1 Introduction

In the classical Lidstone expansion theorem [1], an entire function \(f(x)\) may be expanded with respect to the points 0 and 1 in the form

$$f(x)=\sum_{n=0}^{\infty} \bigl(f^{(2n)}(1)A_{n}(x)-f^{{(2n)}}(0)A_{n}(x-1) \bigr), $$

where \(A_{n}\) is a polynomial of degree \(2n+1\) that satisfies

  1. (i)

    \(A_{0}(x)=x\),

  2. (ii)

    \(A_{n}(0)=A_{n}(1)=0\) for \(n\in\mathbb{N}\),

  3. (iii)

    \(A_{n}''(x)=A_{n-1}(x)\).

The polynomial \(A_{n}\) is called Lidstone polynomial.

Ismail and Mansour [2] introduced a q-analogue of Lidstone’s theorem where the two points are 0 and 1. They expanded the function in q-analogues of Lidstone polynomials which are in fact q-Bernoulli polynomials as in the classical case (see Section 2).

It is the object of this paper to give a q-analogue of the results of [3] using the terminology and results given in [2].

This article is organized as follows. In the next section, we state the q-definitions and present some preliminaries of q-calculus which will play an important role in our main results. In Section 3, we define the Green’s functions of certain q-differential systems which are related to q-Lidstone polynomials, and Section 4 gives q-Fourier expansions of these functions and for q-Lidstone polynomials. Some interesting results and relationships are obtained. In Section 5, we are interested in the existence of solutions to the following boundary value problem:

$$ (-1)^{n} D_{q^{-1}}^{2n}y(x)= \phi \bigl(x,y(x), D_{q^{-1}}y(x), D_{q^{-1}}^{2}y(x), \ldots , D_{q^{-1}}^{k}y(x)\bigr), $$
(1.1)

\(n\in\mathbb{N}\) and \(0\leq k\leq2n-1\), subject to the boundary conditions

$$ D_{q^{-1}}^{2j}y(0)= \beta_{j}, D_{q^{-1}}^{2j}y(1)= \gamma_{j}\quad (\beta _{j}, \gamma_{j} \in\mathbb{C}, j=0,1,\ldots,n-1), $$
(1.2)

with some conditions imposed on y.

2 Preliminaries

In this paper, we assume that q is a positive number less than one with

$$[x]= \frac{1-q^{x}}{1-q}. $$

For \(t>0\), the sets \(A_{q,t}\), \(A^{*}_{q,t}\) are defined by

$$A_{q,t}:=\bigl\{ tq^{n}: n\in\mathbb{N}_{0}\bigr\} , \qquad A^{*}_{q,t} := A_{q,t}\cup \{0\}, $$

where \(\mathbb{N}_{0}:=\{0,1,2,\ldots\}\). Notice, if \(t=1\), we simply use \(A_{q}\) and \(A^{*}_{q}\) to denote \(A_{q,1}\) and \(A^{*}_{q,1}\), respectively.

In the following, we state some of the needed q-notations and results (see [4] and [5]).

The q-shifted fractional is defined by

$$(a;q)_{\infty}= \prod_{j=0}^{\infty} \bigl(1-aq^{j}\bigr)\quad \mbox{and}\quad (a;q)_{n} := \frac{(a;q)_{\infty}}{(aq^{n};q)_{\infty}}\quad \mbox{for } n\in\mathbb{Z}, a\in\mathbb{C}. $$

The q-gamma function is defined by

$$\Gamma_{q}(z) = \frac{(q;q)_{\infty}}{ (q^{z};q)_{\infty}} (1 - q)^{1-z} \quad \mbox{for } z\in\mathbb{C}\setminus\{-n: n\in\mathbb{N}_{0}\}. $$

Let f be a function defined on a q-geometric set A, i.e., \(qx\in A\) for all \(x \in A\). The q-difference operator is defined by

$$ D_{q} f(x):= \frac{f(x)-f(qx)}{(1-q)x} \quad\mbox{if } x\in A-\{0\}. $$

The q-integration by parts rule (see [4]) is

$$\int_{0}^{a} f(qt) D_{q}g(t)\, d_{q}t= (fg) (a)-\lim_{n\rightarrow\infty} (fg) \bigl(aq^{n}\bigr)- \int_{0}^{a} D_{q}f(t) g(t) \,d_{q}t. $$

If X is the set \(A_{q,t}\) or \(A^{*}_{q,t}\), then for \(n>1\), \(C_{q}^{n}(X)\) is the space of all continuous functions with continuous q-derivatives up to order \(n-1\) on X. The space \(C^{n}_{q}(X)\) associated with the norm function

$$\|f\|:= \sum_{k=0}^{n-1} \max _{x\in X} \big| D_{q}^{k} f(t)\big|\quad \bigl(f \in C^{n}_{q}(X)\bigr) $$

is a Banach space (see [4]).

Ismail and Mansour [2] defined a q-analogue of the Bernoulli polynomials \(B_{n}(z;q)\), \(z\in\mathbb{C}\) by the generating function

$$\frac{tE_{q}(zt)}{E_{q}(t/2)e_{q}(t/2)-1}=\sum_{n=0}^{\infty}B_{n}(z;q)\frac {t^{n}}{[n]!}, $$

where the functions \(E_{q}(z)\) and \(e_{q}(z)\) have the series representation

$$e_{q}(z)=\sum_{k=0}^{\infty} \frac{z^{k}}{\Gamma_{q}(k+1)}; \quad|z|< 1 \quad \mbox{and} \quad E_{q}(z)=\sum _{k=0}^{\infty} \frac {q^{k(k-1)/2}z^{k}}{\Gamma_{q}(k+1)}; \quad z\in\mathbb{C}. $$

The q-Bernoulli numbers are defined by

$$\beta_{n}:= B_{n}(0;q). $$

Hence, in terms of the generating function,

$$ \frac{t}{E_{q}(t/2)e_{q}(t/2)-1}=\sum_{n=0}^{\infty}\beta_{n}\frac{t^{n}}{[n]!}. $$
(2.1)

Also, they defined two q-analogues of the Euler polynomials through the generating functions

$$\begin{aligned}& \frac{2 E_{q}(xt)}{E_{q}(t/2)e_{q}(t/2)+1}=\sum_{n=0}^{\infty}E_{n}(x;q) \frac{t^{n}}{[n]!}, \end{aligned}$$
(2.2)
$$\begin{aligned}& \frac{2 e_{q}(xt)}{E_{q}(t/2)e_{q}(t/2)+1}=\sum_{n=0}^{\infty}e_{n}(x;q) \frac{t^{n}}{[n]!}. \end{aligned}$$
(2.3)

Notice, \(E_{0}(x;q)=e_{0}(x;q)=1\), and \(\tilde{E}_{n}:= E_{n}(0;q)=e_{n}(0;q)\) for all \(n\in\mathbb{N}_{0}\).

Proposition 2.1

For \(n\in\mathbb{N}\), the q-Bernoulli and q-Euler polynomials satisfy the following q-difference equations:

$$\begin{gathered}D_{q^{-1}}B_{n}(x;q)=[n]B_{n-1}(x;q); \\ D_{q^{-1}}E_{n}(x;q)=[n]E_{n-1}(x;q) \quad\textit{and} \quad D_{q}e_{n}(x;q)=[n]e_{n-1}(x;q). \end{gathered} $$

Proposition 2.2

The q-Euler polynomials \(E_{n}(x;q)\) and \(e_{n}(x;q)\) are given by

$$E_{0}(x;q)= e_{0}(x;q)=1, $$

and for \(n\in\mathbb{N}\),

$$E_{n}(x;q)=\sum_{k=0}^{n}{n \brack k}_{q} q^{\frac{k(k-1)}{2}} \tilde{E}_{n-k} x^{k},\qquad e_{n}(x;q)=\sum_{k=0}^{n}{n \brack k}_{q} \tilde {E}_{n-k} x^{k}. $$

Recall that (see [6]) an entire function f has a p-exponential growth of order k and a finite type (\(p, k\in\mathbb{R}-\{0\}\) with \(p >1\)) if there exists a real number \(K>0\), α such that

$$\big|f(x)\big|< K p^{\frac{k}{2} (\frac{\log|x|}{\log p} )^{2}} |x|^{\alpha}. $$

The following results from [2] will be needed in the sequel.

Theorem 2.3

Let \(0 < q < 1\) and f be a function of \(q^{-1}\)-exponential growth of order less than or equal to 1. Then

$$f(z)= \sum_{n=0}^{\infty} \bigl( D_{q^{-1}}^{2n} f(1)A_{n}(z)- D_{q^{-1}}^{2n} f(0)B_{n}(z) \bigr), $$

where \(A_{n}\) and \(B_{n}\) are polynomials of degree \(2n+1\) defined by

$$\begin{gathered} A_{n}(z) = \frac{2^{2n+1}}{[2n+1]!} \sum _{j=0}^{2n+1} \left [ \textstyle\begin{array}{c} 2n+1 \\ j \end{array}\displaystyle \right ]_{q} (-z;q)_{j} 2^{-j} \beta_{2n+1-j}, \\ B_{n}(z) = \frac{2^{2n+1}}{[2n+1]!}B_{2n+1} (z/2;q). \end{gathered} $$

Furthermore, the polynomials \(A_{n}\) are defined recursively by \(A_{0}(z)= z\) and, for \(n\in\mathbb{N}\), \(A_{n}\) satisfies the second order q-difference equation

$$ D^{2}_{q^{-1}} A_{n}(z)= A_{n-1}(z),\qquad A_{n}(0)= A_{n}(1)=0 \quad(n\in \mathbb{N}). $$
(2.4)

The polynomials \(B_{n}\) are defined recursively by \(B_{0}(z)= 1-z\) and, for \(n\in\mathbb{N}\), \(B_{n}\) satisfies the second order q-difference equation

$$ D^{2}_{q^{-1}} B_{n}(z)= B_{n-1}(z),\qquad B_{n}(0)= B_{n}(1)=0 \quad(n\in \mathbb{N}). $$
(2.5)

Lemma 2.4

Let \(z\in\mathbb{C}\). Then

$$A_{n}(z):= \varepsilon_{q^{-1}}^{1} B_{n}(z), $$

where \(\varepsilon_{q^{-1}}^{y}\) is a q-translation operator defined by

$$\varepsilon_{q^{-1}}^{y} x^{n} =x^{n} \bigl(-y/x;q^{-1}\bigr)_{n}=q^{-\frac{n(n-1)}{2}}y^{n}(-x/y;q)_{n}. $$

3 The Green’s function of a certain q-differential system

In this section, we consider certain boundary value problems which are related to q-Lidstone polynomials, and then we define these polynomials by using Green’s function.

Consider the following q-differential equation:

$$ D_{q^{-1}}^{2}y(x) - f(x)= 0 \quad\bigl(x\in A^{*}_{q}\bigr), $$
(3.1)

subject to the boundary conditions

$$ y(0)= 0, \qquad y(1)= 0. $$
(3.2)

Theorem 3.1

The boundary value problem (3.1)-(3.2) is equivalent to the basic Fredholm q-integral equation

$$ y(x)= \int_{0}^{1} G\bigl(x,q^{2}t\bigr) f \bigl(q^{2}t\bigr)\, d_{q}t, $$
(3.3)

where

$$ G(x,t)= \left \{ \textstyle\begin{array}{l@{\quad}l} -t(1-x), & 0\leq t< x\leq1; \\ -x (q-t), & 0\leq x< t\leq1. \end{array}\displaystyle \right . $$
(3.4)

Proof

Since \(D^{2}_{q^{-1}} y(x)= \frac{1}{q}(D^{2}_{q} y)(\frac{x}{q^{2}})\), Equation (3.1) can be written as

$$ D_{q}^{2}y(x) - qf\bigl(q^{2}x \bigr)= 0 \quad\bigl(x\in A^{*}_{q}\bigr). $$
(3.5)

By taking double q-integral for (3.5), we obtain

$$ y(x)=q \int_{0}^{x} (x-qt) f\bigl(q^{2}t\bigr)\, d_{q}t+c_{1}x+c_{2}, $$
(3.6)

where \(c_{1}\) and \(c_{2}\) are arbitrary constants. Now, using the boundary conditions, we get

$$c_{1}=-q \int_{0}^{1} (1-qt) f\bigl(q^{2}t\bigr)\, d_{q}t \quad\mbox{and}\quad c_{2}=0. $$

Substituting in (3.6), we obtain the required result. □

Now, consider the following equations:

$$ \begin{gathered} G_{1}\bigl(x,q^{2} t\bigr):= G\bigl(x,q^{2} t\bigr), \\ \begin{aligned}G_{n}\bigl(x,q^{2} t\bigr)&= \int_{0}^{1} G\bigl(x,q^{2}y\bigr) G_{n-1}\bigl(q^{2}y, q^{2} t\bigr) \, d_{q}y \quad(n=2,3, \ldots) \\ &= \int_{0}^{1} \cdots \int_{0}^{1} G\bigl(x,q^{2}t_{1} \bigr) G\bigl(q^{2}t_{1},q^{2}t_{2} \bigr)\cdots G\bigl(q^{2}t_{n-1},q^{2}t\bigr) \, d_{q}t_{1}\,d_{q}t_{2}\cdots d_{q}t_{n-1}.\end{aligned} \end{gathered} $$
(3.7)

Corollary 3.2

The q-Lidstone polynomials \(A_{m}\) and \(B_{m}\) are given by

$$ \begin{gathered} A_{0}(x) = x, \\ A_{m}(x) = \int_{0}^{1} G\bigl(x,q^{2}t\bigr) A_{m-1}\bigl(q^{2}t\bigr) \,d_{q}t = q^{2} \int_{0}^{1} t G_{m}\bigl(x,q^{2}t \bigr)\,d_{q}t, \end{gathered} $$
(3.8)

and

$$ \begin{gathered} B_{0}(x) = 1-x, \\ B_{m}(x) = \int_{0}^{1} G\bigl(x,q^{2}t\bigr) B_{m-1}\bigl(q^{2}t\bigr) \,d_{q}t = \int_{0}^{1} G_{m}\bigl(x,q^{2}t \bigr) \bigl(1-q^{2} t\bigr)\,d_{q}t. \end{gathered} $$
(3.9)

Proof

The proof follows immediately from Theorem 3.1, Equation (3.7), Equation (2.4) and Equation (2.5). □

Theorem 3.3

Let \(0 < q < 1\) and \(g \in C^{2n}(A^{*}_{q})\). Then

$$g(x)= \sum_{m=0}^{n-1} \bigl[ D_{q^{-1}}^{2m} g(1)A_{m}(x)- D_{q^{-1}}^{2m} g(0)B_{m}(x) \bigr]+ \int_{0}^{1} G_{n}\bigl(x,q^{2}t \bigr) D_{q^{-1}}^{2n}g\bigl(q^{2}t\bigr)\, d_{q}t, $$

where \(A_{m}\) and \(B_{m}\) are q-Lidstone polynomials of degree \(2m+1\).

Proof

From Theorem 3.1 we can verify that, for \(q\in(0,1)\) and \(g\in C^{2n}( A^{*}_{q} )\), the q-integral equation

$$g(x)= \int_{0}^{1} G_{n}\bigl(x,q^{2}t \bigr) f\bigl(q^{2}t\bigr)\, d_{q}t $$

is the solution of the q-differential system

$$\left \{ \textstyle\begin{array}{l} D_{q^{-1}}^{2n}g(x)- f(x)= 0 \quad(x\in A^{*}_{q}), \\ D_{q^{-1}}^{2k} g(0)= D_{q^{-1}}^{2k} g(1)= 0 \quad(k=0,1,\ldots,n-1). \end{array}\displaystyle \right . $$

Furthermore, the unique solution of the system

$$ \left \{ \textstyle\begin{array}{l} D_{q^{-1}}^{2n}g(x) - f(x)= 0 \quad(x\in A^{*}_{q}), \\ D_{q^{-1}}^{2k} g(0)=a_{k}, \qquad D_{q^{-1}}^{2k} g(1)= b_{k} \quad(k=0,1,\ldots,n-1) \end{array}\displaystyle \right . $$
(3.10)

is

$$ \begin{aligned} g(x)={}& a_{0} (x-1)+b_{0} x + \sum_{k=1}^{n-1} a_{k} \int_{0}^{1} \bigl(q^{2}t-1\bigr) G_{k}\bigl(x,q^{2}t\bigr)\,d_{q}t \\ &+\sum_{k=1}^{n-1} b_{k} \int_{0}^{1} q^{2}t G_{k} \bigl(x,q^{2}t\bigr)\,d_{q}t + \int_{0}^{1} G_{n}\bigl(x,q^{2}t \bigr) f\bigl(q^{2}t\bigr)\, d_{q}t. \end{aligned} $$

Replacing \(a_{k}\), \(b_{k}\) and \(f(x)\) by their values in terms of \(g(x)\) as given by the q-differential system (3.10), we get

$$ \begin{aligned} g(x)={}& g(0) (x-1)+g(1) x +\sum _{k=1}^{n-1} D_{q^{-1}}^{2k} g(0) \int_{0}^{1} \bigl(q^{2}t-1\bigr) G_{k}\bigl(x,q^{2}t\bigr)\,d_{q}t \\ &+\sum_{k=1}^{n-1} D_{q^{-1}}^{2k} g(1) \int_{0}^{1} q^{2}t G_{k} \bigl(x,q^{2}t\bigr)\, d_{q}t + \int_{0}^{1} G_{n}\bigl(x,q^{2}t \bigr) D_{q^{-1}}^{2n} g\bigl(q^{2}t\bigr)\, d_{q}t. \end{aligned} $$

Therefore, according to Equations (3.8) and (3.9), we obtain the required result. □

Remark 3.4

By using Theorem 3.3, and from Equations (2.4) and (2.5), we have

$$\begin{gathered} \begin{aligned}D^{2j}_{q^{-1}}g(x)={}& \sum _{m=j}^{n-1} \bigl[ D_{q^{-1}}^{2m} g(1) D_{q^{-1}}^{2j}A_{m}(x)+ D_{q^{-1}}^{2m} g(0) D_{q^{-1}}^{2j} B_{m}(x) \bigr] \\ &+ \int_{0}^{1} G_{n-j}\bigl(x,q^{2}t \bigr) D_{q^{-1}}^{2n}g(t)\, d_{q}t \\ ={}&\sum_{m=j}^{n-1} \bigl[ D_{q^{-1}}^{2m} g(1) A_{m-j}(x) + D_{q^{-1}}^{2m} g(0) B_{m-j}(x) \bigr] \\ &+ \int_{0}^{1} G_{n-j}\bigl(x,q^{2}t \bigr) D_{q^{-1}}^{2n}g(t)\, d_{q}t \\ ={}&\sum_{m=0}^{n-j-1} \bigl[ D_{q^{-1}}^{2(m+j)} g(1)A_{m}(x) + D_{q^{-1}}^{2(m+j)} g(0)B_{m}(x) \bigr] \\ &+ \int_{0}^{1} G_{n-j}\bigl(x,q^{2}t \bigr) D_{q^{-1}}^{2n}g(t)\, d_{q}t,\end{aligned} \\ \begin{aligned}D^{(2j+1)}_{q^{-1}}g(x)={}& \sum_{m=0}^{n-j-1} \bigl[ D_{q^{-1}}^{2(m+j)} g(1)D_{q^{-1}} A_{m}(x) + D_{q^{-1}}^{2(m+j)} g(0)D_{q^{-1}} B_{m}(x) \bigr] \\ &+ \int_{0}^{1} D_{q^{-1},x}G_{n-j} \bigl(x,q^{2}t\bigr) D_{q^{-1}}^{2n}g(t)\, d_{q}t.\end{aligned} \end{gathered} $$

4 Certain q-Fourier expansions

The purpose of this section is to obtain the q-Fourier series expansions of the following q-integrals:

$$\int_{0}^{1} \bigl(q^{2}t \bigr)^{k} G_{n}\bigl(x,q^{2}t\bigr) \, d_{q}t, \quad k=0,1, n\leq4, $$

and then to compute the series expansions of some of q-Lidstone polynomials which will be used to solve the boundary value problem (1.1)-(1.2).

First, recall that the q-trigonometric functions \(C_{q}(z)\) and \(S_{q}(z)\) are defined for \(z\in\mathbb{C}\) by

$$\begin{gathered} C_{q}(z) = \sum _{n=0}^{\infty}(-1)^{n} \frac {q^{n(n-1/2)}z^{2n}}{(q;q)_{2n}}= \frac{z}{1-q} {_{1}\phi _{1}}\bigl(0;q;q^{2},q^{1/2}z^{2} \bigr), \\ S_{q}(z) = \sum_{n=0}^{\infty}(-1)^{n} \frac {q^{n(n+1/2)}z^{2n+1}}{(q;q)_{2n+1}}= \frac{z}{1-q} {_{1}\phi _{1}} \bigl(0;q^{3};q^{2},q^{3/2}z^{2}\bigr). \end{gathered} $$

The Fourier series expansion for any function defined on the q-linear grid \(\mathcal{A}_{q}\) is the following (see [7, 8]):

$$S_{q}(f):= \frac{a_{0}}{2}+\sum_{k=1}^{\infty} \bigl[a_{k}C_{q}\bigl(q^{1/2}w_{k}z \bigr)+b_{k}S_{q}(qw_{k}z) \bigr], $$

where \(a_{0}=\int_{-1}^{1} f(t)\, d_{q}t\) and, for \(k=1,2,\ldots\) ,

$$\begin{gathered} a_{k} =\frac{1}{\mu_{k}} \int_{-1}^{1} f(t)C_{q} \bigl(q^{1/2}w_{k}t\bigr)\, d_{q}t, \quad b_{k}=\frac{\sqrt{q}}{\mu_{k}} \int_{-1}^{1} f(t)S_{q}(qw_{k}t) \, d_{q}t, \\ \mu_{k} = (1-q)C_{q}\bigl(q^{1/2}w_{k} \bigr)S'_{q}(w_{k}) \end{gathered} $$

on the q-linear grid \(\mathcal{A}_{q}\), where \(\{w_{k}: k\in\mathbb {N}\}\) is the set of positive zeroes of \(S_{q}(z)\).

One can verify that

$$D_{q,z}C_{q}(wz) = - \frac{w}{1-q} S_{q}(wz \sqrt{q}) \quad\mbox{and} \quad D_{q,z}S_{q}(wz) = \frac{w}{1-q} C_{q}(wz\sqrt{q}). $$

Lemma 4.1

Let \(x\in A^{*}_{q} \) and \(n\in\mathbb{N}\). Then

$$ \int_{0}^{1} G\bigl(x,q^{2}y\bigr) S_{q}\bigl(q^{n}w_{k}y\bigr) \, d_{q}y = \frac {(1-q)^{2}}{q^{2n-5/2}w_{k}^{2}} \bigl(xS_{q}\bigl(q^{n-1}w_{k} \bigr)-S_{q}\bigl(q^{n-1}w_{k} x\bigr) \bigr). $$

Proof

By using Equations (3.1) and (3.3), the q-integral

$$y(x)= \int_{0}^{1} G_{1}\bigl(x,q^{2}y \bigr) S_{q}\bigl(q^{n}w_{k}y\bigr) \, d_{q}y $$

is the solution of the q-differential system

$$ \left \{ \textstyle\begin{array}{l} D_{q^{-1}}^{2}y(x) - S_{q}(q^{n-2}w_{k}x)= 0 \quad(x\in A^{*}_{q}), \\ y(0)= 0, \qquad y(1)= 0. \end{array}\displaystyle \right . $$
(4.1)

Therefore,

$$ \begin{gathered} D_{q} y(x)= \frac{-(1-q)}{q^{n-\frac{3}{2}}w_{k}} C_{q}\bigl(q^{n-\frac {1}{2}}w_{k} x \bigr)+c_{1}, \\ y(x) = \frac{-(1-q)^{2}}{q^{2n-\frac{5}{2}}w_{k}^{2}} S_{q}\bigl(q^{n-1}w_{k} x\bigr)+c_{1} x+c_{2}. \end{gathered} $$
(4.2)

From the boundary conditions, we get

$$c_{1}= \frac{(1-q)^{2}}{q^{2n-\frac{5}{2}}w_{k}^{2}} S_{q}\bigl(q^{n-1}w_{k} \bigr) \quad \mbox{and} \quad c_{2}=0. $$

Substituting the values of \(c_{1}\) and \(c_{2}\) into Equation (4.2), we obtain the required result. □

Lemma 4.2

For \(x\in A^{*}_{q} \), the following q-Fourier series expansion holds:

$$ \int_{0}^{1} G\bigl(x,q^{2}t\bigr)\, d_{q}t=-2\sqrt{q}(1-q)^{2} \sum _{k=1}^{\infty} \frac{L_{k}}{w_{k}^{2}} S_{q}(w_{k}x), $$
(4.3)

where

$$L_{k}:= \frac{1-C_{q}(q^{1/2}w_{k})}{ w_{k} C_{q}(q^{1/2}w_{k})S'_{q}(w_{k})}. $$

Proof

By computing the q-Fourier series expansion of the function \(f(x)= 1\) for \(0< x<1\), we get

$$ 1= 2 \sum_{k=1}^{\infty} \frac{1-C_{q}(q^{1/2}w_{k})}{ w_{k} C_{q}(q^{1/2}w_{k})S'_{q}(w_{k})} S_{q}(qw_{k}t), \quad t\in A^{*}_{q}. $$
(4.4)

Multiplying (4.4) by \(G_{1}(x,q^{2}t)\) and integrating with respect to t from zero to unity, we get

$$ \int_{0}^{1} G_{1}\bigl(x,q^{2}t \bigr) \, d_{q}t= 2 \sum_{k=1}^{\infty} L_{k} \int_{0}^{1} G_{1}\bigl(x,q^{2}t \bigr) S_{q}(w_{k}qt) \, d_{q}t, $$
(4.5)

where

$$L_{k}:= \frac{1-C_{q}(q^{1/2}w_{k})}{ w_{k} C_{q}(q^{1/2}w_{k})S'_{q}(w_{k})}, \quad x\in A^{*}_{q}. $$

By using Lemma 4.1, we get

$$ \int_{0}^{1} G_{1}\bigl(x,q^{2}t \bigr) S_{q}(w_{k}qt) \, d_{q}t= \frac{-\sqrt{q}(1-q)^{2}}{ w_{k}^{2}}S_{q}(w_{k}x). $$
(4.6)

Substituting from (4.6) into (4.5), we obtain the required series. □

Lemma 4.3

For \(x\in A^{*}_{q} \), the following q-Fourier series expansion holds:

$$ \int_{0}^{1} G\bigl(x,q^{2}t\bigr) \bigl(q^{2}t\bigr)\, d_{q}t= -2q^{5/2}(1-q)^{2} \sum_{k=1}^{\infty} \frac{\widetilde{L_{k}}}{w_{k}^{2}} S_{q}(w_{k}x), $$

where

$$\widetilde{L_{k}}:= \frac{qw_{k}C_{q}(q^{1/2}w_{k})-(1-q)S_{q}(qw_{k})}{q^{2} w_{k}^{2} C_{q}(q^{1/2}w_{k})S'_{q}(w_{k})}. $$

Proof

Considering the function \(g(t)= t\) for \(0< t<1\) and computing the q-Fourier series of the extension of g as an odd function on \([-1,1]\), we get

$$ t= 2 \sum_{k=1}^{\infty} \widetilde{L_{k}} S_{q}(qw_{k}t) \quad\mbox{for all } 0< t < 1, $$
(4.7)

where

$$ \widetilde{L_{k}}:= \frac{qw_{k}C_{q}(q^{1/2}w_{k})-(1-q)S_{q}(qw_{k})}{q^{2} w_{k}^{2} C_{q}(q^{1/2}w_{k})S'_{q}(w_{k})}. $$
(4.8)

Hence, the proof can be performed by using (4.7) similar to the proof of Lemma 4.2. So, we will omit it. □

Throughout the following results, we define the constants \(L_{k}\) and \(\widetilde{L_{k}}\) as in Lemma 4.3 and Lemma 4.3, respectively.

Note that, by using Equation (3.7), we get

$$ G_{2}\bigl(x,q^{2}t\bigr)= \int_{0}^{1} G\bigl(x,q^{2}y\bigr) G \bigl(q^{2}y,q^{2}t\bigr) \, d_{q}y. $$
(4.9)

Integrating (4.9) with respect to t from 0 to unity and using Lemma 4.2, we obtain

$$ \int_{0}^{1} G_{2}\bigl(x,q^{2}t \bigr)\, d_{q}t= -2\sqrt{q}(1-q)^{2} \sum _{k=1}^{\infty } \frac{L_{k}}{w_{k}^{2}} \int_{0}^{1} G\bigl(x,q^{2}y\bigr) S_{q}\bigl(q^{2} w_{k} y\bigr) \, d_{q}y. $$

Again, using Lemma 4.1, we get

$$\int_{0}^{1} G\bigl(x,q^{2}y\bigr) S_{q}\bigl(q^{2}w_{k}y\bigr) \, d_{q}y = \frac {(1-q)^{2}}{q^{3/2}w_{k}^{2}} \bigl(xS_{q}(qw_{k})-S_{q}(qw_{k} x) \bigr). $$

Hence,

$$ \int_{0}^{1} G_{2}\bigl(x,q^{2}t \bigr) \, d_{q}t = -2\frac{(1-q)^{4}}{q} \sum _{k=1}^{\infty} \frac{L_{k}}{w_{k}^{4}} \bigl(xS_{q}(qw_{k})-S_{q}(qw_{k} x) \bigr). $$

Repeating the process for \(n=3\) and \(n=4\), we obtain the following result.

Theorem 4.4

For \(x\in A^{*}_{q} \) and \(n\leq4\), the following expansion holds:

$$ \begin{aligned}[b] q^{2} \int_{0}^{1} G_{n}\bigl(x,q^{2}t \bigr) \, d_{q}t ={}& \frac{(-1)^{n-1}(1-q)^{2n}}{ q^{n(n-3/2)}} \Biggl[\sum _{k=1}^{\infty} \frac{L_{k}}{w_{k}^{2n}} \bigl(xS_{q} \bigl(w_{k}q^{n-1}\bigr)-S_{q}\bigl(w_{k}q^{n-1} x\bigr) \bigr) \\ &+ 2\sum_{i=1}^{n-2} (-1)^{i} q^{2(n+i-1)} \sum_{k=1}^{\infty} \frac {L_{k}}{w_{k}^{2(n-i)}}S_{q}\bigl(q^{n-i-1}w_{k}\bigr) \\ &\times\sum_{k=1}^{\infty} \frac{\widetilde{L_{k}}}{w_{k}^{2i}} \bigl(xS_{q}\bigl(q^{i-1}w_{k}\bigr)-S_{q} \bigl(q^{i-1}w_{k} x\bigr) \bigr) \Biggr]. \end{aligned} $$
(4.10)

Remark 4.5

In the classical case, Widder [9] concluded a general formula for a Fourier series of the integral of Green’s functions \(G_{n}\) for all \(n\in\mathbb{N}\). Theorem 4.4 gives a formula for the q-Fourier series of \(\int_{0}^{1} G_{n}(x,q^{2}t) \, d_{q}t \) for \(n\leq4\), we could not put it in a closed form for all \(n\in\mathbb{N}\). However, we can verify that

$$ \int_{0}^{1} G_{n}\bigl(x,q^{2}t \bigr) \, d_{q}t = \frac{(-1)^{n-1}(1-q)^{2n}}{ q^{n(n-3/2)}} S_{k,n} \quad(n\in \mathbb{N}), $$

where \(S_{k,n}\) denotes a sum of q-series which converge uniformly on \(A^{*}_{q}\) and depend on the q-trigonometric function \(S_{q}\) and the constants \(L_{k}\) and \(\widetilde{L_{k}}\).

Theorem 4.6

For \(x\in A^{*}_{q}\) and \(n\leq4\), the following expansion holds:

$$ \begin{aligned} \int_{0}^{1} G_{n}\bigl(x,q^{2}t \bigr) t \, d_{q}t ={}& \frac {(-1)^{n-1}(1-q)^{2n}}{q^{n(n-3/2)}} \Biggl[\sum _{k=1}^{\infty} \frac {\widetilde{L_{k}}}{w_{k}^{2n}} \bigl(xS_{q} \bigl(w_{k}q^{n-1}\bigr)-S_{q}\bigl(w_{k}q^{n-1} x\bigr) \bigr) \\ &+ 2\sum_{i=1}^{n-2} (-1)^{i} q^{2(n+i-1)} \sum_{k=1}^{\infty} \frac {\widetilde{L_{k}}}{w_{k}^{2(n-i)}}S_{q}\bigl(q^{n-i-1}w_{k}\bigr) \\ &\times\sum_{k=1}^{\infty} \frac{\widetilde{L_{k}}}{w_{k}^{2i}} \bigl(xS_{q}\bigl(q^{i-1}w_{k}\bigr)-S_{q} \bigl(q^{i-1}w_{k} x\bigr) \bigr) \Biggr]. \end{aligned} $$

Proof

The proof is similar to the proof of Theorem 4.4 and is omitted. □

The following corollary follows immediately from Theorems 4.4 and 4.6.

Corollary 4.7

For \(x\in A^{*}_{q}\) and \(n\leq4\), the following expansion holds:

$$ \begin{aligned} &\int_{0}^{1} G_{n}\bigl(x,q^{2}t \bigr) \bigl(1-q^{2}t\bigr)\, d_{q}t \\ &\quad= \frac {(-1)^{n-1}(1-q)^{2n}}{q^{n(n-3/2)}} \Biggl[\sum_{k=1}^{\infty} \frac {1}{w_{k}^{2n}} \bigl(L_{k}-q^{2}\widetilde{L_{k}}\bigr)\bigl(xS_{q}\bigl(w_{k}q^{n-1} \bigr)-S_{q}\bigl(w_{k}q^{n-1} x\bigr) \bigr) \\ &\qquad{} + 2 \sum_{i=1}^{n-2} (-1)^{i} q^{2(n+i-1)} \sum_{k=1}^{\infty } \frac{S_{q}(q^{n-i-1}w_{k})}{w_{k}^{2(n-i)}} \bigl(L_{k}-q^{2}\widetilde{L_{k}}\bigr)\\ &\qquad{}\times \sum _{k=1}^{\infty} \frac{\widetilde{L_{k}}}{w_{k}^{2i}} \bigl(xS_{q}\bigl(q^{i-1}w_{k}\bigr)-S_{q} \bigl(q^{i-1}w_{k} x\bigr) \bigr) \Biggr]. \end{aligned} $$

Corollary 4.8

For \(x\in A^{*}_{q}\) and \(n\leq4\), the q-Fourier series for the q-Lidstone polynomials \(A_{n}(x)\) and \(B_{n}(x)\) are given by

$$\begin{aligned}& \begin{aligned} A_{n}(x) = {}&\frac{(-1)^{n-1}(1-q)^{2n}}{q^{n(n-3/2)}} \Biggl[\sum _{k=1}^{\infty} \frac{\widetilde{L_{k}}}{w_{k}^{2n}} \bigl(xS_{q}\bigl(w_{k}q^{n-1}\bigr)- S_{q}\bigl(w_{k}q^{n-1} x\bigr) \bigr) \\ &+ 2\sum_{i=1}^{n-2} (-1)^{i} q^{2(n+i-1)} \sum_{k=1}^{\infty} \frac{\widetilde {L_{k}}}{w_{k}^{2(n-i)}}S_{q}\bigl(q^{n-i-1}w_{k}\bigr) \\ &\times\sum_{k=1}^{\infty} \frac{\widetilde{L_{k}}}{w_{k}^{2i}} \bigl(xS_{q}\bigl(q^{i-1}w_{k}\bigr)-S_{q} \bigl(q^{i-1}w_{k} x\bigr) \bigr) \Biggr], \end{aligned} \\& \begin{aligned} B_{n}(x)={} &\frac{(-1)^{n-1}(1-q)^{2n}}{q^{n(n-3/2)}} \Biggl[\sum _{k=1}^{\infty} \frac{1}{w_{k}^{2n}} \bigl(L_{k}-q^{2}\widetilde{L_{k}}\bigr) \bigl(xS_{q}\bigl(w_{k}q^{n-1}\bigr)-S_{q}\bigl(w_{k}q^{n-1} x\bigr) \bigr) \\ & + 2\sum _{i=1}^{n-2} (-1)^{i} q^{2(n+i-1)} \sum_{k=1}^{\infty} \frac {S_{q}(q^{n-i-1}w_{k})}{w_{k}^{2(n-i)}} \bigl(L_{k}-q^{2}\widetilde{L_{k}} \bigr) \\ &\times \sum_{k=1}^{\infty} \frac{\widetilde{L_{k}}}{w_{k}^{2i}} \bigl(xS_{q}\bigl(q^{i-1}w_{k}\bigr)-S_{q} \bigl(q^{i-1}w_{k} x\bigr) \bigr) \Biggr]. \end{aligned} \end{aligned}$$

Proof

It follows immediately from Theorem 4.6, Corollary 4.7, Equations (3.8) and (3.9). □

Proposition 4.9

There exists a constant C such that

$$ 0\leq(-1)^{n} \int_{0}^{1} G_{n}\bigl(x,q^{2}t \bigr)\, d_{q}t \leq\frac {(1-q)^{2n}}{q^{n(n-3/2)}} C. $$

Proof

By using Equations (3.4) and (3.7), we get

$$(-1)^{n} \int_{0}^{1} G_{n}\bigl(x,tq^{-1} \bigr)\, d_{q}t\geq0. $$

Another inequality follows from Theorem 4.4 together with the result that the series in (4.10) converges uniformly at each fixed point \(x\in A^{*}_{q} \). □

Proposition 4.10

There exists a constant C̃ such that

$$ \int_{0}^{1} \big| D_{q^{-1},x} G_{n} \bigl(x,q^{2}t\bigr) \big| \, d_{q}t \leq\frac {(1-q)^{2(n-1)}}{q^{(n-1)(n-5/2)}} \widetilde{C}. $$

Proof

By using (3.7), we have

$$\begin{aligned} \int_{0}^{1} \big| D_{q^{-1},x} G_{n} \bigl(x,q^{2}t\bigr) \big| \, d_{q}t ={}& \int_{0}^{1} \biggl[ D_{q^{-1},x} \int_{0}^{1} G\bigl(x,q^{2}y\bigr) (-1)^{n-1} G_{n-1}\bigl(q^{2}y,q^{2}t\bigr) \, d_{q}y \biggr]\,d_{q}t \\ ={}& \int_{0}^{1} \int_{0}^{x} (-1)^{n-1} \bigl(q^{2}y\bigr) G_{n-1}\bigl(q^{2}y,q^{2}t \bigr) \, d_{q}y \,d_{q}t \\ &- \int_{0}^{1} \int_{x}^{1} (-1)^{n-1} \bigl(q-q^{2}y\bigr) G_{n-1}\bigl(q^{2}y,q^{2}t \bigr) \, d_{q}y \,d_{q}t. \end{aligned} $$

Interchanging the order of the double q-integrations and using Proposition 4.9, we get

$$\begin{aligned} \int_{0}^{1} \big| D_{q^{-1},x} G_{n} \bigl(x,q^{2}t\bigr) \big| \, d_{q}t ={}& \int_{0}^{x} \bigl(q^{2}y\bigr) \biggl[ \int_{0}^{1} \big| G_{n-1}\bigl(q^{2}y,q^{2}t \bigr) \big| \, d_{q}t \biggr] \,d_{q}y \\ &- \int_{x}^{1} \bigl(q- q^{2}y\bigr) \biggl[ \int_{0}^{1} \big| G_{n-1}\bigl(q^{2}y,q^{2}t \bigr) \big| \, d_{q}t \biggr] \,d_{q}y \\ \leq{}&\frac{(1-q)^{2(n-1)}}{q^{(n-1)(n-5/2)}} C \biggl[ \int_{0}^{x} \bigl(q^{2}y\bigr)\, d_{q}y - \int_{x}^{1} \bigl(q- q^{2}y\bigr)\, d_{q}y \biggr] \\ ={}& \frac{(1-q)^{2(n-1)}}{q^{(n-1)(n-5/2)}} C \biggl[ q(1-x)+ \frac {q^{2}}{(q+1)} \biggr] \\ \leq{}&\frac{(1-q)^{2(n-1)}}{q^{(n-1)(n-5/2)}} C \biggl[ q+\frac {q^{2}}{(1+q)} \biggr]. \end{aligned} $$

Hence, if we define the constant C̃ as

$$\widetilde{C}:= \biggl(q+\frac{q^{2}}{(1+q)}\biggr) C, $$

we get the required result. □

We end this section by computing the q-Fourier expansion of the q-Euler polynomials of degree 2. We start by the following lemma.

Lemma 4.11

$$\sum_{k=1}^{\infty} \frac{L_{k}}{w_{k}} = \frac{\sqrt{q}}{2(1-q^{2})} \quad\textit{and} \quad\sum_{k=1}^{\infty} \frac{\widetilde {L_{k}}}{w_{k}}= -\frac{\sqrt{q}}{2[3]!(1-q)}. $$

Proof

By computing the q-Fourier series for the function \(f(x)= |x|\), we obtain

$$f(x)=\frac{1}{1+q}-\frac{2(1-q)}{\sqrt{q}} \sum_{k=1}^{\infty} \frac{ 1-C_{q}(q^{1/2}w_{k})}{ w_{k}^{2} C_{q}(q^{1/2}w_{k})S'_{q}(w_{k})} C_{q}\bigl(q^{1/2}w_{k} x\bigr). $$

In particular, when \(x=0\), this implies

$$0= \frac{1}{1+q}-\frac{2(1-q)}{\sqrt{q}} \sum_{k=1}^{\infty} \frac { 1-C_{q}(q^{1/2}w_{k})}{ w_{k}^{2} C_{q}(q^{1/2}w_{k})S'_{q}(w_{k})}. $$

Therefore,

$$\sum_{k=1}^{\infty} \frac{L_{k}}{w_{k}} = \frac{\sqrt{q}}{2(1-q^{2})}. $$

Similarly, computing the q-Fourier series for the function \(g(x)= |x|^{2}\), we obtain

$$|x|^{2}= \frac{1}{[3]}+\frac{2[2](1-q)}{\sqrt{q}} \sum _{k=1}^{\infty } \frac{\widetilde{L_{k}}}{w_{k}} C_{q} \bigl(q^{1/2}w_{k} x\bigr). $$

At \(x=0\), we have

$$\sum_{k=1}^{\infty} \frac{\widetilde{L_{k}}}{w_{k}}= - \frac{\sqrt {q}}{2[3]!(1-q)}. $$

 □

Theorem 4.12

For \(x\in A^{*}_{q} \), the q-Fourier series for q-Euler polynomials \(e_{2}(x;q)\) is given by

$$ e_{2}(x;q)= \frac{[2]}{q} \Biggl[ -2 \sqrt{q}(1-q)^{2} \sum_{k=1}^{\infty} \frac{L_{k}}{w_{k}^{2}} S_{q}(w_{k} x)+\biggl(\frac{q}{1+q}-\frac {q}{2} \biggr)x \Biggr]. $$

Proof

By using Proposition 2.1, we have

$$1= e_{0}(x;q)=D_{q}e_{1}(x;q). $$

Therefore, for \(x\in A^{*}_{q}\), the q-Fourier expansion of the function \(D_{q}e_{1}(x;q)\) is

$$ D_{q}e_{1}(x;q)= 2 \sum _{k=1}^{\infty} \frac{1-C_{q}(q^{1/2}w_{k})}{ w_{k} C_{q}(q^{1/2}w_{k})S'_{q}(w_{k})} S_{q}(qw_{k}x). $$
(4.11)

Integrating (4.11) from 0 to x, we obtain

$$ e_{1}(x;q)= \frac{-2(1-q)}{\sqrt{q}} \sum _{k=1}^{\infty} \frac{L_{k} }{w_{k}} C_{q} \bigl(q^{1/2}w_{k}x\bigr)+C_{1}, $$
(4.12)

where C is a constant of integration. This constant is obtained by putting \(x=0\) in Equation (4.12) and then using Lemma 4.11 and the result \(e_{1}(0;q)= \widetilde {E}_{1}(0)=-\frac{1}{2}\). We get \(C_{1}= -\frac{1}{2}+\frac{1}{1+q}\). Hence,

$$ e_{1}(x;q)= \frac{-2 (1-q)}{\sqrt{q}} \sum _{k=1}^{\infty} \frac {L_{k}}{w_{k}} C_{q} \bigl(q^{1/2}w_{k} x\bigr)+ \frac{1}{1+q}- \frac{1}{2}. $$
(4.13)

Again, using Proposition 2.1 with \(n=2\), we get

$$ e_{2}(x;q)= [2] \int e_{1}(x,q)\, d_{q}x +C_{2}. $$
(4.14)

Substituting Equation (4.13) into Equation (4.14) gives us

$$ \begin{aligned} e_{2}(x;q)={}& [2] \Biggl[\frac{-2 (1-q)}{\sqrt{q}} \sum_{k=1}^{\infty} \frac{L_{k}}{w_{k}} \int C_{q}\bigl(q^{1/2}w_{k} x\bigr)\, d_{q}x + \int\biggl(\frac {1}{1+q}-\frac{1}{2}\biggr)\, d_{q}x \Biggr]+C_{2}. \end{aligned} $$

This implies

$$ e_{2}(x;q)= [2] \Biggl[\frac{-2 (1-q)^{2}}{\sqrt{q}} \sum _{k=1}^{\infty} \frac{L_{k}}{w_{k}^{2}} S_{q}(w_{k} x)+ \biggl(\frac{q}{1+q}-\frac{q}{2}\biggr)x \Biggr]+C_{2}. $$

In the last equation putting \(x=0\), we get \(C_{2}=0\), and hence the theorem. □

Corollary 4.13

For \(x\in A^{*}_{q}\), the following holds:

$$e_{2}(x;q)= \frac{[2]}{q} \biggl[ \int_{0}^{1} G\bigl(x,q^{2}t\bigr)\, d_{q}t + \biggl(\frac {q}{1+q}-\frac{q}{2}\biggr)x \biggr]. $$

Proof

The proof follows immediately from Lemma 4.2 and Theorem 4.12. □

Remark 4.14

From Equation (3.9), we have

$$B_{n}(x) = \int_{0}^{1} G_{n}\bigl(x,q^{2}t \bigr) \,d_{q}t - q^{2} \int_{0}^{1} t G_{n}\bigl(x,q^{2}t \bigr) \,d_{q}t. $$

Thus, by using Corollary 4.13 and Equation (3.8), we obtain the following relation:

$$ B_{1}(x)+ q^{2}A_{1}(x)=q \biggl[ \frac{e_{2}(x;q)}{[2]}+ \biggl(\frac{1}{2}-\frac {1}{1+q}\biggr)x \biggr]. $$
(4.15)

If \(q\rightarrow1\), Equation (4.15) coincides with the result which is given by Agarwal and Wong [3] in the classical case.

5 An application: q-boundary value problems

The q-difference equations are important in q-calculus. This subject initiated in the first quarter of the twentieth century [10–13], and it has been developed over the years. Recently, many authors have studied the existence and uniqueness of solutions for some problems of q-difference equations, for instance, see [7, 14–20].

The goal of this section is to solve the boundary value problem (1.1)-(1.2) by using the q-Lidstone expansion theorem. The results here attained are the q-analogue of those given by Agarwal and Wong [3], where they studied the existence of solutions for

$$\left \{ \textstyle\begin{array}{l} (-1)^{n} x^{(2m)} (t)=f(t,x(t),x'(t),\ldots,x^{(k)}(t)), \\ x^{(2i)}(0)=a_{i}, \\ x^{(2i)}(1)=b_{i}, \end{array}\displaystyle \right . $$

where \(0\leq k\leq2m-1\) and \(i=0,1,\ldots,m-1\) with some conditions imposed on f and x.

For our purpose, let us define two constants C and C̃ as in Proposition 4.9 and Proposition 4.10, respectively, and we introduce the following assumptions:

\(H_{1}\): \(K_{j}\), \(0\leq j\leq k\) are given real numbers, and define the nonzero constant M to be the maximum of \(|\phi(x,y_{0}, y_{1}, y_{2}, \ldots , y_{k})|\) on the compact set \(A^{*}_{q} \times E\), where

$$\begin{gathered} E=\bigl\{ (y_{0}, y_{1}, y_{2}, \ldots , y_{k}), |y_{j} |\leq2K_{j}, 0\leq j\leq k\bigr\} . \\ H_{2} : \frac{(1-q)^{2(n-j)}}{q^{(n-j)(n-j-3/2)}} M C \leq K_{2j}, \quad j=0,1,2,\ldots,\frac{k}{2}; \\ H_{3} : \frac{(1-q)^{2(n-j-1)}}{q^{(n-j-1)(n-j-5/2)}} M \widetilde {C}\leq K_{2j+1}, \quad j=0,1,2,\ldots,\frac{k-1}{2}; \\ H_{4} : \max\bigl\{ |\gamma_{j}|, |\beta_{j}| \bigr\} + \sum_{i=1}^{n-j-1} \max\bigl\{ |\gamma_{i+j}|, |\beta_{i+j}| \bigr\} \frac {(1-q)^{2i}}{q^{i(i-3/2)}} C \leq K_{2j}; \\ H_{5} : |\gamma_{j} + \beta_{j} |+ \widetilde{C} \sum_{i=1}^{n-j-1} \max\bigl\{ |\gamma_{i+j}|, |\beta_{i+j}| \bigr\} \frac {(1-q)^{2(i-1)}}{q^{(i-1)(i-5/2)}}\leq K_{2j+1}. \end{gathered} $$

The proof of the existence results for boundary value problem (1.1)-(1.2) depends on q-Lidstone polynomials and the Arzela-Ascoli theorem [21].

Theorem 5.1

Let \(q\in(0,1)\) and \(y\in C_{q^{-1}}^{n} ( A^{*}_{q})\) be a real or complex-valued function. Assume that assumptions \(H_{1}\), \(H_{2}\), \(H_{3}\) and \(H_{4}\) hold. Then the boundary value problem (1.1)-(1.2) has a solution in E.

Proof

By using Theorem 3.3, we conclude that the boundary value problem (1.1)-(1.2) is equivalent to the following Fredholm q-integral equation:

$$ y(x)= \sum_{i=0}^{n-1} \bigl[ \gamma_{i} A_{i}(x)+ \beta_{i} B_{i}(x) \bigr]+ \int_{0}^{1} G_{n}\bigl(x,q^{2}t \bigr) \phi\bigl(t,y(t), \ldots , D_{q^{-1}}^{k}y(t)\bigr)\, d_{q}t. $$
(5.1)

Hence, this problem can be interpreted as a fixed point for the mapping \(T: C_{q^{-1}}^{k} (A^{*}_{q})\rightarrow C_{q^{-1}}^{2n} (A^{*}_{q})\) which is defined by

$$ (Ty) (x)= \sum_{i=0}^{n-1} \bigl[\gamma_{i} A_{i}(x)+\beta_{i} B_{i}(x) \bigr]+ \int_{0}^{1} \big|G_{n}\bigl(x,q^{2}t \bigr)\big|\phi\bigl(t,y(t),\ldots,D_{q^{-1}}^{k}y(t)\bigr)\, d_{q}t. $$
(5.2)

We define the set

$$ J\bigl(A^{*}_{q}\bigr):= \Bigl\{ y(x)\in C_{q^{-1}}^{k} \bigl(A^{*}_{q}\bigr): \big\| D_{q^{-1}}^{j}y\big\| = \max _{0\leq x\leq1}\big| D_{q^{-1}}^{j}y(x)\big| \leq2K_{j}, \, 0\leq j\leq k\Bigr\} . $$

Notice that \(J (A^{*}_{q})\) is a closed subset of the space \(C_{q^{-1}}^{k} (A^{*}_{q})\). We prove that T maps \(J(A^{*}_{q})\) into itself.

Let \(y(x)\in J(A^{*}_{q})\). Then, from Equation (5.2), Remark 3.4, Proposition 4.9 and hypotheses \(H_{1}\), \(H_{2}\) and \(H_{4}\), we get

$$ \begin{aligned}[b] \big|D^{(2j)}_{q^{-1}}(Ty) (x) \big| \leq{}&\sum_{i=0}^{n-j-1} \big| \gamma_{i+j} A_{i}(x)+ \beta_{i+j} B_{i}(x) \big|+ M \int_{0}^{1} \big|G_{n-j}\bigl(x,q^{2}t \bigr)\big| \, d_{q}t \\ \leq{}&|\gamma_{j} x|+ \big|\beta_{j}(1-x)\big|+ \sum _{i=1}^{n-j-1} \bigg| \gamma _{i+j} \int_{0}^{1} \bigl(q^{2}t\bigr) G_{i}\bigl(x,q^{2}t\bigr) \, d_{q}t + \beta_{i+j} \\ &\times \int_{0}^{1} \bigl(1-q^{2}t\bigr) G_{i}\bigl(x,q^{2}t\bigr) \,d_{q}t \bigg| + M \int _{0}^{1} \big|G_{n-j}\bigl(x,q^{2}t \bigr)\big| \,d_{q}t \\ \leq{}&\sup_{x\in A^{*}_{q}} \bigl[|\gamma_{j} x|+ \big| \beta_{j}(1-x)\big| \bigr] + \sum_{i=1}^{n-j-1} \max\bigl\{ |\gamma_{i+j}|, |\beta_{i+j}|\bigr\} \\ &\times \int_{0}^{1} \big|G_{i}\bigl(x,q^{2}t \bigr)\big| \, d_{q}t + M \int_{0}^{1} \big|G_{n-j}\bigl(x,q^{2}t \bigr)\big| \, d_{q}t \\ \leq{}&\max\bigl\{ |\gamma_{j}|, |\beta_{j}| \bigr\} + \sum _{i=1}^{n-j-1} \max \bigl\{ |\gamma_{i+j}|, | \beta_{i+j}| \bigr\} \frac{(1-q)^{2i}}{q^{i(i-3/2)}} C \\ &+ \frac{(1-q)^{2(n-j)}}{q^{(n-j)(n-j-3/2)}} M C \leq2K_{2j}, \quad j=0,1,2,\ldots, \frac{k}{2}. \end{aligned} $$
(5.3)

Similarly, from Equation (5.2), Remark 3.4, Proposition 4.10 and hypotheses \(H_{3}\) and \(H_{5}\), we get

$$ \begin{aligned}[b] \big|D^{(2j+1)}_{q^{-1}}(Ty) (x) \big| \leq{}& |\gamma_{j} + \beta_{j} |+ \widetilde{C}\sum _{i=1}^{n-j-1} \max\bigl\{ |\gamma_{i+j}|, | \beta_{i+j}| \bigr\} \frac{(1-q)^{2(i-1)}}{q^{(i-1)(i-5/2)}} \\ &+ \frac{(1-q)^{2(n-j-1)}}{q^{(n-j-1)(n-j-5/2)}} M \widetilde{C} \\ \leq{}& K_{2j+1}+ K_{2j+1}= 2K_{2j+1}, \quad j=0,1,2,\ldots, \frac{k-1}{2}. \end{aligned} $$
(5.4)

This completes the proof of \(T(J (A^{*}_{q}))\subseteq J (A^{*}_{q})\). Furthermore, from the inequalities (5.3) and (5.4) we conclude that the set

$$\bigl\{ D_{q^{-1}}^{j} (T)y(x): y(x)\in J \bigl(A^{*}_{q} \bigr), 0\leq j\leq k\bigr\} $$

is uniformly bounded and equicontinuous on \(J (A^{*}_{q})\). Therefore, from the Arzela-Ascoli theorem \(\overline{T(J (A^{*}_{q}))}\) is compact. It means that we can find a fixed point of T in E which satisfies the boundary value problem (1.1)-(1.2). □

Corollary 5.2

Assume that the function \(\phi(x,y_{0}, y_{1}, \ldots , y_{k})\) satisfies the following condition on \(A^{*}_{q} \times\mathbb{R}^{k+1}\):

$$ \big|\phi(x,y_{0}, y_{1}, \ldots , y_{k})\big|\leq L+ \sum_{j=0}^{k} L_{j} |y_{j}|^{\alpha_{j}}, $$
(5.5)

where L, \(L_{j}\) are nonnegative constants, and \(0\leq\alpha_{j}<1\). Then the boundary value problem (1.1)-(1.2) has a solution.

Proof

By using (5.5), for \(y(x)\in J (A^{*}_{q})\), we get

$$\big|\phi\bigl(x,y(x), D_{q^{-1}}y(x), D_{q^{-1}}^{2}y(x), \ldots , D_{q^{-1}}^{k}y(x)\bigr)\big|\leq N, $$

where \(N:= L+ \sum_{j=0}^{k} L_{j} (2K_{j})^{\alpha_{j}}\). Hence, the result follows by observing that the hypotheses of Theorem 5.1 are satisfied and replacing M by N such that \(K_{j}\) (\(0\leq j\leq k\)) are sufficiently large. □

6 Conclusion

The goal of this paper is to study some properties of q-Lidstone polynomials by using Green’s function of certain q-differential systems and then to solve the following boundary value problem:

$$\begin{gathered} (-1)^{n} D_{q^{-1}}^{2n} y(x)= \phi\bigl(x,y(x), D_{q^{-1}}y(x), D_{q^{-1}}^{2}y(x), \ldots , D_{q^{-1}}^{k}y(x)\bigr), \\D_{q^{-1}}^{2j}y(0)= \beta_{j},\qquad D_{q^{-1}}^{2j}y(1)= \gamma_{j}\quad (\beta _{j},\gamma_{j} \in\mathbb{C}, j=0,1,\ldots,n-1),\end{gathered} $$

where \(n\in\mathbb{N}\) and \(0\leq k\leq2n-1\).