The goal of this section is to solve the following triple q-integral equations:
$$\begin{aligned}& \int_{0}^{\infty}\psi(u) J_{\nu}\bigl(u \rho; q^{2}\bigr) \,d_{q} u = f_{1} (\rho),\quad \rho\in A_{q,a}, \end{aligned}$$
(3.1)
$$\begin{aligned}& \int_{0}^{\infty}u^{-2\alpha} \psi(u) \bigl[1+w(u) \bigr] J_{\nu}\bigl(u\rho; q^{2}\bigr) \,d_{q} u = f_{2} (\rho), \quad \rho\in A_{q,b} \cap B_{q,a}, \end{aligned}$$
(3.2)
$$\begin{aligned}& \int_{0}^{\infty}\psi(u) J_{\nu}\bigl(u \rho; q^{2}\bigr) \,d_{q} u = f_{3} (\rho),\quad \rho\in B_{q,b}, \end{aligned}$$
(3.3)
where \(0< a< b<\infty\), and α, ν are complex numbers satisfying
$$\Re(\nu) > -1\quad \mbox{and} \quad 0< \Re(\alpha)< 1. $$
ψ is an unknown function to be determined, \(f_{i} \) (\(i=1,2,3\)) are known functions, and w is a non-negative bounded function defined on \(\mathbb{R}_{q,+}\).
Clearly from (2.5), a sufficient condition for the convergence of the q-integrals on the left-hand side of (3.1)-(3.2) is that
$$ \psi\in L_{q,\nu}(\mathbb{R}_{q,+})\cap L_{q,\nu-2\alpha}(\mathbb{R}_{q,+}). $$
(3.4)
For getting the solution of the triple q-integral equations (3.1)-(3.3), we define a function C by
$$C(u):= u^{-2\alpha}\psi(u) \bigl[1+w(u) \bigr], \quad u\in \mathbb{R}_{q,+}. $$
This implies
$$\psi(u)= u^{2\alpha} C(u)- u^{2\alpha} C(u) \biggl[\frac {w(u)}{1+w(u)} \biggr], $$
and the triple q-integral equations (3.1)-(3.3) can be represented as
$$\begin{aligned}& \int_{0}^{\infty}u^{2\alpha} C(u) J_{\nu}\bigl(u\rho; q^{2}\bigr) \,d_{q} u- \int _{0}^{\infty}u^{2\alpha} C(u) \biggl[ \frac{w(u)}{1+w(u)} \biggr] J_{\nu}\bigl(u\rho; q^{2}\bigr) \,d_{q} u \\& \quad = f_{1} (\rho), \quad \rho\in A_{q,a}, \end{aligned}$$
(3.5)
$$\begin{aligned}& \int_{0}^{\infty}C(u) J_{\nu}\bigl(u\rho; q^{2}\bigr) \,d_{q} u= f_{2} (\rho), \quad \rho\in A_{q,b}\cap B_{q,a}, \end{aligned}$$
(3.6)
$$\begin{aligned}& \int_{0}^{\infty}u^{2\alpha} C(u) J_{\nu}\bigl(u\rho; q^{2}\bigr) \,d_{q} u- \int _{0}^{\infty}\frac{w(u)}{1+w(u)}u^{2\alpha}C(u) J_{\nu}\bigl(u\rho; q^{2}\bigr) \,d_{q} u \\& \quad = f_{3} (\rho), \quad \rho\in B_{q,b}. \end{aligned}$$
(3.7)
Since equation (3.6) is linear in C, we may assume that \(C:=C_{1}+C_{2}\) and
$$f_{2}=g_{1}+g_{2},\quad \mbox{on } A_{q,b}\cap B_{q,a}, $$
where \(g_{1}\) defined on \(A_{q,b}\) and \(g_{2}\) defined on \(B_{q,a}\). Therefore,
$$\begin{aligned}& \int_{0}^{\infty}C_{1}(u) J_{\nu}\bigl(u\rho; q^{2}\bigr) \,d_{q} u= g_{1} (\rho), \quad \rho\in A_{q,b}, \\& \int_{0}^{\infty}C_{2}(u) J_{\nu}\bigl(u\rho; q^{2}\bigr) \,d_{q} u= g_{2} (\rho), \quad \rho\in B_{q,a}. \end{aligned}$$
So, the triple q-integral equations (3.5)-(3.7) can be rewritten in the following form:
$$\begin{aligned}& \int_{0}^{\infty}u^{2\alpha} \bigl[C_{1}(u)+C_{2}(u) \bigr] J_{\nu}\bigl(u\rho; q^{2}\bigr) \,d_{q} u \\& \qquad {}- \int_{0}^{\infty}u^{2\alpha} \bigl[C_{1}(u)+C_{2}(u) \bigr]\frac{w(u)}{1+w(u)} J_{\nu}\bigl(u\rho; q^{2}\bigr) \,d_{q} u \\& \quad = f_{1} (\rho),\quad \rho\in A_{q,a}, \end{aligned}$$
(3.8)
$$\begin{aligned}& \int_{0}^{\infty}C_{1}(u) J_{\nu}\bigl(u\rho; q^{2}\bigr) \,d_{q} u= g_{1} (\rho), \quad \rho\in A_{q,b}, \end{aligned}$$
(3.9)
$$\begin{aligned}& \int_{0}^{\infty}C_{2}(u) J_{\nu}\bigl(u\rho; q^{2}\bigr) \,d_{q} u= g_{2} (\rho), \quad \rho\in B_{q,a}, \end{aligned}$$
(3.10)
$$\begin{aligned}& \int_{0}^{\infty}u^{2\alpha} \bigl[C_{1}(u)+C_{2}(u) \bigr] J_{\nu}\bigl(u\rho; q^{2}\bigr) \,d_{q} u \\& \qquad {}- \int_{0}^{\infty}\frac{w(u)}{1+w(u)}u^{2\alpha} \bigl[C_{1}(u)+C_{2}(u) \bigr] J_{\nu}\bigl(u\rho; q^{2}\bigr) \,d_{q} u \\& \quad = f_{3} (\rho), \quad \rho\in B_{q,b}. \end{aligned}$$
(3.11)
Proposition 3.1
Let
\(\psi_{1}\), \(\psi_{2}\)
be the functions defined by
$$\begin{aligned}& \psi_{1}(x):= \int_{0}^{\infty}u^{\alpha}C_{1}(u) J_{\nu-\alpha}\bigl(ux; q^{2}\bigr) \,d_{q}u, \quad x\in B_{q,b}, \end{aligned}$$
(3.12)
$$\begin{aligned}& \psi_{2}(x):= \int_{0}^{\infty}u^{\alpha}C_{2}(u) J_{\nu+\alpha}\bigl(ux; q^{2}\bigr) \,d_{q}u, \quad x \in A_{q,a}, \end{aligned}$$
(3.13)
provided that
\(0<\Re(\alpha)<1\), \(\Re(\nu)>-1\), \(\Re(\nu+\alpha )>0\), and
\(C_{1}\in L_{q,\nu}(\mathbb{R}_{q,+})\), \(C_{2}\in L_{q,-t}(\mathbb {R}_{q,+})\)
where
$$\Re(\nu)+2>\Re(t)>-\Re(\nu)+2\Re(1-\alpha). $$
Then, for
\(u\in\mathbb{R}_{q,+}\), we have
$$\begin{aligned}& C_{1}(u)= u^{1-\alpha} \biggl[ \int_{0}^{b} x \Phi_{1}(x)J_{\nu-\alpha} \bigl(ux; q^{2}\bigr) \,d_{q}x + \int_{b}^{\infty}x \psi_{1}(x)J_{\nu-\alpha} \bigl(ux; q^{2}\bigr) \,d_{q}x \biggr], \end{aligned}$$
(3.14)
$$\begin{aligned}& C_{2}(u)= u^{1-\alpha} \biggl[ \int_{0}^{a} x \psi_{2}(x)J_{\nu+\alpha} \bigl(ux; q^{2}\bigr) \,d_{q}x+ \int_{a}^{\infty}x \Phi_{2}(x)J_{\nu+\alpha} \bigl(ux; q^{2}\bigr) \,d_{q}x \biggr] , \end{aligned}$$
(3.15)
where
$$\begin{aligned}& \Phi_{1}(x)= \frac{(1-q^{2})^{\alpha }x^{\alpha-\nu-1}}{\Gamma_{q^{2}}(1-\alpha)} D_{q,x} \biggl[x^{-2\alpha} \int_{0}^{x} g_{1} (\rho) \rho^{\nu+1} \bigl({q^{2}\rho^{2}}/{x^{2}}; q^{2}\bigr)_{-\alpha} \,d_{q}\rho \biggr] \\& \hphantom{\Phi_{1}(x)}=\bigl(1-q^{2}\bigr)^{\alpha}x^{\alpha-\nu-1}D_{q^{2},x}^{\alpha} \bigl((\cdot )^{\nu/2}g_{1}(\sqrt{\cdot}) \bigr) (x), \quad x\in A_{q,b}, \end{aligned}$$
(3.16)
$$\begin{aligned}& \Phi_{2}(x)= -\frac{(1-q^{2})^{\alpha}q^{2\alpha+\nu -2}x^{\alpha+\nu-1}}{\Gamma_{q^{2}}(1-\alpha)} D_{q,x} \int_{x}^{\infty}g_{2}(\rho) \rho^{1-2\alpha-\nu} \bigl({x^{2}}/{\rho^{2}}; q^{2} \bigr)_{-\alpha} \,d_{q} \rho \\& \hphantom{\Phi_{2}(x)}=-q^{\frac{\alpha(1-\alpha)}{2}}\bigl(1-q^{2}\bigr)^{\alpha}x^{\alpha+\nu -1}D_{q^{2},x} \mathcal{K}_{q^{2}}^{(1-\alpha)} \bigl[(\cdot)^{-\nu /2}g_{2}( \sqrt{\cdot}) \bigr]\biggl(\frac{x}{q^{2}}\biggr),\quad x\in B_{q,a}. \end{aligned}$$
(3.17)
Proof
We start with proving (3.16). Let \(x\in A_{q,b}\). Multiplying both sides of (3.9) by \(x^{-2\alpha} \rho^{\nu+1} ({q^{2}\rho ^{2}}/{x^{2}}; q^{2})_{-\alpha}\) and then integrating with respect to ρ from 0 to x, we get
$$\begin{aligned}& \int_{0}^{x} x^{-2\alpha} \rho^{\nu+1} \bigl({q^{2}\rho^{2}}/{x^{2}}; q^{2} \bigr)_{-\alpha} \int_{0}^{\infty}C_{1}(u) J_{\nu}\bigl(u\rho; q^{2}\bigr) \,d_{q} u \,d_{q} \rho \\& \quad = \int_{0}^{x} g_{1} (\rho) x^{-2\alpha} \rho^{\nu+1} \bigl({q^{2}\rho ^{2}}/{x^{2}}; q^{2}\bigr)_{-\alpha} \,d_{q}\rho. \end{aligned}$$
(3.18)
Notice that the double q-integral on the left-hand side of (3.18) is absolutely convergent for \(0<\Re(\alpha)<1\) and for \(\Re (\nu)>-1\) provided that \(C_{1}\in L_{q,\nu}(\mathbb{R}_{q,+})\). So, we can interchange the order of the q-integrations to obtain
$$\begin{aligned}& \int_{0}^{\infty}C_{1}(u) x^{-2\alpha} \int_{0}^{x} \rho^{\nu+1} \biggl( \frac {q^{2}\rho^{2}}{x^{2}}; q^{2}\biggr)_{-\alpha} J_{\nu}\bigl(u \rho; q^{2}\bigr) \,d_{q} \rho \,d_{q} u \\& \quad = \int_{0}^{x} g_{1} (\rho) x^{-2\alpha} \rho^{\nu+1} \biggl(\frac{q^{2}\rho ^{2}}{x^{2}}; q^{2} \biggr)_{-\alpha} \,d_{q}\rho. \end{aligned}$$
(3.19)
By calculating the q-derivative of the two sides of (3.19) with respect to x and using (2.8), we get
$$ \int_{0}^{\infty}u^{\alpha}C_{1}(u) J_{\nu-\alpha}\bigl(ux; q^{2}\bigr) \,d_{q}u = \Phi _{1}(x),\quad x\in A_{q,b}, $$
(3.20)
where
$$\Phi_{1}(x)= \frac{(1-q^{2})^{\alpha}x^{\alpha-\nu-1}}{\Gamma _{q^{2}}(1-\alpha)} D_{q,x} \biggl[x^{-2\alpha} \int_{0}^{x} g_{1} (\rho) x^{-2\alpha} \rho^{\nu+1} \biggl(\frac{q^{2}\rho^{2}}{x^{2}}; q^{2} \biggr)_{-\alpha} \,d_{q}\rho \biggr]. $$
To prove (3.17), let \(x\in B_{q,a}\). Multiplying both sides of (3.10) by \(\rho^{-2\alpha-\nu+1}({x^{2}}/{\rho^{2}}; q^{2})_{-\alpha}\) and q-integrating with respect to ρ from x to ∞, we get
$$\begin{aligned} \begin{aligned}[b] &\int_{x}^{\infty}\rho^{1-2\alpha-\nu} \bigl({x^{2}}/{\rho^{2}} ; q^{2} \bigr)_{-\alpha } \int_{0}^{\infty}C_{2}(u) J_{\nu}\bigl(u\rho; q^{2}\bigr) \,d_{q} u \,d_{q} \rho \\ &\quad = \int _{x}^{\infty}g_{2} (\rho) \rho^{-2\alpha-\nu+1} \bigl({x^{2}}/{\rho^{2}}; q^{2} \bigr)_{-\alpha} \,d_{q}\rho. \end{aligned} \end{aligned}$$
(3.21)
From (2.5), we can prove that \(u^{t} J_{\nu}(u;q^{2})\) is bounded on \(\mathbb{R}_{q,+}\) provided that \(\Re(t+\nu)>-1\). So, if we take t such that \(\Re(\nu)+2>\Re( t)> -\Re(\nu)+2\Re(1-\alpha)\), we can prove that the double q-integral
$$\int_{x}^{\infty}\rho^{1-2\alpha-\nu} \bigl({x^{2}}/{\rho^{2}}; q^{2}\bigr)_{-\alpha } \int_{0}^{\infty}C_{2}(u) J_{\nu}\bigl(u\rho; q^{2}\bigr) \,d_{q} u \,d_{q} \rho $$
is absolutely convergent and then we can interchange the order of the q-integration to obtain
$$\begin{aligned}& \int_{0}^{\infty}C_{2}(u) \int_{x}^{\infty}\rho^{1-2\alpha-\nu} \bigl({x^{2}}/{\rho^{2}}; q^{2}\bigr)_{-\alpha} J_{\nu}\bigl(u\rho; q^{2}\bigr) \,d_{q} \rho \,d_{q} u \\& \quad = \int_{x}^{\infty}g_{2} (\rho) \rho^{-2\alpha-\nu+1} \bigl({x^{2}}/{\rho ^{2}}; q^{2} \bigr)_{-\alpha} \,d_{q}\rho. \end{aligned}$$
(3.22)
Calculating the q-derivative of the two sides of (3.22) with respect to x and using (2.10) yields
$$ \int_{0}^{\infty}u^{\alpha}C_{2}(u) J_{\nu+\alpha}\bigl(ux; q^{2}\bigr) \,d_{q}u =\Phi _{2}(x), \quad x\in B_{q,a}, $$
(3.23)
where
$$\Phi_{2}(x)= -\frac{(1-q^{2})^{\alpha}q^{2\alpha+\nu-2}x^{\alpha+\nu -1}}{\Gamma_{q^{2}}(1-\alpha)} D_{q,x} \int_{x}^{\infty}g_{2}(\rho) \rho ^{1-2\alpha-\nu} \bigl({x^{2}}/{\rho^{2}}; q^{2} \bigr)_{-\alpha} \,d_{q} \rho. $$
By the above argument, if we assume that \(\psi_{1}\) and \(\psi_{2}\) are given by (3.12) and (3.13), then
$$ \int_{0}^{\infty}u^{\alpha}C_{1}(u)J_{\nu -\alpha} \bigl(ux;q^{2}\bigr) \,d_{q}x=\left \{ \textstyle\begin{array}{l@{\quad}l}\phi_{1}(x),&x\in A_{q,b}, \\ \psi_{1}(x),&x\in B_{q,b} \end{array}\displaystyle \right . $$
(3.24)
and
$$ \int_{0}^{\infty}u^{\alpha }C_{2}(u)J_{\nu+\alpha} \bigl(ux;q^{2}\bigr) \,d_{q}x=\left \{ \textstyle\begin{array}{l@{\quad}l} \phi_{2}(x),&x\in B_{q,a}, \\ \psi_{2}(x),&x\in A_{q,a}. \end{array}\displaystyle \right . $$
(3.25)
Hence, (3.14) and (3.15) follow by applying the inverse pair of q-Hankel transforms (2.1) on (3.24) and (3.25). This completes the proof. □
Remark 3.2
From the definitions of \(\psi_{i}\) and \(\phi_{i}\), \(i=1,2\), in Proposition 3.1, one can verify that \(x^{-\nu-\alpha}\phi _{2}\) is a bounded function in \(B_{q,a}\) and \(x^{-\nu-\alpha}\psi_{2}\) is bounded in \(A_{q,a}\). Also, \(x^{-\nu+\alpha}\phi_{1}\) is bounded in \(A_{q,b}\) and \(x^{-\nu+\alpha}\psi_{1}\) is bounded in \(B_{q,b}\).
Proposition 3.3
For
\(\rho\in B_{q,b}\), \(\psi_{1}(\rho)\)
satisfies the Fredholm
q-integral equation of the form
$$ \psi_{1}(\rho)=\tilde{F}_{1}(\rho)+ \frac{q^{-2\alpha^{2}-\alpha+\nu }}{(1-q)^{2}} \int_{b}^{\infty} x \psi_{1}(x) K_{1}(\rho,x) \,d_{q}x, $$
(3.26)
where
$$\begin{aligned}& K_{1}(\rho,x)= \int_{0}^{\infty}\frac{uw(u)}{1+w(u)} J_{\nu-\alpha } \bigl(ux;q^{2}\bigr) J_{\nu-\alpha}\bigl(u\rho;q^{2}\bigr) \,d_{q}u, \\& \tilde{F}_{1}(\rho)= F_{1}(\rho) -\frac{q^{-2\alpha^{2}-\alpha+\nu}}{(1-q)^{2}} \int_{0}^{a} x \psi _{2}(x) \int_{0}^{\infty} \frac{u}{1+w(u)}J_{\nu+\alpha} \bigl(ux;q^{2}\bigr) J_{\nu-\alpha}\bigl(u\rho;q^{2}\bigr) \,d_{q}u \,d_{q}x, \end{aligned}$$
and
$$\begin{aligned} F_{1}(\rho) =&\rho^{\nu-\alpha} \frac{q^{-2\alpha ^{2}-\alpha+\nu}(1+q)(1-q^{2})^{-\alpha}}{(1-q)^{2} \Gamma_{q^{2}}(\alpha )} \int_{\rho}^{\infty}x^{2\alpha-\nu-1}f_{3}(qx) \bigl(\rho^{2}/x^{2}; q^{2}\bigr)_{\alpha-1} \,d_{q}x \\ &{}-\frac{q^{-2\alpha^{2}-\alpha+\nu }}{(1-q)^{2}} \biggl[ \int_{a}^{\infty}x\Phi_{2}(x) \int_{0}^{\infty}\frac {u}{1+w(u)} J_{\nu+\alpha} \bigl(ux;q^{2}\bigr) J_{\nu-\alpha}\bigl(u\rho;q^{2}\bigr) \,d_{q}u \,d_{q}x \\ &{}+ \int_{0}^{b} x \Phi_{1}(x) \int_{0}^{\infty}\frac{uw(u)}{1+w(u)} J_{\nu -\alpha} \bigl(ux;q^{2}\bigr) J_{\nu-\alpha}\bigl(u\rho;q^{2}\bigr) \,d_{q}u \,d_{q}x \biggr]. \end{aligned}$$
Proof
Equation (3.11) can be written in the following form:
$$ \int_{0}^{\infty}u^{2\alpha}C_{1}(u) J_{\nu}\bigl(u\rho; q^{2}\bigr) \,d_{q} u= G( \rho),\quad \rho\in B_{q,b}, $$
(3.27)
where
$$\begin{aligned} G(\rho) =& f_{3} (\rho) - \int_{0}^{\infty}u^{2\alpha} C_{2}(u) \frac {1}{1+w(u)} J_{\nu}\bigl(u\rho; q^{2}\bigr) \,d_{q} u \\ &{}+ \int_{0}^{\infty}u^{2\alpha} C_{1}(u) \frac{w(u)}{1+w(u)}J_{\nu}\bigl(u\rho; q^{2}\bigr) \,d_{q} u. \end{aligned}$$
(3.28)
By using equations (2.3) and (3.27), we get
$$ G(\rho)= -(1-q) \rho^{\nu-1} q^{\nu-1} D_{q,\rho} \rho^{1-\nu} \int_{0}^{\infty}u^{2\alpha-1}C_{1}(u) J_{\nu-1} \bigl(u\rho q^{-1}; q^{2}\bigr) \,d_{q} u. $$
(3.29)
Substituting the value of \(C_{1}(u)\) from (3.14) into (3.29), we obtain
$$\begin{aligned}& D_{q,\rho} \rho^{1-\nu} \int_{0}^{\infty}u^{\alpha} \biggl[ \int_{0}^{b} x \Phi_{1}(x)J_{\nu-\alpha} \bigl(ux; q^{2}\bigr) \,d_{q}x + \int_{b}^{\infty}x \psi _{1}(x)J_{\nu-\alpha} \bigl(ux; q^{2}\bigr) \,d_{q}x \biggr] \\& \quad {}\times J_{\nu-1} \bigl(u\rho q^{-1}; q^{2} \bigr) \,d_{q} u = - \frac{\rho^{1-\nu}q^{1-\nu} G(\rho )}{(1-q)} , \quad \rho\in B_{q,b}. \end{aligned}$$
(3.30)
From (2.5), there exists \(M>0\) such that
$$\bigl\vert J_{\nu-\alpha}\bigl(ux;q^{2}\bigr)\bigr\vert \leq M(ux)^{\Re(\nu-\alpha )} \quad \mbox{for all } u,x \in\mathbb{R}_{q^{2},b,+}. $$
Hence, from Remark 3.2, the double q-integration is absolutely convergent and we can interchange the order of the q-integrations to obtain
$$\begin{aligned} G(\rho) =& - (1-q)\rho^{\nu-1}q^{\nu-1} \biggl[ \int_{0}^{b} x \Phi_{1}(x) \,d_{q}x + \int_{b}^{\infty}x \psi_{1}(x) \,d_{q}x \biggr] \\ & {}\times D_{q,\rho} \rho^{1-\nu} \int_{0}^{\infty}u^{\alpha} J_{\nu-1} \bigl(u\rho q^{-1}; q^{2}\bigr) J_{\nu-\alpha}\bigl(ux; q^{2}\bigr) \,d_{q} u,\quad \rho\in B_{q,b}. \end{aligned}$$
(3.31)
Therefore, applying Proposition 2.1 with \(\Re(\nu-\alpha) > \Re(\nu-1) > -1\) we obtain
$$ G(\rho)= \frac{-(1-q)^{2}(1-q^{2})^{\alpha}}{\Gamma_{q^{2}}(1-\alpha)} \rho^{\nu-1} D_{q,\rho} \int_{\rho}^{\infty}x^{1-\nu-\alpha} \psi_{1}(x) \bigl({\rho^{2}}/{x^{2}}; q^{2}\bigr)_{-\alpha} \,d_{q} x. $$
(3.32)
By using
$$ \int_{x}^{\infty}f(t) \,d_{q}t = \frac{1}{1+q} \int_{x^{2}}^{\infty}\frac {f(\sqrt{t})}{\sqrt{t}} \,d_{q^{2}}t,\qquad D_{q,\rho}\bigl(f\bigl(\rho^{2}\bigr) \bigr)= \rho(1+q) (D_{q^{2}}f ) \bigl(\rho^{2}\bigr), $$
(3.33)
we obtain
$$\begin{aligned} G(\rho)&= \frac{-(1-q)^{2}(1-q^{2})^{\alpha}}{\Gamma _{q^{2}}(1-\alpha)} \rho^{\nu} D_{q^{2},\rho^{2}} \int_{\rho^{2}}^{\infty}x^{\frac{-(\nu+\alpha)}{2}} \psi_{1}( \sqrt{x}) \bigl({\rho^{2}}/{x}; q^{2}\bigr)_{-\alpha} \, d_{q^{2}} x \\ &=-(1-q)^{2} \bigl(1-q^{2}\bigr)^{\alpha}q^{\alpha^{2}-2\alpha-\nu} \rho^{\nu } \bigl(D_{q^{2} } \mathcal{K}_{q^{2}}^{1-\alpha} \bigl((\cdot)^{-\frac{\nu+\alpha }{2}}\psi_{1}(\cdot) \bigr) \bigr) \bigl( \rho^{2}/q^{2}\bigr). \end{aligned}$$
Replacing ρ by qρ yields
$$\begin{aligned}& -{q^{-\alpha^{2}+\alpha}\bigl(1-q^{2}\bigr)^{-\alpha}} {(1-q)^{-2}} \bigl[(\cdot )^{-\nu/2} G(q\sqrt{\cdot}) \bigr] \bigl(\rho^{2}\bigr) \\& \quad = D_{q^{2},\rho ^{2}}\mathcal{K}_{q^{2}}^{1-\alpha} \bigl[( \sqrt{\cdot})^{(\alpha-\nu )}\psi_{1}(\sqrt{\cdot}) \bigr]\bigl( \rho^{2}\bigr). \end{aligned}$$
(3.34)
Thus, applying Proposition 3.3 yields
$$\begin{aligned} \rho^{\alpha-\nu}\psi_{1}(\rho) =& q^{-\alpha ^{2}}(1-q)^{-2} \bigl(1-q^{2}\bigr)^{-\alpha}\mathcal{K}_{q^{2}}^{\alpha} \bigl[(\cdot)^{-\nu/2}G(q\sqrt{\cdot}) \bigr]\bigl(\rho^{2}/q^{2} \bigr) \\ =&\frac{q^{-2\alpha^{2}-\alpha+\nu}(1-q^{2})^{-\alpha }(1-q)^{-2}}{\Gamma_{q^{2}}(\alpha)} \int_{\rho^{2}}^{\infty}x^{-\frac {\nu}{2}+\alpha-1} G(q\sqrt{x}) \bigl({\rho^{2}}/{ x}; q^{2}\bigr)_{\alpha-1 } \, d_{q^{2}} x. \end{aligned}$$
Using \(\int_{x^{2}}^{\infty}f(t) \,d_{q^{2}}t=(1+q)\int_{x}^{\infty }tf(t^{2}) \,d_{q}t\), we obtain
$$\rho^{\alpha-\nu}\psi_{1}(\rho) =\frac{q^{-2\alpha^{2}-\alpha+\nu }(1-q^{2})^{-\alpha}(1+q)}{(1-q)^{2}\Gamma_{q^{2}}(\alpha)} \int_{\rho }^{\infty}x^{2\alpha-\nu-1} G(qx) \bigl({ \rho^{2}}/{ x^{2}}; q^{2}\bigr)_{\alpha- 1}\, d_{q} x. $$
From (3.28), we can write the last equation in the following form:
$$\begin{aligned}& \psi_{1}(\rho)+ \rho^{\nu-\alpha} \frac{q^{-2\alpha^{2}-\alpha+\nu }(1-q^{2})^{-\alpha}(1+q)}{(1-q)^{2}\Gamma_{q^{2}}(\alpha)} \int_{\rho}^{\infty} x^{2\alpha-\nu-1} \bigl( \rho^{2}/x^{2}; q^{2}\bigr)_{\alpha -1} \\& \qquad {}\times \biggl[ \int_{0}^{\infty}\frac{u^{2\alpha}}{1+w(u)}C_{2}(u) J_{\nu}\bigl(qux; q^{2}\bigr) \,d_{q} u \\& \qquad {}- \int_{0}^{\infty}\frac{ w(u)}{1+w(u)}u^{2\alpha} C_{1}(u) J_{\nu}\bigl(qux; q^{2}\bigr) \,d_{q} u \biggr] \,d_{q} x \\& \quad =\rho^{\nu-\alpha} \frac{q^{-2\alpha^{2}-\alpha+ \nu}(1-q^{2})^{-\alpha}(1+q)}{(1-q)^{2} \Gamma_{q^{2}}(\alpha)} \\& \qquad {}\times\int _{\rho}^{\infty}x^{2\alpha-\nu-1} f_{3}(qx) \bigl({\rho^{2}}/{x^{2}}; q^{2}\bigr)_{\alpha-1} \,d_{q}x,\quad \rho\in B_{q,b}. \end{aligned}$$
(3.35)
From the condition on the function \(C_{2}\), we can prove that the double q-integration
$$ \int_{\rho}^{\infty} x^{2\alpha-\nu-1} \bigl( \rho^{2}/x^{2}; q^{2}\bigr)_{\alpha -1} \int_{0}^{\infty}C_{2}(u)\frac{u^{2\alpha}}{1+w(u)} J_{\nu}\bigl(qux; q^{2}\bigr) \,d_{q} u \,d_{q}x $$
is absolutely convergent. Therefore, we can interchange the order of the q-integrations and use Proposition 2.2 to obtain
$$\begin{aligned}& \psi_{1}(\rho)+ \frac{q^{-2\alpha^{2}-\alpha+\nu}}{(1-q)^{2}} \biggl[ \int_{0}^{\infty}\frac{u^{\alpha}}{1+w(u)}C_{2}(u) J_{\nu-\alpha } \bigl(u\rho; q^{2}\bigr) \,d_{q} u \\& \qquad {}- \int_{0}^{\infty}\frac{u^{\alpha} w(u)}{1+w(u)} C_{1}(u) J_{\nu-\alpha} \bigl(u\rho; q^{2}\bigr) \,d_{q} u \biggr] \\& \quad =\rho^{\nu-\alpha} \frac{q^{-2\alpha^{2}-\alpha+\nu }(1-q^{2})^{-\alpha}(1+q)}{(1-q)^{2} \Gamma_{q^{2}}(\alpha)} \\& \qquad {}\times\int_{\rho }^{\infty}x^{2\alpha-\nu-1} f_{3}(qx) \bigl({\rho^{2}}/{x^{2}}; q^{2}\bigr)_{\alpha-1} \,d_{q}x,\quad \rho\in B_{q,b}. \end{aligned}$$
(3.36)
Substituting the value of \(C_{1}(u)\) and \(C_{2}(u)\) from equations (3.15) and (3.14) into equation (3.36), and then interchanging the order of the q-integrations we get
$$\begin{aligned}& \psi_{1}(\rho)+ \frac{q^{\nu-4\alpha}}{(1-q)^{2}} \biggl[ \int_{0}^{a} x \psi_{2}(x) \int_{0}^{\infty} \frac {u}{1+w(u)}J_{\nu+\alpha} \bigl(ux;q^{2}\bigr) J_{\nu-\alpha}\bigl(u\rho;q^{2}\bigr) \,d_{q}u \,d_{q}x \\& \qquad {}- \int_{b}^{\infty}x\psi_{1}(x) \int_{0}^{\infty} \frac{u w(u)}{1+w(u)} J_{\nu-\alpha} \bigl(ux;q^{2}\bigr) J_{\nu-\alpha}\bigl(u\rho;q^{2}\bigr) \,d_{q}u \,d_{q}x \biggr] \\& \quad = F_{1}(\rho),\quad \rho\in B_{q,b}, \end{aligned}$$
(3.37)
where
$$\begin{aligned} F_{1}(\rho) =&\rho^{\nu+\alpha} \frac{q^{\nu-4\alpha }(1+q)(1-q^{2})^{-\alpha}}{(1-q)^{2} \Gamma_{q^{2}}(\alpha)} \int_{\rho }^{\infty}x^{2\alpha-\nu-1}f_{3}(qx) \bigl(\rho^{2}/x^{2}; q^{2}\bigr)_{\alpha-1} \,d_{q}x \\ &{}-\frac{q^{\nu-4\alpha}}{(1-q)^{2}} \biggl[ \int_{a}^{\infty}x\Phi _{2}(x) \int_{0}^{\infty}\frac{u}{1+w(u)} J_{\nu+\alpha} \bigl(ux;q^{2}\bigr) J_{\nu -\alpha}\bigl(u\rho;q^{2}\bigr) \,d_{q}u \,d_{q}x \\ &{}+ \int_{0}^{b} x \Phi_{1}(x) \int_{0}^{\infty}\frac{uw(u)}{1+w(u)} J_{\nu -\alpha} \bigl(ux;q^{2}\bigr) J_{\nu-\alpha}\bigl(u\rho;q^{2}\bigr) \,d_{q}u \,d_{q}x \biggr]. \end{aligned}$$
Equation (3.37) is nothing else but the Fredholm q-integral equation of the second kind (3.26). This completes the proof. □
Proposition 3.4
For
\(\rho\in A_{q,a}\), \(\psi_{2}(\rho)\)
satisfies the Fredholm
q-integral equation of the form
$$ \psi_{2}(\rho)=\tilde{F}_{2}(\rho)+ \frac{1}{(1-q)^{2}} \int_{0}^{a} x K_{2}(\rho,x) \psi_{2}(x) \,d_{q}x, $$
(3.38)
where
$$\begin{aligned}& K_{2}(\rho,x)= \int_{0}^{\infty}\frac{uw(u)}{1+w(u)} J_{\nu+\alpha} \bigl(ux; q^{2}\bigr) J_{\nu +\alpha} \bigl(u\rho; q^{2} \bigr) \,d_{q}u, \\& \tilde{F}_{2}(\rho)=F_{2}(\rho) - \frac{1}{(1-q)^{2}} \int_{b}^{\infty}x\psi_{1}(x) \int_{0}^{\infty }\frac{u}{1+w(u)} J_{\nu-\alpha} \bigl(ux;q^{2}\bigr)J_{\nu+\alpha}\bigl(u\rho;q^{2}\bigr) \,d_{q}u \,d_{q}x, \end{aligned}$$
and
$$\begin{aligned} F_{2}(\rho) =& \frac{(1-q^{2})^{-\alpha}(1+q) \rho^{\alpha-\nu -2}}{(1-q)^{2}\Gamma_{q^{2}}(\alpha)} \int_{0}^{\rho} \bigl({q^{2}x^{2}}/{ \rho ^{2}}; q^{2}\bigr)_{\alpha-1} x^{\nu+1}f_{1}(x) \,d_{q}x \\ &{}+\frac{1}{(1-q)^{2}} \int_{a}^{\infty}x \Phi_{2}(x) \int_{0}^{\infty}\frac{uw(u)}{1+w(u)} J_{\nu+\alpha} \bigl(ux; q^{2}\bigr) J_{\nu+\alpha} \bigl(u\rho; q^{2} \bigr) \,d_{q}u \,d_{q}x \\ &{}-\frac{1}{(1-q)^{2}} \int_{0}^{b} x \Phi_{1}(x) \int_{0}^{\infty}\frac {u}{1+w(u)}J_{\nu-\alpha} \bigl(ux; q^{2}\bigr) J_{\nu+\alpha} \bigl(u\rho; q^{2} \bigr) \,d_{q}u \,d_{q}x. \end{aligned}$$
Proof
The proof is similar to the proof of Proposition 3.3 and is omitted. □
Theorem 3.5
The solution of (3.1)-(3.2) is given by
$$\psi(u) =\frac{u^{2\alpha}}{1+w(u)} \bigl(C_{1}(u)+C_{2}(u) \bigr). $$
The functions
\(C_{1}\), \(C_{2}\), \(\phi_{1}\), and
\(\phi_{2}\)
are given by Proposition
3.1, and
\(\psi_{1}\), \(\psi_{2}\)
satisfies the Fredholm
q-integral equations (3.38) and (3.26) of second kind.
Example 1
1. Take \(b=aq^{-m}\) and assume that \(m\to\infty\). If we assume that \(f_{1}=f\), \(f_{2}=f\), and \(w=0\). Then the system (3.1)-(3.3) is reduced to the dual q-integral equations
$$\begin{aligned}& \int_{0}^{\infty}\psi(u)J_{\nu}\bigl(u \rho;q^{2}\bigr) \,d_{q}u=f(\rho),\quad \rho\in A_{q,a}, \end{aligned}$$
(3.39)
$$\begin{aligned}& \int_{0}^{\infty}u^{-2\alpha}\psi(u)J_{\nu} \bigl(u\rho;q^{2}\bigr) \,d_{q}u=0, \quad \rho\in B_{q,a}. \end{aligned}$$
(3.40)
Hence, from Theorem 3.5,
$$\begin{aligned}& \psi(u)=u^{1+\alpha} \int_{0}^{\infty}x\psi_{2}(x)J_{\nu+\alpha } \bigl(ux;q^{2}\bigr) \,d_{q}x,\quad u\in \mathbb{R}_{q,+}, \\& \psi_{2}(\rho)=\frac{(1-q^{2})^{-\alpha}(1+q)\rho ^{\alpha-\nu-2}}{(1-q)^{2}\Gamma_{q^{2}}(\alpha)} \int_{0}^{\rho}\bigl(q^{2} x^{2}/\rho^{2};q^{2}\bigr)_{\alpha-1}x^{\nu+1}f(x) \,d_{q}x \\& \hphantom{\psi_{2}(\rho)}=\rho^{-\alpha-\nu}\frac{(1-q^{2})^{-\alpha }}{(1-q)^{2}}I_{q^{2}}^{\alpha} \bigl(t^{\nu/2}f(\sqrt{t}) \bigr) \bigl(\rho^{2}\bigr). \end{aligned}$$
Hence,
$$\psi(u)=u^{1+\alpha}\frac{(1-q^{2})^{-\alpha}}{(1-q^{2})} \int _{0}^{\infty}x^{1-\alpha-\nu}I_{q^{2}}^{\alpha} \bigl(t^{\nu /2}f(\sqrt{t}) \bigr) \bigl(x^{2} \bigr)J_{\nu+ \alpha}\bigl(ux;q^{2}\bigr) \,d_{q}x. $$
This coincides with the result in [15], Theorem 4.1, for solutions of double q-integral equations.
2. Let \(a=q^{m}\) and assume that \(m\to\infty\). If we assume that \(f_{2}=0\), and \(f_{3}=f\), we obtain the dual q-integral system of equations
$$\begin{aligned}& \int_{0}^{\infty}u^{-2\alpha}\psi(u)J_{\nu} \bigl(u\rho;q^{2}\bigr) \,d_{q}u=0,\quad \rho\in A_{q,b}, \end{aligned}$$
(3.41)
$$\begin{aligned}& \int_{0}^{\infty}\psi(u)J_{\nu}\bigl(u \rho;q^{2}\bigr) \,d_{q}u=f,\quad \rho\in B_{q,b}. \end{aligned}$$
(3.42)
Hence, from Theorem 3.5,
$$\begin{aligned}& \psi(u)=u^{1+\alpha} \int_{b}^{\infty}x\psi_{1}(x)J_{\nu-\alpha } \bigl(ux;q^{2}\bigr) \,d_{q}x, \quad u\in \mathbb{R}_{q,+}, \\& \psi_{1}(\rho)=-\frac{(1-q^{2})^{-\alpha}q^{-2\alpha}\rho^{\alpha +\nu}}{(1-q)^{2}\Gamma_{q^{2}}(\alpha)} \int_{\rho}^{\infty}\bigl(\rho^{2}/ x^{2};q^{2}\bigr)_{\alpha-1}x^{2\alpha-\nu-1}f(x) \,d_{q}x. \end{aligned}$$
This is a special case of Theorem 5.1 in [15].
Example 2
We consider the triple q-integral equations
$$\begin{aligned}& \int_{0}^{\infty}\psi(u)J_{0}\bigl(u \rho;q^{2}\bigr) \,d_{q}u = 0,\quad \rho\in A_{q,a}, \end{aligned}$$
(3.43)
$$\begin{aligned}& \int_{0}^{\infty}u^{-1}\psi(u)J_{0} \bigl(u\rho;q^{2}\bigr) \,d_{q}u = 1,\quad \rho \in A_{q,b}\cap B_{q,a}, \end{aligned}$$
(3.44)
$$\begin{aligned}& \int_{0}^{\infty}\psi(u)J_{0}\bigl(u \rho;q^{2}\bigr) \,d_{q}u = 0,\quad \rho\in B_{q,b}. \end{aligned}$$
(3.45)
Hence, we have \(\nu=0\), \(g_{1}=1\), \(g_{2}=0\), \(f_{1}=f_{3}=0\), \(w=0\), and \(\alpha=\frac{1}{2}\).
From Theorem 3.5,
$$\psi(u)=u \bigl(C_{1}(u)+C_{2}(u) \bigr), $$
where
$$\begin{aligned}& C_{1}(u)=\frac{(1-q)(1-q^{2})}{\Gamma_{q^{2}}^{2}({1}/{2})}\frac{\sin (\frac{bu}{1-q};q)}{u} \\& \hphantom{C_{1}(u)={}}{}+\frac{\sqrt{1-q^{2}}}{\Gamma_{q^{2}}(1/2)} \int _{b}^{\infty}\sqrt{x}\psi_{1}(x)\cos \biggl(\frac{xu\sqrt{q}}{1-q};q^{2}\biggr) \,d_{q}x, \\& C_{2}(u)=\frac{\sqrt{1-q^{2}}}{\Gamma_{q^{2}}(1/2)} \int_{0}^{a}\sqrt {x}\psi_{2}(x)\sin \biggl(\frac{xu}{1-q};q^{2}\biggr) \,d_{q}x, \\& \psi_{1}(\rho)=\frac{\sqrt{\rho }(1+q)}{q(1-q)\Gamma_{q^{2}}^{2}(1/2)} \int_{0}^{a}x^{3/2}\frac{\psi _{2}(x)}{q\rho^{2}-x^{2}} \,d_{q}x,\quad \rho\in B_{q,b}, \end{aligned}$$
(3.46)
$$\begin{aligned}& \psi_{2}(\rho)=-\frac{(1+q)\sqrt{\rho }}{(1-q)\Gamma_{q^{2}}^{2}(1/2)} \int_{b}^{\infty}\frac{\sqrt{x}\psi _{1}(x)}{qx^{2}-\rho^{2}} \,d_{q}x \\& \hphantom{\psi_{2}(\rho)={}}{}+ \frac{(1+q)^{3/2}}{\sqrt{1-q}\Gamma_{q^{2}}^{3}(1/2)}\sqrt{\rho } \int_{\rho/q}^{b}\frac{\,d_{q}x}{qx^{2}-\rho^{2}}. \end{aligned}$$
(3.47)
We used [19], pp.455-466 or Proposition 2.4 of [15] to calculate \(\psi_{1}\) and \(\psi_{2}\) in equations (3.46) and (3.47), respectively. Substituting from (3.46) into (3.47), we obtain the second order Fredholm q-integral equation
$$\begin{aligned} \psi_{2}(\rho) =&-\frac{q^{-1}\sqrt{\rho}(1+q)}{(1-q)^{2}\Gamma _{q^{2}}^{3}(1/2)} \int_{0}^{a}t^{3/2}\psi_{2}(t)K_{2}( \rho,t) \,d_{q}t \\ &{}+ \frac{(1+q)^{3/2}}{\sqrt{1-q}\Gamma_{q^{2}}^{3}(1/2)}\sqrt{\rho } \int_{\rho/q}^{b}\frac{d_{q}x}{qx^{2}-\rho^{2}}, \end{aligned}$$
(3.48)
where \(\rho\in A_{q,a}\) and
$$K_{2}(\rho,t)= \int_{b}^{\infty}\frac{x}{(t^{2}-qx^{2})(\rho^{2}- q x^{2})} \,d_{q}t. $$