\gamma_{2}=0.0016$, and $\gamma=\max\{\gamma_{1},\gamma_{2}\}=0.002$, and

$$\begin{aligned} \bigl\vert \varphi(x)-\varphi(y) \bigr\vert &= \Biggl\vert \sum _{i=0}^{\infty}\frac {C_{i}\vert x(t_{i})\vert }{1+\vert x(t_{i})\vert }-\sum _{i=0}^{\infty}\frac {C_{i}\vert y(t_{i})\vert }{1+\vert y(t_{i})\vert } \Biggr\vert \\ &< \sum_{i=0}^{\infty}C_{i} \bigl\vert x(t_{i})-y(t_{i}) \bigr\vert \leq \frac{2}{e^{2}}\Vert x-y\Vert , \end{aligned}$$

so that $(H_{2})$ holds with $\ell=\frac{2}{e^{2}}$.

Also, we can show that

$$\mathfrak{S}=\gamma\Omega+\ell\Phi\approx0.542< 1. $$

Hence, by Theorem 3.1 problem (6.1) has a unique solution.

II. The existence of at least one positive solution to problem ( 6.1 ). We can show that

$$\begin{aligned}& \sigma = \min \biggl\{ \frac{\lambda(\eta-\omega_{0})}{q(T-\omega _{0})},\frac{T-\eta}{(T-\omega_{0})-q(\eta-\omega_{0})} \biggr\} \\& \hphantom{\sigma} = \min\{0.0165,0.997\}=0.0165, \\& \frac{1}{\lambda} = 0.75\leq\sigma_{0}< 1, \\& \sigma^{*} = \min \bigl\{ \sigma,\min\{\sigma_{0}\} \bigr\} =0.0165, \\& \diamondsuit = \frac{1}{ \int^{T}_{\omega_{0}}G(qs+\omega,qs+\omega )\,d_{q,\omega}s}=0.146, \\& \Theta = \frac{1}{\sigma^{*} \int^{T}_{\omega_{0}}G(qs+\omega ,qs+\omega)\,d_{q,\omega}s}=8.850, \\& \Upsilon_{1}=\frac{\Lambda}{\lambda(T-\eta)}=0.749,\quad \text{and}\quad \Upsilon_{2} = \frac{\Lambda}{\sigma^{*} [\lambda(T-\eta) ]}=45.389. \end{aligned}$$

Clearly,

$$\begin{aligned} & \bigl\vert f \bigl(t,x(t),D_{p,\theta}x(pt+\theta) \bigr) \bigr\vert \leq \frac{1+\frac{\pi }{4}}{ (\frac{9}{2}+20 )^{2}}=0.0036=0.049\frac{\diamondsuit }{2} \\ &\quad \text{for }0\leq x\leq r_{1} \leq0.049, \\ & \bigl\vert f \bigl(t,x(t),D_{p,\theta}x(pt+\theta) \bigr) \bigr\vert \geq\frac{1}{e (30 )^{2}} \biggl[\frac{\vert x\vert }{\vert x\vert +1} \biggr]> \frac {1}{2e}=0.0002=0.000045\frac{\Theta}{2} \\ &\quad \text{for }0.00000074=0.000045 \sigma^{*}\leq x\leq r_{2} \leq 0.000045, \\ & \varphi(x)\leq\frac{2}{e^{2}}=0.271=3.3457\frac{\Upsilon _{1}}{2} \quad \text{for }0\leq x\leq r_{1} \leq3.3457,\quad \text{and} \\ & \varphi(x)\geq\frac{1}{e^{2}}=0.135=0.0059 \frac{\Upsilon _{2}}{2}\quad \text{for }0.000097=0.0059 \sigma^{*}\leq x\leq r_{2} \leq0.0059. \end{aligned}$$

Therefore, conditions $(A_{1})$-$(A_{3})$ are satisfied. Consequently, by Theorem 5.1 problem (6.1) has at least one positive solution $x^{*}$ such that $r_{2}=0.000097 \leq \Vert x^{*}\Vert _{\mathcal{C}} \leq0.0036=r_{1}$.